MODULE 1
CHANGES IN QUANTITIES, CONSTANT RATE
OF CHANGE, AND LINEAR FUNCTIONS
Part 1: Investigations #1-5
Investigation #1: Reasoning with Quantities
Investigation #2: Changes in Quantities’ Values
Investigation #3: Constant Rate of Change
Investigation #4: Practice with Constant Rate of Change
Investigation #5: Applying Constant Rate of Change to Find New
Values
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REASONING WITH QUANTITIES
Quantity
A quantity is some attribute of an object that you can
imagine as being measured.
Examples:
i) your height
ii) the number of people in this room
iii) the amount of money in your pocket
iv) the distance around a 400-meter track
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1. Sarah decides to ride a Ferris wheel at a local fair. The
animation shown models her ride around the Ferris
wheel. Assume she is seated in the brown cart and starts
at the bottom of the Ferris wheel. Watch the Ferris wheel
animation as Sarah rotates around twice and then respond
to the following questions.
Choose one of the following links to load a Ferris wheel applet.
Ferris Wheel 1
(requires HTML5)
Rational Reasoning applet. This
is a better option, but may
require a browser upgrade.
Ferris Wheel 2
Applet designed by another
group.
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You will be using your fingers to help model how quantities in
this context change.
Your left index finger will move up and
down to represent values of Sarah’s
height above the ground.
Your right index finger will move left
and right to represent values of the time
elapsed since the Ferris wheel began
moving.
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1. Sarah decides to ride a Ferris wheel at a local fair. The
animation shown models her ride around the Ferris
wheel. Assume she is seated in the brown cart and starts
at the bottom of the Ferris wheel. Watch the Ferris wheel
animation as Sarah rotates around twice and then respond
to the following questions.
a. What varying quantities can you identify in this
situation?
Answers may vary. Some possibilities include
• Sarah’s distance above the ground
• Sarah’s distance from different carts
• The elapsed time since the Ferris wheel began
moving
• etc.
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b. How does Sarah’s height above the ground vary?
Initially, Sarah’s height above the ground increases, then
after reaching the top of the Ferris wheel her height
above the ground decreases until she returns to her
starting location. The pattern then repeats.
c. How does the time since the Ferris wheel began moving
vary?
As the animation runs, the time since the Ferris wheel
began moving increases.
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d. How do the quantities time since the Ferris wheel began
moving and Sarah’s height above the ground vary
together?
As the time since the Ferris wheel began moving
increases, Sarah’s height above the ground initially
increases and then decreases.
The pattern then repeats as the time since the Ferris
wheel began moving continues to increase.
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Points and Graphs
A point shows the corresponding values of two quantities
that we are comparing. A graph of the relationship between
the two quantities is a collection of points showing all pairs
of corresponding values.
2a. The given axes are labeled time since the Ferris wheel began
moving and Sarah’s height above the ground. Your teacher
will play the animation again while you represent how these
quantities are changing together using your fingers. Each
time the video stops, plot a point on the axes that represents
the corresponding values of time since the Ferris wheel
began moving and Sarah’s height above ground.
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A graph of the relationship between the two quantities
is really a collection of points.
Each point tells us the corresponding values of two
quantities that we are comparing.
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b. On the axes in part (a), show all of the corresponding values
of Sarah’s height above the ground and the amount of time
since the Ferris wheel began moving as Sarah travels around
the Ferris wheel twice. (Your graph can be a rough sketch of
how these two quantities change together – you do not need to
worry about making your graph exact).
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c. Choose a random point on your graph. Describe the
information that this point conveys about the situation.
(Specific answers will vary.) A point on a graph can be
thought of as a snapshot that represents a specific amount
of elapsed time since the Ferris wheel began rotating and
the corresponding specific height that Sarah is above the
ground.
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Sam rides a different Ferris wheel. He boards at the bottom and
rides around a few times before getting off. The following graph
represents Sam’s height above the ground (in feet) with respect to
the amount of time since the Ferris wheel began moving (in
minutes) for one complete rotation around the wheel.
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3. The point (1, 50) is plotted on the graph. Explain what
information this point conveys in the context of the
problem.
1 minute after the Ferris wheel began moving, Sam was 50
feet above the ground.
4a. How much did the time since the Ferris wheel began
moving change from 1 to 4 minutes?
The time since the Ferris wheel began moving changed by 3
minutes, from 1 second elapsed to 4 minutes elapsed, and
we can calculate this change with the expression 4 – 1 = 3.
Basically we are asking for the change in time that satisfies
(1 minute elapsed) + (what change in time?) = 4 minutes elapsed.
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b. Represent this change on the graph.
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c. Represent the corresponding change in Sam’s height
above the ground from 1 to 4 minutes.
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d. Approximately how much did Sam’s height above the
ground change over this time interval? Explain how you
determined the change.
This change is represented by the length of the vertical line
segment drawn in part (c) and is approximately 240 feet.
Estimates may vary slightly.
e. Approximately how much did Sam’s height above the
ground change over this time interval? Explain how you
determined the change.
Sam is about 290 feet above the ground 4 minutes after the
Ferris wheel began moving. Estimates and methods may
vary.
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5. Plot the point (5, 320) on the graph and explain its
meaning in the context of the problem.
5 minutes after the
Ferris wheel began
moving, Sam was 320
feet above the ground.
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6a. How much did the time since the Ferris wheel began
moving change from 5 to 7 minutes?
The time since the Ferris wheel began moving changed
by 2 minutes, from 5 minutes elapsed to 7 minutes
elapsed. We can calculate this change with the
expression 7 – 5 = 2.
Basically we are asking for the change in time that
satisfies
(5 minutes elapsed) + (what change in time?) = 7 minutes elapsed.
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b. Represent this change on the graph.
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c. Represent the corresponding change in Sam’s height
above the ground from 5 to 7 minutes.
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d. Approximately what was the change in Sam’s height
above ground over this time interval? Explain how you
determined the value.
Sam’s height above the ground changes by about –105 feet.
Estimates may vary slightly.
e. Estimate Sam’s height above the ground 7 minutes since
the Ferris wheel began moving and plot the point
representing this information.
We can use the graph or the information from part (d) to
determine the new height. Sam is about 215 feet above the
ground 7 minutes since the Ferris when began moving.
Estimates may vary slightly.
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Additional videos are linked on the Rational Reasoning
website and are located in the Pathways Dropbox. You can
use these videos for additional practice with the ideas in
this investigation.
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Investigation
#2
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CHANGES IN QUANTITIES’ VALUES
1. In high school Maria grew from 62 inches tall to 74
inches tall. What was Maria’s change in height during
high school?
Her change in height was 12 inches because her height
started at 62 inches and changed to 74 inches. We calculate
the change with the expression 74 – 62.
We can think of the answer as the number that makes the
following statement true.
62 inches + (what change in height?) = 74 inches
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2. Matt started a new diet, and over the last few months his
weight went from 196.7 pounds to 181.2 pounds. What
was the change in Matt’s weight?
His change in weight was –15.5 pounds because his weight
started at 196.7 pounds and changed to 181.2 pounds. We
calculate the change with the expression 181.2 – 196.7.
We can think of the answer as the number that makes the
following statement true.
196.7 pounds + (what change in weight?) = 181.2 pounds
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3. When Cyndi woke up this morning (6:00 am) the
temperature outside was –6oF. When she got home from
school (4:00 pm) the temperature outside was 28oF.
What was the change in temperature from 6:00 am to
4:00 pm?
The change in temperature was 34oF because the
temperature started at –6oF and changed to 28oF. We
calculate the change with the expression 28 – (–6) or
28 + 6 .
We can think of the answer as the number that makes the
following statement true.
–6oF + (what change in temperature?) = 28oF
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Variable
A variable is a letter used to represent the possible varying
values of a quantity’s measure.
Example:
A bathtub, with a maximum volume of 42 gallons, starts
off full and then drains to empty. Let v be the volume of
water remaining in the bathtub (in gallons). Then v can
represent any possible number from 0 to 42 as the water
drains, such as v = 38 or v = 5.7.
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4a. Define variables to represent the quantities’ values in
Exercises #1-3.
(Answers may vary.)
Let h represent Maria’s height (in inches).
Let w represent Matt’s weight (in pounds).
Let t represent the outside temperature (in degrees
Fahrenheit).
Notice that in our variable definitions we describe the
object and the attribute we are measuring as well as
what units we are using to measure the attribute.
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b. What values might each variable take on?
(Answers may vary.)
h might take on any value between about t = 20 and
maybe t = 75 or t = 80 (depending on how tall Maria is
when she stops growing).
w might take on any value between about w = 10 and
maybe w = 200 (depending on Matt’s highest weight).
t might take on any value between about t = –20 and
maybe t = 100 (depending on where Cyndi lives).
Note that we might choose a tighter set of values if we
restrict ourselves to the time periods described in each
context.
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5. In January 2012, Caitlyn began saving money to make a
down payment on a house. Let m be the number of
months she’s been saving and let v represent the amount
of money she’s saved (in dollars).
a. What is the change in v from v = 940 to v = 2,360?
What does this represent?
The change in v is 1420. Caitlyn saved up an additional
$1420 towards her down payment.
b. The change in v described in part (a) occurred as m
changed from m = 3 to m = 7? What is the change in
m and what does it represent?
The change in m is 4. It took her four months to save
the $1420 described in part (a).
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6. Let x, y, and z represent the values of three different
quantities.
a. If the value of x goes from x = 1 to x = 10, what is the
change in x?
The change in x is 9 (represented by the expression 10 – 1).
b. If the value of x changes to x = –5 from x = 3, what is the
change in x?
The change in x is –8 (represented by the expression –5 – 3).
c. If the value of z changes to z = 18.07 from z = 3.15, what is
the change in z?
The change in z is 14.92 (represented by the expression
18.07 – 3.15).
d. If the value of y goes from y = 6 to y = p, what is the
change in y? The change in y is p – 6.
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Changes in Quantities’ Values
Suppose x is the value of Quantity A and y is the value of
Quantity B. Then ∆x represents a change in the value of
Quantity A and ∆y represents a change in the value of
Quantity B.
Example 1:
If we let x change from x = 4 to x = 11, then ∆x = 7 which
we calculate with the expression 11 – 4.
Example 2:
If we let y change from y = 8 to y = 5, then ∆y = –3 which
we calculate with the expression 5 – 8.
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7. Suppose Gerald bought a tree and planted it in his yard
last summer. At the time of planting it was 4.5 feet tall.
Let h represent the height of the tree in feet.
a. What is the difference between saying h = 5 and
∆h = 5?
Saying h = 5 is stating that the height of the tree is 5 feet.
Saying ∆h = 5 is stating that the change in height was 5
feet (so the tree grew 5 feet over some time period).
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7. Suppose Gerald bought a tree and planted it in his yard
last summer. At the time of planting it was 4.5 feet tall.
Let h represent the height of the tree in feet.
b. If we substitute h = 6.8 into the expression h – 4.5,
what are we calculating?
We are calculating the change in the tree’s height (∆h)
from when it was 4.5 feet tall to when it was 6.8 feet tall.
c. If we substitute h = 10.7 into the expression h – 6.8,
what are we calculating?
We are calculating the change in the tree’s height (∆h)
from when it was 6.8 feet tall to when it was 10.7 feet tall.
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
5
21
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
5–2
5
The change in x
from x = 2 to x = 5
is 3.
21
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
3
5
The change in x
from x = 2 to x = 5
is 3.
21
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
3
5
21 – 5
21
The change in x
from x = 5 to x = 21
is 16.
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
3
5
16
21
The change in x
from x = 5 to x = 21
is 16.
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
3
5
16
21
30 – 21
The change in x from
x = 21 to x = 30 is 9.
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
a.
x
∆x
2
3
5
16
21
9
The change in x from
x = 21 to x = 30 is 9.
30
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8. Fill in the following tables showing the appropriate
changes in the variable values.
b.
y
∆y
10
7.4
2.1
–1.3
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8. Fill in the following tables showing the appropriate
changes in the variable values.
b.
y
∆y
10
7.4 – 10
The change in y from
y = 10 to y = 7.4 is – 2.6.
2.1 – 7.4
The change in y from
y = 7.4 to y = 2.1 is – 5.3.
–1.3 – 2.1
The change in y from
y = 2.1 to y = –1.3 is
– 3.4.
7.4
2.1
–1.3
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8. Fill in the following tables showing the appropriate
changes in the variable values.
b.
y
∆y
10
–2.6
The change in y from
y = 10 to y = 7.4 is – 2.6.
–5.3
The change in y from
y = 7.4 to y = 2.1 is – 5.3.
–3.4
The change in y from
y = 2.1 to y = –1.3 is
– 3.4.
7.4
2.1
–1.3
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8. Fill in the following tables showing the appropriate
changes in the variable values.
c.
w
∆w
–7
–2
2
–3
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8. Fill in the following tables showing the appropriate
changes in the variable values.
c.
w
∆w
–7
–2 – (–7)
The change in w from
w = –7 to w = –2 is 5.
2 – (–2)
The change in w from
w = –2 to w = 2 is 4.
–3 – 2
The change in w from
w = 2 to w = –3 is –5.
–2
2
–3
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8. Fill in the following tables showing the appropriate
changes in the variable values.
c.
w
∆w
–7
5
The change in w from
w = –7 to w = –2 is 5.
4
The change in w from
w = –2 to w = 2 is 4.
–5
The change in w from
w = 2 to w = –3 is –5.
–2
2
–3
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9. Each of the following graphs shows two ordered pairs
(x, y). Find ∆x and ∆y from the point on the left to the
point on the right.
a.
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9. Each of the following graphs shows two ordered pairs
(x, y). Find ∆x and ∆y from the point on the left to the
point on the right.
b.
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10. Show how we can illustrate the values of ∆x and ∆y in
Exercise #9 on the graphs.
a.
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10. Show how we can illustrate the values of ∆x and ∆y in
Exercise #9 on the graphs.
a.
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10. Show how we can illustrate the values of ∆x and ∆y in
Exercise #9 on the graphs.
b.
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10. Show how we can illustrate the values of ∆x and ∆y in
Exercise #9 on the graphs.
b.
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11. Do your answers to Exercise #9 change if we instead
want to find ∆x and ∆y from the point on the right to the
point on the left? Explain.
Yes. The magnitude of the changes will be the same, but
the signs will be opposite.
a.
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11. Do your answers to Exercise #9 change if we instead
want to find ∆x and ∆y from the point on the right to the
point on the left? Explain.
Yes. The magnitude of the changes will be the same, but
the signs will be opposite.
b.
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12. Suppose we start with a value of x = –3.
a. What expression represents the change in x from x = –3
to any possible value of x?
b. Repeat part (a) if our initial reference value is x = –7
instead of x = –3.
c. Repeat part (a) if our initial reference value is x = 1.2
instead of x = –3.
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13. Think about the idea of the change in a variable from
one value to another. What do the following expressions
represent when we substitute a value for the variable?
a. x – 34
The change in x from x = 34 to whatever value of x we
substitute.
For example, if we substitute x = 30, then
30 – 34 = –4, telling us that the change in x from x = 34
to x = 30 is –4.
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13. Think about the idea of the change in a variable from
one value to another. What do the following expressions
represent when we substitute a value for the variable?
b. y – (–2) [or y + 2]
The change in y from y = –2 to whatever value of y we
substitute.
For example, if we substitute y = 5, then 5 + 2 = 7, telling
us that the change in y from y = –2 to y = 5 is 7.
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13. Think about the idea of the change in a variable from
one value to another. What do the following expressions
represent when we substitute a value for the variable?
c. r – 4.7
The change in r from r = 4.7 to whatever value of r we
substitute.
For example, if we substitute r = 5, then 5 – 4.7 = 0.3,
telling us that the change in r from r = 4.7 to r = 5 is 0.3.
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13. Think about the idea of the change in a variable from
one value to another. What do the following expressions
represent when we substitute a value for the variable?
d. p – (–13.8) [or p + 13.8]
The change in p from p = –13.8 to whatever value of p we
substitute.
For example, if we substitute p = –8, then –8 + 13.8 = 5.8,
telling us that the change in p from p = –13.8 to p = –8 is
5.8.
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Investigation
#3
I#3
CONSTANT RATE OF CHANGE
You’ve probably heard someone use the phrase “constant
speed” before. For example, you might hear “He was
driving at a constant speed of 45 miles per hour.” In this
investigation we will explore the idea of constant speed
(and more generally constant rate of change) and link this
back to your work with changes in quantities from
Investigation 2.
For Exercises #1-6 use the “Jane Walking” applet.
Jane is walking from her home to work. When she passes
her mailbox she is 25 feet from her house. Between the
mailbox and a tree she walks at a constant speed, covering
40 feet in 8 seconds.
Jane Walking Applet
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1. After Jane passes the mailbox, how long does it take
her to travel 20 feet? Explain your thinking.
It takes her 4 seconds to travel 20 feet because 20 feet is
times as far as 40 feet, so it will take her times as long as
8 seconds to cover this distance.
2. After Jane passes the mailbox, how far does she travel
in 2 seconds? 6 seconds? Explain your thinking.
(Student reasoning might vary. We give only one possible
way of thinking.)
She travels 10 feet and 30 feet respectively. 2 seconds is
times as long as 8 seconds, so she will travel times as far
as 40 feet. 6 seconds is times as long as 8 seconds, so she
will travel times as far as 40 feet.
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3. Let’s draw line segments to represent the total distance
traveled from the mailbox to the tree and the amount of
elapsed time it takes to cover this distance.
Demonstrate how we can use these line segments to
represent the reasoning in Exercises #1-2.
We can partition the line segments to see how these
two quantities correspond.
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3. Let’s draw line segments to represent the total distance
traveled from the mailbox to the tree and the amount of
elapsed time it takes to cover this distance.
It takes her 4 seconds to travel 20 feet because 20 feet is
times as far as 40 feet, so it will take her times as long as
8 seconds to cover this distance.
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3. Let’s draw line segments to represent the total distance
traveled from the mailbox to the tree and the amount of
elapsed time it takes to cover this distance.
She travels 10 feet and 30 feet respectively. 2 seconds is
times as long as 8 seconds, so she will travel times as far
as 40 feet. 6 seconds is times as long as 8 seconds, so she
will travel times as far as 40 feet.
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3. Let’s draw line segments to represent the total distance
traveled from the mailbox to the tree and the amount of
elapsed time it takes to cover this distance.
She travels 10 feet and 30 feet respectively. 2 seconds is
times as long as 8 seconds, so she will travel times as far
as 40 feet. 6 seconds is times as long as 8 seconds, so she
will travel times as far as 40 feet.
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4. Use the line segments in Exercise #3 and the applet to
help you answer the following questions.
a. How far will she travel during any 1-second time
interval between the mailbox and the tree?
5 feet during any 1-second interval
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4. Use the line segments in Exercise #3 and the applet to
help you answer the following questions.
b. How long will it take her to travel any 1-foot
distance between the mailbox and tree?
0.2 seconds to cover any 1-foot distance between
mailbox and tree
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5. How do your answers to Exercise #4 help you find
a. how far Jane travels for any change in the time
elapsed since she passed the mailbox? (For
example, how far does Jane travel for a change in
time elapsed of 2.8 seconds?)
Once we know the unit rate, this allows us to scale to
any multiple we want.
We know that in 1 second she travels 5 feet, so in 2.8
seconds (2.8 times as long as 1 second) she travels 2.8
times as far as 5 feet, or 2.8(5) = 14 feet.
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5. How do your answers to Exercise #4 help you find
b. how long it takes Jane to travel any distance
between the mailbox and tree? (For example, how
long does it take Jane to travel 33 feet?)
Once we know the unit rate, this allows us to scale to any
multiple we want.
We know it takes her 0.2 seconds to travel 1 foot, so to
travel 33 feet (33 times as far as 1 foot) it will take her 33
times as long as 0.2 seconds, or 33(0.2) = 6.6 seconds.
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6. Think about the work you did with the “Jane Walking”
applet. What does it mean for an object to move at a
constant speed? (Note: Please say something more than
“The speed doesn’t change” – be descriptive and
reference specific quantities.)
Explanations will vary. Some important observations could
include the following.
For an object traveling at a constant speed, the same change
in time elapsed will always correspond to equal changes in
distance traveled, and likewise the same change in distance
traveled will correspond to equal changes in time elapsed.
Furthermore, if we know how far the object travels in some
total amount of time, then in some fraction of the total time
it will travel the same fraction of the total distance.
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7. Felipe was walking to school today. Assume that he
walked at a constant speed during the entire trip, and
also suppose that during one part of the trip he walked
70 feet in 16 seconds.
a. Provide at least four conclusions we can draw from
the given information.
Answers will vary. Some examples are given.
He walked 35 feet ( times as far as 70 feet) in times
as long as 16 seconds (or 8 seconds).
If the trip was long enough, he walked 280 feet (4 times
as far as 70 feet) in 64 seconds (4 times as long as 16
seconds).
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b. How far did Felipe travel in 3 seconds?
Felipe travels 4.375 feet over any 1-second interval, so
in 3 seconds he travels 3(4.375) = 13.125 feet.
c. Does your answer to part (b) depend on which 3-second
interval we’re talking about? Explain.
No. Over any 3-minute interval during this part of his trip
he travels 13.125 feet because his speed was constant.
d. How long did it take Felipe to travel any 10-foot
distance during this part of his trip?
It takes him
of a second to travel 1 foot, so to travel
10 feet it takes
seconds.
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8. Assume that a baseball and a tennis ball are both
traveling at constant speeds (but not necessarily the
same speed).
a. What does it mean to say that the tennis ball is
traveling faster than the baseball? Be descriptive, and
reference the specific quantities involved.
For any given amount of time elapsed, the tennis ball
will cover a greater distance than the baseball will in the
same amount of elapsed time.
Similarly, for any given amount of distance, the tennis
ball will cover that distance in less time than it takes the
baseball to cover that distance.
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8. Assume that a baseball and a tennis ball are both
traveling at constant speeds (but not necessarily the
same speed).
b. What does it mean if the baseball and tennis ball have
the same constant speed? Be descriptive, and
reference the specific quantities involved.
For any given elapsed time, the two objects will cover
the same amount of distance.
Likewise, for any specific amount of distance, it will
take the same amount of time for both objects to travel
that distance.
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9. Suppose we want to rank the following objects in order from
fastest to slowest. Assume each object is traveling at a
constant speed (but not necessarily the same constant speed).
Bike:
11 feet in 0.5 second
Runner:
112 feet in 7 seconds
Car:
212 feet in 10 seconds
Football:
52.5 feet in 3 seconds
Bird:
4.24 feet in 0.2 second
Water Balloon: 24 feet in 1.5 seconds
a. Why is it difficult to compare the objects’ speeds based
on the information given?
The distances traveled and the corresponding time intervals are
all different – we can’t determine which objects travel further
in a fixed amount of time or which objects cover the same
amount of distance in less time by simply looking at the given
information.
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b. What strategy can we use to help us complete the task?
We need to either determine how long it takes each object
to travel some fixed distance (such as one foot) or
determine how far each object travels in a given amount of
time (such as one second). This latter technique is the most
common convention for describing speed – we determine
how far an object travels in one unit of time (in this case
one second). We then apply the reasoning from this
investigation. For example, the bird travels 4.24 feet in 0.2
(or of one second), so in one second the bird will travel
5(4.24) = 21.2 feet.
c. Rank the objects in order from fastest to slowest.
Bike (22 ft/sec), Car and Bird (21.2 ft/sec), Football (17.5
ft/sec), Runner and Water Balloon (16 ft/sec)
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10. What are the benefits of stating speeds in terms of
distance traveled in one unit of time, such as 20 mph
(20 miles traveled in one hour), 9.8 m/s (9.8 meters
traveled in one second), and so on? Brainstorm some
ideas with a classmate and summarize your thinking.
One answer based on the results of Exercise #9 is that it
makes speeds easy to compare, which is very important.
For example, a car’s speedometer always reports your
speed in mph (regardless of how far you travel or if you
travel that speed for less than one hour) which allows you
to easily compare your speed to stated speed limits.
Imagine the trouble that could occur if your car instead
constantly told you information such as “You just traveled
56 feet in 3.9 seconds…30 feet in 1.56 seconds…”
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10. What are the benefits of stating speeds in terms of
distance traveled in one unit of time, such as 20 mph
(20 miles traveled in one hour), 9.8 m/s (9.8 meters
traveled in one second), and so on? Brainstorm some
ideas with a classmate and summarize your thinking.
However, there is another important benefit that we can support
with our understanding of multiplication.
If we report speeds in terms of distance traveled in one unit of
time, then as time elapses we can easily scale the distance by
simply using multiplication. If a car travels 52 mph for 2.1
hours, we simply multiply 52 by 2.1 to get the distance traveled
in miles (2.1 times as long as 1 hour corresponds with 2.1 times
as far as 52 miles). If a car travels 45 mph for
of an hour, we
simply multiply 45 by to get the distance traveled ( times as
long as 1 hour corresponds to times as far as 45 miles). The
task becomes harder if a speed is reported as 3.6 miles per 2.7
minutes.
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11. Suppose that Joe is jogging at a constant speed of 0.12
miles per minute. Let d represent the total distance (in
miles) Joe has traveled during his jog and let t represent
the total number of minutes he’s been jogging.
a. Use ∆ notation to represent a 4-minute time period
during Joe’s jog.
∆t = 4
b. How far does Joe travel during any 4-minute time
period? Represent your answer using ∆ notation.
0.48 miles
∆d = 0.12(4) or ∆d = 0.48
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c. Repeat parts (a) and (b) if instead of a 4-minute time
period we look at a
i) 7.8-minute time period
a) ∆t = 7.8
b) ∆d = 0.12(7.8) or ∆d = 0.936
ii) 0.3-minute time period
a) ∆t = 0.3
b) ∆d = 0.12(0.3) or ∆d = 0.036
iii) any x-minute time period
a) ∆t = x
b) ∆d = 0.12x
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d. For any change in t, what happens to the change in d?
Explain, then represent your thinking using ∆ notation.
The change in distance (in miles) is always 0.12 times as
large as the change in time (in minutes).
∆d = 0.12(∆t)
So far we’ve focused on the idea of constant speed. However,
this is just a specific example of the more general concept of
constant rate of change.
Constant Rate of Change
A constant rate of change of one quantity with respect to
another quantity exists when, for any uniform change in
one quantity’s value, the other quantity’s value changes by
equal amounts.
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12. If we know that the total cost of purchasing a bag of
walnuts increases at a rate of $7 per pound, then it’s easy
to determine the change in total cost if we change the
amount of walnuts we purchase. Let c represent the total
cost of purchasing walnuts (in dollars) and let w represent
the total weight of walnuts purchased (in pounds).
a. Suppose we have some walnuts in a bag and we add 3
more pounds of walnuts to the bag. Represent the
change in the number of pounds of walnuts using ∆
notation.
∆w = 3
b. How much does the total cost of the walnuts we
purchase change? Represent your answer using ∆
notation.
$21
∆c = 7(3) or ∆c = 21
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c. Repeat parts (a) and (b) if instead of adding 3 more
pounds of walnuts we
i) add 2.8 pounds of walnuts to the bag.
a) ∆w = 2.8
b) ∆c = 7(2.8) or
∆c = 19.6
ii) add 0.65 pounds of walnuts to the bag.
a) ∆w = 0.65
b) ∆c = 7(0.65) or
∆c = 4.55
iii) remove 1.2 pounds of walnuts from the bag.
a) ∆w = –1.2
b) ∆c = 7(–1.2) or
∆c = –8.4
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d. For any change in w, what happens to the change in c?
Explain, then represent your thinking using ∆ notation.
The change in total cost (in dollars) is always 7 times as
large as the change in weight (in pounds).
∆c = 7(∆w)
e. If your friend gives you $10 to purchase more walnuts,
how many additional pounds of walnuts could you add
to your bag?
Approaches may vary. One example is given.
Any change in 1 dollar spent on walnuts corresponds with a
change of pounds of walnuts purchased. So if ∆c = 10,
then
You can purchase an additional 1.43 pounds of walnuts.
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13. Suppose we want to add some sand to a child’s
sandbox, and that we know an 8-liter bag of sand
weighs 25.6 pounds. Define variables to represent the
quantities in this context, then represent the relationship
described using ∆ notation.
Answers may vary based on how we define our variables.
Two possible answers are given.
Method 1: Let b represent the volume of sand in the
sandbox (in bags) and let w represent the weight of the
sandbox (in pounds). Then ∆w = 25.6(∆b).
In other words, the change in the sandbox weight (in
pounds) is always 25.6 times as large as the change in
volume of sand (in bags).
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13. Suppose we want to add some sand to a child’s
sandbox, and that we know an 8-liter bag of sand
weighs 25.6 pounds. Define variables to represent the
quantities in this context, then represent the relationship
described using ∆ notation.
Method 2: Let v represent the volume of sand in the
sandbox (in liters) and let w represent the weight of the
sandbox (in pounds). Then ∆w = 3.2(∆v).
In other words, the change in the sandbox weight (in
pounds) is always 3.2 times as large as the change in
volume of sand (in liters).
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I#4
PRACTICE WITH CONSTANT RATE OF CHANGE
1. Jorge left his house to go for a jog this morning. Let
d = Jorge’s distance (in miles) from his house. What’s
the difference in meaning between d = 2 and ∆d = 2?
d = 2 means that Jorge is 2 miles from his house.
∆d = 2 describes a change of 2 miles in the distance Jorge
is from his house, such as the distance covered between the
times he was 1 mile from his house (d = 1) and 3 miles
from his house (d = 3).
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2. Student Council is planning to rent a banquet hall for the
post-graduation party this year. In addition to the rental
costs, the hall charges $14 per person for food. Let p
represent the number of people who will attend the party
and c represent the total cost of the party.
a. What does the statement
mean?
The change in the total cost of the party (in dollars) is 14
times as large as the change in the number of people
attending.
b. If p changes by 7, by how much does c change? What
does this represent?
c changes by 98 since ∆p = 7 and ∆c = 14(∆p) = 14(7) = 98.
If 7 additional people attend, the total cost of the party
increases by $98 dollars.
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c. How much does c change by if p changes by –5? What
does this represent?
c changes by –70 since ∆p = –5 and
∆c = 14(∆p) = 14(–5) = –70.
If 5 fewer people decide to attend the party, the total
cost will decrease by $70.
d. If c changed by 182, by how much did p change? What
does this represent?
p changed by 13 since ∆c = 14(∆p) meaning that
∆p = (1/14)(∆c) = (1/14)(182) = 13.
If the total cost of the party increased by $182, then it
means 13 additional people must have decided to attend.
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3. Suppose we know that two variables (x and y) are related
such that ∆y = –3 ∙ ∆x.
a. What does the statement ∆y = –3 ∙ ∆x mean?
The change in the value of y is always –3 times as
large as the change in the value of x.
b. If x changes from x = 1 to x = 9, by how much does y
change?
c. How much does y change by if x changes from x = 4 to
x = –2?
d. If y changes to y = 8 from y = 10, by how much does x
change?
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In Exercise #2-3 we saw examples where the change in one
variable always had to be some constant times as much as the
change in a second variable. This is a very useful way of
understanding constant rate of change.
Constant Rate of Change as a Constant Multiple
There exists a constant rate of change of Quantity B with
respect to Quantity A if the following is always true.
The constant is the number we use to describe the constant rate
of change, and its value can be calculated as follows.
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Example 1:
Suppose x is the value of quantity A and y is the value of
quantity B. Then there is a constant rate of change of m
between the quantities if
where m is a constant, and
Example 2:
Each of the following represents a situation with a
constant rate of change of one quantity with respect to
another.
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4. Use the following graphs and the given information to
answer the questions that follow.
a. For the graph on the left, if x changes from the given
point to x = 0, by how much does y change if
?
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4. Use the following graphs and the given information to
answer the questions that follow.
b. In the graph on the right, what does the point (5.5, 15)
represent?
If a package weighs 5.5
pounds, then it costs $15
to ship.
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4. Use the following graphs and the given information to
answer the questions that follow.
c. Let w be the weight of a package (in pounds) and c be
the cost to ship the package (in dollars). If w changes
from the given point to w = 3, by how much does c
change if
?
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4. Use the following graphs and the given information to
answer the questions that follow.
d.
Explain the meaning of your answer to part (c).
If we reduce the weight of the package by 2.5 pounds, the
cost to ship the package decreases by $3.
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In Exercises #5-6, explain why each context does not have a
constant rate of change. (Hint: You should first identify
what quantities are changing together.)
5. When Marcus was born he was 19 inches tall. When he
turned 16 he was 65 inches tall.
Let h represent Marcus’s height (in inches) and let t
represent his age (in years). Then ∆h is the change in
Marcus’s height (in inches) and ∆t is the change in his
age (in years).
No constant rate of change exists in this context because,
for uniform changes in his age, we do not expect to
observe equal changes in his height. For example, we
don’t expect his height to change the same from 1 to 4
years old as from 10 to 13 years old.
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6. The number of people following one celebrity’s Twitter
feed doubled every day for a week after she appeared
on Saturday Night Live.
Let n represent the total number of people following the
celebrity’s Twitter feed and let d be the number of days since
she appeared on Saturday Night Live. Then ∆n is the change
in the number of people following the celebrity’s Twitter feed
and ∆d is the change in the number of days since she
appeared on Saturday Night Live.
Doubling does NOT produce a constant rate of change. For
example, if she began with 1,000 Twitter followers, her total
number of followers after the next several days would be
2,000 then 4,000 then 8,000, and it’s clear that ∆d isn’t
constant for a given ∆d (such as ∆d = 1).
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Describing a Constant Rate of Change
Sometimes you might see a phrase like “There is a constant
rate of change of y with respect to x.” This just means that
there is a constant m such that
.
7. While performing a backup for her computer, Elise
noticed that during a certain interval of time the
computer was transferring data at a constant rate of 8
megabytes per second. Complete the following
statement and explain what it tells us.
where ∆t represents the change in the elapsed time for
the backup and ∆n the change in the number of
megabytes transferred.
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7. While performing a backup for her computer, Elise
noticed that during a certain interval of time the
computer was transferring data at a constant rate of 8
megabytes per second. Complete the following
statement and explain what it tells us.
where ∆t represents the change in the elapsed time for
the backup and ∆n the change in the number of
megabytes transferred.
The change in the total data transferred (in megabytes) was
8 times as large as the change in time (in seconds) during
any portion of the given time period.
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8. The cost of covering an area with a certain type of
concrete paver increases at a constant rate of $4.50 per
square foot with respect to the size of the area covered.
Complete the following statement and explain what
it tells us.
where ∆a represents the change in the size of the area
covered (in square feet) and ∆c represents the change in
the total cost of covering the area.
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8. The cost of covering an area with a certain type of
concrete paver increases at a constant rate of $4.50 per
square foot with respect to the size of the area covered.
Complete the following statement and explain what
it tells us.
where ∆a represents the change in the size of the area
covered (in square feet) and ∆c represents the change in
the total cost of covering the area.
The change in the total cost of covering the area (in dollars)
is 4.50 times as large as the change in the area covered (in
square feet).
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9. Suppose we know that
for some constant m,
and we are given the information in the following table.
What is the value of m?
x
–2
1
6
14
y
2
6.5
14
26
We just need to take two entries in
the table and calculate ∆x and ∆y.
For example, suppose we select
(–2 , 2) and (6, 14).
Then
and
, so
You will get the same result for any two points you choose
because there is a constant rate of change.
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10. Suppose we know that
for some constant m,
and we are given the information in the following
graph. What is the value of m and what does it tell us
about how x and y change together?
We use the points on the
graph to determine ∆x and
∆y.
So
and
, and thus
The change in the value of y is always –0.8 times as
large as the change in the value of x.
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11. Suppose we know that
for some constant m,
and we know that two ordered pairs are (–5, 1) and
(–3, 9). What is the value of m?
We use the ordered pairs to determine ∆x and ∆y.
So
and
, and thus
The change in the value of y is always 4 times as large
as the change in the value of x.
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12. Given that the x and y axes have the same scale, what is
the approximate rate of change for the linear functions
whose graphs are given? How did you determine your
answer?
a.
About 1/3.
It appears that the change
in y is about 1/3 times as
large as the change in x.
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12. Given that the x and y axes have the same scale, what is
the approximate rate of change for the linear functions
whose graphs are given? How did you determine your
answer?
b.
About –2.
It appears that the change in y is
about –2 times as large as the change
in x.
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Investigation
#5
I#5
APPLYING CONSTANT RATE OF CHANGE
TO FIND NEW VALUES
1. Suppose that we know that ∆y = m ∙ ∆x for some constant
m, and we are given the information in the following
table. What is the value of m and what does it tells us
about how x and y change together?
x
–7
–4
y
15
9
We just need to take two entries in the
table and calculate ∆x and ∆y. For
example, suppose we select (–4, 9) and
(1, –1).
1
9
–1
–17
Then ∆x = 5 and ∆y = –10, so
The change in the value of y is always –2 times as
large as the change in the value of x.
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I#5
For Exercises #2-4, suppose we know that the changes in the
values of two variables are related according to ∆y = 4 ∙ ∆x.
2a. If we start off at x = 7 and let x change to be x = 10,
what is the change in x?
b. By how much does y change for the change in x you
found in part (a)?
c. Suppose we know that y = 15 when x = 7. What is the
value of y when x = 10? How did you find this?
We find this by adding the change in y to the starting
value of y.
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I#5
For Exercises #2-4, suppose we know that the changes in the
values of two variables are related according to ∆y = 4 ∙ ∆x.
3a. If we end at x = 12 after starting at x = 7, what is the
change in x?
b. By how much does y change for the change in x you
found in part (a)?
c. Suppose we know that y = 15 when x = 7. What is the
value of y when x = 12?
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I#5
For Exercises #2-4, suppose we know that the changes in the
values of two variables are related according to ∆y = 4 ∙ ∆x.
4a. If we start off at x = 7 and let x change to be x = 4,
what is the change in x?
b. By how much does y change for the change in x you
found in part (a)?
c. Suppose we know that y = 15 when x = 7. What is the
value of y when x = 4?
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5. The given graph displays the point (7, 15). Show how
we can represent the reasoning from Exercises #2-4
visually on the graph.
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#2
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#3
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5. The given graph displays the point (7, 15). Show how
we can represent the reasoning from Exercises #2-4
visually on the graph.
#4
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6. Apply your understanding of constant rate of change to
complete the following table of values given that
∆y = 4 ∙ ∆x. Hint: Your first step should be to determine
the change in x from one value to another value.
x
–2
5
y
7
9.5
15
For a given x-value, we find the change in
x away from 7.
For example, to get from x = 7 to x = –2
we must change x by –9 (that is, ∆x = –9).
Then ∆y = 4 ∙ ∆x = 4 ∙ (–9) = –36, so we change y by –36
from y = 15 to get y = –21.
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6. Apply your understanding of constant rate of change to
complete the following table of values given that
∆y = 4 ∙ ∆x. Hint: Your first step should be to determine
the change in x from one value to another value.
x
–2
5
y
–21
For a given x-value, we find the change in
x away from 7.
7
9.5
15
For example, to get from x = 7 to x = –2
we must change x by –9 (that is, ∆x = –9).
Then ∆y = 4 ∙ ∆x = 4 ∙ (–9) = –36, so we change y by –36
from y = 15 to get y = –21.
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I#5
6. Apply your understanding of constant rate of change to
complete the following table of values given that
∆y = 4 ∙ ∆x. Hint: Your first step should be to determine
the change in x from one value to another value.
x
–2
5
y
–21
7
7
9.5
15
25
For a given x-value, we find the change in
x away from 7.
For example, to get from x = 7 to x = –2
we must change x by –9 (that is, ∆x = –9).
Then ∆y = 4 ∙ ∆x = 4 ∙ (–9) = –36, so we change y by –36
from y = 15 to get y = –21.
Repeat for all values of x in the table.
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In Exercises #2-6 we thought about how to find a new
value for y by applying our understanding of constant rate
of change.
We started by finding a change in the value of x, then used
our constant rate of change to determine the corresponding
change in the value of y.
Finally we used this change in y to find a final (new) value
for y.
Use this technique and reasoning to complete the
remaining tasks in this investigation.
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7. Suppose you are considering shipping a package with a
certain shipping company. Let w represent the weight of
the package (in pounds) and let c represent the cost of
shipping the package (in dollars). Furthermore, suppose
c = 15 when w = 5.5 and that ∆c = 1.20 ∙ ∆w.
a. What is the value of c when w = 7.5? What does this
tell us?
∆w = 2 from w = 5.5 to w = 7.5, so ∆c = 1.20(2) = 2.40,
and thus c = 2.40 + 15 = 17.40 when w = 7.5.
The cost to ship a package weighing 7.5 pounds is $17.40.
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7. Suppose you are considering shipping a package with a
certain shipping company. Let w represent the weight of
the package (in pounds) and let c represent the cost of
shipping the package (in dollars). Furthermore, suppose
c = 15 when w = 5.5 and that ∆c = 1.20 ∙ ∆w.
b. What is the value of c when w = 0? What does this tell
us?
∆w = –5.5 from w = 5.5 to w = 0, so
∆c = 1.20(–5.5) = –6.60, and thus c = –6.60 + 15 = 8.40
when w = 0.
The fixed cost to ship a package (prior to determining the
weight and adding the corresponding additional costs) is
$8.40.
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6. Given that ∆r = –1.25 ∙ ∆h, complete the following table
of values.
h
–4
4
6
18
r
6
For a given h-value, we find the change in
h away from 4.
For example, to get from h = 4 to h = 6
we must change h by 2 (that is, ∆h = 2).
Then ∆r = –1.25 ∙ ∆h = –1.25 ∙ (2) = –2.5, so we change r
by –2.5 from r = 6 to get r = 3.5.
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6. Given that ∆r = –1.25 ∙ ∆h, complete the following table
of values.
h
r
–4
4
6
18
6
3.5
For a given h-value, we find the change in
h away from 4.
For example, to get from h = 4 to h = 6
we must change h by 2 (that is, ∆h = 2).
Then ∆r = –1.25 ∙ ∆h = –1.25 ∙ (2) = –2.5, so we change r
by –2.5 from r = 6 to get r = 3.5.
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6. Given that ∆r = –1.25 ∙ ∆h, complete the following table
of values.
h
r
–4
4
6
18
16
6
3.5
–11.5
For a given h-value, we find the change in
h away from 4.
For example, to get from h = 4 to h = 6
we must change h by 2 (that is, ∆h = 2).
Then ∆r = –1.25 ∙ ∆h = –1.25 ∙ (2) = –2.5, so we change r
by –2.5 from r = 6 to get r = 3.5.
Repeat for all values of h in the table.
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Linear Function
If there is a constant rate of change of one quantity with
respect to another quantity, then we say that the relationship
is a linear function.
Example 1:
Suppose ∆y = 2 ∙ ∆x. Then we say that “y is a linear
function of x.”
Example 2:
Suppose a car is traveling at 50 mph. Let d represent the
distance the car has traveled from some location (in miles)
and t represent the corresponding elapsed time (in hours)
since leaving the location, then ∆d = 50 ∙ ∆t. Then we say
that “the distance traveled is a linear function of the time
elapsed.”
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9. Use the given graph to answer the questions that follow
given that ∆y = ∆x.
a. What is the value of y
when x = 3.5? Represent
your reasoning on the
graph.
b. What is the value of y
when x = –4? Represent
your reasoning on the
graph.
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9. Use the given graph to answer the questions that follow
given that ∆y = ∆x.
a. What is the value of y
when x = 3.5? Represent
your reasoning on the
graph.
From the reference point
(1, 2), ∆x = 2.5, so ∆y = 2.5,
and thus y = 2.5 + 2 = 4.5.
b. What is the value of y
when x = –4? Represent
your reasoning on the
graph.
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9. Use the given graph to answer the questions that follow
given that ∆y = ∆x.
a. What is the value of y
when x = 3.5? Represent
your reasoning on the
graph.
From the reference point
(1, 2), ∆x = 2.5, so ∆y = 2.5,
and thus y = 2.5 + 2 = 4.5.
b. What is the value of y
when x = –4? Represent
your reasoning on the
graph.
From the reference point (1, 2), ∆x = –5,
so ∆y = –5, and thus y = –5 + 2 = –3.
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
a. What information does the
point (3.5, 8.4) represent in
this situation?
After burning for 3.5 hours, the
candle’s length is 8.4 inches.
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
b. If L represents the length of
candle remaining (in
inches) and t represents the
number of hours the candle
has been burning, write a
statement showing the
relationship between the
changes in these variables.
_____ = ____________
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
c. What is the length of the
candle 5.5 hours since it
began burning? Find the
answer and demonstrate
your reasoning on the
graph.
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
c. What is the length of the
candle 5.5 hours since it
began burning? Find the
answer and demonstrate
your reasoning on the
graph.
5.2 inches long.
We change t from t = 3.5 to
t = 5.5, so ∆t = 2, and thus
∆L = –1.6(2) = –3.2, and
L = –3.2 + 8.4 = 5.2.
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
d. What is the length of the
candle 1.7 hours since it
began burning? Find the
answer and demonstrate
your reasoning on the
graph.
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
d. What is the length of the
candle 1.7 hours since it
began burning? Find the
answer and demonstrate
your reasoning on the
graph.
11.28 inches long.
We change t from t = 3.5 to
t = 1.7, so ∆t = –1.8, and
thus ∆L = –1.6(–1.8) = 2.88,
and L = 2.88 + 8.4 = 11.28.
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
e. What was the original
length of the candle?
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10. The length of a burning candle decreases at a constant
rate of 1.6 inches per hour.
e. What was the original
length of the candle?
14 inches long.
We can find the value of L
when t = 0. We change t
from t = 3.5 to t = 0, so
∆t = –3.5, and thus
∆L = –1.6(–3.5) = 5.6, and
L = 5.6 + 8.4 = 14.
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