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from Exact Rate of Change Functions

It is worth considering the distinction between accumulation and net
accumulation. If *f* is an accumulation function, then *f*(*x*)
for any value of *x* gives the measure of the accumulating
quantity at that value of *x*. All of it. In the situations we
have dealt with, we developed a method for approximating the accumulation
of quantities from some starting point. These approximations are the *net*
accumulation in the quantity -- the accumulation from the reference point
to a value of *x*, not the entire accumulation up to the value of
*x*.

Put another way, the accumulation in a quantity from a starting point *a*
to a value of *x* of its independent variable is the quantity's *net*
accumulation, or net change in accumulation, in relation to its exact
amount. The function *A* that we developed in Section 5.1
approximates the net change in *f* from *a* to *x*.
It does not approximate the value of *f* at *x.*

Put another way, the functions we defined for the rocket basketball can be used in any situation where there is an unknown exact accumulation function

Given:

- The exact rate of change function $r_f$, every value of which is a
rate of change at a moment of the accumulation function
*f*, - The value of
*a*, the value from which we begin recording changes in*f*, - A value of $\Delta x$, the width of the intervals into which the independent variable is partitioned, then

$$r(x)=r_f(\mathrm{left}(x))$$*Equation
5.2.1. The function r approximates the function $r_f$.*

Our purpose for defining

With

It is convenient to let the value of

Equation 5.2.2 puts all this together.

Each value $A(x)$ is an approximation to the net change in the value

However, we are missing two essential details in order to have GC compute values of

- How to define left(
*x*) in GC for any value of*x*when given values of*a*and $\Delta x$.

- How to compute the number of complete $\Delta x$-intervals between
*a*and the current value of*x*. We need this computation so that GC can compute the upper limit to the summation in Equation 5.2.2.

As a reminder, Equations 5.2.1 and 5.2.2 are the equations that defined an approximate net accumulation function

The second key idea is that a quotient of two numbers tells us how many times as large the numerator is than the denominator. The quotient $\frac{15}{3}$ being 5 tells us that 15 is 5 times as large as 3, or that there are five 3’s in 15. Similarly, $\dfrac{33}{7}=4+\dfrac{5}{7}$ tells us that there are four complete 7’s and $\frac{5}{7}$ of 7 in 33. So 33 contains four complete intervals of size 7.

More generally, the expression $\dfrac{x-a}{\Delta x}\text{, }x \ge a$
computes the number of intervals of size $\Delta x$ that are contained in
the interval from *a* to *x*, including fractional parts
of $\Delta x$.

**In GC, type “floor” to get the floor
function.**

$$\left\lfloor\frac{x-a}{\Delta x}\right\rfloor\text{ if } x \ge a$$

Equation 5.2.2 can now be written completely as:

1. Enter $\lfloor \frac{b-a}{\Delta x} \rfloor$ into GC by typing "floor (x - a) / $\Delta x$". For each row in the table below, set the values of

a |
b |
$\Delta x$ |
$\left\lfloor\dfrac{b-a}{\Delta
x}\right\rfloor$ |
What this value means |

2 | 27 | 0.03 | 833 | There are 833 complete intervals of size 0.03 within the interval that extends from 2 to 27 |

-1.4 | 3.7 | 0.002 | ||

2 | -1.5 | 0.07 | ||

17.33 | 35.24 | 0.012 |

The general definition of left(

$$\mathrm{left}(x)=a+{\Delta
x}\left\lfloor\dfrac{x-a}{\Delta x}\right\rfloor\text{ if }x \ge a$$

To enter the statement $\mathrm{left}(x) = a+\Delta x \left\lfloor \dfrac{x-a}{\Delta x} \right\rfloor \text{ if } x \ge a$ as a function definition in GC, type: \left\ ctrl-9 x = a + \option-j x\ * floor (x - a) / \∆x\ if x ctrl-shift -> a (in Windows, type \Dx\ instead of \option-j x\) |

Equation 5.2.5 in words is, “The value of left(

Here is a YouTube video that gives another explanation of the motive for left(

$$\left\lfloor \dfrac{4.72 - 2}{0.01} \right\rfloor = 2.71$$

while it will also say that

$$\left\lfloor \dfrac{2.72}{0.01} \right\rfloor = 2.72$$

The two

The problem is not with GC. The problem is that computers use a finite number of digits to represent any real number. Click here to see the explanation.

Given an exact rate of change function $r_f$ whose values give the rate of change at every moment of an accumulation function

With the system of statements in Equation 5.2.6, we can have GC approximate any net accumulation function just by knowing its exact rate of change function.

$$A(x)=\left(\sum_{k=1}^{\left\lfloor \frac{x-a}{\Delta x} \right\rfloor} r_f(a+(k-1)\Delta x) \right)+r(x)\left(x-\mathrm{left}(x)\right) \text{ if }x \ge a$$

instead of

$$A(x)=\left(\sum_{k=1}^{\left\lfloor \frac{x-a}{\Delta x} \right\rfloor} r\left(a+(k-1)\Delta x \right) \right)+r(x)\left(x-\mathrm{left}(x)\right) \text{ if }x \ge a.$$

Are they correct? Explain. Hint: Examine $r\left(\mathrm{left}(x)\right) = r_f\left(\mathrm{left}\left(\mathrm{left}(x)\right)\right)$ for any value of

$$A(x)=\left(\sum_{k=1}^{\left\lfloor \frac{x-a}{\Delta x} \right\rfloor} r\left( a+(k-1)\Delta x \right) \right)+r_f(x)\left(x-\mathrm{left}(x)\right) \text{ if }x \ge a$$

instead of

$$A(x)=\left(\sum_{k=1}^{\left\lfloor \frac{x-a}{\Delta x} \right\rfloor} r\left( a+(k-1)\Delta x \right) \right)+r(x)\left(x-\mathrm{left}(x)\right) \text{ if }x \ge a.$$

Are they correct? Explain.

Click on a point in GC’s displayed graph. Record its coordinates. Then interpret the point in terms of the meanings of its coordinates.

The displayed graph in Figure 5.2.2 shows a particular highway's rate of change of elevation over a stretch of road beginning at the Utah-Colorado state border. It shows the highway's rate of change of elevation (in meters of elevation per 100 meters of highway) at every distance from the border up to 14 km from the border. This rate of change of elevation map is modeled by the function $r_E(x) = 3 \cos (\frac{x}{2}) e^{\sin(\frac{2x}{3})}$, where a value of

Study the displayed graph of $r_E$. Its behavior by itself tells us about how the elevation changes with respect to distance from the border. Anywhere that $r_E(x)>0$, the highway’s elevation is increasing. Anywhere that $r_E(x)<0$, the highway’s elevation is decreasing. The highway’s elevation increases more rapidly wherever $r_E$ increases. The highway’s elevation decreases more rapidly wherever $r_E$ decreases.

Our goal is to define a function that approximates $E(x)$, the road’s exact elevation

Determining elevation in this example means to compute accumulated changes in elevation, in meters, from the rate at which the elevation is changing with respect to distance along the road (in $\mathrm{\frac{m}{100m}}$). We can use the general method developed in Equation 5.2.6 to define a function that approximates the exact elevation function, which we will call

Equation 5.2.7 (below) is the result of specializing Equation 5.2.6 to the current situation.

- We used $\Delta x=0.01$, which is $\mathrm{\frac{1}{100} km}$, or 10 meters. This seems a safe size for $\Delta x$-intervals because the elevation will change approximately linearly in any 10-meter segment of the highway.
- We used $a=0$ because we are accumulating changes in elevation starting at 0 km from the Utah-Colorado border.
- We multiply by 10 in the definition of
*A*because the unit of $r_E$ is $\mathrm{\frac{m}{100m}}$, but the distance on the roadway is measured in km (1 km is 10 times as long as 100m). So the rate of change of elevation*x*km from the border, in $\mathrm{\frac{m}{km}}$, is $10r_E(x)$. - Finally, we add 1243 because all accumulated changes in elevation are changes from an initial elevation of 1243 meters.

Figure 5.2.3 displays the graph of the approximate accumulation function

The definition of

- If we want to know the change in elevation from $x=1.2$ km to $x=3.7$ km from the border, we just enter the expression $A(3.7)-A (1.2)$ (after entering all other statements).
- GC will report that $A(3.7) - A (1.2) = 60.244$, which means that the net change in elevation from 1.2 km to 3.7 km from the border is 60.244 meters.
- Put another way, $A(3.7)-A (1.2) = 60.244$ means that the highway is 60.244 meters higher at 3.7 km than it is at 1.2 km.

2. Enter $\dfrac {A(2.61)-A(2.6)}{2.61-2.6}$ on its own line in GC. Determine the point on the graph of $r_E$ that has a

3. Explain what the circled expressions mean in terms of the quantities they evaluate. Click here to get a page that you can print.

4.

a) Graph the ball’s rate of change function
in GC. Interpret the graph in terms of the ball’s motion.

b) Click on a point on GC’s display of $y=r_g(x)$. Record the point’s coordinates, then interpret their meanings in regard to the ball's motion.

c) Interpret the graph again, but this time interpret it in terms of what*each point* on the graph represents about the ball's motion.

d) Define a function*A* whose values approximate the values of *g*.
Enter $y’=A(x’)$ in GC to display the graph of *A* alongside the
graph of $r_g$.

b) Click on a point on GC’s display of $y=r_g(x)$. Record the point’s coordinates, then interpret their meanings in regard to the ball's motion.

c) Interpret the graph again, but this time interpret it in terms of what

d) Define a function

i. Is GC’s graph of displacement with
respect to time consistent with what you anticipated? If not, explain the
error in your reasoning.

ii. Is GC’s graph of A consistent with the graph of $r_g$? Explain.

iii. Click on a point on GC’s display of $y’=A(x’)$. Record the point’s coordinates, then interpret each coordinate's meaning in regard to the bouncing ball.

iv. Click on a point on GC’s display of $y=r_g(x)$. Record the point’s coordinates, then interpret each coordinate's meaning in regard to the graph of $y’=A(x’)$.

ii. Is GC’s graph of A consistent with the graph of $r_g$? Explain.

iii. Click on a point on GC’s display of $y’=A(x’)$. Record the point’s coordinates, then interpret each coordinate's meaning in regard to the bouncing ball.

iv. Click on a point on GC’s display of $y=r_g(x)$. Record the point’s coordinates, then interpret each coordinate's meaning in regard to the graph of $y’=A(x’)$.

5. Use your GC file from Exercise 4 for this question. Define the function