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8.5 Applications of Integrals in the
Physical and Social Sciences

It is common for people to think of the physical and social sciences as being in intellectual opposition. They think of the physical sciences as being about inanimate natural objects while thinking of the social sciences as being about interactions among living beings.

However, when you think of quantities changing at some rate with respect to each other, then the ideas of calculus apply whether the quantities are numbers of people having certain characteristics at various moments in time or the force of gravity acting on an object at different distances from Earth's surface.

If you can state a relationship between changes in quantities' values, then the ideas of calculus will apply.

It is imperative that you use GC to follow the examples given below. Define functions in GC, graph them as in the examples, and reflect on the meaning of the graphs' points in terms of the contexts in which they occur.

Example 1: Spread of infectious disease.

Disease A and Disease B spread at different rates. Disease A spreads slowly at first and more rapidly as time passes. Disease B spreads rapidly at first and less rapidly as time passes. Disease A spreads at a rate of $$r_A(t)=2000t^2$$persons/year for each value of $t$ in the first year. Disease B spreads at a rate of $$r_B(t)=1100\sqrt t$$persons/year for each value of $t$ in the first year. Figure 8.5.1 shows GC's graphs of the diseases' rate of change functions.


Figure 8.5.1. GC's graphs of the rate of change functions for Disease A and Disease B.

Question: Which disease infects the most people in the first year?

It seems intuitive that Disease A will infect more people in the first year because its rate of infection becomes so much greater than that of Disease B.

Now we'll take a closer look.

At any moment $t$, Disease A spreads at the rate of $r_A(t)$ persons/year. The number of people infected by Disease A over a small interval of time $dt$ (e.g., 0.01 years) at any value of $t$ is essetially $r_A(t)dt$. So the number of people in total who have been infected by Disease A from the beginning of the year up to $t$ years will be the accumulation of those incremental changes as time varies from $u=0$ to $u=t$, or $$A(t)=\int_0^t r_A(u)du.$$

Similarly, at any moment $t$, Disease B spreads at the rate of $r_B(t)$ persons/year. The number of people infected by Disease B over a small interval of time $dt$ (e.g., 0.01 years) at any value of $t$ is essentially $r_B(t)dt$. So the number of people in total who have been infected by Disease B from the beginning of the year up to $t$ years will be the accumulation of those incremental changes as time varies from $u=0$ to $u=t$, or $$B(t)=\int_0^t r_B(u)du.$$

Figure 8.5.2 shows the graphs of $y=A(x)$ and $y=B(x)$ for $0 \le x \le 1$. We can see that Disease B has infected more people at all moments in time during the first year despite the fact that, as we saw in Figure 8.5.1, Disease A's rate of change is much larger than Disease B's rate of change for most of the latter half of the year.


Figure 8.5.2. GC's display of the graphs of $y=A(x)$ and $y=B(x)$.


Enter definitions of $r_A,\, r_B,\, A,\, \text{ and }B$ into GC, and evaluate $A(1)$ and $B(1)$. You will see that, according to these models, at the end of the first year approximately 667 people were infected by Disease A and approximately 733 people were infected by Disease B.

Example 2: Toy auto racing.

Toy cars 1 and 2 start simultaneously from the same starting line. Car 1 accelerates rapidly early in the race and less rapidly later. Car 2 accelerates slowly at the beginning of the race and then rapidly later.

Car 1 accelerates at each moment of $t$ at the rate of $$a_1(t)=0.7t+\sin\left(\dfrac{3\pi t}{5}\right)\text{ (ft/sec)/sec.}$$

Car 2 accelerates at each moment of $t$ at the rate of $$a_2(t)=3-3\cos\left(\dfrac{2\pi t}{15}\right)\text{ (ft/sec)/sec.}$$

What are the car's relative speeds at each moment in time? Which car is going faster at the end of 8 seconds?
What are the car's relative distances at each moment in time? Which car has gone farther at the end of 8 seconds?

Figure 8.5.3 shows GC's display of graphs for Car 1's and Car 2's acceleration functions with respect to elapsed time. Car 1 accelerates more rapidly than Car 2 in the first part of their race, Car 2 accelerates more rapidly in the second part of their race, and so on.


Figure 8.5.3. GC's display of the acceleration functions for Car 1 and Car 2.

If we think carefully about Figure 8.5.3 in terms of what is happening to cars' velocities, then we can see that since Car 1 starts with higher accelerations, it will have greater velocity than Car 2. Even when they have equal accelerations at about $t=1.9$ seconds, Car 1 will have a higher velocity. So it will take some time after the cars first have equal accelerations that their velocities will be the same, but we cannot tell when. Also, it is not clear from these graphs which car is ahead at the ends of the second, third, and fourth parts of their race.

We need to see their velocities at each moment in time.

Acceleration is a rate of change of velocity with respect to time. To derive velocity from acceleration, we note that at any moment $t$, a car is accelerating at $a(t)$ (ft/sec)/sec over a tiny interval of length $dt$ seconds, so at $t$ seconds a car's velocity will change by $a(t)dt$ ft/sec.

Let values of the function $v_1$ at a value of $t$ be the accumulated velocity $v_1(t)$ of Car 1 after $t$ seconds have elapsed. Values of $v_1$ will therefore be $${\displaystyle v_1(t)=\int_0^t a_1(u)du}.$$

Let values of the function $v_2$ at a value of $t$ be the accumulated velocity $v_2(t)$ of Car 2 after $t$ seconds have elapsed. Values of $v_2$ will therefore be $${\displaystyle v_2(t)=\int_0^t a_2(u)du}.$$

Figure 8.5.4 shows GC's graphs of $y=v_1(x)$ and $y=v_2(x)$. As elapsed time increases, Car 1 goes at a higher velocity than Car 2, then Car 2 goes at a higher velocity than Car 1. The graphs in Figure 8.5.4 allow us to answer the question of which car is going faster at the end of 8 seconds. Car 2 is going faster.


Figure 8.5.4. GC's display of the velocity graphs $y=v_A(x)$ and $y=v_B(x)$.

You can find out how fast each car is going after 8 seconds have elapsed by entering the definitions of $a_1$, $a_2$, $v_1$, and $v_2$ into GC. Then, when you enter $v_1(8)$ on its own line, GC will display its value. When you enter $v_2(8)$ on its own line, GC will tell you its value.

Enter $v_2(8)-v_1(8)$ to have GC tell you how much faster Car 2 is going than Car 1.

Have GC graph $y'=v_2(x')-v_1(x')$ to see their relative speeds at each moment in time in a new graphing pane. (Remember, prime notation to GC just means to display a graph in a new pane. It does not mean "derivative".) You might need to re-scale the y-axis to see the difference more clearly.

Compare Figures 8.5.3 and 8.5.4 in terms of when the cars have equal accelerations and when they have equal velocities. After they start, the cars' accelerations are equal for the first time at about 1.9 seconds, but their velocities are equal for the first time at about 5.6 seconds. Study each figure in terms of what it represents (acceleration at a moment or velocity at a moment) so that this fact makes sense to you.

We still have not determined which car is ahead after 8 seconds have elapsed. We need to see their distances traveled at each moment in time.

Velocity is a rate of change of displacement with respect to time. Since the cars move in one direction, displacement and distance are the same thing. To derive distance from velocity, we note that at any moment $t$, a car is traveling at $v(t)$ ft/sec over a tiny interval of length $dt$ seconds, so at $t$ seconds a car's distance will change by $v(t)dt$ ft.

Let values of the function $s_1$ at a value of $t$ be the accumulated distance $s_1(t)$ after $t$ seconds have elapsed. Values of $s_1$ will therefore be $${\displaystyle s_1(t)=\int_0^t v_1(u)du}.$$

Let values of the function $s_2$ at a value of $t$ be the accumulated distance $s_2(t)$ after $t$ seconds have elapsed. Values of $s_2$ will therefore be $${\displaystyle s_2(t)=\int_0^t v_2(u)du}.$$

Figure 8.5.5 shows GC's display of the graphs of $y=s_1(x)$ and $y=s_2(x)$. Surprisingly, after they start, Car 1 is ahead of Car 2 throughout the race despite Car 2 going faster than Car 1 at the end of the race.

Figure 8.5.5. GC's display of the distances that Car 1 and Car 2 had traveled at each moment of elapsed time.

Study Figures 8.5.3, 8.5.4, and 8.5.5 in terms of what each represents (acceleration at a moment, velocity at a moment, distance at a moment) so that it makes sense to you that, once they start, Car 1 is always ahead of Car 2.

A Digression

You should have entered all the function definitions for Example 2 and generated all the graphs in Figures 8.5.3, 8.5.4, and 8.5.5 to follow the discussions. If you did not do this, then do it now.

Figure 8.5.6 repeats what you saw (assuming you did) when graphing $y=s_1(x)$ and $y=s_2(x)$. It shows that GC takes a long time to display these graphs. Why?

Figure 8.5.6. GC takes a LONG time to construct the graphs of $s_1$ and $s_2$.

Think about that question ("Why?") before reading on.

For GC to calculate a value of, say, $s_1(4.1)$ it must calculate bits of distance $v_1(t)dt$ for many values of $t$ from 0 to 4.1. But to calculate a value of $v_1(t)dt$ for any value of $t$, GC must also calculate many values of $a_1(u)du$ for values of $u$ ranging from 0 to $t$.

GC must repeat this double-variation process to plot the points associated with all (actually, a large sample) of the values of $x$ from $x=0$ to $x=8$. So for GC to calculate values of $v_1(t)$, GC repeatedly summed an ever increasing number of calculations of $a_1(u)du$ as $u$ varied from 0 to the value of $t$. To calculate values of $s_1(x)$, GC repeatedly summed an ever increasing number of calculations of $v_1(t)$ as the value of $t$ varies from 0 to $x$. All this must happen as $x$ varies from 0 to 8.

Figure 8.5.7 illustrates the above explanation graphically. Suppose that GC computes 2000 incremental values to compute the value of $s_1(4.1)$. Suppose further that GC computes 1500 incremental values to compute each incremental value in the computation of $s_1(4.1)$, and that GC computes 3000 incremental values to compute each of those incremental values. This would explode into $2000\cdot 1500\cdot 3000=9\cdot 10^9$, or 9 billion incremental values to compute the value of $s_1(4.1)$. And GC does this for each value of $x$ that it samples as it varies values of $x$ from 0 to 8.


Figure 8.5.7. GC computes an enormous number of incremental values to compute the value of $s_1(4.1)$.

The fact that GC does so many computations to compute values of $s_1$ and $s_2$, and hence takes so long to generate the graphs in Figure 8.5.5, demonstrates why closed form definitions of accumulation functions can be important. Closed form definitions of $s_1$ and $s_2$ would dramatically reduce the number of computations GC must do.

The closed form definition of $s_1$ is derived below, using the FTC twice.

$$\begin{align}s_1(t)
&=\int_0^t v_1(w)dw\\[1ex]
&=\int_0^t \left(\int_0^w a_1(u)du\right)dw\\[1ex]
&= \int_0^t \left(\int_0^w \left( 0.7u+\sin\left(\dfrac{3\pi u}{5}\right)\right)du\right)dw\\[1ex]
&= \int_0^t \left(
    \frac{0.7w^2}{2}
     -\frac{5\cos\left(\dfrac{3\pi w}{5}\right)}{3\pi}+\frac{5}{3\pi}\right)dw\\[1ex]
&=\frac{0.7t^3}{6}
    + \frac{-25\sin\left(\dfrac{3\pi t}{5}\right)}{9\pi^2}+\frac{5t}{3\pi}
\end{align}$$

We do not intend, at this moment, that you derive closed form definitions of integral functions having complicated rate of change functions. That will come in Chapter 9. But we do encourage you to enter the open and closed form definitions of $s_1$ (using a name other than $s_1$ for the closed form) into GC. This should convince you that the closed form definition of $s_1$ generates the same graph as does the open-form definition of $s_1$, and that GC generates the graph much faster.

We will remind you in Chapter 9 about Example 2 and the motivation it gives for finding closed form definitions of accumulation functions. In Chapter 8 we will continue to focus on conceptualizing quantitative situations, using integral functions defined in open form to do so.

Background for Example 3: Torque, or measuring an amount of twist.

Yoda and Luke are on opposite ends of a seesaw (Figure 8.5.8). Yoda's mass is 81.8 kg. Luke's mass is 54.5 kg. Luke sits 1.8 meters from the seesaw's center (pivot point). The seesaw balances when no one sits on it. How far from the seesaw's center must Yoda sit so that he and Luke balance each other?


Figure 8.5.8 Yoda and Luke on a seesaw. How far from the center should they sit to balance?

The physics of a seesaw entails two opposing rotational (twisting) forces. The measure of a rotational force is called torque. So Yoda and Luke will balance each other when the torques on either side of the seesaw have equal magnitudes and opposite rotational directions.

The standard image for torque is a force applied perpendicularly to an object that rotates around a pivot point (Figure 8.5.9). Torque is calculated as $force\cdot distance$. The SI unit for torque is the Newton-meter, or N-m, which is a rotational force of 1 Newton acting at a distance of 1 meter from a pivot point. One Newton of force is the force that accelerates a mass of 1 kg at a rate of 1 $\mathrm{\dfrac {(m/s)}{s}}$.


Figure 8.5.9. A force of F Newtons applied at a distance of m meters from a pivot point
creates a torque of $F\cdot m$ Newton-meters.

Luke's weight is a force pushing down on Luke's end of the seesaw. Yoda's weight is a force pushing down on Yoda's end of the seesaw. Weight at Earth's surface is the force due to a mass being accelerated by gravity. Acceleration due to gravity at Earth's surface, in the SI system, is $9.8 \mathrm{\frac{(m/s)}{s}}$.

Luke's force on the seesaw when it is in balance will therefore be $9.8\,\mathrm{\dfrac{(m/s)}{s}} \cdot 54.5 \text{ kg}$, or 534.10 Newtons. Yoda's force on the seesaw when it is in balance will be 801.64 Newtons.

For Yoda and Luke to balance on the seesaw they must exert torques that are equal in magnitude but opposite in direction. Luke's torque will be $534.10 \cdot 1.8$ N-m, or 961.38 N-m when he and Yoda are in balance. Yoda will sit $k$ meters from the pivot point and exert a torque of $-(801.64 \cdot k)$ N-m when he and Luke are in balance. For the seesaw to balance,
$$\begin{align}-(801.64\cdot k)&=961.38\\[1ex]
    k&=\frac{961.38}{-801.64}\\[1ex]
    &=-1.20\end{align}$$
So, for the seesaw to balance, Yoda must sit approximately -1.20 meters from the pivot point, or 1.20 meters on the opposite side of the pivot point from Luke.

Example 3. Torque of a variable-density beam.

What is the torque exerted at a pivot point by a beam that is 2 cm high, 5 cm wide, 250 cm long, and has uniform vertical density but a horizontal density that varies with distance from the pivot point? The beam's density is $2.2 \mathrm{g/cm^3}$ at the pivot point, $5.4 \mathrm{g/cm^3}$ at the other end, and varies linearly from pivot point to end. (Figure 8.5.10)


Figure 8.5.10.

Example 3 is just like the seesaw when we think of a small slice of the beam exerting, by itself, a rotational force.

Imagine a dot moving away from the pivot point along the beam's axis. Let $x$ be the dot's distance from the pivot point as the dot moves (Figure 8.5.11).

As $x$ varies from 0 to 250, the beam's density varies from 2.2 $\mathrm{g/cm^3}$ to 5.4 $\mathrm{g/cm^3}$. The beam's density at a distance of $x$ cm from the pivot point is therefore given by the function $f$ defined as
$$f(x)=\frac{5.4-2.2}{250}x+2.2 \mathrm{\quad g/cm^3}.$$
We can now reason as follows.
Figure 8.5.12 shows GC's display of a graph of the torque exerted by the part of the beam from 0 cm to $x$ cm along the beam as $x$ varies from 0 to 250. The torque exerted by the entire beam is $T(250)\approx 13\,204.7$ Newton-meters.

Figure 8.5.12. GC's display of accumulated torque of a beam with variable density as point of measurement varies from x=0 to x=250 cm.

Example 4. Population Growth (Ignore this. I'm going to move it to Chapter 9)

Population growth rates are expressed in terms of a percent of the current number of persons in the population by which the population increases per year. The U.S. population grew approximately 0.77% from July 1, 2015 to July 1, 2016. If we let P be the function that gives the U.S. population in relation to number of years CE, then $P(2016.5)=1.0077P(2015.5)$.

Let $t$ be number of years since July 1, 2016. Let $P_0$ be the U.S. population on July 1, 2016. What, approximately, will be the U.S. population at each moment in time after July 1, 2016 if the population continues to grow at an annual rate of 0.77% per year?

If the U.S. population were to continue growing continuously by 0.77% per year, then its rate of growth in number of people per year would be $r_P(t)=0.0077P(t)$, or in prime notation, $P'(t)=0.0077P(t)$. This says that the rate of change of a population in persons/year at any moment is proportional to the number of people in the population at that moment.

We saw in Chapter 6 that the exponential function is the only function that has a rate of change at a moment that is proportional to its value at that moment, so P(t) is an exponential function. Put another way, $e^{kt+c}$ is an antiderivative of $ke^{kt+c}$ (you should check this). Thus, $r_P(t)=kP(t)$ implies that $$\begin{align}P(t)&=\int_0^t ke^{ku+c}du\\[1ex]
&=e^{kt+c}\\[1ex]
&=e^c \cdot e^{kt}.\end{align}$$
At $t=0$, $P(0)$ is the population on July 1, 2016, so $P(0)=P_0$. But $$\begin{align}P(0)&=e^c\cdot e^{k\cdot 0}\\[1ex]&=e^c\end{align}.$$Hence $P_0=e^c$, and therefore $$\begin{align}P(t)&=e^ce^{0.0077t}\\[1ex]&=P_0e^{0.0077t}\end{align}.$$

We hasten to point out that, in the case exponential growth, we were unable to represent the accumulation function as an open form integral. We needed to make the connection that exponential functions are the only functions that have a rate of change at a moment that is proportional to the value of the function at that moment. This connection, between exponential functions and rate of change being proportional to the value of the function, will be key in many applications.

Example 5. Constrained Growth

Any function f that has a rate of change function $r_f$ of the form
$$r_f(x)=L\frac{ke^{kx}}{\left(e^{kx}+1\right)^2}$$
is called a logistic growth function. Figure 8.5.15 shows GC's graph of $y=r_f(x)$ for $L=10\text{ and } k=1.5$.


Figure 8.5.15. Logistic rate function

The logistic rate of change function is useful when modeling situations where a quantity increases slowly at first, reaches a maximum rate of change, and then increases at a slower rate until it approaches a rate of change of 0. Population growth with respect to time typically fits this pattern when the population is constrained to an environment that has a limited capacity to support it. Indeed, Pierre Verhülst introduced the idea of a logistic function in his attempt to model population growth in constrained environments. As we will see in the exercises, the logistic function shows up in many contexts other than population growth.

Since the logistic rate function $r_f$ gives the rate of change at all moments of a function f, we can recover the net accumulated change in f over any interval by accumulating bits of change $r_f(t)dt$ for $t=a$ to $t=x$. Figure 8.5.16 shows the graph of $f(x)=\int_{-10}^xr_f(x)dx$ with the same parameters as in Figure 8.5.15 $(L=10, k=1.5)$.


Figure 8.5.16. Graph of $\displaystyle y=\int_{-10}^xr_f(t)dt$.

(Examine 8.5.15 and 8.5.16 in relation to each other in terms of implications for growth and value.)

Example 6. Blood Alcohol Concentration

We know from experience that consuming alcohol dulls our senses and affects our moods. We can use calculus to model alcohol's effect over time.

Blood alcohol concentration (BAC) is measured in milliliters of alcohol per liter of blood. According to this website, different BAC levels produce these effects:

Euphoria BAC 0.03 to 0.12
Excitement BAC 0.09 to 0.25
Confusion BAC 0.18 to 0.30
Stupor BAC 0.25 to 0.40
Coma BAC 0.35 to 0.50
Death BAC > 0.50

In most states in the U.S., a BAC of 0.05 or greater is cause for a combination of arrest, imprisonment, loss of drivers license, confiscation of your car, or high fine.

One 12-ounce glass of beer, one 5-oz glass of wine, and one 1.5-oz shot of whiskey typically contain 18 milliliters of alcohol. We will call each of these "a drink". So, a drink in any of these forms contains 18 ml of alcohol.

Ron is at a party. He weighs 198 lbs (90 kg). He chugs 3 drinks, 15 minutes apart. What is Ron's BAC over time?

When you consume alcohol, it is simultaneously absorbed into your bloodstream and eliminated by your liver.

Alcohol is absorbed into your bloodstream at a rate (in milliliters of alcohol per liter of blood per hour since consumption) that is proportional to the amount of alcohol that is available to absorb.

Your liver eliminates alcohol from your bloodstream at a rate (in milliliters of alcohol per liter of blood per hour since consumption) that is proportional to the amount of alcohol in your bloodstream.

The rate at which a person's BAC changes with respect to time is the difference between the rate at which alcohol is absorbed and the rate at which it is eliminated.

The rate at which a person's BAC changes in (ml/liter)/hr is also affected by his or her mass and the amount of alcohol consumed.

When a person with mass $M$ kg consumes $V$ ml of alcohol, the rate at which his or her BAC changes with respect to the number of hours since consuming it is modeled by the function $r_B$, defined as
$$r_B(t)=\begin{cases}
0 &\text{if $t\lt 0$}\\[1ex]
\dfrac{73}{M}\dfrac{V}{40}\left(e^{-t}-0.85e^{-0.85t}\right) &\text{if $t \ge 0$}
\end{cases}$$

Ron drank his first drink at $t=0$ hours. He drank his second drink at $t=0.25$ hours. He drank his third drink at $t=0.5$ hours. The rate at which Ron's BAC changes over time is therefore

$$r(t)=r_B(t)+r_B(t-0.25)+r_B(t-0.50).$$

Reflection 8.5.1. Explain how $r_B(t-0.25)$ gives the rate at which Ron's BAC changes that is due to the $2^{nd}$ drink and how $r_B(t-0.50)$ gives the rate at which Ron's BAC changes that is due to the $3^{rd}$ drink. Then explain how the sum $r_B(t)+r_B(t-0.25)+r_B(t-0.50)$ gives the rate of change of Ron's BAC at all moments of t, $0 \le t$.

Figure 8.5.17 shows GC's display of the rate at which Ron's BAC changes with respect to time at each moment since consuming his first drink. Discontinuities in the graph are due to our assumption that he consumes each drink instantaneously.


Figure 8.5.17. Rate of change of Ron's BAC with respect to time. Discontinuities are due
to the assumption that Ron consumes each drink instantaneously.

Reflection 8.5.2.Elaborate on the comment that discontinuities in the graph in Figure 8.5.17 are due to the assumption that Ron consumed each drink instantaneously.

Reflection 8.5.3. Explain how the graph in Figure 8.5.17 makes sense in relation to Ron's consumption and the way that alcohol is absorbed and eliminated.

According to our model, each value $r(t)$ is the exact rate of change of Ron's BAC at each moment of $t$. So the change in Ron's BAC at each moment of $t$ is essentially $r(t)dt$. Ron's BAC over time will be the accumulation of changes in his BAC. So the function $B_3$ (BAC after 3 drinks), defined as $$B_3(t)=\int_0^t r(u)du$$ will, according to our model, give Ron's BAC from $u=0$ to $u=t$ hours since consuming his first drink when he consumes 3 drinks 0.25 hours apart. Figure 8.5.18 shows GC's display of $y=B_3(x)$, which is a model of Ron's BAC over time. His BAC becomes illegal approximately 0.75 hours after his first drink and does not return to a legal level until approximately 2.3 hours after his first drink.


Figure 8.5.18. Ron's BAC over time since consuming his first drink and
consuming 3 drinks 0.25 hours apart.

Reflection 8.5.4. Explain how the graph in Figure 8.5.18 makes sense in relation to the rate at which Ron's BAC changes with respect to time.

Example 7. Work and Pressure


Exercise Set 8.5

  1. Turn the beam around in Example 3, attaching the denser end to the pivot point. Graph the torque exerted by the beam from the pivot point to $x$ cm from the pivot point for all values of $x, 0\le x \le 250$. Why is the torque curve for the beam now different from the original one?

  2. A bacterial culture has an initial mass of 10 grams. Its mass $t$ hours after being placed in a growth medium changes at the rate of $$r_m(t)=\frac{25(12x^2+120)^{1/2}-300x^2(12x^2+120)^{-1/2}}{12x^2+120}\text{ grams/hr}.$$
    1. Graph the function m such that $m(t)$ gives the culture's mass at each moment $t$ of elapsed time, $t \ge 0$. 

    2. Compare the graph of $y=m(x)$ to the graph of $y=r_m(x)$. Explain how they are consistent with each other.

    3. Why does the behavior of $y=m(x)$ make sense in terms of a bacterial culture in a growth medium?

  3. A pool with a rectangular surface is 15m wide and 24m long. The depth of the pool measured $x$ meters perpendicularly from the deep end is given by
    $$d(x)=\begin{cases}
    5 & \text{ if $0\le x\lt 3$}\\[1ex]
    5-\sin\left(\dfrac{\pi(x-3)}{6}\right) & \text{ if $3\le x\lt 6$}\\[1ex]
    4-\sin\left(\dfrac{\pi(x-6)}{40}\right) & \text{if $6\le x\le 24$}
    \end{cases}$$
    1. Explain how you can think of any value of $15d(x), 0 \le x \le 24$ as the rate of change of the pool's volume with respect to distance from the pool's deep end.

    2. Define a function $V$ whose values give the pool's volume at any distance $x$ meters from the pool's end. Graph $y=V(x)$.
    3. What is the pool's total volume?

    4. Approximately how far from the pool's deep end must you measure to get half the pool's volume?


  4. Re-examine Example 5

    1. Explain why it makes sense, in terms of how the logistic rate of change function is defined, that a population with this rate of change at a moment should approach a limiting value. Think about this question in terms of elapsed time growing larger and larger.
    2. Explain why it makes sense, in terms of how the logistic rate of change function is defined, that a population with this rate of change at a moment has an inflection point at $t=0$. Interpret this inflection point in terms of population growth over time.

    3. Verify graphically, using various values of L and k, that a population having a logistic rate of change always approaches a population of L. This is why L is called the carrying capacity of the population's ecological environment.

  5. A more general logistic rate function is$$r_f(x)=L\frac{Cke^{kx}}{\left(e^{kx}+C\right)^2}.$$
    1. Show that the point of inflection for a population having this logistic rate of change is always at $x=\dfrac{\ln C}{k}$.

    2. Interpret the inflection point in terms of the population growth over time.
  6. A ball suspended by a 10-foot long rubber cord is at rest. It is given a sudden push downward; the cord stretches, then retracts, pulling the ball upward.

    The ball bobs up and down with time. (See the animation, below). The ball's rate of change of displacement from rest is given by the function $r_d$, defined as $$r_d(t)=-4\left(-0.0625\sin(t)e^{-0.0625t}4\sin(t)+t^{0.12}\cos(t)e^{-0.625t}\right), 0≤t≤100$$where $r_d(t)$ is in ft/sec and t is in seconds since the ball was pushed.


    1. Define a function $d$ whose graph $y=d(x)$ gives the ball's displacement from initial rest at each moment $x$ seconds after being pushed. Explain how your function in fact gives the ball's net displacement at values of $x$. (Net displacement is current position relative to a comparison position.  In this scenario, higher than resting position is positive, lower than resting position is negative.)

    2. Use your graph to determine the moments during the first 18 seconds that the ball's displacement reaches a local minimum. A local maximum.

    3. Does the ball's displacement from rest have a global maximum? A global minimum?

    4. At what displacement from initial rest does the ball approach a stable displacement? How is the behavior of its rate of change function consistent with its eventual stable displacement?
  7. Re-examine Example 6

    1. Change the model so that Ron consumes his drinks in sips. Rather than gulping drinks in their entirety, Ron consumes each drink in 3 sips that are 0.1 hours apart, and then begins his next drink 0.25 hours after ending the previous drink. Suppose that Ron consumes 5 drinks by this method. When does Ron become legally intoxicated? How long after his first sip may he drive legally?
    2. The National Football League has a policy that a stadium cannot serve beer past the end of the 3rd quarter. Each quarter of football lasts, on average, 0.75 hours, with a 0.25-hour break after the 2nd quarter.

      Katie, who has a body mass of 50 kg, drank a 20-oz glass of beer during the first quarter of the game and another during the third quarter. She consumed each glass of beer in 4 equal amounts spread equally across a quarter of play, and she reached her car 30 minutes after the game ends. Can Katie legally drive home after the game? If not, how long must she wait?

    3. Suppose that James, an 80 kg male, takes a six-pack of beer to a party and consumes it at a steady pace, consuming 1/4 drink per gulp at equal time intervals. Use a slider whose value gives number of hours between gulps to graphically estimate the least number of hours between gulps that will ensure that James'  BAC remains at or below 0.05.
  8. The force of gravity on objects near Earth's surface is often approximated to be F = m⋅g, where $g = -9.8 \mathrm{\frac{m/s}{s}}$. However, for objects that are very large or very far away, you need to calculate the force of gravity using F=$\frac{-Gm_1m_2}{r^2}$, where $G=6.674⋅10^{-11} \frac{Nm}{kg^2}$.

    Imagine that we have a giant flagpole at the North Pole, 2000 km high (that's one-third the radius of the Earth). This flagpole has mass $9⋅10^9$ kg and has uniform density.

    What is the gravitational force on the flagpole from the flagpole's base to a height of $x$ km, $0<x<2000$? (For convenience, assume that the Earth's mass is concentrated at its center.)

  9. (In progress) At Earth's surface, an object in a vacuum would fall at a constant acceleration of $-9.8 \mathrm{\frac{m/s}{s}}$. However, in the atmosphere objects fall with a decreasing acceleration, starting at $-9.8 \mathrm{\frac{m/s}{s}}$ and decreasing to 0 because of air friction. The acceleration of objects falling in air decreases at a rate that is proportional to its current acceleration; that is, $r_a(t)=-k⋅a(t)$. Note that this is independent of an object's mass.
    1. What is an object's acceleration after t seconds?

  10. (In progress) A non-fatal disease starts to spread in a community. A cure is developed and distributed throughout the community. People are infected at the rate of $r_I(t)=0.9e^{.95t}$ thousand people per year, and the rate at which people are cured is $r_C(t)=2t^2$ thousands of people per year.
    1. Use GC to graph both $r_I(t)$ and $r_C(t)$. Looking at the graph, try to predict when the cure will wipe out the disease and write your guess down before continuing to parts b and c.

    2. If 3000 people were infected when the cure was first sent out, how many years does it take to eradicate the disease?

    3. If 4000 people were infected when the cure was first sent out, how many years does it take to eradicate the disease?

  11. The toy cars from Example 2 race again, but this time they have different accelerations: $$a_1(t)=\sin\left(\frac{3\pi x}{5}\right) \text{ (ft/sec)/sec, and }a_2(t)=\cos\left(\frac{2\pi x}{5}\right) \text{ (ft/sec)/sec.}$$

    1. What does it mean about the motion of a car when it has negative acceleration?

    2. When are the cars first next to each other after they are released? (Answer using appropriate graphs.)

    3. How far apart are the cars when their velocities are the same? (Answer using appropriate graphs.)
    4. The cars' odometers were the same when the race started. What are their odometer readings at the moment(s) their velocities are the same?