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# Section 4.9 Exact Rate of Change Functions

Here are three scenarios where all we know about a quantity is its rate of change at a moment. We do not know how much of the quantity there is.

### Scenario I

Companies with fleets of drivers often keep records of each driver’s speed (in km/h) at successive moments of elapsed time during the day. But they do not record the number of km that the car has traveled at any moment during the day. Let r be the function that gives a drivers speed (rate of change of distance traveled with respect to time) at each moment in time during the day. Anyone looking at these records would know the the value of $r(t)$, the rate at which a driver’s distance changed with respect to time, at values of t. Let d be the function that relates a driver’s distance traveled that day since the day began and let t represent the number of hours since the work day began. Then a value of $r(t)$ gives the rate of change of the driver’s distance traveled that day (in km/h) at the moment that t hours has elapsed in the work day.

Question: Could we reconstruct the driver’s total number of km traveled from the beginning of the day up to any moment in time during that day just by knowing the car's speed at each moment during that day and the number of hours that have elapsed up to that moment?

### Scenario II

Objects near Earth’s surface accelerate due to gravity at -9.8 $\frac{m/s}{s}$, meaning that its velocity (in m/s) changes at a constant rate of -9.8 m/s every second. This number ignores air resistance (addressed in a later chapter). Gravitational acceleration being -9.8$\frac{m/s}{s}$ means that while an object falls after being released, its upward velocity increases -9.8 m/s every second. So its velocity will be $v(t) = -9.8t$ m/s, where t is the number of seconds since being released. Let $d(t)$ be the distance that the object has fallen in t seconds after it was released. Then each value of $v(t)$ gives the rate of change of d with respect to time at the moment it has fallen t seconds. That is, after an object is released we can know its rate of change of distance with respect to time at any moment during its fall, but we do not know how far it has fallen up to that moment.

Question: Could we determine how much the object’s distance from its release point has changed at every moment during its fall just by knowing its acceleration due to gravity and the number of seconds that it has fallen up to that moment?

### Scenario III

Let w be a function that gives the total mechanical work accomplished at a moment in time during one cycle of pulling an amusement park’s slingshot sling backward before it is released (Figure 4.9.1). We are given that each value of the function $r_w(t) = 500\cos(\frac{t}{5}), 0 ≤ t ≤ 15.7$ sec, gives the momentary rate of change with respect to time of $w(t)$ in Newton-meter/sec.

Question: Could we determine how much the applied mechanical work changed from the beginning of a cycle up to any moment within that cycle just by knowing how fast this work changed with respect to time at each moment and the number of seconds that have elapsed up to that moment?

Figure 4.9.1. The sling, holding riders, after being pulled back and released. The sling is thrown into the air by the tension in the wires and towers that accrued due to the sling being pulled backward.

### Common Theme Across Scenarios

In each of the three scenarios, we knew the rate at which one quantity changed with respect another at every moment of the independent variable, but we did not know how much there is of either quantity. The question in each scenario is whether it is possible to recover the total change in a quantity up to a particular moment of its independent variable just by knowing its rate of change at every moment of its independent variable. As Chapter 5 will explain, the answer, in principle, is “yes”.

The functions in Scenarios I-III give the rate of change of another function at every moment of their independent variables. They are called exact rate of change functions. Every value of an exact rate of change function gives the rate of change at a moment for another function.

In preparation for Chapter 5, we will take a closer look at what it means that one function gives the rate of change of another function at every moment of the other function’s independent variable.

## The Meaning of “Essentially Equal To …”

The idea of a number L being essentially equal to a number represented by a sequence of numbers is that you can make the difference between L and values of all but a finite number of terms in the sequence as small as you please. In the sequence $1.9, 1.99, 1.999, …$, we can find a term in this sequence so that all terms in the sequence after it are closer to 2 than any difference we set. If we want all numbers in this sequence whose difference with 2 is within 0.0001, pick a term in the sequence that is larger than 1.9999. All terms after that one will be within 0.0001 of 2. If we want all numbers in this sequence whose difference with 2 is within 0.00000001, pick a term in this sequence that is larger than 1.999999999. All terms after that one will be within 0.00000001 of 2. So we say that 1.999… (infinitely repeating sequence of 9’s) is essentially equal to 2. Every term in the sequence 1.9, 1.99, 1.999 etc. is approximately equal to 2, but the number represented by the infinite sequence of terms itself is essentially equal to 2.

When we say that $r(x_0)$ is the momentary rate of change of f at $x_0$, we mean that the value of $f(x_0 + \mathrm{d}x)$ is essentially equal to the value of $f(x_0) + r(x_0)\mathrm{d}x$, and by “essentially equal to” we mean that the difference between $f(x_0 + \mathrm{d}x)$ and $f(x_0) + r(x_0)\mathrm{d}x$ as dx varies is less than any desired difference that could matter to us. If it mattered that the difference be smaller, we would make the interval through which dx varies smaller.

As a technical matter, we cannot allow dx to be 0. This is because we want to be able to be able to say that $\frac{\mathrm{d}y}{\mathrm{d}x}$, the quotient of the differential in x and the associated differential in y, gives us the same information as $\mathrm{d}y=m\mathrm{d}x$. If we allow dx to be 0, then $\frac{\mathrm{d}y}{\mathrm{d}x}$ is meaningless.

We shall use the symbol “≛” to represent the relation “essentially equal to”. Thus, we would write 1.999… ≛ 2. We could also write, “If values of $r(x)$ give the momentary rate of change of $f(x)$ for all values of x, then for $x = x_0$, and for sufficiently small nonzero values of dx, $f(x_0 + \mathrm{d}x) ≛ f(x_0) + r(x_0)\mathrm{dx}$.

## The Meaning of a Value of an Exact Rate of Change Function

Scenario III included a definition of the function $r_w$ as $r_w(t) = 500\cos(\frac{t}{5}), 0 ≤ t ≤ 15.7$ sec. Read “$r_w(t)$” as “r sub w of t”, which is short for “the momentary rate of change (r) of (the function) w at a value of t”. The conceptual definition of $r_w$ was that each value $r_w(t)$ for a specific value of t gives the momentary rate of change of the amount of mechanical work accomplished in pulling the sling backward before releasing it. What does that mean?

First, remind yourself of what it means that a function has a rate of change at a moment $x_0$. It means that the function has a rate of change that is essentially constant over an interval surrounding $x_0$.

Suppose that $t = 1.25$. Then $r_w(1.25) = 500\cos(\frac{1.25}{5})$, or $r_w(1.25) = 484.456$ Newton-meters/sec. Since any value of $r_w(t)$ is w’s rate of change at at the moment of a value of t, we are assured that for some interval containing $t = 1.25$, w changes at a rate that is essentially constant, and in this instance that rate of change is 484.456 Newton-meters/sec. At this rate of change, if dt, the change in t, were to vary from 0 to 0.0001, then dw, the change in the value of w, would vary from 0 to 484.456×0.0001 Newton-meters. Put another way, while we do not know the value of w(1.25), we do know that the actual change in w, $w(1.25 + 0.0001) ‑ w(1.25)$ is essentially equal to $484.456\cdot0.0001$ Newton-meters. So $w(1.25+0.0001) ≛ w(1.25) + r(1.25)\cdot(0.0001)$.

In the previous sentence we were required say “the actual change in w, $w(1.25 + 0.0001) ‑ w(1.25)$ is essentially equal to $r(1.25)\cdot(0.0001)$ Newton-meters”. This is because we are only assured that the rate of change of w with respect to t is essentially constant over some period of time from $t = 1.25$ to $t = 1.25+\mathrm{d}t$ seconds, and this period might be smaller than 0.0001 seconds. We can nevertheless say that if w were to change at a constant rate of $r(1.25)$ Newton-meters/sec over the time period $t = 1.25$ to $t = 1.2501$ seconds, then the value of w would increase by $r(1.25)\cdot(0.0001)$ Newton-meters.

In Section 4.4 (Rate of Change at a Moment) we emphasized that a function f having a rate of change at a moment $x_0$ meant that f’s rate of change over some interval containing $x_0$ is essentially constant, and hence f’s graph over that interval, in Cartesian coordinates, would be essentially linear. In the present situation, since w has $r_w(1.25)$ as its rate of change at the moment that $t = 1.25$, the graph of w would be essentially linear over sufficiently small intervals around $t = 1.25$. But the graph of w being linear over an interval means that the graph of $y = r_w(x)$ would be essentially constant over that same interval. So we can think of the graph of w being made of short, essentially straight line segments, and the graph of $r_w$ as being made of short, essentially flat line segments.

Figure 4.9.2 gives a visual illustration of the ways that all these meanings interact.

Figure 4.9.2. r(1.25) is the rate of change of w at the moment that x = 1.25. It is essentially constant as x varies from 1.25 to 1.2501. Therefore, y = w(x) is essentially linear as x varies from 1.25 to 1.2501.

There is a lot going on in Figure 4.9.2. Here is a description of what Figure 4.12 illustrates:
• The value of $r(1.25)$ is the momentary rate of change of w with respect to x at the moment that $x = 1.25$.
• “Rate of change at a moment” means that w has a rate of change that is essentially constant over some interval containing $x = 1.25$. This means that the function r, being the rate of change function for w, is essentially constant with a value of $r(1.25)$ over that interval.
• The left graph shows the graph of r as having essentially the constant value of $r(1.25)$ as x varies from 1.25 to 1.2501.
• The right graph assumes, for purposes of illustration, that $w(1.25) = 10.1$.
• As dx varies from 0 to 0.0001 the value of x varies from 1.25 to 1.2501. The essential (approximate) value of w varies at the constant rate of $r(1.25)$. Therefore the value of $w(1.25+\mathrm{d}x)$ as dx varies from 0 to 0.0001 is essentially $w(1.25) + \mathrm{d}y$, or $w(1.25) + r(1.25)\mathrm{d}x$.
It will be beneficial to remind ourselves of what w, r, 0.0001, dx, and dy represent about the scenario in which they occurred.
• The value of $w(x)$ gives the total mechanical work accomplished up to the moment of x seconds in one cycle of drawing the slingshot.
• The value of $r_w(x)$ gives the rate at which work changes with respect to time at the moment that x seconds have elapsed from the beginning of a cycle.
• The value of 1.25 seconds was one moment during the cycle.
• The value 0.0001 seconds was the width of the interval we decided was sufficient for $r(1.25)$ to be essentially constant.
• dx was the change in x as x varied through the interval from $x = 1.25$ to $x = 1.2501$. In other words, as the value of dx varied from 0 to 0.0001, $x = 1.25 + \mathrm{d}x$
• dy was the change in w as dx varied from 0 to 0.0001.
It is important to take note of something that can easily become lost. It is that despite not knowing anything about values of w or its definition, because we know w’s rate of change at each moment we have a good idea of how much it changes over every interval of sufficiently small width. In Chapter 5, we will use this important insight to re-create a function by knowing only its rate of change function.

#### We can generalize Figure 4.9.2 in several ways.

• Our assumption that $w(1.25) = 10.1$ was arbitrary. We assumed it simply so that we could start w’s graph somewhere. But the change in w would be essentially equal to $r_w(1.25)\cdot(0.0001)$ regardless of the actual value of $w(1.25)$.
• We could have picked an arbitrary value for x, such as $x = x_0$. The momentary rate of change of w with respect to x at $x = x_0$ would be $r_w(x_0)$.
• We could have picked an interval with a width other than 0.0001, such as a width of ∆x. So, as dx varies through the interval from 0 to ∆x, $x = x_0 + \mathrm{d}x$ for all values of dx.
• Our reasoning would have been the same. We would end with $w(x_0 + \mathrm{d}x) ≛ w(x_0) + r_w(x_0)\mathrm{d}x$ for $0 ≤ \mathrm{d}x ≤ ∆x$.

## Exercise Set 4.9

1. Values of the function f give the volume of concrete being added to a floor of a New York skyscraper during a 45-second pour. Values of the function $r_f$ defined as $r_f(x) = 2e^{\frac{-x}{15}}$ give the rate of change of f in $\mathrm{\frac{m^3}{sec}}$ at each moment of x, $0 ≤ x ≤ 45$ seconds.

a) Examine the graph of $y = r_f(x), 0 ≤ x ≤ 45$ in GC. Use the graph to describe the flow of concrete during a 45-second pour. Type “r ctrl-L f → ctrl-9 x = …” to define $r_f$ in GC. Type $y = r_f(x), 0 ≤ x ≤ 45$ to generate the graph. Scale the horizontal axis to see the entire graph.

b) At what rate is concrete being poured at the moment that 5.2 seconds has elapsed during the pour?

c) At 5.2 seconds of pouring, 8.79 $\mathrm{m}^3$ of concrete had been poured. Approximately how much concrete will have been poured at 5.4 seconds of pouring? Explain your work.

d) Approximately how much concrete will be poured between 5.2 and 6 seconds? Use one change of 0.8 seconds.

e) Revise your estimate in part d by estimating how much concrete is poured every 0.2 seconds from 5.2 seconds to 6 seconds.  Why are the estimates different? Hint: Examine the graph of $y = r_f(x)$.

f) You were told in part c that $f(5.2) = 8.79 \mathrm{m}^3$. Use your work in part e to sketch a graph by hand of $y = f(x), 5.2 ≤ x ≤ 6$. Scale your axes conveniently.

2. Values of the function $r_g$ defined as $r_g (x) = 100 - x^2$ give the rate of change of a function g at each moment of x, $0 ≤ x ≤ 10$.

a) Examine the graph of $y = r_g (x), 0 ≤ x ≤ 10$ in GC. Use the graph to describe behavior of g over that interval. Type “r ctrl-L g → ctrl-9 x = …” to define $r_g$. in GC. Type $y = r_g (x), 0 ≤ x ≤ 10$ to generate the graph. Scale the vertical axis to see the entire graph.

b) What is g’s rate of change when $x = 4.5$? What does this number mean?

c) $g(4.5) = 419.625$. Approximately how much will the value of g change as the value of x changes from $x = 4.5$ to $x = 4.6$?

d) Approximately how much will the value of g change as the value of x changes from $x = 4.5$ to $x = 5.5$? Use one change of 1 second.

e) Revise your estimate in part d by estimating how much g will change from $x = 4.5$ to $x = 4.6$, from $x = 4.6$ to $x = 4.7$, … , from $x = 5.4$ to $x = 5.5$. Why are the estimates different? Hint: Examine the graph of $y = r_g (x)$.

f) You were told in part c that $g(4.5) = 419.625$. Use your work in part e to sketch a graph by hand of $y = g(x), 4.5 ≤ x ≤ 5.5$. Scale your axes conveniently.

3. Define h in GC as $h(t) = 4\cos(2t)$. Graph $y = h(x)$.

a) Define a linear function f whose graph passes through the points $(3.01, h(3.01))$ and $(3.02, h(3.02))$. Graph $y = f(x)$. The graph of $y = f(x)$ should appear to be tangent to the graph of h at $x = 3.01$. Why?

b) Revise your definition of f so that it passes through the points $(n, h(n))$ and $(n + 0.01, h(n + 0.01))$. Graph $y = f(x)$. Click on “n” in the n-slider. Change the lower bound to -3 and the upper bound to 3. Click OK, then click n’s play button. What happens? Why?

c) What do part a and part b have to do with differentials and rates of change at a moment?