AC Modeling with differential equations

Section 7.5 Modeling with differential equations

Motivating Questions

How can we use differential equations to describe phenomena in the world around us?
How can we use differential equations to better understand these phenomena?

We have seen several ways that differential equations arise in the natural world, from the growth of a population to the temperature of a cup of coffee. In this section, we look more closely at how differential equations give us a natural way to describe various phenoma. As we'll see, the key is to understand the different factors that cause a quantity to change.

Preview Activity 7.5.1.

Any time that the rate of change of a quantity is related to the amount of a quantity, a differential equation naturally arises. In the following two problems, we see two such scenarios; for each, we want to develop a differential equation whose solution is the quantity of interest.

Suppose you have a bank account in which money grows at an annual rate of 3%.
1. If you have $10,000 in the account, at what rate is your money growing?
2. Suppose that you are also withdrawing money from the account at $1,000 per year. What is the rate of change in the amount of money in the account? What are the units on this rate of change?
Suppose that a water tank holds 100 gallons and that a salty solution, which contains 20 grams of salt in every gallon, enters the tank at 2 gallons per minute.
1. How much salt enters the tank each minute?
2. Suppose that initially there are 300 grams of salt in the tank. How much salt is in each gallon at this point in time?
3. Finally, suppose that evenly mixed solution is pumped out of the tank at the rate of 2 gallons per minute. How much salt leaves the tank each minute?
4. What is the total rate of change in the amount of salt in the tank?

Subsection 7.5.1 Developing a differential equation

Preview Activity 7.5.1 demonstrates the kind of thinking we will be doing in this section. In each of the two examples we considered, there is a quantity, such as the amount of money in the bank account or the amount of salt in the tank, that is changing due to several factors. The governing differential equation states that the total rate of change is the difference between the rate of increase and the rate of decrease.

Example 7.5.1.

In the Great Lakes region, rivers flowing into the lakes carry a great deal of pollution in the form of small pieces of plastic averaging 1 millimeter in diameter. In order to understand how the amount of plastic in Lake Michigan is changing, construct a model for how this type pollution has built up in the lake.

Solution

First, some basic facts about Lake Michigan.

The volume of the lake is $5\cdot10^{12}$ cubic meters.
Water flows into the lake at a rate of $5\cdot10^{10}$ cubic meters per year. It flows out of the lake at the same rate.
Each cubic meter flowing into the lake contains roughly $3\cdot10^{-8}$ cubic meters of plastic pollution.

Let's denote the amount of pollution in the lake by $P(t)\text{,}$ where $P$ is measured in cubic meters of plastic and $t$ in years. Our goal is to describe the rate of change of this function; so we want to develop a differential equation describing $P(t)\text{.}$

First, we will measure how $P(t)$ increases due to pollution flowing into the lake. We know that $5\cdot10^{10}$ cubic meters of water enters the lake every year and each cubic meter of water contains $3\cdot10^{-8}$ cubic meters of pollution. Therefore, pollution enters the lake at the rate of

\begin{equation*} \left(5\cdot 10^{10} \frac{m^3 \text{ water} }{\text{year} }\right) \left(3\cdot10^{-8} \frac{m^3 \text{ plastic} }{m^3 \text{ water} } \right) = 1.5\cdot 10^3 \text{cubic m of plastic per year}\text{.} \end{equation*}

Second, we will measure how $P(t)$ decreases due to pollution flowing out of the lake. If the total amount of pollution is $P$ cubic meters and the volume of Lake Michigan is $5\cdot 10^{12}$ cubic meters, then the concentration of plastic pollution in Lake Michigan is

\begin{equation*} \frac{P}{5\cdot10^{12}} \text{cubic m of plastic per cubic m of water}\text{.} \end{equation*}

Since $5\cdot10^{10}$ cubic meters of water flow out each year¹, then the plastic pollution leaves the lake at the rate of

\begin{equation*} \left(\frac{P}{5\cdot10^{12}} \frac{m^3 \text{ plastic} }{m^3 \text{ water} } \right) \left(5\cdot10^{10} \frac{m^3 \text{ water} }{\text{year} } \right)=\frac{P}{100} \text{cubic m of plastic per year}\text{.} \end{equation*}

and we assume that each cubic meter of water that flows out carries with it the plastic pollution it contains

The total rate of change of $P$ is thus the difference between the rate at which pollution enters the lake and the rate at which pollution leaves the lake; that is,

\begin{align*} \frac{dP}{dt} =\mathstrut \amp 1.5\cdot10^{3}-\frac{P}{100}\\ =\mathstrut \amp \frac{1}{100}(1.5\cdot10^{5} - P)\text{.} \end{align*}

We have now found a differential equation that describes the rate at which the amount of pollution is changing. To understand the behavior of $P(t)\text{,}$ we apply some of the techniques we have recently developed.

Because the differential equation is autonomous, we can sketch $dP/dt$ as a function of $P$ and then construct a slope field, as shown in Figure 7.5.2 and Figure 7.5.3.

Figure 7.5.2. Plot of $\frac{dP}{dt}$ vs. $P$ for $\frac{dP}{dt} = \frac{1}{100}(1.5\cdot10^{5} - P)\text{.}$

Figure 7.5.3. Plot of the slope field for $\frac{dP}{dt} = \frac{1}{100}(1.5\cdot10^{5} - P)\text{.}$

These plots both show that $P=1.5\cdot10^5$ is a stable equilibrium. Therefore, we should expect that the amount of pollution in Lake Michigan will stabilize near $1.5\cdot10^5$ cubic meters of pollution.

Next, assuming that there is initially no pollution in the lake, we will solve the initial value problem

\begin{equation*} \frac{dP}{dt} = \frac{1}{100}(1.5\cdot10^{5} - P), \ P(0) = 0\text{.} \end{equation*}

Separating variables, we find that

\begin{equation*} \frac1{1.5\cdot10^5-P} \frac{dP}{dt} = \frac1{100}\text{.} \end{equation*}

Integrating with respect to $t\text{,}$ we have

\begin{equation*} \int \frac1{1.5\cdot10^5-P} \frac{dP}{dt}~dt = \int \frac1{100}~dt\text{,} \end{equation*}

and thus changing variables on the left and antidifferentiating on both sides, we find that

\begin{align*} \int \frac{dP}{1.5\cdot10^5-P} =\mathstrut \amp \int \frac1{100}~dt\\ -\ln|1.5\cdot10^5 - P| =\mathstrut \amp \frac1{100}t + C \end{align*}

Finally, multiplying both sides by $-1$ and using the definition of the logarithm, we find that

\begin{equation} 1.5\cdot10^5 - P = C e^{-t/100}\text{.}\label{wXu}\tag{7.5.1} \end{equation}

This is a good time to determine the constant $C\text{.}$ Since $P = 0$ when $t=0\text{,}$ we have

\begin{equation*} 1.5\cdot 10^5 - 0 = Ce^0 = C\text{,} \end{equation*}

so $C=1.5\cdot10^5\text{.}$

Using this value of $C$ in Equation (7.5.1) and solving for $P\text{,}$ we arrive at the solution

\begin{equation*} P(t) = 1.5\cdot10^5(1-e^{-t/100})\text{.} \end{equation*}

Superimposing the graph of $P$ on the slope field we saw in Figure 7.5.3, we see, as shown in Figure 7.5.4

We see that, as expected, the amount of plastic pollution stabilizes around $1.5\cdot10^5$ cubic meters.

Figure 7.5.4. The solution $P(t)$ and the slope field for the differential equation $\frac{dP}{dt} = \frac{1}{100}(1.5\cdot10^{5} - P)\text{.}$

There are many important lessons to learn from Example 7.5.1. Foremost is how we can develop a differential equation by thinking about the “total rate = rate in - rate out” model. In addition, we note how we can bring together all of our available understanding (plotting $\frac{dP}{dt}$ vs. $P\text{,}$ creating a slope field, solving the differential equation) to see how the differential equation describes the behavior of a changing quantity.

We can also explore what happens when certain aspects of the problem change. For instance, let's suppose we are at a time when the plastic pollution entering Lake Michigan has stabilized at $1.5\cdot10^5$ cubic meters, and that new legislation is passed to prevent this type of pollution entering the lake. So, there is no longer any inflow of plastic pollution to the lake. How does the amount of plastic pollution in Lake Michigan now change? For example, how long does it take for the amount of plastic pollution in the lake to halve?

Resetting $t=0$ at this time, we now have the initial value problem

\begin{equation*} \frac{dP}{dt} = -\frac{1}{100}P, \ P(0) = 1.5\cdot10^5\text{.} \end{equation*}

It is a straightforward and familiar exercise to find that the solution to this equation is $P(t) = 1.5\cdot10^5 e^{-t/100}\text{.}$ The time that it takes for half of the pollution to flow out of the lake is given by $T$ where $P(T) = 0.75\cdot10^5\text{.}$ Thus, we must solve the equation

\begin{equation*} 0.75\cdot10^5 = 1.5\cdot10^5e^{-T/100}\text{,} \end{equation*}

\begin{equation*} \frac12 = e^{-T/100}\text{.} \end{equation*}

It follows that

\begin{equation*} T = -100\,\ln\left(\frac12\right) \approx 69.3 \text{years.} \end{equation*}

In the upcoming activities, we explore some other natural settings in which differential equations model changing quantities.

Activity 7.5.2.

Suppose you have a bank account that grows by 5% every year. Let $A(t)$ be the amount of money in the account in year $t\text{.}$

What is the rate of change of $A$ with respect to $t\text{?}$
Suppose that you are also withdrawing $10,000 per year. Write a differential equation that expresses the total rate of change of $A\text{.}$
Sketch a slope field for this differential equation, find any equilibrium solutions, and identify them as either stable or unstable. Write a sentence or two that describes the significance of the stability of the equilibrium solution.
Suppose that you initially deposit $100,000 into the account. How long does it take for you to deplete the account?
What is the smallest amount of money you would need to have in the account to guarantee that you never deplete the money in the account?
If your initial deposit is $300,000, how much could you withdraw every year without depleting the account?

Hint

Small hints for each of the prompts above.

Answer

$\frac{dA}{dt} = 0.05A\text{.}$
$\frac{dA}{dt} = 0.05A - 10000\text{.}$
The only equilibrium solution is $A = 200000\text{.}$
$t = 20 \ln(2) \approx 13.86$years.
At least $200000.
Up to $15000 every year.

Solution

Let $A(t)$ be the amount of money in a bank account in year $t$ that grows by 5% every year.

The rate of change of $A$ with respect to $t$ is $\frac{dA}{dt} = 0.05A\text{.}$
If $10000 is being withdrawn every year, then

\begin{equation*} \frac{dA}{dt} = 0.05A - 10000\text{.} \end{equation*}
Following is a slope field for the differential equation in (b).

Equilibrium solutions occur where $\frac{dA}{dt} = 0$ and so the only equilibriume solution is $A = 200000\text{.}$ This means that if you have $200000 in the account, then the amount of money in the account will remain constant at $200000.
Suppose we initially have $100000 in the account. We will solve the intitial value problem

\begin{equation*} \frac{dA}{dt} = 0.05A - 10000, A(0) = 100000\text{.} \end{equation*}

Using separation of variables, we find that

\begin{equation*} A = 200000 + C e^{t/20}\text{,} \end{equation*}

and since $A(0) = 100000\text{,}$ we have

\begin{equation*} A = 200000 - 100000e^{t/20}\text{.} \end{equation*}

To determine when the account will be depleted, we solve the equation $A = 0$ for $t\text{.}$ Since $200000 - 100000 e^{t/20} = 0\text{,}$ it follows $e^{t/20} = 2$ and thus $t = 20 \ln(2)\text{,}$ so the account will be depleted in about 13.86 years.
You must have at least $200000 in the account to make sure the account will never be depleted.
Now suppose the intitial deposit is $300000 and we want to what amount $w$ we can withdraw every year without depleting the account. So the initial value problem is

\begin{equation*} \frac{dA}{dt} = 0.05A - w, A(0) = 300000\text{.} \end{equation*}

The solution to this initial value problem is

\begin{equation*} A = 20w + (300000 - 20w) e^{t/20}\text{.} \end{equation*}

To make sure the account is never depleted, we need $(300000 - 20W)$ to be greater than or equal to 0. Solving the inequality $300000 - 2w \geq 0\text{,}$ we obtain $w \leq 15000\text{.}$ We can withdraw up to $15000 every year and the account will not be depleted.

Activity 7.5.3.

A dose of morphine is absorbed from the bloodstream of a patient at a rate proportional to the amount in the bloodstream.

Write a differential equation for $M(t)\text{,}$ the amount of morphine in the patient's bloodstream, using $k$ as the constant proportionality.
Assuming that the initial dose of morphine is $M_0\text{,}$ solve the initial value problem to find $M(t)\text{.}$ Use the fact that the half-life for the absorption of morphine is two hours to find the constant $k\text{.}$
Suppose that a patient is given morphine intravenously at the rate of 3 milligrams per hour. Write a differential equation that combines the intravenous administration of morphine with the body's natural absorption.
Find any equilibrium solutions and determine their stability.
Assuming that there is initially no morphine in the patient's bloodstream, solve the initial value problem to determine $M(t)\text{.}$ What happens to $M(t)$ after a very long time?
To what rate should a doctor reduce the intravenous rate so that there is eventually 7 milligrams of morphine in the patient's bloodstream?

Hint

Small hints for each of the prompts above.

Answer

$\frac{dM}{dt} = -kM\text{,}$ where $k$ is a positive constant.
$k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}$
$\frac{dM}{dt} = 3 - kM\text{,}$ where $k$ is a positive constant.
The equilibrium solution $mM = \frac{3}{k}$ is stable.
$M = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}$
About 2.426 milligrams per hour.

Solution

We will use the differential equation $\frac{dM}{dt} = -kM\text{,}$ where $k$ is a positive constant.
With an initial condition of $M = M_0$ when $t = 0\text{,}$ the solution for the initial value problem is

\begin{equation*} M = M_0 e^{-kt}\text{.} \end{equation*}

Using a half-life of 2 hours, we see that $M = \frac{1}{2} M_0$ when $t = 2\text{.}$ This gives $\frac{1}{2}M_0 = M_0 e^{-2k}\text{,}$ from which it follows that $k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}$
Noting that the rate of change of the drug in the body will be the differnce between the positive contributions of the dosage of 3 milligrams per hour and the decay of $kM$ per hour, it follows that $\frac{dM}{dt} = 3 - kM\text{,}$ where $k$ is a positive constant.
To find the equilibrium solutions, we solve the equation $\frac{dM}{dt} = 0$ for $M\text{.}$ This gives $M = \frac{3}{k}\text{.}$ The graph of $\frac{dM}{dt}$ as a function of $M$ is a straight line with a negative slope. (The slope is $-k\text{.}$) So when $M \lt \frac{3}{k}\text{,}$ $\frac{dM}{dt} \gt 0$ and the $M$ is an increasing function of $t\text{.}$ When $M \gt \frac{3}{k}\text{,}$ $\frac{dM}{dt} \lt 0$ and the $M$ is a decreasing function of $t\text{.}$ This means that the equilibrium solution $mM = \frac{3}{k}$ is a stable equilibrium solution.
We use the initial condition $M = 0$ when $t = 0$ and solve the corresponding IVP. Separating variables,

\begin{equation*} \frac{1}{3 - kM} \frac{dM}{dt} = 1\text{,} \end{equation*}

so

\begin{equation*} \int \frac{1}{3 - kM} dM = \int 1 dt\text{,} \end{equation*}

and thus

\begin{equation*} -\frac{1}{k} \ln |3 - kM | = t + c\text{.} \end{equation*}

Solving for $M$ in the usual way gives us

\begin{equation*} M = \frac{1}{k} \left( 3 - Ae^{-kt} \right)\text{,} \end{equation*}

where $A$ is a constant that arises from the integration constant. Using the initial condition, we obtain $0 = \frac{1}{k} ( 3 - A)\text{,}$ so $A = 3\text{.}$ Hence, the solution to the initial condition problem is $M = \frac{1}{k} \left( 3 - 3e^{-kt} \right) = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}$
We let $q$ be the unknown intravenous rate of morphine (in milligrams per hour). Notice that if we replace $3$ with $q$ in the differential equation and obtain $\frac{dM}{dt} = q - kM\text{,}$ the solution process is the same and the equilibriume solution is $M = \frac{q}{k}\text{.}$ So if we want the equilibrium solution to be $M = 7\text{,}$ we then have $\frac{q}{k} = 7\text{,}$ so $q = 7k \approx 2.426\text{.}$ The intravenous rate of morphine should be 2.426 milligrams per hour.

Subsection 7.5.2 Summary

Differential equations arise in a situation when we understand how various factors cause a quantity to change.
We may use the tools we have developed so far—slope fields, Euler's methods, and our method for solving separable equations—to understand a quantity described by a differential equation.

Exercises 7.5.3 Exercises

1. Mixing problem.

2. Mixing problem.

3. Population growth problem.

4. Radioactive decay problem.

5. Investment problem.

6.

Congratulations, you just won the lottery! In one option presented to you, you will be paid one million dollars a year for the next 25 years. You can deposit this money in an account that will earn 5% each year.

Set up a differential equation that describes the rate of change in the amount of money in the account. Two factors cause the amount to grow—first, you are depositing one millon dollars per year and second, you are earning 5% interest.
If there is no amount of money in the account when you open it, how much money will you have in the account after 25 years?
The second option presented to you is to take a lump sum of 10 million dollars, which you will deposit into a similar account. How much money will you have in that account after 25 years?
Do you prefer the first or second option? Explain your thinking.
At what time does the amount of money in the account under the first option overtake the amount of money in the account under the second option?

Answer

$\frac{dA}{dt} = 1 + 0.05A$
$A(25) = 49.80686$ million dollars.
$A(25) = 34.90343$ million dollars.
The first.
$t = 20 \ln(2) \approx 13.86 \ \text{years} \text{.}$

Solution

Let $A(t)$ be the amount of money (in millions) in the account at time $t\text{.}$ We know that the rate of change of $A$ is affected by the addition of $1$ million dollars a year, plus the interest earned, which is $0.05A\text{.}$ Thus,

\begin{equation*} \frac{dA}{dt} = 1 + 0.05A \end{equation*}
If there is no amount of money in the account when you open it, that means $A(0) = 0\text{.}$ Solving the initial value problem, we can find how much money will be in the account after 25 years. Note that $1 + 0.05A = 0.05(20 + A)\text{,}$ so we have

\begin{equation*} \frac{1}{20+A} \frac{dA}{dt} = 0.05 \end{equation*}

and thus

\begin{equation*} \int \frac{1}{20+A} \, dA = \int 0.05 \, dt\text{.} \end{equation*}

Integrating (and noting that $A(t) > 0$ for all $t\text{,}$

\begin{equation*} \ln (20 + A) = 0.05t + C\text{,} \end{equation*}

so solving for $A$ in the standard way,

\begin{equation*} 20 + A = Ke^{0.05t} \end{equation*}

and thus $A(t) = Ke^{0.05t} - 20\text{.}$ Since $A(0) = 0\text{,}$ $K = 20\text{,}$ and we've shown

\begin{equation*} A(t) = 20(e^{0.05t} - 1)\text{.} \end{equation*}

It follows that $A(25) = 49.80686$ million dollars.
For this option, we know that $\frac{dA}{dt} = 0.05A\text{,}$ $A(0) = 10\text{,}$ and thus by the standard solution to this IVP,

\begin{equation*} A(t) = 10e^{0.05t} \end{equation*}

from which we see that $A(25) = 34.90343$ million dollars.
The first option is obviously better because it leads to more money in the long term.
The amount of money in the account under the first option overtakes the amount of money in the account under the second option when $20(e^{0.05t} - 1) = 10e^{0.05t}\text{.}$ Solving for $t\text{,}$ we find that $10e^{0.05t} = 20\text{,}$ so $e^{0.05t} = 2\text{,}$ and thus

\begin{equation*} t = 20 \ln(2) \approx 13.86 \ \text{years}\text{.} \end{equation*}

7.

When a skydiver jumps from a plane, gravity causes her downward velocity to increase at the rate of $g\approx 9.8$ meters per second squared. At the same time, wind resistance causes her velocity to decrease at a rate proportional to the velocity.

Using $k$ to represent the constant of proportionality, write a differential equation that describes the rate of change of the skydiver's velocity.
Find any equilibrium solutions and decide whether they are stable or unstable. Your result should depend on $k\text{.}$
Suppose that the initial velocity is zero. Find the velocity $v(t)\text{.}$
A typical terminal velocity for a skydiver falling face down is 54 meters per second. What is the value of $k$ for this skydiver?
How long does it take to reach 50% of the terminal velocity?

Answer

$\frac{dv}{dt} = 9.8 - kv$
$v = \frac{9.8}{k}$ is a stable equilibrium.
$v(t) = \frac{9.8 - 9.8e^{-kt}}{k}$
$k = 9.8/54 \approx 0.181481\text{.}$
$t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894$ seconds.

Solution

Let the skydiver's downward velocity at time $t$ be given by $v(t)\text{.}$ Since gravity causes her downward velocity to increase at the rate of $g\approx 9.8\text{,}$ there is a factor of $+9.8$ meters per second squared in the differential equation for $\frac{dv}{dt}\text{,}$ while the factor of wind resistence causing the velocity to decrease contributes the negative factor $-kv\text{.}$ Thus, the differential equation is

\begin{equation*} \frac{dv}{dt} = 9.8 - kv \end{equation*}
To find the equilibrium solutions, we set $\frac{dv}{dt} = 0$ and solve for $v\text{.}$ Doing so, $0 = 9.8 - kv\text{,}$ so $v = \frac{9.8}{k}\text{.}$ Moreover, we note that for $v \lt \frac{9.8}{k}\text{,}$ $\frac{dv}{dt} \gt 0\text{,}$ while for $\frac{dv}{dt} \gt \frac{9.8}{k}\text{,}$ $\frac{dv}{dt} \lt 0\text{.}$ Thus, for velocities just above the equilibrium, the velocity will decrease and for velocities just below the equilibrium, the velocity will decrease, making $v = \frac{9.8}{k}$ a stable equilibrium.
We assume that $v(0) = 0$ and solve the IVP with the differential equation we developed. Taking the differential equation

\begin{equation*} \frac{dv}{dt} = 9.8 - kv \end{equation*}

and separating the variables, we have

\begin{equation*} \frac{1}{9.8 - kv} \frac{dv}{dt} = 1\text{.} \end{equation*}

Integrating with respect to $t\text{,}$

\begin{equation*} \int \frac{1}{9.8 - kv} \, dv = \int 1 \, dt\text{,} \end{equation*}

and using a standard substitution, we have

\begin{equation*} -\frac{1}{k} \ln | 9.8 - kv | = t + C\text{.} \end{equation*}

Solving for $v\text{,}$ we have $\ln | 9.8 - kv | = -kt + C_1$ (where $C_1 = -Ck\text{,}$ or

\begin{equation*} |9.8 - kv| = e^{-kt + C_1} = C_2 e^{-kt}\text{.} \end{equation*}

Incorporating the absolute value into the constant $C_2$ in the usual way, we have for some constant $A$ that

\begin{equation*} 9.8 - kv = Ae^{-kt} \end{equation*}

and thus

\begin{equation*} v = \frac{9.8 - Ae^{-kt}}{k}\text{.} \end{equation*}

Setting $v(0) = 0\text{,}$ we can also determine $A\text{,}$ since then

\begin{equation*} 0 = \frac{9.8 - A}{k} \end{equation*}

so $A = 9.8\text{.}$ The solution to the IVP is therefore

\begin{equation*} v(t) = \frac{9.8 - 9.8e^{-kt}}{k} \end{equation*}
If the skydiver's terminal velocity is 54 meters per second, this means that

\begin{equation*} \lim_{t \to \infty} v(t) = 54 \end{equation*}

Using the solution we found in (c), we observe that

\begin{equation*} \lim_{t \to \infty} \frac{9.8 - 9.8e^{-kt}}{k} = \frac{9.8}{k} \end{equation*}

since $e^{-kt} \to 0$ as $t \to \infty\text{.}$ Thus, $54 = \frac{9.8}{k}\text{,}$ so $k = 9.8/54 \approx 0.181481\text{.}$
To find how long does it take to reach 50% of the terminal velocity, we set $v(t) = \frac{9.8 - 9.8e^{-0.181481t}}{0.181481} = 0.5(54)$ and solve for $t\text{.}$ We have

\begin{equation*} 9.8 - 9.8e^{-0.181481t} = 0.5(54)(0.181481)\text{,} \end{equation*}

so

\begin{equation*} 9.8-9.8e^{-0.181481t} = 4.9\text{,} \end{equation*}

and thus $9.8e^{-0.181481t} = 4.9\text{,}$ which tells us

\begin{equation*} e^{-0.181481t} = 0.5\text{.} \end{equation*}

Solving for $t\text{,}$ $t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894$ seconds.

8.

During the first few years of life, the rate at which a baby gains weight is proportional to the reciprocal of its weight.

Express this fact as a differential equation.
Suppose that a baby weighs 8 pounds at birth and 9 pounds one month later. How much will he weigh at one year?
Do you think this is a realistic model for a long time?

Answer

$\frac{dw}{dt} = \frac{k}{w} \text{.}$
$w(t) = \sqrt{7t+64} \text{;}$ $w(12) = \sqrt{148} \approx 12.17$ pounds.
The model is unrealistic.

Solution

Since the baby's weight, $w(t)$ (in pounds at month $t$), grows at a rate proportional to its reciprocal, we have

\begin{equation*} \frac{dw}{dt} = \frac{k}{w}\text{.} \end{equation*}
Given that $W(0) = 8$ and $W(1) = 9\text{,}$ we can both solve the IVP and determine $k\text{.}$ To start, we separate variables and solve the differential equation by writing

\begin{equation*} w \frac{dw}{dt} = k \end{equation*}

and integrating to get

\begin{equation*} \int w \frac{dw}{dt} \, dt = \int k \, dt\text{.} \end{equation*}

It follows that

\begin{equation*} \frac{1}{2}w^2 = kt + C \end{equation*}

and therefore $w(t) = \sqrt{2kt + C_1}$ (where $C_1 = 2C$ and we take the positive square root because we know $w(t) \gt 0$). Applying the initial condition $w(0) = 0\text{,}$ we find that $8 = \sqrt{C_1}\text{,}$ so $C_1 = 64$ and

\begin{equation*} w(t) = \sqrt{2kt + 64}\text{.} \end{equation*}

To find $k\text{,}$ we use the additional information that $w(1) = 9\text{,}$ so

\begin{equation*} 9 = \sqrt{2k + 64} \end{equation*}

which implies that $81 = 2k + 64$ and thus $k = 17/2\text{.}$ Thus,

\begin{equation*} w(t) = \sqrt{7t+64}\text{.} \end{equation*}

At one year, according to the model the baby weighs $w(12) = \sqrt{148} \approx 12.17$ pounds.
While the function $w(t) = \sqrt{7t+64}$ grows without bound as $t$ increases, the model is unrealistic more because the function initially doesn't grow fast enough. For example, the model predicts that at age $10\text{,}$ $w(120) \approx 30.06$ pounds, and at age $20\text{,}$ $w(120) = 41.76$ pounds. A realistic model would need to grow much more quickly early in life and then level off around age $20\text{.}$

9.

Suppose that you have a water tank that holds 100 gallons of water. A briny solution, which contains 20 grams of salt per gallon, enters the tank at the rate of 3 gallons per minute.

At the same time, the solution is well mixed, and water is pumped out of the tank at the rate of 3 gallons per minute.

Since 3 gallons enters the tank every minute and 3 gallons leaves every minute, what can you conclude about the volume of water in the tank.
How many grams of salt enters the tank every minute?
Suppose that $S(t)$ denotes the number of grams of salt in the tank in minute $t\text{.}$ How many grams are there in each gallon in minute $t\text{?}$
Since water leaves the tank at 3 gallons per minute, how many grams of salt leave the tank each minute?
Write a differential equation that expresses the total rate of change of $S\text{.}$
Identify any equilibrium solutions and determine whether they are stable or unstable.
Suppose that there is initially no salt in the tank. Find the amount of salt $S(t)$ in minute $t\text{.}$
What happens to $S(t)$ after a very long time? Explain how you could have predicted this only knowing how much salt there is in each gallon of the briny solution that enters the tank.

Answer

The inflow and outflow are at the same rate.
$60$ grams per minute.
$\frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}}$
$\frac{3S(t)}{100} \frac{\text{grams}}{\text{minute}} \text{.}$
$\frac{dS}{dt} = 60 - \frac{3}{100} S \text{.}$
$S = 2000$ is a stable equilibrium solution.
$S(t) = 2000 - 2000e^{-\frac{3}{100}t} \text{.}$
$S(t) \to 2000\text{.}$

Solution

The volume of solution in the tank remains constant because the inflow and outflow are at the same rate.
Salt enters the tank with the inflow, which occurs at a rate of $3$ gallons per minute at a concentration of $20$ grams per gallon. Observing that

\begin{equation*} 20 \frac{\text{grams}}{\text{gallon}} \cdot 3 \frac{\text{gallons}}{\text{minute}} = 60 \frac{\text{grams}}{\text{minute}}\text{,} \end{equation*}

we see that salt is entering the tank at a rate of $60$ grams per minute.
Letting $S(t)$ denotes the number of grams of salt in the tank at time $t\text{,}$ since there are $100$ gallons of solution in the tank at any time $t\text{,}$ the concentration of salt in the tank in grams per gallon is

\begin{equation*} \frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}} \end{equation*}

at any time $t\text{.}$
With water leaving the tank at $3$ gallons per minute, it follows that salt is leaving the tank at a rate of

\begin{equation*} \frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}} \cdot 3 \frac{\text{gallons}}{\text{minute}} = \frac{3S(t)}{100} \frac{\text{grams}}{\text{minute}}\text{.} \end{equation*}
Taking into account the rates at which salt enters and leaves the tank as established in (b) and (d) and noting that the units on $\frac{dS}{dt}$ are grams per minute, we have shown that

\begin{equation*} \frac{dS}{dt} = 60 - \frac{3}{100} S\text{.} \end{equation*}
Setting $60 - \frac{3}{100} S = 0\text{,}$ we find that $S = 2000$ is the only equilibrium solution. This makes sense because the inflow is carrying $20$ grams of salt per gallon and thus when the tank reaches this concentration, there will be $20 \cdot 100$ total grams of salt. This equilibrium is stable (both intuitively and because $\frac{dS}{dt}$ is positive for $S \lt 2000$ and negative for $S \gt 2000\text{.}$
Assuming that $S(0) = 0\text{,}$ we can solve the IVP with $\frac{dS}{dt} = 60 - \frac{3}{100} S\text{.}$ For somewhat simpler computations, we write $60 - \frac{3}{100}S = -\frac{3}{100}(S-2000)\text{.}$ Separating variables,

\begin{equation*} \frac{1}{S-2000} \frac{dS}{dt} = -\frac{3}{100} \end{equation*}

and integrating with respect to $t\text{,}$

\begin{equation*} \int \frac{1}{S-2000} \, dS = -\int \frac{3}{100} \, dt \end{equation*}

so that

\begin{equation*} \ln|S-2000| = -\frac{3}{100}t + C\text{.} \end{equation*}

Solving for $S$ in the usual way,

\begin{equation*} S = 2000 + Ke^{-\frac{3}{100}t} \end{equation*}

and using the initial condition that $S(0) = 0\text{,}$ we find $K = -2000$ so that

\begin{equation*} S(t) = 2000 - 2000e^{-\frac{3}{100}t}\text{.} \end{equation*}
As $t \to \infty\text{,}$ we know $e^{-\frac{3}{100}t} \to 0\text{,}$ and thus $S(t) \to 2000\text{.}$ This makes sense because the inflow carries $20$ grams of salt per liter continuously. As time goes on, the concentration of solution in the tank should tend to that same concentration, and with $100$ gallons of solution, the amount of salt will tend to $20 \cdot 100 = 2000$ grams. This value also matches the stable equilibrium that we found in (f).