Mathspace VCE Units 1&2 Specialist Mathematics Flipbook by Mathspace

VCE

Specialist Mathematics Units 1 & 2 Accreditation period from 2023

© 2025 Mathspace Group Holdings Ltd Copyright Notice This Work is copyright. All rights are reserved. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. For permission to use material from this text or product, please email hello@mathspace.com.au For our full digital offering, visit mathspace.co The Victorian Curriculum F-10 content elements are © VCAA, reproduced by permission. The VCAA does not endorse or make any warranties regarding this resource. The Victorian Curriculum F-10 and related content can be accessed directly at the VCAA website. Title: Mathspace VCE Specialist Mathematics Units 1 & 2 Print run 1 - January 2026 ISBN: 978-1-963022-98-8 Editors and lead authors: Erin Gallagher, Jeremy Hunter, Elisha Templin, Nazeel Ali, Katrina Du, Jessica Purton, Isha Tiwari Writing and development team: Adam Humphreys, Ma Lourdes Ona, Sheramie Castillo, Regine Jara, Lesley Ann Lopez, Adriane Abunda, Jady Amarillas, Joy Ann Antonio, Macy Castillo, Christian Dinglasan, Adrian Doctolero, Dharent Fernandez, Dhave Fernandez, Rona Josue, Jessa Lambinicio, Eric Malabag, Julie Ann Manzano, Kyra Joy Manzano, Mary Joy Tacud, Albert Jara Images and design team: Scott Nolan, Chastine Marquez, Keith Gimeno, Jasper Jumawan, Jemark Orevillo, Jessa Ortega

Contents 1

Proof and number

1.01

Propositions and statements

1.02

Symbols in proof

1.03

Direct proof

1.04

Proofs by contradiction

1.05

Proofs by counter-example

1.06

Proof by induction

Investigation: Repetition in code 1.07

Prove properties of integers

1.08

Rational numbers and their representations

1.09

Prove irrationality

Chapter 1 review

Logic and algorithms

2.01

Propositional logic

2.02

Proofs with truth tables

110

2.03

Boolean algebra and axioms

121

2.04

Karnaugh maps

128

2.05

Binary numbers

137

Investigation: Circuits 2.06

Logic gates

150

2.07

Design and interpret algorithms

161

Chapter 2 review

Sequences and series

176

3.01

Introduction to sequences

178

3.02

Arithmetic sequences

194

3.03

Geometric sequences

210

3.04

Applications of sequences Investigation: Spreadsheets and sequences

228

3.05

Arithmetic series

245

3.06

Geometric series

254

3.07

Applications of sequences and series

265

Chapter 3 review

Contents mathspace.co

iii

Combinatorics 274 4.01

Pigeon-hole principle

276

4.02

Inclusion-exclusion principle

283

4.03

Multiplication and addition principles

300

4.04

Arrangements and factorial notation

312

4.05

Permutations

327

4.06

Permutations with restrictions

336

Investigation: Pascal’s triangle 4.07

Combinations

353

4.08

Combinations with restrictions

368

4.09

Combinatorics mixed problems

379

4.10

Combinatorics and probability

391

Chapter 4 review

Matrices 406 5.01

Matrix notation and representation

408

5.02

Addition and subtraction of matrices

424

5.03

Scalar multiplication

433

5.04

Matrix multiplication

443

5.05

Multiplicative inverses and determinants

459

5.06

Solve matrix equations

468

Investigation: Existence of matrix solutions Chapter 5 review

Graph theory

484

6.01

Introduction to graph theory

486

6.02

Degree and adjacency

499

Investigation: Hand-shaking lemma proof 6.03

Graph isomorphisms

513

6.04

Trees

526

6.05

Planar graphs

542

6.06

Walks, paths, and trails

554

6.07

Algorithms in graph theory

570

6.08

Applications of graph theory

581

Chapter 6 review

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Trigonometry 592 7.01

Radians and parts of circles

594

7.02

Foundations of trigonometry

606

7.03

The sine rule

626

7.04

The cosine rule

644

7.05

Applications of sine and cosine rules

656

7.06

Pythagorean identities

669

Investigation: Prove the sine angle sum identity 7.07

Angle sum and difference identities

679

7.08

Double-angle identities

687

Chapter 7 review

Vectors 698 8.01

Introduction to vectors

700

8.02

Magnitude and direction of a vector

709

8.03

Scalar multiples of vectors

725

8.04

Vector addition and subtraction

736

8.05

Unit vectors and component form

752

8.06

Geometric applications of vectors

769

8.07

Polar form

782

8.08

Scalar (dot) product

795

8.09

Scalar and vector projections

809

8.10

Displacement and velocity

818

8.11

Force equilibrium and work done

827

Investigation: Vector proofs Chapter 8 review

Contents mathspace.co

Complex numbers

838

9.01

Introduction to complex numbers

840

9.02

Modulus and argument of complex numbers

851

9.03

Addition and subtraction of complex numbers

861

9.04

Multiplication of complex numbers

870

9.05

Complex conjugates and division

876

I nvestigation: Linear transformations and rotations on the complex plane 9.06

Cartesian and polar form

885

9.07

Complex arithmetic in polar form

900

9.08

Complex inequalities and regions

909

9.09

Complex roots

927

Chapter 9 review

Transformations 940 10.01

Translations

942

10.02

Dilations

949

10.03

Rotations

958

10.04

Reflections

969

10.05

Compositions of linear transformations

978

10.06

Inverse of linear transformations

987

10.07

Area of linear transformations

995

Investigation: Transformations with code Chapter 10 review

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Functions, relations, and graphs

1004

11.01

Absolute value graphs

1006

11.02

Reciprocal functions

1022

11.03

Rational expressions and partial fractions

1036

11.04

Trigonometric functions

1043

11.05

Transformations of trigonometric functions

1057

11.06

Inverse circular functions

1071

11.07

Reciprocal trigonometric graphs

1079

Investigation: Locus definition 11.08

Locus definitions and the distance formula

1101

11.09

Conic sections from locus definitions

1116

11.10

Parametric representations of relations

1127

11.11

Polar coordinates and polar graphs

1140

Chapter 11 review

Simulation, sampling and sampling distributions

1160

12.01

Discrete random variables

1162

12.02

Centre and spread of discrete random variables

1183

12.03

Sums of independent random variables

1195

12.04

Applications of discrete random variables

1202

Investigation: Simulations 12.05

Simulation

1213

12.06

Sample distributions

1220

12.07

Simulate sampling distributions

1229

Investigation: Generating samples with technology 12.08

Sample size and distributions

1236

Chapter 12 review

Answers

1242

Contents mathspace.co

vii

Big ideas • Mathematical proof is built upon the foundational principles of logic, using precise language and symbols to construct valid arguments. • Diverse and rigorous methods of proof are used to establish the truth of mathematical statements, with the appropriate method depending on the nature of the conjecture. • The principles of logical proof are applied to formalise the properties of number systems, including integers and rational numbers.

1 Proof and number Chapter outline 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09

Propositions and statements Symbols in proof Direct proof Proofs by contradiction Proofs by counter-example Proof by induction Investigation: Repetition in code Prove properties of integers Rational numbers and their representations Prove irrationality Chapter 1 review

4 18 33 41 48 55 65 76 87

Mathematical proofs are like detectives solving mysteries – using logic and evidence to uncover truths!

1.01 Propositions and statements After this lesson, you will be able to… • use propositions and connectives in words • write the negation, converse, inverse, and contrapositive of a statement in words • construct valid arguments in words using propositions and connectives • describe the use of pseudocode for if-then decision structures

Language of proof Proof is a logical argument that demonstrates or establishes the truth of a mathematical statement. The study of these methods is logic. Proposition

Truth value

Any statement with a definite truth value.

The status of a proposition as true (T) or false (F).

In other words, a proposition is a statement that is either true or false, but never both at the same time. It is possible to assign a letter, such as T or F, to indicate the proposition’s truth value. Statements that cannot be assigned a truth value are not considered propositions. For example: Simple statement

Proposition?

Reasoning

The swan is white.

Yes

It cannot be true and false at the same time, so it can have a truth value.

These siblings are twins.

Yes

For a particular set of siblings, this will be true or false, not maybe or kind of.

Siblings look alike.

It is a statement that is sometimes true and sometimes false.

Close the door.

It is a command, so it cannot be evaluated as true or false.

Dogs are better than cats.

Based on an opinion — it is not possible to determine its validity as true or false.

Propositions are typically labelled using lowercase letters. Sometimes the letter selected relates to a key word in the proposition. Other times consecutive letters are used, such as the letters p and q. For example: n : The number is even. m : The moon is made of green cheese. p : Pigs fly.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Propositions can be put together using connectives to create logic statements like “If pigs fly and the number is even, then the moon is made of green cheese” or “If p and n, then m.” A statement that connects two or more propositions or simple statements is called a compound statement. New statements can be formed from propositions or logic statements using these connectives: • Negation is the opposite of a given statement. For the proposition, “an animal is a cat”, the negation is “an animal is not a cat.” not p • A conjunction joins two statements with the word “and”. For example, the conjunction of “a shape is a square” and “the shape is green” would be “the shape is a square and the shape is green”. It can be written in the form: p and q • A disjunction joins two statements with the word “or”. For example, the disjunction of “a shape is a square” and “the shape is green” would be “the shape is a square or the shape is green”. It can be written in the form: p or q • An implication, also called an “if-then” or conditional statement, is a type of compound statement that combines two simple statements. For the simple statements “an animal is a cat” and “the animal purrs”, the implication is “If an animal is a cat, then the animal purrs.” If p, then q • Equivalence is also called an “if-and-only-if” or biconditional statement. It means the implication is true in both directions. For the simple statements “an animal is a cat” and “the animal purrs”, the equivalence is “An animal is a cat if and only if the animal purrs.” p if and only if q, which means if p, then q and if q, then p For implication statements, p is called the hypothesis, or the premise. The hypothesis is the ‘if’ part of the statement. Additionally, q is called the conclusion. The conclusion is the ‘then’ part of the statement, and is true if the premise is true. From an implication, related statements can be formed: • The converse is a statement formed by switching the hypothesis and conclusion of an implication. Consider this implication and its converse: • Implication: If an animal is a dog, then it has four legs. • Converse: If an animal has four legs, then it is a dog. • The inverse is a statement formed by negating the hypothesis and conclusion of an implication. Consider this implication and its inverse: • Implication: If an animal is a dog, then it has four legs. • Inverse: If an animal is not a dog, then it does not have four legs. • The contrapositive is a statement formed by interchanging and negating the hypothesis and conclusion of an implication. Consider this implication and its contrapositive: • Implication: If an animal is a dog, then it has four legs. • Contrapositive: If an animal does not have four legs, then it is not a dog.

1.01 Propositions and statements mathspace.co

Combine propositions using connectives to form compound statements, but logic is also concerned with constructing arguments. An argument uses a set of propositions, called premises, to support another proposition, called the conclusion. A valid argument is one where the conclusion logically follows from the premises. If the premises are true, the conclusion is guaranteed to be true. It’s important to distinguish between simply forming a compound statement and constructing a valid argument. Simple valid argument forms include: • Modus Ponens: If p then q. p, therefore, q. Example: “If the alarm rings, I wake up. The alarm rang. Therefore, I woke up.” • Modus Tollens: If p then q, not q. Therefore, not p. xample: “If it is sunny, Yasmin goes to the park. Yasmin did not go to the park. Therefore, it E was not sunny.” Constructing valid arguments involves arranging propositions and connectives into these recognised valid structures.

Exploration Explore some logical relationships using this Venn diagram and these statements: Animal 2 legs

4 legs Dogs

Cats

Corgi

Birds

Pets

• Let p represent the statement “An animal is a corgi.” • Let q represent the statement “An animal is a dog.” • Let r represent the statement “An animal is a cat.” • Let s represent the statement “A corgi is a dog.” 1. If an animal is a corgi, must it also be a pet? 2. Determine whether the statement is true or false: If p, then q. 3. Fill in the blanks with p, q, or r to make a true statement: If ⬚, then ⬚.

4. Fill in the blanks with p, q, or r to make a true statement: If ⬚ is not true, then ⬚ is not true. 5. What other true statements can be written from the diagram?

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 Consider the implication “If it is sunny, then Yasmin goes to the park.” a Identify the hypothesis of the statement.

Create a strategy

Apply the idea

The hypothesis is the “if” part of the statement.

In the statement, the hypothesis is “it is sunny”.

b Identify the conclusion.

Create a strategy

Apply the idea

The conclusion is the “then” part of the statement.

In the statement, the conclusion is “Yasmin goes to the park.”

c Write the converse of the statement.

Create a strategy

Apply the idea

The converse is formed by switching the place of the hypothesis and conclusion.

The converse is “If Yasmin goes to the park, then it is sunny.”

d Write the equivalence of the statement.

Create a strategy Use the original implication and its converse from part (c). An equivalence is the conjunction of both the original statement and its converse.

Apply the idea An equivalence is “If Yasmin goes to the park, then it is sunny, and if it is sunny, then Yasmin goes to the park.” Another way to write the equivalence is “Yasmin goes to the park if and only if it is sunny.”

Reflect and check The ‘if and only if ’ statement is symmetrical, so the order of the two propositions can be reversed without changing the meaning. It can also be written as: “It is sunny if and only if Yasmin goes to the park.”

1.01 Propositions and statements mathspace.co

e Write the inverse of the statement.

Create a strategy The inverse is formed by negating both the hypothesis and the conclusion.

Apply the idea The inverse is “If it is not sunny, then Yasmin does not go to the park.”

Write the contrapositive of the statement.

Create a strategy The contrapositive is formed by negating both the hypothesis and conclusion, and then switching their positions.

Apply the idea The contrapositive is “If Yasmin does not go to the park, then it is not sunny.”

Example 2 Consider these premises: • Premise 1: If the train is late, I will miss my meeting. • Premise 2: The train is late. Construct a valid conclusion using Modus Ponens.

Create a strategy Identify the propositions p and q in Premise 1 (If p, then q). Premise 2 confirms p is true. The Modus Ponens structure allows the conclusion q.

Apply the idea Let p represent “The train is late.” Let q represent “I will miss my meeting.” Premise 1 is “If p, then q”. Premise 2 is “p”. Using Modus Ponens, the valid conclusion is q. Conclusion: Therefore, I will miss my meeting.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary In mathematics, logic is the study of arguments and proof. A proposition is a statement with a definitive truth value, that is either true or false. A compound statement connects two or more propositions using connectives. For simple statements p and q: • • • • • • • •

Negation: not p Conjunction: p and q Disjunction: p or q Implication: if p, then q Converse: if q, then p Inverse: if not p, then not q Contrapositive: if not q, then not p Equivalence: p if and only if q

For an implication (if p, then q), p is the hypothesis and q is the conclusion. A valid argument guarantees that if the premises are true, the conclusion must also be true. Simple valid argument forms include Modus Ponens and Modus Tollens.

Logical equivalence Statements that have the same truth value under all circumstances are known as logically equivalent. In logic, this is called logical equivalence. Note that this differs from an equivalence statement (if and only if). For simple statements or propositions: p

not p

not (not p)

Truth values

This means that a statement and its double negation are logically equivalent. For compound statements: Implication

Converse

Inverse

Contrapositive

Truth values

Depends on statement

Truth values

Depends on statement

This means an implication and its contrapositive always share the same truth value. Likewise, the converse and the inverse of a statement will also have the same truth value.

1.01 Propositions and statements mathspace.co

Example 3 Consider the true implication: “If a shape is a square, then it is a rectangle.” a Write the converse and determine whether it is true.

Create a strategy The implication is in the form “if p, then q”. Write the converse in the form “if q, then p”.

Apply the idea “If a shape is a rectangle, then it is a square.” This statement is false.

Consider this rectangle: It is a rectangle, but it is not a square. 10 m

10 m

b Write the inverse and determine whether it is true.

Create a strategy The implication is in the form “if p, then q”. Write the inverse in the form “if not p, then not q”.

Apply the idea “If a shape is not a square, then it is not a rectangle.” This statement is false.

Consider this rectangle: It is not a square, but it is a rectangle. 10 m

10 m

Reflect and check An inverse and converse statement will always share the same truth value.

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c Write the contrapositive and determine whether it is true.

Create a strategy The implication is in the form “if p, then q”. Write the contrapositive in the form “If not q, then not p.”

Apply the idea

Reflect and check

“If a shape is not a rectangle, then it is not a square.”

This can be confirmed using the law of contrapositive. Since the implication is true, the contrapositive is also true.

This statement is true. Any shape that is not a rectangle cannot be a square since all squares are rectangles.

Idea summary For simple statements, the negation of the negation is logically equivalent to the original statement. For implications, the contrapositive is logically equivalent to the original implication. The converse and inverse of an implication are always logically equivalent to each other.

If-then structures in pseudocode The “if-then” structure from logical implications is fundamental in computer programming and algorithms. Pseudocode is a way to describe the steps in an algorithm using simple English-like statements, without worrying about the exact syntax of a specific programming language. “If-then” structures allow algorithms to make decisions. The basic structure is “IF-THEN”: 1

IF condition is true THEN

2 (follow these instructions) 3

END IF

This means the instructions inside the block are only executed if the condition evaluates to TRUE. Often, an alternative set of instructions is needed if the condition is FALSE. This uses the “IF-THEN-ELSE” structure: 1

IF condition is true THEN

2 (follow these instructions) 3

ELSE

4 (follow these instructions) 5

END IF

1.01 Propositions and statements mathspace.co

Here, one of the two blocks of instructions will always be executed based on the condition’s TRUTH value. Conditions can also be chained using “ELSE IF”: 1

IF first condition is true THEN

2 (follow these instructions) 3

ELSE IF second condition is true THEN

4 (follow these instructions) 5

ELSE

6 (follow these instructions) 7

END IF

The computer checks the conditions in order. As soon as one is found to be TRUE, its instructions are executed, and the rest of the structure is skipped. The final “ELSE” block runs only if none of the preceding “IF” or “ELSE IF” conditions were TRUE.

Example 4 Write pseudocode to determine the maximum of two input numbers, “a” and “b”.

Create a strategy Use an if-then-else structure. If “a” is greater than “b”, then “a” is the maximum. Otherwise, “b” must be the maximum (or they are equal, in which case printing “b” is still correct).

Apply the idea 1

FUNCTION Maximum(a, b)

2 IF a > b THEN 3 RETURN a 4 ELSE 5 RETURN b 6 END IF 7

END FUNCTION

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary Pseudocode uses “if-then”, “if-then-else”, and “if-then-else-if” structures to represent logical decisions in algorithms. These structures correspond to implications and compound logical statements. Further decisions can be nested inside “if”, “else if”, or “else” blocks to handle more complex conditions. Each nested block is typically indented further to show its relationship to the outer block. The general structure for multiple conditions is: 1

IF first condition is true THEN

2 (instructions for first condition) 3

ELSE IF second condition is true THEN

4 (instructions for second condition) 5

...

ELSE

7 (instructions if no conditions were true) 8

END IF

1.01 Practice questions What do you remember? 1

For the statements: • p: The sun is shining. • q: It is raining. “The sun is shining, or it is not raining.”

Is this a simple statement or a compound statement?

What logical connectives are used in this statement?

Write the negation of each statement: a

The ball is blue.

The dog is not brown.

The water is hot.

Some turtles do not have claws.

For the statements: • p: The pillow is white. • q: The pillow is soft. a

Write the conjunction of p and q in words.

Write the disjunction of p and q in words.

1.01 Propositions and statements mathspace.co

4 Ex 1a, 1b

For each implication, identify the hypothesis and conclusion: a

If two angles are vertically opposite, then they are equal.

If the sum of a polygon’s interior angle measures is 180°, then the polygon is a triangle.

Two triangles are congruent if they have the same side lengths.

When two lines are parallel, the co-interior angles formed by a transversal are supplementary.

Determine whether each statement is an implication, an equivalence, or neither: a

If two lines are parallel, then they will never intersect.

Dogs are mammals.

A triangle is equilateral if and only if it is equiangular.

If a number is divisible by four, then it is even.

A number is composite if and only if it has at least three factors.

Right angles measure 90°.

For each pair of implications, determine whether they are converse, inverse, or contrapositive statements for each other: a

If two numbers are even, then their sum will be even. If the sum of two numbers is even, then the two numbers are even.

If two lines do not meet at a right angle, then they are not perpendicular. If two lines are perpendicular, then they meet at a right angle.

If two triangles are similar, then they must have three equal angles. If two triangles are not similar, then they cannot have three equal angles.

Practice 7

For the given diagrams: • The hypothesis, p, is “∠1 + ∠2 = 180° ” • The conclusion, q, is “∠1 and ∠2 are supplementary.”

Ex 1c, 1d, 1e, 1f

Write the implication in words.

Write the converse statement in words.

Write the equivalence in words.

Write the inverse statement in words.

Write the contrapositive statement in words.

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1 2 Figure 1 1

Figure 2

Determine which of these options is the converse of the statement: If 2a = b, then a

b 2

b If 2a ≠ b, then a . 2

b If a , then 2a = b. 2

b If a , then 2a ≠ b. 2

b If 2a = b, then a . 2

Which of these is the inverse of the statement “If a number is an integer, then it is not an irrational number”? A

If a number is an integer, then it is a rational number.

If a number is not an irrational number, then it is an integer.

If a number is not an integer, then it is an irrational number.

If a number is an irrational number, then it is not an integer.

Consider these premises: • Premise 1: If I study for the test, then I will pass. • Premise 2: I studied for the test.

Ex 2

Using Modus Ponens, what is the valid conclusion that can be drawn? 11

Consider these premises: • Premise 1: If the cat is hungry, it will meow. • Premise 2: The cat did not meow. Using Modus Tollens, what is the valid conclusion that can be drawn?

Consider the statement: “A number is even if and only if it is divisible by 2.” Identify the two implications that make up this equivalence statement.

Consider the two implications: “If a number k is a perfect square, then there exists an integer n such that k = n2.” “If there exists an integer n such that k = n2, then k is a perfect square.” Write an equivalence combining these two implications.

15 Ex 3a

Write a statement that is logically equivalent to each statement: a

If a, then b, and if b, then a

Not (not a)

If not c, then not d

If c, then not d

All given statements are true. For each statement: i

Write the converse.

If a number is a whole number, then it is an integer.

If a shape is a rectangle, then it is a quadrilateral.

If two angles are adjacent and supplementary, then they form a straight angle.

Is the converse true or false?

1.01 Propositions and statements mathspace.co

16 Ex 3b

17 Ex 3c

20 Ex 4

All given statements are true. For each statement: i

Write the inverse.

If a number is odd, then it is not even.

If I can paint a picture, then I am an artist.

Is the inverse true or false?

All given statements are true. For each statement: i

Write the contrapositive.

If a number ends in zero, then it is divisible by ten.

If I have a golden retriever, then I have a dog.

Is the contrapositive true or false?

The conjecture “A five-sided polygon is a pentagon.” is true: a

Write the conjecture in if-then form.

Write the contrapositive of the if-then statement.

If a polygon is not a pentagon, what conclusion can be made?

Determine whether each statement is true or false. If false, write a related true statement using its converse, inverse, or contrapositive: a

If a number is a real number, then it is rational.

If a quadrilateral is a parallelogram, then it is a rhombus.

If a number is not even, then it is not divisible by 2.

A person must be 18 years or older to vote. Write pseudocode that takes a person’s age as input and prints “Eligible to vote” if the input age is 18 or older, and “Not eligible to vote” otherwise. Consider this pseudocode: 1

Algorithm: Temperature Check

input temperature

if temperature > 30 then

4 print ‘It is hot.’ 5

else if temperature < 10 then

6 print ‘It is cold.’ 7

else

8 print ‘It is moderate.’ 9

end if

What will this algorithm print if the input temperature is: a

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking 22

Consider these 3 true facts about mammals and monotremes: i

An animal that gives birth to live young is a mammal.

All mammals produce milk.

iii

Monotremes are mammals that lay eggs.

Write each statement in if-then form.

Write the converse of each of the statements in part (a). Is the converse of each statement true? Explain your reasoning.

Write a true if-then statement about mammals, that uses the given information, but is different from the ones in parts (a) and (b). Is the converse of the statement true or false? Explain your reasoning.

Write an implication that is true but whose converse is false. Give an example to show the converse is not true.

Consider the statement: “If m2 − 8 = m + 5, then m = 3.” Can the statement and its converse form a true equivalence? Explain why or why not.

Felix claims the statement “If it is sunny, then it is not winter” can be written as a true equivalence. Identify his error. Explain your reasoning.

Using this Venn diagram, decide whether each statement is true or false. Justify your answer. Assume no region is empty: Real numbers Rational numbers Integers

Irrational numbers

Whole numbers Natural numbers

If a number is a whole number, then it is also a rational number.

If a number is a rational number, then it is also a whole number.

If a number is an integer, then it is also a natural number.

Some rational numbers are not whole numbers.

If a number is an irrational number, then it is not a rational number.

Some numbers are both integers and natural numbers.

Some numbers are both whole numbers and irrational numbers.

1.01 Propositions and statements mathspace.co

1.02 Symbols in proof After this lesson, you will be able to… • use mathematical symbols for implication, equivalence, and equality correctly • apply quantifiers ‘for all’ and ‘there exists’ appropriately • understand the structure of number systems (ℕ, ℤ, ℚ, ℝ, ℂ) and their properties • apply set notation including intersection, union, and complement operations • visualise statements using Venn diagrams to represent sets and subsets

Set operations and Venn diagrams Set theory provides a way to group objects, called elements, into collections called sets. Special notation is used to describe relationships between sets and operations on them. Set

Element

A collection of distinct objects, called elements.

An object belonging to a set.

Example:

In A = {1, 2, 3}, 2 is an element of set A (2 ∈ A).

Example:

A = {1, 2, 3} Key symbols in set theory: Concept

Symbol

Universal set

The set containing all possible elements relevant to the context.

If discussing vowels, U = {a, e, i, o, u}

Element of

∈

Indicates that an object is an element within a set.

If A = {1, 2, 3}, then 2 ∈ A

Not an element of

∉

Indicates that an object is not an element within a set.

If A = {1, 2, 3}, then 4 ∉ A

Subset

⊆

Set A is a subset of set B if every element of A is also in B.

If A = {1, 2}, B = {1, 2, 3}, then A ⊆ B

Proper subset

⊂

Set A is a proper subset of set B if A ⊆ B and A ≠ B.

If A = {1, 2}, B = {1, 2, 3}, then A ⊂ B

Intersection

∩

The set of elements that are in both set A and set B.

If A = {1, 2, 3}, B = {2, 3, 4}, then A ∩ B = {2, 3}

Union

∪

The set of elements that are in set A or set B (or both).

If A = {1, 2, 3}, B = {2, 3, 4}, then A ∪ B = {1, 2, 3, 4}

Complement

A′ or Ac

The set of elements in the universal set U that are not in set A.

If U = {1, 2, 3, 4, 5}, A = {1, 2}, then A′ = {3, 4, 5}

Empty set

∅ or {}

The set with no elements.

If A = {1, 2}, B = {3, 4}, then A ∩ B = ∅

Meaning

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example

Venn diagrams are used to visualise relationships between sets. They typically use overlapping circles inside a rectangle representing the universal set (U). Different regions represent different combinations of set membership. U A

C This Venn diagram shows three sets, A, B, and C, within a universal set U.

U A

C The region where the circles overlap represents A ∩ B (which is also A ∩ B ∩ C since B is a subset of C).

U A

C The entire area covered by both circles represents A ∪ C (which is equivalent to A ∪ B ∪ C).

U A

C The part of circle A not overlapping with B represents elements in A only (A ∩ B′).

U A

C The part of circle B not overlapping with A represents elements in B only (B ∩ A′).

1.02 Symbols in proof mathspace.co

U A

C The part of circle C not overlapping with A or B represents elements in C only (C ∩ A′ ∩ B′).

U A

C The area outside both circles but inside the rectangle represents elements not in A, B, or C ((A ∪ C)′).

Exploration Consider the sets within the Universal set U = {Integers from 1 to 15 inclusive}: A = {Multiples of 3 less than 15} B = {Even numbers less than 10} C = {Prime numbers less than 15} 1. List the elements of set A, set B, and set C. 2. Determine the elements in A ∩ B. 3. Determine the elements in B ∪ C. 4. Determine the elements in A′ (the complement of A with respect to U). 5. Determine the elements in (A ∪ B) ∩ C. 6. Draw a Venn diagram representing sets A, B, and U. Place the elements in the correct regions. 7. Can sets A, B, and C be easily represented on a single Venn diagram with 3 overlapping circles? Why or why not?

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 Let the universal set be U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 3, 5, 7, 9} (odd numbers) and B = {2, 3, 5, 7} (prime numbers): a Determine A ∩ B.

Apply the idea The intersection A ∩ B contains elements that are in both set A and set B. Comparing the elements: A ∩ B means elements in both A and B: • A = {1, 3, 5, 7, 9}, B = {2, 3, 5, 7} • Common elements: 3, 5, 7 Therefore, A ∩ B = {3, 5, 7}.

b Determine A ∪ B.

Apply the idea The union A ∪ B contains elements that are in set A, or in set B, or in both. All unique elements from both sets are listed. • Elements from A: 1, 3, 5, 7, 9 • Elements from B: 2, 3, 5, 7 • Combining them without repetition: 1, 2, 3, 5, 7, 9 Therefore, A ∪ B = {1, 2, 3, 5, 7, 9}.

c Determine B′.

Apply the idea The complement B′ contains elements that are in the universal set U but not in set B. • U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} • B = {2, 3, 5, 7} • Removing the elements of B from U leaves: 1, 4, 6, 8, 9, 10 Therefore, B′ = {1, 4, 6, 8, 9, 10}.

1.02 Symbols in proof mathspace.co

d Draw a Venn diagram to represent U, A, and B, placing all elements from U in the correct regions.

Apply the idea 1. Draw a rectangle for U and two overlapping circles for A and B. 2. Find the intersection A ∩ B = {3, 5, 7}. Place these numbers in the overlapping region. 3. Find elements in A only (A ∩ B′). A has {1, 3, 5, 7, 9}. Removing {3, 5, 7} leaves {1, 9}. Place these in the part of circle A that does not overlap. 4. Find elements in B only (B ∩ A′). B has {2, 3, 5, 7}. Removing {3, 5, 7} leaves {2}. Place this in the part of circle B that does not overlap. 5. Find elements in U that are not in A or B ((A ∪ B)′). A ∪ B = {1, 2, 3, 5, 7, 9}. U = {1, . . ., 10}. The remaining elements are {4, 6, 8, 10}. Place these outside the circles but inside the rectangle. A

B 3 5

6 4 8 10

Idea summary Set theory uses specific notation to describe collections of objects (sets) and their relationships. Key operations and symbols: • • • • • • •

∩ (Intersection): Elements common to both sets. ∪ (Union): Elements in either set (or both). A′ or Ac (Complement): Elements in U (Universal set) but not in A. ⊆ (Subset): All elements of the first set are in the second. ⊂ (Proper subset): Subset, but the sets are not equal. ∈ (Element of), ∉ (Not an element of). ∅ or {} (Empty set).

Venn diagrams provide a visual way to represent sets and the results of set operations, showing overlaps (intersections) and combined areas (unions) within a universal set.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Symbols in proof To prove properties of numbers or expressions, it can be helpful to use specific notation to simplify longer proofs. When referring to number systems, these symbols can be used: ℕ

Set of natural numbers (positive integers, sometimes including 0 depending on context; here assumed to be {1, 2, 3, . . .} unless specified)

ℤ

Set of integers (. . ., −2, −1, 0, 1, 2, . . .)

ℤ

Set of negative integers {. . ., −3, −2, −1}

ℤ

−

Set of positive integers {1, 2, 3, . . .} (same as ℕ in this context)

ℚ

Set of rational numbers (can be expressed as p, where p, q ∈, ℤ, q ≠ 0) q

ℚ′ ℝ ℂ

Set of irrational numbers (cannot be expressed as p; e.g., π, 2) q

Set of real numbers (includes all rational and irrational numbers) Set of complex numbers (numbers of the form a + bi, where a, b, ∈, ℝ and i = 1)

These number systems form a hierarchy, where each set is a subset of the next (except for irrationals, which are separate from rationals but part of reals): ℕ⊂ℤ⊂ℚ⊂ℝ⊂ℂ

Complex numbers

Real numbers Rational Integers Whole Natural . . .–4 –3 –2 –1 0 1 2 3 4 . . . −3 1 − 112 1.3 2.6 13 − 3.12 3 17 ... 3 2 Irrational − 102 1+ 5 π − 2 5 5

Imaginary numbers

−2

–5i

3 + 4i

–2 – 7i

Understanding these properties is crucial. For example, knowing that an integer (ℤ) is also a n rational number (ℚ) because it can be written as , and also a real number (ℝ), is important in 1 proofs. Symbols from set notation are also frequently used in proofs. These symbols include: Statement

Symbol

Is an element of

∈

Such that

: or ∣

Example x is an element of real numbers x∈ℝ

x ∈ ℝ such that x2 = 4 x ∈ ℝ ∣ x2 = 4

1.02 Symbols in proof mathspace.co

Quantifier A symbol used to indicate the quantity or range over which a statement applies. Quantifiers are used to abbreviate certain common phrases. These symbols include: Statement

Symbol

Example 2

For all

∀

There exist

∃

For all x > 1, x > 0 ∀x > 1, x2 > 0 There exists some x such that 2x = 4 ∃x : 2x = 4

While using the correct symbols when writing a mathematical proof is important, it is also important to use symbols in the proper order. Consider this example: “For all real numbers x, there exists some real number y such that x + y = 1.” Using the correct symbols and in the proper order, this statement can be written as “∀x ∈ ℝ, ∃y ∈ ℝ : x + y = 1”.

This statement is not the same as “∃y ∈ ℝ : ∀x ∈ ℝ, x + y = 1”.

When changing the order, this statement has introduced and determined a fixed element y such that it can be added to all real numbers x to always give the result x + y = 1. This is indeed false.

Example 2 Rewrite the statement using correct mathematical symbols: “For all real numbers x, there exists some real number y such that x + y = 0”.

Create a strategy Recall these symbols and their meaning:

Apply the idea

∀

For all

∈

Is an element of

: or ∣

Such that

ℝ

Real numbers

∀x ∈ ℝ, ∃y ∈ ℝ : x + y = 0

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 3 Consider the statement: “The square of any integer is non-negative.” Write this statement using mathematical symbols and quantifiers.

Create a strategy Identify the quantifier needed (“any integer” suggests “for all”). Identify the number set involved (integers: ℤ).

Translate the condition (“square is non-negative”) into a mathematical inequality. Combine using the correct symbols and structure.

Apply the idea “Any integer” translates to “For all integers x”. The symbol for “for all” is ∀, and for integers is ℤ. So the structure begins ∀x ∈ ℤ. “The square of x” is x2.

“Is non-negative” means greater than or equal to zero, which is ≥ 0. Combining these parts results in: ∀x ∈ ℤ, x2 ≥ 0

Example 4 Is the statement ∃x ∈ ℕ :

x ∈ ℕ true or false? Explain. 2

Create a strategy x Find at least one natural number x (from 1, 2, 3, . . .) such that is also a natural number. Test some 2 x small values of x to check if is a natural number. If such an x exists, the statement is true; if no 2 such x is found after testing, further exploration may be required.

Apply the idea The statement reads: “There exists a natural number x such that x divided by 2 is also a natural number.” The natural numbers (ℕ) are {1, 2, 3, . . .}. To prove existence (∃), just one example needs to be found. 1 is not in ℕ. 2 2 Try x = 2: = 1. Since 1 is in ℕ, such an x has been found. 2 Therefore, the statement is true because an example (x = 2) was found that satisfies the condition. Try x = 1:

1.02 Symbols in proof mathspace.co

Idea summary Mathematical proofs use precise symbols for efficiency and clarity. It is important to use the correct symbols and quantifiers in the correct order. Number system symbols and hierarchy: • ℕ: Natural numbers (e.g., 1, 2, 3, . . .) • ℤ: Integers (. . ., −1, 0, 1, . . .)

1 • ℚ: Rational numbers (e.g., , −3, 0.5) 2 • ℝ: Real numbers (includes ℚ and irrationals like π, 2) • ℂ: Complex numbers (e.g., 2 + 3i) • Hierarchy: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ ⊂ ℂ

Quantifiers and set notation: • • • •

∀: For all ∃: There exists ∈: Is an element of : or ∣: Such that

Understanding the properties of number systems and the meaning of quantifiers is essential for constructing and interpreting proofs.

Translate logic statements A variety of symbols are used when working with logical statements. Using symbols can make writing logical statements more efficient. The table shows useful symbols and abbreviations used in proofs: Logical statement A statement or proposition that can be determined to be either true or false.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Definition

Symbol

Read

Example

“Not” p : The plate is dirty. “Not equal to” ¬ p : The plate is not dirty.

Negation

The opposite of a statement.

Conjunction

Joins two logical statements with the word “and”.

∧

“and”

p : The plate is dirty. q : Dinner is over. p ∧ q : The plate is dirty and dinner is over.

Disjunction

Joins two logical statements with the word “or”.

∨

“or”

p ∨ q : The plate is dirty or dinner is over.

Implication

Joins two simple statements in an “if-then” format.

⇒

“if-then”

p ⇒ q : If the plate is dirty, then the dinner is over.

Equivalence

Combines both directions of an implication.

⇔

“if and only if”

p ⇔ q : The plate is dirty if and only if dinner is over.

Note: Equality (=) means two expressions have the same value (for example, 2 + 2 = 4). Equivalence (⇔) means two logical statements have the same truth value. The equivalence p ⇔ q implies that both of these implications are true: • “If the plate is dirty, then dinner is over.” ( p ⇒ q) • “If dinner is over, then the plate is dirty.” (q ⇒ p)

Example 5 Let: p : The temperature is 30°C. q : The fan is working. r : The apartment is hot. Rewrite these statements in symbolic form: a “The fan is working and the apartment is hot.”

Create a strategy

Apply the idea

Identify which simple statements are combined into this compound statement.

“The fan is working” is equivalent to q.

Use the table of logical connectives to translate the statement into its symbolic form.

The connective “and” is used, so this gives:

“The apartment is hot” is equivalent to r. q∧r

Reflect and check This is logically equivalent to r ∧ q.

1.02 Symbols in proof mathspace.co

b “The temperature is not 30°C or the fan is working.”

Apply the idea “The temperature is not 30°C” is equivalent to ¬ p. “The fan is working” is equivalent to q. The connective “or” is used, so this gives: ¬p ∨ q

c “If the apartment is hot and the fan is working, then the temperature is 30°C.”

Apply the idea Notice that the word “and” is used to join the statements “the apartment is hot” and “the fan is working”. This compound statement is joined with the statement “the temperature is 30°C” using the “if-then” connective. This means that translating the statement into the symbolic form results in: (r ∧ q) ⇒ p

d “The apartment is hot if and only if the fan is working, and the temperature is 30°C.”

Apply the idea The words “if and only if” are used to join the statements “The apartment is hot” and “the fan is working”. The word “and” is used to join this compound statement and the statement “the temperature is 30°C”. This means that translating the statement into the symbolic form results in: (r ⇔ q) ∧ p

Idea summary A variety of symbols are used to make writing logic arguments involving logical statements (propositions that can be determined to be either true or false) more efficient. These are summarised as: ¬ ∧ ∨ ⇒ ⇔

Negation Conjunction Disjunction Implication Equivalence

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

“Not” “And” “Or” “If-then” “If and only if”

1.02 Practice questions What do you remember? 1

List all the elements of these sets: a

{Positive odd numbers less than 10}

{Positive factors of 42}

{x | x is a natural number less than 5}

{x | x is an integer between −8 and −3 (not inclusive)}

If U = {1, 2, 3, 4, 5} is the universal set and set A = {1, 3, 5}, what is the complement of A, denoted A′?

Let A = {2, 4, 6} and B = {4, 8, 12}. Find: a

A∪B

Determine whether these statements are true or false: a

A ∩ B

3 ∈ {x | x is a rational number}

− 0.25 ∈ {x | x is a natural number}

If p represents the statement “The cat is sleeping,” what does ¬ p represent? A

The cat is not sleeping.

The cat and dog are sleeping.

The cat is sleeping, or the dog is sleeping.

If the cat is sleeping, then the dog is sleeping.

Let p represent “You can drive” and q represent “You have a licence”. What is the symbolic form for the equivalence statement “You can drive if and only if you have a licence”? A

p∧q

p∨q

p⇒q

p⇔q

Practice 7

Consider the Venn diagram showing sets P and Q within a universal set U: Q

P a b

e f g

U List the elements of these sets: a

P∩Q

P∪Q

P′

(P ∪ Q)′

1.02 Symbols in proof mathspace.co

8 Ex 1

Let the universal set be U = {Integers from 1 to 12}. Let A = {Even numbers in U} and B = {Multiples of 3 in U}: a

List the elements of these sets: A∩B

i b

A∪B

iii

A′

Draw a Venn diagram representing these sets. Place all integers from 1 to 12 in the correct regions of the diagram.

On the Venn diagrams provided, shade the region corresponding to each set expression: a

A ∩ B′

(A ∪ B)′

A B

Let A and B be two sets. Write the set notation for the elements that are in set A or set B, but not in both.

Complete this table to indicate whether each number belongs in the number set (‘Y’ for Yes, ‘N’ for No):

5 −3 3 4 5 1 12

Ex 2

14 Ex 3

ℤ

ℚ

ℕ⊂ℝ

ℝ

⬚

⬚ ⬚ ⬚

⬚

Determine whether these statements are true or false: a

ℕ

ℤ+ ⊆ ℤ

Express each statement using symbols:

⬚ ⬚ ⬚

ℚ⊆ℝ

There exists an integer x such that x2 = 4.

For all positive integers n, the value of n2 is greater than or equal to n.

For all integers n, the square of n is greater than or equal to 0.

ℤ⊆ℕ

Rewrite these statements using mathematical symbols and quantifiers, specifying the appropriate number set for each variable: a

The cube of every real number is a real number.

There exists a positive integer whose square is greater than 100.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Translate the statement into words.

Determine whether each statement is true or false.

∃x ∈ ℝ | x2 = 5

c 16 Ex 4

∃x ∈ ℝ | x2 < 0

∀n ∈ ℤ+ | 2n + 1 is odd. ∀x ∈ ℝ, x3 ≥ 0

Determine whether the statement ∃x ∈ ℚ : x2 = 2x is true or false. Explain your reasoning. Consider the statement: ∀n ∈ ℤ, n is even. a

Write the negation of this statement in symbols.

Determine whether the original statement or the negation is true.

Given the statements “p: It is snowing” and “q: The school is open”, what does the statement p ⇒ ¬ q mean?

Let p, q, r be defined as: p : It is raining.

Ex 5

q : The sky is cloudy. r : It is windy. Rewrite these statements in symbolic form:

“The sky is cloudy or it is windy.”

“It is not raining and the sky is cloudy.”

“If it is raining, then the sky is cloudy and it is windy.”

“It is windy if and only if it is not raining or the sky is cloudy.”

An argument claims that if a condition p is true, then another condition q follows. It also states that condition r is true. Which of these symbolically represents the argument that “If p, then q is true, and also r is true”? A

p ⇒ (q ∧ r)

p ∧ (q ⇒ r)

( p ∧ q) ⇒ r

( p ⇒ q) ∧ r

Let p represent “The sky is blue” and q represent “The grass is green”. Which of these symbolically represents the argument that “It is not the case that both the sky is blue and the grass is green”? A

¬ ( p ∧ q)

p∨¬q

p⇒q

p∧q

Translate this logic statement into words, where p, q, and r are simple statements: ¬ ( p ∧ q) ⇒ r

Translate the symbolic statement ¬ p ∨ p into words and explain why it is always true.

1.02 Symbols in proof mathspace.co

Extend your thinking 24

Consider the sets A = {x ∈ ℝ | x2 ≤ 4} and B = {x ∈ ℝ | x > 0} within the universal set U = ℝ.

Express A ∩ B in interval notation, and represent the intersection on a number line diagram.

Consider the statement: “There exists a prime number p such that p + 2 is also a prime number.” a

Write the statement using mathematical notation.

Determine whether the statement is true or false. Justify your answer using properties of number sets or give an example.

Consider the statement: ∀x ∈ ℝ, if x > 0 then x2 > x. Determine whether the statement is true or false. Justify your conclusion.

For the statement: ∃x ∈ ℚ : ∀y ∈ ℤ, x + y ∈ ℚ

Translate the statement into words and explain its meaning.

Determine whether the statement is true or false. Justify your answer using properties of number sets or give an example that shows the statement is false.

Let: p : The number n is even. q : The number n is divisible by 3. r : The number n is prime. Consider the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}:

Translate the statement “If a number is even and divisible by 3, then it is not prime” into symbolic form. Then, identify the set of numbers in U for which this implication is true.

Draw a Venn diagram representing the sets P = {n ∈ U∣p}, Q = {n ∈ U∣q}, and R = {n ∈ U∣r}. Place all numbers from U into the correct regions.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1.03 Direct proof After this lesson, you will be able to… • prove statements using propositions and connectives • construct valid direct proofs for simple mathematical claims • prove statements using the contrapositive method • apply appropriate deductive reasoning steps in a logical sequence

Proof structure and deductive reasoning Conjecture

Deductive reasoning

An unproven statement, often in the form of an implication or equivalence.

A logical process where conclusions are derived from agreed-upon premises or previously proven facts.

Proof A logical argument that establishes the truth of a conjecture. A statement can be shown to be false through a disproof, often by finding a counter-example.

A proof relies on deductive reasoning, where conclusions are logically derived from agreed-upon premises or previously proven facts. Proofs are often laid out, line by line in a two-column format, with statements on the left and reasoning on the right. For example: a>b b>c ∴a>c

(Given) (Given) (Conclusion from transitive property)

The reason for moving from one step to the next is often a known mathematical law such as transitive property or commutative property, a definition, or a previously established theorem. Each step must logically follow from the previous steps and accepted facts. Line by line proofs may also be done using logic symbols, showing the application of specific rules of inference: p p→q q q → ¬r ¬r

(Given (Premise 1)) (Given (Premise 2)) (Modus Ponens (from lines 1, 2)) (Given (Premise 3)) (Modus Ponens (from lines 3, 4))

Proofs can also be written in clear, concise paragraphs explaining the logical steps taken to reach the conclusion.

1.03 Direct proof mathspace.co

One common proof strategy is direct proof. This involves starting directly from the hypothesis (the ‘if’ part, or given information) and using a sequence of logical deductions to arrive at the conclusion (the ‘then’ part). The basic steps for a direct proof of an implication “If p, then q” are: 1. Assume the hypothesis p is true. 2. Use definitions, axioms, established theorems, and rules of logic to deduce a chain of statements leading towards q. 3. Conclude that q must be true whenever p is true. Direct proofs are often used to prove universal statements (statements involving “for all” or ∀). A universal statement may also be proven false by finding a single counter-example. More generally, a statement can be disproven by proving its negation is true.

Exploration Consider the two conjectures: • Conjecture 1: ∃n ∈ ℤ+ : 2n – 1 is a prime number.

• Conjecture 2: ∀n ∈ ℤ+, 2n – 1 is a prime number.

1. Compare and contrast the two conjectures. What do the symbols ∃ and ∀ mean? 2. Determine a value of n where 2n – 1 is a prime number. What does this mean for the truth of conjecture 1 and conjecture 2? 3. Determine a value of n where 2n – 1 is not a prime number. What does this mean for the truth of conjecture 1 and conjecture 2? 4. What kind of proof approach (finding an example vs. a general argument) is needed for each type of quantifier (∃ vs ∀)?

Example 1 Given these premises: 1. All squares are rectangles. 2. Shape X is a square. Use deductive reasoning to determine a valid conclusion about shape X and present the argument in a two-column format.

Create a strategy The premises provide a general rule (All squares are rectangles) and a specific case (Shape X is a square). Identify the general rule (‘All squares are rectangles’ as a universal statement) and apply it to the specific case. A two-column format will be used, listing statements and their justifications (reasons).

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea All squares are rectangles

(Premise 1 (Given general rule))

Shape X is a square

(Premise 2 (Given specific case))

Therefore, Shape X is a rectangle

(Modus Ponens (from lines 1, 2))

The valid conclusion is that shape X is a rectangle.

Reflect and check This example illustrates basic deductive reasoning. Accepted facts (premises) were used to derive a conclusion that logically follows. The two-column structure clearly shows each statement and the reason for its validity within the argument.

Idea summary A proof uses deductive reasoning to show a statement (conjecture) is true, starting from a hypothesis and reaching a conclusion through logically justified steps. Proofs can be presented in different formats, like two-column or paragraph style. The validity of each step relies on definitions, axioms, or previously proven results. Proving statements with “for all” (∀) requires a general argument, while statements with “there exists” (∃) only require finding one confirming example. Finding one case where a “for all” statement fails (a counter-example) disproves it.

Direct proof Direct proof A method of proving a conditional statement (“If p, then q”) by assuming p is true and logically deriving q (thus proving p ⇒ q). The strategy is straightforward: Assume that the hypothesis p is true. Then, use definitions, postulates, and previously proven theorems to form logically connected steps to show that the conclusion q must also be true. This method directly connects the starting assumption ( p) to the desired outcome (q) through a sequence of deductive steps. It often involves algebraic manipulation or applying definitions related to the properties being discussed (like even/odd numbers, divisibility, etc.).

1.03 Direct proof mathspace.co

Example 2 Consider the statement: “∀n ∈ ℤ+ : n > 1, 2n is not prime.” a Write the statement in words.

Create a strategy Translate each symbol into its corresponding English phrase.

Apply the idea ∀ means “For all”. ∈ ℤ+ means “in the set of positive integers”. “:” can mean “such that”.

The statement reads: “For all positive integers n such that n is greater than 1, the number 2n is not prime.”

b Prove the statement using direct proof.

Create a strategy The hypothesis is n ∈ ℤ+ and n > 1. The conclusion is 2n is not prime.

A number is not prime (i.e., composite) if it can be written as a product of two integers, both greater than 1. It needs to be shown that for any integer n > 1, 2n can be factored in this way.

Apply the idea Assume the hypothesis: Let n be a positive integer such that n > 1. Since n is an integer and n > 1, this means n ≥ 2. Consider the expression 2n. The goal is to show it is not prime. 2n = 2 × 2n–1

Using index laws (am × ak = am+k, here n = 1 + (n – 1))

The expression 2n has been written as a product of two factors: 2 and 2n–1 Now, it is necessary to check if both factors are integers greater than 1. • The first factor is 2. Since 2 is an integer and 2 > 1, this condition is met. • The second factor is 2n–1. Since n ≥ 2, the exponent n – 1 ≥ 1. Therefore, 2n–1 will be 21 = 2, 22 = 4, 23 = 8, etc. All these values are integers greater than or equal to 2. Therefore, 2n–1 > 1 for all n > 1. Since 2n can be written as the product of two integers 2 and 2n–1 , both of which are greater than 1, 2n is not a prime number by definition. Therefore, the statement 2n is not prime for all positive integers n greater than 1 is proven true by direct proof.

Reflect and check The direct proof started with the assumption n > 1 and used definitions (prime numbers, index laws) to directly show the conclusion (2n is not prime).

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary Direct proof proves a statement “If p, then q” by assuming p is true and logically deducing that q must also be true. It directly follows the implication from hypothesis to conclusion using definitions, axioms, and established theorems.

Proof by contrapositive Proof by contrapositive An indirect method of proving a conditional statement (“If p, then q”) by proving its logically equivalent contrapositive (“If not q, then not p”). Sometimes, proving a conditional statement “If p, then q” ( p ⇒ q) directly can be difficult. An alternative method is proof by contrapositive. The contrapositive of the statement “If p, then q” is “If not q, then not p” (¬ q ⇒ ¬ p). A crucial concept in logic is that a conditional statement and its contrapositive are logically equivalent. This means they always have the same truth value. If one is true, the other is true; if one is false, the other is false. Therefore, to prove p ⇒ q is true, its contrapositive ¬ q ⇒ ¬ p can be proven true instead, using a direct proof. The steps for a proof by contrapositive are: 1. Identify the original statement: p ⇒ q 2. Form the contrapositive: ¬ q ⇒ ¬ p 3. Assume the hypothesis of the contrapositive (¬ q) is true. 4. Use definitions, axioms, established theorems, and rules of logic to deduce that the conclusion of the contrapositive (¬ p) must be true. 5. Conclude that since the contrapositive (¬ q ⇒ ¬ p) is true, the original statement ( p ⇒ q) must also be true due to logical equivalence. This method is particularly useful when the negation of q provides a more concrete starting point for the proof than p itself.

1.03 Direct proof mathspace.co

Example 3 Prove this statement using proof by contrapositive: “If n2 is an even integer, then n is an even integer.”

Create a strategy 1. Identify p and q in the statement “If p, then q”. p: n2 is an even integer. q: n is an even integer. 2. Form the contrapositive statement “If not q, then not p”. Not q: n is not an even integer (i.e., n is an odd integer). Not p: n2 is not an even integer (i.e., n2 is an odd integer). Contrapositive: “If n is an odd integer, then n2 is an odd integer.” 3. Prove the contrapositive using a direct proof. Assume “not q” is true and show “not p” must be true.

Apply the idea The statement to prove is: “If n is an odd integer, then n2 is an odd integer.” Assume the hypothesis of the contrapositive: Let n be an odd integer. By definition, an odd integer can be written in the form n = 2k + 1 for some integer k. Now, n2 can be found: n2 = (2k + 1)2 2

= 4k + 2k + 2k + 1 2

= 4k + 4k + 1 2

= 2 (2k + 2k) + 1

Substitute the definition of odd n Expand the product Combine like terms Factorise 2 from the first two terms

Let m = 2k2 + 2k. Since k is an integer, k2 is an integer, 2k2 is an integer, and 2k is an integer. The sum of two integers (2k2 + 2k) is also an integer. Therefore, m is an integer. So, n2 can be written as 2m + 1, where m is an integer. By definition, this means that n2 is an odd integer. It has been successfully shown that assuming n is odd leads to the conclusion that n2 is odd. Therefore, the contrapositive is true, so the original statement is true by logical equivalence. Conclusion: Since the contrapositive is true, the original statement (“If n2 is even, then n is even”) is also true by logical equivalence.

Reflect and check Proving the contrapositive was easier because assuming n is odd gives a specific algebraic form (2k + 1) to work with directly. Trying to prove the original statement directly by assuming n2 is even (n2 = 2k) is harder to manipulate to show n must be even.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary Proof by contrapositive is an indirect proof technique used to prove a conditional statement “If p, then q” ( p ⇒ q). It works by proving the logically equivalent contrapositive statement: “If not q, then not p” (¬ q ⇒ ¬ p). The steps involve assuming ¬ q is true and using direct proof to show that ¬ p must also be true.

1.03 Practice questions What do you remember? 1

3 4

The statement “If p, then q” is logically equivalent to its contrapositive. What is the contrapositive of this statement? A

If q, then p.

If not p, then not q.

If not q, then not p.

p and not q.

To disprove the statement ∀x ∈ ℝ, if x > 1 then x2 > x, what is required? A

A single example where the statement holds true.

A single counter-example where the premise is true but the conclusion is false.

A formal proof showing the statement is always false.

A proof of the contrapositive.

What is required to prove the existential statement ∃x ∈ ℤ : x2 = 25?

What is required to disprove the universal statement ∀n ∈ ℤ, n2 > n?

Practice 5 Ex 1

Consider these premises: 1.

All multiples of 6 are even numbers.

2. 18 is a multiple of 6. Using deductive reasoning, what is the logical conclusion based on these premises?

1.03 Direct proof mathspace.co

Consider this proof that if n2 is even, then n is even. Proof (by contrapositive): 1

Assume the negation of the conclusion: Assume n is not even (i.e., n is odd).

Then n = 2k + 1 for some integer k.

Squaring both sides gives n2 = (2k + 1)2 = 4k2 + 4k + 1.

This can be written as n2 = 2 (2k2 + 2k) + 1.

Let m = 2k2 + 2k. Since k is an integer, m is an integer.

So, n2 = 2m + 1, which means n2 is odd.

This is the negation of the premise (n2 is even).

Therefore, by contraposition, if n2 is even, then n is even.

What definition or property justifies the statement in Step 2? 7

Prove using direct proof that for all integers n, if n is odd, then n + 1 is even.

Ex 2

Prove using direct proof that for any integer n, if n is a multiple of 4, then n2 is a multiple of 16.

Prove the statement: There exists n ∈ ℤ+ such that 3n2 – 1 is a prime number.

10 Ex 3

11 12 13

Prove using proof by contrapositive: If n2 is a multiple of 3, then n is a multiple of 3 (for any integer n). Prove using proof by contrapositive: For any integer a, if a2 + 1 is odd, then a is even. 1 Prove or disprove the statement: ∀x ∈ ℝ where x > 0, x ≥ . x

Prove or disprove the statement: There exists x ∈ ℝ such that x2 – 8x + 17 = 0.

Extend your thinking 14

Prove that the sum of any two rational numbers is always a rational number.

Prove that for any positive integer n,

Prove the statement: For all integers n, if n3 – n is odd, then n is even.

Consider the quadratic function f (n) = an2 + bn + c, where a, b and c are natural numbers with c ≥ 2. Prove that there exists at least one natural number n such that f (n) is composite (not prime).

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

n 1 n > . n n 1

1.04 Proofs by contradiction After this lesson, you will be able to… • understand the concept of proof by contradiction • identify situations where proof by contradiction is appropriate • construct proofs using contradiction • verify the validity of contradiction proofs

Proof by contradiction Proof by contradiction is an indirect proof method. It works by assuming the statement intended for proof is actually false (assuming its negation is true). Then, it is shown that this assumption leads to something impossible or illogical - a contradiction. Because the assumption leads to a contradiction, the original statement must have been true all along. Contradiction A situation where two statements or pieces of information, derived from a logical argument or assumption, are mutually exclusive and cannot both be true simultaneously. Finding a contradiction often invalidates the initial assumption. A contradiction happens when logical steps lead to the conclusion of two things that can’t possibly be true together (like a number being both even and odd, or a statement being both true and false). Suppose it is necessary to prove that statement p is true. To prove a statement p is true, the method follows these steps: 1. Begin by assuming that p is false, which makes ¬ p true. 2. Then, using logical steps based on that assumption, work towards finding a contradiction. If one is found, it means the initial assumption (that p was false) must be wrong. 3. If successful, then it can be concluded that p is true. To prove an implication, p ⇒ q, is true, the steps are: 1. Assume the negation is true: p and ¬ q 2. Show that this assumption leads to a logical contradiction. 3. Conclude that the assumption p “and” ¬ q must be false, therefore the original statement p ⇒ q must be true.

1.04 Proofs by contradiction mathspace.co

Exploration Consider the statement: “There is no smallest positive rational number.” 1. How might the proof begin if attempted directly? What makes it difficult to show that no such number exists among infinitely many possibilities? 2. Now, consider the opposite: “There is a smallest positive rational number.” Does assuming this opposite give a more solid starting point for the argument? 3. This difficulty in proving non-existence directly often motivates the use of indirect proof methods like proof by contradiction.

Proof by contradiction is particularly useful in certain situations: • Proving something doesn’t exist: Sometimes it’s hard to directly show something isn’t possible everywhere. It can be easier to assume it does exist and then show that this leads to an impossible situation (an absurdity). For example, proving ‘there is no largest prime number’. p • Proving a number is irrational: To show a number can’t be written as a fraction like , it often q starts by assuming it can be written as a fraction, and then showing this leads to a contradiction (like proving 2 is irrational). • When the direct approach is tricky: If proving something directly is difficult, assuming the opposite might provide a clearer starting point with equations or properties to work with. • When negating the conclusion of an implication is helpful: For proving “If P, then Q”, assuming “P and not Q” might be easier to work with than starting with just “P”. To ensure a proof by contradiction is valid, verifying a few key points is important: • Correct negation: Has the opposite (the negation) of the original statement been correctly stated as the starting assumption? • Logical steps: Do all the steps in the argument follow logically from the assumption, using correct and valid mathematical rules? • Real contradiction: Is the final conclusion genuinely impossible? It should contradict the initial assumption, a known mathematical fact (like 1 = 0), or be logically impossible (like a number being both even and odd, or P ∧ ¬ P).

Example 1 Prove by contradiction that there are no integers x and y such that x2 = 4y − 2.

Create a strategy The process starts by writing the negation of the statement, then using valid arguments to show that a contradiction arises. Proving non-existence directly can be difficult, so proof by contradiction is a good approach. It is assumed such integers do exist and a logical inconsistency is sought.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea Suppose that there exist integers x and y such that x2 = 4y −2. Consider the given equation, x2 = 4y − 2. Factorise 2 from 4y − 2: x2 = 2 (2y − 1) This implies that x2 is an even number because it is the product of 2 and another number. Since x is an integer, x2 is also an integer. If x2 has a factor of 2, then x must have a factor of 2. Therefore, x is an even number. Since x is even, it can be written in the form 2M, where M is an integer. This gives: (2M)2 = 2 (2y − 1) Expand and simplify to give: 4M2 = 2 (2y − 1) 2M2 = 2y − 1 Here, the left-hand side is even while the right-hand side is odd. Since an integer cannot be both odd and even, this is a contradiction. Therefore, the original statement is true.

Reflect and check The contradiction reached is that the integer 2M2 (which must be even) equals the integer 2y – 1 (which must be odd). An integer cannot be both even and odd. This confirms the validity of the contradiction and therefore the proof.

Example 2 Prove that the geometric mean of two positive numbers, ab, is always less than or equal to their a b arithmetic mean, . 2

Create a strategy To prove this inequality, use a proof by contradiction. The proof starts by assuming the opposite, that the geometric mean is greater than the arithmetic mean, and then showing that this assumption leads to a contradiction. a b allows algebraic steps (like squaring both sides) 2 that might lead to a clearer contradiction than trying to prove the original inequality directly for all positive numbers. Assuming the opposite inequality

1.04 Proofs by contradiction mathspace.co

Apply the idea Beginning with the assumption, derive a contradiction: ab >

a b 2

Write the assumption

2 ab > a + b

Multiply both sides by 2 2

4 ab > (a + b)

Square both sides

4 ab > a + 2ab + b

Expand the squared binomial

Subtract 4 ab from both sides

0 > a – 2ab + b 2

0 > (a – b)

Factorise the trinomial

This final statement, 0 > (a – b) , contradicts the known property that the square of any real number is always non-negative (greater than or equal to zero). Therefore, the initial assumption a b that ab must be false. 2 Since the assumption led to a contradiction, it is proven that the geometric mean ab is always a b less than or equal to the arithmetic mean for any positive numbers a and b. 2

Example 3 Prove that there is no least positive rational number.

Create a strategy To prove this, use a proof by contradiction. The assumption is made that there exists a least positive rational number, and then this assumption is shown to lead to a contradiction. Trying to directly prove that something does not exist within an infinite set like positive rational numbers is challenging. Assuming the opposite (that a least positive rational does exist) provides a specific element, R, which can be used in the argument.

Apply the idea Beginning with the assumption, derive a contradiction: Suppose that R is the least positive (smallest) rational number. R=

m n

Write the assumption

R m R = Form the number 2 2n 2 R Observe that is a positive rational number that is smaller than R, because 2n > n, which implies 2 that: m m 2n n

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

This result contradicts the assumption that R is the least positive rational number, because a R smaller positive rational number, , was constructed, violating the assumption that R is the 2 smallest. Therefore, the assumption must be false, and there cannot be a least positive rational number.

Reflect and check The contradiction is that R was assumed to be the smallest positive rational number, but another R smaller positive rational number was constructed. This violates the “smallest” condition of the 2 assumption, validating the proof.

Idea summary Proof by contradiction is an indirect method. Prove a statement p by following these steps: 1. Assume ¬ p (the negation of p) is true. 2. Show that this assumption leads to a logical contradiction. 3. Conclude that the assumption (¬ p) must be false, meaning the original statement ( p) must be true. For proving an implication ( p ⇒ q): Assume p is true and ¬ q (the negation of q) is true, then derive a contradiction. This method is often useful for proving non-existence, irrationality, or when assuming the opposite provides a useful starting point. Verifying the proof involves checking that the negation was stated correctly and that the final contradiction is mathematically sound.

1.04 Practice questions What do you remember? 1

Identify the negation of each of these statements: a

The product of three consecutive integers is even.

There is no odd integer that can be expressed as the sum of three even integers.

There is no largest prime number.

If x3 is even, then x is even.

If it is raining, then it is cloudy.

1.04 Proofs by contradiction mathspace.co

In each set, determine the two statements which contradict each other: a

AB + CD = EF

ii iii

△ABC is a right-angled triangle.

AB > EF

△ABC is an isosceles triangle.

CD < AB

iii

△ABC is a scalene triangle.

All dogs are mammals.

i The sum of x and y is even.

The product of x and y is even.

Some dogs are reptiles.

iii

x is even and y is odd.

iii

Golden Retrievers are dogs.

To prove the implication “If p, then q” by contradiction, what is the correct initial assumption? A

Assume p is false and q is true.

Assume p is true and q is false.

Assume p is false and q is false.

ssume the contrapositive, “If not q, then A not p”, is true.

Which of these would be a valid contradiction to reach at the end of a proof? A

A statement that is always false (e.g., 1 = 0).

A statement that contradicts one of the initial assumptions (e.g., proving a number is odd after assuming it was even).

A statement of the form P ∧ ¬ P.

All of the above.

Practice 5

Consider the statement: ‘If two integers are odd, then their product is also odd’. This statement will be proved using contradiction: a

Identify the assumptions that should be made for this proof.

Write a proof by contradiction for the statement.

Consider the statement: If 3n + 7 is even, then n is odd. a

Determine the negation of this statement.

Prove the statement by contradiction.

Prove by contradiction that there exists no integers a and b for which 18a + 6b = 1.

Ex 1

Prove that there are no integers m and n such that 2m + 6n = 7.

Prove that the harmonic mean,

Ex 2

their arithmetic mean,

a b . 2

2ab , of two positive numbers is always less than or equal to a b

Prove that for any two positive numbers, a and b, this inequality holds: a3 + b3 ≥ ab (a + b)

If there are 400 people attending a festival, prove that at least two people must share a birthday.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Prove that the statement “if a solid has four faces, then it is not a prism” is true, using proof by contradiction.

Prove by contradiction that there is no largest negative integer.

Ex 3

Consider these statements: • For all integers n, n2 + n is even. • There are no integers x, y such that 6x + 9y = 1. • The sum of the angles in any triangle is 180°.

Which statement is most suitable for proof by contradiction?

Explain why proof by contradiction is a suitable strategy for the chosen statement.

Analyse this proof that claims to show 4 is irrational using contradiction. Identify if the proof is valid or flawed, and explain why. Proof: p 1 Assume 4 is rational. Then 4 where p, q are integers with no common factors, q and q ≠ 0. 2 Squaring both sides gives 4

p2 2

, so 4q2 = p2.

q 2 3 This means p is a multiple of 4, so p must be a multiple of 2. Let p = 2k for some integer k.

2 2 2 2 2 4 Substituting back gives 4q = (2k) = 4k , so q = k . 5 This means q must also be an integer.

6 Integers p and q are found to satisfy the condition. This is a contradiction. Therefore, 4 must be irrational. 16

Prove by contradiction: a

The square of an even integer is always even.

The sum of two odd numbers is even.

The sum of an even number and an odd number is odd.

The product of two rational numbers is rational.

The sum of two rational numbers cannot be irrational.

There is no largest even positive integer.

Show that there cannot be an integer that is both even and odd.

Extend your thinking 18 19 20

Prove that if a, b ∈ ℤ, then a2 − 4b ≠ 2. Prove that 5 is an irrational number.

For the conjecture ∃x, y ∈ ℤ+: x2 = 4y + 2, use proof by contradiction to show that this is impossible. 1.04 Proofs by contradiction mathspace.co

1.05 Proofs by counter-example After this lesson, you will be able to… • understand the role of counter-examples in disproving statements • construct appropriate counter-examples • evaluate the validity of counter-examples • use counter-examples effectively in mathematical arguments

Use examples to prove existence Using examples is a valid proof method only for ‘there exists’ (∃) statements, or when checking every possibility in a finite set. This is called proof by example. Proof by example A method of proving a statement by providing a specific instance that satisfies the given conditions. This method is only valid for proving “existence” statements or verifying statements over finite domains. Example: Proving “There exists an even prime number” by showing the number 2. For “there exists” statements, providing just one specific instance that satisfies the conditions is sufficient to establish the truth of the statement. For example, to prove “There exists an even prime number,” showing the number 2 is both even and prime is enough. Similarly, for statements about a limited set (a finite domain), checking every single case is a valid proof. For example, ∀n ∈ ℤ+ : n < 5, n – 5 < 0

can be proven by checking n = 1, 2, 3, and 4.

Important limitation: Examples, no matter how many are found, cannot prove a general statement or conjecture that applies universally (often indicated by “for all” or ∀) over an infinite domain. Examples can illustrate or suggest a pattern, but they do not provide logical certainty for all cases. To establish the validity of general statements, formal proof methods like direct proof, proof by contradiction, or proof by induction are required. Using examples to try and prove a general statement is a common logical fallacy.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Exploration Consider the conjecture: “All odd numbers greater than 1 are prime.” 1. Find the first three odd numbers greater than 1. Are they prime? Do these examples support the conjecture? 2. Can an odd number greater than 1 be found that is not prime? (Hint: Think about composite numbers). 3. What does finding this one non-prime odd number indicate about the original conjecture? 4. Discuss: Why can finding examples like 3, 5, and 7 not prove the conjecture is true for all odd numbers, while finding just one example like the one in question 2 disproves it?

Example 1 Prove by example that there exists a positive integer n such that n3 + 3n2 + 3n + 1 is a perfect square.

Create a strategy Test values for n, starting from the smallest positive integer, until an example is found where the expression is a perfect square.

Apply the idea A logical strategy is to start at the smallest positive integer: •

Let n = 1: n3 + 3n2 + 3n + 1 = 13 + 3(12) + 3(1) + 1 = 1 + 3 + 3 + 1 = 8 (not a perfect square)

•

Let n = 2: n3 + 3n2 + 3n + 1 = 23 + 3(22) + 3(2) + 1 = 8 + 12 + 6 + 1 = 27 (not a perfect square)

•

Let n = 3: n3 + 3n2 + 3n + 1 = 33 + 3(32) + 3(3) + 1 = 27 + 27 + 9 + 1 = 64 (82, which is a perfect square)

Therefore, the statement “there exists a positive integer n such that n3 + 3n2 + 3n + 1 is a perfect square” has been proven with one example where n = 3.

1.05 Proofs by counter-example mathspace.co

Idea summary Proof by example is a valid method only for demonstrating the truth of “there exists” (∃) statements or for verifying statements over finite domains by checking all cases. It is not sufficient to prove universal (“for all” or ∀) statements over infinite domains; formal proofs are needed for those.

Disprove by counter-example A conjecture is a statement that is believed to be true, but has not been proven. While some conjectures can be proven true using formal proofs, others can be proven false by finding a single counter-example. Counter-example A specific instance or example that shows that a general statement or conjecture is false. It is used to disprove statements that are invalid. Example: For the statement “All birds can fly,” a penguin is a counter-example. Counter-examples are powerful because while proving a statement true often requires a formal proof, proving it false only requires finding a single instance where it fails. In mathematics, there are many conjectures that seemed true for a very long time, but with the growth in computing power, counter-examples have been found. For example, Leonhard Euler (1707-1783) conjectured that it was impossible to determine integers a, b, c, d and e that satisfied the equation a5 + b5 + c5 + d5 = e5. However, about 200 years later, in 1966, a counter-example was found: 275 + 845 + 1105 + 1335 = 1445 When using a counter-example, it is crucial to ensure it is valid. A valid counter-example must genuinely satisfy the conditions (hypothesis) of the statement but fail to meet the conclusion, or directly demonstrate the falsity of the claim. Careful checking is needed to ensure the counter-example is not based on a calculation error or misunderstanding.

Exploration Prime number of the form 2n – 1, where n is a positive integer, is called a Mersenne prime. Consider the conjecture: “All numbers of the form 2n – 1, where n is a positive integer, are prime.” 1. Test the conjecture for n = 2, 3, 5, and 7. Are the results prime numbers? 2. Test the conjecture for n = 4. What is the result? Is it prime? 3. What does the result for n = 4 indicate about the original conjecture? How does this illustrate the power of a counter-example?

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 The statement “If an animal is a shark, then it is a fish” is true. Prove that the converse is false.

Create a strategy First, determine the converse, then identify a counter-example. Since this is purely a contextual problem, the counter-example will involve selecting an animal that satisfies the hypothesis but not the conclusion. It is important to note that a valid counter-example for “If P then Q” must be a case, where P is true (the animal is a fish) but Q is false (the animal is not a shark).

Apply the idea The converse is “If an animal is a fish, then it is a shark.” Consider an animal such as a goldfish, which is a fish but not a shark. This one counter-example proves the statement “If an animal is a fish, then it is a shark” is false.

Reflect and check The contrapositive statement would be true: “If an animal is not a fish, then it is not a shark”.

Example 3 Consider the conjecture that the function P (n) = n2 – n + 41, where n is a positive integer, outputs values that are all primes. Prove that this is false for n ≥ 41.

Create a strategy If the boundary n ≥ 41 was not given, testing the function for many integers would be needed, possibly using software. Since the boundary of 41 is given, it is only necessary to verify that P (41) is not a prime number to disprove the conjecture.

Apply the idea Evaluate the function for P (41), and show that it is not prime. P (n) = n2 – n + 41 2

P (41) = 41 – 41 + 41

Write the equation Substitute n = 41 into the equation

= 41 (41 – 1 + 1)

Factorise 41 from each term

= 41 × 41

Evaluate the expression in brackets

= 1681

Evaluate the multiplication

This shows that P (41) = 1681, is not prime. This counter-example proves that the conjecture P (n) = n2 – n + 41 outputs prime numbers for all positive integer n is false.

1.05 Proofs by counter-example mathspace.co

Reflect and check To ensure this is a valid counter-example, it was confirmed that for n = 41, the result P (41) = 1681 is indeed not prime (it is 41 × 41). Simply finding a value is not enough; it must be shown to violate the conjecture.

Idea summary A counter-example is an example that shows a statement is false. While numerous examples cannot prove a universal statement, one valid counter-example is sufficient to disprove it.

1.05 Practice questions What do you remember? 1

Under what specific condition is “proof by example” a valid method for proving a mathematical statement is true? A

For proving any statement that involves integers.

For proving a universal statement (indicated by “for all” or ∀).

For proving an existential statement (indicated by “there exists” or ∃).

For proving any statement that has at least one example that works.

What is the minimum requirement to disprove a universal statement such as “∀n ∈ ℤ, P (n) is true”?

To be a valid counter-example for the implication “If p, then q”, complete this statement: The case must make p ⬚ and q ⬚.

Practice 4

Prove by example that there exists a positive integer n such that 5n + 3 is an odd integer.

Consider the statement: ∃n ∈ ℤ+ : n is even ∧ n < 10.

Write this expression in words.

Prove the statement using examples.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Consider this statement: ∀n ∈ ℤ+ : n < 5, n2 + 3 < 20. a

Write this expression in words.

Prove the statement using examples.

Which value for x is a counter-example to this conjecture? “For all positive values of x, x3 ≥ x.” A

– 1.0

– 0.1

0.1

1.0

Prove by example that there exists a positive integer n such that n2 – 1 is divisible by 3.

Ex 1

Determine whether each statement is true or false. If it is false, give a counter-example: a

All triangles are equilateral.

All squares are rectangles.

If I have a part-time job, then I am part of the workforce.

If a person goes into a school, then they are a student.

Disprove the conjecture: “Adding 3 to any even number results in another even number” using two counter-examples.

Disprove the conjecture: “Multiplying any two numbers always results in an even number” using a counter-example.

Disprove the conjecture: “If the product of the digits of a number is even, then the number itself is even” using a counter-example.

Sarah claims the statement ‘All multiples of 3 are odd’ is false. She offers the number 10 as a counter-example. Is Sarah’s counter-example valid? Explain your reasoning.

For the true implication, “If a food contains dairy, then it is not vegan.”:

Ex 2

Determine the converse statement.

Prove the converse is false.

Consider the statement: “If x = 2, then x2 = 4.” a

Determine whether this statement is true. If not, provide a counter-example.

Determine whether the converse of this statement is true. If not, provide a counter-example.

Determine whether the inverse of this statement is true. If not, provide a counter-example.

Determine whether the contrapositive of this statement is true. If not, provide a counter-example.

1.05 Proofs by counter-example mathspace.co

Disprove the statement: “Every number of the form n3 + 2 is prime” using a counter-example.

Ex 3

Prove or disprove these statements by showing justification or providing a counter-example: a

All even numbers greater than 2 are composite.

All numbers ending in 0 are composite.

The sum of any two prime numbers is always even.

The number immediately after a multiple of 5 is always a multiple of 6.

Adding 4 to any multiple of 7 will always result in a multiple of 11.

Prove or disprove these conjectures by showing justification or providing a counter-example: a

If a number is prime, then the number plus 6 is also prime.

If a number is prime, then doubling it will also result in a prime number.

Extend your thinking 19

Write a contextual implication (a ⇒ b) which is true, but has a converse that is not true. Give a counter-example to prove that the converse is not true.

Disprove the statement: “There exists n ∈ ℕ : n2 + 3n + 2 is a prime number.”

Did you know?

When you slice an apple into equal pieces to share, you’re already using mathematical ideas of divisibility and fractions! Proof and number concepts help us understand how to divide things fairly and logically, showing that even everyday actions can demonstrate the power and consistency of mathematics.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1.06 Proof by induction After this lesson, you will be able to… • understand the concept of proof by induction • identify situations where proof by induction is appropriate • construct proofs using induction

Proof by induction Exploration Consider the sum of the first few positive odd integers: • 1=? • 1+3=? • 1+3+5=? • 1+3+5+7=? A proposed formula for the sum of the first n odd positive integers is n2. 1. Verify if this formula holds for n = 1, 2, 3, 4 using the given sums. 2. Consider this visual and explain why this formula is true: 9 7 5 3 1 A₁

A₂ A₃ A₄

A₅

3. Explain why the kth positive odd integer is 2k – 1. 4. Assume the formula is true up to the kth positive odd integer, so: 1 + 3 + 5 . . . + (2k – 1) = k2 Using the assumption, show that it is true for (k + 1)th positive odd integer, so show that: 1 + 3 + 5 + 7 + . . . + (2k – 1) + (2(k + 1) – 1) = (k + 1)2 5. The formula holding for the current value, k, means that it holds for the next value, k + 1, and the formula is true for n = 1. What does this suggest about the formula’s validity for all odd positive integers?

1.06 Proof by induction mathspace.co

Mathematical induction is a direct proof technique used to prove certain mathematical statements involving natural numbers. Base case The first step in induction, proving the statement is true for the smallest value of n (often n = 0 or n = 1). Inductive hypothesis The second step in induction, assuming the statement is true for an arbitrary value n = k. Inductive step The third step in induction, proving that if the statement is true for n = k, it must also be true for n = k + 1.

The general process for proofs using induction is: 1. Initial statement(s) or base case: Prove the conjecture is true for the smallest value(s) of n, often n = 0, or n = 1. There can be more than one value necessary for the initial statement. 2. Assumption statement or inductive hypothesis: Assume that the statement is true for n = k. 3. Inductive step: Prove that the statement is true for n = k + 1, using the assumption that the statement is true for n = k. 4. Conclusion: Conclude that the conjecture is true for all values of n in the domain. Since the inductive step proves the conjecture is true for every “next” natural number (k + 1), this means that since it is true for n = 1, it is true for n = 2. If it is true for n = 2, then it is true for n = 3, and so on, thus it is true for all natural numbers n greater than or equal to the base case. Mathematical induction is particularly powerful because it allows statements about an infinite number of cases (all natural numbers starting from the base case) to be proved without having to check each one individually. It bridges the gap from a starting point to all subsequent cases through a logical chain. When a pattern seems to be true for every natural number n, a proof by mathematical induction can be used. Beyond proving summation formulas, mathematical induction is also commonly used to prove statements about divisibility (for example, showing an expression is always divisible by a certain number), inequalities that hold for all natural numbers greater than some starting value, and properties of recursively defined sequences.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 Sn: A polygon of n sides has

n ( n 3) diagonals 2

a Use the statement Sn, to find the number of diagonals for polygons with 3, 4, 5, 6, and 7 sides.

Create a strategy Substitute the values n = 3, 4, 5, 6, and 7 into the formula

n ( n 3) . 2

Apply the idea For a polygon with 3 sides (n = 3): Sn =

n ( n 3) Write the formula for finding 2 the number of diagonals

S3 =

3 ( 3 3) 2

3 0 2 0 = 2 =

Substitute n = 3 Evaluate the subtraction Evaluate the multiplication Evaluate

A triangle (3 sides) has 0 diagonals. For a polygon with 4 sides (n = 4): S4 = =

4 ( 4 3) 2

Substitute n = 4

4 1 2

Evaluate the subtraction

Evaluate

A quadrilateral (4 sides) has 2 diagonals. For a polygon with 5 sides (n = 5): S5 =

5 (5 3) 2

Substitute n = 5

5 2 2

Evaluate the subtraction

Evaluate

A pentagon (5 sides) has 5 diagonals.

1.06 Proof by induction mathspace.co

For a polygon with 6 sides (n = 6): S6 =

6 ( 6 3) 2

6 3 2 =9 =

Substitute n = 6 Evaluate the subtraction Evaluate

A hexagon (6 sides) has 9 diagonals. For a polygon with 7 sides (n = 7): 7 ( 7 3) 2 7 4 = 2 = 14

S7 =

Substitute n = 7 Evaluate the subtraction Evaluate

A heptagon (7 sides) has 14 diagonals.

b Complete the table by writing T for true and F for false based on whether the formula matches geometric reality. True or false? S3:

A polygon of 3 sides has 0 diagonals.

S4:

A polygon of 4 sides has 2 diagonals.

S5:

A polygon of 5 sides has 5 diagonals.

S6:

A polygon of 6 sides has 9 diagonals.

S7:

A polygon of 7 sides has 14 diagonals.

Create a strategy

⬚ ⬚ ⬚ ⬚ ⬚

Use the results from part (a) and knowledge of polygons to check if the formula gives the correct number of diagonals.

Apply the idea True or false?

S3:

A polygon of 3 sides has 0 diagonals.

S4:

A polygon of 4 sides has 2 diagonals.

S5:

A polygon of 5 sides has 5 diagonals.

S6:

A polygon of 6 sides has 9 diagonals.

S7:

A polygon of 7 sides has 14 diagonals.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reflect and check This demonstrates the formula holds for several small values of n, but it does not prove the conjecture in general. A full proof by induction or another approach would be needed to prove it true for all values of n ≥ 3.

Example 2 Use mathematical induction to prove, for all positive integers n, that: 5 + 10 + 15 + . . . 5n =

5n ( n 1) 2

a Show that the statement is true when n = 1. (Base case)

Create a strategy For the base case, substitute n = 1. The left side becomes just the first term, 5 × 1 = 5. Substitute n = 1 into the right side formula and show it also equals 5.

Apply the idea Left-hand side (LHS) when n = 1 is 5. Right-hand side (RHS) when n = 1 is: First term =

5n ( n 1) 2

Write the equation

5 1 (1 + 1) 2

Substitute n = 1

5 1 2 2 5=5 5=

Evaluate the addition Simplify

Since LHS = RHS = 5, the statement is true for n = 1.

Reflect and check The base case is the anchor of the proof. If P (1) were false, the entire ‘domino chain’ would never start, even if P (k) ⇒ P (k + 1) was shown to be true.

1.06 Proof by induction mathspace.co

b Assume the statement is true for n = k. That is, assume this is true: 5 10 15

5k ( k 1) 2

Prove the statement is true for n = k + 1.

Create a strategy Start with the assumed statement for n = k. Add the next term in the series, 5(k + 1), to both sides. Then, algebraically manipulate the right-hand side to show it matches the required formula for 5 ( k 1)(( k 1) 1) 5 ( k 1)( k 2) n = k + 1, which simplifies from to . 2 2

Apply the idea Start with the inductive hypothesis: 5 + 10 + 15 + . . . + 5k = 5 + 10 + 15 + . . . + 5k + 5(k + 1) =

5k ( k 1) 2

Write the equation

5k ( k 1) + 5(k + 1) Add the next term, 5(k + 1), to both sides 2

RHS =

5k ( k 1) 2 5 ( k 1) Combine terms on the right-hand side 2 over a common denominator

RHS =

5 ( k 1) [ k 2] Factorise the common term 5(k + 1) 2 from the numerator

RHS =

5 ( k 1) ( k 2) 2

Simplify

The left-hand side is the sum up to n = k + 1. The right-hand side matches the formula n = k + 1 substituted. Therefore, the statement holds for n = k + 1.

5n ( n 1) with 2

c It has been shown that if the statement is true for k, then it is true for k + 1. Can it now be concluded that the statement is true for every positive integer n?

Create a strategy Determine the final conclusion based on the principle of induction.

Apply the idea Since the statement is true for n = 1 (base case), and it has been shown that if it is true for n = k then it is also true for n = k + 1 (inductive step), by the principle of mathematical induction, the 5n ( n 1) 5 + 10 + 15 + . . . + 5n = is true for all positive integers n. 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 3 Prove by mathematical induction that n3 – n is divisible by 3 for all integers n ≥ 1: a Step 1: Base case

Create a strategy Show the statement is true for the smallest integer in the domain, n = 1.

Apply the idea Let P (n) be the statement “n3 – n is divisible by 3”. For n = 1: 13 – 1 = 1 – 1 = 0 Since 0 is divisible by 3 (as 0 = 3 × 0), P (1) is true.

b Step 2: Inductive hypothesis

Create a strategy Assume the statement P (k) is true for some arbitrary positive integer k ≥ 1.

Apply the idea Assume that k3 – k is divisible by 3. This means k3 – k = 3m for some integer m.

c Step 3: Inductive step

Create a strategy Show that the statement P (k + 1) is also true, i.e., show that (k + 1)3 – (k + 1) is divisible by 3, using the inductive hypothesis.

Apply the idea Consider (k + 1)3 – (k + 1): (k + 1)3 – (k + 1) = (k3 + 3k2 + 3k + 1) – (k + 1) Expand the cube and distribute the negative = k3 + 3k2 + 3k + 1 – k – 1 3

Simplify

= (k – k) + 3k + 3k Group terms to use the inductive hypothesis = 3m + 3k2 + 3k Substitute k3 – k = 3m from the inductive hypothesis = 3(m + k2 + k)

Factorise 3

Since m, k are integers, m + k + k is also an integer. Therefore, (k + 1)3 – (k + 1) is a multiple of 3. Therefore, P (k + 1) is true.

1.06 Proof by induction mathspace.co

d Step 4: Conclusion

Create a strategy Determine the conclusion based on the principle of mathematical induction.

Apply the idea Since the statement is true for n = 1, and assuming it is true for n = k implies it is true for n = k + 1, by the principle of mathematical induction, the statement “n3 – n is divisible by 3” is true for all integers n ≥ 1.

Idea summary Mathematical induction is a proof technique used to prove statements about natural numbers. The process involves four essential steps: 1. Base case: Show the statement is true for the initial value (for example, n = 1). 2. Inductive hypothesis: Assume the statement is true for an arbitrary natural number k. 3. Inductive step: Prove that if the statement is true for k, it must also be true for k + 1. 4. Conclusion: Conclude that the statement is true for all natural numbers n from the base case onwards, by the principle of mathematical induction.

1.06 Practice questions What do you remember? 1

Is each statement true or false? a

With enough examples, a conjecture can be proven true.

With one counter-example, a conjecture can be proven false.

Proof by induction involves an initial statement, assumption statement, and the inductive step which leads to a conclusion.

If a conjecture is assumed true for n = 100 and proven true for n = 101, then the conjecture is true.

If a conjecture is shown to be true for the initial statement, assumed true for n = p – 1 and proven true for n = p, then the conjecture is true.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Choose and arrange the correct steps to prove a conjecture is true for n ≥ 1. A

Assume the statement is true for n = 1.

Prove the statement is true for n = 1.

Prove the statement is true for n = 2.

Prove the statement is true for n = k.

Assume the statement is true for n = k.

Prove the statement is true for n = k + 1.

Assume the statement is true for n = k + 1.

Prove the statement is true for n = k + 2.

Which of these types of mathematical statements is proof by induction most suitable for proving? A

A statement or formula that is claimed to be true for all natural numbers n greater than or equal to some starting value (for example, for all n ≥ 1).

Finding the specific numerical value of a complex calculation.

Proving a geometric theorem that does not depend on an integer parameter n.

Showing a statement is true for a few specific cases, like n = 1, 2, 3.

Practice 4

To prove 2n > n for n ≥ 1: a

Show that the statement is true for n = 1.

Write out the inductive hypothesis for n = k.

Determine the proposition for the inductive step (n = k + 1). Proof is not required here.

Fill in the blanks: For the statement 13 + 23 + 33 + . . . + n3 a b c d

1 2 n (n + 1)2, 4

1 2 ⬚ (1 + 1)2. 4 1 When n = 2, 13 + ⬚3 = (22) (⬚)2. 4 When n = 1, 13 =

When n = k, ⬚ + 23 + . . . + ⬚3 = ⬚k2 (⬚ + 1)2. When n = k + 1, 13 + 23 + . . . + ⬚3 =

1 2 ⬚ (⬚)2. 4

Consider the statement Sn : 12 + 22 + 32 + . . . + n2 =

1 n (n + 1) (2n + 1). 6

According to the statement, what does 12 + 22 + 32 + . . . + 1002 equal? Ex 1

n ( n 3) diagonals. 2 According to the statement S8, how many diagonals does a polygon of 8 sides (an octagon) have? Verify your answer by drawing an octagon and its diagonals if necessary. The statement Sn: A polygon of n sides has

1.06 Proof by induction mathspace.co

Consider the conjecture:

1 3

3 5

...

1 (2n 1)(2n 1)

n 2n 1

Is the conjecture true for the given values of n? a 9

n = 5

n=7

Consider the statement Sn: 1+2+3+...+n=

1 n (n + 1) 2

Complete this table to see if the conjecture is true for 1 ≤ n ≤ 8: 1+2+3+...+n

1 n ( n 1) 2

True or false

⬚

n=1 n=2

⬚

n=3 n=4

⬚

n=5

Ex 2

Ex 3

Consider the statement: 5 Sn : 1 2 3

⬚

n=8

⬚

n=7

⬚

n=6

⬚

6 n 4 ... 2 3 4 n ( n 1) ( n 2)

Calculate S4 using the summation

Calculate S4 using the formula Sn

Is the conjecture true for n = 4?

n ( 3n 7 ) 2 ( n 1) ( n 2)

6 n 4 ... . 2 3 4 n ( n 1) ( n 2)

5 1 2 3

n ( 3n 7 ) . 2 ( n 1) ( n 2)

A conjecture states that “for an integer n ≥ 1,

...

2n 1

.”

Show that this is true for n = 1.

Assume that for n = k, the conjecture is true. Write out the assumption statement.

Using the assumption statement and laws of exponents, show that the conjecture is true for n = k + 1, the inductive step.

Prove that n2 – n is always even for n ≥ 2, using proof by induction.

Consider Sn : n3 – n for n ≥ 1. Determine the largest integer that Sn is always divisible by. Prove the conjecture using induction.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking 1 1 1 . . . and form a conjecture n( n 1) 1 ( 1 1) 2 (2 1) for its value in terms of n. Prove the conjecture using mathematical induction.

Consider the first few terms of the sum

1 2 Prove by mathematical induction that 15 + 25 + 35 + . . . + n5 = n (n + 1)2 (2n2 + 2n – 1) 12 for n ≥ 1.

Investigation: Repetition in code Investigate online

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1.07 Prove properties of integers After this lesson, you will be able to… • prove divisibility properties of non-negative integers • demonstrate the representation of integers as products of prime powers • determine relationships between highest common factor and lowest common multiple using prime factorisations

Properties of integers: odd, even, and divisibility Integers have specific properties related to calculations, such as being even, odd, prime, or a perfect square. The properties may be proven or disproven using mathematical techniques such as: • Contrapositive: The contrapositive is logically equivalent to the original statement, so proving the contrapositive is true also proves the original statement is true. • Contradiction: To prove a statement p is true, start by assuming its opposite (¬ p) is true. Then, show that this assumption leads to a logical impossibility (a contradiction). If the opposite leads to a contradiction, the original statement p must be true. • Counter-example: To disprove a statement, p, find one example where p is false. • Direct proof or algebraic justification: Start with the given information (premises) and use definitions and algebraic steps to logically arrive at the conclusion that needs to be proven. For example, using definitions for any integer k: • An even number can be represented as 2k • An odd number can be represented as 2k + 1 • The next consecutive integer can be represented as k + 1 Beyond odd and even properties, it is often necessary to prove properties related to divisibility.

1.07 Prove properties of integers mathspace.co

Example 1 Prove that ∀n ∈ ℤ, n2 – n – 1 is odd.

Create a strategy

The strategy is to first understand the statement in words, and then apply the method of proof by contradiction. Recall that an integer that can be written as 2k + 1 or 2k –1 for some integer k is odd.

Apply the idea Proof by contradiction: Suppose, for the sake of contradiction, the statement is false, so n2 – n – 1 is even (not odd). This means it can be written in the form 2k for some integer k. n2 – n – 1 = 2k

Suppose the statement is false

n – n = 2k + 1 2

Add 1 to both sides 2

This means that n – n is odd. Factoring gives n – n = n (n – 1). This expression represents the product of two consecutive integers (n – 1 and n). Note that: • If n is even (2M), then n – 1 is odd (2M – 1), so their product, (2M) (2M – 1) = 2(2M2 – M), is even. • If n is odd (2M + 1), then n – 1 is even (2M), so their product, (2M) (2M + 1) = 2(2M2 + M), is even. This means n2 – n is always even, which is a contradiction, so the assumption was false, and the original statement is true. This proves that ∀n ∈ ℤ, n2 – n – 1 is odd.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reflect and check This could be rearranged to be done as direct proof: The expression n2 – n is always even as shown: • If n is even (2M), then n – 1 is odd (2M – 1), so their product, (2M) (2M – 1) = 2(2M2 – M), is even. • If n is odd (2M + 1), then n – 1 is even (2M), so their product, (2M) (2M + 1) = 2(2M2 + M), is even. This means that n2 – n = 2k for some integer k. Furthermore, this means that n2 – n – 1 = 2k – 1, so n2 – n – 1 must be odd.

Example 2 Prove each statement: a If n is an integer such that when divided by 3 it leaves a remainder of 2, then when n2 is divided by 3, it leaves a remainder of 1.

Create a strategy Use direct proof. The condition that n leaves a remainder of 2 when divided by 3 means n = 3k + 2 for some integer k. Assume this hypothesis is true. Square this expression for n and manipulate it algebraically to show that n2 can be written in the form 3j + 1, which means it leaves a remainder of 1 when divided by 3.

Apply the idea The problem asks to prove: If n = 3k + 2 for some integer k, then n2 = 3j + 1 for some integer j. Assume n = 3k + 2 for some integer k, meaning when n is divided by 3, the remainder is 2. Consider n2: n2 = (3k + 2)2 2

Substitute expression for n 2

= (3k) + 2(3k) (2) + 2 2

= 9k + 12k + 4 2

= 9k + 12k + 3 + 1 2

= 3(3k + 4k + 1) + 1

Expand the square Simplify Rewrite 4 as 3 + 1 to isolate a multiple of 3 Factorise 3 from the first three terms

Let j = 3k + 4k + 1. Since k is an integer, it follows that k2, 4k, and 3k2 are also integers. The sum of integers is an integer, so j is an integer. Therefore, n2 = 3j + 1, meaning when n2 is divided by 3, the remainder is 1. This completes the proof, showing that if n leaves a remainder of 2 when divided by 3, then n2 leaves a remainder of 1 when divided by 3.

1.07 Prove properties of integers mathspace.co

b For all integers n ∈ ℤ+, n is divisible by 4 if and only if the number formed by its last two digits is divisible by 4.

Create a strategy This requires proving two directions due to “if and only if”. • Direction 1 (⇒): Assume n is divisible by 4. Represent n using its last two digits (n = 100k + d, where d is the number formed by the last two digits). Show that d must be divisible by 4. • Direction 2 (⇐): Assume the number formed by the last two digits, d, is divisible by 4. Represent n = 100k + d. Show that n must be divisible by 4.

Apply the idea Let n be a positive integer. Represent n in terms of its last two digits: let d be the integer formed by the last two digits of n (so 0 ≤ d ≤ 99). Then, n = 100k + d for some non-negative integer k. • Direction 1 (⇒): Assume n is divisible by 4. If n is divisible by 4, then n = 4m for some integer m. So, 4m = 100k + d Rearrange: d = 4m – 100k Factorise 4: d = 4 (m – 25k) Since m and k are integers, m – 25k is an integer. Let p = m – 25k. Then d = 4p, meaning d, the number formed by the last two digits, is divisible by 4. • Direction 2 (⇐): Assume the number formed by the last two digits, d, is divisible by 4. If d is divisible by 4, then d = 4q for some integer q. Using n = 100k + d, substitute d = 4q: n = 100k + 4q Factorise 4: n = 4 (25k + q) Since k and q are integers, 25k + q is an integer. Let s = 25k + q. Then n = 4s, meaning n is divisible by 4. Since both directions have been proven, the statement is true.

Idea summary Integers possess properties related to being odd or even. An integer n is even if n = 2k and odd if n = 2k + 1 for some integer k. Divisibility is a key concept: a divides b (a|b) if b = ak for some integer k. Properties of integers, including divisibility rules, can often be proven using direct proof, proof by contradiction, or proof by contrapositive.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Prime factorisation Prime number A natural number greater than 1 that has no positive divisors other than 1 and itself. Example: 2, 3, 5, 7, 11 are prime numbers. Composite number A natural number greater than 1 that is not prime, meaning it has positive divisors other than 1 and itself. Example: 4 is composite because 4 = 2 × 2.

The number 1 is neither prime nor composite. Fundamental theorem of arithmetic The theorem stating that every integer greater than 1 is either a prime number or can be expressed as a product of prime numbers in exactly one way (ignoring the order in which the primes are written). Prime factorisation The process of expressing an integer greater than 1 as a product of prime numbers.

For example, the prime factorisation of 12 is 2 × 2 × 3 or 22 × 3. No matter how factoring 12 is started (for example, 2 × 6 or 3 × 4), the result will always contain the prime factors 2 (twice) and 3 (once). For instance, to find the prime factorisation of 60: • Divide by the smallest prime factor, 2: 60 ÷ 2 = 30 • Divide the result by 2 again: 30 ÷ 2 = 15 • The result 15 is not divisible by 2. Divide by the next smallest prime factor, 3: 15 ÷ 3 = 5. • The result 5 is a prime number, so the process stops. The prime factorisation of 60 is 2 × 2 × 3 × 5 or 22 × 3 × 5.

Interactive exploration Discover this concept in action online

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Example 3 Digital only example Explore in the online lesson

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1.07 Prove properties of integers mathspace.co

Example 4 Prove that there are infinitely many prime numbers.

Create a strategy Use proof by contradiction. Assume that there are only finitely many prime numbers, say p1, p2, . . ., pk. Construct a new number N based on these primes and show that this leads to a contradiction regarding the primality or divisibility of N.

Apply the idea Proof by contradiction: Suppose, for the sake of contradiction, that there are only finitely many prime numbers. Let this complete list of all prime numbers be p1, p2, p3, . . ., pk. Consider the number N obtained by multiplying all these prime numbers together and adding one: N = ( p 1 × p 2 × p 3 × . . . × p k) + 1 Now, consider the number N. Since N is an integer greater than 1, by the fundamental theorem of arithmetic, N must have a prime divisor. Let this prime divisor be p. Since it was assumed that p1, p2, . . ., pk is the complete list of all prime numbers, p must be one of the primes in this list. So, p = pi for some i between 1 and k. This means that p (which is pi) must divide N. Also, pi clearly divides the product ( p1 × p2 × . . . × pk) because it is one of the factors. If pi divides both N and the product ( p1 × . . . × pk), then it must also divide their difference: N − ( p 1 × p 2 × . . . × p k) But N was constructed as N = ( p1 × p2 × . . . × pk) + 1, so the difference is: [( p1 × p2 × . . . × pk) + 1] − ( p1 × p2 × . . . × pk) = 1 So, the assumption leads to the conclusion that the prime number pi must be a factor of 1. This implies that pi must divide 1. However, the only integer that divides 1 is 1 or –1, and prime numbers must be greater than 1. This is a contradiction. The initial assumption that there are finitely many prime numbers must be false. Therefore, there are infinitely many prime numbers.

Reflect and check This classic proof, often attributed to Euclid, elegantly constructs a number, N, which must have a prime factor. However, it’s shown that N cannot be divided by any prime in the assumed finite list, leading to a contradiction.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary A prime number is a natural number greater than 1 divisible only by 1 and itself. Numbers greater than 1 that are not prime are composite. Prime factorisation is the process of expressing an integer greater than 1 as a product of prime numbers. The fundamental theorem of arithmetic states that every integer greater than 1 is either prime or can be uniquely factorised into a product of prime numbers (ignoring order). Proof by contradiction can be used to show that there are infinitely many prime numbers.

Applications of prime factorisation Prime factorisation is a powerful tool for understanding relationships between integers, particularly for finding the highest common factor (HCF), also called the greatest common divisor (GCD), and the lowest common multiple (LCM) of two or more integers. Highest common factor (HCF) Also known as the greatest common divisor (GCD), the HCF of two or more integers is the largest positive integer that divides each of the integers without leaving a remainder. Lowest common multiple (LCM) The LCM of two or more integers is the smallest positive integer that is divisible by each of the integers.

Finding HCF using prime factorisation: 1. Find the prime factorisation of each integer. 2. Identify the common prime factors in all factorisations. 3. For each common prime factor, take the lowest power that appears in any of the factorisations. 4. Multiply these lowest powers together to get the HCF. Finding LCM using prime factorisation: 1. Find the prime factorisation of each integer. 2. Identify all prime factors that appear in any of the factorisations. 3. For each prime factor, take the highest power that appears in any of the factorisations. 4. Multiply these highest powers together to get the LCM.

1.07 Prove properties of integers mathspace.co

There is an important relationship between the HCF and LCM of two positive integers a and b: HCF (a, b) × LCM (a, b) = a × b This relationship holds because of how exponents combine. For any prime factor p, if its highest power in a is px and in b is py, the HCF includes the prime factor raised to the smaller of the powers x and y, and the LCM includes the prime factor raised to the larger of the powers x and y. When multiplied, the power of p in HCF × LCM is the sum of the smaller and larger powers, which equals px+y. This matches the power of p in the product a × b, which is also px+y. This applies to all prime factors.

Exploration Explore HCF and LCM: 1. Determine the prime factorisation of 12 and 18. 2. Identify the common prime factors and their lowest powers to find HCF (12, 18). 3. Identify all prime factors involved and their highest powers to find LCM (12, 18). 4. Calculate the product HCF (12, 18) × LCM (12, 18). 5. Calculate the product 12 × 18. 6. Compare the results from Step 4 and Step 5. What relationship is observed between HCF × LCM and a × b? 7. Repeat Steps 1-6 for another pair of integers, for example, 30 and 45. Does the relationship still hold?

Example 5 Determine the HCF and LCM of 72 and 90 using prime factorisation.

Create a strategy First, find the prime factorisation of each number. Then, apply the rules: for HCF, take the lowest power of common prime factors. For LCM, take the highest power of all prime factors involved.

Apply the idea Step 1: Find the prime factorisation of 72. 72 = 8 × 9 = (2 × 2 × 2) × (3 × 3) = 23 × 32 Step 2: Find the prime factorisation of 90. 90 = 9 × 10 = (3 × 3) × (2 × 5) = 2 × 32 × 5

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Step 3: Find the HCF. The common prime factors are 2 and 3. The lowest power of 2 present in both is 21 from 90. The lowest power of 3 common to both factorisations is 32. HCF (72, 90) = 21 × 32 = 2 × 9 = 18 Step 4: Find the LCM. The prime factors involved are 2, 3, and 5. The highest power of 2 present is 23 from 72. The highest power of 3 present is 32 from both 72 and 90. The highest power of 5 present is 51 from 90. LCM (72, 90) = 23 × 32 × 51 = 8 × 9 × 5 = 72 × 5 = 360 So, HCF (72, 90) = 18 and LCM (72, 90) = 360.

Reflect and check The relationship HCF (a, b) × LCM (a, b) = a × b can be verified. HCF (72, 90) × LCM (72, 90) = 18 × 360 = 6480 72 × 90 = 6480 The relationship holds, which increases confidence in the results.

Idea summary The highest common factor (HCF) of integers is found by multiplying the lowest powers of their common prime factors. The lowest common multiple (LCM) is found by multiplying the highest powers of all prime factors present in any of the integers. For two positive integers a and b, their HCF and LCM are related by the formula: HCF (a, b) × LCM (a, b) = a × b

1.07 Practice questions What do you remember? 1

These steps prove that the sum of any two odd integers is even, but the steps are presented out of order. Determine the correct order of the steps: 1 Since a + b + 1 is an integer, the sum is of the form 2k, which is even. 2 Their sum is (2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1). 3 Let the two odd integers be 2a + 1 and 2b + 1, where a, b are integers.

1.07 Prove properties of integers mathspace.co

Determine the correct order of these statements to prove that the square of any even integer is even: 2 2 2 1 Its square is (2n) = 4n = 2 (2n ). 2 Let the even integer be 2n, where n is an integer. 2 3 Since 2n is an integer, the square is of the form 2k, which is even.

Complete the proof that the sum of any two even integers is even:

Let the two even integers be 2a and ⬚, where a, b are integers. Their sum is ⬚ + ⬚ = ⬚. Since ⬚ is an integer, the sum is of the form ⬚, which is even.

Complete the proof that the product of any two odd integers is odd:

Let the two odd integers be 2a + 1 and ⬚, where a, b are integers. Their product is (2a + 1) (⬚) = 4ab + 2a + 2b + 1 = ⬚. Since ⬚ is an integer, let k = 2ab + a + b. The product is ⬚, which is the form of an odd integer.

Practice 5

Determine whether each statement is true or false. Provide a counter-example to support the answer if the statement is false:

For any two integers a and b, if a2 + b2 is even, then both a and b must be even.

For any non-negative integer n, if n is even, then n2 + 2n is even.

For any two non-negative integers a and b, if a and b are both odd, then a2 + b is odd.

Determine the prime factorisation of these numbers: a

For each pair of numbers, find the HCF and LCM by listing factors and multiples: i

Identify the highest common factor (HCF).

Identify the lowest common multiple (LCM).

12 and 18

24 and 32

Prove that the square of any odd integer is odd.

Prove that the sum of the squares of any two consecutive integers is an odd integer.

Prove that for any integer n, the expression n2 + n is always even.

Prove that the product of two consecutive integers is always even.

Prove that for all integers n, the expression n2 + n + 2 is always even.

Prove that for ∀n ∈ ℤ, the expression 3n2 + 7n + 4 is always even.

Prove that the difference between the squares of any two consecutive integers is equal to the sum of those integers.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

15 Ex 1

16 Ex 2

Using proof by contradiction, prove that for any integer n, the expression n2 – n – 1 is always odd. Prove that if an integer n leaves a remainder of 1 when divided by 4, then n2 also leaves a remainder of 1 when divided by 4.

Prove that 2 is the only even prime number.

Determine the prime factorisation of these numbers: a

Ex 3

252

420

1260

2400

Use prime factorisation to determine if 588 is divisible by 42. Explain your reasoning.

Outline the key steps in Euclid’s proof by contradiction to show that there are infinitely many prime numbers. Explain why the constructed number N = ( p1 × p2 × . . . × pk) + 1 leads to a contradiction.

Ex 4

21 Ex 5

Determine the HCF and LCM of 84 and 120 using prime factorisation: a

LCM (84, 120)

Determine the HCF and LCM of 90, 108 and 126 using prime factorisation: a

HCF (84, 120)

HCF (90, 108, 126)

LCM (90, 108, 126)

Let a = 60 and b = 84: a

Determine HCF (60, 84) using prime factorisation.

Determine LCM (60, 84) using prime factorisation.

Calculate HCF (60, 84) × LCM (60, 84).

Calculate a × b = 60 × 84.

Verify that HCF (a, b) × LCM (a, b) = a × b.

Determine if the expression 2n2 + 5n + 3 is always odd for ∀n ∈ ℤ. Provide proof or a counter-example.

Prove that the difference between any two odd prime numbers greater than 2 is even.

Extend your thinking 26 27

Prove that for ∀n ∈ ℤ, the expression n3 + 2n is always divisible by 3. Prove that for any integer n, n3 – n is always divisible by 6.

Let n be an integer. Prove that if n2 is odd, then n must be odd, using proof by contrapositive.

Prove that for any odd integer n, the number n2 – 1 is divisible by 8.

1.07 Prove properties of integers mathspace.co

1.08 Rational numbers and their representations After this lesson, you will be able to… • describe the properties of rational numbers within number systems • convert between fraction and decimal forms of rational numbers

Rational numbers and recurring decimals Rational number (ℚ) p , where p and q are integers and q ≠ 0. Its decimal q representation either terminates or repeats. A number that can be expressed as

Example: 1 , 0.25, 0.3 . . . , 0.83 2 Properties of rational numbers: • Relationship to other number systems: All integers (ℤ) are rational numbers since any integer n n can be written as . Natural numbers (ℕ) are also rational. Rational numbers are a subset 1 of the real numbers (ℝ). So, ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ. • Closure: Rational numbers are closed under addition, subtraction, multiplication, and division (except division by zero). This means performing these operations on two rational numbers always results in another rational number. • Density: Between any two distinct rational numbers, there is always another rational number. This means the rational numbers are ‘dense’ on the number line. • Decimal representation: Every rational number has a decimal representation that either terminates (ends) or repeats in a predictable pattern. Irrational number p A number that cannot be expressed as . Its decimal representation is non-terminating and q non-repeating. Example: 2, 3 1 Some decimals have a finite number of decimal places. For example, is 0.5 and is 0.3. These 10 2 are called terminating decimals.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Recurring decimals, however, have an infinite number of decimal places with a repeating pattern. When this occurs, a line (upper bar) can be used to show the repeating numbers. The line begins on top of the first repeated number and ends at the last repeated number in the pattern. A dot can be used for a single digit repeating. For example: • 0.333 . . .

0.3 •

• 0.45454545 . . .

0.45 •

0.555 . . .

0.5

0.166666 . . .

0.16

Example 1 Identify whether each number is rational or irrational. Explain your reasoning: i

0.75

iii

0.18

Create a strategy p where p, q are integers and q ≠ 0. q Remember that integers, terminating decimals, and recurring decimals are rational. Determine if the number can be written as a fraction

Non-terminating, non-recurring decimals (like the square roots of non-perfect squares) are irrational.

Apply the idea 7 i. 7: Rational. It is an integer and can be written as . 1 ii. 0.75: Rational. It is a terminating decimal and can be written as

75 3 or . 100 4

18 2 . 99 11 iv. 5: Irrational. 5 is not a perfect square, so its square root cannot be expressed as a simple iii. 0.18: Rational. It is a recurring decimal. It can be converted to

fraction and its decimal representation is non-terminating and non-repeating (2.2360679 . . . ).

1.08 Rational numbers and their representations mathspace.co

Example 2 Express each fraction as a recurring decimal: a 6

2 7

Create a strategy Use technology or short division by hand to divide the numerator of the fractional part by the denominator, then add the whole number. Look for a repeating pattern in the quotient.

Apply the idea 2 to a decimal. 7 After calculating the division 2 ÷ 7 using a calculator, the decimal 0.285714285714 . . . is obtained, where 285714 repeats infinitely. First, convert the fractional part

2 0.285714. 7 Now add the whole number part: 6 0.285714 So,

To write this as a recurring decimal, a bar is placed over the six digits that repeat, 285714. 6

2 6.285714 7

Reflect and check Short division can be used if converting the fraction by hand: 0. 285 7 1 4 28 7 2.2060405010302060 Pattern starts to repeat 0. 285 7 1 4 + 6 = 6 . 285 7 1 4

1 3

Create a strategy

Apply the idea

Divide the numerator (1) by the denominator (3) using technology or short division. Look for a repeating pattern in the quotient.

Using division: 1 ÷ 3 = 0.3333. . .. The digit ‘3’ repeats infinitely. A dot over the number is used for a single repeating digit. A bar can also be used. 1 = 0.3 0.3 3

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary p (where p, q are integers, q q ≠ 0). They include integers, terminating decimals, and recurring decimals. Irrational numbers cannot be expressed as such fractions. Rational numbers can be expressed as fractions

All rational numbers have decimal representations that either terminate or recur. Recurring decimals have an infinite number of decimal places with a repeating pattern, indicated using bar or dot notation. The decimal representation of a fraction can be found using a calculator or division.

Convert recurring decimals to fractions Consider converting a recurring decimal into a fraction. Typing a truncated version of a recurring decimal (e.g., 0.12121212) into a calculator and pressing the button to convert it into a fraction will most often just give the fraction for the terminating decimal, not for the true recurring decimal. 1 Some patterns may already be familiar such as 0.3 . From there, patterns may arise like 3 1 1 0.1 0.3 3 3 . However, it may be necessary to prove or justify the conversion 3 9 algebraically.

Exploration Explore how to convert a recurring decimal like 0.5 into a fraction. 1. Let x 0.5. Write this out as x = 0.5555 . . . . 2. What is the value of 10x? 3. Consider the expression 10x – x. Write down the values you found for 10x and x and perform the subtraction. What happens to the repeating decimal part? 10 x x 9x

5.5555 . . . 0.5555 . . . ?

4. The resulting equation should be in the form of 9x = some integer. Solve this equation for x. What is the fractional form of 0.5? This method of multiplying by a power of 10 to shift the decimal and then subtracting to eliminate the repeating part is key to converting any recurring decimal to a fraction.

1.08 Rational numbers and their representations mathspace.co

These steps can be used to convert any recurring decimal to a fraction: Step 1: Introduce x • Let x represent the recurring decimal to be converted to a fraction, for example: x 0.012 Step 2: Determine the repeating digits. • Look for the repeating digits in the recurring decimal, for example for x 0.012 these are 12. Step 3: Shift the repeating digits for alignment. • Create an equation by multiplying x by a power of 10 (10k) so that one full repeating block of digits is to the left of the decimal point, but the decimal part still consists of the repeating block, for example 1000x = 12.1212121 . . . . Step 4: Create a second equation for alignment. • If the repeating block started immediately after the decimal in the original x, then the original equation x = . . . serves as a second equation. • If the repeating block does not start immediately after the decimal point in the original x, create another equation by multiplying x by a different power of 10 (10 j) to shift only the non-repeating part entirely to the left of the decimal, for example 10x = 0.1212121 . . . . • The decimal part of this new equation should now match the decimal part of the equation from Step 3. Step 5: Subtract the equations and solve for x. • Subtract the equation with the smaller coefficient of x (from Step 4 or the original x) from the equation with the larger coefficient of x (from Step 3). • This should eliminate the repeating decimal part. • Then, solve the resulting equation for x and simplify the fraction, for example solving 1000x − 10x = 12.1212121 − 0.1212121 → 990x = 12. To verify the conversion, the fraction can be evaluated as a decimal using a calculator.

Example 3 Express each decimal as a fraction in simplest form: a x 0.243

Create a strategy Define the recurring decimal as a variable, multiply by an appropriate power of 10 to align the decimal places for subtraction, form new equations, subtract the equations to eliminate the repeating part, and then solve for the variable.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea Let x 0.243. The repeating pattern is three digits long (‘243’). Multiply x by 1000 to shift one full repeating block to the left of the decimal point: 1000x = 1000 0.243 = 243.243

Write the equation Evaluate

Now, there are two equations: 1000 x 243.243 x 0.243

1 2

Subtract the equation 2 from equation 1 to eliminate the recurring decimal: 1000x – x = 243.243 0.243 999x = 243

Write the equation Evaluate

243 999

Divide both sides by 999

9 37

Simplify

Reflect and check Use technology to verify the conversion: 9 ÷ 37 = 0.243243 . . .

b x 4.273

Create a strategy Let x be the decimal. First, multiply x by a power of 10 to move the non-repeating part to the left of the decimal point. Then, multiply x by another power of 10 to move one full repeating block to the left as well. Subtract the two resulting equations.

Apply the idea Let x 4.273. The non-repeating part is ‘2’ after the decimal, and the repeating part is ‘73’. Multiply x by 10 to move the non-repeating digit ‘2’ to the left of the decimal point: 10x = 10 4.273 = 42.73

Equation 1

Now, multiply x by 1000 (since the non-repeating part has 1 digit and the repeating part has 2 digits, total 3 digits after the decimal to shift one full pattern over) to get the same repeating decimal part: 1000x = 1000 4.273 = 4273.73

Equation 2

1.08 Rational numbers and their representations mathspace.co

Now, subtract Equation 1 from Equation 2 to eliminate the recurring decimal: 1000x – 10x = 4273.73 42.73 990x = 4231

Evaluate

4231 x= 990 The fraction

Write the equation

Divide both sides by 990

4231 is in its simplest form. 990

Example 4 Evaluate 2.7 0.83 4.1296 as a fraction without a calculator.

Create a strategy Each decimal can be converted individually to a fraction, then these fractions can be added and subtracted using common denominators.

Apply the idea • Convert 2.7: 1 It is known that 0.1 , so 0.7 9 Therefore, 2.7

7 9

18 9

7 9

1 9

7 . 9

25 . 9

• Convert 0.83: Let y 0.83 Then 10 y 8.3 and 100 y 83.3 100 y 10 y

83.3 8.3 75 5 . 90 y 75, so y 90 6 • Convert 4.1296: Let z 4.1296 Then 10 z 41.296 and 10 000z = 41 296.296 10 000z – 10z = 41 296.296 41.296 41 255 8251 9990z = 41 255, so z (Simplifying by dividing by 5). 9990 1998 This fraction can be simplified further by testing for divisibility by small prime numbers. In this case, both numbers share a common factor of 37. This can be further simplified 8251 37 223 by dividing both the numerator and denominator by 37: . Therefore, 1998 37 54 223 4.1296 . 54

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Now, perform the arithmetic with fractions: 2.7 0.83 4.1296 =

25 9

5 6

150 54

45 54

150 45 223 54

Combine numerators over the common denominator

328 54

Evaluate the operations in the numerator

223 54 223 54

2 27

Substitute the fractions for the decimals Express the fractions with a common denominator

Convert to mixed number

Reflect and check An alternative strategy involves first performing the arithmetic with the expanded decimals first. This process gives a final result of 6.074, which can then be converted to a fraction: • Let w 6.074 • Multiply w by 1000 to get: 1000w 6074.074 Subtract the equations to eliminate the recurring decimal: 1000w – w = 6074.074 6.074 999w = 6068

Evaluate the subtraction

6068 999

Divide both sides by 999

w= To simplify

Write the equation

164 6068 , divide by common factor (37). So, w . 27 999

This gives: 2.7 0.83 4.1296

6.074

164 27

2 27

Both methods yield the same result.

1.08 Rational numbers and their representations mathspace.co

Idea summary To convert recurring decimals to fractions, an algebraic method can be used: 1. Set the recurring decimal equal to a variable, x. 2. Identify the repeating block of digits and any non-repeating decimal part. 3. Multiply x by a power of 10 (10 k) to shift one full repeating block and any non-repeating part to the left of the decimal point. 4. Multiply x by another power of 10 (10 j, where j < k) to shift only the non-repeating part to the left of the decimal point. If there’s no non-repeating part, this second equation is simply x itself. 5. Subtract the equation from step 4 from the equation in step 3. This eliminates the recurring part. Solve the resulting equation for x and simplify the fraction. Recognising patterns e.g., 0.1

1 ab , 0.ab can sometimes provide shortcuts for simple 99 9

recurring decimals.

1.08 Practice questions What do you remember? 1

For each number, classify it by identifying all the number sets it belongs to: Natural (ℕ), Integer (ℤ), Rational (ℚ), Irrational, and Real (ℝ): –5

Explain why any integer n is considered a rational number.

Explain why a terminating decimal like 0.6 is considered a rational number.

Give an example demonstrating that rational numbers are closed under addition.

escribe the relationship between the set of integers (ℤ) and the set of rational D numbers (ℚ).

Explain what it means for the rational numbers to be ‘closed’ under addition.

Give an example of an irrational number and explain briefly why it is irrational.

0.6

0.7

Determine if these fractions result in recurring decimals: a

3 4

15 19

24 7

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

16 17

33 30

Determine if these equations are true or false: 4 32 278 0.444444445 a b 13.0416 9 2475 23 3273 2.09 d e 3.273 11 999 g

9.218 9

197 900

8.32

832 99

c f i

61 111 1.23456 49 500 61 6.7 9 18 0.72 25

Practice Ex 1

Ex 2

Identify whether each number is rational or irrational. Explain your reasoning: a

–5

3.14

0.121212 . . .

Express these fractions as decimals: 17 8 a b 2 27 45 e

9982 999

113 33

Show that 0.26 can be written as

Convert these decimals to fractions:

391 999

4 99

5 33

25 66

4 . 15

Ex 3

0.2

0.06

0.3

5.008

0.16

2.13

17.708

1.3146

0.12

0.075

0.29

3.57

Identify the error in this working. Then, provide the correct solution: x = 18.79 100x = 1879.79 100x – x = 1879.79 18.79 99x = 1861 x=

Ex 4

1861 99

1 0.221, as a decimal. 3

Evaluate 0.46

Evaluate these expressions as fractions: a

3.3 0.16 12.178

2.398 4.3287

19 5 17.8 99

1 7

0.5 1.63 8.92

12.56 36.72 2.78 32.267

1.08 Rational numbers and their representations mathspace.co

Solve for n: 4 n 0.53 a 5

1.45

9 n n 11

Extend your thinking 14

Let 1 ≤ x ≤ 9: x x x , 0.0 x , and 0.00 x . 9 99 999

Prove that 0. x

Without solving, predict what 0.0004 will be as a fraction. 454 x 2 3003

Consider the equation: 0.037296x2 decimal.

Consider the equation:

Grace receives a monthly pay of $1200. She spends

Tracey and her two friends, Martha and Burke, decide to invest in a fund and split the contribution evenly between them. They originally plan to each pay $33 333.3. However, both Tracey and Martha do not have enough money at the time, so Burke invests the full sum on behalf of them all. Martha promises to pay back the original investment plus any returns at the end of the second year. Tracey promises to pay back the original investment plus any returns at the end of the third year.

0.4 0.21x

0. Solve for x, expressing the answer as a

6. Solve for x in exact form.

5 1 of her pay on her rent and of what 11 3 1 1 is left on her utilities. She also spends of her pay on her groceries and on petrol. She 4 15 then lends $10 to her friend, Mary. She wishes to save 10% of her remaining money, and she will spend the rest on her birthday. What percentage of her pay will she spend on her birthday?

The fund gives returns of 5.6% by the end of the first year, 7.8% by the end of the second year, and –2.5% by the end of the third year. Assuming they invest all returns into the fund:

Calculate the amount Martha has to pay back at the end of the second year.

Calculate the amount Tracey has to pay back at the end of the third year.

Luchino is making a cake for Galatea’s birthday. He uses 455.276 g of self-raising flour, 140.137 g of plain flour, 435.4231 g of sugar, 314.68 g of eggs, 279.238 g of salted butter, 306.238 g of milk, and 11.4876 g of vanilla extract. What fraction of the weight of the mix is flour?

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1.09 Prove irrationality After this lesson, you will be able to… • understand the concept of irrational numbers • construct proofs of irrationality using contradiction • verify the irrationality of specific numbers

Prove irrationality Numbers may need to be rewritten to clearly show if they are rational or not: Number

1.5

0.3333 . . .

0.285714

–20.1

Fraction form

6 1

3 2

1 3

2 7

0 1

201 10

All real numbers are either rational or irrational and a convincing argument shows that there are more irrational numbers than rational numbers. Famous examples of irrational numbers include π (pi) and e (Euler’s number), though proving their irrationality is more advanced. Irrational numbers have decimal representations that are non-terminating and non-repeating, unlike rational numbers which terminate or repeat. The most common proof of irrationality is for 2. Before looking at a proof of whether 2 is irrational or not, consider that there are many fractions 239 that are very close to 2. For example, the number is an extremely good approximation for 2, 169 but its square is 1.99996498722 . . ., and therefore it cannot be 2. While there are many rational

numbers that are very close to 2, none are equal to exactly 2. However, providing many examples does not prove a conjecture. A proof by contradiction is necessary to prove irrationality. Proof by contradiction is often used because it can be very difficult to directly prove that a p number cannot be written in the form . Assuming it can be written that way provides a concrete q algebraic equation to work with, which can then be shown to lead to a logical impossibility.

1.09 Prove irrationality mathspace.co

Example 1 Prove that: a

2 is irrational

Create a strategy m , where n m and n are coprime integers (meaning their greatest common divisor is 1). Then, derive a contradiction from this assumption. To prove by contradiction, assume that 2 is rational. This means it can be written as

Apply the idea Start the assumption and derive a contradiction: 2= 2=

m n

Assume 2 is rational and can be written as

m2 n2

m2 = 2n2

m n

Square both sides Multiply both sides by n2

Analyse the implications of this equation. Since 2 is a factor of m2, m must also be divisible by 2. Let m = 2k for some integer k. (2k)2 = 2n2 2

4k = 2n 2

2k = n

Substitute m = 2k into the equation Evaluate the square Divide both sides by 2

This implies that n is also divisible by 2, so n must be divisible by 2. However, this means that both m and n are divisible by 2, which contradicts the initial assumption that they have no common factors other than 1. Since the assumption that 2 is rational leads to a contradiction, this means 2 is irrational.

Reflect and check The fact that if n2 being even means that n is also even may not be evidently true. This part of the proof can be shown in more detail: Suppose that n was odd and n2 is even, then n = 2a + 1 for some integer a. This means that: n2 = (2a + 1)2 2

Substitute n = 2a + 1

= 4a + 4a + 1

Expand using perfect square pattern

= 2(2a + 2a) + 1

Factorise 2 out of the first two terms

= 2b + 1

Let b = 2a2 + 2a

Therefore, n2 is odd since it is of the form 2b + 1 for some integer b. However, this contradicts that n2 was even, so n must be even if n2 is even. This makes sense since the prime factorisation of a squared number must have each prime factor show up an even number of times, so if 2 is a factor of n2, then its prime factorisation must contain an even power of 2, which would make n even.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3 is irrational

Apply the idea To prove by contradiction, assume that 3 is rational. 3= 3=

p q

Assume 3 is rational and can be written as

p2 q2

3q2 = p2

p q

Square both sides Multiply both sides by q2

Analyse the implications of this equation. Since p2 is a multiple of 3, p must also be divisible by 3. Let p = 3k for some integer k. 3q2 = (3k)2

Substitute p = 3k in the equation

Evaluate the square

Divide both sides by 3

3q = 9k q = 3k 2

This implies q is also divisible by 3, so q must be divisible by 3. This means both p and q are divisible by 3, which contradicts the initial assumption that p and q have no common factors other than 1. So by contradiction, the assumption that 3 is rational must be incorrect and 3 is irrational.

Did you know?

Proving numbers like 3 are irrational ensures engineers use accurate measurements! For example, diagonal struts in bridges rely on these calculations for proper stress distribution. Irrationality proofs prevent design errors that could affect stability.

1.09 Prove irrationality mathspace.co

Example 2 Prove that 5 2 is irrational, given that 2 is irrational.

Create a strategy Use proof by contradiction. Assume that 5 2 is rational. This means it can be written in the p form for integers p and q (q ≠ 0). Then, manipulate the equation to isolate 2 and show that this q implies 2 must be rational, which contradicts the given information.

Apply the idea Assume, for the sake of contradiction, that 5 2 is rational. Then, by the definition of rational numbers, there exist integers p and q, with q ≠ 0, such that: 5

p q

Now, rearrange the equation to isolate 2: p 5 Subtract 5 from both sides q p 5q = Find a common denominator q q p 5q = Combine the terms q Since p and q are integers, p – 5q is also an integer. 2=

Since q is a non-zero integer, q is also a non-zero integer. p 5q is a ratio of two integers with a non-zero denominator, which means it is a q rational number. Therefore,

The equation shows 2

p 5q , implying that 2 is rational. q

However, it is given that 2 is irrational. This is a contradiction. Thus, the initial assumption that 5 2 is rational must be false. Therefore, 5 2 is irrational.

Reflect and check This proof demonstrates a common technique: if a rational number is added (like 5) to a known irrational number (like 2), the result is always irrational. The proof structure relies on the properties of rational numbers (closure under subtraction and division by non-zero integers).

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary Rational numbers can be written as fractions and have decimal representations that either terminate or repeat. Irrational numbers cannot be written as fractions and have decimal representations that are non-terminating and non-repeating. To prove a number has irrationality, proof by contradiction is often necessary. Furthermore, adding a rational number to an irrational number results in an irrational number.

1.09 Practice questions What do you remember? 1

a The proof that 6 is irrational begins by assuming 6 is rational, so that 6 : b a Identify three properties of the variables a and b that the proof requires. b

Fill in the blanks to complete the proof by contradiction that 6 is irrational: Assume that 6 is rational: a b where a and b are integers with no common ⬚ other than 1 and b ≠ ⬚. 6

Square both sides of the equation:

= a b

6=⬚

6b2 = ⬚

This implies that a2 is divisible by ⬚.

Therefore, a must be divisible by ⬚. Let a = ⬚ for some integer k.

Substitute a = 6k into the equation:

6b2 = ⬚

6b2 = 36k2 2

⬚=⬚

This implies that b is divisible by 6, which means b is divisible by ⬚.

Since both a and b are ⬚ by 6, this ⬚ the assumptions that a and b have no common ⬚ other than 1. Therefore, the assumption leads to a contradiction, indicating that 6 must be ⬚.

1.09 Prove irrationality mathspace.co

Fill in the blanks to prove that x is irrational given that 7x = 2. Assume that x is rational: x

p q

where p and q are positive ⬚ with no common ⬚ other than 1 and ⬚. Substitute this into the equation:

7⬚ = 2

Raise both sides of the equation to the power q:

p q

= 2q

⬚=⬚

This implies that the left-hand side of the equation is ⬚ and the ⬚ side of the equation is even. 3

This gives a ⬚, and therefore x cannot be ⬚, it must be ⬚.

Fill in the blanks to complete the proof by contradiction that 3 5 is irrational. Assume that 3 5 is rational: 3

a b

where a and b are integers with no common ⬚ other than 1, and b ≠ ⬚. ⬚ both sides of the equation:

= a b

⬚=⬚ This implies that a3 is divisible by ⬚.

⬚=⬚

Therefore, a must be divisible by ⬚. Let a = ⬚ for some integer k.

Substitute a = ⬚ into the equation:

5b3 = (⬚)3 b3 = ⬚

b3 = ⬚

This implies that ⬚ is divisible by 5, which means b is divisible by ⬚.

Since both a and b are ⬚ by 5, this ⬚ the assumptions that a and b have no common ⬚ other than 1. The assumption leads to a contradiction, indicating that 3 5 must be ⬚. 92

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice 4

Prove by contradiction that 5 is irrational.

Ex 1

Determine whether each number is rational or irrational and justify your answer with a proof: a

144

5 3

5 7

Prove 2 19 is irrational.

Ex 2

Prove 7 13 7 is irrational.

Prove by contradiction that each number is irrational: 1 1 2 a b c 6 19 3

Is it possible for 3 2 to be a rational number? Explain your reasoning.

Extend your thinking 10

Can the square root of a prime number be a rational number? Justify your answer with a proof.

Determine whether each statement is true or false and support your answer with a proof: a

Given a and b are real numbers, and a + b is irrational, then at least one of a and b is irrational.

Given ab is an irrational number, then at least one of a and b is irrational.

Chapter 1 review Online review

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1.09 Prove irrationality mathspace.co

Big ideas • Formal logic provides a precise language for evaluating the truth and validity of statements and arguments. Using tools like truth tables and the rules of Boolean algebra allows for the analysis, simplification, and proof of equivalence for complex logical expressions, establishing a rigorous foundation for reasoning. • The abstract principles of logic are the building blocks of modern computation. Digital systems represent information in binary and perform operations using electronic logic gates, which can be optimised using tools like Karnaugh maps. These systems execute algorithms, structured, step-by-step procedures, to solve complex problems in a predictable and efficient manner.

2 Logic and algorithms Chapter outline Chapter outline 2.01 2.02 2.03 2.04 2.05 2.06 2.07

Propositional logic Proofs with truth tables Boolean algebra and axioms Karnaugh maps Binary numbers Investigation: Circuits Logic gates Design and interpret algorithms Chapter 2 review

96 110 121 128 137 150 161

Binary code uses only 1s and 0s–just like light switches, they’re either on or off.

2.01

Propositional logic After this lesson, you will be able to… • understand propositions, connectives, and truth values • construct and interpret truth tables for logical expressions • identify tautologies, contradictions, and contingencies

Propositions and truth values A proposition is a statement that is either true or false, forming the foundation of logical reasoning. Each proposition has a truth value, denoted T (true) or F (false). Proposition A statement that is either true or false, but not both. Example: “The sum of the interior angles of a triangle is 180°.” This is a true proposition. Truth value The status of a proposition as T or F.

Propositions are classified as simple propositions (single statements) or compound propositions (combined using logical connectives). Truth values are critical for evaluating logical arguments. Simple proposition A standalone statement without connectives. Example: “It is raining.” Compound proposition A statement formed by joining simple propositions with connectives. Example: “It is raining and it is cold.”

It is important to note that propositions must be declarative sentences—statements that assert something. Questions (like “How old are you?”), commands (like “Turn off the light.”), or expressions of opinion (like “Chocolate is the best flavour.”) are not propositions because objective assignment of a single truth value, T or F, is impossible for these statements.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Exploration Explore what makes a statement a proposition. Consider these sentences: • Canberra is the capital of Australia. • What time is it? • Pass the salt. • x + 5 = 10 • This statement is false. • Running is fun. 1. Which of these sentences can be confidently assessed as either true or false? 2. Which sentences cannot be assigned a single true or false value? Why? 3. What characteristics do the sentences identified in question 1 share? 4. Can a sentence be true sometimes and false at other times? 5. If a sentence contains a variable (like x or y), can one assign it a truth value? Under what conditions might it become a proposition?

Example 1 Identify which are propositions: A How old are you?

7>2

C Turn off the light.

The moon is full tonight.

Create a strategy Check if each can be assigned a definite T or F. Questions and commands are not propositions.

Apply the idea Option A: “How old are you?” is a question, not a proposition. Option B: 7 > 2 is a true statement, so it is a proposition. Option C: “Turn off the light.” is a command, not a proposition. Option D: “The moon is full tonight.” can be true or false, so it is a proposition. Options B and D are propositions because assignment of a truth value (T or F) is possible for these statements. Options A and C are not propositions. Therefore, the propositions are Options B and D.

2.01 Propositional logic mathspace.co

Example 2 Determine the truth value of each statement: a 3 is odd.

Create a strategy

Apply the idea

Recall the definition of an odd number. An integer is odd if it is not divisible by 2. Check if 3 fits this definition.

The number 3 is not divisible by 2. Therefore, the statement “3 is odd” is true. The truth value is T.

b 52 is odd.

Create a strategy First, calculate the value of 52. Then, determine if that value meets the definition of an odd number (not divisible by 2).

Apply the idea Calculate 52 = 25. The number 25 is not divisible by 2. Therefore, 25 is odd. The statement “52 is odd” is true. The truth value is T. c (n + 1)2 is odd, where n is an odd integer.

Create a strategy Given that n is an odd integer. Consider what type of number n + 1 must be. Then, consider the square of that type of number. Determine if the result is odd or even.

Apply the idea If n is an odd integer, then n + 1 must be an even integer. Let m = n + 1. Since m is even, its square m2 = (n + 1)2 must also be even. (The square of an even number is always even, e.g., (2k)2 = 4k2 = 2(2k2)). Since (n + 1)2 is even, the statement that it is odd must be false. The truth value is F.

Idea summary A proposition is a statement with a truth value of T or F. A simple proposition stands alone, while a compound proposition uses connectives.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Logical connectives and truth tables Logical connectives combine propositions to form compound propositions. Key connectives include: Negation (NOT, ¬) A logical connective that reverses a proposition’s truth value. ∼ is sometimes used as an alternative notation, e.g., ∼ p means the same as ¬p. Conjunction (AND, ∧) A logical connective ∧ for two propositions, where the resulting compound proposition is true only if both propositions are true. Disjunction (OR, ∨) A logical connective ∨ for two propositions, where the resulting compound proposition is true if at least one of the propositions is true. Other important logical connectives include implication (symbol ⇒, representing “if ... then ...”) and biconditional (symbol ⇔, representing “if and only if”). However, this lesson focuses on negation, conjunction, and disjunction. A truth table shows all possible truth values of a compound proposition based on its components. For a simple proposition, p, the truth table for the negation of p, ¬p, is: p

¬p

For propositions p and q, the truth table for the conjunction, p ∧ q is: p

p∧q

For propositions p and q, the truth table for the disjunction, p ∨ q is: p

p∨q

When constructing a truth table for a compound proposition involving n distinct simple propositions, there will be 2n possible combinations of truth values, requiring 2n rows in the table (excluding the header row).

2.01 Propositional logic mathspace.co

Note that some texts use ∼ p instead of ¬p for negation. For example, the statement “It is not raining” could be written as ∼ p or ¬p, where p is “It is raining.” Both notations are equivalent.

Interactive exploration Discover this concept in action online

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Example 3 Construct truth tables for these logical expressions: a ¬j

Create a strategy To construct the truth table for ¬j, list all possible truth values for the variable j. Since ¬j (the negation of j) depends only on j, only consider whether j is true or false. The negation ¬j is true when j is false and false when j is true.

Apply the idea List the possible values for j: • j=T • j=F For ¬j, evaluate each row: • If j = T, then ¬j = F. • If j = F, then ¬j = T. The truth table is: j

¬j

b j∧k

Create a strategy To construct the truth table for j ∧ k, list all possible combinations of truth values for the variables j and k. Since there are two variables, there are 22 = 4 combinations. The logical AND (∧) is true only when both j and k are true; otherwise, it is false.

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Apply the idea List the possible combinations for j and k: • j = T, k = T • j = T, k = F • j = F, k = T

• j = F, k = F

For j ∧ k, evaluate each row: • j ∧ k is T only when both j = T and k = T. • In all other cases, j ∧ k is F. The truth table is: j

j∧k

Reflect and check To confirm, check that j ∧ k is T only in the row where both j and k are T, which it is. In all other rows, at least one variable is F, so the result is F, matching the definition of AND. An alternative approach could be to recall the truth table for AND from memory and compare, but constructing it step-by-step ensures understanding.

c j∨k

Create a strategy To construct the truth table for j ∨ k, use the same combinations of truth values for j and k. The logical OR (∨) is true when at least one of j or k is true; it is false only when both are false.

Apply the idea List the possible combinations for j and k: • j = T, k = T • j = T, k = F • j = F, k = T

• j = F, k = F

For j ∨ k, evaluate each row: • j ∨ k is T if at least one of j or k is T. • j ∨ k is F only when both j = F and k = F. The truth table is: j

j∨k

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Reflect and check To verify, ensure that j ∨ k is T whenever at least one variable is T, and F only when both are F. This matches the definition of OR. Another way to think about this is to confirm that j ∨ k is false only when both j and k are false, which occurs only in the final row of the table.

Example 4 Digital only example Explore in the online lesson

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Example 5 Construct the truth table for the logical expression p ∧ (q ∨ ¬p).

Create a strategy The expression involves two simple propositions, p and q, so the table will need 22 = 4 rows for the combinations of truth values. Intermediate columns are needed to build up the expression: ¬p, then q ∨ ¬p, and finally the full expression p ∧ (q ∨ ¬p).

Apply the idea First, list all combinations for p and q. Next, determine the values for ¬p by negating the p column. Then, calculate q ∨ ¬p using the values from the q and ¬p columns. Remember OR is true if at least one input is true. Finally, calculate p ∧ (q ∨ ¬p) using the values from the p column and the q ∨ ¬p column. Remember AND is true only if both inputs are true. The complete truth table is: p

¬p

q ∨ ¬p

p ∧ (q ∨ ¬p)

Reflect and check Check each step: • For the row (T, T ): ¬p is F, q ∨ ¬p is T ∨ F = T, p ∧ (q ∨ ¬p) is T ∧ T = T. • For the row (T, F ): ¬p is F, q ∨ ¬p is F ∨ F = F, p ∧ (q ∨ ¬p) is T ∧ F = F. • For the row (F, T ): ¬p is T, q ∨ ¬p is T ∨ T = T, p ∧ (q ∨ ¬p) is F ∧ T = F. • For the row (F, F ): ¬p is T, q ∨ ¬p is F ∨ T = T, p ∧ (q ∨ ¬p) is F ∧ T = F. The calculations match the table.

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Idea summary Logical connectives combine propositions to form compound propositions. Key connectives include negation (“NOT”, ¬) (which reverses the truth value), conjunction (“AND”, ∧) (true only if both propositions are true), and disjunction (“OR”, ∨) (true if at least one proposition is true). A truth table systematically lists all possible truth values for a compound proposition based on its components.

Tautologies, contradictions, and contingencies Compound propositions are classified using truth tables: Tautology Always true, regardless of component truth values. Contradiction Always false, regardless of component truth values. Contingency Sometimes true, sometimes false, depending on component truth values. The most basic tautology is p ∨ ¬p: p

¬p

p ∨ ¬p

¬p

p ∧ ¬p

The most basic contradiction is p ∧ ¬p:

An example of a contingency is p ∧ q: p

p∧q

These classifications help test argument validity in logic.

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Example 6 Determine these statements as tautologies, contradictions, or contingencies: a p: “It is sunny.” p ∨ ¬p: “It is sunny or it is not sunny.”

Create a strategy Construct the truth table of p ∨ ¬p and check the final column. If all values are T, it’s a tautology; if all are F, it’s a contradiction; if mixed, it’s a contingency.

Apply the idea List the possible values for p: • p=T • p=F Construct the truth table by evaluating each column: • ¬p: • If p = T, then ¬p = F. • If p = F, then ¬p = T. • p ∨ ¬p: • ∨ is true if at least one of p or ¬p is T. • For p = T, ¬p = F, so T ∨ F = T. • For p = F, ¬p = T, so F ∨ T = T. The truth table is: p

¬p

p ∨ ¬p

Analyse the final column (p ∨ ¬p): • All values are T. Therefore, p ∨ ¬p is a tautology.

Reflect and check The final column is all T, which confirms p ∨ ¬p is a tautology, as expected since this is a classic law of logic (the law of excluded middle). To double-check, consider the meaning: “It is sunny or it is not sunny” must always be true because p and ¬p are opposites, ensuring one is always true. An alternative approach could be to recognise this as a known tautology without constructing the table, but building the table ensures understanding.

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b q: “The door is open.” q ∧ ¬q: “The door is open and not open.”

Create a strategy Construct the truth table of q ∧ ¬q and check the final column. If all values are T, it’s a tautology; if all are F, it’s a contradiction; if mixed, it’s a contingency.

Apply the idea List the possible values for q: • q=T • q=F Construct the truth table by evaluating each column: • ¬q: • If q = T, then ¬q = F. • If q = F, then ¬q = T. • q ∧ ¬q: • ∧ is true only if both q and ¬q are T. • For q = T, ¬q = F, so T ∧ F = F. • For q = F, ¬q = T, so F ∧ T = F. The truth table is: q

¬q

q ∧ ¬q

Analyse the final column (q ∧ ¬q): • All values are F. Therefore, q ∧ ¬q is a contradiction.

Reflect and check The final column is all F, confirming q ∧ ¬q is a contradiction, which aligns with the logical principle that a statement cannot be both true and false simultaneously. To verify, consider the meaning: “The door is open and not open” cannot be true under any circumstance, as q and ¬q are opposites. An alternative approach could be to recognise this as a known contradiction (law of non-contradiction), but constructing the table reinforces the concept.

2.01 Propositional logic mathspace.co

105

c p: “I am tired.” q: “I will rest.” p ∨ q: “I am tired or I will rest.”

Create a strategy To classify the statement p ∨ q, construct its truth table and examine the final column. If all values are T, it’s a tautology; if all are F, it’s a contradiction; if mixed, it’s a contingency.

Apply the idea List the possible combinations for p and q: • p = T, q = T • p = T, q = F • p = F, q = T • p = F, q = F Construct the truth table by evaluating the expression: • Compute p ∨ q: • ∨ is true if at least one of p or q is T. • For p = T, q = T, T ∨ T = T. • For p = T, q = F, T ∨ F = T. • For p = F, q = T, F ∨ T = T. • For p = F, q = F, F ∨ F = F. The truth table is: p

p∨q

Analyse the final column (p ∨ q): • The values are mixed: three T’s and one F. Therefore, p ∨ q is a contingency.

Reflect and check The final column has both T and F, confirming p ∨ q is a contingency, meaning its truth value depends on the values of p and q. To verify, consider the meaning: “I am tired or I will rest” can be true (if either p or q is true) or false (if both are false), which matches the mixed results. An alternative approach could be to evaluate specific scenarios mentally, but the truth table provides a systematic way to confirm the classification.

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Idea summary Tautologies are always true. Contradictions are always false. Contingencies are sometimes true, sometimes false, identified via truth tables.

2.01 Practice questions What do you remember? 1

How many cases (rows) would a truth table for a simple statement p have?

How many cases (rows) would a truth table for a compound statement involving two distinct simple statements (p, q) have?

How many cases (rows) would a truth table for a compound statement involving three distinct simple statements (p, q, r) have?

Determine whether each statement is true or false: a

The symbol ¬p will always have the opposite truth value as p.

The conjunction p ∧ q is true only when both p and q are true.

The disjunction p ∨ q is false only when both p and q are false.

A contradiction is a compound statement that is always false.

A tautology is a compound statement that is always true.

Let p = “It is cold”, q = “It is raining”, and r = “I wear a coat”. Translate these English sentences into logical notation using p, q, and r:

It is cold and raining.

It is not cold or I wear a coat.

It is cold and I do not wear a coat.

It is not raining.

Determine the truth value of each statement: a

If p is true, what is the truth value of ¬p?

If p is false, what is the truth value of ¬p?

If p is true and q is false, what is p ∧ q?

If p is true and q is false, what is p ∨ q?

Given the truth table for p ∧ q, identify the truth value when p = T and q = F: p

p∧q

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Write the negation for each proposition: a

The door is locked.

It will snow tomorrow.

The car is red.

I am tired.

Practice 7 Ex 1

8 Ex 2

9 Ex 3

For each statement, determine if it is a proposition. If it is, determine its truth value: a

“What time is it?”

5<9

“Close the door.”

“Paris is the capital of Spain.”

Determine the truth value of each proposition: a

2 + 2 = 4

January is the first month of the year.

A square has 5 sides.

10 ≥ 12

Complete the truth tables for these compound propositions: a

¬p ∨ q

p ∧ ¬q

p q p ¬p ¬p ∨ q

Ex 5

p ∧ ¬q

⬚

⬚ ⬚

p∨q

¬(p ∨ q)

What is the truth value of p ∨ q when p = F and q = T?

What is the truth value of ¬(p ∨ q) when p = T and q = F?

For which combination(s) of truth values for p and q is ¬(p ∨ q) true?

⬚ ⬚

Construct the truth tables for each compound proposition: a

¬(p ∧ q)

¬p ∨ ¬q

p ∧ (q ∨ ¬r)

Determine the truth value of these statements when j is true, k is false, and l is false: a

108

¬q

Consider the truth table:

Ex 4

j ∨ (k ∧ l)

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

¬(j ∧ k) ∨ l

13 Ex 6

For each statement: i

Construct a truth table.

ii Identify the statement as a tautology, contradiction, or contingency. Explain your reasoning.

¬p ∨ p

p ∧ ¬q

¬p ∧ p

(p ∨ q) ∧ ¬p

¬(p ∧ q) ∨ r

(p ∧ q) ∨ ¬p

Classify each proposition as tautologies, contradictions, or contingencies: a

“A number is odd or even.”

“A number is odd or prime.”

“A number is odd and divisible by 2.”

Consider the statement: “The motorbike is not run on petrol and the heat is generated by wind power, and the factory is not owned by an electrician.” Let: • p = “The motorbike is run on petrol.” • q = “The heat is generated by wind power.” • r = “The factory is owned by an electrician.” a

Write the statement using logical connectives.

Construct a truth table for the statement. Write T for true and F for false.

c 16

¬p

¬r

q ∧ ¬r

¬p ∧ (q ∧ ¬r)

⬚

⬚ ⬚ ⬚ ⬚ ⬚ ⬚

Describe the situation(s) where the compound statement is true in terms of the context.

For the statement: ¬[(¬(p ∨ q)) ∨ (q ∧ r)]: a

Construct a truth table for the statement.

Is the statement a tautology, contradiction, or contingency? Explain your reasoning.

Under what specific truth conditions for p, q, and r is the entire statement false?

If p is true and r is false, what can be concluded about the statement’s truth value, regardless of q?

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Extend your thinking 17

Compare the truth tables for ¬(p ∧ q) and ¬p ∨ ¬q to examine De Morgan’s Law.

A compound proposition involving two simple propositions, p and q, produces this final column in its truth table: p

Final result

Write a possible logical expression using only the connectives ¬, ∧, and ∨ that would produce this result. Justify your answer by constructing the truth table for your expression. 19

A ‘majority function’ with three inputs p, q, and r is a compound proposition that is true if and only if at least two of the inputs are true: a

Write a logical expression for this function using only the connectives ¬, ∧, and ∨.

Construct a truth table to verify that your expression works correctly.

2.02

Proofs with truth tables

After this lesson, you will be able to… • construct complete truth tables for logical expressions • identify tautologies and proof patterns using truth tables • determine logical equivalence between different expressions

Evaluate expressions with truth tables Truth tables are tools used to evaluate logical expressions by listing all possible combinations of truth values for proposition variables and computing the result using logical connectives: AND (∧), OR (∨), NOT (¬), and implication (⇒). Logical connective An operator (e.g., ∧, ∨, ¬, ⇒) that combines propositions to form a new proposition.

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For n variables, a truth table has 2n rows. The rules for connectives are: • p ∧ q is true only if both p and q are true. • p ∨ q is true if at least one of p or q is true. • If p is true then ¬p is false. Alternatively, If p is false then ¬p is true. • If p is true and q is false, p ⇒ q is false. Otherwise, p ⇒ q is true. The conditional rule (⇒) might seem unusual when p is false. Consider p ⇒ q as a promise: “If p happens, then q will happen.” This promise is only broken if p happens ( p is true) but q does not (q is false). If p never happens ( p is false), the promise is never tested and therefore not broken, so the statement p ⇒ q is considered true by default. Here is the truth table for the conditional: p

p⇒q

Two expressions are equivalent statements if the expressions possess the same truth values for all possible combinations of variables. This is denoted by p ⇔ q or p ≡ q. One way to show two statements are equivalent is to demonstrate that each statement implies the other, i.e., ( p ⇒ q) ∧ (q ⇒ p) is true, which corresponds to the biconditional p ⇔ q. The phrase “if and only if” (abbreviated to iff) is used to refer to equivalent statements. That is, p ⇔ q is true if and only if p and q have the same truth value. Here is the truth table for the biconditional (if and only if): p

p⇔q

Example 1 Construct a complete truth table for each statement: a ( p ∨ q) ∧ (¬r), where p is “It is sunny,” q is “It is warm,” and r is “It is raining.”

Create a strategy The strategy is to evaluate expressions within brackets first. Intermediate columns for these sub-expressions should be completed before combining them. In truth tables, complete the columns for these bracketed parts before applying more complex logical connectives. List all 23 = 8 combinations of truth values for p, q, and r. Compute p ∨ q, then ¬r, and finally ( p ∨ q) ∧ (¬r).

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Apply the idea p

¬r

p∨q

( p ∨ q) ∧ (¬r)

Reflect and check This means the statement “it is sunny or it is warm, and it is not raining” is sometimes true and sometimes false. This makes sense as this statement depends on the weather which is variable. The statement is a contingency. For example, it is false if it is not sunny, not warm, and raining.

b ( p ⇒ q) ∨ (q ⇒ r), where p is “I study,” q is “I pass,” and r is “I graduate.”

Create a strategy Create 23 = 8 rows for p, q, and r. Evaluate p ⇒ q, then q ⇒ r, and combine with ∨.

Apply the idea p

p⇒q

q⇒r

( p ⇒ q) ∨ (q ⇒ r)

Reflect and check This means the statement “If I study, then I pass, or if I pass, then I graduate,” is always true.

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Idea summary Logical connectives (such as ∧, ∨, ¬, ⇒, and ⇔) are used with truth tables to evaluate logical expressions. The implication (⇒) follows the conditional rule: the statement is false only when the antecedent is true and the consequence is false. Two statements are equivalent statements (denoted ⇔ or ≡) if the expressions possess the same truth values for all possible combinations of variables.

Identify tautologies and proof patterns While truth tables can be used to verify any logical statement, mathematicians and programmers rely on established, valid argument forms to construct proofs and algorithms efficiently. These reliable structures are known as proof patterns. The validity of these proof patterns rests on the concept of a tautology—a compound statement that is true in every possible scenario. Proof pattern A standard logical structure ensuring a valid argument, verifiable with truth tables.

Law of syllogism Step

Statement

p⇒q

q⇒r

∴p⇒r

Law of detachment (Modus Ponens) Step

Statement

p⇒q

p is true

∴ q is true Modus Tollens

Step

Statement

p⇒q

¬q

∴ ¬p

This proof pattern, the law of syllogism, allows for a chain of implications to be simplified into a single implication. Symbolically: [( p ⇒ q) ∧ (q ⇒ r)] ⇒ ( p ⇒ r)

This pattern, the law of detachment or Modus Ponens, allows for a conclusion to be made when the hypothesis of a true implication is met. Symbolically: [( p ⇒ q) ∧ p] ⇒ q This proof pattern is called Modus Tollens. This pattern allows the conclusion that the hypothesis is false if the conclusion of a true implication is false. Symbolically: [( p ⇒ q) ∧ ¬q] ⇒ ¬p

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Modus Tollens is often referred to as ‘proof by contrapositive’ because it relies on the logical equivalence between an implication ( p ⇒ q) and its contrapositive (¬q ⇒ ¬p). It can be shown that all of these proof patterns are tautologies.

Example 2 Determine if ( p ⇒ q) ⇒ ((q ⇒ r) ⇒ ( p ⇒ r)) is a tautology.

Create a strategy Construct a truth table and check if the final column is all T.

Apply the idea p

p⇒q

q⇒r

p⇒r

(q ⇒ r) ⇒ ( p ⇒ r)

( p ⇒ q) ⇒ ((q ⇒ r) ⇒ ( p ⇒ r))

The final column is all T, so it is a tautology.

Example 3 For “If p ⇒ q and q ⇒ r, then p ⇒ r”: a Identify the proof pattern.

Create a strategy

Apply the idea

Compare the statement to known proof patterns and describe its structure.

The form is [( p ⇒ q) ∧ (q ⇒ r)] ⇒ ( p ⇒ r), matching the law of syllogism, where two implications chain to a third.

b Use the statement to simplify to a single implication: • If Angela bakes cookies, then her children will eat them. • If Angela’s children eat cookies, then they will not eat dinner.

Create a strategy Write out the simple propositions, then use the proof pattern to simplify.

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Apply the idea Let p represent “Angela bakes cookies” Let q represent “Angela’s children eat cookies” Let r represent “Her children will not eat dinner” • “If Angela bakes cookies, then her children will eat them.” becomes p ⇒ q • “If Angela’s children eat cookies, then they will not eat dinner.” becomes q ⇒ r Using the proof pattern “If p ⇒ q and q ⇒ r, then p ⇒ r” gives: “If Angela bakes cookies, then her children will not eat dinner.”

Reflect and check This demonstrates how the law of syllogism allows for a valid logical chain, connecting the initial premise (“Angela bakes cookies”) to the final conclusion (“her children will not eat dinner”) through the intermediate step (“her children eat them”).

Idea summary A proof pattern is a standard logical structure that ensures a valid argument. These patterns are based on expressions that are always true (tautologies). Key proof patters used in logic and proofs include: • The law of syllogism (chaining implications). • The law of detachment, also known as Modus Ponens (if p ⇒ q and p is true, then q is true). • Modus Tollens (if p ⇒ q and ¬q is true, then ¬p is true).

Determine logical equivalence Two expressions are logically equivalent if their truth values match for all variable combinations, denoted by ≡ or ⇔. Truth tables confirm this by comparing final columns. Logically equivalent Two expressions with identical truth values for all input combinations. Example: ¬( p ∨ q) and ¬p ∧ ¬q (De Morgan’s law). Alternatively, two expressions, Expr1 and Expr2, are logically equivalent if and only if the statement Expr1 ⇔ Expr2 (read as “Expr1 if and only if Expr2”) is a tautology. Checking if the biconditional connecting the two expressions results in a tautology provides another way to verify equivalence. A fundamental example of logical equivalence is the law of contrapositive, which states that an implication is always equivalent to its contrapositive. That is: ( p ⇒ q) ⇔ (¬q ⇒ ¬p) For example, p ⇒ q and ¬p ∨ q are also equivalent, as their truth tables match.

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Example 4 Use truth tables to prove or disprove the logical equivalence: a p ⇒ q and ¬p ∨ q

Create a strategy Build a truth table and compare the columns for p ⇒ q and ¬p ∨ q.

Apply the idea p

p⇒q

¬p

¬p ∨ q

The columns match, indicating logical equivalence.

b ( p ∧ q) ∨ r and ( p ∨ r) ∧ (q ∨ r)

Create a strategy Construct a truth table for 23 = 8 rows and compare the final columns.

Apply the idea p

p∧q

( p ∧ q) ∨ r

p∨r

q∨r

( p ∨ r) ∧ (q ∨ r)

The truth values in the columns for ( p ∧ q) ∨ r and ( p ∨ r) ∧ (q ∨ r) are identical, so the expressions are logically equivalent.

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c ¬( p ∧ q) and ¬p ∨ ¬q

Create a strategy Use a truth table to compare ¬( p ∧ q) and ¬p ∨ ¬q.

Apply the idea p

p∧q

¬( p ∧ q)

¬p

¬q

¬p ∨ ¬q

The columns match, confirming equivalence (De Morgan’s law).

Idea summary Truth tables determine logical equivalence by comparing truth values; identical final columns indicate equivalence, as in De Morgan’s law.

2.02 Practice questions What do you remember? 1

Match each connective to its definition: i

∧

True if at least one proposition is true

True only if both propositions are true

True unless the first is true and the second is false

∨

iii

⇒

Determine whether each statement is true or false: a

Equivalent statements are always true.

Equivalent statements have exactly the same truth values in the answer columns of their truth tables.

An argument is valid when the conclusion does not necessarily follow from the given set of premises.

A statement is a tautology if it is always true.

Identify whether the expression p ∧ ¬p is a tautology, contradiction, or neither.

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Consider the conditional statement [( p ∧ q) ⇒ r] used to determine whether an argument is valid. If this statement is a tautology, is the argument valid or invalid?

Complete the truth table for the expression ¬p ∨ q, where p is “The sky is clear” and q is “It is daytime.”

¬p

¬p ∨ q

⬚

Are p ⇒ q and q ⇒ p logically equivalent?

⬚

Practice Ex 1

Ex 2

Construct complete truth tables for each statement: a

(a ∨ b) ∧ c, where a is “I am tired,” b is “I am hungry,” and c is “I will rest.”

¬(a ⇒ b), where a is “It is cold” and b is “I wear a jacket.”

( p ∧ q) ⇒ r, where p is “I work hard,” q is “I focus,” and r is “I succeed.”

p ∨ (q ⇒ r), where p is “It is Monday,” q is “I have class,” and r is “I attend.”

Construct a truth table for ¬( p ⇔ ¬q) to determine if it is a tautology, contradiction, or contingency. Use T for true and F for false.

Ex 3a

Ex 3b

¬q

p ⇔ ¬q

¬( p ⇔ ¬q)

⬚

Determine if each statement is a tautology: a

( p ⇒ q) ∧ ( p ∧ ¬q)

⬚ ⬚

(a ⇒ b) ⇒ ((b ⇒ c) ⇒ (a ⇒ c))

Determine whether each statement is a tautology using a truth table: a

( p ⇒ q) ⇒ (q ∨ ¬p), where p is “I study” and q is “I pass.”

(a ∧ b) ⇒ a, where a is “It is sunny” and b is “It is warm.”

Identify the proof pattern in each argument and verify its validity with a truth table: a

“If I practice ( p), I improve (q). I practice. Therefore, I improve.”

“If I cook (a), I eat (b). If I eat, I am full (c). Therefore, if I cook, I am full.”

Simplify each set of statements to a single implication: a

• “If the team completes the research phase on time, then the presentation will be well-prepared.” • “If the presentation is well-prepared, then the project will receive a high grade.”

• “If the garden is watered regularly, the plants will grow strong.” • “If the plants grow strong, they will produce abundant fruit.”

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Is the argument valid or invalid? Use a truth table if necessary: • p⇒q • q • ∴p

Which is a valid conclusion that can be drawn from these statements? • If a triangle is a right-angled triangle, then it has a 90° angle. • If a triangle has a 90° angle, then it is not equilateral.

Every triangle is a right-angled triangle.

No sides of a right-angled triangle are equal.

The angles of a triangle sum to 180°.

No right-angled triangles are equilateral.

Given statements: • If a shape is a rectangle, then opposite sides are equal. • A square is a rectangle. Which is a logical conclusion from the given statements?

A square has opposite sides that are equal.

The opposite angles of a square are equal.

The diagonals of a square are equal.

A square is a quadrilateral.

Express each logical argument symbolically, then determine whether the argument is valid. Justify your answer: a

If a triangle is equilateral, then it has three equal sides. Triangle ABC does not have three equal sides, therefore, triangle ABC is not an equilateral triangle.

If a quadrilateral is a rhombus, then it is a parallelogram. If a quadrilateral is a parallelogram, then it has opposite sides that are parallel. Therefore, a rhombus does not have opposite sides that are parallel. 1 1 1 1 If b = 5, then 7b = 35. If 7b = 35, then b 2 . Therefore, if b = 5, then b 2 . 2 2 2 2

c d 17

If x4 = 256, then x = 4. If x = 4, then x3 = 64. x4 = 256, therefore, x3 = 64.

Determine what conclusion, if any, can be drawn from these statements. Indicate whether the law of detachment, law of syllogism, or law of contrapositive was used: a

If a parallelogram is a rhombus, then it has four equal sides. Parallelogram ABCD does not have four equal sides.

If a line divides a line segment into two equal line segments, then the line is a segment bisector. CD divides AB into two equal line segments.

If a figure is a trapezium, then the figure is a quadrilateral. A figure is a trapezium.

If |x| > 8, then f(x) = 2x – 3. |x| ≤ 8.

Determine whether each statement is equivalent to p ⇒ q: a

¬p ∨ ¬q

¬p ∨ q

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Determine whether each statement is equivalent to ¬( p ⇒ q): a

Ex 4

p∧q

p ∧ ¬q

¬p ∨ q

¬p ∨ ¬q

Determine whether each pair of expressions is logically equivalent: a

¬( p ∨ q) and ¬p ∧ ¬q

p ∧ (q ∨ r) and ( p ∧ q) ∨ r

p ⇒ (q ∧ r) and ( p ⇒ q) ∧ ( p ⇒ r)

¬p ⇒ q and p ∨ q

Prove or disprove logical equivalence using truth tables: a

¬(a ∨ b) and ¬a ∧ ¬b, where a is “I am happy” and b is “I am relaxed.”

p ⇒ q and q ⇒ p, where p is “I exercise” and q is “I am fit.”

Extend your thinking 22

Determine whether the statements are equivalent: ¬[¬( p ∧ ¬q)] and p ∧ ¬q

Josh claims that ( p ⇒ q) ∧ (q ⇒ p) is a tautology because “if p implies q and q implies p, they must always be true together.” Use a truth table to evaluate this claim.

Prove that ¬( p ⇒ q) is logically equivalent to p ∧ ¬q using a truth table. Explain how this relates to the concept of a conditional statement.

Prove that ( p ⇒ q) ∧ (q ⇒ r) and p ⇒ r are not logically equivalent, where p is “I read,” q is “I learn,” and r is “I succeed.” Use a truth table and explain where they differ.

Did you know?

Truth tables are invaluable in computer programming! They help developers confirm that conditional statements behave correctly under all circumstances. By analysing each row of a truth table, programmers can prevent unexpected bugs in software. Truth tables simplify the process of debugging complex logical operations!

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2.03 Boolean algebra and axioms After this lesson, you will be able to… • apply the laws and axioms of Boolean algebra to simplify logical expressions • connect Boolean concepts to set notation

Boolean algebra Boolean algebra is a mathematical framework developed by George Boole, widely used in digital logic and computer programming to represent and manipulate logical expressions. It underpins the design of circuits and computer systems by providing rules to simplify complex logic. Boolean algebra A mathematical system for manipulating logical propositions using variables that are either true/on (1) or false/off (0).

In Boolean algebra, there are only two possible values: 0 and 1. These values are often interpreted as ‘false’ (or ‘off’) and ‘true’ (or ‘on’), respectively. A variable that can only take one of these two Boolean values is called a Boolean variable. These are combined using basic operators: conjunction (AND), disjunction (OR), and negation (NOT). The order of operations for Boolean operators is as follows: evaluate expressions within brackets first, then NOT, followed by AND, and finally OR. The behaviour of these operators can be defined using truth tables. For two Boolean variables A and B: AND (Conjunction): A ∧ B A

A∧B

A∨B

OR (Disjunction): A ∨ B

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NOT (Negation): ¬ A A

¬A

Boolean algebra has strong connections to set theory: • Conjunction (A ∧ B) is similar to set intersection (A ∩ B). • Disjunction (A ∨ B) is akin to set union (A ∪ B). • Negation (¬ A) corresponds to set complement (A′).

Example 1 Evaluate each expression using Boolean algebra: a 1∨0∨0∨1

Create a strategy Apply the OR operation sequentially. Remember that X ∨ 0 = X and X ∨ 1 = 1.

Apply the idea 1 ∨ 0 ∨ 0 ∨ 1 = (1 ∨ 0) ∨ 0 ∨ 1

Group the first two terms for sequential evaluation

=1∨0∨1

Apply the OR identity property

= (1 ∨ 0) ∨ 1

Group the remaining terms

=1∨1

Apply the OR identity property

The OR operation of 1 and 1 results in 1

The expression evaluates to 1.

b 1∧1∧0∧1

Create a strategy Apply the AND operation sequentially. Remember that X ∧ 1 = X and X ∧ 0 = 0.

Apply the idea 1 ∧ 1 ∧ 0 ∧ 1 = (1 ∧ 1) ∧ 0 ∧ 1

Group the first two terms for sequential evaluation

=1∧0∧1

The AND operation of 1 and 1 results in 1

= (1 ∧ 0) ∧ 1

Group the next two terms

=0∧1

Apply the property of AND where a term AND 0 results in 0

Apply the AND identity property

The expression evaluates to 0.

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Example 2 Evaluate each expression using Boolean algebra: a ¬1 ∨ 0

Create a strategy First, evaluate the negation ¬ 1. Then, apply the OR operation.

Apply the idea ¬1 ∨ 0 = 0 ∨ 0 =0

Since ¬ 1 = 0 Apply the OR operation

The expression evaluates to 0.

b (1 ∨ 1) ∧ (1 ∨ 0)

Create a strategy Evaluate the expressions within the brackets first, then apply the AND operation.

Apply the idea (1 ∨ 1) ∧ (1 ∨ 0) = 1 ∧ 1 =1

Since 1 ∨ 1 = 1 and 1 ∨ 0 = 1 Apply the AND operation

The expression evaluates to 1.

c (1 ∧ ¬ 1) ∨ ¬ (1 ∧ 1)

Create a strategy Evaluate the negations and expressions within brackets first, then apply the OR operation.

Apply the idea (1 ∧ ¬ 1) ∨ ¬ (1 ∧ 1) = (1 ∧ 0) ∨ ¬ (1)

Evaluate ¬ 1 = 0 and 1 ∧ 1 = 1

=0∨0

Evaluate 1 ∧ 0 = 0 and ¬ 1 = 0

Apply the OR operation

The expression evaluates to 0.

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Idea summary Boolean algebra is a system of logic using Boolean variables that can have only two values: 1 (true) or 0 (false). The basic operations are conjunction (AND or ∧), disjunction (OR or ∨), and negation (NOT or ¬ ). Truth tables define the outcomes of these operations for all possible input values. Boolean algebra concepts are closely related to set theory operations like intersection, union, and complement.

Boolean axioms Boolean algebra follows specific axioms (also known as laws or postulates) to simplify expressions, which is crucial for reducing complexity in logical design and circuit implementation. Axiom Commutative

Expression A∧B=B∧A A∨B=B∨A

Associative

(A ∧ B) ∧ C = A ∧ (B ∧ C) (A ∨ B) ∨ C = A ∨ (B ∨ C)

Distributive

A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C)

Identity

A∧1=A A∨0=A

Annulment

A∨1=1 A∧0=0

Complement

A ∧ ¬A = 0 A ∨ ¬A = 1

Absorption

A ∨ (A ∧ B) = A A ∧ (A ∨ B) = A

Idempotent

A∧A=A A∨A=A

De Morgan’s

¬ (A ∧ B) = ¬ A ∨ ¬ B ¬ (A ∨ B) = ¬ A ∧ ¬ B

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Example 3 Simplify the Boolean expression A ∧ (A ∨ B) using the axioms of Boolean algebra.

Create a strategy Apply the distributive axiom, then use the idempotent and absorption axioms to simplify.

Apply the idea A ∧ (A ∨ B) = (A ∧ A) ∨ (A ∧ B)

Distributive axiom

= A ∨ (A ∧ B)

Idempotent axiom: A ∧ A = A

Absorption axiom: A ∨ (A ∧ B) = A

The simplified expression is A.

Reflect and check This simplification can be verified using a truth table. The alternative notation (∧ for AND, ∨ for OR) will be used for practice. It is to be shown that A ∧ (A ∨ B) is equivalent to A. A

A∨B

A ∧ (A ∨ B)

Comparing the column for A with the column for A ∧ (A ∨ B), the two columns are identical for all combinations of A and B, confirming the simplification.

Example 4 Prove each Boolean equivalence using axioms: a The absorption axiom: A ∨ (A ∧ B) = A

Create a strategy To prove the absorption axiom A ∨ (A ∧ B) = A, one can use other fundamental axioms such as the identity and distributive axioms.

Apply the idea A ∨ (A ∧ B) = A ∧ 1 ∨ A ∧ B

Identity axiom: A = A ∧ 1

= A ∧ (1 ∨ B)

Distributive axiom (factoring out A)

=A∧1

Annulment axiom: 1 ∨ B = 1

Identity axiom: A ∧ 1 = A

Therefore, the absorption axiom A ∨ (A ∧ B) = A is proven.

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Reflect and check An alternative method to demonstrate the absorption axiom A ∨ (A ∧ B) = A is by using a truth table:

A∧B

A ∨ (A ∧ B)

Comparing the first column (A) with the fourth column (A ∨ (A ∧ B)), shows that their truth values are identical for every possible combination of A and B. This demonstrates the validity of the absorption axiom A ∨ (A ∧ B) = A.

b That (A ∧ B) ∨ (¬ A ∧ B) is equivalent to B using Boolean axioms.

Create a strategy Use the distributive axiom to factorise and apply the complement and identity axioms.

Apply the idea (A ∧ B) ∨ (¬ A ∧ B) = B ∧ (A ∨ ¬ A)

Distributive axiom (factorise out B)

=B∧1

Complement axiom: A ∨ ¬ A = 1

Identity axiom: B ∧ 1 = B

Therefore, (A ∧ B) ∨ (¬ A ∧ B) = B.

Idea summary Boolean algebra utilises specific axioms (also known as laws or postulates) to manipulate and simplify logical expressions. Key axioms include commutative, associative, distributive, identity, annulment, complement, absorption, idempotent, and De Morgan’s. Applying these axioms systematically allows for the reduction of complex Boolean expressions to simpler, equivalent forms, which is essential in digital logic design and circuit optimisation.

2.03 Practice questions What do you remember? 1

126

According to the standard order of operations in Boolean algebra, list the three main operators (NOT, AND, OR) specifying the order in which these should be evaluated.

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Describe the relationship between the three basic Boolean operators and their corresponding operations in set theory.

Determine the two expressions that define the complement axiom in Boolean algebra.

Practice 4

Simplify the Boolean expression A ∧ (A ∧ B) using the laws of Boolean algebra.

Evaluate each expression using Boolean algebra:

Ex 1

6 Ex 2

1 ∧ 0 ∧ 1 ∧ 1

0∨1∨1∨0

Evaluate each expression using Boolean algebra: a

¬0 ∧ 1

(1 ∧ 0) ∨ (0 ∨ 1)

¬ (0 ∨ 1) ∨ (1 ∧ ¬ 0)

Using a truth table, verify that A ∨ ¬ A is a tautology.

Simplify the Boolean expression B ∧ (B ∨ A) using the laws of Boolean algebra.

Ex 3

9 Ex 4

Simplify each Boolean expression using Boolean algebra laws: a

A ∨ A ∧ B

¬A ∧ B ∨ A ∧ B

A ∧ (B ∨ ¬ A)

(A ∨ B) ∧ (A ∨ B)

F = A ∨ (B ∧ C)

Express each Boolean function in set notation: a

F = ¬A ∧ B ∨ A ∧ ¬B

Convert the set notation expression into a Boolean expression: (A ∪ B′) ∩ C

Extend your thinking 12

Prove that A ∧ B ∨ A ∧ ¬ B ∨ ¬ A ∧ B simplifies to A ∨ B using Boolean algebra laws.

Simplify the expression F = A ∧ B ∧ C ∨ A ∧ B ∧ ¬ C ∨ ¬ A ∧ B ∧ C using the laws of Boolean algebra.

An alarm system L is activated if a door is opened (D) and the system is armed (A), or if a window is broken (W) and the system is armed (A): a

Express the alarm system L as a Boolean expression.

Simplify the expression using Boolean algebra laws.

Express the simplified function using set notation.

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2.04

Karnaugh maps

After this lesson, you will be able to… • construct Karnaugh maps for logical functions involving 2 or 3 variables • use Karnaugh maps to find minimal forms of Boolean expressions with 2 or 3 variables

Karnaugh maps and truth tables A Karnaugh map (K-map) is a graphical tool used to simplify Boolean expressions by visually representing a truth table. It reduces the need for algebraic manipulation by grouping adjacent terms. Karnaugh map A grid used to visually minimise Boolean functions. Each cell in the grid represents a specific combination of input variables (a minterm) from a truth table. The key to a K-map is its special ordering of variables, known as Gray code. In this order, only one variable changes value between any two adjacent cells. This special layout is important because it places terms that can be simplified together right next to each other on the map (including when wrapping around the edges). The standard templates for 2, 3, and 4-variable K-maps are: • 2-variable K-map (A, B):

A=0 A=1

B=0

B=1

⬚

• 3-variable K-map (A, B, C): Note the Gray code order for BC : 00, 01, 11, 10

A=0 A=1

BC = 00

BC = 01

BC = 11

BC = 10

⬚

• 4-variable K-map (A, B, C, D): using Gray code for both row (AB) and column (CD) variables.

AB = 00 AB = 01 AB = 11 AB = 10

128

CD = 00

CD = 01

CD = 11

CD = 10

⬚

⬚ ⬚

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⬚ ⬚

Example 1 Construct a truth table and a Karnaugh map for the function Y = ¬A ∨ B ∧ C.

Create a strategy First, build a truth table for the 3 variables A, B, and C by evaluating the expression for all 8 possible input combinations. Then, transfer the output values (Y ) into the correct cells of a 2 × 4 K-map.

Apply the idea First, construct the truth table: A

¬A

B∧C

Y = ¬A ∨ B ∧ C

Next, populate the 3-variable K-map using the output column Y. For example, when A = 0, B = 0, C = 0, the output is 1, so a 1 goes in the cell for A = 0 and BC = 00. BC = 00 BC = 01

BC = 11

BC = 10

A=0

A=1

Example 2 Derive the truth table for the function represented by the given Karnaugh map: BC = 00 BC = 01

BC = 11

BC = 10

A=0

A=1

Create a strategy Work in reverse. For each cell in the K-map, identify the corresponding combination of variables A, B, and C, and write its output value (0 or 1) into a new row in the truth table.

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Apply the idea A truth table with 8 rows for all combinations of A, B, and C is created. The output column is then filled in by reading from the K-map. For example, the cell at A = 0, BC = 01 corresponds to the input (A = 0, B = 0, C = 1), and its value is 1. The cell at A = 1, BC = 10 corresponds to the input (A = 1, B = 1, C = 0), and its value is 1. Conversely, the cell at A = 0, BC = 00 corresponds to the input (A = 0, B = 0, C = 0), and its value is 0. A

Output

Idea summary Karnaugh maps are graphical tools that map the outputs of a truth table onto a grid. The grid’s special layout (Gray code) ensures that terms that can be simplified together appear next to each other, making the simplification process visual.

Minimise Boolean expressions with Karnaugh maps Karnaugh maps are used to minimise Boolean expressions into a minimal sum-of-products (SOP) form. This is done by visually identifying and grouping adjacent 1s on the map. Prime implicant A group of 1s on a K-map that is as large as possible. A prime implicant cannot be fully contained within any other single, larger group. Essential prime implicant A prime implicant that covers at least one 1 that no other prime implicant can cover. All essential prime implicants must be included in the final minimal expression.

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Steps to minimise a Boolean expression: 1. Construct the K-map from the function’s truth table. 2. Circle all 1s by forming the largest possible rectangular groups of size 1, 2, 4, 8 (powers of 2). Groups can wrap around the edges of the map and can overlap. Each group formed is a prime implicant. 3. Identify the essential prime implicants. 4. Select the smallest collection of prime implicants that covers all the 1s on the map. Always start with the essential ones, then only add other prime implicants if needed to cover any 1s that are still left ungrouped. 5. Write the final minimised expression as the sum (OR) of the terms represented by the selected groups. BC 00 01

10 ¬ A

B∧C

Grouping for the function Y = ¬A ∨ (B ∧ C). The red group covers the entire row where A = 0, which simplifies to ¬A. The green group covers the column where B = 1 and C = 1, which simplifies to B ∧ C. Both are essential prime implicants. The final expression is the sum of these terms: Y = ¬A ∨ (B ∧ C)

The Boolean operators for AND (∧), OR (∨), and NOT (for example, ¬A) correspond to the set operations for intersection (∩), union (∪), and complement (for example, A′).

Example 3 Use a Karnaugh map to find the minimal sum-of-products form for the function represented by this truth table: A

Create a strategy Populate a 2-variable K-map with the output values from the truth table. Identify the largest possible groups of 1s to find the prime implicants and write the minimal sum-of-products expression.

Apply the idea The K-map for the function is:

B=0

B=1

A=0

A=1

One prime implicant is identified: • A vertical group of two covering the cells where B = 1 (A = 0, B = 1 and A = 1, B = 1). In this group, B is always 1. This gives the term B. This group covers all 1s and is essential. The minimal sum-of-products form is: F = B

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Example 4 Use a Karnaugh map to find the minimal sum-of-products form for the function represented by this truth table: A

Create a strategy First, populate a 3-variable K-map with the output values from the truth table. Then, identify the largest possible groups of 1s to find the prime implicants and write the minimal expression.

Apply the idea The K-map for the function is: BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Two essential prime implicants are identified: • A vertical group of four covering the two middle columns (where BC = 01 and BC = 11). In this group, the variable C is always 1, while A and B change. This group simplifies to the term C. • A horizontal group of two in the top row, covering the cells at BC = 11 and BC = 10. In this group, A is always 0 and B is always 1, while C changes. This gives the term ¬A ∧ B. Both groups are essential because each covers at least one not covered by the other. The minimal sum-of-products form is the sum (OR) of these terms: F = C ∨ (¬A ∧ B)

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Example 5 Express the Boolean function from this K-map in set notation: B=0

B=1

A=0

A=1

Create a strategy Identify the groups in the K-map, write the corresponding Boolean expression in sum-of-products form, and then convert it to set notation using intersection (∩), union (∪), and complement (‘).

Apply the idea The two groups are individual cells (minterms): • The cell at A = 0, B = 0 gives the term ¬A ∧ ¬B. • The cell at A = 1, B = 1 gives the term A ∧ B. The Boolean expression is the sum of these terms: (¬A ∧ ¬B) ∨ (A ∧ B). This is the XNOR function. In set notation, this is expressed as: (A′ ∩ B′) ∪ (A ∩ B).

Idea summary Karnaugh maps minimise Boolean expressions into a minimal sum-of-products (SOP) form by visually grouping adjacent 1s. These largest possible groupings are known as prime implicants. The final minimal expression is built from the smallest required set of these groups, starting with all essential prime implicants.

2.04 Practice questions What do you remember? 1

In a standard 3-variable Karnaugh map with variables A, B, and C, what is the specific Gray code sequence used to label the four columns for the variables BC?

When grouping 1s on a Karnaugh map to minimise an expression, what are the valid group sizes?

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Determine whether each of these statements about Karnaugh maps is true or false: a

A valid group can be formed by circling 1s that are arranged diagonally on the map.

When forming the largest possible groups of 1s, it is permissible for the groups to overlap.

Every essential prime implicant must be included in the final minimal sum-of-products expression.

Each individual cell in a Karnaugh map represents exactly one row from the function’s corresponding truth table.

Practice 4

Construct a 2-variable Karnaugh map for the Boolean function F = A ∧ B ∨ ¬A ∧ ¬B.

For the Boolean function F = A ∧ ¬B ∨ C:

Ex 1

Construct a truth table for the function.

Transfer the output values from the truth table into the 3-variable Karnaugh map:

A=0 A=1 6

Ex 3, 4

BC = 11

BC = 10

⬚

BC = 00 BC = 01

BC = 11

BC = 10

A=0

A=1

Function 1:

Function 2:

A B F

Construct a Karnaugh map and find the minimal form for: F = A ∧ B ∨ ¬B ∧ C

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⬚

Use a Karnaugh map to find the minimal sum-of-products expression for each function, given by its truth table. Show your groupings on the map. a

BC = 01

Use the Karnaugh map to construct the corresponding truth table for the function:

Ex 2

BC = 00

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For each K-map, determine the minimal sum-of-products expression: a

B=0

B=1

A=0

A=1

BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Use a Karnaugh map to find the minimal sum-of-products form for the function represented by the K-map. This will require using wrap-around grouping. BC = 00 BC = 01

BC = 11

BC = 10

A=0

A=1

A company is designing a simple home security system. The alarm (S) should sound if the door is open (D = 1) and the system is armed (A = 1), or if the window is broken (W = 1) regardless of the door or arm status: a

Write a Boolean expression for the alarm (S) based on the conditions described.

Construct a truth table for the 3 variables D, A, and W.

Use a K-map to confirm the minimal form of the expression.

For the 4-variable K-map shown: CD = 00

CD = 01

CD = 11

CD = 10

AB = 00

AB = 01

AB = 11

AB = 10

Identify all prime implicants. Write the Boolean term for each implicant.

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13 Ex 5

For the function represented by the K-map, determine the minimal sum-of-products expression and then express it in set notation: BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Extend your thinking 14

A 4-variable Boolean function, G (A, B, C, D), is defined by the truth table. Use a Karnaugh map to find its minimal sum-of-products form. A

A 3-variable function has the minimal sum-of-products form F = A ∨ ¬B ∧ C: a

Construct the Karnaugh map for this function.

The expression F = A ∧ B ∧ C ∨ A ∧ B ∧ ¬C ∨ A ∧ ¬B ∧ C ∨ A ∧ ¬B ∧ ¬C ∨ ¬A ∧ ¬B ∧ C is the full sum-of-products form (disjunctive normal form) for this function. Explain why this expression simplifies to the minimal form F = A ∨ ¬B ∧ C, using your K-map from part (a) as evidence.

The K-map contains a redundant prime implicant: BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Identify all three prime implicants in the K-map.

Which prime implicant is redundant (not essential)? Explain why.

Write the minimal sum-of-products expression.

Convert each Boolean expression into its minimal sum-of-products form using a Karnaugh map: F (A, B, C) = (¬A ∨ ¬B) ∧ (B ∨ ¬C)

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2.05 Binary numbers After this lesson, you will be able to… • understand the structure of binary number systems • convert between binary and base 10 numbers • perform operations with binary number • understand that truth is binary

Binary numbers Decimal

Binary

A base-10 number system using digits 0 to 9, where each place value is a power of 10.

A base-2 number system using digits 0 and 1, where each place value is a power of 2. This system is foundational to computer science.

Counting is a fundamental skill, and while a decimal (base-10) system with digits 0 to 9 is typically used, computers rely on a binary (base-2) system using only 0 and 1. This base-2 system is essential in computer science because it aligns with electronic circuits that operate in two states: on (1) and off (0). In decimal, each place value is a power of 10 (e.g., units, tens, hundreds). In binary, each place value is a power of 2 (for example: 1s, 2s, 4s, 8s). For example, the decimal number 2019 is written as: 2 × 103 + 0 × 102 + 1 × 101 + 9 × 100 Binary counting follows a pattern: the units place (rightmost digit) cycles from 0 to 1. When it reaches 1 and needs to increment further, it resets to 0 and carries over a 1 to the next higher place value (2s) which then increases by 1. This carrying process continues across higher place values, similar to how carrying over occurs in decimal arithmetic. Units increase from 0 to 9 0

50 60

72 82

75 85

76 86

Units increase from 0 to 1

Groups of 2 increase from 0 to 1

Groups of 10 increase from 0 to 9

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Interactive exploration Discover this concept in action online

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Beyond numerical representation, binary numbers connect to Boolean logic, where 1 represents true and 0 represents false. This is because many logical statements are either true or false, with no middle ground. This binary nature is fundamental to computer decision-making. Decimal

Binary

Truth value

False

True

This mapping allows a computer to use binary digits to represent logical statements. If each statement is represented by a binary digit, the truth of a combination of those statements can be determined through binary operations, bridging number systems and computational logic.

Example 1 Consider the binary number 11012: a What is the place value of the leftmost 1?

Create a strategy Identify the position of each digit, starting from 0 on the right. The place value is 2 raised to the power of the position number.

Apply the idea The binary number is 11012. The digits and their positions from right to left are: • 1 is in position 0 (20 = 1s place). • 0 is in position 1 (21 = 2s place). • 1 is in position 2 (22 = 4s place). • The leftmost 1 is in position 3 (23 = 8s place). The place value of the leftmost 1 is the 8s place.

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b What is the next largest binary number?

Create a strategy To find the next largest binary number, add 1 to the given number, following the rules of binary addition (where 1 + 1 = 102, meaning 0 with a carry of 1).

Apply the idea Add 1 to 11012: 1 1 1 0 1 1 1 1 1 0 Starting from the right: • 1 + 1 = 10 (Write down 0, carry over 1.) • 0 + 0 + 1 (Carry) = 1 • 1+0=1 • 1+0=1 The next largest binary number is 11102.

c What is the previous binary number?

Create a strategy To find the previous binary number, subtract 1 from the given number, following the rules of binary subtraction (where borrowing may be required).

Apply the idea Subtract 1 from 11012: 1 1 0 1 1 – 1 1 0 0 Starting from the right: • 1–1=0 • 0–0=0 • 1–0=1 • 1–0=1 The previous binary number is 11002.

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Example 2 Show how binary represents the statement “It is raining” as true or false. Assume that it is currently raining on April 5, 2025, in Melbourne.

Create a strategy To represent the statement “It is raining” in binary, assign a binary value based on the truth of the statement. Use 1 to represent true (the statement is correct) and 0 to represent false (the statement is incorrect). Determine the current truth value of the statement based on the given scenario, and assign the corresponding binary digit.

Apply the idea Identify the statement and the given scenario: • Statement: “It is raining.” • Scenario: It is currently raining on April 5, 2025, in Melbourne. Determine the truth value of the statement: • Since it is currently raining, the statement “It is raining” is true. Assign the binary value based on the truth value: • True corresponds to 1 in binary. Consider the alternative case for completeness: • If it were not raining, the statement would be false, corresponding to 0 in binary. Therefore, since it is raining, the binary representation of the statement “It is raining” is 1.

Reflect and check To verify, the assignment of binary values can be reviewed: • The statement “It is raining” is true because the scenario states that it is raining on April 5, 2025, in Melbourne. • According to the strategy, true is represented by 1, which matches the result. • If the scenario were different (e.g., it is not raining), the statement would be false, and 0 would be assigned, confirming that the method correctly handles both cases. This aligns with how binary is used in digital systems: binary digits (bits) represent logical states, where 1 often means “on” or “true,” and 0 means “off” or “false.” Thus, the binary representation of the statement as true is correctly 1.

Idea summary While the common system is decimal (base-10), the binary system (base-2) uses the digits 0 and 1 and is foundational for computing. Each position in a binary number represents a power of 2. Binary digits are also used to represent truth values in logic, where 1 means true and 0 means false.

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Convert between binary and decimal Converting between binary and decimal is a key skill for understanding how computers process numbers. There are straightforward methods for converting in both directions. Binary to decimal conversion To convert a binary number to its decimal equivalent, multiply each binary digit by its corresponding place value (a power of 2) and sum the results:

Decimal value ( d 2p) d is the binary digit (either 0 or 1) p is the position of the digit, starting from 0 for the rightmost digit Decimal to binary conversion There are two common methods to convert a decimal number to binary: • Method 1: Successive division by 2 This method involves repeatedly dividing the decimal number by 2 and recording the remainder. The process continues until the quotient is 0. The binary number is formed by reading the remainders in reverse order (from last to first). 1 FUNCTION DecimalToBinary(n) 2 IF n = 0 THEN 3 RETURN “0” 4 END IF 5 binaryString ← “” 6 WHILE n > 0 DO 7 remainder ← n MOD 2 8 binaryString ← remainder AS STRING + binaryString 9 n ← n DIV 2 10 END WHILE 11 RETURN binaryString 12 END FUNCTION • Method 2: Place value subtraction Find the largest power of 2 that is less than or equal to the decimal number. Place a ‘1’ in that binary position, subtract that power of 2 from the number, and repeat the process with the remainder. For any place values skipped, place a ‘0’.

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Example 3 Convert each binary number to its decimal form: a 10102

Create a strategy To convert the binary number 10102 to decimal, assign each digit a position starting from the right (position 0). Sum the values of the positions where the digit is 1. Each position n corresponds to a value of 2n.

Apply the idea Write the binary number with its place values (powers of 2): Binary digit

Place value

Value

Multiply each digit by its place value and sum the results: (1 × 23) + (0 × 22) + (1 × 21) + (0 × 20) = (1 × 8) + (0 × 4) + (1 × 2) + (0 × 1) Calculate the value of each power of 2 =8+0+2+0

Evaluate the multiplication

= 10

Evaluate the addition

Therefore, the decimal equivalent of 10102 is 1010.

b 110112

Create a strategy Identify the place value for each digit in the binary number 110112. Sum the place values that correspond to a ‘1’.

Apply the idea Write the binary number with its place values: Binary digit

Place value

Value

Sum the values for the positions with a ‘1’: 16 + 8 + 2 + 1 = 27 Therefore, the decimal equivalent of 110112 is 2710.

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Example 4 Convert each decimal number to its binary form: a 610

Create a strategy Use the method of successive division by 2. Divide 6 by 2, note the remainder, then divide the resulting quotient by 2. Continue until the quotient is 0. Read the remainders from bottom to top.

Apply the idea • 6 ÷ 2 = 3 with a remainder of 0 • 3 ÷ 2 = 1 with a remainder of 1 • 1 ÷ 2 = 0 with a remainder of 1 Reading the remainders from bottom up gives 110. Therefore, 610 = 1102.

b 1310, using division

Create a strategy To convert the decimal number 1310 to binary, use the method of successive division by 2, recording the remainders at each step. Start by dividing the number by 2, noting the quotient and remainder. Continue dividing the quotient by 2 until the quotient becomes 0. The binary representation is the sequence of remainders read from bottom to top.

Apply the idea Perform successive division by 2, recording the quotient and remainder at each step: • Step 1: 13 ÷ 2 = 6, remainder: 1 • Step 2: 6 ÷ 2 = 3, remainder: 0 • Step 3: 3 ÷ 2 = 1, remainder: 1 • Step 4: 1 ÷ 2 = 0, remainder: 1 Stop when the quotient is 0. Read the sequence of remainders from bottom to top: 1101 Therefore, 1310 = 11012.

Reflect and check To verify, convert 11012 back to decimal: (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20) = 8 + 4 + 0 + 1 = 13 The result matches.

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c 92110, using place values

Create a strategy To convert 92110 to binary using place values, start by identifying the largest power of 2 that is less than or equal to 921. Subtract this power from 921, record a 1 for that position, and repeat the process with the remainder. For each power of 2 that is not used, record a 0.

Apply the idea The powers of 2 are: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 . . . The largest power of 2 less than 921 is 512 (29). Working

Powers of 2

Binary digit

921 – 512 = 409

512 (2 )

409 – 256 = 153

256 (2 )

128 (2 )

25 < 64

64 (2 )

25 < 32

32 (2 )

16 (2 )

8 (2 )

1<4

4 (2 )

1<2

2 (21)

153 – 128 = 25

25 – 16 = 9 9–8=1

1–1=0

1 (2 )

Reading the binary digits from top to bottom gives 1110011001. Therefore, 92110 = 11100110012.

Idea summary To convert from binary to decimal, sum the products of each digit and its place value (powers of 2). To convert from decimal to binary, either can be used: • Successive division by 2: Reading the remainders in reverse order • Place value subtraction: Finding the largest powers of 2 that fit into the number

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Operations with binary numbers Just like with decimal numbers, arithmetic operations such as addition and multiplication can be performed with binary numbers. The rules are simpler because there are only two digits. • Binary addition The basic rules for binary addition are: • 0+0=0 • 0+1=1 • 1+0=1 • 1 + 1 = 10 (0, carry over 1) • 1 + 1 + 1 = 11 (1, carry over 1) When adding, work from right to left, column by column, carrying over any values greater than 1 to the next column. • Binary multiplication Binary multiplication is also straightforward. The rules are: • 0 × 0 = 0 • 0×1=0 • 1 × 0 = 0 • 1×1=1 The process is similar to decimal long multiplication: multiply the top number by each digit of the bottom number, shift the results, and then add them all together. Alternatively, for any operation, convert the binary numbers to decimal, perform the operation, and then convert the result back to binary.

Example 5 Evaluate each binary expression: a 112 + 1012

Create a strategy Align the numbers by place value, padding the shorter number with leading zeros if necessary. Add column by column from right to left, carrying over where necessary.

Apply the idea To add the numbers, align them vertically. Pad 112 with a leading zero to become 0112 for clear alignment: 1

0 1 1 + 1 0 1 1 0 0 0 • 1s place: 1 + 1 = 10. (Write 0, carry 1.) • 2s place: 1 + 0 + 1 (carry) = 10. (Write 0, carry 1.) • 4s place: 0 + 1 + 1 (carry) = 10. (Write 0, carry 1.) • 8s place: Bring down the carried 1. The result is 10002.

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Reflect and check Check by converting to decimal: 112 = 310, 1012 = 510; 3 + 5 = 8 The result 10002 is 8 in decimal. b 11012 + 10112

Create a strategy 1 1 0 1 Use column addition, adding from right to left and carrying any sums of 2 or more. × 1 0 1 1 1 0 1 Apply the idea 0 0 0 00 + 1 1 0 1 00 1 1 1 1 0 0 0 0 0 1 1 1 0 1 + 1 0 1 1 1 1 0 0 0 • 1s place: 1 + 1 = 10 (Write 0, carry 1.) • 2s place: 0 + 1 + 1 (carry) = 10 (Write 0, carry 1.) • 4s place: 1 + 0 + 1 (carry) = 10 (Write 0, carry 1.) • 8s place: 1 + 1 + 1 (carry) = 11 (Write 1, carry 1.) • 16s place: Bring down the carried 1. The result is 110002.

Example 6 Evaluate each binary expression: a 1012 × 112

Create a strategy Use long multiplication. Multiply 1012 by the 1s digit of 112, then by the 2s digit (shifting the result left), and finally add the partial products.

Apply the idea 101 1 1 101 + 1010 1 11 1

• Multiply by the rightmost 1: 101 × 1 = 101 • Multiply by the next 1: 101 × 1 = 101 (Shift this result one place to the left, giving 1010.) • Add the partial products: 101 + 1010 = 1111 The result is 11112.

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b 11012 × 1012

Create a strategy Use binary long multiplication. Multiply 11012 by each digit of 1012, shifting each partial product to the left, then sum the results.

Apply the idea 1101 101 1101 0 0 0 00 + 1 10 1 00 1000001

• First partial product (×1): 1101 • Second partial product (×0, shifted): 00000 • Third partial product (×1, shifted twice): 110100 • Add the three partial products: 1101 + 0 + 110100 = 10000012 The result is 10000012.

Idea summary Binary arithmetic follows simple rules for addition and multiplication. Addition involves carrying over values of 2 (written as 102) to the next place value. Multiplication is performed similarly to decimal long multiplication, involving shifting and adding partial products. Operations can also be performed by converting to decimal, calculating, and converting back to binary.

2.05 Practice questions What do you remember? 1

What is the base of the binary number system, and what digits does it use?

How does the base of the binary number system differ from the base-10 system used in everyday counting?

In the context of binary numbers, what do the digits 0 and 1 represent when used to model truth values?

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In binary, what does each place value represent?

Write the first five binary numbers starting from 02 to demonstrate the counting pattern in base-2.

Practice Ex 1

Ex 2

Consider the binary number 10112: a

What is the place value of the leftmost 1?

What is the next largest binary number?

What is the previous binary number?

If “It is sunny” is true on April 6, 2025, in Melbourne, what is the binary representation of this statement?

Represent the statement “The sky is blue” in binary, assuming it is true today, May 7, 2025, in your location.

A computer uses binary to represent the truth of the statement “The light is on.” Assign binary values to each scenario and explain your reasoning: a

Ex 3

Ex 4

The light is off.

100002

100012

11012

1012

11102

100112

10102

111112

910

Convert these decimal numbers to binary (base-2 form): 810

310

1210

Convert each decimal number to binary using successive division: a

Convert these binary numbers to decimal (base-10 form):

a 12

The light is on.

710

1910

Convert each decimal number to binary using place values: a

1010

2510

Use the place value method to explain how the binary number 10112 is converted to its decimal equivalent.

For each number, perform the conversion and show your working as instructed:

148

Convert 2310 to binary using the successive division method.

Convert 110102 to decimal, then verify the answer by converting it back to binary.

Convert 3110 to binary using the place value method.

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Ex 5

Calculate each expression in binary: a

Ex 6

1012 + 102

1112 + 12

10112 + 1102

1012 + 11102

Calculate each using binary multiplication: a

112 × 102

1112 × 102

1102 × 102

Extend your thinking 18

A data set contains n binary values (each either 0 or 1). The mean of the data set is calculated as the sum of the values divided by n. If the mean is 0.6 and n = 10, how many values are 1? Connect this to the concept of binary truth.

Convert the decimal number 4210 to binary. Then, using prior knowledge of Boolean algebra, interpret the binary result as a truth value assignment for a logical expression with 6 variables (e.g., A, B, C, D, E, F), where 1 is true and 0 is false. Suggest a possible logical expression.

Explain why the binary number system is limited to 0 and 1, and how this relates to electronic circuits and truth values.

Convert 10010 to binary using place values and verify by converting back to decimal. How many digits does the binary representation have, and why?

A binary number has 8 digits, all 1s (e.g., 111111112). Convert this to decimal and explain how this relates to the maximum value of an 8-bit binary number.

The octal number system is base-8 and uses the digits 0 – 7. Similar to binary (base-2) and decimal (base-10), each place value represents a power of the base: a

Convert the octal number 238 to its decimal (base-10) equivalent.

Show that each octal digit corresponds to a unique 3-bit binary sequence. Use this relationship to convert 238 to binary and verify your answer. (Hint: 23 = 8).

Investigation: Circuits Investigate online

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2.06 Logic gates After this lesson, you will be able to… • identify and use standard logic gate symbols • interpret circuit diagrams in terms of logical functions • apply Boolean algebra and Karnaugh maps to simplify circuits

Introduction to logic gates Logic gates Electronic components that perform basic logical operations on binary inputs (0 or 1) to produce a binary output. The basic types of logic gates are: • AND gate: Outputs 1 only if all inputs are 1. The logical expression is A ∧ B. • OR gate: Outputs 1 if at least one input is 1. The logical expression is A ∨ B. • NOT gate: Outputs the inverse of the input, 1 becomes 0 and 0 becomes 1. The logical expression is ¬ A. This is sometimes also written as A′ or Ā. Each gate has a standard symbol used in circuit diagrams: Name

AND

Graphical symbol

A B

Logical operator

A∧B

A∨B

Truth table A

A NOT

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¬A

NAND

NOR

XOR

A↑B

A↓B

A⊕B

Logic gates are the building blocks of modern digital devices. These gates are used in everything from microprocessors in computers and smartphones to control systems in traffic lights and home appliances. For example, a simple security system might use an AND gate to trigger an alarm only if a window sensor AND a motion detector are both activated. Truth tables summarise the output of each gate for all possible input combinations.

Example 1 Identify the logic gate symbol and convert it to logic notation: C D

Create a strategy First, identify the type of logic gate shown by its unique symbol. Then, write the corresponding logical expression using the input variables provided, which are C and D.

Apply the idea The symbol shown is for an OR gate. With inputs C and D, the logical expression is C ∨ D.

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Example 2 Construct the truth table for an AND gate with inputs C and D.

Create a strategy List all possible input combinations and determine the output for the gate based on its operation.

Apply the idea The truth table for AND gate: C

C∧D

Idea summary Logic gates, such as the AND gate, OR gate, and NOT gate, are basic components of digital circuits that perform binary operations, represented by standard symbols and defined by truth tables.

Design circuit diagrams Complex circuits are built by combining logic gates to implement logical functions. Inputs flow through gates, with each gate’s output feeding into the next as needed. Rules for connecting logic gates in digital circuits: • Inputs must be binary (0 or 1) or outputs from other gates, with outputs connecting to inputs of subsequent gates. • Each gate processes its inputs according to its type (AND, OR, NOT), following the logical expression’s structure. • Ensure proper order of operations (for example, parentheses in expressions dictate gate sequence). • The final output reflects the combined logical function of the circuit. Logical functions (for example, A ∧ B) are translated into circuit diagrams by identifying the operations, mapping them to appropriate gates, and arranging and connecting the gates in sequence.

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Example 3 Design a circuit diagram for the function A ∧ B and determine the output for inputs A = 1 and B = 0.

Create a strategy Use an AND gate with inputs A and B, then evaluate the output using the gate’s truth table.

Apply the idea A

Output 0

B 1 0 AND gate

For A = 1, B = 0, the AND gate outputs 1 ∧ 0 = 0, since both inputs must be 1 for the output to be 1.

Example 4 Digital only example Explore in the online lesson

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Example 5 Derive the logical expression of the circuit diagram: A B

Create a strategy The strategy is to trace the circuit from the inputs on the left to the final output on the right. Each gate is identified in order, and its output expression is written. These smaller expressions are then combined as they feed into subsequent gates to build the final expression.

Apply the idea Step 1: The top gate is an AND gate with inputs A and B. Its output is A ∧ B. Step 2: The middle gate is a NOT gate with input C. Its output is ¬ C. Step 3: The final gate on the right is an OR gate. It takes the outputs from the first two gates as its inputs. Therefore, the final logical expression for the entire circuit is (A ∧ B) ∨ (¬ C).

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Idea summary Circuit diagrams are designed by combining logic gates (AND, OR, NOT) to implement complex logical functions, connecting them according to the expression’s structure and following connection rules to produce the desired output.

Simplify circuits using Boolean algebra and Karnaugh maps Boolean algebra provides a mathematical system to simplify logical expressions using operators like AND, OR, and NOT, reducing the number of gates needed in a circuit. Its laws enable algebraic simplification, while Karnaugh maps (K-maps) offer a visual method to achieve the same by grouping minterms. Key laws of Boolean algebra include: Law

AND form

OR form

Commutative

A∧B=B∧A

A∨B=B∨A

Associative

(A ∧ B) ∧ C = A ∧ (B ∧ C)

(A ∨ B) ∨ C = A ∨ (B ∨ C)

Distributive

A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C)

A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C)

Identity

A∧1=A

A∨0=A

Annulment

A∧0=0

A∨1=1

Complement

A ∧ ¬A = 0

A ∨ ¬A = 1

Absorption

A ∧ (A ∨ B) = A

A ∨ (A ∧ B) = A

De Morgan’s

¬ (A ∧ B) = ¬ A ∨ ¬ B

¬ (A ∨ B) = ¬ A ∧ ¬ B

Karnaugh maps are a visual tool used to simplify Boolean expressions to their minimal Sum-of-Products (SoP) form. This is achieved by grouping adjacent cells containing a 1 into the largest possible rectangular blocks of size 2n cells (i.e., groups of 1, 2, 4, 8, etc.). The map is arranged using Gray code ordering (00, 01, 11, 10), where only one bit changes between adjacent columns or rows. This special ordering ensures that physically adjacent cells are also logically adjacent. Steps for simplification: 1. The Boolean expression is derived from the initial circuit diagram. 2. The expression is then simplified, either algebraically using Boolean laws or visually by constructing a Karnaugh map. 3. For a K-map, adjacent terms containing a 1 are grouped to find the minimal sum-of-products form.

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Example 6 For the function represented by this truth table: A

Output

a Use a Karnaugh map to find the minimal sum-of-products form for the function.

Create a strategy First, populate a 3-variable Karnaugh map with the output values from the truth table. Then, identify the largest possible groups of adjacent 1s to find the essential prime implicants and form the minimal expression.

Apply the idea The outputs from the truth table are placed into a 2x4 K-map. BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Next, the 1s are grouped into the largest possible blocks. BC

Two groups are required to cover all the 1s: 1. Group 1 (Blue): A group of four 1s occupies the two rightmost columns (where B = 1). Across these columns, A changes (is eliminated) and C changes (is eliminated), while B remains 1. This group simplifies to the term B. 2. Group 2 (Red): A group of two 1s in the top row (where A = 0) covers the columns where C = 1. In this group, A is 0 and C is 1, while B changes (is eliminated). This group simplifies to the term ¬ A ∧ C. The final simplified sum-of-products expression is the OR of the group terms: B ∨ (¬ A ∧ C)

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b Draw the simplified circuit diagram for the minimal expression.

Create a strategy Translate the simplified Boolean expression B ∨ (¬ A ∧ C) into a circuit diagram by identifying the necessary logic gates (NOT, AND, OR) and connecting them according to the expression’s structure.

Apply the idea The expression B ∨ (¬ A ∧ C) requires three logic gates: • A NOT gate to invert input A to produce ¬ A. • An AND gate to combine ¬ A and C, producing (¬ A ∧ C). • An OR gate to combine B and the output of the AND gate. The resulting circuit diagram is: A F B C

Example 7 Digital only example Explore in the online lesson

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Example 8 Consider this logic circuit:

Derive its Boolean expression, simplify it using Boolean algebra, and then draw the more efficient, simplified circuit diagram.

Create a strategy The strategy involves three parts. First, the circuit is analysed to derive its full Boolean expression. Second, the laws of Boolean algebra are applied to simplify this expression into its minimal form. Third, a new circuit diagram is drawn based on the simplified expression.

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Apply the idea Step 1: Derive the Boolean expression The expression is determined by following the path of the inputs: • Input A is split. One path goes to a NOT gate, producing the term ¬ A. • This ¬ A term and input B enter an AND gate. The output of this gate is ¬ A ∧ B. • The original input A and the output of the AND gate (¬ A ∧ B) enter the final OR gate. The full expression for the circuit’s output F is F = A ∨ (¬ A ∧ B). Step 2: Simplify the expression The laws of Boolean algebra are now applied to simplify F. F = A ∨ (¬ A ∧ B)

Initial expression

= (A ∨ ¬ A) ∧ (A ∨ B)

Distributive law

= 1 ∧ (A ∨ B)

Complement law: A ∨ ¬ A = 1

=A∨B

Identity law: 1 ∧ X = X

The simplified expression is F = A ∨ B. Step 3: Draw the simplified circuit The simplified expression only requires a single OR gate. A F B

This new circuit is more efficient as it uses only one gate instead of three, but it produces the exact same output for all possible inputs.

Idea summary Boolean algebra and Karnaugh maps simplify circuits by reducing complex logical expressions into fewer gates, using algebraic laws (e.g., distributive, complement) and visual grouping of minterms to optimise design efficiency.

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2.06 Practice questions What do you remember? 1

Identify the binary values used as inputs in logic gates and recall their meaning.

Complete the truth table for a NOT gate with input A: A

¬A

⬚ ⬚

1 3

Complete the truth table for an OR gate with inputs A and B: A

A OR B

⬚

⬚ ⬚ ⬚

Design a circuit diagram for the logical function ¬ A using a single logic gate.

Practice Ex 1

Identify each logic gate and write its logical expression for the given symbol: a

Ex 2

Ex 3

c Z

Y Z

Construct truth tables for each logic gate: a

OR gate with inputs Q and R

AND gate with inputs P and Q

NOT gate with input R

Construct a complete truth table for the logical function F = A ∧ (B ∨ C).

For each function:

158

Design a circuit diagram.

Determine the output for the given input values.

A ∨ B with inputs A = 1, B = 0

¬ A ∨ B with inputs A = 1, B = 0

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9 Ex 4

Ex 5

For each function: i

Design a circuit diagram.

Determine the output for the given input values.

(P ∨ ¬ Q) ∧ R with inputs P = 1, Q = 1, R = 0

P ∧ (¬ Q ∨ R) with inputs P = 1, Q = 0, R = 1

Determine the Boolean expression that represents the function of this circuit: A B Y C

Simplify these Boolean expressions using algebraic laws: a

Ex 6

13 Ex 7

A ∨ (A ∧ B)

¬ A ∨ (A ∧ B)

For the function represented by this truth table: A

Output

Use a Karnaugh map to find the minimal sum-of-products form for the function.

Draw the simplified circuit diagram for the minimal expression.

For the function represented by this truth table: a

Write the unsimplified sum-of-products expression for the function.

Simplify the expression using the laws of Boolean algebra. Show your working.

Draw the logic circuit diagram for the simplified expression.

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Ex 8

Simplify the Boolean expression using a Karnaugh map: A

Output

Simplify this circuit using Boolean algebra and draw the resulting simplified circuit diagram.

A F

B C

Extend your thinking 16

Design a circuit for ¬ (A ∨ B) ∧ C and simplify it using Boolean algebra. Compare the original and the simplified circuits.

Use a Karnaugh map to simplify the circuit given by this truth table, then draw the simplified circuit:

Output

Simplify the given circuit using a Karnaugh map and design the simplified circuit. If A = 1, B = 0, C = 1, predict the output and explain any redundancy.

A B C D

Output

A circuit has inputs A, B, C, and an unused input D = 1. Design a circuit for A ∨ (B ∧ C), then simplify it assuming C = 0 is fixed. Explain how this constraint affects the design.

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2.07 Design and interpret algorithms After this lesson, you will be able to… • define what constitutes an algorithm • identify the key elements of algorithm design • interpret algorithms written in a flowchart • trace the execution of simple algorithms in flowcharts or pseudocode with given inputs

Key elements of algorithm design Algorithm A process that can be carried out systematically, using a well-defined set of instructions, to perform a task or solve a problem. In computer science and mathematics, algorithms are essential for breaking down complex problems into manageable steps. Every algorithm is built using three fundamental constructs: Sequence The overall order in which instructions are executed one after another. Every algorithm follows a sequence from start to finish. Decision A selection process where the flow of instructions branches, executing different steps based on whether a condition is true or false. Repetition The process of repeating a block of instructions until a condition is met, often implemented with “FOR” or "WHILE" loops.

Algorithms can be represented in several ways. Two common methods are pseudocode, which uses plain language and code-like structures, and a flowchart, which uses symbols and arrows to show the flow of logic.

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Pseudocode A simplified, language-independent way to describe an algorithm using a mix of natural language and programming conventions. Flowchart A diagram that represents an algorithm, showing the steps as boxes of various kinds, and their order by connecting them with arrows.

Pseudocode structures Common keywords include: • INPUT/READ: To get data from the user • PRINT or OUTPUT/DISPLAY: To show a result to the user • SET/INITIALISE or ←: To assign a value to a variable The structure for a decision is often written as: 1

IF condition is true THEN

2 instructions to follow if true 3

ELSE

4 instructions to follow if false 5

END IF

A loop for repetition might look like this: 1

WHILE condition is true

2 instructions to repeat 3

END WHILE

Flowchart symbols • Oval: Marks the start or end of the algorithm • Rectangle: Represents an action or process, like setting a value or performing a calculation • Diamond: Indicates a decision point, where the flow splits based on a yes/no condition • Arrows: Show the direction of flow between steps • Parallelogram: Represents input or output operations (e.g., PRINT, READ)

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Example 1 Digital only example Explore in the online lesson

mathspace.co

Example 2 Write pseudocode to find the largest of three numbers: a, b, and c

Create a strategy The algorithm must first get the three numbers as input. A simple strategy involves assuming the first number is the largest, storing it in a variable such as max. The second number is then compared to max, and max is updated if the second number is larger. This process is repeated for the third number. Finally, the value of max is printed.

Apply the idea 1 START 2 INPUT a, b, c 3 SET max = a 4 IF b > max THEN 5 SET max = b 6 END IF 7 IF c > max THEN 8 SET max = c 9 END IF 10 PRINT max 11 END

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Example 3 Write pseudocode to calculate and print the sum of all whole numbers from 1 to N, where N is a positive integer.

Create a strategy The algorithm requires a loop that repeats N times. A variable, such as counter, can track the numbers from 1 to N. A second variable, such as sum, is needed to store the running total and should be initialised to 0. Inside the loop, the value of counter is added to sum. After the loop finishes, the final value of sum is printed.

Apply the idea 1 START 2 INPUT N 3 SET sum = 0 4 SET counter = 1 5 WHILE counter <= N 6 SET sum = sum + counter 7 SET counter = counter + 1 8 END WHILE 9 PRINT sum 10 END

Idea summary Algorithms are built from sequence, decision, and repetition. Algorithms can be represented using pseudocode (text-based) or flowcharts (visual-based), which have their own conventions and symbols.

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Trace and apply algorithms Interpreting an algorithm requires following its logical flow. A key technique is tracing, which involves tracking the values of variables step-by-step to predict the output for a given input and to identify potential errors, such as infinite loops. Tracing Following an algorithm’s steps to see how variables change with given inputs.

Example 4 Describe the algorithm represented by the flowchart: START SET n = 3

Is n < 5?

Yes

PRINT n

No END What is the logical error?

Apply the idea The algorithm starts by setting a variable n to the value 3. It then enters a repetition loop that checks if n < 5. Since 3 < 5 is true, it prints the value of n. The flow then loops back to the decision. However, the value of n is never updated inside the loop. The condition n < 5 will always be true, causing the algorithm to print “3” indefinitely. This demonstrates an infinite loop, a common logical error where a loop’s termination condition is never met.

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Example 5 Trace this flowchart with n set to 1: START SET sum = 0, i = n No

Is i ≤ 3? Yes

sum = sum + 1, i = i + 1 PRINT sum END

START

SET sum = 0

SET i = n

WHILE i <= 3

5 SET sum = sum + i 6 SET i = i + 1 7

END WHILE

PRINT sum

END

Apply the idea Step

Action

Sum

SET sum = 0, i = 1

Is i ≤ 3? (1 ≤ 3)

sum = 0 + 1, i = 1 + 1

Is i ≤ 3? (2 ≤ 3)

sum = 1 + 2, i = 2 + 1

Is i ≤ 3? (3 ≤ 3)

sum = 3 + 3, i = 3 + 1

Is i ≤ 3? (4 ≤ 3)

PRINT sum

The output is 6, the sum of 1, 2, and 3.

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Output

Example 6 Digital only example Explore in the online lesson

mathspace.co

Example 7 Trace this pseudocode with input a set to 4, b set to 7: 1

START

INPUT a, b

IF a < b THEN

4 PRINT a 5

ELSE

6 PRINT b 7

END IF

END

Apply the idea Step

Action

Output

INPUT a, b

IF a < b THEN

PRINT a (since 4 < 7)

END

The algorithm prints 4, the smaller number.

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Example 8 Trace this pseudocode for a simple sorting algorithm (bubble sort) and determine the output for the input list [3, 1, 4]. Assume list indexing starts from 1. 1 START 2 INPUT list 3 SET n = length of list 4 FOR i from 1 to n-1 5 FOR j from 1 to n-i 6 IF list[j] > list[j+1] THEN 7

SWAP list[j] with list[j+1]

8 END IF 9 END FOR 10 END FOR 11 PRINT list 12 END

Create a strategy A trace table is useful for tracking the values of i, j, and the state of the list. The outer loop (i) controls the number of passes, while the inner loop ( j) performs the comparisons and swaps within each pass.

Apply the idea Initial list: [3, 1, 4]. Here n = 3. Outer loop pass 1 (i = 1): The inner loop runs for j from 1 to 3 – 1 = 2. • j = 1: Compare list [1] and list [2] (3 and 1). Since 3 > 1 is true, swap them. List becomes [1, 3, 4]. • j = 2: Compare list [2] and list [3] (3 and 4). Since 3 > 4 is false, do nothing. List remains [1, 3, 4]. Outer loop pass 2 (i = 2): The inner loop runs for j from 1 to 3 – 2 = 1. • j = 1: Compare list [1] and list [2] (1 and 3). Since 1 > 3 is false, do nothing. List remains [1, 3, 4]. The loops are finished. The final list is printed. Output: [1, 3, 4]

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Example 9 The procedure has been written in pseudocode: 1

DECLARE integer n

DECLARE integer result

DECLARE integer i

result ← 1

n ← 5

FOR i from 1 to n

7 result ← result * i 8

END FOR

PRINT result

The output of the pseudocode is a list of numbers. The final number in the list is: A 24

120

720

Create a strategy To find the final number printed, the value of the result variable must be traced as the for loop executes. The loop runs for i = 1, 2, 3, 4, and 5. In each iteration, the current result is multiplied by i.

Apply the idea Tracing the values of i and result step-by-step: Initially: result = 1 • When i = 1, result = 1 × 1 = 1. • When i = 2, result = 1 × 2 = 2. • When i = 3, result = 2 × 3 = 6. • When i = 4, result = 6 × 4 = 24. • When i = 5, result = 24 × 5 = 120. The loop finishes, and the final value of result is printed. The output is 120. The final number in the list is 120. The correct answer is Option C.

Idea summary Tracing an algorithm by tracking variable changes helps to understand its behaviour, predict its output, and verify its correctness for given inputs.

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2.07 Practice questions What do you remember? 1

Identify the key term described in each statement: a

A simplified, language-independent way to describe an algorithm using a mix of natural language and programming conventions.

The fundamental construct where instructions are executed one after another in a specific order.

A diagram that represents an algorithm using standardised symbols and arrows to show the flow of logic.

The process of stepping through an algorithm with specific inputs to track how variables change and determine the final output.

Determine whether each of these statements about algorithms is true or false: a

The ‘decision’ construct in an algorithm allows a set of instructions to be repeated multiple times.

The pseudocode keyword ‘WHILE’ is used to implement the repetition construct.

Every algorithm must contain all three fundamental constructs: sequence, decision, and repetition.

Match each flowchart symbol to its meaning: i

Oval

iii

Diamond

Action or process

Direction of flow

Start or end

Decision point

Rectangle

Arrow

Practice 4 Ex 1

Identify whether each step in this algorithm for making toast is a sequence, decision, or repetition: Step

Instruction

Plug in the toaster.

If the bread is fresh, slice it; else, use it as is.

Set the timer to 2 minutes.

Toast the bread until the timer finishes.

Ex 2

Write pseudocode for an algorithm that takes a number as input and prints “Even” if the number is even, and “Odd” if the number is odd.

Ex 3

Write pseudocode to count from 1 to 5 using a repetition construct.

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Interpret this flowchart by writing its pseudocode: START SET n = 4

Is n > 2?

Yes PRINT n END Ex 4

Interpret this flowchart and explain what it does: START SET count = 0 Is count < 3?

Yes PRINT count, Count = count + 1 END 9

Interpret this pseudocode and write a description of what it does: 1

START

SET count = 0

SET num = 10

WHILE num > 0

5 IF num MOD 2 = 0 THEN 6 count = count + 1 7 END IF 8 num = num - 1 9

END WHILE

10 PRINT count 11 END

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Interpret this flowchart and describe its steps in words: START SET p = 1, q = 8 No

Is p < q?

Yes

SET r=q–p

SET r=p–q PRINT r END

What is the final value printed by the algorithm in this flowchart? START SET counter = 5 Is counter > 3?

Yes

counter = counter – 1

No PRINT counter END A Ex 5

Trace this flowchart with n = 2: START SET sum = 0, i = n Is i ≤ 4?

Yes sum = sum + i, i = i + 1 PRINT sum END

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Trace this pseudocode that uses a FOR loop to find the output:

Ex 6

START

SET product = 1

FOR count FROM 1 TO 3

4 SET product = product * count

END FOR

PRINT product

END

Write pseudocode to calculate a real root of the quadratic equation 2x2 – 5x + 3 = 0 using the positive square root in the quadratic formula x

Ex 7

b2 4 ac . 2a

Write pseudocode to find the minimum of three numbers using decision constructs.

Trace this pseudocode with inputs x = 6 and y = 2 to determine the output: 1

START

INPUT x, y

IF x < y THEN

4 PRINT x 5

ELSE

6 PRINT y

Ex 8

END IF

END

Interpret this pseudocode for sorting two numbers in ascending order and determine the output for inputs 5, 2: 1

START

INPUT a, b

IF a > b THEN

4 SWAP a with b 5

END IF

PRINT a, b

END

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Interpret this pseudocode for a guessing game and predict the output if the secret number is 4 and the guess is 6: 1 START 2 INPUT secret, guess 3 IF guess = secret THEN 4 PRINT ‘Correct!’ 5 ELSE IF guess < secret THEN 6 PRINT ‘Too low’ 7 ELSE 8 PRINT ‘Too high’ 9 END IF 10 END

Ex 9

The procedure has been written in pseudocode: 1

START

SET total = 0

SET n = 4

FOR i from 1 to n

5 SET total = total + i * i 6 PRINT total 7

END FOR

END

The output of the pseudocode is a list of numbers. The final number in the list is: A

Extend your thinking 20

Trace this algorithm with n = 1 and calculate the mean of the printed outputs: 1

START

SET sum = 0

SET i = n

WHILE i <= 5

5 PRINT i 6 SET sum = sum + i 7 SET i = i + 1

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END WHILE

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Trace this pseudocode with x = 3 and calculate the median of all printed outputs. Suggest an improvement to avoid infinite looping. 1

START

INPUT x

WHILE x < 5

4 PRINT x

END WHILE

END

Interpret this flowchart and design an alternative pseudocode that achieves the same result using a different decision condition. Trace both with n = 4 to verify equivalence: START SET n = input Yes

Is n > 0?

PRINT Non-positive

PRINT Positive END 23

Write pseudocode to calculate the factorial of a number n (e.g., 5! = 5 × 4 × 3 × 2 × 1) using a repetition construct. Test your algorithm with n = 4.

Design a flowchart for an algorithm that determines if a number n is positive, negative, or zero, using decision constructs similar to logic gates. Test your algorithm with n = –2.

Write pseudocode to calculate the median of three numbers input by the user. Apply your algorithm to the inputs 4, 2, 7 and interpret the result in the context of data distribution.

Chapter 2 review Online review

mathspace.co

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Big ideas • Sequences are ordered lists of numbers governed by predictable patterns. Classifying these patterns as arithmetic (constant addition) or geometric (constant multiplication) enables the use of recurrence relations and general formulas to determine any specific term and model linear or exponential change. • The sum of a sequence’s terms, a series, allows for the analysis of accumulated totals over time. Recurrence relations and series are powerful tools for modelling real-world phenomena, particularly in finance, and for determining whether an infinite process converges to a finite limit.

3 Sequences and series Chapter outline 3.01 3.02 3.03 3.04 3.05 3.06 3.07

Introduction to sequences Arithmetic sequences Geometric sequences Applications of sequences Investigation: Spreadsheets and sequences Arithmetic series Geometric series Applications of sequences and series Chapter 3 review

178 194 210 228 245 254 265

Every barcode is a sequence—your groceries are basically maths in disguise.

3.01

Introduction to sequences After this lesson, you will be able to … • describe sequences and patterns in words • generate sequences from general and recursive formulas • write the formula for simple sequences in general and recursive form using appropriate subscript notation

Introduction to sequences in general form Sequence A list of numbers that follow a specific pattern. Each number in a sequence is called a term. Sequences can be finite or infinite. Finite sequence

Infinite sequence

A sequence that has a fixed number of terms.

A sequence with terms that are unending.

Example:

Example: 2, 4, 6, 8, 10, . . .

Exploration Consider these patterns:

Step 1

Step 2

Step 3

Step 1

Step 2

Step 3

Pattern ii

Pattern i

Step 1

Step 2

Step 3

Step 1

Pattern iii

Step 2

Step 3

Pattern iv

1. How could these patterns be described in words? 2. Which patterns could be finite, and which ones could be infinite? 3. If only the number of objects in each step is considered, how could the patterns be represented mathematically?

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While describing how a pattern changes from one step to the next is useful (this is called a recursive pattern, which will be explored more in the next section), it is often desirable to find the value of any term in a sequence directly without needing to calculate all the previous terms. A general formula (or explicit formula) allows for the direct calculation of any term. General (explicit) formula A rule that defines the nth term of a sequence, often denoted tn, directly in terms of its position number n. Example: For the sequence of even numbers 2, 4, 6, 8, . . . , the general formula is tn = 2n. To find the 5th term, substitute n = 5: t5 = 2(5) = 10. A common type of sequence is an arithmetic sequence, where the difference between consecutive terms is constant. This constant difference is called the common difference. Arithmetic sequence A sequence in which each term after the first is found by adding a constant, called the common difference, to the previous term. Common difference The constant value added to any term in an arithmetic sequence to obtain the next term. It is denoted by b. Initial term The term of a sequence that corresponds to position n = 0, denoted t0. Also known as the zeroth term. The general formula for an arithmetic sequence allows any term to be found if the first term and the common difference are known.

tn = t1 + (n − 1) b tn

is the nth term of the sequence (the term to be found)

t1 n b

is the first term of the sequence is the term number or position in the sequence is the common difference of the sequence

Alternatively, the general formula for an arithmetic sequence can be expressed in terms of the initial term, t0, which is the term immediately preceding the first term (t1 ).

tn = bn + t0 tn b n t0

is the nth term of the sequence is the common difference of the sequence is the term number or position in the sequence is the initial term (the value when n = 0)

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In this form, t0 is the value of the sequence when n = 0. The relationship between t1 and t0 is given by t1 = t0 + b. This means t0 = t1 − b. Both general formulas, tn = t1 + (n − 1) b and tn = bn + t0, are equivalent for arithmetic sequences. To write the general formula for an arithmetic sequence using tn = t1 + (n − 1) b: 1. Identify the first term, t1. 2. Calculate the common difference, b, by subtracting any term from its succeeding term (e.g., b = t2 − t1 ). 3. Substitute the values of t1 and b into the formula tn = t1 + (n − 1) b. 4. Simplify the expression if required. Alternatively, to use the form tn = bn + t0: 1. Calculate the common difference, b, using the same method. 2. Calculate the initial term, t0, using t0 = t1 − b or by working backward from t1. 3. Substitute the values of b and t0 into the formula tn = bn + t0. 4. Simplify the expression if required.

Example 1 A sequence is defined by the general formula tn = 3n − 5: a Find the first four terms of the sequence.

Create a strategy To find each term, substitute the term number (n = 1, 2, 3, 4) into the general formula.

Apply the idea For n = 1 (first term, t1): t1 = 3(1) − 5 = 3 − 5 = − 2 For n = 2 (second term, t2): t2 = 3(2) − 5 = 6 − 5 = 1 For n = 3 (third term, t3): t3 = 3(3) − 5 = 9 − 5 = 4 For n = 4 (fourth term, t4): t4 = 3(4) − 5 = 12 − 5 = 7 The first 4 terms are − 2, 1, 4, 7.

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b Find the 10th term of the sequence.

Create a strategy Substitute n = 10 into the general formula.

Apply the idea For n = 10 (tenth term, t10): t10 = 3(10) − 5 = 30 − 5 = 25 The 10th term is 25.

Example 2 Determine the general formula tn for each arithmetic sequence: a The sequence 5, 8, 11, 14, . . .

Create a strategy This is an arithmetic sequence. First, identify the first term (t1). Then, find the common difference (b) by subtracting a term from its subsequent term. Finally, substitute these values into the general formula tn = t1 + (n − 1) b.

Apply the idea The first term is t1 = 5. The common difference b = t2 − t1 = 8 − 5 = 3. Check with other terms: 11 − 8 = 3, 14 − 11 = 3. The common difference is indeed 3. Substitute t1 = 5 and b = 3 into tn = t1 + (n − 1) b: tn = 5 + (n − 1)3

Write the equation

= 5 + 3n − 3

Evaluate the multiplication

= 3n + 2

Simplify

The general formula is tn = 3n + 2.

Reflect and check Check if this formula generates the given terms: • For n = 1, t1 = 3(1) + 2 = 5 (Correct) • For n = 2, t2 = 3(2) + 2 = 8 (Correct) • For n = 3, t3 = 3(3) + 2 = 11 (Correct) The formula is correct.

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b The sequence given by the table: n

−2

Create a strategy From the table, identify the first term (t1). Then, find the common difference (b) by looking at the change in tn values. Substitute these into tn = t1 + (n − 1) b.

Apply the idea From the table, the first term (n = 1) is t1 = 10. The common difference b = t2 − t1 = 6 − 10 = − 4. Check with other terms: t3 − t2 = 2 − 6 = − 4, t4 − t3 = − 2 − 2 = − 4. The common difference is − 4. Substitute t1 = 10 and b = − 4 into tn = t1 + (n − 1) b: tn = 10 + (n − 1)(− 4)

Write the equation

= 10 − 4n + 4

Evaluate the multiplication

= − 4n + 14

Simplify

The general formula is tn = − 4n + 14.

Reflect and check Check if this formula generates the given terms: • For n = 1, t1 = − 4(1) + 14 = 10 (Correct) • For n = 4, t4 = − 4(4) + 14 = − 16 + 14 = − 2 (Correct) The formula is correct.

Idea summary A sequence is a list of terms and can be a finite sequence or an infinite sequence. An arithmetic sequence has a constant common difference (b). Its general (explicit) formula finds any term (tn) using either the first term (t1 ) or the initial term (t0 ): • tn = t1 + (n − 1) b • tn = bn + t0 To use the general formula: • To find a specific term, substitute the term number n into the formula. • To write the formula, identify t1 and b from the sequence and substitute them into the standard form.

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Sequences in recursive notation To describe a pattern, consider what is happening from one step to the next. When a previous step is used to describe what happens in the next step, a recursive pattern is being used. Recursive pattern A pattern which can be described by applying a rule to any term in the sequence to find the one after it. While the Fibonacci sequence uses two previous terms, many common recursive patterns, especially first-order linear recurrence relations, define the next term based only on the immediately preceding term. These can often be written in the form tn+1 = m × tn + c, where tn+1 is the next term, tn is the current term, and m and c are constants. For such rules, an initial term (either t0 or t1) must be specified to start the sequence. For example, an arithmetic sequence with common difference, b, can be written recursively as tn+1 = tn + b. A geometric sequence with common ratio, a, can be written as tn+1 = a × tn.

Exploration Consider an arithmetic sequence that starts with the number 5 and has a common difference of 3: 1. Write down the first 5 terms of this sequence. 2. Describe in words how each term is found from the previous one. This is a recursive description. 3. If the 20th term was to be found using this recursive method, how many calculations would be needed after finding the first term? 4. Recall the general formula for an arithmetic sequence: tn = t1 + (n − 1) b. Use this formula to find the 20th term. 5. Compare the effort required to find the 20th term using the recursive description versus the general formula. When might a recursive rule be convenient, and when might a general formula be more efficient?

One of the most famous recursive patterns is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, . . . To get the next term of the sequence, the two previous terms need to be added. This is how the recursive pattern can be described in words. The mathematical representation will now be shown. The general rule for finding the nth term of the Fibonacci sequence in recursive notation is tn = tn−2 + tn−1, where t1 = 1 and t2 = 1.

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To understand the rule, begin by labelling each of the terms using the notation tn, where n represents the number of the term:

The first two terms are given in the rule, so the next one to find is the third term, t3. To find the third term, let n = 3. Substituting this into the rule: t3 = t3−2 + t3−1

Apply the recursive rule

= t1 + t2

Simplify the subscripts

In other words, the first two terms, t1 and t2, need to be added to find the third term, t3: t3 = t1 + t2

Write the equation

=1+1

Substitute the values

Evaluate

Now, this rule can be used to find the next term, t4. In this case, n = 4: t4 = t4−2 + t4−1

Apply the recursive rule

= t2 + t3

Simplify the subscripts

=1+2

Substitute the values

Evaluate

Writing recursive rules relies on some specific notation:

tn+1 = tn + b, starting with t0 or t1 t0

is the initial term (when n = 0)

is the first term (when n = 1)

is the current term in the sequence

tn++1

is the next term in the sequence

is the common difference

Each term of a sequence can be viewed as an ordered pair, (n, tn), where the number of the term is the input and the value of the term is the output. Since each term (output) is identified by a unique term number (input), every sequence is also a function. The domain of a sequence will be some subset of the integers.

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Example 3 Consider the sequence: n

−7

−3

a Describe the recursive pattern in words.

Create a strategy To find the recursive pattern, determine the operation applied to the first term to obtain the second term. Then, ensure that operation is applied to get all the terms in the sequence.

Apply the idea To get from − 7 to − 3, 4 needs to be added. If 4 is added to − 3, the result is 1. Add 4 again, and the result is 5. Add 4 once more, and the result is 9. Recursive pattern: Add 4 to find the next term.

b Find the next 3 terms of the sequence.

Create a strategy To find the next 3 terms, simply add 4 to the last known term. Keep adding 4 to find the other terms.

Apply the idea 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21 The next 3 terms in the sequence are 13, 17, 21.

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c Write the recursive rule for this sequence, tn+1, including the first term, t1.

Create a strategy This question asks to find the (n + 1)th term. To do that, the previous term, which would be tn, needs to be used.

Apply the idea tn+1 = tn + 4, where t1 = − 7

Reflect and check This can be checked by finding the next terms. The first term is already known, so the second term, t2, can be found: t2 = t1 + 4 = − 7 + 4 = − 3 This is the second term given, so the rule is correct. The third term, t3, can be found next: t3 = t2 + 4 = − 3 + 4 = 1 This also matches the third term given, so the rule is correct.

Example 4 A sequence is defined by the rule Tn+2 = 2 (Tn+1 − Tn) with T1 = − 2 and T2 = 0. Find the first 6 terms of the sequence.

Create a strategy The first two terms are already given, and the rule can be used to find T3, T4, T5, and T6.

Apply the idea For T3, n = 1: T3 = 2(T2 − T1) = 2 [0 − (− 2)] = 4 For T4, n = 2: T4 = 2(T3 − T2) = 2 (4 − 0) = 8 For T5, n = 3: T5 = 2(T4 − T3) = 2 (8 − 4) = 8 For T6, n = 4: T6 = 2(T5 − T4) = 2 (8 − 8) = 0 The first 6 terms of the sequence are − 2, 0, 4, 8, 8, 0.

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Example 5 In a certain sequence, the next term is found by doubling the previous term, then subtracting one: a Write the recursive rule for tn+1 in terms of tn.

Create a strategy The next term tn+1 is found based on the previous term tn. The phrase “doubling the previous term, then subtracting one” needs to be translated into this notation.

Apply the idea tn+1 = 2tn − 1

Reflect and check The first term would usually need to be defined here as well, but no information about where the sequence begins is given. This rule describes the pattern, but it cannot help to find the terms of the sequence until at least one term in the sequence is known.

b If t1 = − 2, find t5.

Create a strategy This question asks to find the 5th term. The second, third, and fourth terms will need to be found before finding the 5th term.

Apply the idea t2 = 2t1 − 1 = 2(− 2) − 1 = − 5 t3 = 2t2 − 1 = 2(− 5) − 1 = − 11 t4 = 2t3 − 1 = 2(− 11) − 1 = − 23 t5 = 2t4 − 1 = 2(− 23) − 1 = − 47 The 5th term, t5, is − 47.

Reflect and check Finding the 5th term recursively was manageable. However, if the 50th term needed to be found, this recursive method would be very time-consuming. This sequence is not simply arithmetic or geometric, so finding a simple general formula like tn = t1 + (n − 1) b is not straightforward without more advanced techniques. This highlights a scenario where a general formula, if known, would be much more efficient for finding distant terms.

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Example 6 Consider this graph: 4

2 0

n 1

–2 –4 –6 –8

Write the recursive rule, tn+1, and the initial term t0.

Create a strategy Begin by identifying the points on the graph and labelling the corresponding term. The initial term will correspond to the leftmost point, (0, − 8). The first term corresponds to (1, 4). The second term corresponds to (2, − 2), and the third term corresponds to (3, 1). Next, determine what is happening to the outputs so that the recursive rule can be built.

Apply the idea

Reflect and check

The initial term is t0 = − 8.

Multiplying by

The outputs are being multiplied by time. The recursive rule is

each

is the same as dividing by

− 2, so the rule could also be written as .

Idea summary A recursive pattern defines each term in a sequence by applying a rule to one or more preceding terms. An initial term (e.g., t0 or t1) must be provided to start the sequence. A common type is a first-order linear recurrence relation, where the next term depends only on the immediately preceding term. The rule can be expressed in different but equivalent ways, such as: • tn+1 = tn + b (the next term equals the current term plus b) • tn = tn−1 + b (the current term equals the previous term plus b) Some recursive rules, like the Fibonacci sequence, require more than one preceding term and multiple initial values to be defined.

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3.01 Practice questions What do you remember? 1

For each sequence: i

Describe the pattern in words.

Determine the next number in the sequence.

− 1, 1, 3, 5, 7, . . .

2, − 4, 6, − 8, 10, − 12, . . .

Consider the pattern shown with the given figures:

Step 1

Step 2

Step 3

Step 4

Describe the pattern in words.

Determine the number of squares there will be in the next step in the pattern.

Consider the pattern:

Step 1

64, 32, 16, 8, 4, 2, . . .

Step 2

Step 3

Step 4

Describe the pattern in words.

Determine the number of circles there will be in the next step in the pattern.

Given the sequence tn and the first term is t1, write the notation used for these terms: a

The fifth term of the sequence

The tenth term of the sequence

The nth term of the sequence

The previous term of the sequence, before the nth term of the sequence

The next term of the sequence, following the nth term of the sequence

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Practice 5

Consider the pattern:

Step 1

Ex 1

Ex 2

Ex 3

Step 3

Determine the constant factor.

Determine the number of triangles the next step in the pattern contains.

Write the recursive formula for each sequence: a

− 1, 4, 9, 14, 19, 24 . . .

15, 13, 11, 9, 7, 5 . . .

1, 5, 25, 125, 625 . . .

3, 6, 12, 24, 48 . . .

Write the first five terms of the sequences described by each recursive process: a

Start with 2 and to go from one term to the next, multiply by 3.

Start with − 10 and to go from one term to the next, add 7.

Start with − 100 and to go from one term to the next, divide by 5.

Start with 2.5 and to go from one term to the next, subtract 0.5 and then multiply by 2.

A sequence is defined by the general formula Tn = 4n + 7: a

Determine the first four terms of the sequence.

Determine the 15th term of the sequence.

Determine the general formula Tn for each arithmetic sequence: a

2, 9, 16, 23, . . .

A sequence where the first term is T1 = 20 and the common difference is b = − 6.

The total volume of water that has dripped from a tap is measured each minute and displayed in the table: a

Describe the relationship between time and volume.

Complete the values in the table over the next two minutes.

Use recursive notation to write an equation showing how to determine the volume of the water after n minutes.

190

Step 2

Determine the volume after 8 minutes.

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Time (minutes)

Volume (mL)

⬚

Hannah saves some money each week and puts it in her piggy bank. The total amount in her piggy bank each week is displayed in the table: Time (weeks) Balance (dollars) a Describe the relationship between time 1 4 and balance. 2 8 b Complete the values in the table over the next two weeks. c

d Ex 4

⬚

Use recursive notation to write an equation showing how to determine the balance after n weeks. Determine the balance after 8 weeks.

⬚

Consider the sequence which has a first term of 11 and a second term of 22, and subsequent terms are found by adding the two previous terms: a

Write the recursive rule for the sequence, letting tn be the nth term.

Determine the first five terms of the sequence.

A population of rabbits is counted in an area, and it is found that the population is doubling each month. The population counts at the end of each of the first three months are shown:

Month 1

Month 2

Month 3

Complete the table, showing the population at the end of each of the first 5 months: n

⬚

If the population continues to grow in this manner, use recursive notation to write an equation representing the population at the end of the nth month. Determine the population at the end of one year.

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Ex 5

Ex 6

Write the recursive rule from each description. Let tn be the nth term: a

To determine the next term, add 5 to the previous term.

To determine the next term, add the two previous terms.

To determine the next term, multiply the previous term by negative 1 and then add 7.

To determine the next term, subtract one and a half from the previous term.

To determine the next term, subtract 6 from the previous term.

To determine the next term, multiply the two previous terms.

To determine the next term, multiply the previous term by negative 5 and then subtract 3.

To determine the next term, add three and a quarter to the previous term.

The graph shows the first three terms of a sequence: a

Write a recursive rule for the sequence, including the first term.

Determine the 5th term of the sequence.

Tn 12 9 6 3 n 0

Each of the patterns was created in steps using matchsticks: i

Complete the table of values for each of the given patterns: Step number (t)

Number of matchsticks (m)

⬚

Write a rule for a sequence that describes the relationship between the number of matchsticks, m, and the step number, t.

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Zuber is a taxi service that charges a $1.50 pickup fee and $1.95 per kilometre of travel: a

Determine the charge for a 10-kilometre journey.

Write a recursive rule for Tn in terms of Tn−1 which defines the price of a n kilometre trip, and an initial condition for T0.

Extend your thinking 18

Triangular numbers count the number of objects that make up an equilateral triangle. The first three triangular numbers are shown using spots:

Write a rule to determine the number of spots in the nth term of the sequence, then use this rule to determine the number of spots needed to represent the 10th triangular number. Write the solution using sequence notation. 19

The cumulative rainfall over a series of 4 days is represented on the graph: a

Describe the relationship between time and volume.

Explain how the points on the graph can be described as a sequence.

Assuming the pattern continues over the next three days, determine whether or not it is a function. If it is, describe the domain.

Determine the cumulative volume on day 7.

36 32 28 24 20 16 12 8 4

Volume (mL)

0 20

Time (days) 4 5

A doctor administered pain medication in these amounts in milligrams over four days after their patient underwent surgery: Day (n)

Dosage (m)

2000

1520

1155.2

877.952

Determine the recursive formula which models this sequence. Round the values to two decimal places if necessary.

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3.02 Arithmetic sequences After this lesson, you will be able to… • determine the initial value and common difference for a given sequence • determine properties of arithmetic sequences and their graphs • express arithmetic sequences in recursive and general forms: tn+1 = tn + b and tn = bn + t0 • solve problems involving arithmetic sequences using recursive and general forms

Arithmetic sequences with recursion An arithmetic sequence is a recursive pattern of addition, where the same number is added to each subsequent term. The first term specified for an arithmetic sequence, often denoted as t0 or t1, is the initial value. The constant is called a common difference, denoted by b. The sequences will increase when b is positive or decrease when b is negative. Since the terms increase or decrease by a constant amount, plotting the values on a graph results in a straight line. An arithmetic sequence is represented in recursive notation by the formula:

tn+1 = tn + b tn++1

is the next term in the sequence (term at position n + 1)

is the current term in the sequence (term at position n)

is the common difference

To fully define an arithmetic sequence using a recursive rule, an initial value is needed. This is the starting point of the sequence. The domain for n in a sequence is typically the set of positive integers (n = 1, 2, 3, . . .), or non-negative integers (n = 0, 1, 2, . . .). If the sequence starts with n = 1, the initial value is t1. If the sequence starts with n = 0, the initial value is t0. The recursive rule must be accompanied by a specific initial value, such as t1 or t0, to define a unique sequence.

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Exploration Consider these sequences and see if a pattern can be found: • Sequence A: 2, 5, 8, 11, 14, . . .

• Sequence C: 1, 2, 4, 8, 16, . . .

• Sequence B: 10, 8, 6, 4, 2, . . .

• Sequence D: 3, 3, 3, 3, 3, . . .

1. For each sequence, calculate the difference between each term and the term that came before it (e.g., for Sequence A, calculate 5 − 2, then 8 − 5, and so on). 2. Which sequences have a constant difference between consecutive terms? What is this constant difference for each of them? 3. The sequences with a constant difference are called arithmetic sequences. The constant difference is called the common difference. Can you describe in words how to get from one term to the next in an arithmetic sequence? 4. Plot the first five terms of Sequence A (let the term number be the x-coordinate and the term value be the y-coordinate). What shape do the points form? 5. Plot the first five terms of Sequence C. What shape do these points form? Is it the same as Sequence A? 6. Given the first term of an arithmetic sequence and its common difference, how are the second and third terms found?

Example 1 Determine whether each sequence is arithmetic: a

Create a strategy A sequence is arithmetic if the terms share a common difference.

Apply the idea

The difference between each of the terms is the same, so this sequence is arithmetic. 1 The initial term is t1 and the common difference is . 3

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Reflect and check When plotted, this sequence forms a linear relationship. This is confirmation that it is an arithmetic sequence. tn n

–1 –2 –3

−3

−2

Create a strategy

Apply the idea

Determine if this is arithmetic by finding the difference between terms.

− 2 − (− 3) = 1

Find the first difference

0 − (− 2) = 2

Find the second difference

3−0=3

Find the third difference

The difference between terms is not the same, so the sequence is not arithmetic. c 16

14 12 10 8 6 4 2

Create a strategy

Apply the idea

An arithmetic sequence will form a linear pattern or a straight line when graphed because there will be a constant rate of change.

Since the points in this sequence form a curve, it is not arithmetic.

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Reflect and check Validate the answer by checking the difference between each output and verifying that it is not constant. 16 − 8 = 8 8−4=4 4−2=2 2−1=1 The differences are not the same, so this confirms the sequence is not arithmetic.

Example 2 Emmanuel is selling raffle tickets to raise money for charity. This table shows the cumulative number of tickets he has sold each hour for the first three hours: Time (hours)

Total ticket sales

a Determine whether Emmanuel’s ticket sales represent an arithmetic sequence.

Create a strategy An arithmetic sequence will have a constant rate of change. Values can be compared in the table and see how much the total ticket sales are increasing by each hour.

Apply the idea Emmanuel sells 14 tickets in the first hour. He then sells 28 − 14 = 14 tickets in the second hour and 42 − 28 = 14 tickets in the third hour. The rate of change is constant and therefore, the ticket sales represent an arithmetic sequence. The initial value is t1 = 14 (sales after 1 hour) and the common difference is b = 14 tickets per hour.

b Determine the recursive rule which relates Emmanuel’s ticket sales and time.

Create a strategy The recursive rule is what happens to one term to get to the next. It uses the common difference b and the initial term t1.

Apply the idea From part (a), the initial value is t1 = 14 and the common difference is b = 14. The recursive rule is tn+1 = tn + b, so for this sequence: tn+1 = tn + 14, t1 = 14

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c If Emmanuel’s ticket sales continue in this way, determine the total number of tickets he will have sold after 6 hours.

Create a strategy Use the recursive rule tn+1 = tn + 14 repeatedly, or use a general formula. Since sales are 14 per hour starting from hour 1, the general formula is tn + 14n.

Apply the idea Although the recursive rule could be used to find the 6th term (t6), the fact that Emmanuel is selling 14 tickets per hour could also be used. Since sales are cumulative and start from Hour 1, the total sales after n hours will be 14n. So after 6 hours, if the pattern stays the same, he will have sold 14 × 6 = 84 raffle tickets.

Reflect and check Expand the table to see all of Emmanuel’s ticket sales up to and including the 6th hour: Time (hours)

Total ticket sales

Idea summary An arithmetic sequence is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by b. Each term is found by adding the common difference to the previous term. The graph of an arithmetic sequence forms a straight line. The recursive formula for an arithmetic sequence, given an initial value (t1 or t0), is:

tn+1 = tn + b tn++1 is the next term tn is the current term b is the common difference An initial value (t1 or t0) must be provided with the recursive rule.

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Arithmetic sequences in general form The recursive formula tn+1 = tn + b is useful for finding the next term in a sequence. However, to find a term far into the sequence, like the 100th term (t100), using the recursive rule would require calculating all 99 preceding terms. This is inefficient. A general form (or explicit formula) allows for the direct calculation of any term, given its position n. The general form can be derived as follows. Consider an arithmetic sequence with initial value t1 and common difference b. • t1 = t1 • t2 = t1 + b • t3 = t2 + b = (t1 + b) + b = t1 + 2b • t4 = t3 + b = (t1 + 2b) + b = t1 + 3b A pattern can be seen: the nth term, tn, is the first term t1 plus the common difference b added (n − 1) times:

tn = t1 + (n − 1) b tn is the nth term t1 is the first term n is the term number (position in the sequence) b is the common difference If the sequence starts with an initial value t0 (i.e., the term at position n = 0), the pattern is: • t0 = t0 • t1 = t0 + b • t2 = t1 + b = (t0 + b) + b = t0 +2b • t3 = t2 + b = (t0 + 2b) + b = t0 + 3b The nth term, tn, is the initial term t0 plus the common difference b added n times. This can be written as tn = t0 + nb, or as specified in the learning objectives, tn = bn + t0.

tn = t0 + nb tn is the term at position n t0 is the initial term (at n = 0) n is the term number (position in the sequence, starting from 0) b is the common difference Notice the similarity to the equation of a straight line, y = mx + c. The common difference b acts like the gradient (m), and if the tn = t0 + nb form will be used, t0 is like the y-intercept (c). This reinforces why the graph of an arithmetic sequence is linear.

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Example 3 An arithmetic sequence has an initial term t1 = 4 and a common difference

a Write the recursive rule for this sequence in the form tn+1 = tn + b, stating the initial term.

Create a strategy The recursive rule relates a term to the next term using the common difference, t1 and b are given.

Apply the idea The common difference is

. The recursive rule is

, with t1 = 4.

b Determine the next 5 terms of the sequence (i.e., t2, t3, t4, t5, t6).

Create a strategy The first term t1 = 4 is known and the recursive rule repeatedly.

Apply the idea

The next 5 terms of the sequence are

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. Therefore, the rule can be applied

c Write the general rule for this sequence in the form tn = t1 + (n − 1) b and use it to find the 50th term, t50.

Create a strategy Given t1 = 4 and n = 50 to find t50.

. Substitute these into the general formula tn = t1 + (n − 1) b. Then, substitute

Apply the idea The general rule is: Write the equation

Substitute the values

To find the 50th term, substitute n = 50: Substitute the values Simplify inside the brackets Evaluate the multiplication

Write the fractions with common denominator

Evaluate The 50th term is

Did you know?

The keys on a piano form a sequence of musical notes! Each note follows a pattern of intervals, creating harmony through mathematical relationships in sound. 3.02 Arithmetic sequences mathspace.co

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Example 4 The first term of an arithmetic sequence is t1 = 2. The fifth term is t5 = 26: a Solve for b, the common difference of the sequence.

Create a strategy Use the general formula tn = t1 + (n − 1) b. Substitute the known values n = 5, t5 = 26, and t1 = 2 to create an equation and solve for b. To get from the first term to the fifth term, add the common difference b four times. So, t5 = t1 + 4b. Use this to set up an equation and solve for b.

Apply the idea t5 = t1 + 4b

Write the equation

26 = 2 + 4b

Substitute t1 = 2 and t5 = 26

24 = 4b

Subtract 2 from both sides

6=b

Divide both sides by 4

The common difference is b = 6.

b Write a recursive rule for tn+1 in terms of tn which defines this sequence, and determine the initial term t1.

Create a strategy The initial term is given as t1 = 2. The common difference in part (a) is b = 6. The recursive formula is tn+1 = tn + b.

Apply the idea The recursive rule is tn+1 = tn + 6, with t1 = 2.

Reflect and check Check the answer by using the rule to verify the 5th term is 26: t1 = 2 t 2 = t1 + 6 = 2 + 6 = 8 t3 = t2 + 6 = 8 + 6 = 14 t4 = t3 + 6 = 14 + 6 = 20 t5 = t4 + 6 = 20 + 6 = 26 This matches the given information.

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c Write the general rule for this sequence in the form tn = t1 + (n − 1) b and use it to find the 10th term, t10.

Create a strategy Given t1 = 2 and b = 6. Substitute these into the general formula tn = t1 + (n − 1) b. Then, substitute n = 10 to find t10.

Apply the idea The general rule is: tn = 2 + (n − 1)(6)

Write the equation

= 2 + 6n − 6

Expand the brackets

= 6n − 4

Simplify

To find the 10th term, substitute n = 10: t10 = 6(10) − 4

Write the equation

= 60 − 4

Evaluate the multiplication

= 56

Evaluate

The 10th term is t10 = 56.

Example 5 An arithmetic sequence is defined with an initial term t0 = 7 and a common difference b = 3. Assume n starts from 0 for this sequence: a Write the general rule for this sequence in the form tn = t0 + nb.

Create a strategy

Apply the idea

Substitute the given t0 and b into the general formula tn = t0 + nb.

Given t0 = 7 and b = 3. The general rule is tn = 7 + n(3), or tn = 7 + 3n (which can also be written as tn = 3n + 7).

b Find the term t14 (which is the 15th term since the sequence starts at n = 0).

Create a strategy Use the general rule tn = 7 + 3n and substitute n = 14.

Apply the idea t14 = 7 + 3(14)

Write the equation

= 7 + 42

Evaluate the multiplication

= 49

Evaluate

Therefore, t14 = 49.

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c Write the recursive rule for this sequence in the form tn+1 = tn + b, stating the initial term t0.

Create a strategy The recursive rule uses the common difference b. The initial term is t0.

Apply the idea The recursive rule is tn+1 = tn + 3, with t0 = 7.

Example 6 A student begins a savings plan. In the first week, the student save $14. Each subsequent week, the student add another $14 to their savings account. The total amount saved at the end of each week forms an arithmetic sequence: a Write the general rule for the total savings, tn, after n weeks, in the form tn = t1 + (n − 1) b.

Create a strategy The first term is the amount saved in the first week, t1 = 14. The common difference is the amount added each week, b = 14. Substitute these known values into the general formula.

Apply the idea Given t1 = 14 and b = 14. tn = 14 + (n − 1)(14)

Write the equation

= 14 + 14n − 14

Substitute the values

= 14n

Evaluate

The general rule is tn = 14n. b How much will the student have saved after 20 weeks if this pattern continues?

Create a strategy Use the general rule tn = 14n and substitute n = 20.

Apply the idea tn = 14n

Write the equation

t20 = 14(20)

Substitute the values

= 280 The student will have saved $280 after 20 weeks.

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Evaluate

c After how many weeks will the student's total savings be at least $210?

Create a strategy Use the general rule tn = 14n. Set tn ≥ 210 and solve for n.

Apply the idea 14n ≥ 210

Write the inequality Divide both sides by 14

n ≥ 15

Evaluate

The student’s total savings will reach (or exceed) $210 after 15 weeks.

Idea summary The general form (or explicit formula) of an arithmetic sequence allows direct calculation of any term tn using its position n. If the first term is t1 (for n = 1, 2, 3, . . .) and the common difference is b, the general form is:

tn t1 n b

tn = t1 + (n − 1) b

is the nth term is the first term is the term number is the common difference

If the initial term is t0 (for n = 0, 1, 2, . . .) and the common difference is b, the general form is tn = t0 + nb (or tn = bn + t0):

tn t0 n b

tn = t0 + nb

is the term at position n is the initial term (at n = 0) is the term number (index) is the common difference

The general form highlights the linear nature of arithmetic sequences.

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3.02 Practice questions What do you remember? Ex 1

Determine whether each sequence is an arithmetic sequence:

a c 2

2, 0, − 2, − 4, . . .

60, 71, 82, 93, . . .

2, 22, 24, 26, . . .

3, − 3, 3, − 3, . . .

Determine the common difference, b, for each arithmetic sequence: a

7.6, 6.4, 5.2, 4, . . .

330, 280, 230, 180. . .

Write down the next two terms for each arithmetic sequence: a

4, 8, 12, 16, . . .

6, 2, − 2, − 6, . . .

2, 3.5, 5, 6.5, . . .

Consider the pattern shown with the given figures:

Let Figure 1 be Term 1. a

Determine the common difference, b, of the sequence.

How many dots will be in Figure 4?

Practice 5

206

For each arithmetic sequence: i

Write down the first four terms of the sequence.

Determine the common difference, b.

tn = tn−1 + 7, t1 = − 2

tn = tn−1 + 1.4, t1 = 3

tn+1 = tn – 4, t0 = 9

Write down the first four terms in each arithmetic sequence: a

The first term is − 10 and the common difference is 4.

The first term is − 8 and the common difference is − 2.

The first term is a and the common difference is b.

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Ex 2

Ex 3

Malia is learning French using an app on her phone. Before using the app, she already knows 30 words. The app gives her 10 new words to learn each day: a

Determine how many words she will know in total after using the app for 6 days.

Write down a recursive rule, tn+1, the total number of words she has learnt after (n + 1) days of using the app, in terms of tn.

Write down a recursive rule, sn+1, the total number of words she has learnt after (n + 1) weeks of using the app, in terms of sn.

For each sequence: i

Identify b, the common difference between consecutive terms.

Write down a recursive rule for tn+1 in terms of tn and the initial term t1.

iii

Write down a general rule for tn in the form tn = bn + t0.

5, 10, 15, 20, 25, ...

e 5 4 3 2 1 0 –1 –2 –3

− 1, − 8, − 15, − 22, − 29, ...

−2

−6

− 10

12 10 n 1

8 6 4 2 0

n 1

In an arithmetic sequence with common difference b, the first term is 32: a

Determine how many times the common difference needs to be added to the first term to determine the fifth term.

Write down a simplified equation for t5, the 5th term.

Write down a simplified equation for the 9th term.

Given that the 9th term is 4 times the 5th term, determine the common difference b.

Vanessa is making a sequence of shapes out of tiles as shown in the diagram: a

Write down a recursive rule to represent the relationship between a shape’s sequence number, n, and the number of tiles, tn.

Determine how many times the common difference must be added to the 3rd term to obtain the 20th term.

Determine how many tiles Vanessa will need to make the 20th shape in the sequence.

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Ex 4

There are 12 seats in the first row of an outdoor theatre. There are a total of 14 rows, and the last row has 77 seats: a

Determine the number of times the common difference must be added to the number of seats in the first row to determine the number of seats in the 14th row.

Using your conclusion from part (a), express the number of seats in the 14th row as an equation.

Determine the common difference, b, between the number of seats in each row.

Use the common difference, b, to determine missing terms in each arithmetic sequence. Explain your process: a c

Ex 5

8, ⬚, 16, 20, ⬚ − 12, ⬚, ⬚, ⬚, ⬚, ⬚, 24

b d

⬚, 0, ⬚, 10, ⬚

25, ⬚, ⬚, ⬚, 5

The first three terms of an arithmetic sequence are 82, 75, 68, . . . a

Determine the number of positive terms in the sequence.

Determine the last positive term in the sequence.

Determine the value of x such that x + 4, 6x + 5, and 9x − 8 form successive terms in an arithmetic sequence.

A sequence is defined by the recursive rule tn+1 = tn + 5 with t1 = 3:

Write down the first four terms of the sequence.

Write down a general rule for the sequence in the form tn = t1 + (n − 1) b.

Simplify your rule from part (b) into the form tn = bn + t0.

A car is purchased for $25 000 and depreciates by $1300 each year. The value of the car is modelled by Vn = Vn−1 − 1300, V0 = 25 000, where Vn is the value of the car, in dollars, after n years: a

Determine the value of the car after 1 year.

Determine the value of the car after 7 years.

Determine how much the value of the car has dropped after 7 years.

Maddox has just accepted an offer to play for the Melbourne football club. He will initially be paid a salary of $500 000 a year, with this to increase by $60 000 each year. Let tn be Maddox’s salary in the nth year:

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Determine the value of t1.

Write down a recursive rule for tn+1 in terms of tn that describes Maddox’s salary in a particular year.

Determine Maddox’s salary in Year 5.

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Ex 6

Farhan has saved up $2540 to spend during his vacation to Lebanon. He plans to spend $310 each week while he is there. Let tn be the amount of savings Farhan has left at the beginning of week n:

Determine the value of t1.

Write down the recursive rule that describes how much Farhan has left to spend at the beginning of each week of his vacation.

Determine the number of weeks Farhan will have enough money to spend during his vacation.

Zoomer is a taxi service that charges a $2.25 flag fall and $2.50 per kilometre of travel: a

Determine the total charge for a 8-kilometre journey.

Write down a recursive rule for tn in terms of tn−1 which defines the price of an n-kilometre trip, and an initial condition for t0.

Extend your thinking 20

The kth term of an arithmetic sequence is tk = − 530. Determine the value of k given that t1 = 28 and b = − 18. Explain your process.

Determine which statement is always, sometimes, or never true. Justify your answer: a

An arithmetic sequence has all terms as positive values.

An arithmetic sequence, if plotted on a Cartesian plane, will form a linear pattern.

An arithmetic sequence can be both increasing and decreasing.

The plotted points represent terms in an arithmetic sequence, tn: 8

7 6 5 4 3 2 1 0

n 1

The points are reflected across the horizontal axis to form a new arithmetic sequence. Write down the recursive rule for sn, the nth term in this new sequence, and determine the initial term s0. Explain how this compares to the equation of the original sequence.

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3.03

Geometric sequences

After this lesson, you will be able to… • determine the initial value and common ratio for a given sequence • determine properties of geometric sequences and their graphs • express geometric sequences in recursive and general forms: tn+1 = atn and tn = t0 (an) • solve problems involving geometric sequences using recursive and general forms

Geometric sequences with recursion A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Geometric sequence A sequence of numbers in which each consecutive pair of numbers has a common ratio. Example: 2, 6, 18, 54, . . . (Common ratio is 3) Common ratio The constant factor between consecutive terms in a geometric sequence. It is denoted by a. Example: In the sequence 100, 50, 25, . . ., the common ratio is

Geometric sequences begin with an initial value, generally denoted as t1 (the first term). Each subsequent term is found by multiplying the previous term by the common ratio, denoted by a. To find the common ratio a, any term can be divided by its preceding term:

is the common ratio

tn++1

is a term in the sequence

is the term immediately preceding tn+1

The value of the common ratio a determines the behaviour of the sequence: • If a > 1, the terms will increase in magnitude (grow exponentially if t1 > 0). • If 0 < a < 1, the terms will decrease in magnitude towards zero (decay exponentially if t1 > 0). • If − 1 < a < 0, the terms will alternate in sign and decrease in magnitude towards zero. • If a < − 1, the terms will alternate in sign and increase in magnitude. • If a = 1, all terms are the same (a constant sequence). • If a = − 1, the terms alternate between t1 and −t1.

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A geometric sequence can be defined recursively. This means each term is defined by one or more preceding terms. The recursive formula for a geometric sequence is:

tn+1 = a × tn tn++1

is the (n + 1)th term

is the nth term (the previous term)

is the common ratio

To fully define a sequence recursively, the initial value must also be stated, usually t1. For example, a sequence might be defined by t1 = 10 and tn+1 = 2 × tn. The domain of a geometric sequence is the set of term numbers, typically positive integers starting from 1 (n = 1, 2, 3, . . .).

Example 1 Determine whether each sequence is geometric: a − 3, 3, − 3, 3, − 3, . . .

Create a strategy A sequence is geometric if the ratio between any two consecutive terms is constant. This constant is the common ratio, denoted by a. The strategy is to calculate the ratio for several pairs of consecutive terms by dividing each term by its preceding term (e.g., geometric.

, etc.). If all calculated ratios are identical, the sequence is

Apply the idea First, identify the terms of the sequence: t1 = − 3, t2 = 3, t3 = − 3, t4 = 3, and so on. Next, calculate the ratio for the available consecutive pairs:

Since the ratio between all consecutive terms is constant and equal to − 1, the sequence is geometric. The common ratio is a = − 1.

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Step 1

Step 2

Step 3

Create a strategy Look for a recursive pattern of multiplication between each step. For this one, count the squares in each step to form a sequence, identify the initial value t1, and determine if there is a common ratio a.

Apply the idea The first step has 2 squares, so t1 = 2. The second step has 4 squares (t2 = 4). The third step has 8 squares (t3 = 8). The sequence for this pattern is 2, 4, 8, . . . Next, look for a common ratio a which can be solved by dividing consecutive terms.

The terms do share a common ratio of a = 2, so this sequence is geometric.

60 55 50 45 40 35 30 25 20 0

n 1

Create a strategy When graphed, the terms of a geometric sequence lie on an exponential curve. However, visual appearance can be misleading. Check for a constant common ratio between the y-values (the terms of the sequence, tn) for consecutive x-values (the term numbers, n).

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Apply the idea The points on the curve give the terms: t1 = 60, t2 = 45, t3 = 35, t4 = 27, t5 = 21. Check for a common ratio a by dividing consecutive terms.

Since

, these terms do not share a common ratio, so the sequence is not geometric. Further

calculations for the remaining terms also confirm this.

These terms do not share a common ratio, so the sequence is not geometric. For a sequence to be geometric, its points must lie on a single exponential curve. Since it is not geometric, the points do not consistently follow such a pattern defined by a single common ratio.

Did you know?

When you line up dominoes or tiles, you’re creating a sequence! The spacing or size can follow a mathematical rule that brings order and pattern to the design, showing how maths can turn simple arrangements into something visually satisfying and precise.

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Example 2 A geometric sequence is defined by tn+1 = − 2 × tn, where the initial value is a = − 2.

. The common ratio

Find the next 4 terms of the sequence.

Create a strategy Know the first term t1 and the recursive rule tn+1 = − 2 × tn. Solve t2 by using n = 1 in the rule, then t3 by using n = 2, and so on.

Apply the idea Given

. Use the recursive rule tn+1 = − 2 × tn to find the next term.

For n = 1:

For n = 2:

For n = 3:

For n = 4: t5 = − 2 × t4 = − 2 × (− 1) = 2 The next 4 terms of the sequence are

Reflect and check The recursive formula tn+1 = − 2 × tn clearly shows the common ratio a = − 2. Each term is found by multiplying the previous term by − 2.

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Example 3 The first term (initial value) of a geometric sequence is t1 = 5. The third term is t3 = 80: a Solve for the possible values of the common ratio, a, of this sequence.

Create a strategy In a geometric sequence, to get from t1 to t3, multiply the number by the common ratio a twice. So, t3 = t1 × a × a = t1 × a2. Then, substitute the given values: 80 = 5 × a2 Use this to set up an equation and solve for a.

Apply the idea 5a2 = 80 2

a = 16

Write the equation Divide both sides by 5 Take the square root of both sides

a = ±4

Evaluate the square root

The common ratio a is either 4 or − 4.

Reflect and check Either of these common ratios will satisfy the sequence with the given information. If a = 4, then t1 = 5, t2 = 20, t3 = 80. If a = − 4, then t1 = 5, t2 = − 20, t3 = 80. To find one specific common ratio, an additional term (e.g., an even-numbered term like t2 ) would need to be known to specify a unique a, particularly if the sign pattern matters. Knowing t2 would specify a unique a.

b Determine the recursive rule, tn+1 = a × tn, that defines the sequence if the common ratio a is positive.

Create a strategy The positive common ratio found was a = 4. Given the initial value t1 = 5, substitute these into the recursive form tn+1 = a × tn.

Apply the idea The recursive rule is tn+1 = 4 × tn, where t1 = 5.

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c Determine the recursive rule, tn+1 = a × tn, that defines the sequence if the common ratio a is negative.

Create a strategy The negative common ratio found was a = − 4. Given the initial value t1 = 5. Substitute these into the recursive form tn+1 = a × tn.

Apply the idea The recursive rule is tn+1 = − 4 × tn, where t1 = 5.

Example 4 A group of students is working on a project that is due in 6 weeks, which involves a total of 324 tasks. To manage the workload, they planned to complete tasks frequently at the beginning, with fewer tasks to do each subsequent week. This table shows the number of tasks left to complete at the end of each week, tn, where n is the week number. This can be represented by a geometric sequence: n (Week) tn (Tasks left)

1 2 324 108

3 36

4 12

The initial value (tasks left at the end of Week 1) is t1 = 324: a Identify the common ratio, a.

Create a strategy The common ratio a can be found by dividing a term by the previous term, e.g.,

Apply the idea Since this is a geometric sequence, any pair of consecutive terms can be used. For example, using t4 and t3:

The common ratio is

Reflect and check Other values could have been chosen to arrive at the same result. For example, and

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b Determine the value of t5 (tasks left at the end of Week 5).

Create a strategy Using the common ratio

found in part (a), the recursive rule tn+1 = a × tn can be applied.

Specifically, t5 = a × t4.

Apply the idea Write the recursive relationship for t5 Substitute t4 = 12 from the table =4

Evaluate

The 5th term is t5 = 4, so there are 4 tasks left to complete at the end of the 5th week.

Reflect and check It can be observed that multiplying each term by the common ratio each term by 3.

is the same as dividing

(or 324 ÷ 3 = 108)

Idea summary A geometric sequence has an initial value (first term, t1) and a common ratio (a). The common ratio a is found by dividing any term by its preceding term:

The recursive formula for a geometric sequence is:

tn+1 = a × tn tn++1

is the (n + 1)th term

is the nth term

is the common ratio

The initial value t1 must be specified along with the recursive rule.

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Geometric sequences in general form While the recursive formula tn+1 = a × tn is useful for finding the next term in a sequence, it’s not efficient for finding a term far into the sequence (e.g., the 50th term). For this, the general form (or explicit formula) of a geometric sequence is used. Let t0 be the initial term of the sequence (the term at n = 0) and a be the common ratio. The terms of the sequence can be written as: • t0 • t1 = t0 × a • t2 = t1 × a = (t0 × a) × a = t0 × a2 • t3 = t2 × a = (t0 × a2) × a = t0 × a3 • And so on... This pattern leads to the general form for the nth term of a geometric sequence, when starting with t0:

tn = t0 × an tn

is the nth term

is the initial term (at n = 0)

is the common ratio

is the term number (starting from n = 0, 1, 2,. . .)

If the sequence is defined starting with t1 (the first term, where n = 1), the general formula is:

tn = t1 × an−1 tn

is the nth term

is the first term (at n = 1)

is the common ratio

is the term number (starting from n = 1, 2, 3,. . .)

It is important to pay attention to whether the sequence starts with t0 or t1 when using the general form. For this course, the primary focus is on using tn = t0 × an, t0 can be found if t1 and a are known using

Graphs of geometric sequences: When the terms of a geometric sequence are plotted as points (n, tn) on a Cartesian plane, their appearance depends signifcantly on the common ratio a and the initial term t0. • If t0 > 0 and a > 1, the points show exponential growth, increasing more rapidly as n increases. • If t0 > 0 and 0 < a < 1, the points show exponential decay, decreasing and approaching zero as n increases. • If a < 0, the terms alternate in sign. • If − 1 < a < 0, the points oscillate above and below the n-axis while approaching zero. • If a < − 1, the points oscillate above and below the n-axis with increasing magnitude. • If a = 1, the points form a horizontal line (tn = t0). • If a = − 1, the points alternate between t0 and −t0. The graph of a geometric sequence consists of discrete points, as n typically represents integer term numbers. These points will lie on the curve of an exponential function y = t0 × ax.

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Interactive exploration Discover this concept in action online

mathspace.co

Example 5 Consider a geometric sequence defined by the recursive rule t1 = 80 and tn+1 = 0.5 × tn: a Identify the initial value t0 and the common ratio a for the general form tn = t0 × an.

Create a strategy The common ratio a is given directly in the recursive rule. The term t1 is provided. To find t0, the term before t1, use the relationship t1 = t0 × a, so

Apply the idea From the recursive rule tn+1 = 0.5 × tn, the common ratio is a = 0.5 Given t1 = 80. To find t0, use

So, the initial value for the general form is t0 = 160 and the common ratio is a = 0.5 b Write the general form tn = t0 × an for this sequence.

Create a strategy Substitute the values of t0 and a found in part (a) into the general formula tn = t0 × an.

Apply the idea Using t0 = 160 and a = 0.5, the general form is: tn = 160 × (0.5)n

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c Use the general form to find the term t5 (the term at index n = 5).

Create a strategy Substitute n = 5 into the general formula derived in part (b).

Apply the idea tn = 160 × (0.5)n n

t5 = 160 × (0.5)

Write the equation Substitute n = 5

= 160 × (0.03125)

Evaluate the power

Evaluate

The term t5 is 5.

Reflect and check Check using the recursive definition starting from t1 = 80: t1 = 80 t2 = 0.5 × t1 = 0.5 × 80 = 40 t3 = 0.5 × t2 = 0.5 × 40 = 20 t4 = 0.5 × t3 = 0.5 × 20 = 10 t5 = 0.5 × t4 = 0.5 × 10 = 5 This matches. In this context, t5 refers to the 5th term in the sequence listed as t1, t2, t3, t4, t5,. . . The general formula tn = t0 × an applied with n = 5 gives this value because t0 is the term “before” t1.

Example 6 A geometric sequence has its 2nd term listed as 12 and its 5th term listed as 324. Using the general form tn = t0 × an, this means t2 = 12 and t5 = 324: a Determine the common ratio a and the initial term t0.

Create a strategy Using the general form tn = t0 × an, two equations can be set up: t2 = t0 × a2 = 12 t5 = t0 × a5 = 324 This system of equations can be solved by dividing the second equation by the first to eliminate t0 and solve for a. Then, substitute a back into one of the equations to find t0.

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Apply the idea Equation 1 : t0 × a2 = 12 Equation 2 : t0 × a5 = 324 Divide Equation 2 by Equation 1 : Write the equation

Evaluate the division

Evaluate the power

Take the cube root

Evaluate The common ratio is a = 3. Now, substitute a = 3 into Equation 1 : t0 × a2 = 12 2

t0 × (3) = 12 t0 × 9 = 12

Write the equation Substitute a = 3 Evaluate the power Divide both sides by 9 Evaluate

The initial term is

b Write the general formula for this sequence using tn = t0 × an.

Create a strategy Substitute the determined values of t0 and a into tn = t0 × an.

Apply the idea With

and a = 3, the general formula is:

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c This graph shows points of a sequence: tn 100 80 60 40 20 n 0

Determine if it could represent the geometric sequence found from the previous part, and explain your reasoning by identifying t0 and a from the graph if possible.

Create a strategy From the graph, identify the value of tn when n = 0; this is t0. Then, calculate the ratio between consecutive terms to find the common ratio a. Compare these values to those found for the sequence in parts (a) and (b).

Apply the idea From the graph: When n = 0, the point is approximately (0, 1.3333). So, t0 ≈ 1.3333 which is The terms from the graph are:

t1 = 4, t2 = 12, t3 = 36, t4 = 108.

Check the ratio using these points:

The common ratio from the graph is a = 3. The graph shows an initial value determined for the sequence

and a common ratio a = 3. These match the values in parts (a) and (b). Therefore, the graph could represent

this geometric sequence. The points show exponential growth as t0 > 0 and a > 1.

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Idea summary The general form (or explicit formula) for a geometric sequence with initial term t0 (at n = 0) and common ratio a is:

tn= t0 × an tn is the nth term t0

is the initial term (at n = 0)

is the common ratio

is the term number (n = 0, 1, 2, . . .)

If starting with t1 (at n = 1), the formula is tn = t1 × an−1 The graph of a geometric sequence consists of discrete points lying on an exponential curve. The behaviour of the graph depends on the value of a: • a > 1: Exponential growth (if t0 > 0) • 0 < a < 1: Exponential decay (if t0 > 0) • a < 0: Oscillating terms

3.03 Practice questions What do you remember? 1

Determine the next two terms in each geometric sequence: a

Ex 1

12, − 48, 192

Determine the common ratio for each geometric sequence: a

9, 36, 144, 576, . . .

2, − 6, 18, − 54, . . .

− 64, − 16, − 4, − 1, . . .

− 6, − 42, − 294, − 2058, . . .

Determine whether each sequence is a geometric sequence, an arithmetic sequence, or neither: a

4, 12, 36

4, − 4, 4, − 4, . . .

2, 0, − 2, − 4, . . .

40 000, 4000, 400, 40

2, 22, 24, 26, . . .

Consider the geometric sequences shown: i

Describe the recursive process in words.

Determine the next two terms.

5, 10, 20, 40

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Practice Ex 2

For each recursive relationship, determine the next 3 terms of the sequence: a c

tn = 2tn−1, where t1 = 2 , where t1 = 80

, where t0 = 1072

b d

tn+1 = − 5tn, where t0

Consider the pattern: a

Describe the recursive process in words.

Determine the number of circles the next step in the pattern contains.

Step 1 7

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Step 2

Step 3

Consider the pattern: a

Describe the recursive process in words.

Determine the number of hexagons the next step in the pattern contains. Step 1

1 20

Step 2

Step 3

Consider the sequences described by each recursive process. For each sequence: i

Determine the first five terms of each sequence.

Identify whether the sequence is arithmetic, geometric, or neither.

Start with 5 and to go from one term to the next, multiply by 2.

Start with 3 and to go from one term to the next, add 4.

Start with 81 and to go from one term to the next, divide by 3.

Start with 1 and to go from one term to the next, add 3, and then multiply by 2.

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Write down a recursive rule for tn in terms of tn−1 and the initial term t1 for each sequence: a c

2, 6, 18, 54, . . .

9000

1800

360

9 tn 8 7 6 5 4 3 2 1 0 –1 –2 –3 –4 –5 –6

40, 20, 10, 5, . . .

− 12

− 192

tn 25 20 15

n 1

10 5 n 0

Determine the missing terms in these geometric sequences: a

− 5, ⬚, − 80, 320, ⬚

⬚, ⬚,

,⬚

Determine the sixth term of the finite geometric sequence:

The average daily growth of a seedling is 6% per day. A seedling measuring 8 cm in height is planted:

Determine the height of the seedling at the end of Day 1.

Determine the height of the seedling 4 days after it is planted. Round your answer to two decimal places.

Write a recursive rule, Hn, defining the height of the seedling n days after it is planted.

The average rate of depreciation of the value of a sports car is 14% per year. A new sports car is bought for $90 000: a

Determine the value of the sports car after 1 year.

Determine the value of the sports car after 3 years.

Write a recursive rule for Vn, defining the value of the sports car after n years.

Use a general rule to determine the value of the sports car after 10 years. Round your answer to the nearest cent.

The average annual rate of inflation in Australia is 2.6%. A loaf of bread cost $3.65 in 2015: a

Determine the cost of the loaf of bread in 2016, rounded to the nearest cent.

At this rate, determine the cost of the loaf of bread in 2018, rounded to the nearest cent.

Write a recursive rule, Vn, defining the cost of the loaf of bread n years after 2015.

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Ex 3

Ex 4

Ex 5 20

Futsal is a modified form of soccer, typically played indoors on a smaller pitch with fewer players. A futsal ball has less bounce than a full size soccer ball and is expected to bounce back off the ground to 25% of the height from which it falls after every subsequent bounce: a

Write a recursive rule to represent the height of the nth bounce if it was dropped from an initial height of 2 m.

Determine the height of the third bounce in centimetres. Round your answer to two decimal places.

The first term of a geometric sequence is 6. The fourth term is 384: a

Determine how many times the first term needs to be multiplied by the common ratio to determine the fourth term.

Use your conclusion from part (a) to solve for the common ratio, a, of this sequence.

Determine the recursive rule, tn, that defines this sequence.

The first term of a geometric sequence is 3. The third term is 48: a

Determine how many times the first term needs to be multiplied by the common ratio to determine the third term.

Solve for the possible values of the common ratio, a, of this sequence.

Determine the recursive rule, tn, that defines the sequence with a positive common ratio.

Determine the recursive rule, tn, that defines the sequence with a negative common ratio.

A company is reducing its inventory over 5 months. They started with 625 items and reduced the inventory each month, following a geometric sequence. This table shows the items left at the end of each month, tn, where n is the month number:

125

Identify the common ratio, a.

Determine t5.

The third term of a geometric sequence is 2500. The seventh term is 4: a

Identify the first term of this sequence.

Determine both recursive rules that could define the sequence.

Consider a geometric sequence defined by the recursive rule t1 = 100 and tn+1 = 0.8 × tn: Determine the initial term t0.

Write the general rule for the sequence in the form tn = t0 × an.

Use the general rule to determine t10. Round to two decimal places if necessary.

The third term of a geometric sequence is 16. The sixth term is 128: a

Solve for the common ratio, a, of this sequence.

Identify the initial term, t0, of this sequence.

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1 625

Ex 6

n (Month) tn (Items left)

Determine the general rule, tn = t0 × an, that defines this sequence.

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Match each geometric sequence rule to its corresponding graph, and justify your choices: Rules: i

tn = 40 × (0.5)n

tn = 5 × (1.5)n

iii

tn = 20 × (− 0.8)n

tn = 2 × (− 2)n

Graphs: A

B 20

40 35

–5

–10

5 0

25 20 15 10 5 n 1

–15

n 1

30 25 20 15 10 5 0 –5 –10 –15

Extend your thinking 23

If the first term of a geometric sequence is 27 and the common ratio is a

Determine the 30th term.

Explain the process used.

Explore the different patterns the terms of a geometric sequence form when the common ratio is: a

Greater than 1

Equal to 1

Less than 0

Between − 1 and 1

Equal to − 1

Less than − 1

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3.04 Applications of sequences After this lesson, you will be able to… • apply first-order linear recurrence relations to financial problems • solve real-world problems using sequences

Model with recurrence relations Many real-world situations involve quantities that change in regular patterns over time. Such situations can often be modelled using sequences. A sequence can be defined by explaining how each term is found from the term before it. This is called a recursive definition. A first-order linear recurrence relation is a rule that defines the next term of a sequence based on the current term, using a linear equation. The general form is:

tn+1 = a × tn + b tn++1

is the value of the next term in the sequence

is the value of the current term in the sequence

a is a constant multiplier. It represents a rate of growth or decay b is a constant value added or subtracted at each step. It represents a fixed amount being added or removed To use a recurrence relation, an initial term is also needed, often denoted as t0 (the starting value) or t1 (the first term after the start). Arithmetic sequences are a special case of first-order linear recurrence relations. For an arithmetic sequence with first term t0 and common difference d, the recurrence relation is tn+1 = tn + d. Here, a = 1 and b = d. Geometric sequences are another special case. For a geometric sequence with first term t0 and common ratio r, the recurrence relation is tn+1 = r × tn. Here, a = r and b = 0. The general form tn+1 = a × tn + b allows for the modelling of situations where a quantity is multiplied by a and then has b added or subtracted at each step. For example, a population might grow by a certain percentage each year (a × tn) but also lose a fixed number of individuals (+ b, where b would be negative). To find the terms of a sequence using a recurrence relation, one starts with the initial value and applies the rule iteratively. While this is useful for finding the first few terms, for finding a term far into the sequence (making long-term predictions), using technology (like a spreadsheet or calculator with sequence capabilities) or working out a direct formula (an explicit formula), if possible, is often quicker. This lesson will focus on how to set up these recurrence relations and use them to find terms step-by-step (iterative calculations).

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Exploration A savings jar begins with $100. Each week, parents contribute an additional $20 to the jar. 1. How much money is in the jar at the start (Week 0)? 2. How much money is in the jar after Week 1? 3. How much money is in the jar after Week 2? 4. Let Mn be the money in the jar after n weeks. Write a rule that connects Mn+1 (the money in the jar next week) with Mn (the money in the jar this week)? 5. What is the value of a and b in the rule if it’s in the form Mn+1 = aMn + b?

Example 1 A racing car starts a race with 150 litres of fuel in its tank. The car consumes fuel at a constant rate of 5 litres per minute of racing: a Define a recurrence relation for Fn, the amount of fuel in litres remaining in the tank after n minutes of racing. Determine the initial condition F0.

Create a strategy Identify the initial amount of fuel. Determine how the fuel changes each minute. Write this in the form Fn+1 = aFn + b.

Apply the idea The initial amount of fuel is 150 litres. So, F0 = 150. Each minute, the car consumes 5 litres of fuel. This means the amount of fuel decreases by 5 litres. So, the amount of fuel next minute (Fn+1 ) is the current amount (Fn) minus 5. The recurrence relation is Fn+1 = Fn − 5, with F0 = 150. (In the form tn+1 = atn + b, here a = 1 and b = − 5).

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b Use the recurrence relation to find the amount of fuel remaining after 1, 2, and 3 minutes of racing.

Create a strategy Start with F0 and apply the rule Fn+1 = Fn − 5 iteratively.

Apply the idea Start with F0 = 150. After 1 minute (n = 1): F1 = F0 − 5 = 150 − 5 = 145 L After 2 minutes (n = 2): F2 = F1 − 5 = 145 − 5 = 140 L After 3 minutes (n = 3): F3 = F2 − 5 = 140 − 5 = 135 L c How long will it take for the car to run out of fuel?

Create a strategy The goal is to find when the fuel Fn becomes 0. This can be done by applying the rule step-bystep (iteration) or by using the fact that this is an arithmetic sequence. The direct formula for an arithmetic sequence is Fn = F0 + n × d, so here it’s Fn = 150 + n(− 5) = 150 − 5n.

Apply the idea Using the explicit formula Fn = 150 − 5n: 0 = 150 − 5n

Set Fn = 0

5n = 150

Add 5n to both sides Divide both sides by 5

= 30

Evaluate

It will take 30 minutes for the car to run out of fuel.

Reflect and check Alternatively, by iteration (using a calculator or spreadsheet for speed): F4 = 130, F5 = 125, . . . Continuing this process until Fn = 0 would also yield n = 30.

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Example 2 A car bought at the beginning of 2009 is worth $1500 at the beginning of 2015. The value of the car has depreciated by a constant amount of $50 each year since it was purchased: a What was the initial purchase price of the car at the beginning of 2009?

Create a strategy Multiply the number of years that have passed by the annual depreciation and add the product to the given price.

Apply the idea The number of years that have passed between 2009 and 2015 is 2015 − 2009 = 6. Depreciation amount = 6 × 50 = $300 Purchase price = 1500 + 300 = $1800

ultiply the number of years that have passed by the M depreciation Evaluate Add the given price to the depreciation amount Evaluate

The value of the car in the beginning of 2009 is $1800. b Plot the value of the car, Vn, on the graph from 2009 (represented by n = 0) to 2015 (represented by n = 6).

Create a strategy Subtract 50 from each value generated starting from the purchase price.

Apply the idea • For n = 0, the value will be the purchase price (1800), so the first point is at (0, 1800). • For n = 1, the value is 1800 − 50 = 1750, so the point is at (1, 1750). • For n = 2, the value is 1750 − 50 = 1700, so the point is at (2, 1700). • For n = 3, the value is 1700 − 50 = 1650, so the point is at (3, 1650). • For n = 4, the value is 1650 − 50 = 1600, so the point is at (4, 1600). • For n = 5, the value is 1600 − 50 = 1550, so the point is at (5, 1550). • For n = 6, the value is 1550 − 50 = 1500, so the point is at (6, 1500).

2000 1900 1800 1700 1600 1500 1400 1300 1200 1100 1000 0

n 1

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c Write an explicit rule for the value of the car after n years.

Create a strategy Substitute the values required into the explicit formula Vn = V0 + nd. Here, V0 is the initial value of the car (when n = 0 years) and d is the common difference (the amount the car’s value changes each year).

Apply the idea The common difference d is the amount the car’s value decreases each year, which is − 50. From the graph in part (b), the value of the car after n = 1 year is V1 = 1750. Using the formula Vn = V0 + nd, we can find V0: V 1 = V0 + 1 × d = V0 + 1 × (− 50) 1750 = V0 − 50

Write the formula for n = 1 Substitute d = − 50 Substitute V1 = 1750

V0 = 1750 + 50

Simplify

V0 = 1800

The initial value of the car (at n = 0 years)

Now, substitute V0 = 1800 and d = − 50 into the explicit formula Vn = V0 + nd. Vn = V0 + nd

Write the formula

= 1800 + n × (− 50)

Substitute V0 = 1800 and d = − 50

= 1800 − 50n

Simplify

The explicit rule for the value of the car after n years is Vn = 1800 − 50n. d Solve for the year n at the end of which the car will be worth half the price it was bought for.

Create a strategy into Vn = 1800 − 50n.

Substitute

Apply the idea Write the formula Substitute Purchase price = 1800 = 900

Evaluate

Solving for n: Vn = 1800 − 50n

Write the equation

900 = 1800 − 50n

Substitute Vn = 900

− 900 = − 50n

Subtract 1800 from both sides

18 = n

Divide both sides by −50

At the end of 18 years, the car will be worth half the price it was bought for.

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Example 3 A fish population in a lake is initially 1000. Each year, the population increases by 5% due to natural reproduction. However, 200 fish are harvested from the lake at the end of each year: a Define a recursive relation for Pn, the fish population at the end of year n after harvesting. Determine the initial condition P0.

Create a strategy The initial population is P0. The population increases by 5%, so multiply by 1.05. Then, 200 fish are subtracted.

Apply the idea The initial population is P0 = 1000. Each year, the population Pn first increases by 5%. This means it is multiplied by 1 + 0.05 = 1.05 Then, 200 fish are harvested. So, the recurrence relation is Pn+1 = 1.05Pn − 200, with P0 = 1000. b Calculate the fish population at the end of Year 1 and Year 2.

Create a strategy Use the recurrence relation Pn+1 = 1.05Pn − 200 with P0 = 1000.

Apply the idea End of Year 1 (n = 1): P1 = 1.05P0 − 200

Write the equation

= 1.05(1000) − 200

Substitute P0 = 1000

= 1050 − 200

Evaluate the multiplication

= 850

Evaluate

The fish population at the end of Year 1 is 850. End of Year 2 (n = 2): P2 = 1.05P1 − 200

Write the equation

= 1.05(850) − 200

Substitute P1 = 850

= 892.5 − 200

Evaluate the multiplication

= 692.5

Evaluate

Since half a fish is not possible, population numbers are usually rounded down (unless specified otherwise). So, the fish population at the end of Year 2 is 692.

Reflect and check It is usually best to keep the decimal for ongoing calculations. This practice minimises the accumulation of small errors that can significantly distort the final result. However, the final answer should be rounded if it is a count of things like fish.

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Idea summary A first-order linear recurrence relation defines the next term of a sequence based on the current term using the rule:

tn+1 = a × tn + b tn++1 is the value of the next term in the sequence tn

is the value of the current term in the sequence

is a constant multiplier. It represents a rate of growth or decay

i s a constant value added or subtracted at each step. It represents a fixed amount being added or removed

Terms are generated by applying the rule iteratively from the initial term.

Financial applications of recurrence relations First-order linear recurrence relations are widely used to model various financial situations, such as investments, loans, and depreciation. •

Simple interest

Simple interest is calculated only on the principal amount. If an amount P (or V0 ) is invested at a simple interest rate, the interest earned each period is constant. Let d be this constant interest amount. The value of the investment Vn after n periods can be modelled by:

Vn+1 = Vn + d Vn++1

is the value of the investment after n + 1 periods

is the value of the investment after n periods

is the constant interest amount added each period

The initial condition is V0 = P (the principal). The interest amount d for each period of the recurrence relation can be found using the simple interest formula I = P × rperiod × 1 (for one period). So, d = P × rperiod, where rperiod is the interest rate for that specific time period (e.g., if interest is calculated monthly, rperiod is the monthly rate). •

Flat-rate depreciation

Flat-rate (or straight-line) depreciation is when an asset loses a fixed amount of value each period. Let V0 be the initial value of the asset and d be the constant amount of depreciation each period.

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The value of the asset Vn after n periods can be modelled by:

Vn+1 = Vn − d Vn++1

is the value of the asset after n + 1 periods

is the value of the asset after n periods

is the constant amount of depreciation subtracted each period

The initial condition is V0 (the initial purchase price). •

Compound interest and reducing balance depreciation

Compound interest is calculated on the current balance, including any accumulated interest. If an amount V0 is invested and compounds interest at a rate i per period, the value Vn after n periods is modelled by:

Vn+1 = R × Vn R is the growth factor per period (1+ interest rate per period) Reducing balance depreciation is similar — an asset loses value by a fixed percentage of its current value each period. If the rate of depreciation is i per period, then R = 1 − i.

Vn+1 = R × Vn R is the decay factor per period (1− depreciation rate per period) •

Annuities and loans (combined operations)

Many financial situations involve both compound interest and regular payments or deposits. These are modelled by recurrence relations of the form Vn+1 = R × Vn ± D. Annuities (investments with regular deposits): An initial investment V0 earns compound interest, and a regular deposit D is made each period:

Vn+1 = R × Vn + D R is the growth factor (1 + i) D is the regular deposit amount Reducing balance loans (loans with regular payments): An initial loan amount V0 is charged compound interest (or gains interest from the lender’s perspective), and makes a regular payment P each period.

Vn+1 = R × Vn − P R is the growth factor (1 + i) P is the regular payment amount

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Exploration A loan of $5000 is taken out. This loan is charged compound interest at a rate of 4% per month. At the end of each month, after interest has been added to the outstanding balance, a payment of $600 is made. Let Ln be the balance of the loan after n months. 1. What is the initial loan amount (L0)? 2. How much interest is added in the first month? What’s the new loan balance before the first payment? 3. What is the loan balance after the first payment is made (L1)? 4. Calculate the interest for the second month (based on the balance after the first payment). What’s the loan balance before the second payment? 5. What is the loan balance after the second payment is made (L2)? 6. Write recurrence relation connecting Ln+1 to Ln.

Example 4 The average rate of depreciation of the value of a sports car is 14% per year (reducing balance). A new sports car is bought for $90 000: a Write a recurrence relation for Fn, the value of the sports car after n years and determine F0.

Create a strategy This is reducing balance depreciation. Find the decay factor R. The initial value is F0.

Apply the idea The initial value is F0 = $90 000. The depreciation rate is 14% = 0.14 per year. The decay factor R = 1 − 0.14 = 0.86 The recurrence relation is Fn+1 = 0.86Fn, with F0 = 90 000.

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b Use the recurrence relation to find the value of the sports car after 1 year and after 3 years.

Create a strategy Apply the recurrence relation iteratively.

Apply the idea Given that F0 = 90 000: After 1 year (F1): F1 = 0.86 × F0 = 0.86 × 90 000 = $77 400 After 2 years (F2): F2 = 0.86 × F1 = 0.86 × 77 400 = $66 564 After 3 years (F3): F3 = 0.86 × F2 = 0.86 × 66 564 = $57 245.04 The sports car is worth $77 400 after 1 year and $57 245.04 after 3 years.

Example 5 Sarah invests $10 000 in an account that pays compound interest at a rate of 5% per annum, calculated annually. At the end of each year, after interest is paid on the current balance, Sarah deposits an additional $500 into the account: a Let Vn be the value of Sarah’s investment at the end of year n (after interest and the additional deposit). Write a recurrence relation for Vn, stating V0.

Create a strategy Identify the initial investment V0. Determine the growth factor R due to interest. Identify the additional deposit D. Combine these into the form Vn+1 = RVn + D.

Apply the idea The initial investment is V0 = $10 000. The interest rate is 5% per annum, so i = 0.05 The growth factor is R = 1 + i = 1 + 0.05 = 1.05 The additional deposit each year is D = $500. The recurrence relation is Vn+1 = 1.05Vn + 500, with V0 = 10 000.

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b Calculate the value of Sarah’s investment at the end of Year 1 and Year 2.

Create a strategy Use the recurrence relation Vn+1 = 1.05Vn + 500 with V0 = 10 000 iteratively.

Apply the idea At the end of Year 1 (V1 ): V1 = 1.05 × V0 + 500

Write the equation

= 1.05 × 10 000 + 500

Substitute V0 = 10 000

= 10 500 + 500

Evaluate the multiplication

= $11 000

Evaluate

At the end of Year 2 (V2 ): V2 = 1.05 × V1 + 500

Write the equation

= 1.05 × 11 000 + 500

Substitute V1 = 11 000

= 11 550 + 500

Evaluate the multiplication

= $12 050

Evaluate

The value of Sarah’s investment is $11 000 at the end of Year 1 and $12 050 at the end of Year 2.

Idea summary Recurrence relations model various financial scenarios: • Simple interest: Vn+1 = Vn + d (where d is constant interest per period, V0 = Principal). • Flat-rate depreciation: Vn+1 = Vn − d (where d is constant depreciation per period, V0 = Initial value). • Compound interest: Vn+1 = R × Vn (where R = 1 + i, i is interest rate per period). • Reducing balance depreciation: Vn+1 = R × Vn (where R = 1 − i, i is depreciation rate per period). • Annuities (investment with regular deposits): Vn+1 = R × Vn + D (where D is the regular deposit added each period). • Reducing balance loans (loan with regular payments): Vn+1 = R × Vn − P (where P is the regular payment made each period). In all cases, V0 represents the initial financial value.

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3.04 Practice questions What do you remember? 1

A paver needs to pave a floor with an area of 800 m2. He can pave 50 m2 a day: a

Complete the table showing the area left to pave at the start of each day: Day

Area of floor left to pave (m2)

800

⬚ ⬚

⬚

5 b

What type of change is this? A

Linear growth

Exponential growth

Linear decay

Exponential decay

A new lounge suite depreciates by a constant amount each year and its value is modelled by the recurrence relation Vn+1 = Vn − 700, V0 = 4800 where Vn is the value of the lounge suite, in dollars, after n years:

Determine the initial cost of the lounge suite.

Determine the amount of value lost each year.

An investment pays simple interest annually and is modelled by the recurrence relation Vn+1 = Vn + 2310, V0 = 30 000 where Vn is the value of the investment after n years: a

Determine the initial amount invested.

Determine the amount of interest paid each year.

Calculate the annual interest rate, expressed as a percentage. Round your answer to one decimal place.

Practice 4

Zuber is a taxi service that charges a $2.50 pick-up fee and $2.10 per kilometre of travel: a

Determine the total charge for a 15 km journey.

Write a recursive rule for tn+1 in terms of tn which defines the price of an n -km trip, and an initial condition for t0.

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Skye is a salesperson and is paid a travel allowance of $40 for each business trip she takes to demonstrate the use of the product she sells. Each week she is also paid a retainer of $220: a

Calculate her total pay for a week with 3 business trips.

Write a recursive rule for un+1 in terms of un which defines Skye’s total pay for a week with n trips, and an initial condition for u0.

Write a recursive rule for vn+1 in terms of vn which defines how much Skye is paid in two weeks where she makes n trips, and an initial condition for v0.

A piece of jewellery appreciates in value by a constant amount each year and its value is modelled by the recurrence relation Vn+1 = Vn + 220, V0 = 3000 where Vn is the value of the jewellery, in dollars, after n years:

Determine the initial value of the piece of jewellery.

By how much does it appreciate each year?

Write an explicit rule for Vn for the value of the jewellery after n years.

Determine the value of the investment after 12 years.

A motorbike depreciates in value by a constant amount each year and its value is modelled by the recurrence relation Vn+1 = Vn − 1200, V0 = 15 000 where Vn is the value of the motorbike, in dollars, after n years:

Determine the purchase price of the motorbike.

At what rate is it decreasing in value each year?

Write an explicit rule for Vn that gives the value of the motorbike after n years.

Determine the value of the motorbike after 9 years.

The balance of a savings account earning simple interest each year is given by the explicit rule Vn = 2200 + 300 (n − 1) where Vn is the balance after n years:

How much interest is the account earning each year?

Calculate the account balance after 1 year.

Determine the original investment amount.

Write a recursive rule for Vn+1 in terms of Vn , and an initial condition V0.

For a photo, the animals at an animal shelter have been arranged such that there are 10 animals in the front row and each subsequent row has 7 more animals than the row in front of it: a

240

Write a recursive rule for an+1, the number of animals in row (n + 1), in terms of an, and determine the initial condition a0.

Write a general formula for an in terms of n.

How many animals are in the 7th row?

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Ex 1

Tobias starts his career with a monthly wage of $3500. At the beginning of each year that follows, he receives a raise and his monthly wage for that year will be $160 greater than the previous year: a

Determine his yearly salary in the second year of his service.

Write down an expression for the total amount earned in m years.

Solve for the number of years, y, that it would take for him to earn a total of $447 120.

A farm’s water tank holds 5000 litres. It is estimated that the tank loses 150 litres each day to evaporation. Let Vn be the volume of water in the tank at the beginning of day n:

Ex 2

Determine the value of V0.

Write a recursive rule for Vn+1 in terms of Vn that describes the volume of water in the tank at the beginning of each day.

For how many full days will there be water in the tank?

A piece of art bought in 2012 is worth $4000 in 2020. Its value has appreciated by a constant amount of $150 each year since it was purchased: a

Calculate the purchase price of the art in 2012.

Write an explicit rule for the value of the art, Vn, after n years.

After how many full years, n, will the art be worth at least double its purchase price?

Megan is learning French using an app on her phone. Before using the app she already knows 30 words. The app gives her 10 new words to learn each day: a

How many words will she know in total after using the app for 6 days?

Write a recursive rule, Tn+1, the total number of words she has learnt after (n + 1) days of using the app, in terms of Tn.

Solve for n, the number of days it will take her to learn 500 words.

Write a recursive rule, Wn+1, the total number of words she has learnt after (n + 1) weeks of using the app, in terms of Wn.

When a new school first opened, a students started at the school. Each year, the number of students increases by the same amount, d: a

At the beginning of its 7th year, it had 361 students. Form an equation for a in terms of d.

At the end of the 11th year, the school had 536 students. Form an equation for a in terms of d.

Solve for d, the number of students who joined the school each year.

How many students started at the school when it first opened?

At the end of the nth year, the school has reached its capacity at 921 students. Solve for the value of n.

A mobile phone depreciates in value by a constant amount (in dollars) per month and its value is given by the explicit rule Vn = 800 − 25n, where Vn is the value after n months: a

By how much does the value of the phone depreciate each month?

Determine the purchase price of the phone.

Write a recursive rule for the value of the phone Vn+1 in terms of Vn, and an initial condition V0.

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A rare figurine was purchased for $60 and ten years later it is worth $460: a

How much did the figurine appreciate by each year if it appreciated in value by a constant amount each year.

Write a recursive rule for Vn+1 in terms of Vn, and an initial condition V0.

Write an explicit rule, Vn, for the value of the figurine after n years.

Determine the value of the figurine in another 10 years time.

A piece of machinery depreciated at a constant rate per hour of use. After 160 hours of use, it was worth $21 040. After 210 hours of use, it was worth $20 740: a

Determine the amount of depreciation each hour.

Determine V0, the initial value of the machinery.

Write a recursive rule for Vn+1 in terms of Vn, and an initial condition V0.

Write an explicit rule, Vn, for the value of the machinery after n hours of use.

Solve for n, the number of hours of use after which the machinery will be worth a quarter of its original value.

Glen decided to start saving by depositing some money in a safe each month. On 1st January 2019, Glen put $10 in the safe. On 1st February and 1st March in the same year, he deposited $13 and $16, respectively:

242

If he continued this pattern, how much did Glen put in the safe on August 1st 2019?

Write a recursive rule which gives the amount Tn+1 that Glen deposits in the safe (n + 1) months after January 1st 2019 in terms of Tn.

How much in total did Glen have in the safe by the end of 2019?

Glen wishes to save $770. Solve for n, the number of months it will take to do this.

A cake is taken out of a 156°C oven and left to cool in a room with a constant temperature of 26°C. As soon as it is taken out, the temperature of the cake drops by 2.5°C each minute: a

Determine the temperature of the cake after 8 minutes:

Write a recursive rule that gives the temperature Tn+1 of the cake after n minutes in terms of Tn with initial temperature T0.

By how many degrees has the temperature of the cake dropped after half an hour?

Determine the number of minutes it will take for the cake to reach room temperature.

A bobcat, initially purchased for $70 000, depreciates at a rate of $12 for every day of use: a

Write a recursive rule, Vn+1, that gives the value of the bobcat, in dollars, after (n + 1) days of use.

How much value is lost over a five-week period, if the bobcat is used 6 days a week?

The owner of the bobcat will replace the tyres when the value drops below $59 209. After how many whole days of operation is he first due to change the tyres?

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An investment of $6000 pays simple interest at a rate of 4.2% per annum and is modelled by the recurrence relation Vn+1 = Vn + 252, V0 = 6000 where Vn is the value of the investment after n years:

Calculate the value of the investment after 5 years.

How much interest has been earned in 5 years?

After how many whole years will the investment double?

A motorbike purchased for $5000 depreciates annually by $300. The value of the motorbike is modelled by Vn+1 = Vn − 300, V0 = 5000 where Vn is the value of the motorbike, in dollars, after n years:

After how many whole years will the value of the motorbike first drop below $1000?

After how many whole years will the motorbike first lose more than half its value?

Julia has $3000 to invest and deposits this amount into an account earning 7% simple interest each year. Her sister, Lisa, has $4000 to invest and deposits it into an account earning 4% simple interest each year. After how many whole years will Julia’s investment first be worth more than Lisa’s?

Ex 3

Ex 4

Ex 5

A family takes out a $15 000 loan to renovate their kitchen. The loan accrues interest at a rate of 12% per annum, calculated monthly. The family makes monthly repayments of $400: a

Show that the monthly interest rate is 1%.

Write a recurrence relation for the balance of the loan, Bn, after n months, including the initial balance B0.

What is the balance of the loan after the first two payments have been made?

A new computer is purchased for $2500 and depreciates in value by 20% each year (reducing balance): a

Write a recurrence relation to model the value of the computer, Vn, after n years, including the initial condition V0.

What is the value of the computer after 3 years?

After how many full years will the value of the computer first be less than $1000?

Leo opens a savings account with an initial deposit of $500. The account pays compound interest at 6% per annum. At the end of each year, after the interest is paid, Leo deposits an additional $200: a

Write a recurrence relation to model the value of Leo’s savings, Vn, at the end of year n, including the initial condition V0.

Calculate the value of the savings at the end of the second year.

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Extend your thinking 27

A TV’s value is modelled by two different depreciation methods. Model A (simple depreciation) is shown in the graph. Model B (reducing balance depreciation) assumes the TV loses 8% of its value each year: a

Write a recurrence relation, including an initial condition V0, to represent the value of the TV under Model B.

Determine the year, n, in which the value of the TV under Model B is less than its value under Model A.

The owner plans to sell the TV when its value first drops below $110. Which model predicts this will happen sooner, and by how many years?

An investment that pays simple interest annually is represented by the graph. A second investment option also starts with $2000, but its value is modelled by a geometric sequence. After 2 years, its value is $2420: a

Determine the annual simple interest rate of the investment shown in the graph.

For the second investment option, determine the common ratio r.

Find the first year in which the value of the second (geometric) investment exceeds the value of the first (arithmetic) investment.

The graph shows an investment (Plan A) earning simple interest. An alternative, Plan B, starts with the same principal but earns 8% p.a. compound interest. However, a fee of $50 is withdrawn at the end of each year after interest is paid: a

Write a recurrence relation for Bn, the value of the investment under Plan B after n years.

Determine in which year the value of Plan B will first be greater than the value of Plan A.

190 Vn (Dollars) 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 1 2

n (Years) 3

n (Years) 4 5

1600 Vn (Dollars) 1500 1400 1300 1200 1100 1000 900 800 700 600 500 400 300 200 100 0 1 2

n (Years) 4 5

Vn (Dollars) 3000 2500 2000 1500 1000 500 0

Investigation: Spreadsheets and sequences Investigate online

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mathspace.co

3.05

Arithmetic series

After this lesson, you will be able to… • define arithmetic series and their partial sums • calculate partial sums of arithmetic series • apply sigma notation to calculate arithmetic series sums

Arithmetic series Suppose one wanted to add all of the terms in an arithmetic sequence together. This process involves finding what’s known as an arithmetic series. If the first term of an arithmetic sequence is t1 and the common difference is b, then the sequence terms are: t1 = t1, t2 = t1 + b, t3 = t1 + 2b, t4 = t1 + 3b, t5 = t1 + 4b, . . . Imagine continuing to write the terms all the way up to t100 = t1 +99b, and then finding a way to add up all of the one hundred terms together. The great German mathematician Carl Friedrich Gauss in 1785, at the age of 8, used the following method: Instead of adding the terms from left to right, he added them in this order: (t1 + t100) + (t2 + t99) + (t3 + t98) + . . . + (t50 + t 51) Every grouped pair of terms adds up to 2t1 + 99b and since there are 50 pairs, their total must be 50 × (2t1 + 99b). Gauss knew there were 50 pairs because there are 100 terms. His answer could be written The sum Sn, representing the sum of the first n terms of an arithmetic sequence, is called the nth partial sum of the series. It can be found using one of two related formulas:

1 Sn is the sum of the first n terms n

is the number of terms to be added

is the first term of the sequence

is the common difference of the sequence

Alternatively, if the first term t1 and the nth term tn are known, the sum can be found using:

Sn is the sum of the first n terms n

is the number of terms to be added

is the first term of the sequence

is the nth term of the sequence

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Writing out long sums like t1 + t2 + . . . + t100 can be tedious. Sigma notation provides a concise way to represent the sum of terms in a sequence. It uses the Greek capital letter sigma, Σ. A sum written in sigma notation has the following general form:

• f (k) is the expression for the general term (the summand) that generates the terms to be added • k is the index of summation (or dummy variable), taking on integer values • m is the lower limit of summation, indicating the starting value for the index k • N is the upper limit of summation, indicating the ending value for the index k This expression indicates to sum the values of f (k) as k goes from m to N inclusive For example,

The number of terms being added is N − m + 1. In the example, there are 4 − 1 + 1 = 4 terms. An arithmetic series, which is the sum of the first n terms of an arithmetic sequence tk = t1 + (k − 1) b, can be written in sigma notation as:

Here, the first term is t1 (when k = 1), and the common difference is b.

Example 1 Find the sum of the first 10 terms of the arithmetic sequence defined by t1 = 6 and b = 3.

Create a strategy , where t1 = 6, b = 3, and n = 10.

Use the arithmetic series formula

Apply the idea Write the formula for the partial sum of the series Substitute t1 = 6, b = 3 and n = 10

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= 5 (12 + 27)

Simplify each term

= 195

Evaluate

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 The first term of an arithmetic sequence is − 5 and the 6th term is − 45: a If b is the difference between terms, solve for b.

Create a strategy Use the formula for the nth term of an arithmetic sequence tn = t1 + (n − 1) b, where t6 = − 45, t1 = − 5, and n = 6.

Apply the idea tn = t1 +(n − 1) b

Write the equation

− 45 = − 5 + (6 − 1) b

Substitute n = 6, t1 = − 5 and t6 = − 45

− 45 = − 5 + 5b

Evaluate the subtraction

− 40 = 5b

Add 5 to both sides Divide both sides by 5

b = −8

Evaluate

b Determine the sum of the first 14 terms.

Create a strategy , where n = 14, t1 = − 5, and b = − 8.

Use the arithmetic series formula

Apply the idea Write the formula for the partial sum of a series Substitute n = 14, t1 = − 5 and b = − 8 Evaluate the subtraction Evaluate the multiplication Simplify the value inside the brackets = 7 × (− 114)

Simplify the fraction

= − 798

Evaluate

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Example 3 Consider the arithmetic sequence 4, − 1, − 6, . . . a Write a simplified expression for the sum of the first n terms.

Create a strategy First identify the first term t1 and the common difference b. Use the arithmetic series formula:

Apply the idea The first term is t1 = 4. The common difference is b = − 1 − 4 = − 5 Substitute t1 = 4 and b = − 5 Simplify inside brackets

b Find the sum of the progression from the 19th to the 27th term, inclusive.

Create a strategy To find the sum of the progression from the 19th term to the 27th term inclusive, the sum of the first 27 terms (S27) can be calculated, and then the sum of the first 18 terms (S18) subtracted from it. This leaves the sum of terms t19 + t20 + . . . + t27. From part (a), the simplified expression for the sum of the first n terms is First, calculate S27. Next, calculate S18. Finally, find the difference, S27 − S18.

Apply the idea Calculating S27: Substitute n = 27, into Simplify inside the brackets = − 1647

Evaluate

Calculating S18: Substitute n = 18, into

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= 9(13 − 90)

Simplify the fraction and inside the brackets

= − 693

Evaluate

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Finding the sum from the 19th to the 27th term: Sum = S27 − S18

Write the equation

= − 1647 − (− 693) Substitute the values of S27 and S18 = − 954

Evaluate

So, the sum of the progression from the 19th to the 27th term is − 954.

Example 4 Evaluate the sum of:

Create a strategy The expression 5k − 2 is linear in k, so this is an arithmetic series. First, find the first term (t1) by substituting k = 1 into the expression 5k − 2. Next, find the 15th term (t15) by substituting k = 15 into the expression 5k − 2. The number of terms (n) is 15 − 1 + 1 = 15. Use the formula

to find the sum.

Another approach would be to find the common difference b. Since the term is 5k − 2, the common difference is b = 5 (the coefficient of k). Then the formula

could

be used. However, for this example, the method with t1 and tn will be used.

Apply the idea Find the first term, t1 (when k = 1): t1 = t1 = 5(1) − 2 = 5 − 2 = 3 Find the 15th term, t15 (when k = 15): t15 = 5(15) − 2 = 75 − 2 = 73 The number of terms is n = 15.

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Now, use the sum formula

: Write the formula Substitute n = 15, t1 = 3 and t15 = 73 Simplify inside the brackets = 15 × 38

Simplify the fraction

= 570 Evaluate Therefore,

Example 5 Write the arithmetic series 3 + 7 + 11 + . . . + 39 using sigma notation.

Create a strategy To write a series in sigma notation, the formula for the kth term, tk = t1 + (k − 1) b, and the number of terms, n, need to be found. Identify the first term (t1). Calculate the common difference (b). Find the number of terms (n) by setting the last term (39) equal to tn = t1 + (n − 1) b and solving for n. Write the sum using sigma notation as:

then simplify the expression for the kth term.

Apply the idea The first term is t1 = 3. The common difference is b = 7 − 3 = 4 The formula for the kth term is: tk = t1 +(k − 1) b = 3 + (k − 1)4 = 3 + 4k − 4 = 4k − 1 To find the number of terms, set the last term tn = 39 and solve for n: 4n − 1 = 39

Write the equation

4n = 40

Add 1 to both sides

n = 10

Divide both sides by 4

There are 10 terms in the series. So, the series can be written in sigma notation as:

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Reflect and check To check if the sigma notation:

generates the given series 3 + 7 + 11 + . . . + 39: • For k = 1 (first term): 4(1) − 1 = 3 (Matches) • For k = 2 (second term): 4(2) − 1 = 7 (Matches) • For k = 10: (last term, as n = 10): 4(10) − 1 = 39 (Matches) The terms match, so the sigma notation is correct.

Idea summary An arithmetic series is the sum of the terms in an arithmetic sequence. The sum of the first n terms is called the nth partial sum, denoted Sn. Key formulas for the partial sum Sn are: n 2t1 + ( n − 1)b 2 n Sn t1 t n 2 Sn =

where t1 is the first term, b is the common difference, n is the number of terms, and tn is the nth term. Sigma notation (∑) is used to write sums concisely. The expression

sums the summand f (k) as the index of summation k goes from a lower limit m to an upper limit N. An arithmetic series is written as:

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3.05 Practice questions What do you remember? 1

Complete the formulas for the sum of the first n terms of an arithmetic sequence, Sn: a

When the first term (t1) and common difference (b) are known: ⬚

When the first term (t1) and the nth term (tn) are known: ⬚

Determine whether each statement is true or false: a

An arithmetic series is the sum of the terms in an arithmetic sequence.

The symbol Σ is used to represent the product of a series of terms.

In the expression

the value N is the lower limit of summation.

Consider the arithmetic series represented by the expression a

Determine the first term of the series.

Determine the last term of the series.

Determine the number of terms in the series.

Practice Ex 1

Determine the sum of the first 10 terms of the arithmetic sequence given that: a

t1 = 5 and b =2

t1 = − 18 and b = 4

t1 = 10 and b = 4

t1 = − 4 and b = − 6

Determine the sum of the first 10 terms of each arithmetic sequence given that: a

T1 = 6 and T10 = 3.5

T1 = − 8 and T10 = − 2.75

T4 = 4 and T7 = 13

T6 = 2 and T7 = 5

Determine the sum of the first 20 terms of the arithmetic series defined by T1 = π and T20 = 20π.

Determine the sum of the first 15 terms of each arithmetic series:

2 + 6 + 10 + . . .

− 17 − 15 − 13 − . . .

35 + 31 + 27 + . . .

1 + 8 + 15 + . . .

Determine the sum of the first 14 terms in the arithmetic series where the first term is 13 and the 14th term is

Determine the sum of the first 40 terms of each arithmetic sequence: a

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Tn = 6n

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Tn = 5 − 3n

Ex 2

The first term of an arithmetic sequence is 4 and the eighth term is 25: a

Determine the value of b.

Determine the sum of the first 20 terms.

Consider the arithmetic series − 9 − 5 − 1 + . . . + 59: a

Determine n, the number of terms in the series.

Determine the sum of the series.

The first term of an arithmetic sequence is 7, the common difference is 3 and the sum of the first n terms is 205. Determine the value of n.

Ex 3

For the arithmetic series 50 + 47 + 44 + . . ., calculate the sum of the terms from the 11th to the 20th term, inclusive.

Ex 4

Evaluate the arithmetic series sum:

Ex 5

Express the arithmetic series 5 + 9 + 13 + . . . + 45 using sigma notation.

Extend your thinking 16

Consider the sum of the odd natural numbers: 1 + 3 + 5 + . . . Show that the sum of the first n odd natural numbers is a perfect square.

Determine the sum of all integers between 20 and 50, inclusive, that are divisible by 6.

The sum of the first 10 terms of an arithmetic series is 150. The sum of the first 20 terms of the same series is 600. Determine the first term t1 and the common difference b.

A theatre has 20 rows of seats. The front row has 15 seats, the second row has 17 seats, the third row has 19 seats, and so on, with each row having two more seats than the one in front of it. Tickets for the first 10 rows cost $50 each, while tickets for the back 10 rows (rows 11 − 20) cost $40 each. Calculate the total revenue for the theatre if every seat is sold.

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3.06 Geometric series After this lesson, you will be able to… • define geometric series and their partial sums • calculate partial sums of geometric series • analyse the behaviour of geometric series based on the common ratio

Finite geometric series Recall a geometric sequence is a list of numbers in which each consecutive pair of numbers has a common ratio: n

f(n)

⤻⤻⤻

Common ratio: The explicit rule for finding the nth term in a geometric sequence is:

tn = t1 (a)n−1

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is the nth term

is the first term

is the common ratio

is the number of the desired term

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Exploration Since his 1st birthday, Manuel has received cash gifts from various family members for each birthday. Each year, his mother takes $50 of the money and puts it into a high-yield savings account earning 3% interest annually. On his first birthday, the account has $50. On his second birthday, the money from his 1st birthday will accrue interest and another $50 would get deposited. On his 2nd birthday, the money from his 1st birthday will accrue interest again, the money from his 2nd birthday will accrue interest, and another $50 would be deposited. This pattern continues as shown in the table: Birthday

Amount in the account after each birthday

50 (1.03) + 50

50 (1.03)2 + 50 (1.03) + 50

50 (1.03)3 + 50 (1.03)2 + 50 (1.03) + 50

50 (1.03)4 + 50 (1.03)3 + 50 (1.03)2 + 50 (1.03) + 50

Suppose Manuel wants a way to easily determine the total amount of money in his account after any given birthday. Let Sn represent the amount of money in the account after the nth birthday. Use the questions to derive a shorter formula for Sn. 1. S5 = 50 (1.03)4 + 50 (1.03)3 + 50 (1.03)2 + 50 (1.03) + 50 represents the amount in the account after Manuel’s 5th birthday. If both sides of the equation for S5 are multiplied by the common ratio, 1.03, how will the new equation compare to the equation for S5? 2. How can the given equation and the equation from Question 1 be used to derive a shorter way of finding the sum of money in the account after 5 birthdays? 3. Use the ideas from the previous questions to derive a formula for finding the total amount of money in Manuel’s account after n birthdays.

The exploration with Manuel’s savings provides a method to find a general formula for the sum of the first n terms of any geometric sequence, Sn. Let the first term be t1 and the common ratio be a. The sum of the first n terms is: Sn = t1 + t1 a + t1 a2 + . . . + t1 an−1 Multiply this entire sum by the common ratio a: aSn = t1 a + t1 a2 + t1 a3 + . . . + t1 an Subtracting the second equation from the first results in: + t1 a +

t1 a 2 + . . . +

t1 an−1

aSn = t1 a +

t1 a 2 + . . . +

t1 an−1 + t1 an

Sn − Sn −

aSn =

− t1 a n

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Therefore, Sn − aSn = t1 − t1 an. Factorising Sn on the left side and t1 on the right side gives: Sn(1 − a) = t1(1 − an) If a ≠ 1, dividing both sides by (1 − a) yields the formula for the sum of the first n terms of a geometric sequence: A finite geometric series is the sum of the first n terms of a geometric sequence. It may also be called a partial sum of the sequence. The sum of the first n terms is found using the formula:

is the first term of the geometric sequence

is the common ratio of the geometric sequence

is the number of terms being added

This can only be applied to geometric sequences where the common ratio is not 1 because a = 1 would make the equation undefined. The value of the common ratio a significantly influences the behaviour of the geometric series. An analysis of how Sn changes for different types of common ratios follows: • When |a| > 1 (e.g., a = 2 or a = − 3): The terms t1 ak−1 grow larger in magnitude as k increases. As a result, the partial sum Sn will also grow larger in magnitude (either positively or negatively) as more terms are added. The sum diverges. • When |a| < 1 (e.g., a = 0.5 or a = − 0.2): The terms t1 ak−1 get smaller in magnitude, approaching zero as k increases. The partial sum Sn will approach a specific finite value as n gets very large. In this case, the series is said to converge. This concept is explored further in the study of infinite geometric series. • When a = 1: Each term in the sequence is t1. The sum of the first n terms is simply Sn = n × t1. The formula

is not used here because it would involve division by zero.

The sum grows linearly with n. • When a = − 1: The terms of the sequence alternate between t1 and −t1 (e.g., t1, −t1, t1, −t1,. . .). The partial sum Sn will oscillate. If n is odd, Sn = t1. If n is even, Sn = 0. The sum does not approach a single value. Understanding these behaviours is crucial for analysing geometric series in various applications. A geometric series can be represented in summation notation by:

This notation makes it possible to identify key information about the individual terms in the sequence, which can then be used in the formula to quickly calculate the final sum if n is large.

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Example 1 Sissa is a mythical character from India. Sissa was invited to meet with the king after inventing the game of Chaturanga, an ancient Indian game that evolved into the modern games of chess, xiangqi, shogi, and others. The king, delighted with his creation, asked Sissa how he wanted to be rewarded. Sissa started placing grain in a pattern on the board as shown in this image:

a Assuming this pattern continues, derive a formula to find the total number of grains of rice on the first n tiles of the board.

Create a strategy The geometric sequence is 1, 2, 4, 8, 16, 32, . . . with a common ratio of 2. Recall that any term in the finite sequence will be in the form tn = t1 (2)n−1. Since the first term of this sequence is 1, the last term will be 2n−1. A geometric series is the sum of all the terms, so the sum is represented as Sn = 1 + 2 + 4 + 8 + 16 + . . . + 2n−1.

Apply the idea Multiplying this series by 2, the common ratio, and subtracting the new series from the original one, will cause most of the terms to eliminate. Sn = 1 + 2 + 4 + 8 + 16 + . . . + 2n−1 2Sn = 2 + 4 + 8 + 16 + 32 + . . . + 2n Sn = 1 + 2 + 4 + 8 + 16 + . . . + 2n−1 − 2Sn = − 2 − 4 − 8 − 16 − 32 − . . . − 2n Sn − 2Sn = 1 − 2n Next, Sn can be factored out from the left side of the equation and solved for by dividing by what is left in brackets. Sn (1 − 2) = 1 − 2n

Write the equation

Sn (− 1) = 1 − 2

Evaluate the subtraction

Sn = 2n − 1

Divide both sides by − 1

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b Use the formula you derived in part (a) to find the total number of grains of rice on the first 2 rows of the board.

Create a strategy There are 16 tiles in the first 2 rows, so use n = 16 with the formula that was derived.

Apply the idea Sn = 2n − 1 16

S16 = 2 − 1 = 65 535

Write the equation Substitute n = 16 Evaluate

Reflect and check Should the king in this situation have agreed to Sissa’s reward proposal? After how many tiles would at least 1 million grains of rice have accumulated?

Example 2 Find the sum of the first 9 terms of the geometric sequence

Create a strategy To use the geometric series formula, it is necessary to know the first term, the common ratio, and the number of terms being added. The first term and the number of terms being added are given. The common ratio can be found by dividing any term by the previous term.

Apply the idea The common ratio is: Divide the third term by the second term To divide by a fraction, multiply by its reciprocal Evaluate

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Now, the sum of the first 9 terms can be found using the geometric series formula. Write the geometric series formula

Substitute n = 9, t1 = 2 and

Evaluate the exponent

Evaluate the addition

Evaluate the multiplication

131 072

Evaluate the division Evaluate the multiplication

Using technology, the sum can be approximated as about 1.229

Reflect and check This geometric sequence has the rule:

To add the first 9 terms, the sigma notation would be:

This notation displays each individual term, but the key information can be identified from it and the geometric series formula used instead.

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Example 3 Erin drinks 35 mg of caffeine every 8 hours. She has coffee in the morning, cola mid-afternoon, and tea before bed. If 20% of the caffeine is still in Erin’s body after 8 hours, determine how much caffeine is her body after 3 days.

Create a strategy To see how the caffeine accumulates during the first day, the first 3 terms of the series can be written out. In the morning, Erin has 35 mg of caffeine in her coffee. After 8 hours, 20% of that caffeine is left in her body and she adds 35 mg more by drinking a cola. 35 (0.2) + 35 In the evening, 20% of the 20% of caffeine from her morning coffee remains, 20% of her mid-afternoon cola remains, and she drinks another 35 mg in her tea. 35 (0.2)2 + 35 (0.2) + 35 This pattern will continue, and it can be seen that the terms form a geometric sequence defined by tn = 35 (0.2)n−1, where n represents the number of 8-hour periods that have passed.

Apply the idea To find out how much caffeine is in Erin’s body after 3 days, n = 9 will be used, since there are nine 8-hour periods in 3 days. Using t1 = 35, a = 0.2, n = 9 with the geometric series formula yields: Write the formula Substitute the values ≈ 43.75 mg

Evaluate the expression

About 43.75 mg of caffeine is left in Erin’s body after 3 days of drinking 35 mg every 8 hours.

Reflect and check In reality, there is likely less than this amount of caffeine in Erin’s body after 3 days. Caffeine can be completely flushed from the body after about 10 hours if enough water is consumed. In this model, the caffeine from the coffee on the first day will never entirely leave Erin’s body.

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Example 4 Analyse and compare the behaviour of the partial sums for these two geometric series by calculating S5, S10, and S20 for each: • Series A: First term t1 = 2, common ratio a = 3 • Series B: First term t1 = 100, common ratio a = 0.5

Create a strategy The formula for the sum of the first n terms of a geometric series,

, will be used for

each series and for each value of n (5, 10, 20). Then, how the sums change as n increases for each series will be compared, noting the effect of the common ratio a.

Apply the idea •

For Series A (t1 = 2, a = 3): The formula for Series A is • For n = 5: S5 = 35 − 1 = 243 − 1 = 242 • For n = 10: S10 = 310 − 1 = 59 049 − 1 = 59 048 • For n = 20: S20 = 320 − 1 = 3 486 784 401 − 1 = 3 486 784 400

•

For Series B (t1 = 100, a = 0.5): The formula for Series B is • For n = 5: S5 = 200(1 − (0.5)5) = 200(1 − 0.03125) = 200(0.96875) = 193.75 • For n = 10: S10 = 200(1 − (0.5)10) = 200(1 − 0.0009765625) ≈ 200(0.9990234375) ≈ 199.80 • For n = 20: S20 = 200(1 − (0.5)20) = 200(1 − 0.0000009536743164) ≈ 200(0.9999990463256836) ≈ 199.9998

Analysis of behaviour • For Series A, where a = 3 (so |a| > 1), the partial sums Sn grow very rapidly as n increases. The sum increases from 242 to nearly 3.5 billion by the 20th term. This indicates the sum is diverging. • For Series B, where a = 0.5 (so |a| < 1), the partial sums Sn also increase, but they approach a specific value. The sum starts at 193.75 for n = 5, then 199.80 for n = 10, and 199.9998 for n = 20. It appears the sum is approaching 200. This indicates the sum is converging.

Reflect and check The common ratio a is the key determinant of this behaviour: • When |a| > 1, the term an in the numerator t1(1 − an) becomes very large, causing Sn to grow without bound. • When |a| < 1, the term an approaches 0 as n becomes large. So, Sn approaches . For Series B, this limit would be

, which matches the

observation. This concept of a limit for |a| < 1 is fundamental to understanding infinite geometric series.

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Idea summary A finite geometric series (or partial sum) is the sum of the first n terms of a geometric sequence. The sum Sn can be calculated using the formula:

Sn is the sum of the first n terms t1 is the first term of the geometric sequence a is the common ratio of the geometric sequence (a ≠ 1) n is the number of terms being added The behaviour of a geometric series depends on its common ratio a: • If |a| > 1, the partial sums Sn grow indefinitely (diverge). • If |a| < 1, the partial sums Sn approach a specific finite value (converge) as n increases. • If a = 1, Sn = n × t1. • If a = − 1, Sn oscillates between t1 and 0.

3.06 Practice questions What do you remember? 1

For each series, write the expression that would be used to calculate the partial sum. Do not evaluate the expression: a

The sum of the first 12 terms of the series: 5 + 10 + 20 + . . .

The sum of the first 7 terms of the geometric series: 64 + 16 + 4 + . . .

The sum of the first 5 terms of the series: 24 − 12 + 6 − . . .

Determine whether each statement about geometric series is true or false: a

A finite geometric series is the sum of the terms in a geometric sequence.

The common ratio a is found by subtracting consecutive terms.

If the common ratio |a| > 1, the partial sum Sn will grow without bound as more terms are added.

The sum formula Sn can be used when the common ratio is exactly 1.

Consider the geometric series represented by the expression Without calculating the sum, identify:

262

The first term, t1

The common ratio, a

The number of terms being added, n

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice 4

Determine the sum of the first 10 terms of the series: 3 + 9 + 27 . . .

Determine the sum of the first 6 terms of the geometric series: 81 + 27 + 9 . . .

Determine the sum of the first 6 terms of the series: 32 − 16 + 8 . . . Round your answer to the nearest whole number.

Determine the sum of the first 5 terms of the geometric sequence defined by each. Round your answers to two decimal places: a

t1 = 2.187 and a = 1.134

t1 = − 4.186 and a = − 2.848

Consider the series a

Determine the common ratio, a.

Determine the sum of the first 10 terms, round your answer to one decimal place.

The sum of the first 9 terms of a geometric series is 436 905, and the common ratio is . Determine t1, the first term in the series.

Ex 1

Ex 2

For each series: i

Determine n, the number of terms in the series.

Determine the sum of the series.

4 − 8 + 16 − . . . − 2048

Consider the series

Determine n, the number of terms in the series.

Show the steps required to derive the sum of the series.

Determine the sum of the series, rounded to three decimal places.

The first 3 terms of a sequence are

Write a simplified expression for the sum of the first n terms of this sequence.

Evaluate the sum when x = 5 and n = 9.

The first term of a geometric sequence is 9 and the fourth term is 243. Determine the sum of the first 13 terms of the corresponding series.

Evaluate: a

Suppose Ronald saves $1 the first day of a month, $2 the second day, $4 the third day, $8 the fourth day, and so on. That is, each day Ronald saves twice as much as he did the day before: a

Determine Ronald’s total savings for the first 13 days.

Determine Ronald’s total savings for the first 29 days.

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Ex 3

A conveyor belt is being used to remove materials from a quarry. Every 45 minutes, the conveyor belt empties out

of whatever material remains in the quarry. The quarry initially

holds 19 500 m3 of materials:

Determine how much material has been pumped out after 135 minutes.

Determine how much material is left in the dam after 135 minutes.

A ball is dropped from a height of 8 m, each subsequent bounce is previous bounce.

of the height of the

Derive an expression for the distance the ball has travelled after n bounces, Dn. Ex 4

Analyse and compare the behaviour of the partial sums for these two geometric series by calculating S5, S10, and S20 for each. • Series A: First term t1 = 5, common ratio a = 2 • Series B: First term t1 = 50, common ratio a = 0.8 Based on your calculations, comment on how the sum of each series behaves as more terms are added. Explain how the common ratio a is responsible for this behaviour.

Extend your thinking 19

Mareike takes out a mortgage of $900 000 that is compounded monthly at a rate of 0.75% per month. Mareike makes a repayment of $M a month and is trying to derive an expression for the amount remaining after n months, tn. 1

t0 = 900 000

Amount after 0 months

t1 = 900 000 (1 + 0.0075) − M

Amount after 1 month

= 900 000 (1.0075) − M

1 + 0.0075 = 1.0075

4 5 6

t2 = (900 000 (1.0075) − M ) (1.0075) − M

Amount after 2 months

= 900 000 (1.0075) − M (1.0075) − M

Distributive property of equality

1.0075 + 1 = 2.0075

= 900 000 (1.0075) − M (1.0075 + 1)

= 900 000 (1.0075) − M (2.0075)

tn = 900 000 (1.0075)n − M (n + 0.0075)

Amount after n months

Correct any mistakes in their working. 20

Lisa invests $30 000 at a rate of 1.5% compounded monthly. Each month, she withdraws $600 from her investment after the interest is paid and the balance is reinvested in the account. Calculate the amount Lisa has saved after 3 years.

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3.07 Applications of sequences and series After this lesson, you will be able to… • differentiate between sequences and series in problem contexts • apply sequence and series concepts to solve practical problems

Applications of sequences and series A sequence is an ordered list of numbers following a specific pattern, while a series is the sum of the terms in a sequence. Recognising whether a problem involves a sequence (individual terms) or a series (total sum) is key to applying these concepts in real-world contexts. Sequences and series can be classified as arithmetic or geometric, depending on their patterns. An arithmetic sequence has a common difference, b, between consecutive terms. Its recurrence relation can be written as tn+1 = tn + b, where tn is the term at position n and t0 often represents an initial value before the first term. A geometric sequence has a common ratio, a, between consecutive terms. Its recurrence relation can be written as tn+1 = atn More generally, many sequences in practical problems can be modelled by a first-order linear recurrence relation of the form tn+1 = atn + b for n ≥ 0, where a and b are constants, and t0 is the initial term. Arithmetic sequences are a special case where a = 1 (so tn+1 = tn + b, where b is the common difference). Geometric sequences are a special case where b = 0 (so tn+1 = atn, where a is the common ratio). A note on notation: • When t0 is used to represent an initial state or value before the first term or period (e.g., an initial deposit in a bank account at time 0), then tn often represents the value after n periods. For an arithmetic sequence, this gives tn = t0 + nb. For a geometric sequence, this gives tn = t0 × an. • When t1 is used to represent the first term of a sequence (e.g., the first savings amount, the first bounce height), the sequence is t1, t2, t3,. . .. In this case, the nth term is tn = t1 + (n − 1) b for an arithmetic sequence, and tn = t1 × an−1 for a geometric sequence. These nth term formulas are commonly used when working with sum formulas for series. For series, which involve summing terms of a sequence, formulas for partial sums are often needed. The partial sum of an arithmetic sequence up to n terms, where t1 is the first term and tn is the nth term, is given by:

Sn is the sum of the first n terms n

is the number of terms

is the first term

is the nth term

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The partial sum of a geometric sequence up to n terms, where t1 is the first term and a is the common ratio, is given by:

Sn t1 a n

is the sum of the first n terms is the first term is the common ratio (where a ≠1) is the number of terms

This formula is used when ∣a∣ > 1, to avoid negative denominators. It is equivalent to the formula:

Real-world applications include savings (arithmetic sequences for fixed deposits, geometric sequences for compound interest), depreciation, and population growth or spread (e.g., virus infections). Arithmetic sequences model linear growth or decay, while geometric sequences model exponential trends.

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Example 1 Sarah saves $50 each month in a jar. How much will she have saved after 12 months?

Create a strategy This problem describes accumulating savings over time. If t0 = 0 represents the initial amount before any savings, and $50 is added each month, the total amount after 12 months is the value to be found. This can be viewed as the value t12 in an arithmetic sequence where the amount increases by b = 50 each month.

Apply the idea Let t0 = 0 be the initial amount saved. Each month, $50 is added. The amount saved after n months can be found using the formula tn = t0 + n × b tn = t0 + n × b

Write the formula

t12 = 0 + 12 × 50

Substitute t0 = 0, n = 12 and b = 50

= $600

Evaluate

After 12 months, Sarah has saved $600.

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Reflect and check The sequence of total savings at the end of each month is 50, 100, 150, . . . , 600 if t1 is defined as savings after 1 month. If t0 = 0, then t1 = 0 + 1 × 50 = 50, t2 = 0 + 2 × 50 = 100, and so on, with t12 = 600. The question asks for the total amount, which is a specific term in this sequence of accumulated savings.

Example 2 A bank account earns 5% compound interest annually on an initial deposit of $1000. Calculate the balance after 3 years. Round your answer to two decimal places.

Create a strategy This is a geometric sequence where the balance is multiplied by 1 + 0.05 = 1.05 each year. Let t0 = 1000 be the initial deposit. The balance after n = 3 years is the value to be determined. The common ratio is a = 1.05

Apply the idea The balance after n years can be found using the formula tn = t0 × an. tn = t0 × an

Write the formula 3

t3 = 1000 × (1.05)

Substitute t0 = 1000, a = 1.05, and n = 3

= 1000 × 1.157625

Evaluate (1.05)3

= $1157.625

Evaluate

The balance after 3 years is $1157.63

Example 3 A virus spreads such that the number of new infections increases by 10 each day. If 5 people are infected on Day 1, determine the total number of people infected after 7 days.

Create a strategy This describes an arithmetic series. The number of new infections each day forms an arithmetic sequence: • Day 1: t1 = 5 new infections • Day 2: 5 + 10 = 15 new infections • Day 3: 15 + 10 = 25 new infections, and so on. The first term is t1 = 5 and the common difference is b = 10. The objective is to find the total number infected after n = 7 days, which is the sum of the first 7 terms of this sequence. First, find the number of infections on day 7, t7. Then, use the arithmetic series sum formula .

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Apply the idea The number of new infections on day n is tn = t1 +(n − 1) b. For Day 7: t7 = 5 + (7 − 1) × 10

Substitute t1 = 5, n = 7 and b = 10

= 5 + 6 × 10

Simplify inside the brackets

= 5 + 60

Multiply

= 65 people

Evaluate

So, 65 people are newly infected on Day 7. Now, sum the series using Substitute n = 7, t1 = 5 and t7 = 65 Simplify inside the brackets = 7 × 35

Divide 70 by 2

= 245 people

Evaluate

The total number infected after 7 days is 245 people.

Reflect and check The daily new infections are: 5, 15, 25, 35, 45, 55, 65. Summing these: 5 + 15 + 25 + 35 + 45 + 55 + 65 = 245 This confirms the result obtained using the series formula. A sum (series) is required here because the phrase ‘total number of people infected’ implies an accumulation of cases over the entire period.

Example 4 A ball is dropped from an initial height of 10 m. Each time it bounces, it reaches 70% of its previous height. Calculate the total distance the ball travels downwards until it hits the ground for the 5th time.

Create a strategy The problem asks for the total downward distance travelled over 5 descents. The height of each descent forms a geometric sequence: • The first downward distance is t1 = 10 m • The common ratio a is 70% = 0.7 • Sum the first 5 terms of this geometric sequence, so n = 5 Use the formula for the sum of the first n terms of a geometric series:

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Apply the idea The downward distances are: • 1st descent: t1 = 10 m • 2nd descent: t2 = 10 × 0.7 = 7 m • 3rd descent: t3 = 10 × (0.7)2 = 4.9 m • 4th descent: t4 = 10 × (0.7)3 = 3.43 m • 5th descent: t5 = 10 × (0.7)4 = 2.401 m Now, calculate the sum S5 using t1 = 10, a = 0.7, and n = 5: Write the formula for the sum of the first n terms Substitute t1 = 10, a = 0.7, and n = 5 Evaluate the power Evaluate the subtraction in the denominator Evaluate the multiplication = 27.731 m

Evaluate

The total distance the ball travels downwards until it hits the ground for the 5th time is 27.731 m.

Reflect and check The solution can be checked by summing the individual terms: 10 + 7 + 4.9 + 3.43 + 2.401 = 27.731 The formula provides an efficient way to sum these terms. Note that the question specifically asked for downward travel; if it asked for total distance travelled (up and down), the problem would be different.

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Idea summary This section demonstrates how to apply sequence and series concepts to solve practical problems. Key skills include: • Differentiating between situations requiring a specific term of a sequence versus the sum of terms (a series). • Using recurrence relations to model situations, including arithmetic (tn+1 = tn + b) geometric (tn+1 = atn), and general first-order linear relations (tn+1 = atn + b). • Applying formulas for the nth term and partial sums of arithmetic series and geometric series

• Solving practical problems related to contexts like savings, investments, loan repayments, population growth, depreciation, and physical phenomena like bouncing balls.

3.07 Practice questions What do you remember? 1

Determine whether each situation describes a sequence or a series: a

The amount of money in a savings account after each month with a fixed deposit

The total number of people infected by a virus over several days

The balance in an account with compound interest each year

Identify each as arithmetic or geometric: a

2, 4, 6, 8, . . .

3, 6, 12, 24, . . .

10, 20, 30, 40, . . .

100, 50, 25, 12.5, . . .

What is the common difference or ratio for each sequence? a

270

5, 15, 25, 35, . . .

1000, 1050, 1102.5, . . .

True or false: a

An arithmetic series sums the terms of an arithmetic sequence.

The recurrence relation for a geometric sequence is tn+1 = tn + b.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice Ex 1

Ex 2

Ex 3

Determine the specified term using the recurrence relation for each arithmetic sequence: a

Tom saves $20 monthly, starting with $0. Determine the total amount saved after 8 months.

A car depreciates by $500 yearly from $20 000. Determine the value of the car after 6 years.

Rent increases by $30 monthly from $800. Determine the cost of rent for the 5th month.

A stack grows by 4 blocks daily from 10. Determine the height of the stack after 10 days.

Determine the specified term using the recurrence relation for each geometric sequence: a

An account with $2000 earns 4% annual compound interest. Determine the balance after 4 years.

A population of 500 grows by 10% yearly. Determine the size of the population after 3 years.

A ball drops from 100 cm, rebounding 0.8 of the previous height. Determine the height of the ball after the 5th bounce.

A debt of $10 000 increases by 6% yearly. Determine the total amount owed after 2 years.

Calculate the total using the arithmetic series formula

A pest increases by 3 daily from 2. Determine the total number of pests counted over the first 6 days.

Daily steps increase by 100 from 500. Determine the total number of steps taken over the first 10 days.

A job pays $10 more daily from $30. Determine the total earnings over the first 5 days.

Fans grow by 15 daily from 20. Determine the total number of members in the club at the end of the 8th day.

Write the recurrence relation and determine the requested term: a

A car loses $200 yearly from $15 000. Determine its value at the end of the 4th year.

A fund grows by 3% yearly from $5000. Determine the balance at the end of the 3rd year.

Identify whether each scenario is arithmetic, geometric sequence or an arithmetic series: a

Total distance run over 5 days, increasing by 2 km daily from 3 km

Height of a tree after 4 years, growing by 0.5m yearly from 1 m

Balance after 3 years with 2% annual interest from $1000

Solve: a

A club gains 5 members daily from 10. Determine the total number of members in the club at the end of the 6th day.

An investment of $3000 earns 6% compound interest yearly. Determine the balance at the end of the 2nd year.

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Ex 4

A ball is dropped from a height of 6m. On each bounce, it rebounds to 80% of its previous height: a

What is the height of the ball’s third rebound?

What is the total distance the ball has travelled downwards when it hits the ground for the 5th time? Round your answer to two decimal places.

Extend your thinking 12

A car worth $25 000 depreciates by $2000 yearly. How many years until the value is below $10 000?

A fund starts at $4000. Plan A adds $300 yearly; Plan B applies 8% compound interest yearly. Which plan yields more after 5 years? By how much?

A leak increases by 2 L daily from 1 L:

Determine the total leakage after 10 days.

If a tank holds 150 L, on which day will it overfow?

Compare two savings plans over 6 months: Plan X adds $40 monthly from $0; Plan Y starts at $100 with 10% monthly compound interest. Which plan has a higher balance after 6 months? By how much (round the answer to two decimal places)?

Chapter 3 review Online review

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“The beauty of mathematics only shows itself to more patient followers.” Maryam Mirzakhani

Big ideas • Combinatorics provides a set of fundamental principles for systematic counting, essential for determining the number of possible outcomes in complex scenarios without direct enumeration. • Permutations and combinations are two primary techniques in combinatorics that distinguish between scenarios where order does and does not matter, respectively. • Complex counting problems involving specific restrictions or multiple stages are solved by adapting and combining fundamental combinatorial principles.

4 Combinatorics Chapter outline 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.09 4.10

Pigeon-hole principle Inclusion-exclusion principle Multiplication and addition principles Arrangements and factorial notation Permutations Permutations with restrictions Investigation: Pascal’s triangle Combinations Combinations with restrictions Combinatorics mixed problems Combinatorics and probability Chapter 4 review

276 283 300 312 327 336 353 368 379 391

Using combinatorics, you can solve puzzles like Sudoku. It’s all about arranging numbers without repetition!

4.01

Pigeon-hole principle After this lesson, you will be able to… • describe the pigeon-hole principle • use the pigeon-hole principle to solve problems

Pigeon-hole principle The pigeon-hole principle is a simple idea that plays a role in mathematical proofs and has a significant influence on the applications of computer design and digital communications. Consider placing four pigeons into three pigeon-holes (boxes). The pigeon-hole principle states that at least one of the boxes must contain two pigeons.

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In this analogy, the pigeons represent the n items and the pigeon-holes represent the k containers. Both n and k are positive integers. Pigeon-hole principle

4 Pigeons

A principle stating that if n items are put into k containers and n > k, then at least one container must contain more than one item. (Here n, k are positive integers). 3 Pigeon-holes

Proof: If it is assumed that every container holds at most 1 object, then the total number of objects distributed among the k containers can be no more than k × 1 = k. However, it is given that there are n items and n > k. This contradicts the assumption, therefore, at least one container must contain more than 1 object (i.e., at least 2 objects).

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Example 1 What is the minimum size of a group of people to ensure at least two have the same birthday?

Create a strategy Determine the number of possible birthdays (containers, k) and add one to this number to find the minimum number of people (items, n) needed to ensure at least two people share a birthday.

Apply the idea There are k = 366 possible birthdays (including leap years). These are the ‘pigeon-holes’ or ‘containers’. To ensure at least two people (items) share the same birthday, n = k + 1 people are needed. Therefore, at least 366 + 1 = 367 people are needed.

Example 2 True or false: If the opera house concert hall is filled to capacity (2679 people), there will be at least one pair of people with the same first and last initials?

Create a strategy Identify the ‘items’ (n people) and the ‘containers’ (k possible initial combinations). Calculate the total number of possible combinations of first and last initials. If n > k, then by the pigeon-hole principle, the statement is true.

Apply the idea The items are the n = 2679 people. The containers are the possible combinations of first and last initials. There are 26 letters in the alphabet. The total number of possible combinations of first and last initials (containers) is: Number of initial combinations (k) = 26 × 26 = 676

Write the equation Evaluate

Since the number of people n = 2679 is greater than the number of possible initial combinations k = 676 (2679 > 676), by the pigeon-hole principle, there must be at least one pair of people with the same first and last initials. Therefore, the statement is true.

Idea summary The pigeon-hole principle states that if n items are distributed into k containers, where n > k (and n, k are positive integers), at least one container has more than one item. This occurs when there are more items than containers.

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Generalised pigeon-hole principle The generalised pigeon-hole principle extends the basic principle. It states: If n items are put into k containers, where n and k are positive integers, then at least one container must hold a minimum number of items. This minimum is found by taking the value of that is greater than or equal to it.

and finding the smallest integer

The term ceiling of a number (e.g., x) refers to the smallest integer that is greater than or equal to x. For example, the ceiling of 3.2 is 4; the ceiling of 7 (which is already an integer) is 7; and the ceiling of 0.5 is 1. If the number is not an integer, this is equivalent to rounding up to the next integer. When dealing with discrete items such as pigeons or people, the ceiling of any fractional average must be used to guarantee the minimum in one container. Proof: Let C be the ceiling of

which is defined as the smallest integer greater than or equal to

Assume, for the sake of contradiction, that every container has fewer than C items. Since the number of items must be an integer, this means every container has at most C − 1 items. Then, the total number of items in all k containers would be at most k(C − 1). From the definition of the ceiling function, it is known that

, which implies

. Therefore,

This shows that the total number of items would be less than n, which contradicts the fact that there are n items. Therefore, the assumption must be false, and at least one container must have at least C items (i.e., at least the ceiling of items). Another useful way to express the generalised pigeon-hole principle, especially when aiming for a specific minimum number of items in one container, uses this relationship:

Nmin = mk + 1 Nmin

is the minimum total number of items that must be distributed to guarantee that at least one of the k containers holds at least m + 1 items

is the number of containers (or pigeon-holes), which must be a positive integer

is a non-negative integer

In this formula, k is a positive integer and m is a non-negative integer. This equation shows the number of items (Nmin) that need to be placed to guarantee that at least m + 1 items are in one of the k containers. For example, if m = 1, then Nmin = (1)k + 1 = k + 1. This means if k + 1 items are placed into k containers, at least one container must hold at least 1 + 1 = 2 items, which is the basic pigeon-hole principle.

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Generalised pigeon-hole principle A principle stating that if n items are distributed into k containers (where n, k are positive integers), at least one container contains at least the smallest integer greater than or equal to

items. Alternatively, if Nmin = mk + 1 items are placed in k containers (where k is a

positive integer and m is a non-negative integer), then some container holds at least m + 1 items.

Example 3 The mega-corporation, OmniCorp, has exactly 8000 employees. What is the minimum number of employees guaranteed to share the same birth month?

Create a strategy Use the generalised pigeon-hole principle. Identify the number of employees as ‘items’ (n) and the number of possible birth months as ‘containers’ (k). Calculate the value of and then find the smallest integer greater than or equal to this value (the ceiling).

Apply the idea Here, the ‘items’ are the n = 8000 employees, and the ‘containers’ are the possible birth months. Since there are 12 months in a year, the number of containers is k = 12. Using the generalised pigeon-hole principle, at least one month must be shared by at least the ceiling of

employees.

First, calculate the fraction:

...

The ceiling of 666.666. . . (the smallest integer greater than or equal to it) is 667. Therefore, the minimum number of employees guaranteed to share the same birth month is 667.

Example 4 Digital only example Explore in the online lesson

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Example 5 Consider a square with side length 2 units. If 5 points are placed anywhere inside or on the boundary of this square, prove that there must be at least two points that are no more than units apart.

Create a strategy The ‘pigeons’ are the 5 points. Suitable ‘pigeon-holes’ within the square need to be defined. The goal is to divide the large square into a number of smaller regions such that any two points within the same region satisfy the distance condition (≤ units apart). If the number of regions (k) is less than the number of points (n), the pigeon-hole principle can be applied to guarantee that at least one region contains more than one point.

Apply the idea Divide the square of side length 2 units into four smaller, non-overlapping squares, each of side length 1 unit. This can be done by drawing one line segment parallel to the x-axis and one line segment parallel to the y-axis, both passing through the centre of the large square. The large square is divided into four smaller squares. n = 5 points (pigeons) are now to be placed in these k = 4 smaller squares (pigeon-holes). Since the number of points (n = 5) is greater than the number of smaller squares (k = 4), by the basic pigeon-hole principle, at least one of the smaller squares must contain at least two points. Now, consider any two points located within the same smaller square of side length 1 unit. The maximum distance between any two points in a square of side length, s, occurs along its diagonal. The length of the diagonal, d, of a square with side length s = 1 can be found using Pythagoras’ theorem: Therefore, if two points lie in the same smaller square, the distance between them is at most units. Since it was established that at least one small square must contain at least two points, it follows that there must be at least two points among the chosen five that are no more than units apart.

Reflect and check The critical step was to define the ‘pigeon-holes’ (the four smaller squares) in such a way that the property of interest (maximum distance of ) holds within each hole. The number of holes (k = 4) being less than the number of pigeons (n = 5) allowed the application of the pigeon-hole principle to guarantee the result.

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Idea summary The generalised pigeon-hole principle provides a more powerful way to determine minimums when distributing items into containers. It can be understood through two main formulations: • If n items are put into k containers (where n, k are positive integers), then there is at least one container with at least the smallest integer greater than or equal to items. This is also known as the ceiling of . • A lternatively, to guarantee that at least one container holds at least m + 1 items (where m is a non-negative integer and k is the positive integer number of containers), at least Nmin = mk + 1 items need to be placed. Both forms are crucial for solving problems where a certain number of items sharing a characteristic or falling into the same category need to be guaranteed.

4.01 Practice questions What do you remember? 1

Determine whether each statement about the pigeon-hole principle is true or false: a

If n items are placed into k containers and n > k, at least one container must have more than one item.

The principle applies even if n = k.

If 5 items are placed into 3 containers, at least one container must have at least 2 items.

The principle guarantees that every container will have more than one item if n > k.

Explain why the pigeon-hole principle does not apply if there are 4 pigeons and 5 pigeonholes.

In the generalised pigeon-hole principle, if n = 20 items are distributed into k = 5 containers, what does

represent, and how is it used to determine the minimum number of items in at

least one container?

Practice 4

If 8 pigeons are placed into 5 pigeon-holes, what is the minimum number of pigeons that must be in at least one of the holes?

A school has 51 students. If there are 50 lockers available, explain why at least two students must share a locker.

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6 Ex 1

7 Ex 2

Ex 3

What is the minimum number of socks that must be drawn from a drawer containing 10 red socks and 10 blue socks to ensure at least two socks of the same colour? A class has 101 students, and there are 100 possible exam scores (from 0 to 99). Is it true that at least two students must have the same score? Justify your answer.

A club has 73 members, and each member owns between 1 and 10 pets inclusive. Prove that at least two members must have the same number of pets.

A concert hall has 1350 attendees and 676 possible combinations of first and last initials (from 26 × 26). Using the generalised pigeon-hole principle, what is the minimum number of people who must share the same initials?

A library has 250 books distributed across 10 shelves. What is the minimum number of books that must be on at least one shelf?

In a group of 200 people, there are 12 possible birth months. What is the minimum number of people who must share the same birth month?

A school has n students using 5 different types of laptops. If at least 20 students use the same type, what is the minimum value of n?

A factory produces 1000 widgets stored in 7 bins. What is the minimum number of widgets in at least one bin?

In a sequence of 101 distinct integers from 1 to 200, show that there must be at least one pair of numbers whose difference is exactly 100.

Ex 4

Ex 5

Extend your thinking 15

A school has n students and 7 possible days for their birthdays (e.g., Monday to Sunday). If at least 50 students share the same birthday day of the week, and the maximum number of students per day is 60, determine the range of possible values for n.

A drawer contains an unsorted collection of socks: 10 red, 12 blue, and 15 green. The socks are not in pairs. Socks are drawn one at a time from the drawer in the dark: a

What is the minimum number of socks that must be picked to guarantee at least one matching pair of any colour?

What is the minimum number of socks that must be picked to guarantee at least one matching pair of blue socks?

Prove that in any group of 6 people, there are either 3 mutual friends or 3 mutual strangers. (Assume that for any two people, the relationship is either friends or strangers).

An equilateral triangle has a side length of 2 cm. If 5 points are placed anywhere inside or on the boundary of this triangle, prove that there must be at least two points that are no more than 1 cm apart.

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4.02 Inclusion-exclusion principle After this lesson, you will be able to… • understand the concept of unions and intersections of sets • use the inclusion-exclusion principle formula to determine the number of elements in the union or intersection of two or three sets • solve problems involving the union or intersection of two or three sets using the inclusion-exclusion principle

Inclusion-exclusion principle for two sets Combinatorics A branch of mathematics that involves counting and arranging objects. Inclusion-exclusion principle A counting technique used to find the number of elements in the union of two or more sets. For two sets, it states that the number of elements in their union is the sum of the number of elements in each set minus the number of elements in their intersection. Example: For two sets A and B: ∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣

When counting objects, it can be helpful to organise them using a Venn diagram. A∩B A

The intersection is the overlap in the two circles. The notation for the intersection A and B is A ∩ B. U

A∪B A

The union of A and B is shown by shading anything contained by the two circles. The notation for the union of A and B is A ∪ B. U

In combinatorics, the inclusion-exclusion principle for two sets helps find the total number of elements in their union. This is achieved by considering the number of elements in each set and in their intersection. The notation for the number of elements in a set is ∣ ∣. For example, ∣A∣ means the number of elements in set A. 4.02 Inclusion-exclusion principle mathspace.co

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Sometimes, the notation n(A) is also used to represent the number of elements in set A. So, n(A) = ∣A∣.

Exploration For the given Venn diagrams, each dot represents an element of each set. Venn diagram 1

Venn diagram 2

1. Complete the table: Venn diagram 1 Venn diagram 2

∣A∣

∣B∣

∣A ∪ B∣

∣A ∩ B∣

⬚

2. How many times is ∣A ∩ B∣ counted in ∣A∣ + ∣B∣?

⬚

3. Explain how to determine ∣A ∪ B∣ given ∣A∣, ∣B∣, and ∣A ∩ B∣.

For two sets A and B, the inclusion-exclusion principle says that:

∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣ ∣A∣

is the number of elements in set A

∣B∣

is the number of elements in set B

∣A ∩ B∣ is the number of elements common to both sets A and B ∣A ∪ B∣ is the number of elements in the combined set of elements of A and B This principle is useful in combinatorial problems where it is important to avoid double-counting when determining the total number of elements in the union of two sets. By subtracting the number of elements in the intersection, the formula ensures that elements common to both sets are counted only once.

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A∩B

B This Venn diagram shows the inclusion-exclusion principle for two sets A and B, their intersection A ∩ B, and their union A ∪ B. • ∣A∣ = 10 • ∣B∣ = 7 • ∣A ∩ B∣ = 5 • ∣A ∪ B∣ = 10 + 7 − 5 = 12

A∪B

If sets A and B represent events, and the dots are individual outcomes from experiments, statements can be made about the proportion of outcomes for each event. This helps in understanding their probabilities. Let U be the universal set for this experiment. The preceding Venn diagram illustrates a scenario where ∣A∣ = 10, ∣B∣ = 7, ∣A ∩ B∣ = 5, and ∣A ∪ B∣ = 12. If there are 5 elements outside of A ∪ B but within U (as implied by the denominators in the proportions that follow), the total number of outcomes in the universal set is ∣U∣ = 12 + 5 = 17. Therefore: • Event A occurred in the proportion of • Event B occurred in the proportion of • Event A and B occurred together in

of the trials

The proportion of trials in which event A or event B occurred is

Example 1 In this Venn diagram, set A has 13 elements and set B has 14 elements. The five purple dots are the elements that are common to both A and B: A

Determine ∣A ∪ B∣.

Create a strategy Use the formula: ∣A ∪ B∣ = ∣A∣ + ∣B∣ −∣A ∩ B∣

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Apply the idea A has 13 elements: ∣A∣ = 13 B has 14 elements: ∣B∣ = 14 5 elements are common to both A and B: ∣A ∩ B∣ = 5 ∣A ∪ B∣ = ∣A∣ + ∣B∣ −∣A ∩ B∣

Write the formula

= 13 + 14 − 5

Substitute the values

= 22

Evaluate

Reflect and check The number of elements in A ∪ B can be verified by counting the dots contained in the circles of A and B in the Venn diagram.

Example 2 A marine biologist creates this diagram of 8 favourite types of fish: Freshwater fish

Saltwater fish

Goldfish Largemouth Barramundi bass Salmon Betta fish

Blue tang Clownfish Seahorse

The set of freshwater fish is represented by F, while the set of saltwater fish is represented by S. a How many fish can live in freshwater? Use the notation ∣ ∣ to express the answer.

Create a strategy Count the number of all fish in the freshwater fish circle. Include the fish only in the freshwater set F and those in both the freshwater and saltwater sets.

Apply the idea There are 5 fish that can live in freshwater, therefore ∣F ∣ = 5.

Reflect and check This number includes all fish in the freshwater fish circle, regardless of whether the fish can also live in saltwater or are restricted to freshwater only.

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b Determine ∣F ∩ S∣ and interpret the result in the context of the problem.

Create a strategy

Apply the idea

Count the number of fish in the overlap of the two circles, which represents those that can live in both freshwater and saltwater.

The number of fish in the overlap of the two circles is 2, therefore ∣F ∩ S∣ = 2. ∣F ∩ S∣ = 2 means that there are 2 fish that can live in both freshwater and saltwater.

c Show that ∣S∣ = 5 using the inclusion-exclusion formula, without counting the fish on the diagram.

Create a strategy The inclusion-exclusion principle for two sets can be used with ∣F ∣ = 5 and ∣F ∩ S∣ = 2. Note that ∣F ∪ S∣ = 8, as this represents the total number of fish that the marine biologist included in the diagram.

Apply the idea ∣F ∪ S∣ = ∣F∣ + ∣S∣ − ∣F ∩ S∣

Write the formula

8 = 5 + ∣S∣ − 2

Substitute the values

8 = 3 + ∣S∣

Evaluate 5 − 2

5 = ∣S∣

Subtract 3 from both sides

Therefore, ∣S∣ = 5.

Example 3 In a survey of students, 60 students are enrolled in mathematics, 45 students are enrolled in science, and 25 students are enrolled in both mathematics and science: a Calculate the total number of students enrolled in either mathematics or science or both.

Create a strategy The inclusion-exclusion principle for two sets can be used to calculate the total number of enrolments.

Apply the idea If ∣A∣ represents the number of students enrolled in mathematics, then ∣A∣ = 60. If ∣B∣ represents the number of students enrolled in science, then ∣B∣ = 45. If ∣A ∩ B∣ represents the number of students enrolled in both subjects, then ∣A ∩ B∣ = 25. ∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣

Write the formula

= 60 + 45 − 25

Substitute the values

= 80

Evaluate

Therefore, 80 students are enrolled in either mathematics or science or both.

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b If 10 students are enrolled in neither mathematics nor science, determine the total number of students surveyed.

Create a strategy To determine the total number of students surveyed, add the number of students enrolled in either mathematics or science or both to the number of students enrolled in neither subject.

Apply the idea Total number of students = ∣A ∪ B∣ + Number of students enrolled in neither

Express the sum

= 80 + 10

Substitute the values

= 90

Evaluate

So, the total number of students surveyed is 90.

c How many students are enrolled only in mathematics and not in science?

Create a strategy To determine the total number of students enrolled only in mathematics but not in science, subtract the number of students enrolled in both subjects from the total number of students in mathematics.

Apply the idea Number of students in mathematics only = ∣A∣ − ∣A ∩ B∣

Write the formula

= 60 − 25

Substitute the values

= 35

Evaluate

Therefore, the total number of students enrolled only in mathematics is 35.

Reflect and check The number of students enrolled only in science can be calculated by subtracting those enrolled in both subjects from the total number of students in science. Adding this number to the count of students only in mathematics, along with those enrolled in both subjects and those enrolled in neither subject, confirms the total number of students surveyed is 90. This Venn diagram can be used to check: Maths

Science

20 10

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Example 4 Digital only example Explore in the online lesson

mathspace.co

Idea summary The inclusion-exclusion principle for two sets, a foundation concept in combinatorics, helps find the number of elements in their union. It considers the elements in each set and their intersection, making sure common elements are only counted once. If ∣A∣, ∣B∣ and ∣A ∩ B∣ represent the number of elements in set A, B, and their intersection respectively, then the number of elements in the combined set that includes all elements from A and B is given by: ∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣

Inclusion-exclusion principle for three sets When managing three overlapping groups, such as students in sports, music, and drama, the inclusion-exclusion principle helps with counting the total by accounting for overlaps between these groups. This principle can be extended from two sets to situations involving three or more sets that share common elements. It accounts for all possible intersections, as illustrated by this Venn diagram: A

B It considers all possible intersections between the sets to avoid double- or triple-counting elements that belong to more than one set. This includes: • Pairwise intersections: The intersections between each pair of sets (e.g. Set A with Set B, Set A with Set C, and Set B with Set C). • Intersection of all sets: The intersection where all sets overlap, ensuring that any elements common to all sets are counted only once.

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For three sets A, B and C, the inclusion-exclusion principle is expressed as:

∣A ∪ B ∪ C∣ = ∣A∣ + ∣B∣ + ∣C∣ − ∣A ∩ B∣ − ∣A ∩ C∣ − ∣B ∩ C∣ + ∣A ∩ B ∩ C∣ ∣A∣∣, ∣B∣∣, ∣C∣∣

are the number of elements in set A, set B, and set C, respectively

∣A ∩ B∣∣

is the number of elements common to both sets A and B

∣A ∩ C∣∣

is the number of elements common to both sets A and C

∣B ∩ C∣∣

is the number of elements common to both sets B and C

∣A ∩ B ∩ C∣∣ is the number of elements common to all three sets A, B, and C ∣A ∪ B ∪ C∣∣ is the number of elements in the union of the three sets, counting each element only once ∣A ∪ B ∪ C∣∣ gives the number of elements in the combined set that includes all elements from A, B and C, ensuring that elements common to all three sets are counted only once. There are 21 dots in total in the union of the sets in this Venn diagram: A

B To verify that ∣A ∪ B ∪ C∣ = 21, the inclusion-exclusion principle for three sets can be used: ∣A ∪ B ∪ C∣ = ∣A∣ + ∣B∣ + ∣C∣ − ∣A ∩ B∣ − ∣A ∩ C∣ − ∣B ∩ C∣ + ∣A ∩ B ∩ C∣ • Set A has 11 elements, ∣A∣ = 11 • Set B has 10 elements, ∣B∣ = 10 • Set C has 10 elements, ∣C∣ = 10 • The number of elements common to both sets A and B is 4, therefore ∣A ∩ B∣ = 4 • A and C have 3 elements in common, ∣A ∩ C∣ = 3 • B and C have 5 elements in common, ∣B ∩ C∣ = 5 • The number of elements common to all three sets A, B and C is 2, ∣A ∩ B ∩ C∣ = 2 Substituting these values into the inclusion-exclusion formula: ∣A ∪ B ∪ C∣ = 11 + 10 + 11 − 4 − 3 − 5 + 2 = 21 This matches the number of dots obtained by direct counting.

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Example 5 In a school, 120 students were surveyed about their participation in extracurricular activities. The activities are divided into three categories: art, music, and sports. The results of the survey are: • 40 students participate in art. • 50 students participate in music. • 60 students participate in sports. • 15 students participate in both art and music. • 10 students participate in both art and sports. • 20 students participate in both music and sports. • 5 students participate in all three activities: art, music, and sports a How many students participate in at least one of these activities?

Create a strategy The inclusion-exclusion principle for three sets can be used to determine the number of students who participate in at least one of the three activities. Let A represent art, M represent music, and S represent sports, then determine ∣A ∪ M ∪ S∣.

Apply the idea If ∣A∣ is the number of students enrolled in art, then ∣A∣ = 40. If ∣M∣ is the number of students enrolled in music, then ∣M ∣ = 50. If ∣S∣ is the number of students enrolled in sports, then ∣S∣ = 60. If ∣A ∩ M∣ is the number of students enrolled in both art and music, then ∣A ∩ M ∣ = 15. If ∣A ∩ S∣ is the number of students enrolled in both art and sports, then ∣A ∩ S∣ = 10. If ∣M ∩ S∣ is the number of students enrolled in both music and sports, then ∣M ∩ S∣ = 20. If ∣A ∩ M ∩ S∣ is the number of students enrolled in the three activities, then ∣A ∩ M ∩ S∣ = 5. ∣A ∪ M ∪ S∣ = ∣A∣ + ∣M ∣ + ∣S∣ − ∣A ∩ M ∣ − ∣A ∩ S∣ − ∣M ∩ S∣ + ∣A ∩ M ∩ S∣ Write the formula = 40 + 50 + 60 − 15 − 10 − 20 + 5

Substitute the values

= 110

Evaluate

Therefore, 110 students participate in at least one of the three activities.

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b Determine the number of students who do not participate in any activity.

Create a strategy Subtract the number of students who participate in at least one of these activities found in part (a) from the total number of students surveyed.

Apply the idea Number of students in none = Total students surveyed − ∣A ∪ M ∪ S∣

Write the formula

= 120 − 110

Substitute the values

= 10

Evaluate

So, the number of students in none of the activities is 10.

Reflect and check Another notation for “Number of students in none” is ∣(A ∪ M ∪ S )′∣ because (A ∪ M ∪ S)′ is the complement of A ∪ M ∪ S. The complement of a set contains all the elements not in the original set.

c How many students participate only in music?

Create a strategy To determine the number of students who participate only in music, exclude those involved in sports and art.

Apply the idea Number of students in music only = ∣M ∣ − ∣A ∩ M ∣ − ∣M ∩ S∣ + ∣A ∩ M ∩ S∣ Write the formula = 50 − 15 − 20 + 5

Substitute the values

= 20

Evaluate

Therefore, the total number of students enrolled only in music is 20.

Reflect and check The number of students participating in all three activities (art, music, and sports), ∣A ∩ M ∩ S∣, is added back because these elements were subtracted more than once during the calculation. Here’s why: 1. When subtracting the number of students involved in both music and sports, and those involved in both music and art, the students who are in all three categories are subtracted twice, once for each pair of activities. 2. To correct this, the number of students participating in all three activities needs to be added back to account for the double subtraction. This ensures that these students are correctly included in the count of those participating in only music. In summary, the addition corrects for the over-counting that occurred due to the double subtraction of students who are involved in all three activities.

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d What is the percentage of students enrolled in at most one activity, rounded to one decimal place?

Create a strategy The number of students enrolled in at most one activity is equal to the number of students enrolled in music only, in sports only, in art only, or not enrolled in any activity. After determining the number of students enrolled in at most one activity, divide it by the total number of students then express the answer as a percentage.

Apply the idea The number of students in none of the activities, found in part (b), is 10. The number of students enrolled only in music, found in part (c), is 20. The same approach followed in part (c) can be used to find the number of students who participate only in sports and only in art. To determine the number of students in sports only: Number of students in sports only = ∣S∣ − ∣A ∩ S∣ − ∣M ∩ S∣ + ∣A ∩ M ∩ S∣

Write the equation

= 60 − 10 − 20 + 5

Substitute the values

= 35

Evaluate

To determine the number of students in art only: Number of students in art only = ∣A∣ − ∣A ∩ M ∣ − ∣A ∩ S∣ + ∣A ∩ M ∩ S∣

Write the equation

= 40 − 15 − 10 + 5

Substitute the values

= 20

Evaluate

The number of students enrolled in at most one activity is equal to the number of students enrolled in art only, in music only, in sports only, or not enrolled in any activity, therefore: Number of students in at most 1 activity = (Art only) + (Music only) + (Sports only) + (None)

Write the formula

= 20 + 20 + 35 + 10

Substitute the values

= 85

Evaluate

Therefore, the total number of students enrolled only in at most one activity is 85. Finally, the percentage of students enrolled only in at most one activity is: Write the formula = 70.8%

Evaluate

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Example 6 Thirty four snacks are reviewed for common allergens - egg (E), milk (M) and nuts (N). These were the results: • 3 contained all three allergens • 2 contained no allergens • 7 contained nuts and eggs, but no milk • 5 contained egg and milk, but no nuts • 4 contained milk and nuts, but no egg • 17 contained milk • 15 contained egg a Show how the inclusion-exclusion principle can be used to determine the number of snacks that contain nuts.

Create a strategy First, find the total number of snacks that contain at least one allergen by subtracting the snacks with no allergens from the total. Next, calculate the total number of elements in each pairwise intersection (e.g., ∣E ∩ M∣). The problem gives the values for regions like “egg and milk, but no nuts,” so add the triple intersection back to find the full intersection value. Finally, substitute all known values into the inclusion-exclusion formula and solve for the unknown value, ∣N∣.

Apply the idea Let: • ∣E∣ = 15 snacks contained egg. • ∣M∣ = 17 snacks contained milk. • ∣N∣ is the unknown number of snacks that contained nuts. There were 34 snacks in total and 2 contained no allergens. The number of snacks with at least one allergen is: ∣E ∪ M ∪ N∣ = 34 − 2 = 32 The given is ∣E ∩ M ∩ N∣ = 3. Find the full pairwise intersections: • ∣E ∩ M∣ = (egg and milk, no nuts) + (all three) = 5 + 3 = 8 • ∣E ∩ N∣ = (nuts and egg, no milk) + (all three) = 7 + 3 = 10 • ∣M ∩ N∣ = (milk and nuts, no egg) + (all three) = 4 + 3 = 7

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Using the inclusion-exclusion principle: ∣E ∪ M ∪ N∣ = ∣ E∣ + ∣M∣ + ∣N∣ − ∣E ∩ M∣ − ∣E ∩ N∣ − ∣M ∩ N∣ + ∣E ∩ M ∩ N∣

Write the formula

32 = 15 + 17 + ∣N∣ − 8 − 10 − 7 + 3

Substitute the known values

32 = (15 + 17 + 3) + ∣N∣ − (8 + 10 + 7)

Group like terms

32 = 35 + ∣N∣ − 25

Simplify

32 = 10 + ∣N∣

Simplify

22 = ∣N∣

Subtract 10 from both sides

Therefore, there are 22 snacks that contain nuts.

Reflect and check The key step was to reconstruct the total intersection values (like ∣E ∩ M∣) from the partial information given. Note that: ∣E ∩ M∣ = ∣E ∩ M ∩ N ′∣ + ∣E ∩ M ∩ N∣ A common mistake is to use the “only” values (like “egg and milk, but no nuts”) directly in the formula, which leads to an incorrect answer.

b Create a Venn diagram to represent this information.

Create a strategy To build the Venn diagram, start by filling in the value for the innermost region: the intersection of all three sets. Then, use the given information for the two-allergen regions (e.g., “nuts and eggs, but no milk”) to fill in the surrounding intersection areas. Finally, calculate the number for the “only” regions by subtracting all intersection values from the total for each set.

Apply the idea Milk

Egg 2

5 1

0 4

8 Nut

• Start with the given value for the centre: 3 snacks contained all three allergens. • Fill in the other intersection regions: 7 (nuts and egg, no milk), 5 (egg and milk, no nuts), and 4 (milk and nuts, no egg). • Calculate the “only” regions. For Egg: 15 − (7 + 5 + 3) = 0. For Milk: 17 − (5 + 4 + 3) = 5. For Nuts: 22 − (7 + 4 + 3) = 8. • Place the 2 snacks with no allergens outside the circles.

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Idea summary The inclusion-exclusion principle for three sets is used to determine the number of elements in the union of those sets. For three sets A, B and C, the principle is expressed as: ∣A ∪ B ∪ C∣ = ∣A∣ + ∣B∣ + ∣C∣ − ∣A ∩ B∣ − ∣A ∩ C∣ − ∣B ∩ C∣ + ∣A ∩ B ∩ C∣

4.02 Practice questions What do you remember? 1

Write the inclusion-exclusion principle for two sets A and B.

This Venn diagram shows the distribution of students who were asked about being left-handed and about entering the Maths contest, based on the recorded responses:

3 Ex 1

296

Entered

How many students were surveyed in total?

How many students entered the Maths contest?

How many students were left-handed and entered the Maths contest?

How many students were left-handed or entered the Maths contest?

How many students were left-handed and did not enter the Maths contest?

Consider this Venn diagram: a

∣A∣

∣B∣

Determine: i

Left-handed

iii ∣A ∩ B∣

Using the inclusion-exclusion principle, determine ∣A ∪ B∣.

Determine ∣U∣.

Calculate: a

∣A ∪ B∣ given that ∣A∣ = 10, ∣B∣ = 25 and ∣A ∩ B∣ = 5

∣A ∩ B∣ given that ∣A∣ = 34, ∣B∣ = 11 and ∣A ∪ B∣ = 40

∣A ∪ B∣ given that ∣A∣ = ∣B∣ = 11, and sets A and B have no common elements

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A survey asked students which subjects they enjoyed: Maths (M), Science (S), or English (E). The results are shown in the Venn diagram:

M 10

Determine: a

Total number of students surveyed

∣M∣

∣S∣

∣E∣

∣M ∩ S∣

∣M ∩ E∣

∣S ∩ E∣

∣M ∩ S ∩ E∣

∣M ∪ S ∪ E∣

4 3

8 S

12 E

Practice 6

An ecologist listed some features of rivers and lakes in this Venn diagram:

Ex 2

Rivers

Lakes

Flows across a slope

Usually situated Support in a basin ecosystems with aquatic life Often supports Can regulate local navigation climate Subject to pollution and environmental concerns Can have rapids Often a habitat for or waterfalls diverse species Can be used for recreation Has a defined Contains still or and human riverbed standing water activity

R represents the set of river features, while L represents the set of lake features. a

How many features are listed for rivers? Use the notation ∣∣ to express your answer.

Determine ∣L∣.

Determine ∣R ∩ L∣.

Show that ∣R ∪ L∣ = 11 using the inclusion-exclusion principle.

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This Venn diagram shows the number of students who own a laptop and who own a smartphone: Owns a laptop

Owns a smartphone

10 S represents the set of students who own a smartphone, while L represents the set of students who own a laptop.

8 Ex 3

Determine ∣L∣ + ∣S∣.

Determine ∣L ∩ S∣.

Calculate ∣L ∪ S∣ using the inclusion-exclusion principle.

How can the correctness of the answer found in part (c) be verified?

In a survey of households, 80 households have a pet dog, 60 households have a pet cat, and 35 households have both a dog and a cat: a

How many households have at least one pet?

If 20 households have neither a dog nor a cat, determine the total number of households surveyed.

How many households have only a pet dog but not a cat?

A survey of 200 people asked about juice and soda preference. The results showed 120 people prefer juice, 100 people prefer soda, and 60 people prefer both: a

How many respondents prefer either juice or soda or both?

How many respondents do not prefer either juice or soda?

In a class of 50 students, 30 students passed the Maths exam, and 25 students passed the Science exam. If 18 students passed both exams: a

How many students passed only the Maths exam?

How many students passed exactly one of the exams?

Sarah is inviting guests to her party and sends out invitations to her friends and colleagues. She invites 40 friends and 35 colleagues, knowing that 15 people are both friends and colleagues. It is known that 5 of the people invited did not attend: a

How many people were invited to the party?

How many people attended her party?

In a school, students must study either Maths or Physics or both. Out of a total of 105 students, there are 80 students studying Maths and 70 students studying Physics. How many students study both?

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13 Ex 4

Let A and B be two sets. The number of elements in set A is n(A) = p, the number of elements in set B is n(B) = q, and the number of elements in their union A ∪ B is n(A ∪ B) = r. Determine an expression for the number of elements that are in set A but not in set B, also written as n(A ∖ B), in terms of p, q, and r.

Given three sets X, Y, and Z with the given information: • ∣X∣ = 25 • ∣Y ∣ = 30 • ∣Z∣ = 20 • ∣X ∩ Y ∣ = 10 • ∣X ∩ Z∣ = 8 • ∣Y ∩ Z∣ = 12 • ∣X ∩ Y ∩ Z∣ = 5 Use the inclusion-exclusion principle for three sets to calculate ∣X ∪ Y ∪ Z∣.

Ex 5

A survey of 150 university students found the given information about their enrollment in three popular elective courses: History (H), Psychology (P ), and Sociology (S) • 55 students are enrolled in History. • 65 students are enrolled in Psychology. • 50 students are enrolled in Sociology. • 20 students are enrolled in History and Psychology. • 15 students are enrolled in History and Sociology. • 25 students are enrolled in Psychology and Sociology. • 10 students are enrolled in all three courses.

Ex 6

How many students are enrolled in at least one of these courses?

How many students are enrolled in none of these three courses?

How many students are enrolled in History only?

In a survey of 200 people, 30% prefer tea, 40% prefer coffee, and 15% prefer both tea and coffee: a

What percentage of people prefer either tea or coffee or both?

Determine the number of people who prefer either tea or coffee or both.

What percentage of people prefer tea, but not coffee?

A survey of 100 pet owners provided information about whether they owned a dog (D), cat (C), or bird (B). The results were: • 10 owned all three types of pets. • 15 owned a dog and a cat, but not a bird. • 12 owned a dog and a bird, but not a cat. • 8 owned a cat and a bird, but not a dog. • 50 owned a dog. • 48 owned a cat. • 5 owned none of these pets. a

How many pet owners owned a bird?

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Extend your thinking Two hundred people participated in a survey regarding TV show A and TV show B. Of the respondents, 120 watch show A, 100 watch show B, and 60 watch both TV shows:

How many respondents watch only TV show B?

How many respondents watch exactly one of the TV shows?

How many integers from 1 to 863 (inclusive) are multiples of 7 or 11?

How many integers from 50 to 600 (inclusive) are multiples of 4 or 7?

How many integers from 1 to 950 (inclusive) are not multiples of 11 or 13?

Determine the sum of all the integers from 1 to 150 (inclusive) that are divisible by 4 or 5. Hint: Use the formula Determine how many integers from 8 to 600 (inclusive) are divisible by 7 or 10 but not both.

4.03 Multiplication and addition principles After this lesson, you will be able to… • understand when to apply the multiplication principle versus the addition principle • use the multiplication principle to solve counting problems • use the addition principle to solve counting problems • solve problems that combine both principles

The multiplication principle Multiplication principle A rule used to find the total number of outcomes when multiple choices or events occur in sequence. If there are a ways for the first event, b ways for the second, and so on, the total number of outcomes is a × b × c × . . .

The multiplication principle is a key concept in combinatorics used to determine the total number of possible outcomes when multiple choices occur in sequence. It simplifies counting by multiplying the number of ways each individual event can happen. This means that for each way the first event can occur, all the ways the second event can occur are considered, and for each of those combined possibilities, all the ways the third event can occur are considered, and so on.

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This principle is essential for solving problems related to sequences of choices and forms the basis for more complex probability calculations.

Exploration Suppose there is a food truck that sells ice cream sundaes. An ice cream sundae can be created by choosing one item from each of the listed categories: • Ice cream flavours: Vanilla, chocolate, or strawberry • Toppings: Cherries, sprinkles, cookie crumbles, or peanuts • Sauces: Hot fudge or caramel 1. How many different flavours of sundaes could be created? Explain the process for finding the answer. 2. What calculations could be used to find the number of outcomes more quickly?

When the number of possible outcomes is manageable, listing them all can be helpful to understand how the multiplication principle works. For example, suppose a choice is made for one flavour of ice cream, one type of sauce, and one topping for a sundae from the given list. This scenario can be represented with a tree diagram and the number of branches counted. • Ice cream flavours: Vanilla, chocolate, or strawberry • Sauces: Hot fudge or caramel • Toppings: Cherries, sprinkles, cookie crumbles, or peanuts Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles

Hot fudge Vanilla Caramel

Hot fudge Chocolate Caramel

Hot fudge Strawberry Caramel

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Each of the ice cream choices would get paired with a sauce to make six options, and then each of those gets one of the four toppings to give 24 possibilities. This shows how the multiplication principle simplifies counting the total number of outcomes without needing to list each combination individually. The multiplication principle can be used, regardless of whether listing all outcomes is practical. Consider a licence plate that requires one number, followed by two letters, then one number, and then two letters:

The total number of possible licence plates can be calculated using the multiplication principle by considering the possible outcomes for each position on the licence plate. For each number, there are 10 possibilities (the digits 0–9), while for each letter, there are 26 possibilities. Therefore, the number of possibilities for each position is shown in the image:

Digit Letter

Letter

Digit

Letter

For the first position (a number), there are 10 possibilities. For the second and third positions (letters), there are 26 possibilities each. For the fourth position (a number), there are 10 possibilities. Finally, for the fifth and sixth positions (letters), there are 26 possibilities each. Therefore, the total number of possible licence plates is 10 × 26 × 26 × 10 × 26 × 26 = 45 697 600. This calculation shows how the multiplication principle efficiently determines the total number of possible licence plates without listing each outcome individually. The multiplication principle states that if an event can occur in a different ways, a second event in b different ways, a third event in c different ways, and so on, then the total number of ways these events can occur in sequence is a × b × c × . . . Scenarios that involve the word ‘and’ typically apply the multiplication principle (e.g., choosing an ice cream flavour and a sauce and a topping).

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This can be visualised as determining the number of options for a sequence of events or choices: Number of ways for Event 1 × Number of ways for Event 2 × Number of ways for Event 3 × . . . If there are a ways for Event 1, b ways for Event 2, c ways for Event 3, and so on, the total number of ways is a × b × c × . . . . This is like filling slots, where each slot represents an event or choice: • Event 1: a options • Event 2: b options (for each of the a options from Event 1) • Event 3: c options (for each of the a × b combinations from the previous events) • And so on ... Leading to a total of a × b × c × . . . combinations. In some cases, there are restrictions, such as digits or letters not being allowed to be repeated. In these scenarios, the number of available choices for subsequent steps decreases. This means the values of b, c, and so on, in the calculation a × b × c × . . . may change depending on previous choices.

Example 1 How many 4-digit PINs can be created using the digits 0-9 if digits can be repeated?

Create a strategy The PIN has four positions to fill. Determine the number of choices for each position. Since repetition is allowed, the number of choices for each position will be the same. Then, use the multiplication principle to find the total number of possible PINs.

Apply the idea There are 10 possible digits to choose from for each position in the PIN (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). The number of possibilities for each of the four digits is:

Using the multiplication principle, multiply the number of possibilities for each position: Number of PINs = 10 × 10 × 10 × 10

Multiply the number of ways for each digit

= 10

Simplify

= 10 000

Evaluate

There are 10 000 possible 4-digit PINs.

Reflect and check In this problem, the choice for one digit does not affect the choices for the other digits because repetition is allowed. This is a key difference from problems where items cannot be repeated, where the number of available choices decreases with each selection.

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Example 2 A password contains six characters, which can be lowercase letters or the numbers 0–9. Calculate the total number of possible passwords if each letter and number can only be used once.

Create a strategy For each character in the password, the number of possible ways it can be filled needs to be found. Once the number of possibilities for each character is known, multiply these numbers using the multiplication principle.

Apply the idea There are 26 letters and 10 numbers, so there are 26 + 10 = 36 possibilities for the first character. After choosing a letter or number, the next character will have one less possibility. This pattern will continue, so the possibilities for each character will be:

Number of passwords = 36 × 35 × 34 × 33 × 32 × 31 = 1 402 410 240

Multiply the number of ways Evaluate

Idea summary The multiplication principle helps to determine the total number of possible outcomes when multiple choices occur in sequence, regardless of whether it is possible to list all the possibilities. The multiplication principle states: If an event can occur in a different ways, a second event in b different ways, a third event in c different ways, and so on, then the total number of ways these events can occur in sequence is a×b×c×... This can be thought of as multiplying the number of options available at each step in a sequence: (Options for Choice 1) × (Options for Choice 2) × (Options for Choice 3) ×. . . If there are a options for Choice 1, b options for Choice 2, c options for Choice 3, and so on, the total is a × b × c × . . .

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The addition principle Addition principle A rule used to find the total number of outcomes when choosing between two or more mutually exclusive options. If there are a ways for one option, b ways for another (mutually exclusive) option, and so on, the total number of outcomes is a + b + c + . . . Mutually exclusive Events that cannot happen at the same time. If one event occurs, the other cannot. Example: When rolling a standard die once, rolling a 3 and rolling a 5 are mutually exclusive events.

The addition principle simplifies counting by adding the number of ways each event can occur. This principle is particularly useful for solving problems where events do not overlap and forms the basis for more complex probability calculations. Scenarios that involve the word ‘or’ typically apply the addition principle. For example, consider choosing a cold drink from a selection of 3 different types ( juice, milk, water) or a hot drink from a selection of 4 different types (coffee, tea, hot chocolate and hot cider). Listing all the possibilities for drink choice would give: juice, milk, water, coffee, tea, hot chocolate and hot cider. There is no overlap between the two types of drinks, so the addition principle can be used to say that the total number of choices is 3 + 4 = 7, as each choice of cold or hot drinks is separate and does not overlap. This concept is particularly clear when thinking about mutually exclusive events. Events are mutually exclusive if they cannot happen at the same time. If Event A has m outcomes and Event B has n outcomes, and A and B are mutually exclusive, then the total number of outcomes for “A or B” is m + n. Visually, if the choices are thought of as two separate groups (or sets) that do not overlap (they are disjoint), the total number of choices is found by simply adding the number of choices in each group. For example, if Set A represents the cold drink choices and Set B represents the hot drink choices, and these sets have no drinks in common (A ∩ B = ∅, meaning their intersection is empty), then the total number of choices is the number of items in Set A plus the number of items in Set B: ∣A∣ + ∣B∣ Another example: Suppose a student can choose to travel to school by one of 3 different train lines, or one of 4 different public bus routes, or one of 2 school bus routes. Since the student will only choose one mode of transport, these choices are mutually exclusive. The total number of different ways the student can travel to school is 3 + 4 + 2 = 9 ways. The multiplication and addition principles may be used together.

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Example 3 Dalia won a prize and has the option to choose between a movie voucher with 5 different movies to choose from or a meal voucher with a choice of 3 different types of restaurants. Each option is distinct and does not interfere with the other choices. What is the total number of prize voucher choices Dalia has?

Create a strategy To determine the total number of choices, the addition principle can be used.

Apply the idea There are 5 different movie tickets and 3 different restaurant vouchers. Applying the addition principle, the total number of choices is calculated by adding the number of options in each category 5 + 3 = 8. Therefore, Dalia has 8 possible choices.

Example 4 A restaurant offers two separate set menus. A meal consists of an entree, main dish and dessert: • Menu 1 has a choice of 4 entrées, 3 mains, and 2 desserts. • Menu 2 has a choice of 6 entrées, 2 mains, and 5 desserts. Determine the total number of meal combinations that can be ordered.

Create a strategy Use the multiplication principle to calculate the number of possible combinations for both Menu 1 and Menu 2. Then, apply the addition principle to sum the meal combinations from both menus.

Apply the idea Apply the multiplication principle: For Menu 1: Meal options for Menu 1 = 4 × 3 × 2 = 24

Multiply the number of options for each course Evaluate

For Menu 2: Meal options for Menu 2 = 6 × 2 × 5 = 60

Multiply the number of options for each course Evaluate

Apply the addition principle: Total meal options = 24 + 60 = 84

Add the number of combinations from both menus Evaluate

There are 84 meal options available.

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Idea summary The addition principle helps to determine the total number of outcomes when there are two or more mutually exclusive choices (choices that cannot occur at the same time). It involves adding the number of ways each individual choice can be made. The addition principle states: If a first choice can be made in a different ways, a second, mutually exclusive choice can be made in b different ways, a third, mutually exclusive choice can be made in c different ways, and so on, then the total number of ways one of these choices can be made is a + b + c + . . . This principle is typically used when situations involve the word ‘or’, indicating that one option is being chosen from several distinct groups of possibilities.

4.03 Practice questions What do you remember? 1

A new restaurant offers an $8 lunch combo, which includes one salad, one side, and one drink. The choices are displayed in a tree diagram: Mashed potato Vegetable salad

Potato wedges Fruit cup Mashed potato

Fruity pasta salad

Potato wedges Fruit cup Mashed potato

Caesar salad

Potato wedges Fruit cup

Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea

Count the number of outcomes on the right-hand side of the tree diagram to determine how many different combo meals are possible.

Use the counting principle to determine how many different combo meals are possible.

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Angie has 3 tops, 2 bottoms, and 2 types of footwear: Shorts T-shirt Skirts

Shorts OUTFITS

Polo shirt Skirts

Shorts Sweater Skirts

Boots Sneakers Boots Sneakers Boots Sneakers Boots Sneakers Boots Sneakers Boots Sneakers

Count the number of outcomes on the right-hand side of the tree diagram to determine how many different outfits Angie has.

Use the counting principle to determine how many different outfits are possible.

For each of these scenarios, identify whether the addition principle or the multiplication principle should be used to find the total number of options. No need to calculate the answer: a

Selecting a president, a vice-president, and a treasurer from a group of 20 candidates.

Choosing to travel from Melbourne to Sydney by either plane or train.

Creating a custom sandwich by choosing one type of bread, then one type of meat, then one type of cheese.

A student deciding to enrol in either a Certificate IV in Engineering or a Certificate IV in IT.

A library shelf contains 120 fiction books and 80 non-fiction books: a

How many options are there for choosing one fiction book and one non-fiction book?

How many options are there for choosing just one book, either fiction or non- fiction?

Practice 5

Ex 1

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Ben has 3 shirts, each in a different colour: crimson (C), pink (P) and white (W), and 4 ties, each in a different colour: blue (B), grey (G), red (R) and yellow (Y) a

Construct a table to show all the possibilities of shirt and tie colours that Ben could wear.

How many different outcomes are possible?

A quiz has 5 multiple choice questions, and each question has 4 possible answers (A, B, C, or D). How many different ways are there to answer the entire quiz?

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A company is designing a new employee ID badge. The badge consists of: • A choice of 4 different background colours • A choice of 3 different badge shapes (circle, square, rectangle) • A choice of 5 different font styles for the employee’s name How many different ID badge designs can be created?

A diner offers a special breakfast combo where customers can choose: • 1 type of bread: White, whole wheat, or rye • 1 type of spread: Butter, jam, or peanut butter • 1 type of drink: Coffee, tea, or juice How many different breakfast combos can a customer choose?

A hobby shop offers a model-building kit where customers can choose: • One type of model: Aeroplane, car, or ship • One type of paint set: Primary, metallic, or pastel colours • One type of tool set: Basic, advanced, or deluxe

Ex 2

How many different model-building kits can a customer create?

Each model kit costs $50, each paint set adds $10, and each tool set adds $15. What is the total cost of a model kit with an aeroplane, a metallic paint set, and a deluxe tool set?

All passwords to log into a particular website consist of any four characters, which can be lowercase letters or the numbers 0 – 9. Calculate the total number of possible passwords if each letter and number can only be used once.

Consider the set A which contains three letters a, b and c, and the set B which contains two letters x and y. Deanna is forming a 3-letter code. Determine the number of codes she can create if: a

The first letter is from set A, the second letter is from set B, the third letter is from set A, and the same letter can be used more than once.

The first two letters are from set A without repeating any letter, and the third letter is from set B.

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if the number must be even and repetition is not allowed?

Talal is preparing for a party and wants to choose a stylish outfit. The outfit includes: • n types of shirts, where n is a positive integer • 3 types of pants • 2 types of shoes After considering all options, Talal finds there are 24 different possible outfits. Determine the number of types of shirts, n, he has.

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A school cafeteria is offering a special lunch combo where students can choose one item from each of these categories: • Main course: The options are pizza, burger, or pasta • Side dish: The options are salad or fries • Drink: The options are juice or soda However, due to dietary restrictions, some combinations are not allowed: • Students cannot choose pasta with fries. • Students cannot choose a burger with soda.

How many different lunch combos could a student choose if there are no dietary restrictions?

Considering the dietary restrictions, how many valid lunch combos are available?

James can select either a t-shirt from 4 different brands or a long-sleeved shirt from 5 different brands. Each choice is independent of the other. Should James use the addition or multiplication principle to determine the total number of choices?

Ruth is choosing an elective. She can pick one subject from a list of 3 humanities subjects or one subject from a list of 2 science subjects. To find the total number of elective choices, should the addition or multiplication principle be used? Explain your reasoning.

17 Ex 3

Sarah can choose either a book from 6 different genres or a gift card from 4 different stores. Each choice is independent of the other. How many total choices does Sarah have? A student has the option to join one of three different clubs at school: • The Science Club, which has 15 different activities to choose from • The Art Club, which offers 10 different activities • The Sports Club, which offers 8 different activities How many different activities can the student choose from?

A teacher is organising a field trip and gives students the choice of visiting one of these locations: • A museum, where there are 5 different exhibits to explore • A zoo, which has 7 different animal shows • A botanical garden, which features 4 different guided tours How many different activities can a student choose from?

A local community centre is organising a day of workshops and activities. Participants can choose to attend one of these options: • Cooking classes, with 6 different recipes to learn • Painting workshops, offering 8 different art styles to explore • Fitness sessions, featuring 5 different types of exercises How many different activities can a participant choose from?

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Ex 4

A 4-character password can consist of either four non-repeating letters or four non-repeating numerical digits. Determine the total number of unique passwords that can be formed.

Sarah is travelling from Gold Coast to Cairns. There are 4 trains or 3 buses to take her from Gold Coast to Brisbane. There are 2 coaches and 5 flights to take her from Brisbane to Cairns. Determine the number of ways Sarah can travel from Gold Coast to Cairns via Brisbane.

A person wants to travel from City X to City Z. They can fly directly from City X to City Z (3 airlines). Alternatively, they can travel from City X to City Y by bus (2 companies) and then from City Y to City Z by train (4 lines). How many different ways can they travel from City X to City Z?

Extend your thinking 24

A debate team can be formed by selecting either (one junior and one senior) or (one junior, one different junior as a reserve, and one senior). There are 8 juniors and 5 seniors available. Determine the total number of unique teams that can be created.

A new board game involves creating a character by choosing: • One character type: Knight, Wizard, or Archer • One weapon: Sword, Staff, Bow, or Dagger • One magical ability: Fireball, Healing, Shield, or Invisibility • One mount: Horse, Dragon, or Griffin However, there are special rules: • Wizards cannot choose the Sword as their weapon. • Knights cannot choose Invisibility as their magical ability.

How many different character combinations would be possible if there were no special rules?

Considering the special rules, how many valid character combinations are possible?

Describe one possible road network configuration that connects City A to City B, allowing travel directly or via either Town C or Town D, such that there are exactly 20 possible routes in total.

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4.04 Arrangements and factorial notation After this lesson, you will be able to. . . • understand the concept of factorial notation and its role in counting • use factorial notation to represent arrangements • apply pseudocode to model arrangement computations

Arrangements and permutations of items Arrangements The different ways of organising or ordering a set of distinct items. Example: The arrangements of the letters A and B are AB and BA.

The multiplication principle can determine the number of possible arrangements when there are multiple steps or choices involved. It states that if there are a ways to select an option and b ways to select another, the total number of ways to select both options in sequence is a × b. Permutation A specific arrangement of a group of objects where the order matters.

Start with

Possible permutations when choosing 2 symbols

For example, with a red star, blue circle, and green square, selecting any two and arranging them can result in different permutations. The possible permutations of selecting and arranging two dots from the three are shown in the diagram. Note that in permutations, the order is important. For instance, red star and green square are considered different permutations from green square and red star.

The number of possible permutations is determined through counting techniques. For small sets, such as the example with red, blue, and green symbols, listing each permutation and counting the total number is manageable. For larger sets, where different colours or multiple objects are involved, the multiplication principle offers a more efficient method.

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For example, if there are 7 different coloured dots and the goal is to determine the possible permutations of selecting and arranging 4 dots, using a diagram can help visualise the arrangement. The multiplication principle helps in calculating the total number of permutations by multiplying the number of choices available at each stage. A diagrammatic calculation begins with drawing lines or shapes such as boxes, where the number of lines or boxes corresponds to the number of objects being selected. In this case, since 4 dots out of 7 are being selected, four boxes should be used. Choose the first dot:

Inside the first box, the number of possible choices is 7, as there are 7 dots to choose from. Choose the second dot:

Inside the second box, the number of possible choices is 6, as there are 6 dots to choose from (one dot has already been used).

Choose the third dot:

Inside the third box, the number of possible choices is 5, as there are 5 dots to choose from (two dots have already been used).

Choose the fourth dot:

Inside the fourth box, the number of possible choices is 4, as there are 4 dots to choose from (three dots have already been used).

To calculate the total number of possible permutations, multiply these together using the multiplication principle.

So, there are 840 possible permutations for selecting and arranging 4 dots out of 7. It might be noticed that when arranging all items from a distinct set, like arranging n items in n positions, the calculation often involves multiplying a sequence of decreasing whole numbers, such as n × (n − 1) × (n − 2) × . . . × 1. This common type of calculation has a special mathematical notation called a factorial, which will be explored in the next section. It provides a concise way to represent the number of ways to arrange all items in a set.

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Example 1 Lucy’s ATM pin is a 4-digit number. She remembers the first digit, but cannot remember the remaining three digits. How many different possibilities are there?

Create a strategy Determine the number of possibilities for the next three digits and multiply the numbers together.

Apply the idea A 4-digit pin can have repeated values, so for each unknown value there are 10 options. Therefore, the possibilities for each digit will be:

Multiply the number of possibilities: Number of possibilities = 1 × 10 × 10 × 10 = 1000

Write the equation Evaluate

Example 2 At a gathering of political leaders, there are 3 delegates from European countries, 5 delegates from African countries, and 2 delegates from South American countries. In total, there are 10 delegates: a How many ways can all 10 delegates be seated with no restrictions?

Create a strategy Use the multiplication principle to arrange all 10 distinct delegates, computing the factorial of the total number.

Apply the idea There are 10 distinct delegates and 10 seats. Determine the number of ways to arrange these 10 delegates. For the first seat, there are 10 choices. For the second, 9 choices, and so on, down to 1 choice for the last seat. Number of ways = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3 628 800

Write the equation Evaluate

There are 3 628 800 ways to seat the delegates. This product, 10 × 9 × . . . × 1, is also known as 10! (10 factorial), a notation that will be explored further in the next section.

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b If the 3 European delegates leave, how many ways can the remaining 7 delegates be seated?

Create a strategy Use the multiplication principle to arrange the remaining 7 distinct delegates, computing the factorial of the reduced number.

Apply the idea There are 5 + 2 = 7 distinct delegates remaining (5 African and 2 South American). Determine the number of ways to arrange these 7 delegates in 7 seats. For the first seat, there are 7 choices, for the second 6, and so on. Number of ways = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Write the equation Evaluate

There are 5040 ways to seat the remaining delegates.

c How many ways can all 10 delegates be seated if the 2 South American delegates must sit together on the right?

Create a strategy Treat the 2 South American delegates as a single block, arrange the block and the remaining delegates, and multiply by the arrangements within the block.

Apply the idea First, consider the arrangement of the 2 South American delegates in their fixed block of seats on the right. Then, consider the arrangement of the other 3 + 5 = 8 delegates in the remaining seats to the left. The 2 South American delegates are seated together on the right. The number of ways to arrange these 2 delegates among themselves is 2 × 1 = 2 ways. The remaining 8 delegates (3 European and 5 African) can be arranged in the 8 seats to the left. The number of ways to arrange these 8 delegates is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320 ways. Using the multiplication principle, the total number of ways is: Total number of ways = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) × (2 × 1)

Write the equation

= 40 320 × 2

Evaluate inside brackets

= 80 640

Evaluate

There are 80 640 different ways that the delegates can be seated under this condition.

Reflect and check When arranging items with special conditions, such as seating certain delegates together or in specific positions, it is helpful to address those conditions first. Treat the grouped items as a single block if they must stay together, and then arrange the items within that block if their internal order matters. Then, arrange the block(s) and any remaining individual items.

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Idea summary Arrangements refer to the ways of organising or ordering a set of items. A permutation is a specific arrangement of a group of objects where the order matters. The number of possible arrangements and permutations is determined through counting techniques, with the multiplication principle helping to calculate the number of possible arrangements.

Factorial expressions In 1808, the French mathematician Christian Kramp wrote a very simple notation n! to designate the product of numbers decreasing from n to unity, that is n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 where n is a non-negative integer. This notation is particularly useful in counting. As seen in the previous section, when all n distinct items from a set are arranged, the number of ways to do this is n × (n − 1) × (n − 2) × . . . × 1. This is precisely what n! represents. So, n! is the number of permutations of n distinct items. Today, the product is referred to as “n factorial”. So 5! = 5 × 4 × 3 × 2 × 1 = 120 and 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Factorials grow quickly. For example, 18! = 6 402 373 705 728 000. The number 100! has 24 zeros at the right-hand end of the number. In fact, the number 59! is larger than the current estimate of atoms in the universe. Since 8! is 8 × 7!, then an expression like 8! + 7! can be simplified as: 8 × 7! + 7! = 7! (8 + 1) = 9 × 7! Similarly,

More generally, an expression like Similarly,

can be simplified to , and this type of simplification can be useful in solving certain

equations. For example, to solve n! = 120 × (n − 3)! for n, the solution process might proceed as follows: n! = 120 × (n − 3)! n × (n − 1) × (n − 2) × (n − 3)! = 120 × (n − 3)! n × (n − 1) × (n − 2) = 120 3

n − 3n + 2n = 120 3

Write the equation Expand the left-hand side to include (n − 3)! Divide both sides by (n − 3)! Expand the left-hand side

Subtract 120 from both sides

Factorise

n − 3n + 2n − 120 = 0 (n − 6) (n + 3n + 20) = 0

The factorisation of the cubic equation was completed using polynomial division techniques. The quadratic factor cannot be broken into linear factors using integers. Therefore, n = 6. However, since n is an integer, it might be noticed, at the third line, that 6 × 5 × 4 = 120, and so the solution is determined as n = 6 even before the factorisation step. 316

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By definition, 0! = 1. The definition 0! = 1 might seem unusual, but it is consistent with combinatorial formulas and patterns. For instance, there is one way to arrange zero objects (which is to do nothing), and this work correctly when r = n

definition ensures formulas like the permutation formula (giving

Factorial expressions are commonly used to calculate the number of arrangements and permutations. It will be shown later that the number of permutations of n distinct items is given by the factorial expression n!. For example, the number of ways to arrange 3 distinct items a, b, and c is 3! = 3 × 2 × 1 = 6. The six permutations are abc, acb, bac, bca, cab, and cba.

Example 3 Evaluate each expression: a

Create a strategy Write each factorial term in its expanded form, then simplify.

Apply the idea Expand each factorial =7×6

Quotient of same terms is 1

= 42

Evaluate

Create a strategy Start by identifying common factors, which may be factorials, in the numerator and denominator. Expand the largest factorial in the numerator until it includes the largest factorial from the denominator, allowing for cancellation.

Apply the idea xpand the numerator to include the largest E factorial from the denominator Remove the common factor of 4! Evaluate the numerator and expand the remaining factorial Simplify the expression = 15

Evaluate

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Example 4

Create a strategy Expand the factorial term in the denominator until it includes the factorial term from the numerator. Then, simplify by cancelling the common factor.

Apply the idea Expand the denominator Remove the common factor (n − 1)!

Example 5 Digital only example Explore in the online lesson

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Example 6 A band has 5 members: a In how many ways can all 5 members be arranged in a line for a photo?

Create a strategy Arrange 5 distinct members. The number of ways to arrange n distinct items is n!.

Apply the idea Here, n = 5. So the number of arrangements is 5!. Number of arrangements = 5!

Expand the factorial

= 120

Evaluate

There are 120 ways to arrange the 5 band members.

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Write the equation

=5×4×3×2×1

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b For a promotional poster, only 4 of the 5 members will be chosen and arranged in a line. In how many ways can this be done?

Create a strategy Choose 4 members from 5 and arrange them. This is a permutation of 5 items taken 4 at a time. Using the multiplication principle, there are 5 choices for the first position, 4 for the second, 3 for the third, and 2 for the fourth.

Apply the idea Using the multiplication principle: Number of arrangements = 5 × 4 × 3 × 2 = 120 There are 120 ways to choose and arrange 4 members from 5.

Reflect and check It is interesting to note that arranging all 5 members (5! = 120) gives the same numerical result as choosing and arranging 4 out of 5 members. This occurs because the number of ways to arrange n items is equal to the number of ways to arrange n − 1 items from a set of n.

Idea summary The notation n! is read as “n factorial” and represents the product of numbers decreasing from n. n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 where n is a non-negative integer. By definition, 0! = 1. Expressions in the form The expression

can be rewritten as

, and so on.

can then be simplified to

n! represents the number of ways to arrange n distinct items.

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Model arrangements with pseudocode When solving complex problems, especially those involving repetitive calculations like arrangements, it’s often useful to plan the steps before writing actual computer code. Pseudocode A way to describe the steps of an algorithm using simple English-like statements, rather than a specific programming language. It helps in thinking through the logic and structure of a solution. Common pseudocode conventions include: • INPUT: Represents data given to the algorithm. • OUTPUT or DISPLAY: Represents the result returned by the algorithm. • SET (or LET, or variableName equals value): Assigns a value to a variable. • FOR ... TO ... ENDFOR: Represents a loop that repeats a block of code a specific number of times. • WHILE ... ENDWHILE: Represents a loop that repeats as long as a condition is true. • IF ... THEN ... ELSE ... ENDIF: Represents a conditional statement that executes different code based on a condition. Consider how to model the calculation of n! using pseudocode. Note that n! = 1 × 2 × 3 × . . . × n. A loop can be used to perform this repeated multiplication: 1

INPUT n (a non-negative integer)

SET factorialResult := 1

IF n = 0 THEN

4 factorialResult := 1 5

ELSE

6 FOR i FROM 1 TO n 7 factorialResult := factorialResult * i 8 ENDFOR 9

ENDIF

10 OUTPUT factorialResult This algorithm works as follows: • It takes an integer n as input. • It initialises a variable, factorialResult, to 1. This is important because 0! = 1, and it’s the starting point for multiplication (1 × 1 = 1, 1 × 1 × 2 = 2, etc.). • It checks if n is 0. If so, it correctly sets the result to 1. • If n is greater than 0, it then loops from i = 1 up to n. • In each step of the loop, it multiplies the current factorialResult by i and updates factorialResult. • Finally, it outputs the computed factorialResult.

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Trace this for n = 3: • INPUT n = 3 • SET factorialResult to 1 • n = 3 is not 0, so go to ELSE. • FOR i FROM 1 TO 3: • i = 1: factorialResult set to factorialResult × 1 (so, 1 × 1 = 1). Now factorialResult is 1. • i = 2: factorialResult set to factorialResult × 2 (so, 1 × 2 = 2). Now factorialResult is 2. • i = 3: factorialResult set to factorialResult × 3 (so, 2 × 3 = 6). Now factorialResult is 6. • ENDFOR • ENDIF • OUTPUT factorialResult (which is 6). This correctly gives 3! = 6. This pseudocode can be adapted for calculating arrangements of r items from n (that is, P (n, r) = n × (n − 1) × . . . × (n − r + 1)). Instead of looping n times, the loop would run r times, multiplying factors starting from n and decreasing. For example, to calculate P (n, r): 1

INPUT n, r (non-negative integers, n >= r)

SET arrangementResult := 1

IF n < r THEN

4 OUTPUT Error: n must be greater than or equal to r 5

ELSE

6 FOR i FROM 0 TO r-1 7 arrangementResult := arrangementResult * (n-i) 8 ENDFOR 9

ENDIF

10 OUTPUT arrangementResult This pseudocode calculates n × (n − 1) × . . . × (n − r + 1).

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Example 7 This pseudocode is designed to calculate the number of arrangements of n distinct items: 1

INPUT n

SET factorialResult := 1

IF n = 0 THEN

4 factorialResult := 1 5

ELSE

6 FOR i FROM 1 TO n 7 factorialResult := factorialResult * i 8 ENDFOR 9

ENDIF

10 OUTPUT factorialResult Trace the algorithm for an input of n = 5 to find the number of ways to arrange 5 distinct books on a shelf.

Create a strategy This is an arrangement of 5 distinct items, which is 5!. The factorial pseudocode algorithm can be used with n = 5.

Apply the idea Tracing the algorithm with an input of n = 5: • INPUT n = 5 • SET factorialResult to 1 • n = 5 is not 0, so the ELSE block is executed. • FOR i FROM 1 TO 5: • When i = 1, factorialResult becomes 1 × 1 = 1. • When i = 2, factorialResult becomes 1 × 2 = 2. • When i = 3, factorialResult becomes 2 × 3 = 6. • When i = 4, factorialResult becomes 6 × 4 = 24. • When i = 5, factorialResult becomes 24 × 5 = 120. • The loop finishes. • OUTPUT factorialResult, which is 120. There are 120 ways to arrange the 5 books.

Example 8 Digital only example Explore in the online lesson

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Idea summary Pseudocode is a tool for planning algorithms using simple, English-like statements before writing actual computer programs. I t helps clarify the logic for computations such as factorials (n!) and permutations (P(n, r)). Factorial (n!) pseudocode typically involves initialising a result to 1 and then using a loop to multiply by integers from 1 up to n. Permutation (P(n, r)) pseudocode can involve initialising a result to 1 and then using a loop r times, multiplying by n, then n − 1, and so on, for r factors.

4.04 Practice questions What do you remember? 1

Describe, in words, what factorial notation (for example, 5!) represents in the context of counting arrangements.

What is the defined value of 0!?

There are 3 shorts and 2 t-shirts in a closet. How many possible outfits are there?

How many outcomes are possible when three dice are rolled?

Practice Ex 1

An exam consists of 8 multiple choice questions, where the correct answer could be A, B, C, or D. If the answer to each question is guessed randomly, how many different ways are there of answering all 8 questions?

How many unique vehicle registration plates can be created if each plate consists of 2 letters followed by 4 digits? Repetition of the digits is allowed.

A librarian is arranging books on a new display shelf. There are 4 distinct novels, 3 distinct science fiction books, and 2 distinct poetry books. In total, there are 9 books:

Ex 2

How many ways can all 9 books be arranged on the shelf if there are no restrictions?

If the 2 poetry books are removed, how many ways can the remaining 7 books be arranged?

How many ways can all 9 books be arranged if all 4 novels must be kept together?

At a theme park ticket booth, four friends, Max, Lily, Sarah, and Evan, are waiting in the line. How many different arrangements are there for them to wait in the line?

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Five cards have different letters written on them. The letters are M, L, I, E, and S. The cards are shuﬄed and then arranged in a row on a table, face up. How many possible arrangements are there?

In a card game, 2 cards are picked from a regular deck of 52 playing cards. The first card is then laid facing one player and the second card facing the other player. How many different outcomes can result from this selection?

How many three-digit odd numbers can be formed using the digits 3, 4, 5, 6, 7, and 8 if: a

The repetition of digits is not allowed

The repetition of digits is allowed

How many numbers are there between 99 and 1000 having 5 in the units place?

A family is going to have a group photograph. If this family consists of a mother, father, grandfather, grandmother, a baby, and three children, what is the number of possible arrangements that begin and end with a grandparent?

Evaluate: a

15 Ex 3

Determine whether these expressions are true or false: 6! − 4! = 2!

Ex 4

21 Ex 5

120 = 5 × 4!

(3!)2 = 3! + 3! + 4!

(n + 1)!

Write in expanded form: a

(3!)!

Evaluate:

Simplify:

a 18

3! + 2!

Evaluate:

a 17

(n − 2)!

(p + 2)!

Simplify: a

If (n + 1)! = 5n!, solve for n.

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(k − 25)!

23 Ex 6

25 Ex 7

Grace and Blake each create cards bearing the letters of respective names to determine the total number of possible arrangements: a

Grace has cards marked G, R, A, C, and E. How many arrangements can Grace make of her name?

Blake has cards marked B, L, A, K, and E. How many arrangements can Blake make of his name? Explain why his result is the same as Grace’s result.

A newspaper editor is deciding which 2 of 6 articles to print on the front page:

If their order on the front page matters, how many different arrangements are possible for the front page?

She finds an error in one of the 6 articles and cannot print it. How many different arrangements for the front page are now possible, given that the order of the articles on the front page still matters?

The given pseudocode is intended to calculate n!: 1

INPUT n (a non-negative integer)

SET factorialResult := 1

IF n = 0 THEN

4 factorialResult := 1 5

ELSE

6 FOR i FROM 1 TO n 7 factorialResult := factorialResult * i 8 ENDFOR 9

ENDIF

10 OUTPUT factorialResult Trace the execution of this pseudocode for n = 4, showing the value of factorialResult and i at each step of the loop. What is the final output?

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26 Ex 8

Consider this pseudocode designed to calculate the number of permutations P (n, r), which is the number of ways to arrange r items from a set of n: 1

INPUT n, r (non-negative integers, n >= r)

SET arrangementResult := 1

IF n < r THEN

4 OUTPUT Error: n must be greater than or equal to r 5

ELSE

6 FOR i FROM 0 TO r-1 7 arrangementResult := arrangementResult * (n-i) 8 ENDFOR 9

ENDIF

10 OUTPUT arrangementResult a

Trace the algorithm for an input of n = 5 and r = 2. Show the value of arrangementResult and i at each step of the loop.

What is the final output of the algorithm for an input of n = 6 and r = 3?

Explain the purpose of this line of code from the loop: 1

arrangementResult := arrangementResult * (n-i)

Extend your thinking 27

How many different five-digit numbers can be formed from the set of digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, if digits cannot be repeated and the number must be greater than 30 000?

A code is made using 5 characters chosen from the digits 0 to 9, and the vowels A, E, I, O, and U. Each character can only be used once. How many codes can be made in each of these cases:

There are no restrictions on the choice of characters.

The first character must be a digit, then letters and digits alternate.

The code must have three digits followed by two letters.

In how many different ways can this diagram be coloured with these conditions: • Each of the smaller triangles should be painted with one of three colours: green, pink, purple • No two adjacent regions have the same colour.

Calculate the value of n, if

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4.05 Permutations After this lesson, you will be able to… • define permutations and understand their significance in counting problems • use factorial notation to calculate permutations • apply permutation notation to represent arrangements where order matters • solve simple problems involving permutations

Permutations Permutations A specific arrangement of a group of items where the order matters.

Exploration To explore how arrangements work with a small set of items, consider the set of three distinct letters: {A, B, C} List all the different ways 2 letters chosen from this set can be arranged. It is important to remember that in permutations, the order of the letters is important (so AB is different from BA). For this initial look, ensure that the same letter is not used more than once in any single arrangement. 1. How many distinct two-letter arrangements were found (where letters are not repeated)? 2. Is the arrangement ‘AB’ different from ‘BA’? Explain why or why not in the context of permutations. 3. What method can be used to ensure all possible arrangements are found? 4. How many choices are there for the first letter in an arrangement? Once the first letter is chosen, how many choices remain for the second letter (assuming no repetition)? 5. Is it possible to derive a general rule or method to find the number of two-letter arrangements from a set of n distinct letters if repetitions are not allowed? 6. How can these arrangements be represented using a tree diagram or a table?

The number of different ways to arrange r items chosen from a larger set of n distinct items (where the order of arrangement is important) can be found using the multiplication principle: n × n−1 × n−2 ×...× n−r+1 1st

2nd

3rd

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Permutation calculations often use factorials. The factorial of a non-negative whole number n (shown as n!) is the result of multiplying all whole numbers from n down to 1. For instance, 5! = 5 × 4 × 3 × 2 × 1 = 120. An important rule is that 0! = 1. This can be simplified using the permutation notation and formula:

n is the total number of elements in the set r is the number of elements being selected and arranged The relationship between

and the multiplication principle is shown by:

This simplifies to nPr = n × (n − 1) × (n − 2) × . . . (n − r + 1), which matches the earlier use of the multiplication principle. When using the permutation formula, the number of items chosen, r, from a larger set of n distinct items must always be less than or equal to n. When r is equal to n, it means all items from the set of n distinct items are being arranged. The permutation formula simplifies in this situation:

So, when r equals n, the number of permutations is simply n!. This is logical since all n items are being arranged, and there are n! possible ways to arrange n distinct items. Many scientific calculators have a dedicated function for calculating permutations. The general process is as follows: • Typically, the value for n (the total number of items) is entered first. • Then, a specific permutation button should be located, often labelled nPr, nPr, or sometimes just P. This might be a primary button or require pressing a ‘Shift’ or ‘2nd’ function key. • After activating the permutation function, the value for r (the number of items to arrange) is entered. • Pressing the equals key will then display the result for nPr . • Always refer to the calculator’s manual for specific instructions, as button placement and operation can vary. Similarly, for calculating factorials (n!): • Enter the number n for which the factorial is to be calculated. • Locate the factorial button, usually labelled n! or x!. This may also be a secondary function. • Press the factorial button and then the equals key to see the result. • Consult the calculator’s manual for precise steps.

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Did you know?

Musicians use permutations when composing melodies by rearranging notes to create new variations. This helps them explore countless musical patterns, build unique harmonies, and keep their music fresh. By understanding how different sequences shape rhythm and mood, maths helps turn creativity into structured, captivating sound.

Example 1 Consider 7 swimmers in an international race: a Use the permutation formula to determine how many different orders the 7 swimmers can finish, assuming there are no ties.

Create a strategy The permutation formula can be used with n = 7 and r = 7.

Apply the idea Write the permutation formula Substitute n = 7 and r = 7 Simplify the denominator Expand the factorial = 5040

Evaluate

Therefore, 5040 different arrangements of swimmers can be formed.

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Reflect and check This answer could also be found directly with a calculator’s permutation function for 7P7 or by using the multiplication principle: 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

b Check the answer to part (a) using a different approach to determine the number of different possible arrangements of the 7 swimmers.

Create a strategy To find how many ways the 7 swimmers can be arranged, consider placing them in order one by one and using the multiplication principle.

Apply the idea Choose the first swimmer: There are 7 options. Choose the second swimmer: There are 6 remaining options. Choose the third swimmer: There are 5 remaining options. . . . Continue this process until all 7 swimmers are arranged.

Swimmer 2

Swimmer 3

Swimmer 4

Swimmer 5

Swimmer 6

Swimmer 7

Swimmer 1

This diagrammatic representation shows the step-by-step calculation of the total number of possible arrangements for 7 swimmers:

Thus, the total number of arrangements is calculated as 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. Therefore, 5040 different arrangements of swimmers can be formed. This matches part (a).

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Example 2 Four distinct coloured dots (red, blue, green, and yellow) are available. How many different ways can 2 of these dots be selected and arranged in a sequence?

Create a strategy This is a permutation problem because the order of the dots matters. Identify the total number of items to choose from (n) and the number of items to arrange (r). Use the permutation formula to calculate the total number of possible arrangements.

Apply the idea The total number of dots is n = 4. The number of dots to be selected and arranged is r = 2. Write the permutation formula Substitute n = 4 and r = 2 Simplify the denominator Expand the factorials =4×3

Remove the common factor of 2!

= 12

Evaluate

Therefore, there are 12 possible permutations. This matches the number of arrangements shown in the diagram.

Reflect and check The 12 possible arrangements are shown: The result can also be confirmed using the multiplication principle. There are 4 choices for the first dot and 3 remaining choices for the second dot. The total number of arrangements is 4 × 3 = 12. The key concept is that order matters; for instance, an arrangement of (red, blue) is distinct from (blue, red).

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Example 3 From the word SHELF: a How many different ways can 3 letters be selected and arranged?

Create a strategy Since different orders of the selected letters create distinct arrangements, this is a permutation problem.

Apply the idea The permutation formula is used with n = 5 (total letters in SHELF) and r = 3 (letters to arrange). Write the permutation formula Substitute n = 5 and r = 3 Simplify the denominator Expand the factorial Evaluate = 60

Simplify

Therefore, 60 different 3-letter words can be formed from the letters of the word SHELF.

Reflect and check Notice that instead of using the formula, it can be calculated directly using a calculator by typing in 5 P3 or using the multiplication principle of 5 × 4 × 3 = 60.

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b How many different ways can all the letters be arranged?

Create a strategy Since the order of all 5 letters is important because different orders will produce different arrangements, the permutation formula can be used with n = 5 and r = 5.

Apply the idea Write the permutation formula Substitute n = 5 and r = 5 Simplify the denominator Expand the factorial Evaluate = 120

Simplify

Therefore, 120 different words can be formed from the letters of the word “SHELF”.

Reflect and check Since all 5 letters are being arranged, and there are n! possible ways to arrange n distinct items, the number of possible permutations can be directly expressed as 5!.

Idea summary A permutation is an arrangement of items where the order is important. To find the number of different ways to choose and arrange r items from a set of n distinct is used. Here, n and r are whole

items, the permutation formula

numbers, and r must be less than or equal to n (r ≤ n). n elements

r elements

r≤n

If all n items from the set are being arranged (so r = n), the number of permutations is simply n!.

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4.05 Practice questions What do you remember? 1

List all the possible permutations of the letters X, Y, Z taken 2 at a time.

Determine whether these statements are true or false: a

The number of permutations of 4 items chosen from a set of 7 distinct items is given by 4P7.

When arranging all 8 books on a shelf, the number of possible arrangements is given by 8!.

The permutation formula

can be used to calculate the number of ways to

arrange 5 items out of a set of 10. d 3

Evaluate: a

can be used even if r > n.

The permutation formula

Pn−2

P12

Determine, in terms of n, a simplified form of: a

n+5

Pn+2

Practice 5 Ex 1

A puzzle has 5 distinct, numbered puzzle pieces (1, 2, 3, 4, 5): a

Determine, using the permutation formula, the number of different sequences that can be made by arranging these pieces in a row.

Verify the answer to part (a) using a different approach.

How many different ways can the letters of the word FOUR be arranged?

Seven athletes are participating in a race. In how many ways can the first 3 prizes be won?

Ex 2

Ex 3

A music playlist with 12 songs has been downloaded to a phone. Four songs are to be selected and played in order. How many different ways can this be done?

A yard has 3 gates. How many ways can someone enter then exit the yard using two different gates?

For the word LOGARITHMS:

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How many four-letter arrangements can be formed?

How many ways can all the letters be arranged?

A bookshelf has 7 distinct novels: a

How many ways can 3 of these novels be arranged on a shelf?

If one specific novel must be in the first position, how many arrangements of 3 novels are possible?

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

A company has invited 5 speakers to a conference. For the schedule, 3 of these speakers need to be selected to give presentations in a specific order: a

How many different ways can the company arrange the 3 selected speakers in the schedule?

Later, 2 of the 3 selected speakers are chosen to present two awards (a first prize and a second prize, presented in that order). How many different ways can the prizes be presented by these two speakers?

A local bookstore is organising a reading event with 8 different books available. For the event, 4 of these books must be selected and presented in a specific order: a

Determine the number of ways to select and arrange 4 books out of the 8 available.

From the 4 books already chosen, the bookstore will feature 2 in a special showcase, arranged in a specific order. How many ways can these 2 featured books be selected and arranged?

A tech company is developing a new app and has 7 different features to choose from. The product team needs to select and prioritise 3 features for the initial release, and the order in which these features are prioritised is crucial: a

Determine the number of ways to select and arrange 3 features out of the 7 available for the initial release.

Once the 3 features are selected, the team decides to highlight 2 of these features in a promotional campaign. How many different ways can the 2 highlighted features be arranged for the campaign?

After the initial release and promotional campaign, the company decides to run a user feedback survey featuring all 3 of the initially selected features. How many different ways can the 3 features be arranged for the survey presentation?

Determine how many numbers between 10 and 10 000 can be formed by using the digits 1, 2, 3, 4, 5, under these conditions: a

No digit is repeated in any number

Digits can be repeated

Solve for n: a c

Pn = 8!

P2 = n + 3

Pn−2 = 5 × 4 × 3

Extend your thinking 17

Explain why the permutation formula (nPr) is not suitable for the given scenario: Determine how many different ways there are to answer 6 questions in an exam that consists of multiple choice questions where the answer could be A, B, or C.

If 56Pr+6 : 54Pr+3 = 30 800 : 1, calculate the value of r.

Prove that nPr = n−1Pr + r (n−1Pr−1 ).

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4.06 Permutations with restrictions After this lesson, you will be able to… • understand how restrictions affect permutation calculations • solve complex problems involving permutations with repeated items, groups, and circular arrangements

Restrictions on repeated objects Different types of restrictions can apply to permutations, affecting how objects are arranged and counted based on specific conditions. These restrictions can affect the way objects are arranged and counted, depending on the specific conditions involved.

Exploration Consider the words “ALL” and “AIL”: 1. For either word, is it possible to rearrange some letters and still get the original word? 2. Which word would have more possible arrangements? 3. List all possible unique arrangements for each word. Describe any observations.

When objects like letters, digits, or colours are repeated in a permutation, the number of distinct arrangements decreases because the repeated objects are indistinguishable. To account for this, the total number of permutations is divided by the factorial of the number of repeated objects. For example, when considering the word “SUCCESS”, if all the letters were different, the number of possible arrangements would be calculated as 7! = 5040 (since there are 7 letters). However, since the letter “S” repeats three times, and “C” repeats twice, the calculation must be adjusted to SUCCESS.

. Therefore, there are 420 distinct ways to arrange the letters in

This example demonstrates how considering repeated objects in permutations ensures that each distinct arrangement is accurately counted. To determine how to adjust for repeated objects in permutations, follow these steps to ensure that each distinct arrangement is not double or triple counted: 1. Calculate the total number of permutations by taking the factorial of the total number of objects. 2. Identify the repeated objects and count how many times each one appears. 3. Divide the total number of permutations by the factorial of the number of times each repeated object occurs. 336

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In general, if there are n objects where there are n1 identical objects of type 1, n2 identical objects of type 2, ..., nk identical objects of type k, then the number of distinct permutations of these n objects is given by the formula:

is the number of distinct permutations

is the total number of objects

n1, n2, . . . , nk is the number of identical objects of each type, where n1 + n2 + . . . + nk = n

Example 1 Consider two boxes, A and B, each containing 7 balls: • Box A: All 7 balls are different colours. • Box B: The 7 balls include 4 red, 2 blue, and 1 green. a Calculate the total number of distinct arrangements for Box A, where all 7 balls are different.

Create a strategy Use the permutation formula with n = 7 and r = 7.

Apply the idea Write the permutation formula Substitute n = 7 and r = 7 = 5040

Evaluate

Therefore, 5040 different arrangements of balls can be formed for Box A.

Reflect and check This could also be calculated using the multiplication principle as 7! = 5040.

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b Calculate the total number of distinct arrangements for Box B.

Create a strategy To find the number of distinct arrangements of 7 balls, where some balls are of the same colour, divide the total number of arrangements for all 7 balls by the factorial of the number of times each colour is repeated.

Apply the idea 1. Calculate the total permutations if all balls were unique. For 7 balls, this is 7! = 5040. 2. Identify the repeated balls and count how many times each type is repeated: Red balls = 4! Determine how many ways to arrange 4 identical red balls Blue balls = 2! Determine how many ways to arrange 2 identical blue balls Green ball = Not repeated It does not contribute to repeating arrangements. Its factorial is 1!, which is 1 and does not affect the count. 3. Determine the total number of permutations: Divide by the factorial of the number of repetitions for each type of ball to account for identical arrangements = 105

Evaluate

Idea summary When objects are repeated in a permutation, the number of distinct arrangements decreases because the repeated objects are indistinguishable. To account for this, the total number of permutations is divided by the factorial of the number of times each distinct object is repeated. If there are n objects in total, with n1 identical objects of type 1, n2 identical objects of type 2, ..., nk identical objects of type k, the number of distinct permutations is:

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Restrictions on grouping or adjacent objects When specific objects in a permutation need to stay together or be next to each other, adjust the calculation by treating these objects as a single unit or block. This approach simplifies the arrangement process under the given restriction.

Exploration Consider a 400 m swimming relay team with four swimmers: Alex, Ben, Chloe, and David 1. If there are no restrictions, how many different orders can they swim in? 2. Suppose Ben and Chloe are siblings and want to swim one after the other (in either order, Ben then Chloe, or Chloe then Ben). List all possible swimming orders where Ben and Chloe swim consecutively. How many are there? 3. When Ben and Chloe must swim consecutively, how can the pair be conceptually treated as a single “unit” to simplify counting? How does the internal order of Ben and Chloe affect the total count?

In general, when certain objects in a permutation must be grouped together or be adjacent, the calculation can be adjusted as follows: 1. Treat grouped or adjacent objects as a single unit: Consider the objects that need to be grouped or adjacent as one block or unit. This reduces the problem to arranging fewer units. 2. Calculate permutations of the units: Determine the total number of permutations of these units or blocks, treating each as distinct. 3. Multiply by internal permutations of the block: If the block contains multiple objects, calculate the number of ways to arrange the objects within the block and multiply the result from step 2 by this number. 4. Divide by the factorial of identical objects: If any of the objects (either inside or outside the block) are identical, divide the total by the factorial of the number of identical objects to adjust for over-counting identical arrangements. By using this approach, the permutations accurately reflect the restrictions of grouping or adjacency, ensuring that each arrangement meets the specified conditions. There are also situations where certain objects must not be together. For example, arranging people where two specific individuals must not sit next to each other. This type of problem is often best solved using complementary counting. The principle is: Number of arrangements where items are not together equals (Total possible arrangements) minus (Number of arrangements where items are together)

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For example, to find the number of ways to arrange the letters A, B, C, D such that A and B are not together: • Total unrestricted arrangements of A, B, C, D is 4! = 24. • Arrangements where A and B are together: Treat (AB) as a block. Arrange (AB), C, D. This is 3! = 6 ways. The block (AB) can be AB or BA, so 2! = 2 internal arrangements. Therefore, 3! × 2! = 6 × 2 = 12 ways for A and B to be together. • The number of arrangements where A and B are not together is equal to the total number of arrangements minus the number of arrangements where A and B are together. That is, 24 − 12 = 12 ways. This approach simplifies problems that would otherwise require considering many separate cases.

Example 2 How many ways can the letters of the word MATH be arranged if the letters A and T must be adjacent (sit together)?

Create a strategy To solve problems with an adjacency condition, a grouping strategy is used: 1. Treat the letters that must be together (A and T) as a single block or unit. 2. Calculate the number of ways to arrange this new block along with the other remaining letters. 3. Calculate the number of ways the letters can be arranged within the block. 4. Use the multiplication principle to multiply the results from steps 2 and 3 to find the total number of arrangements.

Apply the idea Step 1: Group the adjacent letters The letters A and T must be adjacent, therefore, the letters are treated as a single conceptual block: (AT). The items to be arranged are now the block (AT), the letter M, and the letter H. Step 2: Arrange the block and other letters There are now 3 distinct units to arrange: (AT), M, H. The number of ways to arrange these 3 units is given by 3!. Arrangements of units = 3!

Write the equation

=3×2×1

Expand the factorial

Evaluate

So, there are 6 ways to arrange the block and the other letters.

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Step 3: Arrange letters within the block The letters inside the block (A and T) can also be arranged. They can be in the order AT or TA. The number of ways to arrange these 2 letters is 2!. Internal arrangements = 2!

Write the equation

=2×1

Expand the factorial

Evaluate

This means for each of the 6 arrangements from Step 2, there are 2 possible internal arrangements of the AT block. Step 4: Calculate the total number of arrangements Using the multiplication principle, the total number of ways is the product of the arrangements of the units and the internal arrangements of the block. Total arrangements = (Arrangements of units) × (Internal arrangements)

Write the formula

= 3! × 2!

Substitute the values

= (3 x 2 x 1) x (2 x 1)

Expand the factorial

=6×2

Simplify each product

= 12

Evaluate

Therefore, there are 12 distinct permutations of the word MATH where the letters A and T are adjacent.

Reflect and check The 12 possible arrangements are shown: AT as a block:

TA as a block:

ATMH

2. ATHM

TAMH

8. TAHM

MATH

HATM

9. MTAH

10. HTAM

6. MHAT

11. HMTA

12. MHTA

5. HMAT

This problem demonstrates a key technique for handling restrictions in permutation problems. When items must be kept together, treating them as a single block simplifies the initial counting process. It is crucial to remember to also account for the possible arrangements of the items within the block and combine these using the multiplication principle.

Example 3 Digital only example Explore in the online lesson

mathspace.co

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Example 4 At a school, there are 9 students. Five students are wearing red shirts and 4 students are wearing blue shirts. How many arrangements are there if red and blue shirts must alternate?

Create a strategy Understand the alternating pattern of red and blue shirts, and then calculate the arrangements for each group separately using the factorial formula n!. Consider starting and ending with a red shirt to maintain the alternating sequence.

Apply the idea Following the alternating pattern RBRBRBRBR, calculate the arrangements for red and blue shirts separately and then multiply these arrangements: For the 5 red shirts, the arrangements are 5!. For the 4 blue shirts, the arrangements are 4!. The total arrangements are given by 5! × 4! which calculates to 120 × 24 = 2880. Therefore, there are 2880 possible arrangements.

Reflect and check This can be verified with the multiplication principle:

Idea summary When certain objects in a permutation need to stay together or be next to each other, treat these objects as one unit or block. irst, find the number of ways to arrange the blocks. Then, multiply this by the F number of ways the objects within the block can be arranged.

Restrictions on selection from multiple groups When selecting objects from different groups with specific rules or limitations, adjust the calculation by treating each group independently while applying the restrictions. Such restrictions often require combining permutations from different groups while respecting the given conditions. It’s important to note that these scenarios often involve permutations because the order of selection matters (for example, selecting for specific roles like president, vice-president, etc.). If the order of selection did not matter (for example, just selecting a committee of 3 people with no specific roles), use combinations. This lesson focuses on permutations where the arrangement or the specific role assigned to a selection is significant.

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For example, imagine selecting three books to read from three different genres: 3 fiction, 4 nonfiction, and 2 science. For example, suppose two fiction books are to be selected and read in a specific order, followed by one science book, the fiction and science groups are treated separately and the non-fiction is not relevant. First, calculate the number of ways to choose from the fiction books, and then from the science books. Finally, combine these selections. In this case, the total number of valid selections is 3P2 × 2P1 = 12. To determine the number of possible selections: • Identify the groups: Determine the distinct groups from which selections need to be made and the number of objects available in each group. • Apply the selection rules: Consider the specific restrictions or rules that dictate how selections must be made from each group and calculate the number of valid selections based on the rules. • Combine the selections: Multiply the number of possible choices from each group according to the rules. The product gives the total number of valid combinations. This method ensures accurate accounting for the restrictions while calculating the total number of possible selections.

Example 5 Mr. Ti needs to select 5 students for a student council from three different groups: • 7 students from the Science club • 5 students from the Maths club • 8 students from the Debating club a Determine how many possible ways he can form a student council if it is made up of: • A president from the Science club • A vice president from the Science club • A secretary from the Science club • A treasurer from the Maths club • A student-rep from the Maths club

Create a strategy First, determine the number of possible selections from the Science club, then determine the number of possible selections from the Maths club, and finally combine the permutations.

Apply the idea The total number of students in the Science club is 7, the number of students to be selected from this group is 3, and order matters because they all have different positions: Write the permutation formula Substitute n = 7 and r = 3 = 210

Evaluate

The number of possible selections of 3 students from the Science club is 210.

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The total number of students in the Maths club is 5, the number of students to be selected from this group is 2, and order matters because they have different positions: Write the permutation formula Substitute n = 5 and r = 2 = 20

Evaluate

The number of possible selections of 2 students from the Maths club is 20. Since the selection from each group is independent of the other, the permutations must be multiplied. The number of possible student councils if Mr. Ti must select exactly three students from the Science club and two students from the Maths club, considering the position for which they are selected, is: Number of student councils = Selections from Science club × Selections from Maths club

Write the formula

= 210 × 20

Substitute the values

= 4200

Evaluate

Therefore, 4200 different student councils can be formed under the given condition.

b How many ways can Mr. Ti form a student council of 5 students if the president or the studentrep must be selected from the Debating club, and that the remaining 4 positions are selected from the Science club and Maths club?

Create a strategy First, determine the number of ways to select a student from the Debating club. Then, determine the number of ways to assign this student to one of the specified roles (president or student-rep). Next, calculate the number of ways to select and arrange the remaining students for the other positions from the combined Science and Maths clubs. Finally, combine these possibilities using the multiplication principle.

Apply the idea The number of possible selections of a student from the Debating club is 8P1 = 8 Since there are two possible positions (president or student-rep) for the student selected from the Debating club, there are 2P1 = 2 ways to assign this student to one of these positions. The total number of students in the Science club and Maths club is 12, the number of students to be selected from this combined group for the remaining 4 positions is:

Write the permutation formula

Substitute n = 12 and r = 4

= 11 880

Evaluate

The number of possible selections of the remaining 4 students is 11 880.

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To find the total number of possible student councils, these possibilities are multiplied together based on the multiplication principle: • Choose 1 student from the 8 in the Debating club: 8P1 = 8 ways • Assign this student to one of the 2 specified positions (president or student-rep): 2P1 = 2 ways • Choose and arrange the remaining 4 students for the remaining 4 positions from the 12 students in the Science and Maths clubs combined: 12P4 = 11 880 ways Total number of student councils: Number of student councils = (Ways to choose debating student) × (Ways to assign position) × (Ways to choose other 4 students)

Write the formula

= 8 × 2 × 11 880

Substitute the values

= 190 080

Evaluate

Therefore, 190 080 different student councils can be formed under the given condition.

Idea summary When selecting objects from different groups with specific rules or limitations, adjust the calculation by treating each group independently while applying the restrictions. Such restrictions often require combining permutations from different groups while respecting the given conditions.

Circular permutations Permutations of objects in a line have been considered, known as linear permutations. However, sometimes objects are arranged in a circle, such as people around a table or charms on a bracelet. These are called circular permutations. Linear permutation An arrangement of objects in a line or row, where there is a distinct start and end. Circular permutation An arrangement of objects in a circle, where there is no distinct start or end, and only the relative positions of the objects matter.

In a linear permutation, arrangements like ABCD and DABC are different because the start and end positions are distinct. In a circular permutation, there is no fixed starting or ending point. If A, B, C, D are arranged in a circle, the arrangement ABCD is the same as BCDA, CDAB, and DABC if the circle is rotated. What matters is the relative position of the objects to each other.

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Exploration Consider arranging 3 people, Alice (A), Bob (B), and Charles (C), around a circular table: 1. If this were a line, how many ways could A, B, C be arranged? 2. Draw or list all the possible linear arrangements (for example, ABC, ACB, ...). 3. Imagine these arrangements are placed around a circle. For example, if A is at the top, B to the right, C to the left for ABC. Which linear arrangements result in the same circular arrangement when rotations are considered? (for example, ABC, BCA, CAB are the same in a circle). 4. How many distinct circular arrangements are there for A, B, and C?

To count distinct circular permutations, the position of one object is fixed. This provides a reference point, and then the remaining n − 1 objects in a line relative to the fixed object. Therefore, the number of distinct circular permutations of n distinct objects is (n − 1)!. Another way to think about this is that for n objects, there are n! linear permutations. However, in a circle, each unique circular arrangement can be rotated n times to produce n different linear arrangements that are considered identical in a circular context. So, the total linear permutations are divided by

Pcircular = (n − 1)! Pcircular is the number of distinct circular permutations n

is the number of distinct objects to be arranged

Reflections (Flipping over) Sometimes, arrangements are considered identical if one can be flipped over to become the other. This is common for items like key rings or necklaces where there isn’t a distinct “up” or “down” side. If clockwise and anti-clockwise arrangements are considered the same (that is, reflections are identical), and n > 2, the number of distinct circular permutations is

. It is important to read

the problem carefully to determine if reflections should be considered identical. If not specified, assume they are distinct.

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Example 6 How many ways can 6 people sit around a circular table?

Create a strategy This is a straightforward circular permutation problem with n = 6 distinct people. Use the formula (n − 1)!. Reflections are considered distinct unless stated otherwise (that is, different relative orderings from a fixed person’s perspective).

Apply the idea The number of ways is (6 − 1)! = 5!. 5! = 5 × 4 × 3 × 2 × 1 = 120

Write the equation Evaluate

There are 120 ways for 6 people to sit around a circular table.

Example 7 How many distinct ways can 5 different charms be arranged on a circular bracelet if the bracelet can be flipped over?

Create a strategy This is a circular permutation where reflections (flipping the bracelet) are considered the same. Use the formula

for n = 5.

Apply the idea The number of ways is Write the equation = 12

Evaluate

There are 12 distinct ways to arrange the 5 charms on the bracelet.

Example 8 Digital only example Explore in the online lesson

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Idea summary Arrangements of objects in a circle are called circular permutations. Unlike linear permutations, there is no fixed start or end point, so rotations of the same relative arrangement are considered identical. The number of distinct circular permutations of n distinct objects is given by the formula: (n − 1)!. If reflections (flipping over) are considered identical (for example, for necklaces or key rings where n > 2), the formula becomes

4.06 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

The arrangement ABC is different from CBA in a linear permutation.

5! means 5 × 4 × 3 × 2 × 1.

If items are repeated in a set, the number of unique arrangements increases.

The order of items does not matter in a permutation.

P2 = 4!

Match each scenario with the correct expression that represents how to calculate the number of arrangements or selections: a

Arranging 3 distinct books on a shelf

Choosing a president and a vicepresident from 5 people

The number of ways the letters in the word SEE can be arranged

iii (4 − 1)!

Seating 4 people around a circular table

iv 3!

Arranging 5 distinct items where two specific items must be together

v 4! × 2! vi 5!

Calculate the value of 5P3.

How many ways can the letters of the word APPLE be arranged?

Practice 5

Determine the number of unique ways the word COMMITTEE can be arranged.

Consider a bag that contains 4 red beads, 3 blue beads, and 2 green beads. Calculate the total number of distinct arrangements of the 9 beads in a straight line.

Ex 1

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Consider a set of 10 different-coloured tokens where there are 4 orange tokens, 3 purple tokens, 2 yellow tokens, and 1 green token: a

Determine the total number of distinct permutations of the 10 tokens.

Determine the number of distinct permutations of the 6 remaining tokens that do not include any of the orange tokens.

A librarian needs to arrange 12 books on a shelf. There are 5 identical mystery novels, 4 identical science fiction novels, 2 identical biographies, and 1 cookbook: a

Determine the total number of distinct permutations of the 12 books on the shelf.

If 3 mystery novels and 2 science fiction novels are sold, determine the number of distinct permutations of the remaining books.

Consider the number 11 223 344, which consists of 8 digits: a

Determine the number of distinct permutations of all 8 digits.

Determine the number of distinct permutations using only the odd digits.

This grid shown consists of unit squares: By travelling only right (R) or down (D) along the grid lines, how many paths are there from point A to point B?

A x y

B 11

A delivery robot in a warehouse navigates a grid of unit squares, moving only right (R) or down (D) along the grid lines. The grid represents a section of the warehouse floor.

How many paths can the robot take from the starting point X (top-left corner) to the destination point Y (bottom-right corner)?

Y 12 Ex 2

How many ways are there to arrange the letters A, B, C and D if C and D must always be adjacent? How does the answer change if, in addition, C must always come before D? In the word MACHINE, how many ways are there to order the letters if the vowels must be grouped together?

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How many different arrangements can be made from the letters of the word CHEESY if the two E’s must always be adjacent?

A group of 6 office workers must stand together for a photo: a

How many ways can they be lined up for the photo?

How many ways can they be lined up if two of the workers, Josh and James, stand beside one another?

How many ways can they be lined up if Josh and James refuse to stand next to each other?

Twenty people are selected to sit along a long banquet table. Of these 20 people, 5 are related and must sit together. How many ways can the people be arranged? Give your answer as a product of factorials.

After returning from her vacation, Sarah has 6 souvenirs to distribute among 6 different family members. The souvenirs include:

Ex 3

• 2 chocolate boxes (truﬄes and caramels) • 3 key rings (ocean, beach, and sunrise) • 1 greeting card a

Calculate the total number of different ways Sarah can distribute the 6 souvenirs among her family.

If the 2 chocolate boxes must be given to her two parents (in any order), how many different ways can she distribute the 6 souvenirs?

What is the total number of possible arrangements if the 2 chocolates are given to her parents and the 3 key rings are given to her three siblings, with both groups arranged in any order?

How many arrangements are possible if the truﬄes must go to her mom and the caramels must go to her step-dad, while the 3 key rings can be distributed freely among her three siblings?

At a school, there are 10 students. How many ways are there to arrange the students if Alice and Bob must be placed on the ends?

In how many ways can 4 tourists and 3 locals board a train, if everyone boards one at a time and the locals board before the tourists?

From a group of 6 boys and 4 girls, a president and a vice-president are to be chosen:

How many ways can this be done if there are no restrictions?

How many ways if the president must be a boy and the vice-president must be a girl?

How many ways if both must be boys?

Emily wants to form 4-letter codes using the letters of the word MISTER. Calculate the number of 4-letter codes Emily can form: a

If letters can be repeated

With at least one repeated letter

With no repeated letters

In how many ways can all the digits 1, 2, 3, 4, 5, and 6 be arranged to form a six-digit number such that the odd digits can only occupy even positions?

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Ex 4

If the letters of the word ORANGE are rearranged, how many different arrangements are possible if the vowels must alternate with consonants?

The digits 1, 2, 3, 4, and 5 must be arranged to form a 5-digit number. If the resulting number must be odd, how many possible numbers can be formed?

Using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, each no more than once, how many three-digit numbers can be formed if:

Ex 5

The number must be even

The first digit must be a 7 and the number must be odd

The digits must be 1, 5, 9 in any order

Ms. Lee needs to select 6 people from three different departments to make a presentation. The order of the presentations is important. There are: • 9 employees from the Marketing department • 6 employees from the Finance department • 10 employees from the Operations department

Determine how many possible ways she can form a presentation order if the first four presentation slots must be filled by members of the Marketing department and the last two presentations slots must be filled by members of the Finance department.

How many ways can Ms. Lee form a presentation order if the first and last presentations must be by members of the Operations department and the middle four slots can be filled with employees either from the Marketing and Finance departments?

In how many ways can 7 distinct guests be seated around a circular dining table?

Ex 6

Ex 7

A group of 7 friends are seated at a table with 7 seats. If two of the friends insist on sitting together, how many ways are there to seat the friends?

How many distinct ways can 7 different beads be arranged on a circular necklace if the necklace can be flipped over?

A committee of 3 boys and 3 girls is to be seated around a circular table:

Ex 8

In how many ways can the committee members be seated if there are no restrictions?

In how many ways can the individuals be seated if all the girls must sit together?

In how many ways can the group be seated if the boys and girls must alternate?

Extend your thinking 31

There are 6 boys who enter a boat with 8 seats, 4 on each side. John and Michael must sit on the port side. How many ways are there for the boys to be seated on the boat?

If the word MATHEMATICS is rearranged, how many different arrangements are possible if the vowels must be arranged in alphabetical order?

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Ms. Smith needs to select 8 students from three different year levels to perform in a school concert. The order of performances is important. There are: • 8 students from Year 10 • 7 students from Year 11 • 5 students from Year 12

Determine how many possible ways she can form a performance order if the first three performance slots will be filled by students from Year 10 and the last three slots will be filled by students from Year 11 and the middle two are filled by any students.

How many ways can Ms. Smith form a performance order if the first and last performances must be from Year 12 students, and the middle six slots can be filled with students from either Year 10 or Year 11?

How many different ways can a performance order be arranged if four Year 10 students perform first, followed by two Year 11 students, and then two Year 12 students?

In chess, a rook can move as many spaces as it wants forwards, backwards, left, or right, but not diagonally. A chessboard is 8 × 8 squares. In how many ways can 8 identical rooks be placed on a chessboard so that no two rooks attack each other?

Investigation: Pascal’s triangle Investigate online

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4.07

Combinations After this lesson, you will be able to… • define combinations and distinguish them from permutations • calculate combinations using the appropriate formula • prove simple combinatorial identities • solve problems involving combinations in various contexts

Combinations Exploration Currently, a school has a president and vice-president for the student council. The school is considering changing this structure to feature two co-presidents: 1. If Xavier was president and Yuri was vice-president, is this the same as Yuri being president and Xavier being vice-president? 2. If Xavier and Yuri were co-presidents, is this the same as Yuri and Xavier being copresidents? 3. If there were 5 students running for the two spots on student council, how would the number of possible student council set-ups compare for the president-vicepresident versus co-president models?

Recall that if the order matters for a context, it is a permutation. If the order does not matter, just the selection, then it is called a combination. Combination A selection of items from a collection where the order of selection does not matter. Example: Choose 3 fruits from a basket of 5 for a fruit salad, where the order of selection does not matter (e.g., apple, banana, cherry is the same as cherry, apple, banana).

A combination solely involves selecting or choosing the items. However, if the selected items are then arranged in a specific order, this becomes a permutation. A permutation can be thought of as an ordered version of a combination.

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Consider an example involving three distinct coloured dots: red, blue, and green. If two dots are to be selected from these three, the possible permutations of dots are as shown: Start with

The possible selections of dots have been identified. The next step is to determine if the order of the dots matters. For example, is a red dot paired with a green dot considered the same as a green dot paired with a red dot? If the answer is no and (red, green) is considered different to (green, red), then the order matters, and it is a permutation and the answer is 6.

Start with If the answer is yes, (red, green) and (green, red) are considered the same, then the order does not matter and so it is a combination and the answer would be 3.

The process of calculating the number of possible combinations is key to counting techniques in mathematics. For smaller sets, it is possible to list all the options and count them manually. However, as the sets grow larger, using manual counting becomes impractical. The process for choosing r items from n distinct items, where order is not important can be done using the combination formula:

is the number of items being selected

is the total number of items to choose from

It is important to note that in the term (n − r)!, the subtraction n − r is performed first, and then the factorial is applied to the result. The notation nCr can be read as “n choose r” and means the number of combinations of r items chosen from a set of n items. For example choose 2 items from a set of 5 distinct items.

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means that there are 10 ways to

Other notations that are used in combinations include C(n, r), nCr, nCr, and the same as nCr .

, which all mean

To determine nCr (the number of combinations of r items from n distinct items) using a calculator, follow these steps: 1. Turn on the calculator and ensure it is in the correct mode (standard or scientific). 2. Enter the total number of items n. 3. Press the combination function on the calculator. This is often labelled as nCr, or C, or

On many scientific calculators, this function is often a secondary option, meaning the Shift or 2nd button may need to be pressed first. 4. Enter the number of items to be chosen r. 5. Press the equals or execute button. The calculator will display the result of nCr.

Example 1 Evaluate the combinations: a

Create a strategy Substitute n = 7 and r = 2 into the combination formula

Apply the idea Write the combination formula Substitute n = 7 and r = 2 Evaluate the subtraction Expand the numerator to remove 5! Remove 5! from numerator and denominator Expand the denominator Evaluate the multiplication = 21

Evaluate

Reflect and check The combination function on a calculator can be used to verify this answer.

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Create a strategy Substitute n = 5 and r = 2 into the combination formula

Apply the idea Write the combination formula Substitute n = 5 and r = 2 Evaluate the subtraction Expand the numerator to remove 3! Remove 3! and expand 2! Evaluate the multiplication = 10

Evaluate

Create a strategy Substitute n = 8 and r = 4 into the combination formula n Cr

n! . r !( n r )!

Apply the idea n

n! r !( n r )!

Write the combination formula Substitute n = 8 and r = 4 Evaluate the subtraction

8 7 6 5 4! 4 !(4 3 2 1)

Expand the numerator to remove 4! and expand the other 4! in the denominator Remove 4! Evaluate the multiplication

= 70

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C5 then compare this to 7C2

Create a strategy Substitute n = 7 and r = 5 into the combination formula part (a).

. Then compare the result to

Apply the idea Write the combination formula Substitute n = 7 and r = 5 Evaluate the subtraction Expand the numerator to remove 5! and expand 2! Remove 5! Evaluate the multiplication = 21

Evaluate

From part (a), 7C2 = 21. Comparing this, 7C5 = 21, so 7C5 = 7C2 . This is an example of the property n Cr = nCn−r, since 7C5 = 7C7−5 = 7C2 . Choosing 5 items from 7 is the same as choosing 2 items to leave out.

Did you know?

Combinations help count the number of different card hands that can be formed from a standard deck. Since the order of cards is irrelevant, each hand is treated as a unique combination. This principle is fundamental in probability and game analysis.

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Example 2 In a lottery game, 3 balls are drawn from a set of 10 balls, numbered 1 to 10. A prize is won if the numbers on a ticket match the 3 numbers drawn. The order in which the balls are drawn does not matter. How many different combinations of 3 numbers are possible?

Create a strategy Since the order of the balls does not matter, this is a combination problem. A direct calculation of permutations would overcount the unique sets of numbers. The strategy involves three main steps: 1. First, calculate the total number of permutations (ordered arrangements) of drawing 3 balls from 10 using the formula 10P3. 2. Next, determine the number of ways any single group of 3 balls can be arranged. This will reveal how many times each unique combination has been overcounted. This is calculated using 3!. 3. Finally, divide the total number of permutations by the number of arrangements for each combination to find the true number of unique combinations. This process is encapsulated by the combination formula, nCr.

Apply the idea Step 1: Calculate the total number of permutations If order mattered, the number of ways to arrange 3 balls from 10 would be a permutation. 10

P3 = 10 × 9 × 8

Calculate the number of ordered arrangements

= 720 Evaluate the multiplication There are 720 possible permutations. Step 2: Determine the overcounting factor The permutation calculation counts every different ordering of the same three numbers as a separate outcome. For example, if the balls drawn are 5, 6, and 7, the 720 permutations count all of the following as distinct: (5, 6, 7), (5, 7, 6), (6, 5, 7), (6, 7, 5), (7, 5, 6), and (7, 6, 5). In the lottery, all of these are the same winning combination. The number of ways to arrange any group of 3 balls is 3!. 3! = 3 × 2 × 1

Write the equation

= 6 Evaluate This means each unique combination of 3 balls has been counted 6 times.

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Step 3: Calculate the number of unique combinations To find the actual number of combinations, divide the total permutations by the number of arrangements for each set. This is formally done using the combination formula. Write the relationship between combinations and permutations Substitute the calculated values = 120

Evaluate the division

Alternatively, using the full combination formula: Write the equation Evaluate the subtraction Simplify the expression by removing 7! = 120

Evaluate

There are 120 different combinations of 3 numbers that can be drawn.

Did you know?

Combinatorics can tell you how many ways you can arrange pizza toppings! With endless choices like cheese, pepperoni, olives, and veggies, combinatorics helps you count all the delicious flavour combinations you can create — just like the variety of pizzas you see here!

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Example 3 A school committee of 5 students is to be chosen from a group of 8 boys and 6 girls. How many different committees can be formed if the committee must consist of exactly 3 boys and 2 girls?

Create a strategy This problem involves two independent selections: choosing the boys and choosing the girls: • Calculate the number of ways to choose 3 boys from 8 using combinations. • Calculate the number of ways to choose 2 girls from 6 using combinations. • Use the multiplication principle to find the total number of ways to form the committee by multiplying the results from the two steps.

Apply the idea Step 1: Choose 3 boys from 8. The number of ways to choose 3 boys from 8 is 8C3.

There are 56 ways to choose the boys. Step 2: Choose 2 girls from 6. The number of ways to choose 2 girls from 6 is 6C2.

There are 15 ways to choose the girls. Step 3: Total number of ways to form the committee. Using the multiplication principle, the total number of different committees is the product of the number of ways to choose the boys and the number of ways to choose the girls. Total committees = (Ways to choose boys) × (Ways to choose girls) 8

= C3 × C2

Write the formula Substitute the values

= 56 × 15 Evaluate the combinations = 840

Evaluate

Therefore, there are 840 different committees that can be formed.

Example 4 Digital only example Explore in the online lesson

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mathspace.co

Example 5 Write an algebraic expression for aCb assuming a and b are non-negative integers and a ≥ b.

Create a strategy replacing n with a and r with b.

Use the combination formula

Apply the idea Substituting n = a and r = b into the formula:

This is the algebraic expression for aCb.

Reflect and check This expression is fundamental in various algebraic proofs and identities involving combinations. The conditions a ≥ b and that a, b are non-negative integers ensure the factorials are well-defined.

Example 6 Digital only example Explore in the online lesson

mathspace.co

Idea summary Combinations involve selecting or choosing a number of items from a set, where the order of the items is not important. The combinations formula is:

is the number of items being selected

is the total number of items to choose from

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Properties of combinations There are some properties of combinations that can help build a deeper understanding of combinatorics.

Exploration Consider the combinations 6C2 and 6C4 : 1. Show the working for calculating each combination. What stands out? 2. Interpret the two combinations. Explain how they are related. 3. Write a combination that would be equivalent to 7C2. Explain your answer. 4. Write a combination that would be equivalent to nC(n-r). Explain your answer.

First, it can be shown that

This means, for example that 12C4 = 12 C8 and 7C1 = 7C6.

Choosing 4 items from 12 (like the 4 blue balls) is equivalent to choosing 8 items to leave out (the 8 red balls). Therefore, 12C4 = 12C8 .

There is a reason behind this clear symmetry. This property arises because choosing r items from n items and leaving out (n − r) items from the set, is the same as choosing (n − r) items to exclude from a set of n item and keeping the remaining r items. Another property is that nC0 = 1. This makes sense contextually, because if no items are selected, there is only one way to do that. To show this is true, recall the definition 0! = 1. Applying the formula shows that This means, for example, 5C0 = 1 and 12C0 = 1. This also means that nCn = 1. Another useful property is nC(n−1) = n. This arises from the fact that selecting n − 1 items from a set of n is equivalent to choosing one item to exclude, and there are n ways to exclude one item.

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Finally, there is a relationship between nCr and the previous nPr. Contextually, a permutation is selecting and then ordering, while a combination is just selecting. This means that dividing out all the duplicates is needed if order is ignored. This gives algebraically:

This can also be shown

Example 7 Determine the value of x: a

C6 = 15Cx

Create a strategy Use the property: nCr = nCn−r

Apply the idea Using the property nCr = nCn−r, 15C6 gives n = 15 and r = 6. To solve for x, write the equation of the part equivalent to the missing value:

x=n−r

Equate x and n − r

= 15 − 6

Substitute the values

Evaluate

Cx = 1

Create a strategy Use the property: nC0 = nCn = 1

Apply the idea For the number of selections to equal 1, either 0 items or all n items must be selected. Therefore, x = 0 or x = 5.

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Example 8 Prove algebraically that nC1 = n for n ≥ 1.

Create a strategy . Substitute r = 1 and simplify the expression. Recall

Use the combination formula that 1! = 1.

Apply the idea Apply the combination formula with r = 1 Since 1! = 1 Expand n! as n × (n − 1)! =n

Remove (n − 1)! from numerator and denominator (since n ≥ 1, (n − 1)! is defined)

Therefore, nC1 = n.

Reflect and check This identity makes contextual sense: the number of ways to choose 1 item from a set of n distinct items is simply n, as each item can be chosen individually

Example 9 Digital only example Explore in the online lesson

mathspace.co

Idea summary These properties may be helpful in justifying or calculating combinations: • • •

Cr = nCn−r C0 = nCn = 1 n C(n−1) = n n

•

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4.07 Practice questions What do you remember? 1

For each scenario, determine whether it involves a permutation or a combination. Explain your reasoning briefly: a

Selecting 3 distinct flavours of ice cream for a triple-scoop cone from 10 available flavours

Awarding gold, silver, and bronze medals to the top three athletes in a race with 8 athletes

Choosing 4 books to read on holiday from a pile of 12 books

For each scenario, write an expression using combination notation (nCr) that represents the number of possible selections. There is no need to calculate the final value: a

A teacher needs to select a 4-person student committee from a class of 12 students.

The number of ways to form a triangle by choosing 3 vertices from a set of 10 available points on a plane, assuming no three points are collinear.

In a lottery game, 6 winning numbers are drawn from a pool of 49 distinct numbers.

A book club has 8 distinct fiction books and 5 distinct non-fiction books to choose from this month. For each scenario, write an expression using combination notation (nCr) for the number of ways the books can be chosen. There is no need to calculate the final value: a

Choosing 3 books from the fiction selection.

Choosing 2 books from the non-fiction selection.

Choosing 4 books from the entire collection of these books.

Choosing all 5 of the non-fiction books.

Practice 4 Ex 1

Evaluate: a

Ex 2

C10

100

C99

Simplify: a

Cn−2

n+1

n+2

A gardener has to plant 3 types of flowers from a selection of 5 types. In how many ways can he choose the flowers to plant?

A fruit basket contains 8 different fruits. How many ways can 3 fruits be selected to make a fruit salad?

A pizza shop offers 7 toppings. How many different pizzas, each with exactly 2 distinct toppings, can be created?

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A dessert shop offers 6 different flavours of ice cream. Determine the number of combinations that can be selected if a customer purchases: a

10 Ex 3

2 flavours

3 flavours

4 flavours

5 flavours

A student council needs to form a subcommittee consisting of 2 junior members and 3 senior members. If there are 6 eligible junior members and 8 eligible senior members, how many different subcommittees can be formed?

From a standard 52-card deck, how many different 4-card hands can be formed?

Write the general algebraic expression for nCr, stating the conditions on n and r for the expression to be valid.

Ex 4

Ex 5

A team of 6 people is to be selected from a group of 10 athletes. How many different combinations of teams can be formed?

During a school field trip, 5 students out of 12 need to be selected for a quiz competition. How many possible selections are there?

A photographer has to select 5 photos out of 10 to include in his portfolio. In how many ways can he make his selection?

Twelve students are divided into two teams, Team A and Team B, each consisting of 6 students. Determine the number of ways this can be done.

In a chess tournament, there are 10 players. Each player plays against every other player exactly once. How many different matches will be played in the tournament?

Determine the value(s) of y:

Ex 6

Ex 7

Cy = 14C3

Cy = 1

C6 = yC14

C6 = 7

Prove algebraically that nCn−1 = n for n ≥ 1.

Ex 8

20 Ex 9

A sports club with 12 members needs to select a group of 4 members to attend a workshop: a

In how many ways can the group of 4 members be selected?

Alternatively, in how many ways can the 8 members who will not attend the workshop be selected?

Explain the relationship between the answers to part (a) and part (b) using a property of combinations.

Show that nCr = nCn−r.

Show that each identity is true for any whole numbers, r and n, where 0 ≤ r ≤ n: a

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Cn = 1

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n+1

Cr = nCr + nCr−1

Extend your thinking 23

At a networking conference, every person shakes hands with every other person. Altogether there are 105 handshakes. How many people are at the conference?

How many ways can a group of 8 letters be selected from the English alphabet (assuming 5 vowels and 21 consonants, all distinct) if each group must consist of exactly 3 vowels and 5 consonants?

How many ways are there to form a triangle in the grid using only three points?

Prove that n−2 Ck−2 × n × (n − 1) = nCk × k × (k − 1), where 2 ≤ k ≤ n.

Show that this identity holds: (nCr+s)(r+sCr) = (nCr)(n−rCs) where n, r and s are positive integers and n ≥ r + s.

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4.08 Combinations with restrictions After this lesson, you will be able to… • apply combination principles to situations with restrictions • solve problems involving ‘at least’ and ‘at most’ conditions using combinations

Combinations with the multiplication and addition principles The notation nCr and its formula are used to determine combinations where the order of items does not matter. The concept of combinations can be extended to more complex scenarios involving the selection of subsets under additional constraints or specific rules, often called restrictions. The addition principle can be used to sum the number of combinations for cases that have no intersection or overlap. In cases where multiple independent choices are involved, the multiplication principle can be applied to calculate the total number of possible combinations.

Exploration Imagine having a group of 10 friends, and needing to choose 3 of the group members to go to a concert. This scenario represents a standard combination problem. Here are some common types of restrictions: • A specific friend, Alex, must be chosen. • A specific friend, Beth, must not be chosen. • You must choose either Charles or Diana, but not both. Consider these questions: 1. How does requiring Alex to be chosen change the number of friends still needed for selection and the pool of friends from which selection occurs? 2. If Beth must not be chosen, how does this affect the total pool of friends available for selection? 3. For the “either Charles or Diana” restriction, consideration of how this problem might be broken into separate cases is helpful.

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Example 1 A selection of 4 people is to be chosen from a group of 9 people:

Oldest

Youngest

How many selections are possible if the youngest person or the oldest person is included, but not both?

Create a strategy The condition “the youngest or oldest is included, but not both” requires splitting the problem into two separate cases. The total number of selections will be the sum of the selections from each case. • Case 1: The youngest is included, and the oldest is excluded. • Case 2: The oldest is included, and the youngest is excluded.

Apply the idea Case 1: Youngest included, oldest excluded One position in the group of 4 is already filled by the youngest person. This leaves 3 more positions to fill. The pool of available people is reduced by two (the youngest, who is already chosen, and the oldest, who is excluded). This leaves 9 − 2 = 7 people to choose from. The number of ways to choose the remaining 3 people from the available 7 is:

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Case 2: Oldest included, youngest excluded This scenario is symmetrical to Case 1. One position is filled by the oldest person, and the remaining 3 must be chosen from the other 7 people. The number of ways is also:

Total selections: Apply the addition principle to combine the results of the two separate cases: Total ways = Ways (Case 1) + Ways (Case 2)

Write the equation

= 35 + 35

Substitute the values

= 70

Evaluate

There are 70 possible selections that meet the condition.

Example 2 A team of 3 is to be chosen at random from a group of 5 Year 11 students and 6 Year 12 students. In how many ways can the team be chosen if: a There are no restrictions.

Create a strategy There are 5 + 6 = 11 possible students to choose from, and 3 of them need to be selected. Substitute n = 11, and r = 3 into

Apply the idea Number of ways = 11C3

Write the notation Write the combination formula Substitute n = 11 and r = 3 Evaluate the subtraction Expand the numerator Remove 8! Expand the denominator

= 165

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Evaluate

b There must be more Year 12 students than Year 11 students.

Create a strategy Use the fact that the only cases in which more Year 12 students than Year 11 students are chosen are: • No Year 11 students and all Year 12 students • 1 Year 11 student and 2 Year 12 students Determine the number of ways each of these cases can happen, and add the results. Remember that the word “and” usually implies multiplication.

Apply the idea To determine the number of ways that all three places on the team are filled by Year 12 students, consider that 3 Year 12 students must be chosen from 6 Year 12 students and 0 Year 11 students must be chosen from 5 Year 11 students. This translates to 6C3 × 5C0 = 20 × 1 = 20. To determine the number of ways that 1 Year 11 student and 2 Year 12 students can be chosen, consider that 2 Year 12 students must be chosen from 6 Year 12 students and 1 Year 11 student must be chosen from 5 Year 11 students. This translates to 6C2 × 5C1 = 75. Number of ways

Write the expression using the addition and multiplication principles = 20 + 75

Substitute the values for each case

= 95

Evaluate

There are 95 possible ways to form the team with more Year 12 students than Year 11 students.

Idea summary The concept of combinations can be extended to more complex scenarios involving the selection of subsets under additional constraints, known as restrictions, and specific rules. The addition principle is used to sum the number of combinations for cases that have no intersection or overlap (separate scenarios that cannot happen at the same time). The multiplication principle can be applied to cases where multiple independent choices are involved (often when selecting items from different groups).

Combinations with restrictions Other common restrictions in combination problems include conditions like ‘at least’ or ‘at most’. While these restrictions still utilise a combination of the multiplication and addition rules, additional reasoning is required to determine the appropriate combinations. These types of restrictions frequently appear in problems where selections are made from distinct groups or categories of items, requiring careful application of combination principles to each group.

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To solve problems with these types of conditions: 1. Identify and list all the scenarios that satisfy the restrictions. 2. Use the combination formula and multiplication principle to calculate the number of combinations for each scenario. 3. Add the results from each scenario to attain the total number of combinations. The cases which satisfy the condition can be added up, or the cases that do not satisfy the condition can be subtracted from the total unrestricted number of combinations (this is known as complementary counting). Recall that combination notation includes nCr, but for more involved contexts, commonly used. Both represent the same calculation.

is more

Example 3 Five Year 11 students and 6 Year 10 students are part of the debate team. Four of them must represent the school at the upcoming debate tournament. In how many ways can the team of 4 be formed if: a It must contain 2 Year 11 students and 2 Year 10 students.

Create a strategy An “and” statement means multiplying the results of choosing 2 Year 11 students from the 5 Year 11 students total and choosing 2 Year 10 students from the 6 Year 10 students total.

Apply the idea Choosing 2 students from the 5 in Year 11 translates to 5C2. Choosing 2 students from the 6 in Year 10 translates to 6C2. 6

C2 × 6C2 = 10 × 15 = 150

Calculate each combination separately: 5C2 = 10, 6C2 = 15 Evaluate using multiplication principle

There are 150 possible ways to form the team. b It must contain at least 1 Year 11 student and 1 Year 10 student.

Create a strategy Since the team must contain at least 1 Year 11 student and at least one Year 10 student, these steps are taken: 1. List all the scenarios (combinations of Year 11 and Year 10 students that sum to 4 team members) that satisfy the condition of having at least 1 Year 11 student and at least 1 Year 10 student. 2. Calculate the number of combinations for each scenario using the multiplication principle. 3. Add the results from each scenario for the total (addition principle).

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Apply the idea There are three scenarios that satisfy the conditions of at least 1 Year 11 student and at least 1 Year 10 student. Each scenario along with its calculation of number of combinations is listed: • Selecting 1 Year 11 student and 3 Year 10 students: 5

C1 × 6C3 = 5 × 20 = 100

• Selecting 2 Year 11 students and 2 Year 10 students: 5

C2 × 6C2 = 5 × 15 = 150

• Selecting 3 Year 11 students and 1 Year 10 student: 5

C3 × 6C1 = 10 × 6 = 60

The total number of combinations is the sum of combinations from each scenario: = 100 × 150 + 60

Sum the combinations from the three valid scenarios

= 310

Evaluate

There are 310 ways of selecting a team with at least 1 Year 11 student and at least 1 Year 10 student.

Reflect and check Another way of calculating this is using complementary counting. There are 5 + 6 = 11 students in total from whom to select 4 students. The total number of unrestricted selections is 11C4. 11

C4 = 330

Evaluate using technology

The cases that do not satisfy “at least 1 Year 11 student and at least 1 Year 10 student” are: • All 4 are Year 11 students (0 Year 10): Selecting 4 of 5 Year 11 students can be translated as 5 C4 = 5 • All 4 are Year 10 students (0 Year 11): Selecting 4 of 6 Year 10 students can be translated as 6 C4 = 15 The sum of these undesired cases is: 5

C4 + 6C4 = 5 + 15

Write the equation

= 20

Evaluate

Subtracting the undesired cases from the total number of ways: Number of ways =

Subtract the sum of undesired cases from the total unrestricted combinations

= 330 − 20 Substitute the combination results = 310 Evaluate This confirms the previous result of 310 ways.

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c It must contain at most 1 Year 11 student.

Create a strategy Consider the scenarios where there is at most one Year 11 student. This means either 0 Year 11 students or 1 Year 11 student. Calculate the individual number of combinations for each scenario and add the results.

Apply the idea The scenarios and number of combinations that satisfy at most 1 Year 11 student: • Selecting 1 Year 11 student and 3 Year 10 students: 5

C1 × 6C3 = 5 × 20 = 100

• Selecting 0 Year 11 students and 4 Year 10 students: 5

C0 × 6C4 = 1 × 15 = 15

The total number of combinations is 100 + 15 = 115.

Example 4 Consider the word MATHEMATICS: a If each of the 11 letter positions in the word MATHEMATICS is treated as distinct, how many ways are there to choose a group of 4 positions?

Create a strategy First, count the total number of letters in MATHEMATICS. Then, consider what “choosing 4 letters” implies in the context of repeated letters.

Apply the idea The word MATHEMATICS has 11 letters. If 4-letter positions were being selected from these 11 positions, without considering if the letters themselves are identical, the number of ways would be: Write the formula = 330

Evaluate using a calculator

b How does the answer to part (a) change if no letter can be chosen more than once (that is, 4 distinct letters must be chosen)?

Create a strategy The question asks for the number of ways to choose 4 letters where no letter is chosen more than once. This means selecting 4 distinct letter types. First, identify the unique letter types available in MATHEMATICS. Then, determine how many ways 4 of these distinct letter types can be chosen.

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Apply the idea The word MATHEMATICS contains these letters: M, A, T, H, E, M, A, T, I, C, S The distinct letter types are M, A, T, H, E, I, C, S. There are 8 unique letter types. 4 distinct letters need to be chosen from these 8 unique letter types. The number of ways to do this is: n r

n r n r

Write the combination formula Substitute n = 8 and r = 4

Evaluate the subtraction in the denominator

Expand the factorial

= 70

Evaluate

So, there are 70 ways to choose 4 distinct letters from the word MATHEMATICS.

Idea summary When solving combination problems, it is helpful to understand the types of restrictions that affect how selections are made and counted. Common restrictions include: • At least condition: Requires including a minimum number of specific items. • At most condition: Limits the number of specific items in the subset. To calculate the total number of combinations with restrictions: 1. Identify all scenarios (cases) that satisfy the restrictions. 2. Calculate the individual combinations for each scenario, often using the multiplication principle if selecting from different groups. 3. Sum the combinations from all valid scenarios using the addition principle for the total. Alternatively, complementary counting can be used by subtracting the number of undesired outcomes from the total number of unrestricted outcomes.

4.08 Practice questions What do you remember? 1

escribe one key similarity between permutations and combinations in the context of D selecting items from a set.

Explain the fundamental difference between permutations and combinations.

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Write the formula for calculating the number of combinations of choosing r items from a set of n distinct items.

Calculate the value of these combinations: a

When is the multiplication principle for counting used in combination problems?

Explain the difference between a problem asking for “at least 2 items” versus “at most 2 items” in the context of combinations.

Practice 6 Ex 1

A company needs to select 2 representatives from a group of 10 employees. If either the CEO or the CFO must be selected, but not both, how many ways can the representatives be chosen?

From a group of 8 people, how many ways can a committee of 3 be chosen if Sarah, one of the 8 people, must be on the committee?

A sports team of 11 players needs to be chosen from 15 available players. If one specific player, John, is injured and must not be chosen, how many different teams can be formed?

How many ways are there to choose 5 people from a group of 12 if the youngest must always be included?

How many ways can 2 students be chosen from a group of 6 to form a class leadership team, if exactly one of the two oldest students is on the team?

An artist has access to 10 buckets of paint where each bucket is a unique colour. The artist must mix 4 colours together. How many ways can she do this if she cannot mix green and red?

A selection of 5 players is to be chosen from a squad of 7 forwards and 9 defenders. How many ways can the team be selected if there must be more defenders than forwards?

A factory has 25 machines, 5 of which are faulty. How many ways can 5 machines be selected such that at most 1 of them is faulty?

There are 12 people to be selected from 20 candidates for jury duty. The candidates are from these age groups: Age group

18 − 25

26 − 35

36−60

60+

Number of people

How many ways can the jury be selected if at most 1 person aged 18−25 can be selected?

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16 Ex 2

In a class of 12, there are 7 students wearing red shirts and 5 wearing blue shirts: a

How many ways can 5 students be chosen so that exactly 2 students are wearing blue shirts?

How many ways can 5 students be chosen so that at least 3 students are wearing red shirts?

A group of 7 senior students and 4 intermediate students are part of a school’s mathematics team. A team of 4 is to be sent to a regional competition. How many ways can this team be formed if: a

There are no restrictions

The team must contain an equal number of senior and intermediate students

The team must contain at most 3 senior students

An art gallery has a total of 10 paintings by a particular artist. Five of these paintings are oil paintings and 5 are watercolour paintings. The art gallery is asked to provide a special exhibition of this artist’s work but is restricted to showing only 5 paintings. Calculate the number of ways in which the 5 paintings can be selected for the exhibition if:

There are no restrictions

One particular painting must be included

There must be exactly 3 watercolour paintings

There must be more watercolour paintings than oil paintings

An ice cream shop offers 10 flavours, which customers can combine into a milkshake: a

How many 3-flavour milkshakes could they put on their menu? Assume that double flavours are not allowed.

How many total milkshakes are possible, with between 1 and 10 combined flavours (but not double flavours) allowed?

An office consists of 10 employees. Three of them are managers and the remaining 7 are workers. How many ways can 5 people be chosen to work on a project if either 1 or 2 of the team members must be managers?

A student is to select 4 books from a shelf containing 10 different novels and 6 different non-fiction books. How many different combinations of books can the student choose if 2 books must be non-fiction?

Ex 3

A fruit basket needs to contain 2 different types of apples and 3 different types of oranges. If there are 5 types of apples and 6 types of oranges available, how many different combinations of fruit baskets can be made?

To promote reading, a teacher decides to feature 3 classics, 4 contemporary novels and 2 non-fiction books at the local library. How many selections can she make from 5 classics, 6 contemporary novels and 4 non-fiction books?

A committee of 4 people is to be selected from a group of 12 candidates, where 5 candidates are from the finance department and 7 candidates are from the marketing department. How many different combinations of committees can be formed if 2 members must be from the finance department? 4.08 Combinations with restrictions mathspace.co

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A teacher has 12 students and wants to select a committee of 3 students. Of the 12 students, 4 were on the committee last year. How many different committees can be formed, if exactly one of the 4 experienced students is on the team?

Calculate the number of ways in which: a

5 children can be divided into groups of 2 and 3

9 children can be divided into groups of 5 and 4

9 children can be divided into groups of 2, 3 and 4

From a standard 52-card deck, a 5-card hand is dealt. How many hands contain exactly 2 aces and exactly 2 kings?

A bag contains 11 Scrabble tiles spelling the word MISSISSIPPI:

Ex 4

If each of the 11 tiles is considered a distinct object, how many ways can a group of 5 tiles be chosen from the bag?

How many ways can a group of 3 distinct letter types be chosen from the word MISSISSIPPI?

Extend your thinking 28

A baker living in Victoria only sells sesame and poppy bagels. On any given day, he has exactly 10 distinct sesame bagels and 10 distinct poppy bagels to sell. How many ways can a student purchase 6 bagels if her purchase must contain at least 1 poppy bagel?

Thirty university students recently completed an exam and the results are shown in this table: Exam results

0 − 20%

21 − 40%

41 − 60%

Number of students

61 − 80% 81 − 100% 7

The professor would like to interview 3 of the students to elicit feedback. However, she would like no more than 2 students to be from the same exam result category. In how many ways can she choose the students to interview? 30

A student needs to choose 1 history book from 4 options and 1 science book from 3 options. How many different pairs of books can the student choose?

How many ways can 9 different books be distributed among 3 students if the first student gets 2 books, the second gets 3, and the third gets 4?

A student council has 10 members. The council need to form a subcommittee of 3 members. One member, Alice, says she will only be on the subcommittee if another member, Bob, is also on it. Bob is willing to be on the subcommittee regardless of Alice. How many different subcommittees can be formed under this condition?

A mathematician uses some basic algebra to prove that interpretation that

gives the number of ways to select k objects from n objects, explain

this result intuitively. 378

. Using the

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4.09 Combinatorics mixed problems After this lesson, you will be able to… • solve mixed counting problems with permutations and combinations • determine appropriate methods for problems with multiple restrictions • model real-world scenarios using combinatorial techniques

Mixed counting problems with restrictions Permutations and combinations are two fundamental counting principles used to determine the number of possible outcomes in complex scenarios. The choice between them depends entirely on the context of the problem: permutations are used for ordered arrangements, while combinations are used for unordered groups. Keywords can help in deciding whether to use permutations or combinations. For permutations, keywords include arrange, order, position, sequence, or specific roles being assigned.

is the total number of items

is the number of items being arranged

For combinations, keywords include select, choose, group, committee, or scenarios where the internal order of the chosen items does not create a new outcome.

is the total number of items

is the number of items being selected

To solve complex counting problems involving a sequence of choices, the multiplication principle (or fundamental counting principle) is used. This principle states that if a task can be broken down into sequential stages, the total number of outcomes is the product of the number of choices at each stage. For example, if a first stage can be done in m ways and a second stage can be done in n ways, the total number of ways to complete the task is m × n. The addition principle is used when there are mutually exclusive choices. If there are m ways for one event to occur and n ways for a second, mutually exclusive event to occur, then there are m + n ways for either event to occur.

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Exploration Consider the difference between situations where order is important and where it is not. A club is selecting students for a project. There are 4 Year 10 students (Y 10A, Y 10B, Y 10C, Y 10D) and 2 Year 11 students (Y 11A, Y 11B). A team of 3 students needs to be chosen from all 6 students: 1. If a group of 3 students is just being formed, does the order of selection change the resultant group? For example, is picking Y 10A, then Y 10B, then Y 11A a different group from picking Y 11A, then Y 10A, then Y 10B? List some possible groups of 3. 2. How many different groups of 3 students can be selected from the 6 students? Which counting method (permutation or combination) would be used here? 3. Suppose the roles in the team matter - selection involves a leader, a recorder, and a presenter from the 6 students. If Y 10A is leader, Y 10B is recorder, and Y 11A is presenter, is this different from Y 11A being leader, Y 10A recorder, and Y 10B presenter? List some possible role assignments for a team of 3. 4. How many different arrangements of 3 students into these specific roles are possible from the 6 students? Which counting method would be used here? 5. Compare the answers and reasoning for question 2 and question 4. When did the order of selection or assignment become important?

Counting problems can be further complicated by restrictions. Common types of restrictions and strategies to handle them include: • Items grouped together: Treat the group of items that must stay together as a single unit. Arrange this unit along with other individual items. Then, consider the number of ways the items within the group can be arranged among themselves. Multiply these counts using the multiplication principle. • Items separated: To ensure certain items are never together, first arrange the items that do not have restrictions. Then, place the restricted items in the available spaces between (or at the ends of) the unrestricted items. • At least one or at most one: Problems asking for ‘at least one’ of a particular item are often best solved using complementary counting. Calculate the total number of arrangements/ selections without any restrictions, and then subtract the number of arrangements/selections where the condition is not met (for example, ‘none’ of the item). For ‘at most one’, sum the cases of ‘none’ and ‘exactly one’. • Identical items (repetitions): If there are identical items within the set being arranged, the standard permutation formula n! overcounts. If n items in total, with n1 identical items of type 1, n2 identical items of type 2, ..., nk identical items of type k, the number of distinct permutations is

. For combinations involving sets with repeated items, direct

application of nCr can be complex, and often case-by-case analysis is needed if selections involve these repeated items. • Fixed positions or alternating patterns: If certain items must be in specific positions, or if items must alternate (for example, boy-girl-boy), place the restricted items first according to the condition, then arrange the remaining items in the remaining positions. 380

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Example 1 Digital only example Explore in the online lesson

mathspace.co

Example 2 Four letters are chosen from the word BANNER. How many selections of 4 letters have at least one N?

Create a strategy This problem asks for selections with ‘at least one N’ from a word with repeated letters (two Ns). The most reliable method is to consider the mutually exclusive cases and then add them up. The letters available are B, A, E, R, and two Ns. A selection of four letters can have either one N or two Ns. • Case 1: Calculate the number of selections with exactly one N. • Case 2: Calculate the number of selections with exactly two Ns. The total number of selections will be the sum of the results from Case 1 and Case 2. Since this is a selection, order does not matter, so combinations will be used.

Apply the idea The word BANNER has the distinct letters {B, A, E, R} and two identical Ns. A group of 4 letters must be formed. The two possible cases for having at least one N are analysed. Case 1: The selection includes exactly one N. First, select one N. Since the Ns are indistinguishable in a selection, there is only 1 way to do this. Then, the remaining 3 letters must be selected from the 4 other distinct letters {B, A, E, R}. The number of ways to choose 3 letters from 4 is:

Case 2: The selection includes exactly two N’s. First, select both Ns. There is only 1 way to do this. Then, the remaining 2 letters must be selected from the 4 other distinct letters {B, A, E, R}. The number of ways to choose 2 letters from 4 is:

Total selections: Using the addition principle, the total number of selections with at least one N is the sum of the ways from both cases. Total selections = Ways from Case 1 + Ways from Case 2

Write the equation

=4+6

Substitute the values

= 10

Evaluate

There are 10 selections of four letters that have at least one N.

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Reflect and check An alternative method for ‘at least one’ problems is complementary counting: (Total Selections) (Selections with no Ns). This method can be tricky in this case. To find the selections with no Ns, 4 letters are chosen from the 4 distinct letters {B, A, E, R}. This is 4 C4 = 1 way (the group {B, A, E, R}). If the total selections are calculated by treating all 6 letters as distinct {B, A, N₁, N₂, E, R}, the result is 6C4 = 15. Using this, the complementary method gives 15 − 1 = 14 selections. This answer of 14 is different from the correct answer of 10. The discrepancy arises because 6 C4 treats the two Ns as distinct items. For example, it counts {B, A, N₁, E} and {B, A, N₂, E} as two different selections. However, in a problem asking for a “selection of letters,” the two Ns are typically considered indistinguishable. Therefore, the direct counting method by cases, which gives 10, is the correct interpretation for this problem.

Example 3 A librarian has 4 distinct history books, 3 distinct science books, and 2 distinct math books: a In how many ways can these 9 books be arranged on a shelf if all the history books must be kept together?

Create a strategy This problem involves an arrangement with a restriction: the history books must be grouped together. Step 1: Treat the 4 history books as a single block or unit. This ‘history block’ is then arranged along with the 3 science books and 2 math books. Step 2: Determine the number of ways to arrange this block and the other individual books. Step 3: Determine the number of ways the 4 history books can be arranged within their block. Step 4: Use the multiplication principle to find the total number of arrangements.

Apply the idea Step 1: Consider the 4 history books as one unit (H). The items to arrange are (H), S1, S2, S3, M1, M2. This means there are 1 + 3 + 2 = 6 items (or units) to arrange. Step 2: The number of ways to arrange these 6 items is 6!. 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 Step 3: The 4 distinct history books can be arranged within their block in 4! ways. 4! = 4 × 3 × 2 × 1 = 24 Step 4: By the multiplication principle, the total number of ways to arrange the books with the history books together is the product of the arrangements from step 2 and step 3. Total arrangements = 6! × 4!

Write the equation

= 720 × 24

Substitute the values

= 17 280

Evaluate

There are 17 280 ways to arrange the books on the shelf with all the history books kept together.

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b If the 9 books are arranged randomly on the shelf, what is the probability that all the history books are kept together?

Create a strategy Probability is calculated by dividing the number of favourable outcomes by the total number of possible outcomes: • The number of favourable outcomes (arrangements with history books together) was calculated in the previous part. • The total number of possible outcomes is the total number of ways to arrange the 9 distinct books without any restrictions.

Apply the idea From part (a), the number of favourable outcomes where all history books are together is 17 280. Next, the total number of possible arrangements for the 9 distinct books is found. This is calculated as 9!. Total arrangements = 9! = 362 880 Now, the probability can be calculated. Write the formula Substitute the values Evaluate The probability that the history books are kept together is approximately 0.0476, or about 4.76%.

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Example 4 A 7-digit number is to be formed by arranging the distinct digits 1, 2, 3, 4, 5, 6, 7 without repetition. How many ways can the 7-digit number be formed if odd numbers and even numbers must alternate?

Create a strategy Identify the odd and even digits available. Determine the possible alternating patterns based on the number of odd and even digits. Calculate the arrangements for each group (odd and even digits) and use the multiplication principle. The available digits are 1, 2, 3, 4, 5, 6, 7. • Odd digits: 1, 3, 5, 7 (4 odd digits). • Even digits: 2, 4, 6 (3 even digits). For a 7-digit number with alternating odd and even digits, and given there are more odd digits, the pattern must start and end with an odd digit: O E O E O E O

Apply the idea The pattern for the 7-digit number must be: Odd, Even, Odd, Even, Odd, Even, Odd There are 4 positions for the odd digits and 3 positions for the even digits. Step 1: Arrange the 4 odd digits (1, 3, 5, 7) in the 4 odd-designated positions. The number of ways to do this is 4!. Ways to arrange odd digits = 4! = 4 × 3 × 2 × 1 = 24 Step 2: Arrange the 3 even digits (2, 4, 6) in the 3 even-designated positions. The number of ways to do this is 3!. Ways to arrange even digits = 3! = 3 × 2 × 1 = 6 Step 3: By the multiplication principle, the total number of 7-digit numbers that can be formed with alternating odd and even digits is the product of these two values. Total alternating numbers = 4! × 3! = 24 × 6

Substitute the values

= 144

Evaluate

There are 144 ways to form such a 7-digit number.

384

Write the equation

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Example 5 Two distinguishable teams, Team A and Team B, each consisting of 5 players are to be selected from a group of 12 players. After selecting the teams, each player on Team A is assigned one of these positions: captain, vice-captain, and three other players. Similarly, Team B assigns its players to the same positions: a Determine the number of ways to select the two teams.

Create a strategy Use combinations to select 5 players for each team from the group of 12, considering that the teams are distinguishable.

Apply the idea First, select 5 players for Team A from a group of 12. This can be done in: 12

C5 = 792 ways

Then, select 5 players for Team B from a group of 7 (the remaining players). This can be done in: 7

C5 = 21 ways

The total number of ways to select both distinguishable teams is the product of these two values (using the multiplication principle): 12

C5 × 7C5 = 792 × 21 = 16 632 ways

Since Team A is selected first, then Team B from the remainder, and the teams are distinguishable (Team A is different from Team B), there is no need to multiply by 2 or divide by 2 for the order of selecting teams. The sequence of choices already accounts for this.

b Determine the number of ways to assign positions after the two teams have been selected.

Create a strategy Use permutations to assign positions within the teams.

Apply the idea There are two specific positions (captain and vice-captain) within each team. The number of ways these positions can be assigned from the 5 selected players is: 5

P2 = 20 ways to assign positions for Team A

P2 = 20 ways to assign positions for Team B

Therefore, the total number of ways to assign positions after the teams have been selected is: 20 × 20 = 400 ways

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c Determine the total number of ways that teams can be selected and assigned to positions.

Create a strategy Combine answers from part (a) and part (b) using the multiplication principle.

Apply the idea The total number of ways to select the two teams and assign positions is the product of the number of ways to select the teams and the number of ways to assign positions for both teams: 16 632 × 400 = 6 652 800 ways

Idea summary When solving counting problems, first determine if the order of selection or arrangement matters to distinguish between permutations (order matters) and combinations (order does not matter). • Permutation formula: • Combination formula: For complex problems or those with multiple stages, the multiplication principle (or fundamental counting principle) (if events are sequential) and addition principle (if events are mutually exclusive) are key. When dealing with restrictions: • Grouped items: Treat the group as one unit, arrange units, then arrange items within the group. • Separated items: Arrange unrestricted items first, then place restricted items in available slots. • ‘At least one’: Often use complementary counting (Total arrangements − Arrangements with ’none’). • Identical items (permutations): Divide by the factorial of the count of each identical item: • Fixed positions or alternating: Address these specific placement rules first. Always carefully read the problem to identify the core task (arranging or selecting) and any conditions or restrictions that modify the approach.

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4.09 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

Order does not matter for combinations.

Combinations are the same as permutations. is the same as nCr.

c d

When finding the number of ways r objects can be selected from a set of n distinct objects, a permutation will result in more outcomes than a combination.

In each situation, determine if the order in which the objects are selected matters: a

A local pizza shop is offering a special on a large pizza with 3 toppings, chosen from 9 available options.

Members need to elect a captain and a vice-captain for their sports team. There are 14 players on the team.

There are 52 runners competing in a marathon. The top 5 runners will receive varying amounts of prize money based on their finishing position.

A student council is organising snacks for an event. They can choose 3 different types of snacks to offer. The local store provides 7 snack options for them to select from.

Write using nCr or nPr notation: a

A team of 5 players is chosen from a squad of 12 players.

Five people are arranged in a line for an interview.

Eight chocolates are chosen from a box of 24 flavours.

Locker codes are created from selecting 3 letters from the alphabet.

The code to open a locker is called a combination. Explain why this name is misleading.

Practice 5

A university has 8 different flags available. In how many ways can 5 flags be chosen to be displayed?

In a bicycle race, a quinella is a bet on the first 2 cyclists that finish the race, but the order in which these 2 cyclists place does not matter. How many different quinella bets are possible for a bicycle race where 14 cyclists are competing?

A school baseball team has 12 players, but a coach can only choose 9 players for the batting lineup. The order in which the players bat is important. How many batting lineups are possible?

If a playlist contains 9 songs, in how many different ways could all 9 songs be played, with no song being repeated?

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11 Ex 1

A local sports club is organising a mini-tournament with 6 players. The team will be arranged for a series of matches, and specific rules apply to each part of the organisation process: a

The club needs to determine the number of unique matches if it is a round-robin tournament where each player plays against every other player exactly once. How many unique matches are there?

From the 6 players, the club needs to select 3 players to be in a special practice group. How many different ways can this practice group be formed?

For each of these problems: i

Identify it as a permutation, combination, or counting principle problem.

Solve the problem.

A book store offers a discount when choosing 6 books from a list of 20. In how many ways can a shopper choose their books?

A car company sells a particular car in 10 different colours and in 3 different styles: hatchback, wagon or sedan. In how many ways can the car be designed?

In the qualifying round of a car race, 8 cars are given random starting positions. In how many different ways can starting positions be allocated?

A corporation has 12 members on its board of directors. In how many ways can it elect a president, vice-president, secretary and treasurer?

A popular brand of pen is available in 3 colours (red, green or blue) and 3 tips (0.3 mm, 0.5 mm and 0.7 mm). How many different choices of pens are available with this brand?

A variety pack of chocolate consists of seven bars, each with a different flavour. If three bars of chocolate are chosen at random, how many different selections are possible?

A music festival organiser is deciding which of several bands to feature in a special performance lineup. She initially plans to select and arrange 4 bands from a pool of 9: a

How many different selections are possible?

If the order in which the 4 selected bands perform matters, how many different arrangements are possible for the performance lineup?

If one of the 9 bands is found to be unavailable before selection, and the order in which the 4 selected bands perform still matters, how many different arrangements are possible for the performance lineup from the remaining bands?

How many distinct arrangements can be made from the letters of the word BANANA?

Four letters are chosen from the word BETTING. How many selections of four letters have at least one T? (Consider the two T’s as indistinguishable for selection of letter types, meaning selecting ‘T’ is one option, and selecting ‘TT’ is another).

Ex 2

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Consider the letters of the word ENVIOUS: a

How many ways are there to choose 1 vowel?

How many ways are there to choose 2 non-vowels?

If 3 letters are randomly chosen from ENVIOUS, how many selections include exactly 1 vowel?

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Amelia is playing a game of cards. She will randomly draw 5 cards from a standard deck of 52 cards, and wants exactly 3 of the cards to have Clubs on them. One quarter of the cards in the deck have Clubs on them. How many different selections would contain the cards she wants?

In how many ways can 4 boys and 3 girls be arranged in a row such that all 3 girls must sit together?

Ex 3

17 Ex 4

A code is formed using the letters A, B, C, D, E. How many 5-letter codes can be formed if vowels (A, E) and consonants (B, C, D) must alternate, and repetition is not allowed?

A combination lock has a 4-digit code with digits from 0 − 9. If the first digit cannot be zero and no repetition is allowed, how many different codes are possible?

In Victoria, standard motor vehicle number plates consist of three digits followed by three letters. How many number plates could end with L or B?

A class of 28 students visit a bowling alley and must organise themselves into groups of 8 to use a lane: a

How many different ways are there for the first group of 8 students to be formed?

How many different ways can the second group of 8 students be formed from the remaining students?

How many different ways can the third group of 8 students be formed from the remaining students?

A manager wants to select 5 people from a group of 25 assistants to help with a specific project: a

In how many ways can the manager choose 5 people?

If the manager needs to fill 5 specific roles within the project, in how many ways can the roles in the project be assigned?

Which way of selecting the 5 people gives the manager fewer options?

In a game of soccer, each team must have 1 goalkeeper and 10 other players on the field. If a soccer team consists of 17 players, 2 of which are goalkeepers. In how many ways can players be chosen to start the game?

A community book club wants to form two discussion groups, Group Alpha and Group Beta, each consisting of 3 members from a pool of 10 members. After selecting the groups, each member in Group Alpha will take on specific roles: leader, note-taker, and discussion facilitator. Group Beta will have members assigned to the same roles: a

Determine the number of ways to select the two groups.

Determine the number of ways to assign roles after the groups have been selected.

Determine the total number of ways that groups can be selected and assigned roles.

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Ex 5

A school needs to create two team names for Team X and Team Y from a selection of the letters A, B, C, D, E, F, G, and H. Each team name will consist of 4 letters: a

Determine the number of ways to select letters for Team X and Team Y from the given group of letters.

Determine the number of ways to arrange the selected letters in Team X and Team Y.

Determine the total number of ways that both team names can be formed and arranged.

Extend your thinking 25

For each of the seven days of the week, Fiona can choose to visit any group (or subset) of her 3 friends. How many different weekly plans are possible?

Ewen is organising a 6-person scientific committee from a pool of 12 scientists. The scientists are categorised into three specialisations A, B and C that include 4, 5 and 3 scientists respectively. How many 6-person committee could include exactly 2 scientists from specialisation A and at least 2 scientists from exactly one other specialisation?

Six men and 3 women join a social evening. There are two round tables available, one consisting of 5 seats and another consisting of 4 seats. In how many ways can the 9 people be seated around the two circular tables if all the women wish to be at the same table? Assume rotations of a table’s arrangement are considered identical.

A group of 8 people, consisting of 5 adults and 3 children, is to be seated at two circular tables for dinner. One table has 4 seats, and the other has 4 seats. In how many ways can the group be seated if all the children wish to sit at the same table? Assume rotations of a table’s arrangement are considered identical.

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4.10 Combinatorics and probability After this lesson, you will be able to… • apply combinations and permutations to probability problems

Apply combinatorics to probability Theoretical probability is defined as the ratio of the number of favourable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely.

Pr( A )

n( A ) n(ε )

n(A) is the number of outcomes that result in event A (favourable outcomes) n() is the total number of possible outcomes in the sample space, denoted by the Greek letter epsilon,  In many scenarios, n(A) and n() can be very large, making direct listing impractical. The counting techniques of permutations and combinations are essential for calculating these values systematically. The key is to determine whether the order of selection matters in the context of the problem: • Permutations are used when the order of arrangement is important. Keywords often include “arrange,” “order,” “sequence,” or specific roles like “president, vice-president.” The formula is:

• Combinations are used when the order of selection does not matter. Keywords often include “select,” “choose,” “group,” or “committee.” The formula is:

It is crucial that both n(A) and n() are calculated consistently—either both as permutations or both as combinations.

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Exploration Consider a bag containing 3 red marbles (R) and 2 blue marbles (B). Two marbles are drawn from the bag without replacement: 1. If order does not matter, how many unique pairs of marbles can be drawn? Calculate 5C2. 2. Consider the event A: “both marbles drawn are red”. How many ways can this happen if order does not matter? Calculate 3C2. 3. What is the probability of drawing two red marbles if order doesn’t matter? 4. Consider if order does matter. How many ordered pairs can be drawn? Calculate 5P2. 5. For the event A: “both marbles drawn are red”, how many ordered pairs are possible? Calculate 3P2. 6. What is the probability of drawing two red marbles if order matters? Compare this to the answer from question 3. What stands out? This exploration illustrates that as long as the counting method (permutations or combinations) is applied consistently to both the sample space and the event, the final probability remains the same.

Example 1 The letters of the word SPACE are to be rearranged: a Determine the number of distinct arrangements of the letters in the word SPACE.

Create a strategy Since the word has 5 letters and all of them are to be arranged, use the permutation formula with n = 5 and r = 5.

Apply the idea Write the permutation formula Substitute n = 5 and r = 5 Evaluate the subtraction in the denominator Expand the factorial = 120

Evaluate

There are 120 different ways the letters of the word SPACE can be arranged. 392

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b What is the probability that the letter ‘E’ will be the first letter?

Create a strategy Determine the number of permutations with ‘E’ in the first position, and then divide this number by the total number of permutations found in part (a).

Apply the idea To determine the number of arrangements where ‘E’ is the first letter, fix ‘E’ in the first position. The remaining 4 letters (S, P, A, C) need to be arranged in the remaining 4 positions. The number of permutations of these 4 letters is: n

Pr = 4P4

Substitute n = 4 and r = 4

= 24

Evaluate using a calculator

The number of permutations of the remaining letters is 24. The probability that ‘E’ will be the first letter is the ratio of the number of favourable arrangements to the total number of arrangements: Write the probability formula Substitute the values Simplify

c What is the probability that the letters are arranged in alphabetical order?

Create a strategy Use the number of possible arrangements found in part (a) and determine the number of arrangements in which the letters are in alphabetical order.

Apply the idea From part (a), there are 120 different ways that the word can be arranged. There is only one arrangement in which the letters are in alphabetical order, A C E P S. So, the probability that the letters are arranged in alphabetical order is:

Reflect and check The probability of any particular arrangement would be letters, it would be more involved.

. However, if there were duplicate

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Example 2 Five people are to be selected from a larger group of 10 candidates. If Amelia is among the candidates, what is the probability that she will be among those selected?

Create a strategy Determine the total number of ways in which 5 people can be selected from 10 without restriction. Then, determine the number of ways 5 people can be selected with Amelia being one of them to determine the total number of desired outcomes, then divide this by the total number of outcomes.

Apply the idea The total number of ways in which 5 people can be selected from 10 without restriction can be translated as 10C5. If Amelia is one of the people in the selection, then there are only 4 places left to fill, and only 9 people left to choose from. This can be translated as 9C4. So, the probability is: Write the probability formula Substitute the known values Evaluate numerator and denominator Simplify

Did you know?

Combinatorics is key in designing secure passwords! By counting the number of possible character arrangements, cybersecurity experts can determine how strong a password is. This helps protect sensitive information from hackers. 394

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Example 3 Consider the digits 4, 5, 6, 7, 8 and 9. A number is created by arranging at least one of the digits. Each digit can only be used once. What is the probability, as a percentage, that the number is greater than 800?

Create a strategy First determine the number of possible outcomes. The number created could be between one and six digits long. Then, determine the number of favourable outcomes that are greater than 800 and divide by the total to get the probability.

Apply the idea 1. Determine the total number of possible outcomes: Since the order matters and there are no duplicate digits, the permutations of one to six digits are added: Total number of outcomes = 6P1 + 6P2 + 6P3 + 6P4 + 6P5 + 6P6

Write the equation

= 6 + 30 + 120 + 360 + 720 + 720

Evaluate each permutation

= 1956

Evaluate

2. Determine the number of favourable outcomes (numbers greater than 800): • For a 3-digit number to be greater than 800, the first digit must be 8 or 9 (2 choices). The remaining 2 digits are chosen from the remaining 5 digits and arranged: 5P2 = 5 × 4 = 20 Total 3-digit numbers: 2 × 5P2 = 2 × 20 = 40. • For a 4-digit number, all such numbers will be greater than 800. Number of 4-digit numbers: 6 P4 = 360 • For a 5-digit number, all such numbers will be greater than 800. Number of 5-digit numbers: 6 P5 = 720 • For a 6-digit number, all such numbers will be greater than 800. Number of 6-digit numbers: 6 P6 = 720 So, the number of favourable outcomes is given by: Number of favourable outcomes = 40 + 6 P4 + 6 P5 + 6 P6 Sum of outcomes for 3, 4, 5, and 6-digit numbers greater than 800 = 40 + 360 + 720 + 720

Evaluate each term

= 1840

Evaluate

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3. Calculate the probability: Write the formula Substitute the values ≈ 0.940695 × 100%

Evaluate the fraction

≈ 94.07% Convert to percentage and round to two decimal places Therefore, the probability of a number created from these digits being greater than 800 is approximately 94.07%.

Reflect and check It is also possible to use the complement and determine the number of outcomes that are 800 or less. Count numbers 800 or less: • 1-digit numbers: These are all less than 800: 6P1 = 6 • 2-digit numbers: These are all less than 800: 6P2 = 30 • 3-digit numbers less than 800: The first digit can be 4, 5, 6 or 7 (4 choices). The remaining 2 digits are chosen from the remaining 5 digits: 4 × 5P2 = 4 × 20 = 80 Total non-favourable outcomes (numbers ≤ 800, with at least one digit): 6 + 30 + 80 = 116. The probability of being greater than 800 is 1 − Pr(number ≤ 800): Pr(number > 800) =

Write the equation

= 1 − 0.0593047...

Evaluate the division

≈ 0.9406953

Evaluate

≈ 94.07%

Multiply by 100% and round

This confirms the previous result.

Example 4 A committee of 3 people is to be chosen from a group of 5 men and 4 women. What is the probability that the committee consists of 2 men and 1 woman?

Create a strategy This is a selection problem where order does not matter, so combinations will be used. First calculate the total number of ways to choose a committee of 3 from the total group of 9 people (n( )). Next, calculate the number of ways to choose 2 men from 5 (n(Men)). Then, calculate the number of ways to choose 1 woman from 4 (n(Women)). The number of favourable outcomes n(A) is n(Men) × n(Women) by the multiplication principle. Finally, calculate the probability

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Apply the idea Total number of people = 5 men + 4 women = 9 people Total number of ways to choose a committee of 3 from 9 people (n()):

Apply the combination formula

Expand and simplify the factorials Evaluate Number of ways to choose 2 men from 5 (n(Men)): Apply the combination formula Expand and simplify the factorials Evaluate Number of ways to choose 1 woman from 4 (n(Women)): Apply the combination formula Expand and simplify the factorials Evaluate Number of favourable outcomes n(A) = n(Men) × n(Women) = 10 × 4

Write the multiplication principle formula Substitute the values

= 40 Evaluate the multiplication Probability: Write the formula Substitute the number of favourable and total outcomes Simplify the fraction The probability that the committee consists of 2 men and 1 woman is

Example 5 Digital only example Explore in the online lesson

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Example 6 From a group of 6 students, 3 are selected for a project. Two of the students, Sarah and Tom, are friends. What is the probability that at least one of Sarah or Tom is selected for the project?

Create a strategy This is a selection problem, so combinations are used. It’s easier to calculate the probability of the complementary event (neither Sarah nor Tom is selected) and subtract this from 1: 1. Calculate the total number of ways to select 3 students from 6 (n()). 2. Calculate the number of ways to select 3 students such that neither Sarah nor Tom is selected. This means selecting 3 students from the remaining 6 − 2 = 4 students (n(A′)). 3. Calculate Pr(A′)

4. The desired probability is Pr(A) = 1 − Pr(A′).

Apply the idea Total number of ways to select 3 students from 6 (n()):

Apply the combination formula

Expand and simplify the factorials Evaluate Number of ways to select 3 students such that neither Sarah nor Tom is selected (n(A′)): This means selecting 3 students from the other 4 students.

Apply the combination formula

Expand and simplify the factorials Evaluate Probability that neither Sarah nor Tom is selected Pr(A′): Substitute the number of favourable and total outcomes Evaluate Probability that at least one of Sarah or Tom is selected Pr(A):

Write the complement rule

Substitute the values Evaluate The probability that at least one of Sarah or Tom is selected is .

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Reflect and check Alternatively, to find the number of ways at least one is selected: • Sarah is selected, Tom is not: Choose 2 more from 4 others: 1C1 × 1C0 × 4C2 = 1 × 1 × 6 = 6 • Tom is selected, Sarah is not: Choose 2 more from 4 others: 1C0 × 1C1 × 4C2 = 1 × 1 × 6 = 6 • Both Sarah and Tom are selected: Choose 1 more from 4 others: 1C1 × 1C1 × 4C1 = 1 × 1 × 4 = 4 Total favourable outcomes n(A) = 6 + 6 + 4 = 16. Probability

, which matches.

This uses the inclusion-exclusion principle implicitly if considering cases for “Sarah or Tom”.

Example 7 A committee of 4 people is to be selected from a group of 7 men and 5 women: a In how many ways can the committee be formed if it must consist of at least 2 men?

Create a strategy This is a selection problem where the order of selection does not matter, so combinations will be used. The condition “at least 2 men” means the committee can have 2 men, 3 men, or 4 men (since the committee size is 4). For each case, calculate the number of ways to select men and women separately, then multiply these numbers (multiplication principle). Finally, sum the results from each case (addition principle). • Case 1: 2 men and 2 women • Case 2: 3 men and 1 woman • Case 3: 4 men and 0 women

Apply the idea Total number of men available: 7 Total number of women available: 5 Committee size: 4 people Case 1: 2 men and 2 women Write the equation = 21 × 10

Evaluate each combination

= 210

Evaluate

Case 2: 3 men and 1 woman Write the equation = 35 × 5

Evaluate each combination

= 175

Evaluate

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Case 3: 4 men and 0 women Write the equation = 35 × 1

Evaluate each combination

= 35

Evaluate

Total number of ways for the committee to have at least 2 men: Total favourable outcomes = 210 + 175 + 35 = 420

Write the equation Evaluate

There are 420 ways to form the committee with at least 2 men.

b If the committee is selected randomly, what is the probability that it consists of at least 2 men?

Create a strategy Probability is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. • The number of favourable outcomes (committee with at least 2 men) was calculated in the previous part. • The total number of possible outcomes is the total number of ways to select 4 people from the entire group of 12 people (7 men + 5 women).

Apply the idea From part (a), the number of favourable outcomes where the committee has at least 2 men is 420. Next, calculate the total number of ways to select 4 people from the total group of 12 (n()):

Apply the combination formula

Expand and simplify the factorials = 495 Evaluate Now, the probability can be calculated. Number of favourable outcomes Total number of possible outcomes

Write the probability formula

Substitute the values Simplify by dividing 15 The probability that the committee consists of at least 2 men is

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Reflect and check Alternatively, using the complementary event (less than 2 men: 0 men or 1 man). • 0 men and 4 women: 7C0 × 5C4 = 1 × 5 = 5 • 1 man and 3 women: 7C1 × 5C3 = 7 × 10 = 70 Total non-favourable outcomes (n(A′)) = 5 + 70 = 75.

. This confirms the result.

Idea summary The probability of an event A is calculated as

, where n(A) is the

number of favourable outcomes and n() is the total number of possible outcomes. Permutations are used to find n(A) and n() when the order of selection or arrangement matters. Combinations are used when the order of selection does not matter. Strategies for counting with restrictions, such as the complementary counting principle (Pr(A) = 1 − Pr(A′)) or the addition principle (for mutually exclusive cases), also apply to probability problems.

4.10 Practice questions What do you remember? 1

Write the formula for theoretical probability in terms of n(A) and n(). What do n(A) and n() represent?

When calculating n(A) and n() for a probability problem, what key consideration determines whether to use permutations or combinations? Give an example for each.

A fair six-sided die is rolled: a

What is n()?

If event A is “rolling an even number”, what is n(A)?

Calculate Pr(A).

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Practice 4 Ex 1

Ex 2

The letters of the word NOTES are rearranged: a

Calculate the number of distinct arrangements of the letters in NOTES.

Determine the probability that the letter N is the first letter.

Determine the probability that the letter N is not the first letter.

Determine the probability that the letters are in alphabetical order.

A permutation of digits 1 to 6 is selected randomly. Determine the probability that the digits are in increasing order.

A standard deck contains 6 cards, each a different colour: red, blue, green, yellow, purple, orange. Three cards are drawn randomly and arranged in a sequence: a

Calculate the total number of possible arrangements of the 3 cards.

Determine the probability that the sequence starts with a red card and ends with a green card.

A bag contains 4 gold, 5 silver, and 6 bronze coins. Three coins are drawn sequentially without replacement. Determine the probability that the sequence is gold, silver, bronze.

A salad bar offers 3 types of lettuce, 4 toppings, and 2 dressings. A customer chooses one of each randomly. Determine the probability that the first lettuce type and second dressing are selected.

A manager needs to select a team of 4 from a group of 12 employees. If one of the employees is named Chloe, what is the probability that she will be on the selected team?

A box contains 6 balls, each a different colour: red, blue, green, yellow, purple, orange. Three balls are selected randomly:

Calculate the total number of possible combinations of the 3 balls.

Determine the probability that the selected balls include blue, red, and green.

A library has 7 distinct books. Three books are selected randomly: a

Calculate the number of ways to select 3 books.

Determine the probability that a specific mathematics book is included, given only one book is on mathematics.

A box contains 10 balls, numbered 0 to 9. Four balls are selected simultaneously: a

Calculate the total number of possible combinations.

Determine the probability that all selected balls have numbers less than 6.

A committee of 3 is formed randomly from 8 teachers. Determine the probability that two specific teachers are included.

A club has 10 members. A board of 4 is elected randomly. Determine the probability that two specific members are both elected.

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Ex 3

Ex 4

How many numbers can Romi create?

What is the probability, as a percentage, that a randomly created number will be less than 350?

A class has 6 boys and 5 girls. A team of 4 students is to be selected: a

Calculate n(), the total number of ways to select the team.

Let A be the event that the team consists of exactly 2 boys and 2 girls. Calculate n(A).

What is the probability that the selected team consists of exactly 2 boys and 2 girls?

Three items are selected from a batch of 10 items, 2 of which are defective. To determine the probability that exactly one selected item is defective: a

Calculate n(), the total number of ways to select 3 items from 10.

Let A be the event “exactly one defective item is selected”. This means 1 defective item is chosen from 2, and 2 non-defective items are chosen from 8. Calculate n(A).

Determine Pr(A).

A bowl contains 5 apples, 7 oranges, and 8 pears. Four fruits are selected randomly. Determine the probability that no apples are included.

A standard 52-card deck is used. A 5-card hand is dealt:

20 Ex 5

Ex 6

Romi wants to create a three-digit number using the 6 digits of 201 563:

Calculate n(), the total number of possible 5-card hands. (Leave as a combination)

Let A be the event that the hand contains exactly 3 Aces. Calculate n(A). (Hint: Choose 3 Aces from 4, and 2 non-Aces from the remaining 48 cards).

What is the probability of being dealt exactly 3 Aces? (Express as a fraction using answers from (a) and (b)).

The letters P, Q, R, S, T, U are arranged randomly in a row: a

Calculate n(), the total number of possible arrangements.

Let A be the event that the letter P is in the first position and the letter U is in the last position. Calculate n(A).

What is the probability that P is first and U is last?

A license plate consists of 3 letters followed by 4 digits, with no repetition. Determine the probability that the second letter is B.

A box contains 10 balls, numbered 0 to 9. Four balls are selected randomly and arranged in a line:

Calculate the total number of possible arrangements.

Determine the probability that the number 1234 appears.

Determine the probability that the first and last balls are even numbers (zero is even).

From a group of 8 employees, 3 are chosen to attend a conference. Two of the employees, Finn and Jake, are best friends: a

Calculate n(), the total number of ways to choose the 3 employees.

Let A′ be the event that neither Finn nor Jake is chosen. Calculate n( A′).

What is the probability that neither Finn nor Jake is chosen?

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Five friends are arranged randomly in a row. Determine the probability that two specific friends are adjacent.

A selection of 4 books is to be made from a shelf holding 3 distinct Math books and 4 distinct Physics books:

Ex 7

In how many ways can the selection be formed if it must consist of at least 2 Math books?

If 4 books are selected randomly, what is the probability that the selection consists of at least 2 Math books?

A bag contains 4 red marbles and 3 blue marbles. Two marbles are drawn without replacement: a

If the order of drawing matters, calculate n() and n(A) for the event A = ‘Both marbles are red’. Find Pr(A).

If the order of drawing does not matter, calculate n() and n(A) for the event A = ‘Both marbles are red’. Find Pr(A).

Compare the answers from part (a) and part (b). Explain the result.

Extend your thinking 27

Six distinct flowers (Rose, Tulip, Lily, Daisy, Orchid, Sunflower) are arranged in a row. Determine the probability that Daisy is adjacent to Orchid and neither Tulip nor Sunflower is at either end.

An ant walks randomly from (0, 0) to (4, 3) using grid lines, moving only up or right. Determine the probability that the path passes through (2, 1).

A group of 5 men and 5 women are arranged in a circle:

What is the total number of distinct circular arrangements?

What is the probability that all the men sit together?

In a lottery, 6 numbers are chosen from 1 to 40 inclusive, without replacement. To win a prize, you must match at least 4 of the 6 winning numbers: a

Calculate n(), the total number of ways to choose 6 numbers. (Leave as a combination)

Calculate the number of ways to match exactly 4 winning numbers. (Hint: Choose 4 of the 6 winning numbers, and 2 of the 34 losing numbers).

Calculate the number of ways to match exactly 5 winning numbers.

Calculate the number of ways to match exactly 6 winning numbers.

What is the probability of winning a prize? (Express as a fraction using your previous answers, simplify if possible, or leave in terms of combinations).

Chapter 4 review Online review

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mathspace.co

“It is better to solve one problem five different ways than to solve five problems one way.” George Pólya

Big ideas • Matrices provide a structured format for representing and manipulating data, with defined operations for addition, subtraction, and multiplication that follow specific algebraic rules. • Advanced matrix concepts, such as determinants and multiplicative inverses, extend matrix algebra to enable the efficient solution of systems of linear equations.

5 Matrices Chapter outline 5.01 5.02 5.03 5.04 5.05 5.06

Matrix notation and representation Addition and subtraction of matrices Scalar multiplication Matrix multiplication Multiplicative inverses and determinants Solve matrix equations Investigation: Existence of matrix solutions Chapter 5 review

408 424 433 443 459 468

GPS technology uses matrices to pinpoint your location anywhere on Earth.

5.01 Matrix notation and representation After this lesson, you will be able to… • identify different types of matrices • determine matrix size and properties • use matrix notation to identify elements • organise and interpret contextual information presented in matrices

Features of a matrix Matrix A rectangular array of numbers, symbols or expressions displayed in rows and columns, enclosed in a set of square brackets and often named with capital letters. Consider this matrix:

1 2 3

13 6 3 10 8 9

Rows run horizontally and are identified from top to bottom.

1 2 13 6 3 10 8 9

Columns run vertically and are identified from left to right.

The order (or size) of a matrix is written in the form m × n, where m is the number of rows and n is the number of columns. The order of matrix A is 3 × 2.

The order of matrix B is 2 × 4.

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Elements are the individual entries of a matrix. The notation aij describes the element in the ith row and the jth column of the matrix.

Example 1 Suppose H is a 3 × 2 matrix: a How many rows does H have?

Create a strategy

Apply the idea

Use the fact that m × n matrix has m rows and n columns.

m×n 3 × 2 = 3 rows × 2 columns So, H has 3 rows.

b How many columns does H have?

Apply the idea m×n 3 × 2 = 3 rows × 2 columns So, H has 2 columns.

Example 2 Consider this matrix:

a What is the entry at a23?

Create a strategy Use the fact that aij denotes the element in the ith row and jth column of the matrix.

Apply the idea A=

−2 −5 −1 1 8 4

5 −7 7

The number at the intersection of row 2 and column 3 is − 7. So, a23 = − 7.

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b Where is the value − 5 located, in terms of i and j?

Apply the idea −2 −5 −1 1 8 4

5 −7 7

The value −5 is located in row 1 and column 2. So, it is the element a12.

Idea summary A matrix is a rectangular array of numbers, symbols or expressions displayed in rows and columns, enclosed in a set of square brackets and represented by capital letters. Rows run horizontally and are numbered from top to bottom. Columns run vertically and are numbered from left to right. The order (or size) of a matrix is written as m × n, where m is the number of rows and n is the number of columns. Elements are the individual entries of a matrix where aij describes the element in the ith row and the jth column of the matrix.

Types of matrices Matrices can be classified based on their shape or the arrangement of their elements. A row matrix has just a single row. For example: T = [3

A column matrix has just a single column. For example:

In a square matrix, the number of rows and columns is equal. For example:

410

2 16 0.1

0 2 3

−1 5 −7

All square matrices have a leading diagonal, which consists of the elements from the top-left to the bottom-right corner. For example, the leading diagonal in matrix J is highlighted.

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A rectangular matrix is a matrix where the number of rows is not equal to the number of columns. If m ≠ n, the matrix is rectangular. Row matrices and column matrices are special types of rectangular matrices, unless they are 1 × 1. For example: 1 2 3 S 4 5 6

7 8

Matrices can also be classified based on their entries or the values of their elements. A zero matrix is a matrix of any size where all the elements are zero. Here are some examples: O

A diagonal matrix is a square matrix where all the elements not on the leading diagonal are zero. Some elements on the leading diagonal can also be zero. For example: D1

2 0 0

0 0 5 0 D2 0 1

7 0 0 0

The identity matrix is any square matrix in which all the elements in the leading diagonal are ones and all other elements are zeros. For example:

A scalar matrix is a diagonal matrix where all the elements on the leading diagonal are equal. If this common diagonal element is k, the matrix can be written as kI, where I is the identity matrix. For example: S1

4 0 0 0 4 0 0 0 4

3 0

0 3

A single-element matrix is a matrix with only one row and one column (i.e., order 1 × 1). It can be considered a row matrix, a column matrix, and a square matrix simultaneously. For example: E = [7]

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Example 3 Classify these matrices in as many ways as possible. For square matrices, also identify the elements on the leading diagonal. a

2 1 0

Create a strategy Observe the number of rows and columns, the placement of any zero elements, and if square, the values on the diagonal to classify the matrix.

Apply the idea Matrix A has 3 rows and 1 column. Since it has only 1 column, it is a column matrix. It is also a rectangular matrix because number of rows is not equal to its number of columns.

b B

1 2 3 4 5 6 7 8 9

Apply the idea Matrix B has 3 rows and 3 columns. Since the number of rows equals the number of columns, it is a square matrix. The elements on its leading diagonal are b11 = 1, b22 = 5, and b33 = 9.

c C

1 0 0 0 1 0 0 0 1

Apply the idea Matrix C is a 3 × 3 square matrix. The elements on its leading diagonal are all 1s, and all other elements are 0s. Therefore, it is an identity matrix. It is also a diagonal matrix and a scalar matrix. The elements on its leading diagonal are c11 = 1, c22 = 1, and c33 = 1.

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d D

0 0 0 0

Apply the idea All elements in matrix D are 0. Therefore, it is a zero matrix. It is also a square matrix, a diagonal matrix, and a scalar matrix. The elements on its leading diagonal are d11 = 0 and d22 = 0.

Idea summary Type of matrix Row matrix (Row vector)

Description

Examples T = [3 4 5]

A single row.

Column matrix A single column. (Column vector)

Square matrix

Identity matrix

Equal number of rows and columns. Has a leading diagonal, which consists of the elements from the top-left to the bottom-right corner. Any square matrix in which all elements in the leading diagonal are ones and all other elements are zeros.

Zero matrix

Every element is zero.

Rectangular matrix

A matrix where the number of rows is not equal to the number of columns.

Single-element matrix

A matrix with only one row and one column.

A square matrix where all the elements not on the leading diagonal are zero. Diagonal matrix Some elements on the leading diagonal can also be zero. A diagonal matrix where all Scalar matrix leading diagonal elements are equal.

O A

1 2 3 4 5 6

T = [1]

1 0 0

0 3 0

4 0 0 4

0 0 N 5

2 0 0 0 0 0 0 0 4

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Represent information in matrices Matrices can be used for storing and displaying information from a variety of applications, such as tables, databases, and connections in social or road networks. Representing information as a matrix allows for useful and efficient mathematical operations. The order of data in a matrix has significance, so it is important to identify what each row and column represents by labelling the matrix rows and columns where necessary.

Exploration Information is often displayed in a format that is easy to interpret, such as a table: Preferred fruit Apples Bananas Oranges

Team Red 10 12 8

Team Blue 15 13 17

This same information can also be represented as a matrix: Red

Blue

Apples

Row 1

Bananas

Row 2

Oranges

Row 3

Column 1 Column 2

1. Compare how the information is presented in the table and in the matrix. 2. What does each row in the matrix represent? 3. What does each column in the matrix represent? 4. Describe how the total number of votes for “Bananas” can be found by looking at the matrix. 5. If another team were added, how would the matrix change? 6. The given matrix is a 3 × 2 matrix (3 fruits, 2 teams). Can the same information be represented using a 2 × 3 matrix? If so, what would the rows and columns represent in this new matrix?

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Example 4 The costs of a train ticket during different periods are: • Weekdays: $7 during peak hours, $4 during off-peak hours • Weekends: $12 during peak hours, $6 during off-peak hours • Public holidays: $18 during peak hours, $10 during off-peak hours a Organise the data into a 2 × 3 matrix, T.

Create a strategy Identify which variable should be represented by rows and which variable by columns.

Apply the idea The two variables are: • Type of day (Weekday, Weekend, or Public holiday) • Time of day (Peak and Off-peak) To construct a 2 × 3 matrix: 1. The variable with two options must be represented by the rows, which in this case is the time of day. 2. The variable with three options must be represented by the columns, which in this case is the type of day. 3. Enter the corresponding ticket costs into the matrix based on time and type of day. The elements in the corresponding location of the matrix: Weekday Weekend T

Peak Off-peak

7 4

12 6

Public holiday 18 10

b If t12 was changed to 8, interpret what that would mean in the scenario.

Create a strategy Identify what t12 represents in the matrix. The matrix rows and columns correspond to the time of day and type of day, respectively.

Apply the idea t12 corresponds to the element in the 1st row and 2nd column which is 12. It is the peak hour cost on weekends. Changing the value of t12 to 8 would mean that the new cost for a peak hour train ticket on weekends would be reduced from $12 to $8.

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Example 5 Represent the road connections shown in this map using a matrix:

Pink Pond

Shell Lake Mountain Range L

Create a strategy Construct a square matrix where rows and columns represent towns, and elements indicate the number of direct road connections.

Apply the idea Construct a blank 4 × 4 matrix to represent the four towns. Label the rows and columns with the ﬁrst letter of each town’s name. This matrix will be labelled as A. Fill in the leading diagonal of the matrix by identifying towns that have a direct road looping back to themselves. These loops should not pass through any other town. If a town has one direct loop, write a 1 in the corresponding diagonal position in the matrix; otherwise, write a 0. J

Pink Pond

The only town that has a loop is M.

So, a11 = 0, a22 = 0, a33 = 0, and a44 = 1.

Shell Lake Mountain Range L

Fill in the remaining elements of the matrix by considering the direct road connections between each town. J

Pink Pond

Shell Lake

There are three direct routes between J and K. So, a12 = a21 = 3.

Mountain Range M

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There are no direct routes between J and L. So, a13 = a31 = 0. J

Pink Pond

Shell Lake

There are three direct routes between J and M. So, a14 = a41 = 3.

Mountain Range L

Pink Pond

Shell Lake

There is one direct route between K and L. So, a23 = a32 = 1.

Mountain Range L

M J

Pink Pond

Shell Lake

There are two direct routes between K and M. So, a24 = a42 = 2.

Mountain Range L

M J

Pink Pond

Shell Lake

There is one direct route between L and M. So, a34 = a43 = 1.

Mountain Range M

Therefore, the matrix that represents the information from the map is:

Reﬂect and check Notice these paths are not one-way connections. Therefore, the square matrix will be symmetrical along the leading diagonal.

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Idea summary Matrices can be used for storing and displaying information in a variety of applications, such as tables, lists of data, and networks. he order of data in a matrix has significance so it is important to identify what T each row and column represents.

5.01 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

For matrices, the terms order, size, and dimensions all mean the same thing.

For matrices, the matrix is usually named with a lowercase letter.

Element a23 of matrix A is the element in row 3 and column 2.

A matrix that has 4 rows and 3 columns has dimensions 4 × 3.

A 1 × 3 matrix is a row matrix.

For the matrix A =

Determine whether these statements are true or false:

3 Ex 1

This matrix is an identity matrix.

This matrix is a square matrix.

a11 = 1

There are exactly 4 elements.

How many columns does K have?

Suppose K is a 4 × 3 matrix: a

How many rows does K have?

This table displays the average temperature (in degrees Celsius) in two cities for each season: Summer

Autumn

Winter

Spring

Perth

Melbourne

This data is organised into the matrix:

418

Do the rows of the matrix represent the temperatures for each season or for each city?

What is the size of the matrix?

Identify the element t32 and interpret its meaning in context.

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Practice 5

Determine the size of each matrix: a

[3 −5

16 − 8]

Identify each element from the given matrix B =

Ex 2a

b14

b22

b32

b21

Identify the location of each value in the form aij, for the matrix A =

Ex 2b

a 8 Ex 3

−6

Classify each matrix using these matrix types: row, column, zero, identity, square, singleelement, diagonal, scalar, rectangular: 10 9 2 8 8 1 10 8 a A b B 9 8 3 5 1 c

0 0 2

E = [ 7 ]

F = [0 0

5 0 0

0 3 0

1 2 3 4 5 6 0]

Determine whether the matrices with these dimensions are square, row, or column matrices: a

1×4

2×2

3×1

3×3

Construct: a

A row matrix consisting numbers − 4, 1, and 4

A column matrix consisting numbers 5, − 2, 2, and 4

A 2 × 2 identity matrix

A 3 × 3 zero matrix

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For the matrices A = a

and B =

Determine the value of a21 − b11.

: b

Determine the value of a13 × b31.

Determine whether these statements are true or false. Explain each answer: a

If two matrices have the same number of elements, then they have the same size.

A square matrix with two or more rows will have more elements than a column matrix with the same number of rows.

Adding a new row to a matrix with size m × n will add the same number of elements as adding another column.

Mia has a part-time job. Each week, she earns money and saves some of it. The matrix represents the amounts earned (E) and saved (S), in dollars, in each of three weeks:

15 Ex 4

420

How much did Mia save in Week 1?

Which week did Mia save the most money?

Determine how much Mia earned in total over the three weeks.

The number of hours worked by Alex (A), Beth (B), Charlie (C), Dana (D), and Ethan (E) across ﬁve days of a week (Monday, Tuesday, Wednesday, Thursday, Friday) is shown in the matrix: a

How many hours did Charlie work on Wednesday?

Who worked 9 hours on Friday?

Calculate the total number of hours worked by Beth during the week.

Who worked the most hours on Monday?

Jack, a chef, is known for these two recipes: Crazy Cookie which contains: Scrumptious Surprise which contains: • 360 g of butter • 420 g of butter • 410 g of sugar • 390 g of sugar • 630 g of ﬂour • 650 g of ﬂour • 185 g of egg • 200 g of egg • 450 g of chocolate chips • 80 g of chocolate chips a

Represent the amounts into a 2 × 5 matrix labelled A.

If the value of a21 was changed to 320, interpret the meaning in the scenario.

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The friendships between ﬁve children are summarised in the graph shown: a

Represent this information in a matrix, using a 0 to indicate no friendship and a 1 to indicate friendship.

The leading diagonal from top left to bottom right contains all zeros. Why is this?

Brad Sharon Adam Joanne Edward

Represent the road connections, alphabetically, in this map using a matrix: A

Ex 5

F E C

The shortest driving distances, in kilometres, between Geelong (G), Ballarat (B), and Daylesford (D), are displayed in the matrix:

G G B D

0 87 101 87 0 44 101 44 0

What is the distance, in kilometres, between Geelong and Daylesford?

Jamie drove 44 km directly between two of the cities. Which two cities did she drive between?

The Martinez family would like to drive from Geelong to Ballarat, and then to Daylesford. Determine the total distance, in kilometres, that they will travel.

A company is reviewing three key expense categories: Training, Equipment, and Marketing, across two departments: Research and Development. The budgeted amounts and the actual expenses for the year are represented in matrices B and C, respectively: T E M R 6000 7000 5000 B D 5500 6000 7500 a

T E M R 4000 7020 6000 C D 6700 5950 8000

For Marketing Research, what were the: i

Budgeted expenses?

Actual expenses?

Was Marketing Research under or over budget?

What does the value at c21 in matrix C represent in the context?

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The Riverdale Soccer Club has ﬁve new players join its team: Jamie (J), Kai (K), Lisa (L), Max (M), and Noah (N).

K L

The graph shows the players who have played soccer together before joining the team. For example, the edge between Jamie and Kai shows that they have previously played soccer together:

N M

Represent this information in a matrix.

How many of these players had Max played soccer with before joining the team?

Who had played soccer with both Kai and Noah before joining the team?

During the season, another new player, Olivia, joined the team. Olivia had not played soccer with any of these players before. Add this information to the matrix.

Extend your thinking 21

Matrix K is a 2 × 3 matrix. The elements of K are determined by the rule kij = (i + j ). Select matrix K: A

2 3 4

3 4 5 D

1 2 1 2 3 2

1 2 3

2 3 4 E

0 1 2

2 3 4

4 5 6

3 4 5 4 5 6

Matrix M is a 2 × 2 matrix. The elements of M are determined by the rule mij = i2 − j. Determine matrix M.

Provide an example of a matrix for each region of the Venn diagram. Identify any regions that are impossible to fill and explain why. Square matrix

Zero matrix 24

422

Identity matrix

For a matrix with 120 elements: a

How many different sizes can this matrix have? Explain your answer.

Which order of matrix will minimise the number of rows and columns combined?

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For the equation:

Determine the value of:

The matrix N displays the number of roads connecting ﬁve towns: Arlington (A), Bristol (B), Camden (C), Dover (D), and Elmont (E):

Construct a road map using the information shown.

Determine whether these statements are true or false:

There is a road loop at Bristol.

Direct travel between Camden and Elmont is possible.

iii

There are two roads connecting Arlington and Camden.

There are three different ways to travel between Bristol and Elmont.

A major ﬂood washes away part of the road connecting Arlington and Camden. Which elements in matrix N will need to be changed to reﬂect the new road conditions between the towns?

The matrix T shows the results of a round-robin chess tournament among six friends: Alan (A), Beth (B), Carlos (C), Dana (D), Evan (E), and Fiona (F ). Each participant played against every other participant exactly once. A 1 in the matrix indicates that the player in that row defeated the player in the corresponding column, and a 0 indicates a loss.

Identify the error in the matrix. Justify using mathematical reasoning.

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5.02 Addition and subtraction of matrices After this lesson, you will be able to… • determine when matrices can be added or subtracted • perform addition and subtraction of matrices • solve mathematical problems involving matrix addition and subtraction

Matrix addition and subtraction Exploration The matrices show the number of goals and behinds for two football teams across four games. Game 1 Team A =

Goals

Behinds

Game 2

Game 3

Game 4

Goals

Behinds

Game 2

Game 3

Game 4

Game 1 Team B =

1. How many goals did Team A score in Game 1? 2. In which game(s) did Team B score more goals than Team A? 3. How many goals were scored in total by both Team A and Team B in Game 4? 4. Determine a way to add the matrices together to create a new matrix that shows the combined goals and behinds for each game. Write the new matrix.

For matrix addition and matrix subtraction, the operations are performed on corresponding elements.

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Formally, matrix addition is represented as: a b c d

e f g h

a e b c g d

f h

Corresponding elements are added 2 1 4

3 0 5

1 6 3

1 1 2

3 7 7

2 1 7

And matrix subtraction is represented as: a b c d

e f g h

a e b f c g d h

Corresponding elements are subtracted 2 1 4

3 0 – 5

1 6 3

1 1 2

1 5 1

4 1 3

Due to the nature of matrix addition and subtraction, where operations are performed element by element, only matrices of the same dimensions can be added or subtracted. The resulting matrix will have the same dimensions as the matrices involved in the operation. Matrix addition follows several important properties, including the existence of an additive identity and additive inverse: Additive identity

Additive inverse

A matrix such that it leaves any matrix unchanged when added. The zero matrix that is the same order (size) as A is the additive identity of A because: A + 0 = A and 0 + A = A.

A matrix that when added results in the zero matrix. For any matrix A, the additive inverse is −A because: A + (−A) = 0 and (−A) + A = 0.

One important law of matrices is that addition of matrices is commutative, that is A + B = B + A. However, subtraction of matrices, like subtraction of real numbers, is not commutative, that is A − B ≠ B − A. Matrices can be added or subtracted using technology by inputting them and using the standard addition and subtraction operations.

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Example 1 Evaluate each matrix operation: a

Create a strategy Add the corresponding elements in each matrix.

Apply the idea 7 −7 −5

8 2 4

7+8 −7 + 2 −5 + 4

Add the elements

15 −5 −1

Evaluate each element

Reflect and check The answer matrix must have the same size as the given matrices.

–

Create a strategy Subtract the corresponding elements in each matrix by hand, or use technology to perform the subtraction.

Apply the idea 9 2 6 5 −7 8

426

−

−3 −5 3 −1 −6 7

9 − (−3) 2 − (−5) 6 − 3 5 − (−1) −7 − (−6) 8 − 7

Subtract the elements

12 7 3 6 −1 1

Evaluate each element

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0 7 7 + 0 0 5

Create a strategy Add the corresponding elements in each matrix.

Apply the idea 7 7 5

0 0 0

7 0 7 0 5 0

Add the elements

7 7 5

Evaluate each element

Example 2 Solve for x in this matrix equation:

Create a strategy Use the elements from the matrices that correspond to x to form an equation.

Apply the idea x+3=8 x=5

Reﬂect and check Write the equation Subtract 3 from both sides

Substitute the value back into the matrix and check the answer. 5+3=8

Example 3 A company tracks its profits and expenses for two of its North and South branches over two quarters, Q1 and Q2. The profits matrix, P, is given by: P

12 000 15 000 10 000 11 000

where the first row represents the North branch, the second row represents the South branch, the first column represents Q1 profits, and the second column represents Q2 profits.

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The expenses matrix, E, for the same period and branches is: E

8000 8500 7000 7200

Determine the net profit matrix for the company, where Net profit = Profits − Expenses.

Create a strategy To find the net profit matrix, subtract the expenses matrix E from the profits matrix P. This involves subtracting corresponding elements.

Apply the idea Let the net profit matrix be N. Then, N = P − E. N=P–E

Write the formula

12 000 15 000 10 000 11 000

12 000 8000 15 000 8500 10 000 7000 11 000 7200

Subtract the corresponding elements

4000 6500 3000 3800

Evaluate each element

The net profit matrix is

8000 8500 7000 7200

Substitute the matrices

4000 6500 . This means: 3000 3800

• North branch Q1 net profit: $4000 • North branch Q2 net profit: $6500 • South branch Q1 net profit: $3000 • South branch Q2 net profit: $3800

Idea summary Matrix addition and matrix subtraction are only possible with matrices of the same size. To add or subtract matrices, add or subtract the corresponding elements. • Addition of matrices is commutative (A + B = B + A). However, subtraction of matrices is not commutative (A – B ≠ B – A). • The additive identity is the zero matrix, 0, which has the same order as matrix A, such that A + 0 = A. • The additive inverse of matrix A, is –A, such that A + (–A) = 0. • Matrix addition and matrix subtraction can be applied in contexts like combining data sets or network analysis.

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5.02 Practice questions What do you remember? 1

For each pair of matrices: i

Identify the size of matrix A.

iii

Is A + B possible?

and B =

Consider the matrices: A =

,B=

Identify the size of matrix B.

and C = [a

and B =

b]:

Identify the size of matrix A.

Identify the size of matrix B.

Identify the size of matrix C.

Is A + B possible?

Is B − C possible?

Is C + A possible?

Consider the matrices:

Identify the size of matrix A.

Identify the size of matrix B.

Identify the size of matrix C.

Is A + B possible?

Is A + B − C possible?

Is B − A possible?

Practice Ex 1a

Determine A + B, if: a

and B =

Determine A − B, if:

Ex 1b, 1c

and B =

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Evaluate: a

Given A = a

Ex 2

A−B+C

,B=

, and C = b

1 4] + [2

2 −1] + [0 −3 −2] − [−1

5 7]

, determine:

B+C−A

C−A+B

Solve these matrix equations for x: a

The number of points scored and assists made by four basketball players were recorded in two different games. The columns represent the players Smith, Johnson, Davis, and Brown, respectively. The ﬁrst row represents the points scored, and the second row represents the assists made in each game: Game 1:

Ex 3

Game 2:

Calculate the matrix which shows the total points scored and assists made by each player over the two games.

Which player scored the highest total points over the two games?

A store received shipments of two types of items (shoes, S, and bags, B) across three branches (X, Y, and Z) on Wednesday:

S X Wednesday : Y Z

On Thursday, they returned some defective items: Assuming there were no sales, determine the current inventory in each branch, after the returns.

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4 7 6 5 3 8 S

X Thursday : Y Z

1 3 2 4 1 2

These tables show the number of fruit and vegetables sold at Mohamad’s three shops over a particular weekend: Saturday: Fruit

Vegetables

Shop 1

Shop 2

Shop 3

Fruit

Vegetables

Shop 1

Shop 2

Shop 3

Sunday:

Represent the sales of fruit and vegetables for Saturday as a 3 × 2 matrix.

Represent the sales of fruit and vegetables for Sunday as a 3 × 2 matrix.

Add the matrices from part (a) and part (b) to determine the total number of sales of fruit and vegetables for each shop over the entire weekend.

Determine the missing matrix X in each equation: X

a 13

Consider this matrix equation:

Determine the value of the unknown variables. 14

Given the matrices A = C=

1 1

and B =

, Angelita calculated A – B and got

3 . Identify and correct her error. 1

Consider matrix A =

and the zero matrix 0 =

. Based on the result of adding

these two matrices, explain what happens when a zero matrix is added to any matrix.

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Extend your thinking 16

Identify these statements as true or false. Justify your reasoning: a

When adding two matrices, the resulting matrix will have the same size as the original matrices.

Two matrices of different sizes can be added together as long as the number of elements are the same.

Subtracting a matrix from itself results in a zero matrix.

Lucas believes that because addition with integers is commutative, matrix addition must also be commutative. Evaluate the reasonableness of Lucas’s assumption.

This table shows information about two matrices, A and B:

Matrix

Order

Rule

3×3

aij = 2i + j

3×3

bij = i − j

Determine A + B. 19

This table shows the number of visitors to a website by country: Australia

New Zealand

Thailand

China

January

February

March

April

May

Show a matrix calculation that determines how many more visitors were from China than from Thailand. 20

In geometry, the centroid of a set of points is the arithmetic mean (average) position of all the points. It is the point where all the coordinate values average out. In a certain city, the coordinates of the three parks A, B, and D are used to determine the centroid: • Park A’s coordinates are represented by the matrix: • Park B’s coordinates are represented by the matrix: • The centroid coordinates are represented by the matrix: Determine the matrix of the coordinates of the third park, D.

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5.03 Scalar multiplication After this lesson, you will be able to… • understand the effect of scalar multiplication on matrix properties • multiply a matrix by a scalar • apply scalar multiplication to solve practical problems

Scalar multiplication A scalar is a quantity described by a single numerical value. Scalar multiplication is the process of multiplying a matrix by a scalar (number). Scalar multiplication can be applied to any matrix and involves multiplying each element within the matrix by the given scalar. In general, the multiplication of a matrix by a scalar can be represented by the equation:

a b c d λ a b c d a b c d

a c

b d

represents the scalar quantity is the original matrix i s the resulting matrix where each element of the original matrix is multiplied by the scalar λ

As an example, the multiplication of a matrix with numerical elements by a scalar will look like this: 3

2 6

2 0

6 18

6 0

Observe that each element was multiplied by 3.

Example 1 If A =

, determine 3A.

Create a strategy Multiply each element of matrix A by 3.

Apply the idea Write the equation and substitute matrix A

Multiply each element by 3

Evaluate each element

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Example 2 Determine

if A =

and B =

Create a strategy Multiply each of the elements in matrix A by

and multiply each of the elements in matrix B by .

Add corresponding elements of the matrices

A and

Apply the idea Write the equation and substitute the matrices A and

Perform the scalar multiplications

Add the corresponding elements and evaluate

Example 3 The cost matrix C shows the current cost, in dollars, per square metre of four different manufacturing materials: C = [50

120 90

70]

Due to an increase in supplier prices, the cost of each material is expected to rise by 8%. Determine the new cost matrix after the price increase.

Create a strategy Use scalar multiplication to change each element in the matrix by the same percentage.

Apply the idea An increase of 8% can be represented as a scalar as 1.08: New cost matrix = 1.08 × [50 = [54

129.6

120 97.2

70]

75.6]

Multiply the original cost matrix by 1.08 Evaluate the scalar multiplication

Reﬂect and check An 8% increase = 100% + 8% = 108% = 1.08. This scalar (1.08) indicates that 100% of the original amount is included plus 8% on top of the original amount. An 8% decrease = 100% − 8% = 92% = 0.92. This scalar (0.92) indicates that the resulting matrix is 92% of the original.

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Idea summary A scalar is a quantity described by a single numerical value. Scalar multiplication is the process of multiplying a matrix by a scalar (number). Scalar multiplication is used to change each element in the matrix by the same factor. In general, the multiplication of a matrix by a scalar will appear similar to

where λ (lambda) represents the scalar quantity.

Associative, commutative, and distributive laws Matrix operations, such as addition, subtraction, and scalar multiplication, have certain properties, similar to operations with real numbers.

Exploration Investigate some properties with specific examples. Technology (like a calculator or software) may be used to assist with the calculations. Consider the scalar k = 2, and the matrices A

1 0 and B –2 3

4 –1 . 2 5

1. Calculate A + B. Then, calculate k (A + B). 2. Calculate kA and kB. Then, calculate kA + kB. 3. Compare your results from question 1 and question 2. What is observed? Consider the scalars k = 3, c = 4, and the matrix A

1 2 . 3 4

1. Calculate kc. Then, calculate (kc) A. 2. Calculate cA. Then, calculate k(cA). 3. Compare your results from question 1 and question 2. What is observed?

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Properties of matrix addition include: • Commutative law: A + B = B + A • Associative law: (A + B) + C = A + (B + C) • Distributive law: kA + kB = k(A + B), where k is a scalar Properties of matrix subtraction include: • Not commutative: A – B ≠ B – A • Not associative: (A – B) – C ≠ A – (B – C ) • Distributive: kA – kB = k(A – B) and kA – cA = (k – c)A, where k and c are scalars Properties of scalar multiplication include: • Commutative: kA = Ak • Associative: (kc)A = k(cA), where k and c are scalars

Example 4 1 2 and B 0 –1

Let the scalar k = 3, and matrices A

4 –2 : 3 5

a Calculate k(A + B).

Create a strategy First, add matrices A and B. Then, multiply the resulting matrix by the scalar k.

Apply the idea First, add A and B: A+B=

1 2 0 –1

5 0 3 4

4 3

2 5

Write the equation and substitute the matrices Add the corresponding elements

Then, multiply the resulting matrix by the scalar 3: k (A + B) = 3 × =

5 0 3 4

15 0 9 12

Write the equation and substitute the matrix Multiply each element by 3

b Calculate kA + kB.

Create a strategy First, perform the scalar multiplications kA and kB. Then, add the resulting matrices.

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Apply the idea Performing scalar multiplications kA and kB: 1 0

kA = 3 =

3 0

Write the product of k = 3 and A

Multiply each element of A by 3

3 4 3

2 5

Write the product of k = 3 and B

12 9

6 15

Multiply each element of B by 3

kB = 3 =

2 1

Adding the resulting matrices: kA + kB =

3 0

6 3

15 9

0 12

12 9

6 15

Write the equation and substitute the matrices Add the corresponding elements

c What property does this demonstrate?

Apply the idea Since k (A +B) =

15 0 15 0 , it can be observed that k (A + B) = kA +kB. and kA + kB = 9 12 9 12

This demonstrates the distributive property of scalar multiplication over matrix addition.

Example 5 Let scalars k = 2, c = − 3, and matrix A =

5 1 : 2 0

a Calculate (kc) A.

Create a strategy First, multiply the scalars k and c. Then, multiply the resulting scalar by matrix A.

Apply the idea (kc)A = 2 × (– 3) = –6 =

30 12

5 1 2 0

5 1 2 0 6 0

Write the equation and substitute the values Multiply the scalars Multiply each element by – 6

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b Calculate k(cA).

Create a strategy First, perform the scalar multiplication cA. Then, multiply the resulting matrix by the scalar k.

Apply the idea Performing the scalar multiplication cA: cA = – 3

5 1 2 0

15 6

3 0

Write the equation and substitute the values Multiply each element by – 3

Multiplying the resulting matrix by k: k(cA) = 2 =

15 6

3 0

Write the equation and substitute the values

30 12

6 0

Multiply each element by 2

c What property does this demonstrate?

Apply the idea Since (kc) A =

30 12

6 and k(cA) = 0

30 12

6 , it can be observed that (kc) A = k(cA). 0

This demonstrates the associative property of scalar multiplication.

Idea summary Laws of matrices can be summarised as: Operation Addition Subtraction Scalar multiplication

438

Commutative Yes No Yes

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Associative Yes No Yes

Distributive Yes Yes N/A

5.03 Practice questions What do you remember? 1

Evaluate these expressions: a

2 0

4 5

1 2

0.5

1 2 7 4 5

0 3 1 2

Let T =

8 6 3 3 1 6

2 4 7 5 4 0

2 7 3

Multiply T by the scalar 0 and explain the result.

Multiply T by the scalar 1 and explain the effect on the matrix and why this is significant.

Determine the scalar that represents these percentage changes: a

20% increase

15% decrease

150% increase

5% decrease

Determine whether each statement is true or false: 1 0 2 1 0 2 1 4 3 3 a 3 1 4 3 0 2 5 0 2 5 b

1 2 3 4 5 6 7 8 9

0 1 0 1 0 1 0 1 0

0 1 2 3 4 5 1 0 1

3 0 2

1 0 1 0 1 0 1 0 1

2 1 1 2 1 0 2 1 0 1 2 1 0 1 2

2 1 5 4 3 1 0 1 2

7 5 2 0 8 4

1 0 2 3 4 5

3 2 1

3 1 2

2 1 1

1 2 3 4 5 6 7 8 9 1 2 0

0 1 2 3 4 5 1 0 1

1 0 1 4 0 1 0 1 0 1

5 1 4 0 2 6 3 0 1 3 3

2 1 0 4 1 2 1 0 1 2

7 2 8

5 3 0 1 4 2

2 1 5 4 3 1 0 1 2

5 1 4 0 2 6 3 0 1

0 4 4 5

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Practice 5 Ex 1

If A = a

, determine:

If multiplying matrix

−A

by a scalar produces

Identify the scalar. Verify your result by performing the scalar multiplication on matrix Q.

Identify the scalar that when multiplied by matrix R, produces matrix Q.

For each expression, determine the value of y:

If A =

and B =

4A + B

If A = a

, determine: b

2A − B

and B =

4A + 2B

Let A =

b ,B=

2A − 2B , and C =

c .

Solve for n: a

Solve these matrix equations for matrix A: a

440

5A + 3B

, determine:

Determine 4A − B + 2C. 12

Ex 2

− 4A

, determine:

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3A + 5B

0.3A + 2.5B

Ex 3

The quarterly revenue (in millions) for three products is given in the table. The company projects a 12% increase in revenue for next year: a

Determine the scalar that represents the percentage increase.

Determine the projected revenue matrix, rounded to two decimal places.

A factory produces three different products with the costs (in dollars) per batch. The factory implements a cost reduction strategy that decreases the cost of each product by 5%: a

Determine the scalar that represents the percentage decrease.

What is the new cost matrix after the reduction?

Product A

Product B

Product C

Product A

200

180

220

Product B

250

230

210

Product C

270

260

240

A pizzeria is about to increase their current prices by 140%. The table shows their current prices: Using scalar multiplication, calculate the new prices. Express your answer as a 3 × 2 matrix.

19 Ex 4

Pizza

Drinks

Small

Medium

$4.50

Large

$12

In a particular town, 30% of households own no pets, 50% of households own one pet, 15% of households own two pets, and 5% of households own more than two pets: a

Organise the percentages into a 1 × 4 row matrix in the same order as stated.

If there are 300 households in the town, multiply your matrix to determine the number of households in each category. Express your answer as a 1 × 4 matrix.

Determine whether these statements as true or false: a

Scalar multiplication by 1 leaves the matrix unchanged.

Scalar multiplication can change the dimensions of a matrix.

Scalar multiplication of a matrix by a negative number reverses the sign of each element in the matrix.

The order in which you multiply a matrix by two different scalars does not matter; that is, k1 (k2 A) = k2 (k1 A).

Given scalar k = 2, A =

1 3 , and B 0 2

1 0 : 4 1

Calculate k(A + B).

Calculate kA + kB.

What property of matrix operations does this demonstrate?

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Ex 5

Let scalars k = 3, c = – 4, and matrix A

2 0 : 1 5

Calculate (kc)A.

Calculate k(cA).

Based on your calculations in parts (a) and (b), what property of scalar multiplication with matrices does this demonstrate? State the property in terms of k, c, and A.

Mia and Ethan completed the same problem but arrived at different answers. Their working out is shown: Mia’s response:

Ethan’s response:

Provide a detailed explanation of any errors in their calculations and identify the correct answer. 22

The time matrix for ﬁve tasks in a project is given by the matrix: T = [3.5 4.2

2.8

5.1 3.9]

(in hours)

The manager estimates that the actual time taken will increase by 50% of the planned time due to potential delays. Determine the estimated additional time required for each of the ﬁve tasks and write them in a 1 × 5 matrix.

Extend your thinking 23

Consider these matrices:

Given the equation A + nB = C, solve for the scalar n. 24

The cost prices of three different electrical items in a store are $230, $290, and $310, respectively. Show a matrix calculation to determine the selling price of each item allowing for a 30% mark up and a $20 commission.

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Given the cost matrix (C) and the sale price matrix (S) for three products across two different stores, calculate the proﬁt matrix for each product in each store, assuming both stores offer a 10% discount on the sale price.

Explain why matrix addition and subtraction are only deﬁned for matrices of the same size, while scalar multiplication is always deﬁned, regardless of the matrix’s dimensions.

If C

2 1 5

1 3 4 0 and D 2 1

1 3 2

0 1 5

2 4 , determine matrices X and Y such that 3

3C − 4X = D and 2C + Y = 3D. 28

The determinant of a matrix A, denoted as det (A), is a scalar value that provides important information about the matrix, such as whether it is invertible. Given a 2 × 2 matrix A

a b , the determinant of A is calculated by: c d

det (A) = ad – bc Prove the property det (αA) = α2 det (A) for a 2 × 2 matrix A.

5.04 Matrix multiplication After this lesson, you will be able to… • determine when two matrices can be multiplied • multiply matrices • apply matrix multiplication to practical situations

Conditions for matrix multiplication Matrix multiplication Matrix multiplication is the process of multiplying one matrix by another matrix. Two matrices A and B can only be multiplied if the number of columns in the first matrix is equal to the number of rows in the second matrix.

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If matrix A is of size m × n and matrix B is of size n × q, then the condition for matrix multiplication is satisﬁed, and the product of A and B can be calculated. Size of 1st matrix

Size of 2nd matrix

m× n

n ×q Must be equal

The size of the resulting matrix will be m × q, or the number of rows in the ﬁrst matrix by the number of columns in the second matrix. Size of 1st matrix

Size of 2nd matrix

m×n

n× q

Size of the answer matrix

Example 1 Consider these matrices: G=

,H=

,I=

Determine whether these matrix multiplications are defined. If defined, determine the size of the resulting matrix: a IGH

Apply the idea I

12 13 14

2 3 4 5

6 7 8 9 10 11

3× 1

2× 2

2 ×3

Not equal

444

Equal

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The matrix multiplication is not possible as the number of columns in matrix I (1) is not equal to the number of rows in matrix G (2).

b GHI

Apply the idea G

2 3 4 5

6 7 8 9 10 11

12 13 14

2×2

2× 3

3× 1

Equal

The matrix multiplication GHI is defined. The resulting matrix will be 2 × 1 in size.

Equal

Answer 2 × 1

Reflect and check The matrix multiplication GHI can be performed in steps: • Consider the size of G (2 × 2) and the size of H (2 × 3). • Since the inner dimensions match (2 = 2), the product GH is possible. • The resulting size of GH is determined by the outer dimensions, which will be 2 × 3. Then, • The size of GH is 2 × 3 and the size of I is 3 × 1. • Since the inner dimensions match (3 = 3), the product GHI is possible. • The resulting size of GHI is determined by the outer dimensions, which will be 2 × 1.

Idea summary Matrix multiplication is the process of multiplying a matrix by another matrix. Two matrices can only be multiplied if the number of columns in the ﬁrst matrix is equal to the number of rows in the second matrix. Must be the same (m × n) (n × p) = (m × p)

order order of A of B

order of AB

If matrix multiplication is possible, the resulting matrix will have the same number of rows as the ﬁrst matrix and the same number of columns as the second matrix.

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Multiplication process Interactive exploration Discover this concept in action online

mathspace.co

The simplest matrix product is the product of a row matrix and column matrix. Each entry in the row matrix is multiplied by the corresponding entry in the column matrix, and the sum of these products is the result. a11

a12

a13

b11 b21 b31

a11b11

a12b21

a13b31

This is why the number of columns of the left matrix must equal the number of rows of the right matrix. This process of multiplying a row by a column is repeated for each row of the first matrix with each column of the second matrix to form the elements of the product matrix. To multiply two matrices: 1. Check if the multiplication is defined and determine the size of the resulting matrix. • Verify that the number of columns in the first matrix matches the number of rows in the second matrix. • The resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix. 2. Represent the resulting matrix using element notation. • When a matrix A (with elements aik) is multiplied by a matrix B (with elements bkj) to get a product matrix C, the element in the ith row and jth column of the resulting matrix C is denoted as cij. 3. Calculate each element by multiplying rows of the first matrix with columns of the second matrix and summing the products. • For each element cij, multiply each element in the ith row of the first matrix (e.g., A) by the corresponding element in the jth column of the second matrix (e.g., B). • Sum these products to obtain the value for cij. If the first matrix A has dimensions m × n and the second matrix B has dimensions n × p, then their product C = AB will be an m × p matrix. The element cij in the ith row and jth column of the product matrix C is calculated using the formula:

cij cij aik bkj n

446

n k 1

aik bkj

is the element in the ith row and jth column of the product matrix C is the element in the ith row and kth column of the first matrix A is the element in the kth row and jth column of the second matrix B is the number of columns in matrix A (which is equal to the number of rows in matrix B)

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 Evaluate this matrix multiplication: A

Apply the idea 1. Check if the multiplication is defined and determine the size of the resulting matrix. 3×2

3×3

3×2

The multiplication is defined, and the resulting matrix will be a 3 × 2. 2. Represent the resulting matrix using element notation.

a11

a12

a21 a31

a22 a32

3. Calculate each element by multiplying rows of the ﬁrst matrix with columns of the second matrix and summing the products. Start with the element a11. A=

a11 a12 a21 a22 a31 a32

Run across row 1 and climb down column 1. 1 2 3 4 5 6 × 7 8 9

10 13 11 14 12 15

a11 = 1 × 10 + 2 × 11 +3 × 12 = 68 Next, consider element a12. A=

68 a12 a21 a22 a31 a32

Run across row 1 and climb down column 2. 1 2 3 4 5 6 × 7 8 9

10 13 11 14 12 15

a12 = 1 × 13 + 2 × 14 + 3 × 15 = 86 Next, consider element a21. A=

68 86 a21 a22 a31 a32

Run across row 2 and climb down column 1. 1 2 3 4 5 6 × 7 8 9

10 13 11 14 12 15

a21 = 4 × 10 + 5 × 11 + 6 × 12 = 167

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Start with the element a22.

Run across row 2 and climb down column 2.

68 86 167 a22 a31 a32

1 2 3 4 5 6 × 7 8 9

10 13 11 14 12 15

a22 = 4 × 13 + 5 × 14 + 6 × 15 = 212 Next, consider element a31. A=

Run across row 3 and climb down column 1. 1 2 3 4 5 6 × 7 8 9

68 86 167 212 a31 a32

10 13 11 14 12 15

a31 = 7 × 10 + 8 × 11 + 9 × 12 = 266 Next, consider element a32. A=

Run across row 3 and climb down column 2. 1 2 3 4 5 6 × 7 8 9

68 86 167 212 266 a32

10 13 11 14 12 15

a32 = 7 × 13 + 8 × 14 + 9 × 15 = 338 Therefore, the resulting matrix is: A

Reﬂect and check Step 3 can be done with the working out in the matrix. 10 13 11 14 12 15

1 2 3 4 5 6 × 7 8 9

a11 a12 a21 a22 a31 a32

1 × 10 + 2 × 11 + 3 × 12 4 × 10 + 5 × 11 + 6 × 12 7 × 10 + 8 × 11 + 9 × 12 68

167

212

266 338

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Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 × 13 + 2 × 14 + 3 × 15 4 × 13 + 5 × 14 + 6 × 15 7 × 13 + 8 × 14 + 9 × 15

Example 3 Frank owns two pizza stores, Pepperoni Palace and Crispy Crust, at which he sells small pizzas for $7, medium-sized pizzas for $15, and large pizzas for $28: Small

Medium

Large

Pepperoni Palace

Crispy Crust

a Organise the prices in a column matrix.

Create a strategy Create a column matrix for the prices, arranging them in the order of small, medium, and then large from top to bottom.

Apply the idea A

7 S 15 M 28 L

b Organise the number of pizzas sold in a 2 × 3 matrix.

Create a strategy Express each numerical element of the table in the corresponding location of the matrix.

Apply the idea The matrix as given in the table: S M L P 21 25 12 B C 26 11 22

c Calculate Frank’s total revenue for each store.

Create a strategy Use the matrices A and B from part (a) and (b), and matrix multiplication.

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Apply the idea To calculate the total revenue for each store, determine the correct order to multiply matrices A and B. Start by considering the information the final matrix should contain. The question asks for the revenue for each store — specifically, the revenue for Pepperoni Palace and Crispy Crust — so the final matrix should be either a 2 × 1 or 1 × 2 matrix. BA AB Next, check if both AB and BA are possible, and determine ( 3 × 1 )( 2 × 3) 2× 3 3× 1 which order gives the correct result. Since AB is not possible, but BA is, multiply B by A to obtain the required 2 × 1 matrix representing the total revenue for each store. 7 21 25 12 BA = 15 26 11 22 28

Not possible

Possible 2×1

Write the equation with the matrices

(21 7 ) (25 15) ( 12 28) (26 7 ) ( 11 15) (22 28)

Determine the calculation for each element

858 963

Evaluate each element

Reflect and check Check the matrix multiplication by considering how the answer could be calculated without matrices. For example, the revenue for Pepperoni Palace can be found by multiplying the number of small pizzas by the cost of a small pizza, the number of medium pizzas by the cost of a medium pizza, and the number of large pizzas by the cost of a large pizza, then adding these products together. Revenue for Pepperoni Palace = 21 × 7 + 25 × 15 + 12 × 28 = 858 This corresponds to the calculation in the matrix multiplication.

Idea summary To multiply two matrices: 1. Check if the multiplication is possible and determine the size of the resulting matrix. 2. Represent the resulting matrix using element notation. 3. Calculate each element by performing a row-by-column multiplication: sum the products of corresponding entries from a row of the first matrix and a column of the second matrix.

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Properties of matrix multiplication Matrix multiplication follows certain properties that allow for simplifications and rearrangements. These laws are crucial for working with matrices in various mathematical and applied contexts. Properties of matrix multiplication include: • Not always commutative: AB ≠ BA (sometimes it is, for example IA = AI, where I is the identity matrix) • Associative: (AB) C = A (BC) • Left distributive: A (B + C ) = AB + AC • Right distributive: (B + C ) A = BA + CA The identity matrix for matrix multiplication is a special matrix that, when multiplied by any matrix of compatible dimensions, leaves the original matrix unchanged. For a 2 × 2 matrix, the identity matrix is: 1 0 0 1

I This means that for a 2 × 2 matrix A: a b c d

a b c d

1 0 0 1

The multiplicative identity for matrices is also commutative. Therefore, for any matrix A: AI = IA = A

Example 4 Determine the matrix that satisfies this equation: 6 1 5 9

6 1 5 9

Create a strategy Recall the multiplicative identity of matrices.

Apply the idea Notice that the result matrix is the same as the original matrix. The multiplicative identity for matrices is the identity matrix. So for any matrix A, AI = IA = A. 6 1 5 9

1 0 0 1

6 1 5 9

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Example 5 Consider these 2 × 2 matrices: A

3 5 ,B 3 2

1 2 ,C 2 3

5 4 5 4

a Calculate A × (B × C).

Create a strategy To multiply two matrices, multiply each row in the first matrix by each column in the second.

Apply the idea A × (B × C) =

3 5 3 2

1 2 2 3

5 4 5 4

3 5 3 2

1 5 2 5 1 4 2 4 2 5 3 5 2 4 3 4

Multiply B and C

3 5 3 2

15 12 25 20

Evaluate the second product

3 15 5 25 3 12 5 20 3 15 2 25 3 12 2 20

Multiply A by the product B × C

170 136 95 76

Evaluate

Substitute the matrices A, B, and C

b Calculate (A × B) × C.

Apply the idea (A × B) × C =

452

3 5 3 2

1 2 2 3

5 4 5 4

Substitute the matrices A, B, and C

3 1 5 2 3 2 5 3 3 1 2 2 3 2 2 3

5 4 5 4

13 21 7 12

13 5 21 5 13 4 21 4 7 5 12 5 7 4 12 4

Multiply the product A × B by C

170 136 95 76

Evaluate

5 4 5 4

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Multiply A and B Evaluate the first product

c Which property do parts (a) and (b) demonstrate? A

(A + B) × C = AC + BC

C × (A + B) = CA + CB

A × (B × C) = (A × B) × C

A + (B + C) = (A + B) + C

Create a strategy Think about the operation used to determine the solutions for parts (a) and (b).

Apply the idea Comparing the results from part (a) A × (B × C) and part (b) (A × B) × C, they are identical. This shows that for these matrices, it does not matter how the matrices are grouped in multiplication; the final answer for the sequence A × B × C is the same. This demonstrates the associative property of matrix multiplication. The correct answer is option C. A × (B × C) = (A × B) × C

Idea summary Properties of matrix multiplication include: • • • •

Not always commutative: AB ≠ BA Associative: (AB)C = A(BC) Left distributive: A(B + C) = AB + AC Right distributive: (B + C)A = BA + CA

The multiplicative identity for matrices is the identity matrix I: AI = IA = A

Did you know?

Matrix multiplication is essential in 3D modelling and digital design, where it is used to rotate, scale, and move objects on a computer screen. By multiplying matrices with coordinate data, software can accurately transform complex models used in areas such as animation, engineering, and product design. 5.04 Matrix multiplication mathspace.co

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5.04 Practice questions What do you remember? 1

For each pair of matrices, determine whether the product AB or BA is possible. If possible, determine the size of the resulting matrix: a

A = [8] and B = [4]

and B = [4 9 − 5]

and B =

4 A = –1 2

and B =

and B = and B =

and B = 5 0 and B = 6 –2 1 8 7 –4 3

For these matrices:

Determine whether these matrix multiplications are possible. If possible, determine the size of the resulting matrix. a 3

Consider A =

. If A can be multiplied by a column matrix B, what is the size of

matrix B? 4

If A is a 4 × 2 matrix and the product AB is a 4 × 5 matrix, what is the size of B?

Matrix P is a 2 × 4 matrix while Q is a 4 × 2 matrix. Explain whether these matrix expressions involving P and Q are possible and justify your answers: a

454

P+Q

P−Q

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice Ex 1

For these matrices:

Determine whether these matrix multiplications are possible. If possible, determine the size of the resulting matrix.

ABC

DEF

BCD

CDE

EFA

FAB

DBA

ADF

A matrix calculation of A = BC resulted in the matrix A =

If B is a 3 × 2 matrix, what is the size of matrix C?

The matrix A

6 2 5

2 1 8

3 3 is used in the matrix multiplication BAC, where BAC is 1

a 3 × 3 matrix: a Ex 2

What is the size of B?

What is the size of C?

and B =

Calculate AB for each pair of matrices:

and B =

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Consider the matrices A =

and B =

What is the size of AB?

Determine the matrix AB.

What is the size of BA?

Determine the matrix BA.

Is AB equal to BA?

Consider the row matrix R = [1

3] and the column matrix C =

Determine these matrices: a Ex 3

Ex 4

Tobias owns two pizza stores, Paris Pizza and Penrith Pizza, at which he sells small pizzas for $5, medium-sized pizzas for $11, and large pizzas for $18. The table shows the number of pizzas sold at each store on a particular day: Small

Medium

Large

Paris Pizza

Penrith Pizza

Organise the prices into a column matrix, A, in increasing size order.

Organise the number of pizzas sold into a 2 × 3 matrix, B, as given in the table.

Calculate Tobias’ total revenue for each store using matrix multiplication.

Consider A =

and I =

Determine the matrix AI.

Determine the matrix IA.

Discuss the signiﬁcance of the identity matrices in matrix multiplication.

Let E =

be the zero matrix, and F =

Determine the product EF.

What happens when any matrix is multiplied by a zero matrix? Explain why this result occurs.

Solve these equations for n: a

456

3 n

7 9

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 9 3 4 2 n

8 6 4

58 32

Ex 5

Consider these matrices: 7 9 7 , 3 6 7

Calculate (A + B) × C.

Calculate A × C + B × C.

6 4 9 4 2 4

9 6 6 , C 7 8 6

What property do parts (a) and (b) demonstrate? A

A + (B + C) = (A + B) + C

C × (A + B) = CA + CB

A × (B × C) = (A × B) × C

(A + B) × C = AC + BC

Given the matrices: X

1 2 , Y 3 4

5 6 , 7 8

2 1

Show that (X + Y) Z = XZ + YZ. 18

A summing matrix is a row or column matrix where all the elements are 1. The purpose of a summing matrix, when used in matrix multiplication, is to calculate the sum of the elements in each row or each column of another matrix. Construct a matrix multiplication to sum the numbers 5, 8, 3, 2, and 6.

Extend your thinking 19

Compare and contrast scalar multiplication with matrix multiplication in terms of what each involves, when each is possible, and the size of the resulting matrix.

The transpose of a matrix A, denoted AT, is formed by swapping the rows and columns of A. For example, if A =

, then AT =

The 2 × 2 matrices A, B, and C are given: 16 18 Determine the value of x, given BT AT = C. 21

Consider this matrix equation:

Identify the two linear equations that this matrix equation represents. Explain how the matrix format corresponds to individual linear equation.

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Determine the values of a and b:

In fields like computer graphics and physics, matrices are often used to represent transformations (like moving or resizing objects). When a matrix A is multiplied by a point P (represented as a column matrix), the result is a new point Q, which is the transformed version of P. This method helps model changes such as rotations, translations (shifts), and scaling (resizing). 3 c , where c and d are constants. 1 d The matrix T maps the point M(3, 4) onto the point N (5, 1). Determine the value of c and d. The 2 × 2 matrix T is defined as: T

In transportation logistics, matrices can be used to model and analyse aspects such as speed, time, and distance for various routes. Suppose a company is analysing the transportation of goods between three routes (Route 1, Route 2, Route 3) using two vehicles (Vehicle 1, Vehicle 2) over two different days (Day 1, Day 2). The following matrices represent the speeds of vehicles (in kilometres per hour) and the time taken (in hours) for these routes. Let S be the speed matrix, where rows represent the two vehicles (Vehicle 1, Vehicle 2) and columns represent the three routes (Route 1, Route 2, Route 3): S= Let T be the time matrix, where rows represent the three routes (Route 1, Route 2, Route 3) and columns represent the two days (Day 1, Day 2): T= Calculate and interpret the distance matrix D, which represents the total distances covered by each vehicle across all three routes for the two different days. Provide a clear explanation of how the three dimensions (routes, vehicles, days) are handled in this calculation.

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5.05 Multiplicative inverses and determinants After this lesson, you will be able to… • calculate and interpret the determinant of a matrix • determine the multiplicative inverse of a matrix, if possible

Multiplicative inverses and determinants Exploration 1. What number, when multiplied by any real number A, gives A as the result? 2. What number, when multiplied by a real number M, gives a result of 1? 3. Is there a number that, when multiplied by any natural number n, results in 1? 4. For a real number D, what value(s) make

1 undefined? D

The multiplicative inverse of a number is the value that, when multiplied by the original number, results in 1. 1 1 For example, 6 , where is the multiplicative inverse of 6 (and vice versa). 6 6 For a square matrix A, the multiplicative inverse, denoted A–1, is a matrix such that when A is multiplied by A–1, the identity matrix I is obtained. Recall that the identity matrix, I, is a square matrix with 1’s in the leading diagonal and 0’s in all other entries. It has the property that for any matrix A: AI = A = IA Therefore, the multiplicative inverse property is expressed as: AA–1 = I = A–1A Note that A 1

1 , as the division of matrices is not defined; instead, the inverse A–1 is used to A

achieve a similar effect to division for matrices.

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Before determining the multiplicative inverse matrix, it is important to first identify the determinant of a square matrix A, written as det (A) or |A|. The determinant is a single number associated with the matrix A and tells us if the inverse of a matrix can be found. For a general 2 × 2 matrix A

a b , the determinant is calculated as follows: c d

det (A) = ad – bc det (A) a, b, c, d

is the determinant of matrix A, also written as |A| are the elements of the matrix A

If the determinant is zero, the multiplicative inverse does not exist. A matrix with no inverse is called a singular matrix. For a 2 × 2 matrix A given by:

a b , provided its determinant is not zero, the multiplicative inverse A–1 is c d

A 1

d 1 det (A ) c

b a

A–1 is the multiplicative inverse of matrix A det (A) is the determinant of matrix A, det (A) ≠ 0 d –b –c a

is the matrix formed by swapping the elements on the main diagonal and changing the signs of the other two elements

1 From this definition, it follows that if det (A) = 0, the scalar multiple would be undefined because 0 is undefined. Therefore, a determinant of zero implies a singular matrix. While this lesson focuses on the manual calculation for 2 × 2 matrices, technology can also be used to find the determinant and inverse of matrices. This is especially useful for larger square matrices, where manual calculation becomes complex. Typically, the matrix is entered into the technology and then a specific function or operator (often looking like A–1 or a command such as inv(A)) is used to find its inverse directly, if it exists.

Did you know?

Mathematicians have found that even music composition can be modelled with matrices — some computer programs use them to generate harmonies and melodies automatically!

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Example 1 8 2 . 1 9

Consider the matrix A

a Calculate the determinant of A.

Create a strategy The determinant of a 2 × 2 matrix is the difference between the products of the diagonals.

Apply the idea det (A)= ad − bc

Write the determinant formula

= − 8 × 9 − [2 × (− 1)]

Substitute a = − 8, b = 2, c = − 1, d = 9

= − 72 − (− 2)

Evaluate the multiplication

= − 70

Evaluate

This means matrix A is invertible.

b Determine the inverse A–1.

Create a strategy For matrix A

a c

b , the multiplicative inverse is defined as: A 1 d

1 det (A)

d c

b a

Apply the idea From part (a), it was determined that det (A) = − 70. A 1 =

1 det (A)

d c

1 9 70 1

2 8

9 = 1

1 70 1 70

b a

Write the equation Substitute the values

2 8

1 70 1 70

Multiply the matrix by the scalar

9 70 1 70

2 70 8 70

Evaluate

9 70 1 70

1 35 4 35

Simplify

1 70

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Reflect and check To verify the inverse, multiply A × A– 1: A A

8 2 1 9

9 70 1 70

1 35 4 35

1 70

1 1 35

9 70

1 70

72 70 = 9 70

2 70 9 70

8 35 1 35

8 35 36 35

70 70

35 35

1 0

Substitute each matrix

1 35

4 35

4 9 35

Compute each element

Evaluate the multiplication

Evaluate the addition and subtraction

0 1

Simplify

This is the 2 × 2 identity matrix, confirming that A– 1 is correct, as the inverse satisfies A × A– 1 = I.

Example 2 Consider the matrix B

2 1 . 4 2

a Calculate the determinant of matrix B.

Create a strategy For a matrix

a b , the determinant is ad – bc. c d

Apply the idea det (B) = ad – bc = (2)(2) – (1)(4)

Substitute a = 2, b = 1, c = 4, d = 2

=4–4

Evaluate

Simplify

The determinant of matrix B is 0.

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Write the equation

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b Determine the multiplicative inverse B–1, if possible.

Create a strategy Recall that the inverse of a matrix A

a b is A 1 c d

d 1 det (A) c

b . The inverse exists only if a

det (A) ≠ 0.

Apply the idea From part (a), we found that det (B) = 0. Since the determinant is 0, the term

1 1 would be , which is undefined. 0 det (B)

Therefore, the matrix B is a singular matrix, and its multiplicative inverse B–1 does not exist.

Example 3 Consider a 2 × 2 matrix

a b : c d

a Write down the formula for the inverse of this matrix. What condition on a, b, c, and d must be satisfied for the inverse to exist?

Create a strategy Recall the formula and the determinant’s role in verifying whether an inverse exists.

Apply the idea The formula is

d ad bc c 1

b provided that ad − bc ≠ 0. a

b Verify that the formula in part (a) is correct by performing the necessary calculations.

Create a strategy Multiply the matrix

a b by the matrix formula in part (a) if it results in the identity matrix I. c d

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Apply the idea A A 1 =

a b c d

d ad bc c

b a

ad bc cd cd

ab ab bc ad

ad bc

ad ad = cd ad

bc bc cd bc

ab ad ad ad

ab bc bc bc

Perform matrix multiplication

Multiply the matrix by the scalar

1 0 0 1

Substitute the matrices

1 ad bc

Simplify

Substitute I

1 0 0 1

c Verify the inverse relationship using matrix multiplication

Create a strategy Use matrix multiplication to confirm the inverse relationship by ensuring that multiplying A by A–1 and A–1 by A both results the identity matrix I.

Apply the idea In part (b), it was shown that AA–1 = I. By the definition of a multiplicative inverse, it must also be true that A–1A = I. Verifying this: d b ad bc c a 1

A 1 A

a b c d

Substitute the matrices

da bc db bd ad bc ca ac cb ad

Perform matrix multiplication

ad bc 0 ad bc 0 ad bc

Simplify terms

1 1

ad bc ad bc = 0 ad bc =

0 ad bc ad bc ad bc

1 0 0 1

This is the identity matrix I. –1

Since both AA = I and A A = I, the inverse relationship is verified.

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1 ad bc

Simplify

=I –1

Multiply each element by

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Idea summary For a square matrix A, the multiplicative inverse, denoted as A–1, is a matrix such that when A is multiplied by A–1, the identity matrix I is obtained. The determinant, det (A) or |A|, is a single number associated with the matrix A and tells us if the inverse of a matrix can be found. If A

a b , then det (A) = ad – bc c d

If the determinant is zero, the multiplicative inverse does not exist. A matrix with no inverse is called a singular matrix. For a 2 × 2 matrix A

a b , the multiplicative inverse is defined as: c d 1 det (A)

A 1

d c

b a

5.05 Practice questions What do you remember? 1

Consider the set of all 2 × 2 matrices with real entries: a

Define the identity element in the context of matrix multiplication.

Identify a 2 × 2 identity matrix and calculate its determinant.

For a 2 × 2 matrix A a

Write the formula for the determinant.

Write the formula for the inverse.

Calculate the determinants of these matrices: a

a b : c d

1 2 2 3

Consider a 2 × 2 matrix A

10 1 0 3

3 4 : 5 6

What condition on 3, 4, 5, and 6 must be satisfied for the inverse of A to exist?

Write down the formula for the inverse of this matrix.

Using the inverse formula you wrote in part (b), verify that it is correct by calculating AA–1 and showing it equals I.

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Practice Ex 1

For each matrix: i

Determine the value of the determinant.

2 3 1 6

Let A a

Ex 2

0 2

4 3

2 7

3 1

Verify that AA–1 = A–1 A = I.

det (–A)

1 2 . Calculate: 3 4

det (A)

Let A

5 2

1 3 : 1 2

Determine A–1.

Let A a

4 5

Determine the multiplicative inverse.

4 1 , B 3 1

det (2A)

1 3

(A–1)–1

1 : 4

Determine AB.

Explain whether B is the multiplicative inverse of A or not. 3 2 is A 1 4 1

1 5

1 n . Solve for n. 4 3

The inverse of A

Let A and B be matrices and assume that A–1 exists. Oliver claims that ABA–1 = AA–1B = B. Explain if Oliver is correct, or if not, provide an example that shows the statement is false.

Consider the matrix A

466

1 0

1 : 0 0 0 . 0 0

Determine a matrix, B, such that AB

Eliza claims that B is the multiplicative inverse of A. Is Eliza’s claim correct? Explain your answer.

Consider the matrix C

6 3 : 4 2

Calculate the determinant of matrix C.

Determine the multiplicative inverse C –1, if it exists. Explain your reasoning.

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Ex 3

Consider a 2 × 2 matrix M =

p q : r s

Write down the formula for the inverse of this matrix. What condition on p, q, r, and s must be satisfied for the inverse to exist?

Using the formula from part (a), verify its correctness by showing that MM–1 = I.

Show that M–1M = I using the formula from part (a). 4 1 and B 3 2

Consider the matrices A

1 2 : 2 5

Calculate det (A) × det (B).

Calculate det (AB).

Compare your results for parts (a) and (b). What do you notice?

Consider the matrix A an inverse.

Consider the matrix B

3 4 . Using only the determinant, determine whether A has 2 6 3 x where x is a real number. For which values of x does x 2

B have no inverse? 16

Let A

k 2

2 2 3

. Determine the value(s) of k for which A–1 exists and determine A–1.

Extend your thinking 17

If A and B are two matrices such that AB = I, determine an equation that relates det (A) and det (B).

A matrix is called self-inverse if A = A–1. Show that if A = A–1, then A2 = I.

Consider the matrix A

1 1 : 0 1

Determine, in terms of the natural number n, the matrix An.

Calculate det (An).

Determine the values of x such that A

x 3 3 is singular. 2 x 2

Let A and B be 2 × 2 matrices and have inverses. Prove that (AB)–1 = B–1A–1.

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5.06 Solve matrix equations After this lesson, you will be able to… • solve matrix equations of the form AX + C = B • represent systems of linear equations using matrices • solve simultaneous equations using inverse matrices • interpret matrix solutions in terms of the original equations

Solve matrix equations algebraically Exploration Solve these equations, making notes of the operations applied to the equation when solving: 1. 2x = 4

2. x × 5 = 10

3. 3x + 5x = 1

Similar to solving algebraic equations, matrix equations can be solved using matrix algebra. Recall that two matrices, A and B, can be added and subtracted element-wise, provided that A and B have the same dimensions. Recall that AA– 1 = A– 1A = I, where A– 1 is the multiplicative inverse of A and I is the identity matrix. Division for matrices is not defined. This is because matrices do not always have an inverse. Instead of matrix division, the pre-multiplication and post-multiplication of inverse matrices is used to solve matrix equations. To solve AX = B, for X, pre-multiply A– 1 (multiply the A– 1 at the front) on both sides: AX = B –1

A AX = A – 1B IX = A– 1B, since A– 1A = I X = A– 1B To solve XA = B, for X, post-multiply A– 1 (multiply the A– 1 at the rear) on both sides: XA = B XAA– 1 = BA– 1 XI = BA– 1, since A– 1A = I X = BA– 1 More complex matrix equations, such as AX + C = B or AX + BX = C, need to first be rearranged to isolate the matrix X, similar to solving a linear equation. The order of these operations is crucial, because matrix multiplication is not commutative, so the order of matrices and operations matters. Once the X matrix is isolated, an inverse matrix can be used to solve for X.

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Example 1 Let A

4 2 ,B 7 6

3 2

1 ,C 3

5 and D = [2 – 4]. 3

Solve for matrix X when: a AX = C

Create a strategy Calculate det (A) to check whether A− 1 exists. If yes, and since A is in front of X, then it indicates the need to pre-multiply both sides by A− 1 and apply the property that AA− 1 = I.

Apply the idea First, calculate det (A): det ( A) = ad – bc

Write the formula

= (4 × 6) – (– 2 × (– 7))

Substitute the values

= 10

Evaluate

Since det ( A) is 10, A– 1 exists and is equal to: A 1

1 det (A)

d b a c

Apply the 2 × 2 inverse matrix formula

1 6 2 10 7 4

Substitute the values

1 1 2 10 10 1 1 4 7 10 10

Multiply each element by the scalar

6 10 7 10

2 10 4 10

Evaluate

3 5 7 10

1 5 2 5

Simplify

1 10

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To solve for X: AX = C –1

Write the matrix equation

–1

Pre-multiply both sides by A– 1

A AX = A C IX = A– 1C

Apply the property that A– 1A = I

X = A– 1C

Simplify

3 5 X= 7 10

1 5 2 5

5 3

12 5 23 10

Substitute the matrices

Evaluate

Reflect and check Verify if AX = C. LHS = AX 4 7

Write the left-hand side 2 6

12 5 23 10

5 3

=C = RHS Since LHS = RHS, the solution is correct.

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Substitute the matrices

Evaluate Substitute C =

5 3

b XA = D

Create a strategy Calculate det (A) to check whether A− 1 exists. If yes, and since A is after X, then it indicates the need to post-multiply both sides by A− 1 and apply the property that A− 1A = I.

Apply the idea From part (a), det (A) is 10. To solve for X: XA = D

Write the matrix equation

= DA

Post-multiply both sides by A– 1

XI = DA– 1

Apply the property that AA– 1 = I

XAA

–1

X = DA– 1

Simplify

3 X = [2 – 4] × 5 7 10 8 6 = 5 5

1 5 2 5

Substitute the matrices

Evaluate

Example 2 Consider the matrices A

4 7

2 ,B 6

3 2

1 and C 3

5 . 3

Solve for matrix X when AX + BX = C.

Create a strategy X is in common, so factorise X at the rear. Then, apply the property that (A + B)– 1(A + B) = I.

Apply the idea For AX + BX = C, first factorise the common matrix X at the rear: AX + BX = C

Write the original equation

(A + B) X = C

Factorise X on the right

To solve for X, pre-multiply by (A + B)– 1 to both sides: (A + B)– 1(A + B) X = (A + B)– 1C

Pre-multiply both sides by the inverse of (A + B)

–1

Apply the inverse property

–1

Apply the identity property

IX = (A + B) C X = (A + B) C

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Now the matrices can be substituted and solved: X = (A + B)– 1C 4 7

2 6

7 5

1 3

( 7 3)

Write the matrix equation 3 2

1 3

5 3

Substitute the matrices

5 3

Add the matrices inside the brackets

3 1 1 ( 1 ( 5) ) 5 7

5 3

Apply the inverse matrix formula

1 12 16 4

Evaluate determinant and matrix multiplication

3 4 1 4

Evaluate

Example 3 Let P

2 1 , Q 3 2

1 , and R 0

7 . 11

Solve the matrix equation PX + Q = R for matrix X.

Create a strategy To solve for X, first isolate the term PX by subtracting matrix Q from both sides of the equation. Then, pre-multiply by P − 1 to solve for X. Calculate P − 1 first.

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Apply the idea The given equation is PX + Q = R. Step 1: Subtract Q from both sides. PX + Q – Q = R – Q PX = R – Q

Subtract Q from both sides Simplify

Substitute the matrices for R and Q: 7 1 11 0 7– 1 11 – 0 6 11

So, PX

Substitute matrices R and Q Subtract the matrices Simplify

6 . 11

Step 2: Find P – 1. First, calculate det (P): det (P) = (2 × 2) – (1 × 3)

Write the equation

=4–3

Simplify

Evaluate

Since det (P) is not zero, the inverse exists. P 1

Step 3: Pre-multiply PX P 1 PX

2 1 det (P) 3 1 1 2 1 3 2 2 1 3 2

1 2

6 by P– 1. 11 6 P 1 11

2 3

1 6 2 11

(2 6) (–1 11) (–3 6) (2 11)

Write the inverse formula Substitute the determinant Simplify

Write the equation Apply the inverse property and substitute matrices Perform matrix multiplication

12 11 18 22

Evaluate

1 4

Simplify

The solution for matrix X is

1 . 4

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Reflect and check Verify the solution by substituting X

1 back into PX + Q = R. 4

LHS = PX + Q

Write the left-hand side of the equation

2 1 1 3 2 4

1 0

( 2 1) ( 3 1)

( 1 4) (2 4)

2 4 3 8

1 0

6 11

1 0

Substitute the matrices P, X, and Q 1 0

Perform the matrix multiplication Evaluate the multiplication within the matrix Simplify the first matrix

6 1 11 0

Perform the matrix addition

7 11

Simplify

This matches the matrix R

= RHS

The left-hand side equals the right-hand side

Since LHS = RHS, the solution is correct.

Idea summary Solving matrix equations often involves using inverse matrices and properties of matrix algebra. The order of multiplication is critical. • To solve AX = B, pre-multiply by A– 1: X = A– 1B. • To solve XA = B, post-multiply by A– 1: X = BA– 1. • To solve AX + BX = C, factorise X first: (A + B) X = C. Then pre-multiply by (A + B)– 1: X = (A + B)– 1C. This only works if A + B can be calculated and has an inverse. • To solve AX + C = B, first isolate AX: AX = B – C. Then pre-multiply by A– 1: X = A– 1(B – C). This only works if matrices have compatible dimensions and A is invertible. Always ensure that matrix dimensions are compatible for the operations being performed (addition, subtraction, multiplication) and that any matrix you are finding an inverse for is square and its determinant is non-zero.

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Applications of matrix equations Matrices and their inverses can be used to solve simultaneous equations. Remember, simultaneous equations are a set of two or more equations that share variables, and solving them means finding values for those variables that make all the equations true at the same time. To do this, the system of equations must first be represented in matrix form. Representing systems of simultaneous linear equations A system of linear equations like ax + by = m cx + dy = n can be rewritten as a single matrix equation: a b c d

x y

m n

This is often expressed in the general form:

AX = B A is the coefficient matrix, which contains the coefficients of the variables (the numbers in front of x and y) X is the variable matrix, a column matrix containing the variables B is the constant matrix, a column matrix containing the constants from the right-hand side of the equations a b , X c d

For the given matrix equation, A

x , and B y

m . n

Solving systems using inverse matrices Once the matrix equation AX = B is obtained, the variable matrix X can be solved by using the inverse of the coefficient matrix, A – 1 (if it exists): AX = B –1

–1

A AX = A B

Write the equation Pre-multiply both sides by A– 1

IX = A– 1B

Since A– 1A = I (the identity matrix)

X = A– 1B

Since IX = X

Remember, a matrix only has an inverse if its determinant is not zero. If det (A) = 0, the matrix A is singular, and the system of equations will not have a unique solution (it will have either no solutions or infinitely many solutions). Systems with unique solutions are typically the focus.

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Steps to solve simultaneous linear equations using matrices: 1. If it is a word problem, define your variables (e.g., let x be... and y be...). 2. Write the simultaneous linear equations based on the information given. Make sure they are in the form ax + by = c. 3. Set up the matrix equation AX = B by identifying the coefficient, variable, and constant matrices. 4. Calculate A– 1 (the inverse of the coefficient matrix). 5. Solve for X using the formula X = A– 1B. 6. Interpret the solution. State the values of the variables and what they mean in the context of the problem, if applicable. For systems with three or more simultaneous equations, these same principles apply, but calculating the inverse of larger matrices (e.g., 3 × 3) by hand can be lengthy, so technology is often used.

Example 4 There are two numbers, x and y. Their sum is 42, and their difference (x – y) is 16. Use matrices to find these two numbers.

Create a strategy First, write the two pieces of information as linear equations. Then, convert this system into a matrix equation AX = B. Solve for X using X = A– 1B. Finally, state the values of x and y.

Apply the idea Step 1: Write the equations From the information given: x + y = 42 x – y = 16 Step 2: Express in matrix form AX = B The coefficient matrix A, variable matrix X, and constant matrix B are: A

x ,B y

1 1 , X 1 1

42 16

So, the matrix equation is: 1 1 1 1

x y

42 16

Step 3: Calculate the determinant of A det (A) = (1 × – 1) – (1 × 1)

Write the equation

= –1 – 1

Simplify

= –2

Evaluate

Since det (A) is not zero, the inverse exists.

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Step 4: Find A–1 A 1

1 1

1 det ( A ) 1 1

1 2 1 2 1 2

1 1

Write the inverse formula Substitute the determinant

1 2 –1 2

Multiply the scalar into the matrix

Step 5: Solve for X using X = A– 1B x y

A 1B 1 2 1 2

Write the solution formula 1 2 –1 2

1 (42) 2 1 (42) 2 21 21 29 13

8 8

42 16 1 (16) 2 – 1 (16) 2

Substitute the matrices

Perform matrix multiplication

Evaluate Simplify

Step 6: Interpret the solution The solution matrix X

29 tells us that x = 29 and y = 13. 13

Reflect and check Checking these values with the original statements: • Sum: 29 + 13 = 42 • Difference: 29 – 13 = 16

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Example 5 At a school canteen, students can order the “Healthy Snack” (which contains 4 pieces of fruit and 2 savoury items) or the “Yummy Snack” (which contains 1 piece of fruit and 3 savoury items). On a particular day, the canteen used a total of 260 pieces of fruit and 140 savoury items. How many of each type of snack were made?

Create a strategy Define variables for the number of Healthy Snacks and Yummy Snacks. Set up two linear equations based on the total fruit used and total savoury items used. Convert this system to a matrix equation AX = B and solve for X using X = A– 1B.

Apply the idea Step 1: Define variables and write equations Let h be the number of Healthy Snacks made and y be the number of Yummy Snacks made. • Equation for total fruit: 4h + y = 260 • Equation for total savoury items: 2h + 3y = 140 Step 2: Express in matrix form AX = B A

4 1 , X 2 3

h ,B y

260 140

The matrix equation is: 4 1 2 3

h y

260 140

Step 3: Calculate the determinant of A det ( A) = (4 × 3) – (1 × 2)

Step 4: Find A A 1

478

Write the equation

= 12 – 2

Simplify

= 10

Evaluate

–1

1 1 3 10 2 4

Write the equation and substitute the values

3 1 – 10 10 2 4 – 10 10

Multiply each element by the scalar

0.3 0.2

Convert each element to decimal

0.1 0.4

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Step 5: Solve for X using X = A– 1B h y

0.3 0.2

0.1 260 0.4 140

Write the equation and substitute the values

(0.3 260) ( 0.1 140) ( 0.2 260) (0.4 140)

Perform matrix multiplication

78 14 52 56

Evaluate the multiplication

64 4

Evaluate

Step 6: Interpret the solution The canteen made 64 Healthy Snacks and 4 Yummy Snacks.

Reflect and check Check the total number of fruit: (64 × 4) + (4 × 1) = 256 + 4 = 260 pieces Check the total number of savoury items: (64 × 2) + (4 × 3) = 128 + 12 = 140 items The solution fits the conditions.

Idea summary To solve a system of simultaneous equations using matrices: • Represent: Write the system as a matrix equation AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. • Solve: If A– 1 exists (i.e., det (A) ≠ 0), find the solution using X = A– 1B. • Interpret: The values in the matrix X are the solutions for the variables in the original equations. Make sure to state these clearly, especially in the context of a word problem.

5.06 Practice questions What do you remember? 1

A and M are two matrices. Given that A– 1 exists, express the solution for X in terms of M and A– 1: a

AX = M

XA = M

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Consider the matrices A

1 2 and B 3 1

4 3 . 7 4

Which matrix X satisfies AX = B? X

3 1 4 3

Let X

3 4 : 1 6

2 1 1 1

Determine X – 1.

Given that A

2 1

5 5 10 5

If XY

1 0 1 1

3 2 , determine Y. 1 5

3 . 5

4 , determine X such that AX 6

Practice 5

Consider the matrices M

4 2 , N 1 3

10 and L 8

40 : 24

Ex 1

a 6

Let P a

Solve for Y in the equation MY = L.

If AP

3 2 : 4 7

Determine P – 1.

Consider the matrices M a

Solve for X in the equation MX = N.

2 4 and Z 8 5

Determine M – 1.

Consider the matrices A

1 : 2 b

2 1 ,B 1 3

3 4 , determine A. 8 6

If MA = Z, determine A.

2 1 and C 3 2

14 2 . 29 5

Which matrix X satisfies AX + BX = C ? A

9 Ex 2

480

5 1 1 0

Consider the matrices A

1 5 0 1

2 4 ,B 1 3

1 0 5 1

3 2 and C 2 1

6 2 . 4 2

Solve for X in the matrix equation AX + BX = C.

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1 0 0 5

10 Ex 3

Solve for X in each matrix equation for the given matrices: A

9 3 , B 4 5

a d

2 1

1 ,C 2

10 15

3 2

XD = B

XA + XB = C

DX – CX = E

AX – BX = A + B

XA + X = B

AX + E = F

Consider the matrices A

2 1 , B 4 3

4 2 ,E 1 3

5 ,C 7

1 , and F 2

9 and D 11

3 . 5

Solve for X in the equation AX + B = 2C – 3D. 1 2 , Q 2 3

2 1 and R 2 3

If P

Consider the matrices A

3 1 , determine X when PQX = R. 2 0

3 1 , B 0 2

1 0 and C 2 1

5 4 : 6 2

Solve for X in the equation AX = B.

Solve for Y in the equation BY = C.

Solve for Z in the equation XZ = Y, using the values of X and Y obtained in parts (a) and (b).

Consider the system of simultaneous linear equations:

Ex 4

2x + 5y = – 4 3x – 2y = 13

Ex 5

Represent this system as a matrix equation of the form AX = B.

Determine A– 1 and use it to solve for X.

State the values of x and y.

A café tracks its sales of coffee and tea over two days. On Day 1, they sold 12 coffees and 15 teas for a total revenue of $201. On Day 2, they sold 8 coffees and 20 teas for a total of $204: a

Represent this system as a matrix equation AX = B, identifying A, X and B. Let x be the price of coffee and y be the price of tea.

Calculate the inverse of matrix A, if exists, and use it to solve for matrix X. Interpret the solution in the context of the problem.

An electronics store sells two types of gadgets, phones and tablets. They recorded the sales: • Day 1: 6 phones and 4 tablets for $2400 • Day 2: 2 phones and 5 tablets for $1900 a

Formulate the matrix equation AX = B for this scenario. Let x1 be the price of a phone and x2 be the price of a tablet.

Calculate the inverse of A and use it to determine matrix X. Interpret the solution in the context of the problem.

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Extend your thinking 17

If A is a 3 × 3 matrix, its inverse is given by: A 1

1 adj (A)T det (A)

where adj (A)T is the transpose of the adjugate matrix, also known as the classical adjoint. To apply this formula for a 3 × 3 matrix, recall: a b For a matrix M = d e g h

c f : i

• The determinant is det (M) = a (ei − fh) − b (di − fg) + c (dh − eg). • The adjugate matrix, adj (M)T, which is the transpose of the cofactor matrix, is: adj(M )T

( ei f h) ( di fg) ( dh eg )

( bi ch) ( ai cg ) ( ah bg )

( bf ce) ( af cd) (ae bd )

1 3 2 0 4 1 ,C 2 3 0

4 2 5 3 6 7 8 1 9

Consider the matrices: A

2 1 4 0 3 5 ,B 1 2 3

Solve for X in the equation AX + B = C. 18

Consider the matrices: T

Solve for Y in the equation BY A = A + C. 19

2 1 3 4 1 2 , B 1 5 3

1 2 1 0 1 3 ,C 4 0 1

3 4 5 1 0 2 2 3 1

Consider the matrices: A

2 1 1 3 1 1 ,B 1 2 3

2 7 1 8 2 8 ,C 1 8 2

4 3 6 7 1 5 9 2 8

Solve for X in the equation (A– 1B + C) X + (B– 1C) = B.

Investigation: Existence of matrix solutions Investigate online

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Chapter 5 review Online review

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“Life is only meaningful when we are striving for a goal.” Aristotle

Big ideas • Graph theory provides a mathematical language to model networks by abstracting objects into vertices and their relationships into edges. The fundamental properties of these components, from vertex degree to overall connectivity, can be described visually, classified into standard forms, and represented numerically, allowing for the systematic analysis of any network’s structure. • The structure of a graph dictates its properties and determines solutions to practical optimisation problems. By applying key theorems and algorithms, we can answer fundamental questions about traversal (Can we cross every bridge once?), connectivity (What is the cheapest way to connect all nodes?), and design (Can this circuit be printed on a flat board?), turning abstract models into powerful problem-solving tools.

6 Graph theory Chapter outline 6.01 6.02 6.03 6.04 6.05 6.06 6.07 6.08

Introduction to graph theory Degree and adjacency Investigation: Hand-shaking lemma proof Graph isomorphisms Trees Planar graphs Walks, paths, and trails Algorithms in graph theory Applications of graph theory Chapter 6 review

486 499 513 526 542 554 570 581

Graphs and networks are behind the delivery apps that find your nearest bubble tea shop.

6.01 Introduction to graph theory After this lesson, you will be able to… • define the basic elements of a graph (vertices and edges) • construct and interpret graphs to model real-world situations

Graphs and networks Network Also known as a graph, is a diagram used to represent relationships between objects, people, or ideas. It consists of a set of points called vertices (or nodes) and a set of lines called edges that connect pairs of these vertices. A key application of networks is to model real-world situations, a process that simplifies complex problems for mathematical analysis.

Exploration To model a situation visually, consider four friends: Alice, Bob, Charlie, and Davina • Alice is friends with Bob and Davina. • Bob is friends with Alice and Charlie. • Charlie is only friends with Bob. • Davina is only friends with Alice. 1. How could dots be used to represent each friend? 2. How could lines between the dots be used to show who is friends with whom? 3. Sketch a representation on paper. Does the drawing clearly show the friendships?

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A point is called a vertex (or node). It is drawn as a dot or circle. The plural of vertex is vertices. An edge (or arc or link) is a line segment that begins and ends at a vertex. An edge represents a relationship between the objects or ideas that it links together, and the kind of relationship depends on the context. Edge Vertex

Vertex

On the left is a single vertex. On the right are two vertices with one edge between them. An edge must be drawn between exactly two vertices. Neither of these are valid edges. An edge must always connect two vertices; it cannot be left dangling from a single vertex. It also cannot connect three or more vertices at once. A network (sometimes called a network graph, or just graph) is a collection of vertices with edges drawn between them. 8

7 9 3 4

These networks demonstrate several features: • Some edges have direction (appearing as arrows), and some do not. • Sometimes a vertex is connected to itself by an edge. • Sometimes a pair of vertices is connected by more than one edge. • One of the networks has numbered edges (weighted edges). These networks use letters or words to label each vertex: B

Athens

A C

Chronos Rhea

Coeus Phoebe

Florence Leto

Zeus E

Berlin

Amsterdam Paris

Palermo

Oxford

Artemis

Vertices are often given vertex labels to identify them. These labels can be specific, such as the name of a city, or abstract, like a capital letter. For example, the first network uses the letters A, B, C, D, and E as labels, while the other two networks use descriptive labels to indicate what the vertices represent.

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Example 1 Which two of these graphs are valid networks? A

Create a strategy A valid network consists of at least one vertex, with all edges connecting exactly two vertices, including the possibility of a vertex connecting to itself via a loop.

Apply the idea Each network in the options has at least one vertex. Option A is invalid because it contains an edge that does not connect two vertices, as it has a dangling end. Option B is invalid because it includes an edge that does not connect two distinct vertices or form a loop, as it has a free end. Option C is valid because all edges connect exactly two vertices. Option D is valid because all edges, including the loop, connect exactly two vertices (a loop connects a vertex to itself). Therefore, Options C and D are valid networks.

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Example 2 This diagram displays newspaper routes between several houses: House B House F

House C House A House E House D

a The vertices are: A

The houses

The space enclosed by the routes

The routes

The connections between the houses

Apply the idea The vertices are the houses since there are house names in each oval where the edges intersect. Therefore, the correct answer is Option A.

b How many vertices are there?

Create a strategy Identify and enumerate the vertices in the network.

Apply the idea The vertices are: House A, House B, House C, House D, House E, and House F. There are 6 vertices.

c How many edges are there?

Create a strategy Count each line segment that connects a pair of vertices.

Apply the idea There are 10 edges.

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Example 3 A small local airline flies direct routes between four cities: Adelaide (A), Brisbane (B), Canberra (C), and Darwin (D). The available two-way direct routes are: • Adelaide to Brisbane • Adelaide to Canberra • Brisbane to Darwin • Canberra to Darwin Construct a graph to represent these cities and direct flight routes.

Create a strategy To represent the situation as a graph, identify the vertices and edges. • The vertices represent the cities: Adelaide (A), Brisbane (B), Canberra (C), and Darwin (D). • The edges represent the direct flight routes between the cities. Represent each city as a labelled vertex and each flight route as an edge connecting the appropriate vertices.

Apply the idea First, draw and label the four vertices to represent the cities: Adelaide (A), Brisbane (B), Canberra (C), and Darwin (D). Next, draw the edges to represent the flight routes: • An edge between A and B • An edge between A and C • An edge between B and D • An edge between C and D

The final graph looks like this: A

Idea summary A network (also called a graph) is a visual diagram used to model relationships. It consists of two fundamental components: • Vertices (or nodes): The points or dots that represent the individual objects (e.g., cities, people). The plural of vertex if vertices. • Edges (or arcs or link): The lines connecting pairs of vertices, which represent the relationships or connections between them. A key rule of a valid network is that every edge must connect exactly two vertices.

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Kinds of edges and networks Edges can be either directed or undirected. • Directed edges (also called arcs) are represented by arrows and symbolise a one-way relationship. • Undirected edges are represented by lines and symbolise a two-way connection. Shark

Eats

France

Spain

Borders Fish

The first network has a directed edge because sharks eat fish and not the other way around - a one-way relationship. The second network has an undirected edge because France and Spain share a border - a two-way relationship.

If a network contains directed edges, it is called a directed network (or digraph). If a network has no directed edges, it is called an undirected network. Considering the networks shown so far, those with arrows are directed, and the rest are undirected. A common question is whether a network can contain both directed and undirected edges. For clarity, networks are typically defined as either fully directed or fully undirected. If a two-way connection is needed in a directed network, it is represented by two separate directed edges. A two-way connection can still be expressed in a directed graph by drawing an edge in each direction. ‘Mixed’ networks are not typically used, as it is clearer if all edges in a graph are of the same type (either all lines or all arrows). A network is considered connected if it is possible to get from any vertex to any other vertex by travelling along the edges (it may take multiple steps). If some vertices or groups of vertices are separate from the rest, the network is disconnected.

The top two networks are connected, the bottom two are disconnected.

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An edge that starts and ends at the same vertex is called a loop. A network is called a simple network if it has: 1. No loops, and 2. No two vertices connected by more than one edge Most networks that will be encountered in this topic are simple. Not simple

Simple

Not simple

The first network is not simple because it has loops. The second network is not simple because there are multiple edges between some vertices. The third network is simple - no vertex is connected to itself, and no vertex is connected to any other in more than one way.

Example 4 Which of these networks are simple networks? Remember that directed graphs can be simple. A

Create a strategy Identify networks that have no loops and no pairs of vertices connected by more than one edge.

Apply the idea There is a loop in the network in Option E. There are two edges between a pair of vertices in the network in Options B and F. The network in Options A, C, and D has no loops, and no “multiple” edges. Therefore, the correct answers are Options A, C, and D.

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Example 5 Model the one-way streets between intersections A, B, and D. • Street 1 goes from A to B • Street 2 goes from B to D • Street 3 goes from A to D

Create a strategy Represent the intersections and one-way streets as a directed graph. The vertices correspond to the intersections: A, B, and D The directed edges, represented as arrows, correspond to the one-way streets, pointing from the starting intersection to the ending intersection.

Apply the idea First, draw three vertices and label them A, B, and D. Next, draw the directed edges: • For Street 1, draw an arrow starting at A and pointing to B. • For Street 2, draw an arrow starting at B and pointing to D. • For Street 3, draw an arrow starting at A and pointing to D. The resulting directed graph: A

Idea summary Edges can be either directed (arrowhead) or undirected (no arrowhead). If a network contains directed edges, it is called a directed network (or digraph). If a network has no directed edges, it is called an undirected network. A network is a connected network when no vertices or groups of vertices are isolated from the rest of the network. If some vertices or groups of vertices are separate from the rest, the network is disconnected. A loop is an edge that starts and ends at the same vertex. A network is a simple network if there are no loops and no “multiple” edges.

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6.01 Practice questions What do you remember? 1

Identify the number of vertices and edges for each network: a

Identify whether each network is directed or undirected: a

For each network, determine the number of vertices that have an odd degree: a

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Practice 4 Ex 1

Determine whether these networks are valid: a

Consider this network: a

Are there more vertices or more edges?

Is this property true for all networks?

For this network, list the pairs of vertices that are connected by an edge. D

C 7

Determine whether these statements are true or false: a

Every network must contain at least two vertices.

In any network, there must be at least one edge.

Each vertex in a network is required to connect to every other vertex.

It is possible for an edge to loop back to its originating vertex.

Edges must always start and end at vertices.

An edge can link more than two vertices simultaneously.

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In the directed network, there is an edge from A to B. List the remaining edges.

A C

B F 9

Identify whether each network contains a pair of vertices that have multiple edges between them: a

This network represents a simplified public transport system in a city:

Ex 2

Public transport map Station A

Station F

Ex 3

Station C

Station B

Station E

Station D

Station G

What do the vertices in this network represent?

What do the edges in this network represent?

How many vertices are there in this network?

How many edges are there in this network?

Station H

Five friends, Amy, Ben, Chloe, David, and Emily, are communicating via text messages: • Amy texts Ben and David. • Ben texts Chloe. • Chloe texts Emily. • David texts Emily. Construct an undirected graph to represent these texting connections, with friends as vertices and texts as edges.

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12 Ex 4

13 Ex 5

Determine whether these networks are simple: a

Five friends, Andrew (A), Betty (B), Cathy (C), Dawson (D), and Ellie (E), are on a social media platform: • Andrew follows Betty and Cathy. • Betty follows Andrew and Dawson. • Cathy follows Andrew, Dawson, and Ellie. • Dawson follows Betty. • Ellie follows Cathy and Andrew.

Represent this ‘follows’ relationship as a directed network.

Is the drawn network connected? Explain your reasoning.

For each network: i

Determine the number of vertices.

Determine the number of edges.

iii

Is the network connected?

B A

C C 15

Determine whether each relationship is best represented by an undirected or directed network: a

Which countries border each other

The results of an elimination-style sports tournament

The animals that different animals eat

Parents and their ancestors

Ways to get from one classroom to another at school (assuming all paths are two-way)

How parts of the body are connected

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Extend your thinking 16

Identify the two changes required to make the network simple: C

A B

F E 17

Deakin University has campuses in Geelong Waterfront, Waurn Ponds, Burwood, Warrnambool, and Melbourne CBD. The university is planning a shuttle service between these locations. They want to visualise their options using a network diagram: • Each campus is represented by a vertex. • A two-way service between campuses is shown by an undirected edge. • A one-way shuttle (only one direction) is shown with a directed edge. a

Draw a simple network diagram showing these services: • A two-way service between Geelong Waterfront and Waurn Ponds • A one-way shuttle from Burwood to the CBD • A two-way service between Burwood and Geelong Waterfront • A one-way shuttle from Warrnambool to Waurn Ponds

Is the network connected? Justify the answer.

Add an edge to make the network fully connected. Explain the meaning of this edge in this context.

Melbourne City Council is mapping scooter paths around key landmarks: Flinders Street Station (F ), Federation Square (S), Melbourne University (M), Queen Victorian Markets (Q), and Docklands (D) They notice that: • There is a scooter loop around D. • Paths go from F to S to M. • There’s a shortcut from S to F. • Students often ride from M to Q, and from there to D. • Some people do a full circle: F - S - Q - D - F. • There is a direct scooter path from M to D.

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Draw a directed network diagram to represent this information.

Does this network contain any loops? Identify them.

Is this a simple network? Explain the reasoning.

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6.02 Degree and adjacency After this lesson, you will be able to… • calculate the degree of vertices in various graphs • represent graphs using adjacency lists and adjacency matrices • convert between different graph representations

Degree of a vertex Degree of a vertex The number of edges that connect to the vertex. The degree of a vertex indicates the number of connections it has to other vertices (or to itself, in the case of a loop) within the network. The degree of a vertex is often written as deg(V ), where V is the vertex label. A useful method involves making a small imaginary ring around the vertex. Count each edge crossing once, with loops counting twice, to find the degree:

Degree 3

Degree 5

Degree 3

Degree 6 Degree 5

Degree 4

Each vertex has a small ring placed around it, and the number of crossings is counted. Using this method ensures that loops are always counted twice.

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Example 1 Consider this network: A B

C a What is the degree of vertex C?

Apply the idea Vertex C is only connected by one edge. The degree of vertex C is 1.

b What is the degree of vertex D?

Apply the idea Vertex D is connected to three edges. The degree of vertex D is 3.

c What is the degree of vertex B?

Apply the idea Vertex B is connected to two edges and has one loop, which is counted twice. The degree of vertex B is 4.

d Which vertices are odd?

Apply the idea Vertex A is connected to two edges, so it has a degree of 2, which is not an odd number. Based on the results from previous parts (a) to (c), vertices C and D have an odd degree.

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Idea summary The degree of a vertex is the number of edges that connect to the vertex. Loops are counted twice.

Adjacency matrices for undirected networks Adjacency matrix A square table or matrix used to represent a graph. It has one row and one column for each vertex, and the entries show the number of edges connecting those vertices. A network can be represented numerically using a table called an adjacency matrix. This is a square array of numbers where each row and column corresponds to a vertex in the graph. Equivalent adjacency matrix:

A B C D E

A B C

D E

0 1

1 0

1 1

0 1

1 0

1 0 0

The number in each cell of the matrix indicates the number of direct edges between the row vertex and the column vertex. For an undirected graph, the matrix is symmetrical.

• Entries: An entry of ‘1’ means there is a single edge connecting the row and column vertices. An entry of ‘0’ means there is no direct edge. For non-simple graphs, an entry can be ‘2’ or higher to represent multiple edges. • Main diagonal: The diagonal from the top-left to the bottom-right represents connections from a vertex to itself. An entry of ‘1’ or more on this diagonal indicates a loop. For simple graphs, the main diagonal is all zeros. • Symmetry: For undirected graphs, the matrix is symmetrical across the main diagonal (e.g., the entry for row A, column B is the same as row B, column A). The adjacency matrix can be used to quickly find the degree of any vertex: • For a simple graph, the degree of a vertex is simply the sum of all the numbers in its corresponding row (or column). • If a graph has loops, the degree is found by summing the entries in the vertex’s row, but the value of the loop (the number on the main diagonal) must be doubled because a loop contributes two to the degree.

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Example 2 Complete the adjacency matrix for this network: A B C a

A B C

A B C D

Create a strategy For each box, count the number of edges that connect the row vertex and the column vertex together. Write 1 if it is connected to the vertex and 0 if it is not.

Apply the idea The given network is an undirected network, therefore each edge connects in both directions. This means that the adjacency matrix is symmetric about the main diagonal. A B C Starting in the top-left, write 0 first because A is not connected to itself. Next, write 1 since A is connected to B and 0 for each other entry in the matrix because A is not connected to C and D.

A B C D

A B C Next, fill out the first column with the same numbers as the first row.

A B C D

0 1

A B C D

0 0 0 1 1 0

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A B C D

0 0 1 0

0 0 A B C

Now, write the remaining connections between C and D in the same way. Therefore, the adjacency matrix for this network is obtained.

0 0

A B C The network shows that B is connected to A and C but not to itself or to D. The completed second row is:

0 0

0 1 1 0 0 1 0 0

0 0 1 0 0 1 1 0

b What is the degree of vertex A?

Create a strategy Count the number of edges connected to the indicated vertex.

Apply the idea The degree of vertex A is 1.

Reflect and check Since the given network is simple, the degree of A will be the sum of the numbers in the first column. Degree = 0 + 1 + 0 + 0

Add the numbers from the first column

Evaluate

Idea summary An adjacency matrix is a square array of numbers, with a row and a column for each vertex, that records the number of connections between vertices. For an undirected network, the entry in row i, column j is the number of edges connecting vertex i and vertex j. This means the matrix will be symmetric.

Adjacency matrices for directed networks Now for a directed network, the entry in a particular row and column indicates the number of edges going from the corresponding row vertex to the corresponding column vertex. A directed, non-simple network with 5 vertices and its corresponding table is shown: A

C D

E A B From C D E

A 0 0 0 1 1

B 1 0 1 0 0

C 0 1 1 0 0

D 0 0 1 0 0

E 0 1 0 2 0

This matrix has different properties: • The matrix is not symmetrical across the main diagonal (an edge from A to B can exist without an edge from B to A). • There is a 1 on the main diagonal (there is a loop at C). • There is a 2 in the table (there are two edges from D to E). For directed networks, the concept of degree is refined to account for the direction of edges. The out-degree of a vertex is the number of edges directed away from it, which can be found by summing the entries in the vertex’s corresponding row in an adjacency matrix. 6.02 Degree and adjacency mathspace.co

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Similarly, the in-degree is the number of edges directed towards a vertex, calculated by summing the entries in its corresponding column. The total degree of a vertex is the sum of its in-degree and out-degree. A loop at a vertex contributes 1 to its in-degree and 1 to its out-degree, consistent with the rule that a loop adds 2 to the total degree. While an asymmetrical matrix means the network is directed, there are directed networks that produce symmetrical tables.

Example 3 A sports tournament between four teams (F, G, H, J) is represented by a directed graph where an arrow from team X to team Y means “X defeated Y”: F

a Create the adjacency matrix for this directed network. F G

F G H J

Create a strategy Create a 4 x 4 matrix with rows and columns for each team. For each row i and column j, place a 1 if an edge goes from vertex i to vertex j, and a 0 otherwise.

Apply the idea Construct the matrix by examining the outgoing edges from each vertex (row) to each corresponding vertex (column). • Row F: Examine vertex F. It has edges going to G and H. Place a 1 in columns G and H. • Row G: Examine vertex G. It has one edge going to H. Place a 1 in column H. • Row H: Examine vertex H. It has no outgoing edges. All entries in this row are 0. • Row J: Examine vertex J. It has one edge going to G. Place a 1 in column G. The completed adjacency matrix is: F G H J

504

F G

0 0 0 0

1 1

0 0 0 0

1 0 0

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0 0

b Determine the degree of team G.

Create a strategy Determine the degree of vertex G by summing its in-degree (number of incoming edges) and its out-degree (number of outgoing edges).

Apply the idea • In-degree: Count the number of edges pointing to vertex G. There are arrows from vertices F and J. The in-degree is 2. • Out-degree: Count the number of edges pointing away from vertex G. There is one outgoing arrow to vertex H. The out-degree is 1. The total degree is the sum of the in-degree and out-degree. Degree of G = 2 + 1 = 3.

Idea summary For a directed network, the adjacency matrix entry in row i and column j shows the number of edges from vertex i to vertex j. Key features can be read directly from the matrix: • • • • •

The out-degree of a vertex is the sum of its corresponding row. The in-degree of a vertex is the sum of its corresponding column. An entry greater than 0 on the main diagonal indicates a loop. An entry greater than 1 indicates multiple edges. An asymmetrical matrix confirms the network is directed.

Adjacency lists and conversions While network diagrams are used to create adjacency matrices, the reverse is also possible. The information in a matrix is sufficient to draw its corresponding graph. A table with 6 rows and columns is shown. So, there are 6 vertices in the network. The network is also considered undirected. 1 1 1 0 1 0 1 0 2 1 1 1 1 2 0 0 0 0 0 1 0 1 1 0 1 1 0 1 0 1 0 1 1 0 1 0

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Drawing the graph from this matrix results in this network: 1 1 1 0

1 0 2 1

1 2 0 0

0 1 0 1

1 1 0 1

0 1 1 0

1 0

1 1

0 1

1 0

0 1

1 0

The process involves: starting with the first row and its corresponding vertex; drawing the edges connected to that vertex; proceeding to the next row, and so on. As well as drawing a graph or creating an adjacency matrix, an adjacency list can also be used to represent the connections in a network. An adjacency list is a collection of lists, where each list corresponds to a vertex in the graph and contains all the vertices that are directly connected to it. For an undirected graph, an edge between A and B is a two-way connection. This means B will be in A’s list, and A will be in B’s list. For a directed graph, an edge from A to B is a one-way connection. This means B will appear in A’s list, but A will not appear in B’s list (unless there is also a separate edge from B to A). Loops and multiple edges can also be shown. A loop at vertex A is shown by including A in its own list. If there are multiple edges between A and B, repeat B in A’s list for each edge. To convert between representations: • Graph to adjacency list: For each vertex, list all the vertices it connects to. • Adjacency list to graph: For each vertex, draw an edge (or arrow for directed) to every vertex in its list. • Adjacency matrix to adjacency list: For each row in the matrix, create a list containing the column-vertex for every non-zero entry. • Adjacency list to adjacency matrix: For each vertex A, look at its list. For every vertex B in A’s list, place the number of connecting edges in the matrix at the position (row A, column B).

Example 4 Create the adjacency list for this undirected graph: A

Create a strategy For each vertex, create a list of all other vertices it is directly connected to by an edge. Since the graph is undirected, each connection is two-way.

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Apply the idea Vertex A is connected to B and D. Vertex B is connected to A and C. Vertex C is connected to B and D. Vertex D is connected to C and A. The final adjacency list is: A: [B, D]

B: [A, C]

C: [B, D]

D: [C, A]

Example 5 The matrix represents an undirected network: P Q P Q R S

R S

1 1 0 0 1 0 2 0 0 2 0 1 0 0 1 0

Convert this adjacency matrix into a graph diagram.

Create a strategy Draw the four vertices P, Q, R, and S. Read the matrix row by row to add the edges. Since the graph is undirected, it is only necessary to read the entries on or above the main diagonal to avoid drawing edges twice.

Apply the idea Start with the vertices P, Q, R, S. • Row P: The entry (P, P) is 1; therefore, there is a loop at P. The entry (P, Q) is 1, so there is an edge between P and Q. • Row Q: The entry (Q, R) is 2; therefore, there are two (multiple) edges between Q and R. • Row R: The entry (R, S) is 1; therefore, there is an edge between R and S. The resulting graph is drawn:

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Example 6 Convert this adjacency list for a directed graph into an adjacency matrix. A: [B] B: [C, D] C: [A]

D: [ ]

Create a strategy Create a 4 × 4 matrix with rows and columns labelled A, B, C, D. For each vertex’s list, place a 1 in the corresponding matrix cell. For example, since B is in A’s list, place a 1 at (row A, column B).

Apply the idea The adjacency list shows the outgoing edges for each vertex: • A → B: Place a 1 at (A, B). • B → C and B → D: Place a 1 at (B, C) and (B, D). • C → A: Place a 1 at (C, A). • D has no outgoing edges, so its row will be all zeros. The resulting adjacency matrix is: A B C D

A B C

0 0

1 0 0 1

0 1

1 0

0 0 0 0

0 0

Idea summary An adjacency list represents a graph by associating each vertex with a list of its adjacent vertices. • In an undirected graph, if A is connected to B, then B is in A’s list and A is in B’s list. • In a directed graph, if there is an edge from A to B, then B is in A’s list. It is possible to convert between graphs, adjacency matrices, and adjacency lists.

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6.02 Practice questions What do you remember? 1 Ex 1

Consider these networks: a

X Y

Determine the degree of vertex X.

Determine the degree of vertex P.

Determine the degree of vertex W.

Determine the degree of vertex S.

Consider the adjacency matrix for an undirected graph with vertices A, B, C, and D: A B C D

A B C

0 1

1 0

0 1

1 0

What is the degree of vertex C?

Create the adjacency list for the graph.

Consider this adjacency list for an undirected graph with vertices P, Q, R, and S: P: [Q]

Q: [P, R, Q]

R: [Q, S]

What is the degree of vertex Q?

Create the adjacency matrix for the graph. P Q

S: [R]

R S

P Q R S

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Practice 4

Create an adjacency matrix for each of these networks: a

Y B

C 5

Consider these networks: i

Create an adjacency matrix for each network.

Determine the degree of each vertex.

S R

S 6

Complete the adjacency matrix for this network: A

Ex 2

A B C

A B C D

C 7 Ex 3

A delivery service operates between four depots (D1, D2, D3, D4). The map shows the oneway routes available. An arrow from one depot to another means a delivery can be made in that direction: a

Create the adjacency matrix for this directed network.

Determine the in-degree, out-degree, and total degree of depot D2.

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Create the adjacency list for this undirected graph: V1

Ex 4

9 Ex 5

1 1

0 0

2 0

0 2

0 0

A B C D A B C D

0 1

1 1

0 0

0 1

0 0

0 1

1 1

1 0

b A B C D d

A B C

0 1 1

1 0 1

1 1 0

P Q P Q R S T

R S T

0 1 2

1 0 0

2 0 0

0 0 0 0 1 0

0 0

1 0

1 1

1 0

Convert this adjacency list into an adjacency matrix, assuming it represents a directed network: S: [T, V ]

A B C D E

Ex 6

Construct a network for each matrix: A B C D E

T: [S]

U: [ ]

V: [U, V ]

The social network Bleeter allows people to share content (in the form of “Bleets”) online. Before anyone can share anything, two people must first make a connection, therefore one person “follows” the other. Whenever a person creates a Bleet, it is shared with everyone who follows them. These people are called “followers”. The given network represents the connections among a group of six users of Bleeter. An arrow from one vertex to another means that the first person follows the second. Therefore, DF follows BK: DF a What kind of network, directed or undirected, represents the connections people form on Bleeter? SS b

Which user has the most followers?

Which user follows the most people?

Is this an example of a simple network? Explain your answer.

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Extend your thinking 12

A group of five friends: Alice (A), Bob (B), Charlie (C), Diana (D), and Edward (E), are working on a project: • Alice can ask Bob for help. • Bob can ask Alice and Charlie for help. • Charlie can ask Diana for help. • Diana can ask Charlie and Edward for help. • Edward can ask Alice for help. Assume these are one-way help channels:

Construct a directed graph to represent these help relationships.

Construct the adjacency matrix for this graph. Rows should represent who is asking for help, and columns should represent who can be asked for help. List vertices in alphabetical order.

Who can ask the most people for help? Who is asked for help by the most people?

The City of Moreland is upgrading its bike lane network. Four key suburbs are being connected: Coburg (C), Brunswick (B), Pascoe Vale (P), and Fawkner (F ). The current network connections are: • Coburg is connected to all three other suburbs. • Brunswick and Pascoe Vale are both connected to Coburg, but not each other. • Fawkner is only connected to Coburg. a

Draw a simple, undirected network diagram to represent this situation.

Construct the adjacency matrix for this network.

What is the degree of each vertex?

If the council wants every suburb to be connected to at least two others, what is the minimum number of new edges they must add? Justify your answer.

Investigation: Hand-shaking lemma proof Investigate online

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6.03 Graph isomorphisms After this lesson, you will be able to… • determine when two graphs are isomorphic • identify and construct subgraphs within a given graph • identify and construct the complement of a given graph • recognise and construct simple, connected, complete, regular, and bipartite graphs

Isomorphic graphs There are many different but equivalent representations of a graph. Rearranging the vertices and edges while maintaining the same connectivity retains the underlying structure of a graph. Isomorphic graphs Two graphs that have the same underlying structure and connectivity, even when drawn with different layouts. Two graphs are isomorphic if there exists a one-to-one correspondence (bijective mapping) between their vertex sets that preserves adjacency. Specifically, these conditions must be satisfied: 1. The graphs have the same number of vertices (necessary). 2. Both graphs have the same number of edges (necessary). 3. The graphs have the same degree sequence, i.e., a matching number of vertices of each degree (necessary). 4. There exists a one-to-one correspondence between vertices such that adjacent vertices in one graph correspond to adjacent vertices in the other (sufficient). These first three conditions are necessary but not sufficient on their own; the fourth condition ensures the graphs are structurally identical. For example, consider these two graphs to determine if they are isomorphic:

Both graphs have 5 vertices and 8 edges. However, the graph on the right has a vertex of degree 2 in the top right, whereas the graph on the left has no vertex of degree 2. The degree sequence of the left graph is (4, 3, 3, 3, 3), while the right graph has (4, 4, 3, 3, 2). Since the degree sequences differ, the graphs are not isomorphic.

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Now, consider a positive example of isomorphic graphs: A

Both graphs have 4 vertices and 5 edges, and have identical degree sequences 3, 2, 3, 2. To confirm isomorphism, a vertex mapping must be tested. For example, the vertices of degree 3 in the first graph are adjacent, as are the vertices of degree 3 in the second graph. This structural similarity suggests a valid mapping exists. Since the mapping preserves adjacency, the graphs are isomorphic.

Example 1 Determine whether Graph G and Graph H shown here are isomorphic: Graph G

Graph H

Create a strategy To determine if two graphs are isomorphic: • Check if the graphs have the same number of vertices. • Check if the graphs have the same number of edges. • Compare their degree sequences. If the sequences are different, the graphs are not isomorphic. • If the above match, attempt to map the vertices of one graph to the other such that all adjacency relationships are preserved. This can sometimes be done by redrawing one graph to look like the other.

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Apply the idea 1. Number of vertices: Graph G has 4 vertices. Graph H has 4 vertices. This matches. 2. Number of edges: Graph G has 5 edges (1-2, 2-3, 3-4, 4-1, 1-3). Graph H has 5 edges (AB, BC, CD, DA, AC). This matches. 3. Degree sequences: • For Graph G: deg(1) is 3, deg(2) is 2, deg(3) is 3, deg(4) is 2. Degree sequence: (3, 3, 2, 2). • For Graph H: deg(A) is 3, deg(B) is 2, deg(C) is 3, deg(D) is 2. Degree sequence: (3, 3, 2, 2). This matches. 4. Mapping vertices and preserving connections: Since the necessary conditions hold, an attempt can be made to map the vertices. Consider the mapping from G to H: • Vertex 1 (degree 3 in G) → Vertex A (degree 3 in H) • Vertex 3 (degree 3 in G) → Vertex C (degree 3 in H) • Vertex 2 (degree 2 in G, connected to 1 and 3) → Vertex B (degree 2 in H, connected to A and C) • Vertex 4 (degree 2 in G, connected to 1 and 3) → Vertex D (degree 2 in H, connected to A and C) Check connections: • In G: (1, 2), (2, 3), (3, 4), (4, 1), (1, 3). • Mapped to H: (AB), (BC), (CD), (DA), (AC). These are precisely the edges in Graph H. Graph H can be redrawn to look like Graph G. Therefore, the graphs are isomorphic.

Idea summary Isomorphic graphs possess these characteristics: 1. The same number of vertices (necessary). 2. The same number of edges (necessary). 3. The same degree sequence, i.e., a matching number of vertices of each degree (necessary). 4. A one-to-one correspondence between vertices that preserves adjacency (sufficient). These conditions ensure the graphs are structurally identical despite appearing different.

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Subgraphs A subgraph is a graph formed by selecting a subset of vertices and edges from a given graph, such that the selected edges connect only the selected vertices. The subgraph must itself be a valid graph, maintaining the connectivity of the chosen edges. For example, consider a graph with four vertices and five edges. A subgraph can be formed by selecting three vertices and three edges connecting them.

Example 2 Which of these graphs is a valid subgraph of the original graph shown? E G

D A

B E

Create a strategy To be a valid subgraph, a graph must only contain vertices and edges that are present in the original graph. A systematic check of each option is required: 1. Verify that all vertices in the option graph exist in the original graph. 2. For each edge in the option graph, verify that an edge connects the same two vertices in the original graph. 3. If an option contains even one vertex or edge not found in the original, it is not a subgraph.

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Apply the idea Option A: This graph consists of vertices A, B, C, D, and edges AB, BC, CD. All these vertices exist in the original graph. Checking the edges: the edge AB exists, BC exists, and CD exists. Since all components are in the original graph, this is a valid subgraph. Option B: This graph contains the edge AF. In the original graph, vertex A is connected to G and B, but not to F. Since the edge AF does not exist in the original graph, this is not a valid subgraph. Option C: This graph contains the edge BE. In the original graph, vertex B is connected to A, C, and F, but not to E. Since the edge BE does not exist in the original graph, this is not a valid subgraph. Option D: This graph contains the edge CF. In the original graph, vertex C is connected only to vertices B and D, and vertex F is connected to B, G, and E. There is no direct edge connecting C and F. Therefore, this is not a valid subgraph. The only option that satisfies the definition of a subgraph is Option A.

Idea summary A subgraph is a graph formed by a subset of vertices and edges from a given graph, where the edges connect only the selected vertices. It retains the structure of the original graph for the chosen components.

Complement of a graph The complement of a graph G is a graph with the same set of vertices as G, but with an edge between two vertices if and only if there is no edge between them in G. For example, consider a graph with four vertices. Three of the vertices are connected to form a triangle, while the fourth vertex is isolated. The complement has the same four vertices, with edges added where none existed in the original graph. This results in edges connecting the previously isolated vertex to each of the other three vertices. Graph

Complement graph

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Example 3 Construct the complement of this graph: A B

Create a strategy Identify all possible edges between the vertices, then include only those edges in the complement that do not exist in the original graph.

Apply the idea A B

The original graph has 5 vertices and 5 edges forming a pentagon. All possible edges are 10 in a complete graph with five vertices. The original graph includes five edges, therefore the complement includes the five edges not present in the original, connecting other vertices.

Idea summary The complement of a graph has the same vertices but includes edges between vertices that are not connected in the original graph, and excludes edges that are present in the original.

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Special classes of graphs Graphs can be classified into various types based on their structure: complete graphs, regular graphs, and bipartite graphs. Complete graph A graph where every pair of distinct vertices is connected by an edge. Denoted Kn, where n is the number of vertices. n ( n 1) with n vertices in complete graph. A complete graph with 2 three vertices forms a triangle. The total number of edges is

These are examples of complete graphs:

Regular graph A graph where every vertex has the same degree. For example, a graph where each vertex connects to 4 others is called a 4-regular graph.

Bipartite graph A graph whose vertices can be divided into two disjoint sets such that every edge connects a vertex in one set to a vertex in the other, with no edges within the same set. For example, a graph with two sets of three vertices each, with edges only between the sets.

K(3,4)

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Example 4 Determine whether this graph is complete, regular, or bipartite: A

Create a strategy Check the definitions: count edges for completeness, check vertex degrees for regularity, and attempt to partition vertices for bipartiteness.

Apply the idea The graph has 4 vertices and 4 edges. A complete graph K4 has complete.

4 ( 4 1) 2

12 2

6 edges, so it is not

Each vertex has degree 2, so it is a 2-regular graph. To check if it’s bipartite, try partitioning vertices into two sets. Place two vertices in one set and two in another. Original graph

Partitioned into two sets

Edges connect between sets, not within, therefore the graph is bipartite. Therefore, it is regular and bipartite.

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Example 5 The school principal wants to allocate 3 teachers to 3 classes, ensuring the allocation is both ‘regular’ and ‘bipartite’. Create a regular bipartite graph that models one possible allocation.

Create a strategy The vertices must be divided into two distinct sets with no edges connecting vertices within the same set. For this problem, the two sets are Teachers and Classes. Edges can only connect a teacher to a class. Every vertex in the graph must have the same degree. This means each teacher must be allocated to the same number of classes, and each class must be assigned the same number of teachers.

Apply the idea Construction plan: First, draw the two sets of vertices. Then, add edges between the sets in a balanced way until every vertex has the same degree. • Teacher 1 is assigned to Class 1 and Class 2. • Teacher 2 is assigned to Class 1 and Class 3. • Teacher 3 is assigned to Class 2 and Class 3. This allocation results in every teacher and every class having a degree of 2. The resulting graph is a 2-regular bipartite graph, as shown: T1

Idea summary Graphs can be classified as: • Complete graph: Every pair of vertices is connected. • Regular graph: All vertices have the same degree. • Bipartite graph: Vertices can be split into two sets with edges only between sets.

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6.03 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

Two graphs are isomorphic if and only if they have the same number of vertices and the same number of edges.

A subgraph can have more edges than the original graph.

The complement of a complete graph (Kn) is a graph with n vertices and no edges.

Every complete graph is also a regular graph.

Match each graph type with its correct definition: i

Complete graph

Regular graph

A graph where every vertex has the same degree.

A graph where vertices are in two sets, and edges only connect vertices from different sets.

A graph where every vertex is connected to every other vertex.

iii

Bipartite graph

List four conditions that must be true for two graphs to be isomorphic.

Practice Ex 1

For each pair of graphs: i

Determine whether the pair of graphs is isomorphic.

If the graphs are isomorphic, briefly explain why or redraw one to show the isomorphism. If the graphs are not isomorphic, explain why.

Graph 1A: Graph 1B:

u2 v1

u1 u5

u6 u4

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v6 u3

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Graph 2B:

Graph 2A:

v1 v7

v3 v5

u6 u5

Ex 2

Consider the graph with vertices A, B, C, D, E, and F:

Graph with vertices A, B, C, D, E, F A

Write the degree sequence for the graph in non-increasing order.

Construct a subgraph of the main graph that contains exactly 3 vertices and forms a cycle. Draw the subgraph and label its vertices.

E D

Ex 3

Construct the complement of this graph: Graph with vertices A, B, C, D C B A

Determine whether these graphs are complete graphs: a

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Ex 4

Determine whether these graphs are bipartite graphs: a

Identify whether the graph is simple, connected, complete, regular, or bipartite. List all properties that apply. Graph with vertices A, B, C, D, E A

C Ex 5

Construct a bipartite graph with 6 vertices and at least 4 edges.

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Determine the degree sequence of the complement of the graph: Graph with vertices A, B, C, D, E A

E 12

For this graph: Graph with vertices A, B, C, D C D

Construct a subgraph.

Construct the complement of the subgraph.

Construct a regular graph with 5 vertices where each vertex has degree 2.

Extend your thinking 14

A graph has degree sequence (4, 3, 2, 2, 1). Construct a graph with this degree sequence and its complement. Are the original graph and its complement isomorphic? Justify your answer.

Prove that any two complete graphs with the same number of vertices are isomorphic.

Construct a bipartite graph with 8 vertices that is also regular. Determine whether it could be isomorphic to its complement.

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6.04 Trees After this lesson, you will be able to… • identify and verify if a graph is a tree • prove equivalent conditions for trees (connected, acyclic, minimal connected, and maximal acyclic) • apply algorithms to find spanning trees in graphs • solve problems involving trees in various applications

Introduction to trees Tree A connected network graph that contains no cycles or loops. These networks have a distinctly tree-like appearance:

A graph is a tree if it meets two conditions: • It is connected: There is a path between any two vertices. • It is acyclic: It contains no cycles (no paths that start and end at the same vertex without backtracking). Leaving a vertex on any of these networks means it cannot be returned to without repeating an edge.

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Example 1 Determine whether each graph is a tree. Justify the answer: a

Create a strategy To be a tree, a graph must satisfy two conditions: it must be connected, and it must be acyclic (contain no cycles). Examine the graph for each of these properties.

Apply the idea This graph is not a tree because it fails both conditions for being a tree. • Condition 1: Connectivity. The graph is disconnected. For example, there is no path from vertex A to vertex D. • Condition 2: Cycles. The graph contains cycles. Specifically, it contains two cycles: A - B - C - A and D - E - F - D. A graph only needs to fail one condition to not be a tree. Since this graph fails both, it is definitively not a tree.

Apply the idea A graph is a tree if it is connected and contains no cycles. This graph is connected, as every vertex can be reached from every other vertex. However, it contains a cycle (for example, the central triangular cycle). Therefore, this graph is not a tree because it has a cycle.

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Apply the idea The graph is connected, as there is a path between any two vertices. It also contains no cycles (it is acyclic). Therefore, it satisfies both conditions and is a tree.

Idea summary A tree is a graph that is both connected and acyclic. • Connected means there is a path between any two vertices in the graph. • Acyclic means the graph has no cycles or loops.

Properties and equivalent conditions for trees While the main definition of a tree is a connected, acyclic graph, there are other ways to test if a graph is a tree. The properties below are all equivalent—if a graph satisfies any one of them, it satisfies all of them. These properties are mathematically equivalent. These properties provide more tools to quickly check if a graph is a tree.

Exploration Discover this property: 1. Draw three different graphs known to be trees. Ensure all resulting graphs are connected and have no cycles. 2. For each tree, count the number of vertices (n). 3. For each tree, count the number of edges (m). 4. What relationship is noticed between n and m for all of your trees? Does it match the formula m = n – 1?

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One of the most important properties is the relationship between the number of vertices and the number of edges.

m=n–1 m is the number of edges in a tree n is the number of vertices in a tree A tree with n vertices will always have exactly n – 1 edges. This simple formula is a quick way to check if a graph could be a tree. Other equivalent conditions for a graph to be a tree include being minimally connected or maximally acyclic. • A graph is minimally connected if it is connected, but removing any single edge would make it disconnected. Trees are minimally connected because there are no redundant edges. • A graph is maximally acyclic if it has no cycles, but adding any single edge between two existing vertices would create a cycle. Trees are maximally acyclic because they have the most edges possible without creating a cycle. For any connected graph with n vertices, these properties are equivalent. If one property is true, then all the other properties are true. The graph is a tree if: • It is connected and acyclic. • It is connected and has n – 1 edges. • It is acyclic and has n – 1 edges. • It is minimally connected. • It is maximally acyclic.

Example 2 A tree has 20 vertices. How many edges does it have?

Create a strategy Use the formula that relates the number of vertices (n) and edges (m) in a tree.

Apply the idea The formula for a tree is m = n – 1. Given n = 20. m = 20 – 1 = 19 The tree has 19 edges.

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Example 3 Verify that the graph shown is a tree by checking its number of vertices and edges. C

Create a strategy First, confirm the graph is connected. Then, count the number of vertices (n) and edges (m). Finally, check if the condition m = n – 1 is met.

Apply the idea 1. Check for connectivity: By inspection, it is possible to travel from any vertex to any other vertex. The graph is connected. 2. Count vertices: The graph has 4 vertices, so n = 4. 3. Count edges: The graph has 3 edges, so m = 3. 4. Check the condition: Check if m = n – 1. 3=4–1 The condition is true. Since the graph is connected and has n – 1 edges, it is a tree.

Idea summary A tree with n vertices always has m = n – 1 edges. A connected graph is a tree if and only if it satisfies any of these equivalent conditions: • • • •

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It is acyclic (the original definition). It has n – 1 edges. It is minimally connected (removing any edge disconnects it). It is maximally acyclic (adding any edge creates a cycle).

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Spanning trees and Prim’s algorithm Spanning tree A subnetwork of a connected network that contains all the vertices of the original network and is also a tree (it has no cycles). Minimal spanning tree For a weighted network, a spanning tree that has the lowest possible total weight of all its edges. The terms ‘minimal spanning tree’ and ‘minimum spanning tree’ (MST) are synonyms and are used interchangeably in most contexts.

Any connected network can have edges removed, taking care not to disconnect the network, until the result is a subnetwork that is a tree - this is then called a spanning tree. Here is an example:

The network in the centre has its eight different spanning trees arranged around it. It is still possible to get from one vertex to any other, but it is not possible to get back without repeating an edge. A connected network that isn’t already a tree can have multiple spanning trees. This network has 11 possible spanning trees:

6 4

5 2

This network has weight 4 + 6 + 5 + 2 + 3 + 10 = 30.

10 3

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Here are all the spanning trees, with their weights: 6 10

3 Weight 23

3 Weight 22 6

2 10 3 Weight 21

5 2

Weight 22

Weight 21 4

2 3 Weight 15

Weight 25

3 Weight 18

3 Weight 24 4

5 2

2 10 3 Weight 20

3 Weight 14

The last spanning tree is special, since it has the lowest weight out of all of them. This is called the minimal spanning tree. Sometimes there can be two or more spanning trees with equal lowest weight, in which case they are all minimal spanning trees. Here is a network with two minimal spanning trees: 3

6 5

Finding both can be a challenge without a systematic approach. Greedy algorithm A problem-solving approach that makes the locally optimal choice at each step with the hope of finding a global optimum. At each stage, the algorithm selects the best option available at that moment, without considering the overall future consequences beyond the immediate step. Prim’s algorithm A greedy algorithm used to find a minimal spanning tree for a weighted, connected graph. It builds the tree one vertex at a time, by adding the cheapest possible connection from the existing tree to a vertex not yet in the tree.

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Prim’s algorithm is an example of a greedy algorithm. While this approach does not guarantee the best solution for all problems, it works effectively for finding minimal spanning trees. Another greedy algorithm for finding minimal spanning trees is Kruskal’s algorithm, though it is not covered here. Prim’s algorithm will be used to find the minimal spanning tree without having to find all the spanning trees. The algorithm was first created by Vojtech Jarník, a Czech mathematician, in 1930. Both Robert Prim (in 1957) and Edsger Dijkstra (in 1959) independently rediscovered it, though Prim’s work popularised its use. Prim’s algorithm (from a network): 1. Pick any vertex to start the spanning tree. 2. Locate the lowest weight edge that connects a vertex currently in the spanning tree to a vertex not yet in the spanning tree. 3. Reject any edge that connects two vertices already in the spanning tree. 4. If there is more than one edge with the lowest weight, pick any of them. 5. Add this edge and the vertex at the other end to the spanning tree. If all vertices are part of the spanning tree, stop. Otherwise, repeat from Step 2.

Example 4 This network represents a mycorrhizal nutrient transfer system connecting the roots of several plants. The weights on the edges represent the distance in metres: A 14 26

B 15 C

12 10

12 D

H 21

F 10

7 G

14 E

To conserve resources during a drought, only the most essential connections can be sustained. Find the minimal spanning tree for this network to determine the shortest total distance required to keep all plants connected, and determine this total distance.

Create a strategy The problem requires finding the minimum spanning tree of the weighted network. Prim’s algorithm will be used to construct this tree. The algorithm begins at an arbitrary starting vertex and iteratively adds the edge with the lowest weight that connects a vertex in the growing tree to a vertex outside the tree. Edges that would form a cycle are rejected. This process continues until all vertices are included in the tree.

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Apply the idea The steps of Prim’s algorithm are applied, starting arbitrarily at vertex F. A 14 26

1. Start at F. The cheapest edge connected to F is FE with a weight of 10. Add edge FE and vertex E to the tree.

15 C

14 12

8 F 12 10 10 14 D E 33

7 G

A 2. Current tree: F - E. The cheapest edge connecting the tree to an outside vertex is ED, also with a weight of 10. Add edge ED and vertex D.

14 26

15 C

8 F 12 10 10 14 D E 33

14 12

A 3. Current tree: F - E - D. The next cheapest edges are DC and DF, both with weight 12. The edge DF must be rejected as it connects two vertices already in the tree and would form a cycle (D - E - F - D). Therefore, select DC. Add edge DC and vertex C.

14 26

15 C

8 F 12 10 10 14 D E 33

14 26

7 G

4. Current tree: F - E - D - C. The next cheapest edge is BF with a weight of 13. Add edge BF and vertex B.

15 C

14 12

10 D

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21 F 10 E

H 8

7 G

A 5. Current tree: F - E - D - C - B. The next cheapest edges are AB, BD, and EG, all with weight 14. Reject BD as it would form a cycle (B - F - E - D - B). Choose one of the others, for example, AB. Add edge AB and vertex A.

14 26

B 15 C

F 10

10 D

7 G

A 6. Current tree: F - E - D - C - B - A. A new, cheaper edge is now available from the tree: AG with weight 8. This is cheaper than the remaining edge of weight 14 (EG). Add edge AG and vertex G.

14 26

B 15 C

F 10

10 D

7 G

A 7. Current tree: F - E - D - C - B - A - G. Only vertex H remains. The cheapest edge connecting H to the tree is GH with weight 7. Add edge GH and vertex H. All vertices are now in the tree.

14 26

15 C

10 D

H 21

F 10

7 G

E A

B The final minimal spanning tree consists of the selected edges:

13 C

10 D

H 8 F

The total weight of the minimal spanning tree is the sum of the weights of its edges: Total weight = 10 + 10 + 12 + 13 + 14 + 8 + 7 = 74 The shortest total distance to connect all plants is 74 metres.

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Reflect and check Prim’s algorithm is a greedy algorithm that is guaranteed to find a minimal spanning tree. Even if a different choice was made during a tie (e.g., choosing edge EG instead of AB in step 5), the algorithm would still find a minimal spanning tree with the same total weight of 74 metres, although the set of edges might be different. The final graph is a tree as it connects all 8 vertices with 7 edges and contains no cycles.

Example 5 B

8 6

A 7

2 D

The rangers of a national park want to clear walking tracks to 6 landmarks in the area. The potential tracks between landmarks are shown in the graph, along with the number of days required to clear each track for use.

To make a network of tracks in the shortest time, the rangers want to find a minimal spanning tree for the graph.

a Which two of these options are possible minimal spanning trees? A

B 7 7 C

F C

A 7

E 2 D 6

Create a strategy Use the Prim’s algorithm to test each option.

536

4 6

B 7 7 C 4 8 A 7

7 8

2 D

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4 8 6

A 7 F

E 2 D 6

Apply the idea Applying Prim’s algorithm, starting from vertex A, allows a choice between edge AB or AF (both weight 7). • Path 1 (choosing AB first): Select edge AB (weight 7). From A or B, the next shortest edge to an unvisited vertex is BE (weight 4). Then, ED (weight 2). Then, DC (weight 6). Finally, to connect F, edge EF (weight 6) is chosen over AF (weight 7). This tree (AB, BE, ED, DC, EF ) corresponds to Option C. Total weight: 7 + 4 + 2 + 6 + 6 = 25. • Path 2 (choosing AF first): Select edge AF (weight 7). From A or F, the next shortest edge to an unvisited vertex is FE (weight 6). Then, ED (weight 2). Then, DC (weight 6). Finally, to connect B, edge EB (weight 4) is chosen over AB (weight 7), CB (weight 9) or DB (weight 8). This tree (AF, FE, ED, DC, EB) corresponds to Option A. Total weight: 7 + 6 + 2 + 6 + 4 = 25. Both Options A and C result in a minimal spanning tree of weight 25. Therefore, the possible minimal spanning trees are in Options A and C.

b What is the minimum time taken to build a network of tracks that connect each landmark?

Create a strategy Add the weights of edges of the minimum spanning trees found in part (a).

Apply the idea In part (a), it was found that the total weight of both minimal spanning trees was 25. The minimum time taken to build a network of tracks is 25 days.

Idea summary A spanning tree is a subnetwork of a connected network that includes all the vertices and is a tree (is connected and has no cycles). A minimal spanning tree is a spanning tree with the lowest possible total weight of edges. Prim’s algorithm is a Greedy algorithm employed to find a minimal spanning tree: 1. Pick any vertex to start. 2. Find the lowest weight edge connecting a vertex in the tree to a vertex outside the tree. 3. Add this edge and the new vertex to the tree, rejecting any edge that would create a cycle. 4. If there is a tie for the lowest weight edge, any of them can be picked. 5. Repeat until all vertices are in the tree.

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6.04 Practice questions What do you remember? 1

In graph theory terms, define: a

A tree

A spanning tree

A minimal spanning tree

Prim’s algorithm

Determine whether these statements about trees are true or false: a

Adding one edge to any tree creates a new tree.

A tree contains no cycles or loops.

If you remove any edge from a tree it creates a disconnected graph.

A tree is a minimally connected graph.

Describe the relationship between the number of edges (m) and vertices (n) in any tree.

Practice 4 Ex 1

Determine whether each graph is a tree. If it is not a tree, explain why: a

D A

B C

B A

d A

C B

D 5

A tree has 15 edges. How many vertices does it have?

Ex 2

Verify that the graph shown is a tree.

Ex 3

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Construct a minimal spanning tree for each weighted network: a

B C 10 6

b A

6 7

8 B3

5 F

4 6

This weighted network shows possible connections (and associated costs, in millions of dollars) for connecting rural towns to a common cabled internet service:

Ex 5

12 9 C

To minimise the cost of connecting each town to the network, determine the minimal spanning tree: a

Ex 4

How many edges will be required for the minimal spanning tree?

E 2

Construct the minimal spanning tree.

Determine the minimum cost of connecting all of the towns to the service.

The direct flight routes between 5 major cities are shown in the graph. Each edge is marked with the distance (in km). To optimise their network, the airline wants to determine a minimal spanning tree for the graph.

B 720

Which two of these options represent a valid minimal spanning tree for this network? A

720

500

520 E

500 720

640 520

520 720

720

560

560 740 720

500

500 720

740 640

640 520 720

520 720

560 b

560

740

640

500

640

740

720 a

3 7

560

What is the minimum total distance required to connect all these major cities?

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Water pipes need to be laid down in a new area that is undergoing construction. This pipe map was suggested:

G F

To minimise the cost, the council wants to connect every house to water, minimising the total length of water pipes used.

9 15 17

Draw the network of pipes that connects all the houses using the minimum total length.

B 19

24 22

A C 11

The bus company decided to redesign its travel routes after some new roads were built. The map shows the routes connecting the towns and the respective times (in minutes) taken to cover the route:

The bus company wants to connect all towns with the most time-efficient route. To be as time-efficient as possible, the company needs to connect all towns using the minimum total travel time. Draw the network of routes that achieves this.

21 15

10 10

H 33 C

29 60

45 31

43 B

Extend your thinking 12

This graph shows the potential monthly profit (in thousands of dollars) from operating flight routes between cities. An $12K airline wants to select a set of routes that connects all JFK cities and maximises the total profit. This is known as finding a maximal spanning tree: a

b c 13

Describe how one could adapt Prim’s algorithm to find a maximal spanning tree instead of a minimal one.

LHR $8K DXB $18K

$7K

Using your adapted algorithm, construct the maximal spanning tree for this network.

$13K

GRU

$15K

SYD

What is the maximum possible monthly profit?

Urban planners are designing a new subdivision with 5 key locations (A, B, C, D, E). The possible direct road connections and their lengths (in metres) are: •

AB : 120 m

• AC : 150 m

•

BE : 180 m

•

CD : 100 m

• AD : 200 m •

CE : 220 m

• BC : 90 m •

DE : 130 m

The urban planners wish to create roads that are most distance efficient, and allow them to connect each location. Represent this information as a weighted graph. Then, construct the minimal spanning tree that is the most distance efficient and state its total length.

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A telecommunications company is planning to lay new fibre optic cables to connect 5 remote data centres (P, Q, R, S, T ). The costs (in thousands of units) for laying cable directly between certain pairs of centres are as follows:

Connection

Cost (thousands)

120

Note: Some pairs like PS and QT have no direct link listed, meaning the cost is prohibitive or not feasible for a direct connection. To make a network of cables in the most cost-effective way, the company wants to find a minimal spanning tree:

Represent this information as a weighted graph. Then, construct the minimal spanning tree to connect all data centres.

Calculate the minimum cost to connect all data centres.

Two park rangers, Alex and Ben, are planning a network of walking trails to connect 6 key landmarks (A, B, C, D, E, F ) in a national park. They want to find a minimal spanning tree (MST) to ensure all landmarks are connected with the minimum total length of cleared trails. The available potential trails and their lengths (in km) are given in the table: Both rangers decide to use Prim’s algorithm as described in their training materials. a

Alex starts Prim’s algorithm from landmark A. List the trails (edges) Alex selects in the order he adds them to form his MST. What is the total length of Alex’s MST? Ben starts Prim’s algorithm from landmark F. List the trails (edges) Ben selects in the order he adds them to form his MST. What is the total length of Ben’s MST?

Trail

Length (km)

Compare the final set of trails selected by Alex and Ben. Are their MSTs identical in terms of the trails included? Explain why this might or might not always be the case when Prim’s algorithm is applied from different starting vertices on the same graph.

After their initial planning, a new survey reveals an additional potential trail between landmarks D and F, with a length of 6 km. If Alex were to re-run Prim’s algorithm (still starting from landmark A) including this new DF trail option, would the DF trail be part of his new MST? Justify your answer by describing the state of Alex’s MST construction at the point the DF trail would be considered.

Considering the original graph (without the DF trail from part (d)), suppose the cost to clear trail DF (originally 9 km) is uncertain. What is the maximum possible length for trail CF such that it could be included in an MST constructed by Prim’s algorithm (for example, starting from A)? Explain your reasoning, detailing how Prim’s algorithm would make its selection.

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6.05 Planar graphs After this lesson, you will be able to… • determine whether a given graph is planar using Euler’s formula (v + f – e = 2) • analyse and solve problems involving planar graphs • prove results about planarity of simple connected graphs

Planar networks A graph may be drawn with edges that cross over one another. However, if the vertices can be rearranged to create a drawing where no edges intersect, that drawing is called a planar representation. A graph that has a planar representation is known as a planar network. No more crossings!

Crossings

Planar?

Move two vertices

Planar!

It is not always obvious at first glance whether a network is planar or non-planar, as achieving a planar representation can require significant rearrangement of the vertices to eliminate all edge crossings. Such rearrangements, which maintain the original vertex connections, result in an isomorphic (equivalent) graph.

The top network is planar, and below it are three of its equivalent planar representations. All four graphs are isomorphic.

Interactive exploration Discover this concept in action online

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Once a planar network is drawn in a planar representation, a face (or region) is defined as an area of the plane bounded by edges. The area of the plane outside the network is also considered a face. In the three planar representations shown previously, each face is shaded with a different colour, revealing that the network has a total of 4 faces.

This octahedral graph (which represents a 3-dimensional octahedron) has 8 faces.

Non-simple networks that are planar also have faces, though some of their faces are bounded by only one or only two edges:

The network on the left has 6 faces, with 4 of them bounded by only 1 edge (loops). The network on the right has 5 faces, with 2 of them bounded by only 2 edges. However, if a network is not planar, then there are no faces to define - the crossing of edges makes it impossible.

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Example 1 Consider this graph: A

a Match the graph with its equivalent planar representation. A

H A

A H

Create a strategy The strategy involves redrawing the crossing edges of the original graph to eliminate intersections. It is essential to verify that the connections between vertices (the edges) are the same in both the original graph and the proposed planar representation.

Apply the idea All the graphs among the options do not contain any intersecting edges, so they are all planar graphs. By checking the edges, it is determined that only the graph in Option A has the same set of edges as the original graph. Therefore the correct answer is Option A.

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b How many faces does this representation have?

Create a strategy The number of faces can be found using Euler’s formula for connected planar graphs, which is: v+f–e=2 The known values for vertices (v) and edges (e) will be substituted into the formula to solve for the number of faces (f ).

Apply the idea The graph has 8 vertices (v = 8) and 10 edges (e = 10). v+f–e=2

Write the formula

8 + f – 10 = 2

Substitute v = 8 and e = 10

f–2=2 f=4

Simplify Add 2 to both sides

Idea summary A planar graph is a graph that can be redrawn into an isomorphic (equivalent) graph where no edges intersect. This non-crossing version is called a planar representation. In a planar representation, a face (or region) is an area of the plane bounded by edges. The area outside the graph is also considered a face. If a graph is non-planar, it cannot be drawn without edge crossings, and therefore does not have faces.

Euler’s formula Euler’s formula A formula that relates the number of vertices (v), edges (e), and faces (f ) of any connected planar graph. For any connected planar graph, the number of vertices (v), edges (e), and faces (f ) are always related by the rule:

v+f–e=2 v is the number of vertices (nodes) e is the number of edges (connections) f is the number of faces (regions, including the outer face)

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This formula holds true for every possible connected planar graph, regardless of how it is drawn. Its main application is to find one of these three quantities when the other two are known. This connected planar graph has v = 6, e = 9, and f = 5. Verifying with Euler’s formula: v+f–e=6+5–9=2 The formula holds true.

Example 2 A connected planar graph has 10 vertices and 15 edges. How many faces does it have?

Create a strategy The number of faces can be found by using Euler’s formula, v + f – e = 2. The known values for vertices (v) and edges (e) are substituted into the formula, which is then solved for f.

Apply the idea Given v = 10 and e = 15. v+f–e=2 10 + f – 15 = 2 f–5=2 f=7

Write Euler’s formula Substitute v = 10 and e = 15 Simplify the expression Add 5 to both sides

The graph has 7 faces.

Example 3 The planar representation of a certain polyhedron has 20 faces and 30 edges. How many vertices does it have?

Create a strategy The number of vertices is found by substituting the known values for faces (f ) and edges (e) into Euler’s formula (v + f – e = 2) and solving for v.

Apply the idea Given f = 20 and e = 30. v+f–e =2 v + 20 – 30 = 2 v – 10 = 2 v = 12

Write Euler’s formula Substitute e = 30 and f = 20 Simplify the expression Add 10 to both sides

The polyhedron has 12 vertices.

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Idea summary A connected planar graph satisfies Euler’s formula:

v f e

v+f–e=2 is the number of vertices is the number of faces is the number of edges Region 1

Region 2 Vertices

Edges

Region 3

Faces (also called Regions)

Test for planarity Euler’s formula can be used in two main ways: to find a missing value if a graph is known to be planar, or to test if a graph with known properties could be planar. If a connected graph’s properties do not satisfy v + f – e = 2, it is proven to be non-planar. A consequence of Euler’s formula provides a powerful inequality for testing if a simple graph is non-planar. For any simple, connected planar graph with 3 or more vertices, this relationship must be true: e ≤ 3v – 6 This inequality provides a simple test: if a graph violates this condition (i.e., if e > 3v – 6), it has too many edges for its number of vertices and is therefore proven to be non-planar. Note that, if a graph satisfies the condition e ≤ 3v – 6, the test is inconclusive; the graph might be planar, but it is not guaranteed.

Example 4 A connected planar graph has 7 vertices, 10 edges, and 5 faces. Verify that it satisfies Euler’s formula.

Create a strategy The properties of the graph are tested by substituting the given values for v, e, and f into the expression v + f – e and checking if the result equals 2.

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Apply the idea The graph has v = 7, e = 10, and f = 5. v+f–e=2 7 + 5 – 10 = 2 12 – 10 = 2 2=2

Write Euler’s formula Substitute v = 7, f = 5, and e = 10 Simplify v + f Verify the formula holds

Since v + f – e = 2, the graph satisfies Euler’s formula.

Example 5 Prove that a simple, connected graph with 8 vertices and 20 edges cannot be planar.

Create a strategy To prove the graph is non-planar, it must be shown that it fails the condition e ≤ 3v – 6.

Apply the idea The graph has v = 8 and e = 20. e ≤ 3v – 6

Write the inequality

20 ≤ 3(8) – 6

Substitute the values

20 ≤ 18

Evaluate

The inequality 20 ≤ 18 is false. Since the graph violates the condition for planar graphs, it is proven to be non-planar.

Idea summary Two key formulas are used to analyse and test the planarity of connected graphs. Euler’s formula: For any connected planar graph, this formula is true: v+f–e=2 If a graph’s properties do not satisfy this formula, it is non-planar. Planarity test for simple graphs: For any simple, connected planar graph, this inequality must be true: e ≤ 3v – 6 • If e > 3v – 6, the graph is definitely non-planar. • If e ≤ 3v – 6, the graph may be planar (the test is inconclusive).

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6.05 Practice questions What do you remember? 1

Determine whether each statement is true or false. a

Euler’s formula (v + f – e = 2) applies to all types of graphs.

A graph is planar if it can be drawn with no edges crossing.

The exterior region of a planar graph is not counted as one of its faces.

Identify whether each graph is planar or non-planar: a

Determine whether a connected planar graph could have these parameters: a

v = 5, f = 6, and e = 9

v = 5, f = 5, and e = 10

v = 8, f = 6, and e = 13

v = 12, f = 10, and e = 20

Practice Ex 1

Draw an equivalent planar representation for each graph: a

For each graph: i

Determine the total number of edges, E, of the planar graph.

Calculate D, the sum of the degrees of the vertices of the planar graph.

iii

Write an algebraic equation relating the sum of degrees (D) and the number of edges (E) of the planar graph.

A E

B C

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For each graph: i

Draw an equivalent planar representation.

Determine the number of faces, f, for the representation.

A D

B c

A E

C 7

For each of these graphs: i

Determine v, the number of vertices.

Determine e, the number of edges.

iii

Draw an equivalent planar representation of the graph.

Determine f, the number of faces of the planar graph.

Evaluate v + f – e for the planar graph.

E B

D B

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Consider these graphs: Graph A A

Graph B B

C E

Graph C A

Graph D G

Complete the table:

Vertices (v) Faces (f ) Edges (e) v+f–e Ex 2, 3

Graph A

Graph B

Graph C

Graph D

⬚

For a connected planar graph with v vertices, e edges, and f faces, determine: a

f given v = 7 and e = 14

e given v = 8 and f = 2

v given f = 3 and e = 12

e given v = 4 and f = 5

v given f = 8 and e = 11

f given v = 6 and e = 7

A connected planar graph has 3 edges and 1 vertex: a

Determine f, the number of faces.

Draw a planar graph with the given parameters.

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Consider these graphs: Graph A

Graph B

A B

Graph C

Graph D A

A B

D D B a

Faces (f ) Edges (e) v+f–e Planar? (Y/N) No. of vertices with odd degree

552

Complete this table:

Vertices (v)

Graph A

Graph B

Graph C

Graph D

⬚

⬚ ⬚ ⬚ ⬚

What is observed about the number of vertices with odd degree?

A connected planar graph has 3 vertices and 3 faces: a

Determine the number of edges, e.

Draw a planar graph with the given parameters.

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⬚ ⬚ ⬚ ⬚

13 Ex 4

Consider the planar graph shown, which represents the vertices and edges of a cube: a

Count the number of vertices (v), edges (e), and faces (f ).

Verify that the graph satisfies Euler’s formula.

A graph has 6 vertices and 9 edges and is drawn with 7 distinct faces. Determine whether this graph can be a connected planar graph. Explain your reasoning.

E Ex 5

A simple, connected graph is described as having 10 vertices and 25 edges: a

Determine whether this graph can be planar. Justify your answer.

Based on your result, can this graph satisfy Euler’s formula? Explain why or why not.

Extend your thinking 16

For these non-planar graphs, determine the minimum number of edges that can be removed to make them planar: a

A connected planar graph has 4 vertices of degrees 2, 2, 3, and 3: a

Determine the number of edges, e.

Determine the number of faces, f.

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6.06 Walks, paths, and trails After this lesson, you will be able to… • distinguish between different types of walks, paths, trails, and circuits in graphs • determine whether a graph contains Euler paths or circuits or Hamiltonian paths/cycles • apply theorems about path existence to solve practical problems • analyse the historical Konigsberg Bridge Problem using graph theory

Traversal: Walks, trails, and paths Walk The most general type of traversal. A sequence of vertices where each consecutive pair is connected by an edge. Both vertices and edges can be repeated. Trail A walk where no edge is repeated. Vertices can be repeated. Path A walk where no vertex is repeated (except possibly the start and end). This is the strictest type of traversal.

In network theory, moving through a graph is described by different types of traversal, each with specific rules. The most fundamental is a walk. A walk is simply a sequence of connected vertices, where both vertices and edges can be revisited. A walk that starts and ends at the same vertex is closed, and if it starts and ends at different vertices, it is open. C

B Start A

G A - B - A - E - F - D - E is an open walk but repeats the edge AB

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G A - B - A - D - F - C - G - D - E - A is a closed walk that starts and ends at the same vertex.

By adding restrictions to a walk, more specific types of traversal can be defined. If a walk cannot use the same edge more than once, it is a trail if it’s open, and a circuit if it’s closed. Vertices can still be repeated in trails and circuits. If a walk cannot use the same vertex more than once (except for the start and end), it is called a path. A closed path is called a cycle. C

End

D Start A End

B Start D

A F

The blue walk D - G - C - F - E - A on the left is an open path, since no vertex is used more than once, and it starts and ends at different vertices. The orange walk B - C - F - D - E - A - B on the right is a closed path (cycle), since no vertex is used more than once, except for the vertex at the start/end. These definitions form a hierarchy, which can be summarised as follows: Walks Open start ≠ end

Closed start = end

Trail no repeat edges

Circuit

Path no repeat vertices

Cycle

A walk is the base type of traversal. A trail is a walk with no repeated edges. A closed trail is a circuit. A path is a walk with no repeated vertices. A closed path is a cycle. This means all paths are also trails, and all cycles are also circuits.

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Example 1 For the graph shown, classify each sequence of vertices as a walk, trail, path, circuit, or cycle. Explain your reasoning. B A D

C G

a Sequence: A - C - H - I - G

Create a strategy Check the sequence against the hierarchy of definitions. First, confirm it is a valid walk. Then, check for repeated edges and repeated vertices to classify it further.

Apply the idea • It is a walk because every consecutive pair of vertices (AC, CH, HI, IG) is connected by an edge. • It is a trail because no edges are repeated. • It is a path because no vertices are repeated. The walk starts at A and ends at G, therefore it is open. Since it is an open path, it cannot be a cycle or circuit. Final classification: Walk, trail, path

b Sequence: B - C - D - E - G - H - F - H - C - B

Apply the idea • It is a walk as all edges exist. • It is a trail because no edges are repeated. • It is not a path because the vertex H is repeated. The walk starts and ends at B, therefore it is closed. Since it is a closed trail, it is also a circuit. Final classification: Walk, trail, circuit.

c Sequence: A - D - C - B

Apply the idea The sequence begins with the step from A to D. There is no edge connecting A and D in the graph. Therefore, the sequence is not a walk.

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d Sequence: G - E - D - C - H - G

Apply the idea • It is a walk as all edges exist. • It is a trail as no edges are repeated. • It is a path as no intermediate vertices are repeated. The walk starts and ends at G, so it is closed. Since it is a closed path, it is a cycle. Since it is a closed trail, it is also a circuit. Final classification: Walk, trail, path, circuit, cycle

Idea summary Traversal types are classified by rules about repeating vertices and edges, forming a hierarchy from general to specific. • A walk is any sequence of connected vertices. Repeats are allowed. • A trail is a walk with no repeated edges. A closed trail is a circuit. • A path is a walk with no repeated vertices. A closed path is a cycle. All paths are trails, and all cycles are circuits, but the reverse is not always true.

Traversability: Eulerian and Hamiltonian graphs Traversable

Hamiltonian graph

Describes a graph that can be drawn completely without lifting the pen from the page or retracing any edge.

A graph that contains a Hamiltonian cycle: a closed path that visits every vertex exactly once (except for returning to the start).

Eulerian graph A connected graph that contains an Eulerian circuit: a closed trail that uses every edge exactly once. A graph is Eulerian if and only if all its vertices have an even degree.

Semi-Hamiltonian graph A graph that contains a semi-Hamiltonian path: an open path that visits every vertex exactly once.

Semi-Eulerian graph A connected graph that contains an Eulerian trail: an open trail that uses every edge exactly once. A graph is semiEulerian if and only if it has exactly two vertices of odd degree.

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In network theory, two key questions relate to traversing a graph. The first asks if it is possible to travel along every edge exactly once, and the second asks if it is possible to visit every vertex exactly once. A graph is considered traversable if it is possible to draw it completely without lifting the pen or retracing an edge. This involves finding a trail or circuit that includes every single edge. To determine if a connected graph is traversable, examine the degree of a vertex (the number of edges connected to it). • A graph is Eulerian if all vertices have an even degree. It contains an Eulerian circuit (a closed trail covering all edges) that can start and end at any vertex. • A graph is Semi-Eulerian if exactly two vertices have an odd degree. It contains an Eulerian trail (an open trail covering all edges) that must start at one odd-degree vertex and end at the other. A different kind of problem involves finding a path or cycle that visits every vertex in the graph exactly once. • A Hamiltonian cycle is a closed path that visits every vertex once before returning to the start. • A semi-Hamiltonian path is an open path that visits every vertex once. Unlike Eulerian graphs, there is no simple rule based on vertex degrees to determine if a Hamiltonian path or cycle exists. The only method is to search for one by inspection (trial and error). However, it is sometimes possible to prove that one cannot exist: • If a graph has a vertex of degree 1, it cannot have a Hamiltonian cycle. • If a graph has more than two vertices of degree 1, it cannot have a semi-Hamiltonian path. A graph can be Eulerian, Hamiltonian, both, or neither. The two properties are independent.

Not Hamiltonian Hamiltonian

Not Hamiltonian

Hamiltonian

Not Euler

Not Euler Euler

Euler

The first graph is both Eulerian and Hamiltonian, the second is only Hamiltonian, the third is only Eulerian, and the fourth is neither.

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Example 2 Consider this network: A

a Classify the network as Eulerian, semi-Eulerian, or neither. Explain your reasoning.

Create a strategy Determine the degree of every vertex in the network. Use the count of odd and even degree vertices to classify the graph.

Apply the idea Calculate the degree of each vertex: The vertices B and C have an odd degree. All other vertices have an even degree. Since the graph is connected and has exactly two vertices of odd degree, the network is semi-Eulerian. An Eulerian trail exists, and it must start at either B or C and end at the other.

Vertex

Degree

b Classify the network as Hamiltonian, semi-Hamiltonian, or neither. Explain your reasoning.

Create a strategy Search for a path that visits every vertex exactly once. If such an open path exists, the graph is semi-Hamiltonian. If a closed path (cycle) exists, the graph is Hamiltonian.

Apply the idea By inspection, it is possible to find an open path that visits all six vertices. One such semi-Hamiltonian path is: A - B - C - D - F - E. Since a semi-Hamiltonian path exists, the network is semi-Hamiltonian. A Hamiltonian cycle does not exist. For example, any cycle that includes vertex A must go through B and C. If a path reaches D, it cannot return to the starting vertex without repeating a vertex.

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Reflect and check This graph is an example of a network that is both semi-Eulerian (due to its vertex degrees) and semi-Hamiltonian (determined by inspection). The existence of one type of path does not guarantee or prevent the existence of the other.

Idea summary Traversal problems ask if every edge or every vertex in a graph can be visited exactly once. The concept of traversable relates to whether such a path or circuit exists. • Eulerian graphs focus on traversing every edge. This is determined by analysing the degree of the vertices: • All even degrees: The graph is Eulerian (contains a closed circuit). • Exactly two odd degrees: The graph is semi-Eulerian (contains an open trail). • Hamiltonian graphs focus on visiting every vertex. There is no simple degree rule; paths and cycles must be found by inspection. • A graph containing a Hamiltonian cycle (a cycle visiting every vertex exactly once) is a Hamiltonian graph. • A graph containing a Hamiltonian path (a path visiting every vertex exactly once but not necessarily starting and ending at the same vertex) is a semi-Hamiltonian graph.

Applications of Eulerian graphs The Konigsberg bridge problem is a famous mathematical problem. A river divided the city of Konigsberg into four landmasses, which were connected by seven bridges as shown:

Euler was asked if it was possible to walk around the city crossing all of the bridges exactly once. A start or end can be made in any area.

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Interactive exploration Discover this concept in action online

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The Seven Bridges of Konigsberg is often cited as the origin of graph theory as a mathematical discipline. This problem of “can all of the bridges be crossed exactly once during a single trip?” is really asking “Is this network traversable?” or “does it have an Eulerian trail or circuit?” B

The Seven Bridges of Konigsberg can be represented using a network graph. The graph shows the degrees of the vertices are: deg(A) = 5, deg(B) = 3, deg (C) = 3, and deg(D) = 3. C

Since all four vertices have odd degrees, there is no Eulerian trail or circuit, therefore the network is not traversable. This means that all of the seven bridges cannot be crossed exactly one time.

Example 3 What would happen if a new bridge was built between the northern region (B) and southern region (D) in the Seven Bridges of Konigsberg problem?

Create a strategy One edge would need to be added to the network between B and D.

Apply the idea B The Seven Bridges of Konigsberg can be represented using a network graph. The graph shows that the degrees of the vertices are: deg (A) = 5, deg (B) = 4, deg (C) = 3, deg (D) = 4. C

Since there are now exactly two vertices with odd degrees and two with even degrees, there is an Eulerian trail, therefore the network is traversable. This means that all eight bridges can be crossed exactly one time.

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Example 4 A postman needs to deliver mail along all streets in a small neighbourhood, as shown in the network graph. The postman starts at the post office (P) and wants to travel along each street exactly once. The vertices A and B represent street junctions: B

P a Can the postman travel along each street exactly once and return to the post office (P)? Explain why or why not.

Create a strategy To determine if such a route is possible, the network needs to be checked for the existence of an Eulerian circuit. This involves finding the degree of each vertex. If all vertices have an even degree, an Eulerian circuit exists, and the postman can complete the route as required.

Apply the idea First, the degree of each vertex in the graph must be determined: • deg (P ) = 2 • deg ( A) = 4 • deg (B) = 2 All vertices have an even degree. Since all vertices in the graph have an even degree, the graph is Eulerian. This means an Eulerian circuit exists. Therefore, the postman can travel along each street exactly once and return to the post office (P).

b Suggest one possible route the postman could take, starting and ending at P.

Create a strategy Since an Eulerian circuit exists, any vertex can be a starting point. A route that traverses every edge exactly once needs to be found.

Apply the idea One possible Eulerian circuit starting and ending at P is: P → A → loop at A → B → P This can be written as the trail: P - A - A - B - P (where A - A represents traversing the loop at A). Other routes are also possible.

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Idea summary Many real-world routing problems can be solved by modelling them as network graphs. A network’s traversability is determined by analysing the degree of its vertices—a technique pioneered to solve the famous Seven Bridges of Konigsberg problem.

6.06 Practice questions What do you remember? 1

Determine the condition for a connected graph to be: a

Eulerian (have an Eulerian circuit)

Semi-Eulerian (have an Eulerian trail)

What is the key difference between an Eulerian trail and a Hamiltonian path?

What is the difference between an open walk and a closed walk?

Practice 4

Consider the given network: F

K R P

Determine if it is a valid walk. If it is a walk, state whether it is open or closed. a

K - F - P - K

K-P-F-K-P

K - P - F - R - M

R-P-F-M-R

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For each graph, determine whether each sequence of vertices defines a valid path: a

A C

B D

E b

E-D-C

D-A-C

iii

E-F-C

A-B-D

A-B-C

C-D-D

D-C-D

iii

A-B-C

B-C-A

A-D-C

A-B-C

E-D-C

iii

B-C-C

D-E-A

B-A-B

E-D-B

A-E-C

iii

E-C-B

A-D-C

B-A-E

B c

C E d

E D

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For each network, determine if the sequence of vertices forms a valid circuit: a

C-E-F-D-B-A-C-G

C-E-F-G-H-D-C

iii

A-B-D-H-G-C-A

E-F-G-H-D-B-E

A B

F-C-D-B-E-G-F

E-B-A-C-D-G

iii

G-D-C-F-E-B-G

A-B-D-C-F-A

H G F

B A b

D E F G 7

Using this diagram, list all valid paths: A

C D

E a 8

From A to D

From A to F

Determine the number of unique paths from A to D. B

E D

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For this traversable network:

Is it possible to draw an Eulerian circuit on the network? Explain your answer based on the degrees of the vertices.

Write an Eulerian trail starting at vertex B.

D 10

For each network, determine a path that meets the given requirements: a

The path starts at B and ends at X

he path starts at H, goes through 4 T other vertices, and then ends at B

B X Y Y W

B X

H c

he path starts at Q, has length 3, T and ends at P

The path starts at P and ends at Q

P U

S T

R 11 Ex 1

For each sequence of vertices, select all applicable classifications: walk, trail, path, circuit, cycle: B X Y

W Q H a

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Sequence: Y - B - H - W - X - Q - Y

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Sequence: Y - B - H - W - X - Q

Wendy wishes to traverse the network shown:

Is the network Eulerian or semi-Eulerian?

From which vertex or vertices can an Eulerian trail start?

Wendy starts to trace an Eulerian trail through the graph, by moving along the vertices D, C, B. Write the sequence of vertices that completes Wendy’s Eulerian trail.

B 13 Ex 2

For each graph, determine if it is Eulerian, semi-Eulerian, Hamiltonian or semi-Hamiltonian. Explain your reasoning: a

The graph shown is semi-Eulerian:

Which two vertices must be the start and end points of the Eulerian trail?

Write down one possible Eulerian trail.

The graph shown has a Hamiltonian cycle. Write down one possible Hamiltonian cycle starting from vertex A.

D 16

For each of these networks, determine a valid path that visits each vertex exactly once: a

B A

C S

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17 Ex 3

A group of engineers is tasked with inspecting six bridges that cross a wide canal in town, as illustrated in the diagram:

After inspecting each bridge (which involves crossing it), the engineers must wait for 18 hours before the bridge can be crossed again. Starting from the car at the bottom-centre of the town, can the engineers complete inspections on all six bridges within one day without retracing over any bridge closed due to inspection? 18

Provide an example from the graph. If not possible, explain why. a

A walk that is not a trail

A trail that is not a path

A cycle

C D E F G

19 Ex 4

A delivery driver must cover every road in a small town exactly once, starting and ending at the depot (D), as illustrated in the network diagram. The intersections are marked as X and Y: Y

568

Is it possible for the delivery driver to travel along each road once and return to the Depot (D)? Explain your reasoning.

Determine one route for the delivery driver, starting and ending at D.

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Extend your thinking 20

For each of these lists, select the odd one out and explain the choice: Walk a • b • Path • Path • Closed walk • Cycle • Cycle • Trail • Closed trail

This graph is not Eulerian because it has 4 vertices of odd degree (B - H - E - G): A

D F

What is the minimum number of edges that need to be duplicated (re-traced) to make the graph traversable (Eulerian)? Which pairs of vertices should be connected by duplicating the shortest paths between them? 22

Draw a simple, connected graph for each of these conditions: a

The graph is Eulerian but not Hamiltonian.

The graph is Hamiltonian but not Eulerian.

A chess knight is on a 3 × 3 chessboard. A knight’s move is ‘L-shaped’ - 2 squares in one direction (horizontal or vertical) and then 1 square in a perpendicular direction.

Model the chessboard as a graph where each square is a vertex and a possible knight’s move is an edge. Can a knight start at a corner square, visit every other square exactly once, and end at the centre square (a Hamiltonian path)? Justify your answer by drawing the graph and determining a path if one exists.

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6.07 Algorithms in graph theory After this lesson, you will be able to… • apply algorithms to solve practical problems with graphs and networks • implement graph traversal methods (depth-first and breadth-first search) • solve optimisation problems using algorithmic approaches

Graph traversal algorithms Graph traversal algorithms are methods for visiting every vertex in a graph in a structured way. This allows for solving problems like finding a path between two points or checking if the graph is fully connected. Depth-First Search (DFS) An algorithm that explores a graph by going as deep as possible along each branch before backtracking. Breadth-First Search (BFS) An algorithm that explores a graph by visiting all neighbours of a vertex before moving to the next level. Example: In a social network, BFS can find all friends of a person before exploring friends of friends.

The two main methods are Depth-First Search (DFS) and Breadth-First Search (BFS). DFS explores as far as possible along each branch before backtracking, while BFS explores all neighbours at the current depth before moving deeper. DFS uses a stack (often implemented recursively), while BFS uses a queue. These algorithms are key in applications like maze-solving (DFS) and shortest path finding in unweighted graphs (BFS). BFS is guaranteed to find the shortest path in an unweighted graph because it explores in layers. It checks all the direct neighbours of the start vertex first (layer 1), then all their neighbours (layer 2), and so on. This ensures it finds the path with the fewest edges. Depth-First Search (DFS) steps: 1. Start at the source vertex, add it to a stack, and immediately mark it as visited. 2. While the stack is not empty, pop a vertex and process it. 3. For each neighbour of the popped vertex, check if it has been visited. 4. If a neighbour has not been visited, add it to the stack and mark it as visited. 5. Repeat until the stack is empty.

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Breadth-First Search (BFS) steps: 1. Start at the source vertex, add it to a queue, and immediately mark it as visited. 2. While the queue is not empty, dequeue a vertex and process it. 3. For each neighbour of the dequeued vertex, check if it has been visited. 4. If a neighbour has not been visited, add it to the queue and mark it as visited. 5. Repeat until the queue is empty.

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Discover this concept in action online

Example 1 A maze is represented as a graph with vertices A, B, C, D, E: A

D B

E Use DFS starting at A to find a path to E.

Create a strategy Start at A and use DFS to explore each branch fully, backtracking when necessary, until E is reached.

Apply the idea Begin at A: A

D B

1. From A, go to B (only neighbour). 2. From B, options are C or D. Choose C (arbitrary). 3. C has no unvisited neighbours, backtrack to B. 4. From B, go to D (unvisited). 5. From D, go to E (target reached).

E The final path found is A - B - D - E.

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Reflect and check The path A - B - D - E is a valid path to the end. The algorithm first explored the path to C, hit a dead end, and then backtracked to B to try the other option. This is the key idea of DFS: it will go as far as it can down one path before it returns to explore other options.

Example 2 In a social network graph with vertices A, B, C, D, E, use BFS starting at A to find the shortest path to E. B D

Create a strategy Start at A and use BFS to explore all vertices level by level, tracking the path to E.

Apply the idea Begin at A (queue: A): B A

C Path: A - B - D - E (distance 3).

1. Dequeue A, visit neighbours B, C (queue: B, C). 2. Dequeue B, visit D (queue: C, D). 3. Dequeue C. Its neighbour D is unvisited. Add D to the queue. (Note: In some implementations, vertices are marked ‘visited’ when enqueued to avoid duplicates, but the outcome is the same). 4. Dequeue D, visit E (target reached).

Reflect and check B A

BFS ensures the shortest path in an unweighted graph. A - C - D - E is also distance 3, but A - B - D - E is valid too.

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Idea summary Depth-First Search (DFS) explores deeply first, ideal for pathfinding in mazes, while Breadth-First Search (BFS) explores level by level, perfect for finding shortest paths in unweighted graphs like social networks.

Optimisation algorithms for graphs Optimisation algorithms help find the best possible solution to a problem on a graph, like finding the absolute shortest path or building the cheapest network to connect all points. Dijkstra’s algorithm An algorithm that finds the shortest path from a source vertex to all other vertices in a weighted graph with non-negative weights. Kruskal’s algorithm An algorithm that finds a minimum spanning tree by adding edges in order of increasing weight, avoiding cycles.

Dijkstra’s algorithm works by maintaining a set of visited vertices and always choosing the unvisited vertex that is “closest” to the source. It marks this vertex as visited and then checks its neighbours, updating their tentative distances from the source if a shorter path is found through the newly visited vertex. This process repeats until the destination is reached or all vertices have been visited. Kruskal’s algorithm builds a minimum spanning tree by sorting all edges in the graph from the lowest weight to the highest. It then iterates through the sorted edges, adding each edge to the tree as long as it does not create a cycle with the edges already added. This continues until all vertices are connected in a single tree. While both Kruskal’s algorithm and Prim’s algorithm find a minimum spanning tree, each operates differently. Kruskal’s builds a “forest” of disconnected trees that eventually merge into one, while Prim’s grows a single tree from a starting vertex, always adding the cheapest edge that connects a vertex in the tree to one outside the tree.

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Example 3 A transportation network has vertices A, B, C, D: B 4

1 2

8 C

Use Dijkstra’s algorithm to find the shortest path from A to D.

Create a strategy Start at A, maintain a table of shortest distances, and update it by selecting the vertex with the smallest distance each step.

Apply the idea Track the shortest distance to each vertex from the source A. Initially, the distance to A is 0, and all others are infinity. B

4 5

4 0 A

∞ D

1 8

2 C B

43 5

574

∞ 10 D

1 8 C

• Distance to B becomes 4. • Distance to C becomes 2.

4 0 A

Step 1 (Visit A): Current vertex is A (distance 0). Update its neighbours:

Step 2 (Visit C): The unvisited vertex with the smallest distance is C (distance 2). Update its neighbours: • Distance to B via C is 2 + 1 = 3. Since 3 is less than 4, update B’s distance to 3. • Distance to D via C is 2 + 8 = 10. Update D’s distance to 10.

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0 A

10 8 D

1 8

2 2

Step 3 (Visit B): The next unvisited vertex with the smallest distance is B (distance 3). Check its neighbours: • The path to D through B has a total distance of 3 + 5 = 8. Since 8 is shorter than the current recorded distance to D (which was 10), update D’s distance to 8.

Shortest path: A - B - C - D Total distance: 8 B Step 4 (Visit D): The final vertex is D. The shortest path has been found.

4 1

The shortest path is A - C - B - D, with a total distance of 8.

2 C

A distance table can be used to track the shortest known distance from the source vertex A to every other vertex. Step

Visited vertex

Distance to A

Distance to B

Distance to C

Distance to D

–

∞

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Example 4 A company is planning to connect 7 of its office buildings (A, B, C, D, E, F, G) with a fibre optic cable network. The costs (in thousands of dollars) of laying cable between pairs of buildings are represented by the network: B A

6 3

2 E

9 1

7 F

Design a minimum-cost network that connects all 7 buildings using Kruskal’s algorithm. Determine the edges included in the network and its total minimum cost.

Create a strategy Use Kruskal’s algorithm. First, list all available edges in ascending order of their weight (cost). Sequentially add each edge to the spanning tree, starting with the cheapest, as long as adding the edge does not create a cycle. Stop once all 7 vertices are connected.

Apply the idea First, list the edges sorted by weight from lowest to highest: EF(1), DE(2), BD(3), AB(4), CD(5), BC(6), FG(7), AD(8), EG(9) Now, build the minimum spanning tree step-by-step: B A

4 6

6 3

1. Add EF (weight 1). No cycle formed. 2. Add DE (weight 2). No cycle formed. 3. Add BD (weight 3). No cycle formed.

2 E

9 1

7 F

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B 4

6 3

2 E

9 1

4. Add AB (weight 4). No cycle formed. 5. Add CD (weight 5). No cycle formed. 6. Consider BC (weight 6). This edge would connect vertices B, D, and C, forming the cycle B - D - C - B. Therefore, skip this edge.

7 F

4 6

B Total cost: $22 6 C 3

Add FG (weight 7). No cycle formed. All 7 vertices are now connected. The algorithm can stop.

2 E

9 7

1 F

The edges included in the minimum spanning tree are: EF, DE, BD, AB, CD, and FG. The total minimum cost is the sum of the weights of these edges: Total cost = 1 + 2 + 3 + 4 + 5 + 7 = 22 The minimum cost to connect all buildings is $22 000.

Reflect and check Kruskal’s algorithm is ‘greedy’ because it always chooses the next cheapest edge. This simple strategy is guaranteed to find the minimum spanning tree. Notice that several more expensive edges like AD and EG were not needed. The final tree has 6 edges, which is always the correct number for a spanning tree with 7 vertices (since the number of edges is always v – 1).

Idea summary Optimisation algorithms provide step-by-step methods to find the best solution for a problem on a weighted graph. • Dijkstra’s algorithm finds the shortest path from a single source vertex to all other vertices in the graph. • Kruskal’s algorithm finds the minimum spanning tree (MST), which is the cheapest network of edges needed to connect all vertices.

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6.07 Practice questions What do you remember? 1

Determine the main difference between DFS and BFS in how they explore a graph.

What data structure does DFS typically use, and how does it differ from the one used by BFS?

For each scenario described with the main objective, identify the most appropriate algorithm from the list to solve the problem. Algorithms: Depth-First Search (DFS), Breadth-First Search (BFS), Dijkstra’s algorithm, Kruskal’s algorithm. Scenario

Objective

Algorithm

1. A GPS system needs to find the fastest route between two locations, using a map where roads are marked with travel times.

Find the path with the minimum total weight.

⬚

2. A utility company wants to connect all houses in a new suburb to the power grid. The company Connect all vertices with the has the costs to run a wire between any two minimum total cost. houses. A connection of all houses for the minimum possible total cost. 3. On a social media site, the goal is to find the connection to another user via the smallest number of intermediate friends.

Find the path with the fewest edges.

4. A robot is programmed to navigate a complex maze from a start point to an end point. It needs to find any valid path by exploring one corridor completely before trying another.

Find a path from a start to an end vertex (not necessarily the shortest).

⬚ ⬚ ⬚

Practice Ex 1

A cave system is modelled as a graph with vertices P, Q, R, S, T. Use DFS starting at P to determine a path to T: a b

List the sequence of vertices in the walk explored by the DFS algorithm, including any backtracking. Write the simplified path from P to T.

R Q

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Ex 2

This graph represents a network of locations. Use the Breadth-First Search (BFS) algorithm, starting at vertex A, to find the shortest path to vertex E.

F B 6

Find the shortest path from D to A. D

Use Dijkstra’s algorithm to find the shortest path from vertex A to vertex D in the weighted graph shown:

What is the total weight of the shortest path?

List the sequence of vertices that make up the shortest path from A to D.

5 C

E 8

5 2

This diagram shows a network of paths in a zoo with distances in metres: Bear 60

50 Reptile 30

Aquarium 40 Bird

Lion

45 20 Monkey

Design a minimum-distance network that connects all 6 exhibits using Kruskal’s algorithm.

Determine the total minimum distance.

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Use Kruskal’s algorithm to find the minimum spanning tree for the network shown. The weights represent the cost of connecting the vertices: B a b

List all the edges in the graph in ascending order of their weight.

C 8

3 6

List, in the order of selection, the edges that form the minimum spanning tree. Skip any edges that would form a cycle.

D 1

7 5

Calculate the total weight of the minimum spanning tree.

Extend your thinking 10

A regional council wants to connect a group of rural properties to a communications tower. The council is considering two technologies: a standard cable network and a new wireless relay system. The costs (in thousands of dollars) for each option are shown in the two graphs: Graph 1: Cable network

Graph 2: Wireless relay

P1 20 P3 10 T

15 7

9 P2

Based purely on minimising the initial connection cost, which technology should the council choose? Justify your answer. 11

The network shows the time in hours to build emergency access roads between several remote communities after a flood: C 3

5 4 A

580

D 6

9 10

Determine the minimum time required to build a road network that provides access to all communities.

If the road between communities B and F can not be built, how does this impact the total number of hours required to create the access roads?

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6.08 Applications of graph theory After this lesson, you will be able to… • apply graph theory to model contexts like electrical circuits • prove theorems involving graph properties • model complex systems and relationships using appropriate graph structures

Fundamental graph properties and proofs Graph theory provides a framework for understanding relationships through vertices and edges. Several fundamental theorems underpin this field, offering insights into graph properties and enabling proofs that apply to real-world scenarios. Key proof techniques in graph theory include direct proof and proof by application of a formula. These methods help establish properties such as the number of edges in a complete graph or the planarity of a graph using Euler’s formula. A notable property of the complete graph Kn, where n vertices is the number of edges it contains. This can be derived using combinatorial reasoning.

e e n

n ( n 1) 2

is the number of edges in Kn is the number of vertices

Another important theorem is Euler’s formula for simple, connected planar graphs, which relates the number of vertices, edges, and faces.

v+f–e=2 v is the number of vertices e is the number of edges f is the number of faces (regions, including the outer region)

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Example 1 Prove that the complete graph Kn on n vertices has

n ( n 1) edges. 2

Create a strategy The proof will use the relationship between the sum of the degrees of the vertices and the total number of edges in a graph. The Handshake Lemma states that for any graph, the sum of the degrees of all vertices is equal to twice the number of edges (Σ deg (v) = 2e).

Apply the idea First, determine the degree of a single vertex in a complete graph Kn. In Kn, every vertex is connected to every other vertex. Since there are n vertices in total, each vertex must connect to the other n – 1 vertices. Therefore, the degree of every vertex in Kn is n – 1. Next, find the sum of the degrees of all vertices in the graph. Since there are n vertices, each with a degree of n – 1, the sum is: Sum of degrees = n × (n – 1) According to the Handshake Lemma, this sum is equal to twice the number of edges (2e). 2e = n(n – 1) e=

Apply the Handshake Lemma

n ( n 1) Divide both sides by 2 to solve 2 for the number of edges, e.

Therefore, the number of edges in Kn is proven to be

n ( n 1) . 2

Reflect and check Test this with a small example. For K3 (a triangle), n = 3. The formula gives edges, which is correct. This confirms the formula holds.

3 ( 3 1) 2

3 2 2

Example 2 Draw a planar representation of the complete graph K4, and verify that it satisfies Euler’s formula, v + f – e = 2.

Create a strategy First, draw K4 on a plane such that no edges cross. Then, count the number of vertices (v), edges (e), and faces (f ) from the drawing. Finally, substitute these values into Euler’s formula to confirm it holds.

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Apply the idea For K4 : • Vertices (v): There are 4 vertices by definition. 4 ( 4 1) 6, edges. • Edges (e): Using the formula, e 2 • Faces ( f ): From the planar drawing, there are 3 small triangular faces inside and 1 outer face, for a total of f = 4.

v+f–e=2

Write Euler’s formula

4+4–6=2

Substitute v = 4, f = 4, and e = 6

8–6=2 2=2

Substitute these values into Euler’s formula:

Simplify v + f Verify the formula holds

Since the result is 2, this verifies that the planar representation of K4 satisfies Euler’s formula.

Idea summary Theorems provide foundational tools for proving graph properties in graph theory. • The number of edges in a complete graph Kn is given by the formula, e

n ( n 1) . 2

• Euler’s formula, v + f – e = 2, relates the vertices, edges, and faces of a connected planar graph and can be used to prove a graph is non-planar.

Model and optimisation with graphs Minimum spanning trees are a powerful tool for solving real-world optimisation problems. A minimum spanning tree is used to connect all vertices in a weighted graph with the smallest possible total weight for the included edges. In practical scenarios, the “weight” of an edge can represent various quantities like cost, distance, or time. Calculating the minimum spanning tree translates to determining the most efficient solution, such as: • The lowest cost to build a network. • The shortest distance to connect locations. • The fastest time to establish connections. This principle applies to many different fields and industries. • Infrastructure: Laying electrical cables, water pipes, or internet fibre optics to connect multiple locations. The goal is to use the least amount of material, which minimises cost. • Transportation: Designing a network of roads or railway lines to connect several towns or points of interest with the minimum total length. • Logistics: Planning routes for a delivery service that must have the potential to reach all warehouses in a region using a minimum road network. • Communications: Designing efficient computer networks that connect all devices with the minimum amount of cabling or identifying the most efficient data pathways.

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Example 3 Rhoda maps out the electrical wiring for a unit she is building:

Electrical box Ceiling light Light Fan

Which of these network graphs could be used to represent the electrical wiring? A

Create a strategy First, determine the degree of each component in the wiring diagram. The ceiling fan is connected to 5 other components (degree 5) and the electrical box to 4 (degree 4). Then, examine the graph options to find the one that has vertices with a matching degree sequence.

Apply the idea Option A has a vertex with degree 8, so this is not correct. Option B contains multiple edges between the same two vertices, which does not match the simple connections in the wiring diagram. Option C is correct.

Reflect and check Label the vertices and highlight the regular lights to check.

Ceiling light

Electrical box

Ceiling light

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Ceiling fan

Example 4 An IT technician is setting up a computer network in a new office. The main server (S) must be connected to all workstations (A, B, C, D). The cost of cabling in dollars is shown on the weighted graph:

20 S

A 25 50

The technician has proposed two different layouts, shown: Layout 1: D

Layout 2:

a Determine if both layouts are minimum spanning trees.

Create a strategy Calculate the total cost for each proposed layout. Then, determine the true minimum spanning tree using Prim’s algorithm and compare its cost to the layouts.

Apply the idea First, calculate the cost of each layout. For Layout 1, the edges are SA(20), AB(25), BC(35), and CD(35). Cost = 20 + 25 + 35 + 35 Add the weights of the edges in Layout 1 = $115

Evaluate

The total cost for Layout 1 is $115. For Layout 2, the edges are SA(20), AB(25), AD(45), and CD(35). Cost = 20 + 25 + 45 + 35 Add the weights of the edges in Layout 2 = $125

Evaluate

The total cost for Layout 2 is $125.

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The layouts have different costs; therefore, both cannot be minimum spanning trees. However, it is possible that neither is the actual minimum spanning tree for this network. Determine the true minimum spanning tree using Prim’s algorithm, starting at S. 1. Start at S. The lowest weighted edge is SA with a weight of 20. Add edge SA. 2. From the tree (S, A), the lowest weighted edge is AB with a weight of 25. Add edge AB. 3. From the tree (S, A, B), the lowest weighted edge is BC with a weight of 35. Add edge BC. 4. From the tree (S, A, B, C), the lowest weighted edge to the final vertex D is CD with a weight of 35. Add edge CD. The total cost is 20 + 25 + 35 + 35 = 115. This matches Layout 1. Therefore, only Layout 1 is a minimum spanning tree.

b The cable path A – B is very difficult to install and will prolong the project. Therefore, the technician wants to avoid this connection. What will be the additional monetary cost to the project if this connection is avoided to save time?

Create a strategy If the connection between AB is removed, the network formed by the remaining edges of the spanning tree becomes disconnected. Determine the cheapest way to reconnect the graph.

Apply the idea There are three ways to reconnect the spanning tree. Option 1: Connect SB which has a weight of 50. Option 2: Connect AD which has a weight of 45. Option 3: Connect AC which has a weight of 60. As the weight between AD is the smallest this would add the least amount of money to the project. The original connection between AB was worth 25 and the connection between AD is 45, therefore the project will cost an additional $20.

Idea summary Graph theory models complex systems by representing entities as vertices and relationships as edges. Weighted graphs can be used to solve optimisation problems, such as finding the minimum spanning tree (MST) to connect all points in a network with minimum cost.

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6.08 Practice questions What do you remember? 1

What is the formula for the number of edges in a complete graph Kn?

Describe this graph as complete or not complete. Explain your answer. D

A 3

What information does Euler’s formula v + f – e = 2 provide about a graph?

The network graph shows collaborations among a group of journalists: Derek Xante Neil Tina Xavier Determine whether each of these statements is true or false: a

Neil has collaborated with every other journalist in the group.

The collaboration network represents a complete graph.

There is a path of length 2 between Tina and Derek.

Xavier and Xante have collaborated directly.

Practice 5

Prove that the complete graph K6 has 15 edges.

Ex 1

Show that K3 is planar using Euler’s formula.

Ex 2

Calculate the number of edges in K7.

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Model this electrical circuit as a graph and determine if it is connected:

Ex 3

3 2

10 200 100

4 20 9

10 100

+ − 6

20 9

200

Considering the map, draw a graph that represents the major roads connecting towns.

Young

Temora

A41

Boorowa

B94 Harden Cootamundra

B94

B81

B85 A41

M31

Yass

Junee A25

A41 Lake Burrinjuck

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Consider the image of a train network showing the line colours and stations:

Expo Centre

Willamette River

Airport

Rose Quarter Pioneer Square

Gateway

Hillsboro Gresham PSU Beaverton

Ex 4

Clackamas Town Centre

Draw a graph that represents the train network.

List the stations that would be passed if travelling from the airport to PSU.

The graph shows the distances, in kilometres, of possible tracks between campsites in a national park. The park rangers want to connect all campsites using the shortest possible total length of tracks: A Entrance

2 E 8

7 B

9 10

Determine the minimum total length of tracks required.

Extend your thinking 12

A network has 6 points (A, B, C, D, E, F ) with edges AB, BC, CD, DE, EF, AC. Model it as a graph and use Euler’s formula to check if it’s planar.

Calculate the number of edges in K4 and K5. Then, explain the general pattern for the increase in edges when moving from Kn to Kn+1.

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An electrical circuit can be modelled as a directed graph where vertices represent the components and directed edges show the path of the current. A circuit is designed with a single battery as the source and a single globe where all paths terminate. The layout is described as follows: • From the battery, the current splits into two main branches. • The first branch flows through a switch, then a resistor, and then connects to the globe. • The second branch flows to a resistor. After this component, the path splits again into two smaller branches that both eventually connect to the globe: • One sub-branch goes through a switch, then another resistor, and finally to the globe. • The other sub-branch goes through a switch and then directly to the globe. Draw a single directed graph that models this complex circuit. Label each vertex with its component type.

An internet provider plans to connect five towns with fibre optic cable. The costs of laying cable between the towns, in tens of thousands of dollars, are given in the table: From

Cost ($10 000s)

Draw a weighted network graph to represent this information.

Use Prim’s algorithm to determine the minimum spanning tree for this network. Draw the resulting tree.

What is the minimum cost to connect all five towns?

Chapter 6 review Online review

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“As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.” Albert Einstein

Big ideas • Radian measure and the unit circle provide a foundation for extending trigonometric concepts beyond right-angled triangles to all angles. • The sine and cosine rules generalise trigonometric principles to solve for unknown sides, angles, and areas in any triangle, with applications in two and three-dimensional problems. • Trigonometric identities provide a set of fundamental equations to manipulate, simplify, and solve complex trigonometric expressions.

7 Trigonometry Chapter outline 7.01 7.02 7.03 7.04 7.05 7.06 7.07 7.08

Radians and parts of circles Foundations of trigonometry The sine rule The cosine rule Applications of sine and cosine rules Pythagorean identities Investigation: Prove the sine angle sum identity Angle sum and difference identities Double-angle identities Chapter 7 review

594 606 626 644 656 669 679 687

Trigonometry helps us measure what lies beyond our reach — like the distance across a canyon or the height of a volcano.

7.01 Radians and parts of circles After this lesson, you will be able to… • convert between radian and degree measure • calculate arc length using radian measure • determine the area of sectors and segments of circles • solve problems involving arc length, and sector and segment areas

Degrees and radians Radians are a way of measuring angles based on the radius of a circle. Unlike degrees, which divide a circle into 360 equal parts, radians use the radius as the unit of measurement. One radian is the angle formed when the length of the arc created by the angle is equal to the radius of the circle. r

1 radian r O

Since the circumference of a unit circle is 2π, there are 2π radians in a full circle. This means that 360° is equal to 2π radians, and therefore, 180° is equal to π radians. Radians are often denoted without a specific symbol, or sometimes with rad. These conversion factors are used: • To convert degrees to radians, multiply by

. 180°

• To convert radians to degrees, multiply by

180° . π 180°

Degrees

Radians

180° π

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The most common radian measures and their corresponding degree values are: Radian measure

Degree value

180°

90°

60°

45°

30°

Example 1 Convert –300° to radians in exact form.

Apply the idea –300° = –300° × =

180°

Multiply –300° by

5 radians 3

radians 180°

Simplify

Example 2 Convert

2 radians to degrees. 3

Apply the idea 2 180° 2 radians = 3 3 = 120°

Multiply

2 180° by 3

Simplify

Idea summary Since π rad = 180° and 1 rad =

180°

, to convert between degrees and radians:

• To convert degrees to radians, multiply by

. 180°

180° . Common radian measures and their degree equivalents are:

• To convert radians to degrees, multiply by

Radian measure

Degree value

180°

90°

60°

45°

30°

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Arc length and sector area Recall the formulas for circumference and area of a circle: Circumference : C = 2πr Area : A = πr2 Arc A portion of the circumference of a circle. Sector A region of a circle bounded by two radii and the arc between the radii, similar to a slice of pizza. A

Arc

The length of an arc is a fraction of the full circle’s circumference. When the angle θ is measured in radians, the formula for arc length is:

l = rθ l r θ

is the arc length is the radius of the circle is the central angle, in radians Arc Sector θ

The perimeter of a sector is the arc length plus the two radii. For an angle θ in radians, the formula is: P = rθ + 2r or P = r(θ + 2) The area of a sector A is a fraction of the full circle’s area. When the angle θ is measured in radians, the formula for sector area is: A

1 2 r 2

Note: While formulas exist for degrees, using radians often simplifies calculations, especially in higher-level mathematics.

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Example 3 Consider a circle with radius 10 cm: a Determine the exact length of the arc subtended by the angle 120°.

Create a strategy Convert the angle measurement in degrees to radians and use the formula for arc length.

Apply the idea Step 1: Convert the angle to radians The formula for arc length requires the angle to be in radians. 2 radians 180° 3 Step 2: Calculate the arc length 120˚ ×

l = rθ

Write the formula for arc length 2 3

Substitute r = 10 and θ =

20 cm 3

Simplify and include units

= 10 × =

120°

Convert 120° to radians

2 3

b Determine the exact perimeter of the sector subtended by the angle 120°.

Apply the idea Step 1: Use the angle in radians 2 radians. 3 Step 2: Calculate the perimeter From part (a), 120° =

P = r(θ + 2)

Write the formula for the perimeter of a sector 2 3

= 10

2 3

Substitute r = 10 and θ =

= 10

2 3

6 3

Write with a common denominator

= 10 =

6 3

20 (

3) 3

Add the terms in the brackets Simplify and include units

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c Determine the exact area of the sector subtended by the angle 120°.

Apply the idea Step 1: Use the angle in radians From part (a), 120˚ =

2 rdians. 3

Step 2: Calculate the area 1 2 r θ 2

Write the formula for the area of a sector

1 2 ( 10)2 2 3

Substitute r = 10 and θ =

1 2 ( 100) 2 3

Evaluate the power

100 cm2 3

Simplify and include units

2 3

Example 4 The arc of a circle with radius 13 cm measures 11.7 cm in length. Find the central angle, θ.

Create a strategy 11.7 cm The arc length is given by l = r × θ, where θ is the angle at the centre measured in radians. 13 cm

In this case, the arc length and the radius of the circle are given. The value of θ needs to be determined.

Apply the idea l = rθ

Write the formula for arc length

11.7 = 13θ

Substitute the r = 13 and l = 11.7

θ=

11.7 Make θ the subject of the equation and 13 divide both sides by 13

= 0.9 radians

Simplify and express with proper units

Example 5 Digital only example Explore in the online lesson

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Example 6 The area of a sector of a circle is 32π cm2. If the central angle of the sector is radians, find the 2 radius of the circle.

Create a strategy 1 2 r θ. The area, A, and the 2 angle θ are given, so these values can be substituted into the formula and rearranged to solve for the radius, r. The formula for the area of a sector with an angle in radians is A =

Apply the idea A=

1 2 r θ 2

Write the formula for the area of a sector

32π =

1 2 r 2 2

Substitute A = 32π and θ =

32π =

2 r 4

Simplify the right-hand side

32 =

1 2 r 4

Divide both sides by π

128 = r2 r = 128 = 8 2 cm

Multiply both sides by 4 Take the square root of both sides Simplify and express in proper units

The radius of the circle is 8 2 cm.

Reflect and check Check the answer by substituting the radius back into the area formula: 1 2 r θ 2 2 1 = 8 2 2 2

1 (128) 2 2 = 32π =

Write the formula for the area of a sector Substitute the values Evaluate the square Simplify

The calculated area matches the area given in the problem.

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Idea summary When the central angle θ is measured in radians, use these formulas for calculations involving an arc or a sector: Measurement Arc length Perimeter of a sector

Formula l = rθ P = rθ + 2r

Area of a sector

1 2 r θ 2

Area of a segment Segment A region of a circle bounded by a chord and the arc subtended by that chord.

The area of a segment can be found by calculating the area of the sector created by the central angle θ and then subtracting the area of the triangle formed by the two radii and the chord. Area of segment = Area of sector − Area of triangle 1 The area of the sector is Asector = r2θ, where θ is in radians. The sine rule for area, 2 1 Atriangle = ab sin (C), can be used to find the area of the triangle. In the triangle formed by the two 2 radii and the chord, the two sides adjacent to the central angle are the radii, so a = r and b = r, and 1 the angle C is θ. This gives the area of the triangle as Atriangle = r2 sin (θ ). 2 By subtracting the area of the triangle from the area of the sector, the formula for the area of a segment is obtained: Asegment = Asector – Atriangle =

600

1 2 1 r θ – r2 sin(θ ) 2 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 Factorising out the common term of r2 gives the final formula. Remember, the angle must be in 2 radians to use this formula: A

1 2 r ( 2

sin ( ))

Example 7 A circle has a radius of 8 cm. Find the area of a circular segment subtended by a central angle 2 of radians. Provide an exact answer and an answer rounded to two decimal places. 3

Create a strategy 2 The question gives the radius, r = 8, and the central angle in radians, θ = . Substitute these 3 1 2 values into the formula for the area of a segment, A = r (θ – sin (θ )). 2

Apply the idea A=

1 2 r (θ – sin (θ )) 2

1 2 2 ( 8) 3 2

sin

2 1 (64 ) 3 2

3 2

= 32

2 3

3 2

≈ 39.31 cm2 The exact area is 32

2 3

Write the formula for the area of a segment 2 3

Substitute r = 8 and θ =

2 3

Evaluate the power and sin

2 3

Simplify the expression Evaluate and round

3 cm2, which is approximately 39.31 cm2. 2

Example 8 A circle has a radius of 12 cm. A chord is drawn which creates a minor segment with a central angle of 1.5 radians. Find the area of the major segment, rounded to two decimal places.

Create a strategy The area of the major segment is the total area of the circle minus the area of the minor segment.

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Apply the idea Step 1: Find the exact area of the circle Acircle = πr2 Write the formula for the area of a circle = π (12)2

Substitute r = 12 2

= 144π cm

Simplify and include units

Step 2: Find the exact area of the minor segment 1 2 r (θ – sin (θ )) Write the formula for the area 2 of a segment 1 2 = (12) (1.5 – sin (1.5)) Substitute r = 12 and θ = 1.5 2

Aminor =

= 72 (1.5 – sin (1.5)) cm2

Simplify and include units

Step 3: Subtract the exact areas and find the approximate value Amajor = Acircle – Aminor

Write the formula

= 144π – 72(1.5 – sin (1.5))

Substitute the exact expressions

≈ 416.21 cm

Evaluate and round

The area of the major segment is approximately 416.21 cm2.

Idea summary A segment of a circle is the region bounded by a chord and an arc. The area of a segment is found by subtracting the area of a triangle from the area of a sector. The formula for the area of a segment when the angle θ is in radians is: A

1 2 r ( 2

sin ( ))

7.01 Practice questions What do you remember 1 Ex 1

Convert each angle from degrees to radians, giving your answers in exact form in terms of π: a

135°

–270°

450°

3π

Convert each angle from radians to degrees:

Ex 2

602

60°

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Practice 3

Calculate the arc length for each sector shown, rounded to one decimal place: a

130°

140°

Calculate the arc length, l, given the radius, r, and central angle, θ. Round your answers to two decimal places: 5 a r = 4 cm, θ = b r = 9 cm, θ = 4 3 c r = 20 cm, θ = 2π – d r = 25 cm, θ = 135° 4

Consider the sector shown:

Ex 3

Ex 4

18 cm

Convert the central angle of 120° to radians.

Determine the exact area of the sector in terms of π.

120°

A sector has a radius of 2 cm and an arc length of 3.9 cm. Determine the central angle, θ, in radians. Round your answer to two decimal places.

A large 17 m long sprinkler is placed in a crop field, with one end fixed and the other end free to move. As it turns, it waters everything underneath it.

Ex 5

If the sprinkler has turned an angle of 126°, determine the area of the crop field it has watered.

126˚ 17 m

Determine the radius, r, given the arc length, l, and central angle, θ. Round your answers to two decimal places: 5 a l = 25 cm, θ = b l = 40 mm, θ = 4 6 c

Ex 6

l = 8.2 cm, θ = 60°

l = 1.75 m, θ = 150°

Determine the radius, r, of a sector with a central angle θ and an area of A. Round your answer to two decimal places: a

A = 18π cm2, θ =

rad 4

A = 75 cm2, θ = 120°

A = 50 cm2, θ = 2 rad

A = 40π cm2, θ = 225°

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Determine the sector area created using the angle θ with a radius of r, rounded to two decimal places: 2 a r = 5 cm, θ = b r = 12 cm, θ = 3 3 c

r = 7 cm, θ = 90°

r = 15 cm, θ = 180°

Determine the angle θ, in degrees, that created a sector area of A with radius, r. Round your answers to two decimal places: a

A = 50 cm2, r = 5 cm

A = 100 cm2, r = 10 cm

A = 90 cm2, r = 12 cm

A = 60 cm , r = 8 cm

For each sector shown, determine: i

The perimeter of the sector, rounded to one decimal place

The area of the sector, rounded to one decimal place

b 59.6 mm

30.5 mm

36˚

In this diagram, O is the centre of a circle with a radius 6 cm. Arc JK has a length measuring 2π cm:

Determine the exact area of sector OJK.

K 6 cm O

The given circle, centred at O, has a radius of 18 cm and arc AB has length of 9 cm: Determine the area of the minor sector OAB.

O 18 cm A

Ex 7

9 cm

Calculate the area of the shaded region, rounded to one decimal place.

12 cm

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Ex 8

A circle has a radius of 10 cm. A chord subtends an angle of 150° at the centre. Calculate the area of the major segment formed by this chord, rounded to one decimal place.

Extend your thinking 17

A circle has a centre at O, and a radius of 20 m. Sector AOC has an included angle of θ and an area of 160 m2. Determine the length of arc AC.

A runner runs 120 m along a circular track with a radius of 50 m, from point A to point B:

Calculate the area of the sector AOB.

The runner then returns from B to A in a straight line (along the chord). Calculate the area of the triangle AOB, rounded to two decimal places.

An arc AB subtends an angle of 90° at the centre of a circle of radius 6 2 cm: a

Determine the length of the chord AB. (Hint: Consider Pythagoras’ theorem.)

Determine the exact difference in length between the minor arc AB and the chord AB.

Roald makes a lady beetle cake with a diameter of 24 cm and a height of 10 cm. Roald removes the slice of cake representing the lady beetle’s head with an angle measure of 45°: a

Calculate the exact surface area of the removed slice covered in black icing (top and curved side).

Calculate the exact surface area of the remaining cake covered in red icing (top and curved side).

This diagram shows a piece of jewellery made out of gold: If the gold costs $4 per square millimetre, determine the cost of the piece of jewellery.

6 mm 3 mm

Three circles, each with a radius of 10 cm, are positioned, so the centres of the circles form an equilateral triangle, as shown:

10 cm

Calculate the exact area of the shaded space formed between the three circles.

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7.02 Foundations of trigonometry After this lesson, you will be able to… • define the reciprocal trigonometric ratios • determine exact values for the acute special angles • calculate exact primary and reciprocal trigonometric values using the unit circle

Unit circle and trigonometric ratios Trigonometric ratios Ratios that relate the angles of a right-angled triangle to the lengths of its sides. The primary ratios are sine, cosine, and tangent. Unit circle A circle with a radius of 1 centred at the origin of a Cartesian plane. It is used to define trigonometric functions for all angles, where the coordinates of a point on the circle are given by (cos(θ ), sin(θ )). Coterminal angles Angles in standard position sharing the same terminal side differ by multiples of a full rotation, which is 360° or 2π radians.

The primary trigonometric ratios (sine, cosine, and tangent) relate an angle in a right-angled triangle to the lengths of its sides. A common mnemonic to remember these ratios is SOH CAH TOA.

Hypotenuse

Opposite

θ A

Adjacent

The trigonometric ratios are: sin (θ ) =

Opposite Hypotenuse

Adjacent Hypotenuse Opposite tan (θ ) = Adjacent

cos (θ ) =

Placing a right-angled triangle on the Cartesian plane with its vertex at the origin and its hypotenuse of length 1, traces a unit circle. This representation allows the trigonometric ratios to be expressed using the x- and y-coordinates of the point on the circle.

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y P (x, y) The trigonometric ratios become: 1

−1

θ x

sin (θ ) =

Opposite y y Hypotenuse 1

cos (θ ) =

Adjacent x x Hypotenuse 1

x 1

tan (θ ) = Opposite Adjacent

y , for x x

−1 Unit circle

−1

y P (cos (θ ), sin (θ ))

Terminal arm

1 θ

y x 1

θ is the angle the terminal arm makes with the positive x-axis, measured anti-clockwise. The angle θ can be measured in degrees or radians. The coordinates of the endpoint are

sin ( ) P (cos (θ ), sin (θ )). The slope is tan (θ ) = , cos ( ) for cos (θ ) ≠ 0. −1 Using the unit circle extends the trigonometric ratios to angles in any of the four quadrants. This means that sine, cosine, and tangent can be found for angles of any size. (cos (θ ), sin (θ )) 1

−1

x 1

• In Quadrant 1, 0° < θ < 90°, all ratios are positive. • In Quadrant 2, 90° < θ < 180°, only sin (θ ) is positive. • In Quadrant 3, 180° < θ < 270°, only tan (θ ) is positive. • In Quadrant 4, 270° < θ < 360°, only cos (θ ) is positive.

−1

Angles that share the same terminal side are called coterminal angles. These angles are separated by full rotations (multiples of 360° or 2π radians). Because the terminal side rests in the same position, coterminal angles have the same trigonometric values. For example, 450° and 90° are coterminal, resulting in identical values for sine, cosine, and tangent.

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Example 1 The terminal side of an angle a intersects the unit circle at P

3 4 , : 5 5

a Determine the value of sin (a).

Create a strategy

Apply the idea

On the unit circle, the value of sin (a) is the y-coordinate of the point P.

sin (a) = y =

4 5

b Determine the value of cos (a).

Create a strategy On the unit circle, the value of cos (a) is the x-coordinate of the point P.

Apply the idea cos (a) = x =

3 5

c Determine the value of tan (a).

Create a strategy y The tangent ratio is defined as tan (a) = . x

Apply the idea tan (a) =

Therefore, tan (a)

y x 4 5 3 5 4 3

Write the equation for tangent

Substitute the values of x and y

Simplify the expression

4 . 3

Example 2 Digital only example Explore in the online lesson

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Example 3 Determine if cos (300°) has a positive or negative result.

Create a strategy Consider which quadrant the angle is in and what the sign of the x-coordinate (cosine) will be in that quadrant.

Apply the idea The angle 300° is between 270° and 360°, so it lies in Quadrant 4. In Quadrant 4, the x-coordinate is positive. Since cos (θ ) = x, the value of cos (300°) will be positive.

Reflect and check y

x 300°

If the angle and terminal arm are graphed, the point lies in Quadrant 4, where the x-coordinate is positive. This confirms that cos (300°) is positive.

Idea summary On the unit circle, the primary trigonometric ratios are defined by the coordinates (x, y) of a point on the circle’s edge: y • sin (θ ) = y (cos (θ ), sin (θ )) 1 • cos (θ ) = x y • tan (θ ) = x

Angles that share the same terminal side are called coterminal angles, and these angles share the same trigonometric values.

−1

The sign of each trigonometric ratio depends on the quadrant in which the angle’s terminal side lies. The acronym CAST helps remember which ratios are positive in each quadrant, moving anti-clockwise from Quadrant 4.

C −1

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Reciprocal trigonometric functions The three reciprocal trigonometric functions are cosecant (cosec), secant (sec), and cotangent (cot). These functions are the reciprocals of the primary trigonometric functions sine, cosine, and tangent, respectively. This table shows each primary trigonometric function and its corresponding reciprocal function for an angle θ : Trigonometric function

Reciprocal function

sin (θ )

Cosecant: cosec (θ )

cos (θ )

Secant: sec (θ )

tan (θ )

Cotangent: cot (θ )

1 , sin ( )

1 , cos ( ) 1 tan ( )

sin (θ ) ≠ 0

cos ( )

cos ( ) , sin ( )

sin ( )

Hypotenuse

Opposite

Reciprocal trigonometric ratios can also be expressed in terms of the sides of a right-angled triangle.

Cosecant: cosec (θ ) =

1 sin ( )

Hypotenuse Opposite

Secant: sec (θ ) =

1 cos ( )

Hypotenuse Adjacent

Cotangent: cot (θ ) =

1 tan ( )

Adjacent Opposite

θ Adjacent

To determine the exact value of a reciprocal trigonometric function, find the value of the corresponding primary function (sine, cosine, or tangent) and then calculate its reciprocal. For angles that are significantly larger or smaller than 360° (or 2π radians), using coterminal angles simplifies the process, as their trigonometric function values are identical. Coterminal angles are angles that share the same terminal side when drawn in standard position, this means the angles differ by a number of full rotations (360° or 2π radians). For example, angles of 30°, 390° and − 1050° are coterminal and the values of the reciprocal functions of 390° and − 1050° remain the same as 30° because the trigonometric functions repeat every 360°. 3 The exact values of the sine, cosine, and tangent functions for an angle of 30° are cos (30°) = , 2 3 1 sin (30°) = , and tan (30°) = , therefore: 3 2 • cosec (390°) = cosec (− 1050°) = cosec (30°) = 2 • sec (390°) = sec (− 1050°) = sec (30°) =

2 3 3

• cot (390°) = cot (− 1050°) = cot (30°) = 3

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In radians, 30° =

• cosec

13 radians, 390° = radians, and − 1050° = 6 6

13 6

cosec

35 6

cosec

• sec

13 6

sec

35 6

sec

2 3 3

• cot

13 6

cot

35 6

cot

35 radians, therefore: 6

Example 4 The terminal side of an angle a intersects the unit circle at P

3 4 , : 5 5

a Determine the value of cosec (a).

Create a strategy The cosecant ratio is the reciprocal of the sine ratio. On the unit circle, sin (a) is the y-coordinate. 1 Therefore, cosec (a) . y

Apply the idea 4 Since y , the reciprocal can be found: 5 cosec (a)

1 1 5 y 4 4 5

b Determine the value of sec (a).

Create a strategy

Apply the idea

The secant ratio is the reciprocal of the cosine ratio. On the unit circle, cos (a) is the x-coordinate.

3 Since x , the reciprocal can be found: 5 1 1 5 sec (a) = x 3 3 5

1 Therefore, sec (a) = . x

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c Determine the value of cot (a).

Create a strategy y x The cotangent ratio is the reciprocal of the tangent ratio, which is . Therefore, cot (a) = . x y

Apply the idea 3 x 5 3 cot (a) y 4 4 5

Idea summary For an angle θ, each trigonometric function alongside its reciprocal trigonometric function is represented in this table: Trigonometric function

Reciprocal function

sin (θ )

Cosecant: cosec (θ )

cos (θ )

Secant: sec (θ )

tan (θ )

Cotangent: cot (θ )

1 sin ( )

1 cos ( )

Hypotenuse , for sin ( ) Opposite

Hypotenuse , for cos ( ) Adjacent

1 tan ( )

cos ( ) sin ( )

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Adjacent , for sin ( ) Opposite

To find the exact value of a reciprocal function, first, find the exact value of the primary trigonometric function (sin, cos, or tan) and then, find its reciprocal.

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Exact values Exploration Two special triangles are each drawn in a circle with a radius of 1: 1

θ = 30°

θ = 45°

60° –1

45° 1

–1 0

–1

1. Calculate the lengths of the legs of each triangle, keeping the answer in fraction form. 2. Write the coordinates of the point where each hypotenuse meets the circle. 3. Write the coordinates of the point where each hypotenuse meets the circle in terms of sine and cosine.

Two special right-angled triangles are commonly used to find the exact values of trigonometric ratios for key angles: 30° = π 6

45° = π 4

45° = π 4 1

60° = π 3 1

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sin

1 2

cos

3 2

1 3

60° =

2 2

3 2

2 2

1 2

45° =

30° =

tan

For any multiple of 30°

Angle

2 2 3 3

, 45° or 60° the exact value of each of the trigonometric 6 3 4

functions can be written in surd form.

2 2 , 2 2

−

(–1, 0)

−

2 2

−

y (0, 1) 3π 4

π 2 135°

90°

45°

π 4

180°

315° 225° 5π 270° 7π 4 4 3π 2

 2 2 ,   2 2  

0° 0, 2π

(0, –1)

y 1 3 , (0, 1) 2 2 π 2π 2 π 3 1 , − 3 3 2 2 5π π 120° 90°60° 6 6 150° 30° (–1, 0) π 180° 0° 0 −1 , 3 2 2

(1, 0) x

− 2, 2 − 2 2

π Multiples of 45° 4

7π 6

3 1 ,− 2 2

3 1 − ,− 2 2

3 1 , 2 2 (1, 0) x

210° 330° 11π 240° 300° 6 3 1 270° ,− 4π 5π 2 2 3π 3 3 2 3 1 ,− (0, –1) 2 2

Multiples of 30°

π π and 60° 3 6

The unit circle can be used to determine the exact values of the ratios where the angles are 0° or are multiples of 90. To find the exact value of a trigonometric ratio for an angle greater than 90°: 1. Sketch the angle on the unit circle to find its quadrant. 2. Determine its reference angle (the acute angle it makes with the x-axis). 3. Use the special triangles to find the exact value for the reference angle. 4. Use the quadrant (CAST) to determine if the final answer is positive or negative.

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Example 5 Determine the exact value of: a tan

7 6

Create a strategy 7 is in, and then determine the reference angle. 6 Once the reference angle is determined, use the exact value triangles to find the answer. Identify which quadrant

Apply the idea y

7 is in Quadrant 3. 6 7 The reference angle is . 6 6 The angle

6π = π 6

6 In Quadrant 3, tangent is positive. From the special triangles, tan

7π 6

b sin

Therefore, tan

7 6

3 . 3

2 3

Create a strategy 2 . Then, determine the reference angle (the acute angle 3 the terminal arm makes with the x-axis). Use the reference angle to find the numerical value from the special triangles, and use the quadrant to determine the sign. First, identify the quadrant for the angle

Apply the idea The angle

2 is in Quadrant 2 as 3 2

2 3

. 3 3 From the special triangles, sin . 3 2 In Quadrant 2, the sine ratio is positive. The reference angle is

Therefore, sin

2 3

3 . 2

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c cos

5 4

Apply the idea The angle

5 is in Quadrant 3 as 4

3 . 2

. 4 2 From the special triangles, cos . 4 2 In Quadrant 3, the cosine ratio is negative. The reference angle is

Therefore, cos

5 4

2 . 2

Example 6

π π 2 3 π 4

2 2  

  2  2 ,

(0, 1)

−1, 3 2 2

Consider the diagram shown:

3 1 , 2 2

(1, 0)

a Identify the value of cos

. 6

Create a strategy On the graph, look at the x-coordinate of the point that is at an angle of measure

. 6

Apply the idea The green line on the graph has an angle of measure Therefore, cos

616

3 . 2

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3 , and its x-coordinate is . 2 6

b Determine the value of sec

. 6

Apply the idea sec

= 6

1 cos 6

= 1 =

3 2

2 3

1 cos ( )

Use that sec ( )

Substitute cos

3 2

Multiply 1 by the reciprocal

2 3

2 3 3 3

Multiply the numerator and denominator by 3

2 3 3

Simplify

Reflect and check Since sec (θ ) is the reciprocal of cos (θ ) and from part (a), cos 6 2 3 . which rationalises to 3

3 2 , its reciprocal would be , 2 3

Did you know?

The foundations of trigonometry are essential in designing roofs! For example, architects use trigonometric ratios like sine, cosine, and tangent to calculate the correct slope, height, and length of each section. By applying these principles, they ensure the roof is both structurally stable and visually appealing. Trigonometry allows designers to create roofs that perfectly balance function and form!

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Example 7 Consider this unit circle:

2 2 , 2 2

−

3 1 , 2 2

3 −1 , 2 2

1 3 , 2 2 π 3 π 4 π 6

2π 3π 3 5π 4 6

2 2 , 2 2 3 1 , 2 2

x 7π 6

− 3, −1 2 2 − 2 − 2 , 2 2

11π 3 −1 , 7π 6 2 2 6 5π 2 − 2 3 , 2 2 1 − 3 , 2 2

5π 4 4π 3

−1 , − 3 2 2

a Determine the exact values of cos

and sec 4

. 4

Create a strategy . Subtract or add 2π until an identified angle in the unit circle is 4 reached. Then, use the x-coordinate and reciprocal of the x-coordinate to determine the exact value of cos and sec , respectively. 4 4 Determine a coterminal angle of

Apply the idea 1. Determine the coterminal angle: Coterminal angle = =

2 4 7 4

Add 2π to Evaluate

7 This shows that is coterminal with 4 functions.

618

and therefore share the same trigonometric 4

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2. Determine the exact value of cos cos

7 = cos 4 4

Use the coterminal angle of

2 2

Substitute the x-coordinate of

Therefore, cos

= cos 4 =

= = =

2 , 2

2 2

: 4

1 4

1 2 2 2 2 2 2 2 2 2 2 2

= 2 Therefore, sec

2 . 2

3. Determine the exact value of sec

sec

: 4

Use that sec (θ ) = Substitute cos

1 cos ( ) 4

2 2

Evaluate Multiply both the numerator and denominator by 2 Evaluate the denominator Simplify

b Determine the exact value of cosec

15 . 4

Create a strategy 15 . Subtract or add 2π until an identified angle in the unit circle is 4 15 reached. Then, use the reciprocal of the y-coordinate to determine the exact value of cosec . 4 Determine a coterminal angle of

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Apply the idea 15 2 4 7 = 4

Coterminal angle =

This shows that

15 4

Evaluate

15 7 is coterminal with and therefore share the same trigonometric functions. 4 4

sin

15 7 = sin 4 4

15 = 4

2 2

= =

2 2

2 , 2

2 2

2 2 2 2

1 sin ( )

15 4

Evaluate

2 2

Use that cosec ( )

Substitute the value of sin

= 2 15 4

Substitute the y-coordinate of

1 15 sin 4 1

Therefore, cosec

15 4

2 . 2

15 4

cosec

Use the coterminal angle of

2 2

Therefore, sin

Subtract 2π from

Multiply both the numerator and denominator by 2

Evaluate the denominator Simplify

Reflect and check 7 2 15 , and share the same trigonometric functions. Starting with , which is the 2 4 4 4 2 2 x-coordinate of the point , results in 2 as the final outcome. Therefore, if a calculation , 2 2 2 2. begins with , the y-coordinate of the same point, the expected result is 2

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Idea summary The two special right-angled triangles help find the exact trigonometric values for key angles without a calculator. These values are summarised in the table: sin

cos

tan

0° = 0°

30°

1 2

3 2

3 3

45°

2 2

60°

3 2

1 2

90°

Undefined

To find exact values for angles outside the first quadrant, use the reference angle and the CAST rule to determine the correct sign.

7.02 Practice questions What do you remember? 1

Which of these identities is correct? Adjacent A sec (θ ) Hypotenuse C

cot (θ )

Adjacent Hypotenuse

cosec (θ )

cot (θ )

Hypotenuse Opposite

Opposite Adjacent

Which two of these statements are possible? 1 4

cosec (θ ) =

1 , sin (θ ) = 4 4

cosec (θ ) = 4, sin (θ ) =

sin (θ ) = 4, cosec (θ ) = 4

1 sin (θ ) = , cosec (θ ) = − 4 4

Which two of these statements are possible? 1 3 A sec (θ ) B cosec (θ ) C 2 4

cosec (θ )

5 3

cot (θ )

4 7

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Consider this triangle: a

Determine: i

sec (x)

iii

cot (x)

Determine: i

cosec (x) 1 sec (x)

sec (90° − x)

cosec (90° − x)

Using your answers from parts (a) and (b), describe the relationship between: i

sec (x) and cosec (90° − x)

cosec (x) and sec (90° − x)

90º

The point P on the unit circle has the coordinates (x, y):

In terms of x and y, determine the value of: a

sec (θ )

cosec (θ )

cot (θ )

θ –1

a b c

3 If sin (θ ) = , what is the value of cosec (θ )? 4 2 If cos (θ ) = , what is the value of sec (θ )? 7 8 If tan (θ ) = , what is the value of cot (θ )? 3 Which statement is true of the function sec (θ )? A B C

Which statement is true of the function cosec (θ )? A B C

622

sec (θ ) is undefined when θ = 0. sec (θ ) is undefined when . 2 sec (θ ) is defined for all values of θ. cosec (θ ) is undefined when θ = 0. cosec (θ ) is undefined when . 2 cosec (θ ) is defined for all values of θ.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x 1

–1

P (x, y)

For each angle: i

Which quadrant does this angle lie in?

What is its reference angle? 3 7 b 4 3

5 3

Practice 9 Ex 1

This graph shows an angle a in standard position with its terminal side intersecting the circle at 15 8 P , : 17 17

y P

Determine the value of:

10 Ex 2

sin (a)

cos (a)

tan (a)

sec (a)

cosec (a)

cot (a)

Determine the coordinates of P (x, y) on the unit circle corresponding to an angle of −135°. Provide the exact coordinates and an approximation rounded to two decimal places.

x –135° P

11 Ex 3

12 Ex 4

Determine whether the value of each trigonometric ratio is positive or negative: 4 7 3 a sin b cos c tan d cot 3 4 4 4 5 12 The terminal side of an angle θ intersects the unit circle at the point P , . 13 13 Determine the value of: a

13 Ex 5

sec (θ )

cosec (θ )

cot (θ )

Determine the exact value of these trigonometric ratios. Express answers with a rational denominator where applicable. a cosec b cot c sec d cosec 6 3 4 4

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Use this unit circle to determine:

Ex 6, 7

2 2 , 2 2

−

3 1 , 2 2

3 −1 , 2 2

y 1 3 , 2 2 2 2 , 2 2

π 3 π 4 π 6

2π 3π 3 5π 4 6

3 1 , 2 2

x 7π 6

− 3, −1 2 2 − 2− 2 , 2 2

11π 3 −1 , 7π 6 2 2 6 5π 2− 2 3 , 2 2 1 − 3 , 2 2

5π 4 4π 3

−1 ,− 3 2 2

3 4

The function g (x) is defined as g (x) = sec (x). Determine the exact value of g

53 . 3

The function h (x) is defined as h (x) = cosec (x). Determine the exact value of h

121 . 6

sec

10 3

The function f (x) is defined as f (x) = cot (x). Determine the exact value of f

81 . 4

For each expression: i

Is the expression positive or negative?

Rewrite the expression in terms of its reference angle.

tan

7 6

11 6

sec

sin

4 3

The point P (− 12, − 5) forms an angle, θ, with the positive x-axis: b

Determine r, the distance from the point to the origin. Determine the exact value of: i

sin (θ )

cos (θ )

iii

tan (θ )

sec (θ )

cosec (θ )

cot (θ )

: 2 Determine the value of x where sec (x) = 2.

For 0 a

624

cot

cosec

4 3

tan

5 3

Determine the value of x where cosec (x) = 2.

Determine the value of x where cot (x) = 1.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

cot

2 3

, solve these equations for x: 2 3 sec (x) 2 0 for x

Given 0

0 for x

2 cosec (x)

cot (x)

0 for x

3 3 If cos (θ ) , 7 2

2 , determine the exact value of cosec (θ ).

If cosec (θ ) = 4,

, determine the exact value of cot (θ ).

8 3 If sec (θ ) , 5 2

2 , determine the exact value of cot (θ ).

A point P (x, y) lies in the fourth quadrant on the unit circle. If cosec (θ ) coordinates of P.

9 17

, find the exact

Extend your thinking 22

A unit circle is centred at the origin, O. A point P (x, y) is in the first quadrant on the circle. A tangent to the circle at P intersects the x-axis at U and the y-axis at V. The angle UOP is θ :

cosec (θ )

V P (x, y)

1 1 θ

sec (θ ) 1

Given that the distance OU is 2 units, find the exact values of: a

sec (θ )

The coordinates of point P, with any denominators rationalised

cosec (θ ) with a rational denominator

The angle θ in radians 3 a , where a, b ∈ ℝ+ and 2 b

If sec (x)

Given a sec (x) = 1 and a cosec (x) = 3, determine a and x.

2 , determine cot (x) − cosec (x) in terms of a and b.

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7.03

The sine rule After this lesson, you will be able to… • determine if the sine rule is applicable • apply the sine rule to solve two- and three-dimensional problems • identify if the ambiguous case applies • prove the sine rule for triangle applications

Sine rule to find a side The sine rule, sometimes called the Law of Sines, relates the sides of a triangle to the sines of their opposite angles. It can be used in any triangle, but it is most useful for non-right-angled triangles. The key condition for using the sine rule is that one “pair”—an angle and the length of its opposite side—must be known. To find an unknown side length, it is necessary to know an angle-side pair plus one other angle. This corresponds to the Angle-Angle-Side (AAS) or Angle-Side-Angle (ASA) cases. B a

For a triangle with side lengths a, b, and c, and opposite angles A, B, and C respectively, the sine rule states:

a b c sin(A ) sin(B) sin(C) A, B, C are the measures of the angles in the triangle a, b, c are the lengths of the sides opposite angles A, B, and C In other words, for any angle in a triangle, the ratio of the length of its opposite side to its sine is constant. While the full rule includes all three ratios, it is usually only necessary to use two to solve a problem: a b sin(A) sin(B)

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To prove the sine rule, an altitude (height) can be drawn from one of the vertices and basic trigonometry can be used. Consider triangle ABC. An altitude of length x can be drawn from C to the opposite side AB. This divides the original triangle into two right-angled triangles: C b

x A c

x In the triangle on the left, the sine ratio is sin(A) . Rearranging this gives x = b sin ( A). b x In the triangle on the right, the sine ratio is sin(B) . Rearranging this gives x = a sin (B). a Since both expressions are equal to the altitude x, these can be set equal to each other: b sin ( A) = a sin (B). By dividing both sides by sin ( A) sin (B), the equation can be rearranged into the familiar form of the sine rule: a b sin(A) sin(B) If an altitude were drawn from a different vertex, such as A, it could be similarly shown that b c . Combining these results gives the complete sine rule. sin(B) sin(C)

Interactive exploration Discover this concept in action online

mathspace.co

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627

Example 1 Determine the side length, a, using the sine rule, rounded to two decimal places: a 69° 18

33°

Create a strategy One pair consisting of an angle and its opposite side is given. Therefore, the sine rule can be applied. Use the sine rule:

b a sin( A) sin(B)

Apply the idea b a = sin ( A) sin(B) a 18 = sin(33°) sin (69°) a=

18 sin (69°)

Write the sine rule Substitute b = 18, A = 33°, and B = 69° sin (33°)

≈ 10.50

Multiply both sides by sin (33°) Evaluate and round

Reflect and check Recall the smallest angle in the triangle will be opposite the smallest side length. Since 33° is the smallest angle, expect side a to be smaller than 18.

Example 2 Digital only example Explore in the online lesson

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mathspace.co

Idea summary Use the sine rule to find a missing side when two angles and any side (AAS or ASA) are known.

a b c sin(A ) sin(B) sin(C) A is the angle opposite side a B is the angle opposite side b C is the angle opposite side c

Sine rule to find an angle The sine rule can also be used to find an unknown angle. Just like finding a side, a known angle-and-opposite-side pair is still required. To find an unknown angle, the length of one other side is also needed. This scenario, where two sides and a non-included angle (SSA) are known, can sometimes lead to the ambiguous case, which is covered in the next section. When using the sine rule to determine an unknown angle, it is common to see the formula in the form: sin(A) sin(B) a b The inverted formula provides the same result as the standard form but makes the calculation more direct when determining an angle.

Did you know?

The sine rule is essential for pilots when navigating across long distances! By applying it to the triangles formed between their flight path, destination, and wind direction, pilots can calculate precise headings and course corrections. The sine rule ensures that aircraft stay on the right route, conserve fuel, and reach their destination safely and efficiently!

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Example 3 Calculate the value of the acute angle, x, using the sine rule, rounded to one decimal place: B

A 56°

x C

Create a strategy One pair consisting of an angle and its opposite side is given. Therefore, the sine rule can be applied. Use the sine rule:

sin(A) sin(C) a c

Apply the idea sin(A ) sin(C ) = a c sin(56° ) sin(x) = 9 17 sin (x) =

sin(56°) 9 17

sin (x) = 0.4389 . . . –1

x = sin (0.4389 . . .) = 26.0°

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Write the inverted sine rule Substitute A = 56°, a = 17, c = 9, and C = x Multiply both sides by 9 Evaluate Take the inverse sine of both sides Evaluate and round

Example 4 Solve for θ in degrees, rounded to two decimal places: 15 cm 80°

62° 42°

35 cm

Create a strategy Use the sine rule in the right-hand triangle to find the length of the common side. Then, use this side length and the sine rule in the left-hand triangle to solve for θ.

Apply the idea To calculate θ, an angle-side pair and a second side must be known. To determine the second side, use the sine rule on the triangle on the right. Let a be the side opposite the 62° angle: a b = sin ( A) sin (B)

Write the sine rule

a 15 = sin(62°) sin( 42° ) a=

Substitute A = 62°, B = 42°, and b = 15°

15 sin(62°) sin( 42°)

= 19.7932

Multiply both sides by sin (62°) Evaluate

Solve for θ by using the sine rule for angles in the left triangle: sin( A ) sin( B ) = b a

Write the inverted sine rule

sin( ) sin(80°) = 19.7932 35 sin( 80°) 19.7932 sin (θ ) = 35

θ = sin 1

sin(80°) 35

≈ 33.90°

19.7932

Substitute the known values Multiply both sides by 19.7932 Take the inverse sine of both sides Evaluate and round

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Example 5 This figure shows a pyramid with a triangular base. In triangle ABD, AB = 12 cm, ∠DAB = 90°, and ∠ABD = 30°. In △ BCD, BC = 15 cm, and ∠BCD = 40°: D

40°

90°

A 30°

12 B Find the measure of ∠CDB, rounded to one decimal place.

Create a strategy Use the right-angled triangle ABD and basic trigonometric ratios to find the length of the common side, BD. Once this length is known, apply the sine rule to triangle BCD to calculate ∠CDB.

Apply the idea Step 1: Find the length of BD. In the right-angled triangle ABD, the side adjacent to the 30° angle ( AB = 12) is known and the hypotenuse (BD) needs to be found. Use the cosine ratio. cos (30°) = BD =

Adjacent 12 Hypotenuse BD

Set up the cosine ratio

12 cos( 30°)

Rearrange to solve for BD

≈ 13.856

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Evaluate

Step 2: Use the sine rule to find ∠CDB. Now, consider △ BCD. The known values are BC = 15, BD ≈ 13.856, and ∠BCD = 40°. The goal is to find ∠CDB, which is opposite side BC: sin( CDB) sin( BCD) = BD BC

Write the sine rule

sin( CDB) sin(40°) = 15 13.856

Substitute the known values

sin (∠CDB) =

15 sin( 40°) 13.856

∠CDB = sin 1

15 sin( 40°) 13.856

∠CDB ≈ 44.1°

Multiply both sides by 15 Take the inverse sine Evaluate and round to one decimal place

The measure of ∠CDB is approximately 44.1°.

Reflect and check This problem required combining right-angled trigonometry with the sine rule. The key was identifying the common side BD that connected the two triangles and solving for its length first. This is a common strategy in 3D problems: break the problem down into 2D triangles that can be solved.

Idea summary The sine rule can be inverted to find an angle:

sin(A ) sin(B) sin (C ) a b c A is the angle opposite side a B is the angle opposite side b C is the angle opposite side c

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Ambiguous case of the sine rule The ambiguous case of the sine rule is a special situation that can lead to zero, one, or two possible triangles. It is necessary to check for the ambiguous case when given two side lengths and a non-included angle (SSA), and the side opposite the given angle is shorter than the other given side.

Interactive exploration mathspace.co

Discover this concept in action online

• Case 1: No triangle exists This happens when the side lengths and angle make it impossible to form a closed triangle. When attempting to solve for the unknown angle, a value for sin (B) will be greater than 1. Since the sine of an angle can never be greater than 1, a calculator will show an error, and no triangle can be formed with the given measurements. • Case 2: Exactly one triangle exists This happens in two situations. First, if the side opposite the given angle is longer than the other given side (a ≥ b), only one triangle is possible. Second, even if a < b, the second possible angle (the supplement) might be too large to fit in the triangle with the original given angle. • Case 3: Two possible triangles exist This can only happen if the side opposite the given angle is shorter than the other given side (a < b). If the second potential angle (the supplement) is small enough to fit in the triangle with the given angle, then two valid solutions will exist. • A triangle that includes the calculated solution • A triangle that includes the supplement to the calculated solution C1 b

C2 b

a B1

Possibility 1

B2 = 180° − B1 Possibility 2

To find the second possible solution, subtract the first value of B that was found from 180°.

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Example 6 Determine the value of x in degrees, given that x is obtuse. Round the answer to two decimal places: B

25° 18.8 x A 12.6 C

Create a strategy Use the sine rule:

sin (A) sin (B) a b

Apply the idea sin( A ) sin(B) = a b

Write the inverted sine rule

sin( x ) sin(25°) = 12.6 18.8

Substitute A = x, a = 18.8, b = 12.6, and B = 25°

sin (x) =

sin(25°) 18.8 12.6

sin (x) = 0.630 573 . . . –1

x = sin (0.630573 . . .) = 39.09°

Multiply both sides by 18.8 Evaluate Take the inverse sine of both sides Evaluate and round

To determine the obtuse angle x, subtract the acute angle from 180°: x = 180 – 39.09 = 140.91°

Subtract 39.09 from 180 Evaluate

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Example 7 If possible, solve the triangle where a = 85, b = 93, and ∠A = 61°. Round the answers to two decimal places.

Create a strategy Draw and label a diagram of the triangle; it does not need to be drawn to scale.

Two sides and their non-included angle are given. This is an example of the ambiguous case and solving it may result in one, two, or no solutions to this problem.

A pair of opposite side and angle is given with side a and angle A in addition to side b. The equation can be sin( A) sin(B) set up as to find ∠B. a b

61° A

Apply the idea Step 1: Find the acute angle using the sine rule. sin(A) sin(B) = Write the inverted sine rule b a

93 sin 1 93

sin(61° ) sin(B) = 93 85

Substitute a = 85, b = 93, and A = 61°

sin(61°) = sin (B) 85

Multiply both sides by 93

sin( 61°) 85

Apply the inverse sine to both sides

73.12° = B

Evaluate and round

Step 2: Find the supplementary angle to see if a second solution is valid. ∠B2 = 180° − ∠B1

The supplement of ∠B1

= 180° – 73.12° Substitute ∠B1 = 73.12° = 106.88° Evaluate Adding ∠A and ∠B2 together gives 61° + 106.88° = 167.88°, which does not exceed the maximum 180° possible in a triangle. That means this triangle has two solutions: B1

636

73.12° and

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

106.88°

Step 3: Solve the acute triangle. Solution 1: ∠B1 = 73.12° The known information is a = 85, b = 93, ∠A = 61°, and ∠B1 = 73.12°. Use the triangle angle sum to solve for the missing angle: ∠C1 = 180° − (61° + 73.12°) = 45.88° Use the sine rule to find the missing side, c1:

sin( 45.88°)

c1 a = sin( A) sin(C1 )

Write the sine rule

85 c1 = sin(61°) sin( 45.88 ° )

Substitute a = 85, A = 61°, and C1 = 45.88°

85 = c1 sin(61°)

Multiply both sides by sin (45.88°)

69.77 = c1

Evaluate and round

The three side lengths and angles measures are: a = 85, b = 93, c1 = 69.77 and ∠A = 61°, ∠B1 = 73.12°, ∠C1 = 45.88° Step 4: Solve the obtuse triangle. Solution 2: ∠B2 = 106.88° The known information is a = 85, b = 93, ∠A = 61° and ∠B2 = 106.88°. Use the triangle angle sum to solve for the missing angle: ∠C2 = 180° − (61° + 106.88°) = 12.12° Use the sine rule to find the missing side, c2:

sin ( 12.12°)

c2 a = sin(A ) sin(C2 )

Write the sine rule

85 c2 = sin(61°) sin( 12.12° )

Substitute a = 85, A = 61°, and C2 = 12.12°

85 = c2 sin(61°)

Multiply both sides by sin (12.12°)

20.40 = c2

Evaluate and round

The three side lengths and angles measures are: a = 85, b = 93, c2 = 20.40 and ∠A = 61°, ∠B2 = 106.88°, ∠C2 = 12.12°

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The diagram shows a visual of both solutions: B1 c1

B2 c2

61° A

85 61°

Reflect and check There will be no solution or a calculator error if the inverse sine of a value greater than 1 is attempted. There will be exactly one solution if the solved angle is acute, but its supplement is too large to create a second valid triangle.

Idea summary When determining an angle using the sine rule (with two known sides and a known angle), use the following rule to find one value of B: sin ( B) sin( A) b a If the given angle is acute and the side opposite it is shorter than the other given side, then the ambiguous case applies. Subtract the acute value of ∠B from 180° to find another possible solution. Check that the second solution is a valid solution using the sum of angles in a triangle.

7.03 Practice questions What do you remember? 1

Determine an equation relating the sides and angles using the sine rule: p Q R r P

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Solve for x, rounded to two decimal places: x 50 a sin( 78°) sin( 43°) c

sin( 78°) 20

sin(x) 13

11 sin(66°)

5 sin(x)

x sin( 34°) 7 sin( 70°)

For each scenario: i

Draw a possible illustration of the triangle being described.

Determine whether there is enough information to calculate all the remaining sides and angles in the triangle using only the sine rule.

Three sides are known.

Two of the angles and the side included between them are known.

Two of the angles and a side not included between them are known.

Two of the sides and an angle included between them are known.

Practice Ex 1

For each triangle, determine the side length a, rounded to two decimal places: a

64°

63°

35°

28°

d 104°

4.5 39°

310

25°

a 5

For each triangle, calculate the missing angle and then use the sine rule to determine the length of side x, rounded to one decimal place: a

20 D

47°

41°

39° 64°

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Two lighthouses, P and Q, are 3.5 km apart along a straight coastline. A ship, S, is observed from both lighthouses. The angle ∠SPQ is 55° and the angle ∠SQP is 68°. Calculate the distance of the ship from lighthouse P, rounded to two decimal places.

Calculate the length of y, using the sine rule. Round your answer to one decimal place.

Ex 2

A 11°

60° y

Consider this triangle: a b

Determine the length of side HK, rounded to two decimal places. Determine the length of side KJ, rounded to two decimal places.

46° H

37° 531.36 km

97° J

Consider the given triangle with two interior angles A = 45.32° and B = 64.18°, and one side length c = 7.452 m: A c b B a C

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Determine the size of angle C, rounded to two decimal places.

Determine the value of side a, rounded to three decimal places.

Determine the value of side b, rounded to three decimal places.

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Ex 3

For each triangle, determine the value of the acute angle x using the sine rule, rounded to one decimal place: a

9 B

39°

68°

9 cm 11.2 cm

x C c

62°

A C

34°

11 18

93 B

Consider the given triangle: a

Determine the size of ∠BAC.

Determine the value of c, rounded to two decimal places.

63° B 12

Ex 4

88° C

For each acute-angled triangle, calculate the size of angle B, rounded to the nearest degree: a

△ ABC where ∠A = 57°, side a = 156 cm, and side b = 179 cm

△ ABC where ∠A = 48°, side a = 2.7 cm, and side b = 1.9 cm

For the quadrilateral shown, determine the value of θ, rounded to two decimal places:

18 cm 58° 75° 47°

θ 42 cm

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Ex 5

This figure shows a pyramid with a triangular base. In triangle ABD, AB = 15 cm, ∠DAB = 90°, and ∠ABD = 35°. In triangle BCD, BC = 18 cm and ∠BCD = 45°.

D ?

Determine the measure of ∠CDB, rounded to one decimal place.

90°

45°

A 15

35° 18

Ex 6

In △ PQR, ∠P = 25°, p = 12.6, and r = 18.8. Determine the value of ∠R in degrees, given that ∠R is obtuse. Round your answer to two decimal places.

Ex 7

In △ ABC, a = 50 cm, b = 65 cm, and ∠A = 48°:

How many distinct triangles can be formed with this information? Show your calculations and reasoning.

If one or more triangles can be formed, determine all possible values for ∠B, ∠C, and side c. Round side lengths and angles to two decimal places.

Prove the sine rule using the triangle shown:

In triangle ACD, write an expression for sin ( A) in terms of x and b.

In triangle BCD, write an expression for sin (B) in terms of x and a.

Rearrange both equations to make x the subject.

By setting the two expressions for x equal to each a b other, show that . sin( A) sin(B)

x A

Extend your thinking 18

Mae observes a tower at an angle of elevation of 12°. The tower is perpendicular to the ground. Walking 67 m towards the tower, she determines that the angle of elevation increases to 35°: D

a 35°

12° A

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Determine the length of side a, rounded to two decimal places.

Determine the height, h, of the tower, rounded to one decimal place.

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A surveyor is calculating the height of two identical blocks of flats. From the top of each building ( A and B), the angle of depression is measured to a point C on the ground between them. From A, the angle of depression is 31°, and from B, the angle of depression is 47°:

303 m

31° 47°

Determine the direct distance between A and C, rounded to two decimal places.

Determine h, the height of the buildings, rounded to the nearest metre. C

A radio signal is sent from a transmitter at tower T, via a satellite S, to a town W, as shown in the diagram: S

26°

18° T

526 km

The town is 526 km from the transmitter tower. The signal is sent out from the transmitter tower at an angle of 18°, and the town receives the signal at an angle of 26°. If the satellite is h km above the ground, determine the value of h, rounded to two decimal places. 21

During football training, the coach marks out the perimeter of a triangular course that players need to run around. The diagram shows some measurements taken of the course, where side length a = 14 m:

Each player must sprint one lap and then jog one lap around the triangle. This process is to be repeated 3 times by each player.

If Tara can run 280 m/min, and can jog at half the speed she runs, calculate the time this exercise will take her, rounded to one decimal place. 55° B

87° a = 14 m

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7.04

The cosine rule After this lesson, you will be able to… • determine if the cosine rule is applicable • apply the cosine rule to solve two- and three-dimensional problems • prove the cosine rule for triangle applications

The cosine rule The cosine rule is a formula for finding sides and angles in non-right-angled triangles. Think of it as a more powerful version of Pythagoras’ theorem that works for any triangle, not just right-angled ones. Two situations where the cosine rule is used are:

Case 1 (SAS): Two side lengths and the angle between them are known, to find the side opposite this angle.

36°

x° 8

Case 2 (SSS): Three side lengths are known to find an unknown angle.

10 In a triangle, angles are labelled with capital letters ( A, B, C) and the sides opposite them with matching lowercase letters ( a, b, c). This labelling is important for using the cosine rule correctly. To find an unknown side:

c2 = a2 + b2 – 2ab cos (C) c is the unknown side length opposite the known angle C a, b are the two known side lengths

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To find an unknown angle:

cos(C )

b2 c2 2ab

is the unknown angle

is the side length opposite the unknown angle C

a, b

are the other two side lengths

The cosine rule can be proven using Pythagoras’ theorem and right-angled trigonometry. Consider a △ABC. To prove the cosine rule, introduce a right angle by constructing an altitude from vertex C to the opposite side AB. Call the point of intersection D. Let the length of the altitude CD be h. Let the length of AD be x. Since the full length of AB is c, the length of DB is c – x. C a

b h A

c−x

c Two right-angled triangles are formed, △ADC and △BDC. For △ADC, two relationships are defined: • Using Pythagoras’ theorem: b2 = h2 + x2, which rearranges to h2 = b2 – x2. x • Using trigonometry: cos(A ) , which rearranges to x = b cos ( A). b For △BDC, apply Pythagoras’ theorem: a2 = h2 + (c – x)2 2

Pythagoras’ theorem in △BDC 2

Substitute h2 = b2 – x2 from △ADC

= (b – x ) + (c – x) 2

= b – x + c – 2cx + x

Expand the term (c – x)2

= b2 + c2 – 2cx

Simplify by removing the x2 terms

= b2 + c2 – 2c (b cos ( A))

Substitute x = b cos ( A) from △ADC

= b + c – 2bc cos ( A)

Rearrange the final term to get the cosine rule

This proves the rule for finding side a. The formulas for side b and side c can be proven in the same way by starting with an altitude from a different vertex.

Interactive exploration Discover this concept in action online

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Example 1 For each triangle, determine if the cosine rule can be used to find the value of x. Justify the answer: a

32° 4

Apply the idea Yes. Given two sides and the included angle (SAS), the cosine rule can be used to find the length of the third side, x.

b x°

Apply the idea Yes. Given all three side lengths (SSS), the cosine rule can be used to find the measure of any angle, including x.

68.7°

Apply the idea No, this case requires the sine rule. The cosine rule is not applicable because the given information is two angles and a non-included side (AAS). While the third angle can be found, there is no SAS or SSS information. The sine rule is the appropriate tool to find x.

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Example 2 Determine the length of a using the cosine rule, rounded to two decimal places: C a

33°

Create a strategy Two sides and the included angle are given (SAS), and the third side is to be determined. Therefore, the cosine rule a2 = b2 + c2 – 2bc cos ( A) can be applied.

Apply the idea a2 = b2 + c2 – 2bc cos ( A) 2

Write the cosine rule

a = 13 + 30 – 2 × 13 × 30 cos (33°) 2

a = 13

Substitute b = 13, c = 30, and A = 33° Take the square root of both sides

2 13 30 cos( 33°)

a ≈ 20.37

Evaluate and round

Example 3 Determine the value of w using the cosine rule, rounded to two decimal places:

32 25 w° 19

Create a strategy Three sides are given, and an angle is to be determined. Therefore, the cosine rule can be applied. Use the cosine rule for angles: cos(C )

b2 c2 2ab

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Apply the idea cos (C) =

a2 b2 c2 2ab

322 192 252 2 32 19 5 cos (w) = 8 cos (w) =

w = cos 1

5 8

Write the cosine rule Substitute a = 32, b = 19, c = 25, and C = w Evaluate Take the inverse cosine of both sides

= 51.32° Evaluate and round

Reflect and check Side c needs to be the opposite side to the unknown angle. The order of sides a and b is not important. This substitution could have also been used in the cosine rule: a = 19, b = 32, c = 25, and C = w.

Idea summary The cosine rule is used in non-right-angled triangles for Side-Angle-Side (SAS) and Side-Side-Side (SSS) cases. • To find a side: c2 = a2 + b2 – 2ab cos (C) • To find an angle: cos(C )

b2 c2 2ab

In these formulas, angle C is always opposite side c.

Applications of the cosine rule in 2D The cosine rule is a powerful tool for solving problems involving two-dimensional shapes that are not simple triangles. Many problems in geometry and measurement involve composite shapes, such as quadrilaterals or polygons, which can be broken down into triangles. The strategy for solving these problems is to: 1. Analyse the shape and identify a triangle that contains the unknown length or angle. 2. Determine if there is enough information for that triangle to use the cosine rule (i.e., SAS or SSS). 3. Label the triangle’s sides and angles carefully. Ensure the side opposite the angle is correctly matched as this is the most common place for errors. 4. Substitute the values into the appropriate form of the cosine rule and solve.

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Example 4 A goal has posts that are 2 m apart. Deepa shoots for the goal when she is 2.6 m from one post and 3.1 m from the other post:

2m 2.6 m 3.1 m

x°

Find the measure of the angle, x, in which she can score a goal. Round the answer to two decimal places.

Create a strategy The problem describes a triangle where all three side lengths are known (SSS): 2.6 m, 3.1 m, and 2 m. The angle x needs to be found. Use the cosine rule for finding an angle: a2

cos(C )

b2 c2 2ab

The side opposite the angle x is 2 m.

Apply the idea cos (C) =

a2 b2 c2 2ab

( 3.1)2 (2.6)2 (2)2 2 ( 3.1) (2.6) 9.61 6.76 4 cos (x) = 16.12 cos (x) =

cos (x) =

12.37 16.12

x = cos 1

Write the cosine rule Substitute the values Evaluate the exponents and multiplication Simplify the numerator

12.37 16.12

x ≈ 39.88°

Apply the inverse cosine to both sides Evaluate and round

The angle in which Deepa can score is 39.88°.

Example 5 Digital only example Explore in the online lesson

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Idea summary To solve 2D problems using the cosine rule, first break down the composite shape to identify a key triangle. Once a triangle is identified, determine if Side-Angle-Side (SAS) or Side-Side-Side (SSS) is known. Apply the correct version of the cosine rule to find the unknown length or angle.

Applications of the cosine rule in 3D Solving 3D problems often requires identifying a 2D triangle within the 3D shape. The cosine rule is essential when these 2D triangles are not right-angled. The general strategy is to: 1. Analyse the 3D object to identify the specific triangle in space that contains the required length or angle. 2. Determine the side lengths of this target triangle. It is often necessary to use Pythagoras’ theorem or right-angled trigonometry on other, simpler triangles within the 3D shape (such as the base) to calculate these side lengths first. 3. Apply the cosine rule to the target triangle once the necessary Side-Angle-Side (SAS) or Side-Side-Side (SSS) information is known.

Example 6 A right pyramid has a rectangular base with side lengths of 16 cm and 12 cm. The height of the pyramid is 10 cm: V

θ A B

12 cm

10 cm D

16 cm

C Find the angle θ at the apex of the triangular face that has the shorter base edge. Round the answer to one decimal place.

Create a strategy Determine the lengths of the slant edges using the base dimensions and the pyramid’s height by Pythagoras’ theorem, then apply the cosine rule to the SSS triangle of the face to find the angle θ.

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Apply the idea Step 1: Find the length of the base diagonal BD. Using the base triangle △BCD, where BC = 12 cm and CD = 16 cm: BD2 = BC2 + CD2 2

Apply Pythagoras’ theorem

BD = 12 + 16

Substitute base side lengths

BD = 144 + 256 = 400

Evaluate

BD = 400 20 cm

Take the square root

The length of BO is half of BD, so BO 10 cm. Step 2: Find the length of the slant edge BV. Using the internal right-angled triangle △BOV, where BO = 10 cm and height VO = 10 cm: BV2 = BO2 + VO2 2

Apply Pythagoras’ theorem

BV = 10 + 10 = 200

Substitute and evaluate

BV = 200 10 2 cm

Take the square root and simplify

Step 3: Use the cosine rule to find θ. In the target triangle △BVC, the side lengths are BC = 12, BV 10 2, and CV 10 2. The angle to be found is θ = ∠BVC, which is opposite the side BC. cos (θ ) = BV

cos (θ ) =

CV 2 BC 2 2( BV )(CV )

200 2

200

Write the cosine rule for angles 2

200

122

Substitute side lengths

cos (θ ) =

200 200 144 2 (200)

Evaluate the squares

cos (θ ) =

256 0.64 400

Simplify

θ = cos–1 (0.64)

Apply the inverse cosine

θ ≈ 50.2°

Evaluate and round

Idea summary To solve 3D problems, the key is to isolate the correct 2D triangle within the 3D shape. It is often necessary to first use Pythagoras’ theorem or right-angled trigonometry to find the side lengths of this target triangle. Once SAS or SSS information is known for the target triangle, apply the cosine rule (or simpler trigonometric rules if the triangle is right-angled) to find the required length or angle.

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7.04 Practice questions What do you remember? 1

Which strategy should be used to find the unknown: Pythagoras’ theorem, trigonometric ratios, sine rule, or cosine rule? a

69° 19

33° 11 c

32 25

27°

w 8m

19 Ex 1

For each triangle configuration, identify which rule (sine rule or cosine rule) should be used to find the first unknown value: a

Three sides are known.

Two of the angles and the side included between them are known.

A c c

Two of the angles and a side not included between them are known.

Two of the sides and an angle included between them are known. C

B b A b

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To use the cosine rule to find the length of AC, which angle must be provided?

B 31.05 cm 38.44 cm

C 4

Write the cosine rule for the given triangles used to find side a: a

b 72° 12

14 c A°

Practice Ex 2

For each of these triangles, use the cosine rule to calculate the length of side a, rounded to two decimal places: a

C 17 A

26°

25°

a A

42°

23.3

131

19.4 B

54° 261

8 . 9 Determine the exact length of side AB in centimetres. In △ ABC, cos(C )

A 7 cm B

9 cm

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3 In △QUV, q = 5, u = 6 and cos (V ) . Determine the value of v. 5

Ex 3

For each of these triangles, calculate the value of the variable in degrees, rounded to two decimal places: a

b 25

12 w°

w°

35 18.1

d 5

11.2

A° A°

7 5.2

In △QUV, v = 6, q = 10 and u = 12. Calculate the exact value of cos (U).

A triangle has side lengths of 5 cm, 13 cm, and 15 cm. Determine the size of the largest interior angle, x, rounded to the nearest degree.

The sides of a triangle are in the ratio 4 : 5 : 8. Determine θ, the smallest angle in the triangle, rounded to the nearest degree.

Mae rode her bike 18 km from point A to point B. At point B, she turned and rode 21 km to point C. The angle between her path from A to B and her path from B to C was 126°. Calculate x, the direct distance from A to C, rounded to two decimal places.

Ex 4

21 km

B 18 km

126° x km

A pendulum with a length of 82 cm swings a horizontal distance of 31 cm: Determine the angle, x, of the pendulum’s movement, rounded to the nearest degree.

82 cm

31 cm

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A rhombus has side lengths of 12 cm, and one of its diagonals is 20 cm. Determine the size of all four interior angles of the rhombus, rounded to the nearest degree. A garden, in the shape of a quadrilateral, is represented in the diagram: Determine, rounded to two decimal places:

Ex 6

The length of BD

The length of CD

The perimeter of the garden

19 m

82°

127°

14 m

In a rectangular box with a length of 10 cm, a width of 8 cm, and a height of 6 cm, the vertices P, Q, and R form a triangle as shown. Determine the measure of ∠PQR, rounded to the nearest degree.

Extend your thinking 17

Prove the cosine rule a2 = b2 + c2 – 2bc cos ( A): a

Identify an expression for a2 in terms of c, x, and h by using Pythagoras’ theorem in △CDB.

Identify an expression for h2 by using Pythagoras’ theorem in △ACD.

Identify an expression for x in terms of cos ( A).

Substitute your expressions for h2 and x into your expression for a2 to prove the cosine rule.

C b

D c−x

A teacher is writing an exam question for her maths class. She draws a triangle, labels the vertices A, B, and C and labels the opposite sides a = 5, b = 8, and c = 15, respectively. She asks her students to find the size of ∠A. Explain why this question is impossible to solve.

Two boats, Boat A and Boat B, leave a harbour at the same time. Boat A travels due north at a speed of 15 km/h, while Boat B travels at a speed of 20 km/h in a direction 60° east of north. Determine the distance between the two boats after two hours.

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7.05 Applications of sine and cosine rules After this lesson, you will be able to… • determine which method to use to find missing sides or angles • solve problems using sine and cosine rules • apply trigonometric rules in three-dimensional contexts

Further problem solving with trigonometry To solve problems involving triangles, an appropriate strategy must be chosen based on what is known and what needs to be found. Recall these strategies for non-right-angled triangles: For a triangle with sides a, b, and c, with corresponding angles A, B, and C: The sine rule: a sin ( A)

b sin ( B)

c sin (C)

sin ( A)

sin ( B)

sin (C)

Alternatively, for finding an unknown angle:

The cosine rule: c2 = a2 + b2 – 2ab cos (C) Alternatively, for finding an unknown angle: cos (C)

a2 b2 c2 2ab

This flowchart shows when each strategy can be used to find side lengths and angles. Sine rule a b c sin (A) sin (B) sin (C) Cosine rule c2 = a2 + b2 − 2ab cos (C) No What are you finding?

Is your triangle a right-angled triangle?

Yes What are you finding? Angle θ = sin−1 ( HO ) θ = cos−1 ( HA ) θ = tan−1 ( AO ) No Pythagoras’ theorem a2 + b2 = c2

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Side Do you have an angle?

Angle Do you have another angle?

Side Do you have a “pair”? (angle and opposite side)

No Cosine rule

Yes Sine rule

Yes sin (θ ) = HO cos (θ ) = HA tan (θ ) = AO

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For a triangle with side lengths a and b and an included angle C, the area of the triangle is given by: 1 Area = ab sin (C) 2 For a triangle with side lengths a, b, and c the area of the triangle is given by: Area

s(s – a)(s – b)(s – c), where s

a b c 2

This is also called Heron’s formula.

Example 1 △ ABC consists of angles A, B, and C which appear opposite sides a, b, and c, respectively. Consider the case where the measures of a, c, and A are given: a Which is given? A

SSA: Two sides and a non-included angle

SAS: Two sides and the angle between them

SAA: Two angles and a non-included side

ASA: Two angles and the side between them

SSS: Three sides

Apply the idea Given that the measures of side a, side c, and angle A are known, this corresponds to the two sides and a non-included angle. Therefore, the correct answer is Option A: SSA.

b Which rule should be used to start solving for the measure of the other sides and angles in the triangle? A

The sine rule

The cosine rule

Create a strategy Consider the guidelines from the flowchart: • Use the sine rule when given two sides and a non-included angle (SSA) or two angles and a side (AAS or ASA). • Use the cosine rule when given two sides and the included angle (SAS) or three sides (SSS).

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Apply the idea To solve △ABC where the measures of sides a, c, and angle A are given, the most appropriate rule to use is the sine rule. The sine rule states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. It is expressed as: a sin ( A)

b sin ( B)

c sin (C)

Since the measure of one side and its opposite angle is given, the sine rule can be used to find the measures of the other angles or the other side. This method is particularly useful in this case because it directly relates angles with their opposite sides, making it possible to find unknown angles or sides given at least one side-angle pair. Therefore, the correct answer is Option A: the sine rule.

Example 2 Consider this diagram: P 33 km 62°

40° 20 km Q a Calculate the direct distance, x km, from P to Q. Round the answer to two decimal places.

Create a strategy First, find the angle between the paths OP and OQ. Then, use the cosine rule to find the unknown side x, as two sides and the included angle will be known: c2 = a2 + b2 – 2ab cos (C)

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Apply the idea The bearing of path OP is given as 062°. The bearing of path OQ is measured from the south line. The angle from north to south is 180°, so the true bearing of OQ is 180° – 40° = 140°. The angle between the two paths at point O is the difference between their bearings. ∠POQ = 140° – 62°

Find the difference between the bearings

= 78°

Evaluate

Now, apply the cosine rule to △ POQ: x2 = 332 + 202 – 2 × 33 × 20 × cos (78°) Substitute c = x, a = 33, b = 20, and C = 78° x = 1089 400 1320 (0.2079...)

Evaluate and take the square root

x = 1214.55 ≈ 34.85 km

Simplify inside the square root Evaluate and round

b Calculate the area A, in km2, enclosed by triangle POQ. Round the answer to two decimal places.

Create a strategy Since two side lengths and the included angle are known, use the sine area rule: Area =

1 ab sin (C) 2

Apply the idea 1 ab sin(C) 2 1 = × 33 × 20 × sin (78°) 2 ≈ 322.79 km2

Area =

Write the formula Substitute a = 33, b = 20, and C = 78° Evaluate and round

c Calculate the bearing α, of Q from P. Round the answer to the nearest degree.

Create a strategy To find the bearing of Q from P, first find the bearing of O from P. This is the back bearing of P from O. Next, find the angle inside the triangle, ∠OPQ, using the sine rule. Finally, combine the back bearing and the internal angle to find the total bearing α.

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Apply the idea N P

α 33 km 62°

40° 20 km Q First, find the bearing of O from P. This is found by adding 180° to the bearing of P from O. Bearing of O from P = 62° + 180° = 242°

Calculate the back bearing Evaluate

Next, find the internal angle ∠OPQ using the sine rule and the length of PQ from part (a). sin ( OPQ ) 20

sin (∠OPQ) =

sin (78)° 34.85 20 sin (78°) 34.85

∠OPQ = sin–1 (0.5613…) ≈ 34°

Apply the sine rule Multiply both sides by 20 Take the inverse sine Evaluate and round

From the geometry of the bearings, the bearing of Q from P is found by subtracting ∠OPQ from the bearing of O from P.

α = 242° – 34° = 208°

Subtract the internal angle from the back bearing Evaluate

Example 3 Find the area of a triangle with side lengths 7 cm, 8 cm, and 9 cm. Round your answer to two decimal places.

Create a strategy Since all three sides of the triangle are known (SSS), Heron’s formula can be used to find the area using the formula: Area

660

s (s – a) (s – b) (s – c), where s

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a b c 2

Apply the idea First, calculate the semi-perimeter, s: 7 8 9 2 24 12 = 2

Substitute the side lengths Evaluate the semi-perimeter

Now, substitute s and the side lengths into Heron’s formula: Area = 12 (12 – 7) (12 – 8) (12 – 9) =

Substitute the values into the formula Simplify the terms in the brackets

12 (5) (4) (3)

= 720

Evaluate the multiplication 2

≈ 26.83 cm

Evaluate and round

Idea summary The sine rule is used when an opposite side-angle pair and one other side or angle are known: a sin ( A)

b sin (B)

c sin (C)

sin ( A)

sin (B)

sin (C)

Used to find a side Used to find an angle

The cosine rule is used when two sides and an included angle or all three sides are known: c2 = a2 + b2 – 2abcos (C) cos (C) =

Used to find a side

b c 2ab

Used to find an angle

The sine area formula is used when side lengths a and b are given and ∠C is between the two sides: Area

1 ab sin (C ) 2

Heron’s formula is used to find the area when all sides are given: Area =

s (s – a) (s – b) (s – c), where s

a b c 2

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Applications in three dimensions The sine and cosine rules can also be used to solve problems in three-dimensional (3D) space. The key is to identify and isolate 2D triangles within the 3D object. While some of these triangles may be right-angled (allowing for the use of Pythagoras’ theorem or SOH CAH TOA), many problems will involve non-right-angled triangles. This is where the sine and cosine rules become essential. E

F H

D A

Diagonals across faces and through the solid can form non-right-angled triangles, such as △ AHC shown here.

A general strategy for solving 3D trigonometry problems: 1. Draw a large, clear diagram of the 3D object and label all known information. 2. Identify the specific triangle that contains the side or angle that needs to be found. 3. Redraw this triangle in 2D to make it clearer, labelling the vertices consistently. 4. Calculate any side lengths of this triangle that are not immediately known. This may require using Pythagoras’ theorem or trigonometry on other, simpler triangles within the solid (e.g., on the faces or in right-angled cross-sections). 5. Once the target triangle has sufficient information (e.g., SAS, SSS, ASA), apply the sine or cosine rule to find the unknown quantity. This method is also used to find the angle a line makes with a plane, or the angle between two intersecting planes. To find the angle between two planes, a triangle is constructed that cuts perpendicularly through their line of intersection.

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Example 4 The diagram shows a rectangular prism with a length of 14 cm, and a square base of side length 3 cm: A D

3 cm B

3 cm

θ°

14 cm

H G

a Calculate the size of the angle ∠HFC, where F, H, and C are vertices of the prism. Round your answer to one decimal place.

Create a strategy The angle ∠HFC is part of the internal triangle △ HFC. To use the cosine rule, the lengths of the three sides HF, FC, and HC must first be calculated using Pythagoras’ theorem on the faces of the prism.

Apply the idea Step 1: Calculate the squares of the lengths of the three sides of △ HFC. • Side HF is the diagonal of the top face EFGH (a 14 × 3 rectangle): HF2 = 142 + 32 = 196 + 9 = 205 • Side FC is the diagonal of the right face BCGF (a 14 × 3 rectangle): FC2 = 142 + 32 = 196 + 9 = 205 • Side HC is the diagonal of the back face DCHG (a 3 × 3 square): HC2 = 32 + 32 = 9 + 9 = 18 Step 2: Apply the cosine rule to the isosceles △ HFC. cos (∠HFC) = = =

HF 2 FC 2 HC 2 2( HF ) ( FC ) 205 2

205

392 2(205)

∠HFC = cos 1 ≈ 17.0°

18 205

205

392 410

Write the cosine rule Substitute the squared values

Simplify the expression Take the inverse cosine Evaluate and round

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b Calculate the angle that the space diagonal DF makes with the base plane ABCD. Round your answer to one decimal place.

Create a strategy The angle between a line and a plane is defined as the angle between the line and its projection on the plane. The projection of the space diagonal DF onto the base plane ABCD is the line DB. Therefore, the required angle is ∠FDB. This angle can be found within the right-angled triangle △ FDB using the tangent ratio.

Apply the idea Consider the right-angled triangle △ FDB. The angle at B is a right angle because the height FB is perpendicular to the base plane. • The opposite side to ∠FDB is the height, FB = 3 cm. • The adjacent side is the base diagonal, DB. First, calculate the length of the base diagonal DB: DB2 = DA2 + AB2 2

Use Pythagoras’ theorem on the base

= 3 + 14

Substitute the side lengths

= 9 + 196 = 205

Evaluate

DB = 205

Take the square root

Now, apply the tangent ratio to △ FDB: tan (∠FDB) = =

Opposite Adjacent

Write the tangent ratio

FB 3 DB 205

Substitute the side lengths

∠FDB = tan

≈ 11.8°

3 205

Take the inverse tangent Evaluate and round

Idea summary To solve problems in three dimensions, identify and isolate relevant 2D triangles within the 3D solid. Many of these triangles will not be right-angled, requiring the use of the sine rule or cosine rule. It is often necessary to perform preliminary calculations (using Pythagoras’ theorem or SOH CAH TOA on right-angled faces/cross-sections) to find the side lengths of the target triangle before applying the sine or cosine rule. Always start with a large, clear diagram to help visualise the required triangles and angles.

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7.05 Practice questions What do you remember? Ex 1

For a non-right-angled triangle, determine whether the sine rule or the cosine rule is the most direct method to use given these information: a

Two sides and the included angle (SAS)

Two angles and any side (AAS or ASA)

All three sides (SSS)

Two sides and a non-included angle (SSA)

Practice 2

Ex 2

A triangle has side lengths of 10 cm, 15 cm and 20 cm: a

Which rule (sine or cosine) is the most appropriate to find the largest angle? Explain your reasoning.

Calculate the size of the largest angle, rounded to the nearest degree.

Two boats leave a harbour at the same time. Boat A travels at 15 km/h on a bearing of 070°. Boat B travels at 12 km/h on a bearing of 150°. How far apart are the boats after 2 hours? Round your answer to one decimal place.

A bushwalker starts at point O and walks 15 km on a bearing of 130° to point A. She then walks 12 km on a bearing of 240° to point B: a

Calculate the direct distance from O to B, rounded to one decimal place.

Determine the bearing of B from O, rounded to the nearest degree.

A triangular plot of land is being replaced by a park. Two sides of the park are 250 m and 350 m in length and the angle between them is 60°. Determine the area of the park, rounded to two decimal places.

Determine the area of this triangular paddock, rounded to the nearest square metre:

50° 685 m

575 m

An industrial site in the shape of a triangle is located at the intersection of three roads. Calculate the area of the site, rounded to two decimal places.

44° 50 m 80 m

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Ex 3

Calculate the area of the triangle with sides of length 5 cm, 6 cm, and 5 cm, using Heron’s formula.

Calculate the area of each triangle, rounded to one decimal place: a

18 cm

20 cm

50°

34 cm

13.1 m

12 m

7.4 m 4 m 123°

9 cm

8 cm

5 cm 10

Calculate the area of each triangle, rounded to the nearest square centimetre: a

39° 25° 22 cm

b 107°

33 cm

29° 11.6 cm

23.2 cm

24.7 cm

17 cm

46° 11.6 cm

79° 155 mm

70°

The area of △ ABC is 520 cm2. Given side BC = 48 cm and ∠ACB = 35°: A b cm B

35° 48 cm

Determine the value of b, rounded to the nearest whole number.

666

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Ex 4

The area of △ ABC is 99.3 cm2. Given side AC = 14.5 cm and ∠ACB = 25°. Determine the length of CB, rounded to one decimal place.

For the rectangular prism shown with length 10 cm, width 4 cm, and height 3 cm: a

Calculate the size of the angle ∠GED, formed by the internal diagonals of three different faces. Round your answer to one decimal place. Calculate the angle that the space diagonal DF makes with the base plane EFGH. Round your answer to one decimal place.

A D C

θ°

4 cm B

3 cm F

10 cm G

Extend your thinking 14

An isosceles triangle has a perimeter of 50 cm and its equal sides are of length 17 cm: a

Determine the length of the third side.

Calculate the area of the triangle.

Suppose a rhombus has a perimeter of 80 cm and one of its diagonals is 24 cm: a

Determine the length of each side of the rhombus.

Calculate the area of the rhombus.

Consider △ PQR, where ∠P = 40°, side p = 8 cm, and side q = 10 cm. This is an example of the ambiguous case (SSA): a

Use the sine rule to find the possible value(s) for ∠Q, rounded to one decimal place.

How many distinct triangles can be formed with this information? Explain your reasoning.

Calculate the area of this quadrilateral: 27 cm

42 cm

45 cm 36 cm

39 cm 18

A triangle and a parallelogram have the same area and share a common base of 21 cm. The other two sides of the triangle are 17 cm and 10 cm. Determine the height of the parallelogram.

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A regular pentagonal garden plot has centre of symmetry O and an area of 86 m2. A

Determine the distance OA, rounded to the nearest whole number. 20

The diagram shows the internal triangle AFC within a rectangular prism with dimensions 15 cm × 8 cm × 6 cm: a

Calculate the lengths of the three sides of triangle AFC. Leave your answers in surd form.

Use the cosine rule to find the size of the angle ∠AFC, rounded to one decimal place.

G F

θ 6 cm C 8 cm

D A 21

15 cm

The square pyramid shown has a base with edge lengths of 8 cm. All sloping edges are 12 cm long: a

Determine PM, the perpendicular height of the pyramid, in surd form.

Determine PN, the slant height of a triangular face, in surd form.

Determine the size of ∠PNM, the angle a face makes with the base, rounded to two decimal places.

P 12

C N

M A

A regular n sided polygon is inscribed in a circle of radius r cm. Determine the formula for the area of the n-sided polygon, in terms of n and r.

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7.06 Pythagorean identities After this lesson, you will be able to… • prove Pythagorean trigonometric identities using the unit circle and reciprocals • use Pythagorean trigonometric identities to simplify trigonometric expressions

Pythagorean identities The Pythagorean identities are fundamental relationships in trigonometry that arise from the Pythagoras’ theorem, which applies to right-angled triangles. These identities provide essential tools for solving various mathematical problems by connecting the trigonometric functions. For a given angle, A: • The sine and cosine functions are related by sin2 ( A) + cos2 ( A) = 1 • The tangent and secant functions are related by tan2 ( A) + 1 = sec2 ( A) • The cotangent and cosecant functions are related by cot2 ( A) + 1 = cosec2 ( A) Before proving these identities, it’s important to recall the definitions of the tangent function and the reciprocal trigonometric functions: secant, cosecant, and cotangent, as well as the quotient identities. These will be used in the derivations that follow: sin ( A) • Tangent (tan ( A)) is defined by the quotient identity tan ( A) , for cos ( A) ≠ 0. In a cos ( A) Opposite right-angled triangle, this is the ratio . Adjacent • Secant (sec ( A)) is defined by the reciprocal identity sec ( A) Hypotenuse right-angled triangle, this is the ratio . Adjacent

1 , for cos ( A) ≠ 0. In a cos ( A)

• Cosecant (cosec ( A)) is defined by the reciprocal identity cosec ( A)

In a right-angled triangle, this is the ratio

1 , for sin ( A) ≠ 0. sin ( A)

Hypotenuse . Opposite

1 or the quotient tan ( A) Adjacent cos ( A) , for sin ( A) ≠ 0. In a right-angled triangle, this is the ratio . Opposite sin ( A)

• Cotangent (cot ( A)) is defined by the reciprocal identity cot ( A) identity cot ( A)

The first Pythagorean identity sin2 ( A) + cos2 ( A) = 1 can be obtained by using the trigonometric ratios and Pythagoras’ theorem for a right-angled triangle.

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The Pythagoras’ theorem for right-angled triangles can be described using trigonometric ratios. Consider a right-angled triangle with legs a and b, and hypotenuse c. Let the angle opposite side a be denoted as α.

a From the right-angled triangle, the trigonometric ratios for the angle can be expressed as: a • The sine of α is given by the ratio of the opposite side to the hypotenuse: sin ( ) c • The cosine of α is given by the ratio of the adjacent side to the hypotenuse: cos ( )

b c

To prove that sin2 (α) + cos2 (α) = 1, follow these steps: sin2 (α) + cos2 (α) = a c

b c

Write the left-hand side of the identity and substitute the values of sin (α) and cos (α)

a2 b2 = Express the indices c2 c2 2

a b = Use common denominator c2 2 c =

c2 c2

Substitute a2 + b2 by c2 ( According to the Pythagoras’ theorem a2 + b2 = c2)

= 1 Simplify Therefore, the first (or fundamental) Pythagorean identity, for any angle A is: sin2 ( A) + cos2 ( A) = 1 The proof of the Pythagorean identity sin2 ( A) + cos2 ( A) = 1 is closely linked to the unit circle. In this context, the coordinates of any point on the unit circle for an angle A are given by (cos ( A), sin( A)), which means the x-coordinate is x = cos ( A) and the y-coordinate is y = sin( A). 1

sin (A) –1

0 cos (A)

–1

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x 1

Since the radius of the unit circle is 1, this results in the equation x2 + y2 = 1, which directly reinforces the relationship sin2 ( A) + cos2 ( A) = 1 (by substituting x = cos ( A) and y = sin ( A)), and provides a clear geometric interpretation of these trigonometric functions. The second Pythagorean identity tan2 ( A) + 1 = sec2 ( A) can be derived using the first identity that has already been proven: sin2 ( A) + cos2 ( A) = 1 sin 2 ( A)

cos2 ( A) 2

cos ( A) sin2 ( A)

cos2 ( A)

cos ( A)

cos2 ( A)

sin ( A)

cos2 ( A)

cos ( A)

= =

Write the first Pythagorean identity 1

Divide both sides by cos2 ( A) (for cos ( A) ≠ 0)

cos ( A) 1 cos

1 cos ( A)

Use the property:

2 ( A) 2

Use

sin2 ( A) cos2 ( A)

A C

A B C

B C

sin ( A) cos ( A)

and

1 cos ( A)

cos2 ( A)

tan2 ( A) + 1 = sec2 ( A) Substitute using the quotient identity sin ( A) and the reciprocal identity tan ( A) cos ( A)

sec ( A)

1 cos ( A)

Therefore, the second Pythagorean identity: tan2 ( A) + 1 = sec2 ( A) The third Pythagorean identity cot2 ( A) + 1 = cosec2 ( A) can also be derived using the first identity: sin2 ( A) + cos2 ( A) = 1 2

sin ( A)

cos ( A) 2

sin ( A) sin2 ( A)

cos2 ( A)

2 ( A)

sin ( A)

cos2( A)

sin

sin ( A) sin ( A)

= =

Write the first Pythagorean identity 1

Divide both sides by sin2 ( A) (for sin ( A) ≠ 0)

2 ( A)

sin

Use the property:

2 ( A)

sin

1 = 2 sin ( A) sin ( A)

Use

A B C

sin2 ( A)

sin ( A)

sin2 ( A)

sin ( A)

A C 2

and

B C 1 sin2 ( A)

1 sin ( A)

cot2 ( A) + 1 = cosec2 ( A) Substitute using the quotient identity cos ( A) and the reciprocal identity cot ( A) sin ( A) 1 cosec ( A) sin ( A) Therefore, the third Pythagorean identity: cot2 ( A) + 1 = cosec2 ( A)

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Example 1 12

Given that sin

2 4

, determine the exact value of cos

. 12

Create a strategy Use the first Pythagorean identity: sin2 ( A) + cos2 ( A) = 1

Apply the idea sin2 ( A) + cos2 ( A) = 1 sin2

Write the first Pythagorean identity

cos2

=1 12

Substitute the angle A by

cos2

=1 12

Substitute sin

cos2

=1 12

Expand

2 3 cos2 =1 4 12

Simplify

8 2 12 8 4 3 to 16 16

2 3 from both sides 4

4 8 2 12 16

cos2

= 1 2 3 12 4

Subtract

cos2

2 3 = 4 12

Simplify

cos

6 2 by the given value 12 4

6 2 4

2 3 4

2 3 = Apply square root of both sides taking only the 12 4 positive value since belongs to the first quadrant 12 2 3 = Take the square root of numerator and denominator 4 (2 3) = Evaluate 2

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Reflect and check Substituting the values back into the first Pythagorean identity: sin2 ( A) + cos2 ( A) = 1 12

sin2

=1 12

Substitute the angle A by

6 2 4

cos

Write the first Pythagorean identity

(2 3) 2

Substitute the values for sin

and cos

8 2 12 16

2 3 =1 4

Evaluate the squares

8 4 3 16

2 3 =1 4

Simplify 2 12 to 4 3

8 4 3 16

8 4 3 =1 16

Multiply the second fraction by

4 4

16 =1 16

Add the numerators of the fractions

1=1

Evaluate

Example 2 Given that cot ( )

a Determine the exact value of cosec ( ), where

Create a strategy Use the third Pythagorean identity: cot2 ( A) + 1 = cosec2 ( A)

Apply the idea cot2 ( A) + 1 = cosec2 (α) 3

1 = cosec2 (α)

3 + 1 = cosec2 (α)

Write the third Pythagorean identity Substitute cot (α) by the given value 3 Evaluate

cosec2 (α) = 4

Perform the addition

cosec (α) = ±2

Apply square root to both sides

= –2 Take only the negative value since belongs to the fourth quadrant

Therefore, cosec (α) = –2.

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b Determine, in radians, the exact value of α.

Create a strategy Determine the angle (α) such that cosec (α) = –2.

Apply the idea cosec (α) = –2

Use the solution found in part (a)

1 = –2 sin ( ) 1 2

sin (α) =

α=

Therefore, ( )

Use the definition cosec ( )

1 sin ( )

Take the reciprocal of both sides

Use the unit circle and the specific trigonometric values

Reflect and check Using ( )

For ( )

, it can be verified that cot ( )

, cot 6

cos sin

6 6

3 2 1 2

3. 3 2 1 2

Idea summary The Pythagorean identities are fundamental relationships in trigonometry that are derived from the Pythagoras’ theorem and the definitions of trigonometric and reciprocal trigonometric ratios. The three Pythagorean identities are: sin2 ( A) + cos2 ( A) = 1 tan2 ( A) + 1 = sec2 ( A) cot2 ( A) + 1 = cosec2 ( A)

Applications of Pythagorean identities The Pythagorean identities are essential for simplifying complex trigonometric expressions. By substituting equivalent forms of sin2 ( A) + cos2 ( A) = 1 or its related identities, these expressions can often be reduced, aiding in proofs and problem-solving. Recognising opportunities to replace and simplify terms helps develop critical thinking and reinforces the value of these identities in both theoretical and applied mathematics.

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Example 3 Simplify: a

1 1 1 sin (x) 1 sin (x)

Create a strategy Multiply the fractions, then identify, and substitute any relevant Pythagorean identities.

Apply the idea 1 1 sin (x)

1 1 = Multiply the numerators and multiply 1 sin (x) (1 – sin (x)) (1 + sin (x)) the denominators = =

1 1 sin2 (x) 1 2

cos (x)

Expand the brackets in the denominator

Use an equivalent form of the first Pythagorean identity: 1 – sin2 (x) = cos2 (x)

= sec2 (x)

Evaluate

b tan (θ ) cos2 (θ ) sec (θ )

Create a strategy Use the quotient and reciprocal identities to express the terms in sine and cosine.

Apply the idea Recall basic trigonometric identities: sin ( ) • tan ( ) cos ( ) • sec ( )

1 cos ( )

tan (θ ) cos2 (θ ) sec (θ ) =

sin ( ) 1 cos2 ( ) cos ( ) cos ( )

Substitute tan (θ ) and sec (θ )

= sin (θ ) Evaluate and cancel out the common factors

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1 1 sin2 (x)

Create a strategy Use Pythagorean identity: sin2 (x) + cos2 (x) = 1

Apply the idea 1 1 sin2 (x)

cos2 (x)

Substitute 1 – sin2 (x) with cos2 (x)

= sec2 (x) – 1 Use the identity: sec ( x)

1 cos ( x)

= 1 + tan2 (x) – 1 Use the identity 1 + tan2 (x) = sec2 (x) = tan2(x) Evaluate and combine like terms

Example 4 Prove the identity 5 cos2 (θ ) – 3 = 2 – 5 sin2 (θ )

Create a strategy Use the fundamental Pythagorean identity, sin2 (θ ) + cos2 (θ ) = 1 to manipulate the left side of the equation until it is identical to the right side.

Apply the idea 5 cos2 (θ ) – 3 = 5 (1 – sin2 (θ )) – 3 = 5 × 1 – 5 sin2 (θ ) – 3 2

= 2 – 5 sin (θ )

Substitute cos2 (θ ) with 1 – sin2 (θ ) Expand the brackets Combine the constant terms

Idea summary Pythagorean identities can be applied to other expressions to simplify or derive additional trigonometric identities.

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7.06 Practice questions What do you remember? 1

Complete the three Pythagorean identities: a c

sin2 ( A) + cos2 ( A) = ⬚

1 + cot2 ( A) = ⬚

tan2 ( A) + 1 = ⬚

Express these trigonometric functions in terms of sin (θ ) and/or cos (θ ): a

tan (θ )

sec (θ )

cosec (θ )

cot (θ )

Practice

Ex 1

Starting with the identity sin2 (θ ) + cos2 (θ ) = 1, prove the identity tan2 (θ ) + 1 = sec2 (θ ). Show your working.

Given that cos ( A)

3 3 A 2 , determine the and angle A is in the fourth quadrant 5 2

exact value of sin ( A). Ex 2

Given that tan (B)

12 B , determine the and angle B is in the second quadrant 2 5

exact value of sec (B).

Ex 3

Given that cosec ( A) = 4 and angle A is in the second quadrant, determine the exact value of cot ( A)

Simplify each expression: a

Ex 4

(1 – cos (x)) (1 + cos (x)) 2

sin (θ ) cot (θ )

sec ( A) – 1

cosec2 (β ) – cot2 (β )

cos (x) + sin (x) tan (x)

cos²(β )(1 + tan²(β ))

Simplify the expression (sec (x) – tan (x)) (sec (x) + tan (x)).

Simplify the expression sin2 (θ ) + sin2 (θ ) tan2 (θ ).

Prove these identities: a

(sin (x) + cos (x))2 = 1 + 2 sin (x) cos (x)

sec (θ ) – cos (θ ) = sin (θ ) tan (θ )

3 – 3 sin2 ( A) = 3 cos2 ( A)

tan (β ) + cot (β ) = sec (β ) csc (β )

Prove the identity (1 – cos2 (x)) (1 + cot2 (x)) = 1.

Prove the identity sin4 (x) – cos4 (x) = sin2 (x) – cos2 (x).

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Extend your thinking 13

A point P on the unit circle corresponding to an angle A has coordinates (x, y): 1

sin (A) –1

0 cos (A)

x 1

–1 a

Identify the coordinates of P in terms of A.

The equation of the unit circle is x2 + y2 = 1. Use this information to prove the identity sin2 ( A) + cos2 ( A) = 1.

Explain how this proof works for any quadrant. cos (x)

cos (x)

1 sin (x)

Simplify the expression

Prove the identity

If 2 cos2 (x) – sin2 (x) = 1, determine the possible values of cos (x).

Determine the values of tan (θ ) if 3 sec2 (θ ) + tan (θ ) – 5 = 0.

If sin ( )

Prove the identity (sin (x) + cosec (x))2 + (cos (x) + sec (x))2 = tan2 (x) + cot2 (x) + 7.

tan ( A) cot ( A) sec ( A) cosec ( A)

1 and cos (θ ) > 0, determine the exact values of all six trigonometric functions. 3

Investigation: Prove the sine angle sum identity Investigate online

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7.07 Angle sum and difference identities After this lesson, you will be able to… • apply angle sum and difference identities for sine and cosine to simplify expressions • prove trigonometric identities using these formulas • determine exact values of other angles using angle sums and differences

Angle sum and difference identities Angle sum and difference identities are formulas used to expand the sine or cosine of a combined angle, like A + B, by using the sines and cosines of the individual angles A and B. These identities are: Sine angle sum identity

sin ( A + B) = sin ( A) cos (B) + cos ( A) sin (B)

Sine angle difference identity

sin ( A – B) = sin ( A) cos (B) – cos ( A) sin (B)

Cosine angle sum identity

cos ( A + B) = cos ( A) cos (B) – sin ( A) sin (B)

Cosine angle difference identity

cos ( A – B) = cos ( A) cos (B) + sin ( A) sin (B)

The sine angle difference identity can be found by rewriting the difference A – B as a sum: A + (– B). This allows the sine angle sum identity to be applied. Using the properties of cosine and sine, cos (– B) = cos (B) and sin (– B) = –sin (B), the equation simplifies to: sin ( A – B) = sin ( A) cos (B) – cos ( A) sin (B) Prove the cosine angle sum identity using complementary angles. It is known that (A B) , and the sine difference , so cos ( A + B) can be rewritten as sin cos ( ) sin 2 2 identity can then be applied: cos ( A + B) = sin = sin = sin

2 2 2

( A B) A

A cos ( B) cos

A sin ( B)

= cos ( A) cos (B) – sin ( A) sin (B) Therefore, the cosine angle sum identity: cos ( A + B) = cos ( A) cos (B) – sin ( A) sin (B)

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The cosine angle difference identity can be obtained using the cosine angle sum identity, as any difference can be expressed as a sum. Specifically, the difference A – B can be rewritten as the sum A + (– B), allowing the application of the cosine angle sum identity: cos ( A – B) = cos ( A + (– B)) = cos ( A) cos (– B) – sin ( A) sin (– B) Using the properties of cosine and sine, cos (– B) = cos (B) and sin (– B) = –sin (B), the equation simplifies to: cos ( A – B) = cos ( A) cos (B) + sin ( A) sin (B) Therefore, the cosine angle difference identity: cos ( A – B) = cos ( A) cos (B) + sin ( A) sin (B)

Example 1 Given that sin (A) a sin ( A + B)

24 20 and cos (B) , determine the exact value of: 25 29

Create a strategy Use trigonometric identity: sin ( A + B) = sin ( A) cos (B) + cos ( A) sin (B)

Apply the idea This problem assumes that angles A and B are acute angles (in the first quadrant), meaning their sine and cosine values will be positive. If the quadrants were specified and different, the signs of sine or cosine for A or B might change based on the CAST rule, which would affect the final calculation. 24 is given. In a right-angled triangle, this means the opposite side is 24 and 25 the hypotenuse is 25. Using Pythagoras’ theorem, find the adjacent side is 7. For angle A, sin ( A )

20 . This means the adjacent side is 20 and the hypotenuse is 29. 29 Using Pythagoras’ theorem, the opposite side is 21. Similarly, for angle B, cos ( B)

sin ( A + B) = sin ( A) cos (B) + sin (B) cos ( A) 24 25 24 = 25 =

20 sin ( B) cos ( A ) 29 20 21 7 29 29 25

Write the sine angle sum identity Substitute the values Substitute sin( B) = cos (A)

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Evaluate

A H

7 25

O 21 = and H 29

b cos ( A – B)

Create a strategy Use trigonometric identity: cos ( A – B) = cos ( A) cos (B) + sin ( A) sin (B)

Apply the idea cos ( A – B) = cos ( A) cos (B) + sin ( A) sin (B) Write the cosine angle difference identity 20 24 sin (B) 29 25 7 20 24 21 = 25 29 25 29 = cos (A )

Substitute the values Substitute cos(A ) sin( B)

644 725

O 21 H 29

A 7 and H 25

Evaluate

Example 2 Determine the exact value of cos

11 . 12

Create a strategy Use the identity: cos ( A + B) = cos ( A) cos (B) – sin ( A) sin (B)

Apply the idea 11 First, to find an exact value, express as a sum or difference of familiar angles from the unit 12 2 circle, such as and : 4 3 2 11 cos = cos Write the equation 3 4 12 2 2 cos sin sin = cos Rewrite in expanded form 3 4 3 4 = cos

cos

sin

Use reference angles: 2 and cos cos 3 3 2 sin sin 3 3 Substitute the values

1 2

2 6 4 4

Evaluate the multiplication

2 6 4

Combine into a single fraction

2 2

3 2

2 2

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Idea summary Angle sum and difference identities are formulas which allow expressing the sine or cosine of a combined angle (like A + B) using the sines and cosines of the individual angles ( A and B). These identities are essential for simplifying expressions, finding exact values, and proving other identities. The identities are: Sine angle sum identity Sine angle difference identity Cosine angle sum identity Cosine angle difference identity

sin ( A + B) = sin( A) cos (B) + cos ( A) sin (B) sin ( A – B) = sin ( A) cos (B) – cos ( A) sin (B) cos ( A + B) = cos ( A) cos (B) – sin ( A) sin (B) cos ( A – B) = cos ( A) cos (B) + sin ( A) sin (B)

Applications of angle sum and difference identities The angle sum and difference identities are powerful tools in trigonometry, particularly useful in breaking down expressions involving combined angles. Their applications include: • Simplifying expressions: These identities allow trigonometric expressions that involve sums or differences of angles to be rewritten in expanded forms. This is especially helpful when working with angles that are not common values in trigonometric tables. • Proving trigonometric identities: The sum and difference identities are commonly used to derive or verify other trigonometric identities. This proves helpful in both algebraic and geometric problem-solving contexts where identities need to be established to find other values. Knowing when to use these identities — forwards or backwards — will make solving trigonometry problems much faster and easier.

Example 3 Express sin ( A) cos (2B) + cos ( A) sin (2B) using one trigonometric ratio.

Create a strategy Use the identity: sin (u + v) = sin (u) cos (v) + cos (u) sin (v)

Apply the idea Comparing the given expression with the identity shows that u = A and v = 2B. Therefore, by applying the identity directly to the given expression, the result is: sin ( A) cos (2B) + cos ( A) sin (2B) = sin ( A + 2B) Therefore, the given expression can be expressed using one trigonometric ratio as sin ( A + 2B).

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Example 4 Use the angle sum and difference identities to evaluate tan ( A + B) in terms of tan ( A) and tan (B).

Apply the idea sin ( A B) sin ( A B) Use the fact that tan ( A B) cos ( A B) cos ( A B) sin ( A ) cos ( B) cos ( A) sin ( B) = Substitute the sum identities cos ( A ) cos ( B) sin (A ) sin ( B) Convert this expression in terms of tangent by dividing every term in the numerator and the denominator by cos ( A) cos (B): tan ( A + B) =

sin ( A ) cos ( B) tan ( A + B) =

cos ( A) cos (B)

cos ( A ) sin ( B) cos ( A) cos (B)

cos ( A) cos (B) cos ( A) cos (B)

sin ( A) sin (B) cos ( A) cos (B)

sin ( A ) sin ( B) cos ( A ) cos ( B) = sin ( A ) sin ( B) 1 cos ( A ) cos (B) =

tan ( A ) tan ( B) 1 tan ( A ) tan ( B )

Divide every term by cos ( A) cos (B)

Remove common factors in each term

sin ( ) cos ( ) final identity Substitute

tan ( ) to get the

Did you know?

Scientists use angle sum and difference identities when reconstructing the original shape and posture of fossilised animals! By combining the angles of preserved bones, they can estimate how the creature moved, stood, and even how it hunted or escaped predators.

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Example 5 Prove the identity: sin ( A + B) + sin ( A – B) = 2 sin ( A) cos (B)

Create a strategy Start with the expression on the left-hand side and use the angle sum and difference identities for sine to expand each term. Then, simplify the resulting expression to show it is equal to the expression on the right-hand side.

Apply the idea The angle sum identity for sine is sin ( A + B) = sin ( A) cos (B) + cos ( A) sin (B). The angle difference identity for sine is sin ( A – B) = sin ( A) cos (B) – cos ( A) sin (B). Start by expanding the left-hand side (LHS). Substitute the angle sum and difference identities: sin ( A + B) + sin ( A – B) = (sin ( A) cos (B) + cos ( A) sin (B)) + (sin ( A) cos (B) – cos ( A) sin (B)) Remove brackets: LHS = sin ( A) cos (B) + cos ( A) sin (B) + sin ( A) cos (B) – cos ( A) sin (B) Group like terms: LHS = sin ( A) cos (B) + sin ( A) cos (B) + cos ( A) sin (B) – cos ( A) sin (B) Combine like terms. The cos ( A) sin (B) terms cancel out: LHS = 2 sin ( A) cos (B) + 0 = 2 sin ( A) cos (B)

Write the equation Simplify

The LHS simplifies to 2 sin ( A) cos (B), which is equal to the right-hand side (RHS). Since the LHS = RHS, the identity is proven.

Idea summary The angle sum and difference identities are important tools in trigonometry. As shown in this section, they can be used to simplify trigonometric expressions involving angle sums or differences and to prove other trigonometric identities.

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7.07 Practice questions What do you remember? 1

Write the equation of the four angle sum and difference identities for sine and cosine: a

sin ( A – B)

cos ( A + B)

cos ( A – B)

sin (2x + y)

tan (2α + 3β )

tan (θ – 3σ )

Expand: a

sin ( A + B)

cos (x – 2y)

A and B are two different angles. Given that sin (A) a

Show that sin ( A + B) = 1.

Determine the exact value of sin ( A – B).

3 4 4 3 , cos ( A ) , sin (B) , and cos (B) : 5 5 5 5

Practice 4

A and B are two different acute angles such that sin(A) a

Determine the exact value of cos ( A).

Determine the exact value of sin (B).

Show that sin (A B)

d 5 Ex 1

3 4 15 . 20 Determine the exact value of cos ( A – B).

Suppose sin ( ) exact value of: a

1 3 and cos(B) : 4 5

3 and cos ( ) 5

9 . Given that α and β are acute angles, determine the 41

sin (α – β )

cos (α – β )

Using an angle sum or difference identity, determine the exact value of sin (15°). (Hint: 45° –30° = 15°)

Using

5 12

, determine the exact value of sin

5 . 12

Ex 2

8 Ex 3

Simplify: a

sin ( A + B) cos (B) – cos ( A + B) sin (B)

Determine the exact value of: 5 a cos 12 Show tan (A B)

cos ( A + B) cos (B) + sin ( A + B) sin (B)

cos

7 12

tan (A ) tan (B) . 1 tan (A ) tan(B)

Ex 4

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Prove the identity: cos ( A + B) + cos ( A – B) = 2 cos ( A) cos (B)

Ex 5

Show that sin ( A + B) sin ( A – B) = sin2( A) – sin2(B).

Show that:

cos (45° – α) cos (45° – β ) – sin (45° – α) sin (45° – β ) = sin (α + β )

cos (α) cos (δ – α) – sin (α) sin (δ – α) = cos (δ )

3 Given that sin (A) , where A is an acute angle, and cos (B) 5 angle, determine the exact value of: a

sin ( A + B)

5 , where B is an obtuse 13

cos ( A – B)

Extend your thinking 15

Expand: a

sin ( A + B + C)

cos ( A + B + C)

tan ( A + B + C)

sin ( A B) sin ( B C ) sin (C A ) 0 cos ( A ) cos( B) cos (B) cos( C ) cos (C ) cos (A)

Show that:

Given that sin( A ) sin( B)

3 and cos( A ) cos( B) 10

2 , determine cos ( A – B). 5

Did you know?

Angle sum and difference identities help carpenters cut wood at precise angles using a protractor! This ensures that joints fit perfectly and structures are stable.

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7.08 Double-angle identities After this lesson, you will be able to… • prove the double-angle formulas • apply double-angle identities for sine and cosine to simplify or rewrite expressions • prove trigonometric identities using these double-angle identities • determine exact values of other angles using double-angle identities

Double-angle identities Double-angle identities express trigonometric functions of a double-angle (like 2A) in terms of the trigonometric functions of a single angle (like A). The primary double-angle identities include: • Sine double-angle identity: sin (2A) = 2 sin ( A) cos ( A) • Cosine double-angle identities:

cos (2A) = cos2 ( A) – sin2 ( A)

cos (2A) = 1 – 2 sin2 ( A)

cos (2A) = 2 cos2 ( A) – 1 • Tangent double-angle identity: tan (2A) =

2 tan ( A ) 1 tan2 ( A )

These identities are derived directly from the angle sum identities. To prove the sine double-angle identity, sin (2A) = 2 sin ( A) cos ( A), use the sine angle sum identity, as the angle 2A can be expressed as the sum A + A. Therefore:

sin (2A) = sin ( A + A)

= sin ( A) cos ( A) + cos ( A) sin ( A)

= 2 sin ( A) cos ( A)

To prove the cosine double-angle identity, cos (2A) = cos2 ( A) – sin2 ( A), use the cosine angle sum identity, as the angle 2A can be rewritten as the sum A + A. Therefore:

cos (2A) = cos ( A + A)

= cos ( A) cos ( A) – sin ( A) sin ( A)

= cos2 ( A) – sin2 ( A)

The second and the third cosine double-angle identities can be obtained using the first cosine double-angle identity: cos (2A) = cos2 ( A) – sin2 ( A). Recall that cos2 ( A) + sin2 ( A) = 1 and therefore two equivalent equations can be formed: cos2 ( A) = 1 – sin2 ( A) or sin2 ( A) = 1 – cos2 ( A)

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To obtain the second cosine double-angle identity, substitute cos2 ( A) = 1 – sin2 ( A) into cos (2A) = cos2 ( A) – sin2 ( A): cos (2A) = cos2 ( A) – sin2 ( A)

= 1 – sin2 ( A) – sin2 ( A)

= 1 – 2 sin2 ( A)

To obtain the third cosine double-angle identity, substitute sin2 ( A) = 1 – cos2 ( A) into cos (2A) = cos2 ( A) – sin2 ( A): cos (2A) = cos2 ( A) – sin2 ( A)

= cos2 ( A) – (1 – cos2 ( A))

= cos2 ( A) – 1 + cos2 ( A)

= 2 cos2 ( A) – 1

To prove the tangent double-angle identity, tan (2 A ) identity with 2A = A + A. Therefore:

2 tan ( A ) 1 tan2 (A )

tan (2A) = tan ( A + A)

, use the tangent angle sum

tan ( A ) tan ( A ) 1 tan ( A ) tan ( A ) 2 tan ( A ) 1 tan2 ( A )

Example 1 Given cos ( A ) a sin ( A)

4 and sin ( A) < 0, determine: 5

Create a strategy Use the identity: sin2 ( A) + cos2 ( A) = 1

Apply the idea sin2 ( A) + cos2 ( A) = 1 sin2 ( A )

4 5

sin2 ( A )

Write the Pythagorean identity

Substitute cos ( A )

16 =1 25

Evaluate the square

sin2 ( A) =

9 25

sin ( A) =

Subtract 9 25

4 5

16 from both sides 25

Take the square root of both sides

3 = Evaluate and take the negative solution given that 5 sin ( A) < 0

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b sin (2A)

Create a strategy Use the identity: sin (2A) = 2 sin ( A) cos ( A)

Apply the idea sin (2A) = 2 sin ( A) cos ( A) =2

3 5

24 25

4 5

Write the sine double-angle identity Substitute cos ( A )

4 and sin ( A ) 5

3 5

Evaluate

c cos (2A)

Create a strategy Use the identity: cos (2A) = cos2 ( A) – sin2 ( A)

Apply the idea cos (2A) = cos2 ( A) – sin2 ( A) = 4 5

3 5

Write the cosine double-angle identity Substitute cos ( A )

4 and sin ( A ) 5

16 9 25 25

Evaluate the squares

7 25

Evaluate

3 5

d tan (2A)

Create a strategy Use the identity tan (2 A )

sin (2 A ) , substituting the values found in parts (b) and (c). cos (2 A )

Apply the idea tan (2A) =

sin (2 A ) cos (2 A )

24 25

7 25

24 7

Write the tangent double-angle identity

Substitute cos (2 A )

7 and sin (2 A ) 25

24 25

Evaluate

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Reflect and check Alternatively, use the tangent double-angle identity. First, find the value of tan ( A): 3 5 4 5

sin ( A ) tan ( A) = cos ( A )

3 4

Now, substitute this value into the tangent double-angle identity:

tan (2A) =

2 tan ( A ) 1

tan ( A )

3 4 3 4

6 4 9 1 16

3 2 7 16

3 16 2 7

48 14

24 7

Both methods yield the same result.

Idea summary Double-angle identities express trigonometric functions of a double-angle (like 2A) in terms of the trigonometric functions of a single angle (like A). The primary double-angle identities include: • Sine double-angle identity: sin (2A) = 2 sin ( A) cos ( A) • Cosine double-angle identities: cos (2A) = cos2 ( A) – sin2 ( A) cos (2A) = 1 – 2 sin2 ( A) cos (2A) = 2 cos2 ( A) – 1 • Tangent double-angle identity: tan (2A) =

2 tan ( A ) 1 tan2 ( A )

Applications of double-angle identities The double-angle identities play a crucial role in a variety of applications. Recognising when to apply these identities enhances problem-solving efficiency. Key applications include: • Simplifying complex trigonometric expressions into a more manageable form. • Proving or verifying other, more complex, trigonometric identities. • Finding exact trigonometric values for related angles, such as deriving and using half-angle relationships from the double-angle formulas.

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Example 2 Consider these questions: a Express

x 1 cos ( x ) in terms of cos using the fact that x 2 2

x 2

x . 2

Create a strategy Use the identity: cos ( x ) cos2

x 2

sin2

x 2

Apply the idea 1 cos ( x ) = 2 =

2 cos2

2 cos

2 x 2

= cos2

x 2

x 1 using the Substitute cos ( x ) 2 cos2 2 x double-angle identity for cosine with an angle of 2 Simplify the numerator Remove the common factor of 2

b Determine the exact value of cos ( x ) if sin ( x )

5 and 13 2

Create a strategy Use the identity: sin2 (x) + cos2 (x) = 1

Apply the idea sin2 (x) + cos2 (x) = 1 5 13

Write the Pythagorean identity 5 13

cos2 ( x ) = 1

Substitute sin ( x )

25 cos2 ( x ) = 1 169

Evaluate the square

cos2 (x) =

Since

144 169

Subtract

25 from both sides 169

cos (x) =

144 169

Take the square root of both sides

cos (x) =

12 13

Evaluate

, x is in the second quadrant where cosine is negative, take the negative value: cos ( x )

12 13

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c Determine cos

x 5 if sin ( x ) and 2 2 13

Create a strategy x 1 cos ( x ) is derived. Take the square root of both sides and use the 2 2 value of cos (x) from part (b). From part ( a), cos2

Apply the idea From part ( a), the identity cos2 cos

x 2

cos ( x ) is derived. Taking the square root gives 2

cos ( x ) . 2

x lies. The given domain for x is 2 x x x . Dividing all parts of this inequality by 2 gives the domain for : . This range falls 2 2 4 2 2 within Quadrant 1, where cosine is positive. Therefore, the positive root is used for the calculation. To choose the correct sign, first determine the quadrant in which

cos

x 1 cos ( x ) = 2 2 =

12 13

Write the equation, taking the positive root

Substitute cos ( x )

2 =

1 13 2

Simplify the numerator

1 26

Simplify the fraction

= =

1 26 26 26

Take the square root of the numerator and denominator Rationalise the denominator

Example 3 Express: a sin (x) cos3 (x) – sin3 (x) cos (x) in terms of sin ( 4x )

Create a strategy Use trigonometric and double-angle identities.

692

12 from part (b) 13

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea sin (x) cos3 (x) – sin3 (x) cos (x) = sin (x) cos (x) (cos2 (x) – sin2 (x)) Factorise the common term = sin (x) cos (x) cos (2x) Apply the cosine double-angle identity =

1 Apply the sine sin (2 x ) cos (2 x ) 2 double-angle identity

1 1 Apply the identity sin ( 4 x ) 2 2 1 sin ( A ) cos ( A ) sin (2 A ) 2 with A = 2x

1 sin ( 4 x ) 4

Simplify

b sin (3x) in terms of sin (x)

Apply the idea sin 3x = sin (2x + x)

Rewrite 3x as a sum of angles

= sin (2x) cos (x) + cos (2x) sin (x)

Apply the angle sum identity 2

= 2 sin (x) cos (x) cos (x) + (1 – 2 sin (x)) sin (x) Apply the double-angle identities for sin (2x) and cos (2x) = 2 sin (x) cos2 (x) + sin (x) – 2 sin3 (x)

Expand and simplify

Apply the Pythagorean identity

Expand the brackets

= 2 sin (x) (1 – sin (x)) + sin (x) – 2 sin (x) = 2 sin (x) – 2 sin (x) + sin (x) – 2 sin (x) 3

= 3 sin (x) – 4 sin (x)

Simplify

c cos (4x) in terms of cos (x)

Apply the idea cos (4x) = cos (2x + 2x)

Rewrite 4x as a sum of angles

= cos (2x) cos (2x) – sin (2x) sin (2x) Apply the cosine angle sum identity = cos2 (2x) – sin2 (2x) 2

Simplify 2

= (2 cos (x) – 1) – (2 sin (x) cos (x)) Apply the double-angle identities for cosine and sine = (4 cos4 (x) – 4 cos2 (x) + 1) – (4 sin2 (x) cos2 (x)) 4

Expand the brackets

= 4 cos (x) – 4 cos (x) + 1 – 4(1 – cos (x)) cos (x) Apply the Pythagorean identity sin2 (x) = 1 – cos2 (x) = 4 cos4 (x) – 4 cos2 (x) + 1 – 4 cos2 (x) + 4 cos4 (x) 4

= 8 cos (x) – 8 cos (x) + 1

Expand the final bracket Combine like terms

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Example 4 Verify that

tan2 ( x )

tan2 (x )

cos (2 x ) is an identity.

Create a strategy Use the identity: tan ( A )

sin ( A ) cos ( A )

Apply the idea tan ( A) =

tan2 ( x )

tan (x )

sin ( A ) cos ( A ) sin ( x ) cos ( x )

sin ( x ) cos ( x )

1 = 1 =

Write the trigonometric identity for tan (θ )

Substitute tan2 (x) with sin ( x ) cos ( x )

sin2 ( x ) cos2 ( x ) sin2 ( x )

Use the exponential rule

A B

A2 B2

cos ( x ) 2

cos ( x )

sin2 ( x )

cos2 ( x )

sin2 ( x )

Multiply each term by cos2 (x)

= cos2 (x) – sin2 (x)

Apply the Pythagorean identity

= cos (2x)

Apply the double-angle identity

Idea summary Double-angle identities are applied in various ways: • To simplify complex expressions by rewriting terms like sin (x) cos (x) or cos2 (x) – sin2 (x). • To prove other trigonometric identities, often by first expressing angles like 3x or 4x as sums (e.g., 2x + x). • To derive related formulas, such as half-angle identities, which can be used to find exact trigonometric values.

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7.08 Practice questions What do you remember? 1

Given α is an acute angle: a

Express cos (2α) in terms of cos (α).

Complete these identities by expressing them as a single trigonometric function: a

Express cos (2α) in terms of sin (α).

2 sin (3θ ) cos (3θ ) = ⬚

sin (2A) = sin ( A + A)

Complete the proof for each double-angle identity: a

= sin ( A) ⬚ + cos ( A) ⬚ = 2 sin ( A) cos ( A)

tan (2A) = tan ( A + A) = =

⬚ 1

2 cos2 (5β ) – 1 = ⬚ cos (2A) = ⬚

= ⬚ cos ( A) – ⬚ sin ( A) = cos2 ( A) – ⬚

cos (2A) = cos2 ( A) – sin2 ( A)

tan ( A ) = (1 – ⬚) – sin2 ( A) ⬚

2 tan ( A ) = 1 – ⬚ 1 tan2 ( A )

Practice 4

Using the angle sum identity, sin ( A + B) = sin ( A) cos (B) + cos ( A) sin (B), prove the double-angle identity for sine: sin (2A) = 2 sin ( A) cos ( A) 4 Given that A is an acute angle and sin ( A ) , determine the exact value of: 5 a cos ( A) b sin (2A) 4 , determine the exact value of cos (2α). 5

Given that α is an acute angle and sin ( )

A and B are two different acute angles such that sin( A )

Ex 1

Determine the exact value of cos ( A).

Determine the exact value of sin (2B).

Determine the exact value of cos (2A).

Determine the exact value of cos (165°) given cos ( 330 ° )

Using the value of tan (45°), determine tan (22.5°).

Given tan ( A) = 2, use the identity tan (2 A ) a

tan (2A)

1 3 and cos ( B) : 4 5

3 . 2

2 tan ( A ) , to determine the exact value of: 1 tan2 (A ) b

tan (4A)

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Using the fact that A

A 2

A , express: 2

Ex 2

a 12

3 sin ( A) in terms of sin

sin (2 A ) 1 cos (2 A )

A A and cos 2 2

Simplify: a

A 2

1 + cos ( A) in terms of cos

2 sec2 ( A ) sec2 ( A )

Express

cosec2 ( A ) 2 cosec2 ( A )

sin (6 ) in terms of sin (θ ). (Hint: sin (3x) = 3 sin (x) – 4 sin3 (x)) cos ( 3 )

Ex 3

Show that

sin ( A ) cos ( A ) cos ( A ) sin ( A )

tan (2 A ) sec (2 A ).

Ex 4

Given that (2 cos ( A) + 1) (2 cos ( A) – 1)(2 cos (2A) – 1) = 2x + 1, express x as a single trigonometric identity.

If tan ( )

1 and sin ( ) 7

1 10

, where α and β are acute angles, determine the value of

tan ( 2 ). Use the identity tan ( A B)

tan ( A ) tan ( B) . 1 tan ( A ) tan ( B)

Extend your thinking x.

Determine x given that 3 cosec ( 20 ° ) sec (20 ° )

Evaluate cos4

cos4

5 8

If 2 tan (θ ) = 3 tan (δ ), show that tan (

)

Given that cos ( )

Show that tan (3A) tan (2A) tan ( A) = tan (3A) – tan (2A) – tan ( A).

Show that

cos4

1 a 2

sec (8 A ) 1 sec ( 4 A ) 1

3 8

cos4

7 . 8

sin (2 ) . 5 cos (2 )

1 , express cos (3θ ) in terms of a3. a

tan (8 A ) . tan (2 A )

Chapter 7 review Online review

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mathspace.co

“He who influences the thought of his times, influences all the times that follow. He has made his impress on eternity.” Hypatia of Alexandria

Big ideas • Vectors are geometric objects defined by magnitude and direction, which can be represented algebraically using components. A consistent set of rules for vector addition, subtraction, and scalar multiplication provides a powerful framework for analysing geometry. The scalar (dot) product extends this framework, offering a tool to determine the relationship between vectors, such as the angle between vectors or the condition of perpendicularity. • Vectors are the fundamental language for modelling physical quantities that involve direction, such as displacement, velocity, and force. Through vector addition and scalar products, complex ideas like relative motion and static equilibrium are transformed from physical principles into solvable algebraic equations.

8 Vectors Chapter outline 8.01 8.02 8.03 8.04 8.05 8.06 8.07 8.08 8.09 8.10 8.11

Introduction to vectors Magnitude and direction of a vector Scalar multiples of vectors Vector addition and subtraction Unit vectors and component form Geometric applications of vectors Polar form Scalar (dot) product Scalar and vector projections Displacement and velocity Force equilibrium and work done Investigation: Vector proofs Chapter 8 review

700 709 725 736 752 769 782 795 809 818 827

Pilots use vectors to navigate through the skies, ensuring they reach their destinations safely and efficiently.

8.01 Introduction to vectors After this lesson, you will be able to… • distinguish between scalar and vector quantities • use correct vector notation and representations • represent position vectors using different notations

Introduction to vectors In mathematics, quantities can be classified as either scalar or vector quantities: Scalar quantity

Vector quantity

A quantity that is fully described by its magnitude (or numerical value) alone. It has no associated direction.

A quantity that is fully described by both its magnitude (numerical value) and its direction.

Example: Speed (60 km/h), distance (5 km), mass (10 kg), area 100 m2 and temperature (25 °C).

Example: Velocity (60 km/h North), displacement (10 m East), force/weight (50 N downwards), acceleration (9.8 m/s2 downwards), and momentum (20 kg · m/s forward).

A boat’s speed of 60 km/h is a scalar quantity because it only contains magnitude, while a boat’s velocity of 60 km/h at a bearing of 020° is a vector quantity because it also contains direction. A A vector is a directed line segment. Consider two distinct points on a plane, A and B: B A

When drawing a line segment from A to B and adding an arrow at the end to show direction, it becomes a directed line segment. B

The point B (where the arrow is located) is called the terminal point. Sometimes the arrow is at the end of the line segment, or it can be in the middle:

700

The point A (where the vector begins) is called the initial point.

The length of the directed line segment represents the magnitude of the vector (its size or length), and the arrow indicates its direction.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 Classify these quantities as scalar or vector: a A force of 8 N acting horizontally right

Create a strategy Determine whether each quantity includes both a magnitude (size) and a direction.

Apply the idea Both magnitude (8 N) and direction (horizontally right) are specified, so it is a vector quantity.

b A displacement of 4 m from point A towards point B

Apply the idea Both magnitude (4 m) and direction along a specific path (the line joining A and B) are specified, so it is a vector quantity.

c A mass of 1 kg

Apply the idea A mass of 1 kg only specifies magnitude without any direction, making it a scalar quantity.

d A time of 1 second

Apply the idea A time interval of 1 second only specifies magnitude without any direction, making it a scalar quantity.

Idea summary Scalar quantities only involve magnitude, while vector quantities involve both magnitude and direction. A vector can be represented as a directed line segment, with an initial point and a terminal point.

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Vector notation A vector is represented visually as a directed line segment, with a tail (starting point) and an arrowhead (terminal point) to indicate direction. The vector from point A to point B is shown. B End

This vector can be labelled as AB or a. The length of the line indicates the magnitude and the arrow points towards the direction. If a vector n has length of 1 unit, then it is called the unit vector.

A Start

Two-dimensional vectors are drawn on a Cartesian plane and can be written as an ordered pair x (x, y) or a column vector . The x-coordinate or x-component describes horizontal movement y (negative means left and positive means right), and y-coordinate or y-component describes vertical movement (negative means down and positive means up). A position vector is a vector with its tail at the origin, (0, 0). Note that not all vectors start at the origin.

Interactive exploration Discover this concept in action online

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4 3 is shown on the plane 4 since the arrow is 3 units to the right and 4 units up from the tail of the vector at (0, 0).

The position vector a a

2 1 –2 –1

702

x 1

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5 4

3 is shown since the arrow is 3 3 units to the right and 3 units down from the tail of the vector. The vector d

3 2 1 –1 –1

x 1

Notice that the tail (start) is not at the origin. The size and direction are what matter for vectors.

A zero vector 0 has zero magnitude and points in any direction, therefore it is a point on the plane.

Example 2 Draw a vector to represent a car going 50 km/h southwest.

Create a strategy Use the compass rose to draw a vector pointing southwest, then label it with its magnitude.

Apply the idea

A compass rose can be used to determine what southwest (SW) looks like, then the magnitude of 50 km/h can be labelled on the vector. 50 km/h

Reflect and check Direction can be given in a variety of ways including with angles, true bearings, cardinal directions (north, east, south, west), ordered pair, or column vector.

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Example 3 For the given vector a on the plane: 4 3 2 1 −4 −3 −2 −1 a −1 −2 −3 −4

x 1

2 3 4

a Determine the column vector represented on the plane.

Create a strategy To identify the vector as a column vector, determine its horizontal (x) and vertical ( y) components from the starting point to the ending point as illustrated on the Cartesian plane.

Apply the idea 4 3 2 1 −4 −3 −2 −1 −1 a −2 −3 −4

x 1

2 3 4

The vector starts at (–2, –3) and ends at the point (1, 1). The vector is defined by the horizontal displacement (right or left) and vertical displacement (up or down) from the tail to the tip. Counting horizontally, the end is 3 units to the right from the start. Counting vertically, the end is 4 units upwards from the start.

The column vector can be written by placing the x- and y-displacement into a column vector 3 format: a 4

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b Draw a vector with the same magnitude and direction as a, but starting at the origin.

Create a strategy a is 3 units to the right and 4 units up, so the new vector will be the same but starting from the origin.

Apply the idea 4 3 2 1

−4 −3 −2 −1

−1 −2 −3 −4

2 3 4

Idea summary Two-dimensional vectors are represented visually as a directed line segment on the Cartesian plane, with a tail (starting point) and arrowhead (terminal point). The length of the line indicates the magnitude and the arrow points towards the direction. y 5 4 3 2 1 −1

−1

x 1

x or as an y ordered pair (x, y). In both notations, the x-component represents the horizontal displacement of the vector and the y-component represents the vertical displacement of the vector. Vectors can be denoted as AB or a, and are often written in column vector notation

A unit vector has a length of 1 unit. A position vector is a vector with its tail at the origin, (0, 0). A zero vector has zero magnitude.

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8.01 Practice questions What do you remember? 1

2 Ex 1

Determine whether each statement is true or false: a

Vectors only have magnitude, not direction.

Two vectors are considered equal if the vectors have the same magnitude and direction, regardless of their initial points.

The magnitude of a vector can never be 0.

A vector’s direction is described by its magnitude only.

Classify these quantities as scalar or vector: a

The temperature inside the room is 22 °C.

A time of 12 seconds to run 100 m

The wind blowing towards the north at 15 km/h

The amount of water in a glass is 250 ml.

Identify the vector quantities from this list: • Speed • Displacement • Length • Volume • Acceleration • Force

Practice 4

Draw a vector to represent:

Ex 2

706

4 metres in the direction of east

2.5 Newtons in the direction of northwest

10 m/s in the direction of 170° T

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

• Area • Temperature • Mass • Power • Momentum • Pressure

5 Ex 3a

Determine the column vector represented on the plane: a

y 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4

b 3 2 1

x 2 3 4

2 3

−2 −3

y 4

x 1

−6 −5 −4 −3 −2 −1 −1

y 2

−2 −1 −1

2 3 4 5 6 7

−2

1 x

−3 −5 −4 −3 −2 −1

−4 −5

6 Ex 3b

Sketch these position vectors on a plane using the origin as the starting point: a

(2, – 8)

(3, 5)

Sketch these vectors on a plane: 4 a from the coordinate (1, 9) 5 c

1 −1

6 from the coordinate (1, 3) 9

(– 4, – 2)

(7, 2)

3 from the coordinate (4, – 1) 2

9 from the coordinate (– 8, 5) 2

Consider the vector AB from point A = (1, 3) to point B = (5, 6): a

Express AB as a column vector.

Sketch AB on a Cartesian plane.

Sketch the position vector AB

Onto the same plane, sketch the vector BC of AB.

Express AB as an ordered pair.

4 . 6 7 which starts from the terminal point 2

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Let a be a real number and consider the vector v a

a>0

a<0

a . Draw a possible example of v when: 1 c a=0

Extend your thinking 11

A particle moves from the origin and travels 6 units southwest. It then changes direction and travels north for 9 units. Use a diagram to show the two displacements as vectors.

The turning point of the graph of the parabola y = x2 + 2x – 3 can be written as (p, q). 2 n and b , where m = p and n = q. m 5 Determine the components of a and b and sketch both vectors on a Cartesian plane. Consider the two position vectors a

A drone starts at a launch point L and its initial flight path is represented by the vector 5 . This takes it to point M. From point M, the drone then flies along a second path, f1 3 represented by the vector f2 which brings it back to the exact launch point L. If the launch point L was at coordinates (2, 7), sketch the path of the drone (vectors f1 and f2 on a Cartesian plane.

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8.02 Magnitude and direction of a vector After this lesson, you will be able to… • define and calculate vector magnitude • determine the direction of vectors using the tangent ratio • apply vector equality to solve problems

Components and magnitude of vectors A vector can be broken down into two perpendicular parts: a horizontal component (the change in horizontal position) and a vertical component (the change in vertical position).

Interactive exploration mathspace.co

Discover this concept in action online

For a vector a AB that starts at an initial point A(x1, y1) and ends at a terminal point B(x2, y2), the components are calculated by finding the difference in the coordinates: • Horizontal component: x = x2 – x1 • Vertical component: y = y2 – y1 This means the vector can be written in component form as a

x2 y2

x1 . y1

y a

y units

x units

As shown in the diagram, the vector and its components form a right-angled triangle. The length or magnitude of the vector, denoted |a|, can be found using Pythagoras’ theorem: _a_

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Example 1 Let G and H be the points G(7, 3) and H(8, –2): a Determine the vector HG in component form: HG

Create a strategy The two points can be plotted and the horizontal and vertical displacement will be the x- and y-components respectively. Subtract the x-coordinates and y-coordinates of the two points. The order matters, so the start point is subtracted from the terminal point.

Apply the idea 4 3 2 1 0 −1 −2 −3 −4

y It is important to subtract the initial coordinate from the final coordinate and not the other way around, in order to get the correct vector.

x 1

2 3 4 5 6 7 8 H

HG is the vector from point H to point G: • Start point: H(8, –2) • Terminal point: G(7, 3) To get from H to G, go one unit left and 5 units upwards.

x-component = x2 – x1 =7–8

Substitute x1 = 8 and x2 = 7

= –1

Evaluate

y-component = y2 – y1

Therefore, HG

Write the equation

= 3 – (–2)

Substitute y1 = –2 and y2 = 3

Evaluate

( 1, 5).

Reflect and check In this case, plotting the points and counting squares was possible, but if the points had non-integer coordinates, then this approach would not work and subtracting the coordinates would be necessary.

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b Find the exact length of the vector HG.

Create a strategy Use the Pythagoras’ theorem: _a_

Apply the idea Use the x- and y-components obtained from part (a): _ HG _ = x2 y2

Write the formula

= ( 1)2 52

Substitute x = –1 and y = 5

= 1 25

Evaluate the squares

= 26

Evaluate the addition

Reflect and check Since the question asked for an exact value, it should be given as a surd, not a rounded decimal answer.

Example 2 Determine the magnitude of the given vector: y

4 3 2 B

−2 −1

x 1

−1 −2

Create a strategy First, calculate the horizontal (x) and vertical ( y) components of the vector by subtracting the coordinates of the initial point from the terminal point. Then, use the magnitude formula with the calculated components: _a_

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Apply the idea Let the vector be a. From the graph, identify the initial and terminal points: • Initial point A(x1, y1) = (3, 4) • Terminal point B(x2, y2) = (– 1, 1) Calculate the horizontal component (x): x = x2 – x1

Write the formula for the horizontal component

= –1 – 3

Substitute the x-coordinates

= –4

Evaluate

Calculate the vertical component ( y): y = y2 – y1

Write the formula for the vertical component

=1–4

Substitute the y-coordinates

= –3

Evaluate

4 . 3 Now, use the magnitude formula with x = – 4 and y = – 3: The vector in component form is a |a| = x2 y2

Write the magnitude formula

2 2 = ( 4 ) ( 3)

Substitute the component values

= 16 9

Evaluate the squares

= 25

Evaluate the addition

Evaluate the square root

Therefore, the magnitude of the vector is 5 units.

Reflect and check The calculated components x = – 4 (4 units left) and y = – 3 (3 units down) match the direction of the vector in the graph shown, confirming the result is reasonable.

Idea summary A vector can be described by its perpendicular horizontal component (the change in x) and vertical component (the change in y). For a vector from point A(x1, y1) to B(x2, y2), the components are: • Horizontal component: x = x2 – x1 • Vertical component: y = y2 – y1

y a

The magnitude (or length) of a vector with components x and y is found using Pythagoras’ theorem:

2 2 a x y |a| is the magnitude of the vector x is the horizontal component y is the vertical component

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x units 0

y units x

Direction of a vector Recall for a right-angled triangle: The trigonometric ratios are: Opposite

sin ( )

Hypotenuse

cos ( )

tan ( )

Opposite Hypotenuse Adjacent Hypotenuse Opposite Adjacent

Adjacent S sin positive

A all positive cos (θ ), sin (θ )

Also recall the quadrants for solving trigonometric equations: • In Quadrant 1, all ratios are positive. • In Quadrant 2, only sin (θ ) is positive. • In Quadrant 3, only tan (θ ) is positive. • In Quadrant 4, only cos (θ ) is positive.

x 1

–1

T tan positive –1

C cos positive

The direction of a vector is the angle it makes with the positive x-axis. The anticlockwise direction is positive, while clockwise direction is negative. y 4 3 Consider the vector from (0, 0) to (3, 3).

2 1 –1

θ 1

x x

This vector has horizontal length of 3 units and vertical length of 3 units.

–1

The angle can be calculated using trigonometric ratios: y x 3 tan (θ ) = 3 θ = tan–1 (1) tan (θ ) =

= 45°

Write the formula for the angle using the tangent ratio Substitute the component values x = 3 and y = 3 Simplify the fraction and take the inverse tangent Evaluate

Therefore, the direction of the vector is 45°.

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For a position vector a

x , the direction a makes with the positive x-axis is: y y tan 1 x

y returns a reference angle, typically between –90° and 90°. To find the true x direction in all four quadrants, use the signs of the x- and y-components and the quadrant rules (ASTC) to determine the correct angle.

Note that tan 1

Example 3 Consider the vector a

2 3 , 2 :

a Determine the magnitude of the vector.

Create a strategy Identify the components x and y and use the magnitude formula: _a_

Apply the idea Given the vector a

2 3 , 2 , the components are x 2 3 and y = 2. |a| = x2 y2

Write the magnitude formula

= (2 3 )2 22

Substitute x 2 3 and y = 2

= 12 4

Evaluate the squares

= 16 =4

Evaluate the addition Evaluate the square root

Therefore, |a| = 4 units.

b Determine the direction of the vector in degrees.

Create a strategy Use the direction formula:

y x

tan–1

Apply the idea θ = tan–1 = tan –1

= tan–1 = 30°

714

y x

Write the direction formula 2

2 3 1 3

Substitute x 2 3 and y = 2 Simplify the fraction Evaluate

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reflect and check

The vector has positive x and y components; therefore, it is in the first quadrant. An angle of 30° is in the first quadrant; therefore, the result is reasonable.

To verify the magnitude and direction, the vector can be graphed using technology:

a =4

2 3

2 30°

–1

c Compare the magnitude and direction of a to b

–1

2 3 , 2 .

Create a strategy Calculate the magnitude and direction for vector b using the same formulas and compare the results to vector a.

Apply the idea For vector b

2 3 , 2 , the components are x 2 3 and y = –2.

Calculating the magnitude: |b| = x2 y2

Write the magnitude formula

= (2 3 )2 ( 2)2

Substitute x 2 3 and y = –2

= 12 4

Evaluate the squares

= 16

Evaluate the addition

Evaluate the square root

The magnitude is 4, which is the same as a. Calculating the direction:

θ = tan–1 –1

= tan

–1

= tan

= –30°

y x

Write the direction formula 2

2 3 1 3

Substitute x 2 3 and y = –2 Simplify the fraction Evaluate

The vectors have the same magnitude but opposite directions, with b being a reflection of a across the x-axis.

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Reflect and check 4 3

Plotting both vectors confirms that both have the same length and are reflections of each other across the x-axis.

2 1 –4 –3 –2 –1–1

30°

1 2 3 30°

–2 –3 –4

Example 4 Consider vector v with magnitude 12 and direction angle –60°: a Determine the horizontal component of v, rounded to two decimal places.

Create a strategy Consider the relationship between trigonometric ratios and the components of a vector.

Apply the idea 0

x –60°

The vector forms a right triangle and with the axes: • The hypotenuse is 12. • The adjacent side is x. • The opposite side is y. Given the hypotenuse and looking for the adjacent side, the cosine ratio can be used to determine the horizontal component.

y To determine x: cos (–60°) =

Adjacent Hypotenuse

Write the equation

cos (–60°) =

x 12

Substitute the known values

12 × cos (–60°) = x 1 12 = x 2 6=x

Multiply both sides by 12 Substitute the exact value of the cosine ratio Evaluate

The horizontal component is 6 units.

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b Determine the vertical component of v, rounded to two decimal places.

Create a strategy Consider the relationship between trigonometric ratios and the components of a vector.

Apply the idea 0

x –60°

The vector forms a right triangle and with the axes: • The hypotenuse is 12. • The adjacent side is x. • The opposite side is y. Given the hypotenuse and looking for the opposite side, the sine ratio can be used to determine the vertical component.

y To determine y: sin (–60°) =

Opposite Hypotenuse

Write the equation

sin (–60°) =

y 12

Substitute the known values

12 × sin (–60°) = y 12

Multiply both sides by 12

3 =y 2

Substitute the exact value of the sine ratio

6 3 = y The vertical component is 6 3

Evaluate 10.39 units.

Reflect and check A direction angle of − 60° lies in the fourth quadrant; therefore, the positive horizontal component and negative vertical component are consistent with the direction of the vector. This means that 6 v . 6 3

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Idea summary For a vector a = (x, y) with horizontal component x and vertical component y, where θ is the angle the vector makes with the horizontal:

x units a

tan ( )

y units x

y x , x

Therefore, the direction of a vector is y q = tan–1 . x Trigonometric ratios can be used to solve for missing components, magnitude or angles.

0 A

Vector equality Two vectors are equal if and only if the vectors have the same magnitude and direction. For example, all these vectors are equal because all have the same length and point in the same direction: y 6 5 4 3 2 1 –9 –8 –7

–6 –5 –4 –3

0 –1

–2 –1

–2 –3

718

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x 1

However, these vectors are not equal because the vectors do not have the same length and point in different directions: y 4 3 2 1 –8 –7 –6 –5 –4 –3

–2 –1

x 0

–1

–2 –3 –4

Example 5 Consider the graph shown:

4 3

2 1 d –4 –3 –2 –1–1 a –2 –3 e –4

x 1

2 3 4

Determine whether the statements are true or false by calculating the component form of each vector: a a=b

Create a strategy Two vectors are equal only if every matching component is identical. For each statement, calculate the horizontal (x) and vertical ( y) components of both vectors using the formula (x2 – x1, y2 – y1). Then, compare the components to determine if the vectors are equal.

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Apply the idea Calculate the components for vector a (from (– 2, 1) to (– 4, – 3)): a= =

4 ( 2) 3 1 2 4

Substitute the coordinates Evaluate

Calculate the components for vector b (from (3, 0) to (1, – 4)): b= =

1 3 4 0 2 4

Substitute the coordinates Evaluate

Since both the horizontal and vertical components are identical, the statement is true.

Reflect and check 2 . Since the components are identical, the vectors 4 must have both the same magnitude and the same direction. The graph confirms this, as both vectors are parallel and share the exact same length. The component forms for a and b are both

b c=f

Apply the idea Calculate the components for vector c (from (– 3, 5) to (– 1, 1)): c= =

1 ( 3) 1 5

Substitute the coordinates

2 4

Evaluate

Calculate the components for vector f (from (2, 4) to (4, 1)): 4 2 1 4 2 = 3

Substitute the coordinates Evaluate

The horizontal components are the same (2), but the vertical components are different (– 4 ≠ – 3). Therefore, the statement is false.

Reflect and check Because the components differ, the vectors are not equal. This indicates that either their magnitudes, their directions or both are different. In this case, both are different: _c_

_f_

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Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

( 3)

22 ( 4 )2

20, while

13. A visual check of the graph confirms that vector c is longer than vector f.

c a=d

Apply the idea 2 . 4 Calculate the components for vector d (from (– 1, – 1) to (1, 3)): From previous calculation, a

1 3

2 4

( 1) ( 1)

Substitute the coordinates Evaluate

The components are not identical (– 2 ≠ 2 and – 4 ≠ 4). Therefore, the statement is false.

Reflect and check This is a special case where the components of d are the negative of the components of a.

Therefore, the vectors have the same magnitude _a_ _d_ 20 but point in opposite directions. For vectors to be equal, both magnitude and direction must be identical. A visual check of the graph confirms this relationship.

Idea summary Two vectors are equal if and only if the vectors have the same magnitude and direction.

8.02 Practice questions What do you remember? 1

Complete these formulas for a vector v a

Horizontal component: x = ⬚ – ⬚

Vertical component: y = ⬚ – ⬚

Direction: tan ( )

x that starts at point (x1, y1) and ends at (x2, y2): y

Magnitude: _v_

A vector AB starts at point A(1, 2) and ends at point B(4, 6): a

Determine the horizontal component of the vector.

Determine the vertical component of the vector.

Calculate the magnitude of the vector, _ AB_.

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Determine whether each statement about vectors is true or false: a b c

Two vectors are equal if the vectors have the same magnitude. 5 The vector is equal to the vector from (2, 3) to (7, 2). 1 5 The magnitude of the vector is 13. 12

Practice Ex 1

5 Ex 2

Consider the points P(–3, 5) and Q(4, 1): a

Determine the vector PQ in component form.

Determine the vector QP in component form.

Calculate the exact magnitude of the vector PQ.

Consider this diagram:

y 7

Determine the magnitude of vector a AB.

6 5

4 3 2 1

x 0

For these vectors, determine: i

The exact magnitude of the vector

The angle, in degrees, that the vector makes with the positive x-axis

(6, 8)

(–4, 12)

6, 2 7

(–24, –10)

4 6 , 4 3

(2, –2)

2 3 , 6

5 3 , 5

Determine the exact magnitude of each vector: a

AB (12, 5)

(7, –7)

Show that the vector v

Show that the vector m

For each vector:

722

BC EF

(3, 4) 1, 3 3

(3, –2)

2 2, 2 2

4 2, 2

3 4 is a unit vector. , 5 5 2 3 , is a unit vector. 13 13

Draw the vector with the start point at the origin.

Calculate the magnitude.

(4, 3)

(–1, 4)

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Determine whether each of these statements is true or false: a

The vector (1, –1) is a unit vector.

The vector (4, 4) has a magnitude of 16. x c The magnitude of the vector v is the same as its negative vector, v y 5 Consider the vector v : 5 3 a Determine the exact magnitude of the vector v. b

12 Ex 3

Ex 4

Determine the direction of the vector v in degrees.

How does the magnitude and direction of v compare to the vector w

5 5 3

Consider vector v with magnitude 64 and direction angle 57°: a b

x . y

Determine the horizontal component of v, rounded to two decimal places. Determine the vertical component of v, rounded to two decimal places.

For the given vectors, use their magnitude and angle to: i

Calculate the horizontal component.

Calculate the vertical component. y

24 45°

– 66.4°

17.5

Express these vectors in the form (a, b), leaving a and b as exact values: a

y 42 135°

x 0

– 120°

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Identify equivalent vectors drawn on this graph:

y b

Ex 5

c d f

x2 8 x 16

Assume that

Determine x for each vector: a b c

. Determine all possible values of x and y.

a = (3, x) which has a magnitude of 10. 1 b x, , where b is a unit vector. 6 c = (4, x) , where u = (2, –5) has the same magnitude as c.

Determine two pairs of points A and B such that AB

Determine four vectors that can have a magnitude of 5.

2 . 1

Extend your thinking 21

A tennis player serves the ball with a total velocity of 168 km/h. If the vertical component of the velocity is 98 km/h downwards: a How fast is the tennis ball travelling horizontally? b At what angle below the horizontal does the tennis player serve the ball?

onsider a vector in the third quadrant which begins from the origin of a Cartesian plane. C Its magnitude is 13 2 and has equal horizontal and vertical components. Determine the vector in exact form (a, b). Consider another vector on the first quadrant which begins from the origin with a magnitude 38 3. If the ratio between the vertical to horizontal component is 3, determine the vector in exact form (a, b).

I dentify and plot a relation to represent the terminal point of all possible vectors with the origin as an initial point and a magnitude of 3.

For this relation, determine y when x = 1. Express this as a vector and plot the graph.

A vector F in the Cartesian plane represents force in Newtons. The vector F has an x-component Fx and Fy. The magnitude of F is not known, but the components are constrained by the equation Fx2 + Fy2 = 36: a Determine the magnitude of the vector F using the constraint. b If the angle between F and the positive y-axis is 30°, determine the possible vectors F in the form (Fx, Fy) where Fx and Fy are exact values.

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Did you know?

Scalar multiples of vectors help explain real-world forces in action! In windsurfing, the wind’s push on the sail can be shown as a vector with direction and magnitude. When the wind becomes stronger, the force increases but still points the same way, making it a scalar multiple of the original vector. This is why the rider can speed up or slow down while moving in the same direction!

8.03 Scalar multiples of vectors After this lesson, you will be able to… • represent and interpret scaled vectors geometrically • calculate scalar multiples of vectors algebraically • apply scalar multiplication to solve problems • understand the effect of scalar multiples on vector magnitude and direction

Scalar multiples of vectors on the plane Interactive exploration Discover this concept in action online

mathspace.co

A scalar is a real number (e.g., 5, – 0.5, π) that has magnitude but no direction. Multiplying a vector by a scalar produces a new vector that is parallel to the original. This operation scales the vector’s length and can reverse its direction. For a vector v and a scalar m, the vector mv is affected as follows: • If ∣m∣ > 1, the vector is stretched (its magnitude increases). • If 0 < ∣m∣ < 1, the vector is compressed (its magnitude decreases). • If m > 0, the direction remains the same. v • If m < 0, the direction is reversed.

–v 2v

8.03 Scalar multiples of vectors mathspace.co

1v 2

725

Example 1 The vector v is shown on the plane as a position vector: y 4 3 2 1 v −4 −3 −2 −1

x 1

−1 −2 −3 −4

2 3 4

a Draw the vector 2v on the same plane.

Create a strategy To draw 2v, draw a new vector that starts at the same point, points in the same direction, but is twice as long as v.

Apply the idea 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4

v 1

2v x 2 3 4

The vector v represents a displacement of ‘2 units right and 1 unit up’. Therefore, 2v will represent twice this displacement: ‘4 units right and 2 units up’.

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Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b Draw the vector −1.5v on the same plane.

Create a strategy The scalar is negative; therefore, the new vector will point in the opposite direction. Since the absolute value is 1.5, it will be 1.5 times as long as v.

Apply the idea y 4 3 2 1 v −4 −3 −2 −1 −1 –1.5v −2 −3 −4

x 1

2 3 4

The opposite of ‘2 units right and 1 unit up’ is ‘2 units left and 1 unit down’. Scaling this by 1.5 gives a final displacement of ‘3 units left and 1.5 units down’.

Reflect and check Geometrically, 2v can be constructed by placing two copies of v head-to-tail. The resultant vector from the start of the first to the end of the second is 2v.

Idea summary Multiplying a vector v by a scalar m produces a new vector mv that is parallel to the original. • The magnitude of the new vector is ∣m∣ times the magnitude of the original vector. • If m > 0, the direction is unchanged. • If m < 0, the direction is reversed.

8.03 Scalar multiples of vectors mathspace.co

727

Algebraic properties of scalar multiplication Algebraically, the process of scalar multiplication is straightforward: each component of the vector is multiplied by the scalar. For a vector v

x and a scalar k, the scalar multiplication is defined as: y kx x kv k y ky

This rule is the foundation for several key algebraic properties that ensure vector operations are consistent and predictable. Multiplying a vector by two scalars in succession is the same as multiplying the vector by the product of those scalars. For scalars k and c: k(cv) = (kc)v • Associative property: The associative property indicates that when multiplying a vector by two scalars, the order in which the multiplication occurs does not matter. x • If k and c are scalars and a is a vector, then: y cx x k ( c a) k ( k c) a kc cy y • This means that multiplying a vector by a product of scalars is the same as multiplying the vector by one scalar and then the result by the other. • Identity property: The identity property states that multiplying a vector by the scalar 1 leaves the vector unchanged. x • For any vector a : 1a a y • This property reinforces the idea that it is the multiplicative identity in scalar multiplication, maintaining the vector’s original magnitude and direction.

Example 2 Given the vector v a 3v

4 , calculate these expressions in column vector form: 6

Create a strategy To calculate the scalar multiple of a vector, multiply each component of the vector by the scalar.

Apply the idea 4 6 3 ( 4)

3v = 3 = =

728

3 6 12 18

Substitute the given vector v Multiply each component by the scalar 3 Evaluate

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b –0.5v

Apply the idea – 0.5v = – 0.5

4 6

Substitute the given vector v

0.5 ( 4 ) 0 .5 6

Multiply each component by the scalar – 0.5

2 3

Evaluate

Example 3 Two vectors are defined as a component y if b = ka.

2 and b 5

6 . Find the value of the scalar k and the y

Create a strategy Set up the vector equation b = ka. Equate the corresponding horizontal components to solve for the scalar k. Then, use this value of k in the equation for the vertical components to solve for y.

Apply the idea Write the vector equation using the component forms: 6 y

2 5

2k 5k

Equate the horizontal components to find k: –6 = 2k

Write the equation

–3 = k

Divide both sides by 2

Now, equate the vertical components and substitute k = –3 to find y: y = –5k

Write the equation

= –5 (–3)

Substitute k = –3

= 15

Evaluate

Therefore, the scalar is k = –3 and the component is y = 15.

Reflect and check The relationship b = –3a means that vector b should be parallel to a but point in the opposite direction. Since a has a positive x-component and negative y-component (points down-right), b should have a negative x-component and positive y-component (points up-left). The result 6 b matches this; therefore, the answer is reasonable. 15

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Idea summary Scalar multiplication is performed algebraically by multiplying each component of the vector by the scalar: kv

x y

kx ky

This operation follows key algebraic properties, such as the associative property k (cv) = (kc)v and the identity property 1v = v. These rules allow vector expressions involving scalars to be consistently simplified and rearranged.

Magnitudes of scalar multiples Interactive exploration mathspace.co

Discover this concept in action online

When a vector v is multiplied by a scalar k, the magnitude of the resulting vector kv is ∣k∣∣v∣, which is the absolute value of the scalar multiplied by the magnitude of the original vector. This relationship is defined by the formula: ∣kv∣ = ∣k∣ ∣v∣ The absolute value ∣k∣ is used because magnitude cannot be negative. For example, multiplying by –2 doubles the magnitude, it does not make it negative. This formula can be proven by starting with the algebraic definition of scalar multiplication. For a x vector v : y |kv| =

kx ky

= ( kx )2 ( ky)2

Apply the formula for the magnitude of a vector

= k2 x2

k2 y2

Apply the power of a product rule to the terms

= k2 ( x2

y2 )

Factorise the common term k2

= k2 x2

Separate the square roots using the property ab a b

= |k||v|

730

Apply the definition of scalar multiplication to the vector v

Recognise that k2 = |k| and x

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

y2 = |v|

Example 4 A vector a has a magnitude of 5 units. A second vector is defined as b = –3a. Find the magnitude of b.

Create a strategy Use the formula for the magnitude of a scaled vector, ∣ka∣ = ∣k∣ ∣a∣, with the scalar k = –3 and the given magnitude |a| = 5.

Apply the idea ∣b∣ = ∣–3a∣

Write the equation

= ∣–3∣ × |a|

Apply the magnitude scaling formula

=3×5

Substitute the known values

= 15

Evaluate

The magnitude of vector b is 15 units.

Reflect and check Even though the scalar was negative, the resulting magnitude is positive, as expected. The negative scalar reverses the direction of the vector but scales its length by a positive factor of 3.

Example 5 Given the vector v

3 , find the possible values of the scalar k such that the magnitude of kv is 15. 4

Create a strategy First, calculate the magnitude of the original vector, ∣v∣. Then, set up the equation ∣k∣ ∣v∣ = 15 and solve for ∣k∣. Finally, determine the possible values for k.

Apply the idea Step 1: Find the magnitude of v. v

32 ( 4 )2

9 16

Step 2: Set up and solve the equation for ∣k∣. ∣kv∣ = 15

Write the equation

∣k∣∣v∣ = 15

Apply the magnitude scaling formula

∣k∣ × 5 = 15

Substitute the calculated magnitude of v

∣k∣ = 3

Divide both sides by 5

Step 3: Determine the possible values for k. If ∣k∣ = 3, then k = 3 or k = –3.

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Reflect and check 9 , which has a 12 9 magnitude of 15 and the same direction as v. k = –3 gives the vector 3v , which also has a 12 magnitude of 15 but points in the opposite direction. Both solutions are valid. There are two possible values for the scalar. k = 3 gives the vector 3v

Idea summary The magnitude of a vector multiplied by a scalar k is the absolute value of the scalar multiplied by the original vector’s magnitude.

∣kv∣ = ∣k∣∣v∣

∣kv∣∣ ∣k∣∣ ∣v∣∣

is the magnitude of the scaled vector is the absolute value of the scalar is the magnitude of the original vector

8.03 Practice questions What do you remember? 1

a 2

2 . Calculate these vectors: 3 4v b –2v

Let v

0.5v

–v

Vector a is shown: 4 3 2 1 −4 −3 −2 −1 b

−1 −2 −3 −4

y c a 1

x d

2 3 4

Which of the other vectors represents –2a? A

732

Vector b

Vector c

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Vector d

Complete these statements about scalar multiplication: a b c

If ∣v∣ = 10, then ∣3v∣ = ⬚.

If∣u∣ = 8, then ∣–4u∣ = ⬚.

Multiplying a vector by a negative scalar reverses its ⬚.

A non-zero vector v is multiplied by a scalar k to create a new vector, kv.

Match each scalar value of k with the correct description of its effect on the vector v: a

k=3

The vector is stretched and its direction is reversed.

k = –2

The vector is compressed and its direction is reversed.

k = 0.5 1 k= 4

iii

The vector is stretched and its direction is the same.

The vector is compressed and its direction is the same.

Practice 5

2 : 8 Draw w on the plane, starting from the origin.

Calculate w and draw on the same Cartesian plane as w.

Let w

Ex 1

1 : 2 Draw this vector on the plane, starting from the origin. 1 Draw the vector 5 on the same graph. 2

Consider the plane vector v a

1 2

For the vector a shown, sketch the vectors: a

–4a

4 3 2 1

1 a 2

a −4 −3 −2 −1

−1 −2 −3 −4

Ex 2

2 : 1 Calculate 3v.

Calculate –5v.

Interpret these results geometrically for a vector on the plane.

x 1

2 3 4

Let v

If b is a unit vector, determine the length of 5b.

8.03 Scalar multiples of vectors mathspace.co

733

a b 11

x y

v for a > 0, determine x and y.

Calculate the scalar multiple.

Draw the original vector and the scalar multiple on the Cartesian plane.

iii

Describe the effect of the scalar on the magnitude and direction of the original vector. 3 8 5 1 1 b c d 2 1 5 1 2 6 3 1 5 , determine k: 12

For each scalar multiple of v a

If a

v, determine x and y.

For each product of a scalar number and a vector:

2 : 4 x If 2 y

Let v

10 24

5 12

2 .5 6

8 : 6 Determine ∣v∣.

For each scalar multiple kv on the plane, determine k and ∣kv∣:

y 8 6 4 2 v

0 0

For v

−8 −6 −4 −2 −2 −4 −6 −8 iii

ii v x

2 4 6 8

16 14 12 10 8 6 4 2

8 6 kv 4 2

x 2 4 6 8

−8 −6 −4 −2 −2 −4 −6 −8

−16 −14−12−10 −8 −6 −4 −2

734

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

8 6 4 2

−8 −6 −4 −2 −2 −4 −6 −8

x 2 4 6 8 kv

14 15

2 . Show graphically that 3 × (2 × v) = (3 × 2) × v. 1 1 Consider the vector v : 1 a Calculate the magnitude of v. Let v

b Ex 3

What scalar multiple of v would have ten times the magnitude?

3 k and b , determine the value of k for which b is a scalar 4 5 multiple of a. Interpret the geometric relationship between a and b for this value of k.

Given two vectors a

Consider the vectors u a

2 6 and v , and the scalars m = 3 and n = –2: 7 8 Determine the exact magnitudes of the vectors u and v.

Calculate the exact magnitude of the scaled vector mu.

Calculate the exact magnitude of the scaled vector nv.

Ex 4

A vector v has a magnitude of 25 units. A second vector is defined as w = –4v. Find the magnitude of w.

Ex 5

Given the vector u ku is 52.

12 , find the possible values of the scalar k such that the magnitude of 5

Extend your thinking a be a vector and k and l be scalars. Verify that k × (l × v) = (k × l) × v. b 256 2a . Express v in the form 2k . 32 1

Let v

Let h

A drone’s initial displacement from its base is represented by the vector d

0.4 . Show that the unit vector in the direction of h is twice as long as h. 0.3

100 metres. If 75 the drone flies to a new position that is 1.5 times the magnitude of its initial displacement but in the opposite direction, what is its new displacement vector from the base?

Prove that ∣ka∣ = ∣k∣ × ∣a∣.

Explain why ∣ka∣ ≠ ∣k∣ × ∣a∣.

8.03 Scalar multiples of vectors mathspace.co

735

8.04 Vector addition and subtraction After this lesson, you will be able to… • determine resultant vectors geometrically using the triangle or parallelogram rule • make the connection between algebraic vector addition and the triangle or parallelogram rule • perform vector addition and subtraction algebraically • solve vector problems using multiple operations • verify vector calculations geometrically

Geometric vector addition Adding vectors means finding a single vector, called the resultant vector, that represents the combined effect of the original vectors. This can be visualised as finding the net displacement after a series of journeys. Resultant vector The single vector that represents the sum of two or more individual vectors. It describes the overall or net effect of the vectors acting together. It can be found graphically using the tip-to-tail method or algebraically by adding the components of the vectors.

Interactive exploration Discover this concept in action online

mathspace.co

There are two main geometric methods for adding vectors. The most fundamental method for adding vectors is the triangle rule. To add vector v to vector u, vector v is translated so that its tail (start) is at the tip (head) of u. The resultant vector is drawn from the tail of u to the tip of v. v

736

u+v=w

This figure illustrates the triangle rule. The resultant vector w = u + v completes the triangle.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

When two vectors start from the same point, the parallelogram rule is a useful way to visualise the sum. The two vectors are drawn as adjacent sides of a parallelogram. The resultant vector is the diagonal of the parallelogram that starts from the common initial point.

The parallelogram rule shows that vector addition is commutative (u + v = v + u), as the diagonal represents the sum regardless of which vector is considered first.

u+v

Example 1 Consider the two vectors a and b, as graphed: y a

0 a Plot the result of a + b on the axes.

Create a strategy

Apply the idea

To add two or more vectors graphically, use the triangle rule by translating the tail of the second vector to the tip of the first. The resultant will then start where the first vector starts, and end where the last vector ends.

The resultant vector will be: y

x 0

8.04 Vector addition and subtraction mathspace.co

737

b Plot the result of b + a on the given axes: y

Apply the idea The resultant vector will be: y

x 0

c Are the two resulting vectors a + b and b + a equal?

Create a strategy Compare the graphs of parts (a) and (b).

Apply the idea Two vectors are equal if both their magnitudes and directions are equal. As shown in the graphs, a + b and b + a have the same length and direction. Therefore, the resulting vectors a + b and b + a are equal.

738

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 4 and v 0

Two vectors, u

1 , are shown starting from a common origin: 3 y 5 4 3 2 v

u 1

Using the parallelogram rule, find the resultant vector r = u + v.

Create a strategy The parallelogram rule for vector addition involves these steps: 1. The two vectors, u and v, are drawn from a common origin. 2. A parallelogram is completed by drawing a vector parallel to u from the tip of v, and a vector parallel to v from the tip of u. 3. The resultant vector, u + v, is the diagonal of the parallelogram that starts from the common origin.

Apply the idea The parallelogram is constructed using the two vectors. The resultant vector is the diagonal from the origin to the opposite corner. y 5 4 3 2 1

r=u+v u

x 3

The resultant vector r starts at the origin and ends at the point (5, 3). Therefore, r

5 . 3

8.04 Vector addition and subtraction mathspace.co

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Idea summary Vectors can be added geometrically to find the resultant vector. • The triangle rule (head-to-tail) involves placing vectors one after another. The resultant connects the very start to the very end. • The parallelogram rule involves drawing two vectors from a common origin. The resultant is the diagonal of the completed parallelogram.

Geometric vector subtraction Vector subtraction is defined as adding the negative of a vector. This allows the use of the same geometric rules learned for addition. For any vector v, the negative vector, denoted –v, is a vector that has the same magnitude as v but points in the exact opposite direction. y 1

v x

−3 −2 −1 −v

When a vector and its negative are added, the resultant is the zero vector: v + (–v) = 0. To subtract vector v from vector u, one calculates u – v. This is geometrically interpreted as adding the negative of v to u: u – v = u + (–v) The head-to-tail method can then be applied. First, draw vector u. Then, draw the negative vector –v starting from the tip of u. The resultant vector u – v is drawn from the tail of u to the tip of –v. An alternative way to visualise u – v is to draw u and v from a common origin. The vector that connects the tip of v to the tip of u is the resultant u – v.

740

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 3 The vectors a and b are shown: 4 3 2 a 1 −5 −4 −3 −2 −1 −1 −2 −3

1 2 3 4

−4 Draw the resultant vector a – b.

Create a strategy To find a – b, it is necessary to add –b to a. First, determine the negative vector –b. Then, apply the head-to-tail rule by translating –b to the tip of a.

Apply the idea Step 1: Draw the negative vector –b. It has the same length as b but points in the opposite direction. 4 3 2 1 −5 −4 −3 −2 −1 –b −1 −2 −3 −4

Step 2: Add –b to the tip of a. The resultant vector connects the tail of a to the tip of –b.

y –b b 1 2 3 4

a–b

4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4

x 1

2 3 4

8.04 Vector addition and subtraction mathspace.co

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Reflect and check An alternative method to verify the result is to apply the parallelogram rule to the addition a + (–b). To do this, both vectors a and –b are drawn from a common origin. A parallelogram is then constructed using these two vectors as adjacent sides. The resultant vector, a – b, is the main diagonal of this parallelogram. y

4 3 2

1 a–b −5 −4 −3 −2 −1 –b −1

x 1 2 3 4

−2 −3 −4 The main diagonal of the parallelogram formed by a and –b ends at the point (–5, 2). This 5 corresponds to the resultant vector , which is the same result obtained from the head-to-tail 2 method.

Idea summary Vector subtraction u – v is defined as adding the negative vector: u – v = u + (–v) he vector –v has the same magnitude as v but points in the opposite direction. T The subtraction can then be performed visually using the triangle rule and parallelogram rule. y 1

v x

−3 −2 −1 −v

742

1 1

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Algebraic vector addition and subtraction While geometric methods are useful for visualisation, the most efficient way to add and subtract vectors is algebraically. The head-to-tail method demonstrates that the net horizontal displacement of a resultant vector is the sum of the individual horizontal components, and the net vertical displacement is the sum of the individual vertical components. This leads to the rule for algebraic addition: to add two or more vectors, add the corresponding components. For vectors u

a and v b

c , the sum is: d u v

a b

c d

a c b d

Similarly, algebraic subtraction is performed by subtracting the corresponding components: u v

a b

c d

a c b d

Vector addition follows several important algebraic properties: • Commutative property: u + v = v + u • Associative property: (u + v) + w = u + (v + w) • Identity property: u + 0 = u, where 0 is the zero vector. Note that vector subtraction is not commutative: u – v ≠ v – u. Instead, u – v = –(v – u).

Example 4 Given the vectors a a a+b

2 and b 7

8 , calculate the resultant vectors: 2

Create a strategy

Apply the idea

To add the vectors, add the corresponding horizontal and vertical components.

2 7

a b

8 2

2 8 7 2

6 9

b a–b

Create a strategy To subtract the vectors, subtract the components of b from the corresponding components of a.

Apply the idea a b

2 7

8 2

2 8 7 2

10 5

8.04 Vector addition and subtraction mathspace.co

743

Example 5 If p

5 and q 1

2 , find the vector x such that 2p + x = q. 3

Create a strategy First, rearrange the vector equation to make x the subject. Then, perform the required scalar multiplication and vector subtraction to find the components of x.

Apply the idea Rearrange the equation to solve for x: x = q – 2p First calculate the scalar multiple 2p: 2p 2

5 1

10 2

Now, substitute the vectors back into the rearranged equation: 2 3

10 2

Finally, perform the subtraction: 2 10 3 ( 2)

12 5

Reflect and check To verify the answer, substitute x back into the original equation: 2

5 1

12 5

10 2

12 5

2 3

which is equal to q. The solution is correct.

Idea summary Vector addition and subtraction can be performed algebraically by adding or subtracting the corresponding components. a c For vectors u and v : b d • Addition: u v • Subtraction: u v

a c b d a c b d

Vector addition is commutative and associative. Vector subtraction is not commutative.

744

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

8.04 Practice questions What do you remember? 1

The vectors u and v are shown on the grid: y 3 2 1 −1

u 1

x 3

−1

Which of these diagrams correctly shows the resultant vector u + v using the triangle rule? A

y 4

2 2

1 −1

x 1

−1

2 1

x 1

−1 −1

−1

8.04 Vector addition and subtraction mathspace.co

745

Vector v is shown in the graph:

y 2 1 v −2

−1

x 2

−1 −2

Which vector correctly represents the negative vector, –v? 2 2 2 A B C 1 1 1 3

1 2

Determine whether each statement about vector addition and subtraction is true or false: a

To add two vectors algebraically, add the corresponding components.

Vector addition is commutative, meaning u + v = v + u.

Vector subtraction is also commutative, meaning u – v = v – u.

The negative vector, –v, has the same magnitude as v but points in the opposite direction.

When using the parallelogram rule to find u + v, the resultant vector is the shorter diagonal of the parallelogram.

Subtracting a vector, v, is geometrically equivalent to adding its negative, –v.

Practice 4

Consider this graph with vectors labelled a, b, u, v, and w: a

746

Which vector is the resultant of u + a?

Which vector is the resultant of w + u?

Which two vectors can be added together so that the resultant is a zero vector?

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

y 7 6 5 4 3 2 1 w 0 1 −1 −2 −3

b u x a  2 3 4 5 6 7 8 9 v

5 Ex 1

The vectors a and b are shown. Plot the resultant vector c = a + b: a

x c

y a b

b a

x 6

For each plane, plot the resultant vector: a

x c

b x

8.04 Vector addition and subtraction mathspace.co

747

Draw the vector u for each of these equations: a b c

u=c–d

u=d–b

Sketch these vectors: a b c d

u=a–b

a–b c–a

c–b

2b – a

c b

x 9 Ex 2

2 4 and w : 6 1 Represent these vectors in the plane, starting from a common origin.

Using the parallelogram rule, draw the resultant vector v + w.

Consider the vectors v

c 10

0 1 ,w : 1 0 Represent these vectors in the plane.

Using the triangle rule, draw the resultant vector 2v – w.

Consider the vectors v

c 11

748

Determine v + w in column vector form.

Determine 2v – w in column vector form.

Consider the vectors v

1 ,w 1

0 : 1

Represent these vectors in the plane.

Using the triangle rule, draw the resultant vector 3v – 2w.

Determine 3v – 2w in column vector form.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Three vectors u, v, w are drawn: y

One vector is the sum of the remaining two:

Determine which one is the resultant vector and explain why.

Translate the vectors in the plane to represent the rule.

Three vectors u, v, w are shown starting from a common origin:

y 6 5 4 3 u −3 −2 −1

w 1

x 2

One vector is the sum of the other two: a

Determine which vector is the resultant and explain why.

On the diagram, construct a parallelogram to show how the resultant vector is the sum of the other two.

8.04 Vector addition and subtraction mathspace.co

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14 Ex 3

Sketch the vector: a

a – b

c–e

y y 5 5

3 2

x 0 1 2 3 4 5 6 7

x 0

v – u

m – 3n

y y 7 5 6 4 5 3n 4 3 3 2 v 2 m u 1 1 x x 0 1 2 3 4 5 6 7 0 1 2 3 4 15

Consider the vectors on the 2D plane labelled a, b, c, and d: 10 9 8 7 6 5 4 3 2 1 0

b d a

1 2 3 4 5 6 7 8 9 10

Express all the vectors in column vector form

Determine: ii iii iv

750

a+b

2c – d

a – c + 2d

a+b+c+d

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x . y

16 Ex 4

a 17

Ex 5

a+b

1 . Calculate these resultant vectors: 5 b

a–b

b–a

a+a

Determine the vector u in column form that satisfies each of these equations: a

4 and b 2

Let a

1 1

2 0

Given the vectors a 3a + x = 2b.

5 3

4 and b 2

1 7

6 2

4 0

4 1

5 5

1 , determine the vector x that satisfies the equation 6

If vector v has a magnitude of 5 units and vector u has a magnitude of 3 units, and the angle between them is 90°, what is the magnitude of the resultant vector using the triangle rule?

Extend your thinking 20

Consider the property of vectors that states: If k is a scalar and u, v are vectors, then 0 1 k × (u + v) = k × u + k × v. For the case of k = 2 and u ,v , use the triangle rule to 1 0 illustrate this property graphically.

Consider the vector equation a b

1 u v 1 Use the triangle rule to draw u + v.

3 : 3

Using the illustration given, explain why there is not enough information to determine u and v explicitly.

The mathematics teacher states that there are four unknown plane vectors s, t, u, and v. Assume that all vectors are drawn with the start at the origin, given this information: • When drawn, the end of the vectors s, t lie in the first quadrant, and u lies in the second quadrant. • The vectors u and v are perpendicular. a b

What possible quadrant(s) can the head of v lie in? Is it possible that s + v = 0?

The vectors a and b are given by a c – 4d = 2a and 3c – d = –2b.

5 and b 12

2 . Determine the vectors c and d, if 1

8.04 Vector addition and subtraction mathspace.co

751

8.05 Unit vectors and component form After this lesson, you will be able to… • calculate and use unit vectors • resolve vectors into i and j components • convert between different vector representations • use components to solve vector problems including magnitude

Vectors in component form Unit vector A vector with a magnitude of 1, used to indicate direction. The standard unit vectors are i for the x-direction and j for the y-direction.

y 4

Recall that a position vector is a vector that begins at the origin, and a unit vector, a, is a vector of magnitude 1.

3 2 1

The unit vectors in the direction of the x- and y-axes are named i and j.

−1 0

x 1

Therefore, i

1 and j 0

0 . 1

y 4 4i + 3j

3 2 1 0

752

a j

i 1

x 2

Any two-dimensional position vector can be written as a combination of i and j. For 4 example, a is 4 units to the right (in the i 3 direction) and 3 units up (in the j direction) from the origin, therefore a = 4i + 3j.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Cartesian (component) form A method of representing a vector using its horizontal and vertical components along the x x-axis and y-axis, typically written as xi + yj or as a column vector . y Even if a vector does not begin at the origin, it can still be expressed in component form. The coefficient of i is the horizontal component and the coefficient of j is the vertical component. For example, d starts at (–1, 1) and ends at (3, 4). The horizontal component is 4 units to the right and vertical component is 3 units up, therefore d = 4i + 3j. y

4i + 3j

4 d

3 2 1 –1

x y

4 j

In general, this means that: 1 0

0 1

Example 1 Rewrite each vector in component form: a

0 9

Create a strategy Component form means in terms of the unit vectors i and j. The unit vector i points in the direction of the x-axis, and the unit vector j points in the direction of the y-axis.

Apply the idea Recall that

x can be written as xi + yj: y x = xi + yj Write the equation y = 0i + 9j

Substitute x = 0 and y = 9

= 9j

Simplify the term with coefficient of zero

8.05 Unit vectors and component form mathspace.co

753

Reflect and check 9 8 7 6 5 4 3 2 1 –1

0 as a position vector on the 9 Cartesian plane, it can be seen that there is no horizontal component, giving 0i , and 9 units upwards, giving 9j. Graphing

0    = 9j 9  x 1 2 3 4 5 6 7 8 9

5 2

Apply the idea x = xi + yj y

Write the equation

5 = 5i – 2j 2

Substitute x = 5 and y = –2

Reflect and check 9 8 7 6 5 4 3 2 1 –1 –9–8 –7–6–5–4–3–2–1 –2 –3 –4 –5 –6 –7 –8 –9

754

Graphing 5i – 2j

1 2 3 4 5 6 7 8 9 (5, –2)

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

5 as a position vector on the 2

Cartesian plane, it can be seen that there is a horizontal component of 5 units to the right, giving 5i, and 2 units downwards, giving –2j.

Example 2 Rewrite the vector –8i as an ordered pair.

Create a strategy

Apply the idea

The unit vector i points in the direction in the x-axis, and the unit vector j points in the direction in the y-axis.

Since there is no unit vector j, the vector sum can be written as –8i + 0j. Remember that (x, y) can be written as xi + yj. Therefore, the ordered pair representation of the vector –8i is (–8, 0).

Idea summary Vectors in Cartesian (component) form are represented in terms of the unit vectors, i = (1, 0) and j = (0, 1). Therefore, a vector can be represented in any of the forms shown: • Ordered pairs: (x, y) x • Column vectors: y • Cartesian (component) form: xi + yj

Calculate unit vectors To determine a unit vector that points in the same direction as a given vector, divide each component of the vector by its magnitude. Magnitude (of a vector) The length or size of a vector, representing its scalar value. For a vector v = xi + yj, the magnitude is calculated as _v_

The unit vector in the direction of n is n

y2 .

n . n

For a non-zero vector, n = xi + yj, the magnitude is n the unit vector, n, in the direction of n is: n= = =

n n

0. Therefore, in Cartesian form,

Write the equation

x y Divide each component of the vector n i j n n by its magnitude x x2 y2

y x2 y2

Substitute the formula for the magnitude |n|

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Now, to show that n has unit length: |n| = x2 y2

Write the formula

y x2

y2 x2 y2 x2 y2

Substitute the components of n

Evaluate the square

x2 y2

Add the fractions

x2 y2

= 1

Simplify the fraction, since x2 Evaluate

Example 3 Calculate the unit vector in the direction of each given vector in the required form: a a = (3, 4), as an ordered pair

Create a strategy Calculate the magnitude of the vector and then apply the unit vector formula: a

Apply the idea First, calculate the magnitude of a: |a| = x2 y2

Write the formula

= 32 42

Substitute the values

= 25

Simplify

Evaluate

Secondly, calculate the unit vector: a= = =

a a (3, 4) 5 3 4 , 5 5

Write the formula Substitute |a| = 5 and a = (3, 4) Evaluate

Reflect and check Notice that this is scalar multiplication by

756

1 , the inverse of the magnitude of a. a

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

a a

b c

6 , in component form 3

Apply the idea First, calculate the magnitude of c: |c| = x2 y2

Write the formula

= 62 32

Substitute the values

= 45

Evaluate

=3 5

Simplify

Secondly, calculate the unit vector: c= =

6i 3j

Divide vector c by its magnitude

3 5 1 3 5

6i 3j

Factorise the expression

5 6i 3j 15

Rationalise the denominator

6 5 3 5 i j 15 15

Expand the brackets

2 5 5 i j 5 5

Simplify the fractions

Reflect and check Since c is a unit vector, it should have magnitude of 1: |c| = x2 y2 2

Write the formula 5 5

2 5 5

20 5 25 25

Square each component

25 25

Add the fractions

Substitute the values

= 1

Simplify

Evaluate

Therefore, c is a unit vector.

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Idea summary The unit vector in the direction of n is n

n . n

Resolve vectors into components In physics, particularly in fields such as mechanics and navigation, the need to add and subtract vectors frequently arises. The polar form effectively represents forces because they have a magnitude, often measured in Newtons (N), and a direction. Polar form (of a vector) A method of representing a vector using its magnitude (length) and the angle it makes with a reference direction, typically the positive x-axis. Example: A vector with magnitude 5 and an angle of 30° from the positive x-axis is in polar form.

Interactive exploration Discover this concept in action online

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Performing operations directly in polar form can be challenging. To facilitate the addition or subtraction of vectors, it is a common practice to convert them from polar form (magnitude and angle) to component form (i and j). In the context of vectors, converting or resolving vectors refers to the process of breaking a vector down into its components along specified axes, typically the horizontal i and vertical j directions. This involves determining the magnitude of the vector in each direction, allowing for simpler mathematical operations like addition, subtraction, or analysis of the vector’s effects in various applications. This process involves the application of trigonometric functions to determine the horizontal and vertical components.

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For example, for a vector with magnitude |v| and angle θ : y

Using trigonometric functions in the right-angled triangle shown, where |v| is the hypotenuse, Vx (the horizontal component) is adjacent to the angle θ, and Vy (the vertical component) is opposite the angle θ.

Vy j

Vx i

Therefore, cos (θ ) =

Adjacent Hypotenuse

Write the trigonometric ratio for cosine

cos (θ ) =

Vx _v_

Substitute the vector components and magnitude

|v| × cos (θ ) = Vx Vx = |v| cos (θ )

Multiply both sides by |v| Rearrange the equation to solve for the x-component

and sin (θ ) = sin (θ ) =

Opposite Hypotenuse Vy v

|v| × sin (θ ) = Vy Vy = |v| sin (θ )

Write the trigonometric ratio for sine Substitute the vector components and magnitude Multiply both sides by |v| Rearrange the equation to solve for the y-component

This results in the component form v = Vx i + Vy j. Once vectors are expressed in component form, addition or subtraction can be executed by combining the respective i and j components. After completing vector operations, if a resultant in polar form is desired, conversion back can be accomplished by: • Calculating the magnitude: _v_ • Determining the angle: q = tan–1

Vx2 Vy2 Vy Vx

or arctan

Vy Vx

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Example 4 A child is flying a kite with a force of 50 Newtons (N) pulling the kite string at an angle of 45° above the horizontal: y

50 N F

45°

Given that vector F can be expressed in component form as F = XF i + YF j, calculate, in Newtons, the exact values of Fx and Fy, the horizontal and vertical components of the force exerted on the kite string.

Create a strategy Resolve F into horizontal i and vertical j components using the trigonometric functions in the right-angled triangle, where |F| is the hypotenuse, Fx (horizontal component) is adjacent to the angle θ = 45°, and Fy (vertical component) is opposite the angle θ = 45°.

Apply the idea For the horizontal component Fx: Adjacent Hypotenuse

Write the cosine ratio

Fx F

Substitute the adjacent side and hypotenuse

cos (45°) =

Fx 50

Substitute θ = 45°and |F| = 50

cos (θ ) =

Fx = 50 cos (45°)

Rearrange the equation

50 2 2

Evaluate as an exact value

= 25 2 N

Simplify

The exact value of the horizontal component is Fx 25 2 N.

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For the vertical component Fy: sin (θ ) = = sin (45°) =

Opposite Hypotenuse Fy F Fy

Write the sine ratio Substitute the opposite side and hypotenuse Substitute θ = 45°and |F| = 50

Fy = 50 sin (45°)

Rearrange the equation

50 2 2

Evaluate as an exact value

= 25 2 N

Simplify

The exact value of the vertical component is Fy 25 2 N.

Example 5 Consider two vectors, OA and OB, of magnitudes 11 and 15, as shown in this image: y B

15 30°

11 60°

OA and OB can be expressed in component form as OA

X OA i YOA j and OB

X OB i YOB j.

a Determine the component form of the vector OB.

Create a strategy Resolve OB into horizontal i and vertical j directions using the trigonometric functions in the right-angled triangle, where _OB_ is the hypotenuse, X OB (horizontal component, negative) is adjacent to the angle θ = 30°, and YOB (vertical component, positive) is opposite the angle θ = 30°. Then, write in the form OB

X OB i YOB j.

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Apply the idea For the horizontal component X OB, the magnitude will be positive, but the component should be negative as it is moving to the left: cos (θ ) =

Adjacent Hypotenuse

X OB

cos (30°) =

X OB

_OB_ 15

Write the cosine ratio Substitute the adjacent side and hypotenuse Substitute θ = 30°and _OB_ = 15

X OB = –15 cos (30°)

Rearrange the equation

15 3 2

Evaluate as an exact value

For the magnitude of the vertical component YOB: sin (θ ) = = sin (30°) =

Opposite Hypotenuse YOB _OB_ YOB 15

Write the sine ratio Substitute the opposite side and hypotenuse Substitute θ = 30°and _OB_ = 15

YOB = 15 sin (30°)

Rearrange the equation

15 2

Evaluate as an exact value

Write the vector in component form: OB = X OB i Y OB j =

15 3 15 i j 2 2

Express OB in terms of unit vectors i and j Substitute X OB

15 3 15 and YOB into OB 2 2

Reflect and check 15 3 15 and can be calculated as positive values if they represent the magnitudes of the horizontal 2 2 and vertical components of vector OB. When using these values to express vector OB in terms of unit vectors i and j, the positive or negative signs should be taken into consideration. An alternative method to determine the correct signs is to use the angles measured from the x-axis in an anticlockwise direction (180° – 30°) instead of 30°. This approach allows the positive or negative signs to be obtained directly, eliminating the need for any further adjustments.

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b Determine the component form of vector OA.

Apply the idea For the magnitude of the horizontal component X OA: cos (θ ) = = cos (60°) =

Adjacent Hypotenuse X OA _OA _ X OA 11

Write the cosine ratio Substitute the adjacent side and hypotenuse Substitute θ = 60°and _OA_ = 11

X OA = 11 cos (60°)

Rearrange the equation

11 2

Evaluate as an exact value

For the magnitude of the vertical component YOA: sin (θ ) = = sin (60°) =

Opposite Hypotenuse YOA _OA_ YOA 11

Write the cosine ratio Substitute the opposite side and hypotenuse Substitute θ = 60°and _OA_ = 11

YOA = 11 sin (60°)

Rearrange the equation

11 3 2

Evaluate as an exact value

Write the vector in component form: OA = X OA i YOA j =

11 11 3 i j 2 2

Write the vector in terms of unit vectors i and j Substitute X OA

11 11 3 and YOA into OA 2 2

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c Express OC, the sum of vectors OA and OB, in component form.

Create a strategy Vectors OA and OB are expressed in component form, adding them can be executed by combining the respective i and j components.

Apply the idea OC = OA OB =

11 11 3 i j 2 2

11 2

Write the vector sum 15 3 15 i j 2 2

15 3 i 2

11 3 2

15 j 2

Substitute the components of OA and OB Combine like terms

d Determine the exact magnitude of the resultant vector OC.

Create a strategy Use the magnitude formula: v

Vx2 Vy2

Apply the idea OC =

11 2

_OC _ =

X OC

15 3 i 2

11 3 2

11 2

15 3 2

YOC 2

11 3 2

15 j 2

Write the vector component form of OC Write the magnitude formula

15 2

= 346

Substitute the values Simplify in exact form

e Determine the direction of the resultant vector OC, rounded to two decimal places.

Create a strategy Use the angle formula, if

764

tan 1

Vy Vx

then the angle formed with the x-axis is arctan

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Vy Vx

Apply the idea OC = 11 2

15 3 i 2

θ = tan

= tan

11 3 2

15 j 2

Write the vector component form of OC

YOC

Write the angle formula

X OC 11 3 15 2 2 11 15 3 2 2

Substitute the values

= tan–1 (–2.273 085 88)

Evaluate inside brackets using technology

= −66.25° or 180° + (−66.25°) Evaluate using technology, and express both possibilities = 113.75° Choose the positive angle representing rotation from the positive x-axis

Reflect and check This means that OC in polar form is ( 346, 113.75°). The vectors cannot be added directly in polar form as OA = (11, 60°) and OB = (15, 150°) and adding the magnitudes and angles would not give the correct resultant.

Example 6 Digital only example Explore in the online lesson

mathspace.co

Idea summary Converting vectors from polar to component form makes addition and subtraction simpler and calculations easier. Once vectors are expressed in component form, addition or subtraction can be executed by combining the respective i and j components. A vector v with magnitude |v| and angle θ can be expressed in the component form: v = Vxi + Vy j where Vx = |v| cos (θ ) and Vy = |v| sin (θ ). After completing vector operations, if a resultant in polar form is desired, conversion back can be accomplished by calculating the magnitude: v

Vx2 Vy2 and determining the angle:

tan 1

Vy Vx

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8.05 Practice questions What do you remember? 1

Draw these vectors on a Cartesian plane: a

0i − 4j

3i − 2j

Ex 1

Express each vector in component form: 1 4 a b 3 0

Express each vector as an ordered pair:

Ex 1

− 4i + 5j

− 4i − 5j

6i + 4j

4 7

2 5

− 3j

Express each vector as a column vector: a

3i − 19j

i+j

– i + 3j

Express these given vectors in Cartesian (component) form: a

0 1

–1

–2

–3

1 –4

x 0 c

766

–3

–2

–1

1 –4

1 x

0 –1

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

–4

–3

–2

–1

0 –1

Practice Ex 3

Let g = (−3, 7). Calculate a unit vector in the direction of g.

Consider the position vector h = (2, 1). Calculate the unit vector in the direction of h.

Consider the position vector f = (−6, −4). Calculate the unit vector in the direction of f.

For each vector: i

Determine the scalar that can be used to make a unit vector in the same direction.

Calculate the unit vector in the same direction.

6 8

5 12

8 15

3 1

Ben walks on a hiking trail from point A to B. The trail takes him 3 km north, 2 km west, 4 km south and 5 km east. Determine a single vector in component form representing his journey from A to B (take north to be the positive y-axis).

A(–5, –2), B(7, –1), C(10, 6) are three points on the Cartesian plane. Identify each of these vectors in i, j notation: a

AC BC

y C

6 4 2

−6 −4 −2 2 4 6 8 10 −2 B A −4 −6 −8

The vector, 2i + 3j points from point X to point Y. If point Y has coordinates (–3, 19), determine the coordinates of point X.

Calculate a unit vector which is in the opposite direction to g = (−3, 5).

ABCD is a rectangle where AB 5i and AD 8j. Express these vectors in component form: a

The unit vector a is parallel to the line y = 3x. Determine the possible values for a.

Consider a vector t = (8, −5). Calculate a vector which has a magnitude of 14 in the same direction as t.

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Ex 4

Ex 5

A ball is struck with a velocity of 40 m/s at an angle of 30° to the horizontal: a

Find the exact value of the horizontal component, vx.

Find the vertical component, vy.

Two forces, F1 and F2, act on an object at the origin:

• F1 has a magnitude of 20 N and acts at an angle of 60° to the positive x-axis. • F2 has a magnitude of 30 N and acts at an angle of 150° to the positive x-axis.

a b 19

Find the component form of F1 and F2.

Find the component form of the resultant force, R = F1 + F2.

Three forces, F1, F2, and F3, act on an object at the origin as shown in this diagram:

Ex 6

20 20 N

F2 45°

15 10 5

–20 –15 –10 –5 –5 –10 –15

15 N

x 30° 60° 10 15 20 F3

10 N

–20 Find the magnitude and direction of the resultant force R = F1 + F2 + F3. Round your final answers to one decimal place.

Extend your thinking 20

Calculate the unit vector in the direction of a vector which has a direction of 60°.

The coordinates of points A and B are (12, 15) and (–8, 13), respectively. The point C is the midpoint of A and B. Determine the vector CB in component form.

Point A has position vector 3i + 10j. Point B has position vector –5i + 6j. Point C lies on the interval AB such that AC : CB = 2 : 1. Determine the position vector OC.

ABCD is a parallelogram, with points A(6, 9) and C(33, 18). The vector AB Determine the position vector OD.

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21 i 14 j.

8.06 Geometric applications of vectors After this lesson, you will be able to… • use vectors to represent line segments • construct vector proofs to prove geometric theorems

Represent line segments with vectors Exploration For the vector AB, the points A, B, and O, along with (undirected) line segments between them, are shown:

B (xB, yB)

A (xA, yA)

x O (0, 0)

1. Write the position vectors OA and OB in Cartesian form: OA

2. Write a vector sum where AB is the resultant. Use two of OA, AO, OB, BO as the addends: AB 3. Write the vector sum in Cartesian form: AB 4. Calculate AB in Cartesian form. 5. Is AB BA? If so, justify the equality. If not, explain how they are related.

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Line segment The straight path connecting two distinct points, encompassing all the points between them. It has a specific length (magnitude) but no inherent direction.

11 y B 10 9 8 7 6 5 4 3 2 1 x A 0 1 2 3 4 5 6 7 8 9 10

Midpoint The point on a line segment that divides it into two equal parts. It is equidistant from both endpoints.

M A

A directed line segment can be represented by a vector. The vector from point A to point B, written as AB, describes the displacement—both the distance and direction—required to move from A to B. y A (1, 3.5) B (7, 3) (0, 0)

x (6, –0.5)

This vector can be found using the position vectors of the points A(xA, yA) and B (xB, yB) relative to an origin O. By applying the triangle rule for vector addition (AB AO OB), a simple formula can be derived:

AB OB OA OB

is the position vector of the terminal point B

is the position vector of the initial point A

This allows for the calculation of the components of the vector AB directly from the coordinates of its endpoints. xB xA xB x A • In column vector form: AB yB yA yB y A • In component form: AB

770

( xB

x A ) i ( yB

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

yA ) j

The midpoint of a line segment is a special point that divides the segment in a 1 : 1 ratio. The position vector of the midpoint M is the average of the position vectors of the endpoints A and B.

4 y B

–2 A

–4

1 OA OB 2

This leads to the familiar midpoint formula for coordinates: M

xA 2

xB y A ,

yB 2

More generally, a point P can divide a line segment AB in any ratio m : n. The position vector of such a point can be found using the section formula, which is a weighted average of the endpoints’ position vectors.

nOA mOB m n

OP is the position vector of the point P that divides the line segment AB m : n is the ratio in which the segment AB is divided. Note the “cross-over” of the weights: n multiplies A and m multiplies B

Example 1 Digital only example Explore in the online lesson

mathspace.co

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Example 2 Find the coordinates of the midpoint M of the line segment joining the points A(1, 2) and C(4, 5).

Create a strategy Use the midpoint formula, which averages the x- and y-coordinates of the endpoints: M

Write the formula

xC y A ,

yC 2

Apply the idea M=

xA 2

xC y A ,

1 4 2 5 , 2 2

Substitute the coordinates of points A and C

5 7 , 2 2

Evaluate

The midpoint is M(2.5, 3.5).

Example 3 Consider the line segment with endpoints A(– 5, – 4) and B(1, 8). Determine the point, P, that 1 divides the segment AB in the ratio 1 : 2. (This is the point of the way from A to B). 3

Create a strategy 1 Determine the components of the vector AB. Calculate of this vector and translate point A by 3 the result to find point P.

Apply the idea

8 7 6 5 4 3 2 1

–5 –4 –3 –2 –1–1 –2 –3 –4 A

772

First, find the vector representing the full displacement from A to B: AB x

1 2 3 4 5

1 ( 5) 8 ( 4)

6 12

The vector from A to P is one-third of this total displacement:

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 AB 3

1 6 3 12

2 4

Add the components of the vector AP to the coordinates of the starting point A to find the coordinates of P. OP = OA AP =

5 4

3 0

Write the vector addition formula

2 4

Substitute the vector values Evaluate

The coordinates of the point P are (–3, 0).

Reflect and check This result can be verified using the section formula, which treats the point P as a weighted average of the endpoints A and B. For a ratio of 1 : 2 (m = 1, n = 2), the formula is: OP =

2OA 1OB 1 2

1 2 3

1 9 3 0

5 4

Write the formula 1

1 8

Substitute the position vectors values

3 0

Evaluate the expression

Both methods yield the same result.

Idea summary The vector AB from an initial point A(xA, yA) to a terminal point B(xB, yB) is found by subtracting the position vector of A from the position vector of B: AB

xB yB

OB OA

xA yA

The midpoint M of a line segment AB is the average of the endpoints’ positions. Its position vector is given by: OM

1 OA OB 2

A point P that divides a line segment AB in the ratio m : n can be found using the section formula, which is a weighted average of the position vectors: OP

nOA mOB m n

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Geometric proofs with vectors Vectors provide a useful algebraic method for proving geometric theorems without relying on coordinate geometry, angles, or lengths. By representing points and line segments as vectors, geometric relationships can be translated into algebraic equations. The strategy for most vector proofs involves expressing all relevant vectors in terms of a few base vectors (e.g., a and b) and then demonstrating a specific algebraic property. Key conditions used in vector proofs include: • Parallel vectors: Two vectors u and v are parallel if one is a scalar multiple of the other. That is, u = kv for some non-zero scalar k. • Equal vectors: Two vectors are equal if both vectors have the same magnitude and the same direction. This is a key property for proving sides of a quadrilateral are parallel and equal in length (i.e., proving it is a parallelogram). • Collinear points: Three points A, B, and C are collinear (lie on the same straight line) if the vector AB is a scalar multiple of BC. That is, AB kBC. The shared point B is essential.

Example 4 Digital only example Explore in the online lesson

mathspace.co

Example 5 Use vectors to prove the midpoint theorem, which states that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half its length. A

Consider △ABC. Let D be the midpoint of side AB and E be the midpoint of side AC.

Create a strategy Represent the vertices A, B, and C with position vectors a, b, and c from an origin O. Express the position vectors of the midpoints D and E using the midpoint formula. Finally, find the vector DE and compare it to the vector BC to show it is a scalar multiple.

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Apply the idea Let the position vectors of the vertices be OA a, OB b, and OC c. Since D is the midpoint of AB, its position vector is: 1 OA OB 2

1 (a + b) 2

Since E is the midpoint of AC, its position vector is: 1 OA OC 2

1 (a + c) 2

Now, find the vector DE by subtracting the position vectors: DE = OE OD

Write the formula

1 1 (a + c) (a + b) 2 2 1 1 1 1 a b c = a 2 2 2 2

1 1 = c b 2 2

Substitute the position vectors Distribute the scalar

1 2

Combine like terms

1 (c – b) 2

Factorise

1 2

Next, find the vector BC: BC

OC OB c b

1 BC. 2 This proves that DE is parallel to BC and its length is half the length of BC. By comparing the two results, it is clear that DE

Reflect and check This proof demonstrates the efficiency of using vectors. Complex geometric relationships are simplified into straightforward algebraic manipulations, proving properties of parallelism and length in a few steps without needing coordinates or angles.

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Example 6 Use vectors to prove that the diagonals of a parallelogram bisect each other.

C B

A O

Create a strategy Define the vertices of the parallelogram using position vectors a and b. Find the position vector of the midpoint of the first diagonal, then find the position vector of the midpoint of the second diagonal. Show that these two position vectors are identical.

Apply the idea Let the parallelogram be OACB, with the vertex O at the origin. Define the position vectors of the adjacent vertices as OA a and OB b. By the triangle rule for vector addition, the position vector of the fourth vertex C is OC Since AC OB b, then OC a + b.

AC.

First, find the position vector of the midpoint of the diagonal OC: 1 OO OC 2

Write the midpoint formula

1 0 2

Substitute the position vectors

1 (a + b) 2

M1 =

a b

Simplify

Next, find the position vector of the midpoint of the other diagonal AB: M2 = =

1 OA OB 2

Write the midpoint formula

1 (a + b) 2

Substitute the position vectors

Because the position vectors for both midpoints are identical (M1 = M2), the diagonals intersect at a common midpoint. This shared point confirms that the diagonals bisect each other.

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Idea summary Vector algebra is a powerful tool for proving geometric theorems. The general strategy is to represent points with position vectors and use vector operations to prove relationships. Key properties for proofs include: • Two vectors are parallel if one is a scalar multiple of the other (u = kv). • Two vectors are equal if they have the same magnitude and direction, which is used to prove properties of parallelograms. • Three points A, B, C are collinear if the vector between two of the points is a scalar multiple of the vector between another two points that share a common point (e.g., AB kBC).

8.06 Practice questions What do you remember? 1

Given the points A(1, –2) and B(5, 1): a

Find the vector AB as a column vector.

Express AB in component form using i and j.

Determine the coordinates of the midpoint M of the line segment connecting P(–4, 6) and Q(2, 8).

The points A and B have position vectors OA a and OB b respectively, relative to an origin O: a b

Express the vector AB in terms of a and b.

Express the position vector of the midpoint of AB in terms of a and b .

Complete these statements about vectors: a b

Two non-zero vectors u and v are parallel if u = ⬚ for some non-zero scalar k. Three distinct points P, Q, and R are collinear if PQ

for some scalar k.

Practice 5 Ex 1

Determine the vector AB as a column vector when the coordinates of points A and B are: a

A(1, 2); B(3, 5)

B(3, –5); A(1, 2)

B(3, 0); A(0, 2)

A(1, 2); B(3 3, 5)

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Let E and F be points on the Cartesian plane. Determine the vector EF in component form ⬚ i + ⬚ j: a

E(0, 0) and F(12, 5)

Determine the vector AB as a column vector.

Determine the vector BC as a column vector.

Show that AB BC

Show that _ AB_

AC.

_ BC _ _ AC _.

Consider a Cartesian plane with origin O (0, 0), the points A (1, 2), B (3, 5) and C (0, 4): a

Plot the points A, B, and C on the Cartesian plane.

Determine these vectors in column vector notation:

E(2, 3) and F(5, 7)

Consider the points A(2, 3), B(4, 6) and C(1, 5):

iii

Determine the exact magnitude of vector BC.

Determine one possibility for the coordinates of two points A(xA, yA) and B(xB, yB) given that: 4 a AB b _ AB_ 5 0 3 Consider the vector u and the point B(3, 5). Determine the coordinates of a point 4 A(xA, yA), given that: a

AB u

AB u 0

Determine the coordinates of P, the point described in relation to A and B: 1 1 a of the way from A(–7, 6) to B(5, –3) b of the way from A(–4, 3) to B(2, 0) 3 3 y y 8 5 A 4 6 A 3 4 2 2 1 x x B 0 –8 –6 –4 –2 2 4 6 8 –5 –4 –3 –2 –1 1 2 3 4 5 –1 –2 –2 B –4 –3 –6 –4 –8 –5

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2 3 of the way from A(–3, 1) to B(5, 7) d of the way from A(–7, 9) to B(5, 0) 5 4 y y 10 8 A B 9 7 8 6 7 5 6 4 5 4 3 3 2 A 2 1 x 1

–4 –3 –2 –1 0 –1 –2 12 Ex 2

Ex 3

1 2 3 4 5 6

B x 0 –8–7–6–5–4–3–2–1 –1 1 2 3 4 5 6 –2

Determine the coordinates of each midpoint: a

The midpoint M of the line segment AB, where A and B are the points with coordinates A(2, 5), B(0, 1).

The midpoint N of the line segment DC, where C and D are the points with coordinates C(–2, –15), D(8, 1).

Consider the line segment with endpoints A(2, 5) and B(8, –1). Find the coordinates of the point P that divides the segment AB in the ratio 2 : 1.

Consider a triangle with vertices at A(4, 1), B(2, 2), and C(4, 3). Verify the midpoint theorem by showing that the line segment connecting the midpoints of sides AB and BC is parallel to side AC and half its length:

Ex 4

Determine the coordinates of M, the midpoint of AB, and L, the midpoint of BC.

Determine the vectors ML and AC as column vectors. 1 Show that ML AC. 2

c 15

Consider a Cartesian plane with origin O(0, 0), the points A(4, 1), B(2, 2) and C(4, 3): a

Plot the points A, B and C, and the line segments AB, AC and BC on the Cartesian plane.

Determine the midpoints M, N and L of the line segments AB, AC, and BC, respectively.

Use the results found in part (b) to plot M, N and L on the same Cartesian plane. 1 Show that AM AB and express BL in terms of BC. 2

d 16

The quadrilateral ABCD has vertices with position vectors OA a, OB b, OC c, and OD d relative to an origin O. The points P, Q, R, and S are the midpoints of the sides AB, BC, CD and DA, respectively: a

Write down the position vectors of P and Q in terms of a, b, and c.

Find an expression for the vector PQ in terms of a and c.

Find an expression for the vector SR.

What conclusion can be drawn about the quadrilateral PQRS? Explain your reasoning.

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Ex 5

In △OAB, OA a and OB b. The point P lies on AB such that AP : PB = 3 : 1 a

Find the vector AB in terms of a and b.

Find the vector AP.

Prove that the position vector of P is OP

1 a 3b . 4

Consider triangle XYZ where M is the midpoint of XZ and N is the midpoint of YZ: X

Using a vector proof, show that MN is parallel to XY and MN = Ex 6

1 XY . 2

A parallelogram PQRS has vertices at P(1, 2), Q(3, 5), R(6, 4), and S(4, 1). Show that the diagonals of the parallelogram bisect each other.

This graph shows: B H

4 3

2 A

G 1

x –1

• The rectangle ABCD where the coordinates of A, B, C and D are given by A 1 11 11 B , , C 3, , and D 3, 5 2 3 3 3

1 5 , , 2 3

• The points H, E, F and G which are the midpoints of the line segments AB, BC, CD, and DA, respectively • The point I, the centre of quadrilateral HEFG a

Determine the coordinates of the points H, E, F, and G.

i Determine the exact magnitude of _ HE _. ii Show that the quadrilateral HEFG is a rhombus.

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Determine the coordinates of point I.

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Consider a Cartesian plane with origin O(0, 0), the points A(–2, 0), B(4, –1), and C(4, 2): a b

y 4

1 Determine the coordinates of the point L that is of 3 the way from A to C.

–3 –2 –1 –1

Express CL in terms of AL.

Determine graphically the coordinates of a point N that lies on line segment BC, such that triangle LNC is a right-angled triangle.

x 1

3 4

5 B

–2

Extend your thinking 22

Determine the coordinates of point P that divides the line segment AB in the given ratio: a

Divides AB in the ratio 2 : 3, where A(1, 4) and B(6, –1)

Divides CD in the ratio 3 : 1, where C(–2, –5) and D(10, 3)

Consider a triangle with vertices A(–3, 0), B(1, 8) and C(–7, 4). Classify the △ ABC (e.g., isosceles, right-angled, equilateral) by determining the exact length of each side.

In a triangle ABC, the median from vertex A meets the opposite side BC at point D. The position vectors of the vertices are OA a, OB b, and OC c. Prove that the position vector of the centroid G, which divides the median AD in the ratio 1 a b c. 2 : 1, is given by OG 3

The points A, B, and C have coordinates (–1, 2), (2, 1), and (k, –1) respectively. Find the value of k if the points are collinear.

Let ABCD be a trapezoid with AB parallel to DC, and let DC Let the diagonals AC and BD intersect at P.

k AB for some scalar k > 1.

By expressing AP in two different ways (as a fraction of AC and by following the path AB BP ), show that the diagonals divide each other in the ratio 1 : k.

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8.07

Polar form After this lesson, you will be able to… • understand vector representation in polar form • express vectors using polar notation • interpret the meaning of r and θ in polar form • convert vectors from polar to Cartesian form • use technology appropriately for conversions

Vectors in polar form Interactive exploration mathspace.co

Discover this concept in action online

Recall that vectors on the Cartesian plane can be expressed as an ordered pair (x, y), a column x vector , or Cartesian (component) form xi + yj. y Another method to express vectors is polar form which includes the vector’s length (magnitude) and direction (angle). y The polar form of a vector a = (r, θ ), where r is the magnitude and θ is the angle from the positive x-axis.

a r

θ 0

Note that the angle, given in degrees or radians, is positive when the vector is turned anticlockwise from the positive x-axis, and the angle is negative when the vector is turned clockwise from the positive x-axis.

If an angle α is given with respect to the negative x-axis, the standard polar angle θ is calculated by subtracting α from 180° (or π radians).

θ = 180° - α Similar adjustments must be made for angles given relative to the positive or negative y-axis.

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Example 1 Express the vector b in polar form: y

b 10 40°

x 0

Create a strategy First, identify the magnitude of the vector. Then, calculate the angle the vector makes from the positive x-axis. Finally, express the vector in polar form (r, θ ).

Apply the idea The magnitude of the vector is 10. Therefore, r = 10. The angle the vector makes from the positive x-axis can be calculated using supplementary angles. That is, θ = 180° - 40° = 140°. b = (r, θ )

Write the equation

= (10, 140°)

Substitute the values

Reflect and check y 140° x –220°

The angle is positive because it is measured in an anticlockwise direction from the positive x-axis. An alternative solution for the angle is 140° - 360° = -220°, but the positive angle is typically preferred in polar form.

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Example 2 Sketch the vector d

. 4

Create a strategy Identify the magnitude and angle the vector makes from the positive x-axis. If the angle is positive, turn anticlockwise, and if the angle is negative, turn clockwise from the positive x-axis.

Apply the idea y

. 4 The vector is 5 units long and makes an angle of from the positive x-axis, turning in a 4 clockwise direction. Polar form is (r, θ ); therefore, r = 5 and θ =

x π 4

0 5

Idea summary y The polar form of a vector can be written as a = (r, θ ), where r is the magnitude and θ is the angle from the positive x-axis.

a r

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θ is positive when turned anticlockwise from the positive x-axis, and negative when turned clockwise from the positive x-axis.

Cartesian form to polar form Interactive exploration mathspace.co

Discover this concept in action online

In right-angled triangles, trigonometric ratios can be used to describe the relationship between a given angle and two sides.

Opposite

The trigonometric ratios are:

Hypotenuse

sin ( ) cos ( )

tan ( ) Adjacent

Opposite Hypotenuse Adjacent Hypotenuse Opposite Adjacent

Recall that vectors can be given in terms of their horizontal and vertical components: • Ordered pairs: (x, y) • Column vectors:

x y

• Cartesian (component) form: xi + yj, where i is the unit vector in the horizontal direction and j is the unit vector in the vertical direction. The polar form of a vector can be found from its Cartesian (component) form a = xi + yj as a = (r, θ ), where:

r x2 y2 r is the magnitude, calculated using Pythagoras’ theorem

tan 1

y x

q is the angle, measured anticlockwise from the positive x-axis These conversions can be performed efficiently using technology. Many devices have built-in functions for converting between rectangular (Cartesian) and polar forms. To use these, one typically inputs the (x, y) components to receive the (r, θ ) values directly. Alternatively, technology y can be used to evaluate the trigonometric expressions, such as tan–1 . It is crucial to be aware x of the angle’s quadrant, as technology may provide a reference angle that requires adjustment (e.g., adding 180°).

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y The magnitude r can be calculated using Pythagoras’

xi + yj r

α  −x  r u=  − y   −y tan(θ ) = −x y = x

 x v=  y r x  α θ = α – 180°

theorem, x2 + y2 = r2. Rearranging gives r x2 y2 . The angle can be calculated using trigonometric y identities, specifically the tangent ratio: tan ( ) . x y Rearranging gives tan 1 . x

Since tan is positive in Quadrant 1 and Quadrant 3 and negative in Quadrant 2 and Quadrant 4. Recall that the tan-1 function typically returns an angle in the range (-90°, 90°). If the vector is in Quadrant 2 or Quadrant 3, it will be necessary to add or subtract 180° to determine θ such that it satisfies the stated range for θ. This could be 0° ≤ θ < 360°, -180° ≤ θ < 180°, or even -360° ≤ θ < 0° to ensure the angle corresponds to the vector’s position in relation to the positive x-axis.

Sketching the vector on the Cartesian plane can be helpful to avoid making an error and giving polar form in terms of the acute angle with the negative x-axis. y Similarly, tan (90°) is undefined; therefore, for vectors that have no horizontal component, x = 0, 0 using tan 1

y will result in an error. However, drawing the vector on the plane, it can be seen 0

that θ = 90° or θ = -90° depending on the sign of y.

Example 3 Convert each equation to polar form, where -180° < θ ≤ 180°: a u = 3i − 5j

Create a strategy Identify the horizontal and vertical components of the vector. Apply the magnitude formula y . Then, express in polar form (r, θ ). r x2 y2 and the angle formula tan 1 x

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Apply the idea The horizontal component is x = 3 and the vertical component is y = -5. The magnitude ∣u∣ x2

y2 32 ( 5)2 34

Applying the angle formula:

θ = tan

= tan

y x

Write the angle formula

5 3

Substitute the values

= -59°

Evaluate

Therefore, the polar form of u is ( 34 , −59°).

Reflect and check Sketch u = 3i − 5j and check which quadrant the vector lies in: y

x 0 –5

3 u

The vector is in Quadrant 4, which agrees with the negative angle from the positive x-axis.

b v=

2 2

Create a strategy First, calculate the magnitude r. Then, determine the quadrant by looking at the signs of the y components. Calculate the reference angle α using the formula α = tan 1 . Finally, adjust this x reference angle based on the quadrant to find the correct polar angle θ.

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Apply the idea The horizontal component is x = -2 and the vertical component is y = -2. Since both components are negative, the vector lies in the third quadrant. y

θ = α – 180°

 −2  v =   −2  Calculate the magnitude r: r = ( 2)2 ( 2)2

Write the equation

= 4 4

Evaluate the squares

= 8

Evaluate the addition

=2 2 Evaluate the exact value Next, calculate the reference angle α using the absolute values of the components:

α = tan 1 = tan 1

y Write the reference x angle formula 2 2

Substitute the absolute values of the components

= tan-1(1)

Simplify the fraction

= 45°

Evaluate

Since the vector is in the third quadrant, adjust the reference angle to find θ in the required range of -180° < θ ≤ 180°:

θ = α - 180°

Write the equation

= 45° - 180°

Substitute the values

= -135°

Evaluate

Therefore, the polar form is v = 2 2, 135° .

Reflect and check If the range for θ was 0° ≤ θ < 360°, the angle would be calculated by adding the reference angle to 180°: θ = 180° + 45° = 225°. The polar form would be v = 2 2, 225° . Always check the required domain for the angle.

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Idea summary To convert from Cartesian form to polar form, the vector a = xi + yj can be written as a = (r, θ ), where: x2

• r

y2 is the magnitude.

y is the angle from the positive x-axis. x y For vectors in Quadrant 2 or 3, the angle found from tan 1 must be adjusted by adding or x subtracting 180° to correctly position θ within the specified range (e.g., -180° < θ ≤ 180°). • tan 1

Polar form to Cartesian form a = xi + yj x where x = r cos (θ ), the horizontal component x is found using the cosine of the angle y where y = r sin (θ ), the vertical component y is found using the sine of the angle Trigonometric ratios are applied to determine the horizontal and vertical components of the vector. The horizontal component: Adjacent • cos ( ) or cos ( ) Hypotenuse

x r

• x = r cos (θ ) r

θ 0

The vertical component: Opposite y • sin ( ) or sin ( ) Hypotenuse r • y = r sin(θ )

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Example 4 Express the vector v in Cartesian form: y

v 30 2π 9

Create a strategy Identify the magnitude and angle the vector makes with the positive x-axis. Then, apply the formulas x = r cos (θ ) and y = r sin (θ ) to calculate the Cartesian form xi + yj.

Apply the idea The vector has magnitude 30. Therefore, r = 30. The diagram shows the angle with the negative x-axis is

2 . The required angle θ is measured 9 2

from the positive x-axis, so it is found by calculating the supplementary angle:

θ= =

2 9

Write the equation

7 9

Simplify

Apply the formula for horizontal component: x = r cos (θ ) = 30 cos = -22.98

Write the equation 7 9

Substitute the values Evaluate

Apply the formula for vertical component: y = r sin(θ ) = 30 sin = 19.28

Write the equation 7 9

Substitute the values Evaluate

Therefore, v = −22.98i + 19.28j.

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Reflect and check Sketch v = −22.98i + 19.28j to determine if both Cartesian form and polar form lie in the same quadrant: y v

19.28 x 0

–22.98

Since the Cartesian form and the polar form are both in Quadrant 2, the solution is confirmed to be correct.

Idea summary To convert from polar to Cartesian form, the vector a = (r, θ ) can be written as:

a = xi + yj x y

= r cos (θ ) = r sin (θ )

where r is the magnitude and θ is the angle measured anticlockwise from the positive x-axis.

8.07 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

In the polar form of a vector (r, θ ), r represents the direction and θ represents the magnitude.

The angle θ in polar form is always measured from the positive y-axis.

A negative angle θ means the angle is measured in a clockwise direction.

The vector (5, 40°) has the same magnitude as the vector (5, -40°).

A vector has a magnitude of 10 and a direction of 120°. Which polar form correctly represents this vector? A

(10, 60°)

(120, 10°)

(10, 120°)

(10, -60°)

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3 : 3

Consider the vector v = a

In which quadrant does this vector lie?

Calculate its exact magnitude, r.

Calculate its direction, θ, where 0° ≤ θ < 360°.

Write the vector in polar form (r, θ ).

Practice 4 Ex 1

Express in polar form (r, θ ): a

a x

12 30°

20°

4 b

y 25° 7

0 15°

0 8 c

6 Ex 2

For each vector in polar form: i

Sketch on the Cartesian plane.

Express in polar form with -180° < θ ≤ 180°.

a = (8, 300°)

b = (3, −200°)

b = (2, 115°)

f = (1, π)

25 c = 10, 9

37 18

d = 1,

d = (5, −75°) 5 h = 2, 6

Sketch: a e

792

a = (3, 25°) e = 2, 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

c = (6, −135°) 3 g = 1, 2

A vector w has magnitude 3 and is in the direction of 180° to the positive x-axis: a

Sketch the vector.

Express the vector in polar form.

Express the vector as an ordered pair.

Express p = (3, 60°) in two different polar form notations.

On the Cartesian plane, these points in polar coordinates are plotted: A(2, 60°), B(1, π), C(1, 135°) Which point has the shortest distance to the x-axis?

Which point is the furthest from the x-axis?

2 radians to the positive x-axis. 3 Express this vector in Cartesian component form xi + yj, rounded to two decimal places.

A vector has a magnitude of 20 and is at an angle of

Convert the vector a = (15, 215°) to Cartesian component form (x, y), rounded to two decimal places.

Ex 3

5 Convert the vector b = 8, to Cartesian component form (x, y), rounded to two decimal 6 places.

Convert the vector u = 5i − 12j to polar form (r, θ ), where -180° < θ ≤ 180°. Round r to two decimal places and θ to one decimal place.

Express the column vector

7 in polar form (r, θ ), where 0° ≤ θ < 360°. Round r to two 24

decimal places and θ to one decimal place. Ex 4

For each vector in the diagrams shown, express the vector in Cartesian component form x v= . Give your answers as exact values where possible; otherwise, round to two decimal y places: a

y 6

y –5

–4

–3

3 2

–1

150° x

45° 1

–2

–8 –7 –6 –5 –4 –3 –2 –1

1 2

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240° 1 –5 –4 –3 –2 –1 c

–1

–1 –1 –2

–3 –4 –5 –6 –7

–2 –3 –4 –5

2 3 4 5 6 7 –60° d

Use technology to convert these vectors, rounding your answers to two decimal places: a b

p = (9, 2.5 rad) to Cartesian form (x, y)

q = (−4.5, −8.1) to polar form (r, θ ), where 0 ≤ θ < 2π

Extend your thinking 17

Explain how polar coordinate representations are not unique. Let P be a point on the Cartesian plane which is not the origin. Determine the number of possible polar coordinates representing P.

Let a = (5, 120°). b is three times as long as a and pointing in the opposite direction. Express b in polar form.

Express a in polar form, rounded to two decimal places: y (4, 2) a

x 0 (–3, –2)

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8.08

Scalar (dot) product

After this lesson, you will be able to… • calculate scalar products using the formula • understand the geometric interpretation of scalar product • determine when vectors are parallel or perpendicular using scalar products • solve geometric problems using vector properties

Scalar product and properties The scalar (dot) product is a fundamental concept in vector algebra, particularly in two-dimensional space. It provides a means to calculate a single value that reflects the relationship between two vectors, incorporating both their magnitudes and the angle between them. The scalar product can be defined in two main ways: • Geometric definition: If a and b are non-zero vectors in the plane, and θ represents the angle between the vectors a and b, then the dot product can be defined geometrically as: a ⋅ b = ∣a∣∣b∣ cos (θ ). This formula emphasises the connection between the magnitudes of the vectors and the cosine of the angle separating them. • Component definition: If a =

a1 b1 and b = are non-zero vectors expressed in column vector notation, then: a2 b2 a⋅b=

a1 a2

b1 = a 1 ⋅ b1 + a 2 ⋅ b2 b2

This expression illustrates how to calculate the dot product using the individual components of the vectors. By multiplying the corresponding components and summing the results, the scalar product is derived directly. The value of the scalar product gives insight into the alignment of the vectors:

The scalar product is positive if the vectors are working somewhat together.

a b

The scalar product is zero if the vectors are working independently of each other.

θ b

The scalar product is negative if the vectors are working against each other.

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Several algebraic properties of the scalar product are useful to understand. These algebraic properties of the scalar product are presented without proof but can be verified geometrically or by expressing the vectors in column vector notation and performing the operations: • Commutativity: a ⋅ b = b ⋅ a for any two vectors a and b. • Distributivity: a ⋅ (b + c) = a ⋅ b + a ⋅ c for any two vectors a, b, and c. • Scalar multiplication: (ka) ⋅ b = k (a ⋅ b), where k is a scalar. Note that a direct connection between the length of a vector and the scalar product exists. Specifically, the dot product of a vector with itself equals the squared length of the vector, a ⋅ a = ∣a∣2 because a ⋅ a = a1 ⋅ a1 + a2 ⋅ a2 = a12 + a22 = ∣a∣2.

Example 1 This diagram shows vectors OA , OB, OC, and OD, where OA = 5, OB = 7, and OC = 10: A 5 C 10

α = 60°

β = 30° 7

D a Determine the scalar product OA OB.

Create a strategy Use the scalar product formula: OA OB = OA × OB × cos (α)

Apply the idea OA OB = OA × OB × cos (α) Write the formula = 5 × 7 × cos (60°) = 5 7 =

35 2

= 17.5

1 2

Substitute the values of OA , OB , and α Evaluate cos (60°) Evaluate the multiplication Evaluate

Therefore, the exact value of the scalar product OA OB is 17.5

Reflect and check The scalar product is expected to be positive when two vectors point in generally the same direction, as both vectors are effectively “working together.” In such cases, the angle between them is acute (less than 90°), resulting in a positive cosine value. This positive scalar product reflects the degree of alignment and shared direction between the vectors.

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b Determine the exact value of the magnitude of OD, given that OD OD = 153.

Create a strategy Use the property that the scalar product of a vector with itself equals the square of its magnitude: v ⋅ v = ∣v∣2

Apply the idea OD OD = OD

Write the equation

153 = OD 2

Substitute the given scalar product

OD = 153

Take the square root of both sides

OD = 9 17

Factorise the number under the square root

OD = 3 17

Simplify the square root

Therefore, the exact value of the magnitude of OD is 3 17.

c Given that OA OC = −17, determine θ, the angle between OA and OC, rounded to two decimal places.

Create a strategy Use the formula OA OC = OA × OC × cos (θ ) to form an equation that contains cos (θ ) and solve for it.

Apply the idea OA OC = _OA_ × OC × cos (θ )

Write the formula

–17 = 5 × 10 × cos (θ )

Substitute the values

–17 = 50 × cos (θ )

Evaluate the multiplication

17 cos (θ ) = 50

θ = cos 1 = 109.88°

Divide both sides by 50 17 50

Use the inverse function Evaluate and round

Reflect and check An angle greater than 90° is expected because the scalar product is negative.

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Example 2 Consider the vectors a =

2 6 1 ,b= , and c = , where c2 is a real number: c2 1 2

a Determine the scalar product a ⋅ b.

Create a strategy a1 a2

b1 = a 1 ⋅ b1 + a 2 ⋅ b2 b2

a⋅b=

a1 a2

b1 b2

Write the formula

1 2

6 1

Substitute the components of vectors a and b

Use the formula: a ⋅ b =

Apply the idea

= 1 × 6 + 2 × (– 1)

Apply the formula

=6–2

Evaluate the multiplication

Evaluate

b Determine the value of c2 given that a ⋅ c = − 8.

Create a strategy Use the formula a ⋅ c =

a1 a2

c1 = a1 ⋅ c1 + a2 ⋅ c2 and solve for c2, the equation a ⋅ c = − 8. c2

Apply the idea Express the scalar product in terms of c2: a⋅c=

a1 a2

c1 c2

Write the formula

1 2

2 c2

Substitute the components of vectors a and c

= 1 × 2 + 2 × c2

Apply the formula

= 2 + 2c2

Evaluate the multiplication

Solve the equation a ⋅ c = − 8. For c2: a ⋅ c = –8

2 + 2c2 = – 8 c2 = – 5

798

Write the equation Substitute a ⋅ c = 2 + 2c2 Divide both sides by 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary The scalar (dot) product is a key concept in vector algebra used to measure and express how two vectors relate to each other in terms of their direction and magnitude, in two-dimensional space. For any non-zero vectors a and b with θ representing the angle between them, the scalar product can be defined in two ways: • Geometric form: a ⋅ b = ∣a∣∣b∣ cos (θ ) • Component form: a ⋅ b =

a1 a2

b1 = a 1 ⋅ b1 + a 2 ⋅ b2 b2

Parallel vectors Two non-zero vectors are parallel if both point in the same or opposite directions. There are two main algebraic methods to test for parallelism. •

Method 1: Scalar multiples

Two vectors a and b are parallel if one is a scalar multiple of the other. That is, if a non-zero scalar k exists such that: a = kb • If k > 0, the vectors point in the same direction. • If k < 0, the vectors point in opposite directions. This is often the most direct method for determining if vectors are parallel and in which direction the vectors point relative to each other. •

Method 2: The scalar product The scalar product can also be used to test for parallelism. Recall the geometric definition: a ⋅ b = ∣a∣∣b∣ cos (θ ) • If vectors are parallel and in the same direction, the angle between the vectors is θ = 0°. Since cos (0°) = 1, the scalar product becomes a ⋅ b = ∣a∣∣b∣. • If vectors are parallel and in opposite directions, the angle is θ = 180°. Since cos (180°) = –1, the scalar product becomes a ⋅ b = −∣a∣∣b∣.

Therefore, two vectors are parallel if the scalar product is equal to the product (or negative product) of their magnitudes: ∣a ⋅ b∣ = ∣a∣∣b∣

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Example 3 3 and OB 2

Two vectors are given by OA

1 . Determine: 4

a Vector c, where c = AB

Create a strategy Use the formula: AB

AO OB

Apply the idea c = AB = AO OB

Write the formula

= OA OB Rewrite AO as OA 3 1 = Substitute the components of vectors OA and OB 2 4 =

b Vector e =

4 2

Evaluate

16 , parallel to vector c and points in the opposite direction 8

Create a strategy Use the fact that if the scalar constant k is negative, the vectors point in opposite directions.

Apply the idea e = kc

Write the equation

16 =k 8

4 2

Substitute the components

4k 2k

Expand the expression

Equating the corresponding components, results in 16 = –4k and 8 = –2k. Solving these equations for k results in k = –4. Therefore, vectors e and c point in opposite directions.

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c Vector d, where d has a magnitude of 4 5 and is parallel to c in the opposite direction

Create a strategy Use the magnitude formula to calculate the constant k, where d = kc. Ensure k is negative as d is in the opposite direction of c.

Apply the idea d = kc

Write the formula

4 2

Substitute the components

4k 2k

Evaluate

Determine k by using the magnitude formula: ∣d∣ = x2 y2

Write the formula

4 5 = ( 4 k)2 ( 2k)2

Substitute the known values

16 × 5 = (–4k)2 + (–2k)2 2

80 = 16k + 4k 80 = 20k 4=k

Square both sides Evaluate the multiplication and squares Evaluate the addition

Divide both sides by 20

k = ±2

Take the square root of both sides

Since vector d is parallel to vector c but points in the opposite direction, the scalar constant must be negative, k = –2. Therefore, d = − 2c: d= 2 =

4 2

Multiply vector c by the scalar constant

8 4

Evaluate

Reflect and check Use this formula to check if the vectors are parallel but point in the opposite direction: c ⋅ d = − ∣c∣∣d∣ Solving for the LHS: LHS = c ⋅ d =

4 2

Write the formula 8 4

Substitute the components of vectors c and d

= (–4) × 8 + (–2) × 4 Apply the scalar product using the component definition = –40

Evaluate

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Solving for the RHS: RHS = –∣c∣∣d∣

Write the formula

( 4 )2 ( 2)2

82 42 Substitute the magnitude of vectors c and d

Simplify the expression

= 1600

Multiply the surds

= –40

Evaluate

Since the LHS = RHS, vectors c and d are parallel and point in opposite directions.

Idea summary Vectors a and b are parallel if one is a scalar multiple of the other, meaning a = kb, where k is a scalar constant. If k > 0, the parallel vectors point in the same direction; if k < 0, the parallel vectors point in opposite directions. For parallel vectors pointing in the same direction: a ⋅ b = ∣a∣∣b∣ For parallel vectors pointing in the opposite directions: a ⋅ b = − ∣a∣∣b∣

Perpendicular vectors Two non-zero vectors are perpendicular (or orthogonal) if the angle between them is a right angle 90° or radians . 2

Interactive exploration Discover this concept in action online

mathspace.co

The scalar product provides a simple and powerful test for perpendicularity. Recall the geometric definition of the scalar product: a ⋅ b = ∣a∣∣b∣ cos(θ ) When two vectors are perpendicular, the angle between them is θ = 90°. Since the value of cos (90°) is 0, the entire expression becomes zero: a ⋅ b = ∣a∣∣b∣ × 0 = 0 Therefore, two non-zero vectors are perpendicular if and only if the scalar product is zero. This is the fundamental algebraic test for perpendicularity.

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Example 4 Are the vectors a = 6i + 4j and b = 2i − 3j perpendicular, parallel or neither?

Create a strategy To determine the relationship between two vectors, use the dot product for perpendicularity and the scalar multiples for parallelism.

Apply the idea First, calculate the dot product a ⋅ b:

a ⋅ b = 6 × 2 + 4 × (–3)

Write the equation

= 12 – 12

Evaluate the multiplication

Evaluate

The dot product is zero; therefore, the vectors are perpendicular.

Example 5 Two vectors are given by a = 3i − 4j and b = − 2i + yj, where y is a constant. Determine the value of y if a and b are perpendicular.

Create a strategy Use the component definition of the scalar product and equate to zero for perpendicular vectors. Then, solve for y.

Apply the idea a ⋅ b = a 1 × b1 + a 2 × b2 = 0 0 = 3 × (– 2) + (– 4) × y – 6 – 4y = 0

Substitute the values Evaluate the multiplication

– 4y = 6

Add 6 to both sides

6 y= 4

Write the formula and equate to zero

3 2

Divide both sides by – 4 Simplify

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Reflect and check Sketching the vectors can provide a visual verification of the perpendicularity. However, it is important to note that visual inspection alone may not be reliable for accurately determining right angles, as small deviations might not be easily distinguishable. 5 4 3 2 1 –5 –4 –3 –2 –1–1

b = –2i + yj –2 –3 –4 –5

x 1 2 3 4 5

a = 3i + 4j

Idea summary Two vectors a and b are perpendicular (meet at a right angle) if the scalar (dot) product equals zero: a⋅b=0

8.08 Practice questions What do you remember? 1

Determine whether each of these statements about the scalar (dot) product is true or false: a

The scalar product of two vectors results in another vector.

Two non-zero vectors are perpendicular if the scalar product is zero.

The scalar product of a vector with itself, a ⋅ a, is equal to the magnitude of the vector, ∣a∣.

If the angle between two vectors is obtuse (between 90° and 180°), their scalar product is negative.

Calculate the scalar product u ⋅ v for each pair of vectors: 3 4 5 2 a u= ,v= b u= ,v= 2 1 1 10 c

804

4 6 ,v= 0 3

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 1 ,v= 7 7

Calculate the scalar product a ⋅ b using the given geometric information: a

∣a∣ = 8, ∣b∣ = 5, and the angle between them is 60°.

∣a∣ = 10, ∣b∣ = 4, and the vectors are perpendicular.

∣a∣ = 3, ∣b∣ = 7, and the vectors are parallel and point in opposite directions.

Given a and b are two non-zero vectors with θ representing the angle between them. Determine the exact value of the scalar product a ⋅ b when: 2 2

∣a∣ = 5, ∣b∣ = 7, and cos ( )

∣a∣ = 4, ∣b∣ = 2, and θ = 30°

∣a∣ = 3, ∣b∣ = 3, and θ = 135°

The magnitudes of a and b are reciprocal and cos ( )

1 3

Practice 5

Determine the scalar product OA OB: a

B 5

60° B

B 8

130° O

30°

B 7

55°

35°

Consider the scalar product of two vectors OA OB: B

θ O

Determine, rounded to two decimal places, the value of: a

OA given that OB = 3, θ = 45°, and OA OB = 10

OB given that OA = 5, θ = 60°, and OA OB = 8

The angle θ, given that OA = 3, OB = 6, and OA OB = 6

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Ex 1

This diagram shows vectors OA, OB, OC, and OD where OA = 10, OB = 7, and OC = 11: A 10

α = 120° θ

β = 45°

C a

Determine the scalar product OA OB.

Determine the exact value of the magnitude of OD, given that OB OD

Determine the exact value of the scalar product b ⋅ c if: 2 0 2 1 a b= and c = b b= and c = 3 2 1 2

2 3

and c =

2 5

Consider the vectors a =

Ex 2

a b 10

35 2. Given that OA OC = 16, determine θ, the angle between OA and OC , rounded to two decimal places.

3 1

and c =

1 2

5 b 2 , b = 1 , and c = , where b1 is a real number: 2 3 2

Determine the scalar product a ⋅ c.

Determine the value of b1 given that a ⋅ b = 1.

Consider the position vectors OP

2 and OQ 5

4 : 1

Ex 3

Determine vector PQ .

Determine if vector w = Explain your reasoning.

Ex 4

18 is parallel to PQ and points in the opposite direction. 12

Determine a vector Z that has a magnitude of 6 13 and is parallel to PQ in the same direction.

Determine whether each pair of vectors are perpendicular using the scalar product: 1 3 2 4 a a= and b = b p= and q = 2 5 1 2

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Ex 5

Given vectors u = perpendicular.

k and v = 3

6 , determine the exact value of k if u and v are 4

The magnitude of a is 5 and the magnitude of b is 8. Calculate a ⋅ b if the angle between a and b is 90°.

Vector u has a magnitude of 6 and vector v has a magnitude of 3. Given u ⋅ v = 9 2, determine the angle between u and v.

Consider the vectors c = a b

1 3 and d = : 2 1

Calculate the dot product of c and d.

Calculate the angle between c and d.

Given a = kb where k is a scalar, calculate a ⋅ b.

Determine whether these statements are true or false, given ∣a∣ = 3 and ∣b∣ = 5: a

a ⋅ b ≥ 15

a ⋅ b = 15

a ⋅ b ≤ 15

Extend your thinking 18

Vector OB in the diagram shown can be expressed in column vector notation as OB y

X OB YOB

B x

θ = 30° O

Determine the exact values of X OB and YOB given that X OB 19

YOB

= 64.

b1 2 1 ,b= , and c = , where b1 and b2 are two real numbers. b2 1 2 Determine the values of b1 and b2 given that a ⋅ b = − 1 and b ⋅ c = − 4.

Consider the vectors a =

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Consider two vectors, OA and OB , of magnitudes 7 and 10, as shown in the diagram: y A

B 10

7 65°

35°

OC is the resultant vector of OA and OB .

Determine the value of OC OA OC OB, rounded to two decimal places. 21

Consider two vectors, OA and OB such that OA OB 3: A B 80° O OC is the resultant vector of OA and OB.

Given OC = 13.16 and the angle between OA and OB is θ = 80°, determine the value of OA OB . 22

Charlie and Sonja have gotten lost on a camping trip and are trying to retrace their steps to get back to their tent. From their tent they walked for 100 m northeast. They then turned right and walked for another 50 m. They then walked for another 80 m but cannot remember in which direction. Sonja thinks they walked directly northeast, but Charlie thinks they walked at a true bearing of 030°. A vector to represent their current position from the tent can be given by: OT

808

115 2, 65 2

Are either Charlie or Sonja correct? Explain your reasoning.

Calculate the angle between OT and their most recent walk.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

8.09 Scalar and vector projections After this lesson, you will be able to… • define scalar and vector projections • calculate projections using formulas • interpret the geometric meaning of projections • apply projections to solve problems

Scalar projections Scalar projection The scalar projection of vector a onto vector b is the signed length of the “shadow” that a casts on b. It is a scalar quantity. A scalar projection measures the length of the “shadow” that one vector casts onto another. It is a scalar value, not a vector, because it only represents magnitude (or length), not direction.

|u|

As shown in the diagram, the vector a, the line of b, and the perpendicular line from the tip of a form a right-angled triangle. The scalar projection is the length of the adjacent side, denoted compba. Therefore, the magnitude ∣u∣ shown in the diagram corresponds to the absolute value of the scalar projection. Using the basic trigonometric ratio for cosine: cos( )

Adjacent Hypotenuse

compba a

Rearranging this gives the formula for the scalar projection when the magnitude of a and the angle θ between the vectors are known (i.e., for vectors in polar form):

compba = |a| cos(θ ) compba |a| θ

is the scalar projection of a onto b

is the magnitude of vector a is the angle between the two vectors

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To find a formula for vectors in Cartesian form, the geometric definition of the dot product, a ⋅ b = |a||b| cos(θ ), can be used. Rearranging this formula to isolate the |a| cos(θ ) term yields: a cos( ) =

a⋅b b

This provides the equivalent formula for the scalar projection in Cartesian form:

compba

a⋅b b

compba is the scalar projection of a onto b

a⋅b |b|

is the dot product of the two vectors is the magnitude of the vector being projected onto

Note: The scalar projection can be negative if the angle θ is obtuse (90° < θ ≤ 180°), as cos (θ ) is negative in this range. A negative result indicates the projection points in the direction opposite to b.

Example 1 Let g = 3i – 7j and h = − i + 4j. Determine the scalar projection of g on h.

Create a strategy Use the formula: comphg

g⋅h h

. First, find the dot product g ⋅ h, then find the magnitude |h|, and

finally divide the two results.

Apply the idea First, calculate the dot product of g and h: 3 1 7 4 = (3)(–1) + (–7)(4)

g⋅h=

Substitute the vectors Apply the dot product formula

= –3 – 28

Evaluate the subtraction

= –31

Evaluate

Next, calculate the magnitude of h: |h| = ( 1)2 42

Apply the magnitude formula

= 1 16

Evaluate the squares

= 17

Evaluate

Finally, calculate the scalar projection: comphg = =

g⋅h h 31 17

Substitute the values

31 17 17

Rationalise the denominator

810

Write the formula

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 Consider the vectors a and b in polar form: a

and b 6

Determine the scalar projection of b on a.

2, 3

Create a strategy To find the scalar projection of b on a, use the formula: compab = |b| cos(θ )

Apply the idea Calculate the angle between the vectors:

θ=

Subtract the angles Evaluate

Calculate the scalar projection: Scalar projection = |b| cos(θ ) = 2 cos =0

Write the formula using the appropriate vector labels Substitute the values Evaluate

Reflect and check A scalar projection of 0 means the vectors are perpendicular and neither vector b has any component in the direction of vector a, nor does a have any component in the direction of b.

Idea summary A scalar projection is the signed magnitude, or length, of the projection of one vector onto another. The formulas for the scalar projection of a onto b are: Polar form: compba a cos ( ) Cartesian form: compba

a⋅b b

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Vector projections Vector projection The vector projection of vector a onto vector b is a vector representing the “shadow” of a on b. It has a magnitude equal to the scalar projection and the same direction as b. Example: Formula (Cartesian): projba

a⋅b b b⋅b

Recall that the projection of one vector onto another can be visualised as a ‘shadow’. While the scalar projection measures the length of this shadow, the vector projection is a vector that represents the shadow itself, possessing both magnitude and direction.

θ u

In the diagram, vector u is the shadow of vector a on vector b. First, the length of this shadow is found, called the scalar projection, and then it is used to find the shadow vector, called the vector projection. The vector projection of a onto b, denoted projba, is a vector. It has the magnitude of the scalar projection and the same direction as b.

b b Therefore, the vector projection is the scalar projection (magnitude) multiplied by the unit vector (direction): The direction of b is given by its unit vector:

a⋅b b b projba (compba) b b b Simplifying this gives the formula for vector projection in Cartesian form. Since |b|2 = b ⋅ b, the result is: a⋅b a⋅b b projba 2 b b b⋅b

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Did you know?

Scalar and vector projections are important in biomechanics! They help analyse muscle forces acting on joints by projecting forces along bones. This allows researchers to better understand movement and reduce injury risks.

Example 3 Let a = 5i + 4j and b = –2i + 3j. Determine the vector projection of: a a on b

Create a strategy Use the formula:

a⋅b b b⋅b

Apply the idea Vector projection =

a⋅b b b⋅b

5 4 = 2 3 2 2 = 13 3 4 13 = 6 13 4 6 The vector projection is i j. 13 13

Write the formula 2 3 2 3

2 3

Substitute the values

Evaluate the scalar products in the numerator and denominator Perform the scalar multiplication

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b b on a

Create a strategy Use the formula:

b⋅a a a⋅a

Apply the idea Vector projection =

b⋅a a a⋅a

2 3 = 5 4

5 4 5 4

Write the formula

5 4

Substitute the values

2 5 = Evaluate the dot products in the numerator 41 4 and denominator 10 41 = Perform the scalar multiplication 8 41 The vector projection is

10 8 i j. 41 41

Reflect and check It is important to check which vector is being projected onto which, as the formula will change accordingly.

Example 4 5, and d 7, 6 4 Determine the vector projection of c on d in polar form. Consider the vectors c and d in polar form: c

Create a strategy The vector projection is found by multiplying the scalar projection by the unit vector of the direction. 1. Find the scalar projection of c onto d, using the formula compdc = |c| cos(θ ). d 2. Find the unit vector in the direction of d, which is d d 3. Multiply the results: projdc = (|c| cos(θ )) d

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Apply the idea First, calculate the angle θ between the vectors: 6 4 5 = 12 Next, calculate the scalar projection of c onto d:

θ=

compdc = |c| cos(θ )

= 5

Evaluate Write the formula

5 12

= 5 cos

Subtract the angles

Substitute the values

Evaluate the cosine

5 6 5 2 = 4

Simplify

≈ 0.9412 The vector projection is a vector that has: • A magnitude equal to the scalar projection, which was calculated as 5 6 5 2 . 4 • A direction equal to the direction of the vector it is projected onto, d, which is . 4 Since the polar form of a vector is defined as (magnitude, direction), the vector projection can be written directly by combining these two results: projdc

5 6 5 2 , 4 4

Idea summary The vector projection of vector a onto vector b is a vector that represents the “shadow” of a on b. The formula for the vector projection of a onto b is:

projba

a⋅b b b⋅b

This formula can be derived by multiplying the scalar projection (the length of the shadow) by the unit vector for direction.

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8.09 Practice questions What do you remember? 1

Given the vectors a = 2i + 5j and b = –3i + 4j. Calculate the dot product, a ⋅ b.

Calculate the magnitude of the vector v = 7i – 2j.

Determine the unit vector in the direction of w = –6i + 8j.

Consider the vectors p 4, and q 3, . Calculate the dot product, p ⋅ q. 3 6

Given the vectors r a

5 1 and s . Determine: 3 6 r + s b s–r

Practice Ex 1

Let m = 5i + 2j and n = –3i + 6j. Determine the scalar projection of: a

n on m

q on r

p on q

If a ⋅ b = 5 and b = 4i – 2j, determine the scalar projection of a on b.

Consider the vectors m and n in polar form: m 5, and n 3, . Determine the 2 4 scalar projection of: a

n on m

m on n

Consider the vectors x 7, , y 3, , and z 2, . Determine the scalar projection of: 6 3 2 a y on z b x on z Consider the vectors x 5, , y 3, , and z = 2i – 4j. Determine the scalar 4 3 projection of: a

Ex 3

Consider the vectors p = 6i + 3j, q = –4i + 7j and r = –5i + 2j. Determine the scalar projection of: a

Ex 2

m on n

x on y

y on z

Given the vectors a = 3i + 7j and b = 2i + 1 j, determine the vector projection of: a a on b b b on a

Given the vectors u = 5i –4j, v = –2i + 5j and w = –3i + j, determine the vector projection of: a

816

v on w

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

u on v

Ex 4

1 u and v are two vectors such that u ⋅ v = 10 and u . Calculate the vector projection of v 4 onto u.

Given the vectors p and q in polar form: p 6, and q 5, , determine the vector 4 3 projection of: a

p on q

q on p

Consider the vectors p 6, , q 4, , and r 5, . Determine the vector 4 3 2 projection of: a

q on r

p on q

Consider the vectors a 4i 3j, b , and c –5i 2j. Determine the vector projection of: 6 a

a on b

b on c

For the given vectors, calculate the vector projection of vector r onto vector b:

y 4 2 r

−4

b 2

−2

x 4

−2 −4

Extend your thinking 19

The vector projection of x

a on y 3

The scalar projection of a

x 4 on b is 10 5. Determine the value of the constant x. 10 2

Consider the vectors p = 8i + 6j and q = 5i – 3j:

3 6 is . Determine the value of the constant a. 1 2

Determine a vector r that is parallel and in the same direction as q and has a magnitude of 10.

Determine the scalar projection of i on r.

Two vectors are given by a = 7i – 3j and b = ki + 5j. If the angle between a and b is , 4 determine the possible vector projections of a on b. Show how the definition of a scalar product is related to scalar projection and explain the difference between them.

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8.10 Displacement and velocity After this lesson, you will be able to… • model displacement and velocity using vectors • solve problems involving displacement and velocity

Displacement Kinematics is the study of motion without considering the forces that cause it. In kinematics, the focus is on how objects move, how their positions change, and their speeds. Vectors are useful for mapping motion and analysing the movement of objects. Displacement can be represented as a vector because it has both magnitude and direction. A displacement vector represents the straight-line distance from an object’s starting position to its final position, taking direction into account. This differs from the distance travelled, which measures the entire path an object follows and is a scalar quantity, considering only magnitude. The magnitude of a displacement vector represents the straight-line distance from an object’s starting position to its final position. When referring to this straight-line distance in problems, it is equal to the magnitude of the resultant displacement. The letter commonly used to represent displacement is s or x. The variable d is often used to represent the magnitude of a displacement vector, which corresponds to the straight-line distance between the start and end points.

Example 1 A drone flies 12 km at an angle of 60° anticlockwise from the positive x-axis (East). It then changes direction and flies for an additional 10 km at an angle of 200° anticlockwise from the positive x-axis.

10 8

y (North) s2

6 sR

4 2

−2

x (East) 2

The drone’s journey is modelled by adding two displacement vectors, s1 and s2, head-to-tail. The resultant displacement is the vector sR, which represents the drone’s final position relative to its start.

−2

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Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

a Determine the final displacement of the drone from its starting point as a vector in component form.

Create a strategy First, each part of the drone’s flight is modelled as a displacement vector. Since the magnitude and direction angle are given for each leg, each vector is resolved into its x- and y-components using trigonometry (x = r cos (θ ), y = r sin (θ ). Next, the corresponding components of the two displacement vectors are added to find the resultant displacement vector, sR.

Apply the idea Resolve each displacement vector into components. First leg (s1): Magnitude = 12 km, Angle = 60°. The components are found using x = r cos (θ ) and y = r sin (θ ): s1 =

12 cos (60° ) 12 sin (60° ) 12

1 2

3 2

Write the equation

Substitute the exact values for the trigonometric functions

Evaluate

6 3

Second leg (s2): Magnitude = 10 km, Angle = 200°. The components are kept in their exact trigonometric form: s2 =

10 cos (200° ) 10 sin (200° )

Calculate the resultant displacement vector sR by adding the component vectors: sR = s1 + s2 = =

6 6 3

Write the equation 10 cos (200° ) 10 sin (200° )

6 10 cos (200° ) 6 3 10 sin (200° )

ubstitute the exact component S forms of the vectors Combine the corresponding components

≈

6 9.397 10.392 3.420

valuate the trigonometric E expressions using a calculator

≈

3.40 6.97

Calculate the final components, rounded to two decimal places

The final displacement of the drone is approximately

3.40 km. 6.97

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b How far is the drone from its starting point, rounded to two decimal places?

Create a strategy The distance from the starting point is the magnitude of the resultant vector, sR. This is calculated using Pythagoras’ theorem on the components of the resultant vector found in the previous part.

Apply the idea The components of the resultant vector are x ≈ –3.40 and y ≈ 6.97 Calculate the magnitude of the resultant vector sR: ∣sR∣ = x2 y2

Write the magnitude formula 2

≈ ( 3.40)

(6.97 ) Substitute the components of the resultant vector ≈ 11.56 48.58 Evaluate the squares ≈ 60.14

Evaluate the addition

≈ 7.75 km

Evaluate the square root

The drone is approximately 7.75 km from its starting point.

Idea summary Kinematics is the study of motion without considering the forces that cause it. Vectors are useful for mapping motion and analysing the movement of objects. Displacement can be represented as a vector because it has both magnitude and direction. A displacement vector represents the straight-line distance from an object’s starting position to its final position, taking direction into account. This differs from the distance travelled, which is a scalar quantity, considering only magnitude.

Velocity Velocity is the rate at which displacement changes with time. It is also a vector quantity as it has both magnitude and direction. The average velocity over some interval is the total displacement during that interval, divided by the time. The instantaneous velocity at some moment in time is the velocity of the object at that specific time. In what follows, only average velocity is considered. Speed is the rate at which distance changes with time. It is a scalar quantity, as it only considers magnitude and not direction. Speed is defined as the magnitude of the velocity vector, Speed = ∣Velocity∣ The relationship between displacement and velocity is given by s t where v is the velocity, s is the displacement and t is time. v=

This relationship can be rearranged to determine time or displacement for contextual problems.

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Example 2 Kenneth wants to fly his plane directly north with a true ground speed of 410 km/h. If an 80 km/h wind is blowing from the northwest (towards the southeast), what air velocity should he fly his plane to maintain his northward path? Express your answer in terms of the unit vectors i and j, where north corresponds to the positive y-direction.

Create a strategy Establish a coordinate system, where i represents the positive east direction and j represents the positive north direction. Define the desired ground velocity of the plane (vground). Then, determine the velocity vector of the wind (vwind). Since the wind is blowing “from the northwest (towards the southeast)”, resolve the wind velocity into its x- and y-components. Use the vector equation vground = vair + vwind to find the required air velocity.

Apply the idea Let north be the positive y direction ( j) and east be the positive x direction (i). 1. Define known velocity vectors: • Desired ground velocity of the plane (vground): Since the plane flies directly north at 410 km/h: vground = 410j • Wind velocity (vwind): The wind blows at 80 km/h “from the northwest (towards the southeast)”. Southeast corresponds to an angle of –45° (or 315°) from the positive x-axis: vwind = (80 cos (−45°))i + (80 sin (−45°)) j Write the equation = 80

1 2

i + 80

j Substitute cos ( 45° ) sin ( 45° )

= 40 2 i − 40 2 j 2.

1 2

and

Simplify

Apply vector equation: The relationship between ground velocity, air velocity, and wind velocity is vground = vair + vwind. Rearranging to find the air velocity: vair = vground − vwind

Calculate air velocity (vair): vair = 410j − 40 2 i 40 2 j

Substitute the vectors

= 410j − 40 2i + 40 2j

Distribute the negative sign

= 40 2i + 410 40 2 j

Combine like components

≈ − 56.57i + (410 + 56.57) j

Approximate 40 2 56.57

≈ − 56.57i + 466.57j

Evaluate

Kenneth should fly his plane with an air velocity of approximately −56.57i + 466.57j to maintain a northward path.

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Example 3 A boat is attempting to travel across a river. The boat’s engine propels it with a velocity of 10 m/s at an angle of 70° from the positive x-axis (representing East). The river has a current with a velocity of 5 m/s at an angle of 340° from the positive x-axis: a Sketch a labelled diagram showing the vectors for the boat’s velocity relative to the water (vb), the velocity of the water’s current (vw), and the resulting true velocity of the boat relative to the ground (vg).

Create a strategy

Apply the idea

A vector diagram is drawn starting from a common origin. The boat’s velocity (vb) and the water’s velocity (vw) are added using the parallelogram rule. The resultant vector sum, vg = vb + vw, represents the true velocity of the boat.

y (m/s)

8 6 340°

70°

−2

vg x (m/s) 4

b Determine the true speed and direction of the boat. Round the answers to two decimal places.

Create a strategy First, each velocity vector is resolved into its horizontal (x) and vertical (y) components using trigonometry. The components of the two vectors are then added to find the components of the resultant velocity vector (vg). The magnitude of vg gives the true speed, and its angle gives the true direction.

Apply the idea 1.

Resolve the velocities into components: • Boat’s velocity (vb): 10 cos ( 70° ) 10 sin ( 70° )

Write the equation

3.42 9.40

Evaluate

5 cos ( 340° ) 5 sin ( 340° )

Write the equation

4.70 1.71

Evaluate

vb = ≈ • Water’s velocity (vw): vw = ≈

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Calculate the true velocity of the boat (vg): vg = vb + vw

≈

3.42 9.40

≈

8.12 7.69

Write the formula 4.70 1.71

Substitute the values Evaluate

Determine the true speed (magnitude of vg): ∣vg∣ ≈ (8.12)2 ( 7.69)2 Write the equation ≈ 125.08

Evaluate the squares then the addition

≈ 11.18 m/s Evaluate The true speed of the boat is approximately 11.18 m/s. 4. Determine the true direction (angle of vg):

θ = arctan

y x

Write the formula

≈ arctan

7.69 8.12

Substitute the values

≈ 43.44°

Evaluate

The true direction of the boat is approximately 43.44° from the positive x-axis. Therefore, the true speed and direction of the boat are 11.18 m/s and 43.44°.

Idea summary Velocity is the rate at which displacement changes with time. It is also a vector quantity as it has both magnitude and direction. Speed is the rate at which distance changes with time. It is a scalar quantity, as it only considers magnitude and not direction. Speed is defined as the magnitude of the velocity vector, Speed = ∣Velocity∣ The relationship between displacement and velocity is given by v=

s t

where v is the velocity, s is the displacement, and t is time.

8.10 Displacement and velocity mathspace.co

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8.10 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

Speed is the magnitude of the velocity vector.

If a person walks 5 km east and then 5 km west, their final displacement is 10 km.

The resultant displacement of multiple movements is found by adding the individual displacement vectors.

The vector 10j represents a movement of 10 units due east.

A velocity vector v has a magnitude of 20 m/s and a direction of 60° anticlockwise from the positive x-axis. Calculate the exact components of the velocity vector: a

Horizontal component (vx)

Vertical component (vy)

An object undergoes two displacements. The first is s1 =

5 0 m and the second is s2 = m: 0 12

Find the resultant displacement vector, sR.

What is the total distance from the starting point to the final point?

Practice

Ex 1

A ship sails 40 km due north and then changes direction and sails 60 km due east. What is its final displacement from its starting point? Round the answer to one decimal place.

A delivery truck travels 30 km north, then 20 km east, then 30 km south. What is the truck’s displacement from its starting point?

A car travels 50 km at an angle of 60° east of north, then turns and travels 30 km south. What is the car’s displacement from its starting point?

A boat travels from its starting point in two stages, as shown in the diagram:

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• First, it travels 20 km at an angle of 45° to the positive x-axis (s1). • Then, it travels 15 km at an angle of 180° to the positive x-axis (s2).

Determine the final displacement of the boat from its starting point, sR, as a vector in component form. Give your components as exact values.

How far is the boat from its starting point? Round your answer to one decimal place.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

y (km) s2

12 8 6 sR 2 −4 −2

x (km) 2 4 6 8 10 12 14

A hiker undergoes two successive displacements: • The first displacement, s1, has a magnitude of 8 km at an angle of 45° to the positive x-axis. • The second displacement, s2, has a magnitude of 6 km at an angle of 160° to the positive x-axis.

Ex 2

Find the resultant displacement vector, sR = s1 + s2, in component form. Round your components to two decimal places.

How far is the hiker from their starting point? Round your answer to two decimal places.

What vector represents the path the hiker must take to return directly to their starting point?

A drone undergoes two successive displacements. The first is d1 = 6i + 4j and the second is d2 = −2i − 5j: a

Determine the drone’s final displacement vector, dR, from its starting point.

Calculate the total distance from the starting point to the final point (i.e., the magnitude of the final displacement vector). Leave your answer in exact form.

Find the direction of the final displacement, as an angle measured anticlockwise from the positive x-axis. Round your answer to the nearest degree.

A boat is moving at 9 m/s north across with respect to the ground that is flowing at 6 m/s to the east. What is the boat’s resultant velocity with respect to the river? Round the answer to one decimal place.

A pilot wants to fly their plane directly east. The plane’s true velocity relative to the ground is 350 km/h. A wind is blowing from the northeast (towards the southwest) at 60 km/h. What is the required velocity of the plane relative to the air (vair) for the pilot to maintain their eastward course? Express your answer in component form vair = xi + yj, where i is east and j is north. Round components to two decimal places.

A rocket is launched at a speed of 20 000 km/h at an angle of 75° above the horizontal. If the rocket maintains this speed, what will be its altitude 2 hours after lift-off?

A particle moves with a velocity vector of v = 6 2 i 6 2 j km/h:

Determine the speed of the particle.

Determine the direction of the particle’s motion, as an angle measured anticlockwise from the positive x-axis.

A drone undergoes three successive displacements starting from point P. The angles are measured anticlockwise from the positive x-axis (East): • First, it travels 125 m at an angle of 120° to point Q. • Then, it travels 40 m at an angle of 30° to point R. • Finally, it travels 78 m at an angle of 300° (or –60°) to point S. Calculate the final displacement vector from P to S, PS , and express it in polar form (r, θ ). Round your final answers to two decimal places.

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Ex 3

A swimmer wants to swim 100 m north along a river starting from its riverbank. The swimmer can swim at a speed of 2 m/s and the river flows at 1 m/s east: a

At what acute angle should the swimmer aim relative to the bank?

How long will it take the swimmer to cross the river?

A small plane that has a cruising speed of 241 km/h is planning to fly from airport X to airport Y, where X Y = 324i − 486j km and a steady wind of 24i − 7j km/h is forecast for the region: a

Determine the velocity vector of the plane relative to the air that is required to reach airport Y.

Calculate the expected flight time.

An aeroplane is flying with an air velocity (vair) of 250 km/h at an angle of 120° from the positive x-axis (East). The wind has a velocity (vwind) of 40 km/h at an angle of 210° from the positive x-axis: a

Sketch a labelled diagram showing the vectors for the plane’s air velocity (vair), the wind’s velocity (vwind), and the resulting true velocity of the plane relative to the ground (vground).

Determine the true speed and direction of the airplane. Round answers to two decimal places.

Extend your thinking 18

An aeroplane is flying at a speed of 300 km/h in still air. If there is a wind blowing from the northwest at 50 km/h, at what bearing should the pilot set to maintain a course directly east?

A small boat has a cruising speed of 14 km/h in still water. The boat leaves point A at 6:30 a.m. and travels to point B, 6 km due west of A, where it turns and travels to point C, 5.2 km due north of B. The boat then returns to A. A current of 3.2 km/h runs in a westerly direction throughout the area. Determine the time at which the boat is expected to return to A.

In a sailing race, two boats start from the same point and sail to a buoy located 2 km east and 3 km north of the starting point. Boat A sails directly to the buoy while Boat B sails first north for 1 km and then directly to the buoy. If Boat A travels at 30 km/h and Boat B travels at 40 km/h, which boat will reach the buoy first and by how many minutes?

Point A and B lie on opposite sides of a river so that B is 320 m away from A on a bearing of 290°. A uniform current flows due south in the river between A and B at 0.35 m/s. Riley can swim at a steady speed of 1.4 m/s and plans to swim from A to B and then back to A. Determine the time it will take to complete the swim. Hint: A bearing is an angle measured clockwise from the north direction.

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8.11 Force equilibrium and work done After this lesson, you will be able to… • analyse equilibrium situations using vectors • solve problems involving inclined planes • apply scalar product to calculate work done • model complex force problems

Force and equilibrium Force A push or pull that can change an object’s motion. It is a vector quantity, meaning it has both magnitude and direction. Example: The force of gravity pulling an object downwards is called its weight. Force is typically measured in Newtons (N). The standard international unit of force is measured in Newton. One Newton is defined as the force required to accelerate a 1 kg mass at 1 m/s2. F = 1 kg

1 m s2

kg m

There are many different types of forces, some common ones are: • Weight: An object near the surface of the Earth is subject to the force of gravity. This force is referred to as weight. Weight acts vertically downwards. Weight is often calculated by multiplying the object’s mass by the gravitational field strength, which is approximately 9.8 m/s2 on Earth. However, in some problems, the weight might be provided directly; therefore, multiplying by 9.8 is not always necessary. Weight is typically measured in Newtons (N), but it may also be expressed in kilogram-force (kg wt), where 1 kg wt = 9.8 N. • Normal: Any object with mass placed on a surface, whether horizontal or inclined, experiences a force perpendicular (or normal) to the surface. This is the force exerted by the surface to support the object and prevent it from sinking into or passing through the surface: N

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• Tension: Tension is the pulling force exerted by a rope, string, cable, or similar object when it is pulled tight. It acts along the length of the object and is directed away from the object, trying to pull it apart:

WN When there are multiple forces acting on a body or object at once, their combined effect is called the resultant force. The resultant force is the vector sum of all the forces acting on the object. If the resultant force acting on a body is zero, the body is said to be in equilibrium. The key principle for solving equilibrium problems is that if the overall resultant vector is zero, then the sum of its individual components must also be zero. This principle leads to two separate equations: • The sum of all horizontal components of the forces must be zero. • The sum of all vertical components of the forces must be zero. By resolving each force into its horizontal (i) and vertical ( j) components, two equations can be set up and solved to find any unknown forces. Often, a single force acts at an angle. To analyse its effect and solve problems involving multiple forces, it is useful to break this force down into two perpendicular parts called components. Typically, these components are aligned with the horizontal and vertical axes. 6 5 4 3 2 1 0

4 Fx 6

The force vector F forms the hypotenuse of a right-angled triangle. Its components can be found using trigonometry: • The horizontal component is Fx = ∣F∣ cos (θ ). • The vertical component is Fy = ∣F∣ sin (θ ).

In some problems, such as those involving inclined planes, it is more useful to resolve forces into components that are parallel and perpendicular to the surface, rather than horizontal and vertical. The same trigonometric principles apply, but the coordinate system is rotated to align with the slope.

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Example 1 A still object with a weight of 8 kg wt is suspended by two cables attached to two points on the horizontal ceiling. The two strings make angles of 50° and 25° to the horizontal: a Draw a diagram of the situation, labelling the tension forces in the cables as T1 and T2.

Create a strategy Draw each force with arrows to show the direction of the force, labelling magnitudes and angles.

Apply the idea 25°

50°

T1 kg wt

T2 kg wt

8 kg wt

b Calculate the tension force in each cable, rounded to two decimal places. Express your answer in polar form using force units of kg wt.

Create a strategy Since the forces are in equilibrium, represent the forces as a triangle (vector sum is zero) and use trigonometry to calculate the magnitude and direction of the two tension forces.

Apply the idea Represent the force vectors as triangle: 1. Connect the end of tension force, T1, to the start of the weight force. The angle between the weight and T1 is 90° – 25° = 65°. 2. Connect the end of the weight force to the start of the tension force T2. The angle between the weight and T2 is 90° – 50° = 40°. The vectors form a triangle. Therefore, the angle between T1 and T2 can be calculated by 180° – 65° – 40° = 75°. The forces can be resolved into the i- and j components where the sum of all i-components is zero and the sum of all j-components is zero.

T1 65° 75°

40°

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The resolution can be determined in both i- and j-components: 25°

Find the resolution in both i- and j-components.

50°

T1 kg wt

T2 kg wt

Resolution in i- direction:

−T1 cos (25°) = T2 cos (50°) T2 =

cos (50°)

Resolution in j-direction:

8 kg wt

T1 sin (25°) + T2 sin (50°) − 8 = 0

T1 kg wt

T2 kg wt 50°

25°

T1 cos (25°)

8 kg wt

T2 =

T1 sin (25°) sin(50°)

Equating 1 and 2 : T1 cos (25°)

cos (50°) Using a Classpad or calculator:

T1 sin (25°) sin (50°)

T1 = 5.32 kg wt T2 = 7.51 kg wt

Reflect and check Alternatively, the forces can be solved by applying the sine rule to solve for T1, where a is T1, A is the opposite angle, 40°, b is 8 kg wt and B is the opposite angle, 75°: a b = sin ( A ) sin ( B) 8 T1 = sin ( 40° ) sin ( 75° ) T1 =

8 sin ( 40° ) sin ( 75° )

= 5.32 kg wt

Write the formula Substitute the values Multiply both sides by sin (40°) Evaluate using technology

Apply the sine rule to solve for T2, where a is T2, A is the opposite angle, 65°, b is 8 kg wt and B is 75°: b a = sin ( A ) sin ( B) T2 8 = sin (65°) sin ( 75° ) T2 =

8 sin (65° ) sin ( 75° )

= 7.51 kg wt

Write the formula Substitute the values Multiply both sides by sin (65°) Evaluate using technology

Expressing both vectors in polar form: T1 = (5.32 kg wt, 155°) and T2 = (7.51 kg wt, 50°)

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Example 2 A 30 N box attached to a rope is about to be pulled up a smooth slope on an angle of 20°: N T 20° 30 N Determine the tension in the rope just before the box starts to move and the normal force acting on the box due to the slope, rounded to two decimal places.

Create a strategy Since the box is in equilibrium (about to move), the vector sum of all forces is zero. The most effective strategy is to resolve all forces into components that are parallel and perpendicular to the slope. 1. Define a coordinate system with the x-axis parallel to the slope and the y-axis perpendicular to it. 2. Resolve the weight vector into its components in this new system. 3. Set the sum of forces in the parallel direction to zero to find the tension (T ). 4. Set the sum of forces in the perpendicular direction to zero to find the normal force (N ).

Apply the idea First, the weight vector (30 N) must be resolved into components parallel and perpendicular to the slope. The component of weight parallel to the slope (acting downwards along the slope) is 30 sin (20°). The component of weight perpendicular to the slope (acting into the slope) is 30 cos (20°). Now, apply the equilibrium conditions. 1. Forces parallel to the slope: The tension T acts up the slope, and the parallel component of weight acts down the slope. Since the forces are balanced: T – 30 sin (20°) = 0

Write the equation

T = 30 sin (20°)

Add 30 sin(20°) to both sides

≈ 10.26 N

Evaluate using technology

2. Forces perpendicular to the slope: The normal force N acts outwards from the slope, and the perpendicular component of weight acts into the slope. Since these forces are balanced: N – 30 cos (20°) = 0 N = 30 cos (20°) ≈ 28.19 N

Write the equation Add 30 cos(20°) to both sides Evaluate using technology

The tension in the rope is approximately 10.26 N, and the normal force is approximately 28.19 N.

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Idea summary A force is defined as any push or pull that can change an object’s motion. A force is a vector because it has a magnitude and direction. The units of force are newton (N) and N (kilogram weight). Some common forces include: • Weight: The force of gravity acting on an object near Earth’s surface, directed vertically downward. • Normal: The perpendicular force exerted by a surface to support an object and prevent it from sinking. • Tension: The pulling force exerted by a rope, string or similar object. When multiple forces act on an object, their combined effect is called the resultant force, which is the vector sum of all the forces. If the resultant force is zero, the object is in equilibrium, meaning it will remain at rest or move with constant velocity. Equilibrium problems are often solved by resolving forces into horizontal and vertical components.

Work done Work In physics, work is the energy transferred to or from an object when a force is applied over a certain displacement. It is a scalar quantity, calculated as the dot product of the force and displacement vectors (W = F ⋅ x). Example: The energy used to push a box across a floor is the work done on the box. Work is measured in Joules (J). Energy or work done is measured in Joules (J) where 1 J = 1 Newton-metre. When a constant force is applied to an object, the work done on the object is calculated using the formula:

W=F⋅x W F x

is the work done is the force is the displacement vector

Since the definition of a scalar product is a ⋅ b = ∣a∣∣b∣ cos (θ ), the work done can also be expressed as:

∣F∣∣ ∣x∣∣ cos (θ )

832

W = F ⋅ x = ∣F∣∣x∣ cos (θ ) is the magnitude of the force vector is the magnitude of the displacement (distance) is the angle between the force and displacement

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Note that if the force and displacement are both in the same direction, cos (θ ) = 1 and therefore: W = ∣F∣∣x∣ Work is positive when the force and displacement are in the same direction (0° ≤ θ < 90°), indicating that energy is transferred to the object. Work is negative when the force acts in the opposite direction of displacement (90°< θ ≤ 180°), suggesting energy is used to overcome a force, such as friction. Zero work occurs when there is no displacement or when the force is perpendicular to the displacement.

Example 3 A person pushes a box along a flat surface with a constant force of 50 N for a distance of 8 m. Calculate the work done by the person in moving the box.

Create a strategy The formula for work done, W, when the force and displacement are in the same direction is: W = ∣F∣∣x∣

Apply the idea W = ∣F∣∣x∣

Write the formula

= 50 × 8

Substitute the values

= 400 J

Evaluate

Example 4 How much energy is gained by a mass of 40 kg when it rolls 75 metres down a slope at 12° to the horizontal due to gravity?

75 m 12°

Note that force exerted by gravity is given by F (in Newtons) = Mass (kg) × 9.8 m/s2 acting downwards. Round your answer to two decimal places.

Create a strategy Calculate the force due to gravity and determine the angle between the force and displacement, then use the formula: W = ∣F∣∣x∣ cos (θ )

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Apply the idea The force due to gravity can be calculated using the given formula: F = Mass (kg) × 9.8 m/s2

Write the formula

= 40 × 9.8

Substitute the values

= 392 N

Evaluate

The angle between the force and displacement can be calculated using the fact that angles in a triangle add to 180°: 180° – 90° – 12° = 78° Apply the work done formula: W = ∣F∣∣x∣ cos (θ )

Write the formula

= 392 × 75 × cos (78°)

Substitute the values

= 6112.60 J

Evaluate

Idea summary Work done is defined as the energy transferred by a force acting over a distance. Energy or work done is measured in Joules (J). The formula for work done is:

W = F ⋅ x = ∣F∣∣x∣ cos (θ ) W F x ∣F∣ ∣x∣ cos (θ )

is the work done is the force is the displacement vector is the magnitude of the force vector is the magnitude of the displacement (distance) is the angle between the force and displacement

8.11 Practice questions What do you remember? 1

834

Determine whether each statement is true or false: a

Force is a scalar quantity because it only has magnitude.

An object is in equilibrium if the resultant (net) force acting on it is the zero vector.

Work is a vector quantity, measured in Newtons.

If a force is applied perpendicular to the direction of motion, the work done is zero.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Match each type of force with its correct description: a

Weight

The pulling force exerted by a rope or string.

Normal force

Tension

The force of gravity acting vertically downwards.

iii

The support force exerted by a surface, perpendicular to that surface.

Two forces, F1 =

5 N and F2 = 3

8 N, act on an object. 1

A third force, F3, is applied to keep the object in equilibrium. What is the vector F3?

Practice 4

Consider these forces in equilibrium: • A force of a N acting at an angle of 30° east of the north direction. • A force of 100 N acting vertically downwards. • A force of b N acting 150° anticlockwise from the positive x-axis.

Draw a diagram to represent the situation.

Represent as a triangle of forces and include the internal angles of the triangle.

Determine a.

Determine b.

Two forces, u and w, act on an object, where u = (300 N, 70°) and w = (550 N, 160°): a

Draw a diagram of the situation.

Determine the magnitude of the resultant of the two forces, rounded to one decimal place.

Determine the direction of the resultant force, rounded to two decimal places.

6 An object weighing 35 N is suspended from two ropes of length 13 cm and 15 cm. The end of the ropes are attached to two points on the ceiling 18 cm apart.

18 cm T1 13 cm

T2 15 cm

etermine the tension in each string, rounded to two D decimal places.

35 N Ex 1

7 A block weighing 12 N hangs from two ropes attached to the ceiling. The strings both make angles of 57° with the horizontal: a

Draw a diagram to represent the situation.

Determine the tension in each string, rounded to two decimal places.

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8 A small boat is being pulled by a cable with a tension force of 6000 N towards the west. The current in the water is applying a force of 300 N on the boat towards the northeast direction:

Ex 2

Draw a diagram to represent the situation.

Determine the magnitude and direction of a third force that must act on the boat if it is to remain still. Round your answer to one decimal place, when applicable. N

A 27 N box attached to a rope is about to be pulled up a smooth slope on an angle of 15°: Determine the tension in the rope just before the box starts to move, rounded to two decimal places.

15° 27 N 10 A mass of 15 N is kept stationary on a smooth inclined plane of 35° by a force acting at an angle of 10° to the plane: a

Determine the magnitude of the force in Newtons, rounded to two decimal places.

Determine the magnitude of the normal force in Newtons, rounded to two decimal places.

F 10°

35° 15 N

11 An object weighing 24 kg wt is maintained at rest on a smooth inclined plane by a rope. Determine the tension in the rope, in kg wt, rounded to two decimal places, if: a

The plane is at 20° to the horizontal and the rope is parallel to the plane

The plane is at 5° to the horizontal and the rope is held at an angle of 15° to the plane

12 Two forces, a and b, have equal magnitudes. Force a acts at an angle of 20° and force b acts at an angle of –30°. The angles are measured anticlockwise from the positive x-axis. I f a third force, c, keeps the system in equilibrium, calculate the direction of c as a positive angle measured anticlockwise from the positive x-axis. Ex 3

13 A 5 N force is applied to push a chair 1.5 m across the room along a friction free surface. Calculate the total work done, rounded to one decimal place. 14 A 56 kg person is going skydiving from a plane 3 km above the ground. Calculate the work done by gravity to cause the person to reach the ground from the plane, assuming the acceleration due to gravity is 9.8 m/s2 (Hint: Force = Mass × Acceleration). 15 Calculate the work done on an object when a force of 300 N with direction of 20° moves the object 4 m in the horizontal direction, rounded to two decimal places.

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Ex 4

16 A 100 kg tree has fallen and slid 50 m down a hill at a uniform angle of 30° to the horizontal. Assuming the acceleration due to gravity is 9.8 m/s2 and Force = Mass × Acceleration, calculate the work done by gravity on the tree. 17 A force of 250 N acting due south moves an object 4.1 m in a southeasterly direction. Determine the work done, rounded to one decimal place. 18 A force of 155 N does 270 J of work in moving an object 190 cm. Determine the angle between the force and the direction of the movement.

Extend your thinking 19

A 17 kg photo frame is hung on a wall with a string attached to its top-left and top-right corners. The middle of the string is placed on a nail on the wall, and the angle between each half of the string and the horizontal is 45°. Calculate ∣T∣, the magnitude of the tension in each half of the string, rounded to the nearest tenth of a Newton. Assume the acceleration due to gravity is 9.8 m/s2.

20 This diagram shows forces F1 and F2 acting in the same vertical plane on a hook fixed to a vertical wall. F1 has a magnitude 147 N and acts at an angle of elevation of 22° and F2 has a magnitude of 195 N and acts at an angle of depression of 42°. he wall exerts a force on the hook of equal magnitude T but opposite direction to the resultant force of F1 and F2. Determine the force exerted by the wall on the hook, expressing it using unit vectors.

y F1 x

21 Two forces, F1 = i + 4j and F2 = 3i − 2j, act on a single body P and causes the body P to be displaced by 8i + 4j. Determine the work done on P. 22

3x 2 y as the only force acting on some object where x and y 4y 5 represent the object’s displacement. If F = –11j:

Consider a force F a

Calculate the displacement of the object.

Calculate the total work done on the object.

Investigation: Vector proofs Investigate online

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Chapter 8 review Online review

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8.11 Force equilibrium and work done mathspace.co

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Big ideas • The complex number system extends the real numbers by introducing the imaginary unit, i, creating a powerful two-dimensional system that unifies algebra and geometry. Complex numbers can be represented algebraically in Cartesian form, (a + bi), or geometrically in polar form, (r cis(θ )), with a complete set of arithmetic rules. The choice of form is strategic, as polar form elegantly reveals multiplication and division as geometric operations of scaling and rotation on the Argand diagram. • The algebra of complex numbers provides a powerful language for describing geometry, where equations involving the modulus and argument define specific loci such as lines and circles on the Argand diagram. This system also guarantees solutions to all polynomial equations with real coefficients, where any non-real roots always occur in conjugate pairs.

9 Complex numbers Chapter outline 9.01 9.02 9.03 9.04 9.05

9.06 9.07 9.08 9.09

Introduction to complex numbers Modulus and argument of complex numbers Addition and subtraction of complex numbers Multiplication of complex numbers Complex conjugates and division Investigation: Linear transformations and rotations on the complex plane Cartesian and polar form Complex arithmetic in polar form Complex inequalities and regions Complex roots Chapter 9 review

840 851 861 870 876

885 900 909 927

Complex numbers can model waves, from ocean waves to sound waves!

9.01 Introduction to complex numbers After this lesson, you will be able to… • understand the definition i2 = − 1 and properties of complex numbers • convert a surd with negative radicand to a complex number • identify the real and imaginary parts of a complex number • represent complex numbers on an Argand diagram

Definition of a complex number Complex number A number that can be expressed as the sum of a real part and an imaginary part. It is typically written in the form: z = a + bi where a is the real part and b is the imaginary part of the complex number. The imaginary number, i, is defined by the property: i2

1 or i

The imaginary number, i, extends the real number system. This allows finding solutions for equations that do not have real solutions, like x2 = − 1. The real and imaginary parts of a complex number, z, can be denoted as Re(z) and Im(z). For example, if z = 3 + 2i, then Re(z) = 3 and Im(z) = 2. The set of complex numbers is denoted using the symbol ℂ. It can be formally defined as ℂ = {a + bi|a ∈ ℝ, b ∈ ℝ}.

• When a = 0, the complex number z becomes z = bi, which is a pure imaginary number. • When b = 0, the complex number z becomes z = a, which is a real number.

If both the real part a and the imaginary part b are non-zero (e.g., 2 + 3i), the number is said to be a non-real complex number. Therefore, the set of real numbers, denoted by ℝ, is a subset of the set of complex numbers, ℂ. Recall that ℝ consists of rational and irrational numbers. p , where p, q Z and q q ℚ is a subset of ℂ. Q

0 , the set of rational numbers, is a subset of ℝ. Therefore,

When p = mq, where m is an integer, then x = m, which is an integer. The set of integers, ℤ = {. . ., − 2, − 1, 0, 1, 2, . . .}, is a subset of ℚ. Consequently, ℤ is a subset of ℝ and ℂ.

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Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Recall that ℤ consists of positive integers, negative integers and zero. The set of natural numbers, ℕ = {0, 1, 2, 3, . . .}, consisting of non-negative integers, is a subset of ℤ. Therefore, ℕ is a subset of ℚ, ℝ and ℂ. This is shown in this graph: Complex numbers ℂ

Real numbers ℝ

Rational numbers ℚ

Integers ℤ

Natural numbers ℕ

Negative integers

Fractions

Irrational numbers

Imaginary numbers

Note: The symbol ∈ denotes “is an element of”, while ∉ denotes “is not an element of”. These symbols are used to describe membership of an element in a set. For example, if a ∈ ℤ, it means that a is an integer, and if k ∉ ℝ, it means that k is not a real number.

Example 1 Identify the real and imaginary parts for each complex number: a z = 2 + 5i

Create a strategy Recall that complex numbers are expressed in the form z = a + bi, where a is the real part, and b is the imaginary part.

Apply the idea For the complex number z = 2 + 5i, the real part is Re(z) = 2 and the imaginary part is Im(z) = 5.

b z = − 4i

Apply the idea For the complex number z = − 4i, which can be written as z = 0 − 4i, the real part is Re(z) = 0 and the imaginary part is Im(z) = − 4.

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c z = 10

Apply the idea For the complex number z = 10, which can be written as z = 10 + 0i, the real part is Re(z) = 10 and the imaginary part is Im(z) = 0.

d z = 3i − 6

Create a strategy

Apply the idea

The complex number z = 3i − 6 can be rewritten as z = − 6 + 3i so that it is in the form a + bi.

For the complex number z = − 6 + 3i, the real part is Re(z) = − 6 and the imaginary part is Im(z) = 3.

Example 2 Classify these complex numbers as either real, non-real complex or pure imaginary: a z = −3

Create a strategy

Apply the idea

A complex number is classified as real, pure imaginary, or non-real based on whether its imaginary part, real part, or neither part is zero.

The complex number z = − 3 is real since the imaginary part is zero.

b z

i 2

Apply the idea The complex number z

i is purely imaginary since the real part is zero. 2

c z = 5 + 5i

Apply the idea z = 5 + 5i is a non-real complex number since both its real and imaginary parts are non-zero.

Example 3 Digital only example Explore in the online lesson

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Idea summary Complex numbers can be expressed in the form: z = a + bi

where a is the real part, b is the imaginary part and i is the imaginary number defined as: i2

1 or i

The set notation for numbers classifies different types of numbers, where ℕ is a subset of ℤ, ℤ is a subset of ℚ, ℚ is a subset of ℝ and ℝ is a subset of ℂ. Complex numbers ℂ

Real numbers ℝ

Rational numbers ℚ

Integers ℤ

Natural numbers ℕ

Negative integers

Fractions

Irrational numbers

Imaginary numbers

The symbol ∈ denotes “is an element of”, while ∉ denotes “is not an element of”. These symbols are used to describe membership of an element in a set.

Operations with negative surds 1, negative numbers under the square root can be written as complex Since i2 = − 1 or i numbers in the form z = a + bi using index laws including surds and these properties: •

•

a b

• a c b c

( a b) c

• a c b c

( a b) c

Recall that: a b is not equal to a b. Using these surd properties and the definition i 1, expressions that have negative numbers under a square root can be simplified and written as complex numbers.

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Example 4 Express 100 in terms of i.

Create a strategy Express − 100 as a product of − 1 and a 100, and use the fact that i

Apply the idea 100 =

1 100 Rewrite − 100 as the product of − 1 and a number

Use the rule ab

100

= i 100

Substitute

= 10i

Evaluate 100

Example 5 33

Determine the value of

3 11

Create a strategy Apply the properties of square roots and complex numbers (where i2 = − 1).

Apply the idea 33

3 11

1 33

11 1

= = = =

Rewrite each radicand as a product of − 1 and a number 3

11 i 33 i 3 11 i2 99 11 99

844

1 3

11 99 11

Use the fact that ab Rewrite 1 as i

Evaluate the multiplication Use the fact that i2 = − 1 Use the fact

a b

= 9

Simplify the fraction

= −3

Evaluate

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a b

Example 6 Express

16 in the form a + bi.

Create a strategy First, convert each surd with a negative radicand into its imaginary form using i perform the addition.

1. Then,

Apply the idea 9

16 =

1 9

Rewrite each negative radicand 1 16 as a product involving − 1 9

16 Use the rule ab each term

= (i × 3) + (i × 4) Substitute 1 square roots

b for

i and evaluate the

= 3i + 4i

Simplify each term

= 7i

Add the imaginary terms

In the form a + bi, this is 0 + 7i.

Idea summary Using properties of surds and the fact that i expressions involving negative surds.

1 allows for simplifications of

Surds involving negative numbers can be rewritten as complex numbers in the form of z = a + bi.

Did you know?

Complex numbers help us understand quantities beyond ordinary real numbers! For example, an oscillating compass needle can be modelled with real and imaginary parts to predict its rotation, understand wave patterns, and design stable systems.

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Complex numbers on the Argand diagram The complex plane or Argand diagram is used to plot complex numbers. It is named after the Swiss mathematician Jean Argand (1768 − 1822).

Exploration Consider a standard Cartesian plane with an x-axis and a y-axis. When ordered pairs (a, b) are plotted on this plane, consider how their positions relate to complex numbers of the form a + bi? Rename the horizontal axis as the real axis, Re(z) and the vertical axis as the imaginary axis, Im(z). This new plane is called an Argand diagram or complex plane. 1. If a complex number has only a real part (e.g., z = 3 or z = − 2), where would it be plotted on this Argand diagram? 2. If a complex number has only an imaginary part (e.g., z = 2i or z = − 5i), where would it be plotted on this Argand diagram? 3. How does plotting the complex number a + bi as a point on the Argand diagram compare to plotting the ordered pair (a, b) on a standard Cartesian plane?

The complex plane consists of two perpendicular axes: the real axis (horizontal) and the imaginary axis (vertical). Complex numbers are represented as ordered pairs (a, b), where a is the real part and b is the imaginary part. For any complex number z = a + bi, the ordered pair (a, b) is its corresponding point on the Argand diagram. To plot this point, real part (a) is used for the horizontal position and the imaginary part (b) is used for the vertical position. For example: z1 = 2 + 3i

(a, b) = (2, 3)

z2 = 12 − 5i

(a, b) = (12, − 5)

z3 = − 4 − 4i

(a, b) = (− 4, − 4)

z4 = 7

(a, b) = (7, 0)

z5 = − 6i

(a, b) = (0, − 6)

Im (z) 6 4

2 –6 –4 –2 –2 z3

Re (z)

2 4 6 8 10 12 14

–4 –6 z5

These complex numbers can be plotted on an Argand diagram as points, or represented as vectors from the origin (0, 0) to the point (a, b). The diagram shows z1, z2, z3, z4 and z5 represented both as points and as vectors.

–8

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Example 7 Plot the complex number 6 + 2i on the Argand diagram.

Create a strategy Identify the real part (6) and the imaginary part (2). The real part will be plotted along the horizontal axis, and the imaginary part along the vertical axis.

Apply the idea Im (z) 3 Start by drawing a Cartesian plane with labelled axes: the horizontal axis representing the real part and the vertical axis representing the imaginary part. The point representing 6 + 2i will be placed 6 units right of the origin along the horizontal axis and 2 units up along the vertical axis.

(6, 2)

2 1

Re (z) –1

2 3 4 5 6 7

–1

Example 8 The Argand diagram shows a complex number represented by a vector from the origin to the point (− 2, − 1): Im (z) 1 Re (z) –3

–2

–1

(–2, –1)

1 –1 –2 –3

Identify this complex number in the form z = a + bi.

Create a strategy A complex number z = a + bi has a real part a and an imaginary part b. The coordinates x and y represent the real and imaginary parts of the complex number respectively.

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Apply the idea From the given information: Real part a = x = − 2 Imaginary part b = y = − 1 Therefore, the complex number z = a + bi is: z = − 2 + (− 1) i Simplified, z = −2 − i

Idea summary The complex plane or Argand diagram is a two-dimensional coordinate system used to represent complex numbers graphically. In this plane, the horizontal axis represents the real part of the complex number, and the vertical axis represents the imaginary part. Each complex number, in the form a + bi, is represented as a point (a, b) or as a vector from the origin to the point (a, b).

9.01 Practice questions What do you remember? 1

Determine whether these statements are true or false: a

The imaginary number i is equal to 1.

i2 = − 1

All non-real complex numbers include the imaginary number i when simplified.

Express the complex number with the given real and imaginary parts: a

3 Ex 1

4 Ex 2

5 Ex 3

848

a = 4, b = 2

a = 1, b = − 3

a 0, b 7

a = 8, b = 0

z = 2 − 9i

Identify the real and imaginary parts of these complex numbers: a

z = 1 + 2i

z = 7 − 3i

z = 0 + 3i

Classify these complex numbers as either real, non-real complex or pure imaginary: 1 a z b z = 3 − 4i c z = − 9i d z=0 2 For each number, identify which of the listed number sets it belongs to: Natural (ℕ), Integer (ℤ), Rational (ℚ), Real (ℝ), Complex (ℂ). 4 5 a −7 b 3 − 2i c d 9

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Identify whether these numbers are real or non-real complex: a

2 8

Determine the values of a and b if: a

a bi

i 2 5 100 3

a bi

5 4

a bi

6 3

a bi

1 4

a bi

9 4

Practice 8 Ex 4

Express in the simplest form in terms of i: a

8 25

12 55

Evaluate: a

i 7 8

1 5

4 7 4

4i 14

i 6

9 3

3i 3

i 7

36 3

Evaluate: 10

5 5

15 9

64 16

i 100

1 2 17

–3

Evaluate: 4

a d

Ex 6

Determine whether these equations are true or false:

Ex 5

2 3

6 3 4

9 i

Express in the form a + bi: a

4 3

7 5

4 2 5

49 25

25 9

100

121 + 3

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Ex 7

Plot and label these complex numbers in an Argand diagram: • z1 = 3 + 4i • z2 = − 2 − 5i • z3 = − 4i • z4 = 5

Ex 8

This Argand diagram shows four complex numbers represented by points A, B, C, and D: Im (z)

4 3 2 1

D –4 –3 –2 –1 –1

–2 –3

Re (z) 2 3 4

–4 Express each number in the form a + bi.

Extend your thinking 16

Determine whether these statements are true or false: a

i×i=i

When multiplying two real numbers, the product will always be a real number.

When multiplying a non-zero real number and a non-real complex number, the product will always be a complex number.

When multiplying two non-real complex numbers, the product will always be a non-real complex number.

When multiplying any real number by a purely imaginary number the result will always be a purely imaginary number.

Evaluate: i

i×i×i

i×i×i×i

iii

Determine the general pattern for the result of in.

A square ABCD is drawn on an Argand diagram. Vertex A is at 1 + 2i and vertex B is at 4 + 2i. If the vertices are listed in an anticlockwise direction, find the complex numbers representing vertices C and D.

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9.02 Modulus and argument of complex numbers After this lesson, you will be able to… • calculate and interpret the modulus of a complex number • calculate and interpret the argument of a complex number

Modulus The modulus of a complex number a + bi is its distance from the origin on the Argand diagram. In other words, it is the size of the number. The modulus is denoted using many different notations. Common notations include: • |z| • mod z • |a + bi| In this diagram, the point P has been plotted. It corresponds to the complex number 3 + 4i: Imaginary axis 7 6 5

P (3, 4)

4 3

The vertical distance is b units

2 1 –1

0 –1

Real axis 1 2 3 4 The horizontal distance is a units

For this complex number, the horizontal distance is 3 (Re(z)) and the vertical distance is 4 (Im(z)). The modulus is the length of the vector OP, which can be found by using Pythagoras’ theorem on the right-angled triangle. This distance can be found using Pythagoras’ theorem: |z|2 = (Re (z))2 + (Im (z))2 |z| = (Re ( z))2 (Im ( z))2 = 32 42 =5 The general formula for the modulus of a complex number z = a + bi is given by: |z| = |a + bi| = a2 b2

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It is also useful to note that a complex number and its conjugate have the same modulus. For a complex number z = a + bi and its conjugate z = a − bi, the modulus is calculated as: • z

a2 b2

• z

a2 ( b)2

Therefore, z

a2 b2

Example 1 Calculate the modulus of the complex number − 8 + 3i.

Create a strategy Use the formula: |z| = |a + bi| = a2 b2

Apply the idea For the complex number − 8 + 3i, a = − 8 and b = 3 |z| = a2 b2 Write the formula for modulus of a complex number = ( 8)2 32

Substitute a = − 8 and b = 3

= 64 9

Evaluate the squares

= 73

Evaluate and leave as exact value

Example 2 Calculate the modulus of these complex numbers: a z = −4

Create a strategy First, write the real number in complex form as a + 0i. Then, use the modulus formula.

Apply the idea For the complex number z = − 4, it can be written as z = − 4 + 0i. Here, a = − 4 and b = 0. |z| = a2 b2 Write the formula for modulus of a complex number

852

= ( 4 )2 02

Substitute a = − 4 and b = 0

= 16

Evaluate the squares

Evaluate

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b w = 5i

Create a strategy Write the imaginary number in complex form as 0 + bi. Then, use the modulus formula.

Apply the idea For the complex number w = 5i, write it as w = 0 + 5i. Here, a = 0 and b = 5. |w| = a2 b2 Write the formula for modulus of a complex number = 02 52

Substitute a = 0 and b = 5

= 25

Evaluate the squares

Evaluate

Reflect and check The answers can be checked by considering the position of these numbers on an Argand diagram. The number z = − 4 corresponds to the point (− 4, 0), which is 4 units away from the origin on the negative real axis. The number w = 5i corresponds to the point (0, 5), which is 5 units away from the origin on the positive imaginary axis. The calculated moduli match these distances.

Example 3 If z = 12 − 16i and w = 3 − 4i, determine: a |z|

Apply the idea For the complex number z = 12 − 16i, a = 12 and b = − 16 |z| = a2 b2

Write the modulus formula

= 122 ( 16)2

Substitute a = 12 and b = − 16

= 144 256

Evaluate the squares

= 400

Evaluate the addition

= 20

Evaluate

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b |w|

Apply the idea For the complex number w = 3 − 4i, a = 3 and b = − 4 |w| = a2 b2

Write the modulus formula

= 32 ( 4 )2

Substitute a = 3 and b = − 4

= 9 16

Evaluate the squares

= 25

Evaluate the addition

Evaluate

c |z + w|

Create a strategy Add z and w first, then calculate the modulus of the result.

Apply the idea Add z and w: (12 − 16i) + (3 − 4i) = 15 − 20i

Add the real parts and imaginary parts separately

Calculate the modulus of the result: |z + w| = a2 b2

Write the modulus formula

= 152 ( 20)2

Substitute the values

= 225 400

Evaluate the squares

= 625

Evaluate the addition

= 25

Evaluate

d |z| + |w|

Create a strategy

Apply the idea |z| + |w| = 20 + 5 = 25

Add the moduli calculated in parts (a) and (b).

Reflect and check In the preceding example, it was found that |z + w| = 25 and |z| + |w| = 25, so in this case, |z + w| = |z| + |w|. It is important to recognise that this is a special case and not a general rule. The general relationship between these quantities is known as the triangle inequality, which states: |z + w| ≤ |z| + |w|

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Idea summary The modulus |z|, of the complex number z = a + bi, represents the distance of the complex number from the origin in the complex plane. The modulus, |z|, of the complex number z = a + bi, is calculated using the formula:

Argument Exploration The argument of a complex number depends on which quadrant it is in. Explore how to find the argument for numbers in different quadrants. Imaginary (−2 + 2i) B

A (2 + 2i)

Real

(−2 − 2i) C

D (2 − 2i)

1. Consider the four complex numbers plotted in the diagram: zA = 2 + 2i, zB = − 2 + 2i, zC = − 2 − 2i, and zD = 2 − 2i. For each complex number, form a right-angled triangle with the real axis. What is the acute angle made with the real b axis for each of these triangles? Hint : Use tan( ) . a 2. The argument, arg (z), is measured from the positive real axis. Angles measured anticlockwise are positive, and angles measured clockwise are negative. Based on the quadrant, how could the acute angle from question 1 be used to find the argument of each complex number? 3. The principal argument, Arg (z), must be in the range −π < θ ≤ π. Find the principal argument for each of the four complex numbers.

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The argument of the complex number z = a + bi is the angle the vector makes with the positive real axis. It is denoted using the notation: arg (z) Imaginary axis z = a + bi

θ a

b Real axis

In the diagram, the argument of the complex number z is labelled θ. Since the triangle formed is a right-angled triangle, the argument of the complex number, arg(z), can be calculated using trigonometry. b Therefore, arg(z) = θ, where tan (θ ) = , where a ≠ 0. a

An angle can be expressed in many ways. For example, rotating 30° is the same as rotating a full circle plus 30° (i.e., 390°). For consistency, the principal argument is used, denoted Arg (z), which is always in the range − π < θ ≤ π or − 180° < θ ≤ 180°. Quadrant

Conditions on real (a) and imaginary (b) parts

Formula for principal argument, Arg(z)

a > 0, b > 0

Arg(z) = α

a < 0, b > 0

Arg(z) = π − α

III

a < 0, b < 0

Arg(z) = − π + α

a > 0, b < 0

Arg(z) = − α

Other possible arguments for z can be found by adding or subtracting full rotations from the principal argument. This is given by the formula: arg(z) = Arg(z) + 2πn, n ∈ ℤ

When calculating the argument or principal argument of a complex number, begin by drawing an Argand diagram and label the given complex number. Note that the argument can be equal to zero (for example, the argument of the complex number z = 1 + 0i). However, arg (0) is undefined.

Example 4 Determine the principal argument of each complex number: a x

3 i

Create a strategy Begin by plotting the complex number on an Argand diagram, labelling a and b. Recall that the b principal argument, Arg (z), is calculated using tan (θ ) = , where a ≠ 0. The angle is in the range a − π < θ ≤ π.

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Apply the idea Since a 3 , b 1 Imaginary axis

tan (θ ) =

z = 3+i

tan (θ ) =

Real axis

θ=

b a

Write the formula for the argument θ

1 3 6

Substitute a = 3 and b = 1 Take the inverse tangent to solve for θ

This is the argument of z as it is measured from the positive real axis. Since this angle lies in the range −π < θ ≤ π, it is also the principal argument. Arg(z)

b z

3 i

Apply the idea Imaginary axis

Real axis

3, b = − 1, the complex number lies in the Since a third quadrant. When considering the right-angled triangle formed, the angle labelled α is calculated using: tan (α) =

Substitute the absolute values of a and b

α = Take the inverse tangent to 6 solve for α

z = − 3−i

Since the complex number is in the third quadrant, it is easier to find the principal argument by measuring clockwise (negative angle) from the positive real axis. The angle to the negative real axis is −π, and then add the acute angle α. Arg(z) = − π + α Write the formula for the principal argument in the third quadrant = −π + =

5 6

Substitute the value of α Evaluate

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Example 5 The points P and Q are shown on the Argand diagram: 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4

Im (z)

Re (z) 2 3 4 P 1 − 3i

Q (−4i)

For each point: i. Determine its modulus ii. Determine its principal argument in radians. a Point P, representing the complex number 1 3i

Create a strategy To find the modulus, use the formula |z| = a2 b2 . To find the principal argument, first find the b related acute angle α in the fourth quadrant using tan (α ) = . The principal argument will be a Arg ( z) = −α.

Apply the idea i.

Modulus

For z = 1 − 3i, a = 1 and b = − 3. |z| = a2 b2 = 12

Write the formula for the modulus 3

Substitute the values

= 1 3

Simplify

= 4

Evaluate the addition

Evaluate

ii. Principal argument The complex number is in the fourth quadrant. First, find the related acute angle α: tan (α) =

α = tan–1

Substitute the absolute values of a and b, and simplify

Take the inverse tangent to solve for α 3 For a number in the fourth quadrant, the principal argument is the negative of the related acute angle: 3

Arg(z)

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b Point Q, representing the complex number − 4i

Create a strategy To find the modulus, use the formula or identify the distance from the origin directly from the diagram. To find the principal argument, identify the angle made with the positive real axis directly from the diagram.

Apply the idea i.

Modulus

For z = − 4i, it can be written as z = 0 − 4i. Therefore, a = 0 and b = − 4. |z| = 02 ( 4 )2

Substitute the values of a and b

= 16

Simplify

Evaluate

The point is 4 units from the origin. ii. Principal argument The point (0, − 4) lies on the negative imaginary axis, which corresponds to a principal argument of: Arg(z)

Idea summary The argument of the complex number z = a + bi is the angle the vector makes with the positive real axis. It is denoted using the notation arg(z). It is calculated using b arg(z) = θ, where tan (θ ) = , and a ≠ 0. a The principal argument of z, Arg(z), is in the range − π < θ ≤ π or − 180° < θ ≤ 180°.

9.02 Practice questions What do you remember? 1

Determine the formula for the modulus, |z|, of a complex number z = a + bi.

What is the range of values for the principal argument, Arg (z), in radians?

Identify the quadrant in which the complex number z = − 4 − 9i would be plotted on an Argand diagram.

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Convert these angles to principal angles: a

540°

− 720°

7 radians 3

5 radians 2

Express the formula for the modulus of a complex number, |z|, in terms of Re(z) and Im(z).

Practice 6 Ex 1

7 Ex 2

8 Ex 3

9 Ex 4

Calculate the modulus of these complex numbers: a

Ex 5

3 + 3i

5 − 2i

1 + 4i

5 16

|zw|

|z| + |w|

−1 − i

− 7 − 24i

− 4i

If z = 2 − i and w = 5 − 2i, calculate: a

|z + w|

|z| − |w|

Determine the principal argument for each complex number: a

1+i

3 i

Determine the principal argument for each complex number: 2 + 3i

− 3 + 2i

− 5 + 12i

The principal argument

−4

1 i 1 i

2i 4 2i

For each complex number, determine: i

The modulus

4 + 5i

−3 + i

Calculate the modulus of these complex numbers: a

Calculate the modulus of these complex numbers:

a 11

1+i

(2 − 3i) (4 + i)

(3 + 4i)2

The points A and B are shown on the Argand diagram. For each point, determine its modulus and its principal argument in radians: 3 i

Point A, representing the complex number

Point B, representing the complex number − 3

4 3 2 A − 3, 1 1 B (−3, 0) −4 −3 −2 −1

−1 −2 −3 −4

Re (z) 1

2 3 4

If |z| = 5, determine the modulus of: a

860

Im (z)

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

z 2

(4 + 3i) z

Extend your thinking 15

For a complex number z, show that z z = |z|2.

For any complex number z, show that z

For any real number A and complex number z, prove that |Az| = |A||z|.

For any two complex numbers z and w, show that |zw| = |z||w|.

For any non-zero complex number z, show that z

Determine the complex number z if |z| = 5 and Arg (z) = 135°.

Show that |z + w|2 = |z|2 + |w|2 if and only if

z z2

z is purely imaginary. w

9.03 Addition and subtraction of complex numbers After this lesson, you will be able to… • perform addition and subtraction of complex numbers • classify the sum or difference of complex numbers into their number system • interpret complex number addition geometrically on the Argand diagram

Addition and subtraction of complex numbers Complex numbers can be added and subtracted by considering the real components as like terms and the imaginary components as like terms. The sum or difference of two complex numbers is always a complex number of the form a + bi. Sometimes, the result can be a purely real number (if the imaginary part b becomes zero) or a purely imaginary number (if the real part a becomes zero). This process is similar to combining like terms with surds. For example, just as a x b x ( a b) x , if x = − 1 is considered, then a 1 b 1 ( a b) 1, which means ai + bi = (a + b) i. Therefore, when adding or subtracting complex numbers, the real parts are combined and the imaginary parts are combined.

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In general, for two complex numbers z1 = a + bi and z2 = c + di, where a, b, c, and d are real numbers: • Addition: z1 + z2 = (a + bi) + (c + di) = (a + c) + (b + d)i • Subtraction: z1 − z2 = (a + bi) − (c + di) = (a − c) + (b − d)i Two complex numbers are defined to be equal if and only if their real parts are equal and their imaginary parts are equal. That is, for complex numbers z1 = a + bi and z2 = c + di: a + bi = c + di if and only if a = c and b = d This means a single equation involving complex numbers can be equivalent to two equations involving real numbers.

Example 1 Evaluate: a (3 + 2i) + (6 − 4i)

Create a strategy Add the real components and the imaginary components separately.

Apply the idea Consider the two complex numbers z1 = 3 + 2i and z2 = 6 − 4i.

Real components

First, identify the real and imaginary components of each complex number. 3

Imaginary components Now, add the real components and add the imaginary components. Sum of real components 3+6=9 3+6=9 2i + (− 4i) = − 2i (3 + 2i) + (6 − 4i) = 9 − 2i

Add the real components Add the imaginary components

9 − 2i

Express in the form a + bi 2i + (− 4i) = 2i − 4i = − 2i Sum of imaginary components

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6 7 2i

3 7

Create a strategy Subtract the real components and the imaginary components separately.

Apply the idea 6 7 3 7 = 9 7

Subtract the real components

− 2i − 7i = − 9i 6 7 2i

3 7

Subtract the imaginary components

7i = 9 7 9i

Express in the form a + bi

c Evaluate (5 + 3i) + (− 1 − 3i) and classify the result.

Apply the idea 5 + (− 1) = 4

Add the real components

3i + (− 3i) = 0i

Add the imaginary components

(5 + 3i) + (− 1 − 3i) = 4 + 0i

Express in the form a + bi

The result is 4. Since the imaginary part is 0, this is a real number. (It is also a complex number where b = 0).

d Evaluate (2 − i) − (2 + 4i) and classify the result.

Apply the idea 2−2=0 − i − 4i = − 5i (2 − i) − (2 + 4i) = 0 − 5i

Subtract the real components Subtract the imaginary components Express in the form a + bi

The result is − 5i. Since the real part is 0 and the imaginary part is non-zero, this is a purely imaginary number. (It is also a complex number where a = 0).

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Idea summary Complex numbers can be added and subtracted by considering the real components as like terms and the imaginary components as like terms. Consider the complex numbers: z1

a bi and z2

c di

where a, b, c and d are real numbers. Then: z1 + z2 = (a + c) + (b + d)i z1 − z2 = (a − c) + (b − d)i Adding or subtracting complex numbers yields (a + c) + (b + d)i and (a − c) + (b − d)i, both complex numbers. If the resulting imaginary number is zero, the number is real. If the resulting real part is zero (and the imaginary part is non-zero), the number is purely imaginary.

Vector addition and subtraction on the Argand diagram Interactive exploration Discover this concept in action online

mathspace.co

In the Argand diagram, complex numbers can be added and subtracted visually using vector operations. • Addition: To add two complex numbers, represent them as vectors originating from the origin. Then, place the tail of the second vector at the head of the first vector (head-to-tail method). The resultant vector is drawn from the origin to the endpoint of the second vector. • Subtraction: To subtract one complex number from another, reverse the direction of the vector being subtracted (find its negative) and then add it to the other vector using the head-to-tail method. The resultant vector is drawn from the origin to the endpoint of this second (negative) vector when its tail is placed at the head of the first vector. Addition and subtraction of complex numbers on the Argand diagram can be verified by performing the corresponding arithmetic with the complex numbers.

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Example 2 Determine the sum of the two complex numbers (1 + 2i) and (3 + i) using vectors.

Apply the idea Represent the complex number 1 + 2i as a vector starting at the origin (0, 0) and ending at the point (1, 2).

Im (z) 4

Represent the complex number 3 + i as another vector by moving 3 units to the right and 1 unit up, aligning the tail of this vector at the head of the first vector.

3+i

2 1 + 2i 1

The second vector will then end at the point (4, 3), which is the result of adding the real and imaginary components.

4 + 3i Re (z)

−1

Reflect and check Vector addition can be performed algebraically and compared with the result on the Argand diagram. (1 + 2i) + (3 + i) = 4 + 3i

Add the real parts and imaginary parts

Since the Argand diagram shows a resultant vector of 4 + 3i, it is consistent with the algebraic result.

Example 3 Determine the difference of (1 + 2i) from (3 + 4i) using vectors.

Apply the idea Represent the complex number 3 + 4i as a vector from the origin to the point (3, 4). To subtract the complex number 1 + 2i, first find its negative, which is − (1 + 2i) = − 1 − 2i. Represent − 1 − 2i as a vector. Add the vector for 3 + 4i and the vector for − 1 − 2i using the head-to-tail method. Place the tail of the vector for − 1 − 2i at the head of the vector for 3 + 4i (which is at the point (3, 4)). The head of the vector for − 1 − 2i (when its tail is at (3, 4)) means moving 1 unit to the left and 2 units down. The resultant vector, representing the difference (3 + 4i) − (1 + 2i), is drawn from the origin to the point (2, 2). Therefore, the difference is 2 + 2i.

Im (z) 4 3 2 1 −1 −1

− 1 − 2i

3 + 4i 2 + 2i 1

Re (z) 3

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Example 4 The Argand diagram shows two complex numbers, z1 and z2, represented as vectors from the origin. It also shows the vector representing their sum z1 + z2: Im (z) 4

(6, 4)

(2, 3)

3 2 1 −1

z1 z2 1

z1 + z2 (4, 1) Re (z)

2 3 4 5

−1 a Identify the complex numbers z1 and z2.

Create a strategy For a complex number z = a + bi represented by a vector ending at (a, b), the real part is the x-coordinate and the imaginary part is the y-coordinate.

Apply the idea From the diagram, the vector for z1 ends at the point (2, 3). Therefore, z1 = 2 + 3i. The vector for z2 ends at the point (4, 1). Therefore, z2 = 4 + i.

b Determine the sum z1 + z2 algebraically.

Apply the idea z1 + z2 = (2 + 3i) + (4 + i)

Write the equation

= (2 + 4) + (3 + 1) i

Combine real parts and imaginary parts

= 6 + 4i

Evaluate

c Verify that the algebraic sum matches the vector z1 + z2 shown on the diagram.

Create a strategy Check if the real and imaginary parts of the algebraic sum correspond to the coordinates of the endpoint of the vector z1 + z2 on the diagram.

Apply the idea The algebraic sum is 6 + 4i. On the Argand diagram, the vector labelled z1 + z2 starts at the origin and ends at the point (6, 4). This corresponds to the complex number 6 + 4i, so the algebraic sum matches the geometric representation.

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Idea summary In the complex plane, complex numbers can be added and subtracted graphically by treating them as vectors. Key steps include: • To add two complex numbers (z1 + z2): 1. Represent z1 and z2 as vectors from the origin in the complex plane. 2. Place the tail of the vector for z2 at the head of the vector for z1. (Head-to-tail method) 3. The resultant vector, representing z1 + z2, is drawn from the origin (tail of z1) to the head of z2. • To subtract two complex numbers (z1 − z2): 1. Represent z1 as a vector from the origin. 2. Find the negative of z2, which is −z2. Represent −z2 as a vector (it has the same magnitude as z2 but points in the opposite direction). 3. Add the vector for z1 and the vector for −z2 using the head-to-tail method. Place the tail of −z2 at the head of z1. 4. The resultant vector, representing z1 − z2, is drawn from the origin to the head of −z2.

9.03 Practice questions What do you remember? 1

Let z = a + bi and w = c + di: a

Identify the real and imaginary parts of z and w.

Determine the real and imaginary parts of z + w.

Determine the real and imaginary parts of z − w.

Determine a and b if: a

2 + bi = a − 3i

a + 5i = − 2 + bi

2a − 2i = 4 − bi

6 + 2bi = a − 4i

Evaluate: a

3.7 + 9.1i + 0.3 − 0.5i

2.7 + 8i − 8.5 + 3.2i

5.1 − 3.5i + 0.5 − 4.5i

1.2i − 9.3 − 9.7 − 3.8i

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Practice 4

Evaluate: a

(3 − 4i) + (1 + 7i)

(8 + 8i) − (9 − i)

(7 − 4i) + (9 − 9i) − (2 − 3i)

(5 + i) − (3 + 6i) − (− 9 + 3i)

1 2

1 i 2

Evaluate: a

Ex 1

Ex 2, 3

4 5

1 i 3

5 6

7i 9

a + bi = 2 − 7i − (4 + 2i)

a + bi = (1 − i) + (8 − 8i)

a bi

(2 5i)

4 21

(5 3i)

(9 2i) ( 1 4i)

2 i 3

4 i 6

( 9 i)

Express these in the form a + bi: a

(2 3i)

3 4

Evaluate and classify each result as real, pure imaginary, or non-real complex: a

(7 − 4i) + (− 7 − 2i)

(1 + 5i) + (2 – 7i)

(− 8 − 2i) − (–5 – 6i)

(10 + 2i) + (− 4 − 5i)

If z = 3 − 3i, w = 2 + 8i, v = 5 + 5i, calculate and classify the result as real, pure imaginary, or non-real complex: z+w−v

z−v−w

Let z = 2 + 3i and w = 4 − i: On the Argand diagram provided, a draw vectors to represent z, w, and the sum z + w.

v+z+w

On the Argand diagram provided, draw vectors to represent z, −w, and the difference z − w.

6 5 4 3 2

−2 −1 −1

w−z−v

4 3 2 1

Re 1

2 3 4 5 6

−2

868

Determine a and b if:

a 10

1 6i 4

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Re −4 −3 −2 −1 −1

2 3

Ex 4

The Argand diagram shows vectors for two complex numbers, z1 and z2, and their sum z1 + z2: Im

(4, 6)

6 5

(1, 4)

4 3 2

z1 + z2 z2

(3, 2)

z1 1

Identify the complex numbers z1 and z2 in the form a + bi.

Calculate the sum z1 + z2 algebraically.

Verify that the algebraic sum from part (b) matches the vector z1 + z2 shown on the diagram.

Extend your thinking 12

Determine, in the form a + bi, two complex numbers z and w such that: a

z + w = 2 + 4i

z + w = 7i

z – w = 8 + 4i

z – w = 5i

9 2

7 i, and z w 2

5 2

7 i, determine z and w. 2

If z w

Let z1 = 3 + 2i and z2 = x + yi. If the sum z1 + z2 results in a pure real number, and the difference z1 − z2 has a real part of − 1, find the values of x and y.

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9.04 Multiplication of complex numbers After this lesson, you will be able to… • perform multiplication of complex numbers • calculate powers of i

Multiplication of complex numbers Multiplication of complex numbers is performed by applying the distributive law, similar to expanding algebraic binomials. The resulting expression is then simplified by combining like terms and using the property i2 = − 1. A complex number can be multiplied by a real number (a scalar). If z = a + bi is a complex number and k is a real number, to find their product, multiply both the real and imaginary parts of the complex number by the scalar.

k(a + bi) = ka + kbi k a b

is a real number (scalar) is the real part of the complex number is the imaginary part of the complex number

Consider the complex numbers z1 = a + bi and z2 = c + di, where a, b, c and d are real numbers. When multiplying complex numbers, the distributive law is used as well as the fact that i2 = − 1. To find the product z1 × z2, expand the brackets and simplify the result using i2 = − 1: (a + bi) (c + di) = a × c + bi × c + a × di + bi × di = ac + bci + adi + bdi2 = ac + bci + adi + bd (− 1) = ac − bd + i(bc + ad)

Example 1 Simplify: a − 6(3 − 5i)

Create a strategy Use the distributive law A (B + C) = AB + AC.

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Apply the idea A (B + C) = AB + AC

Write the equation for the distributive law

− 6(3 − 5i) = − 6(3) + (− 6) (− 5i)

Substitute A = − 6, B = 3, and C = − 5i

= − 18 + 30i

Evaluate

b i(5 + 6i)

Create a strategy Use the distributive law A (B + C) = AB + AC, and the fact that i2 = − 1.

Apply the idea i(5 + 6i) = i(5) + i(6i) = 5i + 6i

Apply the distributive law

Evaluate the multiplication

= 5i + 6(− 1)

Substitute i2 = − 1

= 5i − 6

Simplify

= − 6 + 5i

Express in the form a + bi

Example 2 Simplify: a (1 − 3i) (4 + 2i)

Create a strategy Use binomial expansion and the fact that i2 = − 1.

Apply the idea (1 − 3i) (4 + 2i) = (1) (4) + (− 3i) (4) + (1) (2i) + (− 3i) (2i) = 4 − 12i + 2i − 6i = 4 − 10i − 6i

Expand the brackets Evaluate the multiplication Combine like terms

= 4 − 10i − 6(− 1)

Substitute i2 = − 1

= 4 − 10i + 6

Evaluate the multiplication

= 10 − 10i

Combine like terms

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b (2 − 3i)3

Create a strategy First, write the expression as (a + bi) (a + bi) (a + bi). Begin by multiplying two of the factors, for example, (2 − 3i) (2 − 3i). Then, multiply the result by the remaining factor (2 − 3i). Use the distributive law and the property i2 = − 1.

Apply the idea First, calculate (2 − 3i)2: (2 − 3i)2 = (2 − 3i) (2 − 3i) Express the square of a binomial in expanded form = 2(2) + 2(− 3i) − 3i(2) − 3i(− 3i) = 4 − 6i − 6i + 9i

Expand the brackets Evaluate the multiplication

= 4 − 12i + 9(− 1) Combine like terms and substitute i2 = − 1 = 4 − 12i − 9

Evaluate the multiplication

= − 5 − 12i

Simplify

Now, multiply this result by (2 − 3i): (2 − 3i)3 = (− 5 − 12i) (2 − 3i) Rewrite as a product using the result of (2 − 3i)2 = − 5 − 12i = − 5(2) − 5(− 3i) − 12i(2) − 12i(− 3i) = − 10 + 15i − 24i + 36i

Expand the brackets Evaluate the multiplication

= − 10 − 9i + 36(− 1) Combine like terms and substitute i2 = − 1 = − 10 − 9i − 36

Evaluate the multiplication

= − 46 − 9i

Simplify

Idea summary When multiplying complex numbers, use the distributive law and the property i2 = − 1. A complex number a + bi multiplied by a real scalar k results in ka + kbi. The product of two complex numbers, a + bi and c + di, is given by the formula: (a + bi) (c + di) = (ac − bd) + (bc + ad) i

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Powers of i Interactive exploration mathspace.co

Discover this concept in action online

Using the property i2 = − 1, powers of i can be simplified. A clear, repeating pattern emerges: • i1 = i • i2 = − 1 • i3 = i2 × i1 = (− 1) × i = − i • i4 = i2 × i2 = (− 1) × (− 1) = 1 This four-step cycle provides an efficient strategy for simplifying any high power of i. To simplify in, divide the exponent n by 4 and find the remainder. The value of in will be the same as i raised to the power of that remainder. This relationship can be expressed using these rules: • If the remainder is 0: in = i0 = 1 (since n = 4k) • If the remainder is 1: in = i1 = i (since n = 4k + 1) • If the remainder is 2: in = i2 = − 1 (since n = 4k + 2) • If the remainder is 3: in = i3 = − i (since n = 4k + 3) where k is a non-negative integer.

Example 3 Simplify: a (2i)9

Create a strategy First, use the index law (ab)n = anbn to separate the terms. To simplify the resulting power of i, divide the exponent by 4 and use the remainder to determine the value.

Apply the idea (2i)9 = 29i9 = 512i

Apply the index law 9

Evaluate 29

To simplify i9, divide the exponent 9 by 4: 9 ÷ 4 = 2 with a remainder of 1 9

This means i is equivalent to i , which is i. 512i9 = 512(i) = 512i

Substitute i9 = i Simplify

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Reflect and check The result can be verified using the foundational property i2 = − 1. By rewriting i9 in terms of i2, the same answer is obtained: i9 = i4×2+1 = (i2)4 × i1 = (− 1)4 × i = 1 × i = i This confirms that 512i9 = 512i.

Apply the idea 5i

6 6

Apply the index law

= 125i6

Evaluate

Rewrite i6 as a power of i2

= 125(i )

= 125(− 1)

Substitute i2 = − 1

= − 125

Evaluate

Idea summary Simplifying powers of i involves recognising its repeating cycle: i1 = i, i2 = − 1, i3 = −i, i4 = 1 i

= 1, i

and 4k+1

= i, i4k+2 = − 1, i4k+3 = −i

Integrating this with algebraic simplifications, such as combining like terms and applying the distributive law, allows for the simplification of complex expressions.

9.04 Practice questions What do you remember? 1

Let z = a + bi: a

Determine iz.

Determine − iz.

Describe what happens to the real and imaginary parts of z when it is multiplied by i.

Describe in terms of real part and imaginary part what happens to any complex number z when multiplied by − i.

Evaluate these powers of i: a

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Practice 3 Ex 1

4 Ex 2

Simplify: a

2i(8 + 3i)

2i + 3(i + 2)

(8 + i) (2 − 2i)

(2i − 7)2

9 3i (2 2i)

(3 − 4i)2

(3 + 7i)3

Determine the values of a and b in these equations: a

4 + 7i (2 − 4i) = a + bi

(2 + i) (2 − i) = a + bi

Calculate (3 + 3i) (3 − 3i).

What is the imaginary part of the result?

What can be said about the imaginary part of a complex number with the form (a + bi) (a − bi)?

Simplify: 2i2

3i4

− 2i6

9i9

3i2 − 9i4

6i10 − 2i5

i − i2 − i3

i− 2024

3i2025 + 2i2024

(i100)10

(x + 2i)3

(y − 3i)2

(vi)30

Simplify: i3 + i7

Evaluate: a

− i(2 + 3i) = a + bi (i + 2) = a + bi

a 11

(4 − 2i)4

a 10

2 3i

(− 8 + i)2

If (a + 2i)3 = 144 + 208i, determine the value of a.

i − 9i (4 − 5i)

Use binomial expansion to evaluate these expressions:

Simplify:

Ex 3

− 7(9 − i)

i1000

Simplify: a

(xi)4

For (x + iy)3, let (x + iy)3 = − 9 + 46i, and y = 2. Determine the value of x.

Let z a

i 1

3, w

2i 3i, determine:

iz + 3w

z − i2w

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Extend your thinking 15

Prove the commutative property for multiplication of complex numbers. That is, if z = a + bi and w = c + di, show that zw = wz.

Given that zw = 5 + 2i, determine the values of a and b if z = 2 + 2i and w = a + bi.

Sketch the graph of y = (x − (3 − 3i)) (x − (3 + 3i)).

Factorise the polynomial x2 + 2x + 5 into the form (x + (a + bi)) (x + (a − bi)).

Solve the equation x2 + 2x + 5 = 0.

9.05 Complex conjugates and division After this lesson, you will be able to… • prove that the product of a complex number and its conjugate is a real number • calculate complex conjugates and perform division • use conjugates to simplify complex expressions

Complex conjugates For any complex number, there exists a complex conjugate where the imaginary part has the opposite sign. a + bi Conjugates of each other a − bi Consider these conjugate pairings: 6

876

1 i 4

− 15 − 3i

− 15 + 3i

1−i

1+i

26i

− 26i

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The symbol used to denote the conjugate of a complex number, z, is z. It is pronounced z-bar. That is, for z = a + bi, z

a bi.

An important property of complex conjugates is that the product of a complex number and its conjugate is always a real number. To prove this: Let z = a + bi, where a and b are real numbers. Then, its conjugate is z

a bi.

The product z z is: z z = (a + bi) (a − bi)

Substitute z = a + bi and z

= a(a − bi) + bi(a − bi) 2

2 2

a bi

Expand the brackets using the distributive law

= a − abi + bai − (bi)

Distribute further

= a − abi + abi − b i

Simplify bai to abi and (bi)2 to b2i2

= a2 − b2i2

Combine like terms (− abi + abi = 0)

Substitute i2 = − 1

= a − b (− 1) = a 2 + b2

Simplify 2

Since a and b are real numbers, a and b are also real numbers. Therefore, their sum a2 + b2 is a real number.

z z z z a b

( a bi)(a bi)

is a complex number a + bi is the complex conjugate a − bi is the real part of the complex number is the imaginary part of the complex number

The use of complex conjugates to create a real denominator is directly analogous to the technique of using conjugate surds to rationalise a denominator. In algebra, an expression such as a b has the conjugate a b. The product of these terms, a2 − b, removes the surd term. Similarly, the product of a complex number and its conjugate removes the imaginary part, resulting in a real number (a2 + b2), which is the key to simplifying division. Consider the example,

2 3i : 4 5i This is the conjugate of the denominator

2 + 3i 4 + 5i

4 − 5i 4 − 5i This fraction is equivalent to 1 and multiplying by 1 does not change the value of the initial expression

Therefore, when dividing by a complex number: 1. Multiply by a fraction that is equivalent to 1. 2. Use the conjugate of the denominator to construct this fraction. 3. Multiply the binomials in the numerator and denominator, then simplify. 4. The denominator will always simplify to a2 + b2 (for the expansion (a + bi) (a − bi)), which is a real number.

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Example 1 Express in the form a + bi: a

4 7i 2 i

Create a strategy Multiply both the numerator and denominator by the conjugate of the denominator. Recall that the conjugate of a complex number in the form a + bi is a − bi.

Apply the idea The conjugate of 2 + i is 2 − i. 4 7i 4 7i 2 i = 2 i 2 i 2 i = =

8 4i 14i 7i2 4 i2 8 10i 7i2 2

4 i

ultiply both the numerator and the M denominator by the conjugate of the denominator, 2 − i Evaluate the multiplication implify by combining like terms S in the numerator

8 10i 7 ( 1) 4 1

Substitute i2 = − 1

8 10i 7 4 1

Evaluate the multiplication

15 10i 5

Combine like terms

= 3 + 2i

implify, expressing the answer in S the form a + bi

Reflect and check A good way to verify a division is by using multiplication. If the answer 3 + 2i is correct, then multiplying it by the original denominator (2 + i) should yield the original numerator (4 + 7i). Performing the multiplication (2 + i) (3 + 2i) confirms this, providing confidence in the result.

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2 3i 2 3i

Apply the idea The conjugate of 2 − 3i is 2 + 3i. ultiply both the numerator and M the denominator by the conjugate of the denominator, 2 + 3i

2 3i 2 3i 2 3i = 2 3i 2 3i 2 3i =

4 12i 9i2

Expand the brackets and evaluate

4 9i2

4 12i 9 4 9

Substitute i2 = − 1

5 12i 13

Combine like terms

5 13

implify, expressing the answer in S the form a + bi

12 i 13

Reflect and check Notice that this problem involved dividing a complex number z = 2 + 3i by its own conjugate. 5 12 i. A key property to check is that when z is divided by its conjugate, the The result was 13 13 modulus of the result should always be 1. Calculating the modulus of the answer confirms that it is indeed 1, which verifies the calculation.

Idea summary For the complex number, z = a + bi, its conjugate, denoted by z, is z the sign of the imaginary part is reversed.

a bi, where

Complex conjugates can be used to remove the imaginary number from the denominator when dividing complex numbers. The product of a complex number z = a + bi and its conjugate z z z

a bi is always a real number:

a2 b2

Complex numbers and four operations When adding, subtracting, multiplying, and dividing complex numbers, consider these points: • To add or subtract complex numbers, add or subtract corresponding real and imaginary components. • To multiply complex numbers, use standard algebraic expansion rules (like for binomials), remembering that i2 = − 1. • To divide complex numbers, multiply by a fraction equivalent to 1, constructed using the conjugate of the denominator. This ensures the denominator is a real number, since (a + bi) (a − bi) = a2 + b2. 9.05 Complex conjugates and division mathspace.co

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Example 2 Express in the form a + bi: 2i ( 4 3i)2 1 i

Create a strategy Square the binomial and use the distributive law to expand the numerator, and then multiply both the numerator and denominator by the conjugate of the denominator.

Apply the idea Begin by considering the square of the binomial (4 − 3i)2: (A − B)2 = A2 − 2AB + B2 2

Write the formula for the square of a binomial 2

(4 − 3i) = 4 − 2(4) (3i) + (3i)

Substitute A = 4 and B = 3i

= 16 − 24i − 9

Evaluate the multiplication and squares

= 7 − 24i

Combine like terms

Now, multiply the result by − 2i: − 2i(7 − 24i) = − 14i + 48i2 = − 14i + 48 × (− 1) = − 48 − 14i

Use the distributive law Substitute i2 = − 1 Simplify, expressing the answer in the form a + bi

Finally, divide the result by 1 + i and use the complex conjugate of 1 + i to remove the imaginary number from the denominator. The complex conjugate of 1 + i is 1 − i. 48 14i ( 48 14i) ( 1 i) = ( 1 i) ( 1 i) 1 i =

48 14i 48i 14i2 12 ( i)2

Expand the brackets

48 34i 14 ( 1) 1 ( 1)

Combine like terms and substitute i2 = − 1

48 34i 14 ( 1 1)

Evaluate the multiplication

62 34i 2

Combine like terms

= − 31 + 17i

880

ultiply the numerator and the denominator by M the conjugate of the denominator, 1 − i

implify, expressing the answer in the S form a + bi

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 3 Simplify the expression and express it in the form a + bi: ( 1 i)2 2 i

1 i (2 i)2

Create a strategy Follow the order of operations by first evaluating the exponents, then rewriting the division as multiplication by the reciprocal, and finally simplifying the resulting complex fraction using the conjugate of the denominator.

Apply the idea Step 1: Evaluate the powers. (1 + i)2 = 12 + 2(1) (i) + i2

Expand the square

= 1 + 2i − 1

Evaluate and substitute i2 = –1

= 2i

Simplify

(2 + i)2 = 22 + 2(2) (i) + i2

Expand the square

= 4 + 4i − 1

Evaluate and substitute i2 = –1

= 3 + 4i

Simplify

Therefore, the expression becomes

2i 2 i

1 i . 3 4i

Step 2: Rewrite division as multiplication by the reciprocal. 2i 2 i

1 i 2i 3 4i = 3 4i 2 i 1 i

Step 3: Multiply the numerators and denominators. 2i ( 3 4i) 6i 8i2 = (2 i) ( 1 i) 2 2i i i2

Expand numerator and denominator

6i 8 ( 1) Substitute i2 = − 1 and combine like terms 2 3i ( 1) in denominator

8 6i 1 3i

Simplify numerator and denominator

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Step 4: Simplify by multiplying by the conjugate of the denominator (1 + 3i). Multiply the numerator and denominator 8 6i 1 3i 8 6i = with the conjugate of the denominator 1 3i 1 3i 1 3i =

( 8) ( 1) ( 8) ( 3i) (6i) ( 1) (6i) ( 3i) 2

( 3i)

8 24i 6i 18i2

1 9i

xpand numerator and use E (a − b) (a + b) = a2 − b2 for denominator valuate the multiplication in the numerator E and the powers in the denominator

8 18i 18 ( 1) 1 9 ( 1)

Combine like terms and substitute i2 = − 1

8 18i 18 1 9

Evaluate the multiplication

26 18i 10

Simplify

Step 5: Express in the form a + bi. eparate the fraction into its real and S imaginary parts

26 18i 26 18 i = 10 10 10 = Therefore,

( 1 i)2 2 i

13 9 i 5 5 1 i 2

( 2 i)

Simplify the fractions 13 5

9 i. 5

Idea summary For complex numbers, each operation has specific rules and procedures: • Addition and subtraction: Combine real parts and imaginary parts separately. • Multiplication: Apply algebraic rules and use i2= − 1 to simplify. • Division: Multiply the numerator and denominator by the conjugate of the denominator to make the denominator a real number.

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9.05 Practice questions What do you remember? 1

Which of these statements about complex conjugates is correct? a

The complex conjugate of a real number is always zero.

Multiplying a complex number by its conjugate always yields a negative real number.

The complex conjugate of a complex number has the same real part, but the imaginary part has the opposite sign.

The complex conjugate of a complex number is found by changing the sign of its real part.

Determine the conjugate of: a

− 7 − 12i

2 2 31i

16i − 12

18 6 2i 3

Multiply these complex numbers by their conjugate: a

3 + 4i

9 + 10i

− 3 + 7i

5 3i

Express in the form a + bi: a

(5 − 4i) + (− 6 + 9i)

(− 4 − i) − (7 − 3i)

(6i + 5) − (2 + 5i) + (10 − 9i)

(4 − 14i) + (− 20 − i) − (i + 4)

6 12i i

8 5i 7 3i

(4 + 6i)2

− 12i(5 − 7i)2

i(− 3 + 9i) (2i + 3)

3i ( 4 5i)

z2 − 2w

3w z

z w

Re(5z + w− 1)

Practice 5 Ex 1

Express in the form a + bi: a

12i

3i(7 + 2i)

Express in the form a + bi: a

Ex 2

3 4i 3 4i

2i ( 4 5i)2 5 2i

For z = 3 − 2i and w = − 6 + 8i, calculate: z a 2z + 3w b w

2 3i 2 5i

For z = 1 + 8i and w = − 5 − 7i, calculate: a

Re(z2)

Im(w2 + z)

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14 i

Re(z − w)

a 11

Ex 3

If z = 6 − 2i and w the form a + bi:

, determine each expression, expressing any complex numbers in zw

1 z

1 w

Determine the values of a and b that satisfy these equations: 3 5i 3 a (5 − 4i) (a + bi) = 1 − 4i b a bi 50 (u v)2

z2 w

29 i 50

the form a + bi, if u = 4 + 3i and v = − 2 + i.

Determine w

Simplify the expression and express it in the form a + bi:

Let z = a + bi. Show that the product of z and its conjugate z is always a real number.

(u v)3

( 3 i)2 1 2i

2 i ( 1 2i)2

Extend your thinking 15

If z = a + bi, determine all possible values of a and b such that: z i z i 2i z 2 a b z 2 z 1

Consider a complex number z such that the square of z minus twice the conjugate of z equals 16 + 16i multiplied by Im(z). If the real part of z is three times its imaginary part, determine the complex number z.

Explain why it is necessary to multiply both the numerator and denominator by the complex conjugate of the denominator when dividing complex numbers.

Investigation: Linear transformations and rotations on the complex plane Investigate online

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9.06 Cartesian and polar form After this lesson, you will be able to… • convert between Cartesian and polar form of complex numbers • interpret the polar form of a complex number arithmetically and geometrically

Cartesian and polar form Complex numbers in the form z = a + bi are said to be in Cartesian form. This is because numbers in this form can easily be plotted on the Cartesian plane where: • The real part, a, is represented by the x-coordinate. • The imaginary part, b, is represented by the y-coordinate. Alternatively, a polar coordinate system can be used. In this system: • r represents the direct distance from the origin to the point. • θ represents the principal angle between the positive horizontal axis (known as the polar axis) and the point. The complex number z = r cis (θ ) in polar form is represented as the point (r, θ ) in the polar coordinate system. π 2

2π 3 5π 6

π 3 Im(z)

π 6

Re(z) 2π 2 3 11π 6

7π 6 4π 3

3π 2

5π 3

Complex numbers can be written in polar form, where the number is expressed in terms of r and θ . Consider the complex number z = a + bi, represented as point P (a, b) on the Cartesian plane: Im(z)

r=z θ

P (a, b) b = r sin (θ ) Re(z)

a = r cos (θ )

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From the diagram, the measurement, r, is the modulus of the complex number calculated by: r = |z| = |a + bi| = a2 b2 The angle, θ, is the principal argument of the complex number calculated by: b where a ≠ 0 and − π < θ ≤ π a Since a and b can be written in terms of r and θ, the complex number z = a + bi can be written as:

θ = tan− 1

z = a + bi

Write the complex number

= r cos (θ ) + ir sin (θ )

Substitute a = r cos (θ ) and b = r sin (θ )

= r (cos (θ ) + i sin (θ ))

Factorise r

= r cis (θ )

Abbreviate cos (θ ) + i sin (θ ) to cis (θ )

The abbreviation cis (θ ) uses the first letter of cos, the imaginary unit i, and sin. Typically, radians are used in polar form. Therefore, a complex number in Cartesian form z = a + bi, can be converted to polar form, z = r cis (θ ), using these relationships: • Modulus (r) from Cartesian form: r

• Argument (θ ) from Cartesian form:

θ = atan2 (b, a) or tan− 1

b with quadrant adjustments a

b , it is important to consider the quadrant in which the complex a number a + bi lies to ensure the correct value of θ is chosen, as the tan− 1 function on most Note: When using θ = tan− 1

, . Sketching the complex number on the Argand 2 2 diagram can help determine the correct principal argument. The atan2 (b, a) function, available in many programming languages and some calculators, directly returns the principal argument in the range (− π, π]. calculators returns a value in the range

A complex number in polar form, z = r cis (θ ), which is z = r (cos (θ ) + i sin (θ ), can be converted to Cartesian form, z = a + bi, using these relationships: • Real part (a) from polar form: a = r cos (θ ) • Imaginary part (b) from polar form: b = r sin (θ )

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Complex numbers in polar form can be represented on the polar plane. The polar plane consists of concentric circles corresponding to different magnitudes (r) and equally spaced rays representing angles (θ ). π 2

0 Re

3π 2 To plot a complex number on the polar plane: 1. Determine the modulus (r): Locate the appropriate concentric circle corresponding to the magnitude. 2. Determine the argument (θ ): • For positive arguments, measure the angle anticlockwise from the positive real axis. • For negative arguments, measure the angle clockwise from the positive real axis. 3. Plot the point: Place the point at the intersection of the circle (determined by r) and the ray (determined by θ ). While complex numbers in polar form are commonly expressed in radians, the polar plane may sometimes use degrees. It is crucial to identify the increments of the rays on the plane and ensure the point is plotted accurately. Plotting the polar coordinates of complex numbers after multiplication or division makes it easier to observe transformations like rotations and dilations. This is because, in polar form, multiplication and division of complex numbers directly correspond to geometric operations on the complex plane: • Multiplying complex numbers results in a dilation (scaling) and a rotation. The magnitudes are multiplied, and the angles (arguments) are added. • Dividing complex numbers results in a dilation (scaling) and a rotation in the opposite direction. The magnitudes are divided, and the angles are subtracted. If there are two complex numbers in polar form, z1 = r1 cis (θ1) and z2 = r2 cis (θ2), their product is: z1 z2 = r1 r2 cis (θ1 + θ2) And their quotient is: z1 r1 cis (θ1 − θ2) z2 r2 By comparing the polar coordinates before and after these operations, the effects of scaling and rotation become visually clear, making it easier to understand the transformations geometrically. It is also important to note that a complex number and its conjugate pair are reflections of each other across the x-axis.

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Example 1 Express each complex number in polar form, where − π < θ ≤ π: a 4 − 4i

Create a strategy Use the formulas r

b2 and θ = tan− 1

b and express as z = r cis (θ ). a

Apply the idea For the complex number 4 − 4i, a = 4 and b = − 4. To determine the value of r: r = a2 b2 = 4

Write the modulus formula 2

( 4)

Substitute a = 4 and b = − 4

= 32

Evaluate the squares and the addition

=4 2

Simplify

To determine the value of θ :

θ = tan− 1

b a

Write the formula for principal argument

= tan− 1

4 4

Substitute a = 4 and b = − 4

= tan− 1(− 1)

Simplify the fraction

Evaluate

Therefore, 4 − 4i = 4 2 cis

Reflect and check Sketch the complex number on a Cartesian plane to confirm the principal angle: 6

Im(z)

π 2 −4 –6 –4 –2 –2 –4 –6

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Re(z) 2

4 – 4i

b − 6i

Create a strategy Sketch the complex number on the Cartesian plane and then observe the values of r and θ.

Apply the idea Since a = 0, the formula for θ cannot be used. Instead, a diagram can be used to determine the values of r and θ. 6

Im(z)

4 2 –6 –4 –2 –2

Re(z) 2

–4 –6 – 6i The distance from the origin to the point, r, is simply 6 since it is a vertical line of 6 units. The principal angle, θ, is

as observed in the diagram. This results in an undefined value when

using the formula θ = tan− 1

b . a

Therefore, − 6i = 6 cis

c 2 2 3i

Apply the idea For the complex number − 2 − 2 3i, a = − 2 and b = − 2 3. This number is in Quadrant 3. To determine the value of r: r = a2 b2 2

= ( 2)

Write the modulus formula 2

( 2 3 )

Substitute a = − 2 and b

= 4 12

Evaluate the squares

= 16

Evaluate the addition

Simplify

2 3

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To determine the value of θ : First, find the reference angle α = tan− 1

α = tan− 1 = tan− 1

b . a

2 3 2

Substitute a = − 2 and b

Simplify the fraction

2 3

Evaluate 3 Since the complex number is in Quadrant 3, the principal argument is θ = −π + α. =

θ= = Therefore, 2 2 3i

Substitute the reference angle

3 2 3

4 cis

Evaluate 2 . 3

Reflect and check π 2

Im(z)

Re(z)

π –4

–2

Sketch the complex number on an Argand diagram 2 to confirm it lies in Quadrant 3 and the angle 3 corresponds to this position. The modulus is 4.

3π 2 While the manual method of finding a reference angle and adjusting for the quadrant is essential for understanding, many scientific calculators and software packages offer a more direct approach. Functions like atan2 (b, a) or Arg (a + bi) are specifically designed to return the correct principal argument in the range (− π, π] without needing manual adjustments. For this example, inputting atan 2 2 3 , 2 or Arg 2 2 3i into a capable calculator would have directly yielded the 2 answer θ = − , confirming the result in a single step. It is a valuable tool for efficiently checking 3 answers.

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Example 2 Express 4 cis

in Cartesian form.

Create a strategy Use the formulas: a = r cos (θ ) and b = r sin (θ )

Apply the idea , r = 4 and θ = − . 3 3 To determine the value of a in the complex number in the form a + bi: For the complex number 4 cis

a = r cos (θ ) = 4 cos

Write the formula 3

Substitute r = 4 and θ = −

Evaluate

To determine the value of b in the complex number in the form a + bi: b = r sin (θ ) = 4 sin

Write the formula 3

= 2 3 Therefore, 4 cis

Substitute r = 4 and θ = −

Evaluate

2 2 3i.

Example 3 Digital only example Explore in the online lesson

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Example 4 The complex number z is represented in this diagram:

π 2 Im(z) 4 2

π –4

Re(z) –2

–2 –4 3π 2 Express z exactly in the form z = a + bi.

Create a strategy Determine the polar coordinates from the diagram and express z in polar form. Convert z to Cartesian form using the formulas: a = r cos (θ ) and b = r sin (θ )

Apply the idea

π 2 Im(z) 4 Using the diagram, r = 4 since the complex number 2

π –4

–2

Re(z) 2

–2 –4

892

sits on the ring 4 units away from the origin and θ =

2π 3

since each division is the positive real axis.

8 and the point sits 12 12

Therefore, the polar coordinates are 4, 3π 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 . 3

2 3

2 from 3

Converting these polar coordinates to Cartesian form using r = 4 and θ = a = r cos (θ ) = 4 cos

Write the formula

2 3

Substitute r = 4 and θ =

= −2

Write the formula

2 3

Substitute r = 4 and θ =

=2 3 Therefore, z

2 3

Evaluate

b = r sin (θ ) = 4 sin

2 , these results to: 3

2 3

Evaluate

2 2 3i.

Reflect and check Ensure z in both polar form and Cartesian form lie in the same quadrant to check the answer. z

2 3

is in Quadrant 2 as shown.

z 2 2 3i is also in Quadrant 2 since Re(z) is negative and Im(z) is positive and this only occurs in Quadrant 2.

Example 5 Consider the complex numbers z = 1 cis a Plot z on the polar plane.

and w

Create a strategy Plot the point on the ring with a radius of 1 that 4 is radians clockwise from the positive 3 12 real axis.

2i :

Apply the idea 2π 3 5π 6 π –2

π 2 Im(z) π 2 3 1

π 6 Re(z)

–1

7π 6

1 –1

4π 3

–2 3π 2

2 11π 6

5π 3

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b Plot w on the same polar plane.

Create a strategy Convert the complex number to polar form, then plot the point on the polar plane.

Apply the idea For the complex number w

2i, a

2 and b 2.

To determine the value of r: r = a2 b2 =

Write the modulus formula

Substitute the values Evaluate

To determine the value of θ :

θ = tan− 1 = tan− 1 = Therefore, w = 2 cis

b a

Write the formula for principal argument 2

3 4

Evaluate, considering correct quadrant

3 . 4

Plotting this on the polar plane, where r = 2 and θ = 2π 3 w 5π 6 π –2

3 : 4

π 2 Im(z) π 2 3 1

1 –1

4π 3

π 6 Re(z)

–1

7π 6

894

2 and b 2

Substitute a

–2 3π 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 11π 6

5π 3

c Plot zw on the same polar plane.

Create a strategy Use the polar forms of z and w to multiply the complex numbers using: z1 z2 = r1 r2 cis (θ1 + θ2), then plot the result on the polar plane.

Apply the idea z × w = 1 cis

= 1 × 2 cis = 2 cis

× 2 cis

3 4

Write the multiplication in polar form

3 4

Multiply the moduli and add the arguments

5 12

Evaluate

Plotting this on the polar plane, where r = 2 and θ = 2π 3 w 5π 6 π –2

5 : 12

π 2 Im(z) π 2 3 zw 1

Re(z)

–1

7π 6

1 –1

4π 3

π 6

–2 3π 2

2 11π 6

5π 3

d Describe the transformation of z when it is multiplied by w.

Create a strategy Use the polar plane from part (c) to observe the transformation of z.

Apply the idea The length of z has been scaled by a factor of 2 and z has been rotated

3 radians anticlockwise. 4

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Idea summary A complex number in Cartesian form, z = a + bi, can be converted to polar form, z = r cis (θ ), where: r = |z| = |a + bi| = a2 b2

and

θ = arg(z), where − π < θ ≤ π The principal argument θ is found using tan− 1 or by inspection for points on the axes.

b (if a ≠ 0) and adjusting for the correct quadrant, a

A complex number in polar form, z = r cis (θ ), can be converted to Cartesian form, z = a + bi, where: a = r cos (θ ) and b = r sin (θ ) Complex numbers in polar form can be represented on the polar plane. The polar plane consists of concentric circles corresponding to different magnitudes (r) and equally spaced rays representing angles (θ ). Complex numbers are plotted on the intersection of the appropriate circle and ray. Remember that in polar form z = r cis (θ ), r represents the distance from the origin (the modulus) and θ represents the angle measured anticlockwise from the positive real axis (the principal argument). Multiplying complex numbers in polar form involves multiplying their moduli and adding their arguments. Dividing involves dividing their moduli and subtracting their arguments. These operations correspond to scaling and rotation on the complex plane.

9.06 Practice questions What do you remember? 1

What does the expression cis (θ ) abbreviate?

Complete the formulas used to convert a complex number from polar form, z = r cis (θ ), to Cartesian form, z = a + bi:

896

• a=⬚ • b=⬚

Given z1 = r1 cis (θ 1) and z2 = r2 cis (θ2), complete the rules for their product and quotient by filling in the blanks: z1 r1 cis (θ1 ⬚ θ2) a z1 z2 = r1 r2 cis (θ1 ⬚ θ2) b z2 r2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice Ex 1

Express these complex numbers in polar form z = r cis (θ ): a

Ex 2

4 cis

3 4

− 4 + 4i

(r, θ ) = 4,

(r, θ ) = 2,

5 4

Identify the argument and modulus of these complex numbers: z

(7 − 7i)2

7 cis

Simplify in the form of r cis (θ ): a

4 cis

Identify the Cartesian coordinates (x, y) of these polar coordinate points:

a 9

− 2 + 2i

Identify which quadrant each of these complex numbers lie in: 7 a 4 cis b 3 cis c 7 − 3i 3 6

a 8

Express these complex numbers in Cartesian form: a

1 3i

(− 1 + i) (− 4 − 4i)

Convert these expressions to polar form: 1 a 3 i

3 1 3i

Determine the equation of the circle in polar form.

Determine the equation of the circle in Cartesian form.

y 2 x −2

2 −2

If z = 2 + 3i and w a

3 3i, express these in polar form: z z + 2w b w

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Ex 3

Consider the point 8,

in polar coordinates: π 2 Im 10 A E

B F

π –10

0 10 Re

G H –10

D 3π 2

Determine the distance from the point to the origin in the polar plane.

Determine the angle that the point makes with the polar axis.

Which of these points shows the location of 8,

on the polar grid?

Convert the equation x2 + y2 = 9 into polar form.

Convert the equation θ =

etermine the equation of this graph in Cartesian D form.

Determine the equation of the line in polar form.

7 into Cartesian form. 6 y (0, 2)

3 −3 −2 −1

−1

(1, 0) 1 2

−2 −3

Ex 4

The complex number z is represented by the point P on the polar grid. Each concentric circle represents an increase of 1 unit in the modulus, and each ray is separated by radians: 12 a Identify the polar coordinates (r, θ ) of the point P, where − π < θ ≤ π. b

898

Convert the polar coordinates to exact Cartesian form a + bi.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 1 −3 −2 −1 P

−1

−2 −3

Re 2

Ex 5

Consider the complex numbers z1 = 2 cis

and z2 = 1 + i: Im

2 1 Re −2

−1

−1 −2

On the polar grid provided, plot the points representing z1 and z2.

Calculate the product z1 z2 and express it in polar form.

Plot the result z1 z2 on the same polar grid.

Describe the geometric transformation that maps the point for z1 to the point for z1 z2.

Consider the complex number z = 3 cis

Find the complex conjugate z and express it in polar form.

On an Argand diagram, describe the geometric relationship between the points representing z and z. If a general complex number is given by w = r cis (θ ), what is its conjugate w in polar form?

Extend your thinking x3 as a polar equation. 3

Write the equation y

Write r = sin (2θ ) in Cartesian form. Express your answer in terms of x and y with one side of the equation equal to zero.

Consider the polar equation r = A sin (3θ ). Determine this equation in Cartesian form. (Hint: sin (3θ ) = 3 sin (θ ) − 4 sin3 (θ ))

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9.07 Complex arithmetic in polar form After this lesson, you will be able to… • perform multiplication and division of complex numbers in polar form • interpret the product and quotient of complex numbers geometrically • determine the sum and difference of complex numbers given in polar form • compare the complex number arithmetic in polar and Cartesian form

Multiplication and division in polar form When performing arithmetic with complex numbers, the form of the number can make certain operations much simpler. While addition and subtraction are most easily done in Cartesian form, multiplication and division are often much simpler when the numbers are in polar form.

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To derive the rule for multiplying two complex numbers in polar form, suppose there are two complex numbers, z1 = r1 cis (θ1) and z2 = r2 cis (θ2). z1 z2 = r1 cis (θ1) × r2 cis (θ2) = r1(cos (θ1) + i sin (θ1)) × r2(cos (θ2) + i sin (θ2)) = r1 r2(cos (θ1) + i sin (θ1)) (cos (θ2) + i sin (θ2)) = r1 r2((cos (θ1) cos (θ2) − sin (θ1) sin (θ2)) + i (sin (θ1) cos (θ2) + cos (θ1) sin (θ2)) Using the trigonometric sum identities, cos (A + B) = cos ( A) cos (B) − sin ( A) sin (B) and sin (A + B) = sin ( A) cos (B) + cos ( A) sin (B), the expression can be simplified:

z1 z2 = r1 r2 cis (θ1 + θ2) z1, z2 r1, r2 θ1, θ2

are the two complex numbers being multiplied are the moduli of the two complex numbers are the arguments of the two complex numbers

Geometrically, multiplying a complex number z1 by z2 results in a rotation of z1 by the angle θ2 about the origin and a scaling of its modulus by a factor of r2.

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A similar rule applies for division:

z1 z2 z1, z2 r1, r2 θ1, θ2

r1 cis( 1 r2

are the two complex numbers being divided are the moduli of the two complex numbers are the arguments of the two complex numbers

Geometrically, dividing a complex number z1 by z2 results in a rotation of z1 by the angle − θ2 1 (a clockwise rotation by θ2) about the origin and a scaling of its modulus by a factor of . r2

Example 1 Consider the product 5 cis

2 3

3 cis

: 6

a Determine the value of the product in Cartesian form with exact values.

Create a strategy To multiply two complex numbers in polar form, multiply their moduli and add their arguments using the rule: z1 z2 = r1 r2 cis (θ1 + θ2). Then, convert the result to Cartesian form.

Apply the idea Multiply the moduli and add the arguments: 5 cis

2 3

3 cis

= 6

3 cis

= 15 cis

4 6

= 15 cis

3 6

= 15 cis

2 3

Multiply moduli and add arguments

Simplify moduli and write arguments 6 with common denominators Combine the arguments Simplify

Convert the complex number from polar form to Cartesian form: 15 cis

= 15 cos 2

i sin

Expand the cis notation

= 15 (0 i ( 1 ))

Evaluate the cosine and sine values

= 15i

Simplify to Cartesian form

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Reflect and check To verify the result, convert both complex numbers to Cartesian form first, then multiply. 2 : 3

First, convert 5 cis 2 3

5 cos

i sin

2 3

= Next, convert 3 cis 3 cos

1 3 Evaluate the cosine i 2 2 and sine values

= 5

5 15 i 2 2

Distribute 5

: 6 i sin

= 3

3 2

1 Evaluate the cosine 2 and sine values

3 3 i 2 2

Distribute 3

Now, multiply the Cartesian forms: 5 2

15 i 2

3 2

3 3 5 15 3 15 45 2 i = i i i 2 4 4 4 4 =

3 5 4

4 15 3 5 i 4 4

= 15i

Expand the brackets Simplify Combine like terms

This method is more calculation-intensive but will produce the same result of 15i. This comparison highlights the efficiency of multiplying in polar form.

b Describe the geometric effect on 5 cis

2 of multiplying it by 3 cis . 6 3

Apply the idea results in two geometric effects: 6 • A rotation about the origin by an angle of 6

Multiplying by 3 cis

• A scaling of the modulus by a factor of 3

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Example 2 Consider the division

3 3 3i : 5i

a Express 3 3 3i in polar form, r cis (θ ), with r ≥ 0 and − π < θ ≤ π.

Apply the idea First, find the modulus r: r = ( 3 3 )2 ( 3 )2 = 27 9

Write the formula for the modulus Substitute the values of a and b

= 36

Simplify

Evaluate

Next, find the argument θ. The complex number is in the fourth quadrant:

θ = tan

= tan

= Therefore, 3 3 3i

3 3 3 1 3

6 cis

Write the formula for the argument Substitute a and b and simplify Evaluate

. 6

b Express − 5i in polar form, r cis (θ ), with r ≥ 0 and − π < θ ≤ π:

Apply the idea The complex number − 5i can be written as 0 − 5i. It lies on the negative imaginary axis in the complex plane. The modulus, r, is the distance from the origin, which is 5. The argument, θ, is the angle measured from the positive real axis. For the negative imaginary axis, this angle is . 2 Therefore, 5i 5 cis . 2

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c Use the results from parts (a) and (b) to evaluate

3 3 3i , in Cartesian form. 5i

Apply the idea Divide the moduli and subtract the arguments: 3 3 3i = 5i

6 cis 5 cis

6 2

Substitute the polar forms from parts (a) and (b)

6 cis 5

3 6

6 2 cis 5 6

Evaluate the addition

6 cis 3 5

Simplify

Divide the moduli and subtract the arguments Write arguments with common denominators

Convert the complex number from polar form to Cartesian form: 6 6 = cis cos 3 3 5 5

i sin

Expand the cis notation

6 1 3 i 5 2 2

Evaluate the trigonometric functions

6 6 3 i 10 10

Distribute the modulus

3 3 3 i 5 5

Simplify

Reflect and check To verify the answer, perform the division directly in Cartesian form by multiplying the numerator and denominator by the conjugate of the denominator. 3 3 3i 3 3 3i = 5i ( 5i )

904

5i 5i

Multiply by the conjugate of the denominator

15 15 3i 25

Expand the numerator and denominator

3 3 3 i 5 5

Separate into real and imaginary parts and simplify

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary To multiply two complex numbers in polar form, multiply their moduli and add their arguments. z1 z2 = r1 r2 cis (θ1 + θ2) To divide two complex numbers in polar form, divide their moduli and subtract their arguments. z 1 z2

r1 cis ( 1 2 ) r2

Geometrically, multiplication and division correspond to a rotation and a scaling in the complex plane: • Multiplying complex numbers involves multiplying their moduli (scaling) and adding their arguments (rotation). • Dividing involves dividing moduli (scaling) and subtracting arguments (rotation).

Addition, subtraction, and comparison of forms While polar form is ideal for multiplication and division, it is not convenient for addition and subtraction. For these operations, Cartesian form is far simpler. To add or subtract complex numbers given in polar form, first convert them to Cartesian form. 1. Convert each complex number from polar form r cis (θ ) to Cartesian form a + bi using the conversion formulas: a = r cos (θ ) and b = r sin (θ ). 2. Perform the addition or subtraction on the Cartesian forms by adding or subtracting the real and imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d) i 3. If required, convert the resulting complex number back into polar form. Comparison of forms • Multiplication Multiplying in polar form involves one multiplication and one addition. Multiplying in Cartesian form involves four multiplications, one subtraction, and one addition. •

Polar form: (r1 cis (θ1)) (r2 cis (θ2)) = r1 r2 cis (θ1 + θ2)

•

Cartesian form: (a + bi) (c + di) = (ac − bd) + (ad + bc)i

• Addition Adding in Cartesian form is direct. Adding in polar form requires converting both numbers to Cartesian form first, which involves evaluating four trigonometric functions and performing four multiplications before the addition can even begin. •

Cartesian form: (a + bi) + (c + di) = (a + c) + (b + d)i

•

Polar form: To find r1 cis (θ1) + r2 cis (θ2), it is necessary to first calculate (r1 cos (θ1) + r2 cos (θ2) + i(r1 sin (θ1) + r2 sin (θ2))

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Example 3 Let z1

4 cis

and z2 3

2 cis

5 . Expressed in Cartesian form a + bi, find: 6

a z1 + z2

Create a strategy First, convert both z1 and z2 to Cartesian form. Then, add the real parts and the imaginary parts together.

Apply the idea Convert z1 to Cartesian form: 3

z1 = 4 cos =4

i sin

1 3 i 2 2

Expand the cis notation Evaluate the trigonometric functions

= 2 2 3i

Distribute the modulus

Convert z2 to Cartesian form: z2 = 2 cos

5 6

3 2

=2 =

i sin

5 6

1 2

Expand the cis notation Evaluate the trigonometric functions

Distribute the modulus

Now, add the two Cartesian numbers: z1 + z2 = 2 2 3i = 2

3 i 2 3 1 i

Substitute the Cartesian forms Group the real and imaginary parts

b z1 − z2

Create a strategy Using the Cartesian forms from part (a), subtract the real and imaginary parts.

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Apply the idea Subtract the Cartesian numbers: z1 − z2 = 2 2 3i

3 i

= 2

2 3 1 i

= 2

3 3

2 3 1 i

Substitute the Cartesian forms Group the real and imaginary parts Simplify

Reflect and check The results for addition and subtraction can be verified using a scientific calculator in complex number mode. Alternatively, converting the final answers back to polar form can serve as a check.

Idea summary To add or subtract complex numbers given in polar form, follow these steps: 1. Convert the complex numbers to Cartesian form (a + bi). 2. Perform the addition or subtraction. 3. Convert back to polar form if required. A summary of which form is best for which operation: • Polar form is best for multiplication, division, and finding powers. • Cartesian form is best for addition and subtraction.

9.07 Practice questions What do you remember? 1

Which form (Cartesian or polar) is generally more efficient for each of these operations? a

Addition

Subtraction

Multiplication

Division

Geometrically, what two transformations occur when a complex number z1 is multiplied by z2 = r2 cis (θ2)?

For each operation, identify which form (Cartesian or polar) is more efficient to use and explain your reasoning: 2 cis a (2 + 4i) + (1 − 5i) b 5 cis c (3 − 2i) × (1 + i) 3 6

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Practice 4

Evaluate, leaving your answer in polar form: a

Evaluate, leaving your answer in polar form: 2 cis (60°) a 4 cis (30°)

cis

6 cis

2 cis (80°) 8 cis (120°)

6 cis ( ) 2 8 cis 3

10 cis

d 2 cis

3 5

2 cis

3 2 5

Evaluate 4 cis

Evaluate, leaving your answer in Cartesian form, rounded to three decimal places: 3 5 cis 5 a 2 cis (45°) b 3 cis ( 60°) 3 cis 2

Describe the geometric transformation that occurs to z1 = 3 cis (50°) when it is multiplied by z2 = 2 cis (− 20°).

Ex 1b

Ex 2

Identify the quadrant each of these complex numbers is in: 7 a 3 cis (40°) × 2 cis (70°) b 5 cis 6 c

Ex 1a

4 cis (45°) × 3 cis (30°)

, leaving your answer in polar form.

Express 7 + 3i in polar form.

Express 2 + 3i in polar form.

Using your answers from parts (a) and (b), calculate polar form.

11 Ex 3

7 3i and express the answer in 2 3i

2 and z2 = 3 cis . Determine each in Cartesian form a + bi, leaving your 3 4 answers with exact values: Let z1 = 6 cis a

z1 + z2

a b

Express 3 i in polar form. Express 2 − 2i in polar form.

Calculate

z1 − z2

3 i ( 2 2i ) and express your answer in polar form.

Evaluate (1 − i)− 2, leaving your answer in polar form.

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By first expressing the numerator and denominator in polar form, simplify the expression: cos (4x) + i sin (4x) cos (5x) + i sin (5x)

Extend your thinking 15

By considering that (r cis (θ ))2 = r2cis (2(θ )), show that: a

cos (2x) = cos2(x) − sin2 (x)

sin (2x) = 2 sin (x) cos (x)

Let z = r cis (θ ) be a complex number. Determine a complex number w that, when multiplied by z, results in each geometric transformation: a

Rotate z by 90° anticlockwise, while leaving the length of z unchanged

Scale the length of z by 2 while leaving the argument of z unchanged

Scale the length of z by 0.5 and rotating z by 45° clockwise

Rotate z by 180° anticlockwise

Use polar form to determine a formula for zn.

For complex numbers z1 = r1 cis (θ ) and z2 = r2 cis (φ), show that

z1 z2

r1 cis ( r2

9.08 Complex inequalities and regions After this lesson, you will be able to… • describe regions in the complex plane using a combination of lines, rays, circles and ellipses • represent complex inequalities geometrically with bound regions

Rays and lines in the complex plane In the complex plane, rays and lines can be represented using complex numbers and their properties. A ray starts from a point and extends infinitely in one direction, while a line extends infinitely in both directions. These geometric objects can be described using inequalities and equations involving complex numbers.

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A ray can be described by the angle it makes from a fixed point. Im (z) 1 Re (z) –1 0 –1

–2

6 A ray starting from the point z0 and making an angle θ with the positive real axis.

π 8

–3

arg (z − (2 − 3i)) =

arg (z − z0) = θ z is a variable point on the ray z0 is the fixed starting point of the ray, which is excluded from the locus θ is the constant angle the ray makes with the positive real axis The principal value of the argument, often denoted Arg(z), is the unique value of the angle θ such that − π < θ ≤ π. A line can be described as the locus of points that are an equal distance from two fixed points. This locus forms the perpendicular bisector of the line segment connecting the two points. Im (z) A

4 2

–4

–2 0 –2

Midpoint Re (z) 2

4 B

Perpendicular bisector {z : |z − (− 2 + 4i)| = |z − (4 − 2i)|}

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The line of points z that are equidistant from two fixed points a and b. This line is the perpendicular bisector of the segment connecting a and b.

|z − a| = |z − b| z a b

is a variable point on the line is a fixed point in the complex plane is a second fixed point in the complex plane

The expression |z − a| can be thought of as the distance from the locus to point a, which can help geometrically interpret the locus for various equations. Horizontal and vertical lines in the complex plane are defined using the real and imaginary parts of z. • A vertical line is formed by fixing Re (z) = c, where c is a constant, representing all points with the same real part. • A horizontal line is formed by fixing Im (z) = k, where k is a constant, representing all points with the same imaginary part.

Example 1 For the equation {z : |z − 4i| = |z − 2|}: a Represent the given equation on the complex plane.

Create a strategy Sketch the perpendicular bisector of the line segment between 4i and 2.

Apply the idea The line segment between 4i and 2 has a gradient of − 2. Therefore, the perpendicular line must 1 1 have a gradient of , since − 2 × = − 1. 2 2 The midpoint of the line segment is (1, 2). 5 4 3 2 1 –5 –4 –3 –2 –1 0 –1 –2 –3 –4 –5

Im (z)

(1, 2) Re (z) 1 2 3 4 5

1 A line is sketched with a gradient of that passes 2 through (1, 2).

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b If z = x + yi, determine the Cartesian equation for the set of points.

Create a strategy Substitute z = x + yi into the given equation and use the definition of the magnitude to determine a Cartesian equation.

Apply the idea |z − 4i| = |z − 2|

Write the equation

|x + yi − 4i| = |x + yi − 2|

Substitute z = x + yi

|x + (y − 4)i| = |(x − 2) + yi|

Group real and imaginary parts

Use the definition |z|2 = a2 + b2

x + (y − 4) = (x − 2) + y

x2 + y2 − 8y + 16 = x2 − 4x + 4 + y2 − 8y + 16 = − 4x + 4 − 8y = − 4x − 12

Expand the brackets Subtract x2 and y2 from both sides Subtract 16 from both sides

4 12 y = x Divide both sides by − 8 and simplify 8 8 the signs of the terms y=

1 3 x 2 2

Simplify

Reflect and check Compare this equation to the line sketched to ensure the y-intercept and gradient match. 1 From the graph, the y-intercept of 1.5 matches the y-intercept in the equation. The gradient is for 2 both the graph and equation.

Example 2 Sketch arg (z − 1) =

in an Argand diagram. 3

Create a strategy The argument is an angle measured from the positive real axis. An equation in the form arg (z − z0) = θ describes a ray on an Argand diagram from z0.

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Apply the idea Im (z) 3 A ray is drawn from the point (1, 0) such that the angle made with the positive real axis is . The point (1, 0) is 3 excluded since arg (0) is undefined.

2 1 (1, 0) 0

60° 1

Re (z) 3

Idea summary Rays are described using an equation that defines a starting point and an angle, like arg(z − z0) = θ. The locus of points representing a line can be described as the perpendicular bisector of two points, using an equation like |z − a| = |z − b|.

Circles in the complex plane Interactive exploration Discover this concept in action online

mathspace.co

The equation |z| = r represents all the points, z, that are r units from the origin. This can be defined as a circle, centred at the origin, with radius r. In Cartesian form, this circle has equation x2 + y2 = r2. The equation for a circle can be shifted away from the origin:

|z − z1| = r z is a variable point on the circle’s circumference z1 is the fixed centre of the circle, z1 = a + bi r is the radius of the circle, where r > 0 This represents all points, z, which are r units away from a fixed point z1. If z1 is expressed as the complex number z1 = a + bi, then the Cartesian equation of this circle is (x − a)2 + (y − b)2 = r2, representing a circle centred at the point (a, b) with radius r.

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Example 3 Sketch the circle described by |z − 1 + 2i| = 2 on an Argand diagram.

Create a strategy Rearrange the equation into the standard form |z − z1| = r to identify the circle’s centre, z1, and its radius, r.

Apply the idea Manipulate the given equation to ensure the centre point is correct: |z − 1 + 2i| = |z − (1 − 2i)|

Factorise − 1

Therefore, this is a circle with a radius of 2 and centre (1 − 2i): Im (z) –1

Re (z)

–1 –2

(1, –2)

–3 –4

Idea summary The equation |z − z1| = r, where z1 = a + bi, describes a circle with radius r and centre z1, which corresponds to the point (a, b) in the Cartesian plane.

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Ellipses in the complex plane An ellipse is defined as the locus of points z in the complex plane such that the sum of the distances from two fixed points, called foci ( f1 and f2), is a constant value k. Im (z) 4 2 Re (z) –4

–2

–2 –4

This definition translates into the standard equation for an ellipse in the complex plane:

|z − f1| + |z − f2| = k z is the variable point on the ellipse f1 is the first fixed focus in the complex plane f2 is the second fixed focus in the complex plane k is the constant sum of distances from the foci For an ellipse to exist, the constant sum k must be strictly greater than the distance between the two foci, |f1 − f2|. If k = |f1 − f2|, the locus degenerates into a line segment between the foci. If k < |f1 − f2|, no such locus exists.

Interactive exploration Discover this concept in action online

mathspace.co

Example 4 Consider an ellipse defined by the equation |z − 2| + |z + 2| = 8 in the complex plane. Identify its foci and constant sum, and sketch the resulting ellipse.

Create a strategy Identify the foci f1 and f2 by comparing to |z − f1| + |z − f2| = k. Determine the constant sum k. Calculate the distance between the foci and verify the condition for an ellipse. Sketch the ellipse, locating the foci and estimating the vertices.

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Apply the idea Comparing |z − 2| + |z + 2| = 8 to the standard form |z − f1| + |z − f2| = k: • The foci are f1 = 2 (or 2 + 0i) and f2 = − 2 (or − 2 + 0i). • The constant sum is k = 8. Calculate the distance between the foci: |f1 − f2| = |2 − (− 2)| = |4| = 4 Verify the condition for an ellipse: k > |f1 − f2|. Since 8 > 4, the equation defines an ellipse. The centre of the ellipse is the midpoint of the foci, which is at the origin (0, 0). The major axis length is equal to k = 8. Since the foci are on the real axis, the major axis is k , 0 , so (±4, 0). horizontal. The vertices are at 2 f1 f2 4 2. The distance from the centre to each focus is c 2 2 2 k 2 c2 bminor, where bminor is the semi-minor axis length. For an ellipse, 2 k 2 42 = 22 bminor Substitute the semi-major axis = 4 2 and focal distance c = 2 2 16 = 4 bminor

Evaluate the squares

2 = 12 bminor

2 Rearrange the equation to solve for bminor

bminor = 12

Take the square root of both sides

=2 3 The minor axis endpoints are at 0, 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

Simplify 2 3 (approximately (0, ±3.46)).

Im (z)

Re (z) 2 3 4

The ellipse with foci at 2 and − 2, and a constant sum of distances 8.

Reflect and check Confirm that the foci are correctly identified and that the sum of distances from any point on the ellipse to the foci equals the constant k. The ellipse should be symmetric about the real and imaginary axes (since the foci are symmetric about the origin).

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Example 5 Determine the Cartesian equation for the ellipse defined by the equation |z − 3| + |z + 3| = 10.

Create a strategy To find the Cartesian equation, substitute z = x + yi into the given locus equation. Isolate one of the square root terms, square both sides to eliminate the root, simplify, and then square both sides again to eliminate the remaining square root. Rearrange the resulting equation into the standard Cartesian form of an ellipse.

Apply the idea |z − 3| + |z + 3| = 10

Write the equation

|(x − 3) + yi| + |(x + 3) + yi| = 10 Substitute z = x + yi ( x 3)2

( x 3)2

y2 = 10 Apply the modulus definition

( x 3)2

y2 = 10

( x 3)2

Isolate one square root y2 term

(x − 3)2 + y2 = 10

( x 3)2

Square both sides y2

2 x2 − 6x + 9 + y2 = 100 20 ( x 3)

( x 3)2

y2 Expand both sides

x2 − 6x + 9 + y2 = 100 20 ( x 3)2

x2 6 x 9

− 6x = 100 20 ( x 3)2

6 x Remove the common terms

y2 Expand the remaining squared term

20 ( x 3)2 y2 = 12x + 100 Isolate the square root term 5 ( x 3)2 y2 = 3x + 25 Divide both sides by 4 25((x + 3)2 + y2) = (3x + 25)2 Square both sides again 25(x2 + 6x + 9 + y2) = 9x2 + 150x + 625 Expand the expressions 25x2 + 150x + 225 + 25y2 = 9x2 + 150x + 625 Distribute the 25 16x2 + 25y2 = 400 Rearrange and simplify 16 x2 25 y2 =1 400 400

Divide both sides by 400

x2 y2 =1 25 16

implify to the standard S Cartesian form

x2 y2 1, which is an ellipse centred at the origin with a semi-major 25 16 axis of 5 and a semi-minor axis of 4. The Cartesian equation is

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Idea summary An ellipse in the complex plane is the locus of points z where the sum of the distances from two fixed points (foci, f1 and f2) is a constant k. Its standard complex equation is:

|z − f1| + |z − f2| = k z f1 f2 k

is the variable point on the ellipse is the first fixed focus in the complex plane is the second fixed focus in the complex plane is the constant sum of distances from the foci

It is important to distinguish this complex plane representation from the Cartesian equation of an ( x h)2 ( y k)2 ellipse, which typically involves x and y variables, like 1. a2 b2

Hyperbolas in the complex plane A hyperbola is the locus of points z where the absolute difference of the distances from two fixed foci ( f1 and f2) is a constant value k. Im (z) |z – f 1| 2

|z – f 2| z

f1 −2

−1

Re (z) 2

−2

This definition gives the standard equation for a hyperbola in the complex plane:

||z − f1| − |z − f2|| = k z is the variable point on the hyperbola f1 is the first fixed focus f2 is the second fixed focus k is the constant difference in distances, where 0 < k < ||f1 − f2| The equation |z − f1| − |z − f2| = k (without the outer absolute value) describes only one branch of the hyperbola. The sign of k determines which branch it is.

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Example 6 Identify and sketch the locus defined by the equation ||z − 5| − |z + 5|| = 6.

Create a strategy Compare the equation to the standard form ||z − f1| − |z − f2|| = k to find the foci and the constant difference. This equation describes a full hyperbola with two branches. The vertices lie on the line segment connecting the foci.

Apply the idea From the equation ||z − 5| − |z + 5|| = 6: • The foci are f1 = 5 and f2 = − 5. • The constant difference is k = 6. The distance between the foci is |5 − (− 5) | = 10. Since 0 < 6 < 10, the locus is a hyperbola. The centre is the midpoint of the foci, which is the origin. The length of the transverse axis is 2a = k = 6, so a = 3. The vertices are at (±3, 0). The distance from the centre to each focus is c = 5. For a hyperbola, c2 = a2 + b2. Substituting the values to find b gives 52 = 32 + b2, which becomes 25 = 9 + b2. This gives b2 = 16; therfore, b = 4. The asymptotes of the hyperbola are given by the lines y 4

b x, which are y a

4 x. 3

Im (z)

3 2 1 −6−5−4 −3 −2−1 −1

Re (z) 1 2 3 4 5 6

−2 −3 −4

The sketch shows a hyperbola with foci at ±5, vertices at ±3, and asymptotes y

4 x. 3

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Idea summary A hyperbola is the locus of points z where the absolute difference of distances from two foci, f1 and f2, is a constant k. The standard complex equation is:

||z − f1| − |z − f2|| = k z is the variable point on the hyperbola f1 is the first fixed focus f2 is the second fixed focus k is the constant difference in distances, where 0 < k < ||f1 – f2| The condition for a hyperbola is 0 < k < ||f1 − f2|.

Regions and inequalities in the complex plane Exploration Consider the diagram: Im (z) 2

Re (z) 0

1. Describe the shape of the region and consider the relationship that would describe this shape in the complex plane. 2. Describe the inequalities used to shade this region. 3. How would the inequalities change if the boundary lines were dashed instead of solid?

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When rays, lines, and circles are combined with inequalities, these boundaries define shaded regions or areas in the complex plane: • For circles, < refers to the region inside the circle while ≤ refers to the region inside the circle and on the circumference. • > refers to the region outside the circle while ≥ refers to the region outside the circle and on the circumference. Inequalities can also define other fundamental shapes: • An inequality of the form |z − a| < |z − b| describes the half-plane of points closer to a than to b. The boundary is the perpendicular bisector of the segment connecting a and b. • An inequality of the form θ 1 < arg(z − z0) < θ2 describes the region between two rays starting from z0, forming a sector. When the inequality symbols < or > are used, the line or curve is drawn as a dashed line to indicate that points on the line are not included. When shading regions, points can be tested to help determine the correct region in the plane. Regions can be combined using set notation. The intersection symbol, ∩ (meaning ‘and’), represents the overlapping area of two regions. The union symbol, ∪ (meaning ‘or’), represents the total area covered by both regions.

Example 7 Sketch the region defined by 0 ≤ Im(z) ≤ 2 ∩ |z| ≤ 2 in an Argand diagram.

Create a strategy For a complex number z = x + yi, Im(z) refers to the y-value. An inequality in the form |z − z1| ≤ r describes the inside of a circle, centred at z1, with radius r.

Apply the idea 2

Im (z)

Re (z) –2

The first inequality describes the region 0 ≤ y ≤ 2. The second inequality describes the inside of the circle centred at (0, 0) with a radius of 2. The required region is where both intersect (overlap).

–2

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Example 8 Sketch the region defined by the inequality z + z > Im(z).

Create a strategy For a complex number z = x + yi, Im(z) is y. The complex conjugate of z is z = x − iy.

Apply the idea Begin by rewriting the inequality in Cartesian form: z z > Im(z) x + iy + x − iy > y 2x > y

Write the inequality Substitute z = x + yi, z = x − yi and Im(z) = y Simplify

The line y = 2x divides the plane into two regions. A point can be tested to help determine the correct region to shade. Test a point, for example (1, 1). Substitute this point into the inequality and determine whether the resulting statement is true or false. If the statement is true, then the chosen point lies in the required region. If the statement is false, the point does not lie in the required region. Testing the point (1, 1) gives the inequality 2 > 1. This is true, therefore the region containing the point (1, 1). is shaded. The line y = 2x is a dashed line as it does not include points on the line.

4 3 2 1 –4 –3 –2 –1 0 –1 –2 –3 –4

Im (z)

Re (z) 1 2 3 4

Example 9 Digital only example Explore in the online lesson

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Example 10 Shade the region defined by z : z

0 arg(z)

on an Argand diagram.

Create a strategy First, identify the individual regions described by each inequality: • |z| ≤ 2 represents the region on or inside a circle. • 0 ≤ arg (z) ≤ represents a sector between two rays starting from the origin. 2 The symbol ∩ means intersection, so the final region is where these two regions overlap. The boundaries will be solid lines because the inequalities are ≤.

Apply the idea The inequality |z| ≤ 2 describes all points on or inside the circle with centre at the origin and radius 2. The inequality 0 ≤ arg(z) ≤ describes all points in the first quadrant, including the positive real and 2 imaginary axes, forming a sector with an angle of from the origin. 2 The intersection of these two regions is the part of the first quadrant that is on or inside the circle of radius 2. This forms a quarter-circle sector. Im (z) 2 The shaded region satisfies both conditions: it is inside the circle |z| = 2 and within the sector defined by 0 ≤ arg(z) ≤ . 2 Re (z) 0

Idea summary Inequalities define regions in the complex plane. The type of inequality determines the boundary: • < or >: The boundary is not included and is drawn as a dashed line. • ≤ or ≥: The boundary is included and is drawn as a solid line. Common regions include the interior or exterior of circles (|z − z1| < r), half-planes (|z − a| ≤ |z − b|), and sectors (θ1 ≤ arg(z) ≤ θ2). Use the intersection symbol ∩ (‘and’) for the overlap between regions, and the union symbol ∪ (‘or’) to combine the total area of regions.

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9.08 Practice questions What do you remember? 1

Match each relationship to its geometric description: a

The distance between the origin and z is always 3.

Re(z) = − 2

The distance from z and 1 is equal to the distance from z to − i.

|z − 1| = |z + i|

The real part of z is always − 2.

iii

arg(z) =

The argument of z from the origin is . 4

|z| = 3

Classify each expression as either vertical line or horizontal line: a

Re(z) = 3

Im(z) = − 2

Re(z + 2 − 3i) = 1

Im(z − 2i) = − 2

|z − (3 − i)| = 2

|z + i| = 1

For each circle, determine its centre and radius: a

|z| = 3

|z − 3| = 5

Determine an equation for a circle in the complex plane that has: a

Centre (2 + i) and radius 2

Centre (− 1 + 0i) and radius 3

Centre (3 − 4i) and radius 5

Centre (− 4 − 7i) and radius 4

Practice 5 Ex 1

6 Ex 2

For each equation: i

Represent the locus of points on an Argand diagram.

Determine the Cartesian equation.

|z − 3| = |z + 6i|

|z − 1 + i| = |z + 1 − i|

|z − 2 + 4i| = |z + 2 + i|

Sketch each expression in the complex plane: a

arg(z)

arg(z (2 2i))

5 6

arg(z)

arg(z ( 1 3i))

7 4

Sketch the region defined by the inequality Re(z) > Im(z).

Let z = x + iy. Sketch the graph of arg( z )

Sketch the graph of Re((3 − i) z) = 2.

Sketch the graph of Im((2 + 2i) z) = 0.

The graphs from parts (a) and (b) intersect at (1, − 1). Determine the Cartesian equation for each graph and show these equations form a pair of simultaneous equations with the solution x = 1 and y = − 1.

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10 Ex 3

Sketch the circle: a

|z| = 5

|z − 6| = 1

|z + i| = 2

Determine an equation of this circle in the complex plane. 4

Im (z)

2 Re (z) 0 –2

13 Ex 4, 5

etermine an equation of a circle in the complex plane that has a centre 0 + 0i and D radius 3.

Convert the answer into a Cartesian equation.

Consider the equation of an ellipse |z − 5| + |z + 5| = 14 in the complex plane: a

Identify the complex numbers representing the foci, f1 and f2, and determine the constant sum of distances, k.

Calculate the distance between the foci and verify that the condition for an ellipse (k > |f1 − f2|) is met.

Determine the coordinates of the vertices on the major axis and the minor axis.

Sketch the ellipse on a complex plane, labelling the foci and vertices.

Determine the Cartesian equation of the ellipse.

Sketch the region |z| ≤ 1 in the complex plane.

Sketch the region defined by the inequality |z| > 2.

Consider the region defined by |z + (3 − 4i)| ≤ 6. Determine whether each complex number lies within the region: a

17 Ex 6

2 + 2i

9 − 8i

Consider the equation of a hyperbola ||z − 4i| − |z + 4i|| = 4 in the complex plane: a

Identify the complex numbers representing the foci, f1 and f2, and determine the constant difference of distances, k.

Determine the coordinates of the vertices on the transverse axis.

Sketch the hyperbola on a complex plane, labelling the foci and vertices.

Sketch the region defined by |z − i| ≤ 1 ∩ Re(z) ≥ 0 on an Argand diagram.

Sketch the region defined by z

arg(z)

3 . 4

Ex 7

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Sketch the region defined by |z − 3| > |z + i| ∩ Im(z) > 0.

Sketch the region defined by the inequality Im(z z ) ≤ Re (z).

Ex 8

Determine an inequality of this region in the complex plane. Im (z) 4

Re (z) 0 23

a b

Determine an equation of a circle in the complex plane with radius 5 and centre 3 − 2i. 1 Re (z) 0 and the circle in part (a). Determine the intersection points of the line Im(z) 2

etermine an equation of a circle in the complex plane with centre z = 2 + 2i and D radius 1.

Determine an equation of a circle in the complex plane with centre z = − 1 + 3i and radius 3.

Determine if these two circles intersect, justifying your answer algebraically.

Sketch the region in the complex plane defined by the inequality |z − 3i| + |z + 3i| ≤ 10.

Sketch the region defined by z : z

Ex 9

arg( z)

on an Argand diagram.

Ex 10

Extend your thinking 27

Determine an equation of the circle in the complex plane for which the line segment connecting w = 2 − 5i and u = 3 − 3i is a diameter.

For the locus |z − 2i| = 1, identify the exact maximum value for |z + 2|.

Consider the circle with equation |z| = 3. Determine an equation of a line tangent to this circle at the point, where Re(z) = − 2 and Im(z) > 0. Provide the answer in the complex form Re(wz) = k, where w is a complex number and k is a real number.

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9.09

Complex roots

After this lesson, you will be able to. . . • determine when a quadratic equation has complex roots • solve quadratic equations over complex numbers • prove the complex conjugate theorem • identify conjugate roots in quadratic solutions

Complex roots of quadratics Quadratic equations can be expressed in the form P(x) = ax2 + bx + c, where a, b and c are real coefficients and a ≠ 0. When solving a quadratic equation, ax2 + bx + c = 0, a key method is using the quadratic formula: x

b2 4 ac 2a

Other methods for solving quadratic equations include factorising and completing the square, which will be discussed later for complex roots. When using the quadratic formula, the discriminant, Δ = b2 − 4ac, plays a key role in determining the nature of the roots of an equation: • If Δ > 0, the quadratic equation has two unequal real roots. • If Δ = 0, the quadratic has one real root. • If Δ < 0, the quadratic equation has no real roots. When Δ < 0, the two solutions can be found using complex numbers. Therefore, when the discriminant is negative, the quadratic equation is said to have complex solutions. Complex roots refer to solutions of an equation that involve complex numbers, which consist of a real and an imaginary part. These roots arise when an equation has no real solutions but still has valid solutions in the complex number system. For example, consider the quadratic equation x2 + 1 = 0. Using the fact that i

1, this equation can be solved: x2 = − 1 x=

x=±i Therefore, i and − i are roots (solutions) of the equation x2 + 1 = 0. The quadratic is said to have two complex solutions which in this case, are pure imaginary and form a conjugate pair (0 + i and 0 − i).

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The complex conjugate root theorem states: For a polynomial P(x) with real coefficients, if P(x) = 0 has a complex root a + bi, then it also has the root a − bi. This theorem arises from the requirement that the coefficients of a polynomial remain real, which ensures that the terms involving i cancel out during expansion. For example, in the polynomial P(x) = x2 − 2x + 5, the roots are 1 + 2i and 1 − 2i. Their conjugate relationship preserves the polynomial’s real coefficients. This property is particularly useful when solving polynomials or factorising them over the set of real numbers, as it guarantees that complex roots appear in conjugate pairs. It is also important to note that a polynomial of degree n has n solutions or roots. So a quadratic equation, which is of degree 2, has exactly two roots, while a cubic equation, which is of degree 3, has exactly 3 roots. For a quadratic function with real coefficients that has complex solutions, the solutions will always form a conjugate pair. Similarly, for a cubic function with real coefficients that includes complex solutions, there will be one real solution and a conjugate pair of complex solutions.

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Example 2 Determine a quadratic equation with real coefficients so that 1 + 4i is a root.

Create a strategy Use the complex conjugate root theorem to determine both roots, write as factors and expand the expression.

Apply the idea Since 1 + 4i is a root, by the complex conjugate root theorem, 1 − 4i is also a root. The quadratic equation can be formed by multiplying the factors corresponding to these roots: 0 = (x − (1 + 4i)) (x − (1 − 4i))

Write the formula

= (x − 1 − 4i) (x − 1 + 4i)

Expand each factor

= (x − 1) − (4i)

Expand the brackets

Simplify

= x − 2x + 1 + 16 = x − 2x + 17

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Express as difference of two squares

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Example 3 Determine the roots of P(x) = 2x3 − 5x2 − 2x + 15, given one root is 2 + i.

Create a strategy To determine the roots, use the complex conjugate root theorem. Then, perform polynomial division to determine the remaining real root.

Apply the idea If 2 + i is one solution, then by the complex conjugate root theorem, 2 − i must also be a root. Determine the quadratic expression formed by the complex roots: 0 = (x − (2 + i)) (x − (2 − i))

Write the formula

= (x − 2 − i) (x − 2 + i)

Expand each factor

= (x − 2)2 − (i)2

Express as difference of two squares

Expand the brackets

Simplify

= x − 4x + 4 + 1 = x − 4x + 5

Perform polynomial division to determine the remaining factor: 2x + 3 x2 – 4x + 5 2x3 – 5x2 – 2x + 15 –(2x3 – 8x2 + 10x) 3x2 – 12x + 15 –(3x2 – 12x + 15) 0 Equate the factor to zero: 2x + 3 = 0

Equate factor to zero

2x = − 3

Subtract 3 from both sides

3 x= 2

Divide both sides by 2

3 Therefore, the roots of the polynomial are 2 + i, 2 − i and . 2

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Idea summary Complex roots of a quadratic equation ax2 + bx + c = 0 (where a, b, c are real) occur when the equation has no real solutions. These roots involve complex numbers. The discriminant, Δ = b2 − 4ac, determines the nature of the roots: • If Δ > 0, there are two distinct real roots. • If Δ = 0, there is one real repeated root. • If Δ < 0, there are two complex conjugate roots. The complex conjugate root theorem states that for a polynomial P(x) with real coefficients, if a + bi is a root, then its conjugate a − bi is also a root. The quadratic formula to find roots is x

b2 4 ac . 2a

Solve quadratic equations with complex roots All strategies that were used to solve quadratic equations with real roots can also be used for equations that have complex roots. The key strategies are using the quadratic formula, completing the square, and factorisation: • Solving the sum of two squares (ax2 + c = 0): A special case of quadratic equations occurs when the b term is zero, resulting in the form ax2 + c = 0 (where a and c have the same sign). These can be solved directly by isolating x2 c to get x2 and then taking the square root of both sides. This will yield a pair of purely a c i. This form is also known as a ‘sum of two squares’ and imaginary conjugate roots: x a can be factorised directly. • Using the quadratic formula: The quadratic formula, x

b2 4 ac , can be used to determine real or complex 2a

solutions of any quadratic equation ax2 + bx + c = 0 with real coefficients. When the discriminant Δ = b2 − 4ac is negative, the term b2 4ac will involve 1 i, leading to complex roots. The ± sign in the formula ensures that if complex roots exist, these roots will appear as a conjugate pair. For example, if b2 − 4ac = − k where k > 0, then b2 4 ac be x

930

ki 2a

b 2a

k i , which are complex conjugates. 2a

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i k. The roots would then

• Completing the square: Completing the square can also be used to solve quadratic equations with complex roots. The process transforms the equation ax2 + bx + c = 0 into the form (x + p)2 = q. If q is negative, taking the square root will introduce complex numbers. The steps are generally: 1. If a ≠ 1, divide all terms by a. Let the equation be x2 + Bx + C = 0 where B

b c , C . a a

2. Move the constant term C to the right side: x2 + Bx = − C 3. Add

B 2

to both sides to complete the square on the left: x2

B 2

4. Factorise the left side as a perfect square: x

B 2

5. Take the square root of both sides. If the term under the square root is negative, the roots will be complex. x

B 2

6. Solve for x: x

B 2

• Factorisation: To solve a quadratic equation with complex solutions using factorisation, express the quadratic expression as a product of two linear factors, introducing imaginary numbers where necessary. For more general quadratics ax2 + bx + c = 0, if the roots are z1 and z2, the factorised form is a (x − z1) (x − z2) = 0. Finding these factors directly can be challenging if b ≠ 0. Sometimes, completing the square first can help reveal a difference of two squares involving complex numbers. For example, (x + p)2 − (qi)2 = (x + p − qi) (x + p + qi). While the quadratic formula and completing the square always work, factorisation for complex roots might require insight or be combined with other methods.

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Example 5 Solve the equation, x2 + 4x + 5 = 0, using the quadratic formula.

Create a strategy b

Identify a, b, and c and substitute into the quadratic formula: x

b2 4 ac . 2a

Apply the idea For the equation x2 + 4x + 5 = 0, the coefficients are a = 1, b = 4, and c = 5.

= =

4 (1) (5)

2 (1) 4

16 20 2

= =

2 4

2i 2

4 2

2i 2

= −2±i

Substitute a = 1, b = 4, c = 5 into the formula Evaluate the squares and products Simplify the radicand Evaluate

Separate into real and imaginary parts Simplify

The solutions are x = − 2 + i and x = − 2 − i.

Reflect and check The solutions − 2 + i and − 2 − i are a complex conjugate pair, as expected for a quadratic equation with real coefficients and complex roots.

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Example 6 Solve the equation, x2 + 2x + 2 = 0, by completing the square.

Create a strategy Manipulate the equation to form a perfect square on the left-hand side, then solve for x.

Apply the idea Given equation: x2 + 2x + 2 = 0 1. The coefficient of x2 is already 1. 2. Move the constant term to the right side: x2 + 2x = − 2 3.

To complete the square for x2 + 2x, x2 + 2x + 1 = − 2 + 1 2

(x + 1) = − 1 Take the square root of both sides: x+1=

x+1=±i 5.

2 2

1 is added to both sides:

Write the equation Simplify Write the equation Evaluate

Solve for x: x = −1 ± i

The solutions are x = − 1 + i and x = − 1 − i.

Example 7 Solve 4x2 + 9 = 0 by factorising into a product of linear factors.

Create a strategy Rewrite the equation as a difference of two squares using the fact that a2 + b2 = a2 − (ib)2, then factorise and solve.

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Apply the idea Given equation: 4x2 + 9 = 0 4x2 + 9 = 0 2

Write the equation

Rewrite terms as squares

Use A2 + B2 = A2 − (Bi)2

(2x) + 3 = 0 2

(2x) − (3i) = 0 (2x − 3i) (2x + 3i) = 0

Factorise using difference of squares

Set each factor to zero to find the solutions: Solution 1: 2x − 3i = 0

Write the equation

2x = 3i x=

Add 3i to both sides

3 i 2

Divide both sides by 2

Solution 2: 2x + 3i = 0

The solutions are x

3 i and x 2

Write the equation

2x = − 3i

Subtract 3i from both sides

3 x= i 2

Divide both sides by 2

3 i. 2

Reflect and check This equation could also be solved using inverse operations: 4x2 + 9 = 0

Write the equation

4x = − 9 x2 =

9 4

Divides both sides by 4 9 4

= =

Subtract 9 from both sides

4 3i 2

Take the square root of both sides Rewrite Evaluate

The results match.

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Example 8 Solve x2 + 2x + 2 = 0 by factorisation, after first completing the square.

Create a strategy First, complete the square for x2 + 2x + 2 to rewrite it in the form (x + p)2 + q. Then, rewrite (x + p)2 + q as a difference of two squares involving i, factorise, and solve.

Apply the idea Given equation: x2 + 2x + 2 = 0. Step 1: Complete the square for x2 + 2x + 2. x2 + 2x + 2 = (x2 + 2x + 1) + 2 − 1 = (x + 1)2 + 1 Therefore, the equation becomes (x + 1)2 + 1 = 0. Step 2: Rewrite as a difference of two squares. Since 1 = − (− 1) = −(i2), the expression becomes (x + 1)2 − (i)2 = 0. Step 3: Factorise using A2 − B2 = (A − B) (A + B). ((x + 1) − i) ((x + 1) + i) = 0 (x + 1 − i) (x + 1 + i) = 0

Write the equation Simplify

Step 4: Solve by setting each factor to zero. Solution 1: x+1−i=0

Write the equation

x = − 1 + i Subtract 1 from both sides and add i to both sides Solution 2: x+1+i=0 x = −1 −i

Write the equation Subtract 1 and i from both sides

The solutions are x = − 1 + i and x = − 1 − i. This matches the result from solving by completing the square directly.

Idea summary Quadratic equations ax2 + bx + c = 0 with complex roots can be solved using several methods: b b2 4 ac • Quadratic formula: x . If b2 − 4ac < 0, the roots are complex 2a conjugates. • Completing the square: Transform the equation into (x + p)2 = q. If q < 0, taking the square root yields complex solutions. • Factorisation: Express the quadratic as a product of linear factors, possibly involving complex numbers (e.g., A2 + B2 = (A − Bi) (A + Bi)). This may sometimes involve completing the square first for more general quadratics.

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Proof of the complex conjugate root theorem The complex conjugate root theorem is a fundamental result concerning polynomials with real coefficients. • Statement of the theorem If P (x) = anxn + an−1 xn−1 + . . . + a1x + a0 is a polynomial where all coefficients an, an−1, . . ., a0 are real numbers, and if z = a + bi is a complex root of P(x) = 0 (where b ≠ 0), then its complex conjugate z a bi is also a root of P (x) = 0. The proof involves showing that if P (z) = 0, then P ( z ) must also be zero. This relies on the properties of complex conjugates: • The conjugate of a sum is the sum of the conjugates: z1 z2 z1 z2 . • The conjugate of a product is the product of the conjugates: z1 z2 z1 z2 (and by extension zn (z)n). • A real number is its own conjugate: if k is real, then k k.

Example 9 Prove the complex conjugate root theorem for a general polynomial P(x) with real coefficients.

Create a strategy Let P(x) = anxn + an−1 xn−1 + . . . + a1 x + a0, where all ai are real. Assume z is a complex root, so P(z) = 0. The objective is to show that P ( z ) 0 by taking the conjugate of the equation P(z) = 0 and using properties of conjugates.

Apply the idea Let P(x) = anxn + an−1xn−1 + . . . + a1x + a0, where a0, a1, . . ., an are real numbers. Suppose z is a complex root of P(x) = 0. Then, by definition: P(z) = an zn + an−1 zn−1 + . . . + a1 z + a0 = 0 Take the complex conjugate of both sides of this equation: an zn

an 1 zn 1

a1 z a0

Since 0 is a real number, 0 0. Using the property that the conjugate of a sum is the sum of the conjugates ( A B an zn

an 1 zn 1

a1 z a0

A B) :

Using the property that the conjugate of a product is the product of the conjugates (AB AB): an zn

an 1 zn 1

a1 z a0

Since all coefficients ai are real numbers, their conjugates are themselves (ai ai). Also, the conjugate of a power is the power of the conjugate zk an ( z )n

an 1 ( z )n 1

( z )k . a1 z a0

This last expression is precisely P ( z ). P( z )

Therefore, if z is a root of P(x) = 0 (where P(x) has real coefficients), then its complex conjugate z is also a root.

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Idea summary The complex conjugate root theorem states that for any polynomial with real coefficients, if a complex number z = a + bi is a root, then its conjugate z a bi must also be a root. The proof relies on the properties of complex conjugates, specifically that conjugation distributes over addition and multiplication, and that real numbers are their own conjugates. If P(z) = 0, then by conjugating the entire equation and applying these properties, it can be shown that P ( z ) 0. Geometrically, this means that complex roots of polynomials with real coefficients always appear in pairs that are reflections of each other across the real axis in the Argand diagram.

9.09 Practice questions What do you remember? 1

Match each value of the discriminant, Δ = b2 − 4ac, to the nature of the roots for a quadratic equation with real coefficients: a

Δ>0

One real root

Δ=0

Two complex conjugate roots

Δ<0

iii

Two distinct real roots

Determine the complex conjugate root theorem by completing this statement:

For a polynomial with ⬚ coefficients, if a + bi is a root, then ⬚ must also be a root.

Express the quadratic formula used to solve the equation ax2 + bx + c = 0.

A cubic polynomial with real coefficients has roots at x = 5 and x = 2 − i. What must the third root be?

Practice 5

Calculate the discriminant for each equation and classify the nature of its roots: a

Ex 1

c 6

P(x) = x2 + 4x + 5 = 0 2

P(x) = x − 4x + 4 = 0

P(x) = x2 + 2x − 1 = 0

P(x) = 2x2 + 1 = 0

The roots of a quadratic equation with real coefficients are 3 3i. Determine the equation.

Ex 2

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Determine a quadratic equation with real coefficients given that 1 + i is one of its roots.

Consider the polynomial P(x) = 5x2 + 5x + 5: a

Determine the roots of this polynomial.

Identify the relationship between these roots.

A complex root of the equation x2 + 2x + 4 = 0 is given by x = A + Bi. Determine the value of A.

Solve these quadratic equations: a

Ex 3

(x + 4)2 = − 16

Given that 3 − i is a root of the polynomial P(x) = x3 − 4x2 − 2x + 20, determine all the roots of P(x).

Solve these equations for x, expressing your answers in terms of i: a

Ex 4

c 13

14 Ex 5

Ex 6

16 Ex 7

17 Ex 8

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4x + 25 = 0

x2 + 5 = 0

2x2 + 36 = 0

x2 + 9 = 0

(x + 1)2 + 25 = 0

P(x) = 5x2 + 2x + 1 = 0

P(x) = 2x2 − 4x + 3 = 0

2x2 − 4x + 5 = 0

Solve using the quadratic formula: a c

x2 + 9 = 0

Solve by factorising: a

Ex 9

(x − 3)2 = − 49

P(x) = x2 + 3x + 4 = 0 2

P(x) = x + x + 4 = 0

Solve by completing the square: a

x2 + 6x + 10 = 0

Solve these equations by factorising into a product of linear factors: a

x2 + 16 = 0

9x2 + 25 = 0

4x2 + 3 = 0

x + 7 = 0

For each equation, complete the square to form a difference of two squares, then factorise to solve for x: a

x2 + 2x + 5 = 0

x2 − 8x + 25 = 0

x2 − 10x + 26 = 0

x + 4x + 13 = 0

Consider the quadratic equation, ax2 + bx + c = 0, where a, b, c ∈ ℝ. If z is one of the complex solutions of the quadratic equation: a

Explain why az2 + bz + c = 0.

By considering properties of conjugates, show that a( z )2 b z c 0.

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Extend your thinking x2 2 Ax

5 2

For which values of A does P(x)

Given x = 3 + 4i is a root of P(x) = x2 + Ax + B, where A and B are real numbers, determine the values of A and B.

The quadratic equation x2 + Ax + 20 = 0 has a complex root x = B + 2i, where A and B are real numbers. Determine the possible values of A.

Prove that for a quadratic equation ax2 + bx + c = 0 with real coefficients, if z is a complex root, then its conjugate z must also be a root.

Solve: x4 + 4x2 + 4 = 0

0 have complex roots?

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Did you know?

Complex roots play a crucial role in understanding real-world systems that involve oscillations and rotations! For example, a spinning bicycle wheel can be modelled using equations whose solutions include complex roots. By analysing these roots, engineers can predict the wheel’s wobble, determine when it’s most stable, and design mechanisms that minimise vibration. This helps ensure the ride feels smoother, safer, and more efficient on all types of terrain.

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Big ideas • Geometric transformations that shift, scale, turn, and flip shapes in the plane can be represented algebraically using vectors and matrices. While a translation is defined by vector addition, linear transformations such as dilations, rotations, and reflections are each defined by a unique 2 × 2 matrix, allowing complex geometric operations to be performed through systematic matrix multiplication. • The algebraic framework of matrices provides powerful insights into the properties of transformations. A sequence of transformations corresponds to the product of their matrices, and the inverse matrix represents the ‘undoing’ of a transformation. Furthermore, a single number, the determinant, provides a direct geometric interpretation of a transformation’s effect on area and orientation.

10 Transformations Chapter outline Chapter outline 10.01 10.02 10.03 10.04 10.05 10.06 10.07

Translations Dilations Rotations Reflections Compositions of linear transformations Inverse of linear transformations Area of linear transformations Investigation: Transformations with code Chapter 10 review

942 949 958 969 978 987 995

In nature, butterfly wings show nearly perfect reflection symmetry. That’s maths in flight.

10.01

Translations

After this lesson, you will be able to… • understand translations and their representation as column vectors • apply translations to points in the plane using coordinate notation • identify invariant points under translations

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Transformation A function that maps an original (pre-image) point P (x, y) to a transformed (image) point P′ (x′, y′). Translation A transformation that shifts each point of a figure or a graph the same distance in the same direction.

y P′ (x′, y′) T P (x, y) 0

Transformations of points on a Cartesian plane can be represented using matrices. For a translation, this is done by adding a translation vector to the point’s coordinates. Other transformations, such as reflections and rotations, involve matrix multiplication.

A translation can be represented using matrix addition. The translation matrix equation defines how a pre-image point P (x, y) is mapped to an image point P′ (x′, y′):

P′ = P + T P′′ is the matrix representing the image coordinates,

x y

P is the matrix representing the pre-image coordinates, T is the translation vector,

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a b

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x y

Expanding this gives the mapping from the pre-image coordinates to the image coordinates. x y

x y

a b

x a y b

This can also be written using coordinate notation as (x, y) → (x + a, y + b). T

a is called the translation vector. b • If a is negative, the point is moved |a| units left. • If a is positive, the point is moved |a| units right. • If b is negative, the point is moved |b| units down. • If b is positive, the point is moved |b| units up. Invariant point An invariant point is a point that remains unchanged after a transformation. For a point P to be invariant, its image P′ must be identical to P.

For a translation, an invariant point exists if P′ = P. P′ = P x y

x a = y b a x = b y

x y

a 0 = b 0

0 This shows that invariant points only exist if the translation vector is the zero vector, . 0 In this case, every point is an invariant point because no translation occurs. For any non-zero translation, there are no invariant points.

Example 1 Determine the image coordinates of triangle ABC with vertices A (2, − 3), B (0, 1) and C (− 1, 2) 5 under the translation vector T . 1

Create a strategy First, convert the pre-image coordinates into column matrices, then apply the translation matrix equation to determine the image coordinates.

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Apply the idea The vertices of the triangle can be written as a single 2 × 3 pre-image matrix. Then, the image of all vertices can be found at once. Pre-image matrix: P

2 0 3 1

1 2

To find the image matrix P′, the translation vector is added to each column of the pre-image matrix. P′ =

2 0 3 1

1 2

2 5 0 5 3 1 1 1

7 5 4 4 0 1

5 5 5 –1 –1 –1

Add the translation vector to each point

1 5 2 1

Add the corresponding elements Evaluate

Therefore, the image coordinates are A′ (7, − 4), B′ (5, 0) and C′ (4, 1).

Reflect and check The translation vector T right and 1 unit down.

5 means to move 5 units to the 1

Graph the original triangle ABC and the image triangle A′B′C′. This graph shows that triangle ABC has been shifted 5 units to the right and 1 unit down to A′B′C′, so the image coordinates are correct.

4 3

C 2 1 −1 −1 −2 −3

C′

x B′ 1 2 3 4 5 6 7

A A′

−4

Example 2 Determine the equation of the curve y = x2 under the translation T

1 . 4

Create a strategy Use the translation equation to write the original coordinates, x and y, in terms of the new coordinates, x′ and y′. Then, substitute these expressions into the curve’s original equation.

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Apply the idea Substitute the translation vector into the translation matrix equation: P′ = P + T

Write the translation matrix equation

x x = y y

1 4

x x 1 = y y 4

Substitute the matrix values Evaluate

Applying equivalent matrices, two simultaneous equations are obtained: x′ = x + 1

y′ = y − 4

Rearrange 1 and 2 to make x and y the subject: x = x′ − 1

y = y′ + 4

Substitute 3 and 4 into the curve to be translated: y = x2

Write the equation of the curve 2

y′ + 4 = (x′ − 1) Substitute the expressions for x and y from equations 3 and 4 y′ = (x′ − 1)2 − 4

Subtract 4 from both sides 2

Therefore, the image equation is y′ = (x′ − 1) − 4.

Reflect and check 1 means to move 1 unit to the right and 4 units down. 4 This translation can also be represented using coordinate notation: (x, y) → (x + 1, y − 4) The translation vector T

This gives the image coordinates (x′, y′) in terms of the pre-image coordinates (x, y). So, x′ = x + 1 and y′ = y − 4. To find the equation of the translated curve, it is necessary to express x and y in terms of x′ and y′. Rearranging gives x = x′ − 1 and y = y′ + 4. Substituting these into the original equation y = x2 gives y′ + 4 = (x′ − 1)2. This simplifies to y′ = (x′ − 1)2 − 4, which matches the answer found using matrix methods. This can also be verified graphically:

y 4 3 2 y = x2 1 −4 −3 −2 −1 −1 −2 −3 −4

x 1

2 3 4

y′ = (x′ − 1)2 − 4

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Example 3 Show that the translation defined by the vector T

3 has no invariant points. 2

Create a strategy For an invariant point, the image point P′ (x′, y′) must be the same as the pre-image point P (x, y). Set up the translation equation with this condition and show that it leads to a contradiction.

Apply the idea An invariant point must satisfy the condition P′ = P, which means x′ = x and y′ = y. The translation matrix equation is: x y

x y

3 2

x x = y y

3 2

Substitute x′ = x and y′ = y:

x y

3 x = 2 y 3 0 = 2 0

This results in the statement

0 0

3 , which is false. 2

Therefore, there are no points that satisfy the condition for invariance under this translation. The translation has no invariant points.

Reflect and check As shown in the theory, the only translation that has invariant points is the zero translation, where 0 . Since the given translation vector is non-zero, it can be concluded immediately that there T 0 are no invariant points.

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Idea summary A transformation is a function that maps an original (pre-image) point P (x, y) to a transformed (image) point P′ (x′, y′). A translation is a transformation that shifts every point the same distance in the same direction. A translation can be represented using the matrix equation: P′ = P + T x x = y y

a b

x x = y y

a b

Or by using coordinate notation: (x, y) → (x + a, y + b) a is the translation vector. b

• If a is negative/positive, the point is moved |a| units left/right. • If b is negative/positive, the point is moved |b| units down/up. • For a translation, there are no invariant points (points that remain unchanged) unless the 0 translation vector is , in which case all points are invariant. 0

10.01 Practice questions What do you remember? 1

Calculate: 3 5 a 4 2

2 0

1 6

Write the translation vector for each transformation: a

5 units left and 3 units down

2 units up and 3 units right

6 units up

7 units down

For each point, find its image after the given translation: 6 a Point A (− 5, 3) with T b Point B (4, − 2) with T 1 c

Point C (− 10, − 4) with T

8 7

Point D (5, 2) with T

1 5 –8 4

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Practice 4

Describe in words the translation represented by the vector T

A translation is defined by the mapping (x, y) → (x − 4, y + 6):

Write this translation as a column vector T.

Find the image of the point P (3, − 2) under this translation.

Determine the translation vector that transforms the pre-image point to the image point: a

7 Ex 1

7 . 4

B (5, 6) to B′ (7, 8)

H (− 5, − 8) to H′ (0, − 10)

Determine the image coordinates of the parallelogram ABCD with vertices A (− 1, 3), B (4, 3), 3 C (1, 7) and D (− 4, 7) under the translation vector . 4 1 and 4

Determine the image of the point D (5, − 7) under successive translations T1 5 . T2 1

A point, F (3, − 4), undergoes these translations in order: 3 units right, 5 units down, 2 units left and 1 unit up. Use matrix calculations to find the coordinates of its final image.

Which translation has invariant points? A

11 Ex 2

2 2

0 4

The equation of a circle is x2 + y2 = 4. It is translated by the vector T a

Use matrix transformations to find the equation of the image.

Verify your result with a sketch.

A parabola with the equation y = 3x2 − 4 is translated by the vector T a

Find the equation of the image.

Verify your result with a sketch.

Show that the translation defined by the vector T

Ex 3

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0 0 4 : 2

3 : 4

5 has no invariant points. 1

Extend your thinking 14

A point A (3, − 9) is translated by successive translations T1

10 and T2 such that the 5

image point is A′ (3, − 7). Determine the second translation vector, T2. 15

Determine the translation vector that transforms the parabola y = x2 to the image parabola y = x2 + 6x + 15.

Determine the translation vector that transforms the circle x2 + y2 = 15 to the image circle x2 + y2 − 4x + 6y − 2 = 0.

10.02

Dilations

After this lesson, you will be able to… • apply dilations of the form (x, y) → (ax, by) to transform shapes in the plane • identify invariant properties under dilation transformations • represent dilation with a factor parallel to the x-axis and a factor parallel to the y-axis with 2 × 2 matrices

Dilations Interactive exploration Discover this concept in action online

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Linear transformation Any transformation that can be represented by a 2 × 2 matrix. Linear transformation matrix equation The equation P′ = TP, which relates the pre-image points (P) to the image points (P′) using a transformation matrix (T ). Dilation A linear transformation that changes the size of a pre-image by enlarging or reducing it.

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The linear transformation matrix equation is:

P′ = TP P′ is the matrix of image coordinates T is the 2 × 2 transformation matrix P is the matrix of pre-image coordinates Note that the transformation matrix T is always written before the pre-image matrix P. This order is important because matrix multiplication is not commutative, meaning TP ≠ PT. A dilation of factor a parallel to the x-axis and factor b parallel to the y-axis transforms any point (x, y) to an image point (x′, y′) using the mapping rule (x, y) → (ax, by). This is equivalent to the matrix operation: x y

a 0 0 b

x y

ax by

The transformation for a dilation of factor a parallel to the x-axis is T stretched or compressed horizontally. y 4 3 2 1 0

4 3 2 1

1 2 3 4 5 6

Pre-image

4 3 2 1

1 2 3 4 5 6

Pre-image

950

2 0 0 1

Image

The transformation for a dilation of factor b parallel to the y-axis is T stretched or compressed vertically.

a 0 . The pre-image is 0 1

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 0 . The pre-image is 0 b T

x 1 2 3 4 5 6 Image

1 0 0 2

a 0 0 b

T is the dilation transformation matrix a is the dilation factor parallel to the x-axis b is the dilation factor parallel to the y-axis 6 5 4 3 2 1 0

x 1 2 3 4 5 6 Pre-image

6 5 4 3 2 1 0

2 0 0 3

x 1 2 3 4 5 6 Image

Note that: • When the dilation factor is greater than 1, the image is enlarged. • When the dilation factor is between 0 and 1, the image is reduced. • When the dilation factor, a, is less than 0, the image is reflected across the y-axis. • When the dilation factor, b, is less than 0, the image is reflected across the x-axis. An invariant property is a feature of a shape or a set of points that does not change after a transformation is applied. The properties that are invariant under a dilation of the form (x, y) → (ax, by) include: • The origin (0, 0). A point at the origin is a fixed point under this transformation. • Collinearity: Points that are on a straight line in the pre-image remain on a straight line in the image. • Parallelism: Lines that are parallel in the pre-image remain parallel in the image. • Orientation: The order of vertices (e.g., clockwise) is preserved as long as a and b are positive. If either factor is negative, the shape is reflected, and the orientation is reversed. Properties that change under this type of dilation (unless the factor is 1) include: • Side lengths: A segment of length L parallel to the x-axis becomes length ∣a∣ L. A segment parallel to the y-axis becomes length ∣b∣ L. • Area: The area of a shape is scaled by a factor of ∣ab∣. • Angle measures are generally not preserved unless the dilation is uniform (a = b). For example, a square can be dilated into a non-square rectangle.

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Example 1 Determine the image of point A (3, 5) under a dilation of factor 3 parallel to the y-axis.

Create a strategy First, identify the dilation factors a and b from the description. Then, apply the coordinate mapping rule (x, y) → (ax, by) to find the new coordinates.

Apply the idea A dilation parallel to the y-axis does not change the x-coordinate. This means the dilation factor parallel to the x-axis is a = 1. The dilation factor parallel to the y-axis is given as b = 3. The mapping rule is (x, y) → (1x, 3y). Substitute the coordinates of point A(3, 5) into the rule: (3, 5) → (1 × 3, 3 × 5) (3, 5) → (3, 15) Therefore, the image point is A′ (3, 15).

Reflect and check The image point can also be determined using the matrix method. The transformation matrix is 3 1 0 T , and the pre-image point is A . 5 0 3 A′ = TA

Write the formula

1 0 3 = Substitute the values of the matrices 0 3 5 =

3 15

Evaluate

Both methods give the same image point, A′ (3, 15).

Example 2 Digital only example Explore in the online lesson

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Example 3 Determine the equation of the image of the circle x2 + y2 = 4 after it is dilated by a factor of 2 parallel to the x-axis.

Create a strategy To find the equation of the image, expressions for x and y need to be substituted into the original equation. x x 1. Set up the matrix equation . T y y 2. Use this to write equations for x′ and y′. 3. Rearrange these equations to make x and y the subjects. 4. Substitute the expressions for x and y into the circle’s equation, x2 + y2 = 4.

Apply the idea The transformation matrix for a dilation of factor 2 parallel to the x-axis is T

2 0 . 0 1

Apply the linear transformation matrix equation: Write the linear transformation matrix equation

P′ = TP x 2 0 = y 0 1

x y

Substitute the values of the matrices

x 2x = y y

Evaluate

By comparing the elements in the matrices, this gives two simple equations: x′ = 2x

y′ = y

Rearrange 1 to make x the subject: x=

x 2

Substitute 3 and 2 into the curve to be transformed: x 2

x2 + y2 = 4

Write the equation

ubstitute the expressions for x and y S into the original circle’s equation

( y )2 = 4

( x )2 ( y )2 = 4 4 Therefore, the equation of the image is

Expand the brackets ( x )2 4

( y )2

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Reflect and check The pre-image equation is a circle of radius 2 and centre (0, 0). A dilation of factor 2 parallel to the x-axis means to stretch the circle horizontally by a factor of 2.

This can be verified graphically:

y x2 + y2 = 4 4 x

−4

−2

−2 −4

Example 4 A rectangle has vertices P (1, 1), Q (4, 1), R (4, 3), and S (1, 3). 2 0 : 0 0.5

It is transformed by the matrix T

y 4 3

S (1, 3)

R (4, 3)

P (1, 1)

Q (4, 1)

2 1

0 1

2 3

4 5 6 7

x 8

a Find the coordinates of the image vertices P′, Q′, R′, and S′.

Create a strategy Combine the pre-image vertices into a single 2 × 4 matrix. Then, apply the linear transformation matrix equation P′ = TP to find all image vertices at once.

Apply the idea The pre-image matrix is P

1 4 4 1 . 1 1 3 3

P′ = TP 2 0 1 4 4 1 = 0 0.5 1 1 3 3 =

954

2 8 8 2 0.5 0.5 1.5 1.5

Write the equation Substitute the values of the matrices Evaluate

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The image vertices are P′ (2, 0.5), Q′ (8, 0.5), R′ (8, 1.5), and S′ (2, 1.5). y 4 3

S (1, 3)

R (4, 3) R′ (8, 1.5)

S′ (2, 1.5)

Q (4, 1) P (1, 1) P′ (2, 0.5) 0 1 2 3 4 5 6 7

Q′ (8, 0.5) x 8

b Identify one property that is invariant and one property that has changed under this dilation.

Create a strategy Analyse the pre-image and image shapes. To find an invariant property, look for a feature that stayed the same (like parallel sides). To find a changed property, compare measurements like side lengths or angle measures.

Apply the idea

Reflect and check

An invariant property is parallelism. In the pre-image, PQ is parallel to SR. In the image, the corresponding side P′Q′ is also parallel to S′R′.

Note that the area of the pre-image (3 × 2 = 6) and the image (6 × 1 = 6) are the same. This is a special case that happens because the area is scaled by the factor ∣ab∣ = ∣2 × 0.5∣ = 1. In this unique case, multiplying the area by 1 leaves it unchanged. For any other dilation where ∣ab∣ ≠ 1, the area would change.

A property that has changed is side length. The length of side PQ is 4 − 1 = 3 units. The length of the corresponding image side P′Q′ is 8 − 2 = 6 units. The horizontal side length was multiplied by the dilation factor a = 2.

Idea summary A dilation is a linear transformation that changes the size of a pre-image. It can be represented by the linear transformation matrix equation P′ = TP. The transformation matrix for a dilation of factor a parallel to the x-axis and a 0 factor b parallel to the y-axis is T . This corresponds to the mapping 0 b (x, y) → (ax, by). Invariant properties are features that do not change. Under dilation, the origin, collinearity, and parallelism are invariant. Side lengths and area generally change under dilation. Angle measures are generally not preserved, except in the case of a uniform dilation (a = b), where the image is a similar shape.

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10.02 Practice questions What do you remember? 1

Determine the transformation matrix, T, that represents each dilation: a

Dilation of factor 5 parallel to the x-axis

Dilation of factor 0.2 parallel to the y-axis

Dilation of factor − 5 parallel to the y-axis

Dilation of factor 1.5 parallel to the x-axis and 3 parallel to the y-axis

Stretched vertically to make the image twice as large

Compressed by a factor of 0.5 parallel to the x-axis and by a factor of 0.5 parallel to the y-axis

A mirror distorts an image to look 2 times taller and 3 times wider. Write a transformation matrix for this distortion.

Describe the physical transformation represented by T

5 0 . 0 2

Describe the physical transformation represented by T

4 0 . 0 1

Practice 5

Determine the image of point P (10, − 2) under a dilation of factor 6 parallel to the y-axis.

Ex 1

Determine the image of point M (− 3, 4) under a dilation of factor − 3 parallel to the x-axis.

Determine the image of point L (− 1, − 4) under a dilation of factor 3 parallel to the y-axis and a dilation of factor 2 parallel to the x-axis.

Determine the image points of the triangle with vertices A (2, − 5), B (7, 1) and C (2, − 10) under a dilation with a factor of 4 parallel to the x-axis.

Determine the image points of the triangle with vertices A (4, − 5), B (8, 3) and C (− 4, − 2) under a dilation with a factor of 0.25 parallel to the x-axis and a factor of 3 parallel to the y-axis.

Determine the equation of the image of the line 3x − 4y = 6 under a dilation of factor 3.5 parallel to the x-axis.

Ex 2

Ex 3

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11 Ex 4

A rectangle has vertices A (0, 0), B (5, 0), C (5, 2), and D (0, 2). It is transformed by the matrix 3 0 : T 0 1.5 a

Find the coordinates of the image vertices A′, B′, C′, and D′.

Identify one property that is invariant and one property that has changed under this dilation. Justify your answer.

Determine the equation of the image of the curve y = 5x2 under a dilation of factor 1.5 parallel to the y-axis.

Determine the equation of the image of the circle x2 + y2 = 9 under a dilation of factor 3 parallel to the x-axis and 0.5 parallel to the y-axis.

1 Determine the equation of the image of the hyperbola y after it is dilated by x 2 2 0 . T 0 3 a 0 A point A (− 2, 0) is dilated by T so that its image is A′ (− 3, 0). Calculate the value of a. 0 2 1

is transformed to the image equation y

The equation y

Which geometric property is not always invariant when a shape is transformed by the 3 0 dilation T ? 0 2 A Collinearity of points B Parallelism of lines

x T. Determine a possible matrix for T.

C 18

Angle measures

4( x )2

under a dilation matrix,

The origin point (0, 0)

True or False: A square with vertices at (1, 1), (3, 1), (3, 3), and (1, 3) remains a square after 2 0 being transformed by T . Justify your answer. 0 1

Extend your thinking 19

A quadrilateral ABCD has vertices at A (− 1, 0), B (2, − 2), C (5, 1) and D (2, 7). It is dilated by a factor of 1.5 parallel to the y-axis. Calculate the area of A′ B′ C′ D′. b The circle x2 + y2 = r2 is transformed by a dilation of factor parallel to the x-axis. Prove that r 2 the image has an area of πbr units .

Prove algebraically that matrix multiplication is commutative for dilations. That is, for any two dilation matrices T1 and T2, prove that T1T2 = T2T1.

Consider two parallel lines, y = mx + c1 and y = mx + c2. Show that their images under the a 0 dilation T are also parallel. 0 b

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10.03 Rotations After this lesson, you will be able to… • apply rotations of an angle (θ ) anticlockwise about the origin to transform shapes in the plane • identify invariant properties under rotation transformations • represent rotations with 2 × 2 matrices

Understand and apply rotations Interactive exploration Discover this concept in action online

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Rotation

A type of transformation that turns a figure around a fixed point, known as the centre of rotation, through a given angle and in a specified direction.

P′ (x′, y′) P (x, y) θ

Rotations are rigid transformations, or isometries, which means that the shape and size of the object remain unchanged. In other words, the image is congruent to the pre-image. A rotation in the plane is described by: • A centre of rotation (for this lesson, it will always be the origin). • An angle of rotation. • A direction of rotation (clockwise or anticlockwise).

cos ( ) sin ( )

sin ( ) cos ( )

T is the transformation matrix for rotation about the origin θ is the angle of rotation. A positive angle indicates an anticlockwise rotation from the positive x-axis. A clockwise rotation uses a negative value for θ Recall that the linear transformation matrix equation is P′ = TP, where P is the matrix of the pre-image coordinates, P′ is the matrix of the image coordinates and T is the transformation matrix. To rotate a polygon, this transformation is applied to each of its vertices.

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To understand why the rotation matrix has this specific form, it can be derived by considering the rotation of a single point on the unit circle. This helps connect the geometry of rotation with trigonometry. y

P′ (x′, y′) θ

–1

P (x, y) φ

–1

Let P (x, y) be a point on the unit circle which makes an angle φ with the positive x-axis. P′ (x′, y′) is obtained by rotating P (x, y) by an angle θ in an anticlockwise direction. Therefore, P′ (x′, y′) makes an angle θ + φ with the positive x-axis. Using trigonometry, P (x, y) = (cos (φ), sin (φ)) and P′ (x′, y′) = (cos (θ + φ), sin (θ + φ)). By applying the trigonometric sum identities: cos (A + B) = cos (A) cos (B) – sin (A) sin (B) sin (A + B) = sin (A) cos (B) + cos (A) sin (B) The image point can be obtained as shown: x′ = cos (θ + φ) = cos (θ ) cos (φ) – sin (θ ) sin (φ) = x cos (θ ) – y sin (θ ) y′ = sin (θ + φ) = sin (θ ) cos (φ) + cos (θ ) sin (φ) = x sin (θ ) + y cos (θ ) cos ( ) x = sin ( ) y

sin ( ) cos ( )

cos ( ) sin ( )

sin ( ) cos ( )

∴T =

x y

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Example 1 Determine the image of point A(2, 3) under a rotation about the origin by 30° anticlockwise.

Create a strategy First, build the transformation matrix, T, for a 30° anticlockwise rotation. Then, write the point’s coordinates as a column matrix A. Finally, use the formula A′ = TA to find the image coordinates.

Apply the idea The transformation matrix for a rotation of 30° anticlockwise about the origin is: cos ( 30° ) sin ( 30°)

sin ( 30°) cos ( 30°)

3 2

1 2

3 2

2 . 3 Apply the linear transformation matrix equation: The pre-image matrix is A

A′ = TA x = y

Write the transformation matrix equation

3 2

1 2

3 2

3 3 1 2 2 3 3 2 2 3 3 2

Therefore, the image point is A

960

2 3

Substitute the matrices

Evaluate

Combine terms

2 3 3 2 3 3 , . 2 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 A triangle has vertices at A(2, 1), B(4, 1), and C(4, 3). Determine the coordinates of the image triangle A′B′C′ after a rotation of 90° anticlockwise about the origin.

Create a strategy First, determine the transformation matrix for a 90° anticlockwise rotation. Write the vertices of the pre-image as a single 2 × 3 matrix. Then, multiply the transformation matrix by the pre-image matrix to find the coordinates of the image vertices.

Apply the idea The transformation matrix for a 90° anticlockwise rotation is: cos (90° ) sin (90° )

sin (90° ) cos (90° )

0 1

1 0

The pre-image matrix P containing the vertices A, B, and C is: 2 4 4 1 1 3

Multiply the transformation matrix by the pre-image matrix to find the image matrix P′. P′ = TP

Write the transformation matrix equation

0 1

1 2 4 4 0 1 1 3

Substitute the matrices

1 2

1 4

Evaluate

3 4

The coordinates of the image triangle are A′(−1, 2), B′(–1, 4), and C′(−3, 4). A′

B′

y 4 A

3 C′

2 1

–4 –3 –2 –1

C 1

B x

2 3 4

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Example 3 Determine the equation of the image of the line y = 2x – 1 under a rotation about the origin by 60° anticlockwise.

Create a strategy The goal is to substitute expressions for the original coordinates (x and y) into the line’s equation. To find these expressions, start with the transformation equation P′ = TP and rearrange it to make the pre-image matrix P the subject. This requires using the inverse matrix, T –1.

Apply the idea The transformation matrix for a rotation of 60° anticlockwise about the origin is: T

cos (60° ) sin (60° )

sin (60° ) cos (60° )

1 2

3 2

1 2

Apply the linear transformation matrix equation and inverse matrices: P′ = TP –1

Write the formula

–1

T P′ = T TP

Pre-multiply T –1 on both sides

T –1P′ = IP

Apply the rule T –1T = I

∴P = T –1P′

Simplify

The determinant is 1. Therefore, the inverse of T is: T

1 2

3 2

1 2

Using the inverse of T, this result is obtained: P = T –1P′ x = y

x = y

Write the formula

1 2

3 2

1 2

1 x 2

3 y 2

x y

Substitute the matrices

Perform matrix multiplication 1 3 x y 2 2 Applying equivalent matrices, two simultaneous equations are obtained: x= y=

962

1 x 2 3 x 2

3 y 2

1 y 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Substitute 1 and 2 in y = 2x – 1: y = 2x – 1

Write the equation of the line

3 x 2

1 1 y =2 x 2 2

3 y 2

3 x 2

1 y =x 2

Expand the brackets

1 y 2

3y = x

3 x 2

Rearrange to combine like terms

1 2

1 2 3 2 3 =x 2 2

3 2

3 =x 1

y′ = x

y′ =

Factorise x′ and y′

2 Make all coefficients over a common 2 denominator of 2

2 2

x 2

Substitute the expressions for 1 x and y from 1 and 2

Multiply both sides by

1 2 3

Evaluate

1 2 3

Therefore, the equation of the image is y

3 x

1 2 3

Reflect and check The rotation about the origin by 60° anticlockwise can be verified graphically: 4 y′ =

2 + 3 x′ − 2 1−2 3

3 2 1

–5 –4 –3 –2 –1 –1

x 1

2 3 4

–2 y = 2x – 1 –3 –4

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Idea summary A rotation is a transformation that turns a pre-image around a fixed point (the origin) by a certain angle. To rotate a polygon, apply the rotation transformation to each of its vertices. cos ( ) sin ( ) . sin ( ) cos ( ) The angle of rotation, θ, is positive when rotated anticlockwise, and negative when rotated clockwise. The angle is measured from the positive x-axis. The transformation matrix for an angle of rotation θ is T

Invariant properties of rotations Invariant properties Features of a shape or object that do not change after a transformation is applied. Invariant point A point that remains in the same position after a transformation.

Since rotation is an isometry (a rigid transformation), several key properties are invariant. A′

B′

A Triangle A′B′C′ is the image of triangle ABC after a 90° anticlockwise rotation.

C′ C

B x

The properties that are invariant under rotation about the origin include: • The origin: The centre of rotation, (0, 0), is an invariant point, meaning it does not move. • Distance between points: The lengths of the sides of a shape do not change. • Angle measures: All angles within the shape remain the same. • Area: The area of the shape is preserved. • Orientation: The order of the vertices (e.g., clockwise) remains the same. • Parallelism: Lines that are parallel in the pre-image remain parallel in the image.

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While many properties are invariant, some properties do change: • Coordinates of points: Unless a point is the centre of rotation, its coordinates will change. • Slope of lines: The slope of any line will change, unless the rotation is by 180°.

Example 4 A triangle has vertices at A(2, 1), B(4, 1), and C(4, 3). It is rotated 90° anticlockwise to produce the image with vertices A′(−1, 2), B′(−1, 4), and C′(−3, 4). Show that the length of side AB is equal to the length of side A′B′.

Create a strategy ( x2 x1 )2 ( y2 y1 )2 , to find the length of side AB and the Use the distance formula, d corresponding side A′B′. Compare the results to demonstrate that side length is an invariant property.

Apply the idea First, calculate the length of side AB using A(2, 1) and B(4, 1). Length of AB

( 4 2)2 ( 1 1)2

22 02

2 units

Next, calculate the length of side A′B′ using A′(− 1, 2) and B′(− 1, 4). Length of A B

( 1 ( 1))2 ( 4 2)2

02 22

2 units

The length of AB is equal to the length of A′B′. This demonstrates that distance is an invariant property of rotation.

Example 5 A square has an area of 25 square units. If the square is rotated 270° anticlockwise about the origin, what is the area of its image? Justify your answer.

Apply the idea Rotation is a rigid transformation, which means it preserves the size and shape of the pre-image. Area is an invariant property under rotation. Therefore, the area of the image will be the same as the area of the pre-image. The area of the image is 25 square units.

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Example 6 A point P(5, 0) is on a circle centred at the origin. If the point is rotated 45° anticlockwise about the origin to a new point P′, what is the distance of P′ from the origin?

Create a strategy Rotation is a type of transformation that preserves certain geometric properties. Determine the distance of the original point P from the origin, and consider how a rotation affects this distance.

Apply the idea The distance of point P(5, 0) from the origin (0, 0) is 5 units. This is the radius of the circle. One of the invariant properties of rotation is that the distance from any point to the centre of rotation is preserved. Therefore, the distance of the image point P′ from the origin will also be 5 units.

Idea summary Invariant properties are features of a shape that do not change after a transformation. Rotation about the origin is classified as an isometry (or rigid transformation) because it preserves distance and angle measure. The invariant properties observed during this transformation are: • • • • •

The origin (0, 0) is an invariant point Distance between points (side lengths) Angle measures Area Orientation

Since these properties are preserved by the isometry, the image of a rotated shape is necessarily congruent to the pre-image.

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10.03 Practice questions What do you remember? 1

Determine the transformation matrix that represents a rotation about the origin by: a

90° anticlockwise

45° clockwise

210° anticlockwise

300° clockwise

Describe the rotation represented by each matrix: a

1 3 2 2 3 2

1 2

2 2

Under a rotation about the origin, one point in the plane is always invariant (it does not move). What are the coordinates of this point?

Practice 4

Determine the image of the point H(5, 1) after a rotation of 90° anticlockwise about the origin.

Determine the image of the point P(– 1, 2) after a rotation of 150° clockwise about the origin.

Ex 1

3 Determine the image of the point P(3, – 4) after a rotation of radians anticlockwise about 2 the origin.

5 Determine the image of the point R(5, 2) after a rotation of radians clockwise about the 4 origin.

8 Ex 2

Determine the image of the quadrilateral P(2, 3), Q(5, 3), R(5, – 1) and S(2, – 1) after a rotation of 90° anticlockwise about the origin.

Determine the image of the triangle with vertices A(1, 1), B(4, 4) and C(4, – 2) after a rotation of 120° anticlockwise about the origin. Sketch the pre-image and the image on the same axes.

A triangle has vertices X(–3, 2), Y(0, 5) and Z(–2, –4). Determine the coordinates of the 5 image after a rotation of radians clockwise about the origin. 6

Determine the equation of the image of the line y = 4 – x under a rotation of 210° anticlockwise about the origin. Verify your results graphically.

Ex 3

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Ex 4

Ex 5

Determine the equation of the image of the parabola y = x2 + 1 under a rotation of 90° anticlockwise about the origin. Verify your results graphically.

The line segment with endpoints A(– 2, 5) and B(3, – 1) is rotated about the origin by 245° clockwise:

15 Ex 6

Determine the coordinates of the image endpoints, A′ and B′.

Show that the line segments AB and A′B′ have the same length.

A rectangle has vertices P(1, 1), Q(5, 1), R(5, 3), and S(1, 3): a

Calculate the area of rectangle PQRS.

The rectangle is rotated 60° anticlockwise about the origin. Find the coordinates of the image P′Q′R′S′.

Calculate the area of the image rectangle P′Q′R′S′.

What property of rotation is demonstrated by this comparison of areas?

A point Q(3, –4) lies on a circle centred at the origin. The point is rotated 120° clockwise about the origin to become Q′. Without calculating the coordinates of Q′, determine its distance from the origin and explain your reasoning. A triangle with vertices A(1, 2), B(4, 2), and C(1, 6) is rotated 45° anticlockwise about the origin to form triangle A′B′C′: a

Find the coordinates of the image vertices.

Calculate the measure of angle BAC.

Calculate the measure of angle B′A′C′.

What property of rotation is illustrated by this result?

Extend your thinking 17

A square with vertices A(0, 0), B(1, 0), C(1, 1) and D(0, 1) is rotated about the origin by 45° clockwise: a

Determine the coordinates of the image and sketch the pre-image and image on the same set of axes.

Determine the area of the overlapping region.

A point P(x, y) has been rotated about the origin in an anticlockwise direction to the image 3x 4 y 4 x 3 y point P . Determine the angle of rotation. , 5 5

Determine the transformation matrix needed to reverse the effect of a rotation about the origin by θ anticlockwise.

Prove algebraically that rotations are commutative. That is, for any two anticlockwise rotations T1 (by angle θ1) and T2 (by angle θ2), show that T1T2 = T2T1.

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10.04

Reflections

After this lesson, you will be able to… • apply a reflection across a line through the origin to transform shapes in the plane • identify invariant properties under reflection transformations • represent reflections with 2 × 2 matrices

Reflections Interactive exploration mathspace.co

Discover this concept in action online

Reflection A linear transformation that mirrors a pre-image in a line of reflection to create an image on the opposite side of the line. Each point on the image is the same distance from the line of reflection as its corresponding point on the pre-image. Each point on the image is the same distance from the line of reflection as its corresponding point on the pre-image. The line connecting a point to its image is always perpendicular to the line of reflection. Pre-image points that lie on the line of reflection remain unchanged. A′ B′

y 5 A 4 3 2 C = C′ 1

−5 −4 −3−2 −1 –1 –2 –3 –4 –5

B x

1 2 3 4 5

The blue polygon is the image of the purple pre-image after a reflection across the y-axis (the line x = 0). Notice that point C is on the line of reflection, so it does not move. It is an invariant point, and C = C′.

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Any reflection across a line that passes through the origin can be described with a matrix. The transformation matrix for a reflection across the line y = x tan (θ ) is shown. (Remember that tan (θ ) is the gradient of the line.)

cos(2 ) sin(2 )

sin(2 ) cos(2 )

T is the transformation matrix for a reflection θ is the angle the line of reflection makes with the positive x-axis Some useful standard reflection matrices: Component

Reflected across x-axis

Reflected across y-axis

Reflected across y=x

Reflected across y = −x

Coordinates

(x, − y)

(− x, y)

(y, x)

(− y, − x)

Orientation

Reversed

1 0 0 1

0 1 1 0

Matrix

1 0

0 1

1 0

Reflections have several important invariant properties, which are features that do not change after the transformation. Property

Invariant under reflection?

Shape and size

Yes

Reflection is an isometry (a rigid transformation).

Side lengths

Yes

Distances between points are preserved.

Angle measures

Yes

Angles within the shape remain unchanged.

Parallelism

Yes

Lines that are parallel stay parallel.

Collinearity

Yes

Points that are on a line remain on a line.

Orientation

The order of vertices is reversed (e.g., from clockwise to anticlockwise).

Coordinates

Coordinates change unless the point is on the line of reflection.

Explanation

The determinant of any reflection matrix is always − 1. This value indicates that the transformation preserves area but reverses the orientation of the shape. An important property of reflections is that applying the same reflection twice returns an object to its original position. If a shape is reflected and then immediately reflected back in the same line, it returns to its starting position. This means the transformation is its own inverse. T 2 = I and T −1 = T

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Example 1 Determine the image of point A(− 3, 2) under a reflection across the line y = − x.

Create a strategy The plan is to: 1. Determine the angle (θ ) that the line y = − x makes with the positive x-axis. 2. Use this angle to create the specific transformation matrix T. 3. Apply the formula A′ = TA to find the image coordinates.

Apply the idea The angle that the reflection line y = − x makes with the positive x-axis is 90° + 45° = 135°. y

90° + 45° = 135°

y=−x The transformation matrix for a reflection across the line y = − x is: T=

cos (2 135° ) sin(2 135° )

cos (270° ) sin(270° )

sin(2 135° ) cos (2 135° ) sin( 270°) cos (270° )

0 1 = 1 (0) =

0 1

1 0

The pre-image matrix is A

1 0

3 2

2 3 Therefore, the image point is A′(− 2, 3). =

Evaluate the trigonometric functions

3 . 2

A′ = TA 0 1

Simplify the angles

Simplify

Apply the linear transformation matrix equation:

ubstitute θ = 135° into the S general formula

rite the linear transformation W matrix equation Substitute the matrices

Evaluate

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Example 2 Determine the equation of the image of the curve y = x2 after it is reflected across the line y 3 x.

Create a strategy To find the equation of the image, the original variables (x and y) must be expressed in terms of the image variables (x′ and y′). 1. First, find the transformation matrix T for the reflection. 2. Use the relationship P = T −1 P′ to find equations for x and y. 3. Substitute these expressions for x and y into the original curve’s equation, y = x2.

Apply the idea The angle that the reflection line y 3 x makes with the positive x-axis is tan( ) tan 1 3 60°. y

3 , therefore, 1

y = 3x

3 3 x

θ 1

The transformation matrix for a reflection across the line y 3 x is: T= =

972

cos (2 60°) sin (2 60°) sin (2 60°) cos (2 60°)

Substitute θ = 60° into the general formula

cos (120°) sin (120°) sin (120°) cos (120°)

Simplify the angles

1 2

3 2

1 2

3 2

1 2

Evaluate the trigonometric functions

Simplify

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the linear transformation matrix equation and inverse matrices: P′ = TP

rite the linear transformation W matrix equation

T −1 P′ = T −1 TP

Pre-multiply T −1 on both sides

T −1 P′ = IP

Apply the rule T −1 T = I

∴P = T −1 P′ The inverse of T is: 1 T −1 = 1

1 2

3 2

1 2

3 2

1 2

3 2

1 2

Notice that the inverse matrix T −1 is the same as the original matrix T. This is a property of all reflection matrices. Therefore, the result is: P = T −1P′ x = y

Write the formula

1 2

3 2

1 2

x y

Substitute the matrices

This means that the system of equations is given by: x=

1 x 2

3 y 2

3 x 2

1 y 2

Substituting 1 and 2 in y = x2, the image equation is: 3 x 2

1 y = 2

1 x 2

3 y 2

ubstitute the x and y-values S from equation 1 and 2

Reflect and check The reflection can be verified graphically: 4 y = x2

3 2 1

−4 −3 −2 −1 −1

x 1 2 3 4

−2 −3 −4

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Example 3 A triangle ABC has vertices A(1, 2), B(3, 4), and C(1, 5). Find the vertices of the image triangle A′B′C′ after a reflection across the line y = x.

Create a strategy First, determine the transformation matrix for a reflection across the line y = x. Then, put all three pre-image vertices into a single 2 × 3 matrix. Finally, multiply the matrices to find the new vertices all at once.

Apply the idea The transformation matrix for a reflection across the line y = x is T

0 1 . 1 0

The pre-image matrix, P, contains the vertices of the triangle: P

1 3 1 2 4 5

Apply the linear transformation matrix equation: P′ = TP

Write the linear transformation matrix equation

0 1 1 3 1 = 1 0 2 4 5

Substitute the matrices

2 4 5 = 1 3 1

Evaluate the matrix product

Therefore, the image vertices are A′(2, 1), B′(4, 3), and C′(5, 1).

Example 4 Consider the reflection of the square with vertices O (0, 0), P (2, 0), Q (2, 2), R (0, 2) across the x-axis: a Find the coordinates of the vertices of the image square O′P′Q′R′.

Create a strategy The transformation matrix for a reflection across the x-axis can be applied to the vertices, or the mapping rule (x, y) → (x, − y) can be used for each vertex.

Apply the idea Using the mapping rule (x, y) → (x, − y): • O(0, 0) → O′(0, 0) • P(2, 0) → P′(2, 0) • Q(2, 2) → Q′(2, − 2) • R(0, 2) → R′(0, − 2) The image vertices are O′(0, 0), P′(2, 0), Q′(2, − 2), and R′(0, − 2).

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b List two geometric properties that are invariant and describe how the orientation has changed.

Create a strategy Compare the pre-image square and the image square. Consider properties like side lengths, angles, and the order of the vertices.

Apply the idea Invariant properties: • Side lengths are invariant. For example, the length of PQ is 2, and the length of P′Q′ is also 2. • Angle measures are invariant. The pre-image is a square with 90-degree angles, and the image is also a square with 90-degree angles. Orientation change: The orientation is reversed. The vertices of the pre-image OPQR are in a anticlockwise order. The vertices of the image O′P′Q′R′ are in a clockwise order.

c Identify any invariant points or lines under this transformation.

Create a strategy Invariant points are points that do not move after the transformation. An invariant line is a line where every point on the line is mapped to another point on the same line.

Apply the idea Invariant points: Any point on the line of reflection (across the x-axis) is an invariant point. For the given square, the vertices O (0, 0) and P (2, 0) are invariant points. Invariant lines: An invariant line is a line that maps onto itself. • The line of reflection (across the x-axis) is an invariant line because every point on it is mapped to another point on the x-axis. • Any line perpendicular to the line of reflection is also an invariant line. For example, the line x = 2 is reflected onto itself, even though most points on the line (like Q) move to a new position on that same line.

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Idea summary A reflection is a linear transformation that mirrors a pre-image in a line of reflection. The general transformation matrix for a reflection across a line making an angle θ with the positive x-axis is: T

cos(2 ) sin(2 )

sin(2 ) cos(2 )

Key properties of reflections: • Points on the line of reflection are invariant points (do not change position). • Reflections preserve distances, angle measures, and area. The shape and size of an object remain the same. • Reflections reverse the orientation of an object. • The determinant of a reflection matrix is always − 1. • A reflection is its own inverse: T −1 = T

10.04 Practice questions What do you remember? 1

Determine the image of point A(4, 1) under a reflection across the line y = x.

Determine the image of point A(3, − 1) under a reflection across the line y = 0.

Determine the image of point A(− 3, − 2) under a reflection across the line y = − x.

Ex 1

Determine the image of point A(7, − 5) under a reflection across the line x = 0.

Determine the transformation matrix that represents a reflection across the line: 2 a y = 2x b y c y=0 x 3 d

x 3

y = −x

For each transformation matrix, find the angle the line of reflection makes with the positive x-axis: 1 0 0 1 a b 0 1 1 0

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x=0

1 2 3 2

3 2 1 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 2

3 2

1 2

Practice 7

Determine the equation of the image of the line y = 2x + 1 after a reflection across the line y = x.

Ex 2

8 Ex 3

Determine the image of the triangle with vertices A(− 2, 1), B(3, 7) and C(− 5, 2) after a reflection across the line y = x.

Determine the image of the triangle with vertices A(− 4, 1), B(− 7, − 3) and C(− 2, − 5) after a reflection across the line y = − x.

Determine the image of the quadrilateral with vertices A(− 5, 3), B(6, − 2), C(2, 5) and D(1, 1) after a reflection across the line y = − x.

Determine the image of point A(1, 2) under a reflection across the line y

Determine the image of point F(9, − 10) under a reflection across the line y = − 4x.

A kite with vertices A(− 5, 4), B(0, 4), C(3, 0) and D(− 4, − 4) is reflected across the line y = 2x.

3 x. 3

Determine the image points and sketch ABCD and A′B′C′D′ on the same axes. 14 Ex 4

A square has vertices at A(1, 1), B(3, 1), C(3, 3) and D(1, 3). The square is reflected across the line y = x: a

Determine the coordinates of the image vertices A′, B′, C′, and D′.

List two geometric properties (e.g., side length, angle measure) that are invariant under this reflection.

The orientation of the vertices ABCD is anticlockwise. What is the orientation of the image vertices A′B′C′D′?

Are there any invariant points for this specific shape under this reflection? Explain why or why not.

Determine the equation of the image of the curve y = (x − 2)2 after a reflection across the y-axis (x = 0).

Determine the equation of the image of the curve y = x3 + 1 after a reflection across the line y = − x.

Extend your thinking 17

A point P(x, y) is reflected across the line y = x tan (θ ) to the image point 3 x 4 y 4 x 3 y . Determine the equation of the reflection line. P , 5 5

Determine the transformation matrix needed to reverse the effect of a reflection across the line y = − 3x.

Let T be the transformation matrix for a reflection across the line y = x tan (θ ). Prove algebraically that T −1 = T, and explain the significance of this result. 10.04 Reflections mathspace.co

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10.05 Compositions of linear transformations After this lesson, you will be able to… • understand and use composition of linear transformations and the corresponding matrix products • apply compositions of linear transformations to plane subsets • describe the combined effect of multiple transformations geometrically

Composition of linear transformations Composition of linear transformations The process of applying two or more linear transformations in sequence, which can be represented by the product of their transformation matrices. Identity transformation A transformation that leaves every point unchanged, represented by the identity matrix, I. Inverse transformation A transformation that reverses the effect of another transformation. When composed with the original transformation, it results in the identity transformation.

A composition of linear transformations is what happens when two or more transformations are applied, one after the other. For example, a shape might be rotated and then reflected. The combined effect of applying transformation T1 followed by T2 is equivalent to a single transformation, which can be found by multiplying their matrices.

P′ = T2 T1 P P′′ is the matrix representing the image coordinates T2 is the matrix for the second transformation applied T1 is the matrix for the first transformation applied P

is the matrix representing the pre-image coordinates

The order of matrix multiplication is important, as transformations are not always commutative. In other words, applying T1 then T2 is not always the same as applying T2 then T1. Identity transformation is represented by the identity matrix, I

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1 0 . 0 1

When a transformation, T, is composed with its inverse, T −1, the result is the identity transformation. TT −1 = T −1T = I For example: • The inverse of a rotation by angle θ is a rotation by angle −θ. • A reflection is its own inverse. Applying the same reflection twice returns the object to its original position.

Exploration What happens when two reflections are combined? Investigate this by considering a reflection TA across the x-axis and a reflection TB across the y-axis. Follow these steps to see their combined effect: 1. Write down the matrix for the reflection TA across the x-axis. 2. Write down the matrix for the reflection TB across the y-axis. 3. Find the matrix for the composite transformation TC = TB TA (reflection TA followed by reflection TB). 4. What single geometric transformation does TC represent? Describe it. 5. Find the matrix for the composite transformation TD = TA TB (reflection TB followed by reflection TA). Is TD the same as TC? What does this indicate about the order of transformations?

Translations are handled differently from other transformations. Unlike rotations, reflections, and dilations, translations are not linear transformations and cannot be represented by a 2 × 2 matrix. Translations are combined using matrix addition, not multiplication.

Example 1 Determine the image of the point P(3, 5) under a dilation of factor 3 parallel to the y-axis followed by a rotation of 60° anticlockwise about the origin.

Create a strategy First, identify the matrix for the dilation (T1) and the matrix for the rotation (T2). Since the rotation is applied after the dilation, the correct order for the matrix multiplication is T2 T1. The image can be found by calculating P′ = T2 T1 P.

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Apply the idea The transformation matrix for a dilation of factor 3 parallel to the y-axis is T1: T1

1 0 . 0 3

The transformation matrix for a rotation about the origin by 60° anticlockwise is T2: cos (60°) sin (60°)

sin (60°) cos (60°)

1 2

3 2

1 2

First, find the single matrix T that represents the composition T = T2 T1. T = T2 T1 =

Write the formula

1 2

3 2

1 2

3 3 2

3 2

1 0 0 3

Substitute the matrices

Evaluate

Now, apply this composite matrix to the pre-image point P P′ = TP

980

Write the formula

1 2

3 3 2

3 2

3 5

3 15 3 2 3 3 15 2

Therefore, the image point is P

3 . 5

Substitute the matrices

Evaluate

3 15 3 15 3 3 . , 2 2

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reflect an check The composition of linear transformations can be verified graphically. y 15

P1 (3, 15) After dilation

P′(P2 After rotation) 10

P(3, 5)

x –10

–5

Example 2 Show that the result of a reflection across the line y = −x followed by a reflection across the line y 3x is a single rotation.

Create a strategy Find the matrices for each reflection, T1 and T2. Then, multiply them in the correct order (T2 T1) to get a single matrix representing the combined transformation. Finally, examine the structure of this new matrix to determine if it matches the form of a known transformation, such as a rotation.

Apply the idea The angle that the reflection line y = −x makes with the positive x-axis is 90° + 45° = 135°. y

90° + 45° = 135° x 0

y = –x

The transformation matrix for a reflection across the line y = −x is: T1

cos ( 2 135°) sin ( 2 135° )

sin (2 135°) cos (2 135°)

0 1

1 0

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The angle that the reflection line y 3x makes to the positive x-axis is: tan (θ ) =

3 1

Relate the gradient of the line to tan (θ ) 3 1

θ = tan 1

Use the inverse tangent function to find θ

= 60°

Evaluate 2

y y = 3x

1 3

θ –2

–1

x 2

–1 –2 The transformation matrix for a reflection across the line y 3x is: cos ( 2 60°) sin ( 2 60°)

sin (2 60°) cos (2 60° )

1 2

3 2

1 2

Determine the single transformation matrix: T = T2 T1 =

Write the formula

1 2

3 2

1 2

0 1

3 2

1 2

3 2

cos ( 210°) sin ( 210° )

1 0

Substitute the matrices

Evaluate

sin (210° ) cos ( 210° )

Convert to trigonometric functions

Therefore, the single transformation matrix is a single rotation of 210° anticlockwise about the origin.

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Example 3 A line segment has endpoints A(1, 2) and B(3, 4). Determine the coordinates of the endpoints of the image segment A′B′ after a rotation of 90° anticlockwise about the origin, followed by a reflection across the x-axis.

Create a strategy First, identify the matrices for the rotation (T1) and the reflection (T2). Then, calculate the composite transformation matrix T = T2 T1. Finally, apply this single matrix T to a matrix containing both points A and B to find their images A′ and B′.

Apply the idea The transformation matrix for a rotation of 90° anticlockwise about the origin is: cos (90°) sin (90°)

sin (90°) cos (90°)

0 1

1 0

The transformation matrix for a reflection across the x-axis is: 0 1

1 0

Next, find the single transformation matrix for the composition, T = T2 T1. T = T2 T1 1 = 0 = =

Write the formula

0 0 1 1

1 0

Substitute the matrices

(1) (0) (0) (1)

(1) (–1) (0) (0)

(0) (0) (–1) (1)

(0) (–1) (–1) (0)

0 1

1 0

Perform matrix multiplication Evaluate

Now, apply the composite transformation matrix T to the pre-image matrix P P′ = TP

Write the linear transformation equation

1 1 3 0 = 1 0 2 4 = =

Substitute the matrices

(0) (1) (–1) (2)

(0) (3) (–1) (4)

(–1) (1) (0) (2)

(–1) (3) (0) (4)

2 1

1 3 . 2 4

4 3

Perform matrix multiplication Evaluate

The coordinates of the image endpoints are A′(−2, −1) and B′(−4, −3).

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Reflect and check The final matrix T represents a reflection across the line y = −x. This can be checked by verifying two key properties: • The line segment connecting a point to its image (e.g., AA′) should be perpendicular to the line of reflection. • The midpoint of that segment should lie on the line of reflection. It can be verified that these properties hold true for both points A and B and their images.

Example 4 Find the image of the point P(4, 1) after a rotation of 90° anticlockwise about the origin, followed by 2 a translation by the vector . 3

Create a strategy First, apply the linear transformation (the rotation) using matrix multiplication. Then, take the result and apply the non-linear transformation (the translation) using matrix addition.

Apply the idea Step 1: Apply the rotation 0 1 Let the intermediate point after the rotation be P∗. The matrix for a 90° anticlockwise rotation is T P∗ = TP =

0 1

1 4

1 . Apply this to the point P 0

Write the transformation equation 1 4 0 1

Substitute the matrices Evaluate

After the rotation, the point is at (−1, 4). Step 2: Apply the translation 2 to the intermediate point, P∗. 3

Now, add the translation vector V P′ = P∗ + V =

1 4

1 7

Write the translation equation 2 3

Substitute the vectors Evaluate

The final image of the point is P′(1, 7).

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4 . 1

Reflect and check Performing these operations in the correct order is very important. If the translation is applied first, followed by the rotation, the result is different: 2 1. Translate P(4, 1) by : (4 + 2, 1 + 3) = (6, 4) 3 0 1 6 4 2. Rotate the new point (6, 4) by 90°: 1 0 4 6 The result is (− 4, 6), which is completely different. This shows that the order of composition matters, especially when mixing linear transformations with translations.

Idea summary A composition of linear transformations is what happens when two or more transformations are applied, one after the other. The combined effect of applying transformation T1 followed by T2 is found using the matrix product T = T2 T1. The resulting image is found using P′ = TP. If a transformation T is followed by its inverse transformation (T −1), the result of the composition is the identity transformation (I), which maps all points back onto themselves. The order of transformations is crucial, as matrix multiplication is not always commutative (T2 T1 ≠ T1 T2). A composition of linear transformations can be applied to points, lines, and shapes, and results in a new, single linear transformation. The overall geometric effect of this new transformation can often be described as a simpler one. For example, two reflections can be equivalent to a single rotation.

10.05 Practice questions What do you remember? 1

A point P is transformed first by TA and then by TB to get the image P′. Which matrix equation correctly represents this composition? A

P′ = PTA TB

P′ = PTB TA

P′ = TA TB P

P′ = TB TA P

Identify the term or property described in each statement: a

The transformation that reverses the effect of another transformation.

The property that the order of transformations does not matter (e.g., T1 T2 = T2 T1). 1 0 The transformation that leaves every point unchanged, represented by . 0 1

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Determine whether these statements about composite transformations are true or false: a

The composition of reflection across any line with a reflection across any second line is always equivalent to a single rotation.

The composition of a transformation T and its inverse T −1 results in the zero matrix.

The order in which transformations are applied does not usually matter.

A reflection is its own inverse.

Practice

Ex 1

Determine the single transformation matrix for a rotation about the origin by 120° anticlockwise followed by a reflection across the line y = x.

Determine the single transformation matrix for a dilation by a factor of 3 parallel to the y-axis followed by a rotation about the origin by 160° clockwise.

Determine the image of the point B(−1, 3) under a dilation by a factor of 2 parallel to the x-axis followed by a reflection across the line y = −2x.

Show that a reflection across the line y = x followed by a reflection across the line y results in a single rotation.

Ex 2

1 3

Find the image of the point M(2, −1) after a rotation of 90° anticlockwise about the origin, followed by a dilation by a factor of 2 parallel to the y-axis.

Determine the image of a triangle with vertices A(1, 2), B(3, 5) and C(5, 2) after a dilation by a factor of 3 parallel to the y-axis and a dilation by a factor of 2 parallel to the x-axis, followed by a rotation about the origin by 90° anticlockwise. Verify the results graphically.

A line segment has endpoints C(2, −1) and D(−3, 5). Determine the coordinates of the image points C′ and D′ after the segment is reflected across the line y = x, and then rotated 180° about the origin.

Ex 3

The line with the equation y = 2x − 1 is reflected across the line y = −x, and then rotated 135° anticlockwise about the origin. Find the equation of the resulting image line.

Show that a rotation about the origin by 45° anticlockwise twice is the same as a rotation about the origin by 90° anticlockwise once.

Find the image of the point A(−2, 5) after a rotation of 180° about the origin, followed by a 4 translation by the vector . 3

Ex 4

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Consider the point P(2, 6) and two transformations: T1 is a reflection across the x-axis, and T2 is a dilation by a factor of 4 parallel to the x-axis: a

Determine the image point after a transformation of T1 followed by T2.

Determine the image point after a transformation of T2 followed by T1.

Does the order of the transformations matter in this case? Explain why or why not.

Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Consider the point P(−5, 10) and two transformations: T1 is a rotation of 90° anticlockwise about the origin, and T2 is a reflection across the line y = x: a

Determine the image point after a transformation of T1 followed by T2.

Determine the image point after a transformation of T2 followed by T1.

Does the order of the transformations matter in this case? Explain why or why not.

Extend your thinking 16

A rotation T1 by an angle of θ ° anticlockwise and a reflection T2 across the line y = x are applied. If the order of these transformations does not matter (i.e., T1 T2 = T2 T1), find all possible values for the angle θ °.

Prove that a reflection across the line y = x tan (θ °), followed by a reflection across the line y = x tan (α°) is equivalent to a single rotation about the origin.

10.06 Inverse of linear transformations After this lesson, you will be able to… • determine the inverse of linear transformations • apply inverse transformations to reverse effects on plane subsets • understand the connection between an inverse matrix and the inverse of linear transformations

Inverse linear transformations A linear transformation, represented by a matrix T, moves a point or shape (P) to a new position (P′). An inverse linear transformation performs the reverse operation: it takes the image (P′) and maps it back to its original pre-image (P). This “undoing” transformation is represented by the inverse matrix, T −1. The inverse of a linear transformation matrix, T −1, will transform the image P′ back to its pre-image P. Inverse linear transformation A transformation, represented by an inverse matrix T −1, that maps an image back to its original pre-image position. It undoes the effect of the original transformation T. Singular matrix A square matrix that does not have an inverse. A matrix is singular if and only if its determinant is equal to 0.

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987

If T

a b , then T 1 c d

1 ad bc

d c

b a

T is the original 2 × 2 transformation matrix T −1 is the inverse transformation matrix a, b, c, d is the elements of the matrix T ad – bc is the determinant of matrix T, often denoted as det (T ). The inverse matrix T −1 exists if and only if ad – bc ≠ 0 If the determinant ad – bc = 0, the transformation matrix T is called a singular matrix, and an inverse transformation does not exist. Geometrically, this means the transformation has squashed the entire 2D plane onto a single line or even a single point. If multiple different points from an original shape all land on the same spot, it becomes impossible to know where each one came from in the reverse process. Because a unique original position cannot be recovered, an inverse transformation doesn’t exist.

Exploration The “undoing” nature of inverse transformations can be explored. Consider the point P(2, 5) and the transformation matrix T this matrix is T 1

2 0 . The inverse of 0 3

0.5 0 1 . 0 3

Follow the steps and answer the questions: 1. Apply the transformation T to the point P to find the image P′ by calculating P′ = TP. 2. Take the result for P′ and apply the inverse transformation T −1 to it by calculating T −1 P′. 3. Calculate the matrix product T −1T. • What point is obtained after applying T −1 to P′? How does it compare to the original point P? • What special matrix is obtained from the product T −1T? • Describe the effect of matrix T −1 in relation to the effect of matrix T. This shows that applying a transformation and then its inverse is the same as applying the identity matrix, which returns the point to its starting position.

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Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Matrix inverses can be applied to the linear transformation matrix equation. P′ = TP T −1 P′ = T −1 TP T −1 P′ = IP ∴P = T −1 P′

Example 1 Point A was rotated 45° anticlockwise about the origin to obtain the image point A′(2, –3). Determine the pre-image point A.

Create a strategy First, write the transformation matrix T for the rotation. Then, calculate its inverse, T −1. Finally, apply the formula A = T −1A′ to find the pre-image coordinates.

Apply the idea The transformation matrix for a rotation about the origin by 45° anticlockwise is: cos ( 45° ) sin( 45° )

sin ( 45° ) cos ( 45° )

2 1

2 . 3 Apply matrix inverses and the linear transformation matrix equation: The image matrix, representing point A′ is

A′ = TA

Write the formula

T −1A′ = T −1TA

Pre-multiply T −1 on both sides

T −1A′ = IA

Apply the rule: T −1T = I

∴A = T −1A′

Simplify

First, calculate the determinant of T: det (T) =

1 1 2 2 =1 =

Use the determinant formula, ad – bc Evaluate the multiplication Evaluate

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Since det (T) ≠ 0, the inverse exists. Now find T −1 using the formula. 1

2 1

1 T −1 = 1

2 1

Substitute into the inverse matrix formula

Evaluate

The pre-image coordinates can now be found by calculating A = T −1A′: A=

2 1

2 3

2 2

2 3

Substitute the matrices

Multiply the matrices

1 =

2 5

Simplify

2 2 2

Rationalising the denominators gives

Therefore, the pre-image point is A

5 2 2

2 5 2 , . 2 2

Reflect and check y

The pre-image and image points can be verified graphically.

2 x –6 –4 –2 A −

1 5 2 –2 ,− 45° 2 2 –4

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A′ (2, –3)

Example 2 A triangle with vertices A′(–1, 8), B′(0, 4), and C′(1, 8) is the image of triangle ABC after a linear 1 2 transformation T . Find the coordinates of the vertices of the pre-image triangle ABC. 2 0

Create a strategy First, calculate the inverse of the transformation matrix, T −1. Then, group the image vertices into a single matrix and apply the formula P = T −1 P′ to find all pre-image vertices at once.

Apply the idea 1 2

The transformation matrix is T

2 . First, find its inverse. 0

Calculate the determinant: det (T ) = (1) (0) – (–2) (2) = 0 + 4 = 4 Calculate the inverse matrix: T 1

1 4

0 2 2 1

0 0.5 0.5 0.25

1 0 1 . Now find the pre-image matrix P: 8 4 8

The image matrix is P P = T −1P′ 0 =

1 2

Write the formula 1 2 1 4

1 0 1 8 4 8

1 ( 8) 2 1 1 ( 1) ( 8) 2 4

0 ( 1) =

4 2 = 1 2 1 2

1 (4) 2 1 1 ( 0) (4) 2 4

0 (0)

Substitute the matrices

1 ( 8) 2 1 1 ( 1) ( 8) 2 4

0 ( 1)

4 1 2 2

Multiply the matrices

Simplify

4 2 4 = 5 3 1 2 2 The pre-image vertices are A 4,

5 3 , B(2, 1), and C 4, . 2 2

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Example 3 For each transformation matrix, determine its inverse or explain why an inverse does not exist: a M

3 1 5 2

Create a strategy Calculate the determinant of the matrix. If it is non-zero, use the formula for the inverse of a 2 × 2 matrix to find M −1.

Apply the idea First, calculate the determinant of M: det (M) = (3)(2) – (1)(5) = 6 – 5 = 1 Since the determinant is 1, which is not 0, the inverse exists. M−1 = = The inverse matrix is M 1

b N

2 1

2 5

1 1

2 5 2 5

1 3 1 3

Substitute the values Evaluate

1 . 3

4 2

Create a strategy Calculate the determinant of the matrix. If it is zero, an inverse does not exist.

Apply the idea First, calculate the determinant of N: det (N) = (2) (2) – (–4) (–1) = 4 – 4 = 0 Since the determinant is 0, the matrix is singular and an inverse transformation does not exist.

Idea summary The inverse linear transformation matrix, T −1, transforms the image P′ back to its pre-image P using the relationship: P = T −1P′. d b a b 1 , its inverse is T 1 . The inverse ad bc c a c d exists if and only if the determinant ad − bc ≠ 0. If the determinant equals zero, the matrix T is called a singular matrix, and the inverse transformation T –1 does not exist.

For a transformation T

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10.06 Practice questions What do you remember? 1

Complete the formula for the inverse of a 2 × 2 matrix T T 1

a b . c d

Determine whether these statements about inverse transformations are true or false: a

A matrix has an inverse if and only if its determinant is non-zero.

A reflection transformation is its own inverse.

The inverse of a rotation by angle θ is a reflection across the line y = x.

Fill in the blanks to complete the key relationship between a pre-image P, its image P′, a transformation T, and its inverse T −1. If P′ = TP, then P = ⬚P′.

Practice 4 Ex 1

Ex 2

A reflection across the line y = x resulted in the image point B′(–5, 7). Calculate the pre-image point B.

A rotation of 90° clockwise about the origin resulted in the image point A′(4, –3). Calculate the pre-image point A.

A dilation with a factor of 2 parallel to the y-axis resulted in the image point P′(6, 10). Calculate the pre-image point P.

A reflection across the line y = –x resulted in the image point C′(2, –4). Calculate the pre-image point C.

Determine the pre-image point P if a reflection across the line y = x followed by a rotation of 90° about the origin in a clockwise direction resulted in the image point P′(–3, 6).

The image points of the vertices of a triangle A′(0, 0), B′(8, 2) and C′(15, 6) were transformed 3 5 by matrix T . 1 2 Determine the pre-image points of the vertices of the triangle ABC. Verify your answer graphically.

A rotation of 180° clockwise about the origin resulted in the image curve y = –2(x – 1)2. Determine the pre-image curve. Verify your answer graphically.

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11 Ex 3

Determine the inverse transformation matrices of: 2 1 a b 1 2

1 0 1 0

Determine: a

The transformation matrix for a dilation with a factor of 3 parallel to the x-axis and a factor of 0.5 parallel to the y-axis

The inverse of the dilation matrix from part (a)

Let T

2 0 be a transformation matrix: 0 3

Describe T in terms of dilations, reflections and rotations.

Determine the inverse transformation matrix and then describe the transformation of T −1.

Determine: a

The transformation matrix of a rotation of 75° anticlockwise about the origin

The inverse of the rotation matrix from part (a)

Determine: a

The transformation matrix for a reflection across the line y = x followed by a rotation of 90° anticlockwise about the origin

The inverse of the transformation matrix from part (a)

Determine: a

The transformation matrix for a dilation with a factor of 2 parallel to the x-axis followed by a reflection across the line y = –x

The inverse of the transformation matrix from part (a)

Determine: a

The transformation matrix for the reflection across the line y = x

The transformation matrix for the rotation of 45° clockwise about the origin

The overall transformation matrix by applying the reflection first, followed by the rotation

The inverse of the combined transformation matrix from part (c)

Explain the geometric effects of the transformation matrix T matrix cannot be inverted.

1 2 and why this means the 2 4

Extend your thinking 19

Determine the transformation matrix that transforms the points A(4, 3) and B(9, 10) to A′(2, 1) and B′(5, 6).

Demonstrate that the determinant of any rotation or reflection transformation matrix is always ±1, and explain why this property holds when finding the inverse of such matrices.

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Prove that a reflection across the line y = x tan (θ ) is its own inverse.

Determine the transformation matrix which rotates the line 4x – 3y = 0 about the y-axis.

Determine the transformation needed before a clockwise rotation about the origin by 90° in order to result in a reflection across the line y = 0.5x.

10.07 Area of linear transformations After this lesson, you will be able to… • determine the effect of linear transformations on area using determinants • understand the relationship between determinants and area scaling

Area of linear transformations Interactive exploration Discover this concept in action online

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The determinant of a transformation matrix indicates how much the area of a shape changes after transformation. The absolute value of the determinant, ∣det (T )∣, is the scale factor for the area. If the determinant is 0, the image area will be 0. Geometrically, this means the transformed shape has collapsed into a line or a point. If the determinant is negative, the transformation includes a reflection. However, the area is still scaled by the absolute value of the determinant, as area is always a positive quantity. When a pre-image shape is transformed by a linear transformation matrix, T, the image area is calculated using the following formula: Areaimage = ∣det (T)∣ × Areapre-image A unit square with vertices A(0, 0), B(0, 1), D(1, 1) and C(1, 0) can be transformed by a linear a b transformation matrix T . c d

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Applying the linear matrix transformation equation, P′ = TP, to each vertex yields the image vertices A′(0, 0), B′(b, d), D′(a + b, c + d) and C′(a, c), as shown: y

y D (1, 1)

B (0, 1)

D′ (a + b, c + d)

c B′ (b, d) d

C′ (a, c)

x A (0, 0)

C (1, 0)

A′ (0, 0)

The pre-image is a square with dimensions 1 by 1, unit, so its area is 1 unit2. The transformed shape is a parallelogram. Its area is calculated by finding the area of the large rectangle, (a + b) (c + d), and then subtracting the areas of the surrounding shapes: two rectangles bd ac with an area of bc, two triangles with an area of , and two other triangles with an area of . This 2 2 leads to: bd ac 2 Set up the area calculation 2 2 = ac + ad + bc + bd − 2bc − bd − ac Expand the brackets and simplify

Areaimage = ( a b) ( c d ) 2bc 2 = ad − bc

Combine like terms 2

= det (T) units

Recognise the determinant

As area is a positive quantity, the absolute value is used to ensure positivity.

Example 1 Triangle ABC with vertices A(2, 1), B(4, 6) and C(6, 1) is transformed by T area of A′B′C′.

2 0 . Determine the 0 3

Create a strategy Sketch the pre-image triangle ABC and calculate its area. Calculate the determinant of T. Then, apply the formula for the area of linear transformations to calculate the image area.

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Apply the idea First, sketch the pre-image triangle: B

6 4 2

A 0

The pre-image triangle has a base length of 4 units and perpendicular height of 5. Therefore, its area is: b h 2

Areapre-image Given that T

4 5 2

10 units2

2 0 , det (T ) = ad − bc = −2 × 3 − 0 × 0 = −6. 0 3

Apply the area of linear transformations formula: Areaimage = ∣det (T )∣ × Areapre-image Write the area of linear transformations formula = ∣ −6∣ × 10

Substitute the values

= 6 × 10

Take the absolute value of −6 2

= 60 units

Evaluate

Therefore, the image area is 60 units2.

Reflect and check Calculate the individual image points using P′ = TP, to find: A′(− 4, 3), B′(− 8, 18), C′(− 12, 3). Sketch the image points: B′ 15 10 5 A′

C′ –15

–10

–5

The image triangle has a base length of 8 units and perpendicular height of 15. Therefore, its area is: Areaimage

b h 2

8 15 2

60 units2

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Example 2 A square with vertices (0, 0), (1, 0), (1, 1), and (0, 1), is transformed by the matrix T the area of the transformed shape?

2 4 . What is 1 2

Create a strategy First, calculate the area of the pre-image square. Then, determine the determinant of the transformation matrix T. Finally, use the formula for the area of linear transformations to find the image area.

Apply the idea The pre-image is a square with a side length of 1 unit. Therefore, its area is 1 unit2. Given T

2 4 , calculate its determinant: 1 2 det (T ) = ad − bc

Write the determinant formula

=2×2−4×1

Substitute the matrix values

=4−4

Evaluate the multiplication

Evaluate

Now, apply the area of linear transformations formula: Areaimage = ∣det (T)∣ × Areapre-image Write the area of linear transformations formula = ∣0∣ × 1

Substitute the known values

=0×1

Take the absolute value 2

= 0 units

Evaluate

Therefore, the area of the transformed shape is 0 units2.

Reflect and check When the determinant of the transformation matrix is 0, the transformation maps the entire shape onto a line segment or a single point, resulting in an area of zero. For example, applying this transformation to the given square results in image vertices: • (0, 0):

2 4 1 2

0 0

• (1, 0):

2 4 1 2

1 0

2 1

• (1, 1):

2 4 1 2

1 1

6 3

2 4 0 4 1 2 1 2 1 All these image points (0, 0), (2, 1), (6, 3), and (4, 2), lie on the line y x, confirming that the 2 square collapses into a line segment. • (0, 1):

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Example 3 A triangle with an area of 15 units2, is transformed by a linear transformation matrix. The area of the image triangle is 75 units2. What is the absolute value of the determinant of the transformation matrix?

Create a strategy Rearrange the formula for the area of linear transformations, Areaimage = ∣det (T)∣ × Areapre-image, to solve for ∣det (T )∣. Then, substitute the given areas and calculate the result.

Apply the idea Given a pre-image area of 15 units2, and an image area of 75 units2. Rearrange the formula: Areaimage = ∣det (T )∣ × Areapre-image ∣det (T )∣ = =

Areaimage Areapre-image 75 15

rite the area of linear W transformations formula Divide both sides by Areapre-image Substitute the given area values Evaluate

The absolute value of the determinant of the transformation matrix is 5.

Reflect and check The absolute value of the determinant acts as an area scaling factor. A value of 5 means the transformed area is 5 times larger than the original area. This matches the problem: 15 units2 × 5 = 75 units2

Did you know?

The area of linear transformations is vital in practical applications! For example, architects use a floor plan to see how shapes change when scaled or stretched, and by analysing the determinant, they can check how each room’s area is affected and keep the layout accurate.

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Idea summary When a pre-image shape is transformed by a linear transformation matrix, T, the image area is calculated using this formula:

Areaimage = ∣det (T )∣ × Areapre-image Areaimage refers to the area of the shape after it has been transformed |det (T )| represents the absolute value of the determinant of the transformation matrix. This value acts as the area scale factor Areapre-image refers to the area of the original shape before transformation The absolute value ∣det (T )∣ is used because area is always a positive quantity. If det (T ) = 0, the area of the transformed shape will be 0, as the shape collapses into a line or a point.

10.07 Practice questions What do you remember? 1

Determine whether these statements about the area of transformed shapes are true or false: a

If the determinant of a transformation matrix is negative, the area of the image will be negative.

The absolute value of the determinant of a transformation matrix acts as the scale factor for the area.

If the determinant of a transformation matrix is zero, the area of the image is also zero.

A shape is transformed by a matrix with a determinant of 0: a

What is the area of the resulting image?

Describe what this implies geometrically about the transformed shape.

Calculate the determinant of each transformation matrix: a

4 2

1 3

3 1

6 2

Practice 4

A kite with an area of 18 units2 is transformed by matrix T Calculate the area of the image.

1000 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 3 . 8 10

A parallelogram with an area of 36 units2 is transformed by matrix T Calculate the area of the image.

Ex 1

7 Ex 2

Ex 3

A triangle A(0, 3), B(2, 8) and C(4, 3) is transformed by T a

Calculate the area of ABC.

Calculate the determinant of matrix T.

Calculate the area of A′B′C′.

4 2

1 . 3

3 1

6 . What is the 2

2 0 : 0 8

A rectangle with an area of 12 units2 is transformed by the matrix T area of the image?

A polygon with an initial area of 20 units2 is transformed by a linear transformation. If the transformed area is 150 units2, what is the absolute value of the determinant of the transformation matrix?

A parallelogram with vertices A(− 3, −1), B(0, 5), C(3, 5) and D(0, −1) is transformed by matrix 2 1 . Calculate the area of the image. T 5 3

A kite with vertices E(− 4, 5), F(2, 8), G(4, 1) and H(− 1, 2) is transformed by a dilation of factor 5 parallel to the y-axis and a dilation of factor 2 parallel to the x-axis:

Determine the area of the pre-image.

Determine the transformation matrix, T.

Calculate the area of the image.

A rectangle with vertices C(1, 1), D(10, 1), E(1, 5) and F(10, 5) is rotated about the origin 60° clockwise: a

Determine the transformation matrix, T.

Calculate the area of the image.

A quadrilateral H(5, 5), I(8, 10), J(11, 5) and K(14, 10) is reflected across the line y = 12x, then it is dilated by a factor of −2 parallel to the y-axis, and finally rotated 45° anticlockwise about the origin. Calculate the area of the image.

Extend your thinking 13

A unit square with vertices A(0, 0), B(0, 1), C(1, 0) and D(1, 1) is transformed by the a 3 transformation matrix T into a parallelogram. The image area is 4 units2. Determine a a all possible value(s) of a. k 2 Determine the values of k where the matrix maps the unit square to a parallelogram 4 3 of area greater than 3.

10.07 Area of linear transformations 1001 mathspace.co

Prove that the area of the image is the same as the area of the pre-image under a rotation of angle θ anticlockwise.

Explain the effect of a singular transformation matrix on the area of a shape. Include in your explanation what this implies about the nature of the transformation and provide a geometric interpretation.

Investigation: Transformations with code Investigate online

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Did you know?

The area of linear transformations is a powerful tool in visualising how shapes change under transformations! For example, when a biologist studies how cells deform under pressure, they can use a matrix to represent the transformation. By analysing the determinant of the matrix, they can determine whether the area of the cell increases, decreases, or stays the same, helping to understand cell behaviour under different conditions.

1002 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.” S. Gudder

Big ideas • The characteristics of complex functions are revealed by understanding their connection to simpler parent graphs. Applying systematic operations like taking the absolute value, reciprocal, or inverse, and transforming the extensive family of circular functions allows for the analysis and sketching of a wide variety of graphs. Algebraic techniques such as partial fraction decomposition are essential for breaking down complex rational expressions to understand their underlying structure. • Curves can be defined not just by a single Cartesian equation, but also as a locus of points satisfying a geometric rule, which gives rise to the family of conic sections. Parametric equations and the polar coordinate system offer alternative frameworks for describing these curves, providing a dynamic perspective of how a path is traced or a more elegant description for shapes with rotational symmetry.

11 Functions, relations, and graphs Chapter outline 11.01 11.02 11.03 11.04 11.05 11.06 11.07 11.08 11.09 11.10 11.11

Absolute value graphs Reciprocal functions Rational expressions and partial fractions Trigonometric functions Transformations of trigonometric functions Inverse circular functions Reciprocal trigonometric graphs Investigation: Locus definition Locus definitions and the distance formula Conic sections from locus definitions Parametric representations of relations Polar coordinates and polar graphs Chapter 11 review

1006 1022 1036 1043 1057 1071 1079 1101 1116 1127 1140

In photography, reciprocal functions help balance light and exposure settings!

11.01 Absolute value graphs After this lesson, you will be able to… • determine the absolute value of a real number • describe the key features of the absolute value function • graph the absolute value function and its transformations

Absolute value Absolute value The distance of a number from zero on the number line The absolute value is denoted by |x|, where x is any real number, and the result is always a non-negative quantity. For any real number x, the absolute value |x| is defined as: x

x, if x 0 x, if x 0

This definition means: • If x is greater than or equal to 0, the absolute value of x is simply x itself. • If x is less than 0, the absolute value of x is − x, which makes it positive. In both cases, the absolute value |x| gives the non-negative distance from zero. For example, x = +3 or x = − 3 implies |x| = 3, as |3| = | − 3| = 3. 3 units

3 units

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 Number line The numbers − 3 and 3 are both 3 units from 0, so both have the same absolute value. Similarly, the distance between two numbers, x and y, is defined as the absolute value of their difference, |x − y|. This formula uses the absolute value to ensure that the distance is always non-negative, reflecting the actual distance between the two numbers regardless of their order. This table outlines key properties of absolute values: Property Non-negativity

|x| ≥ 0

Identity property

If |x| = 0, then x = 0

Multiplicative property

|xy| = |x| × |y|

Division property

x y

1006 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x (for y ≠ 0) y

Example 1 Evaluate: a p = | − 3| + 2 − | − 5|

Create a strategy If x ≥ 0, use |x| = x. If x < 0, use |x| = − x.

Apply the idea p = | − 3| + 2 − | − 5|

Write the equation

= 3 + 2 −5

Substitute | − 3| = 3 and | − 5| = 5

Evaluate

b q = | − 4 + |2 − 8||

Apply the idea q = | − 4 + |2 − 8||

Write the equation

= | − 4 + | − 6||

Evaluate 2 − 8 inside the inner absolute value

= | − 4 + 6|

Substitute | − 6| = 6

= |2|

Evaluate − 4 + 6 inside the absolute value

Substitute |2| = 2

c |2a − b| when a = − 1 and b = 3

Create a strategy Substitute the given values of a and b into the expression. Then, evaluate the expression inside the absolute value before applying the absolute value definition.

Apply the idea |2a − b| = |2(− 1) − 3|

Substitute a = − 1 and b = 3

= | − 2 − 3|

Evaluate the multiplication

= | − 5|

Evaluate

Evaluate the absolute value

Therefore, |2a − b| = 5 when a = − 1 and b = 3.

11.01 Absolute value graphs 1007 mathspace.co

Idea summary The absolute value of a number is defined as its distance from zero on the number line, regardless of direction. It is denoted by |x|, where x is any real number, and the result is always a non-negative quantity. For any real number x:

x, x,

if x 0 if x 0

The distance between two numbers, x and y, is defined as the absolute value of their difference, |x − y|.

Absolute value graphs and transformations The graph of the basic absolute value function y = |x| can be constructed using the definition of the absolute value. Specifically, |x| = x for x ≥ 0 and |x| = − x for x < 0. This results in two line segments: for x ≥ 0, the graph follows the line y = x, and for x < 0, it follows the line y = − x. These segments intersect at the origin (0, 0), forming a V-shape. Consequently, the graph y = |x| is characterised by these two lines, reflecting the absolute value function’s properties of non-negativity and symmetry about the y-axis. Therefore, the graph of y = |x| is: y 3 y = −x

y=x

1 x −3 −2 −1

−1

The graph of y = |x| has several key features: • Vertex of an absolute value function: The point where the graph changes direction. For y = |x|, the vertex is at (0, 0). • Axis of symmetry: A line that divides the graph into two mirror images. For y = |x|, the axis of symmetry is the y-axis (the line x = 0). • Domain: The set of all possible input values. For y = |x|, the domain is all real numbers, which can be written as (− ∞, ∞) or x ∈ ℝ. • Range: The set of all possible output values. For y = |x|, the range is all non-negative real numbers, which can be written as [0, ∞) or y ≥ 0.

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The graph of the absolute value function can be transformed. The general form for transformations is:

y = a|x − h| + k a controls the vertical dilation (vertical stretch or vertical compression) and reflection across the x-axis, where the graph is stretched vertically if ∣a∣ > 1, compressed vertically if 0 < ∣a∣ < 1 and reflected across the x-axis if a < 0 h controls the horizontal shift (a horizontal translation), where the graph shifts right by h units if h > 0 and shifts left by ∣h∣ units if h < 0 k controls the vertical shift (a vertical translation), where the graph shifts up by k units if k > 0 and shifts down by ∣k∣ units if k < 0 (h, k) represents the coordinates of the vertex of an absolute value function of the transformed function The axis of symmetry for the transformed function y = a|x − h| + k is the vertical line x = h. The domain remains all real numbers, (− ∞, ∞). The range depends on a and k. If a > 0, the range is [k, ∞). If a < 0, the range is (− ∞, k]. Visually, these transformations affect the graph as follows: Effect of parameter a (vertical dilation and reflection): 4

y = 2|x| y = |x|

2 y= –4

–2

1 |x| 2 x 4

y = –|x|

11.01 Absolute value graphs 1009 mathspace.co

Effect of parameter h (horizontal shift): 6

4 2

y = |x + 2|

y = |x|

y = |x – 2| x

–6

–4

–2

–2 Effect of parameter k (vertical translation):

y y = |x| + 2

2 y = |x| x –4

–2

2 y = |x| – 2

–2

Therefore, the transition from the basic graph y = |x| to the graph y = a|x − h| + k can be achieved by applying three key types of transformations: • Vertical dilation (stretch/compression) and reflection: The parameter a scales the graph vertically. • If |a| > 1, the graph stretches vertically, whereas if 0 < |a| < 1, it compresses vertically. • If a < 0, it also reflects the graph across the x-axis. • Horizontal translation (shift): The parameter h shifts the graph horizontally. • If h > 0 (e.g., |x − h|), the graph shifts to the right. • If h < 0 (e.g., |x + h| = |x − (− h)|), it shifts to the left. • Vertical translation (shift): The parameter k shifts the graph vertically. • If k > 0, the graph shifts up. • If k < 0, it shifts down. For example, the graph y = 2|x − 3| + 4 can be obtained by starting with the graph of y = |x|, then applying a vertical dilation by a factor of 2, translating right by 3 units, and then translating up by 4 units.

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Example 2 Plot: y = − |x − 1| + 3

Create a strategy This is the basic absolute value graph, y = |x|, reflected across the x-axis, shifted 1 unit to the right and 3 units up. Compare to y = a|x − h| + k to find a = − 1, h = 1, k = 3.

Apply the idea The transformations are: • Reflection across the x-axis (due to a = − 1) • Horizontal translation 1 unit to the right (due to h = 1) • Vertical translation 3 units up (due to k = 3) The vertex of an absolute value function is at (h, k) = (1, 3). The axis of symmetry is x = 1. Since a < 0, the graph opens downwards.

4 3 2 1 −3 −2 −1

Therefore, the graph of y = −|x − 1| + 3 is:

−1 −2 −3 −4

x 1

2 3 4 5

Reflect and check Firstly, ensure the graph opens downwards because a = − 1 is negative. Then, verify the vertex is at (1, 3) due to the combined effects of shifting 1 unit right and 3 units up. Finally, verify that some key points from y = |x| like (0, 0), (1, 1), and (− 1, 1) are transformed as follows: • Original points: (0, 0), (1, 1), (− 1, 1) • After reflection ( y = −|x|): (0, 0), (1, − 1), (− 1, − 1) • After horizontal translation 1 unit right ( y = −|x − 1|): (1, 0), (2, − 1), (0, − 1) • After vertical translation 3 units up ( y = −|x − 1| + 3): (1, 3), (2, 2), (0, 2) These points are on the graph.

Example 3 Describe the sequence of transformations applied to y = |x| to obtain the graph of y = |x| − 4 and sketch the graph. Determine its vertex, axis of symmetry, domain, and range.

Create a strategy Compare the equation y = |x| − 4 to the form y = a|x − h| + k. Identify the values of a, h, and k. A non-zero value for k indicates a vertical translation. The vertex of an absolute value function will be at (h, k). The axis of symmetry is x = h.

11.01 Absolute value graphs 1011 mathspace.co

Apply the idea Comparing y = |x| − 4 to y = a|x − h| + k, this yields: • a=1 • h=0 • k = −4 Since k = − 4, this is a vertical translation of the graph of y = |x| downwards by 4 units. The key features are: • Vertex of an absolute value function : (h, k) = (0, − 4) • Axis of symmetry: x = h which implies x = 0 (the y-axis) • Domain: (− ∞, ∞) • Range: Since a = 1 > 0, the range is [k, ∞) which is [− 4, ∞). Graph of y = |x| − 4: 4 3 2 1 −5 −4 −3 −2 −1

−1 −2 −3

x 1 2 3 4 5

−4

Idea summary The graph of the basic absolute value function y = |x| is V-shaped. Key features include: • • • •

Vertex of an absolute value function at (0, 0). Axis of symmetry is x = 0 (the y-axis). Domain is (−∞, ∞). Range is [0, ∞).

Transformations of y = |x| can be expressed as y = a|x − h| + k: • The parameter a causes a vertical dilation (vertical stretch/vertical compression) and/or a reflection across the x-axis. • The parameter h causes a horizontal shift. • The parameter k causes a vertical translation. For y = a|x − h| + k, the vertex of an absolute value function is at (h, k), and the axis of symmetry is x = h. The domain remains (−∞, ∞). The range becomes [k, ∞) if a > 0, or (−∞, k] if a < 0.

1012 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Graphs of y = f (x), y = | f (x)|| and y = f |x|| Studying the relationships between the graphs of y = f (x), y = |f (x)| and y = f |x| provides insights into how changes in the function’s input and output influence its graphical representation: • y = f (x): The baseline graph of the original function. • y = |f (x)|: The graph of y = f (x) with all negative outputs reflected across the x-axis. This highlights how absolute value affects the function’s range. • y = f |x|: The graph of y = f (x) where the input is transformed to its absolute value, resulting in symmetry about the y-axis. This highlights how absolute value affects the domain and symmetry of the function. Construct or plot the graph of y = |f (x)| given the graph of y = f (x) by following these steps: 1. Graph the original function: Start by plotting the graph of y = f (x) for both positive and negative values of x. 2. Identify the positive side: Focus on the portion of the graph where f (x) ≥ 0 (i.e., where the graph is above or on the x-axis). This part will remain unchanged in the transformation (|f (x)| = f (x)). 3. Reflect the negative side: Identify the portions of the graph where f (x) is negative (i.e., where the graph is below the x-axis). For these regions, |f (x)| = − f (x). Reflect these portions of the graph in the x-axis to make all values non-negative. This involves taking the negative of the function’s values for these regions, which reflects the portions of the graph that lie below the x-axis to lie above it. 4. Combine both parts: Combine the original graph (for f (x) ≥ 0) with the reflected parts (for f (x) < 0) to complete the graph of y = |f (x)|. Ensure that the graph is always above or on the x-axis. To construct or plot the graph of y = f (|x|) given the graph of y = f (x), follow these steps: 1. Graph the original function: Start by plotting the graph of y = f (x) for both positive and negative values of x. 2. Identify the positive side (for input x): Focus on the portion of the graph where x ≥ 0. This part of the graph y = f (x) will remain unchanged because if x ≥ 0, then |x| = x, so f (|x|) = f (x). 3. Reflect the positive x-side: Discard the portion of the graph where x < 0. Then, reflect the portion of the graph for x ≥ 0 across the y-axis. This creates the graph for x < 0. This is because if x < 0, then |x| = − x. Therefore, for a negative x-value, f (|x|) = f (− x). This means the y-value for x = − c (where c > 0) will be the same as the y-value for x = c. 4. Combine both parts: The graph of y = f (|x|) consists of the original graph for x ≥ 0 and its reflection across the y-axis for x < 0. The resulting graph will be symmetrical about the y-axis.

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11.01 Absolute value graphs 1013 mathspace.co

Example 4 Given the function f (x) = x2 − 4, sketch the graph of the transformed function y = |f (x)|.

Create a strategy The transformation y = |f (x)| affects the output of the function. The process involves graphing the original function f (x), then reflecting any part of the graph that is below the x-axis to be above the x-axis. The parts of the graph already on or above the x-axis remain unchanged.

Apply the idea Step 1: Graph the original function. First, sketch the graph of the original function, y = x2 − 4. This is a parabola with its vertex at (0, − 4) and x-intercepts at x = − 2 and x = 2. 5 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4 −5

x 1

2 3 4 y = x2 − 4

Step 2: Reflect the negative portion. To obtain the graph of y = |x2 − 4|, any part of the original graph that is below the x-axis must be reflected above it. • The parts of the graph where y ≥ 0 (for x ≤ − 2 and x ≥ 2) remain unchanged. • The part where y < 0 (for − 2 < x < 2) is reflected across the x-axis. y 5 4 3 2 1 x y = |x2 – 4| −4 −3 −2 −1 1 2 3 4 −1 −2 −3 y = x2 – 4 −4 −5 The final graph has a distinctive ‘W’ shape, with the vertex that was at (0, − 4) now reflected to (0, 4).

1014 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 5 Given the function f (x) = x − 5, sketch the graph of the transformed function y = f (|x|).

Create a strategy The transformation y = f (|x|) can be sketched in three main steps. First, graph the original function f (x). Second, keep only the part of the graph where x ≥ 0 and discard the rest. Finally, reflect this remaining part across the y-axis to create a symmetrical graph.

Apply the idea Step 1: Graph the original function. The original function is f (x) = x − 5, which is a straight line with a slope of 1 and a y-intercept of − 5.

y 4 2 −6 −4 −2

x 2

−2

−4 −6 y=x−5

−8

Step 2: Keep the portion where x ≥ 0. The part of the graph to the right of the y-axis is kept, and the part to the left is discarded.

y 4 2 −6 −4 −2 −2

x 2

−4 −6 −8

11.01 Absolute value graphs 1015 mathspace.co

Step 3: Reflect the remaining part across the y-axis. The remaining portion of the graph is reflected across the y-axis to produce the final graph of y = f (|x|) = |x| − 5. y 4 2 y = |x| – 5 −6 −4 −2 −2

x 2

−4 −6 −8

The final result is a V-shaped graph that is symmetric about the y-axis, with its vertex at (0, − 5).

Example 6 Digital only example Explore in the online lesson

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Idea summary Studying the graphs of y = f (x), y = | f (x)|, and y = f (|x|) shows how absolute values affect a function’s graph: • y = |f (x)|: Reflects any negative outputs of f (x) (parts of the graph below the x-axis) above the x-axis, making all y-values non-negative. The parts of the graph of f (x) that are on or above the x-axis remain unchanged. • y = f (|x|): The part of the graph of f (x) for x ≥ 0 is kept. The part for x < 0 is discarded and replaced by a reflection of the x ≥ 0 part across the y-axis. The resulting graph is symmetric about the y-axis. These transformations reveal how absolute values impact the range and symmetry of a function.

1016 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.01 Practice questions What do you remember? 1

The absolute value of a real number x is defined piecewise. Complete the definition: x

Identify the key features of the basic absolute value function, y = |x|: a

Vertex coordinates

Axis of symmetry

Domain

Range

In the general form y = a|x − h| + k, what transformation does each parameter affect? a

y = |f (x)|

y = f (|x|)

Identify which property is always true for any real numbers a and b: A

Describe how to obtain these graphs from the graph of a function y = f (x): a

x, if x x, if x

|a + b| = |a| + |b|

|a × b| = |a| × |b|

|a − b| = |a| − |b|

Determine whether each statement is true or false: a

| − 3| < 2

|4| ≥ | − 4|

| − 2 − 3| = |2| + | − 3|

|3 − 1| ≤ | − 2| − |5|

Practice 7 Ex 1a, 1b

Evaluate: a

| − 2| + |5|

| − 2 + 5|

|8 − (− 6)|

|8| − | − 6|

| − 3| + | − 7|

| − 3 + (− 7)|

|8 − 9| + | − 6 + 4|

|9 − 11| − | − 8 + 5|

| − 6 − 4| − |2 + 3|

Evaluate: a

| − 3| × |4|

| − 3 × 4|

|10 × (− 2)|

|12| ÷ | − 4|

| − 8| × | − 4|

| − 8 × (− 4)|

|6 − 8| × | − 7 + 4|

5 7 4 6

18 10 3 7

11.01 Absolute value graphs 1017 mathspace.co

9 Ex 1c

Evaluate these expressions for the given values: a

|3x − 2y| when x = − 2 and y = 4

|a| − |b − a| when a = 5 and b = 1

Sketch the number line representation of: a

|x| ≤ 3

|x| ≠ 2

|x| > 6

|x| < 7

|x| = − 7

|x| = 9

|x| ≥ 5

|x| > − 5

Complete this table: x

−2

−1

f (x) = |x|

⬚

f (x) = x − 3

⬚

f (x) = |x − 3| f (x) = 2|x − 3| − 5|

⬚

Complete this statement:

The graph of y = 4|x − 3| + 2 can be obtained by starting with the graph of ⬚, then applying a vertical dilation by a factor of ⬚, translating ⬚ by 3 units, and translating up by ⬚ units.

Plot y = − |x − 3| + 1 without using a table of values.

Ex 2

Plot y = 2|x − 3| + 1 using a table of values.

Plot the graph of: a

Ex 3

f (x) = 1 − 3|x|

f (x) = |1 − 3x|

Consider the functions f (x) = |x| and f (x) = 4|x| − 1: a

Complete this table: x

−2

−1

f (x) = |x|

⬚

f (x) = 4|x| − 1

⬚

Plot y = 4|x| − 1.

What transformations are applied to the graph of y = |x| to obtain the graph of y = 4|x| − 1?

1018 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Given that the blue section represents the function y = |x|, determine the equation of each function in the form of y = a|x − h| + k: a

y 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4

4 3 2 1

x −4 −3 −2 −1

2 3 4

y 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4

2 3 4

−4 −3 −2 −1

x 1

−1 −2 −3 −4

4 3 2 1

−1 −2 −3 −4

2 3 4

x 1

2 3 4

Consider the graph of a function y = f (x): 9 8 7 6 5 4 3 2 1 −3 −2 −1−1 −2 −3

x 1 2 3 4 5 6 7

Determine whether the statement is true or false: a

The domain of f (x) is all x ≥ 0.

f (x) ≥ 0 for all values of x.

The graph has a symmetry about the y-axis.

The curve could be the graph of a function y = − |x − 1| + 1.

11.01 Absolute value graphs 1019 mathspace.co

Consider the equation f (x) = |x| + 3: a

Sketch the graph of the function.

Which of these statements about the symmetry of the graph is true? A

The graph is symmetric about the y-axis.

The graph is rotationally symmetric about the origin.

The graph is symmetric about the x-axis.

The graph has no symmetry.

Determine a simplified expression for f (− x).

Complete the statement:

The result of part (c) shows that f (− x) = ⬚, which verifies that the function is symmetric about the y-axis. 20

Consider the function f (x) = 11 − |x|: a

Sketch the graph of the function.

What is the maximum value of f (x)?

On which interval is the function increasing? A

Ex 4, 5, 6

(− ∞, 0)

(− ∞, 0]

(− ∞, 11]

[11, ∞)

(11, ∞)

[0, ∞)

(0, ∞)

Using the graph of f (x) = − 2x + 3, plot the graph of: a

y = |f (x)|

Using the graph of f (x) = a

On which interval is the function decreasing? A

(− ∞, 11)

y = f (|x|)

1 x − 5, plot the graph of: 2

y = |f (x)|

y = f (|x|)

Consider the function f (x) = x2 − 5, plot the graph of: a

y = f (x)

y = |f (x)|

y = f (|x|)

Extend your thinking 24

The graph of y = − |x − 4| + 2 is translated 2 units down and 3 units left, then reflected across the x-axis. What is the equation of the transformed graph?

Consider the function f (x) = |x + 6| + |x − 6|: a

Complete the table of values to completely define the function in each domain:

|x + 6| |x − 6| |x + 6| + |x − 6|

Simplified expression in domain x < − 6

Simplified expression in domain − 6 ≤ x < 6

Simplified expression in domain x ≥ 6

⬚

1020 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

⬚

Fill in the missing values to define f (x). 2 x for x for 6 for x

f ( x) c d

Evaluate the function when x = − 6 and x = 6. Which of these is the graph of f (x)? A

y 12 9 6 3 −12 −9 −6 −3 −3 −6 −9 −12

Plot: y = |x − 1| + |x + 3|

Plot: y = |x2 − 4| + 2

Plot: |y − 1| = 2|x − 5|

18 15 x

12 9 6 3

−12 −9 −6 −3

3 6 9 12

y −12 −9 −6 −3 −3 −6 −9 −12 −15 −18 −21 −24 −27

3 6 9 12

y 12 9 6 3 −12 −9 −6 −3 −3 −6 −9 −12

y 21

3 6 9 12

y 27 24 21 18 15 12 9 6 3

−12 −9 −6 −3 −3 −6

6 3 −12 −9 −6 −3 −3 −6 −9 −12 −15 −18 −21 −24 −27

x 3 6 9 12

y x 3 6 9 12

11.01 Absolute value graphs 1021 mathspace.co

11.02

Reciprocal functions

After this lesson, you will be able to… • interpret aspects of the reciprocal function based on original function • determine the key features of the rectangular hyperbola • perform simple transformations on reciprocal functions

Simple reciprocal functions Reciprocal function A function defined as the reciprocal of another function, y =

1 . f ( x)

Vertical asymptote A vertical line that the graph of a function approaches but never touches or crosses. It is where the function is undefined. Horizontal asymptote A horizontal line that the graph of a function approaches as the input x approaches positive or negative infinity. Rectangular hyperbola A specific type of hyperbola with asymptotes that are perpendicular to each other, typically described by the equation y = k. x

y y

1 f ( x)

is the output of the reciprocal function

f (x) is the output of the original function for a given input x This definition implies that for each input x where f (x) ≠ 0, the output of the reciprocal function is the reciprocal of the value of f (x). Specifically, if f (x) = a (where a ≠ 0), then the reciprocal function 1 evaluates to y = . a 1 It is important to understand how the graph of y = is affected by the original function, y = f (x). f ( x) A key area of interest is what happens when f (x) gets very close to zero or becomes very large.

1022 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Some properties for the reciprocal function y =

1 are: f ( x)

1 for each x-value f ( x) where f (x) = 0. The domain of the reciprocal function excludes these values. 1 • As f (x) approaches zero from the positive side, approaches positive infinity. f ( x)

• Vertical asymptotes: A vertical asymptote occurs on the graph of y =

• As f (x) approaches zero from the negative side,

1 approaches negative infinity. f ( x)

• Horizontal asymptote: If the original function f (x) approaches ±∞ as x → ±∞, then its 1 reciprocal approaches zero. This results in a horizontal asymptote at y = 0. f ( x) • General shape: The reciprocal’s behaviour is opposite to the original function’s magnitude. 1 • When f (x) is large, is small (close to zero). f ( x) 1 • When f (x) is small (but not zero), is large. f ( x) Consider, for example, the function y =

1 which is the reciprocal function of y = x: x

• It has a vertical asymptote at x = 0, this is because the function is undefined where x = 0. 1 As x approaches zero from the positive side, approaches positive infinity. Conversely, as x 1 x approaches zero from the negative side, approaches negative infinity. x 1 • As x approaches positive or negative infinity, the value of approaches zero. Therefore, x the horizontal asymptote for this function is y = 0. This means that as x becomes very large (positively or negatively), y gets closer and closer to zero but never actually reaches it. 1 • Around the vertical asymptote at x = 0, the graph of y = exhibits hyperbolic behaviour. This x means that the graph has two distinct branches: • For x > 0: The graph approaches the vertical asymptote from the positive side and heads towards the horizontal asymptote as x increases. • For x < 0: The graph approaches the vertical asymptote from the negative side and also approaches the horizontal asymptote as x decreases. 10

y y=

1 x y=x

–10

–5

x 10

–5 –10 Therefore, the graph of

1 is a basic example of a hyperbola. x 11.02 Reciprocal functions 1023 mathspace.co

The graph of y =

1 is the most basic example of a rectangular hyperbola. Its key features are: x

• Asymptotes: It has a vertical asymptote at x = 0 (the y-axis) and a horizontal asymptote at y = 0 (the x-axis). • Domain and range: The domain is all real numbers except 0, x ∈ (− ∞, 0) ∪ (0, ∞). The range is also all real numbers except 0, y ∈ (− ∞, 0) ∪ (0, ∞). • Shape: The graph consists of two separate, and symmetric branches. 1 • Quadrants: For y = , the branches are in Quadrants I and III. x • Axes of symmetry: The graph is symmetric about the lines y = x and y = − x. k The general form of a simple rectangular hyperbola is y = . The value of k dilates the graph. If k is x negative, the graph is reflected across the x-axis (or y-axis) and its branches lie in Quadrants II and IV.

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Example 1 Consider the function y =

1 x , where f (x) = + 4: f ( x) 2

a Solve the equation f (x) = 0 for x, and explain the significance of the solution for the reciprocal 1 function y = . f ( x)

Create a strategy Set the expression for f (x) equal to zero and solve for x. The zero of the original function f (x) 1 corresponds to the vertical asymptote of its reciprocal . f ( x)

Apply the idea x 4 = 0 2

Set the function equal to zero

x = −4 2

Subtract 4 from both sides

x = −8

Multiply both sides by 2

The solution is x = − 8. This value is where the original function is zero, so the reciprocal function is 1 undefined. Therefore, the reciprocal function y = has a vertical asymptote at x = − 8. x 4 2

1024 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b Analyse the behaviour of f (x) as x approaches positive and negative infinity and interpret the result for the graph of the reciprocal function.

Create a strategy As x becomes very large (positive or negative), f (x) will also become very large. The reciprocal of a very large number is a number very close to zero. This determines the horizontal asymptote.

Apply the idea Consider the end behaviour: x + 4 → ∞. 2 x • As x → − ∞, f (x) = + 4 → − ∞. 2 In both cases, as the magnitude of f (x) becomes very large, its reciprocal • As x → ∞, f (x) =

1 approaches zero. f ( x) Therefore, the line y = 0 is the horizontal asymptote for the reciprocal function.

c Sketch the graph of y =

1 , labelling all key features including asymptotes and intercepts. x 4 2

Create a strategy First, simplify the expression for the reciprocal function. Then, draw the vertical asymptote at x = −8 and the horizontal asymptote at y = 0. Find the y-intercept by setting x = 0. Determine the shape of the graph based on the behaviour around the asymptotes and plot the y-intercept to help sketch the curve.

Apply the idea The function can be simplified: y =

1 x 4 2

1 x 8 2

2 x 8

• Vertical asymptote: x = −8 • Horizontal asymptote: y = 0 • y-intercept (let x = 0): y=

2 0 8

1 The y-intercept is at 0, . 4

2 8

4 3 2 1

1 4 −−12 −10 −8 −6 −4 −2

• x-intercept (let y = 0): 0=

2 x 8

−1 −2 −3 −4

x 2

This equation has no solution, which is expected as y = 0 is a horizontal asymptote. There are no x-intercepts.

11.02 Reciprocal functions 1025 mathspace.co

Example 2 For the functions f (x) =

4 x2 and g (x) = 2 : 4 x

a Sketch the graph of f (x).

Create a strategy x2 Recognise f (x) = as a standard parabola with its vertex at the origin, opening upwards. 1 4 The factor of makes it wider than the basic parabola y = x2. 4

Apply the idea 8 7 6 5 2 x f ( x) = 4 4 3 2 1 −4 −3 −2 −1

x 1

2 3 4

b Describe how f (x) is related to g (x).

Apply the idea The function g (x) is the reciprocal of f (x). The relationship is shown by the calculation: 1 1 4 2 2 = g (x) f ( x) x x 4

c Using the graph of f (x), determine the key features and sketch the graph of g (x) =

1 . f ( x)

Create a strategy • Vertical asymptotes: Find where f (x) = 0. • Horizontal asymptotes: Analyse the behaviour of f (x) as x → ±∞. • Sign: Note that f (x) is always positive (or zero), so g(x) must also be always positive. The graph will be entirely above the x-axis. • Invariant points: Find where f (x) = 1. At these points, g(x) = 1 as well. • Sketch the graph using these features.

1026 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea x2 = 0 when x = 0. Therefore, g(x) has a vertical asymptote at x = 0. 4 1 Horizontal asymptote: As x → ±∞, f (x) → ∞. Therefore, g(x) = → 0. The horizontal asymptote f ( x) is y = 0. Vertical asymptote: f (x) =

x2 Invariant points: Set f (x) = 1: = 1, which gives x² = 4, therefore x = ± 2. The graphs intersect at 4 (−2, 1) and (2, 1). The domain of g(x) is x ≠ 0. The range is y > 0.

f ( x) = x 4

−4 −3 −2 −1

8 7 6 5 4 3 2 1

g ( x) =

4 x2

x 1

2 3 4

Idea summary This section covered the key principles for interpreting and graphing reciprocal functions. 1 • The graph of a reciprocal function y = can be determined from the graph f ( x) of y = f (x). 1 • Vertical asymptotes of occur where f (x) = 0. f ( x) 1 • A horizontal asymptote of y = 0 exists for if f (x) → ±∞. f ( x)

• The two graphs intersect where f (x) = ±1 (invariant points). 1 1 • Where f (x) is positive, is positive. Where f (x) is negative, is f ( x) f ( x) negative. k • The graph of y = is a rectangular hyperbola with key features: x • Vertical asymptote at x = 0 and horizontal asymptote at y = 0. • Domain: x ≠ 0. Range: y ≠ 0. • If k > 0, branches are in Quadrants I and III. If k < 0, branches are in Quadrants II and IV. • Axis of symmetry are y = x and y = −x.

11.02 Reciprocal functions 1027 mathspace.co

Transformations of the rectangular hyperbola 1 Transformations can be applied to the basic rectangular hyperbola y = to create more complex x reciprocal functions. The general form of a transformed hyperbola is:

A k, A 0 x h

A is the vertical dilation h is the horizontal translation, vertical asymptote at x = h k is the vertical translation, horizontal asymptote at y = k Key features of the transformed hyperbola: • Vertical asymptote: x = h • Horizontal asymptote: y = k • Domain: All real numbers except h, x ≠ h • Range: All real numbers except k, y ≠ k A + k is: x h 1. Identify the values of A, h, and k.

A clear strategy for sketching f (x) =

2. Draw the vertical asymptote x = h and horizontal asymptote y = k as dashed lines. 3. Determine the location of the branches. If A > 0, the branches are in the top-right and bottom-left quadrants formed by the asymptotes. If A < 0, the branches are in the top-left and bottom-right. 4. Calculate and plot the y-intercept by setting x = 0. 5. Calculate and plot the x-intercept by setting y = 0. 6. Sketch the two branches of the hyperbola passing through the intercepts and approaching the asymptotes. If ∣A∣ > 1, then the graph is dilated by a factor of ∣A∣ from the x-axis. For example, when A = 3: y=

3 1 compared to y = x x 4 y 1 y= x 3 2 1

(–1, –1) –4 –3 –2 –1

–1

–2 (–1, –3) –3 –4

1028 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3 y= x (1, 3) (1, 1)

1 2 3 4 Asymptotes: x=0 y=0

If 0 < ∣A∣ < 1∣, then the graph is dilated from the x-axis by a scale factor of ∣A∣. For example, 1 when A = : 2 4 y 1 1 y= y= 3 2x x 2 1  (1, 1)  − 1, − 2  1   x –4 –3 –2 –1 (–1, –1) –1 –2 –3 –4

1 2 3 4  1  1, 2    Asymptotes: x=0 y=0

If A < 0, then the graph is reflected across the x-axis and then dilation may also take place. For example, when A = − 2: 4 3 (–1, 2)

1 x

2 (1, 1)

1 –4 –3 –2 –1 –1 (–1, –1) –2 Asymptotes: –3 x=0 y=0 –4

x 1

(1, –2) 2 y=− x

Example 3 Consider the function y =

2 x 3

a Describe the transformations applied to y =

1 to obtain this function. x

Apply the idea • Dilation by a factor of 2 from the x-axis. • Horizontal translation of 3 units to the right. • Vertical translation of 1 unit up.

11.02 Reciprocal functions 1029 mathspace.co

b Determine the equations of the asymptotes, the domain, and the range.

Apply the idea A By comparing the function to the general form y = x h identified as A = 2, h = 3, k = 1.

k, the values of the parameters can be

• Vertical asymptote: x = h, which is x = 3 • Horizontal asymptote: y = k, which is y = 1 • Domain: x ≠ 3 • Range: y ≠ 1

c Sketch the graph, showing all key features.

Create a strategy First, find the x- and y-intercepts. Then, plot the asymptotes and intercepts on a Cartesian plane and sketch the curve. Since A = 2 is positive, the branches will be in the top-right and bottom-left sections relative to the asymptotes.

Apply the idea y-intercept (let x = 0): y = x-intercept (let y = 0):

2 0 3

1 1 . Point is 0, . 3 3

2 1 3

2 1 x 3

Write the equation

−1 =

2 x 3

Subtract 1 from both sides

−(x − 3) = 2

Multiply both sides by x – 3

−x + 3 = 2

Distribute –1

−x = −1

Subtract 3 from both sides

x=1

Divide both sides by –1

The x-intercept is at (1, 0). 5 4 3 2 1 −1

−1 −2 −3

x 1

2 3 4 5 6 7

1030 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 4 The graph of a rectangular hyperbola has asymptotes x = 1 and y = − 2, and passes through the point (3, 0). Find its equation.

Create a strategy A k. x h Then, substitute the coordinates of the given point into the equation to solve for A. Use the asymptotes to find the values of h and k in the general equation y =

Apply the idea The vertical asymptote x = 1 gives h = 1. The horizontal asymptote y = −2 gives k = −2. A 2. The equation is of the form y = x 1 Substitute the point (3, 0) into the equation to find A: y=

A 2 x h

Write the equation

A 2 3 1

Substitute x = 3 and y = 0

0 = A 2 2 2=

A 2

A=4

Evaluate the subtraction Add 2 to both sides Multiply both sides by 2

4 2. The equation of the graph is y = x 1

Idea summary 1 The graph of the rectangular hyperbola y = can be transformed into the x A k. form y = x h

• The vertical asymptote is x = h. • The horizontal asymptote is y = k. • The value of A determines the dilation and orientation of the branches. To sketch the graph, first identify and draw the asymptotes. Then, find the x- and y-intercepts. Use these features to accurately draw the two branches of the hyperbola.

11.02 Reciprocal functions 1031 mathspace.co

11.02 Practice questions What do you remember? 1

Complete this table: x

−2

−1

f (x) = x − 3

⬚

1 f ( x)

⬚

Consider a function y = f (x) defined over the domain [− 1, ∞). Determine whether the statement is true or false:

The zeros of f (x) become horizontal asymptotes for the graph of

If f (x) increases positively,

The domain of y =

1 approaches zero. f ( x)

1 is [− 1, ∞). f ( x)

Consider the functions f and g defined by f (x) = 2x − 5 and g (x) = Determine the behaviour of: a c

f (x) when x approaches ∞ 1 when x approaches − ∞ f ( x)

b d

g (x) when x approaches − ∞ 1 when x approaches ∞ g( x )

Determine the behaviour of: b c d 5

1 . x 3

Consider the functions f and g defined by f (x) = − 2x + 4 and g (x) = a

1 . 2x 1

f (x) when x approaches 2 1 g(x) when x approaches from the right 2 1 when x approaches 2 from the left f ( x) 1 1 when x approaches g( x ) 2

Determine the domain of the function g defined by g (x) = a

1 . f ( x)

f (x) = 2 − x

f (x) = x2

1032 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 when: f ( x)

f (x) = 2 + x4

f ( x)

1 x

Practice Ex 1

c d e

b c

10 11

Determine the behaviour of f (x) as x approaches − ∞ and ∞. 1 Plot the graph of y = . f ( x) 1 , where f (x) = ∣x − 5∣: f ( x) Solve for x the equation f (x) = 0.

Consider the function y = a

Ex 2

1 , where f (x) = x2 + 2x − 3: f ( x) Solve for x the equation f (x) = 0.

Consider the function y = a

1 , where f (x) = x2 − 4: f ( x) Solve the equation f (x) = 0 for x, and explain the significance of the solutions for the 1 reciprocal function y = . f ( x) Determine the behaviour of f (x) as x approaches − ∞ and ∞, and interpret the result for 1 the graph of the reciprocal function . f ( x) 1 Determine the behaviour of y = around the lines with equations x = 2 and x = − 2. f ( x) 1 Plot the graph of y = . f ( x) 1 Plot the graph of y = 2 on the same Cartesian plane. x 4

Consider the function y =

Determine the behaviour of f (x) as x approaches − ∞ and ∞. 1 Plot the graph of y = . f ( x)

For the functions f (x) = x2 − 1 and g (x) =

1 2

x 1

Sketch the graph of f (x).

Describe how f (x) is related to g (x).

Using the properties of f (x), determine the key features and sketch the graph of g(x).

1 1 Using the graph of f (x) = 3 + , plot the graph of y = . x f ( x) Using the graph of f (x) = x2 + 3, plot the graph of: 1 1 a y b y f ( x) f ( x)

1 f( x)

11.02 Reciprocal functions 1033 mathspace.co

Ex 3

Consider the function y = a

Determine the zero of f (x) and determine the equation of the vertical asymptote for

Describe the end behaviour of f (x) and determine the equation of the horizontal 1 asymptote for . f ( x) 1 Determine the y-intercept of . f ( x) 1 Sketch the graph of y = . 2x 6

c d

Ex 4

1 , where f (x) = 2x − 6: f ( x)

3 + 2: x 1

Consider the function y = a

Describe the sequence of transformations that maps the graph of y =

Determine the equations of the vertical and horizontal asymptotes.

Determine the domain and range of the function.

Determine the coordinates of the x- and y-intercepts.

Sketch the graph.

1 . f ( x)

1 to this function. x

A rectangular hyperbola has a vertical asymptote at x = − 2 and a horizontal asymptote at y = 4. It passes through the point (0, 3). Determine the equation of the hyperbola.

Given y = f (x) = x2 − k, f (x) and A

x = ±1

Consider the graph of

1 will intersect when: f ( x) C y = ±1 k

1 . f ( x) 0.4 0.3 0.2

0.1 −0.1 −0.2 −0.3 −0.4

x 1 2 3 4 5 6 7 8 9

Determine whether the statement is true or false: a

The domain of f (x) is all x ≥ 0.

f (x) ≥ 0 for all values of x.

The curve of f (x) has a symmetry about the line with equation x = 5.

The curve of f (x) could be the graph of a certain parabola.

1034 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking 1 7 and y = + k intersect at A and B, where A lies on the x-axis, determine x x 2 the value of k and the coordinates of B.

Given y = 3 −

This graph represents a function f (x):

1 Plot the graph of y = on a Cartesian plane. f ( x)

y y = f (x)

x 5

–5

–5 10 19

x=0

6 x2 9 x 5 This graph represents the function f (x) = : 10 x 5 1 a Plot the graph of . f ( x) 1 b Plot the graph of y = on the same Cartesian f ( x ) plane.

8 6 4 2

−8 −6 −4 −2 −2 −4 −6 −8

This graph represents two functions f and g such that f (x) × g (x) = 1: Determine g (x).

f (x) g (x)

4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4

Sketch the graph y = f (x) =

2x 1 2x

x2 8 x 4

2 4 6 8

x 1

2 3 4

11.02 Reciprocal functions 1035 mathspace.co

11.03 Rational expressions and partial fractions After this lesson, you will be able to… • write rational functions as a sum of a reciprocal functions using partial fractions • apply the technique of equating coefficients of polynomials

Partial fraction decomposition Partial fraction decomposition A technique used to rewrite a complex rational function as a sum of simpler rational functions whose denominators are factors of the original denominator. Example: 3x 1 ( x 1) ( x 2)

A x 1

B x 2

Partial fraction decomposition is useful for breaking down complicated rational expressions into simpler pieces. This technique relies on the factors of the denominator. This lesson focuses on denominators that can be factored into distinct linear terms and irreducible quadratic terms. If the numerator’s degree is greater than or equal to the denominator’s, perform polynomial long division first. This is to express the rational function as a polynomial plus a proper rational function (where numerator degree is less than denominator degree). The proper rational function part is then decomposed. •

Case 1: Distinct linear factors

If the denominator h(x) can be factored into distinct linear factors (a1 x + b1)(a2 x + b2) . . . (an x + bn), the decomposition takes the form:

g ( x) h ( x)

A1 a1 x b1

A2 a2 x b2

An an x bn

Ai is a constant that must be determined for each corresponding linear factor ai x + bi are the distinct linear factors of the denominator

1036 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

•

Case 2: Irreducible quadratic factors

If the denominator h(x) contains an irreducible quadratic factor ax2 + bx + c (meaning it cannot be factored further into linear factors with real coefficients, i.e., b2 − 4ac < 0), the decomposition includes a term of the form:

Ax B ax2 bx c A, B 2

ax + bx + c

are the constants that must be determined is the irreducible quadratic factor

If the denominator has multiple factors (linear and irreducible quadratic), the decomposition is a sum of terms corresponding to each factor type. •

Finding the constants (Equating coefficients) After setting up the appropriate form for the decomposition: 1. Multiply both sides of the equation by the original denominator h(x). This clears the fractions. 2. Simplify the resulting equation. This results in a polynomial equation: g(x) = a polynomial in (x) with coefficients involving A, B, C, . . .) 3. Equate the coefficients of corresponding powers of x on both sides of the identity. For example, the coefficient of x2 on the left must equal the coefficient of x2 on the right. 4. This creates a system of linear equations for the unknown constants (A, B, C . . .). Solve this system to find the values of the constants.

Alternatively, for distinct linear factors, substituting the roots of these factors into the cleared equation (step 2) can be a quicker way to find some or all constants.

Example 1 Express

5x 1 in partial fractions. ( x 3) ( x 1)

Create a strategy The denominator has two distinct linear factors, (x − 3) and (x + 1). The degree of the numerator (1) is less than the degree of the denominator (2), so no long division is needed. Set up the partial A B fraction form . Clear the denominators and solve for A and B by equating coefficients or x 3 x 1 substituting strategic values.

11.03 Rational expressions and partial fractions 1037 mathspace.co

Apply the idea Set up the decomposition: 5x 1 A = ( x 3) ( x 1) x 3

B x 1

Multiply both sides by the common denominator (x − 3)(x + 1): 5x − 1 = A(x + 1) + B(x − 3)

Clear the denominators

Method 1: Equating coefficients 5x − 1 = Ax + A + Bx − 3B

Expand the right-hand side

5x − 1 = (A + B)x + (A − 3B)

Group terms by powers of x

Equate coefficients of corresponding powers of x: • Coefficient of x: 5 = A + B 1 • Constant term: − 1 = A − 3B

Solve the system of equations. Subtract 2 from 1 : 5 − (− 1) = (A + B) − (A − 3B) Subtract the corresponding sides of the equations

Substitute B

6 = 4B

Simplify the expression

6 4

Divide by 4 to solve for B

3 2

Simplify the fraction

3 into 1 : 2 5 = A

3 2

Substitute the value of B

A=5

3 2

Rearrange to solve for A

10 2

3 2

Write with a common denominator

7 2

Simplify the fraction

Method 2: Substitution Start from 5x − 1 = A(x + 1) + B(x − 3). Substitute x = 3 (root of x − 3): 5(3) − 1 = A(3 + 1) + B(3 − 3)

Substitute the value of x

14 = A(4) + B(0)

Evaluate the terms

14 = 4A

Simplify the equation

14 4

Solve for A

7 2

Simplify the fraction

1038 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Substitute x = − 1 (root of x + 1): 5(− 1) − 1 = A(− 1 + 1) + B(− 1 − 3)

Substitute the value of x

− 6 = A(0) + B(− 4)

Evaluate the terms

− 6 = − 4B

Simplify the equation

6 B= 4 3 = 2 Both methods yield A

Solve for B Simplify the fraction

7 3 and B . 2 2

Therefore, the partial fraction decomposition is: 7 3 5x 1 2 2 ( x 3) ( x 1) x 3 x 1

7 2 ( x 3)

3 2 ( x 1)

Reflect and check To check, combine the fractions on the right side using the common denominator 2(x − 3)(x + 1): 7 2 ( x 3)

7 ( x 1) 3 ( x 3) 3 = 2 ( x 3) ( x 1) 2 ( x 1) 7 x 7 3x 9 2( x 3) ( x 1) 10 x 2 = 2( x 3) ( x 1) =

2 (5 x 1) 2 ( x 3) ( x 1) 5x 1 = ( x 3) ( x 1) =

Write with a common denominator Expand the numerator Collect like terms Factorise the numerator Remove the common factor

This matches the original expression, confirming the result.

Example 2 Express

3 x2

x 10

( x 2) ( x2 4 )

in partial fractions.

Create a strategy The denominator has one linear factor (x + 2) and one irreducible quadratic factor (x2 + 4) (since 02 − 4(1)(4) = −16 < 0). The degree of the numerator (2) is less than the degree of the denominator A Bx C (1 + 2 = 3). Set up the partial fraction form . Clear denominators and use equating x 2 x2 4 coefficients to find A, B, C.

11.03 Rational expressions and partial fractions 1039 mathspace.co

Apply the idea Set up the decomposition: 3 x2

x 10

( x 2) ( x

A Bx C 2 x 2 x 4

Multiply both sides by (x + 2)(x + 4): 3x2 − x + 10 = A(x2 + 4) + (Bx + C)(x + 2)

Clear the denominators

Expand and group terms: 3x2 − x + 10 = Ax2 + 4A + Bx2 + 2Bx + Cx + 2C = (A + B)x2 + (2B + C)x + (4A + 2C) Equate coefficients: • Coefficient of x2: 3 = A + B

• Coefficient of x: −1 = 2B + C

• Constant term: 10 = 4A + 2C

Equation 3 can be simplified by dividing by 2: 5 = 2A + C 3 a From 1 , B = 3 − A. From ( 3 a), C = 5 − 2A. Substitute these into 2 : −1 = 2(3 − A) + (5 − 2A)

Substitute expressions for B and C

−1 = 6 − 2A + 5 − 2A

Expand the brackets

−1 = 11 − 4A

Collect like terms

4A = 11 + 1 = 12

Rearrange to solve for A

A=3

Solve for A

B=3−A

Write the expression for B

=3−3

Substitute the value of A

Simplify

Find B and C:

C = 5 − 2A

Write the expression for C

= 5 − 2(3)

Substitute the value of A

=5−6

Evaluate the multiplication

= −1

Solve for C

Therefore, A = 3, B = 0, C = −1. The partial fraction decomposition is: 3 x2

x 10

( x 2) ( x

3 x 2

1040 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

0x 1 x

3 x 2

1 x

Idea summary g ( x) (where h ( x) deg (g) < deg (h)) as a sum of simpler fractions based on the factors of h(x). Partial fraction decomposition rewrites a rational function

• For each distinct linear factor (ax + b), include a term

A . ax b

• For each irreducible quadratic factor (ax2 + bx + c), include a term

Ax B

ax2 bx c Find the unknown constants (A, B, C, . . .) by clearing the denominators and either equating coefficients of like powers of x or substituting strategic values of x.

11.03 Practice questions What do you remember? 1

For each polynomial, determine its degree and leading coefficient: a

P(x) = 5x3 − 2x2 + 7x − 1

Q(x) = 4 − 3x + 6x2

R(x) = (2x + 1)(x − 3)

S(x) = 8

x2 − 9

x2 − 6x + 9

Factorise these quadratic expressions: a c

x2 + 5x + 6 2

2x − 5x − 3

Determine whether these statements are true or false: a

The first step for decomposing an improper rational function (where the numerator’s degree is greater than or equal to the denominator’s degree) is to perform polynomial long division.

Partial fraction decomposition is a method for combining multiple simple fractions into a single, more complex fraction.

A distinct linear factor in the denominator, such as (x − 3), corresponds to a term of the A form in the partial fraction decomposition. x 3

An irreducible quadratic factor in the denominator, such as (x2 + 1), corresponds to a term A of the form 2 in the partial fraction decomposition. x 1 The quadratic expression x2 − 4x + 4 is an irreducible quadratic.

11.03 Rational expressions and partial fractions 1041 mathspace.co

Practice 4

For each rational expression, write the form of its partial fraction decomposition. Do not solve for the constants: a

Ex 1

5 x2 1

x ( x 9)

The partial fraction decomposition of

Express

Express each rational expression in partial fractions:

c 8

x 4 ( x 1) ( x 2) ( x 3)

4x 7 A B has the form . Multiplying the ( x 1) ( x 2) x 1 x 2 denominator yields the identity 4x + 7 = A(x + 2) + B(x + 1). By expanding and equating the coefficients of the powers of x, determine the system of linear equations for A and B.

Ex 2

2x 1 ( x 5) ( x 3 )

8x 1 in partial fractions. ( x 2) ( x 3)

2x 3 x2 2 2

x 2

x 4

x 10 x2

x 12 6

x ( x 2 3)

Express these rational expressions in partial fractions: a

4 x2 3 x 5 ( x 1) ( x2 1)

2 x2 3 x 2 ( x 1) ( x2 2)

First perform polynomial division, then find the partial fraction decomposition of

Express

x2 3 x 5 . x 2

4 x2 6 x 10 in partial fractions. ( x 1) ( x 2) ( x 3)

Extend your thinking x 3 2 x2 5 x 1

in the form P ( x )

x 1

, where P(x) is a polynomial.

Express

The decomposition for a rational function with a repeated linear factor, such as g( x ) (x − a)n, includes a term for each power from 1 to n. For example, would have the ( x a)2 2 A B 3x 7 x 3 form . Use this principle to express in partial fractions. x a ( x a)2 ( x 1)2 ( x 2)

1042 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.04

Trigonometric functions

After this lesson, you will be able to… • derive the graphs of sine, cosine, and tangent from the unit circle • graph sine, cosine, and tangent functions over restricted domains • identify and analyse key features of circular function graphs

Sine and cosine functions Period The length of the smallest interval over which the graph of a periodic function repeats. Amplitude The measure of half the vertical distance between the maximum and minimum values of a periodic function.

The graphs of sine and cosine functions can be derived from the unit circle. For any point on a circle with radius 1, at an angle of θ from the positive x-axis, the coordinates are given by (cos (θ ), sin (θ )).

Interactive exploration mathspace.co

Discover this concept in action online

As θ moves from 0 to 2π, a full cycle of each graph is completed. The period is the length of one full cycle, which is 2π for both y = sin (x) and y = cos (x). The amplitude is the distance from the centre line to the maximum or minimum value. Since the unit circle has a radius of 1, the amplitude is 1. To graph sine or cosine functions over a given domain, follow these steps: 1. Identify the key points for one cycle. A cycle of sine or cosine has five key points: the maximums, the minimums, and the x-intercepts. 3 , 1 , and (2π, 0). 2 3 3. For y = cos (x) over the domain [0, 2π], the key points are (0, 1), , 0 , (π, − 1), , 0 and (2π, 1). 2 2 4. Plot these key points on the Cartesian plane and connect them with a smooth, wavelike curve. 2. For y = sin (x) over the domain [0, 2π], the key points are (0, 0),

, 1 , (π, 0),

5. Extend the pattern to the left or right to cover the entire given domain.

Example 1 Digital only example Explore in the online lesson

mathspace.co

11.04 Trigonometric functions 1043 mathspace.co

Example 2 Consider the shown section of the curve y = sin (x): y 1 1 2 −2π

−23π

−π

1π −2

x 1 2π

1 −2

3 2π

2π

−1

a If one cycle of the graph of y = sin (x) starts at x = 0, when does the next cycle start?

Create a strategy

Apply the idea

The graph can be used to determine the next point where the function has the same value and a positive gradient as the intercept at x = 0.

At x = π, the value of the function is the same as x = 0 (i.e. sin (0) = sin (π) = 0) but the function is decreasing. At x = 2π, y = 0 and the function is increasing. Therefore, the next cycle starts at x = 2π.

b In which regions is the graph of y = sin (x) decreasing?

Create a strategy Use the graph to determine the intervals of x for which the function is decreasing.

Apply the idea y 1 1 2 −2π

3π −2

−π

−21 π

x 1 2π

1 −2

3π 2

2π

−1

The graph shows that y = sin (x) is decreasing at

1044 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3 and 2

3 2

c What is the x-value of the x-intercept where 0 < x < 2π?

Create a strategy

Apply the idea

Use the graph to determine the x-value where y = 0.

The graph shows that y = 0 when x = π. This is the only x-intercept between x = 0 and x = 2π.

Example 3 Sketch the graph of the function y = cos (x) over the domain [0, 3π].

Create a strategy Identify the five key points for one cycle of the cosine function over the domain [0, 2π]. This pattern is then extended to complete the graph up to 3π.

Apply the idea Step 1: Find the five key points for one cycle starting at x = 0. • Maximum at x = 0: (0, 1) • Intercept at x

• Minimum at x = π: (π, − 1) 3 3 : ,0 • Intercept at x 2 2 • Maximum at x = 2π: (2π, 1) Step 2: Extend the pattern to cover the domain up to 3π. The next key points will be an intercept at 5 and a minimum at x = 3π. x 2 2 2 5 5 : ,0 • Intercept at x 2 2 • Minimum at x = 3π: (3π, − 1) Step 3: Plot all the points and connect them with a smooth curve. y 1 1 2 0 1 − 2 −1

x 1 π 2

3 π 2

2π

5 π 2

11.04 Trigonometric functions 1045 mathspace.co

Example 4 Sketch the graph of the function y = sin (x) over the restricted domain

, . 2 2

Create a strategy The solution involves identifying the key points of the sine function that fall within the given domain. These points include the values at the endpoints of the domain and any intercepts or stationary points in between.

Apply the idea Step 1: Find the values of the function at the endpoints of the domain. • At x • At x

sin

2 2

1. Point: 1. Point:

Step 2: Find any key points that lie between the endpoints. The x-intercept at x = 0 is in this domain. • At x = 0, y = sin (0) = 0. Point: (0, 0) Step 3: Plot these three points and connect them with a smooth curve. y 1 1 2 –π

1 − π 2

1 2 −1

−

1046 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x 1 π 2

Idea summary Key features of y = sin (x): • • • • •

Domain is all real numbers Range is −1 ≤ y ≤ 1 Period = 2π Amplitude = 1 x-intercepts (solutions) occur at . . ., − π, 0, π, 2π, . . . 5 9 • Local maximums at: , , ,... 2 2 2 3 7 • Local minimums at: , , ,... 2 2 2 y 1

y = sin (x)

−π

1 2 1 − π 2

Amplitude 1 π 2

1 2 −1

−

x 3 π 2

2π

Period Key features of y = cos (x): • • • •

Domain is all real numbers Range is −1 ≤ y ≤ 1 Period = 2π Amplitude = 1

3 • x-intercepts (solutions) occur at . . ., , , ,. . . 2 2 2 • Local maximums at: 0, 2π, . . . • Local minimums at: −π, π, . . . y Period

1 y = cos (x)

−π

1 − π 2

1 2 Amplitude 1 2 −1

−

1 π 2

x π

3 π 2

2π

11.04 Trigonometric functions 1047 mathspace.co

Tangent functions Tangent function A circular function defined as the ratio of sine to cosine, representing the gradient of the line forming an angle with the positive x-axis. Asymptote A line that a curve approaches as it heads towards infinity.

The tangent function is defined as: sin ( ) cos ( )

tan ( )

This also represents the gradient of the line that forms the angle θ to the positive x-axis.

Interactive exploration mathspace.co

Discover this concept in action online

The graph of y = tan (θ ): y 10 5

θ –2π

3 − π 2

–π

−

π 2

–5

π 2

3 π 2

2π

–10

Graphing tangent functions over a given domain involves these steps: 1. Identify the period, which is π for the function y = tan (q ). 2. Locate and draw the vertical asymptotes. These occur where the function is undefined, at + kπ for any integer k. Draw these as dashed vertical lines. x 2 3. Identify the key points for one cycle between a pair of asymptotes. The x-intercept occurs halfway between the asymptotes. The function passes through (intercept + , 1) and 4 (intercept − , − 1). 4 4. Plot the key points and sketch the characteristic curve, ensuring it approaches the asymptotes. 5. Repeat the pattern of asymptotes and curves to cover the entire given domain.

1048 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Key features of the graph of y = tan (θ ) are: • Vertical asymptotes: At θ = + kπ, where k is an integer. For example, on the given graph 2 3 3 , , , . asymptotes are 2 2 2 2 • Axis intercepts: y-intercept: (0, 0), x-intercepts at: θ = kπ, where k is an integer - at each intercept there is a point of inflection • Period: π, the distance between two successive asymptotes • Range: All real numbers ℝ, or (− ∞, ∞) • Domain: θ : θ ∈ ℝ, θ ≠ 2 vertical asymptote

k , where k is any integer, as the function is undefined at each

• Key points: Useful for graphing, the function passes through

, 1 and

• Symmetry: The tangent graph is an odd function, that is tan (− θ ) = − tan (θ ). This means the graph has point symmetry about the origin (or any point of inflection) by 180°. The asymptotes of the function are where the angle θ would cause the terminal arm to be vertical sin ( ) and so the gradient is undefined. This can also be seen through the definition tan . cos ( ) The function is undefined where cos (θ ) = 0 and the graph approaches the vertical lines at these values of θ, forming asymptotes. The tangent function repeats at intervals of π. This can be proven algebraically by looking at the sin ( ) ratio . If π is added to an angle θ, then the diagram shows that sin (θ + π) has the same cos ( ) magnitude as sin (θ ) but opposite sign. The same relation holds between cos (θ + π) and cos (θ ). y 1

θ +π

sin (θ )

cos (θ + π) –1 sin (θ + π)

θ θ

cos (θ )

–1

sin (θ ) = − sin (θ + π) sin (θ + π) = − sin (θ )

cos (θ ) = − cos (θ + π) cos (θ + π) = − cos (θ )

11.04 Trigonometric functions 1049 mathspace.co

Making use of the definition: tan (θ + π) =

sin( cos (

) )

Apply the definition of tangent, tan (q ) =

sin ( ) cos ( )

sin ( ) Use the angle identities sin (q + p) = –sin (q ) and cos (q + p) = –cos (q ) cos ( )

= tan (θ )

Remove the negative signs and simplify back to tan (q )

Example 5 Digital only example Explore in the online lesson

mathspace.co

Example 6 Select the two functions that have the same graph as y = tan (x): A y

tan x

3 2

y = tan (x + π)

tan x

y = tan (x + 2π)

Create a strategy The period represents the interval that the graph repeats itself.

Apply the idea Consider the graph of y = tan (x). Think about by how much each of the curve could be moved to the right so that they land on the graph again or by how much could the asymptote be moved to the right so it lands on another asymptote. The period of y = tan (x) is π, so the graph repeats itself every π units. The graph of y = tan (x) can be shifted horizontally by multiples of π units and the graph will align with itself. y

2 x 1 − π 2

1 π 2

3 π 2

−2

Therefore, the options that have the same graph as y = tan (x) are Options B and D.

1050 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 7 Sketch the graph of the function y = tan (x) over the domain [− π, 2π].

Create a strategy First, identify and draw the vertical asymptotes within the domain [−π, 2π]. Then, find the intercepts and other key points for each cycle within the domain. Finally, sketch the curves approaching the asymptotes.

Apply the idea Step 1: Identify the asymptotes in the domain [−π, 2π]. Asymptotes are at x • For k = −1, x • For k = 0, x • For k = 1, x

k .

2 2 3 2

Step 2: Identify the x-intercepts. These are at x = kπ. • Points are: (−π, 0), (0, 0), (π, 0), and (2π, 0). Step 3: Identify key points passing through y = 1 and y = −1 for each cycle. For the cycle between , 1 . This pattern repeats. , 1 and and , the points are 4 4 2 2 Step 4: Plot the points, draw the asymptotes, and sketch the tangent curves over the domain. 3

2 1 −π

1 − π 2

−1 −2

x 1 π 2

3 π 2

2π

−3

11.04 Trigonometric functions 1051 mathspace.co

Idea summary Key features of the graph of the tangent function, y = tan (x) are: 4 3 2 1 –2π

–π –1

Midline π

2π

–2 –3 –4 Period

k , k is • Domain: x ∈ ℝ, x ≠ 2 any integer • Range: (−∞, ∞) • Asymptotes: x = + kp, k is any 2 integer • x-intercept: x = kπ, k is any integer • y-intercept: (0, 0) • Period: π • Amplitude: None • Midline: y = 0 • Key points:

, 1 and

To graph a tangent function use the key features mentioned.

11.04 Practice questions What do you remember? 1

Consider the graph of y = sin (x) and determine whether these statements are true or false:

Ex 1

y 1 x 1 3 −2π − 2π −π −2π

1π 3π 2π 2 π 2

−1

The graph of y = sin (x) is symmetric about the line x = 0.

The graph of y = sin (x) is symmetric with respect to the origin.

The y-values of the graph repeat after a period of 2π.

1052 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Consider the graph of y = cos (x): If one cycle of the graph of y = cos (x) starts at x = 0, at what value of x does the next cycle start?

Determine whether the graph of y = cos (x) is increasing or decreasing over these domains:

Ex 2

0<x<π

π < x < 2π

iii

− 3π < x < − 2π

1 x

−1

5 . 2

Consider the curve y = sin (x):

Identify the x-intercept over the domain − 2π < x < 0.

If one cycle of the graph of y = sin (x) starts at x = − 2π, at what value of x does the next cycle start?

Determine whether the graph of y = sin (x) is increasing or decreasing over these domains: 3 3 5 i ii x x 2 2 2 2 3 2

3π 1π 2 π 2 2π

1 3 −2π−2π −π −2 π

Identify the x-intercept over the domain

iii

5 2

1 x 3π −π −1 π −2π−2 2

1 3 2π π 2π 2π

−1

3 2

Consider the graph of y = cos (x): y 1 x 3π −π −1 π −2π−2 2

1 3 2π π 2π 2π

−1

Determine the coordinates of the y-intercept.

Determine the range of the function.

Determine the period of the function.

Determine the x-intercepts over the domain 0 < x < 2π.

11.04 Trigonometric functions 1053 mathspace.co

Consider the right-angled triangle containing angle θ and the graph of y = cos (θ ): y 1 Opposite

Hypotenuse

θ –2π− 3 π–π − 1 π 2 2

1 π π 3 π 2π 2 2

–1 Adjacent

a b

As angle θ increases from 0 to , explain what happens to the value of the opposite side 2 in the triangle. Explain what happens to the value of tan (θ ) as angle θ increases from 0 to , given that 2 Opposite tan (θ ) is defined as . Adjacent

Express tan (θ ) in terms of sin (θ ) and cos (θ ).

Determine the values of θ for which cos (θ ) = 0, given − 2π ≤ θ ≤ 2π.

Identify the values of θ between − 2π and 2π for which tan (θ ) is undefined.

Complete the table of values:

− 2π

7 4

5 4

−π

3 4

tan (θ )

⬚

3 4

⬚

tan (θ )

⬚

5 4

⬚

7 4 ⬚

Sketch the graph of y = tan (θ ) over the domain − 2π ≤ θ ≤ 2π.

2π ⬚

Determine the equations of the vertical asymptote(s) of the graph of y = tan (x) for − 2π < x < 2π.

Complete this statement: The domain of the graph y = tan (x) is x : x ∈ ℝ, x ≠ ⬚, where k is any integer.

1054 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice Ex 3

Sketch a graph of each function over the given domain: a

Ex 4

cos ( x ),

3 2

By first sketching each function over the given domain, determine the increasing regions of: a

y = sin (x), − 2π ≤ x ≤ 0

cos( x ),

y = sin (x), 0 ≤ x ≤ π

Use the diagram of the unit circle to explain these properties of the graph of y = sin (x): a

The range of values for y = sin (x) is − 1 ≤ y ≤ 1.

The graph of y = sin (x) repeats after every 2π radians.

y (cos (θ ), sin (θ ))

−1

Consider the graph of y = tan (x) for − 2π ≤ x ≤ 2π: a

Identify the y-intercept of the graph.

Identify the period of the function.

Identify the equations of the vertical asymptotes over the domain 0 ≤ x ≤ 2π.

Does the graph of y = tan (x) increase or decrease between any two successive vertical asymptotes?

If x > 0, determine the least value of x for which tan (x) = 0.

1 –2π− 3π –π − 1 π 2 2 –1

x 1π π 2

3π 2π 2

–2 –3 –4

12 Ex 5

Consider the graph of y = tan (x) for − 2π ≤ x ≤ 2π: a

Which of these best describes the tangent function: A

Power function

Circular function

Periodic function

Linear function

Determine the range of y = tan (x).

As x increases, determine the equation of the next asymptote of the graph after x = −

5 . 2

11.04 Trigonometric functions 1055 mathspace.co

Consider the graph of y = tan (x) for − 2π ≤ x ≤ 2π: 3 a Determine the sign of tan (x) for π ≤ x < . 2 3 b In which quadrant of a unit circle is angle x if π ≤ x < ? 2 c Determine the sign of tan (x) for < x ≤ 0. 2 d In which quadrant of a unit circle is angle x if < x ≤ 0? 2 e Describe the connection between the unit circle and the graph with regard to the sign.

Select the two functions that have the same graph as y = tan (x): 3 A y = tan (x − π) B y tan x C y tan x 2 2

Ex 6

y = tan (x + 3π)

For the graph y = tan (x), identify all the x-intercepts within the interval [2π, 6π].

Ex 7

Consider the graph of y = tan (x): a

How many x-intercepts would occur in the interval (0, π)?

How many x-intercepts would occur in the interval [0, π]?

Extend your thinking 17

Consider the graph of y = sin (x). Write an expression in terms of n to determine the x-intercepts of the function, where the angle x is measured in radians and n is an integer.

Consider the graph of y = cos (x). Write an expression in terms of n to determine the x-intercepts of the function, where the angle x is measured in radians and n is an integer.

Describe the sequence of transformations that maps the graph of f (x) = sin (x) to the graph of g (x) = cos (x).

Determine all the points of intersection between the graphs of y = sin (x) and y = cos (x) in the interval 0 ≤ x ≤ 2π. Provide the x-coordinates of these points in both radians and degrees.

1056 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.05 Transformations of trigonometric functions After this lesson, you will be able to… • apply dilations, reflections, and translations to graphs of sine, cosine, and tangent functions • identify key features of transformed trigonometric graphs

Transformations of y = sin (x) and y = cos (x) The graphs of y = sin (x) and y = cos (x) can be transformed through dilations, reflections, and translations. These transformations are controlled by four parameters, A, n, h, and k, in the general forms:

y = A sin (n(x − h)) + k and y = A cos (n(x − h)) + k A controls the amplitude and vertical reflection n controls the period (horizontal dilation/reflection) h controls the horizontal translation (phase shift) k controls the vertical translation and the midline around which the graph oscillates

Interactive exploration Discover this concept in action online

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The amplitude is half the vertical distance between the maximum and minimum values of the function. It is a measure of vertical dilation and is calculated as ∣A∣. If A < 0, the graph is also reflected across the midline. The period is the length of one complete cycle of the graph. The parameter n causes a horizontal 1 dilation by a factor of . The standard period of sine and cosine is 2π, so the new period is n 2 calculated as Period = . n A horizontal translation is controlled by h. The graph is translated horizontally by h units. Note the form (x − h); a positive value of h (e.g., (x − 3)) shifts the graph to the right. A negative value of h (e.g., (x − (− 3)), written as (x + 3)) shifts the graph to the left. A vertical translation is controlled by k. The graph is shifted vertically by k units. This translation also moves the midline, which is the horizontal line about which the graph oscillates. Its equation is y = k.

11.05 Transformations of trigonometric functions 1057 mathspace.co

Example 1 Identify the amplitude and midline for each trigonometric function: a y = − 4 sin (x)

Create a strategy Compare the equation to the general form y = A sin (n(x − h)) + k to identify the values of A and k. The amplitude is ∣A∣ and the midline is y = k.

Apply the idea For y = − 4 sin (x), the parameters are A = − 4 and k = 0. The amplitude is ∣A∣ = ∣ − 4∣ = 4. The midline is y = 0 (the x-axis).

b y

cos (x ) 1 2

Create a strategy First, rewrite the equation in the standard form y = A cos (n(x − h)) + k. Then, identify A and k to find the amplitude and midline.

Apply the idea Rewrite the equation as y = 1 and k = −1. 2 1 The amplitude is |A|= 2

1 cos (x) −1. 2

Here, A

1 . 2

The midline is y = −1.

Example 2 First describe the sequence of transformations that have occurred from the base graph to obtain the given function, then sketch each graph: a y = cos (2x) over the domain 0 ≤ x ≤ 2π

Create a strategy 2 Identify the parameter n. Calculate the new period using . Note that all other parameters are n unchanged (A = 1, h = 0, and k = 0).

1058 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea 1 For y = cos (2x), the parameter n = 2. This is a horizontal dilation by a factor of . 2 2 The period is = π. This means the graph completes two full cycles within the specified domain 2 of [0, 2π]. y 1 0.5 0 –0.5

x 1π 2

3π 2

2π

–1

b y = cos (2x − π)

Create a strategy The equation must be in the form y = A cos (n(x − h)) + k to correctly identify the horizontal shift. Factorise out the value of n inside the cosine function.

Apply the idea Factorising out n = 2 from the argument gives: y = cos 2 x The transformations are:

1 from the y-axis (period is π). 2 • A horizontal translation of units to the right. 2 y • A dilation of factor

1 0.5 0 –0.5

x 1π 2

3π 2

2π

–1

11.05 Transformations of trigonometric functions 1059 mathspace.co

2 cos 4 x

Create a strategy Identify all four parameters A, n, h, and k and describe the effect of each one.

Apply the idea For y = − 2 cos 4 x

− 1, the parameters are:

• A = −2 • n=4 • h= 2 • k = −1 The transformations are: • Reflection across the midline (due to negative A). • Dilation of factor of 2 from the x-axis (amplitude is 2). 1 2 . from the y-axis period is 4 4 2 • Horizontal translation of units to the right. 2 • Vertical translation of 1 unit down (midline is y = − 1). • Dilation of factor of

y 1 0 –1

x 1π 2

3π 2

2π

–2 –3

Idea summary The graphs of y = sin (x) and y = cos (x) can be transformed into y = Af (n(x − h)) + k, where f is sine or cosine. • Amplitude is ∣A∣. If A < 0, the graph is reflected vertically. 2 • Period is . n • Horizontal translation is h units. • Vertical translation is k units. The midline is y = k.

1060 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Did you know?

A guitar string vibrates in patterns that match transformed sine curves, creating different musical notes.

Transformations of the tangent function The graph of the tangent function can also be transformed. While it does not have an amplitude, its steepness, period, and position can be changed.

y = A tan (n(x − h)) + k A controls the vertical dilation (steepness) and reflection across the midline n controls the period (horizontal dilation/reflection) h controls the horizontal translation k controls the vertical translation and the midline

Interactive exploration Discover this concept in action online

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• The effects of A: • ∣A∣ > 1 vertically stretches the graph. • 0 < ∣A∣ < 1 vertically compresses the graph. • A < 0 reflects the graph across the midline y = k. • The effects of n: • The period is calculated as Period =

. n • ∣n∣ > 1 horizontally compresses the graph. • 0 < ∣n∣ < 1 horizontally stretches the graph.

• The effect of h is a horizontal translation of h units. This translates the asymptotes. For y = tan (x), an asymptote is at x = . For the transformed graph, this asymptote translates to x = + h. 2 2n • The effect of k is a vertical translation of k units. The midline of the graph is the line y = k.

11.05 Transformations of trigonometric functions 1061 mathspace.co

Example 3 Describe the sequence of transformations applied to y = tan (x) that result in the graph of 1 y = 2 tan − 1. Then, determine the key features. x 2 4

Create a strategy Compare the equation to the general form y = A tan (n(x − h)) + k to identify the four parameters. Describe the effect of each one to list the transformations. Use the parameters to calculate the key features.

Apply the idea The parameters are A = 2, n

1 ,h 2

, and k = −1.

The transformations are: • A dilation of factor of 2 from the x-axis. • A dilation of factor of 2 from the y-axis since n • A horizontal translation of units to the right. 4 • A vertical translation of 1 unit down.

1 . 2

The key features are: • Period:

2 . 1 2 • Midline: y = k, so y = −1. • Asymptotes: To find the asymptotes, set the argument of the tangent function equal to the 1 x standard asymptote locations: + mπ. Then solve for x. The new asymptotes 2 4 2 Period 5 2 are at x = + h + m × Period. Therefore, x = + m(2π) = π + + 2mπ = + 2mπ. 2 4 2 4 4 n

Reflect and check To verify the transformations, the location of the point that is usually at the origin can be checked. , Due to the horizontal and vertical shifts, its new location should be at (h, k) which is 4 can be tested by substituting x = into the equation: 4

1 . This

1 −1 Write the equation x 2 4 1 = 2 tan −1 Substitute x = 4 2 4 4 = 2 tan (0) − 1 Simplify the expression inside the tangent function

y = 2 tan

= 2(0) − 1

Evaluate tan (0)

= −1

Evaluate

The point is correct.

1062 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 4 Consider the transformed graph of f (θ ) = tan (θ ): 5 4

3 2 1

1 3 1 –π − π− π− π−1 4 2 4 −2

1 3 1 π π π π 4 2 4

−3 −4 −5

a Determine the equation of the graphed function.

Create a strategy Identify the key features from the graph to determine the parameters A, n, h, k. Observe the vertical shift to find the midline (k). Find the period to determine n. Check for a horizontal shift (h) by looking at the point on the midline that is halfway between the asymptotes.

Apply the idea The midline is at y = 4; therefore, k = 4. The graph passes through the point (0, 4) on its midline, and the next corresponding point on the , 4 . The horizontal distance between these points is the period. Period = . Since midline is at 2 2 Period = , it follows that ; therefore, ∣n∣ = 2. The simplest case, n = 2, is used. n 2 n The point on the midline is at θ = 0; therefore, there is no horizontal translation and h = 0. The graph is not reflected, indicating that A > 0. With no vertical dilation or compression present, it is assumed that A = 1. The equation is y = tan (2θ ) + 4.

11.05 Transformations of trigonometric functions 1063 mathspace.co

b Determine the domain and range.

Apply the idea The range of any transformed tangent function is all real numbers, (−∞, ∞). + mπ. Solving for θ gives θ = 2 4 The domain is all real numbers except for these values. The asymptotes occur when 2θ =

m , where m is any integer. 2

Idea summary The graph of y = tan (x) can be transformed into the general form y = A tan (n(x − h)) + k. • Dilation from the x axis: Factor of ∣A∣. If A < 0, the graph is reflected vertically. • Period: n • Horizontal translation: h units. • Vertical translation: k units. The midline is y = k.

11.05 Practice questions What do you remember?

Ex 1

True or False: The horizontal translation of a function f (x) = cos (x − h) translates the graph h units to the left if h > 0.

What does the parameter k represent in the function y = sin (x) + k?

Identify the amplitude of each function: a c

y = 4 cos (x) 4 y cos (x ) 3

b d

y = − 5 cos (x) sin( x ) y 8

For each function, identify and describe: i

The horizontal translation

The vertical translation

y = sin (x) − 2

y = cos (x + π)

cos x

1064 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

sin x

5 2

Practice 5

Sketch a graph of each function for 0 ≤ x ≤ 2π: a

y = 2 sin (x)

For each graph of the form y = sin (bx) or y = cos (bx), where b is positive: i

Identify the period.

x 1 π 2

3 π 2

–1

y 1

–1

x 1 π 2

–1

Determine the equation of the function.

Ex 2

y = − 3 sin (x)

x π 2π 3π 4π 5π 6π 7π 8π 9π

–1

Sketch a graph of each function showing one full cycle over the corresponding domain: 2 3 sin( 3 x ), 0 x a y = sin (2x), 0 ≤ x ≤ π b y 3 c y = 3 sin (2x) + 1, 0 ≤ x ≤ π d y = cos (2x) + 1, 0 ≤ x ≤ p

11.05 Transformations of trigonometric functions 1065 mathspace.co

Identify the horizontal translation for each graph, and complete the corresponding equation: a

y = sin (x + ⬚)

y = cos (x − ⬚)

y 1

1 x 2 1 −π − π − π 3 3

1 2 π ππ 3 3

1 −π − π 2

–1

1 π 2 –1

y = sin (x − ⬚)

y = cos (x + ⬚)

y 1

1 x 1 3 ππ π 2π 2 2

3 1 –2π − π−π− π 2 2

–1

Calculate the maximum and minimum values for each trigonometric function: 1 sin x 2 2 sin x a y = 2 sin (x) + 3 b y c y 2 4 2

Sketch a graph of each function showing two full cycles over the corresponding domain. Check using technology: a

x 1 2 π π π 3 3

2 1 −π − π − π 3 3

–1

2 sin x

17 4

3 sin x

4 3

Sketch the graph of each function and check using technology: a

3 sin 2 x

5 cos

1 x 2 3

2 cos 3 x

2 sin 2 x

1066 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

6 4

1 3

8 3

This graph shows the average daily temperature (°C) for Tokyo over two consecutive years: Average daily temperature in Tokyo over two consecutive years 40

Temperature (°C)

35 30 25 20 15 10 5 0

a 13

J F M A M J J A S O N D J F M A M J J A S O N D J Time (months)

What is Tokyo’s hottest month?

Identify the period and the amplitude.

A weather station records the temperature variations in a coastal town over a week. The temperature levels are recorded in degrees Celsius (°C). The first reading was taken at 6:00 a.m., and the graph of the readings is shown: Temperature variations over a week 40 Temperature (°C)

35 30 25 20 15 10 0

Ex 3

72 96 Time (hours)

120

144

168

Does the temperature level appear to be a periodic function over time?

At about what time does the temperature level seem to peak?

When does the temperature level appear to be at its lowest?

What is the period?

What is the amplitude?

Describe the sequence of transformations that maps the graph of y = sin (x) to the graph of y

3 sin 2 x

11.05 Transformations of trigonometric functions 1067 mathspace.co

The functions f (x) and g (x) = f(kx) have been graphed on the same set of axes: y y = g (x)

1 x

− 3π 4

−1π 2

y = f (x)

1π 4

−1π 4 −1

Determine the period of f (x).

Determine the period of g (x).

Describe the sequence of transformations required to map the graph of g (x) from the graph of f (x).

Calculate the value of k.

Consider the graph of y = sin (x):

The first maximum point for x ≥ 0 is indicated at

,1 . 2 By considering the transformation that has taken place, determine the coordinates of the first maximum point of each of the given functions for x ≥ 0:

3π 4

1π 2

y = 5 sin (x)

y = − 5 sin (x)

y = sin (x) + 2

y = sin (3x)

y = 5 sin (x) + 2

sin x

1 x

1π 2

3π 2

2π

−1

Consider the graphs of y = sin (x) and y = g (x) graphed on the same set of axes: y y = sin( x )

1 π 4 −1

x 1 π 2

3 π π 4 y = g( x )

5 π 4

3 π 2

7 π 4

2π

9 π 4

Describe the sequence of transformations required to obtain the graph of y = g (x) from the graph of y = sin (x).

Identify the equation of g (x) .

1068 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Consider the function y = tan (2x): a

Complete the table of values: x

−π

⬚

tan (2x)

Ex 4

3 4 3 2

⬚

3 4

⬚

Undefined

⬚

Determine the interval between the asymptotes of y = tan (2x).

Determine the period of y = tan (2x).

Sketch the graph of y = tan (2x) over the domain − π ≤ x ≤ π.

⬚

For each graph: i

Identify the period, midline, and point of inflection.

Describe the transformations from f (x) = tan (x) to the graphed function.

iii

Identify the equation of the graphed function.

Determine the domain and range.

y 4

1 –2π− 3 π –π − 1 π 2 –1 2

y 4

x 1π π 2

–2π − 3 π –π − 21 π 2 –1

3 π 2π 2

–2

–3

–4

y 4

1π π 2

3 π 2π 2

y 4

1 1π –2π − 3 π –π − 2 2 –1

x 1π π 2

3 π 2π 2

1 –2π − 3 π –π − 1 π 2 –1 2

–2

–3

–4

x 1π π 2

3 π 2π 2

11.05 Transformations of trigonometric functions 1069 mathspace.co

Compare the graph of g(θ ) = tan (θ ) to each function: a

g(θ ) = tan (2θ )

g (θ ) = − 3 tan (θ )

g( )

tan

Extend your thinking 21

Consider the functions f (x) and g(x) = jf (kx), graphed on the same set of axes: y 1 2 3 1 3

g ( x) –8π

–6π

–4π

–2π

f ( x)

x 2π

−1 3 −2 3 −1

4π

Determine the equation for f (x).

Determine the values of j and k, and so, the equation of g (x).

6π

8π

Calculate the values of c over the domain − 2π ≤ c ≤ 2π that make the graph of y = sin (x − c) the same as the graph of y = cos (x).

Determine two possible equations of the graphed function. y 3 2 1 1 3 −2π − π −π− π 2 −1 2

x 3 1 π π π 2π 2 2

−2 −3

Using technology, consider the functions y

2 cos x

and y

sin

x . 4

Determine the number of solutions over the domain 0 ≤ x ≤ 4π, for the equation: 2 cos x

1070 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

sin

x 4

f (x) = a sin (bx) + d has a range of [3, 11]. Determine possible values for a, b and d.

Determine the period of each function in terms of b: a

y = 2 tan (bx)

2 tan

2 sin (2 x ) looks like and explain any 4 cos (2 x ) transformations on the graph of y = tan (x) to achieve it.

Without technology, determine what the graph of y

11.06

Inverse circular functions

After this lesson, you will be able to… • graph the inverse sine, cosine, and tangent functions • apply dilations, reflections, and translations to inverse graphs of sine, cosine, and tangent functions • use exact values to evaluate inverse circular functions

Graphs and transformations of inverse circular functions Inverse circular function

Restricted domain

A function that reverses the effect of a circular function (sine, cosine, or tangent) over a restricted domain, providing the angle for a given trigonometric ratio.

A limited interval of input values for a function, chosen to make the function one-to-one so that a valid inverse function can be defined.

The graphs of inverse circular functions are obtained by reflecting the graphs of the sine, cosine, and tangent functions (over their restricted domains) across the line y = x. Restrictions are necessary because the original functions are not one-to-one over their full domains, and inverses require a unique output for each input. • y = sin (x) restricted to

, 2 2 1.5 1

0.5 3 − π 2

−π

1 − π –0.5 2 –1

x 1 π 2

3 π 2

–1.5

11.06 Inverse circular functions 1071 mathspace.co

y 1 π 2 1 π 4 –1 y = sin (x) −1 1 − π 4 1 − π π 2   −1, − 2   

 π  1, 2    y = sin (x) x 1

Reflecting y = sin (x) across the line y = x produces the graph of y = sin−1 (x). • Domain: [− 1, 1] • Range:

, 2 2

• Key points (endpoints):

and 1,

• y = cos (x) restricted to [0, π] y 1.5 1 0.5 3 − π 2

−π

1 − π –0.5 2 –1

x 1 π 2

3 π 2

–1.5

(−1, π)

3π y = cos–1 (x) 4 1π 2 1 π 4 −1

(1, 0) 1 − π 4

y = cos (x) x 1

Reflecting y = cos (x) across the line y = x produces the graph of y = cos−1 (x). • Domain: [− 1, 1] • Range: [0, π] • Key points (endpoints): (− 1, π) and (1, 0)

1072 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

• y = tan (x) restricted to

, 2 2 4 3 2 1

3 − π 2

1 − π 2

−π

x 1 π 2

−1 −2 −3 −4

3 π 2

y y = tan (x)

1 π 2

x −4 −3 −2 −1 y = tan–1 (x) 1 − π 2

2 3 4

Reflecting y = tan (x) across the line y = x produces the graph of y = tan−1 (x) with domain (− ∞, ∞) and range

, . 2 2

Key features of the graphs: Function

Domain

Range

Asymptotes

sin−1 (x)

[− 1, 1]

, 2 2

None

cos−1 (x)

[− 1, 1]

[0, π]

None

tan−1 (x)

(− ∞, ∞)

, 2 2

The graphs of inverse circular functions can be modified by transformations. These transformations are described by the general form y = Af (nx) + k, where f is an inverse circular function.

y = A sin−1 (nx) + k A

is the vertical dilation by a factor of ∣A∣

is the horizontal dilation by a factor of

is the vertical translation

1 n

11.06 Inverse circular functions 1073 mathspace.co

• Vertical dilation factor: If A < 0, the graph is also reflected across the x-axis. This affects the range. • Horizontal dilation: If n < 0, the graph is also reflected across the y-axis. This affects the domain. • Vertical translation: The graph shifts up by k units if k > 0 or down by ∣k∣ units if k < 0. This affects the range. These transformations change the domain and range of the parent function. • Domain: The domain of y = sin−1 (x) and y = cos−1 (x) is [− 1, 1]. For y = f (nx), the new domain 1 1 , becomes . The domain of y = tan−1 (nx) remains (− ∞, ∞) for any n ≠ 0. n n • Range: Vertical transformations alter the range. For y = Af (x) + k, the new range is found by applying the transformations to the endpoints of the parent range. For y = A cos−1 (x) + k, the parent range is [0, π]. The new range will be [k, Aπ + k] if A > 0, and [Aπ + k, k] if A < 0.

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Example 1 Sketch the graph of y = 2 sin−1 (x) − 1, and determine its domain and range.

Create a strategy Start with y = sin−1 (x), apply a dilation of factor 2 from the x-axis, then translate down by 1 unit.

Apply the idea y 1π 2 x −1

1 1 − π 2 −π

Domain: [−1, 1] (unchanged) Range: Original range

, dilated by 2 becomes [−π, π], then shifted down by 1 to [−π −1, π −1]. 2 2

1074 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 Digital only example Explore in the online lesson

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Example 3 Sketch the graph of y = sin−1 (2x) and determine its domain and range.

Create a strategy 1 Start with y = sin−1(x). The parameter n = 2 causes a dilation of factor from the y-axis. This will 2 affect the domain. The range is unaffected.

Apply the idea y

x −0.5

0.5

Domain: The original domain [−1, 1] is horizontally 1 1 , . dilated by a factor of 2, becoming 2 2 Range:

, (unchanged) 2 2

Example 4 Sketch the graph of y = tan−1 (x) + , and determine its domain, range, and asymptotes. 2

Create a strategy Start with the graph of y = tan−1 (x) and apply a vertical translation up by units. This will shift the 2 range and the horizontal asymptotes. The domain is unaffected.

11.06 Inverse circular functions 1075 mathspace.co

Apply the idea y Domain: (−∞, ∞) (unchanged) Range: The original range become (0, π).

1 π 2 x −4 −3 −2 −1

2 2

is shifted up by

Asymptotes: The original asymptotes y up by to become y = 0 and y = π. 2

to 2

are shifted

2 3 4

Idea summary The graphs of the inverse circular functions are reflections of their original functions (over a restricted domain) across the line y = x. The key features of the parent inverse circular functions are: Function

Domain

Range

sin−1 (x)

[−1, 1]

, 2 2

cos−1 (x)

[−1, 1]

[0, π]

tan−1 (x)

(−∞, ∞)

2 2

with horizontal asymptotes at y = ±

These graphs can be transformed using the general form y = Af (nx) + k, where f is an inverse circular function. • Parameter A (Vertical dilation/reflection): Dilates the graph vertically by a factor of ∣A∣. If A < 0, the graph is also reflected across the x-axis. This affects the range. 1 • Parameter n (Horizontal dilation/reflection): Dilates the graph horizontally by a factor of . n This affects the domain. 1 1 , • The new domain for sin−1 (nx) and cos−1 (nx) is . n n

• Parameter k (Vertical translation): Translates the graph vertically by k units. This also affects the range. The new range is found by applying the vertical transformations (y′ = Ay + k) to the endpoints of the original range.

1076 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.06 Practice questions What do you remember? 1

A function is considered one-to-one if it passes the horizontal line test. Determine whether each graph represents a one-to-one function: a

Evaluate using exact values: a

sin

cos (π)

cos

sin

Evaluate these composite functions using exact values: 1 1 a sin sin b cos (cos−1 (− 0.8)) 3 d

sin 1 sin

3 4

cos 1 cos

4 3

tan

tan (0)

tan (tan−1 (10))

tan 1 tan

2 3

Determine the restricted domain and range for these functions so that their inverse functions are defined: a

y = sin (x)

y = cos (x)

y = tan (x)

11.06 Inverse circular functions 1077 mathspace.co

Practice 5

Evaluate each expression using exact values in radians: 1 2

sin 1

sin−1 (0)

2 2

cos 1

3 2

tan−1 (1)

cos−1 (− 1)

tan 1

cos−1 (0)

tan−1 (0)

Determine the domain and range of these inverse circular functions: a

y = sin−1 (x)

y = cos−1 (x)

y = tan−1 (x)

Sketch the graph of y = sin−1 (x). Label the endpoints, domain, and range.

Sketch the graph of y = cos−1 (x). Label the endpoints, domain, and range.

Sketch the graph of y = tan−1 (x). Label the horizontal asymptotes, domain, and range.

For each function, describe the transformations applied to the parent inverse circular function, and determine the domain and range: y = 3 sin−1 (x)

y = cos−1 (x) + π 1

Ex 1

Sketch the graph of y

Ex 2

Consider the function y = − 3 cos−1 (x):

Ex 4

Ex 3

2 cos ( x )

y = − tan−1 (x)

. Label the endpoints and range.

Describe the transformations applied to the parent function y = cos−1 (x).

Determine the domain and range of the transformed function.

Sketch the graph, labelling the endpoints.

x : 3 Describe the transformation applied to the parent function y = sin− 1 (x).

Determine the domain and range of the transformed function.

Sketch the graph, labelling the endpoints.

Consider the function y

2 cos 1 ( x )

sin 1

tan 1 ( x )

Sketch the graph of y

Sketch the graph of y = − cos−1(x) + π. Label the endpoints and range.

. Label the horizontal asymptotes and range.

1078 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking 16

Explain why sin−1 (x) + cos−1 (x) = for all x in [− 1, 1]. (Hint: Consider a right-angled triangle with 2 hypotenuse 1 and one side x.)

Explain why there are no solutions to the equation sin−1 (x) = sin−1 (x − 1). Justify your answer from a graphical perspective.

Derive the equation of the transformed inverse tangent function y = A tan−1 (x) + k whose graph has horizontal asymptotes at y and y . Determine the domain and range. 3 3 x Show that tan (sin−1 (x))= for x ∈ (− 1, 1). Use this to sketch the graph of y = tan (sin−1 (x)) 2 1 x over its domain, labelling key points.

11.07 Reciprocal trigonometric graphs After this lesson, you will be able to. . . • derive the graphs of cosecant, secant, and cotangent from the primary trigonometric functions • identify key features of the cosecant, secant, and cotangent graphs • solve problems involving the reciprocal trigonometric functions • apply dilations, reflections, and vertical translations to the graphs of cosecant, secant, and cotangent functions • draw graphs of reciprocal trigonometric functions from contexts

Parent reciprocal functions and features Three new trigonometric functions, called reciprocal functions, can be defined using the sine, cosine, and tangent functions. 1 • The cosecant function, cosec (θ ), is the reciprocal of the sine function: cosec (θ ) = . sin( ) Key features of y = cosec (x): • Domain: All real numbers except for integer multiples of π (e.g., −π, 0, π, 2π). This is written as x ≠ jπ, where j is any integer. • Range: (− ∞, − 1] ∪ [1, ∞) • Period: 2π • Vertical asymptotes: x = jπ, where j is any integer. • Amplitude: Undefined (since the function does not have a maximum or minimum value). • Midline: y = 0 • x-intercepts: None • y-intercept: None

11.07 Reciprocal trigonometric graphs 1079 mathspace.co

• The secant function, sec (θ ), is the reciprocal of the cosine function: sec (θ ) = Key features of y = sec (x):

e.g., • Domain: All real numbers except for odd multiples of 2 written as x ≠ + jπ, where j is any integer. 2

3 , 2

1 . cos ( )

3 . This is 2 2 2 ,

• Range: (− ∞, −1] ∪ [1, ∞) • Period: 2π

• Vertical asymptotes: x = + jπ, where j is any integer. 2 • Amplitude: Undefined • Midline: y = 0 • x-intercepts: None • y-intercept: y = 1 • The cotangent function, cot (θ ), is the reciprocal of the tangent function: cot (θ ) =

1 tan( )

cos ( ) sin( )

Key features of y = cot (x): • Domain: All real numbers except for integer multiples of π. This is written as x ≠ jπ, where j is any integer. • Range: (−∞, ∞) • Period: π • Vertical asymptotes: x = jπ, where j is any integer • Amplitude: Undefined • Midline: y = 0 x • x-intercepts: Odd multiples of 2 • y-intercept: None

j , where j is any integer

Since division by zero is undefined, these reciprocal functions have vertical asymptotes wherever the original function in the denominator (sin (θ ), cos (θ ), or tan (θ )) equals zero. The corresponding graphs and key features are closely related to the graphs of sine, cosine, and tangent.

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1080 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

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Example 1 Sketch the graph of f (x) = cosec (x) on the interval [−2π, 2π].

Create a strategy To graph y = cosec (x), first sketch its reciprocal function, y = sin (x). • Draw vertical asymptotes where sin (x) = 0. • Plot points where sin (x) = 1 (so cosec (x) = 1) and where sin (x) = −1 (so cosec (x) = −1). These are the local minimums and maximums of the cosecant curves. • Draw the ‘U-shaped’ curves approaching the asymptotes and passing through the points plotted in the previous step. • Extend the pattern over the required interval [−2π, 2π].

Apply the idea First, sketch y = sin (x) from −2π to 2π. The sine function is 0 at x = −2π, −π, 0, π, 2π. Draw vertical asymptotes at these x-values.

y 2

The sine function is 1 at x

1 x 3 1 − π −π − π 2 2 –1

1 π π 2

3 π 2

3 , . The points 2 2

3 , 1 and , 1 will be the bottom points (local 2 2 minimums) of the upward-opening ‘U’ shapes of the cosecant graph. 3 , The sine function is −1 at x . The points 2 2 3 , 1 will be the top points (local , 1 and 2 2

–2

maximums) of the downward-opening ‘U’ shapes. y 2 1 x 3 − π −π − 1 π 2 2 –1

1 π π 2

3 π 2

Draw the cosecant curves approaching the asymptotes and passing through these points. Within the period [0, 2π], the asymptotes are at x = 0, 3 , 1 and , 1 . x = π, x = 2π. The key points are 2 2

–2

Reflect and check Check the graph against the known features of y = cosec (x). The period is 2π. Asymptotes are at multiples of π. The range is (−∞, −1] ∪ [1, ∞). The graph matches these features.

11.07 Reciprocal trigonometric graphs 1081 mathspace.co

Example 2 Sketch the graph of f (θ ) = sec (θ ) on the interval [−2π, 2π].

Create a strategy 1 , the function f (θ ) = sec (θ ) will have vertical cos ( ) asymptotes at the values of θ that make cos (θ ) = 0. Begin by graphing cos (θ ). Since sec (θ ) =

Also, the reciprocal of −1 is −1, and the reciprocal of 1 is 1. The values of θ that make cos (θ ) = ±1 will also make sec (θ ) = ±1.

Apply the idea f (θ ) 2 1

θ 3 1 − π −π − π 2 2 –1

1 π π 2

3 π 2

The asymptotes of the secant function occur where cos (θ ) = 0, which is at odd multiples of . On the 2 3 3 , , , interval [−2π, 2π], these are θ = . 2 2 2 2 The points where cos (θ ) = 1 (at θ = −2π, 0, 2π) become points (−2π, 1), (0, 1), (2π, 1) on the secant graph. These will be the bottom points (local minimums) of the upward-opening ‘U-shapes’.

–2

f (θ ) 2 1

θ 3 1 − π –π − π 2 2 –1 –2

1 π π 2

3 π 2

The points where cos (θ ) = −1 (at θ = −π, π) become points (−π, −1), (π, −1) on the secant graph. These will be the top points (local maximums) of the downward-opening ‘U-shapes’. Sketch the “U-shaped” curves for the secant function passing through these points and approaching the asymptotes.

1082 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reflect and check An alternative method is to create a table of values for y = cos (θ ), then find the reciprocals for sec (θ ), noting where cos (θ ) = 0 leads to undefined values (asymptotes).

3 2

2π

cos (θ )

−1

sec (θ )

Undefined

−1

Undefined

Plotting these points and drawing the asymptotes confirms the graph.

Example 3 Consider the function f (θ ) = cot (θ ): a Determine the values of θ on the interval [0, 2π] for which f (θ ) is undefined. These correspond to the vertical asymptotes.

Create a strategy The function, f (θ ) = cot (θ ) =

cos ( ) , is undefined where the denominator sin (θ ) = 0. sin( )

Apply the idea On the unit circle, sin (θ ) = 0 when θ is an integer multiple of π. The values of θ on the interval [0, 2π] where f (θ ) = cot (θ ) is undefined are θ = 0, π, and 2π. These are the locations of the vertical asymptotes.

Reflect and check 1 . The cotangent function is undefined where tan (θ ) = 0. tan( ) The tangent function is zero at integer multiples of π. This confirms the result. Alternatively, recall that cot (θ ) =

11.07 Reciprocal trigonometric graphs 1083 mathspace.co

b Sketch the graph of f (θ ) = cot (θ ) on the interval [− π, 2π].

Create a strategy • First, draw the vertical asymptotes found in Part (a) and extend them to the interval [− π, 2π]. Asymptotes occur at integer multiples of π. So, draw asymptotes at θ = − π, 0, π, 2π. • Find key points between the asymptotes. The θ -intercepts occur where cot (θ ) = 0, which is where cos (θ ) = 0 at odd multiples of

• Evaluate the function at other points, for example θ = (0, π) to see the shape: cot

= 1 and cot

3 4

3 and θ = , within one period like 4 4

= − 1.

• Sketch the decreasing curve between asymptotes and repeat the pattern.

Apply the idea Asymptotes are at θ = − π, 0, π, 2π 3 θ -intercepts are at θ = , , 2 2 2 3 5 7 ,1 , , 1 , ,1 , , Other points: 4 4 4 4

1 ,

3 ,1 , 4

f (θ ) 2 1

θ 1 − π 2 –1

1 π 2

–2

1084 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3 π 2

c Compare the domain, range, and period of f (θ ) = cot (θ ) to g(θ ) = tan (θ ).

Apply the idea Domain: • cot (θ ): All real numbers except x = jπ, where j is an integer • tan (θ ): All real numbers except x = + jπ, where j is an integer 2 • Comparison: The domains are different because their vertical asymptotes are in different locations Range: • cot (θ ): (− ∞, ∞) • tan (θ ): (− ∞, ∞) • Comparison: The ranges are the same; both cover all real numbers. Period: • cot (θ ): π • tan (θ ): π • Comparison: The periods are the same.

Example 4 Digital only example Explore in the online lesson

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Idea summary The graphs of the parent reciprocal trigonometric functions—the cosecant function (y = cosec x), the secant function (y = sec x), and the cotangent function ( y = cot x) have distinct shapes, periods, domains, ranges, and asymptotes based on their relationship to sine, cosine, and tangent.

Function

Period

y = cosec (x)

2π

y = sec (x)

2π

y = cot (x)

Vertical Domain asymptotes (all x except vertical ( j is any integer) asymptote) x = jπ x

2 x = jπ

x ≠ jπ j

2 x ≠ jπ

Range (− ∞, −1] ∪ [1, ∞)

(− ∞, −1] ∪ [1, ∞) (− ∞, ∞)

Amplitude is undefined for all three reciprocal functions. Graphing the corresponding primary function (sin (x), cos (x), or tan (x)) first can help identify asymptotes and key points.

11.07 Reciprocal trigonometric graphs 1085 mathspace.co

Transformations of reciprocal functions The graphs of cosecant, secant, and cotangent can be transformed by changing the equations using parameters. The general forms involving parameters A, n, h, and k are: • y = A cosec (n(x − h)) + k • y = A sec (n(x − h)) + k • y = A cot (n(x − h)) + k

y = Af (n(x − h)) + k A dilates (stretches or shrinks) the graph vertically; if ∣A∣ > 1 it is a stretch and if 0 < ∣A∣ < 1 it is a compression, while if A < 0 the graph is reflected across the midline; this affects the range of cosec and sec (though amplitude remains undefined) with local minimums or maximums ∣A∣ units from the midline n affects the period by dilating (stretching or shrinking) the graph horizontally; if ∣n∣ > 1 it is a horizontal compression and if 0 < ∣n∣ < 1 it is a horizontal stretch, with the period becoming

Parent period 2 for cosec and sec, n n

for cot , and if n < 0 the graph is reflected across n the y-axis, changing the location of vertical asymptotes h is the horizontal translation (phase shift) that translates the graph h units horizontally; the graph translates right if h > 0 and left if h < 0 (the opposite direction of the sign inside the parentheses), and this parameter also changes the location of vertical asymptotes k is the shift that moves the graph k units vertically; the midline of the graph shifts to y = k, which affects the range of cosec and sec To graph a transformed reciprocal function, it is helpful to first identify the transformations and how these affect the key features of the parent function (period, asymptotes, range, midline, key points). The corresponding transformed primary function (sine, cosine, or tangent) can also be graphed as a guide. The parameters n and h affect the locations of the vertical asymptotes. To find the new asymptote locations (letting j be any integer): 1. Take the expression inside the trigonometric function (the argument), n(x − h). 2. Set this argument equal to the formula for the parent function’s asymptote locations (such as jπ for cosecant and cotangent, or + jπ for secant). 2 3. Solve the resulting equation for x.

1086 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

• For y = A cosec (n(x − h)) + k, the parent asymptotes are at jπ. So, set n(x − h) = jπ and j h solve for x: x n • For y = A sec (n(x − h)) + k, the parent asymptotes are at 2 1 j h solve for x: x n 2

j . So, set n(x − h) =

j and

• For y = A cot (n(x − h)) + k, the parent asymptotes are at jπ. So, set n(x − h) = jπ and j h solve for x: x n

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Example 5 Consider the function f (x) = cosec (3x): a Determine the equations of the asymptotes on the interval [0, 2π]. Find the period of the function.

Create a strategy Compare the function f (x) = cosec (3x) to the general form y = A cosec (n(x − h)) + k to identify the parameters. Here, A = 1, n = 3, h = 0, k = 0. The parameter n affects the period and asymptotes. The parent function y = cosec (x) has asymptotes at x = jπ (where j is any integer). Set the argument of the transformed function equal to jπ and solve for x. Calculate the period using the formula Period

2 . n

Apply the idea The argument is 3x. Set 3x = jπ. j . 3 The vertical asymptotes occur at multiples of . 3 On the interval [0, 2π], the vertical asymptotes are located at x Solving for x, gives x

The period is

2 n

2 3

2 . 3

2 4 5 , , , ,2 . 3 3 3

Reflect and check 1 The transformation is a dilation by a factor of from the y-axis. This compresses the period from 3 2 2π to and makes the asymptotes occur three times as often, at multiples of , which matches 3 3 the findings.

11.07 Reciprocal trigonometric graphs 1087 mathspace.co

b Sketch the graph of f (x) = cosec (3x) on the interval 0,

2 3

(one period).

Create a strategy 2 2 ,x and asymptotes occur at x = 0, x within this interval. Sketch the 3 3 3 corresponding transformed sine function, y = sin (3x), as a guide. The period is

2 : where it crosses the x-axis (these 3 become asymptotes for cosecant), where it reaches its maximum value of 1, and where it reaches its minimum value of −1. 2 • Intercepts (sin (3x) = 0) at x = 0, , (where asymptotes of cosec (3x) will be). 3 3 Find the key points of y = sin (3x) within one period 0,

• Maximum (sin (3x) = 1) occurs when 3 x • Minimum (sin (3x) = −1) occurs when 3 x

giving x

3 giving x 2

. The point is

,1 .

Draw the asymptotes and the “U” shaped curves for cosec (3x) touching the points

1 . 6

, 1 and

1 .

Apply the idea Sketch y = sin (3x) (dashed) and the asymptotes 2 ,x x = 0, x 3 3

Draw the graph of y = cosec (3x) (solid).

π 3

–1

π 3

–1

Reflect and check 2 . It has the correct shape, 3 asymptotes, and touches the points corresponding to the maximum or minimum of sin (3x). The graph shows one full period of cosec (3x) between x = 0 and x

1088 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

c How do the domain, range, and midline of f (x) = cosec (3x) compare to the parent function y = cosec (x)?

Create a strategy Recall the features of the parent function y = cosec (x) from Section 1. Compare them to the features determined by the transformation f (x) = cosec (3x), where n = 3 represents a horizontal compression.

Apply the idea j . The domain is affected 3 (values are excluded more frequently) due to the horizontal compression changing asymptote locations. • Range: Parent range is (−∞, −1] ∪ [1, ∞). The transformation only involves horizontal compression (n = 3) and no vertical stretch or shift (A = 1, k = 0). Therefore, the range remains the same: (−∞, −1] ∪ [1, ∞). • Midline: Parent midline is y = 0. Since there is no vertical shift (k = 0), the midline remains y = 0. • Domain: Parent domain is x ≠ jπ. Transformed domain is x

Example 6 Consider the function f (x) = −2cot (x): a Describe the transformations that map y = cot (x) to f (x). How do these affect the graph’s key features, such as its shape and intercepts?

Create a strategy Compare f (x) = −2 cot (x) to the general form y = A cot (n(x − h)) + k to identify the parameters. In this case, A = −2, n = 1, h = 0, k = 0. The parameter A = −2 indicates a reflection across the x-axis (because A is negative) and a vertical dilation by a factor of 2 (because ∣A∣ = 2). Consider how a reflection combined with a vertical dilation affects the parent function’s characteristics. The parent y = cot (x) is a decreasing function. A reflection will change its direction, and a dilation will make it steeper. Features like asymptotes and intercepts may not be affected.

Apply the idea The transformations are a reflection across the x-axis and a vertical dilation by a factor of 2. • Shape: The parent function y = cot (x) is a decreasing function between its asymptotes. The reflection causes the graph of f (x) = −2 cot (x) to be an increasing function. The vertical dilation by a factor of 2 makes the curve appear steeper. • Intercepts: The x-intercepts occur where the function value is zero. Since − 2(0) = 0, the intercepts j for any integer j. of − 2 cot (x) are in the same location as for cot (x), which is at x 2 • Other features: The transformations do not change the vertical asymptotes (x = jπ), the period (π), the domain (x ≠ jπ), or the range (− ∞, ∞).

11.07 Reciprocal trigonometric graphs 1089 mathspace.co

b Sketch the graph of f (x) = −2 cot (x) on the interval [0, 2π].

Create a strategy First, identify the asymptotes in the given interval. For cot (x), they occur at integer multiples of π, so draw vertical asymptotes at x = 0, x = π, and x = 2π. 3 and x . 2 2 To determine the shape, find the value at key points. For the parent function, cot Next, plot the x-intercepts. These are at x

transformed function, f f

3 4

2 cot

3 4

2 cot

= − 2(1) = − 2. This gives the point

= − 2 (− 1) = 2, giving the point

3 ,2 . 4

1. For the

2 . Similarly,

Draw the increasing curves between the asymptotes, passing through the calculated points.

Apply the idea y 3 2 1 0 –1

x 1 π 2

3 π 2

–2 –3

Reflect and check The graph shows the correct asymptotes at integer multiples of π. Between these asymptotes, the function is increasing, which correctly represents a reflection of the parent cotangent function. The key points at y = −2 and y = 2 confirm the vertical dilation by a factor of 2, and the intercepts are correctly plotted.

1090 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 7 Consider the function f (x) = cot x

. Identify the horizontal translation and find the location of 4 the vertical asymptotes. Sketch one period of the graph.

Create a strategy Compare f (x) = cot x

to y = A cot (n(x − h)) + k. Identify h. 4 The parent function y = cot (x) has asymptotes at x = jπ. To find the new asymptotes, set the j and solve for x. argument equal to the parent asymptotes: x 4 The period is determined by n. Here n = 1, so the period is still π. Find two consecutive asymptotes to define one period. Find the x-intercept within that period by setting the argument equal to Sketch the graph.

(where parent cotangent is 0). 2

Apply the idea Here, A = 1, n = 1, k = 0. The term x

4 Set argument = parent asymptotes: x Solve for x: x

indicates h 4

. This is a horizontal translation right by

j .

. The asymptotes are at . . .,

3 5 ,... , , 4 4 4

5 . 4

One period occurs between consecutive asymptotes, for example, between x and x 4 5 The length of this interval is , which matches the period. 4 4 3 , which gives x Find the x-intercept: Set argument , so x . 4 2 2 4 4 2 3 ,0 . The intercept is at 4 Sketch the graph, showing the decreasing cotangent curve between the asymptotes x x

units. 4

3 5 ,0 . , passing through 4 4

and

y 2 1 0 –1

x π 4

2π 4

3π 4

4π 4

5π 4

–2

11.07 Reciprocal trigonometric graphs 1091 mathspace.co

Reflect and check . The asymptotes are at x 4 3 5 , as calculated. The x-intercept is correctly located at x . 4 4

The graph shows a standard cotangent shape shifted right by x

and

Example 8 Consider the function y = sec (x) + 1. Identify the vertical translation, midline, and range. Sketch the graph over [0, 2π].

Create a strategy Compare y = sec (x) + 1 to y = A sec (n(x − h)) + k. Identify k. The vertical shift determines the new midline y = k. The range of the parent sec (x) is (−∞, −1] ∪ [1, ∞). The vertical shift adds k to the boundary values. Use y = cos (x) + 1 as a guide. Its midline is y = 1 and its amplitude is 1. Asymptotes for sec (x) + 1 are where cos (x) = 0 (unaffected by vertical shift). Find key points based on the shifted cosine graph and draw the secant curves.

Apply the idea Here, A = 1, n = 1, h = 0, k = 1. The vertical translation is k = 1 unit up. The new midline is y = k = 1. The parent secant function has upper curves starting at y = 1 and lower curves starting at y = −1. Translating everything up by 1 (k = 1) means the upper curves now start at y = 1 + 1 = 2 and the lower curves now start at y = −1 + 1 = 0. The new range is (−∞, 0] ∪ [2, ∞). 3 , Asymptotes remain where cos (x) = 0, i.e., at x in 2 2 [0, 2π].

y 3 2 1 x –1

1 π 2

3 π 2

The guide y = cos (x) + 1 has a max value of 1 + 1 = 2 at x = 0, 2π and a min value of −1 + 1 = 0 at x = π. Therefore, y = sec (x) + 1 has points (0, 2), (2π, 2) (local minimum) and (π, 0) (local maximum).

Reflect and check The graph shows the secant shape translated up by 1 unit. The midline is y = 1, and the curves touch y = 2 and y = 0, matching the calculated range.

1092 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 9 Digital only example Explore in the online lesson

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Idea summary The parameters A, n, h, and k in the general forms y = Af (n(x − h)) + k transform the parent reciprocal function graphs: • A: Vertical dilation by factor ∣A∣ from the x-axis and reflection across the midline if A < 0. Affects range (for cosec/sec). 1 • n: Dilation by factor from the y-axis. Affects period and asymptote location. n 2 Period = (cosec, sec ) or (cot ). Reflection across y-axis if n < 0. n n • h: Horizontal translation by h units. Affects asymptote location. • k: Vertical translation by k units. Midline becomes y = k. Affects range (for cosec/sec). To find key features such as asymptotes and range for transformed functions, apply the effects of these parameters to the parent function features. Remember amplitude is not a property of cosecant, secant, and cotangent.

11.07 Practice questions What do you remember? 1

Complete the table of values: x sin (x) cosec (x)

⬚

Undefined

tan (x)

⬚

cot (x)

⬚

⬚ 1

3 ⬚

⬚ 1

⬚

11.07 Reciprocal trigonometric graphs 1093 mathspace.co

Consider the graph of y = sin (x) for −2π ≤ x ≤ 2π: y 1 x –2π− 3 π −π − 1 π 2 2

3 1 π π π 2π 2 2

–1

Complete the table of values: x

cosec (x)

⬚

Solve sin (x) = 0 for − 2π ≤ x ≤ 2π and find the asymptotes for y = cosec (x).

Solve cosec (x) = 1 for −2π ≤ x ≤ 2π.

Determine the period of cosec (x).

Consider the graphs f (x) = cos (x) and g (x) = sec (x) drawn on the same diagram: a

Determine the values of x which satisfy f (x) = g (x).

What is the range of f (x)?

What is the range of g (x)?

What are the asymptotes of g (x) = sec (x)?

What is the period of f (x) and g (x)?

y g (x) = sec (x)

3 2

1 f (x) = cos (x) −2π − 3 π −π − 1 π 2 2 −1

x 3 1 π 2π ππ 2 2

−2 −3

Consider the graphs f (x) = sin (x) and g(x) = cosec (x) drawn on the same diagram:

y g (x) = cosec (x)

Determine the values of x which satisfy f (x) = g (x).

What is the range of f (x)?

What is the range of g (x)?

f (x) = sin (x)

What are the asymptotes of g (x) = cosec (x)?

What is the period of f (x) and g (x)?

−2π − 3 π −π − 1 π 2 2−1

2 1 x

−2 −3

1094 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3 1 π 2π π π 2 2

Consider the graphs f (x) = tan (x) and g(x) = cot (x) drawn on the same diagram: y

g(x) = cot (x)

3 2

f (x) = tan (x)

−2π

3 − π 2

1 1 − π −1 2

−π

x 1 π 2

3 π 2

2π

−2 −3

Determine the values of x which satisfy f (x) = g(x).

What is the range of f (x)?

What is the range of g(x)?

What are the asymptotes of g(x) = cot (x)?

What is the period of f (x) and g(x)?

Practice 6 Ex 1, 2, 3

For each of the parent reciprocal trigonometric functions, sketch the graph on the interval [−π, π]. Label the equations of any vertical asymptotes and the coordinates of key points (turning points or intercepts): a

y = cosec (x)

y = sec (x)

y = cot (x)

The graph of a reciprocal trigonometric function is shown: y

Ex 4

2 1 x 1 − π 2 −1

1 5 3 π π π 2π π 2 2 2

−2

Determine the period of the function.

Determine the range of the function.

State the equations of the vertical asymptotes in terms of an integer, k.

Based on these features, is this the graph of y = cosec (x), y = sec (x), or y = cot (x)? Justify your answer.

11.07 Reciprocal trigonometric graphs 1095 mathspace.co

1 Consider cosec (x) = for the interval (−2π, 2π), sin( x ) as shown: a

y 6

For which values of x will cosec (x) be: i

Undefined

Equal to 1

iii

Equal to 0 (an x-intercept)

x 3 1 π 2π π π 2 2

−2π − 3 π −π − 1 π 2 2−2

What is the lowest possible positive value?

−4 −6

Consider the graph of f (x) = cot (x) given:

Determine whether each statement is true or false: a

f (x) has vertical asymptotes at x = 0 and x = π.

f (x) is periodic with a period of π.

f (x) passes through the origin (0, 0).

f (x) is defined for all x ∈ ℝ.

2 1 x 3 1 –2π − π –π − π 2 2 –1

3 1 π 2π π π 2 2

–2

cos ( x ) Consider cot (x) = for interval [0, 2π], sin( x ) as shown: a

What is the period of cot (x)?

For which values of x will cot (x) be:

Undefined

Equal to 1

cot (x) = 3 when x

y 6 4 2

. Determine:

he next value of x for which T cot (x) = 3

The largest values of x in the interval [0, 2π] for which cot (x) = 3

1096 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

0 −2 −4 −6

x 1 1 1 2 5 7 4 3 5 11 π π π π ππ π π π π π 6 3 2 3 6 6 3 2 3 6

This graph shows y = cosec (x): a

The next positive x-value with an asymptote

The highest negative x value with an asymptote

The function is equal to 2 when x

There is a vertical asymptote at x = π, determine:

2 x

, determine:

6 The next positive x-value where y = 2

−2

The first positive x-value greater than 2π where y = 2

5 4 π π 3 3

1 2 π π π 3 3

−4

iii

The highest negative x-value where y = −2

The lowest positive x-value where y = −2

What is the period of the function?

The graphs of g(x) = sec (x) and h(x) = cosec (x) are shown: y 2 h (x) = cosec (x) 3 −2π − π 2 g (x) = sec (x)

−π

1 − π 2

1 x −1

1 π 2

3 π 2

2π

−2

Determine the coordinates of the points where h(x) = g(x). 13

Describe why f (x) = cosec (x) has vertical asymptotes at certain points. Determine the points in terms of k where k is an integer.

Consider f (x) = 2 sec (πx): a

Calculate: i

Period

Lowest possible positive value

iii

y-intercept

x-intercept(s)

Sketch the graph on the interval [−2, 2].

11.07 Reciprocal trigonometric graphs 1097 mathspace.co

Consider the graphs of y = cosec (x), y = cot (x) and y = sec (x) shown:

y 2

For which intervals over the domain [0, 2π] are these graphs increasing? a

cosec (x)

cot (x)

y = cosec (x) y = cot (x) x

sec (x) −1

1 π π 2 y = sec (x)

3 π 2

2π

−2 Ex 5

Consider the function f (x) = cosec (2x): a

Determine the period of the function. Then, state the equations of the vertical asymptotes on the interval [0, 2π].

Sketch the graph of f (x) = cosec (2x) over one full period, [0, π]. Clearly label any asymptotes and the coordinates of local maximum and minimum points.

Compare the domain, range, and midline of f (x) = cosec (2x) to the parent function y = cosec (x).

What is the period of the function y = cot (2x)? State the equations of the vertical asymptotes in terms of an integer k.

Consider the function g(x) = 2 sec (x):

Ex 6

19 Ex 7

Describe the transformation from y = sec (x).

Determine the range of the function.

Determine the midline of the function.

Sketch the graph of the function on the interval [0, 2π]. Label any asymptotes and the coordinates of local maximum and minimum points.

Consider the function f (x) = cot x

Identify the horizontal translation from the parent function y = cot (x).

Determine the equations for the vertical asymptotes of the function in terms of an integer j. 2 Determine the coordinates of the x-intercept between the asymptotes at x and x . 3 3 2 Sketch the graph of the function over one period, for the interval . Label the , 3 3 asymptotes and the x-intercept.

Ex 8

Sketch the graph of y = sec (x) + 3 for the interval [0, 2π]. Clearly label the equations of any asymptotes and the coordinates of local maximum and minimum points. For the function y = −3 sec 2 x

+ 1, identify the:

Ex 9

Period

Equations of two consecutive vertical asymptotes

Phase shift

1098 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Midline

Range

Determine the largest negative value of x for which cot (x) = 0.

Consider the graphs of f (x) = cosec (x), g(x) = k cosec (x) and h ( x )

1 cosec (x) shown: m

g (x) = k cosec (x) 2

f (x) = cosec (x)

x 1 π 2

1 − π 2

3 − π –π 2 1 h (x) = cosec (x) m

–2π

3 π 2

2π

−2

Determine the values of k and m. 24

Consider the graphs of f (x) = sec (x), g(x) = sec (x + α) and h(x) = sec (x − β ) shown: y 3 2 g (x) 1 − π 2

1 f (x)

h (x)

−1

1 π 2

−2 −3

Determine the values α and β.

Extend your thinking 25

1 cos ( x ) and . Although either may be used as an expression for cot (x), there tan( x ) sin( x ) are certain inputs for which they will not provide the same value.

Consider

Determine the x-intercepts of both for the interval [0, 2π]. Using the results, explain the cos ( x ) 1 difference between and . sin( x ) tan( x )

11.07 Reciprocal trigonometric graphs 1099 mathspace.co

Identify the graph of sec (x) from the graphs. Justify your answer.

y A 2

B C x 1 π 2

1 − π 2

Consider the graph of cosec (x) and m cos (x): The two functions intersect once in the interval [0, π]: a

Determine the value of m that satisfies the graph shown.

There is another value of m for which these two functions would intersect once in the given interval. Determine this value of m.

y 1

x 1 π 2

3 π 2

2π

−1

Consider the graph of f (x) = a sec (x + b) + c given: By considering suitable coordinates, determine the values of a, b and c, given that: • a>0

• 0<b< 2 • c>0

18 16 14 12 10 8 6 4 2 3 1 − π –π − π 2 2 −2

x 3 1 π 2π 5 π3π ππ 2 2 2

Investigation: Locus definition Investigate online

1100 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

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11.08 Locus definitions and the distance formula After this lesson, you will be able to… • apply the distance formula to develop locus definition of lines and curves • construct lines and curves in the plane based on locus conditions • derive the Cartesian form equations of lines and conic sections from their locus definitions

Locus definitions of lines and circles Locus A path or set of points that all follow a specific geometric rule. Geometric shapes can be described using these locus rules.

Interactive exploration Discover this concept in action online

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The locus of a point that is equidistant from two fixed points is a straight line. Specifically, it is the perpendicular bisector of the line segment connecting the two fixed points. Let the two fixed points be A (x1, y1) and B (x2, y2). Let P (x, y) be any point on the locus. The locus condition is PA = PB. Squaring both sides gives the equation PA2 = PB2. The squared version of the distance formula, d2 = (x2 – x1)2 + ( y2 – y1)2, can then be used to find the equation of the line. The locus definition of a circle is the set of all points P (x, y) that are a fixed distance (the radius, r) from a single fixed point (the centre, (h, k)).

11.08 Locus definitions and the distance formula 1101 mathspace.co

y Circle A set of all points in a plane that are equidistant from a fixed point called the centre.

Centre The fixed point from which all points on a circle are equidistant. Radius The constant distance from the centre to any point on the circle.

To translate this definition into an algebraic equation, use the distance formula, which calculates the distance between two points (x1, y1) and (x2, y2) in the Cartesian plane.

( x2

d (x1, y1) (x2, y2)

x1 )2

( y2

y1 )2

is the distance between the two points are the coordinates of the first point are the coordinates of the second point

Applying the locus condition for a circle, the distance between any point P (x, y) on the circle and the centre C (h, k) must equal the radius r. PC = r

Write the locus condition

( x h)2 ( y k)2 = r

Use the distance formula

(x – h) + ( y – k) = r

Square both sides

This gives the standard equation of a circle.

(x – h)2 + ( y – k)2 = r2 (h, k) r (x, y)

are the coordinates of the centre of the circle is the radius of the circle are the coordinates of any point on the circle

1102 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 Determine the equation of the locus of a point P (x, y) that is equidistant from the points A (2, 5) and B (6, 1). Sketch the locus.

Create a strategy Use the squared distance formula d2 = (x2 – x1)2 + ( y2 – y1)2 to set PA2 = PB2. Expand and simplify the equation to find the Cartesian form. Plot the points A and B and draw the resulting line.

Apply the idea The condition for the locus is that the distance from any point P (x, y) to A (2, 5) is equal to the distance from P to B (6, 1). This can be written as PA = PB. To avoid dealing with square roots, work with the squared distances: PA2 = PB2 Now, apply the squared distance formula to each side of the equation. PA2 = PB2 2

Write the equation 2

(x – 2) + ( y – 5) = (x – 6) + ( y – 1) Substitute the coordinates using the distance formula x2 – 4x + 4 + y2 – 10y + 25 = x2 – 12x + 36 + y2 – 2y + 1 Expand the squared terms –4x – 10y + 29 = –12x – 2y + 37 Remove x2 and y2 from both sides and combine constants 8x – 8y = 8 Collect variables on the left and constants on the right y = x – 1 Divide both sides by 8 and rearrange to solve for y The equation of the locus is y = x – 1. This is the equation of a line, which is the perpendicular bisector of the segment AB. To sketch the locus, plot the points A (2, 5) and B (6, 1), the segment connecting them, and the line y = x – 1. 7 6

y A

5 4 3 2 1 –1 –1

B 1

2 3 4 5 6 7

11.08 Locus definitions and the distance formula 1103 mathspace.co

Example 2 Determine the equation of the locus of a point P (x, y) that moves such that its distance from the origin O (0, 0) is always 5 units. Sketch the locus.

Create a strategy The locus of a point that moves at a constant distance (5 units) from a fixed point (the origin) is a circle. The fixed point is the centre and the constant distance is the radius. The circle can be sketched by plotting its key points: the points 5 units above, below, left, and right of the centre, and then drawing a smooth curve through them.

Apply the idea The locus is a circle with centre O (0, 0) and radius r = 5. To sketch the circle, four key points can identified on the circle: • 5 units to the right of the centre: (5, 0) • 5 units to the left of the centre: (–5, 0) • 5 units above the centre: (0, 5) • 5 units below the centre: (0, –5) Here is the sketch of the graph: y 6 5 4 3 2 1 O −6−5−4 −3 −2−1 −1 −2 −3 −4 −5 −6

1 2 3 4 5 6

Reflect and check The Cartesian equation can be verified using the distance formula. Let P (x, y) be a point on the locus and O (0, 0) be the centre. The distance OP must be 5. OP = 5 2

( x 0)

Write the required distance from the origin

( y 0) = 5

Apply the distance formula between P (x, y) and O (0, 0)

x2 y2 = 5

Simplify

x + y = 25 Square both sides The equation is x2 + y2 = 25, which matches the standard equation of a circle centred at the origin with radius 5.

1104 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary The locus of a point P (x, y) that is equidistant from two fixed points A and B is the perpendicular bisector of the segment AB. The condition is PA = PB. The locus of a circle is the set of all points P (x, y) that are equidistant from a single fixed point, the centre C (h, k). The constant distance is the radius r. Using the distance formula, this locus condition leads to the standard equation of a circle: (x – h)2 + ( y – k)2 = r2

Ellipse y Ellipse The set of all points in a plane such that the sum of the distances from two fixed points (the foci) is constant.

x F1

Foci The two fixed points used in the locus definition of an ellipse or hyperbola (singular: focus). Major axis The longest diameter of an ellipse, which passes through its centre and both foci. Minor axis The shortest diameter of an ellipse, which passes through the centre and is perpendicular to the major axis.

An ellipse is defined as the locus of points P (x, y) in a plane such that the sum of the distances from P to two fixed points, called the foci (F1 and F2), is a constant value. Let this constant sum be 2a, where 2a must be greater than the distance between the foci. The locus condition is: PF1 + PF2 = 2a If the foci are at F1(– c, 0) and F2(c, 0), the equation using the distance formula becomes: ( x ( c))2 ( y 0)2 ( x c)2

( x c)2 ( y 0)2 = 2a ( x c)2

y2 = 2a

11.08 Locus definitions and the distance formula 1105 mathspace.co

This equation can be simplified algebraically. The process involves isolating one square root, squaring both sides, simplifying, then isolating the remaining square root and squaring again. This leads to the standard Cartesian equation for an ellipse centred at the origin:

a is half the constant sum of distances; the distance from the centre to a vertex along the major axis b is the distance from the centre to a co-vertex along the minor axis, where b2 = a2 – c2 c is half the distance between the foci, F1(–c, 0) and F2(c, 0) (x, y) are the coordinates of any point on the ellipse

Example 3 A point P (x, y) moves such that the sum of its distances from F1(–4, 0) and F2(4, 0) is always 10. Determine the equation of the locus and then sketch the graph.

Create a strategy This describes an ellipse with foci at (±4, 0) and constant sum 2a = 10. Use the locus definition PF1 + PF2 = 10 and the distance formula. The algebraic simplification involves isolating square roots and squaring twice. Identify a and c (half the distance between foci) to find b2 = a2 – c2 and x2 y2 1. Sketch the ellipse. write the standard equation 2 a b2

Apply the idea The locus condition is PF1 + PF2 = 10. From the problem, 2a = 10, so a = 5. The foci are at (±4, 0), so c = 4. For an ellipse, b2 = a2 – c2. b2 = 52 – 42 = 25 – 16 = 9 Therefore, b = 3. The vertices on the major axis are at (±a, 0) = (±5, 0). The endpoints of the minor axis are at (0, ±b) = (0, ±3). The Cartesian equation is

1, which gives

x2 25

y2 9

Sketch: 4 3 2 1

F1 −5 −4 −3 −2 −1 −1 −2 −3 −4

1106 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

F2 1 2 3 4 5

Reflect and check The full derivation of the equation from the locus definition PF1 + PF2 = 10 is as follows: ( x 4 )2

( x 4 )2

y2 = 10 Write the distance formula for the locus definition

( x 4 )2 y2 = 10

( x 4 )2

Isolate one of the square y2 root terms

(x + 4)2 + y2 = 10

( x 4 )2

x2 + 8x + 16 + y2 = 100 20 ( x 4 )2 16x – 100 = 20 ( x 4 )2

25 – 4x = 5 ( x 4 )2 2

Square both sides x2 8 x 16

y2 Remove common terms and isolate the remaining square root

Divide both sides by – 4

(25 – 4x) = 25((x – 4) + y ) 2

Square both sides

625 – 200x + 16x = 25x – 200x + 400 + 25y 2

y2 Expand both sides

Expand both sides

225 = 9x + 25y

Add 200x to both sides, collect variables on the right and constants on the left

x2 y2 =1 25 9

Divide both sides by 225

Idea summary The locus of an ellipse is the set of all points P (x, y) such that the sum of the distances from P to two fixed foci (F1, F2) is a constant, 2a (the length of the major axis). PF1 + PF2 = 2a x2 For foci at (±c, 0), the standard Cartesian equation is 2 a minor axis and b2 = a2 – c2.

y2 b2

1, where 2b is the length of the

11.08 Locus definitions and the distance formula 1107 mathspace.co

Parabola y Parabola The set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

Focus (of a parabola) The fixed point used in the locus definition of a parabola. Directrix The fixed line used in the locus definition of a parabola.

The locus condition is: PF = Distance (P, directrix). The distance from a point to a line is the perpendicular distance. The vertex of a parabola is the point on the parabola halfway between the focus and the directrix. This vertex is the point where the parabola changes direction. Consider a parabola with its focus at F (0, p) and its directrix the line y = –p. Let P (x, y) be a point on the locus. The distance from P to the focus is PF ( x 0)2 ( y p)2 x2 ( y p)2 . The perpendicular distance from P to the line y = –p is the absolute difference in the y-coordinates, PD = ∣y – (–p)∣ = ∣y + p∣. PF = PD x2 ( y 2

Write the locus condition

p)2 = ∣y + p∣ 2

Substitute distance expressions 2

x + ( y – p) = ( y + p)

Square both sides

x2 + y2 – 2py + p2 = y2 + 2py + p2

Expand the squares

x = 4py

Simplify

This gives the standard equation for a parabola opening upwards with vertex at the origin.

x2 = 4py p is the distance from the vertex to the focus (and from the vertex to the directrix). If P > 0, parabola opens up; if P < 0 opens down (0, p) is the coordinates of the focus y = –p is the equation of the directrix (x, y) is the coordinates of any point on the parabola Other standard forms exist for parabolas opening sideways ( y2 = 4px) or with vertices not at the origin.

1108 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 4 Determine the equation of the locus of a point P (x, y) that is equidistant from the point F (0, 3) and the line y = –3. Sketch the locus.

Create a strategy This describes a parabola with focus (0, 3) and directrix y = –3. Use the locus condition PF = PD, where PD is the perpendicular distance from P to the line y = –3. Apply the distance formula for PF and the formula for vertical distance for PD. Square both sides and simplify to find the equation. Sketch the parabola.

Apply the idea Let P (x, y) be on the locus. The focus is F (0, 3), the directrix is y = –3. 2 2 PF = ( x 0) ( y 3)

x2 ( y 3)2

PD = ∣y – (–3)∣ = ∣y + 3∣ PF = PD

Write the locus condition

( y 3) = ∣y + 3∣

Substitute y = –3 2

x + ( y – 3) = ( y + 3) 2

x + y – 6y + 9 = y + 6y + 9 2

x = 12y

Square both sides Expand the squares Simplify

The equation is x2 = 12y. This is a parabola opening upwards with vertex at the origin. Since 4p = 12, gives P = 3, confirming the focus and directrix. Sketch:

y 8 6 4

F (0, 3)

2 Vertex −8 −6 −4 −2 −2

2 4 6 8

11.08 Locus definitions and the distance formula 1109 mathspace.co

Example 5 This graph shows a parabola:

5 4 3 2 V 1

−7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3

Identify its vertex, focus, and directrix, and determine its locus definition.

Create a strategy Determine the direction the parabola opens. Measure or deduce the distance p from the vertex to the focus (which lies inside the curve) or the directrix (which lies outside). The equation form is ( y – k)2 = 4p (x – h) since it opens sideways. Use the vertex (h, k) and p to find the focus and directrix, then state the locus definition.

Apply the idea From the graph, the vertex is at V (–2, 0). The parabola opens to the left. The standard form is ( y – k)2 = 4p (x – h), with p < 0. The graph shows the focus at F (–4, 0) and the directrix is the y-axis, x = 0. The distance from the vertex (–2, 0) to the focus (–4, 0) is ∣–4 – (–2)∣ = 2. Since it opens left, p = –2. The distance from the vertex (–2, 0) to the directrix x = 0 is ∣0 – (–2)∣ = 2. This confirms ∣p∣ = 2. This analysis gives these features: • Vertex: (–2, 0) • Focus: (–4, 0) • Directrix: x = 0 Locus definition: The parabola is the set of all points P (x, y) such that the distance from P to the focus (–4, 0) is equal to the perpendicular distance from P to the directrix line x = 0.

1110 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary The locus of a parabola is the set of all points P (x, y) equidistant from a fixed focus (of a parabola) F and a fixed directrix line. The midpoint lying on the curve between the focus and the directrix is the vertex of a parabola. PF = Distance (P, directrix) For a focus at (0, p) and directrix y = –p, the equation is x2 = 4py. For a focus at (p, 0) and directrix x = –p, the equation is y2 = 4px.

Hyperbola y

Hyperbola The set of all points in a plane such that the absolute value of the difference of the distances from two fixed points (the foci) is constant.

x F2

The locus condition is: ∣PF1 – PF2∣ = 2a, where 2a is the constant difference. If the foci are at F1 (– c, 0) and F2 (c, 0), the equation using the distance formula becomes: ( x ( c))2 ( y 0)2 ( x c)2

( x c)2 ( y 0)2 = 2a y2

( x c)2

y2 = 2a

Similar to the ellipse, complex algebraic manipulation leads to the standard equation for a hyperbola centred at the origin opening horizontally:

a is half the constant difference of distances; the distance from the centre to a vertex b is related to the asymptotes, where b2 = c2 – a2 c is half the distance between the foci, F1 (− c, 0) and F2 (c, 0); and note that for a hyperbola, c > a (x, y) are the coordinates of any point on the hyperbola If the hyperbola opens vertically (foci on y-axis), the equation is

11.08 Locus definitions and the distance formula mathspace.co

1111

Example 6 Determine the equation of the locus of a point P (x, y) such that the absolute difference of its distances from F1(– 5, 0) and F2(5, 0) is 8. Sketch the locus.

Create a strategy This describes a hyperbola with foci at (±5, 0) and constant difference 2a = 8. Use the locus definition ∣PF1 – PF2∣ = 8 and the distance formula. The algebraic simplification is similar to the ellipse but involves ensuring the difference leads to the correct hyperbola equation. Identify a and x2 y2 c (half the distance between foci) to find b2 = c2 – a2 and write the standard equation 2 1. 2 a b Sketch the hyperbola, including its asymptotes.

Apply the idea First, define the distances using the distance formula: PF1 = ( x ( 5))2 ( y 0)2

( x 5)2

PF2 = ( x 5)2 ( y 0)2

( x 5)2

Now, derive the equation: ∣PF1 – PF2∣ = 8 ( x 5)2

( x 5)2

Write the locus condition

Substitute distance y2 = ±8 expressions

( x 5)2 y2 = 8

( x 5)2

Isolate one of the square y2 root

(x + 5)2 + y2 = 64 16 ( x 5)2

( x 5)2

x2 + 10x + 25 + y2 = 64 16 ( x 5)2

x2 10 x 25

Square both sides y2 Expand terms

20x – 64 = 16 ( x 5)2

Isolate the square root term

5x – 16 = 4 ( x 5)2

Divide both sides by 4

(5x – 16) = 16((x – 5) + y )

Square both sides again

9x2 – 16y2 = 144

Group like terms

25x – 160x + 256 = 16(x – 10x + 25 + y ) 25x – 160x + 256 = 16x – 160x + 400 + 16y

Expand terms

x y =1 16 9

Distribute the 16 on the right-hand side

Divide both sides by 144

This is the equation of a hyperbola centred at the origin opening horizontally. Here, a2 = 16 implies a = 4 and b2 = 9 implies b = 3. The foci are at (±5, 0) where c = 5. Check: b2 = c2 – a2 implies 9 = 25 – 16, which is true.

1112 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The asymptotes are given by y Sketch:

b x a

3 x. 4 y 6 4

2 F1 −6 −4 −2 −2

F2 x 4 6

−4 −6

Idea summary The locus of a hyperbola is the set of all points P (x, y) such that the absolute value of the difference of the distances from P to two fixed foci (F1, F2) is a constant, 2a: ∣PF1 – PF2∣ = 2a For foci at (±c, 0), the standard Cartesian equation is

1, where b2 = c2 – a2:

11.08 Practice questions What do you remember? 1

Calculate the distance between these pairs of points. Give your answer in simplest radical form where necessary: a

A (1, 2) and B (4, 6)

C (– 1, 3) and D (2, – 1)

E (0, 0) and F (5, 12)

G (2, – 5) and H (2, 3)

Match the locus definition to the correct geometric shape: i

The set of all points equidistant from a fixed point.

The set of all points where the sum of the distances from two fixed points is constant.

iii

The set of all points equidistant from a fixed point and a fixed line.

The set of all points where the absolute difference of the distances from two fixed points is constant.

Parabola

Circle

Hyperbola

Ellipse

11.08 Locus definitions and the distance formula 1113 mathspace.co

The standard equation of a circle with centre (h, k) and radius r is given by (x – h)2 + ( y – k)2 = r2: a

Determine the equation of a circle with centre (2, – 3) and radius 4.

Identify the centre and radius of the circle given by the equation x2 + ( y – 1)2 = 9.

Practice 4 Ex 1

5 Ex 2

6 Ex 3

Determine the equation of the locus of a point P (x, y) that moves such that its distance from the point C (2, 1) is always 3 units. Sketch the locus. A point P (x, y) moves such that the sum of its distances from F1(– 3, 0) and F2(3, 0) is always 10. Determine the equation of the locus and sketch the locus.

x2 Sketch the graph of the ellipse whose equation is 16 co-vertices.

This graph shows an ellipse:

9 Ex 4

Ex 5

Determine the equation of the locus of points P (x, y) such that the distance from P to A (1, 2) is equal to the distance from P to B (5, 4). Describe this locus.

Determine the coordinates of the vertices and co-vertices.

Determine the equation of the ellipse.

If the points F1 and F2 are the foci, determine the locus definition of this ellipse.

y2 4

1. Label the vertices and

y 3 2 CV1 1 V2 −5 −4 −3 −2 −1 −1 −2 −3

V1 x 1 2 3 4 5

CV2

Determine the equation of the locus of a point P (x, y) that is equidistant from the point F (0, 2) and the line y = –2. Sketch the locus. This graph shows a parabola. Identify its vertex, focus, and directrix, and determine its locus definition.

1114 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11 Ex 6

Determine the equation of the locus of a point P (x, y) such that the absolute difference of its distances from F1 (– 13, 0) and F2 (13, 0) is 10. Sketch the locus. x2 Sketch the graph of the hyperbola whose equation is 9 equations of the asymptotes.

y2 16

1. Label the vertices and the

Extend your thinking 13

An ellipse has foci at (0, ±3) and the sum of the distances from any point on the ellipse to the foci is 10. Determine the equation of the ellipse.

A parabola has its vertex at (1, 2), and its focus at (1, 4). Determine the equation of the parabola and its directrix.

A point P (x, y) moves such that it is always twice as far from the point A (3, 0) as it is from the y-axis (x = 0). Show that the locus of P is a hyperbola and determine its equation.

Did you know?

The spray of water in a fountain creates a perfect real-life example of a locus! Every point along the water’s arc is the same distance from the centre, forming a circle you can actually see. Using the distance formula, mathematicians can calculate that distance and explain why the fountain’s pattern looks so smooth and symmetrical. It’s a great reminder that maths isn’t just on paper — it’s all around us in everyday places.

11.08 Locus definitions and the distance formula 1115 mathspace.co

11.09 Conic sections from locus definitions After this lesson, you will be able to… • write the equation for parabolas, ellipses, and hyperbolas using locus definitions • identify and analyse key features of conic sections (foci, directrices, axes) • graph conic sections from their equations in standard form

Parabolas Conic sections These are curves formed by the intersection of a plane and a double cone. These curves include circles, ellipses, parabolas, and hyperbolas. A parabola is the locus of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). For a parabola with its vertex at the origin opening upwards or downwards:

x2 = 4py p is the directed distance from the vertex to the focus. Its sign determines the direction of opening (0, 0) is the vertex of the parabola (0, p) is the focus of the parabola y = –p is the equation of the directrix For a parabola with its vertex at the origin opening left or right: y2 = 4px If the vertex is at (h, k), the equations become (x – h)2 = 4p (y – k) (opens up or down) and (y – k)2 = 4p (x – h) (opens left or right).

Example 1 Consider the parabolic equation (y – 2)2 = 8 (x + 1): a Identify its key features: vertex, focus, directrix, and axis of symmetry

Create a strategy The equation is in the form (y – k)2 = 4p (x – h), which represents a parabola opening horizontally. Identify h, k, and p.

1116 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea Matching (y – 2)2 = 8 (x + 1) with (y – k)2 = 4p (x – h): • k=2 • h = –1 • 4p = 8 implies p = 2 Vertex: (h, k) = (–1, 2) Since p = 2 > 0 and the y term is squared, the parabola opens to the right. Focus: (h + p, k) = (–1 + 2, 2) = (1, 2) Directrix: x = h – p = –1 – 2 = –3 Axis of symmetry: y = k implies y = 2

b Sketch the graph of the parabola.

Apply the idea 6 5 4 3 2 1

−3 −2 −1

−1 −2

F x 1

2 3 4 5

Idea summary In the context of conic sections, the standard equations for a parabola with vertex at (h, k) are: • Opens up or down: (x – h)2 = 4p (y – k) • Focus: (h, k + p) • Directrix: y = k – p • Opens left or right: (y – k)2 = 4p (x – h) • Focus: (h + p, k) • Directrix: x = h – p The sign of p determines the direction of opening.

11.09 Conic sections from locus definitions 1117 mathspace.co

Circles A circle is the locus of points in a plane that are at a fixed distance (the radius) from a fixed point (the centre).

(x – h)2 + (y – k)2 = r2 (h, k) is the coordinates of the centre of the circle r is the length of the radius of the circle

Example 2 This graph of a circle is shown: 2

1 −1 −1

x 1 2 (2, –1)

4 5 (5, –1)

−2 −3 −4

Determine its equation.

Create a strategy First, identify the coordinates of the centre (h, k) from the graph. Next, determine the radius r by finding the distance from the centre to any point on the circle. Finally, substitute these values into the standard equation of a circle, (x – h)2 + (y – k)2 = r2.

Apply the idea From the graph, the centre of the circle is at (2, – 1); therefore, h = 2 and k = –1. The radius is the distance from the centre (2, – 1) to a point on the circle, for example, (5, – 1). The distance is 3 units. Substitute h = 2, k = –1, and r = 3 into the standard equation: (x – h)2 + (y – k)2 = r2 2

Write the standard equation of a circle

(x – 2) + (y – (–1)) = 3 2

Substitute h = 2, k = –1, and r = 3

(x – 2) + (y + 1) = 9 2

Simplify 2

The equation of the circle is (x – 2) + (y + 1) = 9.

1118 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary A circle is the set of points equidistant (radius) from a centre. The standard equation is (x – h)2 + (y – k)2 = r2.

Ellipses An ellipse is the locus of all points P (x, y) such that the sum of the distances from P to two fixed points, called the foci, is a constant value, 2a. Distance (P, F1) + Distance (P, F2) = 2a In the standard equations for an ellipse, a represents the length of the semi-major axis, and b represents the length of the semi-minor axis. Therefore, a > b. The distance from the centre to a focus is c, with the relationship between these values being a2 = b2 + c2. The orientation of the ellipse is determined by the location of the larger denominator (a2). If it is under the x term, the major axis is horizontal. If it is under the y term, the major axis is vertical. For a centre at (h, k), the standard equations are: ( x h)2

( y k)2

( x h)2

( y k)2

• Horizontal major axis: • Vertical major axis:

Example 3 Determine the equation of the ellipse with foci at F1 (–3, 0) and F2 (3, 0), and for which the sum of the distances from any point on the ellipse to the foci is 10.

Create a strategy The centre of the ellipse is the midpoint of the foci, which is (0, 0). The sum of the distances is 2a = 10, so a = 5. The distance from the centre to each focus is c = 3. Use the relationship a2 = b2 + c2 to find b2. Then, write the standard equation.

11.09 Conic sections from locus definitions 1119 mathspace.co

Apply the idea The given foci (±3, 0) implies c = 3. Constant sum 2a = 10 which gives a = 5. b2 = a2 – c2 2

Make b the subject of the distance formula equation

=5 –3

Substitute a = 5, c = 3

= 25 – 9

Evaluate the squares

= 16

Evaluate

Therefore, b = 4. The major axis is horizontal since the foci are on the x-axis. The equation is

1. x2

x2 1 implies 25

y2 16

Example 4 ( x 1)2 ( y 2)2 1: 9 25 a Identify its key features: centre, foci, lengths of major and minor axes, and vertices. Consider the elliptical equation

Apply the idea The equation has a larger denominator under the y-term, so the major axis is vertical. The standard ( x h)2 ( y k)2 form for a vertical ellipse is 1. b2 a2 • Centre: (h, k) = (1, –2) • Semi-major axis: a2 = 25; therefore, a = 5 • Semi-minor axis: b2 = 9; therefore, b = 3 • Length of major axis: 2a = 2(5) = 10 • Length of minor axis: 2b = 2(3) = 6 Find the focal distance, c: c2 = a2 – b2

Write the formula

c = 25 – 9

Substitute a2 = 25 and b2 = 9

c2 = 16

Evaluate

c=4

Take the square root of both sides

Foci are on the major axis (vertical), c units from the centre: (h, k ± c) = (1, –2 ± 4) • Foci: F1 (1, –6) and F2 (1, 2) Vertices are the endpoints of the major axis: (h, k ± a) = (1, –2 ± 5) • Vertices: V1 (1, –7) and V2 (1, 3)

1120 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b Sketch the graph of the ellipse.

Apply the idea 3

y V (1, 3) 2

F2 (1, 2)

1 −2 −1 (–2, –2)

−1 −2

x 1

C (1, –2)

4 (4, –2)

−3 −4 −5

F1 (1, –6)

−6 −7

V1 (1, –7)

Idea summary An ellipse is the locus of points where the sum of distances to two foci is constant (2a). The key relationship is a2 = b2 + c2 where a is the semi-major axis. Standard equations (centre (h, k)): ( x h)2

( y k)2

( x h)2

( y k)2

• Horizontal major axis: • Vertical major axis:

Hyperbolas A hyperbola is the locus of all points P (x, y) such that the absolute value of the difference of the distances from P to two fixed points (the foci) is constant, 2a. ∣Distance (P, F1) − Distance (P, F2)∣ = 2a The transverse axis is a line segment that passes through the centre of the hyperbola and has vertices as its endpoints. The vertices are always located on the transverse axis. The conjugate axis is the line segment perpendicular to the transverse axis and passing through the centre of the hyperbola.

11.09 Conic sections from locus definitions 1121 mathspace.co

The semi-transverse axis is half the length of the transverse axis while the semi-conjugate axis is half the length of the conjugate axis. The key relationship between the semi-transverse axis (a), semi-conjugate axis (b), and distance from centre to focus (c) is c2 = a2 + b2. For a centre at (h, k), the standard equations are: ( x h)2 ( y k)2 • Opens horizontally (transverse axis is horizontal): a2 b2 b y k ( ) x h Asymptotes: a ( y k)2 ( x h)2 • Opens vertically (transverse axis is vertical): 1 a2 b2 a ( x h) Asymptotes: y k b

Example 5 Determine the equation of the hyperbola with foci at F1 (0, –5) and F2 (0, 5), and for which the absolute difference of the distances from any point on the hyperbola to the foci is 6.

Create a strategy The foci are on the y-axis, so the hyperbola opens vertically. The centre is the midpoint of the foci, (0, 0). The constant difference is 2a = 6 which gives a = 3. The distance from the centre to each focus is c = 5. Use c2 = a2 + b2 to find b2. Then, write the standard equation for a vertical hyperbola.

Apply the idea Calculate b²: b2 = c2 – a2 2

Write the formula

=5 –3

Substitute c = 5 and a = 3

= 25 – 9

Evaluate the squares

= 16

Evaluate

From b² = 16, it follows that, b = 4. Write the equation for the vertical hyperbola: y2

y2 9

x2 16

Write the equation

Substitute a = 3 and b = 4

Evaluate

1122 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 6 ( x 2)2 ( y 1)2 1: 9 16 a Identify the key features: centre, foci, vertices, and equations of asymptotes. Consider the hyperbolic equation

Apply the idea The x-term is positive; therefore, the hyperbola opens horizontally. Matching the equation to the standard form gives the parameters: • Centre: (h, k) = (–2, 1) • a2 = 9 implies a = 3 • b2 = 16 implies b = 4 Find the focal distance, c: c 2 = a 2 + b2

Write the formula

c = 9 + 16

Substitute a2 = 9 and b2 = 16

c2 = 25

Evaluate

c=5

Take the square root of both sides

The key features are: • Vertices: Located a units horizontally from the centre at (h ± a, k) = (−2 ± 3, 1). The vertices are V1 (−5, 1) and V2 (1, 1). • Foci: Located c units horizontally from the centre at (h ± c, k) = (−2 ± 5, 1). The foci are F1 (−7, 1) and F2 (3, 1). • Asymptotes: y k

b ( x h) a

Write the standard formula for the asymptotes

y 1

4 ( x ( 2 )) 3

Substitute h = –2, k = 1, a = 3 and b = 4

4 ( x 2) 3

Simplify

11.09 Conic sections from locus definitions 1123 mathspace.co

b Sketch the graph of the hyperbola.

Apply the idea

C (–2, 1) V1 (–5, 1)

F1 (–7, 1) −9 −8 −7

−6 −5 −4 −3 −2 −1

8 7 6 5 4 3 2 1 −1 −2 −3 −4 −5 −6

V2 (1, 1) 1

F2 (3, 1) 3

x 5

Idea summary A hyperbola is the locus of points where the absolute difference of distances to two foci is constant (2a), where a is the length of the semi-transverse axis. The key relationship is c2 = a2 + b2. Standard equations (centre (h, k)): ( x h)2 • Opens horizontally: a2 • Opens vertically:

( y k)2

( x h)2

The key features of a hyperbola are its centre, foci, vertices, transverse axis and conjugate axis, and asymptotes.

1124 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.09 Practice questions What do you remember? 1

Determine the formula for the distance d between two points (x1, y1) and (x2, y2).

Match the standard form of the equation to the conic section it represents:

Parabola

Circle

Ellipse

iii

Hyperbola

1 a2 b2 (x – h)2 + (y – k)2 = r2

a2 b2 x2 = 4py x2

Define these terms related to conic sections: a

Focus (of a parabola)

Directrix (of a parabola)

Foci (of an ellipse/hyperbola)

Asymptotes (of a hyperbola)

Practice

Ex 1

Ex 2

Determine the equation of the parabola with focus at F (2, 0) and directrix x = –2. Identify its vertex and axis of symmetry, and sketch the graph.

Consider the parabolic equation (x + 1)2 = –4 (y – 3):

Identify its key features: vertex, focus, directrix, and axis of symmetry.

Sketch the graph of the parabola.

Determine the equation of the circle shown:

(–1, 3)

−5 −4 −3 −2 −1

Ex 3

7 6 5 4 3 2 1 −1

(3, 3) x 1

2 3

An ellipse has foci at F1 (0, –4) and F2 (0, 4). The sum of the distances from any point on the ellipse to the foci is 10. Determine the equation of the ellipse.

11.09 Conic sections from locus definitions 1125 mathspace.co

Ex 4

Ex 5

( x 3)2 ( y 1)2 1: 16 7 Identify its key features: centre, foci, lengths of major and minor axes, and vertices.

Sketch the graph of the ellipse.

Consider the elliptical equation

A hyperbola has foci at F1 (–10, 0) and F2 (10, 0). The absolute difference of the distances from any point on the hyperbola to the foci is 16. Determine the equation of this hyperbola.

Consider the hyperbolic equation

Ex 6

( y 2)2 ( x 1)2 1: 9 16 Identify its key features: centre, foci, vertices, and equations of asymptotes.

Sketch the graph of the hyperbola.

A hyperbolic mirror has the property that light rays directed towards one focus are reflected x2 y2 1. Find the coordinates of towards the other focus. The equation of a hyperbola is 9 16 its foci.

Extend your thinking 12

A point P (x, y) moves such that its distance from the point A (4, 0) is equal to its distance from the line x = –2. Determine the equation of this locus.

Determine the equation of an ellipse if its vertices are at (±5, 0) and it passes through the point (3, 3.2).

A point P (x, y) moves such that the line segment joining P to A (0, 0) is perpendicular to the line segment joining P to B (4, 0). Show that the locus of P is a circle and find its centre and radius.

Did you know?

The curved arches of a bridge often follow a parabolic shape, which is a conic section defined using locus rules. Every point on the arch is positioned so that forces are spread evenly across the structure. This clever use of distance and shape helps make bridges strong, stable, and safe — showing how conic sections appear in real-world engineering.

1126 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.10 Parametric representations of relations After this lesson, you will be able to… • express relations in parametric form using appropriate parameters • convert between parametric and Cartesian representations of curves • graph relations from their parametric equations • analyse the key features of parametrically defined curves

Parametric equations Relations in the Cartesian plane are typically represented by equations linking x and y (for example, y = f (x) or F (x, y) = 0). Another powerful way to describe curves is by using parametric equations. Parametric equations A set of equations that expresses the coordinates of points on a curve as functions of an independent variable, called a parameter (often denoted by t). Example: x = t + 1, y = t2 Parameter An independent variable in terms of which the coordinates of a curve are expressed in parametric equations. It can often be thought of as time or an angle.

x = f (t) and y = g(t) x, y t f (t), g (t)

are the coordinates of a point on the curve is the parameter, often representing time or an angle are the functions defining how x and y change with t

As the parameter t varies over a given interval, the point (x(t), y(t)) traces out the curve.

11.10 Parametric representations of relations 1127 mathspace.co

Exploration Explore a set of parametric equations. Consider x = 2 cos (t) and y = 3 sin (t). Complete the table for values of t from 0 to 2π, and then plot the resulting (x, y) points. t

x = 2 cos (t)

y = 3 sin (t)

(x, y)

2 cos (0) = 2

3 sin (0) = 0

(2, 0)

2 cos

3 sin

(0, 3)

2 cos (π) = −2

3 sin (π) = 0

(−2, 0)

3 2

2 cos

2π

2 cos (2π) = 2

2 3 2

3 sin

2 3 2

(0, −3)

3 sin (2π) = 0

(2, 0)

After plotting the points, consider these questions: 1. What shape does the curve appear to form? 2. How does the curve behave as t increases from 0 to 2π? What is the direction of motion? 3. Determine a possible Cartesian equation that might describe this shape.

Advantages of parametric representations: • Representing curves that are not functions of y in terms of x (or vice-versa), such as circles or curves that loop. • Describing the motion of an object along a path, where the parameter t can represent time. This includes information about direction and speed. • Defining curves in three-dimensional space (by adding a third equation z = h (t)). • Simplifying the representation of complex curves.

1128 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 Sketch the curve defined by the parametric equations x = t + 1 and y = 2t − 3 for −2 ≤ t ≤ 3.

Create a strategy Create a table of values for t, x, and y by choosing several values of t within the given interval. Plot the (x, y) points and connect them to sketch the curve. Note the direction of the curve as t increases.

Apply the idea Table of values: t

x=t+1

y = 2t − 3

(x, y)

−2

−2 + 1 = −1

2(−2) − 3 = −7

(−1, −7)

−1

−1 + 1 = 0

2(−1) − 3 = −5

(0, −5)

0+1=1

2(0) − 3 = −3

(1, −3)

1+1=2

2(1) − 3 = −1

(2, −1)

2 + 1 = 3

2(2) − 3 = 1

(3, 1)

3 + 1 = 4

2(3) − 3 = 3

(4, 3)

Plotting these points: 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

y x 1 2 3 4 5 6

The graph is a line segment starting at (−1, −7) (when t = −2) and ending at (4, 3) (when t = 3), traced from bottom-left to top-right as t increases.

11.10 Parametric representations of relations 1129 mathspace.co

Example 2 Describe the curve represented by the parametric equations x = 4 cos (t) and y = 4 sin (t) for 0 ≤ t ≤ 2π.

Create a strategy Recognise the trigonometric identity cos2 (t) + sin2 (t) = 1. Try to manipulate the given parametric equations to use this identity, which might reveal a familiar Cartesian equation.

Apply the idea From the parametric equations, this information is obtained: cos (t )

x y and sin(t ) 4 4

Substitute these into the Pythagorean identity cos2 (t) + sin2 (t) = 1:

x 4

y 4

x y = 1 Substitute cos (t) = and sin (t) = 4 4 into the identity

x2 y2 =1 16 16

Evaluate the squares

x2 + y2 = 16

Multiply both sides by 16

Write in standard form of a circle equation

x +y =4

This is the equation of a circle centred at the origin (0, 0) with radius r = 4. As t goes from 0 to 2π, the point (x, y) traces the circle once in a anticlockwise direction, starting from (4, 0) when t = 0. y 5 (0, 4) 4 3 2 1

x (–4, 0) (4, 0) −5−4 −3 −2 −1 1 2 3 4 5 −1 −2 −3 −4 (0, –4) −5

1130 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary Parametric equations define the coordinates x and y of points on a curve as functions of a third variable, the parameter t. x = f (t), y = g(t) Parametric representations are advantageous for curves that are not simple functions, for describing motion (direction and orientation), and for complex shapes.

Convert between parametric and Cartesian forms It is often useful to convert between parametric and Cartesian representations of a curve. Converting from parametric to Cartesian form involves eliminating the parameter t. Converting from Cartesian to parametric form involves introducing a parameter. •

Converting parametric to Cartesian

The general strategy is to eliminate the parameter t from the two equations x = f (t) and y = g(t) to obtain a single equation relating x and y. • Solve one equation for t: If possible, solve either x = f (t) for t in terms of x, or y = g(t) for t in terms of y. • Substitute: Substitute this expression for t into the other parametric equation. This will result in an equation involving only x and y. • Use identities: For equations involving trigonometric functions (for example, for circles, ellipses, hyperbolas), use trigonometric identities (like cos2 (t) + sin2 (t) = 1 or sec2 (t) − tan2 (t) = 1) to eliminate t. It is important to consider the domain of t, as this may restrict the resulting Cartesian curve. •

Converting Cartesian to parametric

This process often has multiple solutions, as a single Cartesian curve can have many different parametric representations. • Simple substitution: For y = f (x), a common choice is x = t, which then gives y = f (t). • For x = g(y), a common choice is y = t, which then gives x = g(t). • Trigonometric substitutions: For conic sections, specific trigonometric parametrisations are standard: • Circle (x2 + y2 = r2): x = r cos (t), y = r sin (t) • Ellipse

• Hyperbola

1 : x = a cos (t), y = b sin (t)

1 : x = a sec (t), y = b tan (t)

These can be shifted if the centre is not at the origin (for example, x = h + a cos (t)).

11.10 Parametric representations of relations 1131 mathspace.co

Did you know?

Parametric equations let us describe motion in a more natural way by giving each coordinate its own rule. This is perfect for modelling curved paths like a gymnast’s swing!

Example 3 Convert the parametric equations x = 3t − 1 and y = 2t + 5 to Cartesian form.

Create a strategy Solve one of the equations for t, then substitute this expression for t into the other equation.

Apply the idea Solve the first equation for t: x = 3t − 1

Write the given equation for x

x + 1 = 3t

Add 1 to both sides

x 1 t= 3

Divide both sides by 3

Substitute this expression for t into the second equation: y = 2t + 5 = 2

Write the given equation for y

x 1 3

Substitute t

x 1 into the equation 3

2 (x 1) 5 3 3

Combine fractions over a common denominator

2 x 2 15 3

Distribute and simplify the numerator

2 x 17 3

Combine like terms in the numerator

The Cartesian equation is y

2 x 3

17 , which is an equation of a straight line. 3

1132 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 4 Derive parametric equations for the parabola y = (x − 3)2 + 2.

Create a strategy A simple way to parametrise an equation of the form y = f (x) is to let x = t. Then, substitute t for x in the equation to find y in terms of t.

Apply the idea Let x = t. Substitute x = t into the Cartesian equation: y = (t − 3)2 + 2 Therefore, the parametric equations are: x=t y = (t − 3)2 + 2 The parameter t can typically range over all real numbers.

Reflect and check Another possible parametrisation could involve letting x − 3 = t. Then x = t + 3, and y = t2 + 2. This represents the same parabola.

Idea summary To convert parametric equations to Cartesian form, eliminate the parameter t by substitution or by using algebraic/trigonometric identities. To convert Cartesian equations to parametric form, choose a suitable substitution for x or y in terms of t (for example, x = t for y = f (x)), or use standard parametrisation for known curves like conic sections.

Graph parametric equations Graphing parametric equations x = f (t), y = g(t) involves understanding how the (x, y) point moves as the parameter t varies over its domain. Techniques for graphing • Plotting points: Create a table of values for t, x, and y. Plot the resulting(x, y) points and connect them in order of increasing t. This helps visualise the curve and its direction (orientation). • Eliminating the parameter: Convert the parametric equations to Cartesian form. Graph the Cartesian equation, keeping in mind any restrictions on x and y imposed by the domain of t. • Recognising standard forms: If the parametric equations match standard forms for conic sections (circles, ellipses, parabolas, hyperbolas), their general shape is known, which aids graphing.

11.10 Parametric representations of relations 1133 mathspace.co

Analysing key features • Domain and range: The domain of t determines the portion of the curve traced. The range of x = f (t) and y = g(t) over this domain of t gives the Cartesian range of x and y values for the curve. • Intercepts: • x-intercepts: Set y (t) = 0 and solve for t. Substitute these values of t into x (t) to find the x-coordinates. • y-intercepts: Set x (t) = 0 and solve for t. Substitute these values of t into y (t) to find the y-coordinates. • Symmetry: Analyse if x (t) or y (t) are even or odd functions, or if specific transformations of t lead to symmetries in x and y. • Direction or orientation: Indicate with arrows on the curve the direction in which the point (x, y) moves as t increases. • Asymptotes (for hyperbolas): If the parametric equations represent a hyperbola, convert to Cartesian form to find the equations of the asymptotes, or analyse the behaviour of x (t) and y (t) as t approaches values that make denominators zero (for example, for sec (t) and tan (t)). • Foci (for ellipses and hyperbolas): After converting to Cartesian form and identifying a, b, and c, the foci can be located.

Example 5 Sketch the graph of the parametric ellipse defined by x = 5 cos (t) and y = 3 sin (t) for 0 ≤ t ≤ 2π. Identify its domain, range, intercepts, and lines of symmetry.

Create a strategy Convert the parametric equations to Cartesian form using the trigonometric identity cos2 (t) + sin2 (t) = 1 to identify the curve as an ellipse. Then, determine the domain and range by analysing the possible values of x and y. Find the intercepts by setting x = 0 and y = 0 and solving for corresponding points. Identify lines of symmetry based on the Cartesian equation’s form. Finally, sketch the graph of the ellipse, marking key points.

Apply the idea •

Convert to Cartesian form: From the parametric equations: cos (t )

x y , sin(t ) 5 3

Substitute into the identity cos2 (t) + sin2 (t) = 1:

x 5

y 3

x2 y2 =1 25 9

Substitute cos (t )

x y and sin(t ) into the identity 5 3

Evaluate the squares

This is the Cartesian form of an ellipse centred at the origin (0, 0), with semi-major axis 5 (along the x-axis) and semi-minor axis 3 (along the y-axis).

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•

Domain and range: Since −1 ≤ cos (t) ≤ 1, it follows that −5 ≤ 5 cos (t) ≤ 5. Therefore, the domain is [−5, 5]. Since −1 ≤ sin (t) ≤ 1, it follows that −3 ≤ 3 sin (t) ≤ 3. Therefore, the range is [−3, 3].

•

Intercepts:

To find the x-intercepts, set y = 0. This gives 3 sin (t) = 0, which simplifies to sin (t) = 0. For 0 ≤ t ≤ 2π, the solutions are t = 0, π, and 2π. • At t = 0: x = 5 cos (0) = 5, point: (5, 0) • At t = π: x = 5 cos (π) = −5, point: (−5, 0) • At t = 2π: Same as t = 0, point: (5, 0) x-intercepts: (5, 0) and (−5, 0) To find the y-intercepts, set x = 0. This gives 5 cos (t) = 0, which simplifies to cos (t) = 0. 3 For 0 ≤ t ≤ 2π, the solutions are t = and t = . 2 2 • At t • At t

3 sin

3 :y 2

3 sin

3, point: (0, 3)

2 3 2

3, point: (0, −3)

y-intercepts: (0, 3) and (0, −3) •

Lines of symmetry: The Cartesian equation

x2 25

• Substituting x with −x:

y2 9

( x )2 25

1 is symmetric about both axes: y2 9

x2 25

y2 9

1, unchanged, so symmetric about the y-axis.

x2 ( y)2 x2 y2 1, unchanged, so symmetric about the x-axis. 25 9 25 9 The ellipse is symmetric about both the x-axis and y-axis. • Substituting y with −y:

•

Graph: y 3 (0, 3) 2 1

(–5, 0) –5–4–3–2–1 –1 –2 –3

(5, 0) x 1 2 3 4 5

(0, –3)

11.10 Parametric representations of relations 1135 mathspace.co

Example 6 Plot the parametric hyperbola defined by x = 2 sec (t) and y = 3 tan (t) for t ∈ [0, 2π], t Identify its asymptotes and foci.

3 . 2 2 ,

Create a strategy Convert the parametric equations to Cartesian form using the trigonometric identity sec2 (t) − tan2 (t) = 1. Analyse the resulting equation to confirm it is a hyperbola and determine its orientation. Identify the asymptotes using the standard form of the hyperbola equation. Calculate the foci using the relationship between the semi-axes and the distance to the foci. Consider the domain and range based on the parametric equations and the restricted t-values. Finally, graph the hyperbola, marking the vertices, foci, and asymptotes.

Apply the idea •

Convert to Cartesian form: From the parametric equations: sec (t )

x y , tan (t ) 3 2

Substitute into the identity sec2 (t) − tan2 (t) = 1:

x 2

y 3

Substitute sec (t) =

x2 y2 =1 4 9

y x and tan (t) = into the identity 3 2

Evaluate the squares

This is the Cartesian form of a hyperbola centred at the origin (0, 0), with a2 = 4, so a = 2, and x2 y2 b2 = 9, so b = 3. The form 2 1 indicates the hyperbola opens horizontally. a b2 •

Domain and range: 1 , sec (t) ∈ (−∞, −1] ∪ [1, ∞) excluding t cos (t ) therefore, x ∈ (− ∞, −2] ∪ [2, ∞). For x = 2 sec (t), since sec (t) =

For y = 3 tan (t), tan (t) ∈ (−∞, ∞) with vertical asymptotes at t •

3 ; 2

3 ; therefore, y ∈ (−∞, ∞). 2

Vertices: The vertices are at (±a, 0) = (±2, 0), so (2, 0) and (−2, 0): • For the vertex at x = 2: The condition 2 = 2 sec (t) gives sec (t) = 1. This occurs when t = 0, 2p. At these values, y = 3 tan (t) = 0, giving the point (2, 0). • For the vertex at x = −2: The condition −2 = 2 sec (t) gives sec (t) = −1. This occurs when t = p. At this value, y = 3 tan (p) = 0, giving the point (–2, 0).

•

Asymptotes: For a hyperbola

1, the asymptotes are y

1136 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b x. Here, a = 2, b = 3, so y a

3 x. 2

•

Foci: The distance to the foci is given by c, where c2 = a2 + b2: c2 = a2 + b2 =4+9

Substitute a2 = 4 and b2 = 9

= 13

Evaluate

c = 13

Take the square root of both sides

≈ 3.606 The foci are at (±c, 0) = •

Write the formula for the foci of a hyperbola

Evaluate

13 , 0 ≈ (±3.606, 0).

Graph: y 6 5 4 3 2 x F2 F1 V 1 1 V 2 −6−5−4−3−2−1 1 2 3 4 5 6 −1 −2 −3 −4 −5 −6

Idea summary To graph parametric equations, create a table of values, plot points indicating direction, or convert to Cartesian form. Key features like domain/range (of x and y), intercepts, symmetry, and orientation (direction of tracing) can be analysed from parametric equations. For conic sections, asymptotes, and foci can be found after converting to Cartesian form or using standard parametric properties.

11.10 Parametric representations of relations 1137 mathspace.co

11.10 Practice questions What do you remember? 1

Is each statement about parametric equations true or false? a

In parametric equations, both x and y are expressed as functions of a third variable called a parameter.

A curve represented by parametric equations must also be a function of the form y = f (x).

Eliminating the parameter from x = t + 2 and y = 3t results in the Cartesian equation y = 3x − 6.

The equations x = t and y = 5t + 1 are a possible parametrisation of the line y = 5x + 1.

Complete this table of values for the parametric equations x = t2 − 1 and y = t + 2 for the given values of t: t

x = t2 − 1

y=t+2

(x, y)

−2

⬚

−1 0 1 2 3

⬚

⬚ ⬚ ⬚

Match each set of parametric equations to its corresponding Cartesian equation: a

x = t, y = 4t − 3

y=x−5

x = t + 5, y = t

y = x2 + 1

x = 2 cos (t), y = 2 sin (t)

iii

y = 4x − 3

x2 + y2 = 4

x = t, y = t + 1

Which of these is the Cartesian equation for the parametric equations x = 2t and y = 4t + 3? A

y = 2x + 3

y = 4x + 3

y = 8x + 3

y=x+3

Which trigonometric identity is commonly used to convert parametric equations of the form x = a cos (t), y = b sin (t) to Cartesian form?

If x = t2 and y = t + 1, determine the Cartesian equation by expressing y in terms of x. Consider any restrictions on x or y.

Determine a set of parametric equations for each Cartesian equation: a

y = x2 − 5

1138 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

(x − 1)2 + y2 = 25

Practice Ex 1

Sketch the curve defined by the parametric equations x = 2t and y = t − 1 for −1 ≤ t ≤ 4. Indicate the direction of the curve.

Ex 2

Describe the curve represented by the parametric equations x = 3 cos (t) and y = 3 sin (t) for 0 ≤ t ≤ π. Sketch the curve.

Convert the parametric equations x = 4t + 1 and y = t − 2 to Cartesian form.

Convert the parametric equations x = t + 2 and y = t2 − 1 to Cartesian form.

Ex 3

Convert the parametric equations x = 2 cos (t) and y = 5 sin (t) to Cartesian form.

Ex 4

Derive parametric equations for the circle (x − 2)2 + (y + 1)2 = 9.

Ex 5

Sketch the graph of the parametric ellipse defined by x = 4 cos (t) and y = 2 sin (t) for 0 ≤ t ≤ 2π. Identify its domain, range, and intercepts.

Ex 6

Plot the parametric hyperbola defined by x = 3 sec (t) and y = 2 tan (t) for t in the interval [0, 2π], excluding values where the functions are undefined. Identify its Cartesian equation and asymptotes.

For the curve defined by x = t2 − 1, y = t + 1: a

Determine the Cartesian equation of the curve.

Sketch the curve for −2 ≤ t ≤ 2, indicating the direction.

Extend your thinking 17

A curve is defined by the parametric equations x = sin (2t) and y = cos (t) for 0 ≤ t ≤ 2π: a

Use the identity sin (2t) = 2 sin (t) cos (t) and sin2 (t) + cos2 (t) = 1 to determine the Cartesian equation of the curve.

What are the restrictions on x and y values?

Determine parametric equations for the line segment from P (− 1, 3) to Q (4, − 2) such that t = 0 corresponds to point P and t = 1 corresponds to point Q.

The cycloid is a curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slipping. If the wheel has radius r and rolls along the x-axis, and the point starts at the origin, the parametric equations are x = r (t – sin (t)) and y = r (1 – cos (t)), where t is the angle through which the wheel has rotated: a

If r = 1, determine the coordinates of the point on the cycloid when the wheel has rotated through t radians. 2

At what value of t does the point first touch the x-axis again after starting at the origin? What are the x-coordinates of this point?

11.10 Parametric representations of relations 1139 mathspace.co

11.11 Polar coordinates and polar graphs After this lesson, you will be able to. . . • convert between Cartesian and polar coordinate systems • graph basic polar equations and identify their key features • express conic sections in polar form • analyse symmetry and periodicity in polar graphs

Polar coordinates The Cartesian coordinate system locates points in a plane using a horizontal distance (x) and a vertical distance (y) from an origin. An alternative system for locating points in a plane is the polar coordinate system.

4 2 −4

−2

2 −2

A polar coordinate system uses concentric circles and radial lines emanating from the pole.

−4

Polar coordinates A two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

P(r,θ )

θ O

Pole The fixed reference point in the polar coordinate system, equivalent to the origin in the Cartesian system. Polar axis A fixed ray (half-line) starting from the pole, usually drawn horizontally to the right, from which the angle θ is measured. It corresponds to the positive x-axis in the Cartesian system.

1140 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

In the polar coordinate system, a point P is described by an ordered pair (r, θ ), where: • r is the directed distance from the pole (origin O) to the point P. r can be positive, negative, or zero. • θ is the angle measured from the polar axis to the ray OP. The angle is positive if measured anticlockwise and negative if measured clockwise. Note that a point in polar coordinates does not have a unique representation. For example, (r, θ ) is the same point as (r, θ + 2nπ) for any integer n. Also, (r, θ ) is the same as (–r, θ + π). •

Relationship between polar and Cartesian coordinates

Conversion between polar and Cartesian coordinates uses trigonometry, by considering a rightangled triangle formed by the point P (x, y), the origin, and the projection of P onto the x-axis. P (x, y) r = |z|

y = r sin (θ )

θ x = r cos (θ )

The relationship between Cartesian (x, y) and polar (r, θ ) coordinates. The angle θ is measured from the positive x-axis (polar axis) to the line segment connecting the pole to the point P.

From polar (r, θ ) to Cartesian (x, y):

x = r cos (θ ) x

is the Cartesian x-coordinate

is the polar radial coordinate

is the polar angular coordinate

y = r sin (θ ) y

is the Cartesian y-coordinate

From Cartesian (x, y) to polar (r, θ ):

y2 implies r

r is the polar radial coordinate (usually the non-negative root is taken)

tan( )

y ( if x x

θ is the polar angular coordinate; and the quadrant of (x, y) must be considered to find the correct value of θ When finding θ, if x = 0, then θ =

3 if y > 0, θ = or if y < 0, and θ is undefined if y = 0 (at the pole). 2 2 2

11.11 Polar coordinates and polar graphs 1141 mathspace.co

Example 1 2, 2 3 to polar coordinates (r, θ ), where r > 0 and 0 ≤ θ < 2π.

Convert the Cartesian coordinates

Create a strategy Use the conversion formulas r based on the signs of x and y.

y y2 and tan (θ ) = . Determine the correct quadrant for θ x

Apply the idea Given x = –2 and y 2 3. Calculate r: r = x2 y2

Write the conversion formula

= ( 2)2 (2 3 )2

Substitute x = –2 and y 2 3

= Evaluate the squares 4 4 ( 3) = Evaluate the multiplication 4 12 = Evaluate the addition 16 = 4 Evaluate Calculate θ : 2 3 Write the equation 2 = Evaluate 3

tan(θ ) =

Since x < 0 and y > 0, the point 2, 2 3 is in the second quadrant. The reference angle α for which tan(α) = 3 is α = . 3 2 In the second quadrant, θ = π – α = π – . 3 3 2 Therefore, the polar coordinates are 4, . 3

Reflect and check Check the answer by converting the polar coordinates 4,

x = r cos (θ ) = 4 cos

2 3

y = r sin (θ ) = 4 sin

2 3

The resulting Cartesian coordinates are

2 3 1 2

4 4

back to Cartesian coordinates.

3 2

2 2 3

2, 2 3 , which match the original coordinates.

1142 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 Convert the polar coordinates 3,

5 6

to Cartesian coordinates (x, y).

Create a strategy Use the conversion formulas x = r cos(θ ) and y = r sin(θ ).

Apply the idea Given: r = 3 and θ =

5 6

Calculate x: x = r cos(θ ) 5 = 3 cos 6

Write the equation for x-coordinate Substitute r = 3 and θ =

5 6

3 5 3 = Evaluate cos 2 6 3 3 = Simplify 2 Calculate y: y = r sin(θ ) = 3 sin

5 6

Write the equation for y-coordinate Substitute r = 3 and θ =

5 6

5 1 3 = Evaluate sin 6 2 3 = Simplify 2 Therefore, the Cartesian coordinates are

3 3 3 , . 2 2

Reflect and check 5 is in Quadrant 2, cos(θ ) is negative while sin(θ ) is positive. Reference angle is , 6 6 3 5 3 1 5 1 cos ; cos , sin ; sin 6 2 6 2 6 2 6 2

11.11 Polar coordinates and polar graphs 1143 mathspace.co

Idea summary Polar coordinates (r, θ ) represent points by a distance r from the pole and an angle θ from the polar axis. Conversions: • Polar form to Cartesian form: x = r cos(θ ), y = r sin(θ ) y • Cartesian form to polar form: r x2 y2 , tan(θ ) = (considering the correct x quadrant for θ ).

Graph polar equations A polar equation is an equation that expresses a relationship between r and θ. Graphing polar equations involves plotting points (r, θ ) that satisfy the equation, or converting to Cartesian coordinates if that simplifies the graphing process. Common polar graphs include circles, roses, and limaçons. •

Techniques for graphing polar equations • Plotting points: Create a table of values by choosing values for θ and calculating the corresponding r. Plot these points in the polar coordinate system. • Converting to Cartesian form: If possible, convert the polar equation to its Cartesian equivalent. This can often reveal a familiar shape (e.g., line, circle, conic). • Testing for symmetry: • Symmetry about the polar axis (x-axis): Replace θ with –θ. If the equation remains unchanged (or r(– θ ) = r(θ ) or r(– θ ) = –r(π – θ )), it’s symmetric about the polar axis. (y-axis): Replace θ with π – θ. If the equation remains 2 unchanged (or r(π – θ ) = r(θ ) or r(π – θ ) = –r(– θ )), it’s symmetric about the line θ = . 2 • Symmetry about the pole (origin): Replace r with –r, or replace θ with θ + π. If the equation remains unchanged, it’s symmetric about the pole. • Symmetry about the line θ =

ote: These tests are sufficient but not always necessary. A graph may have symmetry even N if a test fails, due to non-unique polar representations. • Periodicity: For functions involving sin(nθ ) or cos(nθ ), the graph may repeat. The period of 2 sin(θ ) and cos(θ ) is 2π. The period of sin(nθ ) or cos(nθ ) is . n • Maximum |r| value: Find the maximum value of |f(θ )|. This helps determine the scale of the graph. • Zeros: Find values of θ for which r = 0. These are angles where the graph passes through the pole.

1144 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

•

Basic polar graphs: Circles The simplest circle is r = a, which is a circle centred at the pole with radius |a|. 2π 3 5π 6

π 3

π 2

π 6

2 r=3

The graph of r = 3, a circle centred at the pole with radius 3.

–4

–2

7π 6

2 –2

4π 3

11π 6

3π 2

–4

5π 3

Circles passing through the pole can be represented by equations of the form r = a cos (θ ) or r = a sin (θ ). Here, |a| is the diameter of the circle. 2π 3 5π 6

π 6

5π 6

d=2

π –2

2π 3

π 3

π 2

7π 6

–1 4π 3

–2

3π 2

11π 6

d=2

π 6 0

–2

–1

7π 6

1 –1

4π 3

5π 3

r = 2 cos (θ ) has diameter 2, centre on the polar axis at (1, 0)

Equation form

π 2

0 –1

π 3

–2

3π 2

2 11π 6

5π 3

r = 2 sin (θ ) has diameter 2, centre on the line θ = at 1 , 2 2

Centre (Cartesian)

Symmetry

r = a cos (θ ), a > 0

On positive x-axis:

a ,0 2

Polar axis (x-axis)

r = a cos (θ ), a < 0

On negative x-axis:

a ,0 2

Polar axis (x-axis)

r = a sin (θ ), a > 0

On positive y-axis: 0,

a 2

Line θ =

(y-axis) 2

r = a sin (θ ), a < 0

On negative y-axis: 0,

a 2

Line θ =

(y-axis) 2

11.11 Polar coordinates and polar graphs 1145 mathspace.co

•

Basic polar graphs: Roses

Equations of the form r = a cos(nθ ) or r = a sin(nθ ), where n is an integer greater than 1, produce rose curves. • The value |a| is the maximum distance from the pole (length of each petal). • If n is odd, the rose has n petals. • If n is even, the rose has 2n petals. • Symmetry for r = a cos(nθ ): Polar axis • Symmetry for r = a sin(nθ ): If n is odd, symmetric about θ = . If n is even, symmetric 2 about θ = , polar axis, and pole. 2 2π 3 5π 6

2 1

π –2

–1

7π 6

1 –1

4π 3

–2

3π 2

2π 3

π 3

π 2

π 6

5π 6

–2

11π 6 5π 3

r = 2 cos(3θ ) has 3 petals, maximum length 2 •

π 3

π 2

π 6

0 –1

7π 6

1 –1

4π 3

–2

3π 2

2 11π 6

5π 3

r = 2 sin(3θ ) has 3 petals, maximum length 2

Basic polar graphs: Limaçons

Equations of the form r = a ± b cos(θ ) or r = a ± b sin(θ ) (where a > 0, b > 0) produce limaçons. a The shape depends on the ratio . b a • 1 (i.e., a < b): Limaçon with an inner loop b a • 1 (i.e., a = b): Cardioid (heart-shaped) b a • 1 2 (i.e., b < a < 2b): Dimpled limaçon b a • 2 (i.e., a ≥ 2b): Convex limaçon (no dimple) b

1146 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Symmetry for r = a ± b cos(θ ) is about the polar axis. Symmetry for r = a ± b sin(θ ) is about the line θ = . 2 2π 3 5π 6

2π 3

π 3

π 2

π –10

–5

7π 6

5 –5

4π 3

–10

π 6

5π 6

–10

–5

5 –5

4π 3

5π 3

–10

3π 2

Cardioid:

a 3 1 b 5

r = 5 + 5 cos (θ )

2π 3 5π 6

10 11π 6

5π 3

a 1 b

π 3

π 2

π 6

π –10

π 6 0

Limaçon with inner loop: r = 3 + 5 cos (θ )

π 3

π 2

7π 6

11π 6

3π 2

0 –5

7π 6

5 –5

4π 3

–10

3π 2

11π 6 5π 3

Dimpled Limaçon: a 6 r = 6 + 5 sin (θ ) 1 2 b 5

Interactive exploration Discover this concept in action online

mathspace.co

Example 3 Digital only example Explore in the online lesson

mathspace.co

11.11 Polar coordinates and polar graphs 1147 mathspace.co

Example 4 Consider the polar equation r = 4 cos (θ ). Identify its Cartesian equation and sketch its graph.

Create a strategy This equation is of the form r = a cos (θ ), which represents a circle. The graph can also be identified by converting the equation to Cartesian coordinates. To do this, multiply the polar equation by r to introduce terms that can be substituted, such as r2 and r cos (θ ).

Apply the idea The equation r = 4 cos (θ ) is a circle with diameter 4, symmetric about the polar axis, passing through the pole and the point (4, 0) in polar coordinates (which is (4, 0) in Cartesian). Converting to Cartesian: Multiply by r: r2 = 4r cos(θ ) Substitute r2 = x2 + y2 and x = r cos(θ ): x2 + y2 = 4x 2

x – 4x + y = 0 2

Write the equation Rearrange

(x – 4x + 4) + y = 4 Complete the square for 4 2 x: Add  = 4 to both sides 2 2 2 2 (x – 2) + y = 2 Factorise and simplify This is a circle with centre (2, 0) (Cartesian) and radius 2. y

3 2 1

–2 –1 0 –1

Centre (2, 0)

–2 –3 –4 Polar graph of r = 4 cos (θ )

Cartesian graph of (x – 2)2 + y2 = 4

Reflect and check To formally test for symmetry about the polar axis, replace θ with –θ in the equation. r = 4 cos (–θ )

Write the equation

= 4 cos (θ ) Since cos (–θ ) = cos (θ ) (cosine is an even function) Because the equation remains unchanged, the graph is symmetric about the polar axis, as observed in the sketch.

1148 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 5 Sketch the graph of the polar equation r = –4 sin (3θ ).

Create a strategy This is a rose curve of the form r = a sin(nθ ) with a = –4 and n = 3. • Number of petals: Since n = 3 (odd), there are n = 3 petals. • Petal length: The maximum length of a petal is |a| = | –4 | = 4 units. • Symmetry: Since it’s sin(nθ ) with n odd, it is symmetric about the line θ = . 2 2 • Spacing of petals: Petals are spaced radians apart. 3 • First petal tip: For r = a sin (nθ ), tips occur when sin (nθ ) = ±1. 3 5 , Here, sin (3θ ) = ±1.3θ = , ,... 2 2 2 5 Therefore, θ = , , ,... 6 2 6 4 sin 3 4 sin At θ = , r = –4(1) = –4. This point is 4, , which is 6 2 6 6 7 4, plotted as 4, . 6 6 At θ = , r 2 5 ,r 6 5 as 4, 6

At θ =

4 sin 3

4 sin 3 4,

5 6

= –4(–1) = 4. This point is 4, 4 sin

5 2

= –4(1) = –4. This point is

5 , plotted 6

11 . 6

• Zeros (points at pole): Setting r = 0 gives the equation −4 sin (3θ ) = 0, which simplifies to sin (3θ ) = 0. Therefore, 3θ = 0, π, 2π, . . . 2 θ = 0, , , π, . . . 3 3 The graph passes through the pole at these angles. Plotting these key points and using the graph’s symmetry allows for an accurate sketch. The graph completes its full pattern as θ goes from 0 to π (since this interval for θ means 3θ covers 0 to 3π). 2π 3

Apply the idea The petal tips are at 4,

, 4,

11 7 , and 4, . 6 6

The graph is a rose curve with 3 petals, each of length 4. The 7 11 tips are located at angles θ = , . The curve passes , 2 6 6 2 through the pole at θ = 0, , . 3 3

π 2

π 3

5π 6

π 6

–5

7π 6 4π 3

–5

3π 2

11π 6 5π 3

11.11 Polar coordinates and polar graphs 1149 mathspace.co

Example 6 Sketch a graph of the equation r = 4 – 3 cos (θ ).

Create a strategy The equation is r = 4 – 3 cos (θ ). This is a limaçon of the form r = a – b cos (θ ) with a = 4 and b = 3. Analyse the ratio

a : b

• |a| = 4, |b| = 3 a 4 • The ratio b 3 4 • Since 1 2 (i.e., b < a < 2b), the graph is a dimpled limaçon (it has a “dent” but no inner 3 loop). Key features: • Symmetry: Since the equation involves cos (θ ), it is symmetric with respect to the polar axis (x-axis). • Maximum |r| value: Occurs when cos (θ ) = –1 (i.e., θ = π), giving r = 4 – 3(–1) = 7. • Minimum positive |r| value: Occurs when cos (θ ) = 1 (i.e., θ = 0, 2π), giving r = 4 – 3(1) = 1. • Points at pole (r = 0): The equation becomes 0 = 4 − 3 cos (θ ), which simplifies to 4 4 4 3 cos (θ ) = 4, or cos (θ ) = . Since 1, cos (θ ) = has no solution. Therefore, the graph 3 3 3 does not pass through the pole. 3 Create a table of values for key angles, typically 0, , π, , to find specific points. 2 2

cos (θ )

r = 4 – 3 cos (θ )

4 – 3(1) = 1

4 – 3(0) = 4

–1

4 – 3(–1) = 7

3 2

4 – 3(0) = 4

2π

4 – 3(1) = 1

1150 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea From the table, the key points in (r, θ ) form are: (1, 0), 4, , (7, π), 4, 3 , and (1, 2π) (which is the 2 2 same point as (1, 0)). Plot these points on a polar grid: π 2

2π 3

π 3

5π 6

π 6

π –8

–6

0 –4

7π 6

–2 –2

11π 6

–4 4π 3

5π 3

3π 2

Connect the points smoothly, keeping in mind that it’s a dimpled limaçon symmetric about the polar axis. The “dimple” occurs towards the pole at θ = 0 where r = 1. π 2

2π 3

π 3

5π 6

π 6

π –8

–6

0 –4

–2

7π 6

11π 6

–4 4π 3

3π 2

5π 3

11.11 Polar coordinates and polar graphs 1151 mathspace.co

Example 7 Sketch a graph of the equation r = 1 + 2 cos (θ ).

Create a strategy a 1 = < 1, this is a limaçon b 2 with an inner loop. It will be symmetric about the polar axis. The key points can be plotted for θ = 0, 3 , π, , 2π and points where r = 0 to find the loop. 2 2 This is a limaçon of the form r = a + b cos (θ ) with a = 1 and b = 2. Since

Apply the idea Key points:

cos (θ )

1 + 2 (1) = 3

1 + 2 (0) = 1

2 3

–

–1

1 + 2 (–1) = –1

4 3

–

1 2

1 + 2

3 2

1 + 2 (0) = 1

2π

1 + 2 (1) = 3

1 2

r = 1 + 2 cos (θ )

1 + 2

1 = 1 (r = 0, enters pole) 2

1 = 0 (r = 0, leaves pole) 2

1 2 The r = 0 values occur when 1 + 2 cos (θ ) = 0 it follows that cos (θ ) = – . This happens at (θ ) = 3 2 4 and θ = . These define the inner loop. 3

2π 3 5π 6 π –3 –2 7π 6

π 2

π 3

π 6 (0, 3) 0 –1 1 2 3(2π, 3) 3 π   –1  2 , 1  11π   6 4π –2 5π 3π 3 2 3 π  1  , 1 2 

1152 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The graph is a limaçon with an inner loop. The outer extent is r = 3 at θ = 0. The inner loop crosses the negative polar axis at r = –1 (when θ = π).

Idea summary To sketch the graph of a polar equation r = f (θ ), a combination of these techniques can be used: plotting points, converting to Cartesian form, analysing for key features, or recognising a standard shape. Common forms: • Circles: r = a, r = a cos (θ ), r = a sin (θ ) • Roses: r = a cos(nθ ), r = a sin(nθ ). Number of petals depends on n being odd or even. • Limaçons: r = a ± b cos (θ ), r = a ± b sin (θ ). Shape (inner loop, cardioid, a dimpled, convex) depends on ratio. b

Conic sections in polar form Conic sections (parabolas, ellipses, hyperbolas) can also be represented by a single general polar equation when one focus is at the pole and the directrix is either horizontal or vertical. π 3

3π 2π 4 6 5π 6 Directrix

π 2

Focus

π 4π 6 0, 2π

3π 2π 4 6 5π 6 Directrix

3π 2

11π 6 5π 7π 3 4

π 3

π 4π 6

Focus 0, 2π

3π 2π 4 6 5π 6 Directrix π p

1 2 3 4 5

1 2 3 4 5 7π 6 5π 4π 4 3

π 2

7π 6 5π 4π 4 3

3π 2

π 3

π 4 π 6

Focus

0, 2π

1 2 3 4 5

11π 6 5π 7π 3 4

7π 6 5π 4π 4 3

3π 2

11π 6 5π 7π 3 4

If e > 1, hyperbola.

If e = 1, parabola.

If e < 1, ellipse.

π 2

The general polar form for a conic section with a focus at the pole is:

ed or r e cos ( )

ed e sin( )

r is the distance from the pole to a point on the conic θ is the angle from the polar axis to the point e is the eccentricity of the conic section (e > 0) d is the distance from the focus at the pole to the directrix (d > 0) The eccentricity e determines the type of conic section: • If e = 0, it is a circle (though this specific polar form is usually for e > 0; a circle centred at the pole is simply r = constant). • If 0 < e < 1, it is an ellipse. • If e = 1, it is a parabola. • If e > 1, it is a hyperbola.

11.11 Polar coordinates and polar graphs 1153 mathspace.co

The choice of cos (θ ) or sin(θ ) and the sign (±) in the denominator determine the orientation of the conic and its directrix relative to the pole: • 1 + e cos (θ ): Directrix x = d (vertical, to the right of the pole). Conic opens left. • 1 – e cos (θ ): Directrix x = –d (vertical, to the left of the pole). Conic opens right. • 1 + e sin (θ ): Directrix y = d (horizontal, above the pole). Conic opens down. • 1 – e sin (θ ): Directrix y = –d (horizontal, below the pole). Conic opens up. Vertical directrix

5π 6

Directrix right of pole

Directrix left of pole

ed r 1 e cos ( )

2π 3π 6 4

π 2

π 3 π Directrix 4 π 6

5π 6

π 2

Focus

5π 4 4π 3

3π 2

π 4 π 0, 2π

1 2 3 4 5

1 2 3 4 5 7π 6

π 3

Directrix π p

p 0, 2π

Focus

2π 3π 6 4

11π 7π 6 5π 4 3

7π 6

5π 4 4π 3

7π 5π 4 3

3π 2

11π 6

Horizontal directrix Directrix above of pole r 1

5π 6

2π 3π 6 4

Focus

Directrix below of pole

ed e sin ( ) π 2

π 3

Directrix p

π 4 π 6

5π 6

0, 2π

2π 3π 6 4

5π 4 4π 3

3π 2

7π 5π 4 3

11π 6

1154 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

π 2

π 3

π 4 π 6 0, 2π

Focus

1 2 3 4 5 7π 6

ed e sin ( )

r 1

p 7π 6 5π 4 4π 3

1 2 3 4 5 11π 6 7π 5π 4 3

Directrix 3π 2

•

Rotated conic sections

The standard forms assume the directrix is perfectly horizontal or vertical. If the directrix is rotated, the conic’s axis of symmetry also rotates. The general form for a conic with its axis of symmetry rotated by an angle β from the polar axis is:

ed e cos(

)

or r

ed e sin(

)

b is the angle of rotation of the conic’s major axis from the polar axis For b = 0, ed r 1 e cos ( ) Directrix x = −p

•

For

ed 1 e sin ( )

Directrix y = −p

For b = π, ed r 1 e cos( ) Directrix x = p

For

3 ed ,r 2 1 e sin ( )

Directrix y = p

Converting between Cartesian and polar forms for conics

To convert a conic’s polar form to Cartesian, use r x2 y2 , x = r cos (θ ), y = r sin (θ ). Multiply out the denominator, substitute, and rearrange. This can lead to the standard Cartesian forms of conics, though often shifted or rotated if the focus was at the pole. Converting a general Cartesian conic to polar form with a focus at the pole can be more involved and often requires translating the Cartesian equation so one focus is at the origin before converting.

11.11 Polar coordinates and polar graphs 1155 mathspace.co

Example 8 6 , find its 2 cos ( ) eccentricity e, and the equation of its directrix x = d or y = d. Sketch the conic. Identify the type of conic section represented by the polar equation r

Create a strategy ed ed or r by dividing the 1 e sin( ) 1 e cos ( ) numerator and denominator of the given fraction by the constant term in the denominator (in this case, 2). Then identify e and ed to find d. Rewrite the equation in one of the standard forms r

Apply the idea r=

6 2 cos ( )

Write the equation

6 Divide numerator and 2 = (2 cos ( )) denominator by 2 2 3 1 = Simplify 1 cos ( ) 2 Comparing with r

ed : e cos ( ) Eccentricity e

1 2

1 1 Since ed = 3 and e , it follows that d = 3, which gives d = 6 2 2 Since e

1 (which is 0 < e < 1), the conic is an ellipse. 2

The form 1 + e cos (θ ) in the denominator indicates the directrix is vertical and to the right of the pole, so its equation is x = d. The directrix is x = 6. To sketch, find points at key angles: 3 3 • θ = 0: r = 2. The point is (2, 0). 1 3 1 2 2 3 3. The point is 3, . • θ= :r 2 2 1 0 3 3 • θ = π: r 6. The point is (6, π). 1 1 1 2 2 3 3 3 :r 3. The point is 3, • θ= . 2 2 1 0

1156 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x=6

7 5  π  3,  3  2 (6, π)

–7 –5 –3 –1 –1

(2, 0) 1

–3  π   3,  –5  2  –7

Example 9 Express the circle x2 + y2 = 9 in polar form.

Create a strategy Use the relationships x2 + y2 = r2.

Apply the idea Given x2 + y2 = 9. Substitute r2 = x2 + y2: r2 = 9

Write the equation

r = 3 Since r usually represents a non-negative distance, take the positive root The polar form is r = 3. This is a circle centred at the pole with radius 3.

Idea summary Conic sections with a focus at the pole can be represented by the general polar equation: r

ed or r e cos ( )

ed e sin( )

• Eccentricity e determines the type: ellipse (0 < e < 1), parabola (e = 1), hyperbola (e > 1). (The case e = 0 corresponds to a circle. However, a circle centred at the pole is more simply described by r = k. A circle passing through the pole is a special case of a limaçon (r = a cos (θ )). • d is the distance from the focus at the pole to the directrix. • The term (cos (θ ) or sin (θ )) and sign (±) indicate directrix orientation. Conversions use x = r cos (θ ), y = r sin (θ ), r2 = x2 + y2.

11.11 Polar coordinates and polar graphs 1157 mathspace.co

11.11 Practice questions What do you remember? 1

Match each basic polar equation to its corresponding graph description: a b c

r = 4 θ = 3 r = 6 cos (θ )

A line passing through the pole.

A circle with its centre at the pole.

iii

A circle passing through the pole.

Identify the type of curve represented by each polar equation (e.g., circle, cardioid, rose, parabola): 6 a r = 2 + 2 sin (θ ) b r = 3 cos (4θ ) c r = 6 sin (θ ) d r 1 sin( )

Match each conic section description to its standard Cartesian equation: a b c d

x2 + y2 = r2 2

Parabola (vertex at origin, opens up)

Ellipse (horizontal major axis)

iii

Hyperbola (opens horizontally)

Circle

x2 = 4py

Practice 4

Convert these Cartesian coordinates to polar coordinates (r, θ ) where r ≥ 0 and 0 ≤ θ < 2π: a

6 Ex 1

(1, 1)

(0, – 3)

3, 1

Convert these polar coordinates (r, θ ) to Cartesian coordinates (x, y): 7 2, 2, a b (4, π) c d 3 4

Convert the polar coordinates 6,

2 3

to Cartesian coordinates (x, y).

Sketch the graph of the polar equation r = 5.

Ex 3a

(5, 0)

Convert the Cartesian coordinates (–3, –3) to polar coordinates (r, θ ), where r > 0 and 0 ≤ θ < 2π.

Ex 2

Sketch the graph of the polar equation θ =

2 . Describe the graph. 3

Ex 3b

1158 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Ex 4

Sketch the graph of the polar equation r = 6 sin (θ ). Identify its Cartesian equation.

Ex 5

Sketch the graph of the polar equation r = 3 cos (2θ ) (a four-petal rose). Indicate the range of θ needed to trace the full curve.

Consider the equation r = 2 + 2 cos (θ ). First, identify the type of curve. Then, test the equation for symmetry about the polar axis, the line θ = , and the pole. 2

Sketch the graph of the polar equation and identify the type of limaçon:

Ex 6, 7

r = 3 – 2 cos (θ )

r = 2 + 4 sin (θ ) 3

10 , and 2 cos ( )

4 , and sin( )

Ex 8

Identify the type of conic section represented by the polar equation r determine its eccentricity and the equation of its directrix.

Ex 9

Convert the Cartesian equation x2 + y2 = 25 to its simplest polar form.

Identify the type of conic section represented by the polar equation r determine its eccentricity and the equation of its directrix.

Find the polar equation of a parabola with its focus at the pole and its directrix at x = –4.

Extend your thinking 3 to Cartesian form and identify the conic section. 1 2 sin (θ )

Convert the polar equation r

Consider the graphs of r = 2 and r = 4 cos (θ ). Determine their points of intersection in polar coordinates.

Sketch the graph of r2 = 4 cos (2θ ) (a lemniscate). Describe the graph’s symmetry with respect to the polar axis, the line θ = , and the pole. 2

Chapter 11 review Online review

mathspace.co

11.11 Polar coordinates and polar graphs 1159 mathspace.co

Big ideas • Discrete random variables provide a mathematical framework for modelling situations with countable, random outcomes. A probability distribution completely defines the variable's behaviour, which can be summarised by its expected value (the theoretical long-run average) and its variance (a measure of spread). This framework allows to analyse and predict the outcomes of random phenomena through calculation and simulation. • Since analysing entire populations is often impractical, statistics calculated from a random sample are used to estimate unknown population parameters. This process is governed by sampling variability, the natural variation between samples. The Central Limit Theorem is a cornerstone of statistics, stating that the distribution of sample means will be approximately normal for a large enough sample size, allowing to quantify the reliability of our estimates.

12 Simulation, sampling and sampling distributions Chapter outline 12.01 12.02 12.03 12.04 12.05 12.06 12.07 12.08

Discrete random variables Centre and spread of discrete random variables Sums of independent random variables Applications of discrete random variables Investigation: Simulations Simulation Sample distributions Simulate sampling distributions Investigation: Generating samples with technology Sample size and distributions Chapter 12 review

1162 1183 1195 1202 1213 1220 1229 1236

Archaeologists use random sampling to find ancient sites. Digging is expensive—statistics helps them guess right.

12.01 Discrete random variables After this lesson, you will be able to… • determine if a situation can be modelled with a discrete random variable • understand the properties of discrete random variables • use multiple representations for probability distribution for a discrete random variable • apply discrete random variables to solve basic probability problems

Discrete random variables (DRV) A random variable is a fundamental concept in probability theory and statistics. For each possible outcome of a random phenomenon, it assigns a numerical value. Random variables can be classified into two main categories: discrete and continuous Discrete random variable (DRV) A random variable that takes on a countable number of distinct values. Example: The number of heads in three flips of a fair coin, with possible values 0, 1, 2, and 3. Continuous random variable A random variable that takes on any value within a given range. Example: The weights of marbles produced by a machine, measured in mg.

The focus will be on discrete random variables (DRV), defined by these characteristics: 1. Discrete values: The random variable can only take on a countable number of distinct values. 2. Random occurrence: The outcomes occur randomly within the context of the experiment or situation. 3. Varied outcomes: There are multiple possible outcomes, ensuring that the results vary. Examples of situations that can be represented by discrete random variables: • Rolling two dice and finding the sum of the topmost faces, the possible outcomes are: {2, 3, . . . , 12} • Determining the number of days that exceed 20°C during March next year, the possible outcomes are: {0, 1, . . . , 31} • A lemonade trolley is selling lemonade for $3.50. Let M represent the change in the amount of money in someone’s pocket. A customer may purchase a lemonade for $3.50 or make no purchase, so the set of possible outcomes for M is –$3.50, $0. Note that discrete random variables do not always represent counts.

1162 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 The quality control manager of the installation of a fibre-optic network is monitoring the faults found in the cable being used: a Can the metres between successive faults in the fibre-optic cable being analysed be modelled by a discrete random variable?

Apply the idea No, it cannot be represented by a discrete random variable.

b What is the reason this cannot be represented by a discrete random variable?

Create a strategy

Apply the idea

Identify the nature of the outcomes in terms of continuity or discreteness.

The possible outcomes are continuous as the distance can take any value within a range and therefore, are not discrete.

c Can the number of faults found in a randomly chosen m length of the fibre-optic cable be modelled by a discrete random variable?

Create a strategy

Apply the idea

Determine if the number of faults is countable and therefore, can be discrete.

The number of faults is countable and can take integer values, making it suitable to be modelled by a discrete random variable.

Example 2 A player is interested in the number of heads obtained when flipping a fair coin 5 times: a What are the possible outcomes of this situation?

Create a strategy

Apply the idea

Determine all possible numbers of heads that can be obtained.

The possible number of heads that can be obtained when flipping a fair coin 5 times is 0, 1, 2, 3, 4, and 5.

b Can this situation be modelled by a discrete random variable?

Create a strategy Check whether the properties of the discrete random variable are satisfied.

12.01 Discrete random variables 1163 mathspace.co

Apply the idea

Reflect and check

Yes, this situation can be modelled by a DRV because it meets the required criteria: the outcomes are discrete countable values (0, 1, 2, 3, 4, 5), which occur randomly, and the results vary.

Continuous random variables represent outcomes that can take any value within a given range, such as the weights or heights of people. In contrast, this situation is modelled by a discrete random variable, as it involves counting whole numbers — the number of heads obtained when flipping a coin 5 times.

Idea summary Random variables are variables whose possible values are numerical outcomes of a random phenomenon. These variables can be classified into two main categories: continuous and discrete. • Continuous random variable: This type of random variable can take on any real value within a given range. • Discrete random variables (DRV): The values are discrete, countable, occur at random and vary. 1. Discrete values: The outcomes of a random variable are a countable number of distinct possible values. 2. Random occurrence: The outcomes occur randomly within the context of the experiment or situation. 3. Varied outcomes: There are multiple possible outcomes, ensuring that the results vary.

Probability function for a DRV For a discrete random variable DRV, the probability function assigns a probability to each possible value that the random variable can take. If X is a discrete random variable, the probability function is p(x) = P (X = x) where p(x) is the probability that X takes the value x. Properties of a probability function for a DRV: • The probability p(x) must be between 0 and 1 for each value: 0 ≤ p(x) ≤ 1 • The sum of the probabilities for all possible values of x must equal 1 : p ( x)

For example, when a fair six-sided die is rolled, the probability function can be written as: P(X = x) =

1 for x = 1, 2, 3, 4, 5, 6 6

1164 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 3 Morning Oats announced a new promotion: there is a 3% chance of winning a holiday trip with every box of oats purchased. Let X represent the number of purchases that wins. Find the probability that Joanne wins if she buys 2 boxes: a Illustrate the 4 possible outcomes of Joanne’s purchase.

Apply the idea Outcome

Value of X

b Calculate the probability of each value of X.

Apply the idea Value of X Probability

0.97 × 0.97 = 0.9409 (0.97 × 0.03) + (0.03 × 0.97) = 0.0582 0.03 × 0.03 = 0.0009

c Find the probability that Joanne wins at least 1 trip.

Apply the idea P(X ≥ 1) = 1 – P(X = 0) = 1 – 0.9409 = 0.0591

Example 4 Consider the function P(X = x) =

x for x = 1, 3: 4

a Complete the given table of values:

Create a strategy

P(X = x)

⬚

Calculate the probability P(X = x) for x = 1 and x = 3 and complete the table by organising the values and the probabilities.

12.01 Discrete random variables 1165 mathspace.co

Apply the idea For x = 1: P(X = 1) =

1 4

P(X = 3) =

3 4

P (X = x)

1 4

3 4

For x = 3:

Therefore, the table is as follows:

b Identify whether the table represents a discrete random variable.

Create a strategy Check whether the properties of a probability function for a DRV are satisfied.

Apply the idea Yes, the table represents a discrete random variable because X is discrete numerical: p( x ) x

p ( 1)

p( 3)

1 4

3 4

1, and 0 ≤ p(x) ≤ 1

Idea summary The probability function of a DRV assigns a probability to each possible value that the random variable can take. If X is a discrete random variable, the probability function is p(x) = P(X = x) where p(x) is the probability that X takes the value x. Properties of a probability function for a DRV: • The probability p(x) must be between 0 and 1 for each value: 0 ≤ p(x) ≤ 1 • The sum of the probabilities for all possible values of x must equal 1: p ( x) x

1166 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Discrete probability distribution of a DRV The discrete probability distribution of a discrete random variable (DRV) provides a complete description of all possible values the random variable can take along with the associated probabilities. It can be presented in various forms, such as a table, graph, or function. A common way to represent a probability distribution is through a table listing all possible outcomes and the corresponding probabilities. If X is a discrete random variable and P(X = x) = p(x) represents the probability that X takes the value x, the probability distribution for X must adhere to these conditions: 1. All possible outcomes of x are numerical and discrete. 2. The probability p(x) must be between 0 and 1 for each value x: 0 ≤ p(x) ≤ 1 3. The sum of the probabilities for all possible values of x must be 1:

p( x )

For the first condition, “numerical” refers to outcomes that are represented by numbers, while “discrete” specifically refers to the nature of these numbers being distinct and countable. For example, consider a scenario where a coin is tossed twice, with the focus being on the number of tails that appear: • A DRV can be defined as “Let X be the number of tails in a two-coin toss.” • The possible values of X are 0, 1, or 2. • The probability distribution for X shows the probabilities for each outcome. A clear way to calculate the probabilities for this example is to use a tree diagram: Tails 1 2 1 2 Heads

1 2 1 2

Tails

1 2 1 2

Tails

Heads TH HT

Heads HH

If X is the number of tails in a two-coin toss, p(x) = P(X = x) represents the probability that outcome x occurs: 1 • x = 0 means that the number of tails in the two-coin toss is 0 (HH), P(X = 0) = 4 • x = 1 means that the number of tails in the two-coin toss is 1 (HT or TH), P(X = 1) = • x = 2 means that the number of tails in the two-coin toss is 2 (TT), P(X = 2) =

2 4

1 4

12.01 Discrete random variables 1167 mathspace.co

A discrete probability distribution can be created by representing all the possible values of x and the corresponding probabilities p(x) in a table: x

P(X = x)

1 4

2 4

1 4

The table consists of all the outcomes and corresponding probabilities. Other than the probability function and the table of values explored previously, additional ways to represent a probability distribution for a DRV include: 1. A graph, such as a column graph, with outcomes on the horizontal axis and height of the bars reflecting the probability: 1

P (X = x)

3 4 1 2 1 4 0

x 0

2. A cumulative probability function, F(x), is defined as the probability that the random variable X will take a value less than or equal to x. It is written as F(x) = P(X ≤ x). 3. A cumulative probability table of F(x) = P(X ≤ x) represents the probability of the outcome being less than or equal to x. For example: x

P(X ≤ x)

1 4

3 4

4 4

P(X ≤ 1) is also equivalent to 1 − P(X > 1) = 1 − P(X = 2) = 1 −

1 3 4 4

P(X ≤ 1) = P(X = 0) + P(X = 1) =

1 4

2 4

3 4

Example 5 A cat is about to give birth to three kittens. The exact combination of male and female kittens remains unknown until birth: a Represent the possibilities using a tree diagram.

Create a strategy Construct a tree diagram for the outcomes of three kittens with each kitten having an equal possibility of being male or female.

1168 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea Here is the tree diagram representing the possibilities:

Female

FFF

Male

FFM

Female

FMF

Male

FMM

Female

MFF

Male

MFM

Female

MMF

Male

MMM

Female

Male Male

b If X is defined as the number of female kittens born, determine the possible values of X.

Create a strategy

Apply the idea

Count the possible number of female kittens born.

The possible values of X (the number of female kittens) are: X = 0, 1, 2, 3

c Is X a discrete random variable?

Create a strategy

Apply the idea

Check if the values of X are discrete, occur at random, and are countable.

Based on the answer in part (b), the values of X are 0, 1, 2, and 3 represent the number of female kittens born. These values are discrete, occur randomly, and take distinct values.

d Determine the discrete probability distribution of X.

Create a strategy Each value of X has a particular probability of occurring. Calculate the probability of each value of X and summarise these probabilities in a table.

Apply the idea

Reflect and check

Using the tree diagram from part (a):

Sum the probabilities to verify that the total is 1:

P (X = x)

1 8

3 8

1 8

1 3 3 1 =1 8 8 8 8

12.01 Discrete random variables 1169 mathspace.co

e Represent the situation with a cumulative probability table.

Create a strategy Create a table that represents the cumulative probabilities for each possible outcome.

Apply the idea x

P(X ≤ x)

1 8

4 8

7 8

8 8

Example 6 A fair six-sided die is rolled 100 times. The outcomes are recorded: • Number of times 1 appeared: 18 • Number of times 2 appeared: 15 • Number of times 3 appeared: 17 • Number of times 4 appeared: 20 • Number of times 5 appeared: 14 • Number of times 6 appeared: 16 a If the discrete random variable X is defined as the number that appears on the topmost face, determine all possible values of X.

Create a strategy Find all possible numbers that can appear.

Apply the idea The possible values of X (number that appears on the topmost face) are: X = 1, 2, 3, 4, 5, 6

b Determine the relative frequency of each outcome and represent it using a relative frequency column graph.

Create a strategy Use relative frequencies obtained from data to determine point estimates of probabilities associated with the discrete random variable, X.

1170 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea 1. Calculate the relative frequency of each outcome by dividing the number of times each outcome appears by the total number of rolls, then represent these frequencies using a bar chart. 18 2. Relative frequency of 1 = 100 = 0.18 Relative frequency of 2 =

15 100

= 0.15 Relative frequency of 3 =

17 100

= 0.17 Relative frequency of 4 =

20 100

= 0.20 Relative frequency of 5 =

14 100

= 0.14 Relative frequency of 6 =

16 100

= 0.16

Relative frequency

3. Use the calculated relative frequencies to create a column graph: Rolling a 6-sided die experiment

0.25 0.20 0.15 0.10 0.05 0.00

3 4 5 Outcome

Reflect and check Verify that the relative frequencies add up to 1: 0.18 + 0.15 + 0.17 + 0.20 + 0.14 + 0.16 = 1.00 The sum confirms that the relative frequencies are correct.

12.01 Discrete random variables 1171 mathspace.co

c Compare the experimental relative frequencies obtained in part (b) with the theoretical probabilities for a fair six-sided die. Discuss any observed differences.

Create a strategy Establish the theoretical probability for each outcome of a fair six-sided die. Then, tabulate these alongside the experimental relative frequencies. Comment on the discrepancies, considering the nature of experimental data.

Apply the idea For a fair six-sided die, each of the six outcomes is equally likely. The theoretical probability for 1 each outcome x is P (X = x) = ≈ 0.1667 6 The comparison between theoretical probabilities and experimental relative frequencies is as follows: Outcome (x)

Theoretical probability P(X = x)

Experimental relative frequency

1 0.167 6

0.18

1 0.167 6

0.15

1 0.167 6

0.17

1 0.167 6

0.20

1 0.167 6

0.14

1 0.167 6

0.16

The experimental relative frequencies are generally close to the theoretical probability of approximately 0.167. Some outcomes, like ‘4’ (relative frequency 0.20), occurred more often than theoretically expected, while others, like ‘5’ (relative frequency 0.14), occurred less often. These differences are expected due to random variation (or sampling variability), which is inherent in probability experiments. With a finite number of trials (in this case, 100 rolls), the experimental results will rarely match the theoretical model perfectly. According to the law of large numbers, if the number of rolls were significantly increased, the experimental relative frequencies would be expected to converge more closely to the theoretical probabilities.

1172 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reflect and check This comparison illustrates that experimental probabilities (relative frequencies) are estimates of theoretical probabilities: • Differences between experimental and theoretical values are normal due to random variation, especially with a limited number of trials. • Increasing the number of trials typically leads to experimental frequencies that more closely align with theoretical probabilities, reflecting the law of large numbers. This distinction is vital for interpreting results from real-world experiments and simulations.

Idea summary The discrete probability distribution of a DRV provides a complete description of all possible values the random variable can take along with their associated probabilities. It can be presented in various forms, such as a table, graph, or a function. If X is a discrete random variable and p(x) = P(X = x) represents the probability that X takes the value x, the probability distribution for X must adhere to these conditions: 1. All possible outcomes x are numerical and discrete. 2. The probability p(x) must be between 0 and 1 for each value x: 0 ≤ p(x) ≤ 1 3. The sum of the probabilities for all possible values of x must equal 1: p( x )

There are a number of ways to represent a probability distribution: 1. A probability function, which is a mathematical function that gives the probability for each specific outcome. 2. A table of values created by representing all the possible values x and the corresponding probabilities p(x) in a table. 3. A graph such as a column graph or spike graph with outcomes on the horizontal axis and height of the bars reflecting the probability. 4. A cumulative probability table where P(X ≤ x) represents the probability of the outcome being less than or equal to x. The relative frequency can be found by collecting data on the various outcomes of the random variable and calculating the number of times the outcome occurs divided by the total number of observations. This process is repeated to estimate the probability of each possible outcome, these values are known as point estimates.

12.01 Discrete random variables 1173 mathspace.co

Did you know?

A train turnstile counts passenger entries as discrete random variables, creating a clear record of peak and off peak times. Transport planners use this data to improve timetables and passenger flow.

Types of discrete probability distributions Two types of discrete distributions are fundamental in probability theory: • Uniform discrete distributions refer to situations where all outcomes are equally likely. This means that each outcome has the same probability of occurring. For example, when rolling a fair six-sided die, each face (1 through 6) has an equal probability 1 of . 6 If X is a uniform discrete random variable taking values x1, x2, . . . , xn:

P ( X xi )

1 n

i = 1, 2, . . . , n n is the number of possible outcomes • Non-uniform discrete distributions refer to situations where outcomes have varying probabilities. Unlike uniform distributions, where each outcome has an equal chance of occurring, non-uniform distributions reflect varying likelihoods among different outcomes. For example, a coin may be weighted to favour one side over the other (biased coin), resulting in a non-uniform distribution of outcomes (e.g. 70% heads, 30% tails). Computing probabilities in a non-uniform distribution involves using the assigned probabilities for each outcome.

1174 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 7 Identify the situations that can be modelled using uniform discrete distributions: A Rolling a fair six-sided die B Flipping a biased coin C Selecting a student according to their mathematics grade D Flipping a fair coin E Choosing a card from a well-shuffled standard deck of 52 cards

Create a strategy Identify the situations where all outcomes are equally likely.

Apply the idea Analyse each situation to determine if all outcomes are equally likely: • A. Rolling a fair six-sided die: This is a uniform distribution because each of the six faces 1 has an equal probability of landing face up P ( x ) . 6 • B. Flipping a biased coin: This is a non-uniform distribution because a biased coin implies the probabilities of heads and tails are not equal. • C. Selecting a student according to their mathematics grade: This is a non-uniform distribution, as it is unlikely that each grade has an equal number of students, making the probability of selecting a student with a certain grade unequal across different grades. • D. Flipping a fair coin: This is a uniform distribution because heads and tails each have an 1 equal probability of occurring P (Heads) P ( Tails ) . 2 • E. Choosing a card from a well-shuffled standard deck of 52 cards: This is a uniform distribution because each of the 52 cards has an equal probability of being selected 1 . P (Card) 52 Therefore, the situations that can be modelled by a uniform discrete distribution are Options A, D, and E.

Example 8 Digital only example Explore in the online lesson

mathspace.co

12.01 Discrete random variables 1175 mathspace.co

Example 9 A biased spinner has three sections with the following probabilities: P(Red) = 0.6, P(Blue) = 0.3, and P(Green) = 0.1. The spinner is spun twice. A random variable Y is defined as the number of times the spinner lands on Red: a Verify that this is a valid probability distribution for a single spin.

Create a strategy A valid probability distribution must satisfy two conditions: each probability must be between 0 and 1, and the sum of all probabilities must equal 1.

Apply the idea • First, check if each probability is between 0 and 1. • P(Red) = 0.6, which is between 0 and 1. • P(Blue) = 0.3, which is between 0 and 1. • P(Green) = 0.1, which is between 0 and 1. • All individual probabilities are valid. Next, check if the sum of probabilities is 1. Sum of probabilities = P(Red) + P(Blue) + P(Green) = 0.6 + 0.3 + 0.1

Write the equation Substitute P(Red) = 0.6, P(Blue) = 0.3 and P(Green) = 0.1

=1 Evaluate • The sum is 1, so this is a valid probability distribution for a single spin.

b Construct the probability distribution table for the random variable Y, the number of times the spinner lands on Red in two spins.

Create a strategy The random variable Y can take the values 0, 1, or 2. Calculate the probability for each value. It is helpful to first find the probability of the spinner not landing on Red. Then, use the multiplication rule for independent events to find the probabilities for Y = 0 (Not Red, Not Red), Y = 2 (Red, Red), and Y = 1 (Red then Not Red, or Not Red then Red).

1176 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea • Let R be the event of landing on Red, and N be the event of not landing on Red. First, find the probability of not landing on Red, P(N). P(N) = 1 – P(R)

Write the equation

= 1 – 0.6

Substitute P(R) = 0.6

= 0.4

Evaluate

• Y = 2: This corresponds to the outcome RR (Red on both spins). P(Y = 2) = P(R) × P(R)

Write the equation

= 0.6 × 0.6

Substitute P(R) = 0.6

= 0.36

Evaluate

• Y = 1: This corresponds to the outcomes RN or NR (Red on one spin). P(Y = 1) = P(RN) + P(NR)

Write the equation

= (P(R) × P(N)) + (P(N) × P(R)) Substitute P(RN) = P(R) × P(N) and P(NR) = P(N) × P(R) = (0.6 × 0.4) + (0.4 × 0.6)

Substitute P(R) = 0.6 and P(N) = 0.4

= 0.24 + 0.24

Evaluate the multiplication

= 0.48 Evaluate • Y = 0: This corresponds to the outcome NN (Not Red on both spins). P(Y = 0) = P(N) × P(N)

Write the equation

= 0.4 × 0.4

Substitute P(N) = 0.4

= 0.16 Evaluate • The probability distribution for Y is shown in the table. This is a non-uniform discrete probability distribution since the probabilities for the different values of Y are not equal. y

P (Y = y)

0.16

0.48

0.36

Reflect and check Check that the sum of the probabilities for Y is 1. ∑ P(Y = y) = P(Y = 0) + P(Y = 1) + P(Y = 2)

Write the equation

= 0.16 + 0.48 + 0.36 Substitute P(Y = 0) = 0.16, P(Y = 1) = 0.48 and P(Y = 2) = 0.36 = 1.00

Evaluate

The sum is 1, which confirms the distribution is correct.

12.01 Discrete random variables 1177 mathspace.co

Idea summary Two types of discrete distributions are fundamental in probability theory: • Uniform discrete distributions refer to situations where all outcomes are equally likely. If X is a uniform discrete random variable taking values x1, x2, . . . , xn:

P ( X xi ) i n

1 n

= 1, 2, . . . , n is the number of possible outcomes

• Non-uniform discrete distributions refer to situations where outcomes have varying probabilities. Unlike uniform distributions, where each outcome has an equal chance of occurring, non-uniform distributions reflect varying likelihoods among different outcomes. Computing probabilities in a non-uniform distribution involves using the assigned probabilities for each outcome.

12.01 Practice questions What do you remember? 1

Determine whether the random variable X is discrete or continuous: a

X is a random variable for the length of a crocodile.

X is a random variable for the number of games won by a football team in a season.

X is a random variable for the time taken by a driver to complete one lap of a circuit.

Does the table represent a valid discrete probability distribution? Explain why or why not. x

P(X = x)

10%

20%

25%

45%

1178 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

For each column graph: i

Identify which conditions for a discrete probability distribution are evident in the graph.

Determine whether the graph represents a discrete probability distribution.

P (x) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 2 3 4

x 5

P (x) 0.4

0.3 0.2 0.1 0

x 1

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

P (x)

x 1

2 3 4 5 6 7 8

P (x)

x 1

Practice Ex 1

5 Ex 2

6 Ex 3

A photocopier in a busy school breaks on a regular basis: a

Can the number of breakdowns occurring in a fortnight be represented by a discrete random variable?

Can the time between breakdowns be represented by a discrete random variable? Explain your answer.

A coin is tossed three times and the total number of tails is recorded: a

Can this experiment be represented by a discrete random variable?

If X represents the number of tails in the three coin tosses, list all possible values of X.

An archer has a 70% chance of hitting the bullseye with each arrow. Let X represent the number of bullseyes she hits in two shots: a

Calculate the probability that she hits exactly one bullseye.

Calculate the probability that she hits at least one bullseye.

12.01 Discrete random variables 1179 mathspace.co

7 Ex 4

8 Ex 5

9 Ex 6

x for x = 1, 2, 3: 6 Present this probability distribution in a table.

Verify that this is a valid discrete probability distribution.

Consider the function P(X = x) =

A dog has four puppies. Let F represent the number of female puppies in this litter: a

Construct the probability distribution table for F, assuming the probability of a puppy being male or female is equal.

Construct the cumulative probability table for F.

A standard fair six-sided die is rolled 6000 times: a

List the possible outcomes of each roll.

What is the expected shape of the distribution resulting from the experiment? Explain your answer.

The results of the experiment are tabulated: x

Frequency

996

1005

994

1005

996

1004

Complete the probability distribution table. Express the probabilities as fractions with a denominator of 6000:

P (X = x)

⬚

Using the results, calculate the experimental probability P(X < 3).

Compare the experimental and theoretical probabilities for P(X < 3) by finding the percentage error.

The number of fruit in boxes delivered to an office was sampled. The frequency of each observation is recorded in this table: Number of fruit

Frequency

Let X be the number of fruit in a box. Construct a probability distribution table for X using the observations.

Use the experimental probabilities to estimate: i

The probability that the next box delivered contains at least 40 fruit

The probability that the next box delivered contains less than 41 fruit

1180 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The probability function for a uniform discrete random variable is shown in this graph: P (x)

2 3 4

5 6

Calculate the value of k. 12

Given the probability distribution table for X, determine the value of k. x

P(X = x)

A random variable X has the probability distribution: x

P (X = x)

1 16

3 16

1 4

5 16

1 8

Determine these probabilities:

14 Ex 7

P(X = 8)

P(X is even)

P(X > 8)

P(X ≤ 7)

P(6 < X < 8)

P(6 ≤ X < 12)

For each scenario, determine if the described discrete random variable would most likely have a uniform or a non-uniform probability distribution: a

The random variable X represents the outcome of rolling a standard fair six-sided die once.

The random variable Y represents the number of heads obtained when flipping two fair coins.

The random variable Z represents the colour of a randomly selected traffic light signal (red, yellow, green) at a typical intersection, assuming the cycle times for each colour are different.

The random variable W represents the suit (Heart, Diamond, Club, Spade) of a single card drawn from a standard, well-shuffled 52-card deck.

The random variable K represents the number of defective items found when inspecting a batch of 5 items, where the probability of an item being defective is very low but not zero.

12.01 Discrete random variables 1181 mathspace.co

15 Ex 8

16 Ex 9

The probability that a new smartphone has a defective battery is 4%. A quality control inspector randomly selects two phones from the production line for testing: a

Let X be the number of phones with a defective battery in the sample of two. Construct the probability distribution table for X.

Determine the probability that at most one phone has a defective battery.

Is this probability distribution uniform or non-uniform? Explain your answer.

The probabilities of an archer hitting different sections of a target on a single shot are given: P(bullseye) = 0.2, P(outer ring) = 0.5, and P(miss) = 0.3. The archer takes two independent shots: a

Verify that the probabilities for a single shot form a valid probability distribution.

Let X be the number of bullseyes the archer hits in two shots. Construct the probability distribution table for X.

Determine the probability that the archer hits at least one bullseye.

Extend your thinking 17

The cumulative distribution for a discrete random variable is given:

P(X ≤ x)

0.1

0.3

0.65

0.95

P(X = x)

⬚

Complete the given table.

Calculate these probabilities: i

P (1 < X ≤ 3)

P (X ≤ 3∣X > 1)∣

A six-sided die is weighted such that the probability of landing on a number is proportional to that number. Let X be the number that appears uppermost: a

Determine the probability distribution of X.

Calculate the probability that the number rolled is greater than 4.

The probability of being placed in a queue when calling a service centre is given by P(X = x) = k × (0.2)x for x = 0, 1, 2, 3, . . . , where x is the number of customers in the queue ahead of the caller: a

Given that this is a valid probability distribution, the sum of this infinite geometric series k is k (0.2) x 1. Determine the value of k. x 0 1 0.2 Calculate the probability of not being placed in a queue.

1182 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

12.02 Centre and spread of discrete random variables After this lesson, you will be able to… • calculate the mean, variance, and standard deviation of discrete random variables • interpret the centre and spread of discrete random variable distributions • apply the mean and standard deviation to solve basic probability problems

Expected value or mean When analysing the outcomes and probabilities of a discrete random variable, the focus is on measures of central tendency and dispersion. Measures of central tendency include the mean, median, and mode. Mean The average of a probability distribution, calculated as a weighted sum of outcomes by their probabilities, also called the expected value. Expected value The average outcome expected from a discrete random variable over many trials, also called the mean. It is denoted E(X) or μ.

For a discrete random variable X with outcomes xi = {x1, x2, . . ., xn} and associated probabilities pi = P(X = xi), the formula for the expected value or mean is:

E( X ) xi pi i

xi is the outcome of the discrete random variable pi is the associated probability The expected value is not the most likely outcome. The most likely outcome is the mode, which has the highest probability.

12.02 Centre and spread of discrete random variables 1183 mathspace.co

Example 1 Consider the probability distribution of the discrete random variable given in this table: x

P(X = x)

0.2

0.15

0.4

0.1

0.15

a Calculate the expected value of the distribution, E(X).

Create a strategy To calculate the expected value, multiply each possible outcome by its probability and then sum the products.

Apply the idea E (X) =

pi xi

Write the formula

= (0.2 × 1) + (0.15 × 2) + (0.4 × 3) + (0.1 × 4) + (0.15 × 5)

Substitute the values

= 2.85

Evaluate

b Interpret the value obtained.

Create a strategy Explain the significance of the value 2.85 within the given context.

Apply the idea In the given context, it is expected that the long-term average of all outcomes is 2.85.

Reflect and check Clearly, an outcome of 2.85 is not possible since all outcomes are integers, as listed in the table. This is similar to how the average number of children per family can be a decimal, even though a family cannot have a fraction of a child. The mean value gives an idea of what is expected on average. The expected value is not to be confused with the most likely outcome. In the distribution mentioned, the most likely outcome is 3 since P (X = 3) has the highest probability of 0.4. This is also the mode in the distribution. The mode can be used to check the reasonableness of the expected value. The value 2.85 is between 2 and 3 and closer to 3, which seems reasonable as the highest probability is for X = 3 and the probability of X < 3 is greater than the probability of X > 3, so the expected value is slightly less than 3.

1184 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary The expected value or mean of a distribution, is a measure of central tendency, distinct from the mode, is calculated as a weighted average: E(X)

pi xi

where E(X) or μ is the expected value or mean of the random variable X, and pi is the probability of xi, the outcome of the discrete random variable

Variance and standard deviation The variance and standard deviation measure the spread of outcomes in a discrete random variable’s probability distribution, indicating how much outcomes vary from the expected value. Variance is calculated similarly to data variance, incorporating probabilities as weights. Two equivalent formulas exist for variance:

Var ( X ) = xi E ( X ) pi i

pi ( xi

xi is the outcome of the discrete random variable, X pi is the associated probability E (X) is the expected value or mean µ is the mean (expected value) of X, calculated as pi xi E(X)

Var ( X ) ( x i2 pi ) [E( X )]

xi is the outcomes of the discrete random variable pi is the associated probabilities E (X) is the expected value or mean The standard deviation, denoted σ, is the square root of the variance: ( X )

Var ( X )

12.02 Centre and spread of discrete random variables 1185 mathspace.co

Example 2 The probability distribution for a discrete random variable is given in this table: x

P( X = x)

0.2

0.15

0.4

0.1

0.15

Calculate the variance and standard deviation of the distribution, given the expected value E(X) = 2.85.

Create a strategy Use the variance formula Var ( X ) =

( xi2 pi ) [E( X )] 2 to calculate the variance, then find the

standard deviation as σ ( X ) = Var ( X ).

Apply the idea Var (X) = i

( xi2 pi )

[E( X )]

rite the variance W formula

= (12 × 0.2 + 22 × 0.15 + 32 × 0.4 + 42 × 0.1 + 52 × 0.15) – 2.852 Substitute the outcomes, probabilities, and expected value = 1.6275 Evaluate

σ (X ) =

Var ( X )

rite the standard W deviation formula

= 1.6275 Substitute the values ≈ 1.2757 Evaluate the square root The variance is 1.6275, and the standard deviation is approximately 1.2757.

1186 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 3 A carnival game costs $5 to play. A player spins a wheel with four possible outcomes for a cash prize. The probability of winning each prize is given in this table: Prize

Probability

$20

0.05

$10

0.15

0.30

0.50

a Calculate the expected net profit for a player. Should a player expect to make money in the long run?

Create a strategy First, define a discrete random variable, X, for the player’s net profit. The net profit is the prize money won minus the cost to play. Then, create a probability distribution table for X. Finally, calculate the expected value, E(X), and interpret its meaning.

Apply the idea The cost to play is $5. Let X be the net profit. • If the prize is $20, the net profit is $20 – $5 = $15. • If the prize is $10, the net profit is $10 – $5 = $5. • If the prize is $5, the net profit is $5 – $5 = $0. • If the prize is $0, the net profit is $0 – $5 = –$5. The probability distribution for the net profit X is: x (Net profit)

–5

P(X = x)

0.05

0.15

0.30

0.50

Now, calculate the expected value E(X): E(X) = (15 × 0.05) + (5 × 0.15) + (0 × 0.30) + (–5 × 0.50) Multiply each net profit by its probability and sum = 0.75 + 0.75 + 0 – 2.50 Evaluate the multiplication = –1.00

Evaluate

The expected net profit is –$1.00. This means that, on average, a player can expect to lose $1.00 each time they play the game. In the long run, a player should not expect to make money.

12.02 Centre and spread of discrete random variables 1187 mathspace.co

b Calculate the standard deviation of the net profit. What does this value indicate about the game?

Create a strategy First, calculate the variance, Var (X), using the formula Var (X) = i( xi2 pi ) [E( X )] 2. Then, find the standard deviation by taking the square root of the variance. The standard deviation will measure the risk or volatility of the game.

Apply the idea Given that E(X) = –1.00, the first step is to calculate

( xi2 pi ).

( xi2 pi ) = (152 × 0.05) + (52 × 0.15) Substitute the outcomes and probabilities i + (02 × 0.30) + (( – 5)2 × 0.50) = (225 × 0.05) + (25 × 0.15) + (0 × 0.30) + (25 × 0.50)

Evaluate the squares

= 11.25 + 3.75 + 0 + 12.5

Evaluate the multiplications

= 27.5

Evaluate

Now, calculate the variance: Var (X) = i

( xi2 pi ) [E( X )]2

Write the formula for variance

= 27.5 – (–1.00)2

Substitute the values

= 27.5 – 1

Evaluate the square

= 26.5

Evaluate

Finally, calculate the standard deviation:

σ (X) =

Var ( X )

Write the formula

= 26.5

Substitute the values

≈ 5.15

Evaluate

The standard deviation of the net profit is approximately $5.15. A higher standard deviation indicates a greater spread of outcomes. This means the game is quite risky or volatile; while the average loss is $1, the actual outcome for a single game can vary significantly from this average.

Reflect and check A “fair game” is one where the expected net profit is $0. Since the expected value here is –$1.00, the game is biased in favour of the carnival. The standard deviation gives a measure of the risk involved. A player might win $15 or lose $5 in a single game, which are both far from the average loss of $1.

1188 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary The variance of a distribution, denoted as Var (X), is a measure of the spread of the outcomes in a probability distribution: Var ( X ) i

( xi2 pi ) [E( X )]2 or

Var ( X )

E X

The standard deviation, denoted as σ, is calculated by extracting the square root of the variance: (X)

Var ( X )

12.02 Practice questions What do you remember? 1

Consider this table: x

P(X = x)

0.1

0.45

0.15

0.05

0.25

Determine whether the table represents a discrete probability distribution. Explain your answer.

Determine the most likely outcome for the random variable X.

Determine the expected value, E(X).

For each of these probability distribution tables for the random variable X: i a

Determine the most likely outcome.

Determine the expected value, E(X).

–1

P(X = x)

0.05

0.1

0.25

0.6

–6

–4

–2

P(X = x)

2 32

4 32

6 32

8 32

6 32

4 32

2 32

P(X = x)

0.04

0.03

0.07

0.35

0.11

0.08

0.19

0.13

12.02 Centre and spread of discrete random variables 1189 mathspace.co

X is a uniform discrete random variable that takes on the values 1, 2, 3 or 4: a

b 4

Complete the probability distribution table for X. x

P(X = x)

⬚

Determine E(X).

Consider the probability function defined by: P( X a

b 5

x , x 3, 4, 5, 6, 7 25 0, Otherwise

Complete the probability distribution table for X. x

P(X = x)

⬚

Calculate the expected value of the distribution.

For each of the probability distribution graphs: i

List the possible outcomes of X.

Classify the distribution as uniform or non-uniform.

iii

Calculate the expected value of the distribution.

P(x) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 2 3 4

2 5

P(X = x)

1 5

x 5

x 1

Practice 6

Let X be a random variable with this probability distribution table: x P(X = x)

2 0.2

3 0.6p

Determine the possible values of p.

For each possible value of p, determine: i

E(X)

1190 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

0.1

1–p

0.1

P(X ≥ E(X))

Consider this probability distribution table for X: 3

P(X = x)

0.1

0.08

0.2

0.07

0.55

Determine E(X).

Determine E(X2).

Determine E(2X + 5). 1 Determine E , rounded to two decimal places. X Determine Var (X).

This table represents a discrete probability distribution: x

P(X = x)

0.05

0.1

0.15

Use a property of probability distributions to express m in terms of n.

Use the fact that E(X) = 3.1 to express m in terms of n.

Solve for n.

Solve for m.

Calculate the standard deviation, rounded one decimal place.

Consider these discrete probability distribution graphs: Distribution A P(x) 0.5 0.5

0.4

0.3

0.2

0.1

x 1

3 4

5 6

P(x)

Distribution B

x 1

3 4

5 6

Calculate the expected value for each distribution.

Describe the shape of each distribution.

For each distribution, determine whether the median is greater than, less than, or equal to the mean.

Calculate the variance of distribution A.

Calculate the standard deviation of distribution A.

Determine if the standard deviation of distribution B is likely greater than, less than, or equal to that of distribution A. Explain your answer.

12.02 Centre and spread of discrete random variables 1191 mathspace.co

Ex 1

Consider this probability distribution table for X:

a Ex 2

P(X = x)

0.14

0.24

0.13

0.23

0.13

Determine E(X).

Determine Var (X).

Consider the table: x

P(X = x)

0.12

0.15

0.22

0.23

0.28

Explain why the table represents a discrete probability distribution.

Determine E(X).

Determine the variance of X.

Determine the standard deviation, rounded two decimal places.

Consider this probability distribution table for X: a

Determine the value of k.

Determine E(X).

Determine Var (X).

P(X = x)

1 12

1 2

1 20

1 5

For each of these probability distribution tables: i

Determine the value of k.

Determine the mode.

iii

Determine the expected value.

Determine the variance.

Determine the standard deviation, rounded to two decimal places.

P(X = x)

0.15

0.2

0.1

0.3

P(X = x)

For each of these probability functions described: i

Construct the probability distribution table for X.

Determine the expected value of the distribution.

iii

Determine the standard deviation of the distribution. Round your answer to two decimal places where necessary.

P( X

P(X

x , x 1, 2, 3, 4 10 0, Otherwise

P(X

7 x , x 0, 1, 3, 5 19 0, Otherwise

x2 1 x 2, 3, 4, 5 , 50 0, Otherwise

P(X

2x 1 x 1, 2, 3, 4 , 16 0, Otherwise

1192 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

The Vaucluse Vigilantes and the Woollahra Weasels play a series of basketball matches and the first to win four games wins the series. This distribution shows the probabilities of the total number of games that will be played in the series: X

P(X = x)

0.11

0.34

0.33

0.22

Determine the expected number of games that will be played in the series. 16

A game uses a spinner with numbers from 1 to 12, each with an equal probability of occurring. A player wins $6 if the outcome is greater than 9, $4 if the outcome is less than 5 and loses $2 for any other outcome: a

Let X be the return from one game. Construct the probability distribution table for X.

Determine the expected return in dollars.

Calculate the exact standard deviation.

A standard deck of cards has cards numbered 8 through 16. A player draws a card at random. A player wins $3 if the card is odd and loses $3 if it’s even. Let X be the winnings of the player. Determine the expected return in dollars.

A fair die is rolled. If it lands on an odd number, then the score is the number. But if it lands on an even number, then the score is 0. Let X be the score when the die is rolled once: a

Determine E(X).

Determine Var (X).

A cat has a litter of three kittens. Each kitten is equally likely to be born male or female: a

Construct a tree diagram to represent all possible combinations of the gender of these three kittens.

Let M be the number of male kittens born to this cat. Construct the probability distribution table for M.

Determine the expected number of male kittens born.

Xanthe is sitting a multiple choice quiz consisting of 3 questions, each with 4 possible answers. Xanthe has not studied for the quiz so she will guess the answer to each question at random: a

Construct a probability tree diagram to represent all possible combinations of which questions Xanthe got right and wrong.

Let X be the number of correctly answered questions. Construct a probability distribution table for X.

Determine the most likely number of questions Xanthe guessed correctly.

Calculate how many questions Xanthe was expected to answer correctly by guessing.

12.02 Centre and spread of discrete random variables 1193 mathspace.co

Ex 3

To open a tutoring business, Quentin does some market research into the profit or loss of similar businesses in his area in the first year. His findings are summarised in this table shown: Earnings

Probability

–$1000

0.15

0.1

$3000

0.5

$5000

0.25

Does this table represent a discrete probability distribution? Explain your answer.

Calculate Quentin’s expected profit.

Calculate the variance in his expected profit.

Extend your thinking 22

Xavier tosses two coins. He wins $10 for two tails, $5 for two heads, and wins nothing if one head and one tail are tossed. If the game costs $5 to play, is Xavier likely to win or lose in the long run? Explain your answer.

Consider the discrete probability distribution graph:

0.5

Distribution A P (X = x)

0.4 0.3 0.2 0.1 0

x 1

Construct a graph of discrete probability distribution B that has a mean and median of 3, and has the same standard deviation as distribution A.

1194 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

12.03 Sums of independent random variables After this lesson, you will be able to… • calculate the mean and variance of linear combinations of random variables • apply properties of linear combinations to solve probability problems • determine the distribution of sums of independent random variables • calculate probabilities involving sums of discrete random variables

Sums of random variables Independent random variables Random variables where the outcome of one does not affect the other. Example: The results of two separate die rolls. For any two random variables X and Y, the expected value of the sum of random variables E(X + Y ), will be equal to the sum of their individual expected values E(X ) + E(Y ).

Interactive exploration Discover this concept in action online

mathspace.co

For two independent random variables X and Y that are identically distributed (same probability distribution), the distribution of the sum X + Y differs from doubling a single random variable, 2X.

Example 1 1 A fair six-sided die has outcomes {1, 2, 3, 4, 5, 6}, each with probability . Let X be the outcome 6 of one roll. Compare the probability distributions of 2X (doubling one roll) and X + Y (sum of two independent rolls).

Create a strategy List the possible values and probabilities for 2X by scaling the outcomes of X. For X + Y, calculate the probabilities of all possible sums of two dice rolls.

12.03 Sums of independent random variables 1195 mathspace.co

Apply the idea For 2X: 1 Multiply each outcome of X by 2: 2X = {2, 4, 6, 8, 10, 12}, each with probability . 6 For X + Y: The possible sums range from 1 + 1 = 2 to 6 + 6 = 12. Calculate probabilities by counting outcomes: Sum

Probability

1 36

2 36

3 36

4 36

5 36

6 36

5 36

4 36

3 36

2 36

1 36

The distribution of X + Y is triangular, peaking at 7, while 2X is uniform over even numbers.

Reflect and check Verify by checking total probabilities: For 2X, 6

1 6

For X + Y, sum the probabilities: 1 2 3 4 5 6 5 4 3 2 1 36

Idea summary The distribution of the sum of n independent random variables that are identically distributed differs from scaling a single random variable. For two variables, X + Y has a wider spread than 2X.

Mean and variance of linear combinations Linear combination A sum of random variables, each multiplied by a constant, e.g., aX + bY, where a and b are constants. Example: Total cost of 3 items at price X each: 3X. A linear combination of random variables combines them with constant coefficients. The mean and variance of such combinations follow specific properties.

E(aX + bY ) = aE(X ) + bE(Y ) E a, b

is the expected value (mean) of the linear combination are the constants multiplying the random variables

1196 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

For independent random variables X and Y, the variance of their linear combination is given by: Var (aX + bY ) = a2Var (X ) + b2Var (Y ) A special case of this is the variance of the difference between two independent random variables, X – Y. Since X – Y can be written as 1 X + (–1)Y, the formula becomes: Var (X − Y ) = 12Var (X ) + (−1)2Var (Y ) = Var (X ) + Var (Y ) It is important to note that the variances are still added. This is because variability increases regardless of whether the random variables are being added or subtracted.

Exploration Consider two random variables representing daily sales at two stores. Discuss how scaling one store’s sales by a factor (e.g., 2X ) or summing sales from both stores (X + Y ) affects the expected total and variability.

Linear combinations of random variables model real-world scenarios, such as total costs, weights, or scores, where multiple random quantities are summed or scaled. • Sum of identical items: The total weight of n packages, each with weight X, is X1 + X2 + . . . + Xn. • Combined outcomes: The sum of scores from multiple tests or games, e.g., X + Y for two dice rolls. These properties allow for the efficient calculation of the mean and variance of combined random events.

Example 2 Digital only example Explore in the online lesson

mathspace.co

12.03 Sums of independent random variables 1197 mathspace.co

Example 3 Let X and Y be independent random variables with these properties: • E(X ) = 10 and Var (X ) = 4 • E(Y ) = 3 and Var (Y ) = 2 a Calculate E(X + Y ) and Var (X + Y ).

Apply the idea For the mean of the sum: E(X + Y ) = E(X ) + E(Y )

Write the formula

= 10 + 3

Substitute the values

= 13

Evaluate

For the variance of the sum: Var (X + Y ) = Var (X ) + Var (Y )

Write the formula

=4+2

Substitute the values

Evaluate

b Calculate E(X – Y ) and Var (X – Y ).

Apply the idea For the mean of the difference: E(X – Y ) = E(X ) – E(Y )

Write the formula

= 10 – 3

Substitute the values

Evaluate

For the variance of the difference: Var (X – Y ) = Var (X ) + Var (Y )

Write the formula

=4+2

Substitute the values

Evaluate

Reflect and check Note that while the mean of the difference is the difference of the means, the variance of the difference is the sum of the variances. This is because the variability of both random variables contributes to the overall variability of their difference.

1198 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 4 A company ships packages, each with a weight X that is a random variable with mean E(X ) = 2 kg and variance Var (X ) = 0.5 kg2. Find the mean and variance of the total weight of 3 packages, assuming independence.

Create a strategy Model the total weight as X1 + X2 + X3. Use E(X1 + X2 + X3) = E(X1) + E(X2) + E(X3) and Var (X1 + X2 + X3) = Var (X1) + Var (X2) + Var (X3) for independent variables.

Apply the idea For the mean: E(X1 + X2 + X3) = E(X1) + E(X2) + E(X3)

Add the expected values

=2+2+2

Substitute E(X ) = 2 for each package

= 6 kg

Evaluate

For the variance: Var (X1 + X2 + X3) = Var (X1) + Var (X2) + Var (X3) = 0.5 + 0.5 + 0.5 2

= 1.5 kg

Add the variances for independent variables Substitute Var (X ) = 0.5 for each package Evaluate

The mean total weight is 6 kg and the variance is 1.5 kg2.

Example 5 Two fair six-sided dice are rolled independently, with outcomes X and Y, each uniformly distributed over {1, 2, 3, 4, 5, 6}. Calculate the probability that their sum X + Y exceeds 8.

Create a strategy List all possible sums X + Y greater than 8 (i.e., 9, 10, 11, 12). Count the number of outcomes and divide by the total number of outcomes (36).

12.03 Sums of independent random variables 1199 mathspace.co

Apply the idea Possible sums exceeding 8: • 9: Outcomes (3, 6), (4, 5), (5, 4), (6, 3) (4 outcomes) • 10: Outcomes (4, 6), (5, 5), (6, 4) (3 outcomes) • 11: Outcomes (5, 6), (6, 5) (2 outcomes) • 12: Outcome (6, 6) (1 outcome) Number of favourable outcomes = 4 + 3 + 2 + 1 = 10 Total outcomes = 36 Number of favourable outcomes Total number of possible outcomes

Write the probability formula

10 36

Substitute the values

5 18

Simplify

P(X + Y > 8) =

The probability is

5 . 18

Reflect and check Verify by checking the complementary event: P(X + Y ≤ 8) = 1 – P(X + Y > 8). Count outcomes for 26 13 13 5 , and 1 sums 2 to 8 (26 outcomes), so P(X + Y ≤ 8) , confirming the result. 36 18 18 18

Idea summary The mean of a linear combination aX + bY is aE(X ) + bE(Y ), and the variance, for independent variables, is a2Var (X ) + b2Var (Y ). Linear combinations of random variables model practical scenarios like total weights or scores. Their properties enable calculation of means, variances, and probabilities in real-world problems.

12.03 Practice questions What do you remember? 1

Define a linear combination of random variables and give an example.

Determine the formulas for the mean and variance of a linear combination aX + bY, where X and Y are independent random variables.

Explain why the distribution of X + Y (sum of two independent random variables) differs from 2X (doubling a single random variable) when X and Y are identically distributed.

1200 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice 4

Ex 1

A random variable X has mean E(X ) = 5 and variance Var (X ) = 3. Calculate the mean and variance of 2X + 10.

A random variable X has mean E(X ) = 10 and variance Var (X ) = 4. Calculate the mean and variance of –2X + 3.

The scores of two independent tests, Test 1 (X ) and Test 2 (Y ), are random variables with properties:

Ex 2

Ex 3

1 A fair four-sided die has outcomes {1, 2, 3, 4}, each with probability . Let X be the outcome 4 of one roll. Compare the possible values and probabilities of 2X and X + Y (sum of two independent rolls).

• E(X ) = 50, Var (X ) = 16 • E(Y ) = 65, Var (Y ) = 9 a

Calculate the mean and variance of the total score, X + Y.

Calculate the mean and variance of the difference in scores, X – Y.

A game involves rolling two fair six-sided dice, with scores X and Y. The total score is X + Y. Determine the mean and variance of the total score, given each die has mean, E(X ) = 3.5 35 and variance, Var (X ) = . 12

Ex 4

A store sells items whose weights are independent random variables with a mean of 1.5 kg and a variance of 0.2 kg2. Find the mean and variance of the total weight of 4 such items.

Ex 5

Two fair six-sided dice are rolled independently, with outcomes X and Y. Calculate the probability that their sum X + Y is at least 10.

Extend your thinking 11

A factory produces two types of items: • Items of Type 1 have a weight, X, where E(X ) = 100 g and Var (X ) = 16 g2 • Items of Type 2 have a weight, Y, where E(Y ) = 120 g and Var (Y ) = 25 g2 A special package is assembled containing two items of Type 1 and one item of Type 2. Assuming the weights of all items are independent, calculate the mean and variance of the package’s total weight. Then, explain how these values could help in planning the packaging process.

A game involves rolling three fair six-sided dice. The total score is the sum of the outcomes, 35 S = X + Y + Z. Given that a single die roll has a mean of 3.5 and a variance of , calculate 12 the mean and variance of the total score, S.

12.03 Sums of independent random variables 1201 mathspace.co

12.04 Applications of discrete random variables After this lesson, you will be able to… • create a distribution from a context • calculate and interpret the mean and variance of random variables • perform probability calculations given a context

Application of discrete probability distributions When modelling a situation using a discrete probability distribution, the experiment is used to construct the probability distribution, which is represented as a table or a function. For discrete random variables, tables are primarily constructed. To construct a discrete probability distribution: 1. Define what the random variable X represents. Typically, X is used for the random variable. 2. Identify the possible outcomes X can take. 3. Determine a method to calculate probabilities for each outcome, often using a sample space (e.g., a tree diagram or table) or counting techniques (e.g., combinations). 4. Calculate the probabilities and present them in a table with each outcome. After constructing a probability distribution, key measures can be calculated to describe it. The two most important measures are the mean and the variance. The expected value, or mean, of a discrete random variable is the long-run average value of the variable. It is denoted by E(X) or μ.

E ( X ) E(X) x P(X = x) ∑

x P ( X x )

is the expected value (or mean) of the random variable X a possible value of the random variable X is the probability that the random variable X takes the value x is the sum over all possible values of x

The variance measures the spread or dispersion of the outcomes around the mean. A small variance means the outcomes are clustered close to the expected value, while a large variance means they are more spread out. It is denoted by Var (X) or σ 2. Var (X) = E(X2) − [E(X)]2 The standard deviation is the square root of the variance, denoted by SD(X) or σ. It is often easier to interpret because it is in the same units as the random variable itself.

1202 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 1 A fair, standard die is rolled, and the number of dots on the uppermost face is noted. Let X be the number of dots on the uppermost face: a Construct the probability distribution for X.

Create a strategy Determine the possible outcomes for X and their probabilities. Since it is a standard die, each outcome is equally likely.

Apply the idea 1 The outcomes for X are 1, 2, 3, 4, 5, 6. Each has a probability of . 6 x

P (X = x)

1 6

b Calculate the expected value, E(X), and interpret its meaning.

Create a strategy Use the formula E(X) = ∑ x × P(X = x). Multiply each outcome by its probability and sum the results.

Apply the idea E(X) = 1

1 6

1 Substitute the values into the 6 formula

1 2 3 4 5 6 6

Simplify

21 6

Evaluate the addition

= 3.5

Evaluate

The expected value is 3.5. This means that if the die were rolled many times, the average of all the outcomes would be close to 3.5

12.04 Applications of discrete random variables 1203 mathspace.co

c Calculate the variance, Var (X).

Create a strategy First, calculate E(X2) using the formula E(X2) = ∑ x2 × P(X = x). Then, use the variance formula Var (X) = E(X2) − [E(X)]2.

Apply the idea First, find E(X2) : E(X2) = 12

1 6

1 Substitute values into the 6 formula E(X²) = Σx²P(X = x)

1 4 9 16 25 36 Evaluate the squares and 6 combine the fractions

91 6

Simplify the numerator

Now calculate the variance: Var (X) = E(X2) – [E(X)]2 =

91 ( 3.5)2 6

Write the formula Substitute the values

91 Evaluate the square 12.25 6 ≈ 15.167 – 12.25 Convert the fraction to decimal

= 2.917

Evaluate

The variance is approximately 2.917. The standard deviation is 2.917 ≈ 1.708. This means that a typical result from a single roll is expected to be about 1.7 units away from the mean of 3.5.

1204 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 A pencil case contains 5 blue pens and 3 red pens. Four pens are drawn randomly without replacement. Let X represent the number of blue pens drawn: a Describe the probability distribution for X.

Create a strategy Use combinations to calculate probabilities for each possible number of blue pens drawn (x = 1, 2, 3, 4).

Apply the idea 5 x

Calculate probabilities using the formula P(X = x) =

1 5 1

P(X = x)

3 4 x . 8 4

2 3 3

5 2

8 4

3 3 2

5 3

8 4

4 3 1

5 4

8 4

3 0 8 4

Simplify the probabilities: x

P(X = x)

1 14

6 14

1 14

b Calculate the probability of between 1 and 4 blue pens being drawn.

Create a strategy Sum the probabilities for X = 2 and X = 3, as “between 1 and 4 ” excludes 1 and 4.

Apply the idea P(X = 2) + P(X = 3) =

6 14

12 14

12.04 Applications of discrete random variables 1205 mathspace.co

c Calculate P(X > 1∣X < 4).

Create a strategy Use the conditional probability formula: P(X > 1∣X < 4) = This is

P(X

P ( X 2) P ( X 3) 1) P ( X 2) P ( X

P(X

1 P(X

X 4)

Apply the idea P(X > 1∣X < 4)∣ = =

P(X

X 4)

1 P(X

Write the conditional probability formula

P ( X 2) P ( X 3) P ( X 1) P ( X 2) P ( X

Express events as sums of discrete outcomes

6 6 14 14 = 1 6 6 14 14 14

Substitute the values

12 14 = 13 14

Simplify

12 13

Evaluate

d Calculate the expected number of blue pens drawn.

Create a strategy Use the formula E(X) = ∑ x × P(X = x) with the probabilities calculated in part (a).

Apply the idea E(X) = ∑ x × P(X = x) 1 =1 14 =

6 2 14

Write the formula for expected value 6 3 14

1 4 14

Substitute the values into the formula

1 12 18 4 Evaluate the multiplication in the 14 numerator

35 14 = 2.5 =

Evaluate the addition Convert the fraction to a decimal

The expected number of blue pens drawn is 2.5. This means if this experiment were repeated many times, the average number of blue pens drawn would be 2.5

1206 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

e Calculate the variance of the number of blue pens drawn.

Create a strategy First, calculate E(X2). Then, use the formula Var (X) = E(X2) – [E(X)]2.

Apply the idea First, find E(X2) : E(X2) = 12

1 14

6 14

1 14

ubstitute the values into the S formula E(X²) = Σx²P(X = x)

1 ( 1) 4 (6) 9 (6) 16 ( 1) Evaluate the squares and 14 combine the fractions 1 24 54 16 = Evaluate the multiplication 14 in the numerator 95 = Simplify the numerator 14 =

Now, calculate the variance: Var (X) = E(X2) – [E(X)]2 =

95 – (2.5)2 14

Write the formula for variance Substitute the values

95 – 6.25 Evaluate the square of the mean 14 ≈ 6.786 – 6.25 Convert the fraction to its decimal approximation =

= 0.536

Evaluate

The variance is approximately 0.536. The standard deviation is 0.536 ≈ 0.732. This indicates that the number of blue pens in a typical draw is expected to be about 0.73 pens away from the mean of 2.5

Idea summary A discrete probability distribution lists all possible outcomes in the sample space of a discrete random variable X and their associated probabilities. • The expected value (mean), E(X) = ∑ x × P(X = x), is the long-run average outcome. • The variance, Var (X) = E(X2) – [E(X)]2, measures the spread of the outcomes around the mean. • The standard deviation, Var (X ) , measures the typical distance of an outcome from the mean.

12.04 Applications of discrete random variables 1207 mathspace.co

12.04 Practice questions What do you remember? 1

A student designs a spinner for a game with the probabilities for landing on each colour: Colour (x)

Red

Blue

Green

P(X = x)

0.4

0.3

0.2

Explain why this table does not represent a valid discrete probability distribution. 2

A small bakery records the number of cheesecakes sold each day. The probability distribution is shown: Number of cheesecakes (x)

P( X = x)

0.1

0.2

0.4

0.3

Which expression correctly shows the calculation for the expected number of cheesecakes sold per day?

0.1 0.2 0.4 0.3 4

(0 × 0.1) + (1 × 0.2) + (2 × 0.4) + (3 × 0.3)

(0 + 1 + 2 + 3) × (0.1 + 0.2 + 0.4 + 0.3)

0 1 2 3 4

A survey asks households about the number of pets they own. The results are summarised in the probability distribution table: Number of pets (x)

P(X = x)

0.25

0.45

0.20

0.10

What is the most likely number of pets a household owns?

What does the expected value, E(X), of this distribution represent in the context of the survey?

Practice 4

5 Ex 1

The probability that a particular biased coin lands on tails is 0.7. Let X be the number of tails when the coin is tossed twice. Complete the given probability distribution table for X.

P(X = x)

⬚

A biased die is weighted so that rolling an odd number is twice as likely as rolling an even number. Let X be the number the die lands on: a

Construct a probability distribution table for X.

Determine E(X).

Determine Var (X).

1208 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

To be allowed to leave class and go to lunch, Mrs Ammar gives her class a 3-question multiple choice quiz, with each question consisting of 4 possible answers. David, who has not listened to a word Mrs Ammar has said all lesson and will have to guess each of the three questions: a

Determine the probability David will guess none of the questions correct.

Determine the probability David will guess all of the questions correctly.

Determine the probability David will guess just one of the questions correctly.

Let X represent the number of questions David guesses correctly. Construct a probability distribution table for X.

Determine whether the table represents a discrete probability distribution. Explain your answer.

Two fair six-sided dice are rolled and the sum of the numbers on the uppermost faces is recorded. Let Y represent the value of the sum of the two dice: a

Complete the table for this discrete probability distribution: y

P(Y = y)

1 36

3 ⬚

4 3 36

⬚

Identify the most likely sum to occur.

Describe the shape of the distribution.

Determine the expected value.

Determine the variance.

7 6 36

⬚

⬚ ⬚ ⬚

Two standard dice are rolled. The absolute difference between the numbers on their uppermost faces is recorded:

Complete the sample space in the given table.

Let X be the absolute difference. Construct the probability distribution table for X.

⬚

Determine P(X < 3)

⬚

Determine P(X ≤ 4∣X ≥ 2)

4 5 6

⬚

In a game of two-up, a person called the “Spinner” tosses two coins:

12 1 36

⬚

⬚ 2

⬚

⬚ ⬚

• If the coins land with two heads up, then the Spinner wins and the gamblers lose. • If the coins land with two tails up, the Spinner loses and the gamblers win. • If the coins land one head up and one tail up, the Spinner tosses the coins again and the gamblers break even. a

Construct a tree diagram to represent all possible outcomes of tossing two coins.

A gambler bets $3 on a game. If the spinner tosses two tails, the gambler wins $3 (and gets their original $3 bet back). Construct a probability distribution table for the gambler’s profit from one game.

12.04 Applications of discrete random variables 1209 mathspace.co

Ex 2

A pencil case contains 6 blue pens and 5 green pens. 4 pens are drawn randomly from the pencil case without replacement: a

Determine the probability of drawing one blue pen from the pencil case.

Determine the probability of drawing three blue pens from the pencil case.

Let X be the number of blue pens drawn. Complete the probability distribution table: x

P(X = x)

1 66

1 ⬚

2 5 11

⬚

Determine the expected number of blue pens drawn.

Determine the variance of the number of blue pens drawn.

Determine the probability that at least 2 blue pens are drawn, given that fewer than 4 are drawn.

An investment scheme advertises the percentage returns after 2 years based on historical probabilities: Percentage return (x)

10%

15%

25%

P(X = x)

0.7

0.15

Determine the expected percentage return on an investment.

Determine how much an investment of $50 000 is expected to be worth after 2 years.

A salesperson is starting work in a new region and analyses the probability of how many sales he is likely to make in the next month:

a b

Number of sales (x)

P (X = x)

0.35

0.25

0.2

0.15

0.05

Given that he makes at least one sale, state the probability that he will make 2 sales. The salesperson is offered 2 payment schemes: • Option A: Flat monthly income of $1800 • Option B: $1000 flat fee per month plus $500 per sale If he chooses Option B, what is his expected monthly income?

Which option should he choose to maximise his income?

1210 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

At a city car park, the all-day parking fees are as follows: • Cars with just a driver pay $25 • Cars with a driver and one passenger pay $15 • Cars with a driver and at least two passengers pay $12 This table summarises the number of people per car on a typical day:

Number of people

Number of cars

4500

3500

1100

600

300

Determine the probability a randomly selected car is carrying 3 people.

Given that a car was carrying at least 2 people, determine the probability it was carrying 4.

Let X represent the parking fee paid by a randomly selected car. Construct the probability distribution table for X.

Determine the expected revenue per car in this car park.

Determine the standard deviation of the revenue per car in this car park.

A child randomly selects 3 balls without replacement from a box containing 8 red balls and 2 green balls. For every red ball chosen, the child receives 1 chocolate. For every green ball chosen, the child receives 5 chocolates: a

Let X be the number of green balls chosen. Construct the probability distribution table for X.

Let T be the number of chocolates the child receives. Construct the probability distribution table for T.

Determine the most likely number of chocolates that the child receives.

Determine the expected number of chocolates that the child receives.

Extend your thinking 15

In a game, a standard six-sided die is rolled. The player’s score is determined by these rules: • If the result is any number from 2 to 6, that number is the score. • If the result is a 1, a second four-sided die (with faces numbered 3, 4, 5, and 6) is rolled, and that result is the score. Let X be the score of a player in this game:

Construct a probability distribution table for X.

Determine E(X).

Determine Var (X).

Two spinners numbered from 0 to 4 are spun. Let X be the product of the two numbers that come up: a

List all the possible outcomes of X.

Construct a probability distribution table for X.

Determine E(X).

Determine Var (X).

12.04 Applications of discrete random variables 1211 mathspace.co

Two dice are rolled and the difference between the largest number and smallest number is calculated. A player wins $1 if the difference is 3, $2 if the difference is 4, $3 if the difference is 5 and $0 otherwise: a

Complete the sample space in the given table.

Let X be the winnings from one game. Construct a probability distribution table for X.

c d

Determine the expected winnings.

⬚

If it costs $2 to play each game, calculate the player’s expected return.

⬚

4 5 6

⬚ ⬚

⬚ ⬚ ⬚

⬚

Two fair spinners, A and B, are spun. The total score, X, is calculated using the numbers from the spinners according to the rules: Spinner A Spinner B A B; if A B X A B; if A B B A; if A B 4 1 4 1 a List all the possible outcomes of X. b

⬚

Construct a probability distribution for X.

When a student completes a task on Spacemaths, the system monitors the number of hints used: • The probability of using at least 1 hint is 0.6 • The probability of using 2 hints is the same as using 3 hints. • The probability of using 1 hint is the same as using 4 hints. • At most students can use 4 hints. • The probability they use 2 hints is half the probability that they use 0 hints. a

Let X represent the number of hints the student used. Construct a probability distribution table for X.

Determine the expected number of hints a student will use.

Given that a student used at least 2 hints, determine the probability the student used 4 hints.

Investigation: Simulations Investigate online

1212 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

mathspace.co

12.05 Simulation After this lesson, you will be able to… • explain how simulations relate to experimental and theoretical probability concepts • analyse given simulation data • implement simulations to generate random samples • interpret simulation results to solve probability problems

Analyse simulation data Simulation A method of modelling random experiments to estimate probabilities by generating data that mimics real-world processes. Example: Rolling a die 100 times in a computer program to estimate the probability of rolling a 6. Frequency table A table that counts the occurrences of each outcome in a dataset. Population parameter A numerical value that describes a characteristic of an entire population. Example: The theoretical probability of rolling a 6 on a fair die p

1 or the true mean 6

height of all students in a school (μ). Sample statistic A numerical value calculated from a sample, used to estimate a population parameter. Example: The estimated probability of rolling a 6 from a simulation (p) or the mean height of students in a sample (x).

A simulation generates data to model random experiments, helping estimate the true probabilities (which are population parameters) when theoretical calculations are complex or impossible. It is important to use the correct notation to distinguish between the true values of a population and the estimated values from a sample. Population parameters are fixed, often unknown values that describe the entire population. • p: The theoretical probability (or population proportion) of an event. • μ: The true population mean.

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Sample statistics are calculated from a sample (such as the data from a simulation) and are used to estimate the population parameters. • p (read as “p-hat”): The estimated probability (or sample proportion) calculated from the simulation data. • x– (read as “x-bar”): The sample mean calculated from the simulation data. To summarise simulation data: 1. Create a frequency table to count occurrences of each outcome. 2. Use graphs (e.g., column graphs or histograms) to visualise the distribution. 3. Calculate the mean to find the average outcome, or the median to identify the middle value.

Example 1 A simulation models flipping a fair coin 200 times. The theoretical probability (population parameter) of flipping a Heads is p = 0.5. The simulation results are: Heads (108 times), Tails (92 times). Analyse the data by creating a frequency table, calculating the relative frequencies (a sample statistic), and finding the mean outcome (assign Heads = 1, Tails = 0).

Create a strategy Construct a frequency table with outcomes and the counts, compute relative frequencies by dividing each frequency by the total number of flips, and calculate the mean using the assigned values.

Apply the idea First, create the frequency table: Next, calculate the mean outcome: Use the formula for the ( 108 1) (92 0) Mean = mean, with Heads = 1 200 and Tails = 0 =

108 Simplify the numerator 200

= 0.54

Outcome

Frequency

Relative frequency

Heads

108

108 0.54 200

Tails

92 0.46 200

Evaluate

The relative frequencies show Heads occurs 54% of the time. The estimated probability of Heads from this simulation is p 0.54, which is a sample statistic. The mean outcome is also 0.54, close to the theoretical probability (population parameter) of p = 0.5 for a fair coin.

Reflect and check The estimated probability of Heads ( p 0.54), which is a sample statistic, is slightly higher than the theoretical probability (p = 0.5), which is a population parameter. This difference is expected in a finite simulation and should decrease with more trials, per the law of large numbers.

1214 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary A simulation generates sample data to estimate true, real-world probabilities. The key is to distinguish between the true value and the estimated value. • A population parameter is the true value for an entire population, like the theoretical probability ( p) or mean (μ). • A sample statistic is an estimate calculated from the simulation data, like the sample proportion ( p) or sample mean (x). The simulation data is typically organised in a frequency table to help calculate these sample statistics.

Implement simulations Algorithm A finite set of step-by-step instructions for solving a problem or completing a task. Pseudocode A simplified, informal way of describing an algorithm using a combination of plain English and programming language elements.

Implementing a simulation involves designing an algorithm to generate a random sample that mimics a real-world random process. This often requires: • Defining the event space (possible outcomes). • Using a random number generator to produce outcomes according to given probabilities. • Repeating the process many times to create a large sample. Algorithms for simulations can be described using pseudocode, which outlines the steps clearly. 1 Algorithm: Simulate a coin flip 2 Input: Number of trials (n) 3 Output: Frequency of Heads and Tails 4 1. Initialise Heads_count = 0, Tails_count = 0 5 2. For i from 1 to n: 6 a. Generate random number r between 0 and 1 7 b. If r < 0.5 8

Heads_count = Heads_count + 1

9 c. Else: 10

Tails_count = Tails_count + 1

11 3. Return Heads_count, Tails_count

12.05 Simulation 1215 mathspace.co

Example 2 Design a simulation to estimate the probability of winning a game where a player rolls two fair six-sided dice and wins if the sum is 7. Use 1000 trials.

Create a strategy Write a pseudocode to simulate rolling two dice, check if the sum is 7, and calculate the proportion of wins.

Apply the idea The event space is all possible sums of two dice (2 to 12). Count how often the sum is 7. 1 Algorithm: Simulate dice game 2 Input: Number of trials (1000) 3 Output: Estimated probability of winning 4 1. Initialise win_count = 0 5 2. For i from 1 to 1000: 6 a. Generate random integer die1 between 1 and 6 7 b. Generate random integer die2 between 1 and 6 8 c. If die1 + die2 = 7: 9 win_count = win_count + 1 10 3. Calculate probability = win_count / 1000 11 4. Return probability Suppose the simulation yields 166 wins in 1000 trials: P(win) ≈

166 1000

= 0.166

Divide the number of wins by the total trials Evaluate

The estimated probability of winning is 0.166, or about 16.6%.

Reflect and check 6 1 0.167 (since there are 6 favourable 36 6 outcomes out of 36 possible sums). The simulated probability (0.166) is very close, suggesting the simulation is reliable. The theoretical probability of rolling a sum of 7 is

1216 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary To implement a simulation, design an algorithm (often written as pseudocode) that models a random experiment. The key steps of a simulation algorithm are: 1. Initialise: Set up counters for the outcomes 2. Loop: Repeat the experiment for a large number of trials. 3. Generate: Inside the loop, use a random number generator to produce an outcome. 4. Check and count: Use an IF statement to check if the desired outcome occurred and increment the counter if it did. After the loop finishes, calculate the estimated probability: p

Count of successful outcomes Total number of trials

Interpret simulation results Law of large numbers As the number of trials in a simulation increases, the estimated probability approaches the theoretical probability. Example: After 10 coin flips, the proportion of Heads might be 0.7, but after 1000 flips, it is likely closer to 0.5. Interpreting simulation results involves comparing the sample statistics (e.g., estimated probabilities) to the population parameters (e.g., theoretical probabilities) and understanding the law of large numbers. This law ensures that with more trials, the simulation’s sample statistics converge to the true population parameters.

Example 3 A weather model simulates whether it rains each day, with a theoretical probability (population parameter) of rain at p = 0.3. A simulation of 500 days yields rain on 142 days. Interpret the results to estimate the likelihood of rain and assess the simulation’s reliability.

Create a strategy Calculate the estimated probability (a sample statistic), compare it to the theoretical probability (a population parameter), and use the law of large numbers to evaluate reliability.

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Apply the idea Calculate the estimated probability: P(rain) ≈

142 500

= 0.284

Divide the number of rainy days by the total days Evaluate

The estimated probability of rain is p 0.284, which is a sample statistic. This is slightly below the theoretical probability (population parameter) of p = 0.3. The law of large numbers suggests that with 500 trials, the estimate is reasonably reliable, but increasing the number of trials would likely bring the sample statistic p closer to the population parameter p = 0.3 In context, the simulation suggests a 28.4% chance of rain on any given day, which aligns closely with the expected weather pattern.

Reflect and check To check reliability, the simulation could be run for 1000 days. If the estimated probability (p) moves closer to the theoretical probability (p = 0.3), it supports the law of large numbers. The small difference (0.3 – 0.284 = 0.016) between the sample statistic and the population parameter is typical for a finite number of trials.

Idea summary Interpreting simulation results involves comparing sample statistics (e.g., estimated probabilities) to population parameters (e.g., theoretical probabilities), considering the law of large numbers, and applying the results to solve probability problems.

12.05 Practice questions What do you remember? 1

Determine whether these statements about simulations are true or false: a

A simulation is used to calculate the exact theoretical probability of an event.

The estimated probability calculated from a simulation is known as a population parameter.

A frequency table is used to count how many times each outcome occurs in a simulation.

The law of large numbers states that as the number of trials decreases, the simulation results become more accurate.

Determine the formula used to estimate the probability of an event in a simulation.

Explain how a frequency table is used to analyse simulation data.

1218 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice Ex 1

Ex 2

Ex 3

A simulation models spinning a fair spinner with 3 equal sections (labelled A, B, C) 150 times. The results are: A (52 times), B (48 times), C (50 times) a

Create a frequency table for the outcomes, including relative frequencies.

Calculate the mean outcome, assigning A = 1, B = 2, C = 3.

Design a simulation to estimate the probability of drawing a heart from a standard deck of 52 cards (with 13 hearts) using 400 trials: a

Write pseudocode for the simulation.

If the simulation yields 98 hearts, calculate the estimated probability.

A simulation models rolling a fair six-sided die 300 times. The results show 47 rolls of a 6: a

Calculate the estimated probability of rolling a 6.

Compare the estimated probability to the theoretical probability and comment on the result.

Design a simulation to estimate the probability of winning a game where a player rolls two dice and wins if the sum is 9. Use 800 trials: a

Write pseudocode for the simulation.

If the simulation yields 88 wins, calculate the estimated probability.

A simulation models a traffic light with a 0.25 probability of being red, observed 600 times. The results show 146 red lights: a

Calculate the estimated probability of the light being red.

Interpret the result, comparing to the theoretical probability and referencing the law of large numbers.

Explain the difference between experimental probability (from a simulation) and theoretical probability, using the coin flip example from the lesson (200 flips, 108 heads).

Extend your thinking 10

A simulation models a game where a player spins a spinner with 4 sections (1 to 4) twice and records the sum. Design a simulation with 1000 trials to estimate the probability of a sum of 6: a

Write pseudocode for the simulation.

If the simulation yields 122 sums of 6, calculate the estimated probability and compare to the theoretical probability.

Compare two simulations: Simulation A (roll two dice, win if sum = 8, 1000 trials, 138 wins) and Simulation B (weather model, 0.3 chance of rain, 1000 days, 292 rainy days): a

Calculate the estimated probabilities for each simulation.

Compare to their theoretical probabilities and explain why differences occur.

12.05 Simulation 1219 mathspace.co

12.06 Sample distributions After this lesson, you will be able to… • distinguish between population parameters and sample statistics • explain how sample statistics can be used to estimate population parameters • calculate and interpret sample means as estimates of population means • describe the concept of sampling variability

Populations and samples In statistical experiments and surveys, it is important to be precise in identifying the group of subjects that are being studied. To draw accurate conclusions about a population, such as finding the average height of students in Year 12 at a particular school, each and every member of the population could be surveyed. This process is called a census. Population The set of all eligible members of the group intended to be studied. Sample A subset of the population used to infer information about the greater population. Census The procedure of systematically acquiring and recording information from every member of a given population. Population mean The average value of a specific characteristic for all members of a population, denoted by μ. Example: The average height of all Year 12 students in a school is a population mean (μ) Sample mean The average value of a specific characteristic for a sample, denoted by x. It is used to estimate the population mean (μ). Example: The average height of a sample of Year 12 students in a school

While conducting a census provides complete information about a population, it is often impractical due to cost and time. Instead, data is collected from a sample. If this sample is representative of the whole group, its characteristics can be used to make reliable inferences about the entire population.

1220 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Numerical information derived from a population is called a population parameter. For example, the true average height of all Year 12 students in a country is a population mean, denoted by the Greek letter μ. In contrast, any numerical value calculated from a sample is called a sample statistic. The average height of a randomly selected group of 50 of those students is a sample mean, denoted by x. The formulas used to calculate the population mean and the sample mean are very similar, but they use different notation to distinguish between the population size (N) and the sample size (n).

xN N

μ is the population mean x1, x2, . . ., xN are the individual values in the population N is the total number of members in the population

x x x1, x2, . . ., xn n

xn n

is the sample mean are the individual values in the sample is the number of members in the sample

The primary goal of inferential statistics is to use a sample statistic as an estimator for an unknown population parameter. For instance, the sample mean ( x ) is a powerful estimator for the true population mean (μ). Population

Sample

The diagram shows how a sample is a smaller group selected from the entire population. The properties of the sample are used to estimate the properties of the population.

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Example 1 A statistics student wants to estimate the average number of hours per week that students in their school spend on extracurricular activities. They take a sample of five students and record their reported hours: 5, 8, 4, 7, 6 Calculate the sample mean of hours spent on extracurricular activities and explain how it can be used to estimate the population mean.

Create a strategy To find the sample mean ( x ), sum all the values in the sample and divide by the number of values in the sample. This sample mean will then serve as an estimate for the average hours spent by all students in the school (the population mean, μ).

Apply the idea The given sample data is 5, 8, 4, 7, 6. Sum of the values: 5 + 8 + 4 + 7 + 6 = 30 Number of values in the sample: 5 Calculate the sample mean ( x ): x

30 6 5

The sample mean of hours spent on extracurricular activities is 6. This sample mean of 6 hours is an estimator for the population mean (μ) of all students in the school. It suggests that, on average, students in the school spend approximately 6 hours per week on extracurricular activities.

Reflect and check A single sample mean provides an estimate, but it’s important to remember that different samples might yield slightly different sample means. This concept is known as sampling variability.

Idea summary A population is the entire group studied, while a sample is a subset used to infer information. A census collects data from every member, while population parameters (e.g., population mean) are estimated by sample statistics or estimators.

1222 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Sampling variability The reason samples are taken from a population is to approximate, at a smaller scale, what is happening statistically in the larger population. For example, if a random sample of 16-year-old students in a particular school is measured for height, and the mean and standard deviation of the sample are calculated, these are expected to be statistically representative of the population of 16-year-old students in the school as a whole. To be more certain, another random sample might be taken, and the mean and standard deviation calculated once again. The sample mean ( x ) and standard deviation will vary between samples but provide estimators for the population parameters (e.g., population mean, μ) which can be further assessed. Sampling variability The natural variation or differences that occur between the sample statistics (like the mean or standard deviation) calculated from different random samples drawn from the same population. Example: If one sample of students’ heights has a mean of 160 cm and another sample from the same population has a mean of 162 cm, this difference is due to sampling variability. To illustrate sampling variability, consider estimating the average height of all 16-year-old students (μ) in a school. If one random sample of 20 students yields a sample mean ( x ) of 165 cm, and a different random sample of 20 students from the same school yields a sample mean ( x ) of 167 cm, these different values reflect sampling variability. Generally, the larger the sample size, the more likely the sample statistics (such as the sample mean, x ) are to be close to the true population parameters (such as the population mean, μ). This is because larger samples tend to be more representative of the overall population.

Example 2 A biologist wants to estimate the average weight (μ) of a specific species of fish in a large lake. It is impractical to weigh every fish (the entire population). Instead, the biologist takes two different random samples on two different days. • Sample 1: 4 fish are caught with weights of 2.1 kg, 2.5 kg, 2.3 kg, and 2.7 kg. • Sample 2: 4 different fish are caught with weights of 2.4 kg, 1.9 kg, 2.2 kg, and 2.3 kg. a Calculate the sample mean ( x ) for each of the two samples.

Create a strategy For each sample, calculate the mean by summing the individual weights and dividing by the number of fish in that sample.

12.06 Sample distributions 1223 mathspace.co

Apply the idea For sample 1: x1 =

x1 x2 x3 x4 n

Write the formula for the sample mean

2 .1 2 .5 2 .3 2 .7 Substitute the weights from 4 sample 1 and the sample size n = 4

9. 6 4

= 2.4 kg

Evaluate the addition Evaluate

For sample 2: x2 = =

2 .4 1.9 2 .2 2 .3 Substitute the weights from 4 sample 2 and the sample size n = 4 8.8 4

= 2.2 kg

Evaluate the addition Evaluate

b Explain why the two sample means are different.

Apply the idea Recall that different random samples from the same population will likely contain different individuals, leading to different sample statistics. The two sample means, 2.4 kg and 2.2 kg, are different because these values were calculated from two separate, random samples. Each sample, by chance, captured a slightly different collection of fish from the overall population. This natural, expected variation between sample statistics calculated from different samples is known as sampling variability.

c The biologist decides to take a much larger third sample of 100 fish, and the sample mean is found to be 2.31 kg. Which of the three sample means (2.4, 2.2, or 2.31) is likely to be the most reliable estimator for the true population mean weight (μ)? Explain why.

Create a strategy Consider how the sample size affects the reliability of a sample statistic as an estimator for a population parameter.

Apply the idea The sample mean of 2.31 kg from the sample of 100 fish is likely to be the most reliable estimator. The reason is that larger sample sizes tend to reduce the effect of sampling variability and provide estimates that are closer to the true population parameter. A sample of 100 fish is much more representative of the entire lake’s fish population than a sample of only 4 fish.

1224 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary Samples approximate population characteristics. Larger samples provide better approximations of population parameters (e.g., population mean, μ) through sample statistics (e.g., sample mean, x). Sampling variability refers to the natural differences in statistics obtained from different samples from the same population.

12.06 Practice questions What do you remember? 1

Describe what these terms mean with regard to statistical analysis: a

Sample

Population

For each population, give an example of a group of people that make up a sample of the population: a

The population is all the students who attend the local high school.

The population is all people aged under 18, who live in Sydney.

The population is all people in a city who play in any organised sporting competition.

The population is all people who own a pet dog.

For each sample, give an example of a population that the sample could have been chosen from: a

A sample containing 50 people who drive white cars.

A sample of 50 people drawn from a population. In this sample, the youngest is 18 years old, and the oldest is 64.

A sample containing the first 50 people to enter a train station on a given day.

At a certain chocolate factory, 30% of products contain nuts. 400 chocolates are tested to check if they meet the required quality for sale. Of those tested, 61% contained nuts: a

What is the population?

Determine the value of the population proportion.

Determine the value of the sample proportion.

Do the 400 chocolates tested represent a simple random sample?

12.06 Sample distributions 1225 mathspace.co

Practice Ex 1

A librarian wants to estimate the average number of books read per month by high school students in her town. She takes a sample of 10 students and records their reported number of books read last month: 4, 7, 3, 5, 8, 4, 6, 5, 7, 3

Ex 2

Calculate the sample mean number of books read.

Explain how this sample mean can be used to estimate the average number of books read by all high school students in her town.

A survey asked 144 randomly chosen students about attending the school play, and 18 students said yes. A total of 204 students from the school ended up attending the play. Use the sample proportion to estimate the total number of students in the school.

A school careers counsellor surveys a random sample of 50 students from the 840 students that attend the school. 75% claimed to have a university offer and would attend university the following year. From the survey, estimate the number of students from the school that are going to university.

A teacher wants to estimate the average test score (as a percentage) for all Year 12 students in their school. They take two small, random samples of 4 students each: • Sample A: 75, 85, 90, 80 • Sample B: 82, 78, 88, 76

Calculate the sample mean score ( x ) for each of the two samples.

Explain why the two sample means are different.

The teacher then takes a larger third sample of 100 students, and the sample mean is found to be 81.7. Which of the three sample means (82.5, 81, or 81.7) is likely the most reliable estimator for the true population mean score? Explain why.

X is a discrete uniform distribution across the integers 1, 2, 3, 4, 5, 6 and 7: a

Calculate the mean and standard deviation of the distribution.

A sample of size 25 is taken from this distribution and the graph and table of results are shown:

10 9 8 7 6 5 4 3 2 1 0

Sample A

1 2 3 4 5 6 7

Value

Frequency

Calculate the mean and standard deviation of this sample, rounded to two decimal places.

1226 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

A sample of size 100 is taken from this distribution and the graph and table of results are shown:

Sample B 25 20 15 10 5 0

3 4

Value

Frequency

Calculate the mean and standard deviation of this sample, rounded to two decimal places. d

Another sample of size 100 is taken from this distribution and the graph and table of results are shown:

Sample C 25 20 15 10 5 0

3 4

5 6

Value

Frequency

Calculate the mean and standard deviation of this sample, rounded to two decimal places. e

Using the results from the previous parts, describe how the sample distribution, mean, and standard deviation are affected as the sample size increases.

12.06 Sample distributions 1227 mathspace.co

Extend your thinking 10

A quality control inspector takes a sample of 5 components from a large production batch. The weights of the components in grams are recorded as: 12.1, 11.9, k, 12.3, 11.7

The sample mean weight ( x ) for this sample is found to be 12.0 grams. Calculate the value of the unknown weight, k.

The true population mean weight (μ) for all components in the batch is known to be 12.05 grams. Comment on how well the sample mean ( x ) estimates the population mean in this case.

A researcher wants to estimate the average weekly study time of all Year 12 students at a large school. To collect data, the researcher surveys 50 students who are found studying in the school library during lunchtime:

Explain why this sampling method is likely to produce a biased estimate of the true average study time for the entire Year 12 population.

Is the sample mean ( x ) from this survey likely to be an overestimate or an underestimate of the true population mean (μ)? Justify your answer.

An analyst draws two independent samples. Sample A, of size nA = 4, is drawn from Population A. Sample B, of size nB = 6, is drawn from Population B. The data is as follows: • Sample A: 10, 14, 15, 17 • Sample B: 18, 20, 21, 22, 23, 24 a

Calculate the sample mean ( x A ) for Sample A and the sample mean ( xB ) for Sample B.

If both samples are combined to form a new single sample of size 10, calculate the mean of this combined sample.

Under what condition would this combined sample mean be a good estimator for the mean of a new population formed by merging all of Population A and Population B? Explain your reasoning.

1228 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

12.07 Simulate sampling distributions After this lesson, you will be able to… • simulate sampling from a population to generate distributions • display variation in sample means using dot plots and other visual displays • interpret and describe patterns observed in simulated sampling distributions • use spreadsheet formulas or code to generate and display multiple samples • calculate summary statistics from simulated samples with technology

Simulate and analyse sampling distributions Sampling distribution The distribution of a statistic, such as the sample mean, calculated from multiple samples of the same size taken from a population. Sampling simulation involves repeatedly drawing samples from a population to observe the behaviour of sample statistics and explore the relationship to the population parameter, such as the population mean (μ). The purpose of this process is to model how sample statistics behave without repeated real-world data collection. Technology, such as spreadsheets or statistical software, allows for efficient generation of multiple random samples from a population and the calculation of sample statistics. This automation enables large-scale simulations to study sampling distributions. Common spreadsheet functions include: • RANDBETWEEN (a, b): Generates a random integer between a and b. • AVERAGE (range): Calculates the mean of a range of values. Analysing a simulated sampling distribution involves calculating summary statistics and interpreting patterns such as shape, centre, and spread. Common observations include: • The mean of the sample means (which is the expected value of the sample mean, E( x)) approximates the population mean: Mean ( x) ≈ µ • Larger sample sizes reduce the spread (standard deviation) of the sampling distribution. The standard deviation of the sample means, known as the standard error, is given by SD (x ) where σ is the population standard deviation and n is the sample size. • With a sufficiently large sample size, the distribution of sample means often resembles a normal distribution (Central Limit Theorem).

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Frequency

Histogram of sample means (100 samples)

This histogram shows a sampling distribution of sample means (x), with a centre near the population mean (μ) and a spread reflecting sample variability.

10 5 0

4.0

4.6 5.2 5.8 Sample mean

6.4

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 A bag contains 10 marbles with values 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Simulate drawing 3 samples of size 2 with replacement and calculate the sample mean for each.

Create a strategy For each of the three samples, randomly select two values from the population with replacement. Calculate the mean for each pair by summing the values and dividing by the sample size of 2.

Apply the idea Sample 1: Draw marbles with values 3 and 7. x1 =

3 7 2

Add the values and divide by the sample size Evaluate

1 2 = 1.5 2 9 10 Sample 3: Draw marbles with values 9 and 10. The sample mean is x 3 = 9.5 2 The three simulated sample means are 5, 1.5, and 9.5

Sample 2: Draw marbles with values 1 and 2. The sample mean is x2

Reflect and check This small simulation demonstrates the variability in sample means. The population mean is 1 2 10 μ= = 5.5. Notice that the sample means vary around this value. With many more 10 samples, the distribution of these means would become centred at 5.5.

1230 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 2 Use a spreadsheet to simulate 5 samples of size 4 from a population of values 1, 2, 3, 4, and 5. Calculate the sample mean for each.

Create a strategy Use a spreadsheet function like RANDBETWEEN (1, 5) to generate four random values for each of the five samples. Then, use the AVERAGE function to compute the mean for each sample.

Apply the idea A spreadsheet is set up with 5 rows (one for each sample) and 4 columns (for the values in each sample). In each cell, the formula RANDBETWEEN (1, 5) is used. The results of one such simulation are shown, along with the calculated mean for each sample. Sample

Value 1

Value 2

Value 3

Value 4

Sample mean

3.5

2.75

3.5

2.75

3.75

The resulting sample means are 3.5, 2.75, 3.5, 2.75, and 3.75

Example 3 A simulation is conducted by drawing samples of size 5 from a bag of marbles with values 1, 2, . . . , 10. a Using technology, 10 samples were generated. The calculated sample means are: 5.2, 5.7, 5.3, 5.2, 5.9, 5.5, 5.5, 5.9, 5.5, 5.4. Create a dot plot to display the distribution of these sample means.

Create a strategy To create a dot plot, draw a horizontal axis that covers the range of the sample mean values. Then, place a dot above the axis for each of the 10 calculated sample means, stacking dots for repeated values.

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Apply the idea The horizontal axis should range from at least 5.0 to 6.0. Plotting a dot for each of the 10 sample means results in this distribution: Distribution of stimulated sample means

5.0

5.2

5.4 5.6 5.8 Sample mean

6.0

This dot plot shows the distribution of the 10 sample means. Notice that the values are tightly clustered around 5.5

b Using the data from the 10 simulated sample means, calculate the mean of this sampling distribution. Compare this value to the true population mean (μ).

Create a strategy The mean of the sample means is found by summing the 10 sample mean values and dividing by 10. The true population mean (μ) for the values 1 to 10 is 5.5. The two values can then be compared to see how well the simulation estimates the true parameter.

Apply the idea The 10 sample means from the simulation are: 5.2, 5.7, 5.3, 5.2, 5.9, 5.5, 5.5, 5.9, 5.5, 5.4. Calculate the mean of these sample means: x– =

xn n

Write the formula for sample mean

5.2 5.7 5.3 5.2 5.9 5.5 5.5 5.9 5.5 5.4 10

Substitute the values

55.1 10

Evaluate the addition

= 5.51

Evaluate

The mean of the sample means is 5.51 The true population mean is μ = 5.5 Comparison: The mean of the sample means (5.51) is very close to the true population mean (μ = 5.5). This demonstrates that the centre of the sampling distribution is an excellent estimator for the true population mean.

Reflect and check The small difference between the simulated mean (5.51) and the true mean (5.5) is a perfect example of sampling variability. It is expected that the mean of a limited number of samples will be close to, but not exactly equal to, the true population mean. Increasing the number of samples would likely result in an even closer estimate.

1232 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Example 4 Fifty samples of size 10 were simulated from a population with mean μ = 20 and standard deviation σ = 5. The sample means are stored in a spreadsheet. Calculate the mean and standard deviation of the sample means and interpret the results.

Create a strategy Use spreadsheet functions to compute the mean (using AVERAGE) and standard deviation (using STDEV.S) of the 50 collected sample means. Compare these calculated values to the theoretical population parameters.

Apply the idea After calculating the statistics from the 50 sample means in the spreadsheet, suppose the results are: Statistic of the sample means

Value

Mean of sample means

19.8

Standard deviation of sample means

1.6

Interpretation of the mean: The mean of the sample means (19.8) is very close to the true population mean (μ = 20). This demonstrates that the sampling distribution of the mean is centred around the population mean. Interpretation of the standard deviation: The standard deviation of the sample means (1.6), also known as the standard error, is significantly smaller than the population standard deviation (σ = 5). The simulation result is consistent with this theory.

Reflect and check Increasing the number of samples (e.g., from 50 to 500) would likely result in the mean of the sample means moving even closer to the true population mean of 20.

Idea summary Sampling simulation is a tool used to understand how sample statistics behave. By generating multiple random samples from a population, it is possible to create a sampling distribution. • Technology like spreadsheets can efficiently generate many samples and calculate their statistics (e.g., sample mean, x ). • The mean of the sampling distribution provides a good estimate of the true population mean (μ). • The spread of the sampling distribution (its standard deviation) is smaller than the population’s spread and decreases as the sample size increases.

12.07 Simulate sampling distributions 1233 mathspace.co

12.07 Practice questions What do you remember? 1

Define a sampling distribution and explain how it differs from a population parameter.

A population has values 2, 4, 6, 8. Calculate the population parameter for the mean.

Explain why the spread of a sampling distribution decreases as the sample size increases.

List two spreadsheet functions used to simulate sampling and calculate sample statistics.

Practice Ex 1

A simulation was run for a bag containing 8 marbles with values 1, 2, 3, 4, 5, 6, 7, and 8. Three samples of size 3 were drawn with replacement and results are recorded. Calculate the mean for each sample: a

Ex 2

Sample 1: 2, 4, 6

Sample 2: 1, 3, 5

A spreadsheet simulation was used to generate 5 samples of size 4 from a population with values {10, 20, 30, 40}. The results are shown in the table: Sample

Value 1

Value 2

Value 3

Value 4

Mean

⬚

Complete the table by calculating the mean for each sample. 7

Sample 3: 4, 6, 8

⬚ ⬚ ⬚

⬚

A spreadsheet was used to simulate 6 samples of size 5 from a population of values {1, 3, 5, 7, 9}. The results are shown in this table: Sample

Value 1

Value 2

Value 3

Value 4

Value 5

Calculate the mean for each of the 6 samples.

Using your calculated sample means, create a dot plot to show the sampling distribution.

1234 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Ex 3

A simulation is conducted by spinning a fair spinner with five equal sections, numbered 1 to 5. For the simulation, the sample size is 4: a

Using technology, 8 samples were generated. The calculated sample means are: 2.75, 3.0, 3.25, 2.5, 3.0, 3.5, 2.75, 3.25 Create a dot plot to display the distribution of these sample means.

Ex 4

Calculate the mean of the population (μ) and the mean of this sampling distribution (Mean (x )). Compare the two values and comment on the result.

A simulation of 100 samples of size 8 was performed on a population with a known mean μ = 15 and standard deviation σ = 4. After the simulation, the mean of the 100 sample means was calculated to be 14.9, and their standard deviation was 1.4: a

Compare the mean of the sample means (14.9) to the population mean (μ = 15) and explain what this demonstrates.

Calculate the theoretical standard error and compare it to the standard deviation of the sample means (1.4). What does this show?

Describe the expected shape of a sampling distribution of sample means for 200 samples of size 20 from a population with mean μ = 50.

Extend your thinking 11

A population consists of test scores with mean μ = 75 and standard deviation σ = 10. Simulate 100 samples of size 15 using a spreadsheet and create a histogram of the sample means. Describe the shape, centre, and spread of the sampling distribution.

A researcher simulates 50 samples of size 10 and 50 samples of size 30 from the same population. Compare the expected spread of the two sampling distributions.

A population has values 1, 2, 3, 4, 5. Simulate 10 samples of size 3 sampling with replacement, and calculate the sample mean and sample proportion of values greater than 3 for each sample. Compare the sampling distributions of these two statistics.

Investigation: Generating samples with technology Investigate online

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12.07 Simulate sampling distributions 1235 mathspace.co

12.08 Sample size and distributions After this lesson, you will be able to… • compare sampling distributions for different sample sizes • understand the centre and spread of sample mean distributions • analyse the effect of sample size on sampling distributions using simulations

Sample size in sampling distributions Standard error A measure of the variability of a sample statistic (such as the sample mean, x–) across multiple random samples. It is the standard deviation of the sampling distribution.

A sampling distribution describes how a statistic, like the sample mean, x, varies across multiple samples of the same size from a population. The size of the sample, denoted n, affects the centre and spread of this distribution.

The centre of the sampling distribution of x, denoted μ or E(x), is approximately the population x mean, μ. The spread is measured by the standard error (denoted σ ), which is the population x standard deviation, σ, divided by the square root of the sample size:

σ is the population standard deviation n is the sample size As n increases, the standard error decreases, indicating less variability in the sample means.

Example 1 A population has a mean of μ = 50 and a standard deviation of σ = 10. Compare the standard error of the sampling distribution of the sample mean for sample sizes n = 25 and n = 100.

Create a strategy Use the formula for the standard error, SE sample size.

, to calculate the standard error for each n

1236 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Apply the idea For n = 25:

n 10 = 25

SE =

Write the formula of standard error Substitute σ = 10 and n = 25 Evaluate

For n = 100:

n 10 = 100

SE =

Write the formula of standard error Substitute σ = 10 and n = 100 Evaluate

The standard error decreases from 2 to 1 as the sample size increases from 25 to 100, indicating less variability in the sample means.

Reflect and check The standard error is halved when the sample size is quadrupled, as 100 10 is twice 25 5.

Idea summary Larger sample sizes reduce the standard error, leading to sampling distributions with less variability. The mean of the sampling distribution of x remains approximately equal to the population mean.

Central limit theorem The shape of the sampling distribution of the sample mean changes with sample size. For small samples, the shape may resemble the population distribution, but as sample size increases, the sampling distribution becomes more symmetric and bell-shaped, regardless of the population’s shape.

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12.08 Sample size and distributions 1237 mathspace.co

Central Limit Theorem (CLT) States that for a sufficiently large sample size, the sampling distribution of the sample mean (x) will be approximately normally distributed, regardless of the shape of the population distribution. The mean of this sampling distribution will be equal to the population mean (μ), and its standard deviation (standard error) will be . n This behaviour shows that larger samples produce sampling distributions that are approximately normal, making it easier to predict the behaviour of sample means. A common rule of thumb is that a sample size of n ≥ 30 is sufficiently large for the central limit theorem to apply. However, if the parent population is already symmetric or close to normal, a smaller sample size may be adequate.

Example 2 A population has a right-skewed distribution with mean μ = 100 and standard deviation σ = 20. Describe the expected shape of the sampling distribution of the sample mean for samples of size n = 10 and n = 50.

Create a strategy Consider how sample size affects the shape of the sampling distribution of the sample mean, focusing on the tendency toward normality with larger samples.

Apply the idea For n = 10, the sample size is small, so the sampling distribution of the sample mean is likely to retain some right-skewness, similar to the population distribution.

Frequency

For n = 50, the sample size is larger, so the sampling distribution is expected to be more symmetric and approximately normal, regardless of the population’s skewness.

Sample mean n = 10

Sample mean n = 50

Reflect and check Simulations would confirm that the distribution for n = 50 is closer to normal than for n = 10, as larger samples reduce the influence of the population’s shape.

1238 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Idea summary The Central Limit Theorem (CLT) is a fundamental principle stating that for a sufficiently large sample size, the sampling distribution of the sample mean (x) will be approximately normal, regardless of the original population’s distribution. Key properties of this sampling distribution are: • Shape: Approximately normal (bell-shaped) • Mean: The mean of the sample means is equal to the population mean ( x ). • Standard deviation (standard error): The standard deviation of the sample means is x . n A sample size of n ≥ 30 is generally considered large enough for the CLT to apply.

12.08 Practice questions What do you remember? 1

Define a sampling distribution and explain how it is used to estimate population parameters.

What is the standard error of the sample mean, and how does it relate to sample size?

A population has a mean of μ = 75 and a standard deviation of σ = 12. Calculate the standard error of the sampling distribution of the sample mean for a sample size of n = 36.

Describe how the shape of the sampling distribution of the sample mean changes as sample size increases.

Practice Ex 1

A population has a mean of μ = 60 and a standard deviation of σ = 8. Compare the standard error of the sampling distribution of the sample mean for sample sizes n = 16 and n = 64.

Ex 2

A population has a uniform distribution with mean μ = 20 and standard deviation σ = 5. Describe the expected shape of the sampling distribution of the sample mean for samples of size n = 5 and n = 40.

12.08 Sample size and distributions 1239 mathspace.co

A simulation generates 1000 samples of size n = 25 from a population with μ = 30 and σ = 6. Calculate the expected mean and standard error of the sampling distribution of the sample mean.

A population has a mean of μ = 120 and a standard deviation of σ = 15. For a sample size of n = 9, calculate the standard error and describe the expected centre of the sampling distribution.

Use a spreadsheet to simulate 500 samples of size n = 20 from a population with μ = 50 and σ = 10. Describe how you would display the variation in sample means.

A left-skewed population has μ = 80 and σ = 12. Compare the expected shape of the sampling distribution for n = 15 and n = 60.

For a population with μ = 40 and σ = 4, calculate the standard error for sample sizes n = 4, n = 16, and n = 64. Comment on the trend.

A simulation generates 1000 samples of size n = 30 from a uniform population with μ = 25 and σ = 7. What is the expected mean of the sampling distribution, and how could the results be visualised?

A population has μ = 90 and σ = 18. For n = 81, calculate the standard error and predict the shape of the sampling distribution.

Extend your thinking 14

A researcher simulates 2000 samples of sizes n = 10, n = 50, and n = 200 from a right-skewed population with μ = 70 and σ = 14. Compare the expected centre, spread, and shape of the sampling distributions. Use a dot plot to illustrate your answer.

A population has an unknown distribution with μ = 45 and σ = 9. Design a simulation to compare the sampling distributions for n = 25 and n = 100. Calculate the standard errors, predict the shapes, and explain how to verify the results using summary statistics.

Explain why the standard error decreases as sample size increases, and discuss the implications for estimating the population mean in a real-world context, such as surveying student heights. Use a numerical example to support your explanation.

Chapter 12 review Online review

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“The joy of discovering something new is the greatest joy in life.” Sofia Kovalevskaya

Answers

8 B

1.01 Propositions and statements What do you remember?

10 Therefore, I will pass. 11 Therefore, the cat is not hungry.

1 a Compound statement

12 If a number is even, then it is divisible by 2.

b Disjunction and negation

If a number is divisible by 2, then it is even.

2 a The ball is not blue.

13 A number k is a perfect square if and only if there exists an integer n such that k = n2.

b The dog is brown. c The water is not hot.

14 a a if and only if b

d All turtles have claws.

c If d, then c

3 a The pillow is white and soft.

b a d If d, then not c

15 a i I f a number is an integer, then it is a whole number.

b The pillow is white or soft. 4 a Hypothesis: Two angles are vertically opposite. Conclusion: They are equal. b Hypothesis: The sum of a polygon’s interior angle measures is 180°. Conclusion: The polygon is a triangle. c Hypothesis: Two triangles have the same side lengths. Conclusion: They are congruent. d Hypothesis: Two lines are parallel. Conclusion: The co-interior angles formed by a transversal are supplementary. 5 a Implication

9 C

b Neither

c Equivalence

d Implication

e Equivalence

Neither

6 a Converse statements b Contrapositive statements c Inverse statements Practice 7 a If ∠1 + ∠2 = 180°, then ∠1 and ∠2 are supplementary. b If ∠1 and ∠2 are supplementary, then ∠1 + ∠2 = 180°. c ∠1 and ∠2 are supplementary if and only if ∠1 + ∠2 = 180°. d If ∠1 + ∠2 ≠ 180°, then ∠1 and ∠2 are not supplementary. e If ∠1 and ∠2 are not supplementary, then ∠1 + ∠2 ≠ 180°.

1242 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

ii False b i I f a shape is a quadrilateral, then it is a rectangle. ii False c i I f two angles form a straight angle, then they are adjacent and supplementary. ii True 16 a i If a number is not odd, then it is even. ii True b i I f I cannot paint a picture, then I am not an artist. ii False 17 a i I f a number is not divisible by ten, then it does not end in zero. ii True b i I f I do not have a dog, then I do not have a golden retriever. ii True 18 a If a polygon is five-sided, then it is a pentagon. b If a polygon is not a pentagon, then it is not five-sided. c The polygon is not five-sided. 19 a False. The converse is true: “If a number is rational, then it is a real number.” b False. The converse, “If a quadrilateral is a rhombus, then it is a parallelogram” is true. c True.

20 1 Algorithm: Voting Eligibility 2 input age 3 if age >= 18 then 4 print ‘Eligible to vote’ 5 else 6 print ‘Not eligible to vote’ 7 end if 21 a It is hot.

b It is cold.

c It is moderate. Extend your thinking 22 a i I f an animal gives birth to live young, then it is a mammal. ii If an animal is a mammal, then it produces milk. iii If a mammal is a monotreme, then it lays eggs. b i I f an animal is a mammal, then it gives birth to live young. False. Monotremes are mammals but lay eggs. ii If an animal produces milk, then it is a mammal. True. Milk production is a defining characteristic of mammals. iii If a mammal lays eggs, then it is a monotreme. True. This is the definition of a monotreme within mammals. c Answers may vary. Example: Statement: If an animal is a monotreme, then it produces milk. (True, as monotremes are mammals, and all mammals produce milk). Converse: If an animal produces milk, then it is a monotreme. (False. Many mammals like cows produce milk but are not monotremes). 23 Answers may vary. Possible answer: If a polygon is a square, then it has four right angles. The converse “If a polygon has four right angles, then it is a square” is not true because a non-square rectangle also has four right angles.

24 N o. For an equivalence to be true, both the original implication (‘if p then q’) and its converse (‘if q then p’) must be true. The implication ‘If m2 – 8 = m + 5, then m = 3’ is false because the solutions to m2 – m – 13 = 0 are not m = 3. The converse ‘If m = 3, then m2 – 8 = m + 5’ is also false because 32 – 8 = 1 and 3 + 5 = 8, and 1 ≠ 8. Since both (or even just one) are false, a true equivalence cannot be formed. 25 Felix’s error is assuming the converse statement is also true. For the statement to be a true equivalence (“It is sunny if and only if it is not winter”), both the original implication and its converse must be true. Original: “If it is sunny, then it is not winter.” (This is false - it can be sunny in winter). Converse: “If it is not winter, then it is sunny.” (This is false - it can be cloudy in summer/spring/autumn). Since both the original statement and its converse are false, it cannot be written as a true equivalence. Felix needs to understand that an ‘if-then’ statement does not automatically imply its converse is true. 26 a True. The ‘Whole numbers’ circle is entirely contained within the ‘Integers’ circle, which is entirely contained within the ‘Rational numbers’ circle. b False. The ‘Rational numbers’ circle contains regions outside the ‘Whole numbers’ circle 1 (e.g., fractions like , or negative integers like 2 − 3). c False. The ‘Integers’ circle contains regions outside the ‘Natural numbers’ circle (e.g., zero or negative integers like − 2). d True. There is a region within ‘Rational numbers’ but outside ‘Whole numbers’. e True. The ‘Irrational numbers’ and ‘Rational numbers’ circles are disjoint, meaning they do not overlap. f T rue. The ‘Natural numbers’ circle is entirely contained within the ‘Integers’ circle, meaning all natural numbers are also integers. g False. The ‘Whole numbers’ circle is contained within ‘Rational numbers’, and ‘Rational numbers’ and ‘Irrational numbers’ are separate, so there is no overlap between ‘Whole numbers’ and ‘Irrational numbers’.

Answers 1243 mathspace.co

1.02 Symbols in proof

10 (A ∪ B) \ (A ∩ B) or (A ∩ B′) ∪ (A′ ∩ B)

What do you remember?

1 a {1, 3, 5, 7, 9} b {1, 2, 3, 6, 7, 14, 21, 42} c {1, 2, 3, 4} d {− 7, − 6, − 5, − 4} 2 A′ = {2, 4} 3 a {4}

b {2, 4, 6, 8, 12}

4 a True

b False

ℕ Y

ℤ

ℚ Y

ℝ

−3

3 4

12 a True

5 A

b True

13 a ∃x ∈ ℤ : x2 = 4

6 D

c ∀n ∈ ℤ, n ≥ 0

Practice

14 a ∀x ∈ ℝ, x3 ∈ ℝ

7 a P ∩ Q = {c} c P′ = {d, e, f, g}

iii A′ = {1, 3, 5, 7, 9, 11}

9 a

d i F or all real numbers x, x3 is greater than or equal to 0.

ii False

3 1 7 5 11

c i T here exists a real number x such that the square of x is less than 0. ii False

b ∃n ∈ ℤ+, n2 > 100

ii True

ii A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}

b ∀n ∈ ℤ+, n2 ≥ n

b i For all positive integers n, 2n + 1 is odd.

8 a i A ∩ B = {6, 12}

4 8 10 2

d False

ii True

d (P ∪ Q)′ = {f, g}

c True

15 a i T here exists a real number x such that x2 equals 5.

b P ∪ Q = {a, b, c, d, e}

16 True. The statement says “there exists” a rational number that satisfies the condition, so only one needs to be found. If x = 2 is used (which is a rational number), then x2 is 4 and 2x is also 4. Since an example exists, the statement is true. 17 a Negation: ∃n ∈ ℤ : n is not even (is odd).

b The original statement is false because a counter-example exists (e.g., n = 3). The negation is true because an example exists that satisfies it (e.g., n = 3).

U b

18 If it is snowing, then the school is not open. 19 a q ∨ r c p ⇒ (q ∨ r) 20 D 21 A

1244 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b ¬p ∨ q d r ⇔ (¬p ∨ q)

22 If it is not the case that both p and q are true, then r is true.

integer, 1 + 2y is also an integer. The denominator is 2, which is a non-zero integer. Therefore, the result is a fraction of two integers, which is the definition of a rational number. The statement holds.

Alternatively: If not (p and q), then r. 23 The statement translates to “not p or p”. This statement is always true because of the law of the excluded middle; a statement p must be either true or false. If p is true, the second part “(p)” makes the ‘or’ statement true. If p is false, then “not p” is true, and the first part makes the ‘or’ statement true. It is a tautology. Extend your thinking 24 First, solve the inequalities for each set: A = {x ∈ ℝ| x2 ≤ 4} ⇒ − 2 ≤ x ≤ 2 ⇒ [− 2, 2]. B = {x ∈ ℝ| x > 0} ⇒ (0, ∞). The intersection contains elements in both sets. A ∩ B = {x ∈ ℝ| − 2 ≤ x ≤ 2 and x > 0} = (0, 2]. A

B C

–3

–2

–1

–3

–2

–1

–3

–2

–1

28 a Symbolic form: (p ∧ q) ⇒ ¬r. An implication is true if the premise is false, or if both the premise and conclusion are true. The premise p ∧ q (even and divisible by 3) is true only for n = 6. For n = 6, the conclusion ¬r (not prime) is also true. So the implication is true for n = 6. For all other numbers in U, the premise p ∧ q is false. An implication with a false premise is always true. Therefore, the implication is true for all numbers in U, so the set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. b

8 10 4

6 2

26 The statement is false. Let x = 0.5. Here, x > 0 is true. However, x2 is (0.5)2 = 0.25. Since 0.25 is not greater than 0.5, a counter-example is found. 27 a “There exists a rational number x such that for all integers y, the sum x + y is rational.” This means the task is to determine if at least one specific rational number can be found such that adding any integer to it will always result in another rational number. b The statement is true. Choose a rational 1 number, for example, x . Now, show that 2 for any integer y, the sum x + y is rational. y An integer y can be written as . The sum is: 1 1 y 1 2y 1 2 y . x y Since y is an 2 1 2 2 2

25 a ∃p ∈ {primes} : (p + 2) ∈ {primes} b The statement is true. Let p = 3, which is a prime number. Then p + 2 = 3 + 2 = 5. Since 5 is also a prime number, an example has been found.

1.03 Direct proof What do you remember? 1 C 2 B 3 Find a single example that satisfies the condition. For instance, letting x = 5 gives 52 = 25, so the statement is proven true. 4 To disprove a universal statement, find a single counter-example where the statement is false. For the statement ∀n∈ ℤ, n2 > n, let n = 1. Then n2 = 12 = 1. The claim 1 > 1 is false. So n = 1 is a counter-example. Practice 5 18 is an even number. 6 The definition of an odd integer. 7 Ask your teacher for worked solutions.

Answers 1245 mathspace.co

8 Ask your teacher for worked solutions.

12 Ask your teacher for worked solutions.

9 Ask your teacher for worked solutions.

13 Ask your teacher for worked solutions.

10 Ask your teacher for worked solutions.

14 a There are no integers x, y such that 6x + 9y = 1.

11 Ask your teacher for worked solutions. 12 Ask your teacher for worked solutions. 13 Ask your teacher for worked solutions. Extend your thinking 14 Ask your teacher for worked solutions. 15 Ask your teacher for worked solutions. 16 Ask your teacher for worked solutions. 17 Ask your teacher for worked solutions.

1.04 Proofs by contradiction What do you remember? 1 a The product of three consecutive integers is odd. b There exists an odd integer that can be expressed as the sum of three even integers. c There is a largest prime number. 3

b Proof by contradiction is useful for proving non-existence. Assuming that such integers do exist (6x + 9y = 1) gives a concrete equation to work with, from which a contradiction can be derived (for example, by considering divisibility by 3). 15 The proof is flawed. Step 6 claims a contradiction was reached, but no contradiction was actually demonstrated. The argument p correctly shows that if 4 , then p2 = 4q2, q leading to p being even (p = 2k) and q2 = k2. This does not contradict the initial assumption or any known mathematical fact (for example, p = 2, q = 1 satisfies the initial assumption and the derived conditions without contradiction). The error lies in falsely identifying a contradiction. 16 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions.

d x is even and x is odd.

c Ask your teacher for worked solutions.

e It is raining and it is not cloudy.

d Ask your teacher for worked solutions.

2 a i and ii

b ii and iii

e Ask your teacher for worked solutions.

c i and iii

d i and ii

f Ask your teacher for worked solutions.

3 B

17 Ask your teacher for worked solutions.

4 D

Extend your thinking

Practice

18 Ask your teacher for worked solutions.

5 a The assumption is that if two integers are odd, then their product is not odd (i.e., is even).

19 Ask your teacher for worked solutions. 20 Ask your teacher for worked solutions.

b Ask your teacher for worked solutions.

1.05 Proofs by counter-example

6 a Negation: 3n + 7 is even and n is even. b Ask your teacher for worked solutions. 7 Ask your teacher for worked solutions.

What do you remember? 1 C

9 Ask your teacher for worked solutions.

2 The minimum requirement is to find a single valid counter-example, which is a specific value of n for which the statement P (n) is false.

10 Ask your teacher for worked solutions.

3 The case must make p true and q false.

8 Ask your teacher for worked solutions.

11 Ask your teacher for worked solutions.

1246 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice

12 False.

4 5 (1) + 3 = 8 (even)

The conjecture can be disproven using the number 251 as an example. The digits of 251 are 2, 5, and 1. The product of these digits is 2 × 5 × 1 = 10, which is even. However, the number 251 itself is odd, clearly disproving the conjecture.

5 (2) + 3 = 13 (odd) ∴ when n = 2, 5n + 3 is an odd integer. (Other answers may also exist.) 5 a There exists a positive integer n such that n is even and n is less than 10. b The statement is true. Valid examples are n = 2, 4, 6, 8 because these numbers are even and are less than 10. 6 a For all positive integers n less than 5, n2 + 3 is less than 20. b For n = 1, n2 + 3 = 4 For n = 2, n2 + 3 = 7 For n = 3, n2 + 3 = 12 For n = 4, n2 + 3 = 19 All are less than 20. 7 C 8 12 – 1 = 0 (divisible by 3) 22 – 1 = 3 (divisible by 3) ∴ when n = 1 and 2, n2 – 1 is divisible by 3. (Other answers may also exist.) 9 a False. For example, a triangle with side lengths 3 cm, 4 cm, and 5 cm is not equilateral. b True c True d False. Many different people go to schools for work or to visit. For example, a person could go into school and be a teacher.

13 Sarah’s counter-example is not valid. A valid counter-example must satisfy the hypothesis (the condition) of the statement but not the conclusion. The hypothesis is that the number is a ‘multiple of 3’. The number 10 is not a multiple of 3. Therefore, it cannot be used as a counter-example for this statement, even though the statement itself is false. (A valid counter-example would be 6, which is a multiple of 3 but is not odd). 14 a If a food is not vegan, then it contains dairy. b False. Counter-examples are chicken, fish, egg, and honey. They are not vegan and do not contain dairy. 15 a True b False. The converse statement, “if x2 = 4, then x = 2” is not true because x = – 2 is also possible. c False. The inverse statement, “if x ≠ 2, then x2 ≠ 4” is not true because x = – 2 is possible. d The contrapositive statement, “if x2 ≠ 4, then x ≠ 2. This is true. 16 To disprove the statement, a single counterexample must be found. Integer values for n should be tested.

10 False.

Let n = 4. Then n3 + 2 = 43 + 2 = 64 + 2 = 66.

For instance, adding 3 to 2 results in 5, and adding 3 to 4 results in 7, both of which are odd. These examples clearly disprove the conjecture.

The number 66 is not prime because it is divisible by 2, 3, 6, etc. Since n = 4 provides a counter-example, the statement is disproven.

11 False.

Because any even number greater than 2 can be expressed as 2n for some integer n > 1. Such numbers are divisible by both 2 and n, confirming the existence of divisors other than 1 and the number being considered, which makes the number composite. Examples like 4, 6, and 8 illustrate this without exception.

For example, multiplying the odd number 3 by the odd number 5 results in 15, which is odd. This serves as a counter-example.

17 a True.

Answers 1247 mathspace.co

b False. A counter-example is the number 0. The number 0 ends in 0, but it is not a composite number. By definition, composite numbers are positive integers greater than 1. As 0 does not meet this criterion, the statement is false. c False. The statement is not always true. The exception is the prime number 2, the only even prime. Adding 2 to an odd prime results in an odd sum. For example, adding 2 (prime) to 5 (prime) gives 7 (odd), serving as a counter-example. d False. Consider the multiple of 5, which is 10. The number immediately after is 11, which is not a multiple of 6. This provides a counter-example. e False. Consider the multiple of 7, which is 14. Adding 4 gives 18, which is not a multiple of 11. This provides a counter-example. 18 a False. The conjecture that adding 6 to a prime number always results in another prime number is disproven by the example of 19. While adding 6 to the prime number 5 results in 11, another prime, and adding 6 to the prime number 7 results in 13, also a prime, this pattern does not hold universally. For instance, adding 6 to the prime number 17 yields 23, which is indeed prime. However, when 6 is added to the prime number 19, the result is 25, which is not a prime as it can be divided by 5. Thus, 19 is a counter-example to the conjecture. b False. Take the prime number 3, double it to get 6, and 6 is not a prime number because it can be divided by 2 and 3.

1248 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking 19 If a polygon is a square, then it has four right angles. The converse of this statement “If a polygon has four right angles, then it is a square” is not true because a rectangle has four right angles and is not necessarily a square. 20 T o disprove this, prove the negation is true. “For all n ∈ ℕ : n2 + 3n + 2 is not a prime number” n2 + 3n + 2 = (n + 1) (n + 2)

For any n ∈ ℕ, both factors (n + 1) and (n + 2) are greater than or equal to 2. Since for any n ∈ ℕ, both n + 1 and n + 2 are integers greater than 1, their product is always a composite number. Since the negation is true, the original statement is false.

1.06 Proof by induction What do you remember? 1 a False

b True

c True

d False

e True 2 B, E, F 3 A Practice 4 a 21 > 1. Yes, since 21 = 2 > 1. b 2k > k

c 2k+1 > k + 1

1 2 (1 ) (1 + 1)2. 4 1 2 b When n 2, 13 23 (2 ) ( 2 1)2 . 4 1 2 c When n = k, 13 23 k3 k ( k 1)2 . 4 d When n = k + 1, 13 + 23 + . . . + (k + 1)3 = 1 (k + 1)2 (k + 2)2. 4 6 338 350 5 a When n = 1, 13 =

7 Substitute n = 8 into the formula: 8(8 3) S8 = 2

1 1 . 21 2 21 1 2 1 1 . Since LHS = RHS, RHS 2 2 21 the conjecture is true for n = 1.

11 a For n = 1, LHS

= 20 An octagon has 20 diagonals. 8 a LHS for n = 5 is: 1 1 1 1 3 3 5 5 7 RHS for n = 5 is:

b Assume for an integer k ≥ 1, 1

7 9

9 11

5 2(5) 1

5 11

1 3

13 15

12 Ask your teacher for worked solutions.

Extend your thinking

1.07 Prove properties of integers

7 . 15

1+2+3+...+n

1 n( n 1) 2

True or false

n=1

True

n=2

True

n=3

True

n=4

True

n=5

True

n=6

True

n=7

True

n=8

True

5 1 2 3 19 = 15

6 2 3 4

7 3 4 5

8 4 5 6

What do you remember? 1 The correct order is 3, 2, 1. 2 The correct order is 2, 1, 3. 3 Let the two even integers be 2a and 2b, where a, b are integers. Their sum is 2a + 2b = 2(a + b). Since a + b is an integer, the sum is of the form 2k, which is even. 4 Let the two odd integers be 2a + 1 and 2b + 1, where a, b are integers. Their product is (2a + 1) (2b + 1) = 4ab + 2a + 2b + 1 = 2 (2ab + a + b) + 1. Since 2ab + a + b is an integer, let k = 2ab + a + b. The product is 2k + 1, which is the form of an odd integer. Practice 5 a False. Counter-example: If a = 1, and b = 1, then a2 + b2 = 12 + 12 = 2, which is an even number, but a and b are odd.

b Substitute n = 4: 4( 3( 4 ) 7) S4 = 2( 4 1)( 4 2)

b True. If n is even, n = 2k. Then n2 + 2n = (2k)2 + 2 (2k) = 4k2 + 4k = 2 (2k2 + 2k). Since 2k2 + 2k is an integer, the expression is even.

19 15

c Yes, both calculations yield

15 Ask your teacher for worked solutions.

Since LHS is equal to RHS, it is true for n = 7.

14 Ask your teacher for worked solutions.

Using the result for n = 5, this is 5 1 1 7 11 11 13 13 15 15

10 a S4

13 Ask your teacher for worked solutions.

b LHS for n = 7 is: 1 1 1 3 (2( 7) 1)(2( 7) 1)

c Ask your teacher for worked solutions.

5 11

7 2( 7) 1

Since LHS is equal to RHS, it is true for n = 5.

RHS for n = 7 is:

19 . 15

c False. Counter-example: Let a = 1 and b = 3, both odd. Then a2 + b = 12 + 3 = 4, which is even, not odd.

Answers 1249 mathspace.co

6 a 22 × 7 3

b 23 × 7 2

c 2 ×3

d 24 × 5

21 a HCF = 12

b LCM = 840

22 a 90 = 2 × 32 × 5

7 a i HCF = 6

ii LCM = 36

108 = 22 × 33

b i HCF = 8

ii LCM = 96

126 = 2 × 32 × 7

8 Ask your teacher for worked solutions.

HCF = 21 × 32 = 18

9 Ask your teacher for worked solutions.

b LCM = 3780

10 Ask your teacher for worked solutions. 11 Ask your teacher for worked solutions. 12 Ask your teacher for worked solutions. 13 Ask your teacher for worked solutions. 14 Ask your teacher for worked solutions. 15 Ask your teacher for worked solutions. 16 Ask your teacher for worked solutions. 17 Ask your teacher for worked solutions. 18 a 22 × 32 × 7 2

c 2 ×3 ×5×7

b 22 × 3 × 5 × 7 d 25 × 3 × 52

19 Prime factorisation of 588 = 22 × 3 × 72. Prime factorisation of 42 = 2 × 3 × 7. All prime factors of 42 (2, 3, 7) are present in the prime factorisation of 588, and the powers in 588 (22, 31, 72) are greater than or equal to the powers in 42 (21, 31, 71). Therefore, 588 is divisible by 42. 20 Steps: 1. Assume there is a finite list of all prime numbers: p1, p2, . . . , pk. 2. Construct N = (p1 × p2 × . . . × pk) + 1. 3. Argue N must have a prime divisor, p. 4. p must be one of p1, . . . , pk (by assumption). Let p = pi. 5. Show pi divides N and pi divides (p1 × . . . × pk). Thus, pi must divide their difference, which is 1. 6. This leads to a contradiction, as no prime number can divide 1. Therefore, the initial assumption that there is a finite list of primes must be false, proving that there are infinitely many prime numbers. Why N leads to contradiction: N leaves remainder 1 when divided by any prime pi in the assumed list. But N must have a prime divisor. This divisor cannot be in the list (as remainder is 1), contradicting the assumption that the list contained all primes.

1250 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

23 a HCF = 12

b LCM = 420

c 12 × 420 = 5040

d 60 × 84 = 5040

e 5040 = 5040. The relationship holds. 24 The statement is false. Consider cases or factorisation: 2n2 + 5n + 3 = (2n + 3)(n + 1). • Case 1: n is even (n = 2k). Then (2(2k) + 3) (2k + 1) = (4k + 3)(2k + 1). This is Odd × Odd = Odd. • Case 2: n is odd (n = 2k + 1). Then (2(2k + 1) + 3) ((2k + 1) + 1) = (4k + 2 + 3)(2k + 2) = (4k + 5)(2 (k + 1)). This is Odd × Even = Even. Since the expression is even when n is odd, the statement “always odd” is false. Counter-example: If n = 1 (odd), 2 (1)2 + 5 (1) + 3 = 2 + 5 + 3 = 10, which is even. 25 Ask your teacher for worked solutions. Extend your thinking 26 Ask your teacher for worked solutions. 27 Ask your teacher for worked solutions. 28 Ask your teacher for worked solutions. 29 Ask your teacher for worked solutions.

1.08 R ational numbers and their representations What do you remember? 1 a ℤ, ℚ, ℝ d ℚ, ℝ

b ℚ, ℝ

e ℕ, ℤ, ℚ, ℝ

g Irrational, ℝ h Irrational, ℝ

c ℚ, ℝ

f Irrational, ℝ

n 2 a Any integer n can be written as the fraction , 1 where n and 1 are integers and 1 is not zero. b A terminating decimal can be written as a fraction with a power of 10 in the 6 denominator for example, 0.6 , which 10 fits the definition of a rational number.

c Example:

1 2

1 3

3 6

2 6

1 5 1 . Both and are 3 6 2

5 is also rational. 6 3 a All integers are rational numbers because n any integer n can be written as . So, ℤ is a 1 subset of ℚ.

10 18.79 has been misread as 18.79 Correct solution: x = 18.79

rational, and their sum

10x = 187.9 100x = 1879.9 100x – 10x = 1879.9 187.9 90x = 1692 1692 94 x= 90 5

b Closure under addition means that adding any two rational numbers always results in another rational number.

c Example: 2. It is irrational because it cannot p be expressed as a simple fraction , and its q decimal representation is non-terminating and non-repeating. 4 a Yes

b Yes

c Yes

b False

c False

d False

e False

f True

g True

h False

i False

1 879 661 111 100

49 109 594

4 5

c Rational. It’s a recurring decimal 0.12

8309 135

Extend your thinking

6 a Rational. It is an integer and can be written as 5 the fraction . 1 314 b Rational. It’s a terminating decimal . 100 12 . 99

3 d Rational. 9 3, which is an integer . 1 e Irrational. 10 is not a perfect square, so its square root is a non-terminating, nonrepeating decimal. f I rrational. It is a non-terminating, non-repeating decimal that cannot be expressed as a fraction of two integers. 7 a 0.629

b 2.17

c 0.391

d 16.15

e 9.991

f 3.42

g 0.04

h 0.378

8 Ask your teacher for worked solutions. 1 c 3

15 197 990

13 a 0.6

Practice

1 b 15

12 a

d No

5 a True

2 9 a 9

11 0.352

1127 d 225

1 6

211 99

5897 333

2603 1980

4 33

17 225

29 99

161 45

14 a Ask your teacher for worked solutions. 4 b 9999 15 x ≈ – 9.625, x ≈ 5.571 16

33 6 7

17 3.4772% 18 a $37 945.60 19

b $36 996.96

33 042 139 900 107 797 321 777

1.09 Prove irrationality What do you remember? 1 a • a and b are integers. • a and b have no common factors other than 1. • b ≠ 0 b Ask your teacher for worked solutions. 2 Ask your teacher for worked solutions. 3 Ask your teacher for worked solutions. Practice 4 Ask your teacher for worked solutions.

Answers 1251 mathspace.co

12 it can be 1 represented as a fraction, so it is rational.

5 a It is rational. Since 144 12

b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. d Ask your teacher for worked solutions. p , where p and q are q coprime integers and q ≠ 0. p 19 = 2 q 4 p p2 19 = 4 q q2 2 2 19q = 4q – 4pq + p2

6 Suppose 2

0 = p2 – 4pq – 15q2 ( 4 q)2 4( 1)( 15q2 ) p = 2( 1) 2q q 19 q(2 19 ) which is divisible by q. 4q

This contradicts the statement that p and q are coprime. Therefore, by contradiction, 2 19 is irrational. 7 Ask your teacher for worked solutions. 1 p 8 a Suppose , where p and q are coprime q 6 integers and q ≠ 0. 6p2 = q2 q2 6 which implies q2 is divisible by 6, which implies q is divisible by 6. p2 =

q = 6a

where a is some integer

m and n are coprime integers. Cubing both m3 sides of the equation yields 2 3 , which n implies 2n3 = m3. This suggests that m3 is an even number, and therefore m must be even. If m is even, then m can be written as 2k, which leads to 2n3 = (2k)3 = 8k3, so n3 = 4k3. But this implies n is even as well, contradicting that m and n are coprime. Therefore, the cube root of 2 cannot be rational. Extend your thinking 10 Ask your teacher for worked solutions. 11 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions.

2.01 Propositional logic What do you remember? 1 a 2

b 4

c 8

2 a True

b True

c True

d True

e True

3 a p∧q c p ∧ ¬r

b ¬p ∨ r d ¬q

4 a False

b True

c False

d True

5 F (second row shows p = T, q = F, and p ∧ q = F) 6 a The door is not locked.

6p2 = (6a)2

b It will not snow tomorrow.

p2 = 6a2

c The car is not red.

p2 a2 = 6

d I am not tired.

which implies p2 is divisible by 6, which implies p is divisible by 6. p and q are both divisible by 6 contradicts the statement that p and q are coprime. 1 Therefore, by contradiction, is irrational. 6 b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. d Ask your teacher for worked solutions. 9 Assume that the cube root of 2 is a rational m number, which can be expressed as , where n

1252 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice 7 a Not a proposition (it is a question, no definite T or F) b A proposition, truth value T (since 5 < 9 is true) c Not a proposition (it is a command, no definite T or F) d A proposition, truth value F 8 a True (T)

b True (T)

c False (F)

d False (F)

9 a

¬p

¬p ∨ q

p T

13 a Tautology

¬p

¬p ∨ p

¬q

p ∧ ¬q

b Contingency

b F (second row) c Only when p is false and q is false (fourth row)

p ∧ ¬q

¬q

c Contradiction

10 a T (third row)

11 a

p∧q

¬p

¬p ∧ p

d Contingency

¬(p ∧ q)

p∨q

¬p

(p ∨ q) ∧ ¬p

¬p

¬q

¬p ∨ ¬q

e Contingency

p∧q

¬(p ∧ q)

¬(p ∧ q) ∨r

¬r

q ∨ ¬r

p ∧ (q ∨ ¬r)

12 a True (T ∨ (F ∧ F) = T ∨ F = T) b True (¬ (T ∧ F) ∨ F = ¬F ∨ F = T ∨ F = T)

f Contingency

p∧q

¬p

(p ∧ q) ∨ ¬p

Answers 1253 mathspace.co

14 a Proposition, tautology (equivalent to p ∨ ¬p)

Extend your thinking

b Proposition, contradiction (equivalent to p ∧ ¬p)

c Proposition, contingency (equivalent to p ∨ q) 15 a ¬p ∧ (q ∧ ¬r) b p

¬p

¬r

q ∧ ¬r

¬p ∧ (q ∧ ¬r)

p∧q

¬(p ∧ q)

¬p

¬q

¬p ∨ ¬q

The final columns of both truth tables are identical (F, T, T, T). This means the two statements ¬(p ∧ q) and ¬p ∨ ¬q are logically equivalent.

c The compound statement is true only when p is false, q is true, and r is false (Row 6: F, T, F ⇒ T). In context: The motorbike is not run on petrol, and the heat is generated by wind power, and the factory is not owned by an electrician. 16 a

p∨q

¬(p ∨ q)

q∧r

(¬(p ∨ q)) ∨ (q ∧ r)

¬[(¬(p ∨ q)) ∨ (q ∧ r)]

b Contingency. The final column of the truth table contains both true and false values. c The statement is false when (p, q, r) are (T, T, T), (F, T, T), (F, F, T), or (F, F, F). d If p = T and r = F, examine Rows 2 (T, T, F) and 4 (T, F, F). In both rows, the statement evaluates to true. Therefore, the statement is true if p is true and r is false, regardless of q.

1254 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

18 A possible logical expression is ¬p ∨ q. Justification via truth table:

¬p

¬p ∨ q

The final column of the truth table for ¬p ∨ q matches the required final result, proving it is a valid expression. 19 a A possible expression is (p ∧ q)∨ (p ∧ r) ∨ (q ∧ r). b

p∧q

p∧r

q∧r

(p ∧ q) ∨ (p ∧ r) ∨ (q ∧ r)

The final column is true only when at least two of the inputs (p, q, r) are true, which matches the definition of a majority function.

2.02 Proofs with truth tables What do you remember? 1 a ii

Practice

b i

c iii

7 a

a∨b

(a ∨ b) ∧ c

3 It is a contradiction because it is not possible for both p and ¬p to be true, so the statement is always false.

4 Valid

a⇒b

¬(a ⇒ b)

2 a False

b True

c False

¬p

¬p ∨ q

d True

6 No, they are not logically equivalent. For example, the statement “If a shape is a square, then it is a rectangle” is true, but its converse “If a shape is a rectangle, then it is a square” is false.

Answers 1255 mathspace.co

p∧q

(p ∧ q) ⇒ r

p ⇔ ¬q

¬(p ⇔ ¬q)

q⇒r

p ∨ (q ⇒ r)

¬q

Contingency. The truth table shows ¬(p ⇔ ¬q) is true in some cases (TT, FF) and false in others (TF, FT), so it is neither always true nor always false.

9 a No, it is a contradiction (all F).

p⇒q

p ∧ ¬q

(p ⇒ q) ∧ (p ∧ ¬q)

¬q

b Yes, it is a tautology (all T).

a⇒b

b⇒c

a⇒c

(b ⇒ c) ⇒ (a ⇒ c)

(a ⇒ b) ⇒ ((b ⇒ c) ⇒ (a ⇒ c))

1256 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

10 a

12 a “If the team completes the research phase on time, then the project will receive a high grade.”

p⇒q

¬p

q ∨ ¬p

( p ⇒ q) ⇒ (q ∨ ¬p)

13 Invalid. This is the fallacy of the converse.

b “If the garden is watered regularly, then the plants will produce abundant fruit.”

All T in the last column, so it is a tautology. a

a∧b

(a ∧ b) ⇒ a

14 D

11 a Modus Ponens: p ⇒ q, p, therefore q.

15 A

p⇒q

16 a Let p represent a triangle that is equilateral.

p⇒q

Statement

Symbolic representation

If a triangle is equilateral, then it has three equal sides

p⇒q

Triangle ABC does not have three equal sides

¬q

Therefore, triangle ABC is not an equilateral triangle

∴¬ p

Let q represent a triangle that has three equal sides.

When p and p ⇒ q are T (Row 1), q is T, so valid. b Hypothetical syllogism: a ⇒ b, b ⇒ c, therefore a ⇒ c .

The truth table shows when q = T and p = F, p implies q is true, but p is false.

All T in the last column, so it is a tautology.

a⇒b

b⇒c

a⇒c

The original statement is true and this uses the law of contrapositive, so is a valid argument.

When a ⇒ b and b ⇒ c are T, a ⇒ c is T, so valid.

Answers 1257 mathspace.co

b Let p represent a quadrilateral that is a rhombus.

d Let p represent x4 = 256. Let q represent x = 4.

Let q represent a quadrilateral that is a parallelogram.

Let r represent x3 = 64.

Let r represent a quadrilateral with opposite sides that are parallel.

Statement If x4 = 256, then x = 4.

Symbolic representation

Statement If a quadrilateral is a rhombus, then it is a parallelogram

p⇒q

If a quadrilateral is a parallelogram, then it has opposite sides that are parallel

q⇒r

Therefore, a rhombus does not have opposite sides that are parallel

∴p⇒¬r

If x = 4, then x = 64. x = 256.

Therefore, x3 = 64.

∴r

Justification: Law of contrapositive b Conclusion: CD is a segment bisector of AB.

Let q represent 7b = 35.

Justification: Law of detachment (Modus Ponens)

1 1 b 2 . 2 2

c Conclusion: The figure is a quadrilateral. Justification: Law of detachment (Modus Ponens)

Symbolic representation

Statement If b = 5, then 7b = 35

p⇒q

1 1 If 7b = 35, then b 2 2 2

q⇒r

Therefore, if b = 5, then 1 b 2 1 2 2

d Conclusion: No conclusion can be drawn. Justification: The argument is of the form p ⇒ q, ¬p, which is an invalid argument form known as the fallacy of the inverse (denying the antecedent).

∴p⇒r

The statements are true and uses the law of syllogism. The argument is valid.

18 a Not equivalent

b Equivalent

19 a Not equivalent

b Equivalent

c Not equivalent 20 a Yes, the pairs are equivalent (De Morgan’s law).

q⇒r

17 a Conclusion: Parallelogram ABCD is not a rhombus.

c Let p represent b = 5.

Let r represent

p⇒q

The argument is valid because it uses the law of syllogism followed by the law of detachment (Modus Ponens). However, the conclusion is not guaranteed to be true because the first premise, “If x⁴ = 256, then x = 4,” is a false statement (since x could also be –4).

The first and second statements are true. The conclusion does not logically follow from the premises using the law of syllogism. Therefore, the argument is invalid.

Symbolic representation

p∨q

¬(p ∨ q)

¬p

¬q

¬p ∧ ¬q

1258 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

d Not equivalent

b No, the columns, Column 5 and Column 7, are not equivalent.

q∨r

p ∧ (q ∨ r)

p∧q

(p ∧ q) ∨ r

c Yes, the pairs are equivalent.

q∧r

p ⇒ (q ∧ r)

p⇒q

p⇒r

(p ⇒ q) ∧ (p ⇒ r)

d Yes, the pairs are equivalent.

¬p

¬p ⇒ q

p∨q

21 a a

a∨b

¬(a ∨ b)

¬a

¬b

¬a ∧ ¬b

Column 4 and Column 7 match, confirming logical equivalence (De Morgan’s law).

p⇒q

q⇒p

The last two columns differ (e.g., Row 2); therefore not logically equivalent.

Answers 1259 mathspace.co

Extend your thinking 22 Equivalent. Using De Morgan’s law: ¬( p ∧ ¬q) = ¬p ∨ ¬¬q = ¬p ∨ q, so ¬[¬( p ∧ ¬q)] = ¬(¬p ∨ q) = ¬¬p ∧ ¬q = p ∧ ¬q, matching the second expression.

24 The pair of expressions are equivalent; the columns (Column 4 and Column 6) match. This shows that negating a conditional p ⇒ q means p is true and q is false, the only case where the conditional fails.

23 The claim is false; it is not a tautology (for example, row 2 is F ). This expression is equivalent to p ⇔ q.

p⇒q

q⇒p

(p ⇒ q) ∧ (q ⇒ p)

¬(p ⇒ q)

p⇒q

q⇒r

(p ⇒ q) ∧ (q ⇒ r)

p⇒r

¬q

p ∧ ¬q

By comparing the final two columns of the completed truth table, the specific rows where they differ are: • Row 3 ( p = T, q = F, r = T ): • The statement ( p ⇒ q) ∧ (q ⇒ r) evaluates to False. • The statement p ⇒ r evaluates to True. • Row 6 ( p = F, q = T, r = F ): • The statement ( p ⇒ q) ∧ (q ⇒ r) evaluates to False. • The statement p ⇒ r evaluates to True. Therefore, they are not logically equivalent.

2.03 Boolean algebra and axioms

2 The relationships are as follows:

What do you remember?

• Conjunction (AND, ∧) corresponds to set intersection (∩).

1 The order of operations is: 1. NOT (¬) 2. AND (∧) 3. OR (∨) Expressions within brackets are always evaluated first.

1260 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

• Disjunction (OR, ∨) corresponds to set union (∪). • Negation (NOT, ¬) corresponds to set complement (Ac or A′). 3 The complement axiom is defined by these two expressions: •

A ∧ ¬A = 0

•

A ∨ ¬A = 1

9 a A ∨ A ∧ B = A (Absorption law: X ∨ X ∧ Y = X)

Practice 4 A ∧ (A ∧ B) = (A ∧ A) ∧ B (Associative law)

(A ∧ A) ∧ B = A ∧ B (Idempotent law: A ∧ A = A)

¬A ∧ B ∨ A ∧ B = B ∧ (¬A ∨ A)

5 a 1 ∧ 0 ∧ 1 ∧ 1 = 0. The AND operation results in 0 if any operand is 0. b 0 ∨ 1 ∨ 1 ∨ 0 = 1. The OR operation results in 1 if any operand is 1.

c ¬(0 ∨ 1) ∨ (1 ∧ ¬0) = ¬(1) ∨ (1 ∧ 1) = 0 ∨ 1 = 1 ¬A

A ∨ ¬A

(Complement law)

(Identity law)

(A ∧ (B ∨ ¬A) = (A ∧ B) ∨ (A ∧ ¬A) (Distributive law)

b (1 ∧ 0) ∨ (0 ∨ 1) = 0 ∨ 1 = 1

=B∧1 c

6 a ¬0 ∧ 1 = 1 ∧ 1 = 1

(Distributive law)

=A∧B∨0

(Complement law)

=A∧B

(Identity law)

d (A ∨ B) ∧ (A ∨ B) = A ∨ B (Idempotent law: X ∧ X = X) 10 a (A′ ∩ B) ∪ (A ∩ B′). b The equivalent set notation is A ∪ (B ∩ C).

It is a tautology because the output column is always 1 (Complement law).

11 The equivalent Boolean expression is F = (A ∨ ¬B) ∧ C.

8 Using the absorption law, X ∧ (X ∨ Y) = X. Therefore, B ∧ (B ∨ A) = B. Extend your thinking 12 Ask your teacher for worked solutions. 13 F = (A ∧ B ∧ C ∨ A ∧ B ∧ ¬C) ∨ ¬A ∧ B ∧ C

(Group terms)

= A ∧ B ∧ (C ∨ ¬C) ∨ ¬A ∧ B ∧ C

(Distributive law)

= (A ∧ B ∧ 1) ∨ ¬A ∧ B ∧ C

(Complement law)

= A ∧ B ∨ ¬A ∧ B ∧ C

(Identity law)

= B ∧ (A ∨ (¬A ∧ C))

(Distributive law)

= B ∧ (A ∨ C)

(Absorption law: X ∨ ¬XY = X ∨ Y )

= (B ∧ A) ∨ (B ∧ C)

(Distributive law)

= (A ∧ B) ∨ (B ∧ C)

(Commutative law)

The minimal sum-of-products form is (A ∧ B) ∨ (B ∧ C). 14 a L = (D ∧ A) ∨ (W ∧ A) b L = (D ∧ A) ∨ (W ∧ A) = A ∧ (D ∨ W)

(Distributive law)

c In set notation, this is A ∩ (D ∪ W).

2.04 Karnaugh maps What do you remember? 1 The column sequence is: 00, 01, 11, 10 2 The groups must be rectangular and contain a number of cells that is a power of 2. Valid sizes are 1, 2, 4, 8, etc. 3 a False

b True

c True

d True

Answers 1261 mathspace.co

Practice

B=0

B=1

A=0

A=1

5 a

BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

The expression F = A ∧ B ∨ ¬B ∧ C is already in its minimal sum-of-products form. The groups on the map are A ∧ B (covering minterms m6 and m7) and ¬B ∧ C (covering minterms m1 and m5). There is no simpler form.

9 a F=A∨B

b F = ¬A ∧ ¬C ∨ ¬B ∧ C

c F = ¬C ∨ A ∧ B

d F=B

b BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

10 The four 1s in the columns where BC = 00 and BC = 10 can be combined into a single group of four using wrap-around grouping. In this group, the variable A changes, the variable B changes, but the variable C is always 0. Therefore, the minimal expression is F = ¬C. 11 a S = (D ∧ A) ∨ W b

Output

7 a Minimal form: ¬A∨B. The K-map shows a group for the row where A = 0 and a group for the column where B = 1. b Minimal form: ¬C. The K-map shows a group of four 1s covering the columns where C is always 0 (while A and B change), which are the columns for BC = 00 and BC = 10.

c The K-map will have a group of 4 for the variable W, and a group of 2 for the term D ∧ A. This confirms the minimal form S = W ∨ D ∧ A. 12 There are two prime implicants: • A vertical group of two in the first column, covering cells where AB = 00, CD = 00 and AB = 10, CD = 00. In this group, B = 0, C = 0, and D = 0. The term is ¬B ∧ ¬C ∧ ¬D. • A 2x2 group in the centre, covering cells where AB is 01 or 11 and CD is 01 or 11. In this group, B = 1 and D = 1. The term is B ∧ D.

1262 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

13 The minimal sum-of-products expression is F = (¬A ∧ ¬C) ∨ (A ∧ C). The set notation for this would be (A′ ∩ C′) ∪ (A ∩ C). Extend your thinking

17 First, expand the expression using the distributive law: F = (¬A ∧ B) ∨ (¬A ∧ ¬C) ∨ (¬B ∧ B) ∨ (¬B ∧ ¬C) The term ¬B ∧ B is always 0. This gives the sum-of-products form:

14 The minimal sum-of-products form is G = (¬A ∧ ¬C) ∨ (C ∧ ¬D).

F = ¬A ∧ B ∨ ¬A ∧ ¬C ∨ ¬B ∧ ¬C

The K-map for the function is: CD = 00

CD = 01

CD = 11

CD = 10

AB = 00

AB = 01

AB = 11

AB = 10

Constructing the K-map and simplifying gives the minimal form F = ¬A ∧ B ∨ ¬B ∧ ¬C. In this case, the term ¬A ∧ ¬C is the redundant prime implicant as its 1s are already covered by the other two groups.

2.05 Binary numbers What do you remember?

There are two essential prime implicants: • The group for C ∧ ¬D is a group of four covering the last column (where CD = 10). • The group for ¬A ∧ ¬C is a 2x2 group of four covering cells where A = 0 and C = 0 (the intersection of the first two rows and the first two columns). Together, these two groups cover all the 1s in the minimal way. 15 a BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

b Each term in the long expression corresponds to a single ‘1’ on the K-map. The term A in the minimal form represents the group of four ‘1s’ in the row where A = 1. The term ¬B ∧ C represents the group of two ‘1s’ where B = 0 and C = 1. Together, these two groups cover all the individual ‘1s’ from the long expression, providing a simpler, logically equivalent function. 16 a The three prime implicants are ¬A ∧ ¬B, ¬B ∧ C, and A ∧ C. b The prime implicant ¬B ∧ C is redundant. Its two 1s are already covered by the other two essential prime implicants ( ¬A ∧ ¬B and A ∧ C). c The minimal sum-of-products expression is the sum of the essential prime implicants: F = ¬A ∧ ¬B ∨ A ∧ C

1 Base-2, digits 0 and 1 2 The base-10 system, or decimal system, has a base of 10, using digits from 0 to 9. 3 In binary, 0 represents “false” and 1 represents “true,” aligning with Boolean logic where these values denote the truth or falsity of a statement. 4 Powers of 2 (e.g., 1s, 2s, 4s) 5 02, 12, 102, 112, 1002 (Pattern: rightmost digit flips 0 to 1, then resets to 0 as next digit increases by 1). Practice 6 a It is the 8s place. b 11002 c 10102 7 1 8 Since the statement “The sky is blue” is true today, it is represented as 1 in binary (True = 1, False = 0). 9 a 1 (True, because the statement matches the scenario). b 0 (False, because the statement does not match the scenario). 10 a 1610

b 1710

c 1310

d 510

e 1410

f 1910

g 1010

h 3110

11 a 10002

b 112

c 11002

d 10012

Answers 1263 mathspace.co

12 a 7 ÷ 2 = 3 remainder 1 3 ÷ 2 = 1 remainder 1 1 ÷ 2 = 0 remainder 1 Binary: 1112 (bottom to top) b 19 ÷ 2 = 9 remainder 1 9 ÷ 2 = 4 remainder 1 4 ÷ 2 = 2 remainder 0 2 ÷ 2 = 1 remainder 0 1 ÷ 2 = 0 remainder 1 Binary: 100112 (bottom to top) 13 a Largest powers of 2: 23 = 8 (10 – 8 = 2, use 1) 22 = 4 (2 < 4, use 0) 21 = 2 (2 – 2 = 0, use 1) 20 = 1 (0 < 1, use 0) Binary: 10102 b Largest powers of 2: 24 = 16 (25 – 16 = 9, use 1) 23 = 8 (9 – 8 = 1, use 1) 22 = 4 (1 < 4, use 0) 21 = 2 (1 < 2, use 0) 20 = 1 (1 – 1 = 0, use 1) Binary: 110012 14 T he binary number 10112 has place values from right to left: 20, 21, 22, 23. So: 10112 = 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 =1×8+0×4+1×2+1×1 =8+0+2+1 = 1110 15 a 23 ÷ 2 = 11 remainder 1 11 ÷ 2 = 5 remainder 1 5 ÷ 2 = 2 remainder 1 2 ÷ 2 = 1 remainder 0 1 ÷ 2 = 0 remainder 1 Reading the remainders from bottom to top gives 101112. b Conversion to decimal: 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20 = 16 + 8 + 0 + 2 + 0 = 2610 Verification (converting back to binary): 26 ÷ 2 = 13 remainder 0 13 ÷ 2 = 6 remainder 1 6 ÷ 2 = 3 remainder 0 3 ÷ 2 = 1 remainder 1 1 ÷ 2 = 0 remainder 1

1264 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Reading the remainders up gives 110102, which matches the original. c The largest power of 2 less than 31 is 16(24). Remainder: 31 – 16 = 15. The next largest power of 2 is 8(23). Remainder: 15 – 8 = 7. The next largest power of 2 is 4(22). Remainder: 7 – 4 = 3. The next largest power of 2 is 2(21). Remainder: 3 – 2 = 1. The next largest power of 2 is 1(20). Remainder: 1 – 1 = 0. Using a ‘1’ for each power of 2 used gives the binary number 111112. 16 a 1112 17 a 1102

b 10002

c 100012

b 11102

d 100112

c 11002

Extend your thinking 18 Let k be the number of 1s. The sum of the values is k (since each 0 contributes 0). The k k mean is 0.6, so 0.6. Solving, n 10 k = 0.6 × 10 = 6. Therefore, there are 6 values of 1. In binary truth, these 1s represent “true” outcomes, suggesting 6 instances where a condition is satisfied. 19 First, convert 4210 to binary: 42 ÷ 2 = 21, remainder: 0 21 ÷ 2 = 10, remainder: 1 10 ÷ 2 = 5, remainder: 0 5 ÷ 2 = 2, remainder: 1 2 ÷ 2 = 1, remainder: 0 1 ÷ 2 = 0, remainder: 1 So, 4210 = 1010102. As a 6-bit sequence for variables A, B, C, D, E, F, it’s A = 1, B = 0, C = 1, D = 0, E = 1, F = 0 (true, false, true, false, true, false). A possible logical expression could be A ∧ ¬B ∧ C ∧ ¬D ∧ E ∧ ¬F, which is true when these conditions match. 20 Binary uses only 0 and 1 because it aligns with electronic circuits, which have two states: off (0) and on (1). This simplicity ensures reliable signal processing in computers. In Boolean logic, these correspond to false (0) and true (1), enabling binary to represent logical statements efficiently.

2.06 Logic gates

21 Place values: 26 = 64 (100 – 64 = 36, use 1), 25 = 32 (36 – 32 = 4, use 1), 24 = 16 (4 < 16, use 0),

What do you remember?

23 = 8 (4 < 8, use 0),

1 B inary values: 0 and 1. 0 means off or false, 1 means on or true.

22 = 4 (4 – 4 = 0, use 1),

21 = 2 (0, use 0),

2 = 1 (0, use 0). Binary: 11001002 Verify: 11001002 = (1 × 26) + (1 × 25) + (0 × 24) + (0 × 23) + (1 × 22) + (0 × 21) + (0 × 20)

¬A

A OR B

= 64 + 32 + 0 + 0 + 4 + 0 + 0

= 10010

Digits: 7, because 26 = 64 ≤ 100 < 27 = 128, requiring 7 bits to represent all values up to 100. 22 111111112 = (1 × 27) + (1 × 26) + (1 × 25) + (1 × 24) + (1 × 23) + (1 × 22) + (1 × 21) + (1 × 20)

4 Use a NOT gate with input A and output ¬A.

= 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 This is the maximum value for an 8-bit binary number because it uses all 8 bits as 1s, representing 28 – 1 = 256 – 1 = 255, the largest unsigned integer possible with 8 bits (range 0 to 255). 1

23 a 238 = (2 × 8 ) + (3 × 8 ) = (2 × 8) + (3 × 1)

Practice 5 a This is an OR gate. Logical expression: X ∨ Y b This is a NOT gate. Logical expression: ¬Z c This is an AND gate. Logical expression: Y∧Z 6 a

= 16 + 3 = 1910 3

b Since 2 = 8, each octal digit corresponds to a 3-bit binary group. For the number 238: • The digit 28 is 0102. • The digit 38 is 0112. Combining these gives 238 = 010 0112, or 100112.

To verify, convert the binary result to decimal: 100112 = (1 × 24) + (0 × 23) + (0 × 22) + (1 × 21) + (1 × 20)

Q∨R

P∧Q

= 16 + 0 + 0 + 2 + 1 = 1910 This matches the decimal value of 238 found in part (a), confirming the conversion is correct.

¬A

= 25510

¬R

Answers 1265 mathspace.co

b A

B∨C

A ∧ (B ∨ C)

13 a F = (¬A ∧ B ∧ ¬C) ∨ (A ∧ ¬B ∧ ¬C) ∨ (A ∧ B ∧ ¬C)

8 a i

b F = (¬A ∧ B ∧ ¬C) ∨ (A ∧ (¬B ∨ B) ∧ ¬C) Distributive law on last two terms

= (¬A ∧ B ∧ ¬C) ∨ (A ∧ 1 ∧ ¬C) Complement law: ¬B ∨ B = 1

A∨B

= (¬A ∧ B ∧ ¬C) ∨ (A ∧ ¬C) Identity law

= (¬C) ∧ ((¬A ∧ B) ∨ A)

ii 1 b i

Distributive law

= (¬C) ∧ ((A ∨ ¬A) ∧ (A ∨ B))

¬A ∨B

Distributive law

= (¬C) ∧ (1 ∧ (A ∨ B))

ii 0

Complement law

9 a i

=¬ C ∧ (A ∨ B)

Output

Identity law

The simplified expression is F = (A ∨ B) ∧ ¬C. ii 0 b i

c A

P ∧ (¬Q ∇ R)

ii 1 C

10 The expression is Y = ¬((A ∧ B) ∨ (¬C)). 11 a A ∨ (A ∧ B) = A. This is a direct application of the Absorption law. b ¬A ∨ (A ∧ B) = (¬A ∨ A) ∧ (¬A ∨ B) (Distributive law). Since ¬A ∨ A = 1 (Complement law), this becomes 1 ∧ (¬A ∨ B). This simplifies to ¬A ∨ B (Identity law). 12 a The Karnaugh map for the given truth table is:

B=0

B=1

A=0

A=1

Groups: 1. The column where B = 0 forms a group (simplifies to ¬B). 2. The row where A = 1 forms a group (simplifies to A).

BC = 00

BC = 01

BC = 11

BC = 10

A=0

15 The simplified expression is (A ∧ B) ∨ C.

A=1

The simplified circuit diagram is:

The minimal sum-of-products expression is F = ¬C ∨ (A ∧ ¬B).

Simplified: A ∨ ¬B

A B

Output AND C

1266 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking

18 Expression: (A ∧ B) ∧ (B ∨ C). K-map:

16 Original circuit: An OR gate for A ∨ B, a NOT gate for ¬(A ∨ B), and an AND gate combining the result with C. Simplification using De Morgan’s law: ¬(A ∨ B) ∧ C = (¬A ∧ ¬B) ∧ C. Using the associative law, this is ¬A ∧ ¬B ∧ C. Simplified circuit: Two NOT gates for ¬A and ¬B, and a single 3-input AND gate for the final output. Comparison: The original circuit uses 3 gates (OR, NOT, AND). The simplified circuit also uses 3 gates (two NOTs, one AND). While the gate count is the same, the simplified circuit may be more efficient in some architectures by removing one level of gate delay. 17 The Karnaugh map for the given truth table is: BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Identify the largest possible groups of 1s to cover all the 1s.

BC = 00

BC = 01

BC = 11

BC = 10

A=0

A=1

Group: Cells where A = 1, BC = 11 and A = 1, BC = 10 form a group (simplifies to A ∧ B). Simplified: A ∧ B Circuit: One AND gate with inputs A and B. Output: For A = 1, B = 0, C = 1, 1 ∧ 0 = 0. Redundancy: The original circuit’s OR gate (B ∨ C) and second AND gate are redundant. The output depends only on A ∧ B, as the term B ∨ C does not affect the output when A ∧ B = 0. The simplified circuit with a single AND gate is minimal. 19 Original circuit: One AND gate for (B ∧ C) and one OR gate to combine the result with A (2 gates total). With the constraint C = 0, the expression is simplified step-by-step: A ∨ (B ∧ C) = A ∨ (B ∧ 0) Substitute C = 0

• Group 1: The two cells in the column BC = 11. Here, B = 1 and C = 1 are constant. This group simplifies to B ∧ C. • Group 2: The two cells at (A = 1, BC = 01) and (A = 1, BC = 11). Here, A = 1 and C = 1 are constant. This group simplifies to A ∧ C.

= A ∨ 0 Annulment Law (since B ∧ 0 = 0) =A

Identity Law (since A ∨ 0 = A)

Simplified form: A single wire for input A (0 gates).

• Group 3: The two cells at (A = 0, BC = 00) and (A = 0, BC = 10) are adjacent. Here, A = 0 and C = 0 are constant. This group simplifies to ¬A ∧ ¬C.

Constraint effect: The constraint C = 0 makes the AND gate’s output permanently 0, rendering it and the subsequent OR gate redundant. The entire circuit’s logic reduces to just the input A. The unused input D has no impact as it is not part of the expression.

The simplified expression is (B ∧ C) ∨ (A ∧ C) ∨ (¬A ∧ ¬C).

2.07 Design and interpret algorithms

Simplified circuit: The circuit requires two NOT gates (for ¬A and ¬C), three 2-input AND gates (for each term), and one 3-input OR gate to combine the results. A B

A’C’ AC Output

What do you remember? 1 a Pseudocode

b Sequence

c Flowchart

d Tracing

2 a False

b True

c False

3 a ii

b iv

c i

d iii

BC C

Answers 1267 mathspace.co

Practice 4

Step

Instruction

1 START 2 SET n = 4 3 IF n > 2 THEN 4 PRINT n 5 END IF 6 END

Type

Plug in the toaster.

Sequence

If the bread is fresh, slice it; else, use it as is.

Decision

Set the timer to 2 minutes.

Sequence

Toast the bread until the Repetition timer finishes.

8 The flowchart sets count to 0. It checks if count < 3. If true, it prints count and increments it, repeating until count reaches 3. It prints 0, 1, and 2, then ends.

9 The algorithm starts by setting count to 0 and num to 10. It then enters a loop that continues while num is greater than 0. In each iteration, it checks if num is even (divisible by 2 with no remainder). If true, it increments count by 1. Then, it decreases num by 1. When num reaches 0, it prints count, which is the number of even numbers from 10 down to 1 (i.e., 10, 8, 6, 4, 2 gives 5).

1 START 2 INPUT number 3 IF number MOD 2 = 0 THEN 4 PRINT ‘Even’ 5 ELSE 6 PRINT ‘Odd’ 7 END IF 8 END 6

10 Start the algorithm. Set p to 1 and q to 8. Check if p < q. If yes, set r to q – p; if no, set r to p – q. Print r. End the algorithm.

1 START 2 SET count = 1 3 WHILE count <= 5 4 PRINT count 5 SET count = count + 1 6 END WHILE 7 END 12

Step

Action

11 C

sum

SET sum = 0, i = 2

Is i ≤ 4? (2 ≤ 4)

sum = 0 + 2, i = 2 + 1

Is i ≤ 4? (3 ≤ 4)

sum = 2 + 3, i = 3 + 1

Is i ≤ 4? (4 ≤ 4)

sum = 5 + 4, i = 4 + 1

Is i ≤ 4? (5 ≤ 4)

PRINT sum

The output is 9 (sum of 2, 3, 4).

1268 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Output

13 The algorithm calculates the factorial of 3(1 × 2 × 3). Loop iteration Count

product (Start of loop)

Action: product = product * count

product (End of loop)

Start

N/A

SET product = 1

product = 1 * 1

product = 1 * 2

product = 2 * 3

The final output is 6. 14

16 1 START 2 SET a = 2 3 SET b = -5 4 SET c = 3 5 SET discriminant = (b * b) (4 * a * c) 6 SET root = (-b + SQRT(discriminant)) / (2 * a) 7 PRINT root 8 END

15 1 START 2 INPUT a, b, c 3 SET min = a 4 IF b < min THEN 5 SET min = b 6 END IF 7 IF c < min THEN 8 SET min = c 9 END IF 10 PRINT min 11 END

Action

INPUT x, y

IF x < y THEN (6 < 2)

PRINT y (since 6 < 2 is false)

END

Step

Output

Output: 2 17 Inputs: a is 5, b is 2. 5 > 2 is true, so swap: a becomes 2, b becomes 5. Output: 2,5. 18 With secret 4 and guess 6:6 = 4 is false, 6 < 4 is false, so it prints ‘Too high’. 19 C

Answers 1269 mathspace.co

Extend your thinking 20 Tracing with n set to 1: Step

Action

sum

Output

SET sum = 0, i = 1

Is i ≤ 5? (1 ≤ 5)

sum = 0 + 1, i = 1 + 1

Is i ≤ 5? (2 ≤ 5)

sum = 1 + 2, i = 2 + 1

Is i ≤ 5? (3 ≤ 5)

sum = 3 + 3, i = 3 + 1

Is i ≤ 5? (4 ≤ 5)

sum = 6 + 4, i = 4 + 1

Is i ≤ 5? (5 ≤ 5)

sum = 10 + 5, i = 5 + 1

Is i ≤ 5? (6 ≤ 5)

Outputs are 1, 2, 3, 4, 5. Mean is 3 21

Step

Action

Output

INPUT x

WHILE x < 5 (3 < 5)

Alternative: Use n less than or equal to 0 and reverse outputs. 1 START 2 INPUT n 3 IF n <= 0 THEN 4 PRINT ‘Non-positive’ 5 ELSE 6 PRINT ‘Positive’ 7 END IF 8 END

Infinite loop: Outputs ‘3’ repeatedly since x never changes. Median of outputs (conceptually infinite 3s) is 3. Improvement: Add x incremented by 1 inside the loop to increase x, stopping at x = 5 after printing 3, 4. 22 O riginal: Checks if n is greater than 0, prints ‘Positive’ if true, ‘Non-positive’ if false. Trace (n set to 4): 4 is greater than 0, prints ‘Positive’. 1 START 2 INPUT n 3 IF n > 0 THEN 4 PRINT ‘Positive’ 5 ELSE 6 PRINT ‘Non-positive’ 7 END IF 8 END

1270 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Trace (n set to 4): 4 less than or equal to 0 is false, prints ‘Positive’. Both yield ‘Positive’, confirming equivalence. 23 1 START 2 INPUT n 3 SET factorial = 1 4 SET i = 1 5 WHILE i <= n 6 SET factorial = factorial * i 7 SET i = i + 1 8 END WHILE 9 PRINT factorial 10 END

3.01 Introduction to sequences

For n = 4, the steps are: • When i = 1, factorial becomes 1 × 1 = 1. • When i = 2, factorial becomes 1 × 2 = 2. • When i = 3, factorial becomes 2 × 3 = 6. • When i = 4, factorial becomes 6 × 4 = 24. 24 Flowchart: Start leads to Input n, then a decision ‘Is n > 0?’ If yes, leads to Print ‘Positive’. If no, leads to a decision ‘Is n < 0?’ If yes, leads to Print ‘Negative’. If no, leads to Print ‘Zero’. All paths lead to End. START

b i T he sequence starts at 64 and each term is subsequently halved. ii 1 c i T he sequence starts at 2 and each term goes up by 2 in magnitude, while the sign of the terms alternate. ii 14 d i The value in the next term is obtained by

INPUT n

adding to

No Is n > 0?

Is Yes n < 0? No

PRINT Positive

1 a i T he sequence starts at −1 and each term increases by 2. ii 9

The final output is 24.

Yes

What do you remember?

PRINT Zero

PRINT Negative

END Test with n = – 2: – 2 > 0 is false, – 2 < 0 is true, prints ‘Negative’. 25

to the previous term.

ii 2 a The number of squares in the next step is 3 more squares than the previous step. b 16 3 a The number of circles in the next step is 4 more circles than the previous step. b 17 4 a t5

b t10

d tn-1

e tn+1

c tn

Practice 1 START 2 INPUT a, b, c 3 IF a > b THEN 4 SWAP a with b 5 END IF 6 IF b > c THEN 7 SWAP b with c 8 END IF 9 IF a > b THEN 10 SWAP a with b 11 END IF 12 SET median = b 13 PRINT median 14 END

For 4, 2, 7: First, 4 > 2, swap: 2, 4, 7. Then, 4 < 7, no swap. Then, 2 < 4, no swap. Median is b = 4. In context, 4 is the central value, with 2 and 7 showing a range of 7 – 2 = 5.

5 a 5

b 125 triangles

6 a tn = tn−1 + 5, where t1 = −1 b tn = tn−1 − 2, where t1 = 15 c tn = 5tn−1, where t1 = 1 d tn = 2tn–1, where t1 = 3 7 a 2, 6, 18, 54, 162 b −10, −3, 4, 11, 18 c −100, −20, −4, −0.8, −0.16 d 2.5, 4, 7, 13, 25 8 a 11, 15, 19, 23

b 67

9 a Tn = 7n − 5

b Tn = −6n + 26

Answers 1271 mathspace.co

10 a The volume increases by a factor of 3 each minute. b

Time (minutes)

Volume (mL)

b 14 16 a i

243

729

c Let Vn be the volume after n minutes. Vn = 3Vn−1, where V1 = 3. d 6561 mL 11 a The balance increases by a factor of 2 each week. b

15 a The first term is T1 = 2. The recursive rule is Tn = Tn−1 + 3.

Time (weeks)

Balance (dollars)

128

c Let Bn be the balance after n weeks. Bn = 2Bn−1, where B1 = 4.

Step number (t)

1 2 3 4

5 10

Number of 4 8 12 16 20 40 matchsticks (mt) ii mt = 4t b i

Step number (t)

1 2 3 4

5 10

Number of 3 6 9 12 15 30 matchsticks (mt) ii mt = 3t c i

Step number (t) Number of matchsticks (mt)

1 2 3 4

5 10

4 7 10 13 16 31

ii mt = 3t + 1 d i

Step number (t)

1 2 3 4

5 10

Number of 3 5 7 9 matchsticks (mt)

11 21

ii mt = 2t + 1 17 a $21 b Tn = Tn−1 + 1.95, T0 = 1.5

d $512 12 a The recursive rule is tn = tn−1 + tn − 2, where t1 = 11 and t2 = 22. n

b The recursive equation is Pn = 2Pn−1, where P1 = 2. The population at the end of one year (12 months) is 4096 rabbits. 14 a tn = tn−1 + 5

18 a One possible rule is Then, S10 = 55 spots

b 11, 22, 33, 55, 88 13 a

Extend your thinking

b tn = tn−1 + tn−2

c tn = −tn−1 + 7

e tn = tn−1 − 6

f tn = tn−1 tn−2

g tn = −5tn−1 − 3

1272 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

19 a The volume increases by a factor of 2 each day. b The y-values (cumulative volumes) of the points form a sequence. The sequence of cumulative volumes is: 4, 8, 16, 32, . . . This is a geometric sequence where the first term is 4 and the common ratio is 2. c Yes, the relationship is a function. The domain is {1, 2, 3, 4, 5, 6, 7}. d 256 mL 20 m1 = 2000, mn = (0.76) mn−1

10 a tn+1 = tn + 2, t1 = 3

3.02 Arithmetic sequences

b 17 times What do you remember?

c 41

1 a No

b No

2 a b = 11

b b = −1.2 c

3 a 20, 24

b 8, 9.5

4 a b=2

c Yes

d No d b = −50

c −10, −14 d b 7

b i

ii b = 7 c i 3, 4.4, 5.8, 7.2

9, 5, 1, −3

ii b = −4 d i

ii b = 1.4

b t14 = 12 + 13b c b=5 12 a 8, 12, 16, 20, 24 The common difference between any two terms in an arithmetic sequence can be determined. So, b = 20 − 16 = 4.

Practice 5 a i −2, 5, 12, 19

11 a 13 times

11, 10.5, 10, 9.5

From there, the common difference can be added to determine terms after the given ones and subtracted to determine previous terms. b −5, 0, 5, 10, 15 To go from t2 to t4, add b two times, so

6 a −10, −6, −2, 2 b −8, −10, −12, −14

. From there, the common

c a, a + b, a + 2b, a + 3b 7 a 90 words b tn+1 = tn + 10, t0 = 30

difference can be added to determine terms after the given ones and subtracted to determine previous terms. c −12, −6, 0, 6, 12, 18, 24

c sn+1 = sn + 70, s0 = 30

To go from t1 to t7, add b six times, so

8 a i b=5

. From there, the

ii tn+1 = tn + 5, t1 = 5

common difference can be added to determine terms after the given ones and subtracted to determine previous terms.

iii tn = 5n b i b = −7 ii tn+1 = tn − 7, t1 = −1

d 25, 20, 15, 10, 5 To go from t1 to t5, add b four times, 5 – 25 so b = = –5. From there, the common 4 difference can be added to determine terms after the given ones and subtracted to determine previous terms.

iii tn = −7n + 6 c i b=3 ii tn+1 = tn + 3, t1 = 2 iii tn = 3n − 1 d i b=−4 ii tn+1 = tn − 4, t1 = 2

13 a 12

iii tn = − 4n + 6

14 x = −7

e i b = 1.5 ii tn+1 = tn + 1.5, t1 = − 2 iii tn = 1.5n − 3.5

c tn = 5n − 2

ii tn+1 = tn − 3, t1 = 12

16 a $23 700

iii tn = −3n + 15 c t9 = 32 + 8b

15 a 3, 8, 13, 18 b tn = 3 + (n − 1)5

f i b=−3

9 a 4 times

b 5

b t5 = 32 + 4b d b = −12

b $15 900

c $9100

17 a $500 000 b tn+1 = tn + 60 000, t1 = 500 000 c $740 000

Answers 1273 mathspace.co

18 a t1 = 2540

3 a Geometric sequence

b tn+1 = tn − 310, t1 = 2540

b Geometric sequence

c 8 weeks

c Arithmetic sequence

19 a $22.25

b tn = tn−1 + 2.50, t0 = 2.25

d Neither

Extend your thinking

4 a i M ultiply by 2 to get from one term to the next

20 k = 32

ii 80 160

Answers may vary. An example explanation: To get the kth term, the common difference was subtracted from the first term a certain number of times. To solve this like an equation, begin by subtracting the first term from the kth term:

b i D ivide by 10 to get from one term to the next ii 4, 0.4 Practice

−530 − 28 = −558

5 a t2 = 4, t3 = 8, t4 = 16

Then, divide by −18 to determine how many times the common difference was subtracted:

b t1 = −268, t2 = 67,

−558 ÷ −18 = 31 This means the common difference was added 31 times to the first term, so k = 1 + 31 = 32. 21 a Sometimes — all positive if t1 > 0 and b > 0, a mix if t1 and b have different signs or if t1 > 0 and b < 0 but not large enough to make terms negative, and all negative if t1 < 0 and b < 0. b Always — the common difference means that it has a constant rate of change, which creates a linear pattern. c Never — if b > 0, then it will always be increasing, if b < 0, then it will always be decreasing, if b = 0 then it will be constant. It cannot be both increasing and decreasing. 22 s n+1 = sn − 3, s0 = 1. The original sequence has rule tn = 3n − 1. Reflecting across the n-axis negates the term value, so the new sequence sn has the rule sn = −(tn) = −(3n − 1) = −3n + 1. This corresponds to a recursive rule with a common difference of −3 and an initial term s0 = 1.

3.03 Geometric sequences

c t2 = 120, t3 = 180, t4 = 270 d 6 a Multiplying by 3

b 108 circles

7 a Multiplying by 4

b 128 hexagons

8 a i Sequence: 5, 10, 20, 40, 80 ii Geometric b i Sequence: 3, 7, 11, 15, 19 ii Arithmetic c i Sequence: 81, 27, 9, 3, 1 ii Geometric d i Sequence: 1, 8, 22, 50, 106 ii Neither 9 a tn = 3 × tn−1, t1 = 2 b t

c t

d tn = − 4 × tn−1, t1 = 3 e tn = −2 × tn−1, t1 = −1 f t

10 a −5, 20, −80, 320, −1280 What do you remember? 1 a 108, 324 c

b −768, 3072 d

2 a 4

b −3

d 7

1274 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b 11 a 12 a 8.48 cm b 10.10 cm c Hn = 1.06 × Hn−1, H0 = 8

13 a $77 400 b $57 245.04 c Vn = 0.86Vn−1, V0 = 90 000 d $19 917.14

• Graph D matches Rule (iv): It shows a diverging oscillating sequence (common ratio is less than −1) with a starting value of 2. Rule (iv) is tn = 2 × (−2)n. Extend your thinking

14 a $3.74 b $3.94 c Vn = 1.026 × Vn−1, V0 = 3.65 15 a ttn+1 n 1 =

tn, t0 = 2

b 3.13 cm

23 a b The process involves using the general rule for a geometric sequence, tn = t1 × an−1 . The first term is t1 = 27 and the common ratio is . Substituting the values to find the 30th

16 a 3 times term gives

b a=4 c tn = 4 × tn−1, t1 = 6

. To solve, 27 is rewritten as 33,

17 a 2 times

leading to the expression

b a=±4

equals

c tn = 4 × tn−1, t1 = 3 d tn = − 4 × tn−1, t1 = 3 18 a

b 1

19 a 62 500 b t

. This simplifies

t t

20 a 125 n

b tn = 125 × (0.8) c 13.42 21 a a = 2 b 2 c tn = 2 × (2)n

22 • Graph A matches Rule (i): It shows exponential decay (common ratio is between 0 and 1) with a starting value (t0) of 40. Rule (i) is tn = 40 × (0.5)n. • Graph B matches Rule (iii): It shows an oscillating sequence (common ratio is negative) that starts at 20. Rule (iii) is tn = 20 × (−0.8)n. • Graph C matches Rule (ii): It shows exponential growth (common ratio is greater than 1) with a starting value of 5. Rule (ii) is tn = 5 × (1.5)n.

, which

24 a The terms increase more and more rapidly towards infinity, each time the common ratio is applied it produces a proportionately large number to then again apply the common ratio to. b The terms will remain constant, always the same value as the initial term, since multiplying by 1 results in the same number. c The terms alternate between positive and negative. This is because multiplying a term by a negative common ratio reverses its sign. d The terms get closer to zero but never quite reach it. This is because each time the common ratio is applied the result is a number that has a smaller magnitude than the number before it. If this continues, the terms will get smaller and smaller but will never reach zero because nothing is ever multiplied by zero. e Similar to the case when equal to 1, but instead, the terms alternate between the positive and negative version of the initial term. f S imilar to the case when greater than 1, but instead, the terms alternate between positive and negative with the magnitude increasing more and more rapidly with each new term.

Answers 1275 mathspace.co

3.04 Applications of sequences

11 a Vn = 5000 b Vn+1 = Vn − 150, V0 = 5000

What do you remember? 1 a

c 33 days 2

12 a $2800

Day

Area of floor left to pave (m )

800

b Vn = 2800 + 150n

750

c 19 years

700

13 a 90 words

650

b Tn+1 = Tn + 10, T0 = 30

600

c n = 47

b C 2 a $4800 b $700 3 a $30 000 b $2310 c 7.7% Practice 4 a $34.00 b tn+1 = tn + 2.10, t0 = 2.50 5 a $340 b un+1 = un + 40, u0 = 220 c vn+1 = vn + 40, v0 = 440 6 a $3000 b $220 c Vn = 3000 + 220n d V12 = $5640 7 a $15 000 b 8% c Vn = 15 000 − 1200n d $4200 8 a $300 b $2200 c $1900 d Vn+1 = Vn + 300, V0 = 1900 9 a an+1 = an + 7, a0 = 10 b an = 10 + 7n c 52 animals 10 a $43 920 b c y=9

1276 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

d Wn+1 = Wn + 70, W0 = 30 14 a a = 361 − 6d c d = 35

b a = 536 − 11d d a = 151

e n = 22 15 a $25 b $800 c Vn+1 = Vn − 25, V0 = 800 16 a $40 b Vn+1 = Vn + 40, V0 = 60 c Vn = 60 + 40n d $860 17 a $6 b V0 = $22 000 c Vn+1 = Vn − 6, V0 = 22 000 d Vn =22 000 − 6n e n = 2750 hours 18 a $31 b Tn+1 = Tn + 3, T0 = 10 c $318 d n = 20 19 a 136°C b Tn+1 = Tn − 2.5, T0 = 156 c 75°C d 52 minutes 20 a Vn+1 = Vn − 12, V0 = 70 000 b $360 c 900 days 21 a $7260 b $1260 c 24 years 22 a 14 years

b 9 years

23 21 years

Practice

24 a The annual rate is 12%. Since interest is calculated monthly, the monthly rate is

4 a S10 = 140 c S10 = 280 5 a S10 = 47.5 c S10 = 85

b Bn+1 = 1.01Bn − 400, B0 = 15000 25 a Vn+1 = 0.80Vn, V0 = 2500

7 a S15 = 450

b S15 = −45

c S15 = 105

d S15 = 750

9 a S40 = 4920

b S40 = −2260

10 a b = 3

b S20 = 650

11 a n = 18

b S18 = 450

b $1280 c 5 years

26 a Vn+1 = 1.06Vn + 200, V0 = 500 b $973.80 Extend your thinking 27 a The recurrence relation is Vn+1 = 0.92Vn, with V0 = 180. b The value under Model B is first less than Model A in the first year (n = 1). c This happens in the 6th year. Model B predicts it will happen sooner, by 2 years. 28 a The interest is $200 per year on a principal or 10% p.a.

b The common ratio is 1.1 c The geometric investment first exceeds the arithmetic investment in the 2nd year. 29 a The growth factor is R = 1.08. The withdrawal is $50. The recurrence relation is Bn+1 = 1.08Bn − 50, with B0 = 1000. b It will be in the 28th year

3.05 Arithmetic series

12 n = 10 13 65 14 738 15 Extend your thinking 16 The sum of an arithmetic series is calculated as n Sn = (2t1 + (n – 1)b) 2 where n is the number of terms, t1 is the first term and b is the difference between consecutive values. Substituting t1 = 1 and b = 2:

What do you remember? 1 a n−1 b tn 2 a True b False

b S10 = −53.75 d S10 = 5

6 S20 = 210π

c $14 497.50

of $2000.

b S10 = 0 d S10 = −310

n (2 × 1 + (n – 1) × 2) 2 n = (2 + 2n – 2) 2 n = (2n) 2

Sn =

= n2

17 180 18 First term: t1 = 1.5, common difference: b = 3 19 The total revenue is $29 600.

c False 3 a The first term is 7. b The last term is 64. c There are 20 terms.

Answers 1277 mathspace.co

3.06 Geometric series 12 a

What do you remember?

b 2555 1 a

13 7 174 449 14 a 242 b

c 2 a True

b False

c True

d False

15 a $8191 b $536 870 911 16 a 9516 m3

3 a The first term is 5. b The common ratio is 2. c The number of terms is 10.

b 9984 m3 17 Dn = 120 − 112

Practice

18 For Series A:

4 88 572

• S5 = 5(25 − 1) = 155

• S20 = 5(220 − 1) = 5 242 875

6 21

For Series B:

7 a 14.29

• S10 = 5(210 − 1) = 5115

b − 204.92

b 10.0 t1 = 327 680

10 a i n = 10 b i n=6

• S10 = 250(1 − 0.810) ≈ 223.16 • S20 = 250(1 − 0.820) ≈ 247.12

8 a

• S5 = 250(1 − 0.85) = 168.08

ii S10 = − 1364 1365 ii S6 16

11 a n = 9 b

For Series A, the partial sums grow very quickly and without bound. This is because the common ratio a = 2 has an absolute value greater than 1 (∣a∣ > 1), causing the terms of the series to become larger and the sum to diverge. For Series B, the partial sums increase but appear to be approaching a fixed value. This is because the common ratio a = 0.8 has an absolute value less than 1 (∣a∣ < 1). As n increases, the term an gets closer to zero, causing the sum to converge to a limit of .

c 85.333

1278 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Extend your thinking 19 M areike incorrectly generalises the expression for the amount after n months too soon. Instead, they should have stopped at step 6 , and then found an expression for the third month so that they could have noticed a pattern. The correct working is given:

t0 = 900 000

Amount after 0 months

t1 = 900 000 (1 + 0.0075) − M

Amount after 1 month

= 900 000 (1.0075) − M

1 + 0.0075 = 1.0075

t2 = (900 000 (1.0075) − M ) (1.0075) − M

= 900 000 (1.0075)2 − M (1.0075) − M

Amount after 2 months Distributive property of equality

6 7

Amount after 3 months

= 900 000 (1.0075)3 − M (1.0075)2 − M (1.0075) − M

Distributive property of equality Distributive property of equality

Geometric series formula

7 a 57 20

is the

expression for the amount in her account after n months. Using that to solve for the amount in her account after 3 years we get t36 = $22 908.61, where n = 36 since it is compounded monthly.

3.07 Applications of sequences and series What do you remember? 1 a Sequence

b Series

c Sequence

c $250

b 9500 d 580

8 a tn+1 = tn − 200, t0 = 15 000, $14 200 b tn+1 = 1.03 tn, t0 = 5000, $5463.64 9 a Arithmetic series b Arithmetic sequence c Geometric sequence 10 a 40

b $3370.80

11 a 3.072 m

b 20.17 m

Extend your thinking 12 8 years

2 a Arithmetic

b Geometric

c Arithmetic

d Geometric

3 a b = 10

b a = 1.05

4 a True

b False

13 Plan B, by $377.31 14 a 100 L

b Day 13

15 Plan X: $240, Plan Y: $177.16, Plan X by $62.84

4.01 Pigeon-hole principle Practice 5 a $160 c $920 6 a $2339.72 c 32.768 cm

b $17 000 d 50 b 666 d $11 236

What do you remember? 1 a True b False. It requires n > k. c True d False. It only ensures at least one container has more than one item, not all.

Answers 1279 mathspace.co

2 T he pigeon-hole principle requires n > k to guarantee that at least one container has more than one item. Here, n = 4 and k = 5, so 4 < 5. This means it is possible to place one pigeon in each of four holes, leaving one hole empty, and no hole needs more than one pigeon.

Extend your thinking 15 Minimum: 7 × (50 − 1) + 1 = 344 Maximum: 7 × 60 = 420 Range: 344 ≤ n ≤ 420 16 a 4 socks b 27 socks

represents the average number of

items per container. The principle states that at least one container must have a number of items greater than or equal to this average. Therefore, the minimum is the ceiling of 4, which is 4.

5 Since 51 > 50, the pigeon-hole principle states that with 51 students (items) and 50 lockers (containers), at least one locker must contain more than one student. 6 3 7 Y es, since 101 > 100, the pigeon-hole principle ensures at least one score (container) has more than one student. 8 There are 10 possible pet counts (containers) and 73 members (items). Since 73 > 10, the pigeon-hole principle guarantees at least one pet count has more than one member, so at least two members share the same number of pets. , so at least 2 people must share the same initials.

2 a 30

b 13

c 4

d 15

e 2

3 a i 8

ii 5

b 10

c 13

4 a 30

b 5

c 22

5 a 55

b 22

c 22

d 25

e 8

f 7

g 9

h 3

i 48

iii 3

Practice 6 a ∣R∣ = 7

b ∣L∣ = 7

c ∣R ∩ L∣ = 3 d Ask your teacher for worked solutions. 7 a 170

b 50

c 120

d Compare the result 120 with the total count from the Venn diagram regions that make up the union: 30 + 50 + 40 = 120

10 , so at least 17

12 n = 19 × 5 + 1 = 96 13

4.02 I nclusion-exclusion principle

1 ∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣

4 2

18 Ask your teacher for worked solutions.

What do you remember?

Practice

17 Ask your teacher for worked solutions.

, so at least 143

14 Let the selected integers be the “pigeons”. Define the “pigeon-holes” as 100 sets: {1, 101}, {2, 102},. . .,{100, 200}. Each of these sets contains two integers whose difference is 100. Since we are selecting 101 distinct integers (pigeons) and there are only 100 pigeon-holes (sets), by the pigeon-hole principle, at least two of the selected integers must belong to the same set. As the numbers in each set have a difference of 100, this guarantees there is a pair of numbers whose difference is exactly 100.

1280 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

8 a 105

b 125

c 45

9 a 160

b 40

10 a 12

b 19

11 a 60

b 55

12 45 13 n(A ∖ B) = r − q 14 50 15 a 120

b 30

c 30

16 a 55%

b 110

c 15%

17 a 52 pet owners

6 1024

7 60 8 27 Dog

Cat

15 10 12

9 a 27 10 1 413 720 passwords 11 a 18 codes

b 12 codes

12 24

Bird

13 n = 4

14 a 12 5

b 8

15 Addition principle

Extend your thinking 18 a 40

b $75

b 100

19 190 20 196

16 Addition principle. The choices are mutually exclusive (the student picks one humanities subject OR one science subject, not both types simultaneously). When choices are ‘OR’ options from distinct sets, the addition principle is used. 17 10

21 797

18 33

22 4577

19 16

23 128

20 19

4.03 M ultiplication and addition principles

21 363 840 22 49 ways

What do you remember?

23 11

1 a 18

b 3 × 3 × 2 = 18

2 a 12

b 3 × 2 × 2 = 12

Extend your thinking 24 320 unique teams

3 a Multiplication principle b Addition principle

25 a 144

c Multiplication principle

26 • Direct routes: 2 roads directly from City A to City B

d Addition principle 4 a 9600

b 200

b 120

• Routes via Town C: 2 roads from City A to Town C and 3 roads from Town C to City B • Routes via Town D: 3 roads from City A to Town D and 4 roads from Town D to City B

Practice 5 a

C, B

P, B

W, B

C, G

P, G

W, G

C, R

P, R

W, R

C, Y

P, Y

W, Y

This configuration provides exactly 20 possible routes from City A to City B.

b 12

Answers 1281 mathspace.co

4.04 Arrangements and factorial notation What do you remember? 1 Factorial notation, n!, represents the product of all positive integers from 1 up to n. In the context of counting arrangements, n! gives the total number of ways to arrange n distinct items in a sequence.

23 a 120 b Blake can make 5! = 120 arrangements of his name. His result is the same as Grace’s because both names consist of 5 distinct letters, and the number of arrangements for n distinct items is n!. 24 a 30

b 20

25 Trace for n = 4: • INPUT n = 4

2 1

• SET factorialResult := 1

3 6

• IF n = 0 (4 = 0 is False) THEN go to ELSE

5 65 536

• FOR i FROM 1 TO 4: • i = 1 : factorialResult := 1 ∗ 1 = 1 • i = 2 : factorialResult := 1 ∗ 2 = 2 • i = 3 : factorialResult := 2 ∗ 3 = 6 • i = 4 : factorialResult := 6 ∗ 4 = 24

6 262 × 104 = 6 760 000

• ENDFOR

4 216 Practice

• ENDIF

7 a 362 880 ways

• OUTPUT factorialResult (which is 24)

b 5040 ways

The final output is 24.

c 17 280 ways

26 a Trace for n = 5, r = 2:

8 24

• INPUT n = 5, r = 2

9 120

• SET arrangementResult := 1

10 2652

• FOR i FROM 0 TO 1: • i = 0 : arrangementResult := 1 ∗ (5 − 0) = 5 • i = 1 : arrangementResult := 5 ∗ (5 − 1) = 20

11 a 60

b 108

12 90

• ENDFOR

• OUTPUT arrangementResult (which is 20)

13 1440 14 a 120

b 8

15 a 210

b 132

16 a 6

b 720

17 a

b 46

18 a False

b True

c True

d True

19 a (n − 2)(n − 3)(n − 4) . . . × 3 × 2 × 1

b (p + 2)(p + 1) p (p − 1)(p − 2) . . . × 3 × 2 × 1 c (n + 1) n (n − 1) . . . × 3 × 2 × 1

d (k − 25)(k − 26)(k − 27) . . . × 3 × 2 × 1

20 a n(n − 1)

b n2 + n

21 a

b –

22 n = 4

1282 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b The algorithm calculates P (6, 3) = 6 × 5 × 4 = 120. The final output is 120. c This line multiplies the current result by a decreasing sequence of numbers starting from n. In the first loop step (i = 0), it multiplies by n. In the second step (i = 1), it multiplies by n − 1, and so on. This carries out the calculation for a permutation, such as n × (n − 1) × (n − 2) × . . . Extend your thinking 27 21 168 28 a 360 360 29 24 30 n = 16

b 14 400

c 14 400

4.05 Permutations

The correct approach is to use the multiplication principle. Since there are 3 possible answers for each of the 6 questions, the total number of different ways to answer the exam is 3 × 3 × 3 × 3 × 3 × 3 = 36 = 729.

What do you remember? 1 XY, XZ, YX, YZ, ZX, ZY 2 a False

b True

c True

d False

3 a 60

b 1

c 39

d 12!

4 a n (n − 1)

b n!

18 41 19 Ask your teacher for worked solutions.

4.06 Permutations with restrictions What do you remember?

Practice

1 a True

b True

c False

d False

e False

5 a 120 b Multiplication principle or 5! = 120

2 a iv b ii

6 24

c i

7 210

d iii

8 11 880

e v 3 60

9 6 10 a 5040

b 10! = 3 628 800

4 60

11 a 210

b 30

Practice

12 a 60

b 6

5 45 360

13 a 1680

b 12

14 a 210

b 6

15 a 200 16 a n = 8

6 1260 c 6

b 775 b n=3

c n=5

7 a 12 600

b 60

8 a 83 160

b 630

9 a 2520

b 6

Extend your thinking

10 210

17 The permutation formula, nPr, is used for arranging a selection of distinct items from a set without replacement.

11 20

This scenario does not meet these conditions for two main reasons:

12 12 ways. If C is before D, then 6 ways. 13 720 14 120

• Repetition is allowed: The same answer choice (e.g., ‘A’) can be selected for multiple questions. Permutations assume that once an item is chosen, it cannot be chosen again.

15 a 720

17 a 720

b 48

• Items are not from a single set: The scenario does not involve selecting and arranging 6 items from a set of 3. Instead, choices are made independently for each of the 6 questions.

c 12

d 6

b 240

c 480

16 16! × 5!

18 80 640 19 144

Answers 1283 mathspace.co

20 a 90

b 24

c 30

8 21

21 a 1296

b 360

c 936

9 a 15

22 36

10 840

23 72

11 270 725

24 72

25 a 224

b 28

26 a 90 720

c 6

d 6

where n, r are non-negative

13 210

27 720

14 792

28 240

15 252

29 360

16 924 b 36

c 15

integers and n ≥ r.

b 2 948 400

30 a 120

b 20

c 12

17 45 18 a y = 3 or y = 11

Extend your thinking

c y = 20

31 4320

b y = 0 or y = 9 d y=7

32 415 800 33 a 12 841 920

b 72 072 000

c 1 411 200 34 40 320

4.07 Combinations 20 a 12C4 = 495

What do you remember?

c The answers are identical. This illustrates the property nCr = nCn−r. Selecting 4 members to attend is equivalent to selecting 8 members not to attend from the group of 12.

1 a Combination. The order of selecting flavours does not change the final set of flavours. b Permutation. The order of awarding medals (gold, silver, bronze) matters and results in a different outcome.

b 12C8 = 495

c Combination. The order in which the books are chosen to be read does not change the selected set of books. 2 a 12C4 3 a 8C3

b 10C3 b 5C2

c 49C6 c 13C4

d 5C 5

5 a

b Ask your teacher for worked solutions. Extend your thinking

Practice 4 a 6

22 a Ask your teacher for worked solutions.

b 286

c 100

d 18

b n+1

23 15 24 203 490 25 410

6 10 7 56

1284 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

26 Ask your teacher for worked solutions. 27 Ask your teacher for worked solutions.

4.08 Combinations with restrictions What do you remember?

17 a 252 ways

b 126 ways

c 100 ways

d 126 ways

18 a 120 possible 3-flavour milkshakes

1 a A key similarity is that both are counting techniques used to determine the number of ways to select a subset of r items from a larger set of n distinct items. b The fundamental difference is that order matters in permutations, but order does not matter in combinations.

b 1023 possible milkshakes 19 210 ways 20 675 21 200 combinations 22 900 ways 23 210

24 112 different committees 3 a 35 c 1

210

4 The multiplication principle is used when a selection process involves multiple independent stages or choices, and the total number of combinations is found by multiplying the number of options at each stage. 5 “At least 2 items” means 2 or more items must be selected, requiring summing combinations for 2 items, 3 items, and so on, up to the maximum possible. “At most 2 items” means 0, 1, or 2 items can be selected, requiring summing combinations for these cases.

25 a 10 ways

b 126 ways

c 1260 ways

26 1584 hands 27 a 462 ways b 4 ways Extend your thinking 28 38 550 ways 29 3566 ways 30 12 pairs 31 1260 ways 32 92 ways

Practice

33 Imagine that among the n objects there is a single-distinguished object. In any selection, an item is either included or not. If it is not included, k objects must be selected from n − 1

6 16 ways 7 21 ways 8 364 ways

choices, giving

possibilities. If it is

9 330 ways

included, then the remaining k − 1 objects from

10 8 combinations

n − 1 choices, giving

11 182 ways

any selection either contains the special distinguished object or not, there are in total

12 2772 ways

possible ways to select k

13 39 729 ways

objects from n choices. Therefore,

14 43 316 ways 15 a 350 ways

possibilities. Since

b 546 ways

16 a 330 possible team selections b 126 possible team selections c 295 possible team selections

Answers 1285 mathspace.co

4.09 Combinatorics mixed problems What do you remember?

22 6006 23 a 4200

b 36

c 151 200

24 a 70

b 576

c 40 320

1 a True

b False

c True

d True

Extend your thinking

2 a No

b Yes

25 2 097 152

c Yes

d No

3 a 12C5

b 5P5

c 24C8

d 26P3

26 240 27 3024

4 The order of the numbers in the code matter, so it is actually a permutation calculation. If the order of the numbers did not matter, like 123 and 312, then a combination calculation would be used.

28 180

4.10 Combinatorics and probability What do you remember? 1 The formula for theoretical probability is . n(A) represents the number of ε favourable outcomes for event A, and n( ) represents the total number of possible outcomes in the sample space,  .

Practice 5 56 6 91 7 79 833 600 8 362 880 9 a 15

b 20

10 a i Combination

C6 = 38 760

b i Counting principle

ii 30

c i Counting principle

ii 8! = 40 320

d i Permutation

e i Counting principle

ii 9

f i Combination

11 a 126

b 3024

2 The key consideration is whether the order of selection or arrangement matters. If order matters, use permutations (e.g., arranging letters in a word). If order does not matter, use combinations (e.g., selecting a committee). 3 a n( ) = 6 c

P4 = 11 880

C3 = 35 c 1680

Practice 4 a 5P5 = 120 c

12 60 13 20 14 a 4

b n(A) = 3

b d

5 b 3

c 12

15 211 926

6 a 6P3 = 120

16 720 8

17 12 18 4536

19 1 352 000

10 a 6C3 = 20

11 a 7C3 = 35

12 a 10C4 = 210

20 a 3 108 105 21 a 53 130

b 125 970

c 495

b 6 375 600

c Choosing 5 people with no specific roles

1286 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

29 a n( ) = 362 880

b 14 15 a 100

b 52%

16 a n( ) = 330

b n(A) = 150

30 a n( ) = 40C6 b n(match 4) = 8415 c n(match 5) = 204 d n(match 6) = 1

c 17 a n( ) = 120

b n(A) = 56

e n(A) = 8620, 40C6 = 3 838 380

5.01 Matrix notation and representation

18 19 a n( ) = 52C5

What do you remember?

b n(A) = 4512 c

1 a True

b False

d True

e True

2 a False

20 a n( ) = 720

b True

c False

b n(A) = 24

3 a 4

b 3

4 a Season

b 4×2

c t32 = 15, representing 15°C in Melbourne during winter.

21 22 a 10P4 = 5040

b Practice

c 23 a

d False

b d

5 a 3×2

b 1×5

c 4×1

d 3×3

6 a 10

b 0

c −2

d 12

7 a a33

b a13

c a23

d a21

8 a Square matrix

24 25 a 22 b 26 a n( ) = 7P2 = 42, n(A) = 4P2 = 12, b n( ) = 7C2 = 21, n(A) = 4C2 = 6, c The probabilities are the same. This illustrates that as long as n( ) and n(A) are calculated consistently (both considering order or both not considering order), the probability will be the same.

b Column matrix, rectangular matrix c Square matrix, diagonal matrix d Rectangular matrix e Single-element matrix, row matrix, column matrix, square matrix, diagonal matrix f Row matrix, zero matrix, rectangular matrix 9 a Row

10 a [− 4

b Square

c Column d Square

Extend your thinking c

11 a − 3

b 2

27 28

Answers 1287 mathspace.co

12 a False. A matrix with size m × n and n × m have the same number of elements, but different sizes. b True. The number of elements in a square matrix with n rows is n2 and the number of elements in a column matrix with n rows is n. When n ≥ 2, n2 > n. c False. Adding a new row to an m × n matrix adds n elements. Adding a new column adds m elements. These are only equal if m = n (a square matrix). 13 a $250

b Week 1

14 a 6 hours

b 2

c Jamie

c $1430

b Both Dana and Ethan

c 39 hours 15 a

20 a

d Dana 630 650

185 450

b If a21 is changed from 420 to 320, it means that the amount of butter used in the ‘Scrumptious Surprise’ recipe is reduced by 100 g, from 420 g to 320 g.

Extend your thinking 2 3 4 3 4 5 0 –1 22 M 3 2

21 A: K

23 Answers may vary. Possible answer:

16 a

Square matrix 23 5 1 00 00

b A person cannot be friends with themselves. 17

0 0 0 0 0 0 Zero matrix

Identity matrix

The region ‘Only identity matrix’ is not possible because, by definition, all identity matrices are square matrices. Therefore, any identity matrix must lie in the intersection of the ‘Identity’ and ‘Square’ circles.

18 a 101 km b Daylesford and Ballarat c 131 km 19 a i $5000

10 01

ii $6000

b Since the actual expenses ($6000) exceed the budgeted expenses ($5000), Marketing Research is over budget by $1000. c c21 represents the actual expenses for Training in the Development department, which is $6700.

1288 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

24 a All the factor pairs of 120 can be used as the order of a matrix with 120 elements. The factor pairs are: 1 × 120, 2 × 60, 3 × 40, 4 × 30, 5 × 24, 6 × 20, 8 × 15, and 10 × 12. Since each order can be ﬂipped, the matrix has 16 possible orders. b The matrix with order 10 × 12 or 12 × 10 will have the least number of rows and columns combined.

25 a t = − 1

b u=7

e x=2

f y=0

26 a

c v=3

d w=8 d A+B=

5 a A−B=

b A−B= 3

A c A−B=

B b i False

E ii True

6 a

iii True

iv False

c If one road is washed away, the values need to decrease by 1 on elements nAC and nCA. 27 Each game must have a clear winner and a clear loser, meaning that for any pair of players, if tij = 1 (indicating Player i defeated Player j), then tji must equal 0 (indicating Player j lost to Player i). For example, if A defeats B, then t12 = 1 and t21 = 0. The error in the matrix involves the game between Dana (D) and Fiona (F). The matrix indicated that t46 = 0 and t64 = 0 meaning both Dana and Fiona lost to each other. In a chess tournament, each game must have a winner, therefore this scenario is impossible. One of the elements should be a 1.

b Operation not possible due to mismatched size c d [5

f g h [2 −5

b B+C−A= c C−A+B=

What do you remember?

9 a

ii 2 × 2

iii Yes

b i 3×2

ii 2 × 3

iii No

b 2×1

e No

f No

3 a 3×3

b 3×3

e No

f Yes

c 1×2

b x = −1

b Smith 10 Branch X: 3 shoes and 4 bags Branch Y: 4 shoes and 1 bag Branch Z: 2 shoes and 6 bags

c 2×2

d Yes 11 a

Practice 4 a A+B=

d No

− 6]

7 a A−B+C=

8 a x=5

1 a i 2×2

11]

e Operation not possible due to mismatched size

5.02 Addition and subtraction of matrices

2 a 2×2

d A−B=

12 a X = b A+B=

c b X=

13 u = 10, v = 4, w = − 3, x = 7, y = 2

c A+B=

Answers 1289 mathspace.co

14 Angelita’s error in calculating A − B for A=

2 4 3 1 and B = was that she 6 8 5 7

computed B − A instead of A − B. She got 1 3 2 1 4 3 C= , which is B – A = , 1 1 5 6 7 8 when she should have calculated: A–B=

2–3 6–5

4–1 –1 3 = 8–7 1 1

15 When these two matrices are added: A 0

This result shows that adding a zero matrix to A leaves A unchanged. General rule: When you add a zero matrix to any matrix A (of the same size), the result is the matrix A itself. Mathematically, this can be expressed as: A+0=A Extend your thinking 16 a This statement is true. When two matrices are added, the operation is performed element by element. If the matrices A and B have the same size, the resulting matrix C will also be the same size. b This statement is false. Even if two matrices have the same number of elements, they cannot be added together unless they have the same size (i.e., the same number of rows and columns). c This statement is true. When a matrix is subtracted from itself, each element in the matrix is subtracted from itself, resulting in a zero for each element. 17 Lucas’s assumption is correct. Matrix addition is commutative. The reason is that the operation of adding corresponding elements in two matrices mirrors the commutative property of integer addition. Therefore, for any two matrices A and B of the same size: A + B = B + A.

19 Visitors =

Now, add all these differences together: 12 + 32 + 9 + 13 + 4 = 70 visitors 20 D =

5.03 Scalar multiplication What do you remember? 1 a

2 a 0T = Multiplying by zero results in a zero matrix b 1T = Multiplying by one leaves the matrix unchanged. This is significant as it acts like an identity element for scalar multiplication with matrices. 3 a 1.20

c 2.50

d 0.95

4 a True

b True

c False

d True

e False

f False

Practice 5 a 2A =

b − 4A =

6 a 3A =

b −A =

7 a The scalar is 3. b

18 A + B = c

1290 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b 0.85

8 a y=2

b y=7

20 a

9 a 4A + B =

24 12

0 60

24 12

c The calculations show that 24 0 and k( cA) ( kc) A 12 60

b 2A − B =

0 60 24 12

0 . 60

Since (kc)A = k(cA), this demonstrates the associative property of scalar multiplication with matrices.

c 5A + 3B = d

21 First scalar multiplication: 10 a 4A + 2B = b 2A − 2B =

This was calculated correctly by Mia but not by Ethan. 3 × 3 = 9, but Ethan writes 6.

c 3A + 5B =

Second scalar multiplication:

d 0.3A + 2.5B =

This was calculated correctly by both Mia and Ethan.

12 a n = 10

b n = 16

13 a A =

b A=

Subtraction:

14 a 1.12 11.20 13.44 12.32 b 16.80 17.92 15.68 14.56 15.68 16.80

Mia’s error: 15 − (−2) = 17 but Mia incorrectly calculated this as 13. Ethan’s subtraction is impacted by his error in the ﬁrst multiplication. So he calculated 6 − 0 = 6 instead of 9 − 0 = 9.

15 a 0.95 190 171 209 b 237.5 218.5 199.5 256.5 247 228

Both Mia and Ethan made mistakes, so neither is correct. The correct answer is:

16 22 T = [1.75 17 a [30% b [90

50% 150

18 a True 19 a k( A

15%

2.1

1.4

2.55

1.95]

5%] Extend your thinking

15]

b False

c True

0 6 8 6

b kA

d True kB

0 6 8 6

c This demonstrates the distributive property of scalar multiplication over matrix addition: k(A + B) = kA + kB.

23 n = 4 24

25 Profit =

Answers 1291 mathspace.co

26 Answers may vary. Possible answer: Matrix addition and subtraction involve adding or subtracting corresponding elements from two matrices. For these operations to be possible, both matrices must have the same size. If the matrices have different sizes, there would be elements in one matrix that do not have corresponding elements in the other, making the operation undeﬁned. Scalar multiplication involves multiplying every element of a matrix by a single number (scalar). This operation does not depend on the matrix’s dimensions because it applies uniformly to each element, regardless of the size of the matrix. 27 X

1.25 1 .5 4.25

0.75 2.75 3.25 1 2.75 0

1 11 16

2 11 19

12 12 7

a b c d det (αA) = (αa)(αd) – (αb)(αc)

= α2(ad) – α2(bc) = α2(ad – bc) = α2det (A)

5.04 Matrix multiplication What do you remember? 1 a The product AB is possible and the resulting matrix is of size 1 × 1. The product BA is also possible and the resulting matrix is of size 1 × 1. b The product AB is not possible. The product BA is also not possible. c The product AB is possible and the resulting matrix is of size 2 × 3. The product BA is not possible. d The product AB is possible and the resulting matrix is of size 3 × 1. The product BA is not possible. e The product AB is possible and the resulting matrix is of size 2 × 3. The product BA is not possible. f The product AB is not possible. The product BA is also not possible. g The product AB is possible and the resulting matrix is of size 3 × 2. The product BA is not possible. h The product AB is possible and the resulting matrix is of size 3 × 3. The product BA is also possible and the resulting matrix is of size 2 × 2.

1292 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 a Not possible

b Not possible

c Not possible

d Possible. 3 × 3

3 3×1 4 2×5 5 a No. Addition is only possible for matrices of the same size. Since P is 2 × 4 and Q is 4 × 2, their sizes do not match. b No. Subtraction is only possible for matrices of the same size. Since P is 2 × 4 and Q is 4 × 2, their sizes do not match. c Yes. The number of columns in P (4) matches the number of rows in Q (4), allowing for multiplication. The resulting matrix will be 2 × 2. d Yes. The number of columns in Q (2) matches the number of rows in P (2), allowing for multiplication. The resulting matrix will be 4 × 4. Practice 6 a Not possible

b Not possible

c Not possible

d Not possible

e Not possible

f Not possible

g Not possible

h Possible. 2 × 1

7 2×3 8 a 3×3 9 a AB =

b AB = c AB =

d AB =

e AB =

f AB =

b 3×3

10 a 2 × 2

Extend your thinking

b AB = c 3×3 d BA = e No 11 a [32]

12 a A

b 5 11 18

b B

26 11 28 19 24 35

755 989

c BA

20 21 x + 2y = 5 and 3x + 4y = 11

13 a AI =

b IA =

c The identity matrix, I, acts as the multiplicative identity in matrix multiplication, similar to how the number 1 acts in scalar multiplication. This means that multiplying any matrix by I (provided size is appropriate) results in the original matrix, demonstrating that I preserves the elements of the matrix it multiplies.

5 3 370 490 395 535

23 c = – 1 and d =

b When any matrix is multiplied by a zero matrix (provided size is appropriate), the result is always a zero matrix. This occurs because each element in the product is formed by multiplying elements of the original matrix by the corresponding zero elements in the zero matrix, leading to sums of zeros. 15 a n = 4

b n = −3 257 176 212 148 257 176 B C 212 148

B) C

b A C c D

17 Ask your teacher for worked solutions.

The matrix format corresponds to these equations by encoding the coefficients of x and y in the first matrix, the variables in a column matrix, and the constants in the result matrix. Each row of the coefficient matrix, when multiplied by the variable matrix, generates one of the linear equations, making the matrix equation a concise representation of the system. 22 a = 9 and b = 10

14 a EF =

16 a ( A

19 Scalar multiplication involves multiplying every element of a matrix by a single number, while matrix multiplication involves combining two matrices to produce a new matrix. Scalar multiplication is always possible, regardless of the matrix’s size, and the resulting matrix retains the same size as the original matrix. In contrast, matrix multiplication is only possible when the number of columns in the ﬁrst matrix matches the number of rows in the second matrix. The resulting matrix from this operation will have the same number of rows as the ﬁrst matrix and the same number of columns as the second matrix.

1 1 1 1 1

5 1 1 8 3 or 1 1 2 6 1

5 8 3 2 6

24 D

5.05 M ultiplicative inverses and determinants What do you remember? 1 a Let I be a matrix. We say that I is the identity matrix if for any matrix A, we have AI = A = IA. b The identity matrix for 2 × 2 matrices with real numbers as entries consists of the matrix with 1’s along the diagonal and 0’s 1 0 everywhere else, i.e. . Its determinant 0 1 is (1) (1) – (0) (0) = 1. 2 a det (A) = ad – bc b A 1

1 det (A)

d c

b a

d 1 ad bc c

b a

Answers 1293 mathspace.co

3 a –1

b 30

4 a The condition for the inverse to exist is that the determinant (3) (6) − (4) (5) = 18 − 20 = − 2 ≠ 0. 3 4 b Let the matrix be A . The formula for 5 6 its inverse is A– 1 =

1 (3)(6)

(4)(5)

6 5

4 . 3

10 Oliver is not correct. In general, it is known that multiplication of matrices is not commutative. It cannot be assumed that BA–1 = A–1B. A counterexample is when: 2 1 and B 1 3

A 1 then:

2 1 1 1 1 3 0 3

A 1B

c Ask your teacher for worked solutions. d Ask your teacher for worked solutions.

BA 1

Practice 5 a i det = 15 2 1 5 5 ii 2 1 15 15

11 a B

4 9

1 0 1 0

= 12 – 12

a b c

=0 The determinant of matrix C is 0. b From part (a), det (C) = 0. The formula for the inverse of a matrix A is d b 1 . Since det(C) = 0, the A 1 a det (A) c 1 term would involve division by zero det (C) 1 , which is undefined. Therefore, the 0 matrix C is a singular matrix, and its multiplicative inverse C –1 does not exist. 13 a The formula for the inverse is s 1 q . The condition that M 1 ps qr r p

3 1

b Ask your teacher for worked solutions. 7 a –2

b –8

c –2

d A I

3 3

det

d i det = – 19 1 3 19 19 ii 7 2 19 19

8 a AB

1 1 2 1 0 3 1 3

12 a det (C) = (6)(2) – (3)(4) Using the formula

c i det = 7 3 4 7 7 ii 2 5 7 7

1 2 5 1

2 5 and 1 10

b No, AB is equal to the additive identity, not the multiplicative identity.

b i det = – 8 1 0 4 ii 5 1 8 2

6 a A 1

1 1 0 3

1 2 3 4

1 0 0 1

b B is indeed the multiplicative inverse of A, since the product AB yields the multiplicative identity. 9 n = –2

1294 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

must be satisfied for the inverse to exist is that the determinant ps – qr ≠ 0. b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. 14 a det (A) × det (B) = 5 b det (AB) = 5 c The answers to parts (a) and (b) are the same: both equal to 5.

Therefore, det (AB) = det (A) × det (B).

15 a A matrix has an inverse if and only if its determinant is non-zero. In this case, compute the determinant of A as 3 × 6 − 2 × 4 = 10 ≠ 0. Therefore, A has an inverse. b Calculate det (A) = 6 – x2. Hence, x det (A) 0 6. So, if x B has no inverse.

6 , then

16 The values of k for which A–1 exists are all k, 1 except k and k = 1. 2 Therefore, the inverse matrix is: 3 2 1 A 1 2 2 ( ) k 2 k 1 2k 3 k 1

Practice 5 a X

1 .4 2 .2

b Y

7.2 5 .6

6 a P 1

1 13

7 4

b A

5 13 32 13

6 13 2 13

7 a M

Extend your thinking 17 Given AB = I. Taking the determinant of both sides: det (AB) = det (I). We know that det (AB) = det (A) det (B) and that det (I) = 1.

19 a A

1 n 0 1

b 1

8 C 9 X

10 a X

b X

5.06 Solve matrix equations What do you remember? 1 a X = A–1M

b X = MA–1

c X

2 B 3 a X

1 14

1 b Y

4 X

19 8 7 16

6 1

4 7 13 14

4 3

4 2

1 14 2 7

b A

c x = 4 and x = – 3 20 Ask your teacher for worked solutions.

5 1 42 8

det (A) × det (B) = 1 This is the equation relating det (A) and det (B). (This also implies that neither det (A) nor det (B) 1 can be zero, and det (A) .) det (B) 18 Ask your teacher for worked solutions.

2 3

d X

e X

f X

0 1

2 2 7 10 1 2 18 71 23 71

4 5

5 71 37 71

4 7 3 7 21 34

22 39

1 3 7 24

1 3 23 48

2 11 27 11

Answers 1295 mathspace.co

Extend your thinking

11 X

6 8

12 X

5 4 9 14

1 4 1 14

17 X

1 6 1 2

0 13 a X 1

18 Y

b Y

5 4

4 6

c Z

11 30

6 24

14 a

2 3

5 2

7 3

0.95 1.66 0.94

0.15 1.43 0.93

6.01 Introduction to graph theory 4 13

What do you remember?

3 2 c x = 3, y = – 2 X

15 a A

12 15 ,X 8 20

So,

12 15 8 20

x y

–0.380 952 0.428 571 0.285 714 0.506 494 0.731 602 –0.450 216 1.251 082 –0.532 468 –0.324 675

1.61

x y

x ,B y

201 204

1 a V = 5, E = 3

b V = 6, E = 9

2 a Directed

b Directed

c Directed

d Undirected

e Undirected

f Undirected

g Undirected

h Directed

i Directed

j Directed

k Undirected

l Undirected

3 a 0

b det (A) = 120 8 X 7

b 0

c 2

d 2

4 a No

b No

c Yes

d Yes

e Yes

f Yes

g Yes

h Yes

Practice

Therefore, the price of coffee is $8 and the price of tea is $7. 16 a A

6 4 ,X 2 5

x1 ,B x2

So,

x1 x2

2400 1900

6 4 2 5

25 9 19 9 5 3

19 X 2.27 3.12

b det (A) = –19

10 3 13 3

2400 1900

b det (A) = 22 200 X 300 Therefore, the price of one phone is $200 and the price of one tablet is $300.

5 a More vertices

b No

6 • B and C • A and C • D and E 7 a False

b False

c False

d True

e True

f False

8 • B to C • C to A • C to D • D to F • E to C • F to B 9 a Yes

1296 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b No

c Yes

d No

10 a The public transport stations b The transport lines/routes between stations c 8 vertices d 10 edges 11

b No, as Warrnambool (W) only connects to Waurn Ponds (WP) with a directed edge. Since there is no return path from WP to W, W cannot be reached from the rest of the network. c

B C

E W

12 a No

b Yes

13 a

c No

d Yes An additional one-way shuttle bus has been added between campuses in CBD and Warrnambool.

18 a

C Q D

b Yes, the network is connected. It is possible to trace a path between any two friends. All friends are part of a single component. 14 a i 5

ii 9

iii Yes

b i 7

ii 10

iii No

15 a Undirected

b Directed

c Directed

d Directed

e Undirected

f Undirected

Extend your thinking 16 Remove the loop at B and remove an edge between C and A 17 a

b D–D c No, as the network contains a loop.

6.02 Degree and adjacency What do you remember? 1 a i 2

ii 2

b i 4

ii 3

2 a The degree of vertex C is 4. b A : [B, D] B : [A, C] C : [B, C, D] D : [A, C] 3 a The degree of vertex Q is 4.

C GW

WP W

P Q P Q R S

R S

0 1 0 0 1 1 1 0 0 1 0 1 0 0 1 0

Answers 1297 mathspace.co

Practice

9 a

4 a

A B C A B C

X X Y Z

0 1 0 1 0 1 0 1 0

5 a i

P Q P Q R S

0 1 0 0 0 0 1 0 0

R S

0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0

A B

ii P : 2, Q : 1, R : 1, S : 0 b i

P Q P Q R S

0 0 1 1

R S

0 1 1 0 2 0 2 0 1 0 1 0

C D c

ii P : 2, Q : 2, R : 4, S : 2 c i

P Q P Q R S

R S

1 1 1 0 1 0 0 0 1 0 0 1 0 0 1 1

B D

ii P : 4, Q : 1, R : 2, S : 3 6

A B C A B C D

0 1 1 1 1 0 1 0 1 1 0 1 1 0 1 0

7 a

D1 D2 D1 D2 D3 D4

D d

Q R

D3 D4

1 0

0 1

0 0

b In-degree of D2 : 2 (from D1, from D4) Out-degree of D2 : 1 (to D3) Total degree of D2 : 3

S T 10

S T U V S T U V

0 1 0 1 1 0 0 0 0 0 0 0 0 0 1 1

8 V1 : [V2, V3] V2 : [V1, V3, V4, V4] V3 : [V1, V2, V5] V4 : [V2, V2, V5] V5 : [V3, V4]

1298 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11 a Directed

b AR

c DF

d No, because there are multiple edges between the same two vertices. In this network, vertices DF and KJ are connected by two edges (one in each direction).

6.03 Graph isomorphisms

Extend your thinking 12 a

What do you remember?

1 a False 2 a ii

A B C A B C D E

0 1 0 0 1

D E

• Corresponding vertices connect in the same way. Practice

In-degrees: A : 2, B : 1, C : 2, D : 1, E : 1. Alice and Charlie are asked by the most people (2).

B P

4 a i Yes, the two graphs are isomorphic. ii The graphs are isomorphic because both graphs are simple cycles of length 4 (C4). • Vertices: Both graphs have 4 vertices. • Edges: Both graphs have 4 edges. • Degree sequence: In both graphs, every vertex is connected to two other vertices. Therefore, both graphs have the same degree sequence: (2, 2, 2, 2). b i No, the two graphs are not isomorphic.

F C

C B P F

c i

• The graphs have a matching number of vertices of each degree.

1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0

d True

• The graphs have the same number of edges.

c Out-degrees: A : 1, B : 2, C : 1, D : 2, E : 1. Bob and Diana can ask the most people (2).

13 a

b iii

c True

3 • The graphs have the same number of vertices.

b False

0 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0

c Coburg (C): 3 Brunswick (B): 1 Pascoe Vale (P): 1 Fawkner (F): 1 d Currently: • B: 1 → needs 2 → + 1 connection • P: 1 → needs 2 → + 1 connection • F: 1 → needs 2 → + 1 connection A minimum of 3 new edges must be added: • Connect Brunswick to Pascoe Vale • Connect Pascoe Vale to Fawkner • Connect Fawkner to Brunswick

ii The degree sequences are not the same, the graphs cannot be isomorphic. • Vertices: Both graphs have 4 vertices. • Edges: Both graphs have 3 edges. • Degree sequence: For Graph 2A, the central vertex has degree 3, and the three outer vertices each have degree 1. The degree sequence is (3, 1, 1, 1). For Graph 2B, the two inner vertices have degree 2, and the two end vertices have degree 1. The degree sequence is (2, 2, 1, 1). c i Yes, the two graphs are isomorphic. ii The graphs are isomorphic because they have the same structure. This can be verified by checking several properties: • Vertices and edges: Both graphs have 6 vertices and 7 edges. • Degree sequence: Both graphs have the same degree sequence: (3, 3, 2, 2, 2, 2).

Answers 1299 mathspace.co

• Structural adjacency: The connectivity is identical. In both graphs, the two vertices of degree 3 are connected to each other by two distinct paths of length 2 (forming a 4-cycle) and one path of length 3. The isomorphism can be demonstrated by redrawing Graph G to match the layout of Graph H, based on this vertex mapping:

6 A

C B

D 7 a Yes

b No

c Yes

d No

e No

f Yes

g Yes

h No

i Yes

u1 → v1, u2 → v5, u3 → v4, u4 → v3, u5 → v2, u6 → v6

8 a No

b Yes

c Yes

d Yes

d i Yes, the two graphs are isomorphic.

e Yes

f No

g Yes

h Yes

ii The graphs are isomorphic because they are both different visual representations of the same underlying graph: a cycle graph with 7 vertices (C7). • Vertices and edges: Both Graph G and Graph H have 7 vertices and 7 edges. • Degree sequence: In both graphs, every vertex is connected to exactly two other vertices. Therefore, both graphs have the identical degree sequence: (2, 2, 2, 2, 2, 2, 2).

i Yes

Graph G is clearly drawn as a 7-cycle. Graph H, although drawn as a star, is also a single 7-cycle. This can be seen by tracing the path of its edges: v1 – v3 – v5 – v7 – v2 – v4 – v6 – v1. Since both graphs have the exact same structure (a cycle of length 7), they are isomorphic. 5 a The degree sequence is (3, 3, 2, 2, 2, 2).

9 Simple, connected, complete, regular (all vertices degree 4). 10 Connected bipartite graph with 6 vertices C

11 Original degree sequence: (2, 2, 1, 1, 0). Complement edges: AC, AD, AE, BD, BE, CE, DE. Complement degree sequence: (4, 3, 3, 2, 2). 12 a D

b D

b The only subgraph that is a cycle of length 3 is the triangle formed by vertices A, B, and C. B

1300 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Not isomorphic: A self-complementary graph with 8 vertices requires each vertex to have 8 1 degree 3.5, which is impossible as 2 degrees must be integers.

Original graph (Degrees 2)

Extend your thinking 14 Graph: 5 vertices, edges connecting vertex 1 to vertices 2, 3, 4, 5, and vertex 2 to vertices 3 and 4. This gives degrees 4 (vertex 1), 3 (vertex 2), 2 (vertex 3), 2 (vertex 4), 1 (vertex 5). Complement: Edges are vertex 2 to vertex 5, vertex 3 to vertices 4 and 5, and vertex 4 to vertex 5.

Complement graph (Degrees 2)

Complement degree sequence: (0, 1, 2, 2, 3).

Not isomorphic: different degree sequences.

Complement graph Original graph (Degrees: 4, 3, 2, 2, 1) (Degrees: 0, 1, 2, 2, 3) 1 1 5 2 2 5

6.04 Trees What do you remember?

1 a A connected network graph that contains no cycles or loops. b A subnetwork of a connected network that contains all the vertices of the original network and is also a tree (it has no cycles).

15 Proof: A complete graph with n vertices has n ( n 1) n vertices, edges, and each vertex has 2 degree n – 1. For two complete graphs with n vertices, both have the same number of vertices, edges, and degree sequence (n – 1, n – 1, . . ., n – 1) (n times). Since every vertex connects to all others in both graphs, any bijection between vertex sets preserves connectivity, making them isomorphic.

2 a False

16 Example: Two sets of 4 vertices, with edges connecting each vertex in one set to two vertices in the other set, making each vertex degree 2 (regular).

3 In any tree, the number of edges is always one less than the number of vertices. This relationship is described by the formula m = n – 1.

c A spanning tree in a weighted graph that has the minimum total weighting possible. d A greedy algorithm used to find a minimal spanning tree for a weighted, connected graph. b True

c True

d True

Complement: Edges connect each vertex to the two vertices in the other set not connected in the original, forming another bipartite graph with degree 2.

Answers 1301 mathspace.co

Practice

4 a Yes

b No, because it has a cycle.

c Yes

d No, because it has a loop.

6 The graph has 7 vertices (so n = 7) and 6 edges (so m = 6). For the condition m = n – 1, 6 = 7 – 1, which is true. Since the graph is connected and satisfies this condition, it is a tree. B

C 3

b A

C 3

9 C

3 E

7 2

6 F c $21 000 000

12 a Instead of choosing the edge with the lowest weight at each step, one would choose the edge with the highest weight that connects a vertex in the tree to a vertex outside the tree, without creating a cycle. LHR $12K

SYD

c $58 000 13 The graph representation would show 5 vertices (A, B, C, D, E) with edges and weights as listed. One possible MST (using Prim’s starting at A, for example): AB (120 m), BC (90 m), CD (100 m), DE (130 m). Total length: 120 + 90 + 100 + 130 = 440 m.

200

13 D

150

9 15 11

17 B 19

14 A

120

b The minimum total distance is 2300 km.

$13K

$15K

DXB

JFK

9 a A and C

$18K

GRU

8 a 5 b

H 17

4 2

Extend your thinking 6

E 15

5 The tree has 16 vertices.

7 a

C 21

1302 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

180 D

100 130

220 E

14 a The graph representation would show 5 vertices (P, Q, R, S, T) with edges and weights as listed in the table.

Practice 4 a

Q 40 50

80 60

70 120

b 180 000 15 a 19 km b 19 km c Yes, MSTs identical. Trails: {( AB, 4), (BC, 3), (CD, 5), (DE, 2), (EF, 5)}. This graph has a unique MST, which Prim’s finds regardless of start. Not always guaranteed; multiple MSTs can exist, leading to different results based on start/ties. d No, DF (6 km) not included. From A, after A, B, C, D, E connected (cost 14 km), to connect F, options are AF (10), CF (9), DF (6), EF (5). Prim’s selects EF (5) (cheapest). MST remains 19 km. e Max length for CF is 5 km. When A, B, C, D, E are connected (cost 14 km), edges to F are AF (10), CF (x), EF (5). For CF (x) to be chosen, x ≤ 5 km (to be ≤ EF (5)). If CF = 5 km, it’s an alternative MST path.

5 a i E=7

ii D = 14

iii D = 2E

b i E=8

ii D = 16

iii D = 2E

6 a i A

6.05 Planar graphs What do you remember? 1 a False

ii f = 4

b True

b i

c False 2 a Non-planar c Non-planar 3 a Yes

b No

b Planar

d Non-planar c No

d Yes

ii f = 4

Answers 1303 mathspace.co

b i v=6

c i A

ii e = 6

iii F

D B C B

ii f = 3 d i

A E

C ii f = 2 7 a i v=5

ii e = 6 A

v 2

iv f = 2

iii

Graph A

Graph B

Graph C

Graph D

Vertices (v)

Faces (f )

Edges (e)

v+f–e

9 a f=9

b e=8

c v = 11

d e=7

e v=5

f f=3

10 a f = 4

b A

C iv f = 3

11 a

v 2

Graph A

Graph B

Graph C

Graph D

Vertices

Faces

Edges

v+f–e

Planar? (Y/N)

No. of vertices with odd degree

b The number of vertices is always even.

1304 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Practice

12 a e = 4 b Answers may vary. Possible answer:

4 a No

b Yes. Open walk.

c Yes. Open walk. 5 a i Yes

d No

ii No

iii No

iv Yes

ii No

iii Yes

iv No

ii No

iii No

iv No

ii No

iii Yes

iv Yes

6 a i No

ii No

iii Yes

iv Yes

b i Yes

ii No

iii No

iv No

v No b i No 13 a Vertices: v = 8, edges: e = 12, faces: f = 6 (including the outer face) b Substitute the values into Euler’s formula: v + f – e = 8 + 6 – 12 =2 Since the result is 2, the graph satisfies Euler’s formula. 14

v+f–e=6+7–9 =4

Since the result is 4, Euler’s formula is not satisfied. 15 a

e ≤ 3v – 6 25 ≤ 3(10) – 6 25 ≤ 30 – 6 25 ≤ 24

The statement 25 ≤ 24 is false. Therefore, the graph cannot be planar. b No, it cannot satisfy Euler’s formula. Euler’s formula only applies to connected planar graphs. Since we have proven this graph is non-planar, the formula does not apply.

v No c i Yes v No d i No v No

7 a A - B - D and A - C - D b A - B - D - F and A - C - D - F and A - B - D - E - F and A - C - D - E - F 8 5 unique paths 9 a No, it is not possible to draw an Eulerian circuit. An Eulerian circuit requires all vertices to have an even degree. Vertices A and B have an odd degree (deg(A) = 3, deg(B) = 3), and all other vertices (C, D, E) have an even degree making it semi-Eulerian, not Eulerian. b Example trail: B - A - E - C - B - E - D - A 10 a B - Y - Q - X or B - H - W - X b H-X-Y-Q-W-B c Q - S - T - P or Q - R - T - P d P - U - T - S - Q or P - R - T - S - Q

Extend your thinking 16 a 1

b 1

17 a e = 5

b f=3

6.06 Walks, paths, and trails What do you remember? 1 a All vertices must have an even degree. b Exactly 2 vertices must have an odd degree. 2 An Eulerian trail must traverse every edge exactly once. A Hamiltonian path must visit every vertex exactly once. 3 An open walk starts and ends at different vertices. A closed walk starts and ends at the same vertex.

11 a Walk, trail, path, circuit, cycle b Walk, trail, path 12 a The network is semi-Eulerian (it can be traversed but starts and ends at different vertices). b Vertices D and E c Answers may vary. Possible answers: • A - C - E - D - F - E • A - C - E - F - D - E 13 a • Eulerian: The graph is Eulerian. This is because all vertices in the graph have an even degree. An Eulerian circuit, which traverses every edge exactly once before returning to the starting vertex, is therefore possible.

Answers 1305 mathspace.co

• Hamiltonian: The graph is also Hamiltonian. A Hamiltonian circuit, which is a path that visits every vertex exactly once before returning to the start, can be traced (for example, by following the outer edges of the graph). b • Semi-Eulerian: The graph is semi-Eulerian. This is because it has exactly two vertices of odd degree (degree 3), while the remaining vertices have even degrees. This allows for a semi-Eulerian trail, which traverses every edge exactly once but starts and ends at different vertices. • Hamiltonian: The graph is also Hamiltonian. By inspection, it is possible to find a circuit that visits every vertex exactly once before returning to the starting vertex.

Extend your thinking 20 a Cycle, this term is the only one which must start and end with the same vertex. b Path, this term is the only one which is not required to start and end with the same vertex. 21 Two edges need to be duplicated, connecting B and H, and E and G. 22 a Two triangles joined at a single vertex.

b A square with its diagonals.

14 a Vertices B and C. They are the only two vertices with an odd degree. b Answers may vary. For example, B - A - C - E - D - B - C. 15 A - B - C - D - E - A 16 a Example answers: • E - D - C - B - A • A - B - C - E - D • D - A - B - C - E b Example answers: • P - R - Q - S - T • T - S - P - R - Q • Q - P - R - T - S 17 Yes 18 a Example: A - B - D - B - E. The edge BD is used twice (B to D, then D to B). b Example: A - B - D - G - E - B. Vertex B is repeated; no edges are repeated. c Example: A - B - D - C - A. 19 a Yes, since all vertices (D, X, Y) have even degrees, the graph is Eulerian. Therefore, the delivery driver can traverse each road exactly once and return to the depot (D). b D-X-Y-Y-D

1306 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

23 No. The graph of possible knight’s moves on a 3x3 board consists of a cycle of the 8 outer squares, with the centre square being completely isolated (it has a degree of 0). It is impossible to visit an isolated vertex as part of a path, so no Hamiltonian path can exist.

6.07 Algorithms in graph theory What do you remember? 1 DFS explores as deep as possible along each branch before backtracking, while BFS explores all neighbours at the current depth before moving deeper. 2 DFS uses a stack (often recursively), while BFS uses a queue.

3 Scenario

Objective

1. A GPS system needs to find the fastest route between two locations, using a map where roads are marked with travel times.

Algorithm

Find the path Dijkstra’s with the Algorithm minimum total weight.

9 a DE(1), AB(2), BC(3), AF(4), FE(5), CF(6), FD(7), CD(8) b The edges are added in this order: DE, AB, BC, AF, FE. The edge CF is skipped as it would create a cycle. c 15 Extend your thinking

2. A utility company wants to connect all houses in a new suburb to the power grid. They have the costs to run a wire between any two houses. They need to connect all houses for the minimum possible total cost.

Connect all vertices with the minimum total cost.

Kruskal’s Algorithm

3. On a social media site, the goal is to find the connection to another user via the smallest number of intermediate friends.

Find the path with the fewest edges.

BreadthFirst Search (BFS)

4. A robot is programmed to navigate a complex maze from a start point to an end point. It needs to find any valid path by exploring one corridor completely before trying another.

Find a path from a start to an end vertex (not necessarily the shortest).

DepthFirst Search (DFS)

10 The council should choose the cable network. Its minimum connection cost of $29 000 is less than the $34 000 cost for the Wireless relay system. 11 a 25 hours b The new minimum spanning tree has a weight of 27 hours, therefore the total time increases by 2 hours.

6.08 Applications of graph theory What do you remember? n ( n 1) , where n is the 2 number of vertices.

1 The formula is e

2 Complete as the graph is simple and each vertex is joined to every other vertex. 3 It provides a necessary condition for a connected graph to be planar. If the formula does not hold (i.e., v + f – e ≠ 2), the graph is non-planar. If it does hold, the graph may be planar.

Practice 4 a True

4 a P-Q-R-Q-S-T b P-Q-S-T

6 D-C-A 7 a The total weight of the shortest path is 6. b A-B-D 8 a 50 40

30 Lion

c True

d False

Practice

5 A-C-E

Reptile

b False

Bear

5 For K6, the number of vertices is n = 6. 6 (6 1) e= Substitute the values 2 6 5 = Evaluate the values in the numerator 2 = 15 Evaluate Therefore, K6 has 15 edges.

60 Aquarium Bird 20

70 Monkey

b The total minimum distance is 200 m.

Answers 1307 mathspace.co

6 For K3 (a triangle): • Vertices: v = 3 • Edges: e = 3 • Faces: f = 2 (the inner face and the outer face)

Extend your thinking 12 The graph has: • Vertices: v = 6 • Edges: e = 6 (AB, BC, CD, DE, EF, AC) • Faces: f = 2 (one bounded region A - B - C - A, and the outer region).

Substitute into Euler’s formula: v+f–e=3+2–3

Substitute into Euler’s formula:

v+f–e=6–6+2

Since the formula holds, K3 is planar.

7 K7 has 21 edges. 8

The formula holds, therefore the graph is planar. 13 For K4, where n = 4: 4 ( 4 1) e= Substitute the values 2 4 3 = valuate the values in the numerator E 2 =6 Evaluate

5 4

9 5

For K5, where n = 5: 5 (5 1) e= Substitute the values 2

6 10

5 4 2 = 10

The electrical circuit can be modelled as the graph shown. The graph is connected because a path exists between any two of its vertices. 9 Temora

Young

Evaluate the values in the numerator Evaluate

The increase in edges is 10 – 6 = 4. General pattern: When a new vertex is added to Kn to form Kn + 1, an edge must be added from this new vertex to each of the n existing vertices. Therefore, the number of edges increases by n.

Boorowa

Harden

Cootamundra Junee

Yass

10 a Hillsboro

Expo centre

Airport

Beaverton

Gateway Rose quarter

Pioneer square PSU

Clackamas town centre

b Airport - Gateway - Rose Quarter - Pioneer Square - PSU 11 21 km

1308 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Gresham

9 a 12.00 cm

Resistor

Switch

Resistor

Switch

Battery

Globe

Resistor

15 a

Switch

B 3

4 A 7

D b

10 a 13.09 cm

b 150.80 cm2

c 38.48 cm2

d 353.43 cm2

11 a 229.18°

b 114.59°

c 107.43°

d 71.62°

12 a i 108.9 mm

ii 730.6 mm2

b i 156.6 mm

ii 1115.9 mm2

14 81 cm2

5 9

d 8.00 cm

13 6π cm2

b 8.46 cm

50 7.07 cm

15 41.1 cm2

E 2

16 208.3 cm2

Extend your thinking 17 AC = 16 m

C A E

18 a 3000 m2

b 844.33 m2

19 a AB = 12 cm

b (3

12) cm

b 336π cm2

20 a 48π cm

21 $282.74

c $140 000

22 ( 100 3 50 ) cm2

7.01 Radians and parts of circles 7.02 Foundations of trigonometry What do you remember? 1 a 3

1 B

5 d 2

3 c 2 2 a 30°

What do you remember?

3 b 4

b 135°

c –90°

2 B, D d 540°

3 C, D 4 a i

Practice 3 a 11.3 units 4 a 3.14 cm c 109.96 cm 2 5 a 3

13 5

13 12

iii

5 12

d 58.90 cm

13 13 ii 5 12 c i The expressions are equal: sec (x) = cosec (90° – x)

b 108π cm2

ii The expressions are equal: cosec (x) = sec (90° – x)

b 19.5 units b 47.12 cm

6 θ = 1.95 radians

5 13

b i

5 a

1 x

1 y

x y

4 3

7 2

3 8

7 317.77 m2 8 a r = 31.83 cm

b r = 15.28 mm

6 a

c r = 7.83 cm

d r = 0.67 m

7 a B

b A

Answers 1309 mathspace.co

4 ii 3 ii 6 ii 3

8 a i Quadrant 2

20 a cosec (θ ) =

b i Quadrant 1 c i Quadrant 4 d i Quadrant 1

b cot (θ ) = 15

17 15

15 b 17

8 c 15

17 8

2 , 2 coordinates: P (–0.71, –0.71).

10 Exact coordinates: P

b Negative

c Positive

d Negative

13 5

13 a

2 3 3

b 1

d 2

13 12

14 a 3

15 a 1

2 3 3 b 2

5 12

5 13

12 13

13 12

13 5

of a.

1 3 , 2 2

a b

23 cot (x) – cosec (x) =

3 10 . The value of x depends on the sign 10

• If a

3 10 , then x 10

arctan

1 (Quadrant 1). 3

3 10 , then 10 1 + π (Quadrant 3). arctan 3

7.03 The sine rule

3 ii –cot 3

b i

2 3 3

What do you remember?

ii –sin

17 a r = 13 units

6 ii sec 6

d i Negative

22 a 2

d –2

ii tan

c i Negative

17 9

24 a

c 2

b i Positive

8 ,y 9

• If a

c 1

16 a i Positive

5 39 39

Extend your thinking

15 8

2 . Approximate 2

11 a Positive

12 a

c cot (θ ) = 21 x

Practice 8 9 a 17

7 10 20

p q r sin (P ) sin (Q ) sin (R)

2 a x = 71.71

b x = 6.73

c x = 39.48°

d x = 42.16°

3 a i

5 12 12 vi 5

18 a

19 a

1310 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

iii

c ii No

b i

16 a Two distinct triangles can be formed. Ask your teacher for worked solutions.

b Triangle 1: ∠B1 ≈ 75.04°, ∠C1 ≈ 56.96°, c1 ≈ 56.40 cm

Triangle 2: ∠B2 ≈ 104.96°, ∠C2 ≈ 27.04°, c2 ≈ 30.58 cm x x b sin (B) b a c x = b sin (A) and x = a sin (B)

17 a sin (A)

ii Yes c i B

Extend your thinking

b ii Yes d i

18 a a = 35.65 m

b h = 20.4 m

19 a b = 226.55 m

b h = 117 m

20 h = 102.57 km

C b

d Equating the expressions: b sin (A) = a sin (B). Dividing both sides b a . by sin (A) sin (B) gives sin (B) sin (A)

21 1.8 minutes

7.04 The cosine rule What do you remember? ii No

1 a Sine rule c Cosine rule

Practice

b Pythagoras’ theorem d Trigonometric ratio

4 a a = 13.40

b a = 6.85

2 a Cosine rule

b Sine rule

c a = 3.02

d a = 711.73

c Sine rule

d Cosine rule

5 a x = 13.1

b x = 12.9

3 ∠ ABC 4 a a2 = b2 + c2 – 2bc cos (A)

6 3.87 km

b a2 = 142 + 122 – 2 × 14 × 12 cos (72°)

7 y = 4.4 m 8 a HK = 733.18 km

b KJ = 444.55 km

Practice

9 a C = 70.50°

b a = 5.621 m

5 a a = 10.73

b a = 10.00

c a = 14.14

d a = 212.34

c b = 7.116 m 10 a x = 49.3°

b x = 30.4°

6 AB = 3 2 cm

c x = 32.7°

d x = 20.4°

7 v=5

11 a ∠BAC = 29°

b c = 37.11

8 a w = 41.49°

b w = 31.71°

12 a B = 74°

b B = 32°

c A = 27.33°

d A = 33.56°

13 θ = 28.68°

9 cos (U ) =

14 ∠CDB ≈ 44.0°

10 x = 104°

1 15

15 ∠R = 140.91°

Answers 1311 mathspace.co

11 θ = 24°

11 b = 38

12 x = 34.78

12 CB = 32.4 cm

13 x = 22°

13 a θ = 27.2°

14 The angles are 113°, 67°, 113°, and 67°. 15 a BD = 23.88 m

b CD = 22.82 m

b 15.6°

Extend your thinking 14 a 16 cm

b 120 cm2

15 a 20 cm

b 384 cm2

c 62.82 m 16 60°

16 a There are two possible values: 53.5° and 126.5°

Extend your thinking 17 a a2 = h2 + c2 – 2cx + x2 2

b h =b –x

c x = b cos (A) d a2 = b2 + c2 – 2bc cos (A)

b Two distinct triangles. Both possible values for ∠Q are valid because for each one, the sum of angles with the given ∠P is less than 180° (40° + 53.5° < 180° and 40° + 126.5° < 180°).

18 There is no solution because cos (A) = 1.1 > 1. A triangle cannot be formed as the triangle inequality is not satisfied (5 + 8 is not greater than 15).

19 OA = 6 m

19 36.06 km

20 a AF 261, FC = 10, AC = 17

7.05 A pplications of sine and cosine rules

17 1242 cm2 18 4 cm

b 77.1° 21 a PM 112 cm

b PN 128 cm

c 69.30° What do you remember? 1 a Cosine rule

b Sine rule

n 360° 22 r 2 sin cm2 2 n

c Cosine rule

d Sine rule

7.06 Pythagorean identities

Practice

What do you remember?

2 a The cosine rule because all three side lengths are known (SSS), which is a condition for using the cosine rule.

2 a

sin ( ) cos ( )

cos ( ) sin ( )

b 104° 3 35.0 km 4 a 15.7 km

b 176°

5 37 888.61 m2

b sec2 (A)

1 a 1

1 cos ( )

c cosec2 (A) c

1 sin ( )

Practice 3 Ask your teacher for worked solutions.

6 150 863 m2 7 1389.32 m2 8 12 cm2 9 a 60.2 m2

b 144.0 cm2

c 12.4 m2

d 19.9 cm2

10 a 326 cm2

b 100 cm2

c 110 cm2

d 109 cm2

1312 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

4 sin ( A)

4 5

5 sec (B)

13 5

6 cot (A)

7 a sin2 (x)

b cos (θ )

c tan2 (A)

d 1

e sec (x)

f 1

8 1 2

9 tan (θ ) 10 a Ask your teacher for worked solutions.

tan (2 ) tan ( 3 ) 1 tan (2 ) tan ( 3 )

tan ( ) tan ( 3 ) 1 tan ( ) tan ( 3 )

b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. d Ask your teacher for worked solutions. 11 Ask your teacher for worked solutions. 12 Ask your teacher for worked solutions.

3 a Ask your teacher for worked solutions. 7 b sin ( A B) 25 Practice 4 15 b sin ( B) 5 4 c Ask your teacher for worked solutions.

4 a cos ( A)

Extend your thinking 13 a P (cos (A), sin (A))

d cos ( A

3 15 4 20

5 a sin (

)

133 205

b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. 14 2 tan (x) 15 Ask your teacher for worked solutions. 6 3

16 cos (x) 17 tan ( ) 18

)

156 205

6 2 4

7 sin

2 or tan (θ ) = – 1 3

b cos (

5 12

2 4

8 a sin (A)

b cos (A)

sin (θ )

1 3

5 9 a cos 12

cos (θ )

2 2 3

b cos

7 12

tan (θ )

cosec (θ )

–3

12 Ask your teacher for worked solutions.

sec (θ )

3 2 4

13 a Ask your teacher for worked solutions.

cot (θ )

2 2

2 4

2 4 6 4

10 Ask your teacher for worked solutions. 11 Ask your teacher for worked solutions.

b Ask your teacher for worked solutions. 14 a

33 65

16 65

19 Ask your teacher for worked solutions.

Extend your thinking

7.07 Angle sum and difference identities

15 a sin (A) cos (B) cos (C) + cos (A) sin (B) cos (C) + cos (A) cos (B) sin (C) – sin (A) sin (B) sin (C)

What do you remember? 1 a sin (A + B) = sin (A) cos (B) + cos (A) sin (B) b sin (A – B) = sin (A) cos (B) – cos (A) sin (B) c cos (A + B) = cos (A) cos (B) – sin (A) sin (B) d cos (A – B) = cos (A) cos (B) + sin (A) sin (B) 2 a cos (x) cos (2y) + sin (x) sin (2y) b sin (2x) cos (y) + cos (2x) sin (y)

b cos (A) cos (B) cos (C) – sin (A) sin (B) cos (C) – sin (A) cos (B) sin (C) – cos (A) sin (B) sin (C) c tan ( A) tan (B) tan (C ) tan ( A) tan ( B) tan (C ) 1 tan (A) tan ( B) tan ( A) tan (C ) tan ( B) tan (C ) 16 Ask your teacher for worked solutions. 17 cos ( A

7 8

Answers 1313 mathspace.co

14 6 sin (θ ) – 8 sin3 (θ )

7.08 Double-angle identities

15 Ask your teacher for worked solutions.

What do you remember?

16 x = cos (4 A)

1 a cos (2α) = 1 – 2 sin2 (α)

17 tan (α + 2β ) = 1

b cos (2α) = 2 cos2 (α) – 1 2 a sin (6θ )

b cos (10β )

Extend your thinking

3 a sin (2A) = sin (A + A) = sin (A) cos (A) + cos (A) sin (A) b cos (2A) = cos (A + A) = cos (A) cos (A) – sin (A) sin (A) = cos2 (A) – sin2 (A) c tan (2A) = tan (A + A) tan ( A) tan ( A) = 1 tan ( A) tan ( A)

What do you remember?

= 1 – 2 sin2 ( A) Practice 4 Let B = A: sin (2A) = sin (A + A) = sin (A) cos (A) + cos (A) sin (A) = 2 sin (A) cos (A)

6 cos (2 ) 7 a

1 a False

b True

c False

d False

2 a Scalar quantity

b Scalar quantity

c Vector quantity

d Scalar quantity

3 Displacement, acceleration, force and momentum Practice

24 25

4 a

7 25

15 4

8 cos ( 165° )

7 8

b 2

4 metres

24 25

6 4

9 tan (22.5 ° )

10 a tan (2 A)

4 3

b tan ( 4 A)

11 a 1 cos ( A)

2 cos2

A 2

b 3 sin (A)

1 a3

8.01 Introduction to vectors

= (1 – sin2 ( A)) – sin2 ( A)

1 3 a 2

23 Ask your teacher for worked solutions.

2 tan ( A)

3 5

21 cos ( 3 )

22 Ask your teacher for worked solutions.

1 tan2 ( A) d cos (2A) = cos2 ( A) – sin2 ( A)

5 a

3 2 20 Ask your teacher for worked solutions. 19

= 2 sin (A) cos (A)

18 x = 4

6 sin

2.5 Newtons 24 7

A A cos 2 2

12 a tan (A)

b cot (A)

13 a cos (2 A)

b cos (2 A)

1314 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

y 1 x –4

–3

–2

–1

1 –1 –2

10 m/s

5 a

6 8

6 2

6 a 1 –1

–3

7 4

5 2

y 2

y x 1

–1

x –2 –1

–2 –3

2 3 4 5 6 7

–1

–4 –2

–5 –6 –7 –8

7 a

y 14

1 –1

x 1

–1

–4

–2

x 2

–2

3 b

y 2 1 x 1

–1 –2

Answers 1315 mathspace.co

9 a

–2

–4

x 2

–6

y 2

–4

–2 –2

–4 –6

–18–16–14–12–10 –8 –6 –4 –2 4 3

10 a For a > 0, the vector has a positive horizontal component and a negative vertical component; therefore, it should extend rightward and downward. A possible answer is shown for a = 3.

b (4, 3)

y B

–2

8 a AB

5 4

AB 2

x –2

x 0

–2

b For a < 0, the vector has a negative horizontal component and a negative vertical component; therefore, it should extend leftward and downward. A possible answer is shown for a = –3.

1316 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 , b 1

12 a

4 5 y

1 –5 –4 –3 –2 –1 –1

x 1

3 b

–2

–3 c For a = 0, the horizontal component is zero; therefore, the vector should extend downward.

–2

9 8

7 6

x –2

–4 –3 –2 –1 –1

f1 M

3 –2

2 1

x 0 1

Extend your thinking 11

10 8 9 units N 6 4 2 –10–8 –6 –4 –2 –2 –4 6 units SW –6 –8 –10

8.02 Magnitude and direction of a vector y

What do you remember? 1 a x = x2 – x1 c | v | = x2 x

b y = y2 – y1 y d tan( ) x

2 a Horizontal component = 3

2 4 6 8 10

b Vertical component = 4 c 5 3 a False

b True

c True

Practice 4 a PQ 5

7 4

b QP

7 4

c PQ

Answers 1317 mathspace.co

ii θ = 53.13°

6 a i |v| = 10 units

21 a x = 136.46 km/h

ii θ = 108.43°

b i |v| = 4 10 units c i |v| = 8 units

ii θ = 138.59°

d i |v| = 26 units

ii θ = 202.62°

e i |v| = 12 units

ii θ = 35.26°

f i |v| = 2 2 units g i |v| = 4 3 units

ii θ = –45° ii θ = 120°

h i |v| = 10 units

ii θ = –30°

b θ = 35.69° below horizontal

23 a x2 + y2 = 3

13 units

b BC

5 units

c CD

13 units

d DE

7 2 units

e EF

2 7 units

1 − 3, 0 0

–1

9 Ask your teacher for worked solutions. 17

c 6

11 a False

b False

c True

–2 b y

12 a | v | = 10 c The magnitude of w is | w | = 10, which is the same as the magnitude of v.

The direction of w is – 60° or 300°.

1, a

b F

3, 3 3 N or

3, 3 3 N

13 a x = 34.86 units

b y = 53.67 units

2 C

14 a i x = 16.97

ii y = 16.97

3 a 30

b i x = 7.01

ii y = –16.04

4 a iii

6, 6 3

1 a

21 2, 21 2

8 12

4 6

b 32

4 or (x, y)

13 ,

2 3

c direction

b i

c iv

5 a 13 ,

1 1 .5

d ii

Practice

16 a = c = d, e = f 4

35 13 c x 6 19 Answers may vary. Possible answer: Any two pairs of points A (x1, y1) and B (x2, y2) such that x1 – x2 = 2 and y2 – y1 = 1 are correct. For example: A = (0, 0), B (2, 1) and A′ = (4, 3), B = (6, 4).

18 a x = ±9.54

8.03 Scalar multiples of vectors What do you remember?

0, − 3

2 when x

The vectors have the same magnitude but different directions. They are opposite vectors, meaning w = – v.

17 ( x, y)

24 a 6 N

b The direction is 120°.

15 a

–1

8 Ask your teacher for worked solutions.

a 3, 0

–2

4 units

10 a 5

0, 3

7 a AB

b 19 3 , 57

22 a (–13, –13)

b x

20 Any vector where the x- and y-components satisfy 25 = x2 + y2 is correct. Answers may vary. Possible answers: (5, 0), (0, 5), (3, 4), (–3, 4)

1318 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

6 w

5 4 3 2 1

−4 −3 −2 −1

x 1

7 a

1 4

1 w 2

3 2a

2 a

5 4

−6 −5 −4 −3 −2 −1

3 2

−1

x 1

b 4

6 a

x 1

−2

1 w 2 1 −4 −3 −2 −1

2 a −4 −2 −2

−6

−8

y x 2

8 10 12

−4 −4a

−10

−12

x 1

5 c

2.5 2

y 10

6 5×v

2 v

x 1

1.5

1 1 0.5 a x 2 –2.5 –2 –1.5 –1 –0.5 0.5 1 1.5 2 2.5 –0.5 –1 –1.5 –2 –2.5

8 a

6 3

10 5

c The vector 3v is in the same direction as v but three times longer, whereas –5v is in the opposite direction to v and five times longer. This shows how scalar multiplication affects the vector’s magnitude and direction on the plane. 9 5

Answers 1319 mathspace.co

b x

10 a x = –1, y = 2 11 a i

2 ,y a

4 a

5 5

c i

6 2

y 6

4 Scaled

3 2 1 −1

−1

Original 1

Scaled 3

5 3

x 1

−1

iii The scalar 2 doubles the magnitude of the original vector and does not change its direction. b i

−6 −5 −4 −3 −2 −1

Original

iii The scalar –5 reverses the direction of the vector and increases its magnitude by a factor of 5. 4 d i 3 ii

y 6

1 −5 −4 −3 −2 −1 −1 Scaled

Original

2 Original 1

3 2

x Scaled

2 3 4 5

−8 −7 −6 −5 −4 −3 −2 −1

−2 −3

x −1

1 reduces the magnitude of the 2 vector to half, while keeping the direction unchanged.

iii The scalar iii The scalar –1 reverses the direction of the original vector but keeps its magnitude unchanged.

12 a k = 2

b k = –1

c k

13 a | v | = 10 b i k = –1 and | kv | = 10 1 ii k and | kv | = 5 2 iii k = 2 and | kv | = 20 iv k

1320 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 and | kv | = 2.5 4

1 2

d k=0

14 As shown in the graphs, both vectors 3 × (2 × v) and (3 × 2) × v point to the same endpoint, (12, 6) from the origin. This visually confirms that the expressions 3 × (2 × v) and (3 × 2) × v are indeed equivalent.

17 a The magnitudes are | u | = 10 and | v | = 53. b The magnitude of mu is 30. c The magnitude of nv is 2 53. 18 The magnitude of vector w is 100 units. 19 k = 4 and k = –4

y 12

Extend your thinking

20 Ask your teacher for worked solutions.

21 v 25

22 Ask your teacher for worked solutions.

4 2

23 1

3 × (2 × v)

2v 2

23 The new displacement vector from the base is 150 metres. 112.5 24 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions.

y 12

8.04 Vector addition and subtraction

What do you remember?

1 A

2 C

4 2 0

(3 × 2) × v

v 2

3 a True

b True

c False

d True

e False

f True

12 Practice

15 a

b The possible scalar multiples are 10 and –10. 16 If b is a scalar multiple of a, then b = ma for some scalar m. k 3 3m m 5 4 4m From the vertical components: 5 –5 = 4m ⇒ m = 4 Substituting this into the horizontal components: 5 15 k = 3m = 3 4 4 Geometric interpretation: Since the scalar is negative m

4 a b 5 a

b a

c u and v

y c

5 , the vector b points in the 4

opposite direction to a. Since the magnitude of the scalar is greater than 1, b is stretched to be 5 times the length of a. 4

Answers 1321 mathspace.co

y 8 7

6 5

4 c

a+b

2 1

0 c

y c

x 5

7 6

5 4

a+b

3 2

1 d

0 a

y 8 7

6 b

5 4 3

x 6 a

2 1

8 7 6 5

4 3 2 1 0

a+b

x 1

1322 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

a a+b+c x 1 2 3 4 5 6 7 8 9 10 11

7 a

y –b

y 4

u 3

–a

a 2

c–a

x 0

–d

c–b

1 x

0 d u

y 2b

3 d

–b

2 x 8 a

–b

x c

y 3

2b – a

–a

x 0

–b a–b

1 x 0

Answers 1323 mathspace.co

9 a 7

w 2

6 5 4 3

2 w −4 −3 −2 −1

–1

x 0

–2

x 2

−1

–1

2v – w

–2

c 2v − w =

1 2

11 a

5 4 v+w 3 2 w −4 −3 −2 −1

c v+w=

x 1

−1

2 7

10 a

y 5 4

v 0

x w

–2w

3 3v – 2w 2

1 x 0

3 5 12 a It is not possible to have u + v = w since the y-components of u, v are respectively greater than the y-component of w. Similarly, it is not possible to have v + w = u since both v, w have positive x-components while u has a negative x-component. Therefore, u + w = v. c 3v − 2w =

1324 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

y 5

c–e

0 2 e

x 0 13 a Vector v is the resultant, meaning v = u + w.

Explanation: Vector v is purely vertical (its horizontal component is zero). Vectors u and w have horizontal components that are equal in magnitude but opposite in direction (one goes left, one goes right). When added, the horizontal components will cancel each other out, resulting in a purely vertical vector, which matches the direction of vector v. b

5 v–u

3 2 1

–u

3 u −3

−2 −1

w 1

x 2

–3n m

m – 3n

3 2

14 a

−1

5 −1 4

a–b

1 x 0

Answers 1325 mathspace.co

4 2 ,b= ,c= 2 4

4 0 ,d= 1 4

6 6

15 a a = b i 3 3

16 a

8 6

iii

6 13

2 9

5 7

8 4

4 b u= 10

1 17 a u = 1 c u= 18 x = 19

21 a

10 2

u+v −3

−2

 −3  2    3 1 x −1

3 3

d u=

−1

14 18

4 provides only the 2 sum of u and v. Without any additional information about u or v individually, there are infinitely many pairs that satisfy the equation. Therefore, u and v cannot be determined uniquely.

b The equation u + v =

Extend your thinking 20 First, use the triangle rule to work out 1 0

3  1 −   1

−4

34 units

0 : 1

22 a Since v is perpendicular to u and u lies in the 2nd quadrant, then v lies in either the 1st or 3rd quadrant.

y u+v

2(u + v) v

b Yes, consider the example set up satisfying the conditions imposed in the questions. y

x u

t s

v Now, 2

1 0

0 : 1 y 2u + 2v 2v x 0

1326 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

26 34 11 11 23 c = ,d= 16 70 11 11

8.05 Unit vectors and component form

y 7

What do you remember?

1 a

x –2

–1

–2

2 1

–3

x 0

–4 –5

0 2

b 4i

3 a (3, –19)

b (–4, 5)

4 a

–3

d 2i − 5j

–2

c 4i − 7j

c (0, –3)

–1

1 1

1 3

2, 0

5 0

5 a 4i + 2j

b 2i − 3j

c −2i

d 1.5i − 1.5j

0 2

Practice 6 A unit vector in the direction of

–4

g is

y c

2 a i + 3j

x 1

0 –5

–4

–3

–2

7 h=

2 5 5 , 5 5

8 f=−

2 13 3 13 i− j 13 13

–1 –1

3 58 7 58 . , 58 58

–2 1 9 a i Scalar 10

–3 –4 –5

3 5 ii a = 4 5

1 b i Scalar 13

ii b =

1 17

ii c =

c i Scalar

d i Scalar

10 10

ii d =

5 13 12 13 8 17 15 17 3 10 10 10 10

Answers 1327 mathspace.co

10 3i − j

6 a EF = 12i + 5j

11 a 12i + j

b 15i + 8j

c 3i + 7j

2 3

7 a AB

12 (–5, 16) 13

3 34 , 34

2 2 = 3 3

1 5

b BD = − 5i + 8j

c DB = 5i − 8j 10 3 10 , , 10 10

3 1

b BC

c AC = AB + BC

5 34 34

14 a AC = 5i + 8j

b EF = 3i + 4j

1 1 = 2 2

d CD = − 5i 10 , 10

3 10 10

3 1

Therefore, proved.

d AB = 13

112 89 70 89 , 16 v = 89 89

BC = 10 AC = 5 13 + 10 ≠ 5

17 a vx = 20 3 m/s b vy = 20 m/s

8 a

18 a F1 = 10i + 10 3j

F2 = 15 3i + 15j

b R = 10 15 3 i + 15 + 10 3 j

y B

19 The resultant force has a magnitude of approximately 13.5 N and a direction of 73.5°.

Extend your thinking

1 3 20 i + j 2 2

–2

–1

ii BC

3 1

iii OB

–1

21 −10i − j

2 3

22 7 22 i + j 3 3

b i

23 12i + 32j

c ∣BC∣ = 10

8.06 Geometric applications of vectors What do you remember? 4 b AB = 4i + 3j 3 2 The midpoint is M(–1, 7). 1 a AB

3 a AB = b − a

b OM

4 a u = kv

b QR

1 (a + b) 2

9 a Answer may vary. Possible answer: A (5, 3), B (9, 3) b Answer may vary. Possible answer: A (–2, 4), B (2, 1) 10 a A (0, 1)

b A (6, 9)

11 a P (–3, 3)

b P (–2, 2)

11 c P 3, 2 12 a M (1, 3)

Practice

d P

11 27 , 5 5

b N (3, – 7)

13 The coordinates of point P are (6, 1).

5 a AB

2 3

b AB

c AB

3 2

d AB

2 7 2

1328 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3 5

5 3 14 a The midpoints are M 3, and L 3, . 2 2 0 0 b ML = , AC = 1 2 c Using the vectors from the previous part: 1 1 0 AC = 2 2 2

0 = 1 Since ML

0 1 , it follows that ML AC . 2 1

This verifies that the line segment connecting the midpoints is parallel to the third side and half its length. 15 a

y C

16 a OP

1 1 (a + b), OQ (b + c) 2 2

1 (c − a) 2 1 c SR = (c − a) 2 b PQ =

d Since PQ = SR, the side PQ is parallel to and equal in length to the side SR. A quadrilateral with one pair of opposite sides that are parallel and equal in length is a parallelogram. Therefore, PQRS is a parallelogram. 17 a AB OB OA = b − a b Since P divides AB in the ratio 3:1, 3 the vector AP is of the vector AB. 4 3 Therefore, AP (b − a). 4 c Ask your teacher for worked solutions. 18 Ask your teacher for worked solutions.

19 Ask your teacher for worked solutions.

20 a H

A x

65 4

5 8 , 4 3

22 a P(3, 2)

B A x

2 b N 4, 3

c CL

2 AL

Extend your thinking N

ii HE

2 21 a L 0, 3

65 4

c I

b i

3 5 b M 3, , N(4, 2), L 3, 2 2 c

1 8 5 11 8 5 5 , , E , , F 3, and G , . 2 3 4 3 3 4 3

d Ask your teacher for worked solutions.

b P(7, 1)

23 Sides are ∣AB∣ = 80 = 4 5, ∣BC∣ = 80 = 4 5, and ∣AC∣ = 32 = 4 2. The triangle is isosceles. Since ∣AB∣2 + ∣AC∣2 ≠ ∣BC∣2, it is not rightangled. 24 Ask your teacher for worked solutions. 25 k = 8 26 Ask your teacher for worked solutions.

Answers 1329 mathspace.co

8.07 Polar form

c i

What do you remember? 1 a False

b False

c True

d True

500°

2 C b r

3 a Quadrant 2 c θ = 135°

3 2

d 3 2 135°

Practice 4 a a = (12, 30°)

b b = (4, 200°)

c c = (8, 285°)

d d = (7, 115°)

5 a i

ii c = (10, 140°)

d i

x 300°

–370°

x 0 8

1 a

ii a = (8, −60°) b i

ii d = (1, −10°)

6 a

y a

b 3

25°

0 –200°

b ii b = (3, 160°)

b 2

1330 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

y 115°

−

–135°

3π 2

c d

–75° 5

5π 6

y e 2

π 2

x 7 a

180°

w f

f 1

b w = (3, 180°)

c (–3, 0)

8 p = (3, −300°)

p = (3, 420°)

9 a Point B

b Point A

10 v = − 10.00i + 17.32j 11 (– 12.29, – 8.60) 12 (– 6.93, – 4.00) 13 (13.00, – 67.4°) 14 (25.00, 106.3°)

Answers 1331 mathspace.co

15 a a =

6 cos ( 45° ) 6 sin ( 45° )

b b=

9 cos ( 150° ) 9 sin ( 150° )

4 .5 3 4 .5

c c=

5 cos (240° ) 5 sin (240° )

2 .5 3

d d=

7 cos ( 60° ) 7 sin ( 60° )

16 a (– 7.21, 5.39)

6 4 b Yes, because w = 3PQ . Since the scalar multiple k = –3 is negative, the vectors are parallel and point in opposite directions. 18 c Z= 12

3 2

10 a PQ

3 2

2 .5 3 .5

11 a Perpendicular

3 .5 3

12 k = 2

b (9.27, 4.21 rad)

13 0

b Not perpendicular

14 45°

Extend your thinking 17 Polar coordinate representations are not unique since many different values of θ can represent the same angle. There are countably infinite polar coordinate representations. In particular, P(r, θ ) is equivalently P(r, θ + 2πn) and P(–r, θ + (2n + 1)π) .

15 a – 1

b 98.13°

16 k∣b∣2 17 a False

b False

c True

Extend your thinking 4 3; YOB 4

18 b = (15, 300°)

18 X OB

19 a = (8.06, 29.74°)

7 6 19 b1= ; b2= 5 5

8.08 Scalar (dot) product

20 OC OA OC OB = 173.31

What do you remember?

21 OA

1 a False

b True

c False

d True

2 a 14

b 0

c –24

d 50

3 a 20 4 a a⋅b=

b 0 35 2 2

c a⋅b=

d a⋅b=

5 a OA OB = 9

1 3

b OA OB

10 3

d OA OB = 0

E S

100 m

c OA OB = 48 cos (130°)

50 m b

c 80 m

a+b

∣a∣ = 2 x2 (since a is in the northeast direction, the magnitude of its x-component will equal the magnitude of its y-component) 100 = 2 x2

b OB = 3.20 c θ = 70.53°

7 a OA OB = − 35

b OD = 10

c θ = 81.64° c b ⋅ c = 2 2 5 3

9 a a ⋅ c = −16

c – 21

Practice

8 a b⋅c=6

22 a

b a⋅b=4 3

9 2 2

6 a OA = 4.71

OB = 70

b b ⋅ c = −4

d b⋅c=4 2 3 b b1 = 1

1332 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x = 50 2 a = 50 2, 50 2

∣b∣ = 2 x2 (since b is in the southeast direction, the magnitude of its x-component will equal the magnitude of its y-component)

13 a

50 = 2 x2 x = 25 2

b = 25 2, 25 2

33 10 11 10

10 u 17 3

15 a

a + b = 75 2, 25 2

75 2, 25 2

b 15.52°

What do you remember?

3 i 5

3 3,

2 3

5 4

or 1.29,

5 4

or 2.83, or 5.20,

3 4

2 b 4.019i − 1.608j

23 3

20 x = 20

4 j 5

21 a

4 3

3 4

or 1.55,

3 18 b 5

19 a

4 6 3 5 a

2 2,

2 3

Extend your thinking

1 14 53

16 a

17 a 4.30i + 2.48j

8.09 Scalar and vector projections

= 40 2, 40 2 Since their last walk has a positive x-component equal to its y-component, Sonja is right that their last walk was directly northeast.

c = t − (a + b) = 115 2, 65 2

60 29 150 29

6 9

25 34 15 34 i j 17 17

11 34 17

22 5i + 2j 23 Ask your teacher for worked solutions.

Practice 6 a

5 5

34 29 7 a 29 8

3 29 29

3 65 b 65

5 2

3 2 9 a 2 10 a

3 3 2

11 a –1.294 12 a

6 7 b 5

5 2 b 2 b

7 2

b 2.995 b

6 7 a 16

8.10 Displacement and velocity What do you remember? 1 a True

b False

2 a vx = 10 m/s 3 a sR =

5 m 12

c True

d False

b vy = 10 3 m/s b ∣sR∣ = 13 m

Practice 4 Magnitude: ∣sR∣ ≈ 72.1 km 40 Direction: θ = tan–1 ≈ 33.7° from the East 60 5 20.0 km east 6 Magnitude: ∣sR∣ ≈ 43.6 km Direction: θ ≈ – 6.6° (or 353.4°) from the East

Answers 1333 mathspace.co

7 a sR =

10 2

b ∣sR∣ ≈ 14.2 km

10 2

8.11 Force equilibrium and work done What do you remember?

0.02 8 a sR ≈ 7.71 0.02 c −sR = 7.71

b ∣sR∣ ≈ 7.71 km

1 a False

b True

2 a ii 3 F3 =

9 a dR = 4i − j b Distance: ∣dR∣ = 17 units

c False

b iii

d True c i

3 N 2

Practice

c Direction: θ ≈ 346°

4 a

10 10.8 m/s at an angle of 56.3° from the East 11 The required air velocity is approximately 392.43i + 42.43j km/h.

30° 150°

12 38 637 km

13 a Speed: ∣v∣ = 12 km/h b Direction: θ = 315° 14 The final displacement in polar form is (61.71, 79.59°). 15 a The swimmer should aim at an angle of 30° west of north.

100 N

b It would take approximately 57.7 seconds. 16 a 120i − 209j km/h

30°

b 2.25 hours

17 a

y (km/h) 200

90°

150 vair

100

vground

−150

−100

−50

50 x (km/h) vwind

60°

100 N c a = 86.60 N 5 a

d b = 50.00 N y 300 N

b The true speed is approximately 253.18 km/h and the direction is approximately 129.09°.

550 N

160° 70°

Extend your thinking 18 83.2° 19 7 : 56 a.m. 20 Boat B will reach faster by 1.46846 minutes (1 minute and 28.11 seconds) 21 8 minutes

1334 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

b 626.5 N

c 131.39°

6 T1 = 25.04 N, T2 = 20.37 N 7 a

57°

9.01 Introduction to complex numbers

57° T2

What do you remember? 1 a False

b True

2 a 4 + 2i

c True c 0 7i

b 1 – 3i

d 8 + 0i

3 a Re( z) = 1, Im( z) = 2 b Re( z) = 7, Im( z) = −3 c Re( z) = 0, Im( z) = 3

12 N

d Re( z) = 2, Im( z) = −9

b T1 = T2 = 7.15 N

4 a Real

8 a

b Non-real complex c Pure imaginary 300 N 6000 N

45°

b The magnitude of the third force needed to keep the boat stationary is 5791.8 N, directed approximately 2.1° below the positive x-axis. 9 6.99 N 10 a F = 8.74 N

b N = 10.77 N

11 a T = 8.21 kg wt

b T = 2.17 kg wt

d Real 5 a ℤ, ℚ, ℝ, ℂ

b ℂ

6 a Non-real complex

b Real

c ℝ, ℂ

c Real

d Non-real complex

e Non-real complex

f Non-real complex

g Real

h Real b a = 0, b

5 2

c a = 1, b 4 5

d a = 5, b

10 3

e a = −9, b 2

f a

3 2

7 a a = 5, b

12 175°

Practice

13 7.5 J

8 a 5

14 1 646 400 J

d 3i

15 1127.63 J

9 a 16

16 24 500 J 17 724.8 J 18 θ = 23.5°

21 40 J 22 a s =

x y

8 3 4

b 44 J

b i 2 e 10i 2 7 i

2 13

3, b

c i 42 f 2i 165 4i

c 1 + 2i

e 3 7

g 5

h –6

10 a False

b True

c False

d False

e False

f True

5 5

b i

c –2i

i 15 3

1 i 2

12 a

2i 2 3

3i 2 4

19 ∣T∣ = 117.8 N 20 281.21i + 75.41j

d 4 – 14i

11 a i

Extend your thinking

d ℚ, ℝ, ℂ

d 4i 2

e –10

f –18

f –i 3 3 3 c 6 10 4 f –12

Answers 1335 mathspace.co

13 a 4 i 3

b 1 + 7i

c 1 i 34 d

e 5 + 5i

f 0 + 12i

g 6 – 15i

7 i 5

h 5 + 2i

i 7 + 11i

8 a

5 29

145

5 29

5 b 4 6 10 a 0.983 radians 9 a

14 5 4 3 2 1

Im(z) z1 Re(z) z4

–5 –4 –3 –2 –1 1 2 3 4 5 –1 –2 –3 z2 –4 z3 –5

3 d 6 4 b 2.554 radians

c 1.966 radians

d −1.855 radians

11 a i

ii 0.896 radians

b i

ii 2.820 radians

c i 4

ii π radians

d i 5

12 a

221

b 25

radians 2

c 1

5 6 b Modulus: 3, Principal argument: π

5 5

13 a Modulus: 2, Principal argument:

15 • A : – 3 + 2i 14 a

• B : 4 + 4i • C : 2 – 3i

5 2

d 5 5

Extend your thinking

• D : – 4 + 0i

15 Ask your teacher for worked solutions. Extend your thinking 16 a False

b True

d False

e False

17 a i –i

16 Ask your teacher for worked solutions. c True

17 Ask your teacher for worked solutions. 18 Ask your teacher for worked solutions.

iii i

ii 1

19 Ask your teacher for worked solutions.

b in = −1 or i or −i or 1 18 C is at 4 + 5i, D is at 1 + 5i

9.02 Modulus and argument of complex numbers What do you remember? 1

1 a The real part of z is a and the imaginary part of z is b.

3 Quadrant 3

The real part of w is c and the imaginary part of w is d.

b 0° d radians 2

b The real part of z + w is a + c and the imaginary part of z + w is b + d.

(Re(z)) 2+ (Im(z)) 2

5 z

c The real part of z – w is a – c and the imaginary part of z – w is b – d.

Practice 6 a 7 a 2

9.03 A ddition and subtraction of complex numbers What do you remember?

2 The range is −π < Arg ( z) ≤ π.

4 a 180° radians c 3

5 2 5 2 i 2 2 21 Ask your teacher for worked solutions.

20 z

b 3 2

b 4

c 5

1336 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

2 a a = 2, b = –3

b a = –2, b = 5

c a = 2, b = 2

d a = 6, b = –2

3 a 4 + 8.6i

b –5.8 + 11.2i

c 5.6 – 8i

d –19 – 2.6i

Practice 4 a 4 + 3i

b –1 + 9i

11 20

5 a

19 i 3

c 14 – 10i d 11 – 8i

4 23 i 3 18

6 a a = –2, b = –9

c 8

4 i 3

b a = 9, b = –9

c a = 13, b = –1

b z vector from (0, 0) to (2, 3). –w vector from (0, 0) to (– 4, 1). z – w vector from (0, 0) to (– 2, 4). Head-to-tail addition should be shown.

d a = –16, b = –12

4 3

7 a 7 + 3i

b 9

2 3

6 i

c –19 + i

d 9

3 i

z− w

8 a –6i. This is a purely imaginary number.

b 3 – 2i. This is a non-real complex number.

−w

c –3 + 4i. This is a non-real complex number.

−4 −3 −2 −1

d 6 – 3i. This is a complex non-real number.

−1

9 a 0 The result is a real number. b –4 – 16i The result is a complex number. c 10 + 10i The result is a complex number. d –6 + 6i The result is a non-real complex number. 10 a z vector from (0, 0) to (2, 3). w vector from (0, 0) to (4, – 1). z + w vector from (0, 0) to (6, 2). Head-to-tail addition should be shown. 6

11 a z1 = 3 + 2i, z2 = 1 + 4i b (3 + 2i) + (1 + 4i) = 4 + 6i c The algebraic sum is 4 + 6i. The vector z1 + z2 on the diagram ends at the point (4, 6), which corresponds to the complex number 4 + 6i. The results match. Extend your thinking

12 a Answers may vary. Possible answer: z = 1 + i, w = 1 + 3i

5 4

b Answers may vary. Possible answer: z = –1 + 4i, w = 1 + 3i

c Answers may vary. Possible answer:

2 1 −2 −1 −1 −2

z 1

z+ w 2 3 4 w

z = 9 + 5i, w = 1 + i Re 5

d Answers may vary. Possible answer:

z = 1 + 6i, w = 1 + i 13 z

7 2

7 i, w = 1 2

14 F ind the sum: z1 + z2 = (3 + x) + (2 + y)i. For the sum to be purely real, the imaginary part must be 0, so 2 + y = 0, which means y = –2. Find the difference: z1 – z2 = (3 – x) + (2 – y)i. The real part is 3 – x. Since the real part is –1, it follows that 3 – x = –1, so x = 4. Therefore, x = 4 and y = –2.

Answers 1337 mathspace.co

9.04 Multiplication of complex numbers

y 25

What do you remember? 1 a –b + ai

b b – ai

c The original real part becomes the new imaginary part, and the original imaginary part multiplied by –1 becomes the new real part.

b –i

c 1

d –1

3 a –63 + 7i

b –6 + 16i

c 6 + 5i

d –45 – 35i

b 4 2

y = x2 – 6x + 18 18 a (x + (1 + 2i)) (x + (1 – 2i))

What do you remember? 12 5 2 i

1 C

c 45 – 28i 5 a –7 – 24i

b –414 – 154i

c 63 – 16i

d –112 – 384i

6 a=6 7 a a = 3, b = –2

b a = 32, b = 14

c a = 3, b = 4

d a = 5, b = 0

8 a 18

b 0 b 3

c 2

10 a –2i

b –12

c –6 – 2i d 1 + 2i

11 a 1

b 1

c 3i + 2

d 9i

d 1

b x – 12x + i(6x2 – 8)

12 a x

c y2 – 6yi – 9

c 2 2 31i 3 a 181

d –v30

12i i 4

Extend your thinking

c 28 b –11 + 2i

c 13 – 8i

d –20 – 16i

5 a c

7 25

3 4

1338 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

24 i 25

41 59 i 58 58

6 a –6 + 21i c –840 + 288i 364 250 i 29 29

d 44

b 12 + 6i d

1 2

b –20 + 48i d –21 – 27i b

454 344 i 29 29

8 a –12 + 20i

c 17 – 28i

15 Ask your teacher for worked solutions. 7 , b 4

b 58

4 a –1 + 5i

7 a

13 x = 3 2

d –12 – 16i

Practice

c 0

9 a –2

1 4 2

b –7 + 12i

2 a 3 – 4i

d 168 + 168i

16 a

9.05 Complex conjugates and division

4 a 18 – 14i

b 3

b x = –1 + 2i, x = –1 – 2i

Practice

14 a

(3, 9)

d The original imaginary part becomes the new real part, and the original real part multiplied by –1 becomes the new imaginary part. 2 a –1

(0, 18)

9 a –63 33 c 74

17 3 i 50 25

102 36 i 13 13

b 78 365 d 74

10 a

1168 197

19 c 20 11 a a

86 22 i 197 197

Practice 3

4 a 2 cis

d 424 – 368i

21 , b 41

2 1 i 5 5

14 48 i 5 5

16 41

b a = 8, b = 6

2 2i

6 a Quadrant 1

b Quadrant 3

c Quadrant 4

d Quadrant 2 2,

b arg (z) 3 3 mod ( z) = 2 mod ( z) = 7 a bi.

2 3 b cis 2 3 b x2 + y2 = 4

b 98 cis

9 a 8 cis (0) 1 cis 6 2 11 a r = 2

= a2 – (bi)2

10 a

= a2 – b2 i2 = a2 – b2 (–1) = a2 + b2

25 8 3cis (–0.50)

12 a

Since a and b are real numbers, a2 + b2 is also a real number.

13 2 3

cis (2.03)

Extend your thinking 13 a 8

6 3 ,b 5 5 1

2 2

8 a arg ( z)

14 Let z = a + bi and z

b a

3 4

5 a 2 2 3i

7 a (0, 4)

z z = (a + bi) (a – bi)

15 a a

b 2 2 cis

2 1 ,b 2

radians 6

c Point B

14 r = 3 ∓

2 1 2

16 z = 0 17 M ultiplying the denominator by its conjugate eliminates the imaginary part, resulting in a real number (since (a + bi) (a – bi) = a2 + b2). This process is called ‘realising the denominator’. Once the denominator is a real number, the fraction can be easily split into its real and imaginary parts, expressing it in the standard form x + yi.

9.06 Cartesian and polar form What do you remember?

15 y 3 x 3 16 a y = –2x + 2

17 a

2 2 cos ( ) + sin ( )

b r

3 4

18 a

3 2 3 2 i 2 2

Im 2 z2

z1 Re

−2

−1

1 cos (θ ) + i sin (θ ) 2 a = r cos (θ ), b = r sin (θ ) 3 a z1 z2 = r1 r2 cis (θ1 + θ2) z r1 cis ( 1 – 2 ) b 1 z2 r2

−2

z1 is at approximately (1.73, 1). z2 is at (1, 1). 5 b 2 2cis 12

Answers 1339 mathspace.co

c Cartesian form is more efficient as the numbers are already in that form. While polar form is good for multiplication, the effort to convert both numbers to polar form first makes it less efficient for this specific problem.

z1 z2

2 z2

z1 Re

−2

−1

Practice

−1 −2

d The point z1 is rotated anticlockwise by

radians and its modulus is scaled (dilated) by a factor of 2. 3 cis

b The point representing z is a reflection of the point representing z across the real axis. c w

b Quadrant 4 d Quadrant 2

2 8 a 3.346 – 0.897i

10 a

58 cis (0.405)

58 cis (–0.578) 13 3 2

3 2

3 3 i 2

3 2

3 3 i 2

12 a 2 cis

c 4 2 cis b Cartesian d Polar

2 1. A rotation about the origin by the angle θ2. 2. A scaling (dilation) of its modulus by a factor of r2. 3 a Cartesian form is more efficient. Addition in Cartesian form is a single step of adding real and imaginary parts. In polar form, it would require converting both numbers to Cartesian first, which is inefficient. b Polar form is more efficient. Multiplication in polar form is a single step of multiplying moduli and adding arguments. In Cartesian form, it would require converting both numbers first.

1340 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

13 cis (0.983)

3 2

What do you remember?

c Polar

11 a

22 x4 + 2x2y2 + y4 – 3 Ax2y + Ay3 = 0

1 a Cartesian

b 0.709 + 0.230i

9 T he complex number z1 is rotated by –20° (or 20° clockwise) about the origin and its modulus is scaled by a factor of 2.

21 (x2 + y2)3 – 4x2y2 = 0

9.07 Complex arithmetic in polar form

7 64 cis

r cis (– )

3 tan ( ) sec ( )

3 cis 4 3

c Quadrant 4

Extend your thinking 20 r

17 20

1 cis (30°) 2 6 a Quadrant 2 5 a

The product z1 z2 is at approximately (0.73, 2.73).

19 a z

b 6 cis

4 a 12cis (75°)

b 2 2 cis

1 cis 2 2

14 cos (x) – i sin (x) Extend your thinking 15 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 16 a w cis i 2 1 cis c w 2 4

b w = 2 cis (0) = 2 d w = –1

17 zn = rn cis (nθ ) 18 Ask your teacher for worked solutions.

9.08 Complex inequalities and regions

c i 2

What do you remember? 1 a iv

Im(z)

b ii

c i

Re(z)

d iii –3 –2

–1 0 –1

2 a Vertical line

b Horizontal line

c Vertical line

d Horizontal line

–2

3 a Centre: 0 + 0i

b Centre: 3 + 0i

–3

Radius: 3 Radius: 5 c Centre: 3 – i

b ∣z + 1∣ = 3 d ∣z + 4 + 7i∣ = 4

ii y

4 x 3

5 2

6 a

Practice

5 a i 2 1 –4 –3 –2 –1 0 –1 –2

–5

Radius: 2 Radius: 1 c ∣z – (3 – 4i)∣ = 5

–4

d Centre: 0 – i

4 a ∣z – (2 + i)∣ = 2

Im(z)

Im(z) Re(z) 1

Re(z)

3 4 –5

–3 –4 –5 –6

–5

–7

b 5

1 x 2

ii y

Im(z)

9 4

b i

Im(z) Re(z)

3 –5

1 Re(z) –3

–2

–1 0 –1

–5

–2 –3

ii y = x

Answers 1341 mathspace.co

c 5

Im(z)

Re(z)

Re(z) 0

–5

d 5

–5

Im(z)

9 a

Im(z)

Re(z) 0

–5

Re(z)

–5

–5 7 4

Im(z) b 5

3 2 1 –4 –3 –2 –1 0 –1

Im(z)

Re(z) 1

Re(z)

–2

–5

–3 –4 –5 c 3x + y = 2, y + x = 0

1342 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

10 a

10 Im(z) 5

Re(z) –10

–5

–10 b

f1 Re(z)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7 −1 −2 −3 −4 −5

Im(z)

5 4 3 2 1

x2 49

e The Cartesian equation is

Im(z) 14

y2 24

Im(z)

| z| ≤ 1 Re(z) –10

–5

Re(z)

10 –1

–5

1 + 0i

–i

–10 c

Im(z)

Im(z) 4 3

2 Re(z) –10

–5

1 –4 –3 –2 –1 0 –1

Re(z) 1

3 4

–2 –3

–10

–4

11 ∣z – (2 + i)∣ = 2 12 a ∣z∣ = 3

16 a Yes 2

b x +y =9

13 a Foci: f1 = 5, f2 = –5 Constant sum: k = 14 b Distance between foci: ∣5 – (–5)∣ = 10. Condition: 14 > 10, so the condition is met.

b No

17 a Foci: f1 = 4i, f2 = –4i Constant difference: k = 4 b The vertices are at (0, ±2) on the imaginary axis.

c Major axis vertices are at (±7, 0). Minor axis vertices are at 0 , 2 6 .

Answers 1343 mathspace.co

Im(z) 4

1 −4 −3 −2 −1

Re(z) 1

−1

−2

−3

−4

Re(z)

−4 −3 −2 −1 −1

−2

Im(z)

22 ∣z∣ ≤ 3, Re ( z) ≥ 0, Im ( z) ≥ 0 or in polar form,

Im(z)

∣z∣ ≤ 3, 0 ≤ arg z ≤

23 a ∣z − (3 − 2i)∣ = 5 Re(z) 2

–2

16 4 6 8 2 6 and , 5 5 16 4 6 , 5

24 a ∣z – (2 + 2i)∣ = 1

–2 19

b ∣z – (–1 + 3i)∣ = 3 c Distance between centres:

Im(z)

Re(z) 2

Im(z)

= 32

= 10

Re(z) –5

1344 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

(–1)

The sum of the radii is 1 + 3 = 4. The distance between the centres is 10 ≈ 3.16. Since the distance between the centres is less than the sum of the radii, the circles intersect. 5 4 3 2 1 –5 –4 –3 –2 –1 –1

–10

∣(2 + 2i) – (–1 + 3i)∣ = ∣3 – i∣

–15

8 2 6 5

–2 –3 –4 –5

Im(z) f1 Re(z) 1 2 3 4 5 f2

10 a x = 3 + 7i, x = 3 – 7i

26 Im(z) 4

b x = –4 + 4i, x = –4 – 4i 11 The roots are 3 – i, 3 + i, and –2.

12 a x = ± 3i 5 i c x 2 13 a x = 3i, –3i

–4

–2

27 z

b x

2i 5

2i 2

b i

x i 7

0 x

5 i 3

0 x

i 7

x i 7

d 2x i 3 2x i 3

What do you remember? 1 a iii

3i 2

6 6 i, x 1 i 2 2 16 a (x – 4i) (x + 4i) = 0 ⇒ x = ±4i

9.09 Complex roots

c ii

2 For a polynomial with real coefficients, if a + bi is a root, then a – bi must also be a root. 2

4 ac b b 2a 4 The third root must be the conjugate of the complex root, which is 2 + i. 3 x

b (3x 5i)(3x 5i)

2 i 5 z

d x

15i d x 2 15 a x = –3 + i, x = –3 – i b x

5 2

28 2 2 + 1 29 Re

i 5

b x = –1 + 5i, – 1 – 5i

c x

Extend your thinking 5 2

14 a x

Re(z)

b x

Practice 5 a Δ = –4. Therefore, there are two complex roots. b Δ = 8. Therefore, there are two real roots. c Δ = 0. Therefore, there is one real root. d Δ = –8. Therefore, there are two complex roots. 6 x2 – 6x + 12 = 0 7 x2 – 2x + 2 = 0 1 3 1 3 8 a x i, x i 2 2 2 2 b The two roots are complex conjugates.

0 x

3 i 2

17 a (x + 1)2 + 4 = 0 ⇒ (x + 1)2 – (2i)2 = 0 ⇒ x = –1 ± 2i b (x – 4)2 + 9 = 0 ⇒ (x – 4)2 – (3i)2 = 0 ⇒ x = 4 ± 3i c (x + 2)2 + 9 = 0 ⇒ (x + 2)2 – (3i)2 = 0 ⇒ x = –2 ± 3i d (x – 5)2 + 1 = 0 ⇒ (x – 5)2 – i2 = 0 ⇒ x = 5 ± i 18 a Since z is a root of the quadratic equation, it satisfies the equation. Therefore, substitute and let x = z to give az2 + bz + c = 0. b Ask your teacher for worked solutions. Extend your thinking 5 2 20 A = –6, B = 25 19 A

21 A = ±8 22 Ask your teacher for worked solutions. 23 x i 2 (multiplicity 2) and x

i 2 (multiplicity 2)

9 A = –1

Answers 1345 mathspace.co

12 a y′ = 3 (x′ − 3)2 − 8

10.01 Translations

What do you remember? 2 6

1 a

3 6

2 a T

5 3

b T

3 2

c T

0 6

d T

0 7

−2

c C′ (− 2, − 11)

−2 y = 3x2 − 4 −4 −6

b B′ (3, − 7)

3 a A′ (1, 2)

−8

d D′ (– 3, 6)

y′ = 3 (x′ − 3)2 − 8

Practice 4 The translation is a shift of 7 units to the left and 4 units upwards. 5 a T

4 6

b P′ (− 1, 4)

6 a T

2 2

b T

13 For an invariant point, P + T = P, which means 0 5 0 T . Since , there are no invariant 0 1 0 points. Extend your thinking

5 2

7 A′ (− 4, − 1), B′ (1, − 1), C′ (− 2, 3) and D′ (− 7, 3) 8 D′ (− 1, − 4)

14 T2

10 3

15 T

3 6

16 T

2 3

9 F′ (4, − 8) 10 C 11 a (x′ − 4)2 + ( y′ + 2)2 = 4 b

6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

10.02 Dilations

What do you remember? 1 a

5 0 0 1

1 0 0 0.2

1 0

1 .5 0 0 3

1 0 0 2

0.5 0 0 0.5

x2 + y2 = 4 x 1 2 3 4 5 6

0 5

3 0 0 2 3 The pre-image is dilated by a factor of 5 parallel to the x-axis and a factor of 2 parallel to the y-axis.

2 T (x′ − 4)2 + ( y′ + 2)2 = 4

4 The pre-image is dilated by a factor of 4 parallel to the x-axis and reflected across the y-axis. Practice 5 P′ (10, − 12) 6 M′ (9, 4) 7 L′ (− 2, − 12)

1346 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

8 A′ (8, − 5), B′ (28, 1), C′ (8, − 10) 9 A′ (1, − 15), B′ (2, 9), C′ (− 1, − 6)

10.03 Rotations What do you remember?

10 The image equation is 3x′ − 14y′ = 21. 11 a The image vertices are A′ (0, 0), B′ (15, 0), C′ (15, 3), and D′ (0, 3).

1 a

0 1

1 0

2 2

1 2

3 2

1 2

b Answers may vary. Invariant: Parallelism. In the pre-image, AB is parallel to DC. In the image, A′B′ is still parallel to D′C′. Changed: Side lengths. The length of side AB is 5, while the length of image side A′B′ is 15. 12 The image equation is y′ = 7.5 (x′)2. 13

( x )2 81

14 y

4 ( y )2 9

1 2

3 2

2 a 120° anticlockwise

b 45° anticlockwise

3 The invariant point is the origin, (0, 0).

4 H′ (− 1, 5)

5 P

15 a = 1.5 16 T

1 2

Practice

6 (x

3 2

1 2

6 P′ (− 4, − 3)

0.5 0 0 1

7 R

17 C

7 2 3 2 , 2 2

8 P′ (− 3, 2), Q′ (− 3, 5), R′ (1, 5), and S′ (1, 2)

18 False. The horizontal sides are stretched to a length of 4 units, while the vertical sides remain 2 units long. The image is a rectangle, not a square.

• A

Extend your thinking

• B

2 2 3, 2 3

19 40.5 square units

• C

20 T he image is an ellipse with equation ( x )2 ( y )2 1 . The semi-axes are b and r. b2 r2 The area of an ellipse is π × (Semi-axis1) × (Semi-axis2), so the area is πbr. 21 Let T1

a 0 and T2 0 b

c 0 . Then 0 d

ca 0 ac 0 and T2T1 . Since 0 bd 0 db scalar multiplication is commutative (ac = ca and bd = db), T1 T2 = T2 T1. T1 T2

9 The coordinates of the image vertices are: 1

3 2

3 1 2

3, 2 3

2 1

C′ y 4

3 B′

1 A′ −6 −5 −4 −3 −2 −1 −1 −2

A 1

2 3 4 C

−3 −4

bm 22 The gradients of the image lines are both , a proving they are parallel.

Answers 1347 mathspace.co

10 The coordinates of the image vertices are: • X

3 3 2 3 2 3 , 2 2

• Y

5 , 2

5 3 2

• Z

2, 1 2 3

11 y

2 x

15 The distance of Q from the origin is 32 ( 4)2 9 16 25 5 units. Distance from the centre of rotation is an invariant property under rotation. Therefore, the distance of the image point Q′ from the origin is also 5 units. 16 a A′ (− 0.71, 2.12), B′ (1.41, 4.24), C′ (− 3.54, 4.95). b 90°

4 4 3

c 90°

d Angle measures are invariant under rotation.

y=4−x

−2

10 12

−2

Extend your thinking

−4 −6 y −8

2 , 2

17 a A′ (0, 0), B D 3 2 x

4 4 3

2 ,C 2

2, 0 , and

2 2 , 2 2 y

−10

1.5 1

12 x′ = –( y′)2 − 1

C D′

0.5 y = x2 + 1

A′ −0.5

x′ = − ( y′) − 1

−1

B 0.5

C′ x 1.5

−0.5

1 −4 −3 −2 −1

x 1

B′

4 b

−2 −3 −4

13 a Line segment A′B′ coordinates: A′ (− 3.69, − 3.93) and B′ (− 0.36, 3.14) b The length of both segments is 61 7.81 units. 14 a 8 square units b P′ (− 0.37, 1.37), Q′ (1.63, 4.83), R′ (− 0.10, 5.83), S′ (− 2.10, 2.37). c 8 square units d The areas are the same, which shows that area is invariant under rotation.

1348 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 0.414 square units

cos 1

3 5

53.13°

19 The reverse transformation is a clockwise rotation by θ, or an anticlockwise rotation by –θ. cos ( ) sin ( ) cos ( ) sin ( ) T sin ( ) cos ( ) sin ( ) cos ( ) 20 Let the rotation matrices for angles θ1 and θ2 be T1 and T2. By multiplying T1 T2 and T2 T1 and applying trigonometric sum identities, both products simplify to the same matrix: cos ( 1 2 ) sin ( 1 2 ) sin ( 1 2 ) cos ( 1 2 ) Since T1 T2 = T2 T1, the order of rotation does not matter. This property is known as commutativity.

14 a A′ (1, 1), B′ (1, 3), C′ (3, 3), D′ (3, 1)

10.04 Reflections

b Side lengths and angle measures are invariant. (Other possible answers include area, parallelism, and collinearity).

What do you remember? 1 A′ (1, 4)

c The orientation is reversed to clockwise.

2 A′ (3, 1)

d Yes. Any point on the line of reflection is invariant. For this square, the vertices A (1, 1) and C (3, 3) lie on the line y = x, therefore these points are invariant points.

3 A′ (2, 3) 4 A′ (− 7, − 5) 3 5 4 5

5 a

4 5 3 5

1 0 0 1

6 a 45°

5 13 12 13

12 13 5 13

1 2

3 2

1 2

b 90°

1 0

0 1

1 0

1 x 2

8 A

( 1, 2), B

c 30°

d 120°

(2, 5)

13 A D

3 2

2 a Inverse transformation b Commutativity c Identity transformation

3 a False

222 17

b False

c False

d True

Practice

31 8 , ,B 5 5 4 28 , 5 5 A

19 Ask your teacher for worked solutions.

1 D

10 A′ = (− 3, 5), B′ = (2, − 6), C′ = (− 5, − 2), D′ = (− 1, − 1)

12 F

0.6 0.8

What do you remember? ( 7, 3), C

9 A′ = (− 1, 4), B′ = (3, 7), C′ = (5, 2)

55 , 17

0.8 0.6

1 2

17 y = −2x

10.05 C ompositions of linear transformations

7 y

1 2

16 x = y3 − 1 Extend your thinking

Practice

11 A

15 y = (x + 2)2

16 12 , ,C 5 5

y 5 4 B C′ 3 2 1

9 12 , , 5 5

B′

x C −5 −4 −3−2 −1 1 2 3 4 5 6 −1 −2 A′ −3 −4 D −5 −6 D′

4 T

3 2

1 2

3 2

5 T

cos ( 160° ) 3 sin ( 160° ) sin ( 160° ) 3 cos ( 160 ° )

6 B

6 17 , 5 5

7 Ask your teacher for worked solutions. 8 M′ (1, 4)

Answers 1349 mathspace.co

9 A′ (− 6, 2), B′ (− 15, 6), C′ (− 6, 10)

8 P (– 3, – 6) y

C′

B′

−10

−5

4 2 A′ −2 A

1 2 x 3 3 12 Ask your teacher for worked solutions. 11 y

13 The final image of the point is A′(6, − 8).

C′

C x

10 The coordinates of the image endpoints are C′ (1, − 2) and D′ (− 5, 3).

14 a (8, − 6)

3 11 y

A′ −15

60 , 11

B′

6 C 8 10 12 14

10 Pre-image curve: y = 2x2 + 4x + 2 4

c No, the order of transformation does not matter in this scenario; both sequences of transformations result in the same image point.

−4 −3 −2 −1

Extend your thinking

10.06 Inverse of linear transformations What do you remember? d ad bc c 1

2 a True

b a

b True

c False

3 P = T P′ Practice 4 B (7, – 5) 5 A (3, 4) 6 P (6, 5) 7 C (4, – 2)

1350 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

−3 −4

2 1 3 3 11 a 1 2 3 3 b The inverse does not exist. 12 a

–1

−1

x 1

−2 y = −2(x − 1)2

17 Ask your teacher for worked solutions.

y = 2(x + 1)2

b (5, 10)

16 The angle is either θ = 0° or θ = 180°.

c Yes, the results are different, indicating that the order of transformations T1 and T2 does matter in this scenario. In general, transformations are not commutative.

1 T 1

b (8, − 6)

15 a (− 5, − 10)

26 2 , , 11 11

9 Pre-image points: A(0, 0), B

3 0 0 0.5

1 0 3 0 2

13 a It stretches by a factor of 2 along the x-axis. It stretches by a factor of 3 along the y-axis. There are no reflections or rotations. 1 0 2 b T 1 1 0 3 1 It shrinks by a factor of along the x-axis. 2 1 Also, it shrinks by a factor of along the y-axis. 3

14 a

2 4

10.07 Area of linear transformations

What do you remember?

1 a False 2

2 4

6 4

15 a

16 a

0 2

1 0 0 1

2 1

Practice 4 792 units2 1

5 504 units2

2 1

6 a 10 units2

2 1

8 7.5

2 1

9 18 units2

10 a 30 units2

This essentially squashes a 2D pre-image to a 1D image. This effect is irreversible, therefore, the determinant of the matrix is 0. Extend your thinking

2 13 15 13

20 Ask your teacher for worked solutions. 21 Ask your teacher for worked solutions.

4 5 3 5

3 5 4 5

4 5 3 5

3 5 4 5

c 80 2 units2

b 8 2

7 0 units2

18 T he points transformed by this matrix essentially all map onto a single line since: 1 2 x x 2y 2 4 y 2x 4 y

5 13 8 13

b 0

1 b

3 a 14

1 2 0

0 1

0 1 1 0

17 a

c True

b It implies the shape has been collapsed into a lower dimension, such as a line segment or a single point.

1 0 0 1 1 0

b True

2 a The area of the image is 0 square units.

2 4

11 a T

b T

1 2

3 2

1 2

2 0 0 5

c 300 units2

b 36 units2

12 60 units2 Extend your thinking 13

− 1 and 4

14 k

5 11 or k 3 3

15 Ask your teacher for worked solutions. 16 A singular transformation matrix is a matrix with a determinant of 0. When a singular matrix is applied to a shape, it transforms the shape such that the area is scaled by a factor equal to the determinant of the matrix. Since the determinant is 0, this implies that the area of the transformed shape becomes 0. Geometrically, a singular transformation matrix flattens the shape into a lower-dimensional subspace. Since the determinant is zero, the transformation cannot be reversed. There is no unique way to recover the original shape from its flattened version.

Answers 1351 mathspace.co

11.01 Absolute value graphs

11 x

What do you remember? x, if x 0 x, if x 0

1 x

b x = 0 (the y-axis)

2 a (0, 0) c (− ∞, ∞) or x ∈ ℝ

d [0, ∞) or y ≥ 0

3 a Horizontal shift (translation) b Vertical shift (translation) c Vertical dilation (stretch/compression) and reflection across the x-axis 4 a Keep the part of the graph on or above the x-axis. Reflect the part of the graph below the x-axis across the x-axis.

c True

b 3

c 14

e 10

f 10

g 3

h −1

i 5

8 a 12

b 12

c 20

d 3

e 32

f 32

g 6

h 1

−3

−2

−1

f (x) = ∣x − 3∣

f (x) = 2∣x − 3∣ − 5

−1

−3

−5

12 The graph of y = 4 ∣x − 3∣ + 2 can be obtained by starting with the graph of y = ∣x∣, then applying a vertical dilation by a factor of 4, translating right by 3 units, and translating up by 2 units. 13

4 y 3 2 1 −4−3−2−1 −1 −2 −3 −4 −5 −6 −7 −8 −9

−5−4 −3 −2−1 0 1 2 3 4 5 b −4 −3 −2 −1 0 1

2 3 4

1 2 3 4 5 6 7 8 9

9 y 8 7 6 5 4 3 2 1

b 1

10 a

−4−3−2−1 −1 −2 −3 −4

1 2 3 4 5 6 7 8 9

15 a

−8 −6 −4 −2 0 2 4 6 8

e No solution f

−4 −3 −2 −1 −12 −9 −6 −3 0 3 6 9 12

−1 −2

g −8 −6 −4 −2 0 2 4 6 8

−3 −4

h −8 −6 −4 −2 0 2 4 6 8

1352 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

i 2 9 a 14

−4

Practice

d 2

d False

7 a 7

−5

5 B b True

−1

f (x) = x − 3

b Keep the part of the graph for x ≥ 0. Discard the part for x < 0 and replace it with a reflection of the x ≥ 0 part across the y-axis.

6 a False

−2

f (x) = ∣x∣

x 1

b 4

19 a

9 y 8 7 6 5 4 3 2 1

3 2 1 −4 −3 −2 −1

x 1

−1

−2

−4 −3 −2 −1 −1 −2 −3 −4

−3 −4 16 a

b A

−2

−1

f (x) = ∣x∣

f (x) = 4∣x∣ − 1

−1

−3

−2

−1

12 10 8 6 4 2 −4 −3 −2 −1

−2 −4 −6 −8 −10 −12

c y = ∣x∣ − 2

d The result of part (c) shows that f (− x) = f (x), which verifies that the function is symmetric about the y-axis. 20 a

18 16 14 12 10 8 6 4 2

x 1

b y = ∣x∣ + 2 d y = ∣x − 1∣ − 3

18 a False

b True

c False

d False

−14−12−10−8−6−4 −2 2 4 6 8 10 12 14 −2 −4

b 11

c The graph of y = ∣x∣ is transformed into y = 4∣x∣ − 1 by first applying a vertical stretch by a factor of 4, making the graph narrower as the y-values are scaled up. This is followed by a downward shift of 1 unit, moving the entire graph down and repositioning the vertex from (0, 0) to (0, − 1). 17 a y = ∣x − 2∣

c f (− x) = ∣x∣ + 3

21 a

c C 9 8 7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3

d C

x 1 2 3 4 5 6 7

Answers 1353 mathspace.co

b 4

23 a

4 y 3 2 1

3 2 1 −5 −4 −3 −2 −1

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

x 1 2 3 4 5

−1

−2 −3 −4

22 a

y 6 4 2

−12 −10 −8 −6 −4 −2 −2

2 −4 −3 −2 −1

−4 f (x) = −0.5x − 5 −6 −8 b

y 8 6 4 2 −4 −3 −2 −1 f (x) = −0.5x − 5

−2 −4 −6

x 1

y = f (|x|)

−8

1354 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

9 y 8 7 6 5 4 3 2 1

8 y = | f (x)|

x 1

−1 −2 −3 −4

4 y 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

x 1

Extend your thinking 24 y = ∣x − 1∣ 25 a

Simplified expression in domain x < − 6

Simplified expression in domain − 6 ≤ x < 6

Simplified expression in domain x ≥ 6

∣x + 6∣

−x − 6

x+6

∣x − 6∣

−x + 6

x−6

∣x + 6∣ + ∣x − 6∣

− 2x

b f ( x)

2 x for x 6 12 for 6 x 2x for x 6

c f (− 6) = 12, f (6) = 12 d B 26

9 y 8 7 6 5 4 3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

9 y 8 7 6 5 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

12 10 8 6 4 2 −2 −2 −4 −6 −8 −10 −12

x 2

8 10 12

Answers 1355 mathspace.co

11.02 Reciprocal functions What do you remember? 1

−2

−1

f (x) = x − 3

−5

−4

−3

−2

−1

1 5

1 4

1 3

1 2

−1

Undefined

1 f ( x)

2 a False

b True

c False

1 f ( x) will approach ∞. As x approaches − 2 from 1 the right, y will approach − ∞. As x f ( x) 1 approaches 2 from the left, y will f ( x) approach − ∞. As x approaches 2 from the 1 right, y will approach ∞. f ( x)

c As x approaches − 2 from the left, y

3 a f (x) approaches ∞ b g (x) approaches 0 c

1 approaches 0 f ( x)

1 approaches ∞ g ( x)

4 a As x approaches 2, f (x) = − 2x + 4 approaches 0

d 4

1 b As x approaches from the right, 2 1 g(x) = approaches ∞ 2x 1

3 2 1

1 c As x approaches 2 from the left, f ( x) approaches ∞ 1 1 d As x approaches , approaches 0 2 g ( x) 5 a (− ∞, 2) ∪ (2, ∞)

−4 −3 −2 −1

1 ,0 2

x 1

−1

−2 −3

b (− ∞, 0) ∪ (0, ∞) c (− ∞, ∞) 1 d , 2

−4 (0,

)

e 4

Practice

6 a The solutions for the equation f (x) = 0 are x = 2 and x = − 2. For the reciprocal function 1 y= , these values correspond to vertical f ( x) asymptotes, as x approaches these values, 1 tends towards ±∞, resulting in vertical f ( x) asymptotes at x = 2 and x = − 2.

b As x approaches ±∞, f (x) = x2 − 4 approaches 1 ∞. Therefore, approaches 0, indicating f ( x) that y = 0 is the horizontal asymptote for the 1 reciprocal function y . f ( x)

1356 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 −4 −3 −2 −1

−1 −2 −3 −4

x 1

c Key features for g(x):

7 a x = − 3 and x = 1 b As x approaches ±∞, x terms dominates, so f (x) approaches ∞.

• Vertical asymptotes: x = ±1

• Horizontal asymptote: y = 0

• Invariant points: (0, − 1)

2, 1 , and

1 −6 −5 −4 −3 −2 −1

2, 1 and

x 1

−1

2 3 4

−2

−3

−2

−1

−3

−1

−4

−2

−3 8 a x=5 b As x approaches ±∞, ∣x − 5∣ approaches ∞. c

−4 −3 −2 −1

1 2 3 4 5 6 7 8 9

−1

−2

−3

−4

9 a 7

f (x)

11 a

0.4

0.3

0.2

0.1

−4 −3 −2 −1 −0.1

2 1 −3

−4−3−2 −1 −1

−2

−1

x 1

−0.2 −0.3 −0.4

b The function g(x) is the reciprocal of f (x), 1 where g (x) = . f ( x)

Answers 1357 mathspace.co

13 a • Vertical dilation by a factor of 3

0.4

• Reflection across the x-axis

0.3

• Horizontal translation 1 unit to the left

0.2

• Vertical translation 2 units up

0.1

b Vertical asymptote: x = − 1

−4 −3 −2 −1 −0.1

Horizontal asymptote: y = 2

c Domain: x ∈ ℝ, x ≠ − 1

−0.2

Range: y ∈ ℝ, y ≠ 2

−0.3

d y-intercept: (0, − 1)

−0.4

x-intercept: c

0.4

1 ,0 2

7 6 5 4 3 2 1

0.3 0.2 0.1

−4 −3 −2 −1 −0.1

−0.2 −5 −4 −3 −2 −1

−0.3 −0.4

12 a The zero of f (x) is x = 3. The vertical 1 asymptote for is x = 3. f ( x) b As x → ∞, f (x) → ∞. As x → − ∞, f (x) → − ∞. 1 Since f (x) approaches infinity, f ( x) approaches 0. The horizontal asymptote for 1 is y = 0. f ( x)

15 C 16 a False

b True

c True

d True

18 4

1 f ( x)

1 x 1

17 k = − 3, B(− 1, − 4)

1 . 6

−1

Extend your thinking

c The y-intercept is at 0, d

x 1

−1 −2 −3

2 x 2

14 The equation is y

−1

–1

–3

−3

–4

5 f (x)

–2

−2

1358 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 f ( x)

19 a 8

2 a (x + 2) (x + 3)

6 4 2

−8 −6 −4 −2 −2

−6

b False

d False

e False

Practice

4 2

−8 −6 −4 −2 −2

x 5

x 3

A Bx C 2 x x 9

−8

23 5 ( x 3)

7 a

7 3 ( x 2)

1 3 ( x 1)

2 x 4

1 2 ( x 2)

2 x

8 a

3 x 1

1 3 ( x 1)

−6

x 2 x2

1 x 3 2x

3 5x 4 3 ( x2

3 9 x 1 x 2

1 x2 1

2 2 x 2 x 3

Extend your thinking

0.5

11 x 2

0.4 0.3

0.2 0.1 −4 −3 −2 −1 −0.1

17 5 ( x 2)

−4

A B C x 1 x 2 x 3 5 Equating coefficients of x: A + B = 4. Equating constant terms: 2A + B = 7

c True

−8

d (x − 3)2

3 a True

4 a

−4

b (x − 3) (x + 3)

c (2x + 1) (x − 3)

x 1

2 x 1

3 2 ( x 1)

9 2 ( x 1)

1 x 2

( x 1)2

11.04 Trigonometric functions What do you remember? 1 a False

b True

c True

2 a x = 2π

11.03 R ational expressions and partial fractions What do you remember? 1 a Degree: 3, Leading coefficient: 5

b i Decreasing

ii Increasing

iii Increasing 3 c 2 3 a −π

b 0

b Degree: 2, Leading coefficient: 6

c i Decreasing

ii Increasing

c Degree: 2, Leading coefficient: 2

iii Decreasing

iv Increasing

d Degree: 0, Leading coefficient: 8

Answers 1359 mathspace.co

b −1 ≤ y ≤ 1 3 d , 2 2

4 a (0, 1) c 2π

Practice 8 a

5 a The opposite side increases

b The value of tan (θ ) increases sin( ) c tan( ) cos( )

1 2

3 , 2

3 2 2 2

3 , 2

3 , , 2 2 2

3 – π 2

–2π

1 − π 2

–π

x 0

−

1 2 –1

− 2π

7 4

5 4

−π

3 4

tan (θ )

−1

3 4

5 4

7 4

2π

tan (θ )

−1

g 4

y 1 1 2 x 3 – π 2

1 − π 2

–π

−

1π 2

1 2

–1

2 1

1 3 – π –π − π 2 –1 2

1π π 2

3π 2

9 a

–2

–3

1 2

–4

x 6 x

3 ,x 2

3 2

7 The domain of the graph y = tan (x) is x : x ∈ ℝ, x ≠ + kπ, where k is any integer. 2

1 − π 2

−

1 2

1π 2

–1

The function is increasing at 0 ≤ x ≤ . 2

1360 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

11.05 T ransformations of trigonometric functions

y 1

What do you remember?

1 2

1 False

x 1 − π 2

1 − 2

2 The parameter k represents the vertical translation of the graph.

1π 2

4 3 4 a i No horizontal translation 3 a 4

–1

b 5

1 8

ii Down by 2 units The function is increasing at 0

b i Left by π units

10 a The radius of the unit circle is 1, therefore the maximum value for the function is 1 and the minimum value for the function is − 1. b 2π = 360° which represents a complete revolution of the circle, therefore the values of the function repeat every 2π radians. 11 a (0, 0) d Increase 12 a B and C

b π

c x

e x=π b −∞ < y < ∞

c x

13 a Positive

b Quadrant 3

c Negative

d Quadrant 4

3 2

ii No vertical translation c i Right by units 3 ii Up by 5 units d i Left by units 2 5 ii Up by units 2 Practice 5 a

y 2

y = 2 sin (x)

y for the x unit circle, the graph of y = tan (x) will be positive where x and y have the same sign. This occurs in Quadrant 1 and 3, where 3 0 < x < and where π < x < . 2 2

e Since the tangent ratio is defined as

1 π 2

−1

3 π 2

2π

−2

14 A and D 15 x = 2π, 3π, 4π, 5π, 6π 16 a No x-intercepts

b b Two x-intercepts

y 3 2

Extend your thinking

17 nπ

18 n

−1

2 19 g (x) = cos (x) is the result of f (x) = sin (x) translated units to the left without reflection. 2 5 , 20 x or x = 45°, 225° 4 4

x 1 π 2

2π

−2 −3

y = –3 sin (x)

Answers 1361 mathspace.co

6 a i π b i 2

ii y = sin (2x) ii y = cos (4x)

c i 8π

ii y

d i 3π

ii y

7 a

d 3

1 cos x 4 2 sin x 3

y = sin (2x)

1 π 4

1 π 2

3 π 4

1 π 4

−1

1 π 2

3 π 4

8 a Left by y

−2

y = cos (2x) + 1

units 3

sin x

units 4

y cos x 3

3 c Right by units d Left by units 6 2 3 y sin x y cos x 2 6

y 3 2

3 2 Minimum: 1 5 Minimum: 2 c Maximum: 3 9 a Maximum: 5

1 x

1 π 3

−1

2 π 3

b Maximum:

Minimum: –1 10 a

−2 y = –3 sin (3x)

−3

b Right by

y 2

π  y = 2 sin  x +  4  

1 c

y = 3 sin (2x) + 1

−1 −2

2 1 0 −1

x 1 π 4

1 π 2

3 π 4

−2

1362 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

x 1 3 5 7 π π π 2π π 3π π 4π 2 2 2 2

b 4

1 2 π 3

−3 −4

11 a 7 6

4 π 2π 3

1 π  y = 5 cos   x −   + 2 2 3   

4 2 − π − π −1 3 3 −2

8 π 3

3 2 1

π  y = –3 sin  x +  3 

1 2 4 5 7 8 10 11 13 π π π π π 2π π π 3π π π 4π π 3 3 3 3 3 3 3 3 3

−1

  π  y = 3 sin  2  x −   + 4 5   

−2 −3

5 4

y 1

2 1 0 −1

x 1 2 3 4 π π π π π 5 5 5 5

−1

6 π 5

1 π 4

1 π 2

3 π 4

−2 −3

b 2 1

−4

  π  y = –2 cos  3  x +   − 1 6   

−5

  π  y = 2 sin  2  x +   − 3 4   

x 1 − π 6 −1 −2 −3

1 π 6

1 π 3

1 π 2

12 a Around June b Period: 12 months Amplitude: 10°C 13 a Yes, the temperature level appears to be a periodic function, as it repeats in a regular pattern over time. b The temperature level seems to peak around 12:00 p.m. each day. c The temperature level appears to be at its lowest around 12:00 a.m. each day. d The period is 24 hours, as the temperature pattern repeats every day. e 10°C

Answers 1363 mathspace.co

1 14 • A horizontal dilation by a factor of from the 2 y-axis (period becomes π). • A horizontal translation of units to the right. 4 • A reflection across the x-axis. • A vertical dilation by a factor of 3 from the x-axis (amplitude becomes 3).

16 a d

3 ,5 2

3 ,1 4

b g ( x)

2 1 c Horizontal dilation by a factor of . 4 d k=4

sin 3 x

or g ( x )

–π

3 4

–2π

3 2

2π

tan (2x)

Undefined

d 4

2 2

,3 ,7

1 from the 3 y-axis and a horizontal translation (phase shift) of units left. 12

18 a

17 a A horizontal dilation by a factor of

• A vertical translation of 2 units up (midline becomes y = 2). 15 a 2π

sin 3 x

2 b i Period: π

Midline: y = 3

Point of inflection: (0, 3)

ii Vertical translation of +3 (shifting the entire graph up by 3 units)

1 1 −π − 43π− 2π− 41 π −1

x 1 π 1 π 3π π 4 2 4

−2

iii y = tan (x) + 3 iv Domain: x

2 Range: (–∞,∞)

−3

c i Period: π

−4

Midline: y = 2

k , where k is any integer

Point of inflection: (0, 2) 19 a i Period: 2π Midline: y = 0 Point of inflection: (0, 0)

ii Reflection across the x-axis (flipping the function). Vertical translation of +2 (shifting up by 2 units)

ii Dilation parallel to the x-axis by a scale factor of 2 (the function stretches horizontally) x iii y tan 2 iv Domain: x ≠ kπ, where k is any odd integer (as vertical asymptotes occur at these points)

iii y = –tan (x) + 2

Range: (–∞, ∞)

ii Vertical dilation by a factor of 3 (stretching the graph vertically)

1364 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

iv Domain: x

2 Range: (– ∞,∞)

k , where k is any integer

d i Period: π Midline: y = 0 Point of inflection: (0, 0)

iii y = 3 tan (x) iv Domain: x

2 Range: (– ∞,∞)

k , where k is any integer

20 a

4 3 2 1 3 −2π − π −π 2

1 − π −1 2 −2 −3 −4

1 π 2

3 π 2

2π

1 2 • The graph oscillates twice as fast, meaning there are more vertical asymptotes within the same interval. • Dilation from the y-axis by a factor of

4 3 2 1 3 −2π − π 2

−π

x 1 π 2

1 − π −1 2 −2 −3 −4

3 π 2

2π

3 π 2π 2

5 π 2

• Dilation from the x-axis by a factor of 3 • Reflection across the x-axis c

4 3 2 1 1 5 3 − π −2π − π −π − π −1 2 −2 2 2 −3 −4

x 1 π π 2

• Dilation from the y-axis by a factor of 4 • Vertical shift up by 1 unit Extend your thinking 21 a f (x) = cos (x) 1 1 1 x b j , k , g ( x ) cos 3 4 3 4 1 3 , 22 c 2 2 23 y

3 cos(x), y

3 sin x

Answers 1365 mathspace.co

24 4

25 a = 4, b = Any positive value, d = 7 26 a

π 2

b b

2 sin(2 x ) : 4 cos(2 x ) 2 sin(2 x) y = Write the equation 4 cos(2 x) 2 1 sin(2 x) y = Simplify 4 2 cos(2 x) sin 1 y = × tan (2x) Substitute = tan cos 2 From y = tan (x), there will be: 1 • Dilation from the x-axis by a factor of 2 1 • Dilation from the y-axis by a factor of 2 • No horizontal and vertical translations

27 Simplifying y

−1, −

π 2

Domain: [−1, 1], Range: 8

, 2 2

y (−1, π)

11.06 Inverse circular functions What do you remember? 1 a No

b Yes

1 2 d −1

2 a

3 a

1 3

c No

d Yes

b 0

c 1

3 e 2

f 0

b −0.8

c 10

(1, 0)

Domain: [−1, 1], Range: [0, π]

d 4

2 e f 3 3 , , Range: [−1, 1] 4 a Domain: 2 2 b Domain: [0, π], Range: [−1, 1] c Domain: , , Range: (−∞, ∞) 2 2

Practice 5 a

e π

f 3

d 4

g 0

i 0 6 a Domain: [−1, 1], Range:

, 2 2

b Domain: [−1, 1], Range: [0, π] c Domain: (−∞, ∞), Range: , 2 2

1366 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Domain: (−∞, ∞), Range: Asymptotes: y 2

, 2 2

10 a Vertical dilation by a factor of 3. 3 3 Domain: [−1, 1], Range: . , 2 2

13 a A horizontal dilation by a factor of 3 from the y-axis. b Domain: [−3, 3], Range: , 2 2

b Vertical translation up by π units. Domain: [−1, 1], Range: [π, 2π].

c Reflection across the x-axis. , . Domain: (−∞, ∞), Range: 2 2

π 2

d Vertical dilation by a factor of 2, followed by a vertical translation down by Domain: [−1, 1], Range: 11

−1,

3π 2

units. 2

3 . , 2 2

−3, −

π 2

x 1, −

Range:

π 2

3 , 2 2

12 a A reflection across the x-axis, followed by a vertical dilation by a factor of 3. b Domain: [−1, 1] (unchanged by vertical transformations). Range: [−3π, 0]. (The parent range [0, π] is dilated by 3 to [0, 3π] and then reflected to [−3π, 0]).

Range:

3 , , Asymptotes: y 4 4

3 and 4

4 y (1, π)

(1, 0)

(−1, 0)

(−1, −3π) Range: [0, π]

Answers 1367 mathspace.co

Extend your thinking −1

16 Let θ = sin (x). Then sin (θ ) = x. In a right-angled triangle with hypotenuse 1 and opposite side x, . the angle is θ. The other acute angle is 2 For this angle φ, cos (φ ) = cos = sin (q ) = x. 2 −1 Since φ is in the range [0, π], φ = cos (x). Therefore, cos 1 (x) sin 1 (x), giving the 2 identity.

Key points: (0, 0), 0.5,

y 2 1

18 Parent asymptotes at y

−1 −0.5 −0.5, − 3 3

11.07 Reciprocal trigonometric graphs What do you remember? 1

Vertical dilation by A and translation by k give

Add equations: 2k = 0, so k = 0 A 2

2 , so A . 3 3 2 1 tan (x) Equation: y 3 Then

Domain: (−∞, ∞), Range:

sin (x)

2 2

1 2

3 2

cosec (x)

2 3 3

tan (x)

Undefined

cot (x)

−1

19 Let θ = sin (x), then sin (θ ) = x. In a right triangle, opposite = x, hypotenuse = 1, and x adjacent . 1 x2 . Therefore, tan ( ) 1 x2 x The key features for sketching y are: 1 x2 Domain: (−1, 1) Range: (−∞, ∞) Vertical asymptotes: x = –1 and x = 1

1368 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 3

2 a x

, 3 3

−1 −2

k new asymptotes: A , 2 3 A k 2 3 A A k k Solving: , 2 3 2

0.5, 3 x 3 1 0.5

(0, 0)

17 Solving sin−1 (x) = sin−1 (x − 1) would imply x = x −1, which simplifies to 0 = −1, an algebraic contradiction. The graph of y = sin−1(x) is a strictly increasing function. The graph of g(x) = sin−1(x − 1) is the same graph translated 1 unit to the right. Because the function is always increasing, for any given y-value in the shared range, the x-value on the graph of g(x) will always be greater than the x-value on the graph of y = sin−1(x). Therefore, the two graphs can never intersect, and there are no solutions.

3 3 , 0.5, 3 3

cosec (x)

–1 2 Undefined

2 3 Undefined 3

b x = –2π, –π, 0, π, 2π The asymptotes for y = cosec (x) are the vertical lines with equations x = –2π, –π, 0, π, 2π. 3 c x , 2 2 d 2π 3 a x = –2π, –π, 0, π, 2π b –1 ≤ f (x) ≤ 1 c g(x) ∈ (–∞, –1] ∪ [1, ∞)

3 3 , , , 2 2 2 2 e Period of f (x) = 2π

d x

y 3 2

Period of g(x) = 2π π, 0 −2

3 3 , , , 2 2 2 2 b –1 ≤ f (x) ≤ 1

4 a x

− 21 π

c g(x) ∈ (–∞, –1] ∪ [1, ∞)

π, 0 2 1π 2

−1

d x = –2π, –π, 0, π, 2π

−2

e Period of f (x) = 2π

−3

Period of g(x) = 2π 7 5 3 3 5 7 , , , , , , , 4 4 4 4 4 4 4 4 b f (x) ∈ ℝ

5 a x

c g(x) ∈ ℝ

Asymptotes at x = –π, 0, π 7 a π b (–∞, ∞) c The asymptotes are at x = kπ, where k is an integer.

d x = –2π, –π, 0, π, 2π e Period of f (x) = π

d y = cot (x)

Period of g(x) = π

The function has a period of π, a range of all real numbers, and vertical asymptotes at integer multiples of π (e.g., x = 0, x = π). This combination of features uniquely matches the cotangent function. The cosecant and secant functions both have a range of (–∞, –1] ∪ [1, ∞) and a period of 2π.

Practice 6 a

y 3 2

π ,1 2

1 − 2π

1π 2

−1 π − 2 , −1 −2

8 a i x = –π, 0, π 3 ii x , 2 2 iii No x-intercept is equal to 0. b cosec (x) is lowest when sin (x) is highest, therefore sin (x) = 1 implies cosec (x) = 1.

−3

9 a True b

b i x = 0, π, 2π

1 (−π, −1)− 2π

7 6 11 a i x = 2π 5 b i x 6 c i x

2 1

(0, 1)

x 1π 2 (π, –1)

−1

iii x

−2

c False

d False

ii x

5 , 4 4

11 6 ii x = –π 13 ii x 6 ii x

iv x

7 6

c 2π

−3 Asymptotes at x

b True

10 a π

Asymptotes at x = –π, 0, π

, 2 2

Answers 1369 mathspace.co

7 9 , 2 , , 2 , , 2 , 4 4 4 3 5 , 2 , , 2 4 4 1 13 f (x) cosec (x) . Vertical asymptotes sin (x) occur when the denominator of a fraction is zero. Identifying them requires finding where sin (x) = 0. This occurs at all integer multiples of π. Therefore, the asymptotes are at the points x = kπ for any integer k.

14 a i 2

ii f (x) = 2

iii f (x) = 2

iv There are none.

5 4 3 2 1 −2 −1.5 −1 −0.5

−1 −2 −3 −4 −5

15 a

3 , and , 2 2

x 0.5

1.5

, and 2 2

c The cotangent function is never increasing on its domain. 2 . _ n_ For f (x) = cosec (2x), n = 2, so the period is 2 . 2 The asymptotes occur where sin (2x) = 0, which is when 2x = kπ for any integer k. k Solving for x, gives x . 2 On the interval [0, 2π], the asymptotes are at 3 x 0, , , , 2 . 2 2 16 a The period is calculated using

y 2 1 x 1π 4

−1

1 π 2

3π 4

π 3π, −1 4

−2

c The domain of the parent function is k x ≠ kπ. The domain of f (x) is x . 2 The horizontal compression changes the locations of the asymptotes, therefore, changing the domain. The range of the parent function is (–∞, –1] ∪ [1, ∞). Since there is no vertical stretch or shift, the range of f (x) remains the same. The midline of the parent function is y = 0. Since there is no vertical shift, the midline of f (x) remains y = 0. 17 The period is

. 2

The asymptotes are at x integer.

k , where k is an 2

18 a A vertical dilation by a factor of 2 from the x-axis. b (–∞, –2] ∪ [2, ∞) c y=0 d

y 3 2

(2π, 2)

(0, 2)

1 x −1 −2 −3

1370 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

π ,1 4

1π 2

3π 2

(π, −2)

units to the left. 3 b The asymptotes are at x

1 , there are no x-intercepts because its tan (x) numerator is never 0.

For

19 a

. 3

, 0

The difference is that at x

expression

is defined and equal to 0, sin (x) whereas 1 is undefined because tan (x) is tan (x) undefined.

3 2 π, 0 6

33333 π 100000

−1

cos (x)

3 and , the 2 2

26 The graph of sec (x) is labelled A. The range of sec (x) is (–∞, –1] ∪ [1, ∞) and sec (0) = 1. 27 a m = 2

−2

28 a

−3

b m = –2 , c 5 4

3, b

11.08 L ocus definitions and the distance formula

What do you remember?

1 a 5

4 (0, 4)

(2π, 4)

2 a iii

c 13

d 8

c iv

d ii

3 a (x – 2) + ( y + 3) = 16 b Centre: (0, 1); Radius: 9 3

1 1π 2

3π 2

3 Asymptotes: x , x 2 2 Local minimum points: (0, 4) and (2π, 4) Local maximum point: (π, 2) 21 a π c y=1 , x e x 2 22 2 23 k = 2, m = 4 24

b i 2

(π, 2)

b 5

, 2

b to the right 4 d (–∞, –2] ∪ [4, ∞)

Practice 4 The locus is the line with equation 2x + y − 9 = 0. This line is the perpendicular bisector of the segment connecting points A and B. 5 Equation: (x – 2)2 + ( y – 1)2 = 9 The locus is a circle with its centre at (2, 1) and a radius of 3 units. y

C (2, 1) x

Extend your thinking cos (x)

, x-intercepts occur when cos (x) = 0, sin (x) 3 which are at x , in the interval. 2 2

25 For

Answers 1371 mathspace.co

6 The equation is

x2 25

y2 16 4

10 Vertex: V (2, 0). Focus: F (3, 0). Directrix: x = 1. Locus definition: The set of all points P (x, y) such that the distance from P to F (3, 0) is equal to the distance from P to the line x = 1.

3 2 1

−5 −4 −3 −2 −1 −1

11 This is the locus of a hyperbola. The equation is x2 y2 1. 25 144 y

2 3 4 5

−2 5

−3

−4

F2 x −10

−5

7 Vertices are at (±4, 0). Co-vertices are at (0, ±2).

−5

−10

2 1 x −4 −3 −2 −1

12 Vertices are at (±3, 0). Asymptotes:

−1

b x a

4 x 3 y

−2

8 a Vertices: (±5, 0); Co-vertices: (0, ±3)

x2 y2 b 1 25 9 c The locus definition is the set of all points where the sum of the distances to the foci is 10. 9 This is the locus of a parabola. Equation is x2 = 8y. y

Extend your thinking 13

y2 25

x2 16

14 Equation of the parabola: (x – 1)2 = 8 (y – 2) x

1372 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

Directrix: y = 0

15 PA = 2∣x∣ Write the formula ( x 3)2

y2 = 2∣x∣ Write the distance formula for PA, where A is (3, 0)

(x – 3)2 + y2 = 4x2 Square both sides 2

x – 6x + 9 + y2 = 4x2 Expand the left-hand side 3x2 + 6x – y2 = 9 2

Rearrange and simplify the equation

3 (x + 2x) – y = 9

Factorise the common term from the x-terms

3 ((x + 1)2 – 1) – y2 = 9

Complete the square for the x-terms

3 (x + 1)2 – 3 – y2 = 9

Distribute the 3

3 (x + 1) – y = 12

Move the constant term to the right-hand side

( x 1)2 4

Divide both sides by 12

y2 =1 12

11.09 Conic sections from locus definitions

5 a Vertex: (– 1, 3) p = –1 Opens downwards

What do you remember? 1 d 2 a ii

( x2

x1 ) ( y2 b iv

Focus: (– 1, 2) Directrix: y = 4

y1 )

c i

d iii

Axis of symmetry: x = – 1 b

3 a The fixed point from which all points on the parabola are equidistant to it and the directrix.

y 4 V (–1, 3) 3

b The fixed line from which all points on the parabola are equidistant to it and the focus.

F (–1, 2) 2

c The two fixed points used in the locus definition of an ellipse (sum of distances is constant) or a hyperbola (difference of distances is constant).

1 x

d Lines that the branches of a hyperbola approach as they tend towards infinity. Practice

–4 –3 –2 –1

6 (x + 1)2 + ( y – 3)2 = 16

4 Vertex: (0, 0) p=2

Parabola opens to the right

y2 25

x2 9

Equation: y2 = 8x

8 a Centre: (– 3, 1)

Axis of symmetry: y = 0 (x-axis)

Major axis: Horizontal a = 4, b = 7, c = 3

5 4 3 2 1 –2 –1

–1 –2 –3 –4 –5

Major axis length: 2a = 8 Minor axis length: 2b = 2 7 Foci: F1 (–6, 1), F2 (0, 1)

V (0, 0) F (2, 0) x 1 2 3 4

Vertices: V1 (–7, 1), V2 (1, 1)

Answers 1373 mathspace.co

11.10 P arametric representations of relations

y 3 2 C

What do you remember? V2 x

–7 –6 –5 –4 –3 –2 –1

1 a True 2

x2 64

y2 36

10 a Centre: (1, 2) Transverse axis: Vertical a = 3, b = 4, c = 5 Vertices: V1 (1, 5), V2 (1, –1) The foci : F1 (1, 7), F2(1, –3) 3 Asymptotes: y – 2 = (x – 1) 4 b y 7 6 F1 (1, 7) 5 4 V1 (1, 5) 3 2 1 C(1, 2)

–2

(3, 0)

–1

(0, 1)

–1

(–1, 2)

(0, 3)

(3, 4)

b i

c iv

d ii

5 cos2(t) + sin2(t) = 1 6 x = (y – 1)2, restriction: x ≥ 0 7 a Answers may vary. A possible answer is: x = t, y = t2 – 5 b Answers may vary. A possible answer is: x = 1 + 5 cos(t), y = 5 sin (t) Practice

Extend your thinking 12 y2 = 12 (x – 1) y2 16

(x, y)

4 A

11 The foci are at (–5, 0) and (5, 0).

x2 25

d True

y=t+2

3 a iii

−4 −3 −2 −1 1 2 3 4 5 6 −1 V2(1, −1) −2 −3 F2(1, −3)

c True

x = t2 – 1

–1

b False

14 Equation: (x – 2)2 + y2 = 4, Centre: (2, 0), Radius: 2 Ask your teacher for worked solutions.

1374 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

8 The curve is a line segment starting at (–2, –2) (when t = –1) and ending at (8, 3) (when t = 4). 3 2 1 −2 −1 t = –1

−1 −2

y t=4 x 1

9 A semi-circle centred at the origin (0, 0) with radius 3, traced anticlockwise from (3, 0) (when t = 0) to (–3, 0) (when t = π). The curve lies in the upper half-plane (y ≥ 0). 4 3

−1

y2 4

(3, 0) 1

−2

2 x 3

1 ; Asymptotes: y y

5 4 3 2 1

(0, 3)

−4 −3 −2 −1

x2 9

2 (−3, 0)

15 Cartesian:

−7−6−5−4−3−2−1 −1 −2 −3 −4 −5

1 2 3 4 5 6 7

−3 −4

10 x = 4y + 9 or y

16 a x = (y – 1)2 – 1 or x = y2 – 2y b The curve is a segment of a parabola opening to the right with vertex at (–1, 1) (when t = 0). It starts at (3, –1) (when t = –2) and ends at (3, 3) (when t = 2). As t increases, the path moves from the starting point, left and up to the vertex, and then right and up to the ending point.

x 9 4

11 y = (x – 2)2 –1 x2 y2 1 4 25 13 x = 2 + 3 cos (t), y = –1 + 3 sin (t)

x2 y2 16 4 Domain: [–4, 4]

14 Cartesian:

t=2

3 2

Range: [–2, 2] x-intercepts: (±4, 0)

1 t = 0 (Vertex)

y-intercepts: (0, ±2)

y (0, 2)

−1

t = –2

(−4, 0)

(4, 0) 1

−1 −2

−1

1 −4 −3 −2 −1

3 4

x Extend your thinking 17 a x2 = 4y2(1 – y2)

(0, –2)

b Domain (x-values): [–1, 1] Range (y-values): [–1, 1] 18 x = –1 + 5t, y = 3 – 5t for 0 ≤ t ≤ 1 1, 1 2 b t = 2π. Point: (2πr, 0)

19 a Point:

Answers 1375 mathspace.co

11.11 Polar coordinates and polar graphs What do you remember? 1 a ii

b i

c iii

5π 6 11π 12

2 a Cardioid (a type of Limaçon) b Rose c Circle

b iii

c ii

d i

Practice 4 a

5 a 1, 3

3 3, 2

b (–4, 0)

5 3 2, 4

3, 3 3

5 2, 6

d (5, 0)

c (1, –1)

d (0, 5)

5π 12

π 3

π 4

π 6

13π 12 7π 6

23π 12 5π 4 4π 3 17 π 3π 12 2

5π 19π 3 12

7π 4

11π 6

10 Cartesian equation: x2 + ( y – 3)2 = 9 (Circle centre (0, 3), radius 3).

8 A circle centred at the pole with radius 5. 7

90° 135°

π 2

45°

3 1

180°

π 12

d Parabola (a type of conic section) 3 a iv

3π 4

2π 3

π 2

7π 12

O 1 2 3 4 56 7

0°

–7 –5 –3 –1 –1

0 1

–3 –5

315°

225°

–7

270° 9 The graph is a straight line passing through the 2 pole with an angle of with the positive polar 3 axis. Its Cartesian equation is y 3 x.

3π 2

11 Range for full curve: 0 ≤ θ < 2π 3 π 2 2 1 π –3

0 –2

–1

1 –1 –2 –3

1376 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

3π 2

12 The curve is a cardioid. It is symmetric about the polar axis. Tests for symmetry about the line θ = and the pole fail. 2 13 a Dimpled limaçon. Key points are (1, 0), 3, , 2 3 (5, π), 3, . It does not pass through the 2 pole. π 2

16 Conic section: Parabola Eccentricity: e = 1 Directrix: y = 4 17 r

4 1 cos ( )

Extend your thinking 18 Cartesian equation: x2 – 3y2 + 12y = 9 Conic section: Hyperbola 19 Points of intersection: 2,

5 and 2, 3 3

20 Symmetric about the polar axis, the line and the pole. π

0 3

1 2 3 4 5 6

–6 –5 –4 –3 –2 –1

, 2

π 2

2 1 3π 2

π –3

0 –2

–1

–1 b Limaçon with inner loop. Symmetric about the line . Passes through the pole at 2 7 11 and . Key points include (2, 0), 6 6 , and (2, π). 6, 2

–2 –3 3π 2

12.01 Discrete random variables What do you remember? 1 a Continuous b Discrete

1 2 3 4 5 6 –6 –5 –4 –3 –2 –1

14 Conic section: Ellipse 2 Eccentricity: e 3 Directrix: x = –5 15 r = 5

c Continuous

2 Yes. All probabilities are between 0 and 1, and the sum is 10% + 20% + 25% + 45% = 100% = 1. 3 a i • 0 ≤ P(x) ≤ 1 • ΣP(x) = 1 • X is discrete numerical. ii Yes b i • 0 ≤ P(x) ≤ 1 • X is discrete numerical. ii No c i • 0 ≤ P(x) ≤ 1 • ΣP(x) = 1 • X is discrete numerical. ii Yes d i • 0 ≤ P(x) ≤ 1 • X is discrete numerical. ii No

Answers 1377 mathspace.co

Practice

1 11 1 13 a 4

12 k

4 a Yes b No, outcomes are continuous. 5 a Yes

b 0, 1, 2, 3

6 a P(X = 1) = 0.42 b P(X ≥ 1) = 0.91 7 a

7 16

3 16

13 16

b Non-uniform

c Non-uniform 1

P( X = x)

1 6

2 6

3 6

d Uniform

e Non-uniform 15 a

b Yes, because all probabilities are between 0 and 1, and the sum of probabilities is 1 2 3 6 . 1 6 6 6 6 f

P(F = f )

1 16

4 16

6 16

4 16

1 16

P(F ≤ f )

1 16

5 16

11 16

15 16

16 16

7 16

14 a Uniform

8 a

5 16

P(X = x)

0.9216

0.0768

0.0016

b 0.9984 c The distribution is non-uniform because the probabilities for each outcome (P( X = 0), P( X = 1), P( X = 2)) are not equal. 16 a The probabilities are all between 0 and 1. The sum of the probabilities is 0.2 + 0.5 + 0.3 = 1.0. Therefore, it is a valid probability distribution. b

P(X = x)

0.64

0.32

0.04

c 0.36 9 a 1, 2, 3, 4, 5, 6 b Uniform, because the outcomes are equally likely. c x

996 1005 994 1005 996 1004 P(X = x) 6000 6000 6000 6000 6000 6000

667 2000

17 a x

P(X ≤ x)

0.1

0.3

0.65

0.95

P(X = x)

0.1

0.2

0.35 13 ii 14

0.3

0.05

b i 0.65 18 a

e 0.05%

2 21

3 21

4 21

5 21

6 21

P(X = x)

1 21

P(X = x)

5 70

7 70

24 70

8 70

26 70

b P( X > 4) =

11 21

10 a

29 35

b i 11 k

Extend your thinking

18 35

1 6

1378 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

19 a k = 0.8

b P( X = 0) = 0.8

12.02 C entre and spread of discrete random variables What do you remember?

b 1

c 1.9

2 a i 8

ii 6.2

b i 0

ii 0

c i 8

ii 10.22

d 1 e 1 f E qual, Distribution B is a mirror image of Distribution A over the same range, so the values will have the same average distance from the mean value. Therefore, Distribution B will have the same standard deviation.

10 a 2.36

P(X = x)

1 4

11 a Yes, because ΣP(x) = 1, X is discrete numerical, and 0 ≤ P(x) ≤ 1.

b 2.5904

b 2.4

5 2

c 1.82

4 a

b Distribution A: Negatively skewed, Distribution B: Positively skewed c Equal for both distributions.

1 a Yes, because ΣP(x) = 1, X is discrete numerical, and 0 ≤ P(x) ≤ 1.

3 a

9 a Distribution A: 4, Distribution B: 3

d 1.35

P(X = x)

3 25

4 25

5 25

6 25

7 25

12 a k

27 5

5 a i 1, 3, 4, 5

ii Non-uniform iii 3.85

b i 1, 2, 3, 4, 5 ii Uniform

iii 3

Practice

7 a 6.78

b 53.8

d 0.24

e 7.8316

v 2.87

b i k

1 17

2344 289

b m = 1 – 2n

c n = 0.3

d m = 0.4

iii 7.4

ii 5

v 2.85

P(X = x)

1 10

2 10

3 10

4 10

iii 1

b i

P(X = x)

7 19

6 19

4 19

2 19

28 19

c i

iii 1.63 x

P(X = x)

3 50

8 50

15 50

24 50

e 1.2

79 17

iii

ii 3

4811 900

iv 8.24

c 18.56

8 a m = 0.7 – n

157 30

ii 11

14 a i

2 58 b i For p and for , E ( X ) 3 15 16 p 1, E ( X ) 5 2 8 , P ( X E ( X )) ii For p and for 3 15 1 p 1, P ( X E ( X ) ) 5

13 a i k = 0.25

2 6 a p , 1 3

1 6

21 5

iii 0.92

Answers 1379 mathspace.co

d i

P(X = x)

1 16

3 16

5 16

7 16

P(X = x)

27 64

9 64

1 64

3 4

21 a Yes. 0 ≤ P(x) ≤ 1, X is discrete numerical, and ΣP(x) = 1

16 a

–$2

b $2600

5 12

4 12

3 12

c 4 140 000

P(X = x) b $2

Extend your thinking

c $2 3

22 X avier will lose. The cost of the game to play is more than the expected value to be won.

17 –$0.33 3 2

19 a

43 12

23 0.5 Male

Male

Female Male

Male Female

Female

Male

Male Female Male

Female

iii 0.93

15 5.66

c 0, 1

25 ii 8

18 a

P(M = m)

1 8

3 8

1 8

Right 1 4 3 4 Wrong

Right 1 4 3 4 Wrong Right 1 4 3 4 Wrong

1 4 3 4 1 4 3 4 1 4 3 4 1 4 3 4

Distribution B

0.4 0.3 0.2 0.1 0

x 1

12.03 Sums of independent random variables What do you remember?

3 2

20 a

P (X = x)

Right Wrong Right Wrong Right Wrong Right Wrong

1380 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

1 A linear combination of random variables is a sum of random variables, each multiplied by a constant, e.g., aX + bY. Example: The total cost of 2 items at price X each is 2X. 2 For aX + bY, the mean is E(aX + bY) = aE(X) + bE(Y) and the variance is Var(aX + bY) = a2Var(X) + b2Var(Y). 3 T he distribution of X + Y combines outcomes of two independent variables, leading to a wider range and triangular shape (e.g., dice sums from 2 to 12). 2X scales the outcomes of one variable, maintaining a uniform distribution over scaled values (e.g., even numbers 2, 4, . . ., 12 for a die).

12.04 Applications of discrete random variables

Practice 4 For 2X: Values {2, 4, 6, 8}, each with 1 probability . 4

What do you remember? 1 For a valid discrete probability distribution, the sum of all probabilities must equal 1. Here, the sum is 0.4 + 0.3 + 0.2 = 0.9, which is not equal to 1.

For X + Y: Possible sums are 2 to 8. Sum

Probability

1 16

2 16

3 16

4 16

3 16

2 16

1 16

2 B 3 a One pet (as it has the highest probability of 0.45).

2 X is uniform; X + Y is triangular, peaking at 5. 5

E(2X + 10) = 20 Var(2X + 10) = 12

E(–2X + 3) = –17 Var (–2X + 3) = 16

b The expected value represents the long-run average number of pets per household if the survey were conducted many times. Practice

7 a The mean of the sum is 115 and the variance is 25.

b The mean of the difference is –15 and the variance is 25. 8

0.09

0.42

0.49

P(X = x)

2 9

1 9

2 9

1 9

2 9

1 9

5 a

E(X + Y) = 7 35 Var(X + Y) = 6 E(X1 + X2 + X3 + X4) = 6 kg

x P(X = x)

Var(X1 + X2 + X3 + X4) = 0.8 kg2

1 6

10 3

6 a

27 64

10 P(X + Y ≥ 10) =

Extend your thinking

E(S) = 10.5 35 Var(S) = 4

7 a

y P(Y = y)

b 7 d 7

26 9

1 64

P(X = x)

27 64

9 64

1 64

11 The mean is 320 g and the variance is 57 g2. The mean helps estimate the average packaging material needed, while the variance indicates the variability, aiding in selecting containers that accommodate most daily weights.

27 64

e Yes, because ΣP(x) = 1, X is discrete numerical, and 0 ≤ P(x) ≤ 1.

1 2 3 4 5 6 5 4 36 36 36 36 36 36 36 36

3 36

2 36

1 36

c The distribution is symmetrical. 35 e 6

Answers 1381 mathspace.co

8 a

b $56 500

12 a

4 13

b $1650

13 a

11 100

x P(X = x)

6 10 8 36 36 36

2 3

9 10

P(X = x)

0.2

0.35

0.45

e $5.62

P(X = x)

7 15

1 15

P(T = t)

7 15

1 15

b H

6 55

14 a

6 4 2 36 36 36

9 a

c Option A

d $18.90

11 a 13%

T c 3, 7

–$3

P(X = x)

1 4

1 2

1 4

2 10 a 11

10 b 33

Extend your thinking 15 a

P(X = x)

1 66

2 11

5 11

10 1 33 22

84 121

24 11

d 5.4

P(X = x)

1 6

5 24

49 12

50 63

16 a 0, 1, 2, 3, 4, 6, 8, 9, 12, 16 b

c 4

P(X = x)

9 25

1 25

2 25

3 25

2 25

1 2 25 25

1 25

d 20

1382 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

275 144

12.05 Simulation

What do you remember?

17 a

P(X = x)

24 36

6 36

4 36

2 36

c $0.56

1 a False 2 P(A)

4 a

P(X = x)

3 16

3 2 2 2 2 1 1 16 16 16 16 16 16 16

d False

Number of favourable outcomes Total number of trials

Practice

d –$1.44

c True

3 A frequency table counts the occurrences of each outcome in a dataset, allowing calculation of relative frequencies to estimate probabilities.

18 a 1, 2, 3, 4, 5, 6, 7, 8 b

b False

Outcome

Frequency

Relative frequency

52 0.347 150

48 0.32 150

50 0.333 150

b Mean ≈ 1.987 19 a

P(X = x)

0.4

0.1

0.2

0.1

b 1.5

1 5

5 a 1 Algorithm: Simulate card draw 2 Input: Number of trials (400) 3 Output: Estimated probability of drawing a heart 4 1. Initialise heart_count = 0 5 2. For i from 1 to 400: 6 a. Generate random integer r between 1 and 52 7 b. If r <= 13: 8 heart_count = heart_count + 1 9 3. Calculate probability = heart_count / 400 10 4. Return probability b P (Heart) ≈ 0.245 6 a P(6) ≈ 0.157 1 0.167. Estimated probability: 0.157. The estimated probability is close to the 6 theoretical, suggesting reliability, with slight variation due to the finite 300 trials.

b Theoretical probability:

Answers 1383 mathspace.co

7 a 1 Algorithm: Simulate dice game 2 Input: Number of trials (800) 3 Output: Estimated probability of winning 4 1. Initialise win_count = 0 5 2. For i from 1 to 800: 6 a. Generate random integer die1 between 1 and 6 7 b. Generate random integer die2 between 1 and 6 8 c. If die1 + die2 = 9: 9 win_count = win_count + 1 10 3. Calculate probability = win_count / 800 11 4. Return probability b P(win) ≈ 0.11 8 a P(red) ≈ 0.243 b Estimated probability: 0.243. Theoretical probability: 0.25. The estimated probability is very close, suggesting reliability. The law of large numbers indicates that with more trials, the estimate would likely converge closer to 0.25. 108 0.54 for heads. Theoretical probability 200 is based on the mathematical model: 0.5 for a fair coin. The experimental probability (0.54) differs slightly from the theoretical (0.5) due to the finite number of trials, but the law of large numbers suggests convergence with more flips.

9 Experimental probability is calculated from simulation data:

Extend your thinking 10 a 1 Algorithm: Simulate spinner sum 2 Input: Number of trials (1000) 3 Output: Estimated probability of sum = 6 4 1. Initialise sum_six_count = 0 5 2. For i from 1 to 1000: 6 a. Generate random integer spin1 between 1 and 4 7 b. Generate random integer spin2 between 1 and 4 8 c. If spin1 + spin2 = 6: 9 Sum_six_count = sum_six_count + 1 10 3. Calculate probability = sum_six_count / 1000 11 4. Return probability 122 3 0.122 Theoretical probability: 0.1875 (3 favourable outcomes: (2, 4), 1000 16 (3, 3), (4, 2) out of 16 possible pairs). The estimated probability is lower, likely due to the finite 1000 trials. The law of large numbers suggests more trials would bring the estimate closer to 0.1875.

b Estimated probability:

1384 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

138 0.138 1000 292 0.292 Simulation B: 1000

11 a Simulation A:

b Simulation A theoretical probability: 5 0.139 (5 favourable outcomes for 36 sum = 8). Simulation B theoretical probability: 0.3. Both estimated probabilities (0.138, 0.292) are close to theoretical (0.139, 0.3), with slight differences due to finite trials. The law of large numbers suggests more trials would reduce these differences.

12.06 Sample distributions What do you remember? 1 a A selection of some of the population b The entire group being considered in a statistical analysis 2 a Answers may vary. Possible answers: • All the seniors. • The first 30 students to arrive at school that day.

3 a Answers may vary. Possible answers: • Drivers at a drive-in theatre. • People stuck in traffic due to roadworks. b Answers may vary. Possible answers: • People stuck in traffic due to roadworks. • Employees at a bank. c Answers may vary. Possible answers: • All people who catch the train in the morning on a given day. • All people who catch public transport on a given day. 4 a All products in the factory b

Practice 5 a x = 5.2 b The sample mean of 5.2 books per month is a sample statistic that serves as an estimator for the population mean number of books read by all high school students in the town. It suggests that, on average, students in the town read approximately 5.2 books per month.

• All the students who were in the library at lunch time.

6 1632 students

b Answers may vary. Possible answers:

7 630 students

• All students of a primary school in Sydney. • All students in Year 7 in a particular suburb. • 100 children chosen from the Sydney under 15s baseball tournament. c Answers may vary. Possible answers: • The members of 3 teams chosen from the local hockey tournament. • All students from a local school who compete in organised sport. • All the active members of a local football club. d Answers may vary. Possible answers: • 30 people walking their dog at the local park. • All the students from a local school who own a dog.

30%

c 61% d No

8 a The sample mean for Sample A is 82.5, and for Sample B is 81. b The two sample means are different due to sampling variability. Each random sample, by chance, contains a different group of students from the population, leading to slightly different sample statistics. c The sample mean of 81.7 from the sample of 100 students is the most reliable estimator. Larger samples are more representative of the population and reduce the effect of sampling variability, meaning their sample statistics are generally closer to the true population parameters. 9 a μ = 4, σ = 2 c x = 4.05, s = 1.91

b x = 3.8, s = 1.76 d x = 3.83, s = 2.05

e The larger the sample, the more likely the graph, the mean and the standard deviation are close to the graph, mean and standard deviation of the population distribution.

Answers 1385 mathspace.co

Extend your thinking 10 a k = 12.0 grams b The sample mean (x = 12.0 g) is a very good estimator of the population mean (μ = 12.05 g). The sample statistic is very close to the population parameter, suggesting the sample is quite representative, despite its small size. 11 a This sampling method is biased because it is not a random sample of the entire Year 12 population. It only includes students who choose to be in the library at lunchtime, a group that is more likely to be diligent or dedicated to studying than the general student population. Students who study at home, do not study, or participate in other lunchtime activities are excluded. b The sample mean is likely to be an overestimate of the true population mean. Students who use the library for studying at lunch are likely to study more, on average, than the general student population. Because the sample over-represents these diligent students, the calculated sample mean will likely be higher than the true average for all Year 12 students. 12 a xA = 14 xB ≈ 21.33 184 = 18.4 10 c The combined sample mean would only be a good estimator if the original populations, A and B, were of roughly equal size (or more accurately, if the sample sizes were proportional to the population sizes). Because the sample from B (n = 6) was larger than the sample from A (n = 4), the combined mean is weighted more towards the values from Population B. If Population A was much larger than Population B in reality, this combined sample mean would be a poor, biased estimator of the merged population’s true mean. b xcombined

1386 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

12.07 Simulate sampling distributions What do you remember? 1 A sampling distribution is the distribution of a sample statistic, such as the sample mean, calculated from repeated samples of the same size from a population. A population parameter is a numerical measure, like the mean, that describes the entire population. The sampling distribution shows the variability of the sample statistic, while the population parameter is a fixed value. 2 μ=5 3 The spread of a sampling distribution, measured by the standard error, is approximately , where σ is the population n standard deviation and n is the sample size. As n increases, the denominator n grows, reducing the standard error and thus the spread. 4 RANDBETWEEN (a, b) to generate random values and AVERAGE (range) to calculate the sample mean. Practice 5 a x1 = 4

b x2 = 3

c x3 = 6

6 Sample Value 1 Value 2 Value 3 Value 4 Mean 1

17.5

7 a Sample 1 mean: 5.0 Sample 2 mean: 5.0 Sample 3 mean: 4.2 Sample 4 mean: 5.8 Sample 5 mean: 4.2 Sample 6 mean: 3.8

Extend your thinking 3.5 4.0 4.5 5.0 5.5 6.0

The dot plot shows the sampling distribution of the six sample means. The population mean is μ = 5. The sample means are clustered around this value, demonstrating sampling variability.

11 The sampling distribution is approximately normal, centred at 74.8, close to μ = 75. The spread, with a standard deviation of 2.6, matches 10 the expected standard error ≈ 2.58 n 15 30

2.0

2.5

3.0

3.5

4.0

The dot plot shows the distribution of the 8 sample means, clustered between 2.5 and 3.5.

Frequency

8 a

b Population mean: 3 Mean of the sample means: 3 The mean of the sample means (3) is exactly equal to the true population mean (μ = 3). This demonstrates that the centre of the simulated sampling distribution is an excellent estimator for the true population mean. 9 a The mean of the sample means (14.9) is very close to the true population mean (15). This demonstrates that the sample mean is a good estimator for the population mean, as the centre of the sampling distribution aligns with the population parameter. b The theoretical standard error is 4 ≈ 1.414 n 8 The standard deviation of the sample means (1.4) is very close to the theoretical standard error (1.414). This shows that the spread of the sampling distribution is significantly smaller than the population standard deviation (σ = 4) and matches the expected value. 10 With 200 samples of size 20 , the sampling distribution of sample means is expected to be approximately normal (bell-shaped) and centred around the population mean, μ = 50. This is due to the Central Limit Theorem.

80 75 Sample means

The histogram shows a normal-like distribution, confirming the Central Limit Theorem. 12 The spread of a sampling distribution is measured by the standard error, . For n samples of size 10, the standard error is

. Since 30 10, the 30 standard error for size 30 is smaller, so the sampling distribution for larger samples has less spread. For size 30, it is

13 The sampling distribution of sample means will be centred around the population mean (μ = 3). The sampling distribution of sample proportions of values greater than 3 will be centred around 2 the population proportion p = = 0.4 . 5 To compare the spread, calculate the theoretical variances. The variance of the 2 2 sample mean is Var ( x ) d ≈ 0.667. The n 3 variance of the sample proportion is p ( 1 p) 0.4 (0.6) = 0.08. Var ( p̂) 3 n Since 0.667 > 0.08, the sampling distribution of the sample mean is expected to have a significantly larger spread (more variability) than that of the sample proportion.

Answers 1387 mathspace.co

12.08 Sample size and distributions What do you remember? 1 A sampling distribution is the distribution of a statistic, such as the sample mean, calculated from multiple samples of the same size from a population. It is used to estimate population parameters by providing information about the variability and expected value of the statistic, with the sample mean estimating the population mean μ. 2 The standard error is the standard deviation of the sampling distribution of the sample mean, given by SE = , where σ is the population n standard deviation and n is the sample size. As n increases, the standard error decreases, indicating less variability. 3 SE = 2 4 As sample size increases, the sampling distribution of the sample mean becomes more symmetric and approximately normal, regardless of the population’s original distribution shape, with reduced spread. Practice 5 For n = 16, SE = 2. For n = 64, SE = 1. The standard error decreases from 2 to 1 as sample size increases.

10 For n = 15, the sampling distribution may retain some left-skewness. For n = 60, it is expected to be approximately normal due to the larger sample size. 11 For n = 4: SE =

9 Calculate the sample mean for each of the 500 samples. Create a dot plot or histogram with the x-axis labelled ‘Sample Mean’ and y-axis labelled ‘Frequency’ to display the variation. The plot will show the spread and approximate normality of the sampling distribution.

1388 Mathspace VCE Specialist Mathematics Units 1 & 2 mathspace.co

4 16

= 1.

4 = 0.5. The standard error 64 decreases as sample size increases, indicating less variability. 12 The expected mean is μ = 25. Visualise using a histogram with x-axis ‘Sample Mean’ and y-axis ‘Frequency’, showing an approximately normal distribution centred at 25. 13 Standard error: SE = 2 The sampling distribution is expected to be approximately normal due to the large sample size. Extend your thinking 14 Expected centre for all: μ = 70. Spread (standard 14 error): n = 10: SE = ≈ 4.43; n = 50: 10 SE ≈ 1.98; n = 200: SE ≈ 0.99. Shape: For n = 10, right-skewed; n = 50, more symmetric; n = 200, approximately normal. A dot plot would show sample means clustering tighter around 70 with larger n, with fewer outliers for n = 200. Distribution of 2000 sample means from a skewed population n = 10

Density

8 Standard error: SE = 5 The expected centre is the population mean, μ = 120.

= 2. For n = 16: SE =

For n = 64: SE =

6 For n = 5, the sampling distribution may resemble the uniform population distribution, with some variability. For n = 40, the sampling distribution is expected to be approximately normal due to the larger sample size. 7 Expected mean: μ = 30; Standard error: SE = 1.2

67.5

72.5

77.5

67.5

72.5

77.5

67.5

70 72.5 75 Sample mean

77.5

n = 50

65 n = 200

15 Simulate 1000 samples for each sample size. Calculate sample means. Standard errors: 9 n = 25: SE = = 1.8; n = 100: 25 9 SE = = 0.9. Shapes: n = 25 may reflect 100 population shape; n = 100 is approximately normal. Verify by calculating the mean (should be near 45) and standard deviation (should approximate the SE) of the sample means, and visualise with histograms. 16 The standard error, SE =

, decreases as n n increases because the denominator grows, reducing variability. For example, if student heights have σ = 10 cm, then for n = 25, 10 SE = = 2 cm, but for n = 100, SE = 1 cm. In 25 surveying, larger samples yield sample means closer to the true population mean, improving accuracy and confidence in estimates.

Answers 1389 mathspace.co