Math SL

Mathematics SL S. Budd Lamar High School pdf January 12, 2011

ii

Mr. Budd, compiled January 12, 2011

Contents I

Calculus

1

1 Area and Slope 1.1 Using Graphs to Multiply: Definite Integrals . . . . . . . .

1.1.1 What is Calculus? . . . . . . . . . . . . . . . . . . .

1.1.2 What is a Definite Integral? . . . . . . . . . . . . . .

1.1.3 Approximating Area: Counting Squares . . . . . . .

1.1.4 What is “Signed� Area . . . . . . . . . . . . . . . .

1.1.5 Using Known Shapes to Evaluate Definite Integrals .

1.1.6 Using Symmetry . . . . . . . . . . . . . . . . . . . .

1.2 Repeatable Approximation of Definite Integrals . . . . . . .

1.2.1 Using Tables of Data . . . . . . . . . . . . . . . . . .

1.2.2 Rectangular Approximation Method (RAM) . . . .

1.2.3 Trapezoidal Approximation: Quasi-RAM . . . . . .

1.2.4 Streamlining Calculations for Equal Widths . . . . .

1.2.5 Finding a Range of Values . . . . . . . . . . . . . . .

1.2.6 Midpoint RAM . . . . . . . . . . . . . . . . . . . . .

1.2.7 Unequal Subdivisions . . . . . . . . . . . . . . . . .

1.3 Approximating Definite Integral from Formulas . . . . . . .

1.3.1 Definite Integrals from Known Shapes . . . . . . . .

1.3.2 Approximating Definite Integrals . . . . . . . . . . .

1.3.3 Using Symmetry . . . . . . . . . . . . . . . . . . . .

1.4 Slope and Rate of Change . . . . . . . . . . . . . . . . . . .

1.4.1 Instantaneous Rate of Change . . . . . . . . . . . . .

1.4.2 Definition and Notation . . . . . . . . . . . . . . . .

1.4.3 Approximating Derivatives from Tabular Data . . .

1.5 IROC as a limit . . . . . . . . . . . . . . . . . . . . . . . . .

1.5.1 Approximating Rate of Change from a Formula . . .

1.5.2 Kinematics: Displacement, Velocity, Acceleration . .

1.6 Slope and Area: Pulling It Together . . . . . . . . . . . . .

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3 5 5 6 7 8 8 10 15 15 19 20 21 22 23 25 33 33 34 36 39 39 40 41 49 49 50 55

2 Limits 57 2.1 Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . . . 59 2.1.1 Graphic Introduction to Limits . . . . . . . . . . . . . . . 59 2.1.2 Step Discontinuities & One-Sided Limits . . . . . . . . . . 61 iii

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CONTENTS . . . . . . . . . . . . . . . . . .

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91 95 95 95 97 98 99 103 103 106 113 121 121 127 127 128 129 130 130 135

4 Transcendentals 4.1 Antidifferentiation of Trigonometric Functions . . . .

4.1.1 Antidifferentiating Trigs . . . . . . . . . . . .

4.1.2 The Anti-Chain Rule . . . . . . . . . . . . . .

4.2 u-Simplification . . . . . . . . . . . . . . . . . . . . .

4.2.1 u-Simplification with Trigonometrics Outside

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137 139 139 141 143 143

2.2

2.3

2.4

2.1.3 Limits from a Table . . . . . . . . . .

2.1.4 Limits from an Expression . . . . . . .

2.1.5 Substitution and Properties of Limits Limits from a Formula . . . . . . . . . . . . .

2.2.1 Limits at Cancelable Discontinuities .

2.2.2 De-rationalizing with Conjugates . . .

2.2.3 De-denominatorizing with LCDs . . .

2.2.4 Derivative at a Point . . . . . . . . . .

Limit Definition of Derivative as a Function .

2.3.1 Derivative as a Function . . . . . . . .

2.3.2 Tangent Lines . . . . . . . . . . . . . .

2.3.3 Normal Lines . . . . . . . . . . . . . .

Basic Calculus of Polynomials . . . . . . . . .

2.4.1 Notation . . . . . . . . . . . . . . . . .

2.4.2 Basic Properties of Derivatives . . . .

2.4.3 Power Rule . . . . . . . . . . . . . . .

2.4.4 Higher Order Derivatives . . . . . . .

2.4.5 Kinematics . . . . . . . . . . . . . . .

3 Basic Differentiation 3.1 Antidifferentiation of Polynomials . . . . .

3.1.1 Notation of Antiderivatives . . . .

3.1.2 Anti-Power Rule . . . . . . . . . .

3.1.3 Kinematics Revisited . . . . . . . .

3.1.4 General vs. Particular Solutions .

3.1.5 Introduction to Slope Fields . . . .

3.2 Product and Quotient Rules . . . . . . . .

3.2.1 Product Rule . . . . . . . . . . . .

3.2.2 Quotient Rule . . . . . . . . . . . .

3.3 Chain Rule . . . . . . . . . . . . . . . . .

3.4 Differentiation of Trigonometric Functions 3.4.1 Trigonometric Derivatives . . . . .

3.5 Tangent Lines . . . . . . . . . . . . . . . .

3.5.1 Tangent Lines . . . . . . . . . . . .

3.5.2 Horizontal Tangents . . . . . . . .

3.5.3 Vertical Tangents . . . . . . . . . .

3.5.4 Normal Lines . . . . . . . . . . . .

3.5.5 Tangent Line Approximations . . .

3.6 Basic Differentiation Review Problems . .

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Mr. Budd, compiled January 12, 2011

CONTENTS

4.3

4.4

v

4.2.2 Anti-Chain Rule with Trigonometrics Inside .

4.2.3 Linear u-Simplification . . . . . . . . . . . . .

Basic Calculus of Exponential Functions . . . . . . .

4.3.1 Differentiating the Exponential Function . . .

4.3.2 Antidifferentiating the Exponential Function Natural Logarithm . . . . . . . . . . . . . . . . . . .

4.4.1 Differentiating the Logarithmic Function . . .

4.4.2 Antidifferentiating the Reciprocal Function .

4.4.3 Antidifferentiating Rational Functions . . . .

5 Extrema and Optimization 5.1 Relating Graphs of f and f 0 . . . . .

5.1.1 Direction of f . . . . . . . . .

5.1.2 Relative Extrema . . . . . . .

5.1.3 First Derivative Test . . . . .

5.2 Relating the Graphs of f , f 0 , and f 00 5.2.1 Concavity . . . . . . . . . . .

5.2.2 Second Derivative Test . . . .

5.2.3 Points of Inflection . . . . . .

5.2.4 Connecting f , f 0 , and f 00 . .

5.3 Interval Testing . . . . . . . . . . . .

5.3.1 Direction . . . . . . . . . . .

5.3.2 Concavity . . . . . . . . . . .

5.3.3 More Second Derivative Test 5.4 Optimization . . . . . . . . . . . . .

5.4.1 Absolute Extrema . . . . . .

5.4.2 Optimization . . . . . . . . .

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144 145 147 147 149 153 153 155 157

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161 163 163 163 164 171 171 172 173 174 179 179 180 181 185 185 186

6 Definite Integrals 6.1 Evaluating Definite Integrals . . . . . . .

6.1.1 Definite Integral . . . . . . . . . .

6.1.2 Riemann Sums . . . . . . . . . . .

6.1.3 Fundamental Theorem of Calculus 6.1.4 Fundamental Theorem of Calculus 6.2 Area . . . . . . . . . . . . . . . . . . . . .

6.2.1 Definite Integral . . . . . . . . . .

6.2.2 Riemann Sums . . . . . . . . . . .

6.2.3 Area: Slicing dx . . . . . . . . . .

6.2.4 High and Low y Switch . . . . . .

6.2.5 Area: Slicing dy . . . . . . . . . .

6.2.6 Total Distance . . . . . . . . . . .

6.3 Volume . . . . . . . . . . . . . . . . . . .

6.3.1 Slicing into Discs . . . . . . . . . .

6.3.2 Slicing into Washers . . . . . . . .

6.4 Evaluating Definite Integrals . . . . . . .

6.4.1 Fundamental Theorem of Calculus

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189 191 191 191 195 197 201 201 201 202 203 204 204 211 211 213 219 222

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Mr. Budd, compiled January 12, 2011

vi

CONTENTS 6.4.2

II

Fundamental Theorem of Calculus . . . . . . . . . . . . . 224

I.B. Topics

7 Statistics 7.1 Discrete Data . . . . . . . . . . .

7.1.1 Central Tendency . . . . .

7.1.2 Dispersion . . . . . . . . .

7.1.3 Frequency . . . . . . . . .

7.1.4 Cumulative Frequency . .

7.1.5 Population vs. Sample . .

7.2 Continuous Data . . . . . . . . .

7.2.1 Discrete vs. Continuous .

7.2.2 Groups . . . . . . . . . .

7.2.3 Histograms . . . . . . . .

7.3 Cumulative Frequency . . . . . .

7.3.1 Continuous Data . . . . .

7.3.2 Percentiles and Quartiles

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229 231 232 233 235 237 238 245 245 246 249 255 255 258

8 Vectors and Matrices 8.1 Introduction to Matrices . . . . . . . . . .

8.1.1 Matrix Addition . . . . . . . . . .

8.1.2 Matrix Multiplication . . . . . . .

8.1.3 Matrix Division . . . . . . . . . . .

8.1.4 Determinants . . . . . . . . . . . .

8.2 Vector Basics . . . . . . . . . . . . . . . .

8.2.1 Vectors . . . . . . . . . . . . . . .

8.2.2 Representation . . . . . . . . . . .

8.2.3 Simple Arithmetic . . . . . . . . .

8.2.4 Magnitude . . . . . . . . . . . . .

8.2.5 Making Unit Vectors . . . . . . . .

8.2.6 Scalar Product . . . . . . . . . . .

8.2.7 Angle . . . . . . . . . . . . . . . .

8.3 Vector Form of a Line . . . . . . . . . . .

8.3.1 Lines in Two Dimensions . . . . .

8.3.2 Applications . . . . . . . . . . . .

8.4 Lines in Space . . . . . . . . . . . . . . . .

8.4.1 Lines in Three Dimensions . . . .

8.4.2 Angle Between Lines . . . . . . . .

8.4.3 Intersection of Lines . . . . . . . .

8.4.4 Applications in Three Dimensions 8.5 Matrices Reloaded . . . . . . . . . . . . .

8.5.1 Solving Systems of Equations . . .

8.5.2 Singularity . . . . . . . . . . . . .

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273 275 275 276 276 277 281 281 282 282 284 284 284 285 289 289 291 297 297 297 298 298 303 303 305

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Mr. Budd, compiled January 12, 2011

CONTENTS

vii

9 Probability 9.1 Introduction to Series . . . . . . . . . . . . . . . . .

9.1.1 Arithmetic Series . . . . . . . . . . . . . . . .

9.1.2 Geometric Series . . . . . . . . . . . . . . . .

9.2 Introduction to Probability . . . . . . . . . . . . . .

9.2.1 The Basics . . . . . . . . . . . . . . . . . . .

9.2.2 Probability and Set Theory: Venn Diagrams 9.3 Conditional Probability . . . . . . . . . . . . . . . .

9.3.1 Conditional Probability . . . . . . . . . . . .

9.3.2 Sampling Without Replacement . . . . . . .

9.3.3 Counting Methods . . . . . . . . . . . . . . .

9.4 Independent Events . . . . . . . . . . . . . . . . . .

9.4.1 Statistical Independence . . . . . . . . . . . .

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309 311 311 312 317 317 321 327 327 329 331 337 337

10 Probability Distributions 10.1 Discrete Random Variables . . . .

10.1.1 Discrete Random Variables 10.1.2 Variance . . . . . . . . . . .

10.1.3 Other Measures . . . . . . .

10.2 Binomial Expansion . . . . . . . .

10.3 Binomial Distribution . . . . . . .

10.4 Normal Distribution . . . . . . . .

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343 345 345 346 348 353 359 367

11 Review 11.1 Limits Involving Infinity . . . .

11.1.1 Infinite Limits . . . . .

11.1.2 Limits at Infinity . . . .

11.1.3 Horizontal Asymptotes .

11.1.4 Problems . . . . . . . .

11.2 Introduction to Series . . . . .

11.3 Probability and Statistics . . .

11.4 Graphing and Calculus . . . . .

11.5 Retired Problems . . . . . . . .

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375 377 377 378 380 381 383 385 391 397

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Mr. Budd, compiled January 12, 2011

viii

CONTENTS

Mr. Budd, compiled January 12, 2011

Part I

Calculus

1

Unit 1

SL Introduction to Calculus: Area and Slope 1. The Definite Integral as Area 2. Approximating Definite Integral by Riemann Slicing 3. Rate of Change 4. Approximating Rate of Change from Graph, Table, or Equation 5. Slope and Area: Pulling It Together

Advanced Placement Concept of the derivative.

• Derivative presented geometrically, numerically, and analytically.

• Derivative interpreted as an instantaneous rate of change.

Interpretations and properties of definite integrals.

• Computation of Riemann sums using left, right, and midpoint evaluation points.

• Basic properties of definite integrals (Examples include additivity and linearity.)

3

4

SL Unit 1 (Area and Slope)

Numerical approximations to definite integrals. Use of Riemann and trapezoidal sums to approximate definite integrals by functions represented algebraically, geometrically, and by tables of values.

International Baccalaureate (MM 8.6) The estimation of the numerical value of a definite integral using the trapezium rule. Included: an appreciation of the effect of doubling the number of sub-intervals.

Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 1: Using Graphs to Multiply: Definite Integrals

1.1

5

Using Graphs to Multiply: Definite Integrals

International Baccalaureate 7.5 Definite integrals. Area between a curve and the x–axis in a given interval.

Rb Included: a y dx.

7.6 Kinematic problems involving displacement, s, velocity, v, and acceleration, a.

Included: Area under velocity-time graph represents distance.

Textbook We won’t be following the book too closely the first couple weeks, so that there is limited correspondence to a section in the book. The closest section in content would be §4.3 Area or §4.4 The Definite Integral. [15] Resources §5.1 Areas and Integrals in Ostebee and Zorn [16]. §1-3 One Type of Integral of a Function in Foerster [10]. Explorations 1-3a:“Introduction to Definite Integrals” and 1-4a:“Definite Integrals by Trapezoidal Rule” in [9].

1.1.1

What is Calculus?

Ostebee and Zorn describe the focus of calculus as follows: “The tangent-line problem and the area problem are the two main geometric problems of calculus.”

[16] The tangent-line problem is an issue of slope, so that our main concerns in calculus are slope and area. Our interest in slope and area is not purely geometric, however. Slope is our codeword for rate of change, which can be rate of people entering AstroWorld, or the rate at which oil leaves a gash in an oil tanker. Likewise, area can be a whole range of things from people who have entered Super Happy Fun Land in a six-hour time period to the amount of oil that has bled out of a shipwrecked tanker. Area, as we shall see, can even be used to represent the volume of an object.

Foerster, whose materials we will see much of this year, describes calculus as consisting of four things: limits, derivatives, integrals, and integrals [10]. These things probably have no meaning for you, and you are probably wondering why integrals is listed twice. Understanding these things is what we will seek to do over the next nine months. As a brief introduction, I will tell you that derivatives are related to slope, and one of the integrals (definite integrals) is related to area. Both derivatives and definite integrals are limits, and the other type of integral (indefinite) is also related to both derivatives and indefinite integrals.

Mr. Budd, compiled January 12, 2011

6

SL Unit 1 (Area and Slope)

1.1.2

What is a Definite Integral?

Example 1.1.1 Begin with Exploration 1-3: Introduction to Definite Integrals.

Figure 1.1: Velocity, v(t), as a function of the number of seconds, t, since you started slowing. [9]

Distance is Velocity times Time. Looking at the graph in Figure 1.1, we can see that after 30 seconds, the velocity is basically constant at 60 feet per second.

If we were to find the distance traveled between 30 and 50 seconds, we would multiply 20 seconds by 60 feet per second, yielding 1200 feet.

Notice that the same product (20 seconds Ă— 60 feet per second) is represented by the area under the curve of v(t) from 30 seconds to 50 seconds. The shape of this region is a rectangle, and the formula for finding the area of a rectangle is A = l Ă— w. If you look at the region under the curve of v(t) from 30 seconds to 50 seconds, you should notice that it is a rectangle with width of 20 seconds and height of 60 feet per second. The area of this rectangle is 1200 feet, or 20 seconds Ă— 60 feet per second.

The distance traveled from a starting time to a stopping time is the area under the velocity curve between the two times. This area is called the definite integral. This is true for the simple case between 30 and 50 seconds, but it is also true for the less simple case between 0 and 20 seconds. To find the distance traveled between 0 and 20 seconds, we would need to multiply the time difference (20 seconds) by the velocity, except this is not so straightforward because the velocity is changing with time. So instead of finding the product by multiplying two numbers, I need a different approach.

Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 1: Using Graphs to Multiply: Definite Integrals

7

When we multiplied two constants, we were doing the same arithmetic as finding the area of a simple rectangle of constant height. In the less straightforward case of multiplying a varying v(t) by t, the distance is still the area under the curve of v(t) from the starting time to the ending time. But since we don’t have a simple formula (because we don’t have a simple shape) we need a different approach to finding the area.

Definition 1.1 (Definite Integral). The definite integral of the function f from Rb x = a to x = b [written a f (x) dx] gives a way to find the product of (b − a) and f (x), even if f (x) is not a constant. [10] This definition of the definite integral tells us why we need the definite integral.

It doesn’t tell us how to get it. Here’s a geometric explanation from a different author: Definition 1.2 (The Integral as Signed Area). Let f be a function defined for a ≤ x ≤ b.

Z b f (x) dx a

denotes the signed area bounded by x = a, x = b, y = f (x), and the x-axis. [16]

1.1.3

Approximating Area: Counting Squares

Rb If f (x) is always positive, and a is less than b, then a f (x) dx is the area under f between a and b. We will talk more about the qualifier “signed” in a bit.

One way to approximate an area is the “Counting Squares” approach.[10] This approach is fairly basic; you superimpose a grid on your graph and lightly shade the area you are approximating (or imagine the shading in your head).

You count the number of whole squares that are shaded. Sometimes I like to count all the whole squares in vertical strips, and this approach will be helpful to us in the future. Then you count the partial squares, rounding the shaded portion of each square to the nearest tenth, 0.1. After you have added the total number of whole and partial squares, you multiply the number of squares by the area of each square, being conscious of your units. The number of squares multiplied by the area of each square is the area of the shape.

Looking at the example Figure 1.1, we can see that the area under the curve from 30 to 50 seconds gives 24 squares. At 50 feet per square (10 feet per second times 5 seconds), that gives a displacement of 1200 feet. How many squares are between 0 and 20 seconds? Therefore what is the change in your position in the first twenty seconds?

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SL Unit 1 (Area and Slope)

1.1.4

What is “Signed” Area

To find the definite integral of f (x) from x = a to x = b, you are basically looking for the signed area under the curve of f between a and b. What do we mean by signed area? If the area is above the x-axis, it is counted as positive; if the area is below the x-axis it is counted as negative (i.e., a negative amount above the axis). Think about why this is important. What’s happening when the velocity is negative? If the velocity were negative, how should you count the area/ distance?

[Another way to make the signed area negative is if b is less than a, so that taking you from a to b means that you go right to left on the graph, i.e., the change in x, ∆x ≈ dx, is negative.]

Figure 1.2: [10]

Example 1.1.2 The graph in Figure 1.2 shows v(t) centimeters per second as a function of t seconds after an object starts moving. At what time does the object change direction? How far is the object from its starting point when t = 9 sec? What is the total distance traveled by the object? [adapted from [10]] [Ans: 5 sec, 7.1 cm] Note the distinction between displacement and total distance.

1.1.5

Using Known Shapes to Evaluate Definite Integrals

Example 1.1.3 Several areas are shown in Figure 1.3, labeled as integrals. Use familiar area formulas to evaluate each integral. [16]

Ans: 6; k (b − a); 94 π + 32 Mr. Budd, compiled January 12, 2011

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Figure 1.3: [16]

Example 1.1.4 (adapted from AB ’03) Let f be a function defined

Figure 1.4: From 2003 AP Calculus AB exam

on the closed interval −3 ≤ x ≤ 4. The graph of f 0 , a function that is related to f , but different from f , known as the derivative of f , consists of one line segment and a semicircle, as shown in Figure 1.4 Find (a)

R0

(b)

R4

−3

0

f 0 (x) dx

f 0 (x) dx

Ans: - 32 ; −8 + 2π Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope)

1.1.6

Using Symmetry

Example 1.1.5 (adapted from AB ’01) A car is traveling on a straight road with velocity 40 ft/sec at time t = 0. For 0 ≤ t ≤ 18 seconds, the car’s acceleration a(t), in ft/sec2 , is the piecewise linear function defined by the graph in Figure 1.5.

Figure 1.5: A car’s acceleration

(a) How fast is the car going at time t = 0? (b) How much does the car’s velocity increase during the first second? During the first two seconds? (c) What is the car’s velocity at time t = 2? At time t = 6? (d) What change takes place to the car’s velocity at time t = 6? (e) At what time does the velocity of the car return to 40 ft/sec?

[Ans: 40 (ft/sec); 15, 30 (ft/sec); 70, 100 (ft/sec); v decreases; 12 s]

Problems big giant blue-green Calculus book p. 380: Writing Exercises # 1,2; # 41-44 1.A-1 The online supplement for Mathematics SL during the academic year has been migrated to Lamar’s new Moodle site, so take the following steps to enroll. Go to http://moodle.houstonisd.org/lamarhs/ and follow the instructions for creating a new account. You will then need to search for Mathematics SL to enroll. When asked for it, the enrollment key for this class will be LamarMathSL11p# where # is the period of your class. This is a vital online supplement to what happens in class.

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Figure 1.6: [10]

1.A-2 In Figure 1.6, a car is slowing down from a speed of v = 60 ft/sec. Estimate the distance it goes from time t = 5 sec to t = 25 sec by finding the definite integral. [10] [Ans: about 680 feet] 1.A-3 In Figure 1.7, a car speeds up slowly from v = 55 mi/hr during a long trip.

Figure 1.7: [10]

Estimate the distance it goes from time t = 0 hr to t = 4 hr by finding the definite integral. [10] [Ans: about 266 miles] 1.A-4 In the previous two problems, you found a distance using a definite integral. Suppose you use the formula d = vt, rearranged to v = dt . What velocities do you get in the previous two problems when you divide the distance by the change in time? What do you think this represents?

1.A-5 The rate at which people enter an amusement park on a given day is modeled by the function E of time t. E(t) is measured in people per hour and time t is measured in hours after midnight. When the parkR opens at 9 17 a.m., there are no people in the park. Explain the meaning of 9 E(t) dt.

Is this equal to the number of people in the park? Why or why not?

[Ans: The number of people who entered the park by 5 p.m.; no] 1.A-6 A blood vessel is 360 millimeters (mm) long with circular cross sections of varying diameter. If x represents the distance from one end of the blood Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope) vessel, and B(x) is a function that represents the diameter at that point,

2 R 360 B(x) then using correct units explain the meaning of 0 π dx, and 2 2

R 275 B(x) dx. π 125 2

1.A-7 Let g be the function shown graphically in Figure 1.8. When asked to Figure 1.8: Graph of g [16]

R2 estimate 1 g(x) dx, a group of calculus students submitted the following answers: −4, 4, 45, and 450. Only one of these responses is reasonable; the others are “obviously” incorrect. Which is the reasonable one? [16] [Ans: 45] 1.A-8 The graph of a function f is shown in Figure 1.9. [Adapted from [16]] Figure 1.9: Graph of f [16]

R6 (a) Which of the following is the best estimate of 1 f (x) dx: −24, 9, 20, 38? Justify your answer.

R8 (b) 6 f (x) dx ≈ 4. Does this approximation overestimate or underestimate the exact value of the integral? Justify your answer.

R7 (c) Explain a quick way to tell that 12 ≤ 3 f (x) dx.

[Ans: 20; underestimate] Mr. Budd, compiled January 12, 2011

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Figure 1.10: Graph of f [16]

1.A-9 The graph of a function f (shown in Figure 1.10) consists of two straight lines and two one-quarter circles. Evaluate each of the following integrals.

R2 (a) 0 f (x)dx R5 (b) 2 f (x)dx R5 (c) 0 f (x)dx R9 (d) 5 f (x)dx R4 (e) 4 f (x)dx R 15 (f) 0 f (x)dx R 15 (g) 0 |f (x)| dx

Ans: 4,

9π 4 ,

4+

9π 4 ,

−4π, 0, −8 −

7π 4 ,

16 +

25π 4

1.A-10 Suppose Mr. Budd is driving to the Utah Shakespearean Festival in Cedar City, UT. Once he gets on the road, he sets his cruise control for 55 mph.

Let t be the number of hours since he started driving on cruise control. (a) How far has he gone during the first half hour on cruise control? the first hour? the first two hours? (b) Write an equation for the velocity, i.e., v(t) =(something). (c) Graph the velocity versus time. (d) Find the area under the curve of v(t) from t = 0 to t = 0.5. Also, find the area from t = 0 to t = 1 and also to t = 2. (e) What shape are these areas in? If I look at the area from t = 0 to t = tstop , what is the width of the figure (as an expression with tstop in it)? the height? the area (as an expression of tstop )? Call your expression for area A(tstop ). (f) Plot a graph of distance traveled versus time. Use the points (0.5, distance for 0.5), (1, distance for 1), and (2, distance for 2). Look for a pattern, and use your result for A(tstop ) to connect the dots.

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SL Unit 1 (Area and Slope) (g) On your graph of distance versus time, what is the slope at t = 0.5? at t = 1? at t = 2? Indicate units.

1.A-11 (from Explorations 1-3a [9]) As you drive on the highway you accelerate to 100 feet per second to pass a truck. After you have passed, you slow down to a more moderate speed. Table 1.1 shows your velocity, v(t), as a function of the number of seconds, t, since you started slowing.

Table 1.1: Your velocity after passing a truck t (s) 0 5 10

v(t) ft/s 100 77.3755 67.5477

(a) How fast are you going at t = 0? How fast are you going at t = 5?

Why is it not so straightforward to ask how fast you were going for 0 ≤ t ≤ 5? (b) If your speed is constantly decreasing, give an upper estimate of how far you traveled in the first 5 seconds. Give a lower estimate of your displacement in the first 5 seconds. (c) If you had to give one number for your distance in feet for the first 5 seconds, what might it be? Give a reason for how you obtained your answer. (d) How far did you travel for 5 ≤ t ≤ 10? For 0 ≤ t ≤ 10?

[Ans: 100, 77.3755, ; 500, 386.878 (ft); 443.439 ft; 362.308, 805.747 ft] 1.A-12 Read the handout “How to Succeed in Calculus.” (a) Give examples of three things on the list that you already do. (b) Name one thing on the list that you will try to improve this year.

Describe specifically what actions you will take this week. (c) Submit your answer in the appropriate place on the moodle site.

1.A-13 Start Exploration 1-4a: “Definite Integrals by Trapezoidal Rule”; do problems 1 through 3.

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15

Repeatable Approximation of Definite Integrals

International Baccalaureate 7.5 Definite integrals. Area between a curve and the x–axis in a given interval.

Rb Included: a y dx.

7.6 Kinematic problems involving displacement, s, velocity, v, and acceleration, a. Included: Area under velocity-time graph represents distance. (MM 8.6) The estimation of the numerical value of a definite integral using the trapezium rule.

Included: an appreciation of the effect of doubling the number of sub-intervals.

Textbook §4.7 Numerical Integration [15] Resources §5.1 Areas and Integrals in Ostebee and Zorn [16]. §1-4 Definite Integrals by Trapezoids, from Equations and Data in Foerster [10]. Exploration 1-4a:“Definite Integrals by Trapezoidal Rule” in [9].

1.2.1

Using Tables of Data

Example 1.2.1 (from Explorations 1-3 [9]) As you drive on the highway you accelerate to 100 feet per second to pass a truck. After you have passed, you slow down to a more moderate speed. Table 1.2 shows your velocity, v(t), as a function of the number of seconds, t, since you started slowing.

Table 1.2: Your velocity after passing a truck t (s) 0 5 10 15 20 25 30

v(t) (ft/s) 100 77.3755 67.5477 63.2786 61.4242 60.6187 60.2687

(a) What’s the fastest you went in the first five seconds? The slowest? Give an upper and lower range for the displacement in the first 5 seconds.

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SL Unit 1 (Area and Slope) (b) How might one obtain a single best estimate for the change in position for the first 5 seconds? (c) On Figure 1.11, show that each estimate, upper and lower, is represented graphically by a rectangle with a width of 5 seconds. Graphically visualize why the upper-estimate rectangle includes too much area, and the lower-estimate rectangle does not include enough area. (d) If, instead of rectangles that are either too big or too small, suppose we represent the area with one trapezoid, with a width of 5 seconds. The formula for the area of a trapezoid is

h1 + h2 AT = b 2 i.e., the base times the average of the two heights. For a trapezoid that best represents the area of the graph between t = 0 and t = 5, what are the two heights, and what is the area? (e) Estimate your change in position for each of the subintervals [5, 10], [10, 15], [15, 20], and for the overall interval [0, 20].

Figure 1.11: Velocity, v(t), as a function of the number of seconds, t, since you started slowing. [9]

Example 1.2.2 (adapted from Finney, et al. [8]) Try this in your mighty, mighty groups of four. A power plant generates electricity by burning oil. Pollutants produced by the burning process are removed by scrubbers in the smokestacks. Over time the scrubbers become less efficient and eventually must be replaced when the amount of pollutants released exceeds government standards. Measurements Mr. Budd, compiled January 12, 2011

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taken at the end of each month determine the rate at which pollutants are released into the atmosphere as recorded in the Table 1.3.

Table 1.3: [8] Month Pollutant Release Rate (tons/day)

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

0.20

0.25

0.27

0.34

0.45

0.52

0.63

0.70

0.81

0.85

0.89

0.95

(a) What is an upper estimate for the total tonnage of pollutants released in the month of January? February? June? (b) Suppose you plotted the data on a graph of Pollution Rate (tons/day) vs. Time (day). Describe how the total tonnage released for each of those months represents a rectangle, one for each month. (c) What are the lower estimates for these months? (d) Give an upper estimate of the total tonnage of pollutants released from the beginning of January to the end of June. Assuming that new scrubbers allow only 0.05 ton/day released, what is a lower estimate? Why would this problem be easier if the scrubbers didn’t decline, and the pollution rate stayed at 0.05 tons/day?

[Ans: 61.32, 47.04] (e) In the best case, approximately when will a total of 125 tons of pollutants have been released into the atmosphere? [Ans: Oct 27] (f) The upper and lower approximations give a range of reasonable values for the definite integral, but neither one of them is necessarily very reliable. Graphically, instead of having a rectangle at the highest possible y-value for each subinterval, or the lowest possible y-value, what might be a more reasonable approach. Numerically, rather than using the upper or lower approximations, what might be a more reasonable approach?

For this problem, the Upper Rectangular Approximation also happens to be a Right-Endpoint Rectangular Approximation. An Upper RAM will be the Right RAM so long as the function is always increasing. For this problem, the Lower Rectangular Approximation also happens to be the Left-Endpoint Rectangular Approximation. The Left RAM will happen to be the Lower RAM whenever the function is increasing. Were the function decreasing, the left RAM would be the upper approximation.

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SL Unit 1 (Area and Slope)

Terms In using a rectangular approximation to estimate

Rb a

f (x) dx:

• The interval starts at a and ends at b. For this problem, it is the beginning of January to the end of June. The interval width is b − a, e.g., 181 or 182 days depending on leap-hood.

• The interval is divided into subintervals. For the pollution problem, the subintervals are the months. The subinterval widths would be 31 days, 28 (or 29) days, etc.

• We will pretend that Riemann sum is German for RAM. With a rectangular approximation method, the actual area for each subinterval is replaced with the area of a rectangle. The rectangle will have the same width as the subinterval, and the height is determined by whichever Rectangular Approximation Method is chosen. There are countless types of Rectangular Approximation Methods, but five which you need to know: • Left endpoint Rectangular Approximation Method (RAM)[8] - the height of each approximating rectangle is the height of the left side of the corresponding subinterval (e.g., beginning of the month).

• Right endpoint RAM - the height of each approximating rectangle is the height of the right side of the corresponding subinterval (e.g., end of the month).

• Midpoint RAM - the height of each approximating rectangle is the height in the middle of the subinterval (e.g., the sixteenth of the month).

• Upper RAM - the height of each approximating rectangle is the maximum height in the corresponding subinterval.

• Lower RAM - the height of each approximating rectangle is the minimum height in the corresponding subinterval.

In addition to the Rectangular Approximation Methods, there is also: • Trapezoidal Approximation Method, in which the actual area of each subinterval is replaced by the area of a trapezoid. The base of the trapezoid is the same as the subinterval width, just as for rectangles. For each trapezoid, the two heights used are the two heights on the left and right of each subinterval.

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Example 1.2.3 Why did I not ask for the Midpoint Approximation for the pollution problem? In what cases could I ask for a Midpoint RAM?

1.2.2

Rectangular Approximation Method (RAM)

There are many physical situations where we must multiply two quantities, one of which is not a set constant, but a continuously changing variable. In order to multiply two things, one of which is changing, we utilize the definite integral, which is nothing more than the signed area under a curve. Previously, we have estimated the area under a curve via the “counting squares” approach. Another way to estimate the area of a funky shape is to approximate the shape with a series of vertical rectangles, which, together, form a blocky or pixelized version of the original shape.

Our approach here is to divide the shape or region into a number of funky strips that have three straight sides and a curved top (or bottom if below the x-axis) that follows the function whose definite integral we are finding, i.e., that we are integrating.

Once you have divided the shape into a number of strips, the next thing to do is replace the strip with a rectangle of approximately the same size. The idea is that we are approximating the area of the strip, which we don’t know (since we don’t have a formula to find the area of a funky strip) with the area of a rectangle, for which we do have a formula. The area of a rectangle is base times height.

• The width of each rectangle is the subinterval width, which is usually determined to some extent either by the way the problem is asked, or by the data itself.

• The constant height of each approximating rectangle is based on the varying height of the funkily-shaped strip.

There are several different rules for assigning a height to each rectangle. Here are two, but we will discuss others a little later: • Left endpoint Rectangular Approximation Method (RAM)[8] - the height of each approximating rectangle is the height of the left side of the corresponding strip.

• Right endpoint RAM - the height of each approximating rectangle is the height of the right side of the corresponding strip.

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SL Unit 1 (Area and Slope)

Figure 1.12: Left endpoint and right endpoint rules for the Rectangular Approximation Method [8]

The left endpoint rectangular approximation is designated by Ln , where n is the number of rectangles (or strips or slices or subintervals). The right endpoint rectangular approximation is designated by Rn .

1.2.3

Trapezoidal Approximation: Quasi-RAM

What is usually better than taking a constant height at the right endpoint, or a constant height at the left endpoint is joining the left and right endpoints of each funky strip with a line segment, creating a trapezoid instead of a rectangle. The Trapezoidal Approximation Method finds the funky area by adding up the areas of multiple replacement trapezoids, just like the Rectangular Approximation Method added the areas of multiple rectangles.

Example 1.2.4 (adapted from AB ’98) A table of values for the velocity v(t), in ft/sec, of a car traveling on a straight road, at 5 second intervals of time t, for 0 ≤ t ≤ 50, is shown in Table 1.4.

Table 1.4: Velocity of a car traveling on a straight road t (seconds) 0 5 10 15 20 25 30 35 40 45 v(t) (ft/sec) 0 12 20 30 55 70 78 81 75 60

50 72

R 50 (a) Approximate 0 v(t) dt with a left and right rectangular and trapezoidal approximations, each with five subintervals. (b) Draw rectangles or trapezoids on the graphs in Figure 1.13 to demonstrate each of the three methods of estimation. From the graphs, which seems to be the most accurate? (c) Find L10 , R10 and T10 .

R 50 (d) Using correct units, explain the meaning of 0 v(t) dt. (e) How could you estimate the average velocity of the car?

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Figure 1.13: Draw the appropriate rectangles or trapezoids for L5 , R5 , and T5

1.2.4

Streamlining Calculations for Equal Widths

If all of the subinterval widths are the same, the calculations for rectangular and trapezoidal approximations can be simplified.

Left RAM b

Z

f (x) dx ≈ Ln = ∆x (y0 + y1 + y2 + · · · + yn−1 ) a

where ∆x =

b−a n .

Right RAM

Z

b

f (x) dx ≈ Rn = ∆x (y1 + y2 + · · · + yn−1 + yn ) a

Trapezoidal Approximation

Z

b

f (x) dx ≈ Tn = ∆x a

y

0

2

+ y1 + y2 + · · · + yn−1 +

yn ∆x = (y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn ) 2 2

Example 1.2.5 Refer again to Table 1.2. (a) What is the width of each subinterval?

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SL Unit 1 (Area and Slope) (b) Use the above formulas to find T4 , R4 , and L4 to approximate R 20 v(t) dt on your calculator in one input step, using values 0 from the table.

Example 1.2.6 R 40 Refer once again to the data in Table 1.4 on page 20. Estimate 5 v(t) dt. What does this represent?

1.2.5

Finding a Range of Values

There may be a situation in which you want to find a range of values for the definite integral, i.e., what is the best case scenario, and what is the worst case scenario.

Here are two more rules for approximating the height of each funky shape.

• Upper RAM - the height of each approximating rectangle is the maximum height in the corresponding strip.

• Lower RAM - the height of each approximating rectangle is the minimum height in the corresponding strip.

The upper Riemann sum is designated Un , where n is the number of subintervals.

The lower sum is designated Ln . That’s right, the lower and left sums have the same designation. You will have to tell which is which from context. If you’re asked to find L8 and U8 , you should find a lower and upper Riemann sum, not a left and upper.

The importance of these rules are in giving a range of values. The upper RAM is always an overestimate, whereas the lower RAM always underestimates the actual integral. If I is the actual value of the definite integral, then Ln ≤ I ≤ Un

Example 1.2.7 Refer once again to the data in Table 1.4 on page 20. Find lower and upper rectangular approximations, using five subintervals. Then find U10 and L10 . What’s happening to the range as you increase the number of subintervals? Draw rectangles on Figure 1.14 to demonstrate these approximation methods.

Key Questions Mr. Budd, compiled January 12, 2011

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Figure 1.14: Draw the appropriate rectangles or trapezoids for U5 and L5 .

1. When is a lower sum always the same as a left sum?

2. What happens to the range between the upper and lower rectangular approximations as the number of subintervals increases?

1.2.6

Midpoint RAM

Recall that we have discussed the left and right rectangular approximation methods. A third, similar, method is the midpoint approximation, which typically gets confused with the trapezoidal approximation. • Midpoint RAM - the height of each approximating rectangle is the height of the strip in the middle. If you were to take the funky strip and fold it so that the right side and the left side touch, the creased side would be the midline, and the length of that folded side would be used as the height of the approximating rectangle. This would be done for each funky strip.

Figure 1.15: Left endpoint, right endpoint, and midpoint rules for the Rectangular Approximation Method[8]

Example 1.2.8 Refer yet again to the data in Table 1.4 on page 20. Estimate the area under the graph by using the Midpoint Rectangular Approximation with n = 5. Draw rectangles on the graphs Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope) in Figure 1.16 to show that you understand the midpoint approximation.

Figure 1.16: Draw the appropriate rectangles for M5

Key Questions 1. What is the difference between Mn and Tn ?

Accuracy Upper and Lower are obviously the worst methods in terms of accuracy, as they give us extreme values.

Right and Left cannot be considered much more reliable than upper and lower, and frequently give upper and lower. (Why?) Trapezoidal approximations can be considered more accurate than right, left, upper, or lower approximations. But what about the midpoint rule?

If you consider a trapezoidal and a midpoint approximation with the same number of subintervals, then generally the midpoint approximation is about twice as accurate. However, think about using a table of data. If I have nine equally spaced data points, i.e., eight subintervals, then I can use all eight data points to calculate T8 . I would not, however, be able to find M8 . (Why not?)

The best I could do would be M4 . For midpoint and trapezoidal approximations, doubling the number of subintervals generally quadruples your accuracy. So that: • Tn is accurate • Mn is roughly twice as accurate as Tn Mr. Budd, compiled January 12, 2011

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• T2n is roughly four times as accurate as Tn and therefore roughly twice as accurate as Mn It is worth noting that if Mn overestimates the actual integral, then Tn underestimates, and vice versa. Since the midpoint approximation is about twice as accurate as the trapezoidal approximation, we can make a super approximation that is a weighted average of the midpoint and trapezoid, with the midpoint being weighted twice as much as the trapezoid: S2n =

2Mn + Tn 2+1

Why does the number of subintervals double?

1.2.7

Unequal Subdivisions

Recall that the data you have may be such that it is not evenly spaced. In these cases, you must calculate each rectangle or trapezoid separately, and add them at the end.

Example 1.2.9 (adapted from AB ’03) The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by function R of time t. A table of selected values of R(t), for the time interval 0 ≤ t ≤ 90 minutes is shown.

Table 1.5: Rate of fuel consumption of a plane t (minutes) 0 30 40 50 70 90

R(t) (gallons per minute) 20 30 40 55 65 70

(a) Approximate the amount of fuel consumed in the first 30 minutes using left and right endpoint and trapezoidal methods.

Indicate units. (b) Approximate the amount of fuel consumed in the time interval 30 ≤ t ≤ 40 minutes, using left and right endpoint and trapezoidal methods. Indicate units.

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SL Unit 1 (Area and Slope) (c) If we know that R(t) is what we call strictly increasing, i.e., R always increases and never decreases, then what would be a range for the amount of fuel consumed by the plane in the first 90 minutes? Why is it important that we know that R is strictly increasing? (d) Draw rectangles on Figure 1.17 to demonstrate that you understand the upper and lower approximation methods.

R 90 (e) Approximate the value of 0 R(t) dt using the five subintervals indicated by the data in the table. What is the most appropriate method: left, right, trap, upper, lower, or mid? Draw appropriate polygons on Figure 1.17 to demonstrate which method you used.

Rb (f) For 0 < b â&#x2030;¤ 90 minutes, explain the meaning of 0 R(t) dt in terms of fuel consumption for the plane.

R 90 1 R(t) dt? (g) What do you think is the physical meaning of 90 0 R 1 b Of b 0 R(t) dt

Figure 1.17: from 2003 AP Calculus AB exam

Problems 1.B-1 Refer to the graph which is repeated in Figure 1.18. (a) Find a range of values for the area under the graph using upper and lower rectangular approximation methods, with n = 4. (b) Approximate the area under the graph using the Left, Right, and Midpoint Rectangular Approximation Methods and the Trapezoidal Approximation Method, with n = 4. Are these values within your upper/lower range? (c) Draw appropriate rectangles or trapezoids to demonstrate your understanding of M4 , T4 and R4 .

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Figure 1.18: [20] Draw the appropriate rectangles or trapezoids for M4 , T4 and R4

(d) Approximate the area under the graph using the Left, Right, and Midpoint Rectangular Approximation Methods and the Trapezoidal Approximation Method, with n = 8. (e) What happens to the discrepancies between the various methods as you increased the number of subintervals? (f) How might you make all four methods of approximation get closer and closer to the same number?

1.B-2 Refer to Figure 1.19. By counting squares, find an approximation for the definite integral of f (x) from x = 2 to x = 14. Find an estimate of the definite integral of f (x) from x = 2 to x = 14, using rectangular approximation method with: Figure 1.19: [10]

(a) 3 subintervals and a midpoint method for finding the height of the rectangle. (b) 3 subintervals and a left-point method for finding the height of the rectangle.

Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope) (c) 3 subintervals and a right-point method for finding the height of the rectangle. (d) 6 subintervals and a left-point method for finding the height of the rectangle. (e) 6 subintervals and a right-point method for finding the height of the rectangle. (f) 6 subintervals and an upper-point method for finding the height of the rectangle. (g) 6 subintervals and a lower-point method for finding the height of the rectangle.

[Ans: 308, 252, 356, 280, 332, 346, 266]

1.B-3 (adapted from AB â&#x20AC;&#x2122;04) A test plane flies in a straight line with positive velocity v(t), in miles per minute at time t minutes, where v is a function of t. Selected values of v(t) are shown in Table 1.6.

Table 1.6: Test plane velocities t (minutes) 0 5 10 15 20 25 v(t) (miles per minute) 7.0 9.2 9.5 7.0 4.5 2.4

30 2.4

35 4.3

40 7.3

(a) Use a midpoint Riemann sum and values from the table, what is the R 40 most number of subintervals that could be used to approximate v(t) dt?

0 (b) Find M4 . (c) Find T4 . Does T4 use the same data points as M4 ? (d) Find the super-approximation S8 =

2M4 +T4 .

3

1.B-4 Oil is leaking out of a tanker damaged at sea. The damage to the tanker is worsening as evidenced by the increased leakage each hour, recorded in Table 1.7. [8]

Time (h) Leakage (gal/h)

0 50

1 70

Table 1.7: [8] 2 3 4 97 136 190

5 265

6 369

7 516

8 720

(a) Give an lower and upper estimate of the total quantity of oil that has escaped after 5 hours. (b) Repeat (a) for the quantity of oil that has escaped after 8 hours.

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(c) The tanker continues to leak 720 gal/h after the first 8 hours. If the tanker originally contained 25,000 gal of oil, approximately how many more hours will elapse in the worst case before all the oil has leaked? in the best case?

[Ans: 543–758 gal, 543 gal; 1693–2363 gal; 31.4 more hours, 32.4 hours] 1.B-5 (adapted from Acorn book) Table 1.8 gives the values for the rate (in gal/sec) at which water flowed into Lake Lamar, with readings taken at specific times.

Table 1.8: Water Flow into Lake Lamar Time (sec) 0 10 25 37 46 60 Rate (gal/sec) 500 400 350 280 200 180

(a) Give a range of values for the total amount of water that flowed into the lake during the time period 0 ≤ t ≤ 60. (b) Find a trapezoidal approximation to the amount of water that flowed into the lake during that time period. (c) Does your trapezoidal approximation fall within the range you gave?

[Ans: 16930–20520 gal, 18725 gal, yes] 1.B-6 An object is dropped straight down from a helicopter. The object falls faster and faster but its acceleration (rate of change of its velocity) decreases over time because of air resistance. The acceleration is measured in ft/sec2 and recorded every second after the drop for 5 sec, as shown in Table 1.9.

Table 1.9: Acceleration of a falling object [8] t 0 1 2 3 4 5 a(t) 32.00 19.41 11.77 7.14 4.33 2.63

(a) Use L5 to find an upper estimate for the speed when t = 5. (b) Use R5 to find a lower estimate for the speed when t = 5. (c) Use upper estimates for the speed during the first second, second second, and third second to find an upper estimate for the distance fallen when t = 3.

[Ans: 74.65 ft/sec; 45.28 ft/sec; 146.59 ft] Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope)

Figure 1.20: [20]

R4 1.B-7 Let I = 0 f (x) dx, where f is the function whose graph is shown in Figure 1.20. [20] (a) Use the graph to find L2 , R2 , and M2 . (b) Are these underestimates or overestimates of I? (c) Use the graph to find T2 . How does it compare with I? (d) For any value of n, list the numbers Ln , Rn , Mn , Tn , and I in increasing order.

[Ans: 6, 12, 9.6; L2 : u, R2 : o, M2 : o; 9 < I; Ln < Tn < I < Mn < R] 1.B-8 As the fish and game warden of your Buddville, you are responsible for stocking the town pond with fish before the fishing season. The average depth of the pond is 20 feet. Using a scaled map, you measure the distances across the pond at 200-foot intervals, as shown in the diagram in Figure 1.21. [8] (a) Use the Trapezoidal Rule to estimate the volume of the pond. (b) You plan to start the season with one fish per 1000 cubic feet. You intend to have at least 25% of the opening dayâ&#x20AC;&#x2122;s fish population left at the end of the season. What is the maximum number of licenses the town can sell if the average seasonal catch is 20 fish per license?

[Ans: 26.36 million cubic feet; 988]

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31

Figure 1.21: [8]

Mr. Budd, compiled January 12, 2011

32

SL Unit 1 (Area and Slope)

Mr. Budd, compiled January 12, 2011

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1.3

33

Approximating Definite Integral from Formulas

International Baccalaureate 7.5 Definite integrals. Area between a curve and the x–axis in a given interval.

Rb Included: a y dx.

7.6 Kinematic problems involving displacement, s, velocity, v, and acceleration, a.

Included: Area under velocity-time graph represents distance.

(MM 8.6) The estimation of the numerical value of a definite integral using the trapezium rule. Included: an appreciation of the effect of doubling the number of sub-intervals.

Textbook §4.3 Area, §4.4 The Definite Integral, and §4.7 Numerical Integration [15] Resources §5.9 Approximate Integration in Stewart [20]. §5.1 Estimating with Finite Sums and §5.5 Trapezoidal Rule in Finney, et al. [8]. §1-4 Definite Integrals by Trapezoids, from Equations and Data by Foerster [10]. Exploration 1-4: “Definite Integrals by Trapezoidal Rule” in [9].

1.3.1

Definite Integrals from Known Shapes

We have looked at calculating definite integrals using graphs and using tables.

Many times, however, instead of having a graph or a table, we have a formula or expression which we are integrating.

Sometimes, we can find these definite integrals by looking at a graph of the expression.

Example 1.3.1 Find: Z

50

(a)

60 dt 30

Z

0

(−x − 2) dx

(b) −3

Z (c)

4

q 2 4 − (x − 2) − 2 dx

0

Ans: 1200, - 32 , −8 + 2π Mr. Budd, compiled January 12, 2011

34

SL Unit 1 (Area and Slope) Z

1

Example 1.3.2 Find

q

2

16 − (x + 3) dx.

−3

1.3.2

Approximating Definite Integrals

Unfortunately, very often, when we graph the integrated expression, we are not given a graph with nice shapes, for which we have area formulas. Although we may not be able to calculate the exact area under these curves (yet), we still can use the other techniques which we’ve already learned.

Graphing Example 1.3.3 Let f (x) = 1−x2 . Estimate a value for the integral R2 I1 = 0 f (x) dx. To graph use an xstep of 1 and a ystep of 1. [16] Example 1.3.4 Let g(x) = x3 . Estimate an xstep of 0.5 and a ystep of 0.5. [16]

R1 0

g(x) dx. To graph use

Riemann slicing Remember that when we had data points (or graphs), we approximated definite integrals by dividing the overall interval into subintervals. A definite integral was approximated for each subinterval, using area formulas for rectangles or trapezoids, and then the individual areas were added together. For tabular data, the subintervals were usually predetermined by what data was available.

Graphically, this meant dividing the funky shape into several funky strips, each of which was replaced with a rectangle or trapezoid of similar area. The area of all the funky strips were added together, to get the area of the funky shape. For graphs, the number of subintervals was limited by our resolution to distinguish small changes in height or width.

If we are given an expression to integrate, our approach will be similar. We divide the overall shape into smaller strips, and then replace the smaller strips with rectangles or trapezoids, whose areas we then add together. The advantage of using formulas is that we don’t have restrictions on which or how many subintervals to use. The number of strips can be a few, if we are going to calculate he areas by hand, or infinitely many, in a theoretically ideal case. If you are looking at an interval from t = 0 min to t = 8 min, you might naturally pick 8 strips, each of width 1 min. You might also pick a factor of 8, such as 2 Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 3: Approximating Definite Integral from Formulas

35

or 4, to give subinterval widths of 4 and 2 minutes, respectively. Alternatively, you might choose a multiple of 8, such as 16 (subinterval widths of 30 seconds).

It is usually best to make all the strips of equal width.

When we divide the shape into several strips we determined the width or base of each rectangle. total interval width wslice = number of subintervals or ∆x =

b−a n

When the areas of all the rectangles are added together, the sum of the areas is called a Riemann sum. The Riemann sum is an approximation to the definite integral. It is named after a German guy who apparently invented rectangles.

P Definition 1.3 (Riemann sum). A sum of the form f (x)∆x where each term of the sum represents the area of a rectangle of altitude f (x) and base ∆x. A Riemann sum gives an approximate value for a definite integral.[10]

Example 1.3.5 Return to Exploration 1-4. (You should have already completed problems 1 through 3 for homework.) (a) Use your graphing calculator to make a table of values for the equation v(t) = t3 − 21t2 + 100t + 110 for the even values of t from t = 0 to t = 8. (b) Do problem 4 on Exploration 1-4 (c) Find a way to determine the answer for problem 4 on your graphing calculator in one line of input that doesn’t require you to copy data from a table.

Example 1.3.6 Using a program, calculate rectangular and/or

R 20 2x dx trapezoidal approximations to the integral 0 60 + 40 (0.92) using 4, 8, and 16 subintervals.

R2 Example 1.3.7 Without a program, find M3 for 0 1 − x2 dx.

Check with a program, and then use more and more subintervals to find the actual value.

16 Ans: − 27 ; − 23 Mr. Budd, compiled January 12, 2011

36

SL Unit 1 (Area and Slope) R1 Example 1.3.8 Without a program, find T4 for 0 x3 dx. Check with a program, and then use more and more subintervals to find the actual value.

Ans:

17 1 64 ; 4

Example R π 1.3.9 In your mighty, mighty groups of four: Approximate 0 2 sin2 x dx using: (a) M2 (b) T2 (c) M3 (d) T3 (e) R4

1.3.3

Using Symmetry

Example 1.3.10 Let f (x) = 1 − x2 . Find (or estimate) values for R2 R2 the integrals I1 = 0 f (x) dx and I2 = −2 f (x) dx. [16]

Ans: − 32 , − 43 Note that f (x) is an even function.

Example 1.3.11 Let g(x) = x3 . Find or estimate R1 g(x) dx. [16] −1

R1 0

g(x) dx and

[Ans: 0.25, 0] Note that g(x) is an odd function. (Why?)

Problems big giant blue-green Calculus book p. 368: #33; p. 380 #41-44; p. 413, #5 1.C-1 Quickly draw a graph of the appropriate functions, then calculate each definite integral. [16] Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 3: Approximating Definite Integral from Formulas (a)

R3

(b)

R3

(c)

R3

−3 −3 −3

37

(x + 2) dx |x + 2| dx (|x| + 2) dx [Ans: 12, 13, 21]

Z 1q 2 1.C-2 Evaluate 1 − (x − 1) dx. [Hint: Sketch a graph of the integrand, 0 q

2 i.e., 1 − (x − 1) .] [16] Ans: π4 Z

3

6−

1.C-3 Evaluate

q

2

4 − (x − 3)

[Ans: 12 − π]

dx exactly. [16]

1

Z 1.C-4 Evaluate 0

3

q

2

4 − (x − 1) dx exactly.

h

√

Ans:

3 2

+

4π 3

i

1.C-5 Go to http://math.furman.edu/~dcs/java/NumericalIntegration.html Z 2 2 and estimate dt.

2 −1 1 + 4t (a) Start with a left-hand rule, using four subintervals. While doubling the number of subdivisions, watch what happens to the error (i.e., the difference between the estimate and the actual value). The “successive error ratio” that is reported if you use the “Double” button gives the ratio of the new error to the error from before, i.e., the one with half as many subintervals. What value does the successive error ratio approach as you continue to double the number of subintervals, while using the left-hand rule?

[Ans: 0.5] (b) Now using the trapezoidal rule, what value does the successive error ratio approach as you continue to double the number of subintervals, i.e., as n → ∞?

[Ans: 0.25] (c) Now using the midpoint rule, what value does the successive error ratio approach as you continue to double the number of subintervals?

[Ans: 0.25] (d) For midpoint or trapezoidal methods, if you tripled the number of subintervals, what would you expect to happen to the error?

[Ans: one-ninth of what it was before] (e) For a large number of subintervals, compare the absolute value of the error for the midpoint approximation and for the trapezoidal approximation. Which one is bigger? By roughly what percentage?

(Make sure your approximations use the same number of subintervals) [Ans: error for trapezoidal is roughly double that of midpoint] Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope)

1.C-6 Go to http://math.hws.edu/javamath/config_applets/RiemannSums.html and set f (x) = ex +1. Let xmin be 0 and xmax be 1. Set ymin and ymax so that you can see the graph. As you increase the number of subintervals, what does the sum appear to be approaching? Do you recognize this number?

[Ans: 2.718 = e] 1.C-7 Play around with http://www.plu.edu/~heathdj/java/calc2/Riemann.html Rπ 1.C-8 Estimate 0 sin x dx using trapezoids with 2 subintervals, 3 subintervals, and 4 subintervals. Give exact and decimal answers to three places after the decimal. Do your trapezoidal approximations over- or underestimate this definite integral? What’s happening to the values as you increase the number of subintervals?

Make a√conjecture as to what the exact answeri h √ might be.

Ans: π2 = 1.571, π 3 3 = 1.814, π4 1 + 2 = 1.896; under;

Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 4: Slope and Rate of Change

1.4

39

Slope and Rate of Change

International Baccalaureate 7.1 Derivative interpreted as a gradient function and as rate of change.

Textbook §1.1 A Brief Preview of Calculus and §2.1 Tangent Lines and Velocity [15] Resources §1-1 The Concept of Instantaneous Rate in Foerster [10]. Exploration 1-1: “Instantaneous Rate of Change of a Function” in [9].

1.4.1

Instantaneous Rate of Change

Recall: average velocity and slope.

Example 1.4.1 (a) A car driving due east away from Houston is 20 miles from the city limits at 1 p.m. and 130 miles from the city limits at 3 p.m. What is the car’s average velocity between 1 p.m. and 3 p.m. (b) A ball thrown up into the air has an average velocity, between 3 seconds and 5 seconds after it was thrown, of −14 feet per second. If, 5 seconds after it was thrown, the ball was 20 feet above the ground, how high was it 3 seconds after it was thrown? (c) A ball thrown straight up into the air has a height above the ground of s(t) = −16t2 +96t feet, t seconds after it was thrown.

Find the average velocity of the ball during the time period between 1 and 3 seconds after it was thrown.

Slope = Rate of Change 1. Average Rate of Change. This is the Algebra I version of slope. The slope of a secant line between two points. It is rise over run; change in y over change in x.

2. Instantaneous Rate of Change. This is the Calculus version of slope. It is the slope of the tangent line at one point. In a sense, it is an oxymoron, because there is no change in an instant.

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SL Unit 1 (Area and Slope)

1.4.2

Definition and Notation

Definitions and Notation The derivative is another way of saying instantaneous rate of change. It is denoted by a ‘prime’ after the function, i.e., the derivative of f (x) is written f 0 (x).

Definition 1.4 (Derivative). The derivative of a function at a particular value of the independent variable is the instantaneous rate of change of the dependent variable with respect to the independent variable.[10] We’ve already noted that the rate of change is essentially the slope, so that the instantaneous rate of change is the slope at a point.

Notation The derivative of f is denoted by f 0 . f 0 (3) is the slope of the curve of f at the point where x = 3.

The second derivative is the derivative of the derivative, and is denoted by f 00 .

Approximating the derivative given a graph Example 1.4.2 Graph s(t) = −16t2 + 96t. (a) Draw the line tangent to the graph of s(t) at t = 1. (b) Estimate s0 (1) by finding the slope of your tangent line. (c) What is the physical meaning of s0 (1), the rate of change of height, with respect to time, at t = 1? (d) Write an equation of the line tangent to the graph of s(t) at s = 1, and use it to approximate s(1.1). Compare this approximation to the actual value of s(1.1). (e) Estimate s0 (3). Why is s0 easy to find at t = 3?

[Ans: ; 128; v(3); ; 0] Example 1.4.3 Graph f (x) = x2 + x. (a) Draw the line tangent to the graph of f (x) = x2 + x at x = 2.

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(b) Estimate the slope of this line, i.e., f 0 (2) (c) Write the equation of this line. Plug 2.5 into the formula for your tangent line, and compare it to the actual value of 2.5.

When do you suppose the tangent line approximation is a good approximation?

[Ans: ; 5; y = 5 (x − 2) + 6, 8.5, 8.75]

1.4.3

Approximating Derivatives from Tabular Data

Rate of Change = Difference Quotient • Rate from ratio is a quotient.

• Change is the difference.

Difference Quotients • Forward: an interval to the right of the point of interest. This interval starts at the point at which you are estimating the derivative (i.e., instantaneous rate of change), and ends at some very slightly higher x-value.

• Backward: an interval to the left of the point of interest. This interval starts at some point very slightly lower x-value than where you are estimating the derivative, and ends at the point of interest.

• Symmetric: an interval to the left and right of the point of interest. This interval starts with a very slightly lower x-value than where you are approximating the derivative, and ends at a very slightly higher x-value.

Note that any difference quotient can be a forward, backward, or symmetric difference quotient. The point of interest helps decide which it is.

Example 1.4.4 Use the table of values to answer the following questions [19]. x y

−1.5 2.027

−1 0.632

−0.5 −0.357

0 −1

0.5 −1.399

1 −1.718

(a) What is the best approximation of f 0 (1), the derivative at 1?

Is this difference quotient forward, backwards, or symmetric?

Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope) (b) What is the best estimate of f 0 (−1.5)? Name the type of interval. (c) Find the best estimate of f 0 (−1.25). Name the type of interval. (d) What is the best estimate of f 0 (−1)? Name the type of interval. (e) Extension Find the best estimate of f 00 (−1), i.e., the second derivative at −1, i.e., the rate of change of the rate of change.

[Ans: −0.638, b; −2.79, f ; −2.79, s; −2.384, s; 1.624] Key Questions 1. Before doing any calculations, how can you determine whether the derivative should be positive or negative?

Example 1.4.5 Suppose that f is a function for which f 0 (2) exists.

Use the values of f given in the table to estimate f 0 (1.9), f 0 (2), and f 0 (2.02). Name the type of difference quotient used. [16] x f (x)

1.9 6.6

1.97 6.905

2.0 7

2.02 7.059

2.2 7.5

[Ans: 4.357 f, 2.95 f, 2.95 b] Note that difference quotients might be used for graphs and expressions as well.

Problems big giant blue-green Calculus book p. 155, WE #3; #1-7 odd, 37 1.D-1 The position, s(t) (measured in inches), at any time, t (measured in seconds), of an object is described in Figure 1.22. Use the graph to determine: (a) s(0) (b) s(1) (c) v(2) (d) Is v(3) > 0? (e) Is v(1) > 0?

[Ans: 1, 0, 0, Y, N] Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 4: Slope and Rate of Change

43

Figure 1.22: Displacement, s(t) [14]

Figure 1.23: Displacement, s(t) [14]

1.D-2 The graph of a position function in Figure 1.23 represents the distance in miles that a person drives during a twelve minute drive to school. Make a velocity function. [14]  sketch of the corresponding           Ans:        

                 

1.D-3 The graph of a function f is shown in Figure 1.24. Rank the values of Mr. Budd, compiled January 12, 2011

44

SL Unit 1 (Area and Slope) f 0 (−3), f 0 (−2), f (0), and f 0 (4) in increasing order. [16] [Ans: f 0 (4), f 0 (−3), f 0 (−2), f 0 (0)] Figure 1.24: Graph of f [16]

1.D-4 Suppose that f (x) = x3 − 5x2 + x − 1 and that g(x) = x3 − 5x2 + x + 4.

Explain why f 0 (x) = g 0 (x) for every x. [Hint: How are the graphs of f and g related?] [16] 1.D-5 The graph of the derivative of a function f appears in Figure 1.25. [16] Figure 1.25: Graph of f 0 [16]

(a) Suppose that f (1) = 5. Find an equation of the line tangent to the graph of f at (1, 5). (b) Suppose that f (−3) = −6. Find an equation of the line tangent to the graph of f at (−3, −6).

[Ans: y − 5 = 2 (x − 1); y = −6] 1.D-6 (adapted from AB ’06) The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by the function f , shown in Figure 1.26.

Mr. Budd, compiled January 12, 2011

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45

Figure 1.26: Graph of f , Burning Calories

(a) Find f 0 (22). Indicate units of measure. (b) For the time interval 0 ≤ t ≤ 24, at approximately what time t does f appear to be increasing at its greatest rate? Why? (c) Find the total number of calories burned over the time interval 6 ≤ t ≤ 18 minutes.

Z 18 1 (d) What do you suppose is the meaning of f (t) dt (18 − 6) min 6 [Ans: −3 cal/min/min; t = 2; 132; ] 1.D-7 The graph shows how the price of a certain stock varied over a recent trading day. [16]

(a) For each time interval below, find the total change in price and the average rate of change of price. (Be sure to indicate units used to measure these quantities.) i. ii. iii. iv.

8:00 to 11:00 9:00 to 1:00 9:30 to 2:00 11:00 to 1:00 Mr. Budd, compiled January 12, 2011

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SL Unit 1 (Area and Slope)

Ans: 1, 0, 53 , −1 $/hr (b) Estimate the instantaneous rate of change of the stock’s price at each of the following times. (Be sure to indicate units with your answers.) i. ii. iii. iv.

9:15 a.m.

10:30 a.m.

12:15 p.m.

1:45 p.m.

[Ans: −4, 4, −2, 8 $/hr]

1.D-8 Go to http://math.hws.edu/javamath/basic_applets/SecantTangentApplet. html (a) For f (x), put in the function x2 + x, and hit ‘New Function’. (b) Put Tangent at x=2 (c) Change the window so that you can see the parabola, along with the red dot, the red tangent line, and the green secant line. (d) The green secant line is anchored at the point x = 2. You control the placement of the other point. Drag the green circle along the parabola, and notice how the slope of the secant line changes. (e) Pay careful attention to what happens to the green secant line as you drag the green dot closer and closer to the red dot. Make a table of the ‘Secant at x=’ values with the ‘Secant Slope =’ values. What do you notice? If the ‘Secant at x=’ value could be 2, what would the ‘Secant Slope =’ value be?

1.D-9 (adapted from AB ’06) Rocket A has positive velocity v(t) after being launched upward from an initial height of 0 feet at time t = 0 seconds.

The velocity of the rocket is recorded for selected values of t over the interval 0 ≤ t ≤ 80 seconds, as shown. t (seconds) v(t) (feet per second)

0 5

10 14

20 22

30 29

40 35

50 40

60 44

70 47

80 49

(a) Find the average rate of change of the velocity of Rocket A over the time interval 0 ≤ t ≤ 80 seconds. Indicate units of measure. (b) Approximate the instantaneous rate of change of the velocity of Rocket A at t = 0. Repeat for t = 80 seconds, t = 20, and t = 55 seconds. Indicate units of measure.

R 70 (c) Using correct units, explain the meaning of 10 v(t) dt in terms of the rocket’s flight. Use a midpoint R 70Riemann sum with 3 subintervals of equal length to approximate 10 v(t) dt.

Mr. Budd, compiled January 12, 2011

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(d) What do you suppose is the meaning of

Ans:

11 20

1 (70 â&#x2C6;&#x2019; 10) sec

Z

70

v(t) dt?

10

ft/s2 ; 0.2, 0.75, 0.4 ft/s2 ; , 2020 ft;

1.D-10 A differentiable function f has values shown. Estimate f 0 (1.5). [13] x f (x)

1.0 8

1.2 10

1.4 14

1.6 22 [Ans: 40]

1.D-11 Suppose that f is a function for which f 0 (2) exists. Use the value of f given below to estimate f 0 (1.99), f 0 (2), f 0 (2.01), and f 0 (2.1). Explain how you obtained your estimates. [16] x f (x)

1.9 25.34

1.99 33.97

1.999 34.896

2.0 35

2.001 35.104

2.01 36.05

2.1 46.18

[Ans: 102.889 fdq, 104 (sdq), 105.111 (bdq), 112.556 (bdq)]

Mr. Budd, compiled January 12, 2011

48

SL Unit 1 (Area and Slope)

Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 5: IROC as a limit

1.5

49

Instantaneous Rate of Change from a Limiting Process

International Baccalaureate 7.1 Informal ideas of limit and convergence.

Derivative interpreted as a gradient function and as rate of change; Finding equations of tangents.

Textbook §1.1 A Brief Preview of Calculus and §2.1 Tangent Lines and Velocity [15] Resources §1-2 Rate of Change by Equation, Graph, or Table in Foerster [10]. §2.5 Average and Instantaneous Rates: Defining the Derivative in Ostebee and Zorn [16].

1.5.1

Approximating Rate of Change from a Formula

Graphs Example 1.5.1 In your mighty, mighty groups of four: Do Exploration 1.5: “Instantaneous Rate of Change”

Tables Example 1.5.2 Refer to f (x) = x2 + x. (a) Write a formula for the forward difference quotient used to estimate f 0 (2) using an interval of [2, 2 + h]. (b) Put your formula into the calculator, using a variable step-size.

Evaluate the forward difference quotient for h = 0.1, h = 0.01, and h = 0.001. Continue using smaller step sizes (h) until there is no change in the thousandths place of your slope. (c) What would be the appropriate interval for a backward difference quotient of width h to approximate f 0 (2)? (d) As you did with the forward difference quotient, use smaller and smaller h’s until you see no change in the thousandths place of your slope.

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Preview of a formal definition of derivative using formulas Example 1.5.3 Refer to f (x) = x2 + x. (a) Write an expression for the average rate of change of f (x) from 2 to 2.1. (b) Write an expression for the average rate of change of f (x) from 2 to 2.01. (c) Write an formula for the average rate of change of f (x) from 2 to x. If we want an instantaneous rate of change, we want x to be as close as possible to what? (d) Make a table of values for the AROC from 2 to x for different values of x that get closer and closer to 2 from both sides. (e) Rewrite your formula, using the expression for f (x) and the value for f (2). (f) Factor the numerator in your above expression. Cancel. When would canceling not be allowed? What happens to the remaining expression as x gets closer and closer to 2? (g) Write an expression for the difference quotient for f (x) from 2 to 2 + h. If we want this difference quotient to represent the f 0 (2), what do we want h to approach? (h) Multiply the numerator out and combine like terms. Cancel.

When would canceling not be allowed? What happens to the remaining expression as h get closer and closer to 0?

(2) in the above example. Try to see Look at the work for simplifying f (2+h)â&#x2C6;&#x2019;f h that you can replace the 2 with x and you could still do the problem.

1.5.2

Kinematics: Displacement, Velocity, Acceleration

Velocity is the instantaneous rate of change of displacement, i.e., velocity is the derivative of displacement. v(t) = d0 (t) or v(t) = x0 (t).

Speed is the magnitude of velocity. In one dimension, speed is the absolute value of velocity. speed = |v(t)| Acceleration is the instantaneous rate of change of velocity, i.e., acceleration is the derivative of velocity. a(t) = v 0 (t) Acceleration is the derivative of the derivative of displacement, i.e., acceleration is the second derivative of displacement. a(t) = d00 (t) Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 5: IROC as a limit

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Problems big giant blue-green Calculus book p. 156: # 11, (31 or 33), 37, 39 1.E-1 Visit http://www.slu.edu/classes/maymk/SecantTangent/SecantTangent. html and use the following settings: • Let f (x) be x2 − x.

• Let X0 be 3. This is the reference point at which we will be estimating the derivative.

• Let dX be 1 for now.

• Don’t worry about guessing f 0 (x) for now.

• Make sure the first scroll-down menu is on ‘Right Secant’.

• Change the second scroll-down menu to ‘X1 click’.

• Make sure the two boxes are unchecked.

• Change your window as appropriate. (a) Try to make some sense out of all the information that you’re being given, e.g., what are the meanings of X0, dX, and X1? Can you identify the backward, symmetric, and forward difference quotients?

What do you suppose a negative dX means? (b) Use your mouse pointer or the ‘dX In’ and ‘dX Out’ buttons to move the second point on the secant line closer to and farther away from the reference point at x = 3. What’s happening to the line? What’s happening to the difference between the three slopes (left, balanced, and right)? (c) Write down the left, balanced, and right slope for X0= 3.0, and dX= 0.1, 0.01, and 0.001. (d) Play around: use some different functions, different points, etc. Go nuts! (e) Can you find some other, similar, websites? Anything better that we should know about?

1.E-2 Refer to g(x) = x2 − x. (a) Write an expression for the difference quotient for g(x) from 3 to 3.1.

Find the value. (b) Write an expression for the difference quotient for g(x) from 3 to 3.01. Find the value.

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SL Unit 1 (Area and Slope) (c) Write a formula for the difference quotient for g(x) from 3 to x, using the expression for g(x) and the value for g(3). (d) Factor the numerator in your above expression. Cancel. When would canceling not be allowed? What happens to the remaining expression as x gets closer and closer to 3? h i g(3.01)−g(3) g(x)−g(3) x2 −x−6 Ans: g(3.1)−g(3) = 5.1, = 5.01, , , x + 2, goes to 5 3.1−3 3.01−3 x−3 x−3 (e) Write a formula for the difference quotient for g(x) from 3 to 3 + h. (f) Multiply the numerator out and combine like terms. Cancel. When would canceling not be allowed? What happens to the remaining expression as h get closer and closer to 0?

((3+h)2 −(3+h))−6 5h+h2 = h ; goes to 5 Ans: (3+h)−3

1.E-3 Refer to q(x) = x2 . (a) Write a formula for the difference quotient for q(x) from 1 to x.

Factor the numerator in your difference quotient and cancel. What happens to the remaining expression as x gets closer and closer to 1? h i x1 −1 Ans: x−1 = x + 1, goes to 2 (b) Repeat part (a), replacing 1 with 2, then 3, and then −1. Look for a pattern, and guess a formula for q 0 (x). √ 1.E-4 Let r(x) = x. [16] (a) Graph r(x) on your calculator. Zoom in to the point (1, 1) repeatedly, until the graph looks like a straight line. (b) Use that point, i.e., (1, 1), and one other nearby point on the ‘line’ to find the slope of the ‘line’, which is an estimate of r0 (1). [Ans: 0.5] (c) Use to estimate r0 (1/4), r0 (9/4), r0 (4), r0 (25/4), and r0 (9). zooming

1 1 1 1 Ans: 1; 3 ; 4 ; 5 ; 6 (d) Use these results to sketch a graph of r0 (x) over the interval [−1, 2]. i h 1 (e) Use your results and your graph to guess a formula for r0 (x). Ans: 2√ x 1.E-5 Let f (x) = ln x. [16] (a) Use zooming to estimate f 0 (1/5), f 0 (1/2), f 0 (1), f 0 (2), and f 0 (5).

[Ans: 5, 2, 1, 0.5, 0.2] (b) Use your results to sketch a graph of f 0 over the interval [0, 5].

(c) Use your results to guess a formula for f 0 (x).

Ans:

1 x

1.E-6 Repeat the zooming process for a function of your choosing.

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1.E-7 (AB ’01) The temperature, in degrees Celsius (◦ C), of the water in a pond is a differentiable function W of time t. The table shows the water temperature as recorded every 3 days over a 15-day period. t (days) 0 3 6 9 12 15

W (t) (◦ C) 20 31 28 24 22 21

(a) Use data from the table to find an approximation for W 0 (12). Show the computations that lead to your answer. Indicate units of measure. (b) A student proposes the function P , given by P (t) = 20 + 10te(−t/3) , as a model for the temperature of the water in the pond at time t, where t is measured in days and P (t) is measured in degrees Celsius.

Estimate P 0 (12). Using appropriate units, explain the meaning of your answer in terms of water temperature. h Ans:

21−24◦ C 15−9 days

◦

= − 12 C/day; −0.549◦ C/day: water in pond decreasing at a rate of 0.549◦ C/day

Mr. Budd, compiled January 12, 2011

i

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SL Unit 1 (Area and Slope)

Mr. Budd, compiled January 12, 2011

SL Unit 1, Day 6: Slope and Area: Pulling It Together

1.6

55

Slope and Area: Pulling It Together

Do the following problems neatly on graph paper.

1.F-1 Suppose Mr. Budd is driving to the Utah Shakespearean Festival in Cedar City, UT. Once he gets on the road, he sets his cruise control for 55 mph.

Let t be the number of hours since he started driving on cruise control. (a) How far has he gone during the first half hour on cruise control? the first hour? the first two hours? (b) Write an equation for the velocity, i.e., v(t) =(something). (c) Graph the velocity versus time. (d) Find the area under the curve of v(t) from t = 0 to t = 0.5. Also, find the area from t = 0 to t = 1 and also to t = 2. (e) What shape are these areas in? If I look at the area from t = 0 to t = tstop , what is the width of the figure (as an expression with tstop in it)? the height? the area (as an expression of tstop )? Call your expression for area A(tstop ). (f) Plot a graph of distance traveled versus time. Use the points (0.5, distance for 0.5), (1, distance for 1), and (2, distance for 2). Look for a pattern, and use your result for A(tstop ) to connect the dots. (g) On your graph of distance versus time, what is the slope at t = 0.5? at t = 1? at t = 2? Indicate units.

1.F-2 Let f (t) = 2t (a) Find f (0), f (0.5), f (1), f (1.5), and f (2). (b) Graph f (t).

R1 R 1.5 R0 R 0.5 (c) Using the graph of f (t), find 0 f (t) dt, 0 f (t) dt, 0 f (t) dt, 0 f (t) dt, R2 f (t) dt. Do you see a pattern?

0 (d) RThink of a generic area under f (t) that starts at 0 and ends at x, i.e., x f (t) dt. What shape is it in? Write an expression (with x in it) 0 for the height of the shape. What is the width of the shape? Write an expression in terms of x for the area from t = 0 to t = x. Call this expression A(x) (A for area).

R R R 0 0.5 1 (e) On a separate graph, plot 0, 0 f (t) dt , 0, 0 f (t) dt , 1, 0 f (t) dt ,

R R 1.5 2 1.5, 0 f (t) dt , 2, 0 f (t) dt . Connect the dots using A(x).

1.F-3 Let G(t) = t2 . While doing this problem, keep the previous problem in the back of your mind.

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SL Unit 1 (Area and Slope) (a) Write the expression for the Average Rate of Change of G(t) from 1 to 1 + h. If this slope is to be a better and better approximation of G0 (1), what should be happening to h? Is this a forward, backward, or symmetric difference quotient? (b) In your difference quotient to estimate G0 (1), replace G(1 + h) with 2 (1 + h) , and G(1) with the actual value of G(1). Multiply out and combine like terms. Factor and cancel. (c) If you let h â&#x2020;&#x2019; 0, what happens to 1 + h? What happens to your remaining expression for G0 (1)? (d) Repeat the above process to approximate G0 (2) and G0 (3).

1.F-4 Let f (x) = 3x2 . (a) Find f (1), f (2), and f (3). (b) Use the TRAP program with more and more subintervals to make R1 R2 conjectures for the following definite integrals: 0 f (x) dx, 0 f (x) dx, R3 and 0 f (x) dx. Do you see a pattern?

1.F-5 Let G(x) = x3 . While doing this problem, keep the previous problem in the back of your mind. (a) Use symmetric difference quotients with h = 0.1, h = 0.01, and h = 0.001 to approximate G0 (1). Continue using smaller hâ&#x20AC;&#x2122;s until there is no change in the hundredths place in your estimates of G0 (1). (b) Repeat for G0 (2) and G0 (3). What do you notice?

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Unit 2

Limits and the Definition of the Derivative 1. Limits for Continuous Functions and Removable Discontinuities 2. Limit Definition of the Derivative (at x = c form) 3. Limit Definition of the Derivative (h or ∆x form) 4. One-Sided Limits and Infinite Limits 5. Limits at Infinity

Advanced Placement Limits of functions (including one-sided limits).

• An intuitive understanding of the limiting process.

• Calculating limits using algebra.

• Estimating limits from graphs or tables of data.

Asymptotic and unbounded behavior.

• Understanding asymptotes in terms of graphical behavior.

• Describing asymptotic behavior in terms of limits involving infinity.

57

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SL Unit 2 (Limits) • Comparing relative magnitudes of functions and their rates of change. (For example, contrasting exponential growth, polynomial growth, and logarithmic growth.)

Continuity as a property of functions.

• An intuitive understanding of continuity. (Close values of the domain lead to close values of the range.)

• Understanding continuity in terms of limits.

• Geometric understanding of graphs of continuous functions.

Concept of the derivative.

• Derivative presented graphically, numerically, and analytically.

• Derivative interpreted as an instantaneous rate of change.

• Derivative defined as the limit of the difference quotient.

Derivative at a point.

• Slope of a curve at a point.

• Tangent line to a curve at a point.

• Instantaneous rate of change as the limit of average rate of change.

Derivative as a function.

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SL Unit 2, Day 1: Introduction to Limits

2.1

59

Introduction to Limits

International Baccalaureate 7.1 Informal ideas of a limit and convergence.

Textbook §1.4 Computation of Limits [15] Resources §1-5 Limit of a Function in Foerster [10]. Exploration 1-5: “Introduction to Limits” in [9]. §2.4 Introduction to Limits in Varberg, et al. [23]

2.1.1

Graphic Introduction to Limits

Understanding limits is a fairly intuitive process, and usually the easiest way to understand is to study examples and counterexamples.

Figure 2.1: A piecewise function [13]

The graph in Figure 2.1 is given by the following function:   x + 1 −2 < x < 0    2 x=0  f (x) = −x 0<x<2    0 x=2    x − 4 2 < x ≤ 4 Remember This? Recall the different types of discontinuities.

One thing to notice is that this is one function, not several. It is what is called Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits)

a piecewise function, because it is defined in terms of several pieces, rather than one smooth, nice neat curve. One thing to realize is that functions don’t have to look pretty, they just have to pass the vertical line test.

Make sure you understand what is going on with the function. Notice that f (1) = −1, f (0) = 2 and f (2) = 0. The domain includes 4, but not −2. The function is continuous over (−2, 0), which basically means that I can draw that portion of the graph without lifting my pencil. At x = 0, the function undergoes a step (or jump) discontinuity, because it goes from 1 to 2 all the way down to 0. The function is again continuous over (0, 2) and over (2, 4). At x = 2 there is a removable discontinuity. Removable discontinuities are recognized by a hole in an otherwise continuous portion of the graph. A function has a removable discontinuity at a point if changing that one point would make the function continuous.

As x → −1, f (x) →?

I want to look at the limits of f (x), but first lets go over some notation. When we write lim f (x) = −1, we mean that as x get closer and closer to 1, but not x→1

equal to 1, f (x) (or y) gets closer and closer to −1.

Simple Case: Continuity To find the limit of f (x) as x goes to −1, trace your right forefinger along the function to the right of x = −1, moving left towards the point (−1, 0). Trace your left forefinger along the function, moving right towards the same point (−1, 0). You will see that as the x value gets closer to −1, the y value gets closer to 0.

Example 2.1.1 From the graph in Figure 2.1, find the limit of f (x) as x approaches −1.

As x → −1, f (x) → 0.

Notation lim f (x) = 0

x→−1

This is read: “the limit of f (x) as x approaches −1 is zero”.

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The Important Case: Removable Discontinuity An important point should be made, however, and that is that if I’m looking at the limit as x approaches −1, I don’t actually care what happens at −1, only near −1. So let’s take a look at the limit of f (x) as x approaches 2 (Figure 2.1 still). If I let my right finger approach x = 2 along the function from the right, and let my left finger approach x = 2 from the left, I notice that as the x value gets closer and closer to 2, the f (x) value gets closer and closer to −2. Hence lim f (x) = −2

x→2

Notice that lim f (x) 6= f (2). In other words, the limit is not the same as the x→2

functional value. Remember: I don’t care what’s happening at 2 if I’m taking the limit near 2.

Here are two authors’ informal definitions of limits: Definition 2.1 (Intuitive Meaning of Limit). To say that lim f (x) = L means x→c

that when x is near but different from c then f (x) is near L. [23] Definition 2.2 (Limit). L is the limit of f (x) as x approaches c if and only if L is the one number you can keep f (x) arbitrarily close to just by keeping x close enough to c, but not equal to c.[10] Remember This? Remember the difference between if and if and only if.

The formal definition of limit is quite a bit more complicated. The trick is: how do you objectively, logically, and mathematically reason what close and near mean.

Now You Quickly look at numbers 1-4, 7, and 8 in Foerster §1-5. Find lim f (x). x→c Why didn’t I ask you to find the limits for numbers 5,6 and 9, 10. What types of discontinuities are represented in the problems that do have limits. For problems 5 and 6, what types of discontinuities are represented? For problems 9 & 10?

2.1.2

Step Discontinuities & One-Sided Limits

Notation 1. lim h(x) means the limit as x approaches 2 from the left, i.e., the negative x→2−

side of the axis.

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SL Unit 2 (Limits) 2. lim+ h(x) means the limit as x approaches 2 from the right, i.e., the x→2

positive side of the axis.

Graphically Example 2.1.2 For the function f (x) graphed in Figure 2.1, find lim f (x). x→0

Now let’s take a look at the limit as x goes to 0 of the function graphed in Figure 2.1. As x gets closer and closer to 0 from the right, my y values get closer and closer to 0. As x gets closer and closer to 0 from the left, however, f (x) or y is getting closer and closer to 1. Which is the limit? Neither. The limit does not exist as x approaches 0 because the one-sided limits don’t match. lim f (x) does not exist

x→0

because lim f (x) = 1 6= lim+ f (x) = 0.

x→0−

x→0

lim f (x) is the one-sided limit from the left, or negative side. The one-sided

x→0−

limit from the right, or positive side, is given by using a “+” superscript.

Theorem 2.1. lim f (x) = L if and only if lim f (x) = L and lim f (x) = L. x→c

x→c−

x→c+

If we reexamine lim f (x), we see that it exists (and is equal to −2) because x→2

lim− f (x) = lim+ f (x) = −2. So the overall limit exists because the two one-

x→2

x→2

sided limits match, and the value of that overall limit is the same as that of the one-sided limits.

2.1.3

Limits from a Table

Continuity: The Simple Case Example 2.1.3 Use your calculator to make a table of values for f (x) = x2 + 2x + 4 for values of x near 2. Find lim f (x). Repeat x→2

for lim f (x). x→3

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Removable Discontinuity: The Important Case Example 2.1.4 Make a table of values for g(x) =

x3 − 8 for values x−2

of x near 3. Find lim g(x). Repeat for lim . x→3

x→2

[Ans: 19; 12] Example 2.1.5 What is the difference between the graph of f (x) and the graph of g(x)?

[Ans: g(x) has a hole at (2, 12), representing the removable discontinuity] Remember: in evaluating lim f (x), imagine yourself blind to what is happening x→c to f at x = c. From all the evidence near x = c, what is your best guess as to what f (c) is, or should be?

Another way to think about finding limits is: what would f (c) have to be in order to make f continuous at c? That is the limit of f as x approaches c. In fact the definition of continuity is that the limit of f matches the value of f .

One-Sided Limits, Numerically Example 2.1.6 On your graphing calculator, make a table of x3 −8 near x = 2. Find h(x) = |x−2| (a) the limit of h(x) as x → 2 from the left, aka from below. (b) the limit of h(x) as x → 2 from the right, aka from above. (c) the limit of h(x) as x → 2.

2.1.4

Limits from an Expression

If a function is continuous at x = c, then, by definition lim f (x) = f (c). Therex→c fore, if I know ahead of time that a function is continuous, then I simply plug in the x value that I’m approaching. Polynomial functions and functions involving sine and cosine are continuous, so long as there aren’t any variable expressions in any denominators.

Over the next few class periods, we will learn several different methods for dealing with discontinuities.

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SL Unit 2 (Limits) I. Continuous Functions A. The Simple Case 1. Be on the look out for: a. on a graph: graph you can draw without lifting your pencil b. on a table: no singular points that stick out or are undefined c. with an expression: adding or multiplying polynomials, sine, cosine 2. How we deal with it: plug it in, plug it in

II. Discontinuities A. Removable Discontinuities These are the only types of discontinuities for which limits exist. Limits only exist where a function is continuous, or where there is a removable discontinuity.

1. Be on the look out for: a. on a graph: holes b. on a table: a single point that sticks out c. with an expression: • piecewise functions • Rational Functions that give you 00 . 00 is called an indeterminate form, and could be many things, but are typical candidates for removable discontinuities. Be careful, though, indeterminate forms may be vertical asymptotes, or even step discontinuities.

2. How we deal with it: a. Use Algebra to cancel factors, after • factoring • rationalizing with conjugates • de-denominatorizing with LCDs b. Comparing to known limits, such as the grand-daddy of all known sin θ limits, lim θ→0 θ B. Step Discontinuities 1. Be on the look out for: a. on a graph: y-values that all of the sudden jump to a different value. b. on a table: y-values that all of the sudden jump to a different value c. with an expression: • piecewise functions |x| • variations on lim x→0 x Mr. Budd, compiled January 12, 2011

SL Unit 2, Day 1: Introduction to Limits

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2. How we deal with it: one-sided limits C. Vertical Asymptotes 1. Be on the look out for: a. on a graph: Curves that go up or down off the graph. b. on a table: Numbers that get hugely positive or hugely negative as x approaces a specific value. nonzero c. with an expression: rational expressions where you have 0 2. How we deal with it: a. one-sided limits b. infinity, ∞ D. Infinitesimal Oscillations This is a relatively minor and rare type of discontinuity. Be on the look out for: 1 on a graph: oscillations with periods that get smaller and smaller, eventually getting infinitesimally small near a specific x-value.

2 with an expression: variations on lim sin x1 x→0

2.1.5

Substitution and Properties of Limits

These are the theorems that allow me to simply plug in values if I know that my function is continuous.

Theorem 2.2 (Main Limit Theorem). Let n be a positive integer, k be a constant, and f and g be functions that have limits at c. [23] Then 1. lim k = k x→c

2. lim x = c x→c

3. lim kf (x) = k lim f (x) x→c

x→c

4. lim [f (x) + g(x)] = lim f (x) + lim g(x) x→c

x→c

x→c

5. lim [f (x) − g(x)] = lim f (x) − lim g(x) x→c

x→c

x→c

6. lim [f (x) · g(x)] = lim f (x) · lim g(x) x→c

x→c

x→c

lim f (x) f (x) = x→c , provided lim g(x) 6= 0 lim g(x) x→c g(x) x→c

7. lim

x→c

h in n 8. lim [f (x)] = lim f (x) x→c

x→c

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SL Unit 2 (Limits) 9. lim

x→c

p n

f (x) =

q n

lim f (x), provided lim f (x) > 0 when n is even.

x→c

x→c

An example of the first limit property is that lim 7 = 7. Think about why x→−3

this makes sense numerically and graphically.

The second limit property might be exemplified by lim x = π. x→π

The third property can be seen in action by lim 2x = 2 lim x, which can then x→π x→π be simplified with the second property to 2π.

Example 2.1.7

As an exercise to familiarize yourself with the x3 − 4 , naming the property used at each properties, evaluate lim x→2 2x step.

Theorem 2.3 (Limits of Trigonometric Functions). For every real number c in the function’s domain [23] 1. lim sin t = sin c x→c

2. lim cos t = cos c x→c

3. lim tan t = tan c x→c

4. lim cot t = cot c x→c

5. lim sec t = sec c x→c

6. lim sec t = sec c x→c

Example 2.1.8 Find lim sin x

x→π

[Ans: 0] Theorem 2.4 (Limit of a Composite Function). If f is continuous at b and lim g(x) = b, then lim f (g(x)) = f (b). In other words, [20] x→a

x→a

lim f (g(x)) = f

x→a

lim g(x)

x→a

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Example 2.1.9 Find

π limπ sin 2θ + θ→ 3 3 [Ans: 0]

Problems big giant blue-green Calculus book p. 85: # 1, 2, 3, 5, 15, 17, 21, 23, 25, 29, 31 2.A-1 Go to http://www.calculus-help.com/funstuff/phobe.html Watch: (a) Chapter 1, Lesson 1: What is a Limit? (b) Chapter 1, Lesson 2: When Does a Limit Exist? (c) Chapter 1, Lesson 3: How do you evaluate a limit? (d) Chapter 2, Lesson 1: The Difference Quotient.

What differences in terminology do you notice? What did the tutorial help to clarify?

Figure 2.2: [23]

2.A-2 For the function f graphed in Figure 2.2, find the indicated limit or function value, or state that it does not exist. [23] (a) lim f (x) x→−3

(b) f (−3) (c) f (−1) (d) lim f (x) x→−1

(e) f (1) Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits) (f) lim f (x) x→1

(g) lim− f (x) x→1

(h) lim f (x) x→1+

[Ans: 2; 1; d.n.e.; 2.5; 2; d.n.e.; 2; 1] Figure 2.3: [23]

2.A-3 Follow the directions of problem 2 for the function graphed in Figure 2.3.

[23] [Ans: d.n.e.; 1; 1; 2; 1; d.n.e.; 1; d.n.e.]

Mr. Budd, compiled January 12, 2011

SL Unit 2, Day 2: Limits from a Formula

2.2

69

Limits from a Formula

International Baccalaureate 7.1 Informal ideas of a limit and convergence.

f (x + h) − f (x) Definition of derivative as f 0 (x) = lim . Use of this definition h→0 h for differentiation of polynomials, and for justification of other derivatives.

Textbook §1.3 Computation of Limits and §2.1 Tangent Lines and Velocity [15] Resources §1-5 Limit of a Function in Foerster [10]. Exploration 1-5: “Introduction to Limits” in [9]. §2.4 Introduction to Limits in Varberg, et al. [23]

2.2.1

Limits at Cancelable Discontinuities

Theorem 2.5 (Functions that Agree at All But One Point). Let c be a real number and let f (x) = g(x) for all x 6= c in an open interval containing c. If the limit of g(x) as x approaches c exists, then the limit of f (x) also exists and lim f (x) = lim g(x).

x→c

x→c

In order to find the lim f (x), the value of f (c) is irrelevant. Only the values x→c infinitesimally close to c matter.

x3 −8 x→2 x−2

Example 2.2.1 Find lim

by comparing it to lim x2 + 2x + 4. x→2

Example 2.2.2 Find lim

x→3

3−x x2 − 9

Why are limits important in calculus?

Recall that f 0 (2), the derivative of f (x) at x = 2, may be considered two ways, both of which are average rates of change over an interval that always includes x = 2, but where the interval gets smaller and smaller: Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits) • the average rate of change of f form 2 to 2 + h, where h is getting smaller f (2 + h) − f (2) and smaller, i.e., as h → 0; (2 + h) − 2 • the average rate of change of f from 2 to x, where x is getting closer and f (x) − f (2) closer to 2, i.e., as x → 2. x−2

f (2 + h) − f (2) is a forward (2 + h) − 2 difference quotient, but if h < 0, i.e., h is negative, then you have a backward difference quotient. Also recall that for the limit as h → 0, you would include both positive and negative values of h.

Remember that if h > 0, i.e., h is positive, then

Example 2.2.3 For f (x) = x3 , find lim

∆x→0

f (2 + ∆x) − f (2) (2 + ∆x) − 2

Example 2.2.4 In your mighty, mighty groups of four: Pick a quadratic function f (x). Find (a) f (2) (b) lim f (x) x→2

(c) lim

f (2 + h) − f (2) ; (2 + h) − 2

(d) lim

f (x) − f (2) . x−2

h→0

x→2

Talk about what each of these things means, in terms of the graph of f (x + h) − f (x) f . Put your answers on the board. When you finish, find lim h→0 h

2.2.2

De-rationalizing with Conjugates

Sometimes we have to squeeze the factors out of the expression. One technique for doing that when we have radicals is to use conjugates.

Example 2.2.5 [20] Find lim √

t→0

t2

t2 +9−3 Mr. Budd, compiled January 12, 2011

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Example 2.2.6 [15] Find √ 4− x lim x→16 x − 16 Example 2.2.7 In your mighty, √ mighty groups of four: pick a linear function mx + b. Let f (x) = mx + b. Make sure that f (2) exists. (a) f (2) (b) lim f (x) x→2

f (x) − f (2) . x−2 f (2 + h) − f (2) ; (d) lim h→0 (2 + h) − 2 (c) lim

x→2

Talk about what each of these things means, in terms of the graph of f . When you finish, put your answers on the board. When you f (x + h) − f (x) finish, find lim h→0 h Example 2.2.8 Find √ lim

∆x→0

x + ∆x − ∆x

Note: compare this problem to lim

∆x→0

√

x

f (x+∆x)−f (x) , ∆x

which is the for-

mula for the instantaneous rate of change, √ or... derivative. This problem is finding the derivative for f (x) = x, which means that √ f (x + ∆x) = x + ∆x.

2.2.3

De-denominatorizing with LCDs

Example 2.2.9 Find 1 1 − lim x + 1 4 x→3 x−3 Example 2.2.10 In your mighty, mighty groups of four: Let f (x) = 1 . Pick a value of c 6= − 13 3x + 1 (a) f (c) Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits) (b) lim f (x) x→c

f (x) − f (c) . x−c f (c + h) − f (c) (d) lim ; h→0 (c + h) − c (c) lim

x→c

Talk about what each of these things means, in terms of the graph of f . When you finish, put your answers on the board. After you put your answers on the board: How would you find the point at f (x + h) − f (x) . which the derivative is − 34 ? Also, find lim h→0 h Example 2.2.11 Find 1 1 − x + h x lim h→0 h Note: compare this problem to lim

h→0

f (x+h)−f (x) , h

which is the formula

for the instantaneous rate of change, or... derivative. This problem is finding the derivative for f (x) = x1 , which means that f (x + h) = 1 x+h .

2.2.4

Derivative at a Point

Definition 2.3 (Derivative (at x = c form)). [15] The derivative of f (x), with respect to x, at the point x = c is given by

f (x) − f (c) x→c x−c

f 0 (c) = lim

e.g., f 0 (2) = lim

x→2

f (x) − f (2) x−2

Meaning: The instantaneous rate of change of f with respect to x at x = c.

Graphically: The slope of the line tangent to the graph of f at the point x = c.

Example 2.2.12 Let f (x) = x4 . Find f 0 (2). In your mighty, mighty groups of four: Find Mr. Budd, compiled January 12, 2011

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(a) f 0 (−1); (b) f 0 (3); (c) f 0 (−2).

What is the pattern?

Example 2.2.13 The graph of the function f shown in Figure 2.4 consists of a semicircle and three line segments. Find the following Figure 2.4: Graph of f

limits of difference quotients: (a) lim

x→−2

f (x) − f (−2) x+2

f (x) − f (2.3) x − 2.3 f (x) − f (−3) (c) lim − x+3 x→−3

(b) lim

x→2.3

(d)

lim +

x→−3

(e) lim

x→−3

f (x) − f (−3) x+3

f (x) − f (−3) x+3

Ans: − 31 [Ans: −1] [Ans: 2]

Ans: − 31 [Ans: d.n.e.]

Example 2.2.14 Each of the following limits are derivatives. Tell for which function, and for what point x4 − 81 x→−3 x + 3 √ t−5 (b) lim t→25 t − 25 (a) lim

x2 + x − 6 x→2 x−2

(c) lim

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Using conjugates to fetch factors Example √ of the line tangent to the graph √ 2.2.15 Write an equation of y = 2x + 4 at the point −1, 2 h

Ans: y −

√

2=

√1 2

(x + 1)

i

Problems big giant blue-green Calculus book p. 95: Writing Exercises # 2; #3, 11, 19 p. 156: # 23 x2 − a2 is x→a x4 − a4

Ans:

2.B-1 (adapted from ?) If a 6= 0, then lim

 2 x − 64   x 6= 8 x−8 2.B-2 f (x) =   k x=8 What value of k will make f continuous at x = 8?

1 2a2

[Ans: 16]

2.B-3 [15] Evaluate analytically, then check your answer by making a table on your handy-dandy calculator. x2 + x − 6 x→−3 x2 − 9

2 (x + h) − 2 (x + h) + 1 − x2 − 2x + 1 (b) lim h→0 h (a) lim

2.B-4 Evaluate lim √ v→3

3−v √ . v− 3

[Ans: 2x − 2] √

Ans: −2 3

2.B-5 [15] Evaluate analytically, then check your answer by making a table on your handy-dandy calculator, if possible. √

4− x (a) lim Ans: − 18 x→16 x − 16 √

x+5−3 (b) lim Ans: 16 x→4 x−4 p √ h i 3 (x + h) − 3x (c) lim Ans: 2√33x h→0 h 1 1 − 2.B-6 Find lim 2 − h 2 . h→0 h

Ans:

1 4

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2 2 − 2x − 1 3 . This limit happens to be the derivative of what 2.B-7 Find lim x→2 x−2 h i 2 function, at what point?

Ans: − 94 ; 2x−1 at x = 2 2.B-8 Find lim

x→3

x−3 2 2 − 5x − 8 7

49 Ans: − 10

1 1 − 3 (x + ∆x) + 1 3x + 1 2.B-9 Find lim ∆x→0 ∆x √ √ x x− 8 2.B-10 lim is the derivative of what function for what value of x? x→2 x−2 Evaluate the limit, and check your answer by making a table on your h √ i √ calculator.

Ans: f 0 (2) for f (x) = x x; 3 2 2 2.B-11 Use the limit definition of the derivative to write an equation of the line tangent to f (x) = x2 at the point (−4, f (−4)). [Ans: y − 16 = −8 (x + 4)] 2.B-12 Each of the following limits is a derivative. For each limit, • state for what function it is a derivative, and at what x-value; • approximate the limit numerically, using a table on your calculator; • confirm your answer using algebraic techniques, without using a calculator. x2 − 1 x→−1 x + 1 x3 − 8 (b) lim x→2 x − 2 1 1 − 2 + x 2 (c) lim x→0 x √ √ x+2− 2 (d) lim x→0 x (a) lim

Ans: x2 at −1

Ans: x3 at 2

h Ans:

1 2+x

at 0

i

√ Ans: x + 2 at 0

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Mr. Budd, compiled January 12, 2011

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2.3

77

Limit Definition of Derivative as a Function

International Baccalaureate 7.1 Informal ideas of a limit and convergence.

f (x + h) − f (x) Definition of derivative as f 0 (x) = lim . Use of this definition h→0 h for differentiation of polynomials, and for justification of other derivatives.

Derivative interpreted as a gradient function and as rate of change. Finding equations of tangents and normals.

Textbook §2.1 Tangent Lines and Velocity and §2.2 The Derivative [15]

2.3.1

Derivative as a Function

1 , find f 0 (−1), f 0 (0), f 0 (1), f 0 (2), Example 2.3.1 For f (x) = 3x + 1 etc.

Instead of using the formula for the derivative at a point formula several times, we will find an expression for f 0 (x), and then simply plug in −1, 0, 1, and 2.

Definition 2.4 (Derivative (∆x or h form)). [15] The derivative function of f (x) is given by

f 0 (x) = lim

∆x→0

∆y f (x + ∆x) − f (x) f (x + h) − f (x) = lim = lim h→0 ∆x ∆x→0 ∆x h

This gives me a formula for finding the derivative at any generic point, x, instead of only at a specific point like 2 or −1.

Note that this formula can be adapted to find the derivative at a specific point by replacing x with a specific value, like 2 or −1.

Example 2.3.2 Find the point on the graph of y = 3 5 the tangent line is parallel to y = − x + .

4 4

1 where 3x + 1

First, let’s graph the function on our calculator, with the line − 43 x + 54 . See if you can spot the point where the tangent line is parallel to the line that we’ve graphed. As it turns out, there are two points that will work for this problem.

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SL Unit 2 (Limits) √ Example 2.3.3 Find the point on the graph of y = 7x + 4 where the tangent line is perpendicular to the line 10x + 7y = 5, i.e., where the normal line is parallel to 10x + 7y = 5.

Example 2.3.4 (adapted from [2]) A function g is defined for all real numbers and has the following property: g(a + b) − g(a) = 6a2 b + 6ab2 + 2b3 − 3b. Find g 0 (x).

Ans: 6x2 − 3 Example 2.3.5 (BC89) Let f be a function that is everywhere differentiable and that has the following properties. f (x) + f (h) for all real numbers h and x. f (−x) + f (−h) (ii) f (x) > 0 for all real numbers x. (i) f (x + h) =

(iii) f 0 (0) = −1. (a) Find the value of f (0).

[Ans: 1] 1 (b) Show that f (−x) = for all real numbers x. f (x) (c) Using part (b), show that f (x + h) = f (x)f (h) for all real numbers h and x. (d) Use the definition of the derivative to find f 0 (x) in terms of f (x).

[Ans: f 0 (x) = −f (x)]

Notation dy dy The derivative of a function f (x) might be written as f 0 (x) or . is called dx dx Liebniz notation, and is technically the derivative of y. Other notations include d f (x).

Dx f (x) and dx

Example 2.3.6 Find

2.3.2

dy 1 when y = dx x

Tangent Lines

Example 2.3.7 (adapted from BC97) Refer to the graph in Figure 2.5. The function f is defined on the closed interval [0, 8]. The graph Mr. Budd, compiled January 12, 2011

SL Unit 2, Day 3: Limit Definition of Derivative as a Function

79

of its derivative f 0 is shown. Think about what the graph is telling you. An equation of a line tangent to the graph of f is 3x − y = −1.

Figure 2.5:

(a) What is the x-coordinate of the point of tangency? (b) What is the y-coordinate of the point of tangency?

[Ans: (1, 4)]

Example 2.3.8 (adapted from [2]) If p(x) = (x + 2) (x + k) and if the line tangent to the graph of p at the point (4, p(4)) is perpendicular to the line 2x + 4y + 5 = 0, then k =

[Ans: −8]

Example 2.3.9 (adapted from [2]) If the line 3x − y + 5 = 0 is tangent in the second quadrant to the curve y = x3 + k, then k =

[Ans: 3] Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits)

2.3.3

Normal Lines

Normal lines are perpendicular to tangent lines. The slope of the normal line will be the negative reciprocal of the slope of the tangent line. (Recall that the slope of the tangent line is the . . .)

Example 2.3.10 The line that is normal to the curve y = x2 +2x−3 at (1, 0) intersects the curve at what other point? [19]

13 17 Ans: − , 4 16

Problems 2.C-1 Find g 0 (x) if g(a + b) − g(a) is defined as follows:

Ans: 3x2

(a) 3a2 b + 3ab2 + b3 (b) 2ab + b2 − 2b

[Ans: 2x − 2]

2.C-2 Suppose E(x) is a function for which E(x + h) − E(x) = E(x) (E(h) − 1) E(h) − 1 and lim = 1. Show that E 0 (x) = E(x), i.e., show that, if these h→0 h criteria are met for a function, then that function is its own derivative.

2.C-3 Use the limit definition of the derivative to find: (a) f 0 (x) if f (x) = 12 − x2 ; (b) (c) (d)

dy dx d dx dy dx 0

if y = 14 x3 ;

3x2 − 4x + 1 ; √ if y = x;

(e) f (x) if f (x) = 2x2 + x + k, where k is a constant;

1 d (f) dx x h i 1 12 Ans: −2x; 43 x2 ; 6x − 4; 2√ ; 4x + 1; − x x 2.C-4 The following limit is the derivative of what function: q √ 2 3 2 (x + ∆x) − 1 − 3 2x2 − 1 lim ∆x→0 ∆x √

Ans: 3 2x2 − 1 Mr. Budd, compiled January 12, 2011

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2.C-5 (BC90) Let f (x) = 12 − x2 for x ≥ 0 and f (x) ≥ 0. The line tangent to the graph of f at the point (k, f (k)) intercepts the x-axis at x = 4. What is the value of k?

[Ans: k = 2] 1 2.C-6 (adapted from AB97) At what point on the graph of y = x3 is the 4

tangent line parallel to the line 3x − 4y = 7?

Ans: 1, 41 2.C-7 (adapted from [2]) Let f (x) = 3x3 −4x+1. An equation of the line tangent to y = f (x) at x = 2 is [Ans: y = 32x − 47] √ 2.C-8 (adapted from [2]) Find the point on the graph of y = x between (4, 2) and (9, 3) at which the normal to the graph is perpendicular line to the

5 through (4, 2) and (9, 3).

Ans: 25 , 4 2 2.C-9 (adapted from [3]) If the graph of the parabola y = 2x2 + x + k is tangent to the line 3x + y = 3, then k = [Ans: 5]

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SL Unit 2 (Limits)

Mr. Budd, compiled January 12, 2011

SL Unit 2, Day 4: Basic Calculus of Polynomials

2.4

83

Basic Calculus of Polynomials

International Baccalaureate 7.1 Definition of derivative; Use of this definition for differentiation of polynomials; dy Familiarity with both forms of notation, and f 0 (x), for the first derivative. dx Derivative of xn . Derivative interpreted as a gradient function and as rate of change; Finding equations of tangents and normals.

7.2 Differentiation of a sum and a real multiple. The second derivative; Familiarity d2 y and f 00 (x), for the second derivative. with both forms of notation, dx2 7.6 Kinematic problems involving displacement, x, velocity, v, and acceleration, a.

Textbook §2.3 Computation of Derivatives: The Power Rule [15] Resources Exploration 3-4: “Algebraic Derivative of a Power Function” and Exploration 3-5: “ Deriving Velocity and Acceleration Data from Displacement Data” in [9].

2.4.1

Notation

Notation dy dy The derivative of a function f (x) might be written as f 0 (x) or . is called dx dx Liebniz notation, and is technically the derivative of y. Other notations include d Dx f (x) and f (x). dx Example 2.4.1 Find the f 0 (x) for f (x) = x4

Plug in 2, −1, and 3. Compare your answers to the answers you got by using the at x = c form of the derivative for each point.

Example 2.4.2 Find the derivative of x3 Example 2.4.3 Find the derivative of x3 + 1 Example 2.4.4 Find f 0 (x) for f (x) =

√

x and for f (x) =

√

x3 + 1

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SL Unit 2 (Limits)

Example 2.4.5 Find

2.4.2

dy 1 1 when y = and when y = 3 dx x x +1

Basic Properties of Derivatives

Derivative of a Constant If f (x) = k, then f 0 (x) = 0, i.e.,

d k=0 dx

Derivative of a Line If f (x) = mx + b, then f 0 (x) = m, i.e.,

d (mx + b) = m dx

Derivative of a Sum (or Difference)

If f (x) = g(x) + h(x), then f 0 (x) = g 0 (x) + h0 (x), i.e.,

d du dv (u + v) = + . dx dx dx

Derivative of a Scalar Multiple d ky = If f (x) = k g(x), where k is some constant, then f 0 (x) = k g 0 (x), i.e., dx dy k . dx Theorem 2.6. Linearity of Differentiation d [a f (x) + b g(x)] = a f 0 (x) + b g 0 (x) dx

Differentiation will be intuitive as long as you are adding, or multiplying by a constant. When things get more complicated than that, things will get more complicated than that.

Mr. Budd, compiled January 12, 2011

SL Unit 2, Day 4: Basic Calculus of Polynomials

2.4.3

85

Power Rule Power Rule d n x = nxn−1 dx

To differentiate a power function y = axb : • Multiply the coefficient by the old exponent; • Lower the exponent by one.

Example 2.4.6 (adapted from AB97) If f (x) = x3 + x − f 0 (−1) =

1 , then x

[Ans: 5] √ Example 2.4.7 (adapted from [2]) For f (x) = x, find the point on the graph, between (1, 1) and (4, 2), where the slope of the tangent line is equal to the slope of the line between (1, 1) and (4, 2).

9 3 Ans: , 4 2 √ Example 2.4.8 (adapted from AB97) Let f (x) = x 3 x. If the rate of change of f at x = c is thrice its rate of change at x = 1, then c =

[Ans: 27]

Example 2.4.9 (AB86) Let f be the function defined by f (x) = 7 − 15x + 9x2 − x3 for all real numbers x. Write an equation of the line tangent to the graph of f at x = 2.

[Ans: y − 5 = 9 (x − 2)] Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits) Example 2.4.10 (adapted from AB98) Write an equation (in slopeintercept form) for the line tangent to the graph of f (x) = x4 − x2 at the point where f 0 (x) = 1 [Ans: y = x − 1.055] √ 3 Example 2.4.11 Use the power rule to find lim

h→0

8+h−2 h

1 Ans: 12

2.4.4

Higher Order Derivatives

f 00 (x) means the derivative of f 0 (x), or the second derivative of f (x). f 000 (x) would be the third derivative of f (x).

Example 2.4.12 Find f 000 (x) if f (x) = x3 + 1.

[Ans: 6]

d dy In Liebniz notation, the second derivative is . To abbreviate this, dx dx 2 d y d d is multiplied: . In reality, is very much not we pretend like the 2 dx dx dx being multiplied, but it is a useful notation. (Think of the dx2 on the bottom d6 y 2 as (dx) .) The sixth derivative would be . dx6

Example 2.4.13 Find

d2 2t3 − 6t2 + 5 dt2 [Ans: 12t − 12]

2.4.5

Kinematics

Velocity is the derivative of displacement: v(t) = d0 (t) Mr. Budd, compiled January 12, 2011

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Acceleration is the derivative of the velocity: a(t) = v 0 (t) Acceleration is the derivative of the derivative of displacement. Acceleration is the second derivative of displacement: a(t) = v 0 (t) = s00 (t) Also: speed is the magnitude of velocity. In one dimension, this means that speed is the absolute value of velocity. speed(t) = |v(t)| Example 2.4.14 A particle moves along the x-axis so that its position at time t, where t is in seconds, is given by d(t) = 2t3 − 6t2 + 5, where d(t) is given in feet. What is the velocity when the acceleration is zero?

[Ans: −6 feet/sec]

Problems 2.D-1 (AB89) Let f be the function given by f (x) = x3 − 7x + 6. (a) Find the zeros of f .

[Ans: 1, 2, −3] (b) Write an equation of the line tangent to the graph of f at x = −1.

[Ans: y − 12 = −4 (x + 1)] x4 − 625 is the derivative of what function, at what point? Use the x→5 x − 5 power rule to evaluate the limit, without going through all that factor and cancel shtuff.

[Ans: 500] √ 4+h−2 2.D-3 As with the problem before, lim is a derivative at a point. h→0 h Evaluate the limit, by identifying at which point and for which function this is the derivative, then use the power rule instead of squeezing out an

h using conjugates.

Ans: 14 2.D-2 lim

2.D-4 [10] Misconception Problem Mae Danerror needs to find f 0 (3), where f (x) = x4 . She substitutes 3 for x, gets f (3) = 81, differentiates 81, and gets zero for the answer. Explain why she also gets zero for her grade.

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SL Unit 2 (Limits) x3 â&#x2C6;&#x2019; x2 â&#x2C6;&#x2019; 3x + 5, plot the graphs of f and f 0 on the 3 same screen. Show that each place where the f 0 graph crosses the x-axis corresponds to a high or low point on the f graph.

2.D-5 [10] For f (x) =

2.D-6 [14] The graphs of a function f and its derivative f 0 are given on the same coordinate axes in Figure 2.6. Label the graphs as f or f 0 and state the Figure 2.6: [14]

reasons for your choice.

2.D-7 [14] The graphs of a function f and its derivative f 0 are given on the same coordinate axes in Figure 2.7. Label the graphs as f or f 0 and state the Figure 2.7: [14]

reasons for your choice.

2.D-8 (adapted from [2])

d 2x3 ln e = dx

Ans: 6x2 Mr. Budd, compiled January 12, 2011

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2.D-9 [16] When an oil tank is drained for cleaning, there are V (t) = 100, 000 − 4000t + 40t2 gallons of oil left in the tank t minutes after the drain valve is opened. (a) At what average rate does oil drain from the tank during the first 20 minutes?

[Ans: 3200 gal/min] (b) At what rate does oil drain out of the tank 20 minutes after the drain valve is opened?

[Ans: 2400 gal/min] (c) Explain what V 00 (t) says about the rate at which oil is draining from the tank.

[Ans: rate of change is increasing, i.e., rate is getting less negative, i.e., rate of drainage is decreasing] 2.D-10 Find the derivative of x3 + x2 − 2, x3 + x2 − 1, x3 + x2 , and x3 + x2 + 58.7

2 Ans: 3x + 2x 2.D-11 Work backwards: find the functions, f (x), for the following derivatives: (a) f 0 (x) = 6x5 (b) f 0 (x) = 13x12 (c) f 0 (x) = 10x (d) f 0 (x) = 3x2 + 2x (e) f 0 (x) = m 0 if f 0 is the derivative Make up6 a word: of f , then f is the (blank) of f .

13 2 3 2 Ans: x ; x ; 5x ; x + x ; mx + b;

2.D-12 (MM99(2)) A ball is thrown vertically upwards into the air. The height, h metres, of the ball above the ground after t seconds is given by h = 2 + 20t − 5t2 , t ≥ 0. (a) Find the initial height above the ground of the ball (that is, its height at the instant when it is released.)

[Ans: 2 m] (b) Show that the height of the ball after one second is 17 metres. (c) At a later time the ball is again at a height of 17 metres. i. Write down an equation that t must satisfy when the ball is at a height of 17 metres. ii. Solve the equation algebraically.

[Ans: t = 1 s, t = 3 s] dh . (d) i. Find dt ii. Find the initial velocity of the ball (that is, velocity at the instant when it is released).

[Ans: v(0) = 20 m/s] iii. Find when the ball reaches its maximum height. [Ans: t = 2 s] iv. Find the maximum height of the ball.

[Ans: 22 m] Mr. Budd, compiled January 12, 2011

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SL Unit 2 (Limits)

2.D-13 [10] Find equations for the velocity, v, and the acceleration, a, of a moving

object if y = 5t4 −3t2.4 +7t is its displacement. Ans: v = 20t3 − 7.2t1.4 + 7, a = 60t2 − 10.08t0.4 2.D-14 [10] Car Problem Calvin’s car runs out of gas as it is going up a hill.

The car rolls to a stop, then starts rolling backward. As it rolls, its displacement, d(t) feet, from the bottom of the hill at t seconds since Calvin’s car ran out of gas is given by d(t) = 99 + 30t − t2 . (a) Plot graphs of d and d0 on the same screen. Use a window large enough to include the point where the d graph crosses the positive t-axis. Sketch the result.

(b) For what range of times is the velocity positive? How do you interpret this answer in terms of Calvin’s motion up the hill? [Ans: Velocity is positive for 0 ≤ t < 15. C

(c) At what time did Calvin’s car stop rolling up and start rolling back?

How far was it from the bottom of the hill at this time? [Ans: At 15 seconds, his car stopped. d (d) If Calvin doesn’t put on the brakes, when will he be back down at the bottom of the hill?

[Ans: t = 33 sec] (e) How far was Calvin from the bottom of the hill when the car ran out of gas?

[Ans: 99 feet from the bottom]

Mr. Budd, compiled January 12, 2011

Unit 3

Basic Differentiation 1. Finding Derivatives Using Limits 2. Power Rule - Derivatives and Antiderivatives of Polynomials 3. Product and Quotient Rules 4. Chain Rule 5. Tangent Lines 6. Newton’s Method AB

1. Derivatives Concept of the derivative.

• Derivative presented graphically, numerically, and analytically.

• Derivative interpreted as an instantaneous rate of change.

• Derivative defined as the limit of the difference quotient.

• Relationship between differentiability and continuity.

Derivative at a point.

• Slope of a curve at a point. Examples are emphasized, including points at which there are vertical tangents and points at which there are no tangents.

• Tangent line to a curve at a point and local linear approximation.

91

92

SL Unit 3 (Basic Differentiation) • Instantaneous rate of change as the limit of average rate of change.

• Approximate rate of change from graphs and tables of values.

Derivative as a function.

• Corresponding characteristics of graphs of f and f 0 .

• Relationship between the increasing and decreasing behavior of f and the sign of f 0 .

Applications of derivatives.

• Interpretation of the derivative as a rate of change in varied applied contexts, including velocity, speed, and acceleration.

Computation of derivatives.

• Knowledge of derivatives of basic functions, including power functions.

• Basic rules for the derivative of sums, products, and quotients of functions.

• Chain rule.

2. Integrals Techniques of antidifferentiation.

• Antiderivatives following directly from derivatives of basic functions.

Applications of antidifferentiation.

• Finding specific antiderivatives using initial conditions, including applications to motion along a line.

International Baccalaureate (MM 6.1) Differentiation of x 7→ xn , n ∈ Q Differentiation of sums of functions and real multiples of functions.

The chain rule for composite functions.

(MM 6.2) Applications of the first derivative to tangents, kinematical problems indv ds = v, and acceleration, = a. volving displacement, s, velocity, dt dt (MM 6.3) Indefinite integration as anti-differentiation.

Indefinite integrals of: xn ; n ∈ Q. Application to acceleration and velocity.

(MM 6.4) Anti-differentiation with a boundary condition to determine the constant term.

(MM 8.1) Further differentiation: the product and quotient rules; the second derivative.

(MM 8.5) The solution of equations using the Newton-Raphson method: xn+1 = Mr. Budd, compiled January 12, 2011

93 f (xn ) . Included: understanding of both graphical interpretations, i.e., conf 0 (xn ) f (x) sideration of y = f (x) intersecting y = 0 and y = x â&#x2C6;&#x2019; 0 intersecting y = x. f (x) Included: an appreciation of the rate of convergence. xn â&#x2C6;&#x2019;

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SL Unit 3 (Basic Differentiation)

Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 1: Antidifferentiation of Polynomials

3.1

95

Antidifferentiation of Polynomials

International Baccalaureate 7.4 Indefinite integration as anti-differentiation. Indefinite integral interpreted as a family of curves. Indefinite integral of xn .

7.5 Anti-differentiation with a boundary condition to determine the constant term. dy 1 Example: if = 3x2 + x and y = 10 when x = 0, then y = x3 + x2 + 10. dx 2 7.6 Kinematic problems involving displacement, x, velocity, v, and acceleration, a.

Textbook §4.1 Antiderivatives [15] Resources Exploration 5–2b: “A Motion Antiderivative Problem” in [9].

3.1.1

Notation of Antiderivatives

We need a notation for antiderivative that is as confusing as humanly possible, so we will use a notation already available to us for a completely unrelated concept: the definite integral.

The antiderivative of f 0 (x) is written as Z f 0 (x) dx Notice that the function, f 0 (x), that you are antidifferentiating is surrounded R by two things. On the right side is the integration sign, . The integral sign is used because antiderivatives are also known as indefinite integrals. On the right, is dx, which tells us what the variable is.

How will you be able to tell the difference between a definite integral, and an antiderivative?

3.1.2

Anti-Power Rule

Revisit problem 11 on page 89.

To anti-differentiate, we need to do the opposite of differentiation, in the opposite order.

Theorem 3.1. Anti-Power Rule Z xn+1 xn dx = +C n+1 Mr. Budd, compiled January 12, 2011

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SL Unit 3 (Basic Differentiation)

To antidifferentiate a power function y = axb : • Raise the exponent by one; • Divide by the new exponent; • Add an arbitrary constant C (the antiderivative of 0).

The antiderivative of a scalar multiple is the scalar multiple of the antiderivative.

Z Z k f (x) dx = k f (x) dx As long as you are multiplying a function by a constant, you can pull the constant out. For your own safety, never try to pull a variable out of the antiderivative: 63% of the time when you pull a variable out of the antiderivative, your pencil will explode. If it doesn’t happen the first time, know that you got lucky, and you are tempting the fates by trying it again.

The antiderivative of a sum is the sum of the antiderivatives: Z Z Z (f (x) + g(x)) dx = f (x) dx + g(x) dx Theorem 3.2. Linearity of Antidifferentiation Z Z Z [a f (x) + b g(x)] dx = a f (x) dx + b g(x) dx

Antidifferentiation is intuitive, so long as you are adding(/subtracting) functions, or multiplying by a constant. It will not be intuitive for multiplication or division.

Example 3.1.1 (adapted from AB93)

R

x3 + x

2

dx =

2x5 x3 x7 Ans: + + +c 7 5 3 Notice that we cannot get the correct answer by squaring the antiderivative of 3 4 x x2 4

+ 2 2 4 2 x3 + x, i.e., the answer is not x4 + x2 + C, or even + C. All 3 n we know how to antidifferentiate right now is power functions (ax ) and sums of power functions.

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g(x + h) − g(x) x3 − 2 , = h→0 h x3

Example 3.1.2 (adapted from [3]) If lim then g(x) could be equal to (A) −2x−4 (B) 6x−4 x3 + 1 x2 (D) x + x2 (C)

(E) 1 − 2x−3

x3 + 1 Ans: x2 Notice that we cannot find the antiderivative of a quotient by taking the quotient x4 − 2x +C of the antiderivatives: i.e., the answer is not 4 x4 4

Example 3.1.3 (BC86) For all real numbers x and y, let f be a function such that f (x + y) = f (x) + f (y) + 2xy and such that f (h) = 7. lim h→0 h (a) Find f (0). Justify your answer.

[Ans: 0] 0

(b) Use the definition of the derivative to find f (x). [Ans: 7 + 2x]

(c) Find f (x).

Ans: 7x + x2

3.1.3

Kinematics Revisited

dx Velocity is the derivative of displacement (v(t) = ), so that displacement is dt the antiderivative of velocity: Z x(t) = v(t) dt or

Z x(t) =

or

Z x(t) =

x0 (t) dt dx dt dt Mr. Budd, compiled January 12, 2011

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SL Unit 3 (Basic Differentiation)

dv d2 x Acceleration is the derivative of the velocity (a(t) = = 2 ), so that velocity dt dt is the antiderivative of acceleration: Z v(t) = a(t) dt or

Z v(t) =

or

Z v(t) =

v 0 (t) dt dv dt dt

Example 3.1.4 (adapted from AB93) The acceleration of a particle moving along the x-axis at time t is given by a(t) = 6t − 4. If the velocity is 18 when the t = 3 and the position is 11 when t = 1, then the position x(t) =

Ans: t3 − 2t2 + 3t + 9 Example 3.1.5 Exploration 3-9: “Displacement and Acceleration from Velocity”

3.1.4

General vs. Particular Solutions

t1.6 In antidifferentiating v(t) = 50+6t0.6 [9], the displacement is d(t) = 50t+6 + 1.6 C. Notice the +C, which is adding some arbitrary constant, the antiderivative of +0.

This d(t) with the +C is a family of parallel curves. There are an infinite number of curves in this family, corresponding to the infinite number of possibilities of C. Note that these curves are considered parallel because their slopes are the same for every x value. If your antiderivative includes +C, it is called a general solution.

If you want one function d(t) instead of a family of functions, what you want is a particular solution, instead of the general solution. In order to narrow the general solution to a particular solution, you need to incorporate some more information into your answer. This is done by solving for C, after plugging in some known point (t, d(t)) into the formula for d(t).

For example, if v(t) = 50 + 6t0.6 , then we can antidifferentiate to get the general solution for the displacement, d(t) = 50t + 3.75t1.6 + C. If we want a particular Mr. Budd, compiled January 12, 2011

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solution, we will have to use some more information. In this particular example, for instance, we may know that the displacement at time 0 is 100. Then we solve 1.6 100 = 50 (0)+3.75 (0) +C, to find that C = 100. Thus, our particular solution would be d(t) = 50t + 3.75t1.6 + 100.

If the point used is (0, d(0)), this is said to be an initial value. Antidifferentiation problems that ask you to find a particular solution using an initial value are called IVP’s, or Initial Value Problems.

3.1.5

Introduction to Slope Fields

Slope fields give you a sense of the direction or current of the general solutions.

To find the particular solution, start at the known point, and draw a curve to the left and right that is always parallel to the slopes as seen in the slope field. dy = −2x + 2 at twentydx 1 3 five points using x-coordinates of 0, 2 , 1, 2 , 2, and y-coordinates of −1, − 12 , 0, 12 , 1. Sketch an antiderivative within your slope field.

Example 3.1.6 Draw a slope field for

Example 3.1.7 With a window of [0, 4.7] × [−3.1, 3.1], use the dy BIGSLOPE program to draw a slope field of = −4x + 8. Then dx R use the 5. Draw Solution option to graph (−4x + 8) dx for several values of C (−10, −8, −6, −4).

Problems 3.A-1 [10] Find a function whose derivative is given. That is, write the general equation for the antiderivative. (a) f 0 (x) = 7x6 (b) f 0 (x) = x5 (c) f 0 (x) = x−9 7

(d) f 0 (x) = 36x 2

1 6 1 −8 9 7 Ans: x + C, x + C, − x + C, 8x 2 + C 6 8 3.A-2 [10] Find the particular function f (x) that has the given function f 0 (x) for its derivative and contains the given point.

Mr. Budd, compiled January 12, 2011

100

SL Unit 3 (Basic Differentiation) (a) f 0 (x) = x4 and f (1) = 10 (b) f 0 (x) = x2 − 8x + 3 and f (−2) = 13

113 1 3 1 5 2 Ans: 5 x + 9.8, 3 x − 4x + 3x + 3

3.A-3 [10] Derivative and Antiderivative Problem Let g 0 (x) = 0.6x. (a) Find the general equation

for the antiderivative, g(x).

Ans: g(x) = 0.3x2 + C (b) Find the particular equation for g(x) in each case.

Ans: 0.3x2

Ans: 0.3x2 + 3

Ans: 0.3x2 + 5

i. g(0) = 0 ii. g(0) = 3 iii. g(0) = 5

(c) Plot the graph of g 0 (x) and the three graphs for g(x) on the same screen, then sketch the results. Why are the three graphs of g(x) called a family of functions?

3.A-4 [2] If functions f and g are defined so that f 0 (x) = g 0 (x) for all real numbers x with f (1) = 2 and g(1) = 3, then the graph of f and the graph of g (A) intersect exactly once; (B) intersect no more than once; (C) do not intersect; (D) could intersect more than once; (E) have a common tangent at each point of tangency.

[Ans: C] Z

3.A-5 (adapted from AB93) x2 + 2

4 3 x5 + x + 4x + C Ans: 5 3 Z 2 3.A-6 3 (ice) d (ice)

2

dx =

[Ans: iceberg]

d2 y is the second derivative, what do you suppose would be the meaning dx2 −1 d y of ? dx−1

3.A-7 If

3.A-8 (BC86) For all real numbers x and y, let f be a function such that f (x + f (h) = 7. y) = f (x) + f (y) + 2xy and such that lim h→0 h Mr. Budd, compiled January 12, 2011

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(a) Find f (0). Justify your answer. (b) Use the definition of the derivative to find f 0 (x). (c) Find f (x).

3.A-9 [10] Displacement Problem Ann Archer shoots an arrow into the air. Let d(t) be its displacement above the ground at time t seconds after she shoots it. From physics she knows that the velocity is given by d0 (t) = 70 − 9.8t. (a) Write the general equation for d(t).

Ans: 70t − 4.9t2 + C (b) Write the particular equation for d(t), using the fact that Ann is standing on a platform that puts the bow 6 m above the ground when she shoots the arrow.

Ans: 70t − 4.9t2 + 6 (c) How far is the arrow above the ground when t = 5? When t = 6?

When t = 9? How do you explain the relationship among the three answers?

[Ans: 233.5 m, 249.6 m, 239.1 m] (d) When is the arrow at its highest? How high is it at that time?

[Ans: 256 m, at about t = 7.1 sec] 3.A-10 (adapted from [2]) At t = 0, a particle starts at the origin with a velocity of 6 feet per second and moves along the x-axis in such a way that at time t its acceleration is 24t2 feet per second per second. Through how many feet does the particle move during the first 2 seconds?

[Ans: 44 feet] 3.A-11 (adapted from [2]) The acceleration, a(t), of a body moving in a straight line is given in terms of time t by a(t) = 4 − 6t. If the velocity of the body is 20 at t = 0 and if s(t) is the distance of the body from the origin at time t, what is s(2) − s(1)?

[Ans: 19] 3.A-12 With a window of [0, 4.7] × [−3.1, 3.1], use the BIGSLOPE program to dy 1 1 draw a slope field of = √ , i.e., put √ into Y1 . After the slope dx √ x x field is drawn, then graph 2 x + C for C = −3, −2, −1 using the 5. Draw Solution option.

3.A-13 With a window of [0, 2.35] × [−3.1, 3.1], use the BIGSLOPE program to

dy draw a slope field of = 4 (x − 1). Then draw solution to 2 x2 − 2x +C dx for C = −1, 1, 3.

3.A-14 With a window of [−2.35, 2.35] × [−3.1, 3.1], use the BIGSLOPE program

R 2 dy to draw a slope field of = x2 −1. Then draw solutions to x − 1 dx dx for C = −2, 0, 2.

Mr. Budd, compiled January 12, 2011

102

SL Unit 3 (Basic Differentiation) dy = 2x − 2 at twenty-five points using dx 3 1 x-coordinates of 0, 2 , 1, 2 , 2, and y-coordinates of −1, − 21 , 0, 12 , 1. Sketch an antiderivative within your slope field.

3.A-15 Draw by hand a slope field for

Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 2: Product and Quotient Rules

3.2

103

Product and Quotient Rules

International Baccalaureate 7.2 The product and quotient rules. ds 7.6 Kinematic problems involving displacement, s, velocity, = v, and acceleradt dv tion, = a. dt Textbook §2.4 The Product and Quotient Rules [15] Resources Exploration 4-2: “Derivative of a Product” in [9]. §3.3 The Product and Quotient Rules in [20].

3.2.1

Product Rule

Example 3.2.1 [20] In this exercise we estimate the rate at which the total personal income is rising in the Miami-Ft. Lauderdale metropolitan area. In July, 1993, the population of this area was 3,354,000 and the population was increasing at roughly 45,000 people per year. The average annual income was $21,107 per capita, and this average was increasing at about $1900 per year (well above the national average of about $660 yearly). Use the Product Rule and these figures to estimate the rate at which total personal income was rising in Miami-Ft. (a) What was the total personal income in July, 1993? (b) Specifically what factors contribute to the growth of total personal income. (c) Being mindful of units, quantify the factors you found in part (b), and estimate the rate at which total personal income was rising in Miami-Ft. Lauderdaule in July, 1993.

[Ans: $70.8 billion; ; $7.322 billion per year] Example f (x) = x7 + 1, g(x) = x5 − 4, and p(x) = 3.2.2 Consider 7 5 12 x + 1 x − 4 = x − 4x7 + x5 − 4. (a) Write an equation of the line tangent to f (x) at x = 1. Write it in the form y = k + m (x − 1), i.e., treat (x − 1) as a single entity, which you will leave alone.

Mr. Budd, compiled January 12, 2011

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SL Unit 3 (Basic Differentiation) (b) Write an equation of the line tangent to g(x) at x = 1. Write it in the form y = a0 + a1 (x − 1). (c) Write an equation of the line tangent to p(x) at x = 1. As before, treat (x − 1) as a single entity. (d) Multiply the tangent lines at x = 1 for f (x) and g(x), keeping (x − 1) as a single variable unto itself: don’t worry about distributing it. The only like terms you need to combine are the (x − 1) terms. Your answer should look like y = a0 + 2 a1 (x − 1) + a2 (x − 1) . What do you see?

Example 3.2.3 d 7 d 4 d 11 x 6= x · x dx dx dx (b) How could we squeeze the real derivative out of x4 and x7 .

Think about: how can we get the correct exponents, and the correct coefficients. (a) Show that

Product Rule d du dv (u · v) = v+u dx dx dx If h(x) = f (x)g(x), then h0 (x) = f 0 (x)g(x) + f (x)g 0 (x)

Derivative of First · Second + First · Derivative of Second.

Note that the order can be rearranged: Derivative of First · Second + Derivative of Second · First.

First · Derivative of Second + Derivative of First · Second.

First · Derivative of Second + Second · Derivative of First.

Do whichever way you remember best.

Example 3.2.4 Use linear approximations of f (x) and g(x) at x = c to demonstrate the product rule.

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Example 3.2.5 Use the limit definition of the derivative to show that, if p(x) = f (x)g(x), then p0 (x) = f 0 (x)g(x) + f (x)g 0 (x).

Example 3.2.6 Suppose that f is the function shown in Figure 3.1 and that h(x) = x2 f (x) [16]. Evaluate h0 (2) and h0 (4) Figure 3.1: Graph of f [16]

[Ans: −16, 64] 2 Example 3.2.7 If f (x) = x3 + 1 , find f 0 (x)

Example 3.2.8 Find

3 i d h 3 x +1 . dx

2 This can be done two ways, one using the derivative of x3 + 1 , and splitting

3 2 x3 + 1 into x3 + 1 x3 + 1 .

Another way is to use a variation of the product rule for differentiating the product of three functions. For the product of three functions, u, v, and w: du dv dw d (u · v · w) = ·v·w+u· ·w+u·v· dx dx dx dx For e(x) = f (x)g(x)h(x), e0 (x) = f 0 (x)g(x)h(x) + f (x)g 0 (x)h(x) + f (x)g(x)h0 (x) Mr. Budd, compiled January 12, 2011

106

SL Unit 3 (Basic Differentiation)

Examine this formula and see if you can notice how the pattern for two functions has been expanded for three functions.

Once you get your answer, simplify and factor it. In the back of your mind, be looking for a method to go straight to the answer very quickly.

Example 3.2.9 (adapted from [2]) If p(x) = (x + 2) (x + k) and if the line tangent to the graph of p at the point (4, p(4)) is perpendicular to the line 2x + 4y + 5 = 0, then k = [Ans: −8] Example 3.2.10 Use the product rule to prove the power rule (for n = 1, 2, . . . using mathematical induction.

3.2.2

Quotient Rule Quotient Rule du dv d u v dx − u dx = dx v v2

If h(x) =

f (x) , then g(x) h0 (x) =

or

(v 6= 0)

g(x)f 0 (x) − f (x)g 0 (x) 2

[g(x)]

LoDeHi − HiDeLo LoLo

Because of the subtraction, it is important that you keep things in this order (unlike the product rule, where order is not really important).

Example 3.2.11 Use the limit definition of the derivative to prove the quotient rule.

Example 3.2.12 Use linear approximations of f (x) and g(x) at x = c to demonstrate the product rule.

Mr. Budd, compiled January 12, 2011

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Figure 3.2:

2 x3 + 1 dy Example 3.2.13 If y = , find .

3 x +1 dx

Notice that this problem has a quick and simple answer, if you cancel x3 + 1, and take the derivative of that to be 3x2 . Confirm that the quotient rule works.

Example 3.2.14 Find

d dx

1 . x3 + 1

2 3 Compare your answer with the derivative you get for x3 + 1 and x3 + 1 and be looking for the pattern, while thinking of the derivatives of x2 , x3 , and 1 x.

Example 3.2.15 from 2000 AP Calculus AB-2. Two runners, A and B, run on a straight racetrack for 0 â&#x2030;¤ t â&#x2030;¤ 10 seconds. The graph in Figure 3.2, which consists of two line segments, shows the velocity, in meters per second, of Runner A. The velocity, in meters per second, of Runner B is given by the function v defined by v(t) = 24t . Find the acceleration of Runner B at time t = 2 seconds.

2t + 3 Indicate units of measure.

10 = 3.333 meters/sec2 3

72 72

Runner B: a(2) = v 0 (2) = = = 1.469 meters/sec2 2

49 (2t + 3) t=2

Runner A: acceleration =

Mr. Budd, compiled January 12, 2011

108

SL Unit 3 (Basic Differentiation)

Tangent Lines When writing the equation of a line, use the point-slope form: y − y1 = m (x − x1 ) When talking about m, the slope of the tangent line, you should immediately be thinking about the derivative. m = f 0 (x1 ) y − y1 = y 0 (x1 ) (x − x1 )

Example 3.2.16 (adapted from AB93) An equation of the line 3x − 2 tangent to the graph of y = at the point (−1, −5) is 2x + 3

[Ans: y + 5 = 13 (x + 1)]

Problems 3.B-1 (adapted from AB98) Let f and g be differentiable functions with the following properties: (a) g(x) < 0 for all x (b) f (1) = π − 1 If h(x) = f (x)g(x) and h0 (x) = f (x)g 0 (x), then f (x) =

[Ans: π − 1]

3.B-2 [20] (a) If F (x) = f (x)g(x), where f and g have derivatives of all orders, show that F 00 = f 00 g + 2f 0 g 0 + f g 00 (b) Find a similar formulas for F 000 .

3.B-3 Use the Product Rule to show that d d 2 [f (x)] = [f (x)f (x)] = 2f (x)f 0 (x) dx dx 3.B-4 [20] Mr. Budd, compiled January 12, 2011

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(a) Use the Product Rule twice to prove that if f , g, and h are differentiable, then d (f gh) = f 0 gh + f g 0 h + f gh0 dx (b) Taking f = g = h in part 4a, show that d 3 2 [f (x)] = 3 [f (x)] f 0 (x) dx 3.B-5 [16] Suppose that f (1) = 2 and f 0 is the function shown in Figure 3.3. Let m(x) = x3 f (x).

Figure 3.3: Graph of f 0 [16]

(a) Evaluate m0 (1).

[Ans: 10]

(b) Show that m is increasing at 2. (c) Estimate m00 (1).

3.B-6 (AB85) Let f be the function given by f (x) = (a) Find the domain of f .

[Ans: 34] 2x − 5 . x2 − 4 [Ans: {x|x ∈ R ∩ x 6= ±2}]

(b) Write an equation for each vertical and each horizontal asymptote for the graph of f .

[Ans: y = 0, x = 2, x = −2] h i Ans: −2(x−4)(x−1) (c) Find f 0 (x).

2 2 (x −4) (d) Write an equation for the line tangent to the graph of f at 5the point

(0, f (0)).

Ans: y − 4 = − 21 x 3.B-7 Differentiable functions f and g have the values shown in the table. [13] Mr. Budd, compiled January 12, 2011

110

SL Unit 3 (Basic Differentiation) x 0 1 2 3

f0 1 2 3 4

f 2 3 5 10

g 5 3 1 0

g0 −4 −3 −2 −1

(a) If A(x) = f (x) + 2g(x), then A0 (3) =

[Ans: 2]

0

(b) If B(x) = f (x) · g(x), then B (2) =

[Ans: −7]

Ans: 13 25

f (x) , then K 0 (0) = g(x) 1 (d) If D(x) = , then D0 (1) = g(x) (c) If K(x) =

3.B-8 (adapted from [2]) If g(x) =

Ans:

x+2 , then g 0 (−2) = x−2

3.B-9 (adapted from [2]) Consider the function f (x) = 0.75. The value of a is 3.B-10 (adapted from [2]) If y =

dy 4 , then = 3 + x2 dx

1 3

Ans: − 41

6x for which f 0 (0) = a + x3 [Ans: 8] h i −8x Ans: (3+x 2 )2

1 is called a witch of Maria Agnesi. Find 1 + x2

1 an equation of the tangent line to this curve at the point −1, . On 2 your calculator, graph the curve and the tangent line on the same screen.

Ans: y = 21 x + 1

3.B-11 [20] The curve y =

3.B-12 [20] If f and g are the functions whose graphs are shown in Figure 3.4, let f (x) u(x) = f (x)g(x) and v(x) = . g(x) Figure 3.4:

(a) Find u0 (1).

[Ans: 0] Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 2: Product and Quotient Rules (b) Find v 0 (5).

111

Ans: − 32

3.B-13 [16] Suppose that f is the function shown in Figure 3.1. Let m(x) = f (x) . Evaluate m0 (0).

[Ans: −4] x2 + 1

Mr. Budd, compiled January 12, 2011

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SL Unit 3 (Basic Differentiation)

Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 3: Chain Rule

3.3

113

Chain Rule

International Baccalaureate 7.2 The chain rule for composite functions.

Textbook §2.5 The Chain Rule [15] Resources §3-7 Derivatives of Composite Functions- The Chain Rule in Foerster [10]. Exploration 3-7:“Rubber-Band Chain Rule Problem” in [9]. §3-5 The Chain Rule in Stewart [20]. §3.6 New Derivatives from Old: The Chain Rule in Ostebee & Zorn [16].

Example 3.3.1 f (x) 2 x3 + 1 3 x3 + 1 4 x3 + 1 n x3 + 1 n (g(x))

f 0 (x)

6x2 x3 + 1 = 2 x3 + 1 · 3x2 2 2 9x2 x3 + 1 = 3 x3 + 1 · 3x2

3 3 12x2 x3 + 1 = 4 x3 + 1 · 3x2

Example 3.3.2 Using a generalized product rule for the product n of n things, find the derivative of (g(x)) Power Chain Rule d n n−1 (g(x)) = n (g(x)) · g 0 (x) dx

d n du u = nun−1 dx dx where u = g(x) Derivative of the Outside · Derivative of the Inside Example 3.3.3 Find 4 d • x3 − x + 1 dx d 5 • (2x + 1) dx Mr. Budd, compiled January 12, 2011

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SL Unit 3 (Basic Differentiation)

Product and Quotient Rules did not vanish Example 3.3.4 [18] Differentiate y = (2x + 1)

h

Ans: 2 (2x + 1)

4

x2 − x + 1

5

x3 − x + 1

3

4

17x3 + 6x2 − 9x + 3

i

Example 3.3.5 [18] Find the derivative of the function

g(t) =

t−2 2t + 1

9

h Ans:

45(t−2)8 (2t+1)10

i

Example 3.3.6 f (x) 1 x 1 x3 + 1 1 g(x) √ x √ x3 + 1 p g(x) F (x) F x3 + 1 F (g(x))

−

f 0 (x) 1 − 2 x

3x2 (x3

+ 1)

2

=−

(x3

1 + 1)

2

· 3x2

1 √ 2 x

F 0 (x)

Chain Rule dy dy du = · dx du dx For h(x) = F (g(x)), h0 (x) = F 0 (g(x)) · g 0 (x) Derivative of the Outside · Derivative of the Inside Mr. Budd, compiled January 12, 2011

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It’s called the chain rule because you can chain as many together as you need: dy dy du dv dw = dx du dv dw dx Example 3.3.7 The function F is defined by F (x) = G (x − G(x)) Figure 3.5: from Best & Lux [2]

where the graph of the function G is shown in Figure 3.5. Find F 0 (7) [2]

Ans: − 23 Example 3.3.8 Use the product rule, together with the chain rule, to prove the quotient rule, by finding

d 1 f (x) · dx g(x) Example 3.3.9 Let h(x) = f (g(x)). Use the information about f and g given in the table below to fill in the missing information about h and h0 [16]. x 1 2 3 4

f (x) 1 2 4 3

f 0 (x) 2 1 3 4

g(x) 4 3 1 2

g 0 (x) 3 4 2 1

Example 3.3.10 Given that f 0 (x) = find F 0 (x) if F (x) = f (g(x)) [19].

x2

h(x)

h0 (x)

√ x and g(x) = 3x − 1, +1

Mr. Budd, compiled January 12, 2011

116

SL Unit 3 (Basic Differentiation)

Ans:

1 2x

Problems 3.C-1 [13] Differentiable functions f and g have the values shown in the table. x 0 1 2 3 (a) If H(x) =

f 2 3 5 10

f0 1 2 3 4

g0 −4 −3 −2 −1

g 5 3 1 0

h

p f (x), then H 0 (3) =

(b) If M (x) = f (g(x)), then M 0 (1) =

√2 10

i

[Ans: −12]

0

3

Ans:

(c) If P (x) = f (x ), then P (1) =

[Ans: 6]

3.C-2 [3] Let the function f be differentiable on the interval [0, 2.5] and define g by g(x) = f (f (x)). Use the table to estimate g 0 (1). x f (x)

0.0 1.7

0.5 1.8

1.0 2.0

1.5 2.4

2.0 3.1

2.5 4.4 [Ans: 1.2]

3.C-3 [20] If f and g are the functions whose graphs are shown in Figure 3.6, let u(x) = f (g(x)), v(x) = g(f (x)), and w(x) = g(g(x)). Find each derivative, if it exists. If it does not, explain why.

Figure 3.6: [20]

(a) u0 (1)

Ans:

3 4

(b) v 0 (1)

[Ans: nonexistent, g 0 (2) DNE]

(c) w0 (1)

[Ans: −2] Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 3: Chain Rule

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Figure 3.7: Graph of f [16]

3.C-4 [16] Suppose that f is the function shown in Figure 3.7 and that g(x) = f (x2 ). √

(a) For which values of x is g 0 (x) = 0?

Ans: 2, 0 (b) Is g increasing or decreasing at −1?

[Ans: increasing] √ √

(c) Is g 0 positive or negative over the interval 2, 5 ? [Ans: positive]

3.C-5 [3] The graphs of functions f and g are shown in Figure 3.8. If h(x) = Figure 3.8: from Best & Lux [3]

f (g(x)), which of the following statements are true about the function h?

I. h(2) = 5.

II. h is increasing at x = 4.

III. The graph of h has a horizontal tangent at x = 1.

[Ans: II and III only] 3.C-6 (BC98) If f and g are twice differentiable and if h(x) = f (g(x)), theni h 2 00 00 h (x) = Ans: f (g(x)) [g 0 (x)] + f 0 (g(x))g 00 (x) Mr. Budd, compiled January 12, 2011

118

SL Unit 3 (Basic Differentiation)

3.C-7 [2] A particle moves along the x-axis so that at time t, t >= 0, its position 3 is given by x(t) = (t + 1) (t − 3) . Findha formula for the velocity, v(t), andi the acceleration, a(t), of the particle.

2

Ans: 4t (t − 3) ; 12 (t − 3) (t − 1) 5

4

3.C-8 (adapted from AB93) Find the derivative of f (x) h = (2x − 3) (7x + 2) . i 4 3 0 At how many different values of x will f (x) be 0? Ans: 2 (2x − 3) (7x + 2) (63x − 32); Three

√ 9x − 2 3.C-9 (adapted from AB97) If f (x) = x 6x − 2, then f 0 (x) = Ans: √ 6x − 2 3.C-10 [16] Suppose that f (1) = 2, that f 0 is the function shown in Figure 3.9, and that k(x) = f (x3 ). Evaluate k 0 (−1).

[Ans: 12] Figure 3.9: Graph of f 0 [16]

3.C-11 [2] Suppose that g is a function with the following two properties: g(−x) = g(x) for all x and g 0 (a) exists. [I.e., g is an even, differentiable function.]

Find g 0 (−x).

[Ans: −g 0 (x)] 3.C-12 [20] Use the table to estimate the value of h0 (0.5), where h(x) = f (g(x)).

[Ans: −17.4] x f (x) g(x)

0 12.6 0.58

0.1 14.8 0.40

0.2 18.4 0.37

0.3 23.0 0.26

0.4 25.9 0.17

0.5 27.5 0.10

0.6 29.1 0.05

3.C-13 [20] If f is the function whose graph is shown, let h(x) = f (f (x)) and g(x) = f (x2 ). Use the graph of f to estimate the value of each derivative. (a) h0 (2) 0

(b) g (2) Z 3.C-14 Find f 0 (g(x)) g 0 (x) dx

[Ans: 0.64] [Ans: 9] [Ans: f (g(x)) + C] Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 3: Chain Rule

119

Figure 3.10: [20]

Z 3.C-15 Find

13 x3 + 1

12

3x2 dx

i h 13 Ans: x3 + 1 +C

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SL Unit 3 (Basic Differentiation)

Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 4: Differentiation of Trigonometric Functions

3.4

121

Differentiation of Trigonometric Functions

International Baccalaureate sin θ 7.1 Informal ideas of limit and convergence, including the result lim = 1. θ→0 θ

f (x + h) − f (x) Definition of derivative as f 0 (x) = lim . h→0 h Derivative of sin x, cos x, tan x.

Textbook §2.6 Derivatives of Trigonometric Functions. Oxford book: §9.1 sin x and cos x and §9.2 Differentiating tan x [15]

3.4.1

Trigonometric Derivatives

Derivative of Sine Example 3.4.1 Let f (x) = sin x (a) Use several symmetric difference quotients of smaller and smaller h to estimate f 0 (0) (b) Explain why you are approximating lim

h→0

sin (0 + h) − sin 0 and h

sin x x→0 x (c) Find the same symmetric difference quotients using nDeriv(Y1,X,0,H) cos h − 1 (d) Can you make a difference quotient that would be lim ? h→0 h (e) Estimate f 0 at several other points to make a sketch of f 0 (x) lim

Example 3.4.2 Find the derivative of f (x) = sin x: (a) If f (x) = sin x, sketch f 0 (x). d sin x is. dx (c) Prove or disprove your conjecture:

(b) Make a conjecture about what

(a) Start with the definition of the derivative as the limit of an average rate of change. (b) Use the sum of angle formula for sine: sin (α + β) = sin α cos β+ cos α sin β. (c) Use the known limits for sinh h and cos hh−1 as h approaches 0.

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122

SL Unit 3 (Basic Differentiation) (d) Give graphic evidence that the derivative is the same as your conjecture by comparing the graphs of nDeriv(Y1 , X, X) and your proposed derivative using the happy bouncing ball.

d sin x = cos x dx

Example 3.4.3 Make and test (with nDeriv and the happy bouncing ball) conjectures for the derivatives of (a) sin x + 2: Think about what the 2 does to the graph. How does this effect the slope? (b) 2 sin x (c) sin 2x (d) sin x2

Why should you have predicted the last one?

Using the Chain Rule d sin (g(x)) = cos (g(x)) g 0 (x) dx

Derivative of Cosine Example 3.4.4 Graph the derivative of cos x. Make, test, then prove a conjecture for the derivative of cos x.

d cos x = â&#x2C6;&#x2019; sin x dx Using the Chain Rule d cos (g(x)) = â&#x2C6;&#x2019; sin (g(x)) g 0 (x) dx Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 4: Differentiation of Trigonometric Functions

123

Derivative of Tangent Example 3.4.5 Find

d tan x. dx

d tan x = sec2 x dx Using the Chain Rule: d tan (g(x)) = sec2 (g(x)) g 0 (x) dx

Derivative of Secant Example 3.4.6 Find

d sec x. dx

Use the Quotient Rule: d sec x = sec x tan x dx Using the Chain Rule: d sec (g(x)) = sec (g(x)) tan (g(x)) g 0 (x) dx

Example 3.4.7 (adapted from AB93) A particle moves along a line so that at time t, where 0 ≤ t ≤ π, its position is given by t2 s(t) = −3 cos t − + 10. What is the velocity when its acceleration 2 is zero?

[Ans: 1.597]

Example 3.4.8 Let f (x) = sin(2x). Find the intersection of the π π two lines tangent to the graph of f at x = and at x = .

6 3 Mr. Budd, compiled January 12, 2011

124

SL Unit 3 (Basic Differentiation) h

Ans:

Example 3.4.9 (adapted from AB97)

π π 4 , 12

√

+

3 2

i ≈ (0.785, 1.128)

d cos3 x2 = dx

Ans: −6x cos2 x2 sin x2

Example 3.4.10 (BC97) The position of an object attached to a 1 1 spring is given by y(t) = cos(5t) − sin(5t), where t is time in 6 4 seconds. In the first 3 seconds, how many times is the velocity of the object equal to 0? [You may use a calculator.]

[Ans: Five] Example 3.4.11 (adapted from [2]) Administrators at Massachusetts General Hospital believe that the hospital’s expenditures E(B), measured in dollars, are a function of how many beds B are in use with 2

E(B) = 7000 + (B + 1) .

On the other hand, the number of beds B is a function of time, t, measured in days, and it is estimated that

t B(t) = 25 sin + 50.

10 At what rate are the expenditures decreasing when t = 100?

[Ans: $157/day]

Checking Antiderivatives Example 3.4.12 (adapted from AB97) Which of the following are antiderivatives of f (x) = sin x cos x? sin2 x 2 cos2 x II. F (x) = − 2 cos(2x) III. F (x) = − 4 I. F (x) = −

[Ans: II and III only] Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 4: Differentiation of Trigonometric Functions

125

Problems

3π 3π cos + h − cos 2 2 3.D-1 (adapted from AB acorn ’02) What is lim ? h→0 h [Hint: This is not really a limit problem. It is a derivative problem disguised as a limit problem.]

[Ans: 1] 3.D-2 (adapted from [2]) If g(x) = x2 − sin x, then lim

h→0

[Ans: 2x − cos x] 3.D-3 (adapted from AB98) If f (x) = tan(2x), then f 0

π 8

g(x + h) − g(x) = h =

[Ans: 4]

3.D-4 (adapted from AB98) Find equation of the line tangent to the graph of y = 2x + cos (πx) at the point (1, 1). Graph the function and its tangent line on your calculator. Find the equation of the line tangent to the graph of y = 2x + cos (πx) at the point (−1, −3).

[Ans: y = 2x − 1] 2

3.D-5 (adapted from AB93) If f (x) = (x − 1) cos x, then f 0 (0) =

[Ans: −2]

3.D-6 (adapted from [2]) For x 6= 0, the slope of the tangent to y = x sin x equals zero whenever (A) tan x = −x 1 (B) tan x = x (C) tan x = x (D) sin x = x (E) cos x = x [Ans: tan x = −x]

Ans: 2π 3

3.D-7 (AB93) The period of 2 cos(3x) is 3.D-8 (adapted from [2]) If y = 2 sin x cos x =, then y 0 =

[Ans: 2 cos (2x)]

dy = Ans: −9 cos2 (3x) sin(3x) dx π

3.D-10 (adapted from AB acorn ’02) If f (x) = sin2 − x , then f 0 (0) = 4 [Ans: −1] 3.D-9 (adapted from [2]) If y = cos3 (3x), then

2 2 3.D-11 (adapted h π π i from [2]) For f (x) = cos x and g(x) = −0.5x on the interval − , , the instantaneous rate of change of f is greater than the instan2 2 taneous rate of change of g for which value of x? [Note: you should use a calculator.]

Mr. Budd, compiled January 12, 2011

126

SL Unit 3 (Basic Differentiation) (A) −1.5 (B) −1.2 (C) −0.8 (D) 0 (E) 0.9 [Ans: −0.8]

3.D-12 (adapted from [2]) lim

h→0

sin (x + h) − sin x h

=

[Ans: cos x]

3.D-13 (adapted from [2]) At how many points on the interval −2π ≤ x ≤ 2π π does the tangent to the graph of the curve y = x sin x have slope ?

2 [Ans: Three] 3.D-14 (from Stewart [20]) Very Important Preview of Series If You Ever Plan to Take Elite Ninja Math, or Calc 2: Let f (x) = sin x. (a) Find values of the constants a and b so that the linear function L(x) = a + bx has the properties L(0) = f (0) and L0 (0) = f 0 (0). (b) Find the value of the constant c for which the quadratic function Q(x) = L(x) + cx2 has the properties Q(0) = f (0), Q0 (0) = f 0 (0), and Q00 (0) = f 00 (0). (c) Find the value of the constant d for which the cubic function C(x) = Q(x) + dx3 has the properties C(0) = f (0), C 0 (0) = f 0 (0), C 00 (0) = f 00 (0), and C 000 (0) = f 000 (0). (d) Plot f , L, Q, and C on the same axes over the interval [−4, 4].

Ans: L(x) = x = Q(x); C(x) = x − 61 x3 3.D-15 Repeat the previous problem for f (x) = cos x

Ans: L(x): Q(x) = 1 − 21 x2 = C(x)

Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 5: Tangent Lines

3.5

127

Tangent Lines

International Baccalaureate 7.1 Derivative interpreted as a gradient function and as rate of change. Finding equations of tangents and normals. 7.7 Graphical behaviour of functions; tangents and normals.

3.5.1

Tangent Lines

Lines with known y-intercept: y = mx + b Otherwise, in writing equations for lines, it is best to use the point-slope formula, in which you know a point and a slope: y − y1 = m (x − x1 ) If we are talking about writing the equation of a tangent line, remember that m, the slope of the tangent line is the same as the derivative.

Example 3.5.1 (adapted from AB93) An equation of the line tan3x − 2 at the point (−1, −5) is gent to the graph of y = 2x + 3

[Ans: y + 5 = 13 (x + 1)] Note: if you are simply asked to write an equation for the line, this would be an acceptable answer. If you’re asked to put the equation in standard form, it would be 13x − y = −8. If you’re asked to put the equation in slope-intercept form, the answer would be y = 13x + 8. You should be aware of each of these forms because you may need to spot them on a multiple choice exam.

Example 3.5.2 (adapted from AB98) Write an equation (in slopeintercept form) for the line tangent to the graph of f (x) = x4 − x2 at the point where f 0 (x) = 1

[Ans: y = x − 1.055] Mr. Budd, compiled January 12, 2011

128

SL Unit 3 (Basic Differentiation) Example 3.5.3 (adapted from [2]) If p(x) = (x + 2) (x + k) and if the line tangent to the graph of p at the point (4, p(4)) is perpendicular to the line 2x + 4y + 5 = 0, then k =

[Ans: â&#x2C6;&#x2019;8]

Example 3.5.4 (adapted from [2]) If the line 3x â&#x2C6;&#x2019; y + 5 = 0 is tangent in the second quadrant to the curve y = x3 + k, then k =

[Ans: 3]

3.5.2

Horizontal Tangents

f (x) has a horizontal tangent at (c, f (c)) if the slope of the tangent line is zero, i.e., f 0 (c) = 0. The equation of the tangent line will be y = f (c).

Example 3.5.5 The composite function h is defined by h(x) = f (g(x)), where f and g are functions whose graphs are shown in Figure 3.11. Find the number of horizontal tangent lines to the Figure 3.11: from Best & Lux [2]

graph of h. [2]

[Ans: 6] Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 5: Tangent Lines

129

Example 3.5.6 (AB92) Let f be the function defined by f (x) = 3x5 − 5x3 + 2. Write the equation of each horizontal tangent line to the graph of f .

[Ans: y = 0, y = 2, y = 4]

3.5.3

Vertical Tangents

Definition 3.1 (Vertical Tangent). f (x) has a vertical tangent at the point (c, f (c)) if the following conditions are met: 1. There is point of tangency, i.e., f (c) exists.

2. f 0 (c) is infinite.

The equation of the line of tangency is x = c.

Recall that the usual way to get an infinite answer is to have nonzero over zero.

Note that it is not sufficient to state that the denominator is zero.

Notice: both conditions must be met. Recall the definition of limit and realize that the derivative cannot exist unless the point exists. Frequently we may find f 0 to be nonzero over zero when we plug the number into the formula, but f is as well. It is not sufficient to find an infinite derivative; the point must exist as well.

Also, we cannot use the point-form of a line if the slope does not exist. Simply write the equation as x = c.

Example 3.5.7 Write theqequations of all the vertical (and hori2 3 zontal tangents) to f (x) = (x2 − 4) . √

Ans: x = 2, x = −2, (y = 3 16 Example 3.5.8 Write the equations of all the vertical tangents to 1 f (x) = . x−2 [Ans: none] Mr. Budd, compiled January 12, 2011

130

SL Unit 3 (Basic Differentiation) Example 3.5.9 (adapted from [2]) Which of the following is a function with a vertical tangent at x = 0?

(A) f (x) = x2 √ (B) f (x) = 3 x (C) f (x) =

1 x

(D) f (x) = |x| √ [Ans: f (x) = 3 x] d 1 d √ x and for Note that the derivative is infinite at x = 0 for , but the dx dx x 1 1 function is not defined at x = 0. For f (x) = , there is a vertical asymptote x x at x = 0, which is different from a vertical tangent.

3.5.4

Normal Lines

Normal lines are perpendicular to tangent lines. The slope of the normal line will be the negative reciprocal of the slope of the tangent line. (Recall that the slope of the tangent line is the . . .)

Example 3.5.10 The line that is normal to the curve y = x2 +2x−3 at (1, 0) intersects the curve at what other point? [19]

13 17 Ans: − , 4 16

3.5.5

Tangent Line Approximations

Tangent Lines to Approximate Values of f Example 3.5.11 You are stranded on a deserted island with your best friend, Wilson. He building some right triangles and wants √ to know the square root of 9.3. Without a calculator, estimate 9.3.

Example 3.5.12 Let f (x) = graph of f at x = −

1 . Use the tangent line to the x+2

3 to find approximations for: 2

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SL Unit 3, Day 5: Tangent Lines

131

(a) f (−1.4) (b) f (−1.45) (c) f (−1.501) [Ans: 1.6, 1.8, 2.004] Compare these tangent-line approximations to the actual functional values.

Example 3.5.13 (adapted from BC98) Let f be the function given by x2 − 2x + 3. The tangent line to the graph of f at x = 3 is used to approximate values of f (x). For what range of values is the error resulting from this tangent line approximation less than 0.1? Which of the following is the greatest value of x for which the error is less than 0.1?

(A) (B) (C) (D) (E)

3.1 3.2 3.3 3.4 3.5

[Ans: 3.3] Note that the linear approximation has an error below 0.1 from 2.684 . . . to 3.316 . . .

Tangent Lines to Approximate Zeros of f Example 3.5.14 (adapted from AB97) Let f be a differentiable function such that f (2) = 1 and f 0 (2) = −4. If the tangent line to the graph of f at x = 2 is used to find an approximation to a zero of f , that approximation is [Ans: 2.25]

Problems 3.E-1 99 MM2 The function f is given by f (x) =

2x + 1 x−3

,

x∈R

,

x 6= 3.

Mr. Budd, compiled January 12, 2011

132

SL Unit 3 (Basic Differentiation) (a)

(b) (c) (d)

(e) (f)

i. Show that y = 2 is an asymptote of the graph of y = f (x). ii. Find the vertical asymptote of the graph.

[Ans: x = 3] iii. Write down the coordinates of the point P at which the asymptotes intersect.

[Ans: (3, 2)] Find the points of intersection of the graph and the axes.

Ans: − 21 , 0 , 0, − 13 Hence sketch the graph of y = f (x), showing the asymptotes by dotted lines. −7 Show that f 0 (x) = 2 and hence find the equation of the tan(x − 3) gent at the point S where x = 4.

[Ans: y − 9 = −7 (x − 4)] The tangent at the point T on the graph is parallel to the tangent at S. Find the coordinates of T .

[Ans: (2, −5)] Show that P is the midpoint of ST .

3.E-2 (adapted from BC97) Refer to the graph in Figure 3.12. The function f is defined on the closed interval [0, 8]. The graph of its derivative f 0 is shown. The point (1, 4) is on the graph of y = f (x). An equation of the Figure 3.12:

line tangent to the graph of f at (1, 4) is

[Ans: y − 4 = 3 (x − 1)]

3.E-3 (BC90) Let f (x) = 12 − x2 for x ≥ 0 and f (x) ≥ 0. The line tangent to the graph of f at the point (k, f (k)) intercepts the x-axis at x = 4. What is the value of k?

[Ans: k = 2] 1 3.E-4 (adapted from AB97) At what point on the graph of y = x3 is the 4

tangent line parallel to the line 3x − 4y = 7?

Ans: 1, 41 Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 5: Tangent Lines

133

3.E-5 (adapted from [2]) Let f (x) = 3x3 −4x+1. An equation of the line tangent to y = f (x) at x = 2 is [Ans: y = 32x − 47] √ 3.E-6 (adapted from [2]) Find the point on the graph of y = x between (4, 2) and (9, 3) at which the normal to the graph has the same slope line as the

5 through (4, 2) and (9, 3).

Ans: 25 , 4 2 3.E-7 (adapted from [3]) If the graph of the parabola y = 2x2 + x + k is tangent to the line 3x + y = 3, then k = [Ans: 5] 8x at 1 + x3 the point (1, 4) forms a right triangle with the coordinate axes. The area of the triangle is [Ans: 9]

3.E-8 (adapted from [3]) A tangent line drawn to the graph of y =

5

4

3.E-9 (adapted from AB93) Find the derivative of f (x) = (x − 3) (x + 2) .

At how many different values of x will the tangent line be horizontal?

[Ans: Three] 3.E-10 (adapted from [2]) An equation of the normal to the graph of f (x) = x at (1, f (1)) is [Ans: x − 2y = −1] 3x − 2 3.E-11 (adapted from [3]) An equation of the line normal to the graph of f (x) = x at (3, 3) is x−2 [Ans: x − 2y + 3 = 0] √ 3.E-12 (AB88) Let f be the function given by f (x) = x4 − 16x2 . (a) Find the domain of f .

[Ans: Df : x ≥ 4, x ≤ −4, x = 0] (b) Describe the symmetry, if any, of the graph of f . [Ans: Graph of f is symmetric w.r.t. the y-axis]

3 2x −16x 0 (c) Find f (x).

Ans: √ 2 2 x (x −16)

(d) Find the slope of the line normal to the graph of f at x = 5

3 Ans: − 34 3.E-13 (adapted from AB97) No calculator! Let f be a differentiable function such that f (4) = 1.5 and f 0 (4) = 3. If the tangent line to the graph of f at x = 4 is used to find an approximation to a zero of f , that approximation is [Ans: 3.5] 3.E-14 (adapted from [3]) No √ calculator! Let f be a function with f (2) = 6 and derivative f 0 (x) = x3 + 1. Using a tangent line approximation to the graph of f at x = 2, estimate f (2.05) [Ans: 6.15] √ 3.E-15 (adapted from [3]) No calculator! The approximate value of y = x2 + 3 at x = 1.08, obtained from the tangent to the graph at x = 1, is [Ans: 2.04]

Mr. Budd, compiled January 12, 2011

134

SL Unit 3 (Basic Differentiation)

Mr. Budd, compiled January 12, 2011

SL Unit 3, Day 6: Basic Differentiation Review Problems

3.6

135

Basic Differentiation Review Problems

3.F-1 (AB97) If f (x) = −x3 + x +

1 , then f 0 (−1) = x

[Ans: −3]

3.F-2 (AB97) A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by v(t) = 3t2 − 2t − 1. The position x(t) is 5 for t = 2. (a) Write a polynomial expression for the position of the particle at any time t ≥ 0. (b) For what values of t, 0 ≤ t ≤ 3, is the particle’s instantaneous velocity the same as its average velocity on the closed interval [0, 3]

√ 1 x−1 ?

Ans: 3.F-3 [2] What is lim x→1 x − 1 2 3/4 3.F-4 [2] If h(x) = x2 − 4 + 1, then the value of h0 (2) is [Ans: nonexistent] 3.F-5 [2] Which of the following is a function with a vertical tangent at x = 0?

(A) f (x) = x3 √ (B) f (x) = 3 x 1 (C) f (x) = x [Ans: f (x) =

3.F-6 [2] The derivative of

√

1 is x− √ x3x

√ 3

x]

1 −1/2 4 −7/3 Ans: x + x 2 3

3.F-7 (from [2]) A function f is defined for all real numbers and has the following

property: f (a + b) − f (a) = 3a2 b + 2b2 . Find f 0 (x).

Ans: 3x2 3.F-8 [2] If

d d d2 (f (x)) = g(x) and (g(x)) = f (3x), then f x2 is 2 dx dx dx

(A) 4x2 f (3x2 ) + 2g(x2 ) (B) f (3x2 ) (C) f (x4 ) (D) 2xf (3x2 ) + 2g(x2 ) (E) 2xf (2x2 )

3.F-9 [2] If y =

3 dy , then = 4 + x2 dx

Ans: 4x2 f (3x2 ) + 2g(x2 ) " # −6x Ans: 2 (4 + x2 ) Mr. Budd, compiled January 12, 2011

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SL Unit 3 (Basic Differentiation)

3

3.F-10 [2] lim

h→0

5 5 1 1 +h −3 2 2 = h

15 Ans: 16

3.F-11 [2] An object moves along the x-axis so that at time t, t > 0, its position is given by x(t) = t4 +t3 −30t2 +88t. At the instant when the acceleration becomes zero, the velocity of the object is [Ans: 12] 3.F-12 Using the limit definition of derivative, prove the Product Rule.

3.F-13 Use the Chain and Product Rule to prove the Quotient Rule.

Mr. Budd, compiled January 12, 2011

Unit 4

Transcendentals 1. u-Substitution 2. Derivatives and Antiderivatives of Trigonometric Functions 3. Derivatives and Antiderivatives of Exponential Functions 4. Derivatives of Logarithmic, Antiderivatives of Exponential Functions 5. Separable Differential Equations 6. Slope Fields 7. Integration by Parts Advanced Placement

I. Derivatives Derivative as a function. â&#x20AC;˘ Equations involving derivatives. Verbal descriptions are translated into equations involving derivatives and vice versa.

Applications of derivatives. â&#x20AC;˘ Geometric interpretation of differential equations via slope fields and the relationship between slope fields and derivatives of implicitly defined functions.

Computation of derivatives.

137

138

SL Unit 4 (Transcendentals) • Knowledge of derivatives of basic functions, including exponential, logarithmic, and trigonometric functions.

II. Integrals Techniques of antidifferentiation.

• Antiderivatives following directly from derivatives of basic functions.

• Antiderivatives by substitution of variables.

Applications of antidifferentiation.

• Finding specific antiderivatives using initial conditions, including applications to motion along a line.

• Solving separable differential equations and using them in modeling. In particular, studying the equation y 0 = ky and exponential growth.

International Baccalaureate (MM 6.1) Derivative of x 7→ sin x, x 7→ cos x, x 7→ ex , and x 7→ ln x.

(MM 6.2) Applications of the first derivative to tangents, maximum and minimum ds problem, kinematical problems involving displacement, s, velocity, = v, and dt dv acceleration, = a. dt (MM 6.3) Indefinite integration as anti-differentiation.

Indefinite integral of sin x, cos x, ex and the composites with x 7→ ax + b.

Application to acceleration and velocity.

(MM 6.4) Anti-differentiation with a boundary condition to determine the constant term.

(MM 8.3) Integration by substitution of the form u = g(x).

Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 1: Antidifferentiation of Trigonometric Functions

4.1

139

Antidifferentiation of Trigonometric Functions

International Baccalaureate 7.4 Indefinite integration as anti-differentiation. Indefinite integral of sin x, cos x.

The composites of these with the linear function ax + b.

7.9 Integration by substitution.

4.1.1

Antidifferentiating Trigs

Antiderivative of Sine dy = sin x.

Example 4.1.1 Use a program to draw a slope field of dx Conjecture the antiderivative of sine. Prove your conjecture using u-substitution.

Z sin x dx = â&#x2C6;&#x2019; cos x + C

u-substitution gives: Z

sin (g(x)) g 0 (x) dx = â&#x2C6;&#x2019; cos (g(x)) + C

Antiderivative of Cosine dy Example 4.1.2 Use a program to draw a slope field of = cos x. dx Conjecture the antiderivative of cosine.

Z cos x dx = sin x + C

u-substitution gives: Z

cos (g(x)) g 0 (x) dx = sin (g(x)) + C

Mr. Budd, compiled January 12, 2011

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SL Unit 4 (Transcendentals)

Other Trigonometric Antiderivatives Z

sec2 x dx = tan x + C

Z sec x tan x dx = sec x + C

Example 4.1.3 (adapted from AB97) At time t ≥ 0, the acceleration of a particle moving on the x-axis is a(t) = t + sin t. At t = 0, the velocity of the particle is −3. For what value of t will the velocity of the particle be zero?

[Ans: 1.855] Be careful! C 6= −3.

Problems In your SL textbook: page 268, # 3,6 4.A-1 (AB93) If the second derivative of f is given by f 00 (x) = 2x − cos x, which of the following could be f (x)? x3 + cos x − x + 1 3 x3 (B) − cos x − x + 1 3 (C) x3 + cos x − x + 1

(A)

(D) x2 − sin x + 1 (E) x2 + sin x + 1 h Ans:

x3 3

+ cos x − x + 1

i

4.A-2 (adapted from [2]) A particle moves along the x-axis with velocity at time t given by: v(t) = t + 2 sin t. If the particle is at the origin when t = 0, its position at the time when v = 5 is x = [Ans: 17.277]

Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 1: Antidifferentiation of Trigonometric Functions

141

The Anti-Chain Rule International Baccalaureate 7.9 Integration by substitution.

Textbook §4.6 Integration by Substitution. Oxford book: §7.1 Integrating a function of a function on page 212. [15]

4.1.2

The Anti-Chain Rule

Example 4.1.4 Z

6x2 x3 + 1 dx

h i 2 Ans: x3 + 1 + C

Example 4.1.5 Z

3x2 √ dx 2 x3 + 1

Ans:

√

x3 + 1 + C

Example 4.1.6 (adapted from AB93) Z 4x3 √ dx = x4 − 3 √

Ans: 2 x4 − 3 + C

Completing the Derivative of the Inside Example 4.1.7 Z

x2 x3 + 1

2

dx =

h

Ans:

1 9

x3 + 1

3

+C

i

Mr. Budd, compiled January 12, 2011

142

SL Unit 4 (Transcendentals) Example 4.1.8 x2

Z

(x3 + 1)

2

dx

h i Ans: − 3(x31+1)

Example 4.1.9 (adapted from [2]) Z 4 x x2 − 1 dx =

h

Ans:

1 15

x3 + 1

5

+C

i

Example 4.1.10 [11] r2 − 1

Z

(r3 − 3r + 3)

2

dr

Example 4.1.11 [11] Z

1 x2

r 5

1 + 5 dx x

Problems Z 4.A-3

x Z

4.A-4 Z 4.A-5

p 3

1 − x2 dx =

G0 (v(x)) v 0 (x) dx = H 0 (ax + b) dx =

h i 4/3 Ans: − 38 1 − x2 +C [Ans: G (v(x)) + C]

Ans:

1 a H(ax

+ b) + C

Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 2: u-Simplification

4.2

143

u-Simplification

International Baccalaureate 7.9 Integration by substitution.

Textbook §4.6 Integration by Substitution. Oxford book: §7.1 Integrating a function of a function on page 212. [15]

4.2.1

u-Simplification with Trigonometrics Outside

Example 4.2.1 (adapted from [20]) Find

R

x2 cos 2x3

Ans:

Z Example 4.2.2 (adapted from [18]) Calculate

sec

√

dx.

1 6

sin 2x3 + C

√ x tan x √ dx x

[Ans: 2 sec

Z Example 4.2.3 Calculate

√

x + C]

x dx cos2 x2

Ans:

1 2

tan x2 + C

Motivation for finding the happy function Example 4.2.4 What happens when we try to find

I need a function whose derivative is

R

tan x dx?

1 u.

Suppose I had a function h(x) (the happy function), for which h0 (x) = h(x).

Then suppose I had a function I(x) = h−1 (x), which was the inverse function of h(x). Remember, to get an inverse function, I switch the x’s and y’s Mr. Budd, compiled January 12, 2011

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SL Unit 4 (Transcendentals)

So: y = I(x) I = h−1

x = h(y) d d x= h(y) dx dx dy 1 = h0 (y) dx 1 dy = h0 (y) dx 1 dy = h(y) dx 1 dy = x dx

Differentiate both sides Chain Rule, inside is y

h0 = h h(y) = x from before

So, if I can find a happy function, h(x), such that h0 (x) = h(x), then the inverse of the happy function has a derivative x1 , and I can finally find the antiderivative of tan x. Now, if I only had a happy function...

4.2.2

Anti-Chain Rule with Trigonometrics Inside

R Example 4.2.5 Solve cos x sin x dx two different ways, and then show that the solutions are really the same. Do the same for R sec2 x tan x dx.

Example 4.2.6 (adapted from Stewart [20])

R

sec4 x tan x dx

h

Ans:

sec4 x 4

+C

i

Example 4.2.7 [18] Z

√

sin x dx 1 + cos x

√ Ans: −2 1 + cos x + C Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 2: u-Simplification

4.2.3

145

Linear u-Simplification

If the inside function is linear, i.e.

• u = mx + b; •

du = m or k; dx

• the derivative of the inside is constant;

Then all that u-substitution will accomplish is to divide by the derivative of the inside.

Do not try this unless the derivative of the inside is a constant.

Z

f 0 (mx + b) dx =

f (mx + b) +C m

Example 4.2.8 Prove the above equation.

3

(4 − 7x) dx.

2

Z Example 4.2.9 Find

h

Example 4.2.10 Find

Example 4.2.11 Find

R

R

Ans: − (4−7x) +C 56

i

h

i

4

cos (3x + 2) dx.

Ans:

sin(3x+2) 3

+C

sec (πx) tan (πx) dx.

Ans:

1 π

sec (πx) + C

Mr. Budd, compiled January 12, 2011

146

SL Unit 4 (Transcendentals)

Problems 4.B-1 [11]

R

h

sin7 (θ) cos (θ) dθ

4.B-2 (adapted from [2]) If Z 4.B-3 (adapted from [2]) Z 4.B-4 (adapted from [20]) Z 4.B-5 [18] Find Z 4.B-6 [18] Find Z 4.B-7 [18] Find Z 4.B-8 [18] Find Z 4.B-9 [18] Find

dy = cos3 x sin x, then y = dx 6 cos x sin2 x dx =

x sin3 x2 cos x2 dx. √

sec2 x dx.

1 + tan x

1 + tan2 x sec2 x dx.

R 4.B-10 (adapted from [2]) cos (3x + 2) dx =

R 4.B-11 [20] Find x sin 1 − x2 dx √ R sin x √ 4.B-12 [18] Find dx x R 4.B-13 [18] Find x sec2 x2 dx.

R 4.B-14 [18] Find cos (3x + 1) dx.

R 4.B-15 [18] Find sin (3 − 2x) dx.

i

Ans: 2 sin3 x + C h Ans:

tan3 θ sec2 θ dθ

1 + sin x cos x dx.

+C

Ans: − 41 cos4 x + C

cos4 x sin x dx. √

sin8 θ 8

Ans:

tan4 θ 4 +C

i

Ans: − 51 cos5 x + C h

Ans:

2 3

+C

i

sin4 x2 + C

(1 + sin x)

Ans:

1 8

3/2

√ Ans: 2 1 + tan x + C

Ans: tan x +

Ans:

Ans:

1 3

tan3 x + C

1 3

sin (3x + 2) + C

1 2 +C 2 cos 1 − x

√ [Ans: −2 cos x + C]

Ans:

Ans:

Ans:

1 2

tan x2 + C

1 3

sin (3x + 1) + C

1 2

cos (3 − 2x) + C

Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 3: Basic Calculus of Exponential Functions

4.3

147

The Happy Function

International Baccalaureate 7.1 Derivative of ex . Derivatives of ax . 7.4 Indefinite integration as anti-differentiation.

Indefinite integral of ex . The composites of this with the linear function ax + b. 7.9 Integration by substitution.

Textbook §2.7 Derivatives of Exponential and Logarithmic Functions [15]

4.3.1

Differentiating the Exponential Function d x e = ex dx

Example 4.3.1 Find

d x b dx

x d x d b = eln b dx dx d x ln b = e dx = ex ln b ln b x = eln b ln b = bx ln b Applying the Chain Rule: d g(x) e = eg(x) g 0 (x) dx Example 4.3.2 (adapted from AB acorn ’02) Two particles start at the origin and move along the x-axis. For 0 ≤ t ≤ 10, their respective position functions are given by x1 = cos t and x2 = e−2t − 1. For how many values of t do the particles have the same velocity?

[Ans: Four] Mr. Budd, compiled January 12, 2011

148

SL Unit 4 (Transcendentals) eh − 1 is h→0 3h

Example 4.3.3 (adapted from BC97) lim

Ans:

1 3

Example 4.3.4 (adapted from AB97) Let f be the function given 2 by f (x) = 2e4x . For what value of x is the slope of the line tangent to the graph of f at (x, f (x)) equal to 5?

[Ans: 0.246]

3

Example 4.3.5 (adapted from BC93) If f (x) = etan

h

x

, then f 0 (x) =

3

Ans: 3 tan2 x sec2 x etan

x

i

Example 4.3.6 For ef (x) = x, find f 0 (x) in terms of x only. Note d that your answer is ln x. dx

u-Substitution with exponentials Example 4.3.7 If

dy e−x = 2 , then y = dx (3 + 2e−x )

h Ans:

1 2(3+2e−x )

+C

i

Example 4.3.8 [18] Z

e−x 1 + cos e−x dx

[Ans: −e−x − sin (e−x ) + C] Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 3: Basic Calculus of Exponential Functions

4.3.2

149

Antidifferentiating the Exponential Function

Since the derivative of ex is ex the antiderivative of ex is ex .

Z

ex dx = ex + C

Example 4.3.9 Find the antiderivative of

R

2x dx.

x Change 2x to eln 2 to ex ln 2 . Let u = x ln 2. If the antiderivative of your inside function is just a constant number (such as ln 2), you can skip the whole process of u-substitution and just divide by that constant number.

Z

bx dx =

bx +C ln b

Applying u-substitution to the general antiderivative of ex : Z

eg(x) g 0 (x) dx = eg(x) + C

Example 4.3.10 (BC93) A particle moves along the x-axis so that at any time t ≥ 0 the acceleration of the particle is a(t) = e−2t . If 5 17 at t = 0 the velocity of the particle is and its position is , then 2 4 its position at any time t > 0 is x(t) =

h Ans:

e−2t 4

+ 3t + 4

i

Example 4.3.11 (adapted from [2]) Z

3

ex − x2 dx = ex3

"

3

e−x Ans: x + +C 3

#

Mr. Budd, compiled January 12, 2011

150

SL Unit 4 (Transcendentals)

Problems 4.C-1 (adapted from AB98) Let f be the function given by f (x) = 2e3x and let g be the function given by g(x) = 6x3 . At what value of x do the graphs of f and g have parallel tangent lines?

[Ans: −0.344] 4.C-2 (adapted from AB98) If f (x) = cos (e−x ), then f 0 (x) = [Ans: e−x sin (e−x )]

ex−5 , then f 0 (5) = Ans: 72 2

2 x2 e 2ex (x2 −1) 0 Ans: 4.C-4 (adapted from AB97) If f (x) = 2 , then f (x) = x3 x 3

4.C-3 (adapted from BC97) If f (x) = (x − 1) 2 +

4.C-5 (adapted from AB93)

d (3x ) = dx

[Ans: (3x ) ln 3]

√ 4.C-6 (adapted from [2]) The approximate value of y = 3 + ex at x = 0.04, obtained from the tangent to the graph at x = 0, is [Ans: 2.01] 4.C-7 (adapted from [2]) Two particles move along the x-axis and their positions at time 0 ≤ t ≤ 2π are given by x1 = sin 3t and x2 = e(t−3)/2 − 0.75.

For how many values of t do the two particles have the same velocity?

[Ans: Six] 4.C-8 (adapted from [2]) If y = ekx , then 4.C-9 [11]

R

d6 y = dx6

Ans: k 6 ekx

Ans:

ex cos (πex ) dx

4.C-10 [18] Z

√

ex dx ex + 1

1 π

sin (πex ) + C

√ Ans: 2 ex + 1 + C

4.C-11 (from Stewart [20]) Let f (x) = ex . (a) Find the value of the constants a, b, c, and d cubic function C(x) = a + bx + cx2 + dx3 has the properties C(0) = f (0), C 0 (0) = f 0 (0), C 00 (0) = f 00 (0), and C 000 (0) = f 000 (0). (b) Plot f and C on the same axes over the interval [−4, 4].

Ans: C(x) = 1 + x + 21 x2 + 16 x3 h i R t t 4.C-12 (adapted from AB97) π1 e π dt = Ans: e π + C 4.C-13 (adapted from AB97) Z

etan x dx = cos2 x [Ans: etan x + C] Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 3: Basic Calculus of Exponential Functions

151

4.C-14 (adapted from [2]) The acceleration of a particle at time t moving along the x-axis is given by: a = 9e3t . At the instant when t = 0, the particle is at the point x = 2 moving with velocity v = −2. The position of the

1 particle at t = is Ans: e − 23 3 4.C-15 (adapted from [2]) Z

√ 3

e x √ dx 3 3 x2 h i √ 3 Ans: e x + C

4.C-16 (adapted from [2]) The number of bacteria in a culture is growing at a rate of 1000e2t/3 per unit of time t. At t = 0, the number of bacteria present

was 1500. Find the number present at t = 3.

Ans: 1500e2

R 4.C-17 [11] ex cos (πex ) dx Ans: π1 sin (πex ) + C R 4.C-18 (adapted from [18]) sin x ecos x dx [Ans: −ecos x + C] 4.C-19 [18] Z

e1/x dx x2

Ans: −e1/x + C

Z 4.C-20 Find

√

ex dx using the u-substitution u = ex . [Ans: arcsin ex + C] 1 − e2x

Mr. Budd, compiled January 12, 2011

152

SL Unit 4 (Transcendentals)

Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 4: Natural Logarithm

4.4

153

Natural Logarithm

International Baccalaureate Derivative of x 7→ ln x.

Textbook §2.7 Derivatives of Exponential and Logarithmic Functions [15]

4.4.1

Differentiating the Logarithmic Function

y = ln x ey dy ey dx dy dx dy dx

=x =1 1 ey 1 = x =

d 1 ln x = dx x

x>0

Using the Chain Rule gives: d 1 0 ln (g(x)) = g (x) dx g(x)

d g 0 (x) ln (g(x)) = dx g(x)

g(x) > 0

Example 4.4.1 Find the derivative: d ln (−x) dx

x<0

d 1 ln x = dx x

x>0

Combining

Mr. Budd, compiled January 12, 2011

154

SL Unit 4 (Transcendentals)

and

d 1 ln (−x) = dx x

x<0

d 1 ln |x| = dx x

x 6= 0

into one statement gives

.

Example 4.4.2 (adapted from AB97) If f (x) = ln 3x2 − 1 , then f 0 (x) =

6x

(A)

2 3x − 1

6x (B) 2 |3x − 1| 6 |x| (C) 3x2 − 1 6x (D) 3x2 − 1 1 (E) 3x2 − 1 h

Ans:

6x 3x2 −1

i

Example 4.4.3 Show that d d ln xa = a ln x dx dx Example 4.4.4 Show that d d (ln a + ln x) = ln (ax) dx dx Example 4.4.5 Show that d d ln ex = x dx dx Example 4.4.6 Show that d ln x d e = x dx dx Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 4: Natural Logarithm

155

1 π Example 4.4.7 (adapted from [2]) If y = cos u, u = 2v − + − 1, v 2 dy and v = ln x, then the value of at x = e is dx

Ans: − 3e

Example 4.4.8 Show that y = ln x2 + e is a solution to the dy differential equation = 2xe−y . dx

Antidifferentiating to the Logarithmic Function 4.4.2

Antidifferentiating the Reciprocal Function

Why does the Anti-Power Rule not work for

R

x−1 dx?

But, we recently found a function whose derivative is

Z

1 x.

1 dx = ln |x| + C x

Why is the antiderivative ln |x| and not just ln x?

1 is x 6= 0, but the domain of ln x is only the real numbers x R greater than 0. So, x1 dx = ln x only for x > 0. The problem is that x in x1 can be anything but zero, but ln x is impossible for those x’s below zero. I didn’t have this problem before, when I was differentiating ln x, because the domain of ln x is just those numbers above 0, and my derivative, x1 can handle those numbers.

The domain of

I need an antiderivative of

1 for x < 0, since ln x only works for x > 0. x

So let’s take a look at the real numbers less than 0, For x > 0, we’ve already Mr. Budd, compiled January 12, 2011

156

seen that

SL Unit 4 (Transcendentals) d 1 ln x = . Now let’s look at x < 0: dx x = ln (−x) dy 1 d = (−x) dx −x dx dy 1 = − (−1) dx x 1 dy = dx x

So what we see is that (for x < 0).

1 is the derivative of both ln x (for x > 0) and ln (−x) x

1 d ln x = dx x

(x > 0)

1 d ln (−x) = dx x

(x < 0)

Combining these into one statement for all x 6= 0 gives me d 1 ln |x| = dx x d The typical expression that I’m use to seeing, dx ln x = x1 , is basically the same thing, only your restricting yourself to positive x’s, i.e., those numbers for which x = |x|

1 1 is the derivative of ln |x| (for all x 6= 0, the antiderivative of is ln |x| x x (plus some arbitrary constant).

Since

Z

dx = ln |x| + C x

u-substitution gives: Z

g 0 (x) = ln |g(x)| + C g(x)

Mr. Budd, compiled January 12, 2011

SL Unit 4, Day 4: Natural Logarithm

157

Example 4.4.9 Find Z tan x dx

[Ans: − ln |cos x| + C] Example 4.4.10 Use the substitution u = sec x + tan x to find Z sec x dx

[Ans: ln |sec x + tan x| + C] Example 4.4.11 (adapted from [2]) Find a family of curves that intersect at right angles every curve of the family y = 0.5x2 + k for every real value of k?

[Ans: y = − ln |x| + C]

4.4.3

Antidifferentiating Rational Functions

If the order on top is the same or higher than that on bottom, use division (long or synthetic) to reduce the problem to a polynomial + some remainder over the original denominator. The polynomial is easy to antidifferentiate using the anti-power rule, and the remainder over the original denominator should be easier to antidifferentiate than the original fraction.

Example 4.4.12 (adapted from [2]) Z x−1 dx = x−2 [Ans: x + ln |x − 2| + C] Example 4.4.13 If the velocity of Runner B, in meters per second, 24t is given by v(t) = , find an expression for the runner’s position 2t + 3 if her position at time t = 0 seconds is 0 meters.

[Ans: 12t − 18 ln |2t + 3| + 18 ln 3] Mr. Budd, compiled January 12, 2011

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Problems ln(e + h) − 1 4.D-1 (BC97) lim is h→0 h

1 Ans: e

2 4.D-2 (adapted from BC93) If f (x) = ln e3x , then f 0 (x) =

[Ans: 6x]

4.D-3 (adapted from AB93) The slope of the line normal to the graph of y = π 3 ln (sec x) at x = is [Note: normal lines are perpendicular to tangent 4

lines.]

Ans: − 13 4.D-4 (adapted from [2]) If y = 2u + 3eu and u = 1 + ln x, find [Ans: 5e]

1 dy when x = . dx e

t2 + 1 gives the position 4.D-5 (adapted from [2]) The formula x(t) = 2 ln t + 9 of an object moving along the x-axis during the time interval 1 ≤ t ≤ 5.

At the instant when the acceleration of the object is zero, what is the

velocity?

Ans: 43 4.D-6 (adapted graph1 of √ from [2]) What is the slope of the line tangent to the y = ln 3 x at e3 , 1 ?

Ans: 3e3 4.D-7 [3] There is a point between P (1, 0) and Q (e, 1) on the graph of y = ln x such that the tangent to the graph at that point is parallel to the line through points P and Q. What is the x-coordinate of this point?

[Ans: e − 1] 4.D-8 (adapted from AB acorn ’02) Which of the following are antiderivatives ln3 x ? of x ln4 x 4 ln4 x II. +6 4 3 ln x − ln3 x III. x2 I.

(A) I only (B) III only (C) I and II only (D) I and III only (E) II and III only [Ans: I and II only] Mr. Budd, compiled January 12, 2011

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159 3

4.D-9 (adapted from AB98) Let F (x) be an antiderivative of then F (9) = 4.D-10 (adapted from [3]) If eg(x) = 3x − 1, then g 0 (x) =

(ln x) . If F (1) = 2, x [Ans: 7.827] i h 3 Ans: 3x−1

4.D-11 (MM spec) The function f is given by 2x 1 + x2

R By using an appropriate substitution, find f (x) dx. Ans: x − ln 1 + x2 + C f (x) = 1 −

4.D-12 (adapted from [2]) Find Z

sec2 x dx tan x [Ans: ln |tan x| + C]

4.D-13 (adapted from [2]) A particle starts at (3, 0) when t = 0 and moves along the x-axis in such a way that at time t > 0 its velocity is given by v(t) = 1 . Determine the position of the particle at t = 5.

[Ans: 3 + ln 6] 1+t 4.D-14 The antiderivative of

1 cabin

[Ans: Evan’s ark]

4.D-15 (adapted from [2]) Z

12x2 dx = 1 + x3

Ans: 4 ln 1 + x3 + C

4.D-16 (adapted from AB98) Z

x2 − 1 dx = x h Ans:

x2 2

− ln |x| + C

i

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Mr. Budd, compiled January 12, 2011

Unit 5

Extrema and Optimization 1. Relating Graphs of f and f 0 2. First Derivative Test - Increasing/ Decreasing Functions 3. Relating Graphs of f , f 0 , and f 00 4. Points of Inflection and Second Derivative Test 5. Optimization Advanced Placement Derivative as a function • Corresponding characteristics of graphs of f and f 0 .

• Relationship between the increasing and decreasing behavior of f and the sign of f 0 .

Second derivatives.

• Corresponding characteristics of the graphs of f , f 0 , and f 00 .

• Relationship between the concavity of f and the sign of f 00 .

• Points of inflection as places where the concavity changes.

Applications of derivatives.

• Optimization, both absolute (global) and relative (local) extrema.

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Relating Graphs of f and f 0

5.1

International Baccalaureate 7.1 Identifying increasing and decreasing functions.

7.3 Local maximum and minimum points; Testing for the maximum using change of sign of the first derivative.

Textbook §3.3 Maximum and Minimum Values; §3.4 Increasing and Decreasing Functions; §3.6 Overview of Curve Sketching [15] Resources §4.3 Connecting f 0 and f 00 with the Graph of f in [8]

5.1.1

Direction of f

The sign of f 0 determines the direction of f .

• If f 0 is positive over an open interval, then f is increasing over that interval.

• If f 0 is negative over an open interval, then f is decreasing over that interval.

3

2

Example 5.1.1 If x(t) = (t + 2) (t − 2) , for what values of t is x decreasing? At what value of t does x change from increasing to decreasing? Decreasing to increasing?

Ans:

5.1.2

2 5

<t<2

Relative Extrema

Definition 5.1 (Critical Point). A critical point is a point on the graph of f where f 0 is either 0 or undefined.

Definition 5.2 (Critical Number). A critical number is the x-coordinate of the critical point.

Note: the critical number is sometimes and confusingly referred to as the critical point. Judge from context whether the critical point really means the critical point or actually means the critical number.

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SL Unit 5 (Extrema and Optimization) Example 5.1.2 How many critical points does the function f (x) = 4 5 (x + 2) x2 − 1 have? [2] Example 5.1.3 How many critical points does the function f (x) = q 1 2 3 2 (x − 4) + 1 have? How about f (x) = ? x−2

A point on the graph of f where f 0 is 0 has a horizontal tangent line. A point on the graph of f where f 0 is undefined is either a cusp or a vertical tangent.

An extremum is either a maximum or a minimum.

Definition 5.3 (Relative Extrema). If c is in the domain of a function f , then f (c) is a 1. relative maximum of f at c if and only if f (c) ≥ f (x) for all x in near c.

2. relative minimum of f at c if and only if f (c) ≤ f (x) for all x in near c.

Relative extrema are also called local extrema.

5.1.3

First Derivative Test

Critical points are important because they tell me where relative minima or maxima might occur. Just because the derivative is zero or undefined does not mean that I have a local extrema. But finding the critical points allows me to test only a few points for extremity, as opposed to an infinite number of points in a typical domain.

To find out whether the critical points actually are relative extrema, we test them. There are two tests, the First Derivative Test, and the Second Derivative Test.

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The First Derivative Test checks the sign of the first derivative before and after the critical point. If the sign of the first derivative is changing, then the original function is changing direction, and there is a local extremum.

In order to perform the First Derivative Test, I need to cut the number line at each critical number, so that I know how close I need to be when I check on either side of the critical number.

Theorem 5.1 (First Derivative Test for Relative Extrema). (Notes stolen from [8]) At a critical point c: 1. If f 0 changes sign from positive to negative at c (f 0 > 0 for x < c and f 0 < 0 for x > c), then f has a relative maximum value at c.

2. If f 0 changes sign from negative to positive at c (f 0 < 0 for x < c and f 0 > 0 for x > c), then f has a relative minimum value at c.

3. If f 0 does not change sign at c (f 0 has the same sign on both sides of c), then f has no relative extreme value at c.

Example 5.1.4 (AB98) The graph of f 0 , the derivative of f , is shown in Figure 5.1. Which of the following describes all relative extrema of f on the open interval (a, b)?

(A) One relative maximum and two relative minima.

(B) Two relative maxima and one relative minimum.

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Figure 5.1: Graph of f 0 (x)

(C) Three relative maxima and one relative minimum.

(D) One relative maximum and three relative minima.

(E) Three relative maxima and two relative minima.

[Ans: A]

Example 5.1.5 [2]The function f (x) = x4 − 18x2 has a relative minimum at x =

[Ans: −3 and 3]

Example 5.1.6 Determine at what values of x the function f (x) has a relative minimum for:

• f 0 (x) = x2 x2 − 9 • f 0 (x) = x2 (x − 9) 2

• f 0 (x) = x (x − 3) (x + 3)

2

2

• f 0 (x) = x2 (x − 3) (x + 3)

Example 5.1.7 If the derivative of f is given by f 0 (x) = sin (ln x), at which value of x, x ∈ [0, 1.07], does f have a relative maximum value? a relative minimum?

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Problems 5.A-1 (AB00) Figure 5.2 shows the graph of f 0 , the derivative of the function f , or −7 ≤ x ≤ 7. The graph of f 0 has horizontal tangent lines at x = −3, x = 2, and x = 5, and a vertical tangent line at x = 3.

Figure 5.2: AB ’00

(a) Find all values of x, for −7 < x < 7, at which f attains a relative minimum. Justify your answer. [Ans: x = −1, f 0 (x) changes from negative to positive at x = −1] (b) Find all values of x, for −7 < x < 7, at which f attains a relative maximum. Justify your answer. [Ans: x = −5, f 0 (x) changes from positive to negative at x = −5] (c) Find all values of x, for −7 < x < 7, at which f 00 (x) < 0.

[Ans: on the intervals (−7, −3), (2, 3), and (3, 5),f 00 (x) exists and f 0 is decreasing ] 5.A-2 (AB98) The graphs of the derivatives of the functions f , g, and h are shown in Figure 5.3. Which of the functions f , g, or h have a relative maximum on the open interval a < x < b?

[Ans: f only] Figure 5.3: AB98

5.A-3 (adapted from [3]) Which of the following are (is) true about a particle that starts at t = 0 and moves along a number line if its position at time 3 t is given by s(t) = (t − 1) (t − 5)?

I. The particle is moving to the right for t > 4.

II. The particle is at rest at t = 1 and t = 4.

III. The particle changes direction at t = 1.

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SL Unit 5 (Extrema and Optimization) [Ans: I and II only]

5.A-4 (adapted from [3]) A particle starts at time t = 0 and moves along a number line so that its position, at time t ≥ 0, is given by x(t) = 3 (t − 3) (t − 7) . The particle is moving to the left for what values of t?

[Ans: t < 4] 2

3

5.A-5 (adapted from [3]) If the derivative of the function f is f 0 (x) = −2 (x − 3) (x + 1) (x + 2) , then f has a local minimum at x = [Ans: −2] Figure 5.4: Problem 6: Graph of the derivative of f

5.A-6 (adapted from AB97) The graph of the derivative of f is shown in Figure 5.4. Graph the function f .

[Ans: Figure 5.5] 5.A-7 (adapted from AB93) For what value of x does the function f (x) = 2 (x − 3) (x − 2) have a relative minimum?

Ans: 38 5.A-8 (adapted from AB93) How many critical points does the function f (x) = 5 4 (x − 3) (x + 2) have?

[Ans: Three] 5.A-9 (adapted from AB93) The function f given by f (x) = x3 − 12x + 24 is increasing for what values of x?

[Ans: x < −2, x > 2]

Figure 5.5: Answer to problem 6: Graph of f

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Figure 5.6: Problem 10: Graph of h(x)

Figure 5.7: Answer to problem 10: Graph of h0 (x)

5.A-10 (BC98) The graph of y = h(x) is shown in Figure 5.6. Sketch the graph of y = h0 (x) [Ans: Figure 5.7] 5.A-11 [2] A particle moves along the x-axis so that at time t, t >= 0, its position 3 is given by x(t) = (t + 1) (t − 3) . For what values of t is the velocity of the particle increasing?

[Ans: t < 1 or t > 3] x4 x5 − . The derivative of f attains it 2 10 maximum value at x = [Ans: 3] 2 x 2 5.A-13 The derivative of f is given by f 0 (x) = ex /5 cos . Find the x– 5 coordinates of all relative minima of f on the interval x ∈ [0, 7]. [Ans: 4.854]

5.A-12 [2] Consider the function f (x) =

5.A-14 (adapted from AB98) The first derivative of the function f is given by 3 cos2 x − . How many critical values does f have on the open f 0 (x) = x 20 interval (0, 10)?

[Ans: Five] 5.A-15 (adapted from AB97) If the derivative of f is given by f 0 (x) = 2x2 − ex , at which value of x does f have a relative minimum value? [Ans: 1.488]

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SL Unit 5, Day 2: Relating the Graphs of f , f 0 , and f 00

5.2

171

Relating the Graphs of f , f 0 , and f 00

International Baccalaureate dy and f 0 (x) for the first derivative. dx d2 y and f 00 (x) 7.2 The second derivative; Familiarity with both forms of notation, dx2 for the second derivative.

7.7 Graphical behaviour of functions; Included: both “global” and “local” behaviour.

The significance of the second derivative; Use of the terms “concave up” for f 00 (x) > 0 and “concave down” for f 00 (x) < 0. Points of inflection with zero and non-zero gradients; At a point of inflection f 00 (x) changes sign (concavity change). f 00 (x) = 0 is not a sufficient condition for a point of inflection; for example, y = x4 at (0, 0).

7.1 Familiarity with both forms of notation,

Textbook §3.5 Concavity and the Second Derivative Test; §3.6 Overview of Curve Sketching [15] Resources §4.3 Connecting f 0 and f 00 with the Graph of f in [8]

5.2.1

Concavity

There are four basic curve types... (Under Construction) The graph of f is concave upward if f 0 is increasing, i.e., f 00 is positive. The graph of f is concave downward if f 0 is decreasing, i.e., f 00 is negative.

If the function is concave down, then the curve is below the tangent line. If the function is concave up, then the curve is above the tangent line. If the function is concave up, then the tangent line underestimates the actual function; if the function is concave down, the tangent line overestimates the actual function.

Figure 5.8:

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SL Unit 5 (Extrema and Optimization) Example 5.2.1 The graph of a twice-differentiable function f is shown in Figure 5.8. Which of the following is true?

(A) f (1) < f 0 (1) < f 00 (1) (B) f (1) < f 00 (1) < f 0 (1) (C) f 0 (1) < f (1) < f 00 (1) (D) f 00 (1) < f (1) < f 0 (1) (E) f 00 (1) < f 0 (1) < f (1)

[Ans: D]

5.2.2

Second Derivative Test

Recall the First Derivative Test. The First Derivative Test tells me that: • there is a local maximum if the first derivative changes from positive to negative; • there is a local maximum if the first derivative changes from negative to positive.

Note that, if the first derivative is changing from positive to negative through zero, then it is decreasing. So, for a local maximum, if the second derivative exists, it is negative. This makes sense, because if the second derivative is negative, the function is concave down. If the function is concave down, it has a local maximum at the place where f 0 is zero. Similar arguments can be made for local minima.

Theorem 5.2 (Second Derivative Test, Graphically). Based on f 0 : • If f 0 is decreasing through zero at x = c, then f (c) is a local maximum, since the graph of f is concave up with a horizontal tangent.

• If f 0 is increasing through zero at x = c, then f (c) is a local minimum, since the graph of f is concave down with a horizontal tangent.

Or, based on f 00 : • If f 0 (c) = 0 and if f 00 (c) < 0, then f (c) is a local maximum; • If f 0 (c) = 0 and if f 00 (c) > 0, then f (c) is a local maximum; Mr. Budd, compiled January 12, 2011

SL Unit 5, Day 2: Relating the Graphs of f , f 0 , and f 00

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Figure 5.9: BC Acorn ’02

• If f 0 (c) = 0 and if f 00 (c) is 0 or undefined, then the Second Derivative Test fails to determine whether f (c) is a local extremum.

Note: we only use the Second Derivative Test for f 0 (c) = 0. Why would we not use the Second Derivative Test on those critical points where f 0 (c) is undefined?

5.2.3

Points of Inflection

An inflection point is a point on the graph of f where the concavity changes.

At inflection points on the graph of f , • f changes concavity; • f 0 changes direction, i.e., extrema of f 0 yield points of inflection of f ; • f 00 changes sign, i.e., if f 00 crosses the x-axis, then f has a point of inflection.

For inflection points, the tangent line will overestimate the curve on one side of the point, and underestimate it on the other side.

Be careful: for there to be an inflection point, it is not enough for the second derivative to be zero. The second derivative must change sign. This is similar to finding relative extrema: it is not enough for the first derivative to be 0, the first derivative must change sign for there to be a relative extremum.

Example 5.2.2 (BC Acorn ’02) Let f be a function whose domain is the open interval (1, 5). Figure 5.9 shows the graph of f 00 . Count the number of extrema of f 0 and points of inflection of the graph of f 0.

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SL Unit 5 (Extrema and Optimization) [Ans: one rel. min, two p.i.]

2

Example 5.2.3 (adapted from AB98) If f 00 (x) = x (x − 2) (x + 1) , then the graph of f has inflection points at what value(s) of x?

[Ans: 0 and 2 only]

Example 5.2.4 (adapted from BC98) If f is the function defined by f (x) = 3x5 +10x4 +10x3 −60x+7, what are all the x-coordinates of points of inflection for the graph of f ?

[Ans: 0]

Example 5.2.5 [3] Which of the following statements are true 3 about the function f if its derivative f 0 is defined by f 0 (x) = x (x − a) , a > 0.

I. The graph of f is increasing at x = 2a.

II. The function f has a local maximum at x = 0.

III. The graph of f has an inflection point at x = a.

[Ans: I and II only]

5.2.4

Connecting f , f 0 , and f 00 Figure 5.10: Graph of f

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Figure 5.11: Derivative of f from Figure 5.10

Example 5.2.6 (AB98) The graph of f is shown in Figure 5.10.

Sketch the derivative of f .

Example 5.2.7 The graph of f 0 , the derivative of f , is shown in Figure 5.1. Sketch a graph of f .

Figure 5.12: graph of f .

AB â&#x20AC;&#x2122;98 Note: This is the graph of the derivative of f , not the

Example 5.2.8 (AB96) Figure 5.12 shows the graph of f 0 , the derivative of a function f . The domain of f is the set of all real numbers such that â&#x2C6;&#x2019;3 < x < 5. (a) For what values of x does f have a relative maximum? Why? (b) For what values of x does f have a relative minimum? Why? (c) On what intervals is the graph of f concave upward? Use f 0 to justify your answer. (d) Suppose that f (1) = 0. Draw a sketch that shows the general shape of the graph of the function f on the open interval 0 < x < 2.

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[Ans: x = −2,f 0 (x) changes from positive to negative at x = −2] [Ans: x = 4, f 0 (x) changes from negative to positive at x = 4] [Ans: (−1, 1) and (3, 5), f 0 is increasing on these intervals] [Ans: Figure 5.13] Figure 5.13: AB ’96

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Problems 5.B-1 (BC97) The function f is defined on the closed interval [0, 8]. The graph of its derivative f 0 is shown in Figure 5.14. How many points of inflection Figure 5.14: BC ’97 The function f is defined on the closed interval [0, 8]. The graph of its derivative f 0 is shown.

does the graph of f have?

[Ans: Six]

5.B-2 (adapted from AB98) What is the x-coordinate of the point of inflection 1 on the graph of y = x3 − 5x2 − 13?

[Ans: 5] 3 2

5.B-3 (adapted from AB98) If f 00 (x) = x (x + 2) (x − 1) , then the graph of f has inflection points when x = [Ans: 0 and −2 only] 5.B-4 [3] Suppose a function f is defined so that it has derivatives f 0 (x) = x2 (1 − x) and f 00 (x) = x (2 − 3x). Over which interval is the graph of f both increasing and concave up?

Ans: 0 < x < 32 5.B-5 [3] Which of the following are true about the function f if its derivative is defined by 2 f 0 (x) = (x − 1) (4 − x) ?

I. f is decreasing for all x < 4.

II. f has a local maximum at x = 1.

III. f is concave up for all 1 < x < 3.

[Ans: III only] 2 5.B-6 (adapted from AB97) The graph of y = 3x4 − 8x3 − 24x + 162is concave down for what values of x?

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5.B-7 (adapted from [3]) The graph of y = 5x4 − 3x5 has a point of inflection for what value(s) of x?

[Ans: 1 only] 5.B-8 (adapted from [2]) An equation of the line tangent to the graph of y = x3 + 3x2 + 2 at its point of inflection is [Ans: y − 4 = −3 (x + 1)] 5.B-9 (adapted from AB98) What is the x-coordinate of the point of inflection [Ans: −2] on the graph of y = 32 x3 + 4x2 − 10 3 ?

5.B-10 (adapted from AB97) The graph of the function y = x3 + 6x2 + 7x − 2 sin x changes concavity at x = [Ans: −1.67] 5.B-11 (adapted from [2]) The number of inflection points for the graph of y = 2x + cos x2 in the interval 0 ≤ x ≤ 4 is [Ans: Five] 2

3

5.B-12 (adapted from [2]) The function f for which f 0 (x) = (x − 2) (x + 3) has an inflection point where x = [Ans: 0, 2] 5.B-13 (adapted from [3]) At what point does the graph of y = 5x4 − 3x5 have a point of inflection?

[Ans: (1, 2)] 5.B-14 (BC98) Let f be a function defined and continuous on the closed interval [a, b]. If f has a relative maximum at c and a < c < b, which of the following statements must be true?

I. f 0 (c) exists.

II. If f 0 (c) exists, then f 0 (c) = 0.

III. If f 00 (c) exists, then f 00 (c) ≤ 0.

[Ans: II and III only] 5.B-15 (AB99) Suppose that the function f has a continuous second derivative for all x, and that f (0) = 2, f 0 (0) = −3, and f 00 (0) = 0. Let g be a function whose derivative is given by g 0 (x) = e−2x (3f (x) + 2f 0 (x)) for all x. (a) Write an equation of the line tangent to the graph of f at the point where x = 0.

[Ans: y − 2 = −3 (x − 0)] (b) Is there sufficient information to determine whether or not the graph of f has a point of inflection when x = 0? Explain your answer.

[Ans: No] (c) Given that g(0) = 4, write an equation of the line tangent to the graph of g at the point where x = 0.

[Ans: y = 4] (d) Show that g 00 (x) = e−2x (−6f (x) − f 0 (x) + 2f 00 (x)). Does g have a local maximum at x = 0? Justify your answer. [Ans: Yes; g 0 (x) = 0, g 00 (x) < 0]

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SL Unit 5, Day 3: Interval Testing

5.3

179

Interval Testing

International Baccalaureate 7.1 Identifying increasing and decreasing functions.

7.3 Local maximum and minimum points; Testing for the maximum using change of sign of the first derivative.

7.7 Graphical behaviour of functions; Included: both “global” and “local” behaviour.

The significance of the second derivative; Use of the terms “concave up” for f 00 (x) > 0 and “concave down” for f 00 (x) < 0. Points of inflection with zero and non-zero gradients; At a point of inflection f 00 (x) changes sign (concavity change). f 00 (x) = 0 is not a sufficient condition for a point of inflection; for example, y = x4 at (0, 0).

Textbook §3.3 Maximum and Minimum Values; §3.4 Increasing and Decreasing Functions; §3.5 Concavity and the Second Derivative Test; §3.6 Overview of Curve Sketching [15]

5.3.1

Direction

Example 5.3.1 Find all local extrema of f (x) = the interval (0, 2π).

1 x + sin x over 2

q 2 3 Example 5.3.2 Find all local extrema of g(x) = (x2 − 2x − 8) .

Example 5.3.3 Show that the maximum value of h(x) = bxe−bx , b > 0 is independent of b.

Example 5.3.4 (MM 2004) Consider the function f (x) = 2+

1 . x−1

(a) Sketch f (x) (b) Write down the x-intercepts and y-intercepts of f (x). (c) Write down the equations of the asymptotes of f (x). (d) Find f 0 (x) (e) There are no maximum or minimum points on the graph of f (x). Use your equation for f 0 (x) to explain why.

Example 5.3.5 (MM 2003) Figure blah-blah-blah shows the graph of y = ex (cos x + sin x), −2 ≤ x ≤ 3. The graph has a maximum turning point at C (a, b).

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SL Unit 5 (Extrema and Optimization) dy . dx (b) Find the exact value of a and b. (a) Find

Example 5.3.6 (adapted from MM 2002) Let the function f (x) = 2 , x 6= 1.

1 + x3 (a) Write down the equation of the vertical asymptote of the graph of f . (b) Write down the equation of the horizontal asymptote of the graph of f . −6x2 (c) Show that f 0 (x) = 2.

(1 + x3 ) (d) Determine the x-coordinates for which f (a) (b) (c) (d)

5.3.2

is increasing; is decreasing; has a relative maximum; has a relative minimum.

Concavity

Recall concavity If f is concave up, then: • f is like a cup; • f 0 is increasing (if f 0 exists); • f 00 is positive (if f 00 exists); • the curve will lie above the tangent line.

Example 5.3.7 (AB97) For what interval is the graph of y = x4 − 9x3 + 27x2 − 45x + 36 is concave up?

Ans: x <

3 2

or x > 3

Example 5.3.8 For x2 + y 2 = 16: Remember the Quotient Rule: For what values of x or y is the graph of x2 + y 2 = 16 concave up?

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Example 5.3.9 (MM 2003) Figure blah-blah-blah shows the graph of y = ex (cos x + sin x), −2 ≤ x ≤ 3. The graph has a maximum turning point at C (a, b) and a point of inflection at D. dy . dx (b) Find the exact value of a and b. √ π (c) Show that at D, y = 2e 4 . (a) Find

Example 5.3.10 (adapted from MM 2002) Let the function f (x) = 2 , x 6= 1.

1 + x3 (a) Write down the equation of the vertical asymptote of the graph of f . (b) Write down the equation of the horizontal asymptote of the graph of f . (c) Using the fact that f 0 (x) =

−6x2

(1 + x3 ) 3 12x 2x − 1 derivative f 00 (x) = .

3 (1 + x3 )

2,

show that the second

(d) Find the x-coordinates of the points of inflection of the graph of f . (e) Use the trapezium rule with five subintervals to approximate R3 the integral f (x) dx.

1

(f) Given that

R3

f (x) dx = 0.637599, use a diagram to explain why

1

your answer is greater than this.

5.3.3

More Second Derivative Test

Theorem 5.3 (Second Derivative Test for Local Extrema). Let c be a critical value of f such that f 0 (c) = 0 and f 00 (c) exists.

1. If f 00 (c) > 0, then f (c) is a relative minimum.

2. If f 00 (c) < 0, then f (c) is a relative maximum.

3. If f 00 (c) = 0, then the test fails. In such cases, you can use the First Derivative Test.

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SL Unit 5 (Extrema and Optimization) Example 5.3.11 Find the critical values of y = 5x3 − 3x5 , and test each value to decide whether it corresponds to a relative maximum, a relative minimum, or neither. Use the second derivative test.

[Ans: −1 (rel min); 0 (neither); 1 (rel max)] For this problem, there is one point for which the second derivative test fails.

Can you determine whether it is a local extremum merely from the information you’ve received from the second derivative test? b Example 5.3.12 Let f (x) = x + , where b is a positive number. x Use the second derivative test to find and distinguish any relative extrema. h

i √ √ Ans: − b (rel max); b (rel min)

Example 5.3.13 (BC97) Where are the relative extrema of the function f given by f (x) = 3x5 − 4x3 − 3x.

[Ans: rel max at x = −1, rel min at x = 1]

Example 5.3.14 [3] Which of the following are true about the function f if its derivative is defined by 2

f 0 (x) = (x − 1) (4 − x) I. f is decreasing for all x < 4.

II. f has a local maximum at x = 1.

III. f is concave up for all 1 < x < 3.

[Ans: III only]

Problems 5.C-1 For the function h(x) = xe−x (a) Find the critical values of h(x); Mr. Budd, compiled January 12, 2011

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(b) Find all x-coordinates for which h is increasing; (c) Find the x-coordinates of any relative maxima on the graph of h; (d) Find the x-coordinates of any relative minima on the graph of h.

Justify all answers.

[Ans: 1; x < 1; x = 1; ∅]

5.C-2 Exactly find all critical values of f (x) = x2 ln x and classify each as a relative minimum, a relative maximum, or neither. Justify your answer. i h Ans: √1e , relative min 5.C-3 [3] At x = 0, which of the following is true of the function f (x) = sin x+e−x (a) f is increasing (b) f is decreasing (c) f is discontinuous (d) f is concave up (e) f is concave down [Ans: concave up] 1 1 + 3 2 x x [Ans: −2]

5.C-4 (adapted from AB93) At what value of x does the graph of y = have a point of inflection?

3 5.C-5 (adapted from [2]) Consider the function f (x) = x2 − 5 for all real numbers x. At what x-coordinates are the inflection √ √ for the graph points of f ?

Ans: − 5, −1, 1, 5 5.C-6 (adapted from [3]) The slope of the curve y = 32 x2 − e−x at its point of inflection is [Ans: 3 − ln 27]

Mr. Budd, compiled January 12, 2011

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SL Unit 5 (Extrema and Optimization)

Mr. Budd, compiled January 12, 2011

SL Unit 5, Day 4: Optimization

5.4

185

Local and Absolute Extrema

International Baccalaureate 7.3 Local maximum and minimum points; Testing for the maximum using change of sign of the first derivative and using sign of second derivative.

Use of the first and second derivative in optimization problems; Examples of applications; profit, area, volume.

7.7 Graphical behaviour of functions; Included: both “global” and “local” behaviour. The significance of the second derivative; Use of the terms “concave up” for f 00 (x) > 0 and “concave down” for f 00 (x) < 0. Points of inflection with zero and non-zero gradients; At a point of inflection f 00 (x) changes sign (concavity change). f 00 (x) = 0 is not a sufficient condition for a point of inflection; for example, y = x4 at (0, 0).

Textbook §3.3 Maximum and Minimum Values; §3.7 Optimization [15] Resources §8-3 Maxima and Minima in Plane and Solid Figures in [10]

5.4.1

Absolute Extrema

First of all know the difference between absolute extrema and relative extrema • absolute = global = higher (or lower) than everything • relative = local = higher (or lower) than everything nearby The absolute extremum (of a continuous function over a closed interval) will occur either at an endpoint or at a critical point.

To find absolute, aka global, extrema of f over [a, b]. [15] 1. Find the critical numbers of f in (a, b).

2. Evaluate f at each critical number in (a, b).

3. Evaluate f at the endpoints, i.e., at a and at b.

4. The lowest of these values is the absolute minimum, the highest is the absolute maximum.

5. Answer the question that’s asked. Do you need an x-value, or a y-value?

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SL Unit 5 (Extrema and Optimization) Example 5.4.1 Find the absolute minimum and the absolute maximum values of the following functions, over the following intervals: (a) f (x) = πx2 (12 − 3x) over [0, 3]

3 (b) f (x) = x 200 − x over [5, 146.25] 4

5.4.2

Optimization

Example 5.4.2 Old Mac-Donald had a farm. And on that farm, he had some cows, which fought incessantly. In order to separate the Cowpulets from the Montamoos, he built two identical, adjacent rectangular pens which share a side. Old Mac-Donald hires you to find out the dimensions of the pen which give you the greatest area if Old MacDonald has, say, 600 yards of fencing.

[Ans: 75 yards by 100 yards adjoining] • Draw a (big) picture.

• Label variables.

• Write an expression for what you’re maximizing or minimizing.

• Write an equation for the constraints.

• Use the constraint to reduce your maximizing/ minimizing expression to one variable.

• Use the physical reality of the problem to determine end values.

• Differentiate to find critical values.

• Plug end and critical values into your maximizing/ minimizing expression.

• Answer the question that’s asked, not the question that you want to answer.

Technique: Analysis of Maximum-Minimum Problems [10] 1. Make a sketch if one isn’t already drawn.

2. Write an equation for the variable you are trying to maximize or minimize.

3. Get the equation in terms of one variable and specify a domain.

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4. Find an approximate maximum or minimum by grapher.

5. Find the exact maximum or minimum by seeing where the derivative is zero or infinite. Check any endpoints of the domain.

6. Answer the question by writing what was asked for in the problem statement.

College Board does not recognize maxima or minima calculated on the graphing calculator. Justifications require mathematical (i.e., noncalculator) reasoning.

Example 5.4.3 A 400 meter track is to be built around a field that consists of a rectangle with two semicircles at either end. (The base of each semicircle spans the entire width of the rectangle.) How should the track be built in order to maximize (a) the area of the rectangle; (b) the total enclosed area.

Ans: length 100 m, radius

100 π

m ; circle of radius

200 π

m

Example 5.4.4 [from Acorn ’02] Consider one arch of cos x above the x-axis. Draw a rectangle that lies on the x-axis so that its top two vertices lie on the curve y = cos x. Shade the area between y = cos x and the x-axis that is not in the rectangle. Find the minimum area of the shaded region.

[Ans: 0.878] Example 5.4.5 [10] Barb Dwyer must build a rectangular corral along the river bank. Three sides of the corral will be fenced with a barbed wire. The river forms the fourth side of the corral (Figure 5.15). The total length of fencing available is 1000 feet. What is the maximum area the corral could have? How should the fence be built to enclose this maximum area? Justify your answers.

[Ans: 125,000 square feet, 250 feet ⊥ to river by 500 feet parallel] Example 5.4.6 [10] The part of the parabola y = 4 − x2 from x = 0 to x = 2 is rotated about the y-axis to form a surface. A Mr. Budd, compiled January 12, 2011

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SL Unit 5 (Extrema and Optimization)

Figure 5.15: Figure 5.16: cone is inscribed in the resulting paraboloid with its vertex at the origin and its base touching the parabola (Figure 5.16). At which radius and altitude does the maximum volume occur? What is the maximum volume? Justify your answer. √

Ans: x = 2, y = 2, V =

4π 3

Example 5.4.7 (adapted from BC93) Consider all right circular cylinders for which the sum of the height and the circumference is 30π centimeters. What is the radius of the one with maximum volume?

[Ans: 10 cm]

Problems 2

5.D-1 (BC93) If f (x) = 1 + x 3 , which of the following is NOT true?

(A) f is continuous for all real numbers.

(B) f has a minimum at x = 0.

(C) f is increasing for x > 0.

(D) f 0 (x) exists for all x.

(E) f 00 (x) is negative for x > 0.

[Ans: D]

Mr. Budd, compiled January 12, 2011

Unit 6

Definite Integrals 1. Riemann Sums, Area, and Definite Integrals 2. Fundamental Theorem of Calculus 3. Definite Integrals as Sums: Area 4. Definite Integrals as Sums: Volume

189

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SL Unit 6 (Definite Integrals)

Mr. Budd, compiled January 12, 2011

SL Unit 6, Day 1: Evaluating Definite Integrals

6.1

191

Evaluating Definite Integrals

International Baccalaureate 7.4 Indefinite integration as anti-differentiation.

7.5 Definite integrals.

Areas under curves (between the curve and the x-axis), areas between curves.

7.6 Kinematic problems involving displacement, s, velocity, v, and acceleration, a.

Textbook §5.1 Area between Curves [15]

6.1.1

Definite Integral

The definite integral can be seen as several things: 1. An infinite sum. Area is an infinite sum of infinitesimally thin rectangles.

2. A product. Displacement is velocity times time. How is a product really a sum?

3. An accumulated change. If oil is leaking out of a tank at certain rate Rb R(t), then a R(t) dt represents how much oil has leaked out from t = a to t = b. This is the accumulation of all the oil that has leaked out at a rate R(t). Change in velocity is the accumulation of accleration over a certain time period.

6.1.2

Riemann Sums

Example 6.1.1 (AB 1997) The expression r r r r !

1 1 2 3 50 + + + ... + 50 50 50 50 50 is a Riemann sum approximation for r R1 x (a) 0 dx 50 R1â&#x2C6;&#x161; (b) 0 x dx r x 1 R1 (c) dx 50 0 50 Mr. Budd, compiled January 12, 2011

192

SL Unit 6 (Definite Integrals) 1 R1√ x dx 50 0 1 R 50 √ x dx (e) 50 0

(d)

h i R1√ Ans: 0 x dx

Mean Value Theorem Theorem 6.1 (Mean Value Theorem). If f (x) is a function that is continuous over [a, b] and differentiable over (a, b), then ∃ c ∈ (a, b) 3 f 0 (c) =

f (b) − f (a) b−a

A nice, smooth function has a spot where the tangent line is parallel to the secant line, i.e., where the instantaneous rate of change matches the average rate of change.

Example 6.1.2 The function L(t) = 10 000 e−0.2 − e−0.2t gives the number of gallons that have leaked out of a tanker, where t the time in hours after noon.

L(3) − L(1) . Explain the meaning of this value. [Ans: 1349.596] 3−1 Find the average rate at which oil leaked out of the tanker from 3 p.m. to 9 p.m. Indicate units.

[Ans: 639.188 gal/hr] 0 Find L (t). Using correct units, explain the meaning of L0 (t).

At what time between 1 p.m. and 3 p.m. is the instantaneous rate of leakage the same as the average rate of leakage over that same time interval?

At what time between 3 p.m. and 9 p.m. is the instantaneous rate of leakage the same as the average rate of leakage over that same time interval?

(a) Find (b) (c) (d)

(e)

Ans: 2 000e−0.2t , rate of leakage; t = 1.96671 (1:58); t = 5.70352 (5:42) Example 6.1.3 (adapted slightly from AB ’02) Let f be a function that is differentiable for all real numbers. Table 6.5 gives the values of f and its derivative f 0 for selected points x in the closed interval −1.5 ≤ x ≤ 1.5. The second derivative of f has the property that f 00 (x) > 0 for −1.5 ≤ x ≤ 1.5.

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Table 6.1: AB ’02 x f (x) f 0 (x)

−1.5 −1 −7

−1.0 −4 −5

−0.5 −6 −3

0 −7 0

0.5 −6 3

1.0 −4 5

1.5 −1 7

(a) Find a positive real number r having the property that there must exist a value c with 0 < c < 0.5 and f 00 (c) = r. Give a reason for your answer. ( 2x2 − x − 7 for x < 0 (b) Let g be the function given by g(x) = 2x2 + x − 7 for x ≥ 0 The graph of g passes through each of the points (x, f (x)) given in the table above. Is it possible that f and g are the same function? Give a reason for your answer. (c) Write an equation of the line tangent to the graph of f at the point where x = 1. Use this line to approximate the value of f (1.2). Is this approximation greater than or less than the actual value of f (1.2)? Give a reason for your answer. (d) Write an equation of the line tangent to the graph of f at the point where x = 1.5. Use this line to approximate the value of f (1.2). Which of these approximations do you suppose is more accurate, and why?

[Ans: 6; no; −3 < f (1.2); −3.1 < −3 < f (1.2)]

Example 6.1.4 (from Acorn ’02) Let f be a function such that f 00 (x) < 0 for all x in the closed interval [1, 2]. Selected values of f are shown in Table 6.6. Which of the following must be true about Table 6.2: Acorn ’02 # 18 x f (x)

1.1 4.18

1.2 4.38

1.3 4.56

1.4 4.73

f 0 (1.2)? (a) f 0 (1.2) < 0 (b) 0 < f 0 (1.2) < 1.6 (c) 1.6 < f 0 (1.2) < 1.8 Mr. Budd, compiled January 12, 2011

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SL Unit 6 (Definite Integrals) (d) 1.8 < f 0 (1.2) < 2.0 (e) f 0 (1.2) > 2.0 Example 6.1.5 (adapted slightly from AB ’06B) A car travels on a straight track. During the time interval 0 ≤ t ≤ 60 seconds, the car’s velocity v, measured in feet per second, and acceleration a, measured in feet per second per second, are continuous functions.

Table 6.7 shows selected values of these functions.

Table 6.3: AB ’06B t (sec) v(t) (ft/sec) a(t) (ft/sec2 )

0

15

25

30

35

50

60

−20

−30

−20

−14

−10

0

10

1

5

2

1

2

4

2

Z

60

(a) Using appropriate units, explain the meaning of v(t) dt in Z 60 30 terms of the car’s motion. Approximate v(t) dt using a 30

trapezoidal approximation with the three subintervals determined by the table.

Z 60 (b) Using appropriate units, explain the meaning of |v(t)| dt Z 60 30 in terms of the car’s motion. Approximate |v(t)| dt using 30

a trapezoidal approximation with the three subintervals determined by the table. (c) For 0 < t < 60, must there be a time t when v(t) = −5? Justify your answer. (d) For 0 < t < 60, must there be a time t when a(t) = 0? Justify your answer. (e) For 0 < t < 60, must there be a time t when a0 (t) = 0? Justify your answer.

Example 6.1.6 (adapted slightly from AB ’03) A blood vessel is 360 millimeters (mm) long with circular cross sections of varying diameter. Table 6.8 gives the measurements of the diameter of the blood vessel at selected points along the length of the blood vessel, Mr. Budd, compiled January 12, 2011

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Table 6.4: AB ’04B Distance x (mm) Diameter B(x) (mm)

0

60

120

180

240

300

360

24

30

28

30

26

24

26

where x represents the distance from one end of the blood vessel and B(x) is a twice–differentiable function that represents the diameter at that point.

Z 360 B(x) 1 dx (a) Using correct units, explain the meaning of 360 0 2 Z 360 1 B(x) in terms of the blood vessel. Approximate the value of dx 360 0 2 using the data from the table and a midpoint Riemann sum with three subintervals of equal length. Show the computations that lead to your answer. Is your answer reasonable?

2 Z 360 B(x) dx (b) Using correct units, explain the meaning of π 2 0 in terms of the blood vessel. (c) Explain why there must be at least one value x, for 0 < x < 360, such that B 00 (x) = 0.

[Ans: 14 mm; volume of the b.v. from x = 0 to x = 360; B 0 (c1 ) = B 0 (c2 ) = 0, . . . ]

6.1.3

Fundamental Theorem of Calculus

Given: Let f (x) be the derivative of some function G(x) over the interval [a, b].

G(x) is differentiable over [a, b] and over any subinterval inside [a, b]. Why? G(x) is continuous over [a, b] and any subinterval therein. Why? If G(x) is continuous and differentiable, what conclusions can we draw?

Recall: • We partition the interval [a, b] into n subintervals: a = x0 < x1 < x2 < · · · < xn−1 < xn Mr. Budd, compiled January 12, 2011

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SL Unit 6 (Definite Integrals) Pn • The Riemann sum is given by i=1 f (ci )4xi , where 4xi = xi −xi−1 , and ci ∈ [xi−1 , xi ]. ci represents the sample points, or evaluation points within each and every subinterval, where the height is taken for the rectangular approximation of the funky strip.

• There are multiple rules for how to choose the sample points, e.g., left, right, midpoint, Monte Carlo, etc.

The proof involves setting up a Riemann sum such that the sample points, ci , for each subinterval were chosen as theR points guaranteed by the Mean Value Theorem for the antiderivative G(x) = f (x) dx for each subinterval [xi−1 , xi ].

That is, G(xi ) − G(xi−1 ) G0 (ci ) = xi − xi−1 Recognizing that G0 (ci ) = f (ci ) and xi − xi−1 = 4xi , the conclusion of the MVT becomes G(xi ) − G(xi−1 ) f (ci ) = 4xi Replacing this in the formula for the Riemann sum, we have Rn

= =

n X i=1 n X i=1

=

n X

f (ci )4xi G(xi ) − G(xi−1 ) 4xi 4xi G(xi ) − G(xi−1 )

i=1

=

G(x1 ) − G(x0 ) + G(x2 ) − G(x1 ) + G(x3 ) − G(x2 ) .. . + G(xn ) − G(xn−1 )

Rn

=

G(xn ) − G(x0 )

=

G(b) − G(a)

Note that if we choose our sample points this way, then the value we get for Rn , G(b) − G(a), is independent of n, the number of subintervals. So that we can have three, six, one hundred, one million, or even one subinterval, and we get the same value for the Riemann sum, which is therefore the same value for Mr. Budd, compiled January 12, 2011

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197

the definite integral. b

Z

f (x) dx

=

a

=

lim Rn

n→∞

lim G(b) − G(a)

n→∞

= G(b) − G(a)

Example 6.1.7 (adapted from BC ’97) The closed interval [a, b] is partitioned into n equal subintervals, each of width ∆x, by the numbers x0 , x1 , . . . , xn where a = x0 < x1 < x2 < · · · < xn−1 < n √ P 3 x ∆x? xn = b. What is lim i n→∞ i=1

3 4/3 4/3 Ans: b −a 4

6.1.4

Fundamental Theorem of Calculus

Example 6.1.8 (adapted from AB ’97) (a) Find Z 2 i 1 Z 3 ii 1 Z b iii Z1 x iv

4x3 − 6x2 dx

4x3 − 6x2 dx

4x3 − 6x2 dx

4t3 − 6t2 dt

1

(b) What do you suppose is the meaning of Z 2

1 4x3 − 6x2 dx i 2−1 1 Z 3

1 ii 4x3 − 6x2 dx 3−1 1 Z b

1 iii 4x3 − 6x2 dx b−1 1 Z x

(c) Let A(x) = 4t3 − 6t2 dt. Find A0 (x).

1

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Ans: 1, 28, b4 − 2b3 + 1, ; ; 4x3 − 6x2

Example 6.1.9 (AB97) If

Rb a

f (x) dx = a+2b, then

Rb a

(f (x) + 5) dx =

[Ans: 7b − 4a]

Example 6.1.10 (adapted from Acorn ’05) Oil is leaking from a tanker at the rate of R(t) = 2 000e−0.2t gallons per hour, where t is measured in hours after noon.

R9 (a) Using correct units, explain the meaning of 1 R(t) dt.

R9 (b) Approximate 1 R(t) dt using (a) M4 , a midpoint Riemann sum with 4 equal subintervals; (b) T4 , a trapezoidal approximation with 4 equal subintervals; 2M4 + T + 4 [Ans: 6490.959; 6621.211; 6534.376] (c) S8 = 3 (c) Find the exact amount of oil that leaked out between 1 p.m. and 9 p.m.

Ans: 10 000 e−0.2 − e−1.8 = 6534.319 gallons R9 1 (d) Find 9−1 R(t) dt. Using correct units, explain the meaning 1 of this value. You can think of this as (signed) area divided by width. [Ans: 816.790 gal/hr; average of the rate of leakage from 1 p.m. to 9 p.m.] (e) Let L(T ) be the amount of oil that has leaked out from 1 p.m. to time t = T . Write an integral expression for L(T ). Find a formulah for L(T ) that does not require an integral, then findi

RT L0 (T ) Ans: 1 R(t) dt = 10 000 e−0.2 − e−0.2T ; 2 000e−0.2T

Problems Z 6.A-1 [2]

1

sin (πx) dx = 0

2 Ans: π

6.A-2 What is the area of the region above the x-axis, below the graph of 1 y = sec2 (πx), between the lines x = 0 and x = ? What do you 3

suppose is the significance of dividing this area by the width, 31 − 0 ? h i √ Ans: π3 ; average height π 6.A-3 (adapted from AB ’98) If 0 ≤ k ≤ and the area under the curve y = cos x 2 π from x = k to x = is 0.2, then k = [Ans: 0.927] 2 Mr. Budd, compiled January 12, 2011

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6.A-4 (AB ’96) Let R be the region in the first quadrant under the graph of

1 y = √ for 4 ≤ x ≤ 9.

Ans: 2; 25 4 x 6.A-5 (AB ’02) Let f and g be the functions given by f (x) = ex and g(x) = ln x.

Find the area of the region enclosed by the graphs of f and g between 1 x = and x = 1.

[Ans: 1.223] 2 2

6.A-6 (AB acorn ’02) What is the average value of the function f (x) = e−x on the closed interval [−1, 1]?

[Ans: 0.747] Rxp 6.A-7 [2] The average rate of change of the function f (x) = 0 1 + cos (t2 ) dt over the interval [1, 3] to three decimal places is: [Ans: 0.858] 6.A-8 (adapted The approximate average rate of change of the function R xfrom [2]) f (x) = 0 sin t2 dt over the interval [2, 5] is [Ans: −0.092] 6.A-9 (BC97) following, which is the greatest value of x such R x If 0 ≤ x ≤ 4, ofRthe x that 0 t2 − 2t dt ≥ 2 t dt?

(A) 1.35 (B) 1.38 (C) 1.41 (D) 1.48 (E) 1.59 [Ans: B] 6.A-10 (adapted from AB ’04) Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute. The traffic flow at a particular intersection is modeled by the function F defined by

t F (t) = 80 + 6 cos for 0 ≤ t ≤ 30.

2 where F (t) is measured in cars per minute and t is measured in minutes. (a) To the nearest whole number, how many cars pass through the intersection during the middle twenty minutes, i.e., for 5 ≤ t ≤ 25? (b) What is the average traffic flow during the middle twenty minutes?

Indicate units of measure. (c) What is the average rate of change of traffic flow during the middle twenty minutes? Indicate units of measure.

[Ans: 1592 cars; 79.601 cars per minute; 0.540 cars per minute per minute] Mr. Budd, compiled January 12, 2011

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6.A-11 (AB ’05B) A water tank at Camp Newton holds 960 gallons of water at time t = 0. During the time interval 0 ≤ t ≤ 18 hours, water is pumped into the tank at the rate of

√ t W (t) = 76 t cos2 gallons per hour.

6 During the same time interval, water is removed from the tank at the rate

t 2 R(t) = 220 cos gallons per hour.

3 (a) At what rate is water being removed from the tank at t = 15? Indicate units of measure. (b) Is the amount of water in the tank increasing at t = 15? Why or why not?

[Ans: Y;] (c) How much water has been pumped into the tank during the first fifteen hours? What is the average rate at which water is pumped into the tank during that time period?

[Ans: 933.042 gal; 62.203 gph] (d) How much water has been removed from the tank during the first fifteen hours? What is the average rate at which water is removed from the tank during that time period? (e) To the nearest gallon, how much water is in the tank at t = 15? t = 18?

[Ans: 333; ] (f) Write an expression for the gallons of water in the tank at time t. (g) Write an expression for the rate at hwhich the total gallons of wateri Rt in the tank is changing at time t.

Ans: 960 + 0 W (z) − R(z) dz (h) Find the best candidates for the time, 0 ≤ t ≤ 18 at which the amount of water in the tank is an absolute minimum. [Ans: 2.487, 12.450] (i) Find the best candidates for the time, 0 ≤ t ≤ 18 at which the amount of water in the tank is an absolute maximum.

[Ans: 0, 6.198, 18] (j) How would you choose between the best candidates to determine the actual time in which the water is at an absolute extremum?

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6.2

201

Area

International Baccalaureate 7.4 Indefinite integration as anti-differentiation.

7.5 Definite integrals.

Areas under curves (between the curve and the x-axis), areas between curves.

7.6 Kinematic problems involving displacement, s, velocity, v, and acceleration, a.

Textbook §5.1 Area between Curves [15]

6.2.1

Definite Integral

The definite integral can be seen as several things: 1. An infinite sum. Area is an infinite sum of infinitesimally thin rectangles.

2. A product. Displacement is velocity times time. How is a product really a sum?

3. An accumulated change. If oil is leaking out of a tank at certain rate Rb R(t), then a R(t) dt represents how much oil has leaked out from t = a to t = b. This is the accumulation of all the oil that has leaked out at a rate R(t). Change in velocity is the accumulation of accleration over a certain time period.

6.2.2

Riemann Sums

Example 6.2.1 (AB 1997) The expression r r r r !

1 1 2 3 50 + + + ... + 50 50 50 50 50 is a Riemann sum approximation for r R1 x (a) 0 dx 50 R1â&#x2C6;&#x161; (b) 0 x dx r x 1 R1 (c) dx 50 0 50 Mr. Budd, compiled January 12, 2011

202

SL Unit 6 (Definite Integrals) 1 R1√ x dx 50 0 1 R 50 √ x dx (e) 50 0

(d)

h i R1√ Ans: 0 x dx

6.2.3

Area: Slicing dx

The area between two curves is a sum of an infinite number of infinitesimally thin rectangles.

If you slice the area into vertical strips, then the area of each infinitesimally thin rectangle is given by dA = (highy − lowy) dx, and the area is given by Z b Area = (highy − lowy) dx a

Example 6.2.2 (adapted from Ostebee & Zorn [16]) Find the area of the region R bounded by the curves x = 0, y = 2, and y = ex . (a) Find the area of R, with and without a calculator. (b) If the line x = h divides the region R into two regions of equal area, what is the value of h?

[Ans: 0.386; 0.219 ; 1.683] Example 6.2.3 (adapted from AB ’00) Let R be the region in the 2 first quadrant enclosed by the graphs of y = e−x , y = 1 − cos x, and the y-axis. (a) Find the area of R. (b) If the line x = k divides the region R into two regions of equal area, what is the value of k?

[Ans: 0.591; 0.310] Example 6.2.4 Find the area of the region bounded by the graphs 1 of y = ex/2 , y = 2 , x = 2, and x = 3. Try this with, and without x a calculator.

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203

Ans: 2 e3/2 − e −

1 6

= 3.360

π Example 6.2.5 (adapted from AB ’98) If 0 ≤ k ≤ and the area 2 π under the curve y = cos x from x = k to x = is 0.2, then k = 2 [Ans: 0.927]

6.2.4

High and Low y Switch

At times, the graphs may intersect between the bounds of integration, and the high and low y functions might switch. In this case, you should divide the integral up.

When you get to use the calculator, feel free to use abs When you must use the fundamental theorem instead of a calculator to evaluate the definite integral, the absolute value will be worthless. Why? In that case, you need to separate Example 6.2.6 Find the area of the region bounded by the graphs 4 of y = x, y = , x = 1, and x = 4. Try this with x (a) without a calculator; (b) with a calculator, but without using absolute value; (c) with a calculator, using absolute value.

Ans:

9 2

+ 4 ln 4 − 8 ln 2 =

9 2

= 4.500

Example 6.2.7 (adapted from AB ’83) Do the following problem without a calculator, then with a calculator. Find the area bounded by the curve f (x) = 3x2 − 12x + 9 and the x-axis, between the lines x = 0 and x = 2.

[Ans: 6] Example 6.2.8 Find the area of the region bounded by the graphs 1 of y = 2 , y = x, and y = 2. x Mr. Budd, compiled January 12, 2011

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SL Unit 6 (Definite Integrals)

Ans:

6.2.5

7 2

√ −2 2

Area: Slicing dy

If you slice the area into vertical strips, then the area of each infinitesimally thin rectangle is given by dA = (rightx − leftx) dy, and the area is given by Z

b

(rightx − leftx) dy

Area = a

Example 6.2.9 (adapted from Ostebee & Zorn [16]) Find the area of the region R bounded by the curves x = 0, y = 2, and y = ex . (a) Find the area of R, with and without a calculator. (b) If the line y = k divides the region R into two regions of equal area, what is the value of k?

[Ans: 0.386; 0.219 ; 1.683] Example 6.2.10 Find the area of the region bounded by the graphs 1 of y = 2 , y = x, and y = 2. Try this two ways. x

Ans:

6.2.6

7 2

√ −2 2

Total Distance

Distance : Displacement :: Area : Definite Integral There is a difference between the total distance traveled and the displacement.

When you go backward, distance is counted positively, but displacement is counted negatively. How is that similar to the relationship between area and the definite integral?

Speed is the magnitude (absolute value of velocity), and total distance traveled is the accumulation (i.e., definite integral) of speed. Displacement is the accumulation of velocity.

D=

Rb a

|v(t)| dt Mr. Budd, compiled January 12, 2011

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Just like area, total distance traveled must be positive.

In a graph of velocity, the definite integral yields the displacement, i.e., change in position. The area yields the total distance traveled.

Example 6.2.11 (adapted from AB 1997) A bug begins to crawl up a vertical wire at time t = 0. The velocity v of the bug at time t, 0 ≤ t ≤ 8, is given by the function whose graph is shown in Figure 6.1 Figure 6.1: Vertical velocity of a bug

(a) At value of t does the bug change direction? (b) What is the total distance the bug traveled from t = 0 to t = 8? (c) What is the net displacement of the bug between t = 0 and t = 8? (d) What is the total distance traveled downward by the bug? upward? (e) What is the bug’s velocity at t = 5? The acceleration at t = 5? (f) What other questions could we ask about the bug?

Example 6.2.12 (AB ’03) A particle moves along the x-axis so that its velocity at time t is given by 2 t v(t) = − (t + 1) sin 2 At time t = 0, the particle is at position x = 1. (a) Find the acceleration of the particle at time t = 2. Is the speed of the particle increasing at t = 2? Why or why not? (b) Find all times t in the open interval 0 < t < 3 when the particle changes direction. Justify your answer.

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SL Unit 6 (Definite Integrals) (c) Find the total distance traveled by the particle from time t = 0 until time t = 3. (d) During the time interval 0 ≤ t ≤ 3, what is the greatest distance between the particle and the origin? Show the work that leads to your answer.

Ans: 1.588, yes;

√

2π; 4.334; 2.265

Example 6.2.13 (adapted slightly from AB ’83) A particle moves along the x-axis so that at time t its position is given by x(t) = t3 − 6t2 + 9t + 11. (a) What is the velocity of the particle at time t? (b) During what time intervals is the particle moving to the left? (c) What is the total distance traveled by the particle from t = 0 to t = 2? Do this two ways.

R2 (d) What does 0 v(t) dt represent?

Ans: 3t2 − 12t + 9; 1 < t < 3; 6; x(2) − x(0)

Problems 6.B-1 (AB ’93) Set up a definite integral to find the area shaded regioni h of the Rb in Figure 6.2.

Ans: a (d − f (x)) dx Figure 6.2: AP Calculus AB (1993)

6.B-2 (AB ’96) Let R be the region in the first quadrant under the graph of 1 y = √ for 4 ≤ x ≤ 9. x Mr. Budd, compiled January 12, 2011

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(a) Find the area of R. (b) If the line x = k divides the region R into two regions of equal area, what is the value of k?

Ans: 2; 25 4 6.B-3 (adapted from AB ’97) The area of the region enclosed by the graph of y = x2 − 2x + 2 and the line y = 10 is [Ans: 36] 6.B-4 (adapted from AB ’98) What is the area of the region between the graphs

of y = x3 and y = −x from x = 0 to x = 3?

Ans: 99 4 6.B-5 (AB ’02B) Let R be the region bounded by the y–axis and the graphs of x3 and y = 4 − 2x. Find the area of R. [Ans: 3.215] y= 1 + x2 6.B-6 (adapted from BC Acorn) Find the area of the region R bounded by the curves x = 1, y = 1, and y = e3x . (a) Find the area of R, by slicing dx, then again by slicing dy. (b) If the vertical line x = h divides the region R into two regions of equal area, what is the value of h? (c) If the horizontal line y = k divides the region R into two regions of equal area, what is the value of k?

[Ans: 5.362; 0.814 ; 5.065] x2 y2 6.B-7 Find the area enclosed by the ellipse + = 1. What is area in the 16 9 form kπ? What do you suppose is the area enclosed by any generic ellipse y2 x2 + = 1?

[Ans: 12π] a2 b2 6.B-8 (AB ’02) Let f and g be the functions given by f (x) = ex and g(x) = ln x.

Find the area of the region enclosed by the graphs of f and g between 1 x = and x = 1.

[Ans: 1.223] 2 √ 6.B-9 (AB ’03) Let R be the region bounded by the graphs of y = x and y = e−3x and the vertical line x = 1. Find the area of R. [Ans: 0.443] 6.B-10 (adapted from AB ’03B) Let f be the function given by f (x) = 4x2 − x3 , and let ` be the line y = 18 − 3x, where ` is tangent to the graph of f .

Let R be the region bounded by the graph of f and the x-axis, and let S be the region bounded by the graph of f , the line `, and the x-axis. (a) At what point is ` tangent to f (x)? (b) Find the area of R. (c) Find the area of S. Mr. Budd, compiled January 12, 2011

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Ans: (3, 0);

64 3

;7.917

6.B-11 (AB ’05)

1 Let f and g be the functions given by f (x) = + sin (πx) and g(x) = 4−x .

4 Let R be the shaded region in the first quadrant enclosed by the y–axis and the graphs of f and g, and let S be the shaded region in the first quadrant enclosed by the graphs of f and g, as shown in the figure above. (a) Find the area of R. (b) Find the area of S. [Ans: 0.0648; 0.410] 6.B-12 (AB ’05B) Let f and g be the functions given by f (x) = 1 + sin (2x) and g(x) = ex/2 . Let R be the region in the first quadrant enclosed by the graphs of f and g. Find the area of R. [Ans: 0.429] 6.B-13 (BC Acorn 2000) A particle moves along the x-axis so that at any time t ≥ 0 its velocity is given by v(t) = ln (t + 1) − 2t2 + 4t − 1. (a) What is the total distance traveled by the particle from t = 0 to t = 2?

[Ans: 2.178] (b) What is the net displacement of the particle between t = 0 and t = 2?

[Ans: 1.963] 6.B-14 (AB ’87) A particle moves along the x-axis so that its acceleration at any time t is given by a(t) = 6t − 18. At time t = 0 the velocity of the particle is v(0) = 24, and at time t = 1 its position is x(1) = 20.

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(a) Write an expression for the velocity v(t) of the particle at any time t. (b) For what values of t is the particle at rest? (c) Write an expression for the position of the particle at any time t. (d) Find the total distance traveled by the particle from t = 1 to t = 3.

Ans: 3t2 − 18t + 24; 2, 4; t3 − 9t2 + 24t + 4; 6 6.B-15 (AB ’93) A particle moves on the x-axis so that its position at any time t ≥ 0 is given by x(t) = 2te−t . Find the total distance traveled by the particle from t = 0 to t = 5. Try this using fnInt, then try this on your calculator, without calculating a definite integral.

[Ans: 1.404] 6.B-16 (AB 1997) A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by v(t) = 3t2 − 2t − 1. The position x(t) is 5 for t = 2. (a) Write a polynomial expression for the position of the particle at any time t ≥ 0. (b) For what values of t, 0 ≤ t ≤ 3, is the particle’s instantaneous velocity the same as its average velocity on the closed interval [0, 3]? (c) Find the total distance traveled by the particle from t = 0 until time t = 3. Try it with a calculator, then without a calculator.

Ans: t3 − t2 − t + 3; t = 1.786; 17

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Mr. Budd, compiled January 12, 2011

SL Unit 6, Day 3: Volume

6.3

211

Volumes of Rotation: Sweet, Sweet Loaves of Calculus

International Baccalaureate 7.5 Definite integrals.

Volumes of Revolution.

Textbook §5.2 Volume: Slicing, Disks, and Washers [15] Resources §8.2 Finding Volumes by Integration in Ostebee and Zorn [16]. §6.2 Volumes in Stewart [20]. §7.4 Volumes in Finney [8].

6.3.1

Slicing into Discs

In general, b

Z V =

A(x) dx a

or Z V =

d

A(y) dy c

where A(x) or (A(y)) represents the cross-sectional area of the solid at a particular value of x (or y) 2

For volumes of rotation where cross sections are discs, then A(x) = π [r(x)] or 2 A(y) = π [r(y)] .

Example 6.3.1 (adapted from AB 1997) Let R be the √ region bounded by the y-axis, the line y = 2, and the curve y = x. (a) Find the area of region R. (b) Find the volume of the solid generated when region R is rotated about the y-axis.

Ans:

8 32π 3; 5

Things to keep in mind for volume of rotation problems: • If the axis of rotation is the x-axis, or parallel to the x-axis, slice dx. If the axis of rotation is the y-axis, or parallel to the y-axis, slice dy.

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SL Unit 6 (Definite Integrals) • V =

Rb a

πr2 d

• If slicing dy, the radius will be a high x minus a low x. If slicing dx, the radius will be a high y minus a low y. If the axis of rotation is either the x- or y-axis, one of these values will be zero.

1 Example 6.3.2 [3] A region in the plane is bounded by y = √ , x the x-axis, the line x = m, and the line x = 2m where m > 0. A solid is formed by revolving the region about the x-axis. The volume of this solid (A) is independent of m (B) increases as m increases (C) decreases as m decreases 1 (D) increases until m = , then decreases 2 (E) is none of the above

[Ans: A]

Example 6.3.3 Derive, from scratch, the formula for the volume of a sphere of radius r.

Example 6.3.4 Derive, from scratch, the formula for the volume of a cone with height h and base radius r.

Example 6.3.5 [16] The following table gives the circumference (in inches) of a pole at several heights (in feet).

Height Circumference

0 16

10 14

20 10

30 5

40 3

50 2

60 1

Assuming that cross sections of the pole taken parallel to the ground are circles, estimate the volume of the pole using: (a) T3 (b) M3 (c) T6 (d) S2·3 =

M3 + 2T3 , a weighted average of M3 and T3 .

3 Mr. Budd, compiled January 12, 2011

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Nonstandard axes of rotation Remember that each radius is a high y minus a low y (or high x minus low x).

Example 6.3.6 Let R be the region bounded by the graphs of y = ex/2 , y = 1, and x = ln 2. (a) Set up, but do not solve, a definite integral that could be used to find the area of R. (b) Set up, but do not solve, a definite integral that could be used to find the volume of the region obtained by rotating R about the line y = 1. (c) Set up, but do not solve, a definite integral that could be used to find the volume of the region obtained by rotating R about the line x = ln 2. (d) Preview: what changes if the axes of rotation are y = −1 or x = 0?

6.3.2

Slicing into Washers

For washers, h i h i 2 2 2 2 A(x) = π (R(x)) − π (r(x)) = π (R(x)) − (r(x)) or

i h i h 2 2 2 2 A(y) = π (R(y)) − π (r(y)) = π (R(y)) − (r(y))

Remember that each radius is [high y - low y].

Example 6.3.7 Let R be the region bounded by the graphs of y = ex/2 , y = 1, and x = ln 2. Set up, but do not solve, a definite integral that could be used to find the volume of the region obtained by rotating R about (a) the y-axis; (b) the line y = −1; (c) the line x = −1; (d) the line x = 1; √ (e) the line y = 2 Mr. Budd, compiled January 12, 2011

214

SL Unit 6 (Definite Integrals) Example 6.3.8 ([3]) Let R be the region in the first quadrant bounded above by the graph of f (x) = 2 arctan x and below by the graph of y = x. What is the volume of the solid generated when R is rotated about the x-axis?

[Ans: 7.151] Example 6.3.9 (AB â&#x20AC;&#x2122;02) Let f and g be the functions given by f (x) = ex and g(x) = ln x. Find the volume of the solid generated 1 when the region enclosed by the graphs of f and g between x = 2 and x = 1 is revolved about the line y = 4.

[Ans: 23.609] Example 6.3.10 (AB â&#x20AC;&#x2122;97) Let f be the function given by f (x) = 3 cos x. As shown in Figure 6.3, the graph of f crosses the y-axis at point P and the x-axis at point Q. Figure 6.3: AB Exam 1997

(a) Write an equation for the line passing through points P and Q. (b) Write an equation for the line tangent to the graph of f at point Q. Show the analysis that leads to your equation. (c) Find the x-coordinate of the point on the graph of f , between points P and Q, at which the line tangent to the graph of f is parallel to line P Q. Mr. Budd, compiled January 12, 2011

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(d) Let R be the region in the first quadrant bounded by the graph of f and the line segment P Q. Write an integral expression for the volume of the solid generated by revolving the region R about the x-axis.

h

Ans: y − 3 = − π6 (x − 0); y − 0 = −3 (x − π/2); 0.690; π

R π/2 h 0

2

(3 cos x) − − π6 x + 3

Problems 6.C-1 (AB 1993) [No calculator! ]The region enclosed by the x-axis, the line √ is the x = 3, and the curve y = x is rotated about the x-axis. What

volume of the solid generated?

Ans: 29 π 6.C-2 (adapted from AB ’03B) Let R be the region bounded by the graph of f (x) = 4x2 − x3 and the x-axis. Find the volume of the solid generated when R is revolved about the x-axis.

[Ans: 490.208] 6.C-3 [20] A log 10 m long is cut at 1-meter intervals and its cross-sectional areas A (at a distance x from the end of the log) are listed in the table. x (m) 0 1 2 3 4 5

A (m2 ) 0.68 0.65 0.64 0.61 0.58 0.59

x (m) 6 7 8 9 10

A (m2 ) 0.53 0.55 0.52 0.50 0.48

Estimate the volume of the log using: (a) the Midpoint Rule with n = 5. (b) the Trapezoid Rule with n = 5. (c) the Trapezoid Rule with n = 10. (d) S2·5 =

Ans: 5.8 m3

Ans: 5.7 m3

Ans: 5.75 m3

M5 + 2T10 , a weighted average of M5 and T10 . Ans: 5.767 m3 3

6.C-4 [8] We wish to estimate the volume of a flower vase using only a calculator, a string, and a ruler. We measure the height of the vase to be 6 inches.

We then use the string and the ruler to find the circumference of the vase (in inches) at half-inch intervals. (We list them from starting at the top Mr. Budd, compiled January 12, 2011

2 i

i dx

216

SL Unit 6 (Definite Integrals) Circumferences 5.4 10.8 4.5 11.6 4.4 11.6 left, moving down) 5.1 10.8 6.3 9.0 7.8 6.3 9.4 (a) Sketch the vase. (b) Find the areas of the cross sections that correspond to the given circumferences. (c) Express the volume of the vase as an integral with respect to y over the interval [0, 6]. (d) Approximate the integral using the Trapezoidal Rule with n = 12. i h R6 2 1 3 [C(y)] dy; 34.7 in Ans: 2.3, 1.6, 1.5, . . .; 4π 0

6.C-5 (adapted from AB ’00) Let R be the region in the first quadrant enclosed 2 by the graphs of y = e−x , y = 1 − cos x, and the y-axis. (a) Find the volume of the solid generated when the region R is revolved about the x–axis. (b) Find the volume of the solid generated when the region R is revolved about the line y = −1. (c) Suppose R is revolved around the line y = k, where k > 1 so that the line is above the region. Find k if the volume of this solid is the same as the volume of the solid in the previous part, where R is revolved around the line y = −1.

[Ans: 1.747; 5.460; 1.941] √ 6.C-6 (AB ’03) Let R be the region bounded by the graphs of y = x and y = e−3x and the vertical line x = 1. Find the volume of the solid generated when R is revolved about the horizontal line y = 1. [Ans: 1.424] 6.C-7 (AB Acorn ’04-05) Let S be the region enclosed by the graphs of y = 2x and y = 2x2 for 0 ≤ x ≤ 1. Write a definite integral for the volsolid generated when S is revolved about the line y = 3. hume of the i

R1 2 2 2 Ans: π 0 3 − 2x − (3 − 2x) dx 6.C-8 (adapted from AB ’02B) Let R be the region bounded by the y–axis and x3 the graphs of y = and y = 4 − 2x. Find the volume of the solid 1 + x2 generated when R is revolved about Mr. Budd, compiled January 12, 2011

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(a) the x–axis. (b) the line y = 4.

[Ans: 31.885;48.906] 6.C-9 (AB ’05B) Let f and g be the functions given by f (x) = 1 + sin (2x) and g(x) = ex/2 . Let R be the region in the first quadrant enclosed by the graphs of f and g. Find the volume generated when R is revolved about the x–axis.

[Ans: 4.267] 6.C-10 (AB ’05)

1 Let f and g be the functions given by f (x) = + sin (πx) and g(x) = 4−x .

4 Let S be the shaded region in the first quadrant enclosed by the graphs of f and g, as shown in the figure above. Find the volume of the solid generated when S is revolved about the horizontal line y = −1.

[Ans: 4.558]

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Mr. Budd, compiled January 12, 2011

SL Unit 6, Day 4: Evaluating Definite Integrals

6.4

219

Evaluating Definite Integrals

International Baccalaureate 7.4 Indefinite integration as anti-differentiation.

7.5 Definite integrals.

Areas under curves (between the curve and the x-axis), areas between curves.

7.6 Kinematic problems involving displacement, s, velocity, v, and acceleration, a.

Mean Value Theorem Theorem 6.2 (Mean Value Theorem). If f (x) is a function that is continuous over [a, b] and differentiable over (a, b), then ∃ c ∈ (a, b) 3 f 0 (c) =

f (b) − f (a) b−a

A nice, smooth function has a spot where the tangent line is parallel to the secant line, i.e., where the instantaneous rate of change matches the average rate of change.

Example 6.4.1 The function L(t) = 10 000 e−0.2 − e−0.2t gives the number of gallons that have leaked out of a tanker, where t the time in hours after noon.

L(3) − L(1) . Explain the meaning of this value. [Ans: 1349.596] 3−1 (b) Find the average rate at which oil leaked out of the tanker from 3 p.m. to 9 p.m. Indicate units.

[Ans: 639.188 gal/hr] (a) Find

(c) Find L0 (t). Using correct units, explain the meaning of L0 (t). (d) At what time between 1 p.m. and 3 p.m. is the instantaneous rate of leakage the same as the average rate of leakage over that same time interval? (e) At what time between 3 p.m. and 9 p.m. is the instantaneous rate of leakage the same as the average rate of leakage over that same time interval?

Ans: 2 000e−0.2t , rate of leakage; t = 1.96671 (1:58); t = 5.70352 (5:42)

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Table 6.5: AB ’02 x f (x) f 0 (x)

−1.5 −1 −7

−1.0 −4 −5

−0.5 −6 −3

0 −7 0

0.5 −6 3

1.0 −4 5

1.5 −1 7

Example 6.4.2 (adapted slightly from AB ’02) Let f be a function that is differentiable for all real numbers. Table 6.5 gives the values of f and its derivative f 0 for selected points x in the closed interval −1.5 ≤ x ≤ 1.5. The second derivative of f has the property that f 00 (x) > 0 for −1.5 ≤ x ≤ 1.5. (a) Find a positive real number r having the property that there must exist a value c with 0 < c < 0.5 and f 00 (c) = r. Give a reason for your answer. ( 2x2 − x − 7 for x < 0 (b) Let g be the function given by g(x) = 2x2 + x − 7 for x ≥ 0 The graph of g passes through each of the points (x, f (x)) given in the table above. Is it possible that f and g are the same function? Give a reason for your answer. (c) Write an equation of the line tangent to the graph of f at the point where x = 1. Use this line to approximate the value of f (1.2). Is this approximation greater than or less than the actual value of f (1.2)? Give a reason for your answer. (d) Write an equation of the line tangent to the graph of f at the point where x = 1.5. Use this line to approximate the value of f (1.2). Which of these approximations do you suppose is more accurate, and why?

[Ans: 6; no; −3 < f (1.2); −3.1 < −3 < f (1.2)] Example 6.4.3 (from Acorn ’02) Let f be a function such that f 00 (x) < 0 for all x in the closed interval [1, 2]. Selected values of f are shown in Table 6.6. Which of the following must be true about Table 6.6: Acorn ’02 # 18 x f (x)

1.1 4.18

1.2 4.38

1.3 4.56

1.4 4.73

f 0 (1.2)?

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(a) f 0 (1.2) < 0 (b) 0 < f 0 (1.2) < 1.6 (c) 1.6 < f 0 (1.2) < 1.8 (d) 1.8 < f 0 (1.2) < 2.0 (e) f 0 (1.2) > 2.0

Example 6.4.4 (adapted slightly from AB ’06B) A car travels on a straight track. During the time interval 0 ≤ t ≤ 60 seconds, the car’s velocity v, measured in feet per second, and acceleration a, measured in feet per second per second, are continuous functions.

Table 6.7 shows selected values of these functions.

Table 6.7: AB ’06B t (sec) v(t) (ft/sec) a(t) (ft/sec2 )

0

15

25

30

35

50

60

−20

−30

−20

−14

−10

0

10

1

5

2

1

2

4

2

Z 60 (a) Using appropriate units, explain the meaning of v(t) dt in Z 60 30 terms of the car’s motion. Approximate v(t) dt using a 30

trapezoidal approximation with the three subintervals determined by the table.

Z 60 (b) Using appropriate units, explain the meaning of |v(t)| dt Z 60 30 in terms of the car’s motion. Approximate |v(t)| dt using 30

a trapezoidal approximation with the three subintervals determined by the table. (c) For 0 < t < 60, must there be a time t when v(t) = −5? Justify your answer. (d) For 0 < t < 60, must there be a time t when a(t) = 0? Justify your answer. (e) For 0 < t < 60, must there be a time t when a0 (t) = 0? Justify your answer.

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SL Unit 6 (Definite Integrals) Example 6.4.5 (adapted slightly from AB ’03) A blood vessel is 360 millimeters (mm) long with circular cross sections of varying diameter. Table 6.8 gives the measurements of the diameter of the Table 6.8: AB ’04B Distance x (mm) Diameter B(x) (mm)

0

60

120

180

240

300

360

24

30

28

30

26

24

26

blood vessel at selected points along the length of the blood vessel, where x represents the distance from one end of the blood vessel and B(x) is a twice–differentiable function that represents the diameter at that point.

Z 360 1 B(x) (a) Using correct units, explain the meaning of dx 360 0 2 Z 360 1 B(x) in terms of the blood vessel. Approximate the value of dx 360 0 2 using the data from the table and a midpoint Riemann sum with three subintervals of equal length. Show the computations that lead to your answer. Is your answer reasonable?

2 Z 360 B(x) (b) Using correct units, explain the meaning of π dx 2 0 in terms of the blood vessel. (c) Explain why there must be at least one value x, for 0 < x < 360, such that B 00 (x) = 0.

[Ans: 14 mm; volume of the b.v. from x = 0 to x = 360; B 0 (c1 ) = B 0 (c2 ) = 0, . . . ]

6.4.1

Fundamental Theorem of Calculus

Given: Let f (x) be the derivative of some function G(x) over the interval [a, b].

G(x) is differentiable over [a, b] and over any subinterval inside [a, b]. Why? G(x) is continuous over [a, b] and any subinterval therein. Why? If G(x) is continuous and differentiable, what conclusions can we draw?

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• We partition the interval [a, b] into n subintervals: a = x0 < x1 < x2 < · · · < xn−1 < xn Pn • The Riemann sum is given by i=1 f (ci )4xi , where 4xi = xi −xi−1 , and ci ∈ [xi−1 , xi ]. ci represents the sample points, or evaluation points within each and every subinterval, where the height is taken for the rectangular approximation of the funky strip.

• There are multiple rules for how to choose the sample points, e.g., left, right, midpoint, Monte Carlo, etc.

The proof involves setting up a Riemann sum such that the sample points, ci , for each subinterval were chosen as theR points guaranteed by the Mean Value Theorem for the antiderivative G(x) = f (x) dx for each subinterval [xi−1 , xi ].

That is, G(xi ) − G(xi−1 ) G0 (ci ) = xi − xi−1 Recognizing that G0 (ci ) = f (ci ) and xi − xi−1 = 4xi , the conclusion of the MVT becomes G(xi ) − G(xi−1 ) f (ci ) = 4xi Replacing this in the formula for the Riemann sum, we have Rn

= =

n X i=1 n X i=1

=

n X

f (ci )4xi G(xi ) − G(xi−1 ) 4xi 4xi G(xi ) − G(xi−1 )

i=1

=

G(x1 ) − G(x0 ) + G(x2 ) − G(x1 ) + G(x3 ) − G(x2 ) .. . + G(xn ) − G(xn−1 )

Rn

=

G(xn ) − G(x0 )

=

G(b) − G(a)

Note that if we choose our sample points this way, then the value we get for Rn , G(b) − G(a), is independent of n, the number of subintervals. So that we Mr. Budd, compiled January 12, 2011

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can have three, six, one hundred, one million, or even one subinterval, and we get the same value for the Riemann sum, which is therefore the same value for the definite integral.

Z

b

f (x) dx

=

a

=

lim Rn

n→∞

lim G(b) − G(a)

n→∞

= G(b) − G(a)

Example 6.4.6 (adapted from BC ’97) The closed interval [a, b] is partitioned into n equal subintervals, each of width ∆x, by the numbers x0 , x1 , . . . , xn where a = x0 < x1 < x2 < · · · < xn−1 < n √ P 3 x ∆x? xn = b. What is lim i n→∞ i=1

3 4/3 4/3 Ans: b −a 4

6.4.2

Fundamental Theorem of Calculus

Example 6.4.7 (adapted from AB ’97) (a) Find Z 2 i 1 Z 3 ii 1 Z b iii Z1 x iv

4x3 − 6x2 dx

4x3 − 6x2 dx

4x3 − 6x2 dx

4t3 − 6t2 dt

1

(b) What do you suppose is the meaning of Z 2

1 4x3 − 6x2 dx i 2−1 1 Z 3

1 ii 4x3 − 6x2 dx 3−1 1 Z b

1 iii 4x3 − 6x2 dx b−1 1 Mr. Budd, compiled January 12, 2011

SL Unit 6, Day 4: Evaluating Definite Integrals Z (c) Let A(x) =

225

x

4t3 − 6t2 dt. Find A0 (x).

1

Ans: 1, 28, b4 − 2b3 + 1, ; ; 4x3 − 6x2

Example 6.4.8 (AB97) If

Rb a

f (x) dx = a+2b, then

Rb a

(f (x) + 5) dx =

[Ans: 7b − 4a]

Example 6.4.9 (adapted from Acorn ’05) Oil is leaking from a tanker at the rate of R(t) = 2 000e−0.2t gallons per hour, where t is measured in hours after noon.

R9 (a) Using correct units, explain the meaning of 1 R(t) dt.

R9 (b) Approximate 1 R(t) dt using (a) M4 , a midpoint Riemann sum with 4 equal subintervals; (b) T4 , a trapezoidal approximation with 4 equal subintervals; 2M4 + T + 4 [Ans: 6490.959; 6621.211; 6534.376] (c) S8 = 3 (c) Find the exact amount of oil that leaked out between 1 p.m. and 9 p.m.

Ans: 10 000 e−0.2 − e−1.8 = 6534.319 gallons R 9 1 (d) Find 9−1 R(t) dt. Using correct units, explain the meaning 1 of this value. You can think of this as (signed) area divided by width. [Ans: 816.790 gal/hr; average of the rate of leakage from 1 p.m. to 9 p.m.] (e) Let L(T ) be the amount of oil that has leaked out from 1 p.m. to time t = T . Write an integral expression for L(T ). Find a formulah for L(T ) that does not require an integral, then findi

RT L0 (T ) Ans: 1 R(t) dt = 10 000 e−0.2 − e−0.2T ; 2 000e−0.2T

Problems

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Part II

I.B. Topics

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Unit 7

Statistics 1. Descriptive Statistics and Presentation: Discrete Data 2. Descriptive Statistics and Presentation: Continuous Data 3. Cumulative Frequency and Percentiles

229

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7.1

231

Descriptive Statistics: Discrete Data

International Baccalaureate 6.1 Concepts of population, sample, random sample and frequency distribution of discrete .

6.2 Presentation of data: frequency tables and diagrams, box and whisker plots.

Grouped data: mid-interval values, interval width, upper and lower interval boundaries, frequency histogram.

6.3 Mean, median, mode; quartiles, percentiles.

Awareness that the population mean, µ, is generally unknown, and that the sample mean, x, serves as an unbiased estimate of this quantity. Range; interquartile range; variance, standard deviation.

Awareness of the concept of dispersion and an understanding of the significance of the numerical value of the standard deviation.

Obtain the standard deviation (and indirectly the variance) from a GDC and by other methods.

Awareness that the population standard variance, σ 2 , is generally unknown, and that the s2n−1 serves as an unbiased estimate of σ 2 .

6.4 Cumulative frequency; cumulative frequency graphs; use to find median, quartiles.

Recall Mean. Median. Mode. Standard deviation. Frequency.

When we first talk about mean, median, mode, and standard deviation, we usually restrict ourselves to discrete data, and you are probably used to dealing with discrete data. Discrete data are taken from a limited set of values. The number rolled on a die, for example, cannot be any value between 1 and 6, but only one of {1, 2, 3, 4, 5, 6}. Data that are not discrete are continuous.

Example 7.1.1 Give some examples of discrete data. Give some examples of continuous data.

Sometimes the line between discrete data and continuous data gets blurry. While heights can take on any value, there is a limit to our precision of measurement, so that measured values could be considered discrete. Your age, for example, might be 17.247713 . . . years, but you probably just say 17. The way we handle continuous data is basically to make the data discrete by putting the data in groups, or classes.

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SL Unit 7 (Statistics)

Central Tendency

Central Tendency is basically, “Where is the data?” It gives, in some sense, a typical data value.

Mean

The mean, µ, of a population is given by N P

µ=

x1 + x2 + · · · + xN = N

xi

i=1

N

where N is the size of the population. This is the average as you are used to finding it.

Median

As you know from TAKS, the median, µ ˜, is the middle item when the items are placed in order. If there are an even number of items, then it is the average of the middle two.

Sometimes you don’t have all the individual items, as they may be grouped.

Since you don’t know the individual items, the median must be calculated in a different manner. It is calculated as the 50th percentile. Percentiles will be discussed in Section 7.3.2.

Mode

As you also know from TAAS, the mode is the most frequently occurring item.

If a set of data is perfectly symmetric, the mean, median, and mode are the same value. If the data are skewed, then the mean, median, and mode will be different values.

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Other Other measures of central tendency include the geometric mean, v uN uY N t x, i i=1

and the harmonic mean, 1 N

1 N P i=1

, 1 xi

which is the reciprocal of the average reciprocal.

The following example is taken from Probability and Statistics.

Example 7.1.2 [17] Unfair Coin Tossed Thrice This example will be used frequently in this and the next sections. Given a coin for which the probability of head on a single toss is 0.6. Consider the experiment of tossing the coin three times. Let X stand for the number of heads obtained. Suppose the experiment is repeated 40 times, the data, i.e., values of X, in Table 7.1 are obtained. Find the mean, median, and mode.

Table 7.1: Unfair Coin Tosses - Number of Heads 1 2 2 1

7.1.2

0 1 1 2

2 2 2 2

2 1 0 0

3 3 2 2

1 3 1 1

2 2 3 3

1 1 2 2

2 2 1 3

2 3 2 2

Dispersion

Dispersion is, basically, “How spread out is the data?” There is a difference between the set of data {−100, 0, 100} and {−1, 0, 1}, but you would never know that from measures of the central tendency.

Range The range is the difference between the highest and lowest values in the population or sample. It is highly susceptible to outliers.

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Variance and Standard Deviation Example 7.1.3 The mean deviation is the the average difference between each item and the mean value, i.e., N P

(xi − µ)

i=1

N Is the average difference from the mean a good measure of dispersion? Why or why not? How could you improve it?

The variance of a population is given by N P

σ2 =

(xi − µ)

2

i=1

N

It is an average; it is the average square difference between each item and the mean value. The difference is squared to make it positive. (If we didn’t square it, we’d be averaging a bunch of positive and negative numbers, resulting in 0).

The difference between the item and the mean gives you an idea of how far the values are from the mean.

Example 7.1.4 (SL 89) The variance of the numbers 5, 11, and k is equal to 14. Find the possible values of their mean.

[Ans: 6, 10] Example 7.1.5 Show that the variance is the average of the sqaures minus the square of the average.

The standard deviation is the square root of the variance. v u N uP 2 u t i=1 (xi − µ) σ= N It is generally easier to conceptualize because it has the same units as both the sample items and the mean.

Example 7.1.6 Refer to Table 7.1 and find the variance and standard deviation.

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Interquartile Range The interquartile range is the difference between the first and third quartiles, or the 25th and 75th percentiles. Percentiles will be discussed in more detail in Section 7.1.4.

Remember that the median is the middle value, or average of the two middles (once the data is sorted). The first quartile is the median of the bottom half, and the third quartile is the median of the top half.

Example 7.1.7 [4] Find the interquartile range of {2, 6, 5, 4, 1, 0, 5, 2}.

[Ans: 3.5]

Others The mean absolute deviation is the average absolute value of the difference from the mean: N P |xi − µ| i=1

N The standard deviation is preferred to the mean absolute deviation only because there are lots of nice formulas that can be devised with the standard deviation.

7.1.3

Frequency

The frequency is the number of times a particular value occurs in a sample (or population). A frequency distribution for a sample is a tallying and recording of the frequencies for each value.

Example 7.1.8 Refer to Table 7.1 on page 233. Make a frequency chart.

[Ans: The frequency chart is shown in Table 7.2.]

If we’re given the frequency data, where fi is the frequency for each individual value xi , i = 1, 2, . . . k, then we can find the sample (or population) mean by Mr. Budd, compiled January 12, 2011

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SL Unit 7 (Statistics)

Table 7.2: Biased Coin Tossed Thrice Number of Frequency heads, x f 0 3 1 11 2 19 3 7 Total=40

using k

µ or x =

x1 f1 + x2 f2 + · · · + xk fk 1X = xi fi . f1 + f2 + · · · + fk n i=1

Similarly, 2

σ 2 or s2n =

2

k

2

(x1 − x) f1 + (x2 − x) f2 + · · · + (xk − x) fk 1X 2 = (xi − x) fi . f1 + f2 + · · · + fk n i=1

As always, for sn−1 , divide by n − 1 instead of n.

Example 7.1.9 (MM 5/04) Table 7.3 below shows the marks gained in a test by a group of students.

Table 7.3: Marks on a Test Mark Number of students

1 5

2 10

3 p

4 6

5 2

The median is 3 and the mode is 2. Find the two possible values of p.

[Ans: 8, 9] Example 7.1.10 Table 7.4 displays the frequency of occurrences of scores in a competition.

The mean score is 15. Find k.

[Ans: 11] The relative frequency is the frequency divided by the total frequency, n.

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Table 7.4: Problem 8 - Scores in a Competition Score 12 13 14 15 16 17 Frequency 2 4 7 13 k 5

7.1.4

Cumulative Frequency

The cumulative frequency, F (x), of a score x, is basically, ‘What is the combined (or cumulative) frequency of all the scores less than or equal to x’ ?

Cumulative Distribution Function Definition 7.1. [17] The cumulative distribution function [or empirical cumulative distribution function or cumulative frequency distribution] at x0 is defined to be X 1 X fx Fˆ (x0 ) = fˆ (x) = N x≤x0

x≤x0

Example 7.1.11 Refer to the problem where an unfair coin is thrown three times, and the number of heads is recorded. (Page 233). (a) Develop a table for the cumulative frequency in terms of x. (b) Use your table to find the median and interquartile range.

Check your answers with a calculator (c) Develop a table for the cumulative relative frequency in terms of x.

Table 7.5: Cumulative Frequency for Biased Coin Tossed Thrice Cumulative Cumulative No. of heads; x Frequency, f frequency relative frequency 0 3 3 3/40 = 0.075 1 11 3 + 11 = 14 14/40 = 0.350 2 19 14 + 19 = 33 33/40 = 0.825 3 7 33 + 7 = 40 40/40 = 1.000

The answer is given in Table 7.5.

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Figure 7.1: Graph of empirical cumulative distribution function for a discrete random variable.

Graphs If the data are discrete and not grouped, then the function cannot take on values between the isolated data points. Refer to the case of flipping a biased coin three times, counting the number of heads (page 233). Values that are recorded as 2 are 2, and cannot be 1.77893 or 2.431. The frequency of getting less than or equal to 2.73 heads is the exactly the same as the frequency of getting less than or equal to 2.998 heads, which is exactly the frequency of getting less than or equal to 2 heads. (Compare this to continuous data: We would expect more women to be shorter than 67.48 inches than are shorter than 66.82 inches.) In the case of discrete and ungrouped data, we do not connect the dots with lines, we make a series of horizontal lines that start at one data point and end at he next data point. See Figure 7.1 for an example. Note the appropriate use of closed and open circles on the graph.

7.1.5

Population vs. Sample

Sampling Usually, it can be difficult to get statistics for an entire population. Instead, we take a sample, which is a portion of the whole population. CBS cannot determine whom everyone in the state of Florida voted for on election night, so they do exit polling to get a sample of voters. Proper sampling requires that the sample be representative of the whole population, and is a whole field unto itself.

If there are 1376 women at Gates College, then taking the heights of 50 women at Gates college is using a sample. Any statistics will be sample statistics, such as the sample mean, sample variance, and sample standard deviation. Usually we use these sample statistics as estimates of the population statistics, because Mr. Budd, compiled January 12, 2011

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the population is too big to measure, and we don’t have statistics for the entire population.

Sample mean The sample mean, represented by x, is just the mean (average) of the sample: n P

x1 + x2 + · · · + xn x= = n

xi

i=1

n

where n is the size of the sample.

The sample mean happens to give us what is called an unbiased estimator of the population mean.

Sample variance and standard deviation In I.B. world, the sample variance s2n , and sample standard deviation, sn is calculated the same way as the population variance and standard deviation.

That is, n P 2 (xi − x) p s2n = i=1 , sn = s2n . n This gives us an estimate of the variance (and hence standard deviation) for the entire population. The problem is that if a sample size is one (i.e., n = 1) then this gives a standard deviation of 0. In reality, if our sample size is one, we have no way of knowing what the dispersion is like. Mathematically, this estimate is said to be biased, and is not used in formulas that relate the population data to the sample data.

An unbiased estimate of the population variance is given by: n P

s2n−1

=

2

(xi − x)

i=1

n−1

,

sn−1 =

q

s2n−1

The only difference is the division by n − 1 instead of n. If n = 1, then you can’t have an unbiased estimate of the population standard deviation.

Example 7.1.12 (HL 5/03) A teacher drives to school. She records the time taken on each of the 20 randomly chosen days. She finds Mr. Budd, compiled January 12, 2011

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SL Unit 7 (Statistics) that

20 X i=1

xi = 626 and

20 X

x2i = 19780.8 ,

i=1

where xi denotes the time, in minutes, taken on the ith day. Calculate an unbiased estimate of (a) the mean time taken to drive to school; (b) the variance of the time taken to drive to school.

[Ans: 31.3;9.84]

Differences in notation and terminology Now, this wouldn’t necessarily be a big deal, because we have two different formulas, one for the sn , the sample standard deviation (which isn’t very useful mathematically) and one for sn−1 , the unbiased estimate of σ. The problem is in notation and terminology. Table 7.6 summarizes several different notations for the two different standard deviations you get by dividing by n or by n − 1.

Useless Useful

Table 7.6: Notation for Sample Standard Deviation Source What I.B. calls it I.B. TI-83 [7] [17] “Sample Standard Deviation” sn σx s1 s˜ “Unbiased Estimate of σ” sn−1 sx s2 sˆ

[1] σ s

The titles of sn and sn−1 are in quotes because some texts [1],[7] call I.B.’s sn−1 the sample standard deviation. Their point is that if you are dividing by n and not n − 1, then you are treating the data as a population, and what you are finding is σ, the population standard deviation. When you divide by n − 1, you are treating the set of data as a sample, and this standard deviation should be labeled as such. I would tend to agree with these people, but I don’t write the I.B. exams. While I disagree with the I.B. terminology, their notation is perhaps the most clear, distinguishing between n and n − 1.

A very big difference in notation that you need to be aware of is that when I.B. asks for sn , and you do this on your TI-83, you need to use σx and not sx .

It is important for you to understand the differences in notation so that you can get the correct information from the calculator, or from a reference book.

Remember that the unbiased estimate of σ is always larger than sn , because you’re dividing by a smaller number.

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Problems 7.A-1 (MM 5/02) From January to September, the mean number of car accidents per month was 630. From October to December, the mean was 810 accidents per month.

What was the mean number of car accidents per month for the whole year?

[Ans: 675] 7.A-2 (MM 5/01) Given the frequency distribution in Table 7.7, find (a) the median; (b) the mean.

[Ans: 4; 3.8] Table 7.7: Problem 2 - Frequency Distribution Number (x) 1 2 3 4 5 6 Frequency (f ) 5 9 16 18 20 7

7.A-3 (SL 5/07 TZ2) The population below is listed in ascending order.

5, 6, 7, 7, 9, 9, r, 10, s, 13, 13, t The median of the population is 9.5. The upper quartile Q3 is 13. (a) Write down the values of r, s. (b) The mean of the population is 10. Find the value of t.

[Ans: 10, 13; 18] 7.A-4 (MM 5/98) For the population of data x1 , x2 , x3 , x4 , . . . , xi , . . . , x20 , it is 20 20 P P known that xi = 480, and x2i = 12500. i=1

i=1

Find (a) the mean of the population; (b) the standard deviation of the population, s, by making use of the formula P 2 20 20 P x2i xi  i=1   s2 = i=1 −  20  .

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7.A-5 (MM 5/97) Consider the set of scores: 3, 0, 7, 2, 8, 3, 11, a, b. The mean score is 5, the median score is 4, and a < b. Find a and b.

[Ans: 4,7] 7.A-6 (MM 11/95) A set of fifty scores is recorded in the Table 7.8 Table 7.8: Problem 6 - Scores score 1 2 3 4 frequency 20 10 5 15

Calculate the variance of this set.

[Ans: 1.61]

7.A-7 (MM 95) The Mathematical Methods candidates at Trillium College received the grades shown in Table 7.9 in 1994.

Table 7.9: Problem 7 - Trillium College Candidatesâ&#x20AC;&#x2122; Methods Scores Grade 1 2 3 4 5 6 7 Frequency 0 3 7 10 9 5 6

Find (a) the mean grade;

[Ans: 4.6]

(b) the standard deviation of grades.

[Ans: 1.48]

7.A-8 (MM 94) Table 7.4 displays the frequency of occurrences of scores in a competition.

The mean score is 15. (a) Find k.

[Ans: 11]

(b) Calculate the standard deviation of the scores.

[Ans: 1.31]

7.A-9 (SL 91) The mean height of a group of students is 181 cm. Another student whose height is 163 cm joins the group, and the mean height is thereby reduced to 179 cm. What is the number of students in the original group?

[Ans: 8] 7.A-10 (SL 84) Table 7.10 shows the frequencies of the scores when a die is thrown 36 times.

Table 7.10: Problem 10 - Die scores Scores 1 2 3 4 5 6 Frequencies 7 8 9 5 4 3

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Find the mean m and the standard deviation s for this sample, giving the value of s correct to three decimal places. [Note: could you have done this in 1984, when graphing calculators werenâ&#x20AC;&#x2122;t allowed?]

[Ans: 3; 1.53] 7.A-11 (SL 82) A family consists of a mother, father, and three children aged 3, 8, and 10 years. The mother and father are both of the same age. Find the fatherâ&#x20AC;&#x2122;s age if the mean age of the five members of the family is 17 years.

[Ans: 32] 7.A-12 (HL 5/03) A teacher drives to school. She records the time taken on P20 each day of 20 randomly chosen days. She finds that i=1 xi = 626 and P20 2 i=1 xi = 19780.8, where xi denotes the time, in minutes, taken on the ith day.

Calculate an unbiased estimate of (a) the mean time taken to drive to school; (b) the variance of the time taken to drive to school.

[Ans: 31.3; 9.84] 7.A-13 (HL 5/01) A machine produces packets of sugar. The weights in grams of thirty packets chosen at random are shown in Figure 13.

Weight (g) Frequency

Problem 13: Weights of sugar packets 29.6 29.7 29.8 29.9 30.0 30.1 2 3 4 5 7 5

30.2 3

30.3 1

Find unbiased estimates of (a) the mean of the population from which this sample is taken. [Ans: 29.9] (b) the variance of the population from which this sample is taken.

[Ans: 0.0336] 7.A-14 (HL Spec â&#x20AC;&#x2122;00) A machine fills bottles with orange juice. A sample of six bottles is taken at random. The bottles contain the following amounts (in mL) of orange juice: 753, 748, 749, 752, 750, 751.

Find (a) the sample mean; (b) the sample variance; (c) an unbiased estimate of the population variance from which this sample is taken.

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Table 7.11: Problem15 - Hours of Studentsâ&#x20AC;&#x2122; Sleep Hours of sleep 4 5 6 7 8 10 12

Number of students 2 5 4 3 4 2 1

7.A-15 (MM 5/03) The number of hours of sleep of 21 students are shown in the frequency table shown in Table 7.11.

Find (a) the median;

[Ans: 6]

(b) the lower quartile;

[Ans: 5]

(c) the interquartile range.

[Ans: 3]

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7.2

245

Descriptive Statistics: Continuous Data

International Baccalaureate 6.1 Concepts of population, sample, random sample and frequency distribution of continuous data.

6.2 Presentation of data: frequency tables and diagrams, box and whisker plots.

Treatment of both continuous and discrete data.

Grouped data: mid-interval values, interval width, upper and lower interval boundaries, frequency histograms.

6.3 Mean, median, mode; quartiles, percentiles.

Awareness that the population mean, µ, is generally unknown, and that the sample mean, x, serves as an unbiased estimate of this quantity. Range; interquartile range; variance, standard deviation.

Awareness of the concept of dispersion and an understanding of the significance of the numerical value of the standard deviation.

Obtain the standard deviation (and indirectly the variance) from a GDC and by other methods.

Awareness that the population standard variance, σ 2 , is generally unknown, and that the s2n−1 serves as an unbiased estimate of σ 2 .

7.2.1

Discrete vs. Continuous

Discrete Data “Random variables that can only take on isolated values are called discrete random variables.” [7] The difference between discrete and continuous data is probably best expressed by giving examples of the two. The number of heads obtained when tossing three coins is discrete. The number of wins by the Rockets in a season is discrete.

Continuous Data The areas in square miles of the various states is continuous. Height is continuous. However, continuous data is limited by precision of measurement. For example, we can’t tell the difference between someone that is 63.28976 inches tall and someone that is 63.28967 inches tall. We might say that both are 63.29 inches tall. The continuous data is being forcibly grouped by limitations in precision. This grouping can make the continuous data appear or seem discrete.

More on grouping later.

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7.2.2

Groups

These are some definitions associated with grouping of data. mid-interval value is the midpoint of a class interval. Also known as the class mark. interval boundaries are the smallest and largest actual values in the class.

Also outside of the I.B. world as the upper and lower class boundaries. interval width difference between the class boundaries. Also known as the class width.

Grouping due to precision Frequencies for continuous data are a little more sketchy than for discrete data.

For example, no two people have exactly the same height. Instead of assigning a frequency to a specific value, we assign a frequency to an interval of values.

These intervals may be based upon our precision of measurement, but we may also choose to group the data into larger classes.

The following example is taken from Probability and Statistics. [17] Table 7.12: Heights of 50 Women at Gates College, in Inches 61 66 65 66 65

60 63 66 69 70

64 59 68 61 66

63 67 70 64 72

69 65 62 65 62

72 67 64 63 69

65 64 69 67 62

67 71 63 65 69

66 61 64 60 64

66 69 65 64 69

Example 7.2.1 Table 7.12 represents the heights of a sample of 50 women at Gates College, as recorded to the nearest inch. Make a frequency chart.

The frequency chart is seen Table 7.13.

The heights arenâ&#x20AC;&#x2122;t discrete, theyâ&#x20AC;&#x2122;re continuous. Note that the heights are recorded to the nearest inch. That means that a woman recorded with a height of 50 inches might really be 50.472 inches or 49.7 inches. So a height recorded as 50 inches really means a height between 49.5 inches and 50.5 inches.

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Except for the fact that we are assigning a frequency to an interval of values instead of a single individual value, continuous data are treated similarly to discrete data.

Table 7.13: Heights of a Sample of Gates College Women, in Inches Recorded Height Actual Height Frequency 59 58.5-59.5 1 60 59.5-60.5 2 61 60.5-61.5 3 62 61.5-62.5 3 63 62.5-63.5 4 64 63.5-64.5 7 65 64.5-65.5 7 66 65.5-66.5 6 67 66.5-67.5 4 68 67.5-68.5 1 69 68.5-69.5 7 70 69.5-70.5 2 71 70.5-71.5 1 72 71.5-72.5 2 Total=50

[There is some disagreement as to what the range would be. What arguments could you make for giving a range of 13? for 14?]

The relative frequency, as for discrete data, is the frequency divided by the total frequency, n.

Grouping by choice There may be times where we wish to form larger groups than what is allowed by precision of measurement. This consolidates the data, and, in some sense, makes the data more legible. The downside of consolidation, however, is the loss of information.

• In order for the class mark to be an actual recorded value, the class width should be an odd number.

• Class intervals should have the same width.

• The number of classes should be between 5 and 20.

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SL Unit 7 (Statistics) Example 7.2.2 Group the heights of the sample of 50 women from Gates College. (Table 7.12). Nevermind, I’ll do it:

The range is either 13 or 14, depending on whom you ask. We need at least five classes, so that would give them a class interval of 3> 14 5 . Even with six , and one of the classes would be classes we’d need a class interval of 3> 14 6 unused. Seven classes would give a class size of two, which is even and therefore undesirable. (Why?) So: five classes is an excellent choice.

There are two ways I can group the data into five classes. Five classes of width 3 gives fifteen values, and I have fourteen. So I can either start at 57.5 and go up to 72.5, or, I can start at 58.5 and go up to 73.5. It’s just a question of putting the empty value at the end or the beginning.

Table 7.14: Heights of Gates College Women (Grouped Sample Data) Class Class Class Class Label Boundaries Mark Frequency 1 57.5-60.5 59 3 2 60.5-63.5 62 10 3 63.5-66.5 65 20 4 66.5-69.5 68 12 5 69.5-72.5 71 5 Total=50 Did I put the empty value at the end or at the beginning? What are the class boundaries? Find the mean based solely on the grouped data and compare it to the mean for the ungrouped data.

Statistics of Grouped Data To calculate means or standard deviations for the grouped data, treat each data in the class as if it were equal to the class mark. Calculating the median will be described in the Section 7.3.2.

We know that there are 12 women with heights between 66.5 and 69.5 inches, and once the data is grouped, we have lost the information about how the values were distributed within that interval. In determining mean and standard deviation, we will assume that all 12 women have a height of 68 inches. Clearly, that isn’t true, but it will be our operating assumption, unless you got a better idea.

Example 7.2.3 Using the grouped data in Table 7.14, estimate: Mr. Budd, compiled January 12, 2011

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(a) the mean height; (b) the standard deviation and variance of the heights.

Compare your answers to those obtained from the individual, ungrouped heights.

7.2.3

Histograms

A histogram is a bar graph of the frequency distribution. It consists of a “collection of rectangles,” [17] one for each value (for ungrouped, discrete data), or one for each class (grouped, continuous data). The area of the rectangle is proportional to the frequency. If the class widths are equal, then the area is proportional to the height, and therefore the height is proportional to the frequency. That is why we want the class widths to be equivalent.

Figure 7.2: Histogram for Gates Women’s Heights

For grouped, continuous data, the midpoint of the rectangle should be the class mark (see page 246), and the rectangles should start and end at the lower and upper class boundaries, respectively.

Example 7.2.4 See Figure 7.2 for the histogram corresponding to the sample height of Gates women.

Different histograms can be created for different ordinates (y-axes) of frequency ˆ (f ), relative frequency f , and f ∗ = f . Some people choose f ∗ because then Nw

the area of each rectangle is equal to its relative frequency, and the sum of all the areas is 1. Also, some people express the relative frequency in terms of a decimal, and some in terms of a percentage.

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Figure 7.3: Histogram for Gates Womenâ&#x20AC;&#x2122;s Heights

Problems 7.B-1 (SL 11/07) The histogram in Figure 7.4 represents the ages of 270 people in a village.

Figure 7.4: Age of Villagers

(a) Use the histogram to complete the table below.

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Age Range

Frequency

0 ≤ age < 20 20 ≤ age < 40 40 ≤ age < 60 60 ≤ age < 80 80 ≤ age ≤ 100

40

Mid-interval value 10

(b) Hence, calculate an estimate of the mean age.

[Ans: 44.1 yr] Figure 7.5: Lengths of Lab Plants

7.B-2 (MM Spec ’00) Figure 7.5 represents the length, in cm, of 80 plants grown in a laboratory. (a) How many plants have lengths in cm between i. 50 and 60? ii. 70 and 90?

[2 marks]

(b) Calculate estimates for the mean and the standard deviation of the lengths of the plants.

[4 marks] (c) Explain what feature of the diagram suggest that the median is different from the mean.

[1 mark] [Ans: 10, 24; 63, 20.5; assymetric distribution] 7.B-3 (SL 5/07 TZ2) There are 50 boxes in a factory. Their weights, w kg, are divided into 5 classes, as shown in the following Table 7.15. (a) Show that the estimated mean weight of the boxes is 32 kg. (b) There are x boxes in the factory marked “Fragile”. They are all in Class E. The estimated mean weight of all the other boxes in the factory is 30 kg. Calculate the value of x.

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Table 7.15: Fifty Boxes in a Factory Class Weight (kg) Number of boxes A 9.5 ≤ w < 18.5 7 B 18.5 ≤ w < 27.5 12 C 27.5 ≤ w < 36.5 13 D 36.5 ≤ w < 45.5 10 E 45.5 ≤ w < 54.5 8

(c) An additional y boxes, all with a weight in class D, are delivered to the factory. The total estimated weight of all of the boxes in the factory is less than 33 kg. Find the largest possible value of y.

[Ans: 6 (or 14)] 7.B-4 Suppose that the heights of fifty women at Gates College in Table 7.12 was grouped differently. Use five intervals again, each still with an interval of three, but this time let the first group start at 58.5 and the last group end at 63.5. How does the change in grouping effect the mean, standard deviation, and variance (compared with the grouping in Table 7.14?

7.B-5 (MM 11/96) The histogram in Figure 7.6 shows the salary structure of a company which employs 150 people. For example, 28 employees have a salary S such that $15000 ≤ S < $20000. (a) How many employees have a salary S such that $25000 ≤ S < $30000?

[1 mark] (b) Calculate an estimate for the mean salary earned in the company, giving your answer to the nearest $100.

[3 marks] [Ans: 31; $24700] 7.B-6 (adapted from HL 5/04) The heights of 60 children entering a school were measured. The results are shown in Table 7.16. Estimate Table 7.16: Problem 6 - Heights of children Height (m) Number of Students 0.8 < h ≤ 0.9 9 0.9 < h ≤ 1.0 15 1.0 < h ≤ 1.1 15 1.1 < h ≤ 1.2 12 1.2 < h ≤ 1.3 6 1.3 < h ≤ 1.4 3

(a) the mean height

[Ans: 1.05 m] Mr. Budd, compiled January 12, 2011

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Figure 7.6: Salary Structure

(b) the standard deviation and variance of the heights.

Why are the instructions ”Estimate” and not ”Find”?

In Section 7.3.2, we will look at how to estimate the median.

7.B-7 (adapted from HL 5/00) A sample of 70 batteries was tested to see how long they last. The results are shown in Table 7.17 Find (a) the sample mean; (b) the sample variance; (c) an unbiased estimate of the mean of the mean of the population from which this sample is taken.

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Table 7.17: Problem 7 - Battery Lifetimes Time(hours) Number of batteries (frequency) 0 ≤ t < 10 2 10 ≤ t < 20 4 20 ≤ t < 30 8 30 ≤ t < 40 9 40 ≤ t < 50 12 50 ≤ t < 60 13 60 ≤ t < 70 8 70 ≤ t < 80 7 80 ≤ t < 90 6 90 ≤ t < 100 1 Total 70

(d) an unbiased estimate of the variance of the population from which this sample is taken.

[Ans: 49.8; 459 (3 s.d.); 49.8; 466] 7.B-8 [17] A frequency distribution for the weights, in lb, of 80 male college seniors, is given in Table 7.18.

Table 7.18: Problem 8 - Weight of Male College Seniors Class mark, lb 140 155 170 185 200 215 Class frequency, f 17 27 18 12 4 2

(a) Determine i. the interval widths; ii. the upper and lower interval boundaries for each class.

[Ans: 15; 132.5 and 147.5, 147.5 and 162.5, . . .] (b) On graph paper, construct a histogram for these data. (c) Determine i. ii. iii. iv. v.

the the the the the

sample mean. sample standard deviation. sample variance. unbiased estimate of the population mean. unbiased estimate of the population variance.

[Ans: 163] [Ans: 18.8] [Ans: 353] [Ans: 163] [Ans: 358]

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7.3

255

Cumulative Frequency

International Baccalaureate 6.2 Presentation of data: frequency tables and diagrams, box and whisker plots.

6.3 Mean, median, mode; quartiles, percentiles.

Interquartile range.

6.4 Cumulative frequency; cumulative frequency graphs; use to find median, quartiles, percentiles.

Recall Median. Percentile. Quartile.

7.3.1

Continuous Data

Cumulative Distribution Function When dealing with a cumulative distribution function for grouped and/or continuous data, I find the cumulative frequency at each class boundary. For n classes, there will be n+1 boundaries. For each boundary, find the total number of occurrences that are less than that class boundary. The cumulative frequency at the lowest class boundary should always be 0. For a cumulative relative frequency distribution, I simply take each cumulative frequency and divide by the total number of events. The relative cumulative frequency of the uppermost class boundary should always be 1. Cumulative frequencies are given for class boundaries, not for class marks.

Example 7.3.1 Refer to Table 7.14, the frequency distribution for the grouped heights of women at Gates College. Develop a table which contains columns displaying the cumulative frequency and cumulative relative frequency at the class boundaries.

The final cumulative relative frequency table is seen in Table 7.20.

This is sometimes called the empirical cumulative distribution function, because it represents a cdf (cumulative distribution function) that has been empirically determined, i.e., found by experience. This is used to distinguish it from a cumulative distribution function in probability, which is a theoretical and idealized function.

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Table 7.19: Generating Cumulative Relative Frequencies at the Class Boundaries for a Frequency Distribution Class Class Class Cumulative Cumulative mark frequency boundary frequency relative frequency 57.5 0 0 59 3 60.5 0+3=3 3/50 = 0.06 62 10 63.5 3 + 10 = 13 13/50 = 0.26 65 20 66.5 13 + 20 = 33 33/50 = 0.66 68 12 69.5 45 45/50 = 0.90 71 5 72.5 50 50/50 = 1.00

Table 7.20: Cumulative Relative Frequency Table for Sample of Heights of Women at Gates College Class Cumulative boundary, x relative frequency, Fˆ (x) 57.5 0.00 60.5 0.06 63.5 0.26 66.5 0.66 69.5 0.90 72.5 1.00

Example 7.3.2 (MM 95) In Aristia a company has almost 1000 employees. The distribution of employees’ annual salaries is summarized in the cumulative frequency table in Table 7.21. (a) Find the number of employees for whom (a) 20 < S ≤ 25; (b) 30 < S ≤ 40.

[Ans: 334] [Ans: 136]

(b) Calculate an estimate for the mean salary, correct to 3 significant figures.

[Ans: 24.7] (c) Note that we will find the median and quartiles in a way different than assuming all the values are at the class mark. Be aware that for continuous, grouped data, finding the median is not as easy as punching some buttons on the calculator.

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Table 7.21: Annual salaries in Aristia Salary S (in thousands N , the number of of Aristian Crowns) employees not exceeding 10 0 15 45 20 207 25 541 30 829 35 923 40 965 45 985

Cumulative Frequency Polygon Once you have the cumulative frequency distribution, it is a simple matter to create the cumulative frequency polygon. Plot the class boundaries vs. the appropriate cumulative frequencies. You should plot n + 1 coordinates for each of the n + 1 class boundaries. Connect the dots with straight lines.

A cumulative relative frequency polygon is done in a similar manner, simply plotting the relative frequency on the y-axis.

Example 7.3.3 From the tabulated data in Table 7.20, construct a cumulative relative frequency polygon.

Figure 7.7: Cumulative relative frequency polygon for Gates College womenâ&#x20AC;&#x2122;s heights.

The relative cumulative frequency polygon is shown in Figure 7.7.

Example 7.3.4 (HL 5/04) The heights of 60 children entering a Mr. Budd, compiled January 12, 2011

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SL Unit 7 (Statistics) school were measured. The cumulative frequency graph in Figure 7.8 illustrates the data obtained. (a) Determine the interval boundaries, and mid-interval values. (b) Make a frequency chart. (c) Estimate the mean height, and the variance of heights.

Figure 7.8: Heights of Entering Students

7.3.2

Percentiles and Quartiles

Definition 7.2. [17] The kth percentile pk is a value of x [where x denotes a score] such that k FË&#x2020; (pk ) = 100 The first quartile, Q1 , is the 25th percentile, the median is the 50th percentile, and Q3 , the third quartile, is the seventy-fifth percentile.

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Figure 7.9: centiles.

259

Using a cumulative relative frequency polygon to estimate per-

Percentiles from a graph We can use the cumulative relative frequency polygon to help us estimate percentiles. If we are looking for p10 , the tenth percentile, we can draw a horizontal line on the cumulative relative frequency polygon where y = 0.10. The x-value where that horizontal line crosses the polygon is p10 , the score for the tenth percentile. See Figure 7.9, which shows how to estimate the median and third quartile for the sample of 50 women at Gates College.

Example 7.3.5 Using Figure 7.8, estimate the median height of the sixty students entering a school.

Percentiles from a table While the cumulative relative frequency polygon is useful for estimating percentiles, a more precise measurement requires us to do some linear interpolation.

When using grouped data, you cannot use 1-Var Stats on the calculator to determine the median and quartiles. The calculator treats the data discretely, which OK for estimating the mean and variance, but unacceptable in determining percentiles.

Example 7.3.6 Refer to Table 7.20 on page 256, the cumulative relative frequency chart relating the heights of 50 women at Gates College. (a) Calculate p25 , the 25th percentile.

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SL Unit 7 (Statistics) (b) Calculate the median and the third quartile.

In order to do this problem, we must look at the cumulative relative frequency chart (Table 7.20 on page 256). 25%, or 0.25, is between 0.06 and 0.26, so that p25 is somewhere between 60.5 and 63.5. Since 0.25 is much closer to 0.26, we would expect the first quartile to be fairly close to 63.5. To figure out exactly how close, interpolate between the two values. p25 − 60.5 0.25 − 0.06 = 63.5 − 60.5 0.26 − 0.06 In other words, p25 is proportionally the same distance between 60.5 and 63.5 as 0.25 is between 0.06 and 0.26. Since this is a proportion, I can set it up in several different manners, as with all proportions.

63.5 − p25 63.5 − 60.5

=

0.26 − 0.25 0.26 − 0.06

or

63.5 − p25 0.26 − 0.25

=

63.5 − 60.5 0.26 − 0.06

or

p25 − 60.5 0.25 − 0.06

=

63.5 − 60.5 0.26 − 0.06

or flipping:

0.25 − 0.06 p25 − 60.5

=

0.26 − 0.06 63.5 − 60.5

I can also think about it graphically, in terms of slopes of line segments. Since the point (p25 , 0.25) lies on the line segment joining (60.5, 0.06) and (63.5, 0.26), the slope from (p25 , 0.25) to (60.5, 0.06) (or (63.5, 0.26)) is the same as the slope between the two endpoints of the line segment, (60.5, 0.06) and (63.5, 0.26).

Another way to think about it is as follows: There is a 0.20 difference between .19 0.06 and 0.26. 0.25 is .20 = 19 20 of the way between 0.06 and 0.26. So that p25 19 must be 20 of the way from 60.5 to 63.5.

4x 3

=

0.19 0.20

4x

=

3 · 19 20

=

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and, to get the 25th percentile, I need to add 4x to 60.5. p25

=

60.5 + 4x

=

60.5 + 2.85

=

63.35 ≈ 63.4 inches

This is how one author [1] describes it: " Qj/m = b +

j×n m

# − Cf (w) f

“Where Qj/m is the x value below which are j mths of the data, b is the lower boundary of the implied range for the quantile category (the measurement category that contains the quantile), n is the sample size (or N is the population size), Cf is the cumulative frequency from all categories less than the quantile category, f is the frequency in the quantile category, and w is the width of the implied range of the quantile category.” [1] To find the median, we look between the cumulative relative frequencies of 0.26 and 0.66, corresponding to scores between 63.5 and 66.5:

p50 − 63.5 66.5 − 63.5

=

0.50 − 0.26 0.66 − 0.26

p50 − 63.5 3

=

0.24 0.40

p50 − 63.5

=

3×

p50

=

63.5 + 1.8

p50

=

65.3 inches

24 40

or, alternatively, M d = p50

=

63.5 + 4x

=

63.5 + 3 ×

=

65.3 inches

where

0.50 − 0.26 4x = 66.5 − 63.5 0.66 − 0.26

24 40

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To find the third quartile, look between 0.66 and 0.90, corresponding to x values between 66.5 and 69.5: p75 − 66.5 69.5 − 66.5

=

0.75 − 0.66 0.90 − 0.66

p75 − 66.5 3

=

0.09 0.24

p75 − 66.5

=

3×

p75

=

66.5 + 1.125

p75

=

67.625 ≈ 67.6 inches

9 24

or, alternatively, Q3 = p75

=

66.5 + 4x

=

66.5 +

=

67.625 ≈ 67.6 inches

where

4x 0.75 − 0.66 = 69.5 − 66.5 0.90 − 0.66

3·9 24

Interquartile Range The interquartile range is the difference between the first and third quartiles, or the 25th and 75th percentiles. It is used as a measure of dispersion, along with the variance, standard deviation, and range. (See page 233.)

Example 7.3.7 Calculate the interquartile range for the sample of 50 Gates women’s heights.

The interquartile range is Q3 − Q1 = 67.625 − 63.35 = 4.275 ≈ 4.3 inches.

Example 7.3.8 Using Figure 7.8, estimate the interquartile range of the sixty students entering a school.

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Figure 7.10: Symmetric Box-andWhisker Plot

263

Figure 7.11: Skewed Box-andWhisker Plot

Box-and-Whiskers Plot The box extends from Q1 to Q3 with a dividing line at the median, covering the middle two quarters of data. The whiskers cover the upper and lower quarters of data.

Example 7.3.9 The box plot shown in Figure 7.10 shows a distribution summarized by a five-number summary: Q1 , Q2 , Q3 , xs , xl .

Determine from the plot: interquartile range, median, range, and whether the distribution is symmetric or skewed.[1]

The interquartile range is 4 sec, from 13 sec to 17 sec. The median is 15 sec.

The range is 10 sec, from 10 sec to 20 sec. The distribution is symmetric.

Example 7.3.10 The box plot shown in Figure 7.11 shows a distribution summarized by a five-number summary: Q1 , Q2 , Q3 , xs , xl .

Determine from the plot: interquartile range, median, range, and whether the distribution is symmetric or skewed.[1]

The interquartile range is 0.2 kg, from 2.6 kg to 2.8 kg. The median is about 2.65 kg, and the range is 1.0 kg. The distribution is skewed.

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SL Unit 7 (Statistics) Example 7.3.11 (SL 5/07 TZ2) A set of data is 18, 18, 19, 19, 20, 22, 22, 23, 27, 28, 28, 31, 34, 34, 36 The box and whisker plot for this data is shown below.

(a) Write down the values of A, B, C, D, and E. (b) Find the interquartile range.

[Ans: 18, 19, 23, 31, 36; 12] Example 7.3.12 Draw a box and whiskers plot for the height data from the sample of Gates College women.

Problems 7.C-1 (HL 5/02) The 80 applicants for a Sports Science course were required to run 800 metres and their times were recorded. The results were used to produce the cumulative frequency graph in Figure 7.12. Estimate (a) the median; (b) the interquartile range.

[Ans: 135,11] 7.C-2 (MM 5/02) A taxi company has 200 taxi cabs.

The cumulative frequency curve in Figure 7.13 shows the fares in dollars ($) taken by the cabs on a particular morning. (a) Use the curve to estimate i. the median fare; [Ans: $24] ii. the number of cabs in which the fare taken is $35 or less. [Ans: 154] The company charges 55 cents per kilometre for distance travelled.

There are no other charges. Use the curve to answer the following. (b) On that morning, 40% of the cabs travel less than a km. Find the value of a.

[Ans: 40] (c) What percentage of the cabs travel more than 90 km on that morning?

[Ans: 7%]

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Figure 7.12: Problem 1 - Times on 800 metres

Figure 7.13: Cumulative Frequency Curve for Problem 2

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7.C-3 (HL 5/06) Figure 7.14 is the cumulative frequency diagram for the heights of 30 plants given in centimetres.

Figure 7.14: Height of Thirty Plants

(a) Use the diagram to estimate the median height. (b) Complete the following table.

Height (h) 0≤h<5 5 ≤ h < 10 10 ≤ h < 15 15 ≤ h < 20 20 ≤ h < 25

Frequency 4 9

(c) Hence estimate the mean height.

Ans: 11.2; 8, 5, 4; 11.83 7.C-4 (SL 11/06) Figure 7.15 is the cumulative frequency curve for the time, t minutes, spent by 150 people in a store on a particular day. (a) How many people spent less than 5 minutes in the store? (b) Find the number of people who spent between 5 and 7 minutes in the store. (c) Find the median time spent in the store. [Ans: 50; 40; 6 min,15 sec] Mr. Budd, compiled January 12, 2011

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Figure 7.15: Customers in a Store

(d) Given that 40% of the people spent longer than k minutes, find the value of k.

[Ans: 7] (e) On your answer sheet, copy and complete the following frequency table. t (minutes) Frequencey

0≤t<2 10

2≤t<4 25

4≤t<6

6≤t<8

8 ≤ t < 10

(f) Hence, calculate an estimate for the mean time spent in the store.

[Ans: 37, 38, 27; 6 min,15 sec] 7.C-5 (adapted slightly from MM 5/01) Table 7.22 represents the weights, W , in grams, of 80 packets of roasted peanuts. (a) Use the midpoint of each interval to find an estimate for the standard deviation of the weights.

[Ans: 7.41] Mr. Budd, compiled January 12, 2011

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Table 7.22: Problem 5 Weight (W ) Number of (W ) packets 80 < W ≤ 85 5 85 < W ≤ 90 10 90 < W ≤ 95 15 95 < W ≤ 100 26 100 < W ≤ 105 13 105 < W ≤ 110 7 110 < W ≤ 115 4

(b) Copy and complete the cumulative frequency table for the above data.

Table 7.23: Problem 5 Weight (W ) Number of packets

W ≤ 85 5

W ≤ 90 15

W ≤ 95

W ≤ 100

W ≤ 105

W ≤ 110

(c) A cumulative frequency graph of the distribution is shown in Figure 7.16, with a scale 1 cm for 10 packets on the vertical axis and 1 cm for 5 grams on the horizontal axis. [Note: scale may not be accurate.]

Use the graph to estimate i. the median; ii. the upper quartile (that is, the third quartile).

[Ans: 97] [Ans: 101]

Give your answers to the nearest gram. (d) Let W1 , W2 , . . . , W80 be the individual weights of the packets, and let W be their mean. i. What is the value of the sum

W1 − W + W2 − W + W3 − W +. . .+ W79 − W + W80 − W ?

[Ans: 0] ii. What is the value of the sum 2 2 2 2 2 W1 − W + W2 − W + W3 − W +. . .+ W79 − W + W80 − W ?

[Ans: 4390] (e) One of the 80 packets is selected at random. Given that its weight satisfies 85 < W ≤ 110, find the probability that its weight is greater than 100 grams.

[Ans: 0.282] Mr. Budd, compiled January 12, 2011

W ≤ 115 80

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Figure 7.16: Problem 5

7.C-6 (MM 5/00) A supermarket records the amount of money d spent by customers in their store during a busy period. The results are shown in Table 7.24.

Table 7.24: Problem 6 Money in $ (d) Number of customers (n)

0–20 24

20–40 16

40–60 22

60–80 40

80–100 18

100–120 10

(a) Find an estimate for the mean amount of money spent by the customers, giving your answer to the nearest dollar ($).

[Ans: $59] (b) Copy and complete the cumulative frequency table in Table 7.25 and use it to draw a cumulative frequency graph. Use a scale of 2 cm to represent $20 on the horizontal axis, and 2 cm to represent 20 customers on the vertical axis.

Table 7.25: Problem 6b Money in $ (d) Number of customers (n)

< 20 24

< 40 40

< 60

< 80

< 100

< 120

< 140

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120–140 4

270

SL Unit 7 (Statistics) (c) The time t (minutes), spent by customers in the store may be represented by the equation 2 t = 2d 3 + 3 i. Use this equation and your answer to part 6a to estimate the mean time in minutes spent by customers in the store. [Ans: 33.3 min] ii. Use the equation and the cumulative frequency graph to estimate the number of customers who spent more than 37 minutes in the store.

[Ans: 52]

7.C-7 (MM 5/04) The cumulative frequency curve in Figure 7.17 shows the marks obtained in an examination by a group of 200 students.

Figure 7.17: Problem 7 - Marks obtained on an exam

Mark (x) Number of students

0 ≤ x < 20 22

0 ≤ x < 20

0 ≤ x < 20

0 ≤ x < 20

0 ≤ x < 20 20

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(a) Use the cumulative frequency curve to complete the frequency table above. (b) Forty percent of the students fail. Find the pass mark.

[Ans: 50, 66, 42; 43%] 7.C-8 (MM Spec ’00) Refer to problem 2 on page 251. (In this problem, there are 80 plants growing in a laboratory.) Table 7.26 is an extract from the cumulative frequency table: Table 7.26: Cumulative Frequency Table (80 Plants Grown in a Laboratory) length in cm cumulative less than frequency .. .. . .

50 22 60 32 70 48 80 62 .. .. . .

Use the information in the table to estimate the median. Give your answer to two significant figures.

[Ans: 65] 7.C-9 (MM 5/99) One thousand candidates sit an examination. The distribution of marks is shown in Table 7.27.

Table 7.27: Problem 9 Marks Number of candidates

1–10

11–20

21–30

31–40

41–50

51–60

61–70

71–80

81–90

91–100

15

50

100

170

260

220

90

45

30

20

(a) Copy and complete the following table, which presents the above data as a cumulative frequency distribution.

Marks Number of candidates

≤ 10

≤ 20

15

65

≤ 30

≤ 40

≤ 50

≤ 60

≤ 70

≤ 80

905

(b) Draw a cumulative frequency graph of the distribution, using a scale of 1 cm for 100 candidates on the vertical axis and 1 cm for 10 marks on the horizontal axis.

Mr. Budd, compiled January 12, 2011

≤ 90

≤ 100

272

SL Unit 7 (Statistics) (c) Use your graph to answer the parts below. i. Find an estimate for the median score.

[Ans: 46.3] ii. Candidates who scored less than 35 were required to retake the examination. How many candidates had to retake? [Ans: 250] iii. The highest-scoring 15% of candidates were awarded a distinction. Find the mark above which a distinction was awarded.

[Ans: 63.9]

7.C-10 (MM 11/96) Refer to problem 5 on page 252. The histogram in Figure 7.6 on page 253 shows the salary structure of a company which employs 150 people. For example, 28 employees have a salary S such that $15000 â&#x2030;¤ S < $20000. (a) Construct a cumulative frequency table for salaries in the company.

Use these data to draw a cumulative frequency curve. (b) Use the results of part (a) to estimate the median salary in the company, giving your answer to the nearest $100.

[Ans: $23700] 7.C-11 (SL 11/06) The box and whisker diagram shown below represents the marks received by 32 students.

(a) Write down the values of the median mark. (b) Write down the value of the upper quartile. (c) Estimate the number of students who received a mark greater than 6.

[Ans: 3; 6; 8]

Mr. Budd, compiled January 12, 2011

Unit 8

Vectors and Matrices 1. Scalar Product and Angle Between Vectors 2. Projection of a Vector 3. Vector Representation of a Line 4. Distance Between a Vector and a Line

273

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8.1

275

Introduction to Matrices

International Baccalaureate 4.1 Definition of a matrix: the terms element, row, column, and order.

4.2 Algebra of matrices: equality; addition; subtraction; multiplication by a scalar.

Matrix operations to handle or process information.

Multiplication of matrices.

Identity and zero matrices.

4.3 Determinant of a square matrix.

Calculation of 2 × 2 and 3 × 3 determinants.

The inverse of a 2 × 2 matrix.

Conditions for the existence of the inverse of a matrix.

4.4 Solution of linear equations using inverse matrices (a maximum of three equations in three unknowns).

8.1.1

Matrix Addition

In order to add two matrices, they must be of the same order, i.e., have the same number of rows and the same number of columns.

Addition is intuitive.

Scalar multiplication is intuitive.

There is no such thing as subtraction of matrices, but we pretend like there is, just as for vectors.

The Zero Matrices A zero matrix is a matrix full of zeros. There is a zero matrix for every dimension of matrices.

A+0=A

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8.1.2

Matrix Multiplication

The key to matrix multiplication is to multiply rows by columns.

Suppose you are finding the product Am×n Bn×p , which we will designate as Cm×p . The element of C in the ith row and jth column will be the ith row of A times the jth row of B. For this reason each row of A must have the same number of columns as the columns of B have rows.

Example 8.1.1 Use your handy dandy grapher to randomly generate a 3 × 2 matrix, and store it as A. Then store a 2 × 3 matrix into B. By hand, AB and BA, then check your answers.

Matrix multiplication is not necessarily commutative!

AB 6= BA For this reason, we must distinguish the order of multiplication. Pre-multiplication is different from post-multiplication, and if you pre-multiply one side of an equation by A, you should not post-multiply the other side by A

The Identity Matrices The identity matrix is a square matrix with 1s across the main diagonal (top left to bottom right), and 0s everywhere else. It is the matrix form of 1.

AI = IA = A

8.1.3

Matrix Division

There is no such thing as division with matrices.

Instead, you may pre-multiply or post-multiply by the inverse.

A nonsingular matrix A has an inverse matrix, designated A−1 such that AA−1 = A−1 A = I Mr. Budd, compiled January 12, 2011

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Example 8.1.2 [21] From AB = C find a formula for A−1 .

Ans: A−1 = BC−1

The inverse of a 2 × 2 matrix is given by

a c

1 d −b b = d ad − bc −c a

Example 8.1.3 (HL 5/02)

2 (a) Prove using mathematical induction that 0 for all positive integer values of n.

n n 1 2 = 1 0

2n − 1 , 1

(b) Determine whether or not this result is true for n = −1.

The inverse of a product: −1

(AB)

= B−1 A−1

Example 8.1.4 From PA = LU find a formula for A−1 .

Ans: U−1 L−1 P

8.1.4

Determinants det

a c

b = ad − bc d

Example 8.1.5 (SL 11/06) Let A =

3 k

2 4

2 and B = 1

2 .

3

Find in terms of k, (a) 2A − B; (b) det (2A − B).

[Ans: 22 − 4k] Mr. Budd, compiled January 12, 2011

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Problems

a b 1 0 8.A-1 (SL Spec ’06) Let A = and B = . Giving your answers c 0 d e in terms of a, b, c, d, and e, (a) write down A + B; (b) find AB.

8.A-2 (SL 5/06) Let

b 7

3 9 + 8 −2

a+1 Ans: c+d

5 4 8 = .

7 a 15

b a + bd ; e c

(a) Write down the value of a. (b) Find the value of

−4 8.A-3 (SL 5/06) Let 3 2 q.

be 0

[Ans: 5]

b.

8 2 −5 1 q

[Ans: −5]

0 −4

=

−22 −9

a 8.A-4 (MM 5/97) Let the 2 × 2 matrix M = √ 1 − a2 a ≤ 1. Find the matrices

24 . Find the value of 23 [Ans: 3] √

1 − a2 , with −1 ≤ −a

(a) M 2 (b) M 3

1 Ans: 0

0 a , √ 1 1 − a2

√

1 − a2 −a

8.A-5 (MM 11/96) Evaluate the matrix product:  π π √ ! cos − sin  6 6  −2 3 .  π π  2 sin cos 6 6 √ ! −2 3 Then graph the vector , along with your answer.

2 8.A-6 (HL 5/97) Three  1 p A = 1 0 0 0

3 × 3 matrices   q 2 r  , B = 3 2 0

A, B, and C are given by   0 0 −1 −1 1 0 , and C =  2 0 0 1 0 0

Find p, q, and r so that AB = C.

 10 −6 2

[Ans: p = −1, q = 10, r = −6] Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 1: Introduction to Matrices

8.A-7 (HL 5/97) Let A =

2 1

279

−1 .

0

(a) Calculate A2 and A3 .

[2 marks] n

∗

(b) Conjecture a matrix for A , n ∈ N , in terms of n.

[3 marks]

(c) Use mathematical induction to prove your conjecture in part (b).[5 marks] 8.A-8 (HL 5/02) Find the determinant of  1 1 2

the matrix  1 2 2 1 1 5 [Ans: 0]

8.A-9 (MM 5/99) If A =

2p −4p

3 p

and det A = 14, find the possible values of [Ans: −7, 1]

p.

8.A-10 (HL 5/96) Solve the equation  

1 4 1 1   det x 2 2 = det x x2 2 1

3 3

Ans:

5 6,

−1

Mr. Budd, compiled January 12, 2011

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Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 2: Vector Basics

8.2

281

Vector Basics

International Baccalaureate 5.1 Vectors as displacements in the plane and in three dimensions; Distance between points in three dimensions.

Components of a vector; column representation; Components are with respect to the unit vectors i, j, k (standard basis).

The sum and difference of two vectors; the zero vector; the vector, −v.

Multiplication by a scalar, kv.

Magnitude of a vector, |v|.

Unit vectors; base vectors i, j, and k. ~ = a; AB ~ = OB ~ − OA ~ = b − a.

Position vectors OA 5.2 The scalar product of two vectors v · w = |v| |w| cos θ; v · w = v1 w1 + v2 w2 + v3 w3 .

Algebraic properties of the scalar product v · w = w · v; u · (v + w) = u · v + u · w; 2 (kv) · w = k (v · w);v · v = |v| .

Perpendicular vectors; parallel vectors. Included: for non-zero perpendicular vectors v · w = 0; for non-zero parallel vectors v · w = ± |v| |w|.

The angle between two vectors.

8.2.1

Vectors

Scalars • Have magnitude but no direction.

• Just one number (which may be positive or negative.)

• Example: mass, speed.

Vectors • Have direction and magnitude.

• Can be represented by two numbers (two dimensions), three numbers (three dimensions), or more.

• Example: force, weight, velocity, acceleration.

Example 8.2.1 [4] Classify the following situations as needing to be described with a vector or a scalar.

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282

SL Unit 8 (Vectors and Matrices) (a) A classroom chair is moved from the front of the room to the back of the room. (b) The balance in a bank account. (c) The electric current passing through an electric light tube. (d) A dog, out for a walk, is being restrained by a lead. (e) An aircraft starts its takeoff run. (f) The wind conditions before a yacht race. (g) The amount of liquid in a jug. (h) The length of a car. (i) The time that it takes to boil an egg. (j) The number of goals scored in a soccer match.

8.2.2

Representation

Vectors can be represented 1. on a graph as a directed line segment (directed with an arrow at one end).

2. as a sum of components based on the unit vectors i and j(i moves one space to the right, j one space up), e.g., −i + 3j.

−1 3. as a column matrix or column vector, e.g. .

3 4. as an ordered pair, e.g. h−1, 3i.

8.2.3

Simple Arithmetic

Scalar multiples are intuitive.

Example 8.2.2 [4] If a = 2i − j and b = −i + 3j, find (a) −a (b) −2a (c) 3b See the effects graphically.

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283 [Ans: −2i + j; −4i + 2j; −3i + 9j]

Addition is intuitive.

Example 8.2.3 Find a + b. Add analytically and graphically.

[Ans: i + 2j] Subtraction is adding −1 times the vector being subtracted.

Example 8.2.4 Find (a) b − a (b) 3b − 2a

[Ans: −3i + 4j;−7i + 11j]

   −2 3 Example 8.2.5 If p = −1 and q =  0 , find: 3 4 

(a) p + q (b)

3 q−p 2

   −6 1 Ans: −1;  1  7 0.5 



There are two types of multiplication:

• scalar multiplication (dot product), where the answer is a scalar; • vector multiplication (cross product), where the answer is a vector.

There is not really division with vectors. [Note:

q 1 is really q] 2 2 Mr. Budd, compiled January 12, 2011

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SL Unit 8 (Vectors and Matrices)

8.2.4

Magnitude

To get the magnitude of a vector, use the distance formula: p |ha, bi| = a2 + b2 p |ha, b, ci| = a2 + b2 + c2 Example 8.2.6 Find the lengths of the vectors: (a) 3i − 4j (b) −i + 2j − 5k   3 (c) −1 2 √ √

Ans: 5; 30; 14

8.2.5

Making Unit Vectors

To make a unit vector in the direction of v, simply divide v by its own magniv tude, . |v| Example 8.2.7 (MM 5/99) The vectors ~i, ~j are unit vectors along the x-axis and y-axis respectively. The vectors ~u = −~i + 2~j and ~v = 3~i + 5~j are given. (a) Find ~u + 2~v in terms of ~i and ~j.

A vector w ~ has the same direction as ~u + 2~v , and has a magnitude of 26. (b) Find w ~ in terms of ~i and ~j.

8.2.6

Scalar Product

There are at least two ways that vectors can be multiplied. One of the ways is signified with a dot, ·, and the other, with a cross, ×. Dot multiplication yields a scalar answer: Mr. Budd, compiled January 12, 2011

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ha, b, ci · hd, e, f i = ad + be + cf This is called dot multiplication, or, more formally, scalar multiplication. We will discuss cross multiplication, or vector multiplication, later.

Note that the dot product is commutative, i.e., a · b = b · a. This may seem trivial, but the cross product not commutative.

Example 8.2.8 For the following pairs of vectors, find the scalar product.

−1 −1 (a) and 3 2 (b) −5j + 4k and −5i − j − 3k

[Ans: 7; −7]

8.2.7

Angle

Example 8.2.9 Use the law of cosines to show the formula for the angle between two vectors.

cos θ =

a·b |a| |b|

Note that if a and b are unit vectors, then the scalar product is the cosine of the angle between them.

Example 8.2.10 For the following pairs of vectors, find the angle between the vectors, correct to the nearest degree.

−1 −1 (a) and 3 2 (b) −5j + 4k and −5i − j − 3k

[Ans: 8; 101] Mr. Budd, compiled January 12, 2011

286

SL Unit 8 (Vectors and Matrices) Example 8.2.11 If two vectors are perpendicular, what is their dot product?

[Ans: 0] Two vectors are said to be orthogonal if their dot product is 0. This is a fancy word for perpendicular.

Two vectors are parallel if they are scalar multiples of each other, e.g., −i + 3j and 2i − 6j. If v and w are parallel, then v · w = ± |v| |w|, so long as neither is the zero vector.

Example 8.2.12 [4] Three towns are joined by straight roads.

Oakham is the state capital and is considered as the ‘origin’. Axthorp is 3 km east and 9 km north of Oakham, and Buddville is 5 km east and 5 km south of Axthorp.

Considering i as a 1 km vector pointing east and j as a 1 km vector pointing north: (a) Find the position vector of Axthorp relative to Oakham. (b) Find the position vector of Buddville relative to Oakham.

A light rail station (R) is situated two thirds of the way along the road from Oakham to Axthorp. −−→ −−→ (c) Find the vectors OR and BR. (d) Prove that the light rail station is the closest point to Buddville on the Oakham to Axthorp road.

Problems In Mathematics for the International Student: Mathematics SL by Owen, et al., do: Review Exercises: p. 376, 15A: # 2,3,5,7,8; 15B: # 2,4-8; 15C: # 1-8; p.

395, 16A: # 1-5; 16B: # 2 8.B-1 (MM 5/00) The vectors u, v are given by u = 3i + 5j, v = i − 2j. Find scalars a, b such that a (u + v) = 8i + (b − 2) j.

[Ans: a = 2, b = 8] 8.B-2 (MM 5/02) Two boats A and B start moving from the same point P. Boat A moves in a straight line at 20 km h−1 and boat B moves in a straight line at 32 km h−1 . The angle between their paths is 70◦ .

Find the distance between the boats after 2.5 hours.

[Ans: 78.5 km] Mr. Budd, compiled January 12, 2011

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8.B-3 (HL 5/02) Find the angle between the vectors v = i + j + 2k and w = 2i + 3j + k. Give your answer in radians.

[Ans: 0.702]

1 8.B-4 (MM Spec ’99) Find the size of the angle between the two vectors 2

6 and . Give your answer to the nearest degree.

[Ans: 117◦ ] −8 8.B-5 (HL 5/97) The coordinates of P and Q are (3, −1) and (λ, −4 − λ), respectively, where λ is a constant. If O is the origin, find all values of λ for −−→ −−→ which OP is perpendicular to OQ. [Ans: −1] 8.B-6 (MM 11/96) Triangle OAB has one vertex at the origin O. Vertices

A 5 2 ~ = ~ = and B are given by the position vectors OA and OB .

3 −5 Find the size of ∠AOB, giving your answer correct to the nearest tenth of a degree.

[Ans: 99.2◦ ]

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Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 3: Vector Form of a Line

8.3

289

Vector Form of a Line

International Baccalaureate 5.3 Vector equation of a line r = a + tb. Examples of applications: interpretation of t as time and b as velocity, with |b| representing speed.

8.3.1

Lines in Two Dimensions

Forms of a Line from Algebra I: • Point-Slope: y − y1 = m (x − x1 ) • Slope-Intercept: y = mx + b • Standard: Ax + By = C The first two have the advantage of using constants/ parameters with graphic meanings. The graphic advantages of using standard form are not all that apparent.

The problem with these is that if we extend them into three dimensions, e.g.

Ax + By + Cz = D, then they are no longer lines, but planes.

0 Example 8.3.1 Let a be the position vector , and let d be the 28

6 vector . On graph paper, graph the points that correspond to −8 the following position vectors: (a) a (b) a + d (c) a + 2d (d) a + 3d (e) a + 12 d (f) a − d (g) a − 2d (h) a − 3d (i) a − 12 d Mr. Budd, compiled January 12, 2011

290

SL Unit 8 (Vectors and Matrices) (j) a + 32 d (k) a + 0.75d (l) a + 0.875d (m) a plus every single possible multiple of d How can you draw an infinite number of points without drawing an infinite number of points?

A line can be written in the form r = r0 + td  

x x where r = , or y , a generic point on the line, r0 is a specific, known y z point on the line, and d gives the direction of the line. It is very easy to confuse the slope of a line and the direction vector of a line, but remember that slope is a scalar.

Another vector equation of a line (in 2-D) uses the normal vector, n, to the line, instead of the direction vector, d. The normal vector is orthogonal to the line and its direction vector. If r represents a generic point on the line, and r0 is a specific point on the line, then the vector (r − r0 ) must lie in the line and have the same direction as the line. Since (r − r0 ) is in the direction of the line, it must be orthogonal to the normal vector n, so that n · (r − r0 ) = 0, or n · r = n · r0 . This gives way to the standard form of a line.

A line (in 2-D) can also be written in the form ax + by = c

a where gives the normal vector of the line, i.e., a vector that is orthogonal b to the direction vector. To find c, just pick a point (x0 , y0 ) on the line: c = a ax0 + by0 , so ax + by = ax0 + bx0 . If is a normal vector, then you can b obtain a quick direction vector simply by switching a and b, and making one of b −b them negative: or . −a a The descriptions above regarding normal vectors are only valid in two dimensions. In three dimensions, normal vectors are used to describe planes, not lines.

Vector-Based Forms of a Line Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 3: Vector Form of a Line

291

• Vector Form: r = r0 + λd • Parametric Form: x = x0 + λl, y = y0 + λm, z = z0 + λn • Cartesian Form:

x − x0 y − y0 z − z0 = = l m n

λ is considered the parameter. The parametric form is virtually identical to the vector form; each coordinate is listed separately, usually so that it can be written on one line. If each coordinate equation (x, y, and z) is solved for the parameter, λ, you get the Cartesian equation.

0 6 Example 8.3.2 Rewrite r = +t in all of the other 28 −8 forms of a line.

8.3.2

Applications

Differentiate r = r0 +td with respect to time. What is the derivative of position?

One way to think about the line is that r represents the position vector of an object that starts at position r0 and moves with velocity vector d, so that after one unit of time r = r0 + d. After two units of time, r = r0 + 2d. If you look at all the possible points for all the possible times, then you have a straight line (as long as the velocity remains constant).

1 Example 8.3.3 (MM Spec ’99) In this question the vector 0

0 km represents a displacement due east, and the vector km rep1 resents a displacement due north.

Figure 8.1 shows the path of the oil-tanker Aristides relative to the port of Orto, which is situated at the point (0, 0).

The position of Aristides is given by the vector equation

x 0 6 = +t y 28 −8 at time t hours after 12:00. (a) Find the position of the Aristides at 13:00.

[2 marks]

(b) Find (i) the velocity vector; Mr. Budd, compiled January 12, 2011

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SL Unit 8 (Vectors and Matrices)

Figure 8.1: MM Spec

(ii) the speed of the Aristides.

[4 marks]

(c) Find a cartesian equation for the path of the Aristides in the form ax + by = g.

[4 marks] Another ship, the cargo-vessel Boadicea, is stationary, with po18 sition vector km.

4 (d) Show that the two ships will collide, and find the time of collision.

[4 marks] To avoid collision, the Boadicea starts to move at 13:00 with

5 velocity vector km h−1 .

12 (e) Show that the position of the Boadicea for t ≥ 1 is given by

x 13 5 = +t y −8 12 [2 marks] (f) Find how far apart the two ships are at 15:00.

6 6 −1 −1 Ans: ; km h , 10 km h ; 4x + 3y = 84 20 −8 [Ans: 26 km apart]

[4 marks]

[Ans: 15:00]

Speed is the magnitude of velocity. Speed = |v| Example 8.3.4 (MM 5/01) In this question, a unit vector represents a displacement of 1 metre.

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293

A miniature car moves in a straight line, starting at the point (2, 0).

After t seconds, its position, (x, y), is given by the vector equation

x 2 0.7 = +t y 0 1 (a) How far from the point (0, 0) is the car after 2 seconds?

[Ans: 3.94 m]

(b) Find the speed of the car.

Ans: 1.22 ms−1 (c) Obtain the equation of the car’s path in the form ax + by = c.

[Ans: 10x − 7y = 20] Another miniature vehicle, a motorcycle, starts at the point (0, 2), and travels in a straight line with constant speed. The equation of its path is y = 0.6x + 2, x ≥ 0 Eventually, the two miniature vehicles collide. (d) Find the coordinates of the collision point. [Ans: (5.86, 5.52)] (e) If the motorcycle left point (0, 2) at the same moment the car

left point (2, 0), find the speed of the motorcycle. Ans: 1.24 ms−1

Problems 8.C-1 (MM Spec ’99) A line passes through the point (4, −1) and its direction

2 is perpendicular to the vector . Find the equation of the line in the 3 form ax + by = p, where a, b and p are integers to be determined.

[Ans: 2x + 3y = 5] 8.C-2 (MM 5/00) Find a vector equation of the line passing through (−1, 4) and (3, −1). your answer

Give

in the form r = p + td, where t ∈ R. x −1 4 Ans: = +t y 4 −5

3 8.C-3 A line is perpendicular to and passes through (6, 2). −4 (a) Find the equation of the line in the form ax + by = g. Then find

|g| . |n|

[Ans: 3x − 4y = 10; 2] (b) Write an equation of the line in the form r = r0 +td.

6 4 Ans: r = +t 2 3

−−→ (c) Write a vector OR, in terms of t, for the position of a generic point on the line.

[Ans: h6 + 4t, 2 + 3ti] Mr. Budd, compiled January 12, 2011

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SL Unit 8 (Vectors and Matrices) (d) If point S is the closest point on the line to the origin, how is the −→ vector OS related to the direction vector? (e) Find the value of t that gives S, the closest point on the line to the origin.

Ans: − 56

(f) Find the coordinates of S. Ans: 65 , − 85 (g) Confirm the shortest distance from the line to the origin.

8.C-4 (MM 5/02) Three of the coordinates of the parallelogram ST U V are S (−2, −2), T (7, 7), U (5, 15). −→ (a) Find the vector ST and hence find the coordinates of V . [Ans: (−4, 6)] (b) Find a vector equation of the line (U = p+λd where V ) in the form r

x 5 9 λ ∈ R. Ans: e.g., = +t y 15 9

1 (c) Show that the point E with position vector is on the line (U V ), 11

and find the value of λ for this point.

Ans: e.g., − 49

a The point W has position vector , a ∈ R. 17

−−→

√

(d) i. If EW = 2 13, show that one value of a is −3 and find the other possible value of a.

[Ans: 5] −−→ −→ ii. For a = −3, calculate the angle between EW and ET . [Ans: 157◦ ] 8.C-5 (SL 5/08) The point O has coordinates (0, 0, 0), point A has coordinates (1, −2, 3), and point B has coordinates (−3, 4, 2).   −4 −−→   6 . (a) i. Show that AB = −1 ˆ ii. Find B AO. [Ans: 45.8◦ ]       x −3 −4 (b) The line L1 has equation y  =  4  + s  6 . z 2 −1 Write down the coordinates of two points on L1 . [Ans: (−3, 4, 2), (−7, 10, 1)] −−→ (c) The line L2 passes through A and is parallel to OB. i. Find a vector equation for L2 , giving your  answer   in the  form  1 −3 Ans: r = −2 + t  4  r = a + tb.

3 2 ii. Point C (k, −k, 5) in on L2 . Find the coordinates of C. [Ans: (−2, 2, 5)] Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 3: Vector Form of a Line

295

      x 3 1 (d) The line L3 has equation y  = −8 + p −2, and passes z 0 −1 through the point C. Find the value of p at C. [Ans: −5]

1 8.C-6 (MM 5/00) In this question, the vector km represents a displacement 0

0 due east, and the vector km a displacement due north. [Mr. Budd’s 1 note: Do this on a separate sheet, and include a graph on graph paper ] Two crews of workers are laying an underground cable in a north-south direction across a desert. At 06:00 each crew sets out from their base camp which is situated at the origin (0, 0). One crew is in a Toyundai vehicle and the other in a Chryssault vehicle.

18 The Toyundai has velocity vector km h−1 , and the Chryssault has 24

36 velocity vector km h−1 . −16

Ans: 30 km h−1 ;39.4 km h−1

9 18 i. Find the position vectors of each vehicle at 06:30. Ans: ; 12 −8 ii. Hence, or otherwise, find the distance between the vehicles at √ 06:30.

Ans: 481 km

(a) Find the speed of each vehicle. (b)

(c) At this time (06:30), the Chryssault stops and its crew begin their day’s work, laying cable in a northerly direction. The Toyundai continues travelling in the same direction at the same speed until it is exactly north of the Chryssault. The Toyundai crew then begin their day’s work, laying cable in a southerly direction. At what time does the Toyundai crew begin laying cable?

[Ans: 07:00] (d) Each crew lays an average of 800 m of cable in an hour. If they work non-stop until their lunch break at 11:30, what is the distance between them at this time?

[Ans: 24.4 km] (e) How long would the Toyundai take to return to base camp from its lunch-time position, assuming it travelled in a straight line and with the same average speed as on the morning journey? (Give your answer to the nearest minute.)

[Ans: 54 minutes]

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Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 4: Lines in Space

8.4

297

Lines in Space

International Baccalaureate 5.3 Vector equation of a line r = a+tb. Lines in the plane and in three-dimensional space. Knowledge of the following forms for equations of lines.

Parametric form: x = x0 + λl, y = y0 + λm, z = z0 + λn. x − x0 y − y0 z − z0 Cartesian equation of a line in three dimensions = = . l m n Examples of applications: interpretation of t as time and b as velocity, with |b| representing speed.

The angle between two lines.

8.4.1

Lines in Three Dimensions

Example 8.4.1 (HL 5/99) The coordinates of the points P , Q, R, and S are (4, 1, −1), (3, 3, 5), (1, 0, 2c), and (1, 1, 2), respectively. −−→ −→ (a) Find the value of c so that the vectors QR and P R are orthogonal. (b) Find an equation of the line ` which passes through the point −→ Q and is parallel to the vector P R.

[Ans: 1; r = 3 (1 − t) i + (3 − t) j + (5 + 3t) k]

8.4.2

Angle Between Lines

Intersecting lines form two angles (one obtuse and one acute), unless they are perpendicular. One of the angles between two lines is the same as the angle between the two direction vectors. Generally, however, we want the acute angle between two lines, which is given by cos θ =

|d1 · d2 | |d1 | |d2 |

If the dot product of the direction vectors is negative, then it will give the obtuse angle. The absolute value ensures that you will get the acute angle.

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8.4.3

Intersection of Lines

Example 8.4.2 (MM 5/03) The vector equations of two lines are given below.

5 3 −2 4 r1 = +λ , r2 = +t .

1 −2 2 1 The lines intersect at the point P . Find the position vector of P .

Addendum: Find the angle between the lines.

2 Ans: 3 Example 8.4.3 (HL 5/02) The vector equations of the lines L1 and L2 are given by L1 : r

= i + j + k + λ (i + 2j + 3k) ;

L2 : r

= i + 4j + 5k + µ (2i + j + 2k) .

The two lines intersect at point P . Find the position vector of P .

Addendum: Find the angle between the lines.

Example 8.4.4 (HL 5/01) The triangle ABC has vertices at the points A (−1, 2, 3), B (−1, 3, 5), and C (0, −1, 1). Let `1 be the line −−→ parallel to AB which passes through D (2, −1, 0) and `2 be the line −→ parallel to AC which passes through E (−1, 1, 1). (a) Find the equations of the lines `1 and `2 . (b) Hence show that `1 and `2 do not intersect.

[Ans: `1 : x = 2, y = −1 + λ, z = 2λ, `2 : x = −1 + µ, y = 1 − 3µ, z = 1 − 2µ] Non-parallel lines that do not intersect are skew lines.

8.4.4

Applications in Three Dimensions

Example 8.4.5 (Spec ’06) In this question, distance is in kilometres, time is in hours. −1 A balloon is moving at a constant  height with a speed of 18km h , 3 in the direction of the vector 4.

0 At time t = 0, the balloon is at point B with coordinates (0, 0, 5).

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(a) Show that the position vector b of the balloon at time t is given by       x 0 10.8 b = y  = 0 + t 14.4 z 5 0 At time t = 0, a helicopter goes to deliver a message to the balloon. The position vector h of the helicopter at time t is given by       x 49 −48 h = y  = 32 + t −24 z 0 6 (b) (a) Write down the coordinates of the starting position of the helicopter. (b) Find the speed of the helicopter. (c) The helicopter reaches the balloon at point R. (a) Find the time the helicopter takes to reach the balloon. (b) Find the coordinates of R.

Ans: (49, 32, 0), 54 km h−1 ; t =

5 6

hr, (9, 12, 5)

This is a very rare case in which the two parameters are the same for each line.

If two lines do use the same parameter, it should be in a situation like this, where the parameter is time.

Problems 8.D-1 (HL 5/97) (a) The line L1 is parallel to the vector v = 3i+j+3k and passes through the point (2, 3, 7). Find a vector equation of the line. (b) The parametric equations of another line L2 are x = t, y = t, and z = −t, −∞ < t < ∞ Show that i. L1 is not parallel to L2 ; ii. L1 does not intersect L2 .

8.D-2 (SL 5/06) The position vector of point A is 2i + 3j + k and the position vector of point B is 4i − 5j + 21k. −−→ (a) i. Show that AB = 2i − 8j + 20k.

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SL Unit 8 (Vectors and Matrices) −−→ ii. Find the unit vector u in the direction of AB −→ iii. Show that u is perpendicular to OA. Let S be the midpoint of [AB]. The line L1 passes through S and is −→ parallel to OA. (b)

i. Find the position vector of S. [Ans: 3i − j + 11k] ii. Write down the equation of L1 . [Ans: r = (3i − j + 11k) + t (2i + 3j + k)] The line L2 has equation r = (5i + 10j + 10k) + s (−2i + 5j − 3k).

(c) Explain why L1 and L2 are not parallel. (d) The lines L1 and L2 intersect at the point P . Find the position vector of P .

[Ans: 7i + 5j + 13k] (e) Calculate the angle between the lines L1 and L2 .

[Ans: 69.7◦ ]

8.D-3 (SL 5/07 TZ2) In this question, distance is in metres, time is in minutes.

Two model airplanes are each flying in a straight line.

At 13:00 the first model airplane   isatpoint(3, 2, 7). Its position vector x 3 3 after t minutes is given by y  = 2 + t  4 . z 7 10 √

(a) Find the speed of the model airplane. Ans: 5 5 metres per minute At 13:00 the second model airplane is at the point (−5, 10, 23). After two minutes, it is at the point (3, 16, 39).   x (b) Show that its position vector after t minutes is given by y  = z     −5 4  10  + t 3. [Note: “Show that” 6= “Check that”. You should 23 8 solve the problem from scratch, as if you don’t know what the answer already is.] (c) The airplanes meet at point Q. i. At what time do the airplanes meet? ii. Find the position of Q.

[Ans: 13:08] [Ans: (27, 34, 87)]

(d) Find the angle θ between the paths of the two airplanes. [Ans: 0.167 radians] 8.D-4 (SL 5/07) Points P and Q have position vectors −5i + 11j − 8k and −4i + 9j − 5k, respectively, and both lie on line L1 . −−→ (a) i. Find P Q. [Ans: i − 2j + 3k] ii. Hence show that the equation of L1 can be written as r = (−5 + s) i + (11 − 2s) j + (−8 + 3s) k The point R (2, y1 , z1 ) also lies on L1 .

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(b) Find the value of y1 and z1 .

[Ans: −3,13] The line L2 has equation r = 2i + 9j + 13k + t (i + 2j + 3k). (c) The lines L1 and L2 intersect at point T . Find the position vector of T. [Ans: −i + 3j + 4k] (d) Calculate the angle between the lines L1 and L2 .

[Ans: 64.6◦ ]

8.D-5 (adapted from SL 5/08) The point O has coordinates (0, 0, 0), point A has coordinates (1, −2, 3), and point B has coordinates (−3, 4, 2).       x −3 −4 (a) The line L1 has equation y  =  4  + s  6 . z 2 −1 Write down the coordinates of two points on L1 . [Ans: (−3, 4, 2), (−7, 10, 1)] −−→ (b) The line L2 passes through A and is parallel to OB. Find a vector equation for L2 , giving your answer in the form r = a + tb.       3 1 x (c) The line L3 has equation y  = −8 + p −2, and intersects 0 −1 z line L2 at point C. Find the coordinates of point C. [Ans: (−2, 2, 5)] (d) Calculate the acute angle between the lines L2 and L3 . [Ans: 9.76◦ ] 8.D-6 (5/06 TZ2) The following diagram shows a solid figure ABCDEF GH. Each of the six faces is a parallelogram. The coordinates of A and B are Figure 8.2: SL 5/06 TZ 2

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SL Unit 8 (Vectors and Matrices) A (7, −3, −5), B (17, 2, 5). (a) Find 

  10 Ans:  5  10

−−→ i. AB

−−→

ii.

AB

[Ans: 15]

The following information is given.     −6 −2 −−→   −→   6 , AE = −4 AD = 3 4 (b)

−−→

i. Calculate AD .

[Ans: 9]

−→

ii. Calculate AE .

[Ans: 6] −−→ −→ iii. Calculate AD · AE. −−→ −−→ iv. Calculate AB · AD. −−→ −→ v. Calculate AB · AE. vi. Hence, write down the size of the angle between any two inter secting edges.

Ans: π2

(c) Calculate the volume of the solid ABCDEF GH. (d) The coordinates of G are (9, 4, 12).

[Ans: (−1, −1, 2)]

[Ans: 810]

Find the coordinates of H.

(e) The lines (AG) and  (HB)  intersect at point P .

2 −→ Given that AG =  7 , find the acute angle at P .

17

[Ans: 71.6◦ ]

Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 5: Matrices Reloaded

8.5

303

Matrices Reloaded

International Baccalaureate 4.3 Determinant of a square matrix.

Calculation of 2 × 2 and 3 × 3 determinants.

The inverse of a 2 × 2 matrix.

Conditions for the existence of the inverse of a matrix.

4.4 Solution of linear equations using inverse matrices (a maximum of three equations in three unknowns).

8.5.1

Solving Systems of Equations

There is no such thing as division with matrices.

Instead, you may pre-multiply or post-multiply by the inverse.

A nonsingular matrix A has an inverse matrix, designated A−1 such that AA−1 = A−1 A = I The inverse of a 2 × 2 matrix is given by

1 d −b a b = c d ad − bc −c a Example 8.5.1 (barely modified from SL 11/05) Matrices A, B, and C are defined by

5 1 2 4 9 −7 A= ,B = ,C = 7 2 −3 15 8 2 Let X be an unknown 2 × 2 matrix satisfying the equation AX + B = C Find X.

1 Ans: 2

−3 4

Example 8.5.2 (MM 11/99) A and B are 2 × 2 matrices, where

5 2 11 2 A= and BA = . Find B. 2 0 44 8 Mr. Budd, compiled January 12, 2011

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1 Ans: 4

3 12

Example 8.5.3 (barely modified from MM 11/03) Matrices A, B, and C are defined by

5 −2 6 7 −5 0 A= ,B = ,C = 7 1 5 −2 −8 7 Let X be an unknown 2 × 2 matrix satisfying the equation XA + B = C Find X.

2 Ans: −4

−3 1

If there is a unique solution to Ax = b, it is given by x = A−1 b.

Example 8.5.4 (HL 11/99) 

 a −4 −6 7 (a) Find the values of a and b given that the matrix A = −8 5 −5 3 4   1 2 −2 is the inverse of the matrix B =  3 b 1 . −1 1 −3 (b) For the values of a and b you found, solve the system of linear equations x + 2y − 2z = 5 3x + by + z = 0 −x + y − 3z = a − 1 [Ans: 7, 2; (−1, 2, −1)] Example 8.5.5 [22] Consider the system of equations x + 3y − z = 0 3x + 5y − z = 0 x − 5y + z = 8 (a) Show that the determinant is not 0. (b) Find the solution using inverse matrices.

[Ans: (2, −2, −4)] (c) Give a geometric interpretation to the solution.

Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 5: Matrices Reloaded

8.5.2

305

Singularity

A matrix is singular if its determinant is 0. Singular matrices have several properties: • the determinant is zero.

• they do not have inverse matrices.

• one or more rows is a linear combination of the other rows.

3 −2 Example 8.5.6 (HL 5/03) Given that A = and I = −3 4

1 0 , find the values of λ for which (A − λI) is a singular matrix.

0 1

[Ans: 1 or 6] Note: these values of λ are called the eigenvalues of A 

α Example 8.5.7 (HL 5/08) Let M be the matrix  0 −1 Find all the values of α for which M is singular.

2α α −1

 0 1 . α

[Ans: 0, ±1]

Problems 8.E-1 [21] Find the inverses of

0 2 (a) A1 = 3 0

2 0 (b) A2 = 4 2

cos θ − sin θ (c) A3 = sin θ cos θ

Ans:

cos θ − sin θ

sin θ cos θ

8.E-2 [21] If the inverse of A2 is B, show that the inverse of A is AB. (Thus A is invertible whenever A2 is invertible.)

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8.E-3 (SL 11/07) Let A =

0 2

2 .

0

(a) Find i. A−1 ; ii. A2 .

0 .5 4 0 p Ans: ; Let B = .5 0 0 4 0

2 6 (b) Given that 2A + B = , find the value of p and of q.

4 3

2 . q

(c) Hence find A−1 B. (d) Let X be a 2 × 2 matrix such that AX = B. Find X.

0 Ans: 2, 3; 1

8.E-4 (SL 5/08) Let M =

2 2

3 2

1

1 . −1

(a) Find the determinant of M. (b) Find M−1 .

x 4 (c) Hence solve M = . y 8 1

4 1 2

Ans: −4;

a 8.E-5 (MM 5/99) Let M = 2

1 4 − 21

; x = 3, y = −2

2 , where a ∈ Z. −1

(a) Find M2 in terms of a.

5 −4 2 (b) If M is equal to , find the value of a. −4 5 (c) Using this value of a, find M−1 and hence solve the system of equations: −x + 2y 2x − y

= −3 =

2 a +4 Ans: 2a − 2

3

2a − 2 ; −1; x = 1, y = −1 5 Mr. Budd, compiled January 12, 2011

SL Unit 8, Day 5: Matrices Reloaded

307 

1 8.E-6 (SL 5/06 TZ2) [Try this without a calculator.] The matrix A = −3 2   −1 −2 −2 1 1 . has inverse A−1 =  1 a 6 b

2 1 −2

(a) Find the values of a and b. (b) Consider the simultaneous equations x + 2y

=

7

−3x + y − z

=

10

2x − 2y + z

=

12

(c) Write these equations as a matrix equation. (d) Solve the matrix equation.

[Ans: 4, 7; ; x = −3, y = 5, z = 4] 8.E-7 (HL Spec ’99) Solve, by any method, the following system of equations: 3x − 2y + z = −4 x + y − z = −2 2x + 3y = 4 [Ans: x = −1, y = 2, z = 3] 8.E-8 (SL Spec ’05)  1 (a) Write down the inverse of the matrix A = 2 1

−3 2 −5

 1 −1 3

(b) Hence solve the simultaneous equations x − 3y + z = 1 2x + 2y − z = 2 x − 5y + 3z = 3 [Ans: x = 1.2, y = 0.6, z = 1.6] 8.E-9 (adapted from [22]) Consider the system of equations x + y + 2z = 2 2x + y − z = 4 x−y− z=5 Mr. Budd, compiled January 12, 2011

 0 −1 1

308

SL Unit 8 (Vectors and Matrices) (a) Check that the determinant is not 0. (b) Find the solution using an inverse matrix.

8.E-10 (HL 5/02) The matrix A is given by  2 1 A = 1 k 3 4

 k −1 2

Find the values of k for which A is singular.

Ans: 3, − 31

8.E-11 (HL 5/99) Find the value of a for which the following system of equations does not have a unique solution.

4x − y + 2z = 1 2x + 3y

= −6

x − 2y + az =

7 2

[Ans: 1]

1 6 x 8.E-12 (adapted from HL 5/07) Let A = and X = . Given that 4 3 y AX = kX, where k ∈ R, find the values of k for which there is not a unique solution for x.

[Ans: 7,−3]

Mr. Budd, compiled January 12, 2011

Unit 9

Probability 1. Combined Events and Mutually Exclusive Events 2. Conditional Probability 3. Independent Events

International Baccalaureate 5.6 Sample space, U ; the event A. The probability of event A as P(A) =

n (A) . n (U )

The complementary events A and A0 (not A); the relation P(A) + P(A0 ) = 1.

5.7 Combined events, A ∩ B and A ∪ B. The relation P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Mutually exclusive events; the relation P(A ∩ B) = 0.

5.8 Conditional probability; the relation P(A | B) =

P(A ∩ B) .

P(B)

Independent events; the relations P(A | B) = P(A) = P(A | B 0 ).

5.9 Use of Venn diagrams, tree diagrams and tables of outcomes to solve problems.

Applications.

309

310

SL Unit 9 (Probability)

Mr. Budd, compiled January 12, 2011

SL Unit 9, Day 1: Introduction to Series

9.1

311

Introduction to Series

International Baccalaureate 1.1 Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.

Examples of applications, compound interest and population growth.

Sigma notation.

9.1.1

Arithmetic Series

Example 9.1.1 Add: (a) the integers from one to one thousand. (b) the even integers from two to one thousand.

Example 9.1.2 If the first term of an arithmetic series is given by a, and the common difference is given by d, find expressions for: (a) the nth term, tn (b) the sum of the first n terms, Sn

Sn =

n (2a + (n − 1) d) 2

Example 9.1.3 (MM 95) The first three terms of an arithmetic sequence are 100, 96, 92. (a) Find d, the common difference. (b) Find the two values of n for which Sn = 1036 (where Sn represents the sum of the first n terms).

[Ans: −4; 14, 37]

Example 9.1.4 (HL 11/00) Find the sum of the positive terms of the arithmetic sequence 85, 78, 71, . . .

Example 9.1.5 (SL 84) Find the values of x ∈ R for which x, x2 −2, and 2x + 5 form the first three terms of an arithmetic sequence.

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Ans: 3, − 23 Example 9.1.6 (HL 00) The sum of the first n terms of an arithmetic sequence is Sn = 3n2 − 2n. Find the nth term un .

[Ans: 6n − 5]

Example 9.1.7 (HL 5/04) Find an expression for the sum of the first 35 terms of the series ln x2 + ln

x2 x2 x2 + ln 2 + ln 3 + · · · y y y

giving your answer in the form ln

xm , where m, n ∈ N. yn h i 70 Ans: ln yx595

9.1.2

Geometric Series

Be aware of the difference between a geometric sequence and a geometric series.

The sequence is the list of terms, the series is the sum of terms.

Example 9.1.8 Write 0.35 as a fraction.

Example 9.1.9 Write an expression for the first n terms in a geometric series, i.e., the nth partial sum of the geometric series. h Ans: Sn =

Sn =

n X k=0

arn =

a (1 − rn ) 1−r

,

a(1−r n ) 1−r

,

i r 6= 1

r 6= 1

Example 9.1.10 For what values of r will the geometric series S = lim Sn converge? To what will S converge? n→∞

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SL Unit 9, Day 1: Introduction to Series

313

1 1 1 1 − + · · · converges.

Example 9.1.11 [16] The series − + 3 6 12 24 To what limit?

Example 9.1.12 (HL 5/00) The probability distribution of a dis 2 2 , for crete random variable X is given by P(X = x) = k 3 x = 0, 1, 2, . . .. Find the value of k.

1 3

θ 1−cos θ

i

Ans:

Example 9.1.13 (HL 5/04) AOB is a sector on the unit circle, ˆ = θ. where AOB The lines (AB1 ), (A1 B2 ), (A2 B3 ) are perpendicular to OB. A1 B1 , A2 B2 are all arcs of circles with centre O. Calculate the sum to infinity of the arc lengths AB + A1 B1 + A2 B2 + A3 B3 + . . . h Ans:

Example 9.1.14 (HL 5/05) The sum of the first n terms of an arithmetic sequence {un } is given by the formula Sn = 4n2 − 2n.

Three terms of this sequence, u2 , um and u32 , are consecutive terms in a geometric sequence. Find m.

[Ans: 7]

Problems 9.A-1 (MM 5/03) Gwendolyn added the multiples of 3, from 3 to 3750 and found that 3 + 6 + 9 + . . . + 3750 = s .

Calculate s.

[Ans: 2345625]

9.A-2 (MM 5/00) In an arithmetic sequence, the first term is 5 and the fourth

term is 40. Find the second term.

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314

SL Unit 9 (Probability)

9.A-3 (MM Spec ’00) Each day a runner trains for a 10 km race. On the first day she runs 1000 m, and then increases the distance by 250 m on each subsequent day. (a) On which day does she run a distance of 10 km in training? (b) What is the total distance she will have run in training by the end of that day? Give your answer exactly.

[Ans: 37th, 203.5 km] 9.A-4 (MM 5/99) Find the sum of the arithmetic series 17 + 27 + 37 + . . . + 417.

[Ans: 8897] 9.A-5 (MM 5/97) (a) Find the number of terms in the arithmetic series 500 + 505 + 510 + 515 + . . . + 995 + 1000. (b) Calculate the sum of the series in part (a).

[Ans: 101, 75750] 9.A-6 (MM 11/96) In an arithmetic series, S4 = −120 and S8 = 120. Find S6 .

[Ans: −45] 9.A-7 (MM 5/96) (a) How many terms are there in the arithmetic series 258 + 251 + 244 + . . . − 288 − 295? (b) Calculate the sum of the series in part (a).

[Ans: 80, −1480] 9.A-8 (MM 11/95) The sum of the first n terms of a sequence is Sn = t1 + t2 + t3 + . . . + tn .

Given that Sn = n2 − 3n + 5, find (a) S5 ; (b) Sn − Sn−1 , giving your answer in simplified form; (c) t1000 .

[Ans: 15; 2n − 4; 1996] Mr. Budd, compiled January 12, 2011

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9.A-9 (SL 91) Find the sum of the arithmetic series 3 + 7 + 11 + · · · + 399 .

[Ans: 20100]

9.A-10 (SL 90) The sum of the first n terms of an arithmetic progression is given by Sn = 3n2 + 5n.

Find the values of (a) the first term; (b) the common difference.

[Ans: 8, 6] 9.A-11 (SL 89) Find the sum of the arithmetic series 7 + 11 + 15 + . . . + 755.

[Ans: 71628] 9.A-12 (SL 87) The sum of an arithmetic series containing n terms is 750. Its first term is 64 and the common difference is −2. Find the two possible values of n.

[Ans: 15, 50] 9.A-13 (SL 85) Find the sum of the first twenty terms of an arithmetic series whose first term is 17 and second term is 21.

[Ans: 1100] 9.A-14 (SL 82) The sum Sn of the first n terms of an arithmetic series is given by Sn =

1 5n2 + 13n 4

Write down the values of S1 , S2 , and S3 and hence find the first three terms of the series.

[Ans: 4.5, 7, 9.5] Find an expression for the nth term of the series in the form A + Bn for constants A and B. [Ans: 4.5 + 2.5n] 9.A-15 (HL 5/99) The second term of an arithmetic sequence is 7. The sum of the first four terms of the arithmetic sequence is 12. Find the first term, a, and the common difference, d, of the sequence.

[Ans: 15, −8] 9.A-16 (HL 5/98) The first, second, and the nth terms of an arithmetic sequence are 2, 6, and 58, respectively. (a) Find the value of n. (b) For that value of n, find the exact value of the sum of n terms of a geometric sequence whose first term is 2 and common ration is 12 .

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Ans: 15, 4 1 −

1 215

9.A-17 (HL 5/99) The ratio of the fifth term to the twelfth term of a sequence in 6 an arithmetic progression is 13 . If each term of this sequence is positive, and the product of the first term and the third term is 32, find the sum of the first 100 terms of this sequence.

[Ans: 10300] 9.A-18 (MM 5/00) Find the sum of the infinite geometric series 8 16 2 4 − + − + ....

3 9 27 81 9.A-19 (MM 5/98) (a) Find the number of terms in the geometric series 1 + 3 + 9 + 27 + . . . + 177147. (b) Calculate the sum of the series in part (a).

[Ans: 12, 265720] 9.A-20 (MM 5/96) For the geometric series Sn = a + ar + ar2 + . . . + arn−1 , S2 = 3.5, S3 = 11.375, and r < 0. Find a and r.

[Ans: 14, −0.75] 9.A-21 (HL Spec ’00) Find the sum to infinity of the geometric series −12 + 8 −

16 + ....

3

Ans: − 36 5

9.A-22 (HL 5/03) A geometric sequence has all positive terms. The sum of the first two terms is 15 and the sum to infinity is 27. Find the value of (a) the common ratio; (b) the first term.

Ans:

2 3;

9

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317

Introduction to Probability

International Baccalaureate 6.5 Concepts of trial, outcome, equally likely outcomes, sample space U , and event. n(A) The probability of an event A as P(A) = . n(U ) The complementary events A and A0 (not A); P(A) + P(A0 ) = 1.

6.6 Combined events, the formula: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Appreciation of the non-exclusivity of “or”.

P(A ∩ B) = 0 for mutually exclusive events. Use of P(A ∪ B) = P(A) + P(B) for mutually exclusive events.

6.8 Use of Venn diagrams, tree diagrams and tables of outcomes to solve problems.

9.2.1

The Basics

Definitions Definition 9.1. A (probability) experiment E is a specific set of actions, the results of which cannot be predicted with certainty. [17] Definition 9.2. A set S (or U or Ω) is a sample space for an experiment E if each element of S represents a unique outcome of E , and if each outcome of E is represented by a unique element of S. [17] A single experiment may have several different sample spaces.

Example 9.2.1 Suppose we perform the experiment of tossing a coin twice. (Note: in this case, the single experiment is tossing the coin twice.) Indicate appropriate sample space if (a) we wish to record what happens on each toss, or (b) we wish to record the number of heads obtained [17]

If we wish to record what happens on each toss, an appropriate sample space would be S = {T T, T H, HT, HH}. However, in the same experiment, we may use a different sample space. If we wish to record the number of heads, an appropriate sample space would be Sˆ = {0, 1, 2}. Note that the outcome of 1 in Sˆ corresponds to two different outcomes in sample space S. Mr. Budd, compiled January 12, 2011

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Definition 9.3. An event is a subset of a sample space. A simple event is a subset containing a single element. In many discussions, we will liberally interchange the words “event” and “subset.” [17]

Example 9.2.2 Consider the experiment of tossing a coin twice, with sample space S = {T T, T H, HT, HH}. Represent the following verbally described events as subsets of S. [17] (a) exactly one head; (b) head on the second toss; (c) at least one head; (d) head on both tosses. [17] [Ans: A = {T H, HT }, B = {T H, HH}, C = {T H, HT, HH}, D = {HH} (a simple event)] Given the events A, B, and C, determine which of these events are subsets of other events. [17] [Ans: A ⊂ C, B ⊂ C, A ⊂ A, B ⊂ B, C ⊂ C, A 6⊂ B] Definition 9.4. (Complement of a Set) The complement of a set A with respect to S (or, simply, the complement of A), denoted by A0 (some use Ac or A), is the set of all those elements of S that do not belong to A. [17] Definition 9.5. (Intersection of Sets) [17] The intersection of two sets A and B, denoted by A ∩ B, is the set of all those elements which belong to both A and B. In symbols (where “x ∈” means “x is an element of”), we say A ∩ B = {x | x ∈ A and x ∈ B} Definition 9.6. (Union of Sets) The union of two sets A and B, denoted by A ∪ B, is the set of all those elements that are either in A alone, or in B alone, or in both A and B.[17] Often one says that A ∪ B is the set of all elements that belong to A or to B where “or” is used in the inclusive sense of “and/or.” Still another valid statement is “A ∪ B is the set of those elements that belong to at least one of the sets A and B.”[17]

A ∪ B = {x | x ∈ A or x ∈ B}

Example 9.2.3 Consider the experiment of tossing a coin twice, with the associated sample space U = {T T, T H, HT, HH}, with events A = {T H, HT }, B = {T H, HH}, C = {T H, HT, HH}.

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(a) A0 , B 0 , and C 0 ; (b) A ∪ B; (c) A ∩ B; (d) A ∩ C 0 . [17] [Ans: {T T, HH}, {T T, HT }, {T T }; {T H, HT, HH}; {T H}; Ø] Definition 9.7. (Mutually Exclusive Events) The pair consisting of A and B is mutually exclusive (disjoint) if A ∩ B = Ø.[17] Example 9.2.4 Which pairs of events of the following events are mutually exclusive? A = {T H, HT }, B = {T H, HH}, C 0 = {T T }, A0 = {T T, HH}, A ∩ B = {T H}. [17] [Ans: A and C 0 ; A and A0 ; B and C 0 ; C 0 and A ∩ B; A0 and A ∩ B] Definition 9.8. A partition of a set E is a collection of subsets E1 , E2 , . . . , Em of E with the following properties: 1. Ei ∩ Ej = Ø for i 6= j. That is the collection E1 , E2 , . . . , Em is mutually exclusive.

2. E = E1 ∪ E2 ∪ . . . ∪ Em . That is, E equals the union of the Ei ’s.[17] Definition 9.9. The symbol P(A) indicates the probability of an event A of some sample space S. Here, the probability P(A) is a real number. [17]

Axioms Axioms are the basic theoretical assumptions, upon which we prove everything else.

Axiom 9.1. For any event A of a sample space S, 0 ≤ P(A) ≤ 1. [17] Axiom 9.2. P(S) = 1. [17] Axiom 9.3 (Addition Rule). [17] For disjoint (mutually exclusive) events A1 , A2 , A3 , . . ., P(A1 ∪ A2 ∪ A3 ∪ . . .) = P(A1 ) + P(A2 ) + P(A3 ) + . . .

Theorems Theorem 9.1. For the complement A0 of any event A, P(A) + P(A0 ) = 1. [17] Mr. Budd, compiled January 12, 2011

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SL Unit 9 (Probability) Example 9.2.5 A pair of dice is tossed. What is the probability of getting a sum that is at least 3? [17] Ans: 35 36

Theorem 9.2. For the empty set Ø, P(Ø) = 0. [17] Theorem 9.3. If A and B are events in a sample space S, and A ⊂ B, then P(A) ≤ P(B). [17]

Example 9.2.6 A coin, which is not necessarily fair, is tossed twice.

Prove that the probability of getting two heads is less than or equal to the probability of getting a head on the second toss.

Theorem 9.4. For any finite sequence of events (subsets) A1 , A2 , . . . , Am of sample space S, P(A1 ∪ A2 ∪ A3 ∪ . . . ∪ Am ) ≤ P(A1 ) + P(A2 ) + . . . + P(Am )

Examples Example 9.2.7 A particular unbalanced coin is tossed twice. For the sample space S = {T T, T H, HT, HH}, the following probabilities are assigned, based on repeating the double toss several million times, and observing the corresponding relative frequencies: P({T T }) = .16;

P({T H}) = P({HT }) = .24;

P({HH}) = .36

(a) Compute the probability of getting exactly one head. (b) Repeat for the event “head on the second toss.” (c) Repeat for the event “at least one head.” [17] [Ans: .48,.60,.84]

Equally Likely Events: Classic Probability There are certain cases in which sample space is made up of equally likely events.

This is not always the case: rolling a seven on a pair of dice is not as likely as rolling a two. However, the prospect of rolling a six and an ace is as likely as rolling an ace and an ace.

Theorem 9.5 (Equally Likely Simple Events Theorem). Suppose all n of the simple events associated with a finite sample space S = {e1 , e2 , . . . , en } are equally likely. That is, P({e1 }) = P({e2 }) = . . . = P({en }). Then, Mr. Budd, compiled January 12, 2011

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Figure 9.1: Sets A and B [17]

1. P({ei }) = 1/n for i = 1, 2, . . . , n, and 2. P(A) = a/n for any event A to which a elements belong. Put differently, if we let N (A) and N (S) stand for the number of elements in A and S, respectively, then we can write P(A) = N (A) /N (S).

Example 9.2.8 A pair of dice is tossed. Determine the probability that (a) a two appears on at least one die, (b) the sum of the numbers on both dice equals six, and (c) the same number of dots is obtained on both dice.

Ans:

9.2.2

11 5 6 36 , 36 , 36

Probability and Set Theory: Venn Diagrams

Theorem 9.6 (General Addition Rule for Two Events). P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Consider Figure 9.1, which partitions the sample space S into subsets R1 , R2 , R3 , and R4 . Since these are mutually exclusive events, we can use the Addition Rule (Axiom 9.3).

P(A ∪ B)

=

P(R1 ∪ R2 ∪ R3 )

=

P(R1 ) + P(R2 ) + P(R3 )

=

P(R2 ) + P(R1 ) + P(R1 ) + P(R3 ) − P(R1 )

=

P(R2 ∪ R1 ) + P(R1 ∪ R3 ) − P(R1 )

=

P(A) + P(B) − P(A ∩ B)

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Figure 9.2: Venn diagram for Example 9 [17]

Example 9.2.9 The probability that a student will attend the first meeting of a class is .75, the probability that a student will attend the second meeting is .65, and the probability that a student will attend both meetings is .60. (a) What is the probability that a student will attend at least one of the first two meetings of the class? (b) What is the probability that a student will miss both th first and second meetings of the class? (c) What is the probability that a student will attend exactly one of the first two meetings of the class?

[Ans: .80,.20,.20]

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Figure 9.3: Sets A, B, and C [17]

Theorem 9.7 (General Addition Rule for Three Events). P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)

P(A ∪ B ∪ C)

=

P(R1 ∪ R2 ∪ R3 ∪ R4 ∪ R5 ∪ R6 ∪ R7 )

=

P(R1 ) + P(R2 ) + P(R3 ) + P(R4 ) + P(R5 ) + P(R6 ) + P(R7 )

=

P(R1 ) + P(R2 ) + P(R3 ) + P(R4 ) + P(R5 ) + P(R6 ) + P(R7 ) +P(R1 ) − P(R1 ) + P(R1 ) − P(R1 ) + P(R1 ) − P(R1 ) +P(R2 ) − P(R2 ) + P(R3 ) − P(R3 ) + P(R5 ) − P(R5 )

=

P(R1 ) + P(R2 ) + P(R3 ) + P(R4 ) +P(R1 ) + P(R2 ) + P(R5 ) + P(R6 ) +P(R1 ) + P(R3 ) + P(R5 ) + P(R7 ) −P(R1 ) − P(R2 ) − P(R1 ) − P(R3 ) −P(R1 ) − P(R5 ) + P(R1 )

=

[P(R1 ) + P(R2 ) + P(R3 ) + P(R4 )] + [P(R1 ) + P(R2 ) + P(R5 ) + P(R6 )] + [P(R1 ) + P(R3 ) + P(R5 ) + P(R7 )] − [P(R1 ) + P(R2 )] − [P(R1 ) + P(R3 )] − [P(R1 ) + P(R5 )] +P(R1 )

=

P(A) + P(B) + P(C) −P(A ∩ B) − P(A ∩ C) − P(B ∩ C) +P(A ∩ B ∩ C) Mr. Budd, compiled January 12, 2011

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Problems 1 3 9.B-1 (MM 5/04) Let A and B be events such that P(A) = , P(B) = , and 2 4

7 P(A ∪ B) = . Calculate P(A ∩ B).

Ans: 38 8 9.B-2 (MM 99) Two unbiased six-sided dice are thrown; one is red and the other is green. Two numbers, between 1 and 6 inclusive, are shown. Find the probability that (a) the number on the red die is even; (b) the number on the green die is greater than the number on the red die.

5 Ans: 12 ; 12 9.B-3 (MM 97) For the events C, D, P(C) = 0.7, P(D) = 0.3, and P(C ∪ D) = 0.9. Find P(C ∩ D).

[Ans: 0.1] 9.B-4 (MM 97) Two standard six-faced dice are thrown. Let X be the total score obtained. Find (a) P(X = 12); (b) P(X ≤ 3);

Ans:

1 1 36 , 12

9.B-5 (MM 96N) Two fair dice are rolled. Let x be the sum of the two numbers showing uppermost. Find the probability that (a) x = 10; (b) x ≤ 7.

Ans:

1 7 12 ; 12

9.B-6 (MM 96) For the events A and B, P(A) = 0.3 and P(B) = 0.4. Find P(A0 ∩ B 0 ) if A and B are mutually exclusive events.

[Ans: 0.3] 9.B-7 (MM 94) When two standard 6-faced dice are tossed, T is the total score obtained. (a) Evaluate P(T > 8) (b) If the two dice are tossed twice, what is the probability that T exceeds 8 exactly once?

5 65 Ans: 18 ; 162 Mr. Budd, compiled January 12, 2011

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9.B-8 (MM 93S) Two unbiased dice are thrown and the total score is observed.

X is the event that the total score is even, and Y is the event that the total score is a factor of 12. (a) Find P(X). (b) Find P(Y ). (c) Find P(X ∪ Y ).

Ans:

1 1 5 2; 3; 9

9.B-9 (SL 91) An unbiased coin, which can show ‘head’ or ‘tail’, is tossed four times. What is the probability that the fourth toss is a ‘head’

for the 3 second time?

Ans: 16 9.B-10 [17] A probability experiment is to toss a coin three times. Indicate an appropriate sample space if (a) one wishes to record what happens on each toss, or if (b) one wishes to record the total number of heads obtained.

[Ans: U = {T T T, HT T, T HT, T T H, T HH, HT H, HHT, HHH}, {0, 1, 2, 3}] 9.B-11 [17] For sample space U in problem 10a, indicate the following events as subsets of U : (a) obtaining exactly two heads; (b) obtaining exactly three heads; (c) obtaining head on the third toss; (d) obtaining the same result on all three tosses.

[Ans: A = {T HH, HT H, HHT }, B = {HHH}, C = {T T H, T HH, HT H, HHH}, D = {T T T, HHH}] 9.B-12 [17] Two secretaries, Ann and Bobby, work in an office. On any given day, the probability that Ann shows up for work is .9, the probability that Bobby shows up for work is .8, and the probability that at least one of the secretaries shows up for work is .92. (Note that these are not independent events.) Find the probability that, on any given day, (a) both secretaries show up for work, (b) only Bobby shows up, (c) neither secretary shows up, and (d) only one of the secretaries shows up [Ans: .78, .02, .08, .14] 9.B-13 [17] Use Figure 9.1 to prove that for any two events A and B of a sample space U , P(A ∪ B) = P(A) + P(A0 ∩ B).

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9.B-14 [17] Use Figure 9.1 to: 0

(a) prove DeMorgan’s Law, (A ∩ B) = A0 ∪ B 0 ; (b) derive the equation P(A0 ∪ B 0 ) = 1 − P(A ∩ B).

9.B-15 [17] A survey of people who went to the Lamar County Zoo in 1985 showed that 50% visited the Apes building and 40% visited the Reptiles building.

Let A be the event that a person visits the Apes building, and let R be the event that a person visits the Reptiles building. Determine the closest possible lower and upper bounds for P(A ∩ R) and P(A ∪ R).

[Ans: [0, .40],[.50, .90]] 9.B-16 A survey of freshman was conducted at Dashiell University. The letters M , C, and A stand for whether a student takes courses in mathematics, computer science, or anthropology, respectively. Fifty percent of the students take courses in mathematics: that is, P(M ) = .50. Also, P(C) = .30, P(A) = .15, P(M ∩ C) = 0.20, and M ∩ A = Ø and C ∩ A = Ø. The last item indicates that there are no students who take both computer science and anthropology courses. Determine the value of the following probabilities: (a) P(M ∩ C 0 ); 0 (b) P (M ∪ C) ;

(c) P(M 0 ∩ C 0 ); (d) P(M 0 ∩ A); (e) P(M ∪ C ∪ A).

[Ans: .30,.40,.40,.15,.75]

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9.3

327

Conditional Probability

International Baccalaureate 6.7 Conditional probability; the definition P(A | B) =

P(A ∩ B) .

P(B)

6.8 Use of Venn diagrams, tree diagrams and tables of outcomes to solve problems.

9.3.1

Conditional Probability

Definition 9.10. Given that A and B are events of sample space S, and P(A) > 0, then the conditional probability of B given A is defined by the equation P(B | A) =

P(A ∩ B) P(A)

Conditional probability is a probability we use when a given condition is known to exist. If a condition (such as event B has happened) exists, then the sample space has been changed. If I have the condition that event B happens, then there is no probability that B 0 will happen. I need to exclude B 0 from my sample space. My sample space has been reduced from the whole sample space S to the given condition B. The new probability of A given B is equal to the probability of A within the new sample space of B. Therefore, the conditional probability of A given B is the ratio between the overlap between A and B, i.e., that portion of A that is in B, divided by the total probability of my new “sample space”, B. Example 9.3.1 (MM 5/00) In a survey, 100 students were asked ‘do you prefer to watch television or play sport?’ Of the 46 boys in the survey, 33 said they would choose sport, while 29 girls made this choice.

Television Sport Total

Boys

Girls

33 46

29

Total

100

By completing this table or otherwise, find the probability that (a) a student selected at random prefers to watch television; Mr. Budd, compiled January 12, 2011

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SL Unit 9 (Probability) (b) a student prefers to watch television, given that the student is a boy.

Ans:

19 13 50 ; 46

Example 9.3.2 (SL 5/06 TZ2) In a class, 40 students take chemistry only, 30 take physics only, 20 take both chemistry and physics, and 60 take neither. (a) Find the probability that a student takes physics given that the student takes chemistry. (b) Find the probability that a student takes physics given that the student does not take chemistry. (c) Are the events “taking chemistry” and “taking physics” mutually exclusive? Looking ahead: are these events independent?

Example 9.3.3 A medical team has developed a possible vaccine Table 9.1: Example 3 Got cold Got vaccinated Yes No Total Yes 48 32 80 No 52 28 80 Total 100 60 160

for the common cold. They test it on a group of 160 volunteers divided into an 80-person experimental group and an 80 person control group. The members of the experimental group are vaccinated, while the members of the control group are not. After 12 months all 160 people are asked if they got a cold during the past year.

The results are summarized in Table 9.1 (e.g., 48 vaccinated people got a cold.) The probability experiment is to randomly select one of the 160 people. If for this experiment S = {the 160 people}, A = {got a cold}, and B = {vaccinated}, then find: [1] (a) P(A | B) (b) P(A | B 0 ) (c) P(A | B) + P(A0 | B) (d) P(B | A) (e) P(B | A0 ) Mr. Budd, compiled January 12, 2011

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329 0

Example 9.3.4 (HL 5/05) Given that (A ∪ B) = Ø, P(A0 | B) =

6 1 , and P(A) = , find P(B).

Ans: 37 3 7

Multiplication Rules for Two, Three, and k Events Rearranging the definition of conditional probability gives us the multiplication rule.

Theorem 9.8 (Multiplication Rule). If A and B are any two events of a sample space S and P(A) > 0, then P(A ∩ B) = P(A) · P(B | A)

The multiplication rules for three, and for any number of events is as follows.

Theorem 9.9 (Multiplication Rule for Three Events). If A and B and C are any three events of a sample space S and P(A ∩ B) > 0, then P(A ∩ B ∩ C) = P(A) · P(B | A) · P(C | A ∩ B) Theorem 9.10 (General Multiplication Rule). If A1 , A2 , . . . , Ak are any k events (k ≥ 2) of a sample space S, and P(A1 ∩ A2 ∩ . . . ∩ Ak−1 ) > 0, then P(A1 ∩ A2 ∩ . . . ∩ Ak )

=

P(A1 ) · P(A2 | A1 ) · P(A3 | A1 ∩ A2 ) · . . . ·P(Ak | A1 ∩ A2 ∩ . . . ∩ Ak−1 )

Theorem 9.11. Suppose P(A) > 0. Then P(B 0 | A) exists, and is given by P(B 0 | A) = 1 − P(B | A).

9.3.2

Sampling Without Replacement

The key thing to remember is that, if you are picking from n items, after the first pick you have n − 1 items, after the second pick you have n − 2 items, and so on.

Example 9.3.5 Two cards are drawn from a well-shuffled, standard deck of playing cards. If the first card is not replaced between selections, then what is the probability that: (a) both cards will be hearts, Mr. Budd, compiled January 12, 2011

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SL Unit 9 (Probability) (b) both will be queens, (c) one will be a king and the other a queen?

There are at total of 52 card on the first draw, and 51 cards on the second draw.

13 and the probability of getting The probability of getting a first heart will be 52 12 a second hear on the second draw is 51 . The probability of getting two hearts 12 12 would be 13 52 · 51 = 204 = 0.059. In equation form, P(H1 ∩ H2 )

= = =

P(H1 ) P(H2 | H1 ) 13 12 × 52 51 12 = 0.059 204

Note that the probability of drawing a second hear is dependent on drawing a first heart, therefore we have to use P(H2 | H1 ) instead of just P(H2 ). We would use P(H2 ) in a different situation, in which we replaced the first card in the deck and reshuffled, so that the two events were independent of one another.

For drawing two consecutive queens, we’d have a similar equation: P(Q1 ∩ Q2 )

= = =

P(Q1 ) P(Q2 | Q1 ) 4 3 × 52 51 3 = 0.0045 663

For drawing a king and then a queen, we’d have: P(K1 ∩ Q2 )

= = =

P(K1 ) P(Q2 | K1 ) 4 4 × 52 51 4 = 0.0060 663

For a queen and then a king: P(Q1 ∩ K2 )

= = =

P(Q1 ) P(K2 | Q1 ) 4 4 × 52 51 4 = 0.0060 663

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331

Overall, the probability of a king and a queen would be P(K1 ∩ Q2 )+P(Q1 ∩ K2 ) = 0.0060 + 0.0060 = 0.0120. Note that getting a king and then a queen is mutually exclusive from getting a queen and then a king, so that I can find the probability of both simply by adding the two.

Example 9.3.6 What is the probability of getting a spade on the first draw? What is the probability of getting a spade on the second draw, accounting for the fact that the first draw could be a spade, or could not be a spade?

Example 9.3.7 (HL 97) A bag contains 5 white and 7 black balls.

If two balls are drawn at random without replacement, what is the probability that one of them is black and the other is white.

Ans:

9.3.3

35 66

Counting Methods

Tree Diagrams Example 9.3.8 (adapted from SL Nov ’06) (a) A single bag contains 8 red marbles and 7 blue marbles. Two marbles are drawn without replacement. What is the probability that two marbles of different colors are chosen? (b) Bag 1 contains 4 red marbles and 5 blue marbles. Bag 2 contains 4 red marbles and 2 blue marbles. A marble is chosen from Bag 1, followed by a marble from Bag 2. (a) What is the probability that two marbles of different colors are chosen? (b) Suppose a blue marble is drawn. What is the probability that it came from Bag 1?

What is a fundamental difference in the two situations?

Example 9.3.9 Suppose Alice and Bob are playing a tennis match in which the winner is the one who wins three sets first (“best three out of five sets match”). Use a tree diagram to determine all possible ways the match can be decided.

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SL Unit 9 (Probability) (a) Why do teams play best-of series? (b) If Alice has a 60% chance of winning each game, what is the probability that she wins a best of three match? (c) The newspaper reports that Alice won the match, but doesnâ&#x20AC;&#x2122;t give details. What is the probability that she swept?

Figure 9.4: Tree diagram for best three out of five

The tree diagram is seen in Figure 9.4.

Example 9.3.10 (MM 5/04) Dumisani is a student at the Boodle School for International Students. The probability that he will be 7 woken by his alarm clock is . If he is woken by his alarm clock the 8 1 probability that he will be late for school is . If he is not woken by 4 his alarm clock the probability he will be late for school is 53 . (a) Create a tree diagram for this problem. (b) Calculate the probability that Dumisani will be late for school. (c) Given that Dumisani is late for school, what is the probability that he was woken by his alarm clock?

[Ans: 0.294; 0.745] Mr. Budd, compiled January 12, 2011

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2 2 Example 9.3.11 (HL 5/01) Given that P(X) = , P(Y | X) = , 3 5 1 and P(Y | X 0 ) = , find 4 (a) P(Y 0 ); (b) P(X 0 ∪ Y 0 ).

Ans:

13 11 20 ; 15

Fundamental Counting Principle Theorem 9.12 (Fundamental Counting Principle). Given a combined operation consisting of k component operations. Suppose a first operation can be done in n1 ways, and for each of these ways a second operation can be done in n2 ways, and for each of these ways a third operation can be done in n3 ways, and so on, for each of the k component operations.

Then, a combined operation consisting of the k component operations can be done in n1 · n2 · n3 · . . . · nk ways.[17]

Problems 9.C-1 (MM 5/05) The table below shows the subjects studied by 210 students at a college.

History Science Art Totals

Year 1 50 15 45 110

Year 2 35 30 35 100

Totals 85 45 80 210

(a) A student from the college is selected at random.

Let A be the event the student studies Art. Let B be the event the student is in Year 2. i. Find P(A). ii. Find the probability that the student is a Year 2 Art student.

8 Ans: 21 ; 61 (b) Given that a History student is selected at random, calculate the

probability that the student is in Year 1.

Ans: 10 17 (c) Two students are selected at random from the college. Calculate the probability that one of the students is in Year 1, and the other is in Year 2.

Ans: 200 399 Mr. Budd, compiled January 12, 2011

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9.C-2 (MM 11/04) A packet of seeds contains 40% red seeds and 60% yellow seeds. The probability that a red seed grows is 0.9, and that a yellow seed grows is 0.8. A seed is chosen at random from the packet. Create a tree diagram to answer the following problems. (a) Calculate the probability that the chosen seed is red and grows. (b) Calculate the probability that the chosen seed grows. (c) Given that the seed grows, calculate the probability that it is red.

[Ans: 0.36; 0.84; 0.429] 3 1 9.C-3 (MM 5/04) Let A and B be events such that P(A) = , P(B) = , and 2 4

7 Ans: 21 P(A â&#x2C6;Ş B) = . Calculate P(A | B).

8 9.C-4 (MM 5/03) A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the other, without replacement. (a) The first two apples are green. What is the probability that the third apple is red? (b) What is the probability that exactly two of the three apples are red?

693 Ans: 22 23 ; 2300 = 0.301 9.C-5 (MM 5/01) A bag contains 10 red balls, 10 green balls, and 6 white balls.

Two balls are drawn at random from the bag without replacement.

What

is the probability that they are of different color?

Ans: 44 65 9.C-6 (MM Spec â&#x20AC;&#x2122;00) In a survey of 200 people, 90 of whom were female, it was found that 60 people were unemployed, including 20 males. (a) Using this information, complete the table below.

Males

Females

Unemployed Employed Totals

Totals

200

(b) If a person is selected at random from this group of 200, find the probability that this person is i. an unemployed female; ii. a male, given that the person is employed.

Ans: Table, 15 ,

9 14

9.C-7 (HL 99) A bag contains 2 red balls, 3 blue balls, and 4 green balls. A ball is chosen at random from the bag and is not replaced. A second ball is chosen. Find the probability of choosing one green ball and one blue ball in any order.

Ans: 13 Mr. Budd, compiled January 12, 2011

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335

9.C-8 (SL 82) A research worker has a cage containing 3 male animals and 5 female animals. He chooses two animals at random (without replacement).

Find the probability p that the two chosen animals are of different sexes.

15 Ans: 28 9.C-9 (SL 85) Two cards are drawn simultaneously from a normal pack of 52 playing cards. Find, giving answers correct to three significant figures in each case, [10 marks] (a) the probability that both cards are Aces, (b) the probability that one of the cards is the Ace of Clubs, (c) the probability that neither card is an Ace or a Club, (d) the probability that at least one of the cards is an Ace or a Club, and (e) the probability that one of the cards is the Ace of Clubs given that one of the cards at least is a Club.

1 116 17 1 ; 26 ; 105 Ans: 221 221 ; 221 ; 195 9.C-10 (HL 5/04) Robert travels to work by train every weekday from Monday to Friday. The probability that he catches the 8:00 train on Monday is 0.66. The probability that he catches the 8:00 train on any other weekday is 0.75. A weekday is chosen at random. (a) Find the probability that he catches the train on that day. (b) Given that he catches the 8:00 train on that day, find the probability that the chosen day is a Monday.

[Ans: 0.732; 0.180] 9.C-11 (HL 96) Three suppliers A, B, and C produce respectively 45%, 30%, and 25% of the total number of a certain component that is required by a car manufacturer. The percentages of faulty components in each supplierâ&#x20AC;&#x2122;s output are, again respectively, 4%, 5%, and 6%. What is the probability that a component selected at random is faulty?

[Ans: 0.048]

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Mr. Budd, compiled January 12, 2011

SL Unit 9, Day 4: Independent Events

9.4

337

Independent Events

International Baccalaureate 6.7 Independent events; the definition P(A | B) = P(A) = P(A | B 0 ) .

The term “independent” is equivalent to “statistically independent.” Use of P(A ∩ B) = P(A) P(B) for independent events.

6.8 Use of Venn diagrams, tree diagrams and tables of outcomes to solve problems.

9.4.1

Statistical Independence

Definition of Independence Definition 9.11. Events A and B are independent if and only if P(A ∩ B) = P(A) · P(B) This is the mathematical definition. Intuitively, A and B are independent if the probability of A does not change, regardless of whether B happens, or B does not happen. In other words, A and B are independent if(f) P(A) = P(A | B) = P(A | B 0 ), and equally P(B) = P(B | A) = P(B | A0 ).

Example 9.4.1 (MM 95) A and B are events such that P(A) = 0.4 and P(B) = 0.3. (a) If A and B are mutually exclusive events, find P(A ∪ B). [Ans: 0.7] (b) If A and B are independent events, find (a) P(A ∪ B); (b) P(A | B). (c) P(A | B 0 ).

[Ans: 0.58] [Ans: 0.4] [Ans: 0.4]

Example 9.4.2 (MM 11/02) For events A and B, the probabilities 4 3 , P(B) = . are P(A) = 11 11 Calculate the value of P(A ∩ B) if 6 ; 11 (b) events A and B are independent. (a) P(A ∪ B) =

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SL Unit 9 (Probability)

Ans:

1 12 11 ; 121

Example 9.4.3 (HL 11/00) Given that events A and B are independent with P(A ∩ B) = 0.3 and P(A ∩ B 0 ) = 0.3, find P(A ∪ B).

[Ans: 0.8] Example 9.4.4 (HL 5/02) Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for each child is the sum of the two numbers shown on their respective dice. (a) (a) Calculate

the probability that Alan obtains a score of 9.

Ans: 19 (b) Calculate the probability that Alan and Belle both

obtain 1 a score of 9.

Ans: 81 (b) (a) Calculate the probability that Alan and Belle obtain the same score.

Ans: 146 362 (b) Deduce the probability that Alan’s score exceeds Belle’s

score.

Ans: 575 362 Theorem 9.13.

1. Events A and B are independent if and only if P(B | A) = P(B). (Here, assume that P(A) > 0 so that P(B | A) exists.)

2. Events A and B are independent if and only if P(A | B) = P(A). (Here, assume that P(B) > 0 so that P(A | B) exists.)

3. Events A and B are independent if and only if P(B | A0 ) = P(B). (Here, assume that P(A0 ) > 0 so that P(B | A0 ) exists.)

4. Events A and B are independent if and only if P(A | B 0 ) = P(A). (Here, assume that P(B 0 ) > 0 so that P(A | B 0 ) exists.)

Here, the if and only if connectors tell you that the implications of the statements are bidirectional, that is, for part 1, P(B | A) = P(B) implies independence, and independence implies P(B | A) = P(B), as well as P(B | A0 ) = P(B), as well as P(A | B) = P(A), as well as P(A | B 0 ) = P(A).

Theorem 9.14. If the two events A and B are independent, then 1. the two events A and B 0 are also independent, and 2. the two events A0 and B are also independent, and 3. the two events A0 and B 0 are also independent.

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2 Example 9.4.5 Let E and F be events such that P(E) = and 3 1 P(E ∩ F ) = 3 (a) Calculate P(F | E). What must P(F ) be if events E, F are to be independent?

1 (b) Suppose P(E 0 ∩ F 0 ) = . Are E and F independent?

12 5 (c) Suppose instead that P(E ∪ F ) = . Are E and F independent 6 in that case?

Ans:

1 2;

N; Y

Example 9.4.6 One card is drawn, and placed face down, so that its identity is unknown. Accounting for the fact that the first card could be a spade, or might not be a spade, find the probability that the next card drawn is a spade. Are the events “getting a spade on the first draw” and “getting a spade on the second draw” independent of one another?

Definition 9.12 (Independence of Three Events). Three events A, B, and C are independent if and only if the following four equations hold: P(A ∩ B)

=

P(A) · P(B) ;

P(A ∩ C)

=

P(A) · P(C) ;

P(B ∩ C)

=

P(B) · P(C) ;

P(A ∩ B ∩ C)

=

P(A) · P(B) · P(C)

The first three conditions are that the three events are pairwise independent, i.e., each pair is independent. Events A1 , A2 , . . . , Ak are pairwise independent if P(Ai ∩ Aj ) = P(Ai ) · P(Aj ) for every distinct pair i 6= j.

Definition 9.13 (Independence of k Events). Events A1 , A2 , . . . , Ak are independent if and only if the probability of the intersection of any 2, 3, . . . , or k of these events equals the product of their respective probabilities.

Problems 9.D-1 (MM 5/05) The table below shows the subjects studied by 210 students at a college.

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History Science Art Totals

Year 1 50 15 45 110

Year 2 35 30 35 100

Totals 85 45 80 210

Are the events A and B independent? Justify your answer.

[Ans: N]

1 3 9.D-2 (MM 5/04) Let A and B be events such that P(A) = , P(B) = , and 2 4 7 P(A ∪ B) = .

8 (a) Calculate P(A | B). (b) Are the events A and B independent? Give a reason for your answer.

Ans: 12 ; yes 9.D-3 (MM 93S) For events A and B, P(A) =

3 4 6 , P(B) = , and P(A ∪ B) = .

11 11 11

(a) Find P(A ∩ B).

[2 marks]

(b) Find P(A | B).

[2 marks]

0

(c) Find P(A | B ).

[4 marks]

(d) Determine whether the events A and B are independent, giving a reason for your answer.

[4 marks]

1 Ans: 11 ; 14 ; 27 ; N 9.D-4 (MM 5/03) Two fair dice are thrown and the number showing on each is noted. The sum of these two numbers is S. Find the probability that (a) S is less than 8;

[2 marks]

(b) at least one die shows a 3;

[2 marks]

(c) at least one die shows a 3, given that S is less than 8.

[3 marks]

7 1 Ans: 12 ; 11 36 ; 3 9.D-5 (MM 5/98) The events A and B are independent, and P(A ∩ B) = 0.6, P(B) = 0.8. Find (a) P(A | B); (b) P(A | not B).

[Ans: 0.75; 0.75] Mr. Budd, compiled January 12, 2011

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9.D-6 (MM 11/96) A department store has two burglar alarm systems. In the event of an attempted break-in, the systems function properly with probabilities 0.95 and 0.90, respectively. The two systems function independently.

When an attempt is made to break in, what is the probability of at least one of the alarm systems functioning properly?

[Ans: 0.995] 9.D-7 (MM 96) For the events A and B, P(A) = 0.3 and P(B) = 0.4. (a) Find P(A ∪ B) if A and B are independent events. (b) Find P(A0 ∩ B 0 ) if A and B are mutually exclusive events.

[Ans: 0.58, 0.3] 9.D-8 (MM 11/95) A and B are independent events such that P(A) = 0.5 and P(A0 ∪ B 0 ) = 0.7. Find (a) P(B); (b) P(B | A0 ).

[Ans: 0.6, 0.6] 9.D-9 (SL 93) A and B are independent events with P(A ∪ B) = 0.8 and P(B) = 0.2. Find P(A).

[Ans: 0.75] 9.D-10 (SL 91) Two standard unbiased dice are thrown. The random variable X is defined to be the sum of the two values shown on the dice. (a) Write down P(X = 3).

[3 marks]

(b) Calculate P(X is divisible by 4).

[4 marks]

The two dice are thrown twice and X1 and X2 are the two values of X obtained. (c) Show that P(X1 = X2 ) =

146 362

[5 marks]

(d) Hence, or otherwise, determine P(X1 > X2 ).

[3 marks]

(e) Find P(X1 = 3 | X1 > X2 ).

Ans: 9.D-11 (SL 89) If P(A) = calculate

1 3

and P(B) =

2 5

[5 marks]

1 575 2 1 18 ; 4 ; ; 362 ; 575

and A and B are independent events,

(a) P(A ∪ B), (b) P(A0 ∪ B 0 ).

Ans:

3 13 5 ; 15

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9.D-12 (SL 84) There are two roads only from town A to town B; each of the roads is liable to be closed because of landslides. On average the low road is closed on one day in six and on average the high road is closed on one day in five. Assuming that the closures of the roads are independent of one another, find (a) the probability that it is possible to travel from town A to town B on a given day, and (b) the probability that both the roads will be open on a given day.

2 Ans: 29 30 ; 3 9.D-13 (HL 5/03) The independent events A, B are such that P(A) = 0.4 and P(A â&#x2C6;Ş B) = 0.88. Find (a) P(B); (b) the probability that either A occurs or B occurs, but not both.

[Ans: 0.8; 0.56] 9.D-14 (HL 5/00) In a game a player rolls a biased tetrahedral (four-faced) die.

The probability of each possible score is shown in Figure 9.2. Find the probability of a total score of six after two rolls.

Ans: 14 Table 9.2: Problem 14: Score 1 1 Probability 5

Tetrahedral Die 2 3 4 2 1 x 5 10

9.D-15 (HL 5/98) For two independent events A and B, P(A | B) = 41 and 1 P(A â&#x2C6;Š B) = 32 . Find P(A) and P(B).

Ans: 41 , 18

Mr. Budd, compiled January 12, 2011

Unit 10

Probability Distributions

343

344

SL Unit 10 (Probability Distributions)

1. Discrete Random Variables 2. Binomial Distribution Function 3. Normal Distribution

International Baccalaureate 6.9 Concept of discrete random variables and their probability distributions.

Expected value (mean), E (X), for discrete data.

Knowledge and use of the formula E (X) = ÎŁ (xP(X = x)). Applications of expectation, for example, to games of chance.

6.10 Binomial distribution.

Mean of the binomial distribution.

6.11 Normal distribution.

Properties of the normal distribution. Appreciation that the standardized value (z) gives the number of standard deviations from the mean.

Standardization of normal variables. Use of calculator (or tables) to find normal probabilities; the reverse process.

Mr. Budd, compiled January 12, 2011

SL Unit 10, Day 1: Discrete Random Variables

10.1

345

Discrete Random Variables

International Baccalaureate 6.9 Concept of discrete random variables and their probability distributions.

Expected value (mean), mode, median, variance, and standard deviation. Knowledge and use of the formula for E (X) and Var (X).

Applications of expectation, for example, to games of chance.

10.1.1

Discrete Random Variables

Example 10.1.1 (a) Without me telling you how to find it, what do your instincts tell you about how to find the average or mean value from a roll of a die? (b) What changes if the die has two sides with a 1, three sides with a 3, and one side with a 5?

Definition 10.1. [17] The expected value of a discrete random variable X with probability function P(X = x) = f (x) is given by E (X) =

X

x f (x)

x

The expected value of X is calculated essentially the same way as the mean of X, with weighting based on probability instead of frequency.

Weird notation rule: the X stands for what value the random variable is going to take. No mortal knows what this value is, as it will be determined in the future. x is a placeholder for all the different possible values that X might possibly take.

E X 2 would essentially be the average of the squares, and the formula would be X 2 E X2 = x f (x) x

where f (x) is the probability function, f (x) = P(X = x).

Example 10.1.2 Find E (X) and E X 2 where X represents the roll obtained from (a) a normal die; Mr. Budd, compiled January 12, 2011

346

SL Unit 10 (Probability Distributions) (b) a die with two sides with one dot, three sides with three dots, and one side with five dots.

Theorem 10.1. [17] Suppose that X is a discrete random variable with probability function f (x), and that the random variable Y is a function of X given

by Y = g(x). Then, X E(Y ) = E (g(x)) = g(x)f (x) x

where the summation is over all possible values of X. Theorem 10.2. [17] If a and b are constants, then E (aX + b) = aE (X) + b.

Example 10.1.3 Show that E (a g(x) + b h(x)) = aE (g(x))+bE (h(x)).

Do a lot of functions work like that, e.g., is sin (a g(x) + b h(x)) = a sin (g(x)) + b sin (h(x))?

Example 10.1.4 [17] Suppose a pair of fair dice is tossed, and the random variable X is equal to the sum of the numbers that occur on the two dice. Determine E (X), the expected value of X.

[Ans: 7]

10.1.2

Variance

The variance of X is symbolized by Var (X). It is given by

2 Var (X) = E (X − µ) where µ = E (X) In actuality, we usually compute the variance with the following theorem: Theorem 10.3.

2 Var (X) = E X 2 − [E (X)] Example 10.1.5 Find Var (X) both ways where X represents the roll obtained from (a) a normal die; (b) a die with two sides with one dot, three sides with three dots, and one side with five dots.

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347

Check your answer on the calculator.

Example 10.1.6 (extension of MM 5/03) A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the other, without replacement. (a) What is the probability that exactly two of the three apples are red? (b) Create a probability distribution chart. (c) Find the mean, median, and mode.

[Ans: 2.64, 3, 3]

(d) Find the standard deviation and variance.

Example 10.1.7 (SL Spec â&#x20AC;&#x2122;05) Bag A contains 2 red balls and 3 green balls. Two balls are chosen at random from the bag without replacement. Let X denote the number of red balls chosen. (a) Draw a tree diagram to represent the above information, including the probability of each event. Hence create a probability distribution table for X. (b) Calculate E (X), the mean number of red balls chosen. [Ans: 0.8] Bag B contains 4 red balls and 2 green balls. Two balls are chosen at random from Bag B. (c) Draw a tree diagram to represent the above information, including the probability of each event. Hence find the probability distribution

for Y , where is the number of red balls chosen.

Ans: 2,16,12 30 A standard die with six faces is rolled. If a 1 or 6 is obtained, two balls are chosen from bag A, otherwise two balls are chosen from bag B. (d) Calculate the probability that two red balls are chosen. (e) Given that two red balls are chosen, find the conditional prob-

ability that a 1 or 6 was rolled on the die.

Ans: 0.3; 19 Example 10.1.8 (adapted from SL Nov â&#x20AC;&#x2122;99) (a) A single bag contains 8 red marbles and 7 blue marbles. Two marbles are drawn without replacement. What is expected number of red marbles drawn?

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SL Unit 10 (Probability Distributions) (b) Bag 1 contains 4 red marbles and 5 blue marbles. Bag 2 contains 4 red marbles and 2 blue marbles. A marble is chosen from Bag 1, followed by a marble from Bag 2. What is expected number of red marbles drawn?

Example 10.1.9 (MM 5/96) A computer is programmed to generate a sequence x1 , x2 , x3 , . . . of random single digits, each of which takes a value from 0 to 9 inclusive. Each digit is equally likely. Let X be the discrete random variable which represents the value of a digit.

Find the mean and variance of X.

[Ans: 4.5, 8.25] Theorem 10.4. [17] If a and b are constants, then Var (aX + b) = a2 Var (X).

10.1.3

Other Measures

The standard deviation is the square root of the variance.

The median will be the x-value where the cumulative probability distribution function is 0.5.

The mode will be the x-value where the probability function (for a discrete variable) is the largest.

Example 10.1.10 A die has two sides with one dot, three sides

with three dots, and one side with five dots. Find E (X), E X 2 , Var (X), standard deviation, median, and mode.

Example 10.1.11 (MM 11/96) A fair game is a game in which the expected profit of any player is zero.

A player pays $3 to throw a pair of unbiased dice, and then receives a payment, $X, depending on the outcome of the throw, as follows: • X = 0, if the the total score on the dice is less than 7; • X = 3, if the total score is 7; • X = k, if the total score is 11 or 12; • X = 4, otherwise.

The player’s profit is $Y , where Y = X − 3.

Mr. Budd, compiled January 12, 2011

SL Unit 10, Day 1: Discrete Random Variables (a) Show that E (Y ) =

1 12

349

(k − 14).

(b) Given that this is a fair game, find the value of k. [Ans: 14] Hence determine the profit earned by the player if his throw results in a total score of 11.

[Ans: $11] (c) Find (a) (b) (c) (d)

[Ans: −3] [Ans: 0]

Ans: 85 6

5 Ans: 12

the mode of Y ; the median of Y ; the variance of Y ; P(Y > 0).

Example 10.1.12 (adapted from HL 5/02) Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for each child is the sum of the two numbers shown on their respective dice.

Let X denote the largest number shown on the four dice. x 4 , for x = 1, 2, . . . , 6 (a) Show that P(X ≤ x) = 6 (b) Find the median. (c) Copy and complete the following probability distribution table. x P(X = x)

1 1 1296

2 15 1296

3

4

5

(d) Calculate E (X)

6 671 1296

Ans:

575 1296

Problems 10.A-1 (SL 5/06) Three students, Kim, Ching Li, and Jonathan each have a pack of cards, from which they select a card at random. Each card has 0, 3, 4, or 9 printed on it. (a) Kim states that the probability distribution for her pack of cards is as follows x P(X = x)

0 0.3

3 0.45

4 0.2

9 0.35

Explain why Kim is incorrect. (b) Ching Li correctly states that the probability distribution for her pack of cards is as follows. x P(X = x)

0 0.4

3 k

4 2k

9 0.3

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350

SL Unit 10 (Probability Distributions) Find the value of k.

[Ans: 0.1]

(c) Jonathan correctly states that the probability distribution for his x+1 pack of cards is given by P(X = x) = . One card is drawn at 20 random from his pack. i. Calculate the probability that the number on the card drawn is 0. ii. Calculate the probability that the number on the card drawn is

1 ; 19 greater than 0.

Ans: 20 20 10.A-2 (SL 5/06) The probability distribution of the discrete random variable X is given by the following table. x P(X = x)

1 0.4

2 p

3 0.2

4 0.07

5 0.02

(a) Find the value of p. (b) Calculate the expected value of X.

[Ans: 0.31; 2]

10.A-3 (SL 5/06) A game is played, where a die is tossed and a marble selected from a bag.

Bag M contains 3 red marbles and 2 green marbles.

Bag N contains 2 red marbles and 8 green marbles.

A fair six-sided die is tossed. If a 3 or 5 appears on the die, bag M is selected.

If any other number appears, bag N is selected. A single marble is then drawn from the selected bag. (a) Complete a tree diagram. (b) Write down the probability that bag M is selected and a green marble drawn from it. (c) Find the probability that a green marble is drawn from either bag. (d) Given that the marble is green, calculate the probability that it came from Bag M. (e) A player wins $ 2 for a red marble and $ 5 for a green marble. What

2 ; 23 ; 15 ; $ 4 are his expected winnings?

Ans: 15 10.A-4 (adapted from SL 93) An unbiased coin shows H on one side and T on the other. The coin is tossed twice. A random variable X is defined as follows: X = 1 if both tosses have result H, otherwise X = 2 . [Do these without a calculator!] (a) What is the mean of X? Mr. Budd, compiled January 12, 2011

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(b) What is the expected value of X 2 ? (c) What is the standard deviation of X? h Ans:

7 13 4; 4 ;

√

3 4

i

10.A-5 (SL 89) X is a discrete random variable for which r−1 2 1 P(X = r) = 3 3

,

r = 1, 2, 3, . . . .

Calculate P(x ≤ 3), giving your answer as a fraction reduced to its lowest 19 terms.

Ans: 27 10.A-6 (SL 84) A biased die is labeled with 1, 2, 3, 4, 5, and 6 spots on its six faces. When thrown it lands with k spots showing on its upper face with probability proportional to k. Find the mean of the number of spots

shown on the upper face when the die is thrown.

Ans: 13 3 10.A-7 (SL 85) An unbiased die has 3 spots on each of three faces, 2 spots on each of two faces and 1 spot on the remaining face. The die is thrown twice and on both occasions the number of spots on the uppermost face is observed.

These two numbers are added and the total of spots, x is recorded. number 1 5 Find the probability distribution P(x). Ans: 2: 36 ; 3: 19 ; 4: 18 ; 5: 13 ; 6: 14 10.A-8 (HL 5/99) A biased die with four faces is used in a game. A player pays 10 counters to roll the die. The table below shows the possible scores on the die, the probability of each score and the number of counters the player receives in return for each score.

Score Probability Number of counters players receives

1

2

3

4

1 2

1 5

1 5

1 10

4

5

15

n

Find the value of n in order for the player to get an expected return of 9 counters per roll.

[Ans: 30] 10.A-9 (MM 5/97) Let X be the random variable which represents the score when a fair (unbiased) six-faced die is thrown. (a) Calculate E (X), Var (X).

[Ans: 3.5, 2.917]

When a certain biased six-faced die is thrown the score is the random 2 variable Y , and P(Y = y) = k (y − 3.5) + 0.1375. (b) Find k, and hence calculate P(Y = y) for y = 1, . . . , 6. [Ans: .2,.16,.14,.14,.16,.2] (c) Find E (Y ). Explain why, with only this information, the die might be considered fair.

[Ans: 3.5] Mr. Budd, compiled January 12, 2011

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SL Unit 10 (Probability Distributions) (d) Find Var (Y ). By comparing this with Var (X) explain how the die’s bias can now be detected.

[Ans: 3.29]

10.A-10 (MM 11/95) [Try this without a calculator.] A discrete random variable has its probability function completely defined as follows: P(X P(X P(X P(X

= −2) = 0) = 1) = 2)

= 0.20; = 0.35; = 2k; = k.

(a) Find k. (b) If F (x) = P(X ≤ x), find i. F (0); ii. F (3).

[Ans: 0.15; 0.55, 1]

(c) Find the expected value of X [No calculator!]. (d) Find the variance of X. [Still no calculator!] (e) State i. the median of X; ii. the mode of X. [Ans: 0.2; 1.66; 0, 0] 10.A-11 (MM 5/95) [Try this without a calculator.] A bag contains 6 red marbles and 4 yellow marbles. The player of a game draws a sample of three marbles from the bag at the same time. Let R represent the number of red marbles in the sample. (a)

i. Show that P(R = 1) = 0.3. ii. Find the expected value of R.

[Ans: 1.8]

The player wins 10 cents for every red marble drawn and loses 5 cents for every yellow marble drawn. Let W be the profit made by the player. (b) Show that W = 15R − 15. (c) Find i. P(W ≤ 0); ii. the expected value of W .

Ans:

1 3;

12

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Binomial Expansion

International Baccalaureate n 1.3 The binomial theorem: expansion of (a + b) , n ∈ N.

2

(a + b)

3

(a + b)

4

(a + b)

=

(a + b) (a + b)

=

a2 + ab + ba + b2

=

a2 + 2ab + b2

=

(a + b) (a + b)

=

(a + b) a2 + 2ab + b2

=

a3 + 2a2 b + ab2 + a2 b + 2ab2 + b3

=

a3 + 3a2 b + 3ab2 + b3

=

(a + b) (a + b)

2

3

3

=

(a + b) a + 3a2 b + 3ab2 + b3

=

a4 + 3a3 b + 3a2 b2 + ab3 + a3 b + 3a2 b2 + 3ab3 + b4

=

a4 + 4a3 b + 6a2 b2 + 4ab3 + b4

“The calculations are already fairly complex and it is worth looking at these results for the underlying pattern. There are three main features to the pattern.

Looking at the fourth power example above, these patterns are: 1. “The powers of a. These start at 4 and decrease: a4 , a3 , a2 , a1 , a0 .

Remember that a0 = 1.

2. “The powers of b. These start at 0 and increase: b0 , b1 , b2 , b3 , b4 . Putting these two patterns together gives the final pattern of terms in which the sum of the indices is always 4: . . . a4 + . . . a3 b + . . . a2 b2 + . . . ab3 + . . . b4 .

3. “The coefficients complete the pattern. These coefficients arise because there is more than one way of producing most of the terms. Following the pattern begun above, produces a triangular pattern of coefficients known as Pascal’s Triangle. Blaise Pascal (1623-1662) developed early probability theory but is luck to have this triangle named after him as it had been studied by Chinese mathematicians long before he was born.”[4]

Pascal’s Triangle 1 Mr. Budd, compiled January 12, 2011

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SL Unit 10 (Probability Distributions) 1 1 1

1 2

1

1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 “The numbers in bers immediately developed on the produces Pascal’s

3

3

the body of the triangle are found by adding the two numabove and to either side. It is worth looking at the process previous page and comparing it with the same process that Triangle to confirm that the two are effectively the same.

“In using the Pascal’s Triangle method to find the binomial coefficients, all that is necessary is to select the correct row. In the case of the example under discussion, the appropriate row is 1

4

6

4

1

and the complete result follows: 4

(a + b) = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 Example 10.2.1 (a) How many different 3 letter combinations can you make with R’s and G’s, if you are allowed to repeat letters? (b) How many different ways can you order: (a) 3 R’s; (b) 2 R’s and 1 G; (c) 1 R and 2 G’s; (d) 3 G’s 3

(c) Find the expansion of (r + g) (d) How many different 4 letter combinations can you make with R’s and G’s, if you are allowed to repeat letters? (e) How many different ways can you order: (a) 4 R’s; (b) 3 R’s and 1 G; (c) 2 R’s and 2 G’s; (d) 1 R and 3 G’s; (e) 4 G’s.

4

(f) Find the expansion of (r + g) (g) Add up the numbers in each row of Pascal’s triangle.

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Binomial Theorem “An alternative to using Pascal’s Triangle to find the coefficients is to use com4 binatorial numbers. If expanding (a + b) the set of coefficients are:

4 4 4 4 4 = 1, = 4, = 6, = 4, =1 0 1 2 3 4 which is the same as those given by Pascal’s Triangle.[4]

k

(a + b)

k k 0 k k−1 1 k k−r r k 0 k b + ... + b + ... + = a b + a a a b 0 1 r k k k X k n−r r X n−r r = a b = b n Cr a r r=0 r=0

Example 10.2.2 [4] Expand

2x −

2 x

3 = =

2 2x − x

3

0

1

2

3 2 2 2 2 2 1 0 + 3 C1 (2x) − + 3 C2 (2x) − + 3 C3 (2x) − x x x x 8 24 − 3 8x3 − 24x + x x 3

3 C0 (2x)

−

Example 10.2.3 [4] Find the coefficient of the term in x6 in the 7 expansion of 3x2 − 6 3 3 7 The x6 term in 3x2 − 6 is the term containing 3x2 , because x2 = x6 .

3

7−3 The coefficient is 73 , so the term is 7 C3 3x2 (−6) = 35 × 27x6 × 1296 = 6 1224720x . The coefficient of the term is 1224720.

Example 10.2.4 Find the constant term in the expansion of

2 x2 − x 3

9 .

Ans: − 1792 9 Mr. Budd, compiled January 12, 2011

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SL Unit 10 (Probability Distributions) Example 10.2.5 (HL 99) Given that 5

6

(1 + x) (1 + ax) ≡ 1 + bx + 10x2 + . . . . . . + a6 x11 , find the values of a, b ∈ Z∗ .

[Ans: −2, −7] 6

Example 10.2.6 [4] Write the expansion of (1 + x) and hence find 1.016 .

6

6

(1 + x) = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6 . 1.016 = (1 + 0.01) = 2 3 4 5 6 1+6 (0.01)+15 (0.01) +20 (0.01) +15 (0.01) +6 (0.01) +(0.01) = 1+6 (0.01)+ 15 (0.0001)+200.000001+15 (0.00000001)+6 (0.0000000001)+0.000000000001 = 1 + 0.06 + 0.0015 + 0.000020 + 0.00000015 + 0.0000000006 + 0.000000000001 = 1.061520150601.

Problems 7 10.B-1 (MM 5/03) Find the term containing x10 in the expansion of 5 + 2x2 .

Ans: 16800x10 12

10.B-2 (MM 00) Find the coefficient of a5 b7 in the expansion of (a + b) . [Ans: 792] 9 10.B-3 (MM 99S) Determine the constant term in the expansion of x − x22 .

[Ans: −672] 8

10.B-4 (MM 99) Find the coefficient of x5 in the expansion of (3x − 2) . [Ans: −108864] √ 3 10.B-5 (MM 98) Use the Binomial Theorem to express 1 + 5 in the form √ p + q 5 where p and q are integers. (A decimal approximation to the answer will obtain no marks.)

[Ans: 16,8] 5

10.B-6 (MM 96) If (2x − 3) = 32x5 − 240x4 + 720x3 + Ax2 + Bx − 243, find the values of A and B. [Ans: -1080,810] 10.B-7 (MM 94) It is known that, for small values of x, 8

(1 + x) ≈ 1 + 8x + Ax2 + Bx3 . (a) Use the Binomial Theorem to find the values of A and B. [Ans: 28, 56] (b) A calculator gives the value of 1.018 as 1.0828567. Write down the remaining decimal digits of 1.018 .

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10.B-8 (SL 93) Write down the term in which a4 appears in the expansion of 7

(2a − 3b) .

Give your answer in simplest form.

Ans: −15120a4 b3

10.B-9 (SL 87) Find the value of the term independent of x in the binomial expansion of the expression

3 2 . x2 − x [Ans: 12] 10.B-10 (SL 84) Express 2 +

√ 3 √ √

3 in the form a+b 3, where a, b ∈ Z. Ans: 26 + 15 3

10.B-11 (SL 82) Find the coefficient of x3 in the expansion of 5

(2 + 3x) .

[Ans: 1080] 10.B-12 [20] (a) If F (x) = f (x)g(x), where f and g have derivatives of all orders, show that F 00 = f 00 g + 2f 0 g 0 + f g 00 (b) Find similar formulas for F 000 and F (4) . (c) Guess a formula for F (n)

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SL Unit 10 (Probability Distributions)

Mr. Budd, compiled January 12, 2011

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359

Binomial Distribution

International Baccalaureate 6.10 Binomial distribution, its mean and variance. Conditions under which random variables have the binomial distribution.

Characteristics of the Binomial Probability Situation • a fixed number n of independent, identical trials of the same experiment.

• each trial can be classified only as success or failure; • each trial has the same probability p of success, and a probability q = 1−p of failure; • the variable X represents the number of successes out of n trials.

Example 10.3.1 What is the probability in 5 trials of getting SSSFF, i.e., three successes followed by two failures, if the probability of each success is p? What is the probability of SFSFS? How many different words can be made up of the letters SSSFF?

Example 10.3.2 Develop a formula for getting x successes out of n trials, each with probability p of success. How is this related to n the expansion of (p + q) ?

If X ∼ B(n, p), i.e., X is a random variable representing the number of successes in n trials, with each trial having p probability of success, then P(X = x) = n−x n x = nx px q n−x x p (1 − p) In actuality, we do this on the calculator as binompdf(n,p,x).

Example 10.3.3 (MM 93S) In a city with a large population, 65% of the population are vaccinated against infection by a certain disease. However, about 5% of those vaccinated nevertheless eventually acquire the disease, whilst about 20% of those unvaccinated acquire it. (a) If 5 unvaccinated people are selected at random from the city, find the probability that (a) none of them has the disease;

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SL Unit 10 (Probability Distributions) (b) exactly two of them have the disease; (c) at least three of them have the disease.

[Ans: 0.205] [Ans: 0.0579]

(b) Show that the probability that a person has the disease is just over 10%. (c) Show that the probability that an infected person is vaccinated is about 0.32.

Example 10.3.4 (HL 5/01) In a game, the probability of a player 1 scoring with a shot is . Let X be the number of shots the player 4 takes to score, including the scoring shot. (You can assume that each shot is independent of the others.)

9 (a) Find P(X = 3).

Ans: 64 (b) Find the probability that the player will have at least three misses before scoring twice.

Ans: 189 256 (c) Prove that the expected value of X is 4. (You may use the â&#x2C6;&#x2019;2 result (1 â&#x2C6;&#x2019; x) = 1 + 2x + 3x2 + 4x3 . . .)

Example 10.3.5 [4] A fair die is rolled six times. If X denotes the number of fours obtained, find (a) E (X) [What would you expect?] (b) the mode of X (c) the standard deviation and variance

Mean and Variance Example 10.3.6 Using your calculator, (a) Put seq(X,X,0,5) into L1 (b) Put binompdf(5,0.2,L1) into L2 (c) Put binomcdf(5,0.2,L1) into L3 (d) What does binompdf do? What about binomcdf? (e) How many successes out of five would you expect, if the probability of success in each trial is 20%? (f) Use your calculator to find the mean and the variance. (g) Do this again, using a probability other than 0.2. Find a pattern for the expected value and the variance.

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For the binomial probability distribution, E (X) = np Var (X) = np (1 − p) = npq

Example 10.3.7 (a) Show that the binomial distribution is a proper probability distribution in that all the probabilities add to 1. (b) Prove that E (X) = np. You may want to use the fact that sum of all the probabilities is 1 even for n − 1 trials.

(c) Similarly, prove that E X 2 = np ((n − 1) p + 1). Hence show that Var (X) = npq.

Example 10.3.8 [4] The random variable X has a binomial distribution such that E (X) = 8 and Var (X) = 4.8. Find P(X = 3).

[Ans: 0.0123]

Cumulative Distribution and Inequalities The cumulative probability distribution, i.e., P(X ≤ x) is given by binomcdf(n,p,x).

• P(X ≤ x) • P(X < x) = P(X ≤ x − 1) (for x > 0) • P(X > x) = 1 − P(X ≤ x) • P(X ≥ x) = 1 − P(X < x) = 1 − P(X ≤ x − 1) (for x > 0)

Example 10.3.9 (MM 98) A factory produces light bulbs in large batches. In any batch 5% of the light bulbs are defective. The light bulbs are sold in packs of five. (a) Find the probability that a pack contains at least one defective bulb.

[Ans: 0.226] (b) Find the probability that a pack contains more than two defective bulbs.

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SL Unit 10 (Probability Distributions) The quality of the batches is checked in the factory by taking a sample of 50 bulbs from a batch and testing them to find the number of defective bulbs in the sample. (a) Find the mean and standard deviation of the number of defective bulbs in a sample of 50 light bulbs.

[Ans: 2.5, 1.54] (b) An entire batch is rejected if the number of faulty light bulbs found in a sample of 50 exceeds two standard deviations above the mean. Find the least number, n, of faulty bulbs which would justify rejection.

[Ans: 6 > 5.58] (c) What is the probability that the number of faulty light bulbs exceeds two standard deviations above the mean?

Example 10.3.10 (HL 11/06) A bag contains a very large number of ribbons. One quarter of the ribbons are yellow and the rest are blue. Ten ribbons are selected at random from the bag. (a) Find the expected number of yellow ribbons selected. (b) Find the probability that exactly six of these ribbons are yellow. (c) Find the probability that at least two of these ribbons are yellow. (d) Find the most likely number of yellow ribbons selected. (e) What assumption have you made about the probability of selecting a yellow ribbon?

[Ans: 2.5; .0162; .756; 2;] Example 10.3.11 (HL 5/06) Andrew shoots 20 arrows at a target.

He has probability of 0.3 of hitting the target. All shots are independent of each other. Let X denote the number of arrows hitting the target. (a) Find the mean and standard deviation of X. (b) Find (a) P(X = 5); (b) P(4 ≤ X ≤ 8). (c) P(X = 5 | 4 ≤ X ≤ 8)

[Ans: 4.2, 2.05; .179, .780]

Bill also shoots arrows at a target, with probability of 0.3 of hitting the target. All shots are independent of each other. (a) Calculate the probability that Bill hits the target for the first time on his third shot.

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(b) Calculate the minimum number of shots required for the probability of at least one shot hitting the target to exceed 0.99.

[Ans: 0.147; thirteen]

Preview of the Normal Distribution Example 10.3.12 Let X â&#x2C6;ź Bin(10, 0.5). (a) Find the mean and standard deviation. (b) Find the probability that X is (a) within one standard deviation of the mean; (b) within two standard deviations of the mean; (c) more than two standard deviations above the mean. (c) Repeat for n = 20, 40, 80, 160, . . . and/ or for p = 0.4, 0.6, 0.25, 0.75

Problems 10.C-1 (SL 89) A standard unbiased die is thrown 18 times. Find the probability that there are precisely 3 sixes, giving your answer correct to three significant figures.

[Ans: 0.245] 10.C-2 (SL 90) When Alexis plays Boris at tennis, Alexis wins with a probability of 34 . If they play each other seven times, find the probability that Alexis wins five times and Boris twice.

[Ans: 0.311] 10.C-3 (SL 90) A group of 294 people is selected at random. Assuming days of the week are equally likely as birthdays, find (a) the expected number of people in the group born on a Friday, and [Ans: 42] (b) the standard deviation of the number of people in the group born on a Friday.

[Ans: 6] 10.C-4 (SL 82) During a national election 3 votes in every 5 voted for the winner, President X. The voters cast their votes independently. Find the probability that in a random sample of 5 voters, exactly 3 voted for President X. [Ans: 0.346] 10.C-5 (SL 93) An unbiased ten-faced die has the numbers 1, 2, . . . , 10 on the faces. The die is thrown ten times. (a) What is the probability of obtaining a 6 on the fourth throw?

[Ans: 0.1] Mr. Budd, compiled January 12, 2011

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SL Unit 10 (Probability Distributions) (b) What is the probability of obtaining a 6 exactly four times in the ten throws?

[Ans: 0.0112]

10.C-6 (HL 5/03) When a boy plays a game at a fair, the probability that he wins a prize is 0.25. He plays the game 10 times. Let X denote the total number of prizes that he wins. Assuming that the games are independent, find (a) E (X); (b) P(X â&#x2030;¤ 2).

[Ans: 2.5] [Ans: 0.526]

10.C-7 (HL 5/02) When John throws a stone at a target, the probability that he hits the target is 0.4. He throws a stone 6 times. (a) Find the probability that he hits the target exactly 4 times.

[Ans: 0.138] (b) Find the probability that he hits the target for the first time on his third throw.

[Ans: 0.144] 10.C-8 (HL 5/01) X is a binomial random variable, where the number of trials is 5 and the probability of success of each trial is p. Find the value of p if P(X = 4) = 0.12.

[Ans: 0.459] 10.C-9 (HL Spec â&#x20AC;&#x2122;99) In a school, 13 of the students travel to school by bus. Five students are chosen at random. Find the probability that exactly 3 of them travel to school by bus.

[Ans: 0.165] 10.C-10 (HL 5/97) If 30% of college students do not graduate, find the probability that out of 6 randomly selected college students exactly 4 of them will graduate.

[Ans: 0.324] 10.C-11 (HL 5/96) How many times must a pair of dice be thrown so that there is a better than even chance of obtaining a double, that is, the same number on both dice?

[Ans: 4] 10.C-12 (MM 99) A box contains 35 red discs and 5 black discs. A disc is selected at random and its colour noted. The disc is then replaced in the box. (a) In eight such selections, what is the probability that a black disc is selected i. exactly once? ii. at least once?

[Ans: 0.393] [Ans: 0.656]

(b) The process of selecting and replacing is carried out 400 times. What is the expected number of black discs that would be drawn? [Ans: 50] 10.C-13 (MM 97) Refer to problem 9 on page 351. When a certain biased sixfaced die is thrown the score is the random variable Y , and P(Y = y) = 2 k (y â&#x2C6;&#x2019; 3.5) + 0.1375.

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(a) Find k, and hence calculate P(Y = y) for y = 1, . . . , 6. [Ans: .2,.16,.14,.14,.16,.2] (b) Let Z be the random variable which represents the number of sixes obtained when the biased die is thrown 5 times. Calculate i. P(Z = 0); ii. P(Z = 1); iii. P(Z > 1).

[Ans: 0.402] [Ans: 0.402] [Ans: 0.196]

10.C-14 (MM 11/96) Refer to the fair game in problem 10.1.3 on page 348. A fair game is a game in which the expected profit of any player is zero.

A player pays $3 to throw a pair of unbiased dice, and then receives a payment, $X, depending on the outcome of the throw, as follows: • X = 0, if the the total score on the dice is less than 7; • X = 3, if the total score is 7; • X = k, if the total score is 11 or 12; • X = 4, otherwise.

The player’s profit is $Y , where Y = X − 3. (a) Show that E (Y ) =

1 12

(k − 14).

(b) Given that this is a fair game, find the value of k. Hence determine the profit earned by the player if his throw results in a total score of 11.

[Ans: 14; $11] (c) When the player plays this game 8 times, W is the number of times his profit exceeds zero.

Find P(W > 5), giving your answer to 3 decimal places. [Ans: 0.061] 10.C-15 (MM 96) Refer to problem 10.1.2 on page 348. A computer is programmed to generate a sequence x1 , x2 , x3 , . . . of random single digits, each of which takes a value from 0 to 9 inclusive. Each digit is equally likely. Let X be the discrete random variable which represents the value of a digit.

If the computer generates a sequence of 8 random digits, find (a) the probability of obtaining no ‘4’ in the sequence;

[Ans: 0.430]

(b) the probability of obtaining exactly three ‘4’s in the sequence. [Ans: 0.0331] 10.C-16 (SL 93) In an advertising campaign a rail company presents to each child traveling by train an unmarked package. The package contains either a model engine or a model passenger carriage. The company claims that one third of the packages contain engines.

Six packages are chosen at random (a) What is the probability that exactly two of them contain engines?

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SL Unit 10 (Probability Distributions) (b) What is the probability that more than two of them contain engines?

[Ans: 0.320]

10.C-17 (SL 89) If in tossing a supposedly unbiased coin ten times you obtained 8 heads and 2 tails, would you suspect the coin to be biased in favour of heads?

Give reasons for your answer.

[13 marks]

10.C-18 (SL 82) Assume that equal numbers of birthdays occur in each of the twelve months of the year.

[10 marks] (a) What is the probability that a person chosen at random has a birth 1 day in June?

Ans: 12 (b) Three people are chosen at random. Explaining your methods find the probabilities that: h i 1 3 i. all three have birthdays in June Ans: 12 h i 11 3 ii. none of them has a birthday in June Ans: 12 iii. exactly two of them have a birthday in June [Ans: 0.0191]

55 iv. no two of them will have a birthday in the same month. Ans: 72 10.C-19 (SL 82) A manufacturer makes an instrument which goes through a testing procedure. He rejects any instrument which shows a fault. The instruments are tested in batches of three. With 1000 such batches the following results were obtained: No. rejected in batch No. of batches

0 393

1 428

2 155

3 24

(a) Calculate the mean number rejected per batch of three and hence deduce the probability p that an instrument is rejected. [Ans: 0.81, 0.27] (b) Using the value of p from 19a as the basis of a binomial model, the expected numbers of batches with 0 to 3 rejects are given, to the nearest whole numbers, in the table: No. rejected in batch No. of batches

0 389

1 432

2 160

3 20

Explain how the entry â&#x20AC;&#x2DC;432â&#x20AC;&#x2122; was obtained.

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367

Normal Distribution

International Baccalaureate 6.11 Normal distribution.

Properties of the normal distribution. Appreciation that the standardized value (z) gives the number of standard deviations from the mean.

Standardization of normal variables. Use of calculator (or tables) to find normal probabilities; the reverse process.

Standard Normal Distribution 2

Recall that the basic shape of the bell curve is given by e−x /2 . In order to have R∞ a proper probability density function f (x) such that −∞ f (x) dx = 1, we need 2

to scale e−x /2 by a certain amount. The probability density function for the bell curve is given by 2 e−x /2 f (x) = √ 2π This is called the standardized normal probability density function. What is very nice about it is that it has a mean of 0 and a variance of 1.

Example 10.4.1 Z is the standardised normal random variable with mean 0 and variance 1. Write integral expressions for (a) P(Z < a); (b) P(a < Z < b);

Before graphing calculators, there was no good way to quickly find these in−x2 /2 tegrals, has no antiderivative. So, tables were made for F (x) = Z xsince e 1 −x2 /2 √ e dx.

2π −∞ Example 10.4.2 Use the statistical tables to find (a) P(Z < 2); (b) P(−2 < Z < 2); (c) P(−1 < Z < 2); Repeat using fnInt. Repeat again using normalcdf.

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Standardizing the Nonstandard: z-Scores Very rarely are we dealing with situations where the mean is zero and the standard deviation is one. For a general normal distribution, "

2 # 1 x−µ 1 exp − f (x) = √ 2 σ σ 2π where µ is the mean and σ is the standard deviation.

Remember that without a calculator, this is very difficult. And it’s not very feasible to make tables for all the different possibilities of µ and σ. Even on the calculator, this would be a very grim function to have to enter every single time.

Instead, we use a peculiarity of normally distributed functions. All normal distributions have the same probability that X is within one standard deviation of the mean. In other words, P(µ − σ < X < µ + σ) = P(−σ < X − µ < σ) = P(|X − µ| < σ) is the same for all normally distributed functions. Likewise, the probability is the same that X is within two or three or 2.5 standard deviations from the mean for all normal distributions. Also, P(X − µ < 2σ) (X is below two standard deviations above the mean) is the same for any normal distribution, regardless of µ, and regardless of σ.

We use this property of normal distributions to standardize our data, and then we only need to use a single, standard distribution function.

Since P(−σ

< X − µ < σ) X −µ is the same for all normal distributions, so is P −1 < < 1 , just as σ

X −µ P < 2 is constant. σ If X is normally distributed with mean µ and standard deviation σ, then we X −µ can make a new variable Z = which follows the same standard normal σ distribution, regardless of what µ and σ are.

Example 10.4.3 For a normal random variable X with mean of 4 and variance of 9, find the probability that: (a) 1 < X < 7; (b) X < 10; (c) 1.6 < X < 5.

[Ans: 0.683] [Ans: 0.977] [Ans: 0.419]

Example 10.4.4 (MM 5/96) If X is a normal random variable with mean 13.5 and variance 6.25, find Mr. Budd, compiled January 12, 2011

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(a) P(X > 14.95); (b) P(|X â&#x2C6;&#x2019; 13.5| < 1.2).

[Ans: 0.281, 0.369] Example 10.4.5 (SL 90) In a manufacturing process, circular metal cylinders are being produced as components of a certain product.

For a cylinder to by usable, its length must be between 8.44 cm and 8.64 cm and its diameter between 1.52 cm and 1.62 cm. Cylinders are produced which have lengths which are normally distributed about a mean of 8.54 cm with a standard deviation of 0.05 cm while, independently, the diameters are normally distributed about a mean of 1.57 cm with a standard deviation of 0.02 cm. (a) Show that the percentage of cylinders produced whose lengths fall outside the specified limits is 4.5%. (b) Find the percentage of cylinders produced whose diameters fall outside the specified limits.

[Ans: 1.24%] (c) Show that the percentage of cylinders that cannot be used is 5.74%. (d) Find the probability that in a sample of five cylinders taken at random, four are usable and one is not.

[Ans: 0.226] (e) If a cylinder is not usable, what is the probability it has a length in excess of 8.64 cm?

[Ans: 0.397]

Inverse Normal Example 10.4.6 (HL 5/01) Z is the standardised normal random variable with mean 0 and variance 1. Find the value of a such that P(|Z| â&#x2030;¤ a) = 0.75 by (a) using fnInt; (b) using normalcdf; (c) using the statistical tables.

[Ans: 1.15] Example 10.4.7 (MM 5/95) Figure 10.1 shows the probability density function of a normal random variable X. Each of the shaded regions has area 0.08.

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Figure 10.1: MM 5/95

(a) Find the mean of X. (b) Find the standard deviation of X. Example 10.4.8 (MM 11/96) X is a normal random variable with variance 4. Given that P(x > 13.6) = 0.877, find the mean of X. [Ans: 15.9] Example 10.4.9 (MM 5/94) A random variable X is normally distributed with mean 13.5 and variance 1.44. If P(12.3 â&#x2030;¤ X < k) = 0.6055, find the value of k, giving your answer to three significant figures.

[Ans: 14.4] Example 10.4.10 (HL 5/06) The weights in grams of bread loaves sold at a supermarket are normally distributed with mean 200 g .

The weights of 88% of the loaves are less than 220 g . Find the standard deviation.

[Ans: 17.0 g]

Problems 10.D-1 (SL 91) X is a random variable with a normal distribution. If the mean is 1 and the standard deviation is 0.5, find P(0.225 < X < 0.575) [Ans: 0.137] Mr. Budd, compiled January 12, 2011

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10.D-2 (MM Spec ’93) A random variable X is distributed normally with mean 20.2 and standard deviation 4.1.

Find P(16.1 ≤ X ≤ 25.9).

[Ans: 0.759]

10.D-3 (MM 5/97) The normal random variable X has mean 265 and standard deviation 8. Find P(X > 255).

[Ans: 0.894] 10.D-4 (MM 5/00) In a certain country, annual salaries for nurses are normally distributed with a mean of $40 000, and 95% of the nurses can earn between $33 000 and $47 000 per year. Show that the standard deviation of nurses’ salaries is $3570, correct to three significant figures.

10.D-5 (SL 87) X is a normally distributed random variable with mean 5 and standard deviation 3.2. Find the probability, giving your answers to 3 significant figures, that (a) a randomly chosen value of X is negative, and (b) out of two randomly chosen values of X, one is negative and the other is positive.

[Ans: 0.0591;0.111] 10.D-6 (HL 11/06) A certain type of vegetable has a weight which follows a normal distribution with mean 450 grams and a standard deviation 50 grams. (a) In a load of 2000 of these vegetables, calculate the expected number with a weight greater than 525 grams. (b) Find the upper quartile of the distribution.

[Ans: 134; 484 g] 10.D-7 (HL 5/97) A zoologist knows that the lengths of a certain type of tropical snake are normally distributed with mean length L meters and standard deviation 0.12 meters. If 20% of the snakes are longer than 0.70 meters, find the value of L. [Ans: 0.599] 10.D-8 (HL 5/98) The mean test scores for a mathematics class was 60 with a standard deviation of 10. Assuming that the test scores are normally distributed, find the proportion of students scoring more than 80 in a given test.

[Ans: 0.0228] 10.D-9 (HL 5/99) A factory has a machine designed to produce 1 kg bags of sugar. It is found that the average weight of sugar in the bags is 1.02 kg. Assuming that the weights of the bags are normally distributed, find the standard deviation if 1.7% of the bags weigh below 1 kg. Give your answer correct to the nearest 0.1 gram.

[Ans: 9.4 g] Mr. Budd, compiled January 12, 2011

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10.D-10 (HL Spec ’00) The diameters of discs produced by a machine are normally distributed with a mean of 10 cm and a standard deviation of 0.1 cm. Find the probability of the machine producing a disc with a diameter smaller than 9.8 cm.

[Ans: 0.228] 10.D-11 (HL 5/02) The weights of a certain species of bird are normally distributed with mean 0.8 kg and standard deviation 0.12 kg. Find the probability that the weight of a randomly chosen bird of the species lies between 0.74 kg and 0.95 kg.

[Ans: 0.586] 10.D-12 (HL 5/03) The random variable X is normally distributed and P(X ≤ 10) = 0.670 P(X ≤ 12) = 0.937.

Find E (X).

[Ans: 9.19]

10.D-13 (HL 5/04) The weights of adult males of a type of dog may be assumed to be normally distributed with mean 25 kg and standard deviation 3 kg.

Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.

[Ans: 27.5] 10.D-14 (MM 5/00) The lifespan of a particular species of insect is normally distributed with a mean of 57 hours and a standard deviation of 4.4 hours. (a) The probability that the lifespan of an insect of this species lies between 55 and 60 hours is represented by the shaded area in Figure 10.2. This diagram represents the standard normal curve Figure 10.2: Problem 14

i. Write down the values of a and b.

[Ans: −0.455, 0.682] ii. Find the probability that the lifespan of an insect of this species is A. more than 55 hours; [Ans: 0.675] B. between 55 and 60 hours.

[Ans: 0.428] (b) 90% of the insects die after t hours.

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i. Represent this information on a standard normal curve diagram, similar to the one given in part 14a, indicating clearly the area representing 90%. ii. Find the value of t.

[Ans: 62.6 hours] 10.D-15 (MM 5/96) Packets of sugar labeled ‘1 kg’ are filled by a machine which delivers M kg to a packet, where M is a random variable. It is known that M has a normal distribution with a standard deviation of 50 g. A filled packet of sugar is described as ‘underweight’ when it contains less than 1 kg of sugar. (a) In a large batch produced by the machine, the mean mass delivered to each packet is 1.025 kg. Show that almost one-third of the packets are underweight.

[Ans: 0.3085] (b) The sugar supply to each packet is increased in order to reduce the proportion of underweight packets to 10%. What is now the mean mass delivered to each packet?

[Ans: 1.06 kg] (c) Explain briefly why the manufacturer would not attempt to eliminate underweight packages from a batch by adjusting the mean. (d) The regulation is changed to allow no more than 2.5% of the packets to be underweight. If the same mean mass as in part 15b is still delivered to each packet, what standard deviation is required in order to achieve this?

[Ans: 32.7 g] 10.D-16 (HL 5/00) A machine is set to produce bags of salt, whose weights are distributed normally, with a mean of 110 g and standard deviation of 1.142 g. If the weight of a bag of salt is less than 108 g, the bag is rejected.

With these settings, 4% of the bags are rejected.

The settings of the machine are altered and it is found that 7% of the bags are rejected. (a)

i. If the mean has not changed, find the new standard deviation, correct to three decimal places.

[Ans: 1.355 g] The machine is adjusted to operate within this new value of the standard deviation. ii. Find the value, correct to two decimal places, at which the mean should be set so that only 4% of the bags are rejected.

[Ans: 110.37]

(b) With the new settings from part 16a, it is found that 80% of the bags of salt have a weight which lies between A g and B g, where A and B are symmetric about the mean. Find the values of A and B, giving your answers correct to two decimal places.

[Ans: 108.63, 112.11]

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Mr. Budd, compiled January 12, 2011

Unit 11

Review 1. Solving Inequalites 2. First Derivative Test 3. Points of Inflection 4. Second Derivative Test 5. Absolute Extrema 6. Curve Sketching 7. Optimization

375

376

SL Unit 11 (Review)

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SL Unit 11, Day 1: Limits Involving Infinity

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377

Limits Involving Infinity

International Baccalaureate 7.1 Informal ideas of limit and convergence.

7.7 Graphical behaviour of functions; behaviour for large |x|; asymptotes.

Textbook §1.5 Limits Involving Infinity; Asymptotes [15] Resources §2-5 Limits Involving Infinity in Foerster [10].

11.1.1

Vertical Asymptotes: Infinite Limits

Nonexistent, Infinite Limits Example 11.1.1 How many vertical asymptotes does the function 2x2 − x − 6 f (x) = 2 have? What are they? Check your answers on x −x−2 your grapher.

Remember This? How do you distinguish between vertical asymptotes and removable discontinuities?

Example 11.1.2 Using a table or graph, examine

1

lim x→2 x − 2

1

Technically, as x goes to 2, x−2

does not go to one particular number (infinity

1

is not a number). Therefore, we say that lim x−2

does not exist, or that it is x→2

nonexistent. however, we like to treat infinity like a number, so we

Sometimes,

1

write lim x−2

= ∞, and we describe the limit as being infinite. It is true that x→2 the limit is both nonexistent and infinite.

Positive and Negative Zero On a case by case basis, you need to decide the sign (positive or negative) of any zeros in the denominator which yield infinite limits.

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SL Unit 11 (Review) Example 11.1.3 Find

lim tan θ and limπ + tan θ. Confirm your

− θ→ π 2

θ→ 2

answer with a graph and with a table.

Derivatives at a Step Discontinuity Example 11.1.4 (adapted from [18]) Let ( g(x) =

2x − 1, x2 − x

x≤2 x>2

Is g(x) continuous? Where do you expect to have problems? Determine: (a) g(2) (b) lim g(x) x→2−

(c) lim+ g(x) x→2

(d) lim g(x) x→2

(e) lim

x→2−

g(x) − g(2) x−2

(f) lim+

g(x) − g(2) x−2

x→2

g(x) − g(2) x−2 g(x) − g(3) (h) lim x→3 x−3 (g) lim

x→2

11.1.2

Limits at Infinity

If lim f (x) = 0 then lim x→c

x→c

1 is infinite. In other words, 0 in the denominator f (x)

wants an infinite limit.

1 = 0, i.e., ∞ in the denominator x→c f (x)

Similarly, if lim f (x) is infinite, then lim x→c

wants the limit to be zero.

Note also that 0 in the numerator wants a zero limit, and ∞ in the numerator wants an infinite limit.

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 1: Limits Involving Infinity 0 = indeterminate 0 0 =0 nonzero 0 =0 ±∞

nonzero = ±∞ 0 nonzero = answer nonzero nonzero =0 ±∞

379 ±∞ = ±∞ 0 ±∞ = ±∞ nonzero ±∞ = inderterminate ±∞

As x → ∞ or x → −∞, f (x) behaves like its highest order term.

Example 11.1.5 Find the horizontal asymptotes of y =

2x2 − x − 6 x2 − x − 2

2x2 − x − 6 x→−∞ x2 − x − 2 2x6 (b) lim x→−∞ x2 − x − 2 2x2 − x − 6x3 (c) lim x→∞ x2 − x − 2 (a)

lim

Confirm your answers on your handy, dandy calculator with a graph, and with a table.

Example 11.1.6 Find 12 + 7x − 5x3 x→∞ 3x2 − 12x + 9 12 + 7x − 5x2 (b) lim x→−∞ 3x2 − 12x + 9 12 + 7x − 5x2 (c) lim x→∞ 3x2 − 12x3 + 9 (a) lim

Do it the textbook way, and then using highest order terms. What happens if instead of taking the limit to ∞, you take x → −∞?

In using the highest order term, I’m saying that 3x2 − 12x + 9 behaves as 3x2 for larger and larger values of x. Essentially, I’m ignoring the −12x. But if x → ∞, −12x should be extremely massive, not negligible. The difference is in the relative values. Think of how many molecules are in a drop of water.

Conceptually, we will say an infinite amount. Add a drop of water to the Atlantic Ocean. Has the amount of water in the ocean really changed significantly? Even though you’ve added an infinite number of water molecules? So, now imagine that you have three oceans, take out 12 drops, and add nine molecules of water.

You still have three oceans. Which is why we say that 3x2 − 12x + 9 behaves as 3x2 as x → ∞.

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Note that if the degree is higher on top, the limit at infinity is nonexistent (infinite). If the degree on top is lower, then the limit at infinity is zero. If the degree is the same, then the limit at infinity is the ratio of the coefficients.

In determining degrees, ex >> · · · >> xn >> · · · >> x3 >> x2 >> x >> ln x >> sin x.

Example 11.1.7 Find 3x2 − 13x + 7 sin x x→∞ 5x2 + 4x − 9 2 3x − 13ex + 7 sin x (b) lim x→∞ 5x2 + 4 ln x − 9ex 3x2 − 13x183 + 7 sin x (c) lim x→∞ 5x2 + 4 ln x − 9ex (a) lim

You could also see that the sin x becomes irrelevant by using a variation on the squeeze theorem.

1 h2 Example 11.1.8 [18] Find lim 1 h→0 1− h 1−

[Ans: does not exist]

11.1.3

Horizontal Asymptotes

There is a horizontal asymptote at y = L if (and only if) at least one of the following conditions is met: • lim f (x) = L x→∞

•

lim f (x) = L

x→−∞

Note that there are two possibilities for the horizontal asymptote. There might be two horizontal asymptotes, or one, or none. In typical examples of rational expressions, where we just have polynomials over polynomials, the limits at positive and negative infinity are the same. So, to make sure that you know to Mr. Budd, compiled January 12, 2011

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take the limits at both positive and negative infinity, test-makers devise problems where the limits are different, yielding two horizontal asymptotes instead of one. Typically, this is done with radicals, or with absolute values.

Remember This? When is

√

9x2 6= 3x? When is

√

9x4 6= 3x2 ?

Example 11.1.9 Find all horizontal asymptotes of √

−7x + 12 9x2 + 3x + 12

Ans: y = 37 , y = − 73

11.1.4

Problems [Ans: −∞]

11.A-1 lim− cot x is x→π

2

11.A-2 f (x) =

(x − 1) 3x2 − 5x + 2

(a) Is f continuous at x = 1? (b) Give the equation(s) of all vertical asymptotes.

Ans: no; x = 23

(x + 2) x2 − 1 .

11.A-3 Let W be the function given by W (x) = x2 − c (a) Will f be continuous if c = 1? Explain. (b) Will f be continuous if c = 4? Explain. (c) For what positive values of c is f continuous for all real numbers x?

[Ans: no ; no; none] 11.A-4 Find lim− t→1

t , being as specific as possible. ln t

[Ans: −∞]

11.A-5 Do you know your asymptote from a hole in the graph?

11.A-6 Find all the horizontal asymptotes of h(x) = √ 11.A-7 (a) What is lim

x→−∞

x−1 . Ans: y = 21 , y = − 12 2 −3 − x + 4x

x−1 ? Show your process. −3 − x + 4x2

1 (b) Make a single change to the problem so that the answer would be 4

Ans: y = 0; x2 − 1 in numerator Mr. Budd, compiled January 12, 2011

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11.A-8 Let f (x) =

x−1 . −3 − x + 4x2

(a) Create a new function g(x) by making a single change to f (x) so that 1 lim g(x) = − . x→−∞ 2 (b) Show the analysis necessary to evaluate the limit at negative infinity of your new function. A table is not sufficient analysis. (c) Use your calculator to find

lim g(x). Does this confirm your pre-

x→−∞

vious answer? (d) Use your calculator to write a table to find lim g(x). x→+∞

1 h 11.A-9 [18] Find lim 1 h→0 1+ h a t+ t 11.A-10 [18] Find lim b t→0 t+ t

Ans:

1 2

1−

[Ans: −1]

Ans:

a b

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 2: Introduction to Series

11.2

383

Introduction to Series

International Baccalaureate 1.1 Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.

Examples of applications, compound interest and population growth.

Sigma notation.

Problems 11.B-101 Ashley and Billie are swimmers training for a competition. (a) Ashley trains for 12 hours in the first week. She decides to increase the amount of time she spends training by 2 hours each week. Find the total number of hours she spends training during the first 15 weeks.

[3 marks] (b) Billie also trains for 12 hours in the first week. She decides to train for 10% longer each week than the previous week. (i) Show that in the third week she trains for 14.52 hours. (ii) Find the total number of hours she spends training during the first 15 weeks.

[4 marks] (c) In which week will the time Billie spends training first exceed 50 hours?

[4 marks] 11.B-102 The Acme insurance company sells two savings plans, Plan A and Plan B. For Plan A, an investor starts with an initial deposit of $1000 and increases this by $80 each month, so that in the second month, the deposit is $1080, the next month it is $1160 and so on.

For Plan B, the investor again starts with $1000 and each month deposits 6% more than the previous month. (a) Write down the amount of money invested under Plan B in the second and third months.

[2 marks] Give your answers to parts (b) and (c) correct to the nearest dollar. (b) Find the amount of the 12th deposit for each Plan.

[4 marks]

(c) Find the total amount of money invested during the first 12 months (i) under Plan A; (ii) under Plan B.

[2 marks] [2 marks]

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11.B-103 A company offers its employees a choice of two salary schemes A and B over a period of 10 years.

Scheme A offers a starting salary of $11 000 in the first year and then an annual increase of $400 per year. (a)

(i) Write down the salary paid in the second year and in the third year. (ii) Calculate the total (amount of) salary paid over ten years.

[3 marks]

Scheme B offers a starting salary of $10 000 dollars in the first year and then an annual increase of 7% of the previous years salary. (b)

(i) Write down the salary paid in the second year and in the third year. (ii) Calculate the salary paid in the tenth year.

[4 marks]

(c) Arturo works for n complete years under scheme A. Bill works for n complete years under scheme B. Find the minimum number of years so that the total earned by Bill exceeds the total earned by Arturo.

[4 marks] 11.B-104 There were 1420 doctors working in a city on 1 January 1994. After n years the number of doctors, D, working in the city is given by D = 1420+100n. (a)

(i) How many doctors were there working in the city at the start of 2004? (ii) In what year were there first more than 2000 doctors working in the city?

[3 marks]

At the beginning of 1994 the city had a population of 1.2 million. After n n years, the population, P , of the city is given by P = 1 200 000 (1.025) . (b)

(i) Find the population P at the beginning of 2004. (ii) Calculate the percentage growth in population between 1 January 1994 and 1 January 2004. (iii) In what year will the population first become greater than 2 million?

[7 marks]

(c)

(i) What was the average number of people per doctor at the beginning of 1994? (ii) After how many complete years will the number of people per doctor first fall below 600?

[5 marks]

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 3: Probability and Statistics

11.3

385

Probability and Statistics

Example 11.3.5 A company buys 44% of its stock of bolts from manufacturer A and the rest from manufacturer B. The diameters of the bolts produced by each manufacturer follow a normal distribution with a standard deviation of 0.16 mm.

The mean diameter of the bolts produced by manufacturer A is 1.56 mm. 24.2% of the bolts produced by manufacturer B have a diameter less than 1.52 mm.

(a) Find the mean diameter of the bolts produced by manufacturer B. [3 marks]

A bolt is chosen at random from the companyâ&#x20AC;&#x2122;s stock.

(b) Show that the probability that the diameter is less than 1.52 mm is 0.312, to three significant figures.

[4 marks] (c) The diameter of the bolt is found to be less than 1.52 mm. Find the probability that the bolt was produced by manufacturer B. [3 marks] (d) Manufacturer B makes 8000 bolts in one day. It makes a profit of $1.50 on each bolt sold, on condition that its diameter measures between 1.52 mm and 1.83 mm. Bolts whose diameters measure less than 1.52 mm must be discarded at a loss of $0.85 per bolt. Bolts whose diameters measure over 1.83 mm are sold at a reduced profit of $0.50 per bolt.

Find the expected profit for manufacturer B.

[6 marks]

Example 11.3.6

i. The following diagram represents the lengths, in cm, of 80 plants grown in a laboratory.

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(a) How many plants have lengths in cm between (i) 50 and 60? (ii) 70 and 90?

[2 marks] (b) Calculate estimates for the mean and the standard deviation of the lengths of the plants.

[4 marks] (c) Explain what feature of the diagram suggests that the median is different from the mean.

[1 mark] (d) The following is an extract from the cumulative frequency table. length in cm cumulative less than frequency . .

50 22 60 32 70 48 80 62 . .

Use the information in the table to estimate the median.

Give your answer to two significant figures.

[3 marks] ii. Two fair dice are thrown and the number showing on each is noted. The sum of these two numbers is S. Find the probability that (a) S is less than 8; [2 marks] (b) at least one die shows a 3; [2 marks] (c) at least one die shows a 3, given that S is less than 8.

[3 marks]

Example 11.3.7 Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 3: Probability and Statistics

387

i. X is a binomial random variable, where the number of trials is 5 and the probability of success of each trial is p. Find the values of p if P(X = 4) = 0.12.

[6 marks] ii. The mass of packets of a breakfast cereal is normally distributed with a mean of 750 g and standard deviation of 25 g. (a) Find the probability that a packet chosen at random has mass i. less than 740 g; ii. at least 780 g; iii. between 740 g and 780 g.

[5 marks] (b) Two packets are chosen at random. What is the probability that both packets have a mass which is less than 740 g?

[2 marks] (c) The mass of 70% of the packets is more than x grams. Find the value of x.

[2 marks] 10 P iii. Consider the 10 data items x1 , x2 , ...x10 . Given that x2i = i=1

1341 and the standard deviation is 6.9, find the value of x.[6 marks] Example 11.3.8 i. A factory makes calculators. Over a long period, 2% of them are found to be faulty. A random sample of 100 calculators is tested. (a) Write down the expected number of faulty calculators in the sample. (b) Find the probability that three calculators are faulty. (c) Find the probability that more than one calculator is faulty.

[6 marks] 2 1 2 ii. Given that P(X) = , P(Y | X) = , P(Y | X 0 ) = , find 3 5 4 (a) P(Y 0 ); (b) P(X 0 â&#x2C6;Ş Y 0 ).

[6 marks] iii. The table below shows the probability distribution of a discrete random variable X. x 0 1 2 3 P(X = x) 0.2 a b 0.25 (a) Given that E (X) = 1.55, find the value of a and of b. (b) Calculate Var (X) [6 marks]

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Homework 11.C-105 [No calculator] Two standard six-sided dice are tossed. Let X be the sum of the scores on the two dice. (a) Find i. P(X = 6); ii. P(X > 6); iii. P(X = 7 | X > 5).

[6 marks]

(b) Elena plays a game where she tosses two dice.

If the sum is 6, she wins 3 points.

If the sum is greater than 6, she wins 1 point.

If the sum is less than 6, she loses k points.

Find the value of k for which Elena’s expected number of points is zero.

[7 marks] 11.C-106

i. A factory makes switches. The probability that a switch is defective is 0.04. The factory tests a random sample of 100 switches. (a) Find the mean number of defective switches in the sample. (b) Find the probability that there are exactly six defective switches in the sample. (c) Find the probability that there is at least one defective switch in the sample.

[7 marks] ii. In a school with 125 girls, each student is tested to see how many sit–up exercises (sit–ups) she can do in one minute. The results are given in the table below.

(a)

Number of sit–ups

Number of students

15 16 17 18 19 20

11 21 33 q 18 8

(i) (ii) (b) Find (c) Find

Write down the value of p.

Find the value of q. the median number of sit–ups. the mean number of sit–ups.

Cumulative number of students 11 32 p 99 117 125 [3 marks] [2 marks] [2 marks]

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 3: Probability and Statistics 11.C-107

389

i. The probability of obtaining heads on a biased coin is 0.18. The coin is tossed seven times. (a) Find the probability of obtaining exactly two heads. [2 marks] (b) Find the probability of obtaining at least two heads. [3 marks] ii. [No calculators] There are 20 students in a classroom. Each student plays only one sport. The table below gives their sport and gender.

Female Male

Football 5 4

Tennis 3 2

Hockey 3 3

(a) One student is selected at random. (i) Calculate the probability that the student is a male or is a tennis player. (ii) Given that the student selected is female, calculate the probability that the student does not play football. [4 marks] (b) Two students are selected at random. Calcualate the probability that neither student plays football.

[3 marks] 11.C-108 A box contains a large number of biscuits. The weights of the biscuits are normally distributed with mean 7 g and standard deviation 0.5 g. (a) One biscuit is chosen at random from the box. Find the probability that this biscuit (i) weighs less than 8 g; (ii) weighs between 6 g and 8 g.

[4 marks]

(b) Five percent of the biscuits in the box weigh less than d grams. (i) Copy and complete the following normal distribution diagram, to represent this information, by indicating d, and shading the appropriate region.

(ii) Find the value of d.

[5 marks]

(c) The weights of biscuits in another box are normally distributed with mean Âľ and standard deviation 0.5 g. It is known that 20% of the biscuits in this second box weigh less than 5 g.

Find the value of Âľ.

[4 marks]

Mr. Budd, compiled January 12, 2011

390

SL Unit 11 (Review)

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 4: Graphing and Calculus

11.4

391

Graphing and Calculus

Example 11.4.9 Let f (x) = ex 1 − x2 .

(a) Show that f 0 (x) = ex 1 − 2x − x2 .

[3 marks]

Part of the graph of y = f (x), for −6 ≤ x ≤ 2, is shown below. The x–coordinates of the local minimum and maximum points are r and s respectively.

(a) Write down the equation of the horizontal asymptote.[1 mark] (b) Write down the value of r and of s.

[4 marks]

(c) Let L be the normal to the curve of f at P(0, 1). Show that L has equation x + y = 1.

[4 marks] (d) Let R be the region enclosed by the curve y = f (x) and the line L. (i) Find an expression for the area of R. (ii) Calculate the area of R.

[5 marks]

Example 11.4.10 An aircraft lands on a runway. Its velocity v m s−1 at time t seconds after landing is given by the equation v = 50 + 50e−0.5t , where 0 ≤ t ≤ 4. (a) Find the velocity of the aircraft (i) when it lands; Mr. Budd, compiled January 12, 2011

392

SL Unit 11 (Review) (ii) when t = 4.

[4 marks]

(b) Write down an integral which represents the distance travelled in the first four seconds.

[3 marks] (c) Calculate the distance travelled in the first four seconds.[2 marks] After four seconds, the aircraft slows down (decelerates) at a constant rate and comes to rest when t = 11. (d) Sketch a graph of velocity against time for 0 ≤ t ≤ 11. Clearly label the axes and mark on the graph the point where t = 4.

[5 marks] (e) Find the constant rate at which the aircraft is slowing down (decelerating) between t = 4 and t = 11.

[2 marks] (f) Calculate the distance travelled by the aircraft between t = 4 and t = 11.

[2 marks] Example 11.4.11 1 . The diagram below is x−1 a sketch of part of the graph of y = f (x).

(a) Consider the function f (x) = 2 +

Copy and complete the sketch of f (x). (b) (i) Write down the x-intercepts and y-intercepts of f (x). (ii) Write down the equations of the asymptotes of f (x). (c) (i) Find f 0 (x). (ii) There are no maximum or minimum points on the graph of f (x). Use your expression for f 0 (x) to explain why.

The region enclosed by the graph of f (x), the x-axis, and the lines x = 2 and x = 4, is labelled A, as shown below.

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 4: Graphing and Calculus

393

Z (d) (i) Find

f (x) dx.

(ii) Write down an expression that represents the area labelled A. (iii) Find the area of A. (iv) Write down an expression that represents the volume generated when A is rotated about the line y = 4.

Example 11.4.12 Consider the function f (x) = 1 + e−2x . (a) (i) Find f 0 (x). (ii) Explain briefly how this shows that f (x) is a decreasing function for all values of x (i.e., that f (x) always decreases in value as x increases).

[2 marks] 1 Let P be the point on the graph of f where x = − .

2 (b) Find an expression in terms of e for (i) the y–coordinate of P; (ii) the gradient of the tangent to the curve at P.

[2 marks]

(c) Find the equation of the tangent to the curve at P, giving your answer in the form y = ax + b.

[3 marks] (d) (i) Sketch the curve of f for −1 ≤ x ≤ 2.

1 (ii) Draw the tangent at x = − .

2 (iii) Shade the area enclosed by the curve, the tangent and the y–axis. (iv) Find this area.

[7 marks]

Mr. Budd, compiled January 12, 2011

394

SL Unit 11 (Review)

Homework 11.D-109 The following diagram shows the graphs of f (x) = ln (3x − 2) + 1 and g(x) = −4 cos (0.5x) + 2, for 1 ≤ x ≤ 10.

(a) Let A be the area of the region enclosed by the curves of f and g. (i) Find an expression for A. (ii) Calculate the value of A. (b)

[6 marks]

(i) Find f 0 (x). (ii) Find g 0 (x).

[4 marks]

(c) There are two values of x for which the gradient of f is equal to the gradient of g. Find both these values of x.

[4 marks]

1 11.D-110 [No calculator] Consider f (x) = x3 + 2x2 − 5x. Part of the graph of f 3 is shown below. There is a maximum point at M, and a point of inflexion at N. Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 4: Graphing and Calculus

395

(a) Find f 0 (x).

[3 marks]

(b) Find the x–coordinate of M.

[4 marks]

(c) Find the x–coordinate of N.

[3 marks]

(d) The line L is the tangent to the curve of f at (3, 12). Find the equation of L in the form y = ax + b.

[4 marks]

11.D-111

i. [No calculator] A particle moves along a straight line so that its velocity, v m s−1 at time t seconds is given by v = 6e3t + 4. When t = 0, the displacement, s, of the particle is 7 metres. Find an expression for s in terms of t.

[7 marks] ii. [Still no calculator] Z

1 dx.

[2 marks] 2x + 3 Z 3 √ 1 (b) Given that dx = ln P , find the value of P . [4 marks] 0 2x + 3 (a) Find

11.D-112 The equation of a curve may be written in the form y = a (x − p) (x − q).

The curve intersects the x-axis at A (−2, 0) and B (4, 0). The curve of y = f (x) is shown in the diagram below.

Mr. Budd, compiled January 12, 2011

396

SL Unit 11 (Review)

(a)

i. Write down the value of p and of q. ii. Given that the point (6, 8) is on the curve, find the value of a. iii. Write the equation of the curve in the form y = ax2 + bx + c. dy (b) i. Find . dx ii. A tangent line is drawn to the curve at a point P . The gradient of this tangent is 7. Find the coordinates of P . (c) The line L passes through B (4, 0) and is normal to the curve at point B. i. Find the equation of L. ii. Find the x-coordinate of the point where L intersects the curve again.

Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 5: Retired Problems

11.5

397

Retired Problems

Example 11.5.1 Consider functions of the form y = e−kx .

Z 1

1 (a) Show that e−kx dx = 1 − e−kx .

[3 marks] k 0 (b) Let k = 0.5 (a) Sketch the graph of y = e−0.5x , for −1 ≤ x ≤ 3, indicating the coordinates of the y-intercept. (b) Shade the region enclosed by this graph, the x-axis, y-axis, and the line x = 1. (c) Find the area of this region.

[5 marks] dy in terms of k, where y = e−kx . (c) (a) Find dx The point P (1, 0.8) lies on the graph of the function y = e−kx . (b) Find the value of k in this case. (c) Find the gradient of the tangent to the curve at P .[5 marks] Example 11.5.2 (S00–2) Note: radians are used throughout the question. (a) Draw the graph of y = π + x cos x, 0 ≤ x ≤ 5, on millimetre square graph paper, using a scale of 2 cm per unit. Make clear (a) the integer values of x and y on each axis; (b) the approximate positions of the x-intercepts and turning points. (b) Without the use of a calculator, show that π is a solution of the equation π + x cos x = 0. (c) Find another solution of the equation π + x cos x = 0 for 0 ≤ x ≤ 5, giving your answer to six significant figures. (d) Let R be the region enclosed by the graph and the axes for 0 ≤ x ≤ π. Shade R on your diagram, and write down an integral which represents the area of R. (e) Evaluate the integral in part (d) to an accuracy of six significant figures. (If you consider it necessary, you can make use of d (x sin x + cos x) = x cos x.) the result dx (f) Write down an integral which represents the volume generated when R is rotated about the line y = −1.

Example 11.5.3 (00–4) Mr. Budd, compiled January 12, 2011

398

SL Unit 11 (Review) (a) Sketch the graph of y = π sin x − x, −3 ≤ x ≤ 3, on millimetre square paper, using a scale of 2 cm per unit on each axis.

Label and number both axes and indicate clearly the approximate positions of the x-intercepts and the local maximum and minimum points. (b) Find the solution of the equation π sin x − x = 0,

x > 0.

(c) Find the indefinite integral Z (π sin x − x) dx and hence, or otherwise, calculate the area of the region enclosed by the graph, the x-axis, and the line x = 1.

Example 11.5.4 The diagram shows part of the graph of the curve 2 y = a (x − h) + k, where a, h, k ∈ Z.

(a) The vertex is at the point (3, 1). Write down the values of h and of k. (b) The point P (5, 9) is on the graph. Show that a = 2. (c) Hence show that the equation of the curve can be written as y = 2x2 − 12x + 19 Mr. Budd, compiled January 12, 2011

SL Unit 11, Day 5: Retired Problems

399

dy . dx A tangent is drawn to the curve at P (5, 9). (b) Calculate the gradient of this tangent. (c) Find the equation of this tangent.

(d) (a) Find

Homework 11.E-1 Note: Radians are used throughout the question.

Let f (x) = sin (1 + sin x). (a)

(i) Sketch the graph of y = f (x), for 0 ≤ x ≤ 6. (ii) Write down the x-coordinates of all minimum and maximum points of f , for 0 ≤ x ≤ 6. Give your answers correct to four significant figures.

(b) Let S be the shaded region in the first quadrant completely enclosed by the graph of f and both coordinate axes. (i) Shade S on your diagram. (ii) Write down the integral which represents the area of S. (iii) Evaluate the area of S to four significant figures. (c) Give reasons why f (x) ≥ 0 for all values of x.

11.E-2 (a) (03–14) The diagram below shows a triangle and two arcs of circles.

The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the midpoint of [BC].

The arc BDC is part of a circle with centre A. The arc BEC is part of a circle with centre P .

Mr. Budd, compiled January 12, 2011

400

SL Unit 11 (Review) i. Calculate the area of the segment BDCP . ii. Calculate the area of the shaded region BECD. (b) Given that log5 x = y, express each of the following in terms of y. i. log5 x2

1 ii. log5 x iii. log25 x Z 3 g(x) dx = 10, deduce the value of (c) Given that 1

Z

3

i.

1

Z ii.

1 g(x) dx; 2

3

(g(x) + 4) dx.

1

Mr. Budd, compiled January 12, 2011

Bibliography [1] Stephen Bernstein and Ruth Bernstein. Schaum’s Outline of Theory and Problems of Elements of Statistics I: Descriptive Statistics and Probability.

McGraw–Hill, New York, 1999.

[2] George W. Best and J. Richard Lux. Preparing for the (AB) AP Calculus Examination. Venture Publishing, Andover, Massachusetts, 1998.

[3] George W. Best and J. Richard Lux. Preparing for the (BC) AP Calculus Examination. Venture Publishing, Andover, Massachussetts, 1998.

[4] Nigel Buckle, Iain Dunbar, and Fabio Cirrito. Mathematics Higher Level (Core). IBID Press, Camberwell, Australia, second edition, 1999.

[5] Fabio Cirrito, editor. Mathematical Methods. IBID Press, Melton, Australia, second edition, 1998.

[6] Douglas Downing and Jeffrey Clark. Forgotten Statistics: A Self-Teaching Refresher Course. Barron’s Educational Series, Hauppage, New York, 1996.

[7] Douglas Downing and Jeffrey Clark, editors. Statistics: The Easy Way. Barron’s Educational Series, Hauppage, New York, third edition, 1997.

[8] Ross L. Finney, Flanklin D. Demana, Bert K. Waits, and Daniel Kennedy.

Calculus: Graphical, Numerical, Algebraic. Scott Foreesman Addison Wesley, New York, 1999.

[9] Paul Foerster. Calculus: Concepts and Applications; Instructor’s Resource Book. Key Curriculum Press, Berkeley, California, 1998.

[10] Paul A. Foerster. Calculus: Concepts and Applications. Key Curriculum Press, Berkeley, California, 1998.

[11] Nancy Baxter Hastings. Workshop Calculus with Graphing Calculators: Guided Exploration with Review, volume 2. Springer-Verlag, New York, 1999.

401

402

BIBLIOGRAPHY

[12] Melvin Hausner. A Vector Space Approach to Geometry. Dover Publications, Mineola, New York, 1965.

[13] Shirley O. Hockett and David Bock. How to Prepare for the Advanced Placement Examination, Calculus: Review of Calculus AB and Calculus BC. Barronâ&#x20AC;&#x2122;s Educational Series, New York, 1998.

[14] Ann R. Kraus. Test Item File to Accompany Calculus. D. C. Heath and Company, Lexington, Massachusetts, 1994.

[15] Roland E. Larson, Robert P. Hostetler, and Bruce H. Edwards. Calculus with Analytic Geometry. D. C. Heath and Company, Lexington, Massachusetts, 1994.

[16] Arnold Ostebee and Paul Zorn. Calculus From Graphical, Numerical, and Symbolic Points of View. Saunders College Publishing, Fort Worth, 1997.

[17] Ronald I. Rothenberg. Probability and Statistics. Harcourt Brace Jovanovich, Orlando, Florida, 1991.

[18] Salas, Hille, and Garret J. Etgen. Calculus: One and Several Variables.

John Wiley and Sons, Inc., New York, New York, 1999.

[19] Wanda Savage. A. p. summer institute. Bound materials from the summer workshop, 2000.

[20] James Stewart. Calculus:Concepts and Contexts, Single Variable. Brooks/ Cole Publishing Company, Pacific Grove, California, 1998.

[21] Gilbert Strang. Linear Algebra and Its Applications. Harcourt Brace Jovanovich College Publishers, Fort Worth, 1988.

[22] Paul Urban, John Owen, David Martin, Robert Haese, Sandra Haese, and Mark Bruce. Mathematics for the International Student: Mathematics HL (Core). Haese and Harris Publications, Adelaide Airport, Australia, 2004.

[23] Dale Varberg, Edwin J. Purcell, and Steven E. Rigdon. Calculus. Prentice Hall, Upper Saddle River, New Jersey, 2000.

Mr. Budd, compiled January 12, 2011