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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 other--no 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).
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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.
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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 $D=113(t-3.5)+32$ 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. The number of hours that Stan had driven at a constant speed at moments in time since starting for Tuscon is therefore $(t-3.5)$ hours.

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 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.
  1. 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 $dt = t$, $dD = (0.35)t$.

  2. 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?)

  3. 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  $D = 0.35(t-7) + 3.8$.
  4. 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 $(x-x_0)$. So
    $m(x-x_0)$ is the change in y that corresponds to the change $(x-x_0)$.

    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$$y = m(x-x_0) + y_0.$$
    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 $y = m(x-x_0) + y_0$.
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.

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.
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 .
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:
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 $\dfrac{y_2-y_1}{x_2-x_1}$ 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.
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Exercise Set 1.5

  1. 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.



    1. Estimate Rabbit’s speed over and speed back. Zoom the animation to full screen if it is too small as shown.
    2. Estimate Turtle’s speed over and speed back. Zoom the animation to full screen if it is too small as shown.

    3. At what speed must Rabbit travel back so that the two tie?
  2. 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.

  3. 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.
  4. 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.


    1. .
    2. Download this file and perform the requested task.

    3. Download this file and perform the requested task.

    4. Download this file and perform the requested task.

    5. Download this file and perform the requested task.

    6. Download this file and perform the requested task.

    7. Download this file and perform the requested task.
  5. A top-fuel dragster ran a 1/4-mile (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?