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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.

**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?

**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?

**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?

The functions in Scenarios I-III give the rate of change of another function at every moment of their independent variables. They are called

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.

When we say that $r(x_0)$ is the momentary rate of change of

As a technical matter, we cannot allow d

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

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

In the previous sentence we were required say “the actual change in

In Section 4.4 (Rate of Change at a Moment) we emphasized that a function

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

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 d
*x*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 d*x*varies from 0 to 0.0001 is essentially $w(1.25) + \mathrm{d}y$, or $w(1.25) + r(1.25)\mathrm{d}x$.

- 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.

- d
*x*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 d*x*varied from 0 to 0.0001, $x = 1.25 + \mathrm{d}x$

- d
*y*was the change in*w*as d*x*varied from 0 to 0.0001.

- 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 d*x*varies through the interval from 0 to ∆*x*, $x = x_0 + \mathrm{d}x$ for all values of d*x*.

- 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 $0 ≤ \mathrm{d}x ≤ ∆x$.

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.

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

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.

b) What is

c) $g(4.5) = 419.625$. Approximately how much will the value of

d) Approximately how much will the value of

e) Revise your estimate in part d by estimating how much

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

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?

b) Revise your definition of

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