Section
3.15 Constant Rate of Change & Linear Functions
The concept of constant rate of change is central to all of calculus. This
will become evident in Chapters 4 and 5. In this chapter we will review
situations in which the idea of constant rate of change is involved.
Constant rate of change always involves two quantities changing together. Two
quantities change at a constant rate with respect to each other if
changes in one are proportional to changes in the other. If you
are told that two quantities change at a constant rate with respect to each
other, then you know immediately that changes in one are proportional to
changes in the other.
Let x represent the value of Quantity A and let y represent
the value of Quantity B, and let x be the independent variable.
Assume that B changes at a constant rate with respect to A. Then, according
to the meaning of constant rate of change, any change in y will be m
times as large as the change in x that corresponds to it, where m
is a real number and $m \ne 0$.
We can state the above paragraph symbolically. Suppose that y
changes at a constant rate with respect to x. Let dx
represents a change in x. Let dy represents the corresponding
change in y.
The relationship between changes is dy = m dx
for some number $m, m \neq 0$. If the change in the value of x is a
change from 0, then dx = x and dy = mx. The
value of y when $x=0$ is represented by the letter b, thus
$y=b$ when $x=0$. Therefore, $y=mx+b$ is the general form of the
relationship between values of two quantities that change at a constant rate
with respect to each other. The letter m represents the value of
that constant rate of change.
Envisioning Constant Rate of Change
The top bar in Figure 3.15.1 represents the value of y as the value
of x varies. The value of y is b when the value of x
is 0. The bottom bar in Figure 3.15.1 represents the value of x. The
value of y changes at a constant rate of m with respect to
the value of x. As the value of x changes by dx, the
value of y changes by m dx. This is true no matter
the value of m.
Notice also that the value of dx grows within changes. Changes
are not added wholly. Rather, they change smoothly. Therefore values of dy
change smoothly so that $dy = m \cdot dx$ even as dx grows.
Move your pointer away from the animation to hide the control bar.
Figure 3.15.1. Changes in the value of y (dy) are proportional to
changes in the value of x (dx).
Move your pointer away from the animation to hide the control bar.
In Figure 3.15.2., the top bar represents the value of y as the
value of x varies. The bottom bar represents the value of x.
The value of y changes at a constant rate of m with respect
to the value of x. The changes in x aggregate to the value
of x. The corresponding changes in y aggregate to mx.
The value of y is therefore always $y = mx + b$. This is
true no matter the value of m.
Figure 3.15.2. Changes in x aggregate to the value of x, changes in y
aggregate to the value of mx. The value of y is always y = mx + b.
Reflection 3.15.1. Some people think that
Figures 3.15.1 and 3.15.2 say the same thing. Others disagree. Why
might each group of people think as they do about Figures 3.15.1 and
3.15.2?
Alert! How you read the next two examples is important. Read
them with the goal that you become able to repeat their patterns
of reasoning. Do not read them with the goal of
remembering what to write! |
Example 3.15.1
Stan started running errands in Phoenix at 8:00 am. He had driven 32 km at
the end of the last errand. At 11:30 am Stan headed for Tucson at a constant
speed of 113 km/hr; 1.6 hours later he began slowing to a stop.
Let D represent the number of km that Stan drove since 8:00 am, and
let t represent the number of hours since 8:00 am.
While on his trip to Tucson, the rate of change of D with respect to
t was 113 km/hr, so dD = 113dt km, and D,
Stan's distance traveled at the constant speed of 113 km/hr, is
$D=113(t-3.5)+32$ km.
Why must we subtract 3.5 from t? Because it was not until 11:30 am,
when t = 3.5, that Stan began driving at a constant speed.
So it only for $3.5 \le t \le 5.1$ that $D=113(t-3.5)+32$ is a valid model
of the distance Stan drove while driving to Tucson. Stan’s distance driven
with respect to time did not change at a constant rate while he
ran errands nor after he began slowing down in Tucson.
Reflection 3.12.1. Describe similarities and differences between the
story of Stan’s trip and Figures 3.15.1 and 3.15.2.
Example 3.15.2
I am traveling away from home on a straight road at the constant speed of
0.35 km/min. I am 3.8 km from home and I first looked at my watch 7 minutes
ago.
a) How much will my distance from home in
the next 1/10 minute? How much will my distance from home change in the
next 3/10 minute? How much will my distance from home change in the next
1/10 minute?
Let D represent number of km from home and let t represent the number of
minutes since I looked at my watch. Then $dD = 0.35dt$, or $dD =
(0.35)(0.1)$. When $dt = 0.3$, $dD = (0.35)(0.3)$. When $dt = t$,
$dD = (0.35)t$.
b) Was I at home when I looked at my watch?
No, 7 minutes ago I was $(0.35)(-7)$ km from where I am now. Home is
-3.8 km from where I am now. So I was $3.8 + (0.35)(-7)$ km from home
when I first looked at my watch.
c) Define a function that relates my distance from home in km to the
number of minutes since I looked at my watch.
From (b), I was $3.8 + (0.35)(-7)$ km from home when I looked at my
watch. My distance from home increased by 0.35t km in t minutes since
looking at my watch. Therefore my distance from home as a function of
time is $D = 0.35t+(3.8+ 0.35(-7))$, or
$D = 0.35(t-7) + 3.8$.
d) Generalize a-c. The variable y changes at a constant rate m
with respect to x. The value of y is $y_0$ when the value
of x is $x_0$. Define a function f that relates values of
y to their corresponding values of x.
Changes in y and x are proportional.
The change in the value of x from $x_0$ is $(x-x_0)$. So the change in
the value of y that corresponds to the change $(x-x_0)$ is $m(x-x_0)$.
The value of y is $y_0$ when the value of x is $x_0$. Therefore, the
value of y for any value of x is $y = m(x-x_0) + y_0$.
More simply, the value of y for any value of x is equal to the initial
value of y plus the change in y due to the value of x changing from
$x_0$ to x. Define the function f to be
$f(x) = m(x-x_0) + y_0$.
Determining a Constant Rate of Change
Suppose that we are told that Quantity A changed by p kg while
Quantity B changed by q liters, $p, q > 0$, and that they changed
at a constant rate of m with respect to each other. Let y
represent the value of Quantity A and let x represent the value of
Quantity B. What is the value of m?
There are two ways to reason about determining the constant rate of change
of Quantity A with respect to Quantity B. The first one is to reason
quantitatively; the second is to reason algebraically.
Way of Reasoning #1: Look at changes in A relative to a
change of 1 liter in B.
- Since y and x changed at a constant rate with respect
to each other, changes in y are proportional to changes in x.
- The value of x changed by q liters. Thus, for the
value of x to change by 1 liter, it would change by
$\left(\frac{1}{q}\text{ of }q\right)$ liters.
- Since changes in x and y are proportional, y
would change by $\left(\frac{1}{q}\text{ of }p\right)$ kg when x
changes by $\left(\frac{1}{q}\text{ of }q\right)$ liters.
- But $\left(\frac{1}{q}\text{ of }p\right)$ is $\frac{p}{q}$.
- Therefore y changes $\left(\frac{p}{q}\right)$ kg per liter,
or $m = \frac{p}{q}$.
Way of Reasoning #2: Look at total change in A relative to
total change in B.
- Since y and x changed at a constant rate with respect
to each other, all changes in y are proportional to changes in x.
- x changed by p kg. y changed by
q liters.
- So $p = mq$.
- Therefore $m = \frac{p}{q}$.
The quantitative way of reasoning actually explains why you divide the value
of one change by the value of the other change to compute a constant rate of
change. The algebraic way of reasoning is shorter, but it is about algebra,
making it harder to connect with the idea of constant rate of change. You
should become able to think both ways.
Determining Changes and Constant Rate of Change
Let f be a function having a constant rate of change with respect to
its independent variable x. The value of x changes from 7.2
to 9.35. What is the change in the value of x? What is the change in
the value of f? What is f’s constant rate of change with
respect to x?
Change in x: dx = 9.35 - 7.2
Change in f: df = f(9.35) - f(7.2)
Constant rate of change: $\frac{\mathrm{d}f}{\mathrm{d}x} = \frac{f(9.35) -
f(7.2)}{9.35 - 7.2}$
More generally, if x changes from a to b,
$a ≠ b$ then:
Change in x: $\mathrm{d}x = b - a$
Change in f: $\mathrm{d}f = f(b) - f(a)$
Constant rate of change: $\frac{\mathrm{d}f}{\mathrm{d}x} = \frac{f(b) -
f(a)}{b - a}$
Notice that we answered these questions without knowing the definition of f.
Notice also that we assumed nothing about whether $a < b$ or
$b < a$. The statements are true either way.
Connecting Constant Rate of Change to Cartesian Graphs
The animation in Figure 3.15.3 is similar to that in Figure 3.15.2. It shows
an initial value of y (labeled b) and changes in y
that are m times as large as changes in x as the value of x
varies. Figure 3.29 then shows the upper bar rotating so that it rests upon
the value of x as it (the value of x) varies. This is like
plotting the value of y above a value of x in a rectangular
coordinate system. The result is that the graph of $y
= mx + b$ is a line in a rectangular coordinate system.
Figure 3.15.3. Changes in y being proportional to changes in x creates
linear graphs in a rectangular coordinate system.
Exercise Set 3.15
- Turtle and Rabbit ran a race. The graph on the right shows the total
distance they had run at each moment in time while running.
- Estimate Rabbit’s speed over and speed back. Zoom the animation
to full screen if it is too small as shown.
- Estimate Turtle’s speed over and speed back. Zoom the animation
to full screen if it is too small as shown.
- At what speed must Rabbit travel back so that the two tie?
- Type $r = 2θ$ in GC. Print its displayed graph. On the
printed graph, add details that illustrate that r changes at a
constant rate of 2 units per radian with respect to θ.
- A function g has a constant rate of change of 2.25 with
respect to its independent variable. The point (3.25, 4.3125) is on g’s
graph. Use the fact that g has a constant rate of change to give
three other points on g’s graph.
- A function h has a constant rate of change with respect to its
independent variable. The points (1.5, -0.375) and (4.2, 4.35) are on h’s
graph. Use the fact that h has a constant rate of change to give
three other points on h’s graph. How does your strategy build
from your strategy for question 3?
- Define a function that has a constant rate of change of 0.75 with
respect to its independent variable and passes through the point (0, 2)
in the polar coordinate system.
- The animation below illustrates how to answer questions 6a - 6f.
Complete 6a - 6f similarly.
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and perform the requested task.
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and perform the requested task.
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and perform the requested task.
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and perform the requested task.
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and perform the requested task.
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and perform the requested task.
- A function k has a constant rate of change of 2.5 with
respect to values of the function d, where d(x) = 1 + ln(x),
x > 0. Also, k has a value of 3 when d
has a value of 2. Express k as a function of x.