Section 1.5 Constant Rate of Change
The concept of constant rate of change is central to all of calculus.
This will become evident in Chapters 4 and 5. In this section 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
smoothly and continuously.
Two quantities change at a
constant rate with respect to each other if, and only 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 otherno matter how large or small the
changes.
Let x represent the value of Quantity A and let y
represent the value of Quantity B.
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. We will use the letter "d"
preceding a variable to mean "a little change" in that variable's value.
Suppose that y changes at a constant rate with respect to x.
Then dx is a change (large or tiny) in the value of x
and dy is the change in the value of y that occurs as x
changes by dx.
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 over Tiny Intervals
The top bar in Figure 1.5.1 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 1.5.1 represents the value of x.
The value of y changes at a constant rate 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 or the value of dx.
Notice also that the value of dx grows within changes.
Changes are not added in chunks. 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 1.5.1. 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 1.5.2., 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 and no matter how large or small the change in x.
Move your pointer away from the animation to hide the control bar.
Figure 1.5.2. 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.
Move your pointer away from the animation to hide the control bar.
Reflection 1.5.1. Some people think that Figures
1.5.1 and 1.5.2 say the same thing. Others disagree. Why might each
group of people think as they do about Figures 1.5.1 and 1.5.2?
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 1.5.1. Changes in t from an initial value of t
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
$D=113(t3.5)+32$ 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. The number of
hours that Stan had driven at a constant speed at moments in time since
starting for Tuscon is therefore $(t3.5)$ hours.
So it only for $3.5 \le t \le 5.1$ that $D=113(t3.5)+32$ 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 1.5.2. Describe similarities and
differences between this analysis of Stan’s trip and Figures 1.5.1 and
1.5.2.
Example 1.5.2
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.
 How much will my distance from home change 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 t minutes?
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 $dt = t$,
$dD = (0.35)t$.
 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. I am now
3.8 km from home. So I was $3.8 + (0.35)(7)$ km from home when I
first looked at my watch. (Why must we add $0.35(7)$ to 3.8?)
 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
$D = 0.35(t7) + 3.8$.

Generalize a through c: The value of y changes at a constant
rate m with respect to changes in the value of x. The
value of y is $y_0$ when the value of x is $x_0$.
Define a function that relates values of y to their
corresponding values of x.
Changes in y and x are proportional because values of y change at a
constant rate with respect to changes in the value of x.
The change in the value of x from $x_0$ to its current value is
$(xx_0)$. So $m(xx_0)$ is the change in y that
corresponds to the change $(xx_0)$.
When the value of x is $x_0$, the value of y is $y_0$.
Therefore, the value of y for any value of x is the change in y that
is due to the change in x from $x_0$ plus the $y_0$, the initial value
of y. Put another way, the value of y for any value of x when y
changes at a constant rate with respect to x is$$y
= m(xx_0) + y_0.$$
Put another way, 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 its current value. So the function is $y
= m(xx_0) + y_0$.
Again, how you read Examples 1.5.1 and 1.5.2 is important. Read
them again 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! 
Determining a Constant Rate of Change
Let y represent the value of Quantity A and let x represent
the value of Quantity B. Suppose that we are told that Quantity A changed by
dy kg while Quantity B changed by dx liters, $\mathrm{d}x
\neq 0$, and that they changed at a constant rate of m with respect
to each other. 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 (Quantitative): 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 dx liters. Thus, for the
value of x to change by 1 liter, it would change by
$\left(\dfrac{1}{\mathrm{d}x}\text{ of }\mathrm{d}x\right)$ liters.
 Since changes in x and y are proportional, y
would change by $\left(\dfrac{1}{\mathrm{d}x}\text{ of
}\mathrm{d}y\right)$ kg when x changes by
$\left(\dfrac{1}{\mathrm{d}x}\text{ of }\mathrm{d}x\right)$ liters.
 But $\left(\dfrac{1}{\mathrm{d}x}\text{ of }\mathrm{d}y\right)$ is
$\dfrac{\mathrm{d}y}{\mathrm{d}x}$.
 Therefore y changes
$\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)$ kg per liter, or
$m = \dfrac{\mathrm{d}y}{\mathrm{d}x}$.
Reflection 1.5.3. Repeat the above chain of reasoning
with $dx=0.01$ and $dy=0.13$. Use division to evaluate the quotients.
Way of Reasoning #2 (Algebraic): 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 dx kg. y changed by
dy liters.
 Thus, $\mathrm{d}y = m\cdot \mathrm{d}x$.
 Therefore $m = \dfrac{\mathrm{d}y}{\mathrm{d}x}$.
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 Values of Changes and Values of Constant Rates of Change
Suppose that $y$ changes at a constant rate with respect to $x$, and that
the value of $y$ changes from 17.2 to 23.1 while 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 $y$?
 $\mathrm{d}y = 23.117.2$
 What is $y$'s constant rate of change with respect to $x$?
 Remember that $y$ changes at a constant rate with respect to $x$. So
$\mathrm{d}y$ = $m\cdot \mathrm{d}x$. So $(23.1  17.2)=m(9.357.2)$,
and therefore $$m=\frac{(23.1  17.2)}{(9.357.2)}.$$
 The constant rate of change of $y$ with respect to $x$ is
$\dfrac{(23.1  17.2)}{(9.357.2)}$, or approximately 6.93. That is, all
changes in $y$ are approximately 6.93 times as large as changes in $x$.
More generally: Suppose that the value of x
changes from $x_1$ to $x_2$, $x_1 \neq x_2$, while the value of $y$ changes
from $y_1$ to $y_2$, then:
 $\mathrm{d}x=x_2x_1$
 $\mathrm{d}y=y_2y_1$
 Constant rate of change: $$m=\frac{\mathrm{d}y}{\mathrm{d}x} =
\frac{y_2y_1}{x_2x_1}$$
Notice that we answered these questions without knowing how $x$ and $y$ are
actually related. Notice also that we assumed nothing about whether $x_1\lt
x_2$ or $x_2\lt x_1$. The constant rate of change of $y$ with respect to $x$
is computed by the formula $\dfrac{y_2y_1}{x_2x_1}$ either way.
Connecting Constant Rate of Change to Cartesian Graphs
The animation in Figure 1.5.3 is similar to that in Figure 1.5.2. 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 1.5.3 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 1.5.3. Changes in y being proportional to changes in x creates
linear graphs in a rectangular coordinate system. Move your
pointer away from the animation to hide the control bar.
Exercise Set 1.5
 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. Move
your pointer away from the animation to hide the control bar.
 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?

The variable g changes at a constant rate of 2.25 with
respect to the variable h. The point (3.25, 4.3125) is on
the graph of g relative to h. Use the fact that g
changes at a constant rate of with respect to h to give
three other points on the graph of g relative to h.
 A variable k changes at a constant rate respect to the
variable v. The points (1.5, 0.375) and (4.2, 4.35) are on
the graph of k relative to v. Use the fact that k
changes at a constant rate with respect to v to give three
other points on the graph of k relative to v.

The animation below illustrates how to answer questions 6a  6f.
Complete 6a  6f similarly. Move your pointer away from the
animation to hide the control bar.
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and perform the requested task.
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A topfuel dragster ran a 1/4mile (1320 ft) race. It had traveled
1305.48 feet after 3.58 seconds and it traveled the entire 1320 feet
in 3.61 seconds. What was its speed, approximately, in feet per second
at the end of the race? In miles per hour?