Coordinated Calculus

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

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

Since \(5\cdot10^{10}\) cubic meters of water flow out each year⁵and we assume that each cubic meter of water that flows out carries with it the plastic pollution it contains, then the plastic pollution leaves the lake at the rate of

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,

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 Figure8.47 and Figure8.48.

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

Separating variables, we find that

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

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

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

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

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

Using this value of \(C\) in Equation(8.1) and solving for \(P\text{,}\) we arrive at the solution

Superimposing the graph of \(P\) on the slope field we saw in Figure8.48, we see, as shown in Figure8.49

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

1. Small hints for each of the prompts above.

1. \(\frac{dA}{dt} = 0.05A\text{.}\)

2. \(\frac{dA}{dt} = 0.05A - 10000\text{.}\)

3. 

   The only equilibrium solution is \(A = 200000\text{.}\)

4. \(t = 20 \ln(2) \approx 13.86\)years.

5. At least $200000.

6. Up to $15000 every year.

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

1. The rate of change of \(A\) with respect to \(t\) is \(\frac{dA}{dt} = 0.05A\text{.}\)

2. If $10000 is being withdrawn every year, then

   \begin{equation*} \frac{dA}{dt} = 0.05A - 10000\text{.} \end{equation*}

3. Following is a slope field for the differential equation in (b).

   

   Equilibrium solutions occur where \(\frac{dA}{dt} = 0\) and so the only equilibrium 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.

4. Suppose we initially have $100000 in the account. We will solve the initial 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.

5. You must have at least $200000 in the account to make sure the account will never be depleted.

6. Now suppose the initial 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.

1. Small hints for each of the prompts above.

1. \(\frac{dM}{dt} = -kM\text{,}\) where \(k\) is a positive constant.

2. \(k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}\)

3. \(\frac{dM}{dt} = 3 - kM\text{,}\) where \(k\) is a positive constant.

4. The equilibrium solution \(mM = \frac{3}{k}\) is stable.

5. \(M = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}\)

6. About 2.426 milligrams per hour.

1. We will use the differential equation \(\frac{dM}{dt} = -kM\text{,}\) where \(k\) is a positive constant.

2. 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 \(k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}\)

3. Noting that the rate of change of the drug in the body will be the difference 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.

4. 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.

5. 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 \(M = \frac{1}{k} \left( 3 - 3e^{-kt} \right) = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}\)

6. 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 equilibrium 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.

1. Differentiation term by term gives

   \begin{equation*} y' = \sum_{k=1}^{\infty} ka_kx_{k-1}\text{.} \end{equation*}

   We then substitute this series into the differential equation (8.2) to obtain the equation

2. When we write the first few terms of the series on either side of our differential equation we obtain

   \begin{align*} \amp a_1 + (2)a_2x + (3)a_3x^2 + (4)a_4x^3 + \cdots + (k+1)a_{k+1}x^{k} + \cdots\\ \amp= a_0 + a_1x + a_2x^2 + \cdots + a_kx^k + \cdots\text{.} \end{align*}

   Equating the constant terms gives us \(a_1 = a_0\text{.}\)

3. Equating the degree 1 terms gives us \(2a_2 = a_1\) or \(a_2 = \frac{a_1}{2}\text{.}\) Since \(a_1 = a_0\text{,}\) we have \(a_2 = \frac{a_0}{2}\text{.}\)

4. Equating the degree 2 terms gives us \(3a_3 = a_2\) or \(a_3 = \frac{a_2}{3}\text{.}\) Since \(a_2 = \frac{a_0}{2}\text{,}\) we have \(a_3 = \frac{a_0}{3!}\text{.}\)

5. Equating the degree 3 terms gives us \(4a_4 = a_3\) or \(a_4 = \frac{a_3}{4}\text{.}\) Since \(a_3 = \frac{a_0}{3!}\text{,}\) we have \(a_4 = \frac{a_0}{4!}\text{.}\)

6. Equating the degree \(k-1\) terms gives us \(ka_k = a_{k-1}\) or \(a_k = \frac{a_{k-1}}{k}\text{.}\) It appears that \(a_{k-1} = \frac{a_0}{(k-1)!}\text{,}\) so we have

   \begin{equation*} a_k = \frac{a_0}{k!}\text{.} \end{equation*}

7. Since \(a_k = \frac{a_0}{k!}\) we have

   \begin{equation*} y = a_0 \sum_{k=0}^{\infty} \frac{x^k}{k!}\text{.} \end{equation*}

   So the functions that satisfy the differential equation (8.2) are the exponential functions of the form \(y = a_0e^x\text{.}\)