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In Chapter 4 we developed the idea that knowing a function’s rate of change at a moment of its independent variable gives information about how much the function varies during that moment:
If a function f has a momentary rate of change of $r_f(x_0)$ at the moment that $x=x_0$, then the change in f during that moment is $r_f(x_0)dx$ as $dx$ varies through that moment. We do not know the function’s value at any moment, but we have a good approximation of how much it varies during any moment. We will return to this idea repeatedly.
The concept of an accumulation function is quite general. When the measure of Quantity A varies at some nonzero rate of change with respect to the measure of Quantity B, Quantity A accumulates variations at a rate that is essentially constant in relation to sufficiently small variations in B.
This is the meaning of Quantity A having a rate of change at a moment at a value of Quantity B. Deriving a function whose values give the net accumulation of one quantity by knowing its rate of change at every moment of its accumulation addresses the first foundational problem of calculus. Namely,
You know how fast a quantity is varying at every moment; you want to know how much of that quantity there is at every moment.
When we know a function’s rate of change at every moment of its independent variable, we can estimate the function’s net change as its independent variable varies. We do this by accumulating variations in the function’s value as they occur while the value of its independent variable passes through successive moments.
An accumulation function is a function that is generated by accumulating variations in its value over moments of its independent variable.
The accumulation occurs by the function varying at some rate at every moment of its independent variable as its independent variable varies.
The remainder of this chapter will develop the idea of approximating an unknown accumulation function f when all we know about it is $r_f$, f’s exact rate of change function. We will then develop a method for creating these approximations so that GC can compute them automatically. Finally, we will introduce a representation of exact accumulation functions from rate of change functions that GC understands.
Reminder to Windows users: You cannot type "$\Delta x$", "$\Delta t$" etc. in GC. Instead, type "\Dx\", "\Dt\", etc. Windows gives no easy way to type the character "∆".
Reminder to Mac users: To type “$\Delta x$”, "$\Delta t$" etc. in GC, type "\optionj x\", "\optionj t\", etc. The keystroke optionj gives the character "$\Delta$" on a Mac.
A team of engineering students designed a rocketpowered basketball for their course project.
They designed the motor so that within the first 90 seconds after launch, the rocket will have a velocity of $t^{1.3}$ m/s t seconds after the motor starts. That is, they know that the rocket’s velocity function v will be $v(t)=t^{1.3},\, 0\le t \le 90$.
Figure 5.1.1 shows how the team anticipates the rocket’s velocity will vary with time.
The team must also predict the rocket’s altitude at each moment because they have maneuvers to program that are based on the rocket’s altitude.
Unfortunately for them, their rocket will have a clock, but not an altimeter. Their problem, then, is to develop a function h such that $h(t)$ will give the rocket’s height at each moment after the rocket is launched when all they can predict is the rocket’s velocity at each moment.
The solution to defining h would be simple if the rocket rose at a constant velocity k. The rocket’s altitude would be $h(t)=kt$. Unfortunately, the rocket’s velocity is not constant over any interval of time within 90 seconds after launch. On the other hand, its velocity is essentially constant over sufficiently small intervals of time within 90 seconds after launch.
The team cannot determine a definition of $h$. However, they can approximate values of $h(t)$ by drawing on their understanding of rate of change at a moment:
The team’s dilemma at this juncture is that they do not know the widths of the time intervals over which $v(t)$ is essentially constant. Were they to know these widths, they could get the total accumulated change in altitude from $t=0$ to any current value of t by adding all the variations in h that accumulate over these intervals of time.
However, they can experiment with values of $\Delta t$.
Suppose that h has essentially a constant rate of change over time intervals of length $$\Delta t$=0.1$ seconds. Let $A$ be the function whose values $A(t)$ approximate values of $h(t)$.
Over the first two seconds, the rocket would rise $$A(2)=v(0)(0.1)+v(0.1)(0.1)+v(0.2)(0.1)+...+v(1.9)(0.1)$$
A is the accumulation function that approximates the exact accumulation function h. We could write $A(2)$ with summation notation as
$$\color{red}{\text{(Eq. 5.1.1)}}\qquad A(2)=\sum_{k=0}^{19}v(k\Delta t)\Delta t$$
In principle, the team's approach to solving its problem starts by cutting up the time axis into intervals of width $\Delta t$ (generically, “$\Delta t$intervals”).
They will then assume that h has rates of change that are essentially constant over each $\Delta t$interval.
What value of v should they choose as the constant value of $v(t)$ as t varies through an interval?
Actually, there is no “should” in their choice. A better question is what would be a convenient choice for the value of t at which to evaluate $v(t)$ as the essentiallyconstant rate of change of h over the interval. They chose, for convenience, the value of v at the beginning of each $\Delta t$interval.
Figure 5.1.2 uses the ridiculously large value of 0.9 seconds as the width of each $\Delta t$interval. This is just to make it easy for you to visualize how this approach works.
You must examine Figure 5.1.2 in a particular way to understand all that is going on in it.
In Figure 5.1.2 we assume that the rocket’s altitude varies at a constant rate throughout each $\Delta t$interval, starting at $t=0$. We also assume that the constant rate over an interval of t is the value of v at the beginning of the interval.
GC’s graph on the right side of Figure 5.1.2 shows the accumulated variations that happened by assuming that h varies at a constant rate over each of its $\Delta t$intervals.
Our assumption that h varies at a constant rate over each $\Delta t$interval forces us to assume an approximate rate function r (left graph in Figure 5.1.2) which is constant over each $\Delta t$interval. All these assumptions produce a function A (for “approximate accumulation”). Values of A comprise an approximate altitude function that is composed of approximate variations in altitude.
Let r be the approximate rate function illustrated in Figure 5.1.2. This is the function that has a constant value as t varies through each $\Delta t$interval. We can define $r$ piecewise, as:
$$\color{red}{\text{(Eq. 5.1.2)}}\qquad r(t)=\begin{cases} v(0) &\text{if $0\le t \lt 0.9$}\\[1ex] v(0.9) &\text{if $0.9\le t \lt 1.8$}\\[1ex] v(1.8) &\text{if $1.8\le t \lt 2.7$}\\[1ex] &\text{... and so on}\end{cases}$$
Reflection 5.1.1. Let r be defined as in Equation 5.1.2. Complete this table.
t 
r(t) 
0.5  
0.6  
1.2  
1.3  
1.4  
2.71  
2.72  
2.95 
It is worthwhile to note that there are two types of variation happening in Figure 5.1.2: variation within a current $\Delta t$interval (the $\Delta t$interval containing the current value of t) and variation of the completed $\Delta t$intervals (the $\Delta t$intervals through which the value of t has passed). Figure 5.1.3 illustrates the distinction between completed and current $\Delta t$intervals.
Using values of r to compute approximate variations in the exact accumulated altitude function h, we get the function A, the accumulation function that approximates the exact accumulated variations in altitude:
$$\color{red}{\text{(Eq. 5.1.3)}}\qquad A(t)= \begin{cases} r(t)(t0) &\text{if $0\le t \lt 0.9$}\\[1ex] \left(r(0)\Delta t\right)+r(t)(t0.9) &\text{if $0.9\le t \lt 1.8$}\\[1ex] \left(r(0)\Delta t+r(0.9)\Delta t\right)+r(t)(t1.8) &\text{if $1.8\le t \lt 2.6$}\\[1ex] \left(r(0)\Delta t+r(0.9)\Delta t+r(1.8)\Delta t\right)+r(t)(t2.7) &\text{if $2.7\le t \lt 3.6$}\\[1ex]\text{... and so on}\end{cases}$$
There are two things to notice about Equation 5.1.3.
Keep in mind that the current value of t varies as time passes, so both "completed change" and "current change" vary as time passes. "Completed change" varies in chunks while "current change" varies continuously.
We use the letter "d" preceding a variable to mean that the variable's value "varies a little bit".
The way to envision a differential is to first keep in mind that a variable represents the value of a quantity whose value varies. Variables vary, always.
Another key idea in understanding differentials is that they are variables whose values vary "a little bit". To vary a little bit, they must vary from somewhere. We therefore think of differentials of an independent variable as varying through fixed intervals in the variable's domain.
Finally, a differential is a variable by which another variable varies. This surely sounds confusing. Figure 4.1.1 illustrates this idea. The term "left(x)" in the animation means the left end of the interval containing the current value of x.
Variables and Differentials. The value of x varies smoothly through its domain. Differentials in x (dx) are small variations in the variable's value. The symbols "dx" and "$\Delta x$" have differnt meanings.
As the value of t varies, it is always within some $\Delta t$interval. We will use "left(t)" to represent the number at the beginnning of the $\Delta t$interval containing the current value of t.
left(t) = the value of the left end of the $\Delta t$interval containing the current value of t.
Figure 5.1.4 illustrates the concept of left(t). It shows the independent axis partitioned into $\Delta t$intervals of size 0.9.
As t varies from 0, its current value is always within some $\Delta t$interval. The value of left(t) is the value of the left end of the $\Delta t$interval containing the current value of t. The value of $dt$ is the difference between the value of t and the value of left(t). Therefore $dt=t\mathrm{left}(t)$ as the value of t varies in the interval $\mathrm{left}(t)\lt t \le \mathrm{left}(t)+\Delta t$.
Another way to think of left(t) is given in Figure 5.1.4a. It shows that with a = 0 and $\Delta t$ = 0.9, every real number within a ∆ tinterval is mapped to the number that begins that interval. Figure 5.1.4b shows the same thing except that it leaves a record of how values of t are mapped to values of left(t).
Figure 5.1.4a. The top number line shows values of t varying continuously from 0 to 3.5. The bottom number line shows values of left(t) as t varies continuously.
Move your pointer away from the animation to hide its control bar.
Figure 5.1.4b. The top number line shows values of t varying continuously from 0 to 3.5. The bottom number line shows values of left(t) as t varies continuously. The video shows the record of the mapping $t \to \mathrm{left}(t)$.
Move your pointer away from the animation to hide its control bar.
We can now summarize Equation 5.1.2 in one line: $$\color{red}{\text{(Eq. 5.1.4)}}\qquad r(t)=v(\mathrm{left}(t))$$
Now that we have left(x) defined, we can summarize Equation 5.1.3 with one line:
You must understand all that is going on in Figure 5.1.5 in order to represent what is going on symbolically.
Reflection 5.1.2. The definition of A given above defines the upper limit of the sum conceptually. It does not use a formula for actually calculating an upper limit. Does this matter in terms of defining the function A? Explain.
Reflection 5.1.3. The definition of A uses the function named "left" even though left(x) is still not defined computationally. Does this matter in terms of understanding how the definition of A is supposed to work? Explain.
Reflection 5.1.4. Why is it important to use "number of complete ∆tintervals from a to t" as the upper limit of the summation?
Reflection 5.1.5. How does the term $r(t)(t\mathrm{left}(t))$ in Equation 5.1.5 represent "accumulation so far in the current $\Delta x$interval"?
For Questions 1  3. Consider the video below. It shows a ball hanging from a board by a rubber cord. The ball hangs at rest 10 feet below the board. At $x = 0$ seconds, the ball is set in motion with a quick shove downward, causing the rubber cord to stretch and the ball to bounce. The graph shows the rate of change of the ball's displacement from rest with respect to elapsed time, which is the velocity function $v(x) = \mathrm{cos}(x)$.
Value of a 
Value of x 
∆x  No. of complete
∆xintervals between a and x 

a)  2  17  0.03  
b)  3.2  5.23  0.0125  
c)  3  1.2  0.025  
d)  1  5  1.2  
e)  5  128.3  0.2 
We made a mistake on the velocity function! It is actually $$v(t) = 15 \frac{2^{t10}}{2^{t10}+1}.$$
a) What needs to be changed in the system of statements that calculates approximations of the rocket’s height for values of elapsed time?
b) What would the graph of elevation relative to elapsed time look like now?