Multiple Representations for a relationship Mathematical Models

Relationship: As hours worked increases, allowance increases

Equation

This means that one representation may be converted into the other representation.

Words

A = 10 + 2H (On calc: y = 10 + 2x )

Table

Graph Hours Worked 0

Allowance 10

5

20

10

30

15

40

40

90

‚My name’s Bobby. I get a \$10 allowance for the month, & I can make another \$10 for every 5 hours of extra work I do!‛

Answering Questions Example: How many hours should Bobby work in order to buy a \$37 video game?

Words

Equation

Table

Graph Words

Mathematical Models

Equation Variables: Allowance depends on Hours Worked

dependent variable

A = 10 + 2 ● H dep var

start

rate

ind var

independent variable

3.

Graph

6. After plotting at least two points, draw a line through the whole graph

My name’s Bobby. I get a \$10 allowance for the month, & I can make another \$10 for every 5 hours of extra work I do!

1. Identify the variables (what’s changing?) and their relationship 2. Rate of change =

Table

Starting with: a description of some real-world situation

Hours Worked 0

Allowance 10

5

20

10

30

15

40

40

90

5. Each row in the table is an (x,y) point on the graph: Ex: 10 hrs, \$30 is the point: (10, 30)

Dep. var. chg = \$10 = \$2/hr Ind. var. chg 5 hr

3. Starting Allowance = \$10

4. To make a table, pick #’s for H, plug in to find A: Ex: A = 10 + 2H A = 10 + 2*5 (5 hrs) A = 10 + 10 A = \$20

Graph

Table

Equation

Words Words

Mathematical Models

‚The rubber band’s length is 3 inches with no weight hanging on it, and gets 0.5 inches longer for each ounce added onto it.‛

Equation

Variables: Length depends on Weight dependent variable

Starting with: an experiment that measured the length of a rubberband with different weights hung on it.

Graph

L = 3 + 0.5 ● W dep var

start

rate

ind var

independent variable

Starting Length

Table

y

1. Make a scatterplot of the experimental data 2. Draw a ‚best fit‛ line through the points

∆x

Ind. Var (x)

Dep. Var (y)

Weight (oz.)

Length (in.)

0

3

6

6

8

7

6. Plug the two variables, starting length, and rate of change into an equation. 5. Rate of change (slope of the graph) = ∆y = Dep. var. chg = 7-6 = 1 in. = 0.5 in ∆x

Ind. var. chg

8-6

2 oz.

oz

4. Starting Length = 3 in. With no weight (0 oz.), the band is 3 inches long.

The ‚y-intercept‛ on the graph ∆y

X 3. Estimate at least 2 points on the ‚best fit‛ line, including at x=0 (a weight of 0 oz.), and place them into the table. *Estimate points at a crosshairs (+)

Answering Questions with a graph Q1: If Bobby worked 9 hours, what would his allowance be?

Q2: How many hours

y

d a) You’re given 9 hours (independent variable ‘x’) So go to 9 on the Hours axis

Equation: A = 10 + 2H

Graph

should Bobby work in order to buy a \$37 video game?

e

d) You’re given \$37 (dependent variable ‘y’) So go to 37 on the Allow. axis

c

b) Go up till you hit the line c) Go left till you hit the axis

4. Press ‘9’ (X=9), and Enter 5. Y=28 is your answer: So at X = 9 (hours), Y = 28 (allowance of \$28)

f) Go down till you hit the axis

f

You found the answer! “If Bobby works 9 hour, his allowance will be \$28.” Answer Q1 with graphing calculator: 1. In ‘Y=‘, put: y = 10 + 2x 2. Setup the WINDOW (Xmin= 0, Xmax= 16, Xscl= 1, Ymin= 0, Ymax= 40, Yscl= 2) 3. Display the GRAPH

e) Go right till you hit the line

b

X

a

That’s the answer! “Bobby needs to work 13.5 hours to buy a \$37 video game.” Answer Q2 with graphing calc: 1 – 3: same as for Q1 at left 4. Press TRACE, move  and  until Y is close to 37 5. It’s between these two points

(9,28)

6. Try X=13.5 -- type: 13.5, Enter (13.5, 37)

So at X = 13.5 (hours), Y = 37 (allowance of \$37)

Answering Questions with a table (on graphing calculator) Equation: L = 3 + 0.5W Q1: If the band is stretched to 5.4 in., how much weight is on it? (to nearest 0.1 oz.) 1. Enter equation: y = 3 + 0.5x in Y= 2. Setup the table: TBLSET *Always start X at TblStart=0, and increase it by ∆Tbl=1 3. Display the TABLE 4. We’re looking for 5.4 in the Ycolumn. Hmm, it’s in between 5 (X=4) and 5.5 (X=5). Let’s start a new table: TBLSET Start it at TblStart=4, & increase it by only ∆Tbl=.1 (since need the nearest .1 oz.)

5. Display the TABLE again, and scroll down till find Y = 5.4. 6. The X value (4.8) is your answer! The full answer is: “If the band is stretched to 5.4 in., then there’s 4.8 oz. hanging on it”

Words

Table

‚The rubber band’s length is 3 inches with no weight hanging on it, and gets 0.5 inches longer for each ounce added onto it.‛

Weight (oz.) 0

Length (in.) 3

2

4

4

5

Q2: If we attached a 3.7 oz.

6

6

weight to the band, how long would it get? 1. Enter equation: y = 3 + 0.5x in Y=

2. Setup the table: TBLSET *Start X at TblStart=0, and increase it by ∆Tbl=1 3. Display the TABLE 4. We’re looking for 3.7 in the Xcolumn. Hmm, not there! Hey, since we know the X (weight), can’t we just start there? Set TblStart=3.7 in TBLSET 5. Display the TABLE again X = 3.7 will be at the top 6. The Y value (4.85) is your answer! The full answer is: “If you hang a 3.7 oz. weight on the band, it will stretch to 4.85 in.”

IT6710 Job Aid - Math Models

5 20 10 30 15 40 40 90 … … Example: How many hours should Bobby work in order to buy a \$37 video game? Answering Questions (On calc: y = 10...

IT6710 Job Aid - Math Models

5 20 10 30 15 40 40 90 … … Example: How many hours should Bobby work in order to buy a \$37 video game? Answering Questions (On calc: y = 10...