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Section 7.6 Population Growth and the Logistic Equation

The growth of the earth's population is one of the pressing issues of our time. Will the population continue to grow? Or will it perhaps level off at some point, and if so, when? In this section, we look at two ways in which we may use differential equations to help us address these questions.

Before we begin, let's consider again two important differential equations that we have seen in earlier work this chapter.

Preview Activity 7.6.1.

Recall that one model for population growth states that a population grows at a rate proportional to its size.

  1. We begin with the differential equation

    \begin{equation*} \frac{dP}{dt} = \frac12 P\text{.} \end{equation*}

    Sketch a slope field below as well as a few typical solutions on the axes provided.

  2. Find all equilibrium solutions of the equation \(\frac{dP}{dt} = \frac12 P\) and classify them as stable or unstable.

  3. If \(P(0)\) is positive, describe the long-term behavior of the solution to \(\frac{dP}{dt} = \frac12 P\text{.}\)

  4. Let's now consider a modified differential equation given by

    \begin{equation*} \frac{dP}{dt} = \frac 12 P(3-P)\text{.} \end{equation*}

    As before, sketch a slope field as well as a few typical solutions on the following axes provided.

  5. Find any equilibrium solutions and classify them as stable or unstable.

  6. If \(P(0)\) is positive, describe the long-term behavior of the solution.

Subsection 7.6.1 The earth's population

We will now begin studying the earth's population. To get started, in Table 7.6.1 are some data for the earth's population in recent years that we will use in our investigations.

Year 1998 1999 2000 2001 2002 2005 2006 2007 2008 2009 2010
Pop
(billions)		 \(5.932\) \(6.008\) \(6.084\) \(6.159\) \(6.234\) \(6.456\) \(6.531\) \(6.606\) \(6.681\) \(6.756\) \(6.831\)
Table 7.6.1. Some recent population data for planet Earth.
Activity 7.6.2.

Our first model will be based on the following assumption:

The rate of change of the population is proportional to the population.

On the face of it, this seems pretty reasonable. When there is a relatively small number of people, there will be fewer births and deaths so the rate of change will be small. When there is a larger number of people, there will be more births and deaths so we expect a larger rate of change.

If \(P(t)\) is the population \(t\) years after the year 2000, we may express this assumption as

\begin{equation*} \frac{dP}{dt} = kP \end{equation*}

where \(k\) is a constant of proportionality.

  1. Use the data in the table to estimate the derivative \(P'(0)\) using a central difference. Assume that \(t=0\) corresponds to the year 2000.

  2. What is the population \(P(0)\text{?}\)

  3. Use your results from (a) and (b) to estimate the constant of proportionality \(k\) in the differential equation.

  4. Now that we know the value of \(k\text{,}\) we have the initial value problem

    \begin{equation*} \frac{dP}{dt} = kP, \ P(0) = 6.084\text{.} \end{equation*}

    Find the solution to this initial value problem.

  5. What does your solution predict for the population in the year 2010? Is this close to the actual population given in the table?

  6. When does your solution predict that the population will reach 12 billion?

  7. What does your solution predict for the population in the year 2500?

  8. Do you think this is a reasonable model for the earth's population? Why or why not? Explain your thinking using a couple of complete sentences.

Hint

  1. Small hints for each of the prompts above.

Answer

  1. \(P'(0) \approx 0.0755\text{.}\)

  2. \(P(0) = 6.084\text{.}\)

  3. \(k \approx 0.012041\text{.}\)

  4. \(P(t) = 6.084 e^{0.012041t}\text{.}\)

  5. \(P(10) \approx 6.8878\text{.}\)

  6. \(t = \frac{1}{0.012041} \ln \left( \frac{12}{6.084}\right) \approx 56.41\text{,}\) or in the year 2056.

  7. \(P(500) \approx 3012.3\) billion.

Solution

We let \(P(t)\) be the population after year 2000 with \(\frac{dP}{dt} = kP\text{,}\) where \(k\) is a constant of proportionality.

  1. Using the data in the table,

    \begin{equation*} P'(0) \approx \frac{P(1) - P(-1)}{2} \approx \frac{6.159 - 6.008}{2} \approx 0.0755\text{.} \end{equation*}

  2. Since \(t = 0\) corresponds to the year \(2000\text{,}\) \(P(0) = 6.084\text{.}\)

  3. Using \(\frac{dP}{dt} = kP\) and the preceding values at \(t = 0\text{,}\) we have \(0.0755 \approx k (6.084)\text{,}\) so \(k \approx 0.012041\text{.}\)

  4. The solution for the initial value problem is \(P(t) = 6.084 e^{0.012041t}\text{.}\)

  5. The year 2010 corresponds to \(t = 10\text{.}\) So \(P(10) \approx 6.8878\text{.}\) The model predicts that the population in 2010 will be about 6.888 billion.

  6. The population will be 12 billion when \(P(t) = 12\text{.}\) So we solve the equation \(6.084e^{0.012041t} = 12\) for \(t\text{.}\) We see that \(e^{0.012041t} = \frac{12}{6.084}\text{,}\) so \(t = \frac{1}{0.012041} \ln \left( \frac{12}{6.084}\right) \approx 56.41\text{.}\) The population will reach 12 billion during the year 2056.

  7. The year 2500 corresponds to \(t = 500\text{.}\) So \(P(500) \approx 3012.3\text{.}\) The model predicts that the population in 2500 will be about 3012 billion or about about 3 012 300 000 000. This is rather unreasonably large.

Our work in Activity 7.6.2 shows that that the exponential model is fairly accurate for years relatively close to 2000. However, if we go too far into the future, the model predicts increasingly large rates of change, which causes the population to grow arbitrarily large. This does not make much sense since it is unrealistic to expect that the earth would be able to support such a large population.

The constant \(k\) in the differential equation has an important interpretation. Let's rewrite the differential equation \(\frac{dP}{dt} = kP\) by solving for \(k\text{,}\) so that we have

\begin{equation*} k = \frac{dP/dt}{P}\text{.} \end{equation*}

We see that \(k\) is the ratio of the rate of change to the population; in other words, it is the contribution to the rate of change from a single person. We call this the per capita growth rate.

In the exponential model we introduced in Activity 7.6.2, the per capita growth rate is constant. This means that when the population is large, the per capita growth rate is the same as when the population is small. It is natural to think that the per capita growth rate should decrease when the population becomes large, since there will not be enough resources to support so many people. We expect it would be a more realistic model to assume that the per capita growth rate depends on the population \(P\text{.}\)

In the previous activity, we computed the per capita growth rate in a single year by computing \(k\text{,}\) the quotient of \(\frac{dP}{dt}\) and \(P\) (which we did for \(t = 0\)). If we return to the data in Table 7.6.1 and compute the per capita growth rate over a range of years, we generate the data shown in Figure 7.6.2, which shows how the per capita growth rate is a function of the population, \(P\text{.}\)

Figure 7.6.2. A plot of per capita growth rate vs. population \(P\text{.}\)

From the data, we see that the per capita growth rate appears to decrease as the population increases. In fact, the points seem to lie very close to a line, which is shown at two different scales in Figure 7.6.3.

Figure 7.6.3. The line that approximates per capita growth as a function of population, \(P\text{.}\)

Looking at this line carefully, we can find its equation to be

\begin{equation*} \frac{dP/dt}{P} = 0.025 - 0.002P\text{.} \end{equation*}

If we multiply both sides by \(P\text{,}\) we arrive at the differential equation

\begin{equation*} \frac{dP}{dt} = P(0.025 - 0.002P)\text{.} \end{equation*}

Graphing the dependence of \(dP/dt\) on the population \(P\text{,}\) we see that this differential equation demonstrates a quadratic relationship between \(\frac{dP}{dt}\) and \(P\text{,}\) as shown in Figure 7.6.4.

Figure 7.6.4. A plot of \(\frac{dP}{dt}\) vs. \(P\) for the differential equation \(\frac{dP}{dt} = P(0.025 - 0.002P)\text{.}\)

The equation \(\frac{dP}{dt} = P(0.025 - 0.002P)\) is an example of the logistic equation, and is the second model for population growth that we will consider. We expect that it will be more realistic, because the per capita growth rate is a decreasing function of the population.

Indeed, the graph in Figure 7.6.4 shows that there are two equilibrium solutions, \(P=0\text{,}\) which is unstable, and \(P=12.5\text{,}\) which is a stable equilibrium. The graph shows that any solution with \(P(0) \gt 0\) will eventually stabilize around 12.5. Thus, our model predicts the world's population will eventually stabilize around 12.5 billion.

A prediction for the long-term behavior of the population is a valuable conclusion to draw from our differential equation. We would, however, also like to answer some quantitative questions. For instance, how long will it take to reach a population of 10 billion? To answer this question, we need to find an explicit solution of the equation.

Subsection 7.6.2 Solving the logistic differential equation

Since we would like to apply the logistic model in more general situations, we state the logistic equation in its more general form,

\begin{equation} \frac{dP}{dt} = kP(N-P)\text{.}\label{yri}\tag{7.6.1} \end{equation}

The equilibrium solutions here are \(P=0\) and \(1-\frac PN = 0\text{,}\) which shows that \(P=N\text{.}\) The equilibrium at \(P=N\) is called the carrying capacity of the population for it represents the stable population that can be sustained by the environment.

We now solve the logistic equation (7.6.1). The equation is separable, so we separate the variables

\begin{equation*} \frac{1}{P(N-P)}\frac{dP}{dt} = k\text{,} \end{equation*}

and integrate to find that

\begin{equation*} \int \frac{1}{P(N-P)}~dP = \int k~dt\text{.} \end{equation*}

To find the antiderivative on the left, we use the partial fraction decomposition

\begin{equation*} \frac{1}{P(N-P)} = \frac 1N\left[\frac 1P + \frac 1{N-P}\right]\text{.} \end{equation*}

Now we are ready to integrate, with

\begin{equation*} \int \frac 1N\left[\frac 1P + \frac 1{N-P}\right] ~dP = \int k~dt\text{.} \end{equation*}

On the left, observe that \(N\) is constant, so we can remove a factor of \(\frac{1}{N}\) and antidifferentiate to find that

\begin{equation*} \frac 1N (\ln|P| - \ln|N-P|) = kt + C\text{.} \end{equation*}

Multiplying both sides of this last equation by \(N\) and using a rule of logarithms, we next find that

\begin{equation*} \ln\left|\frac{P}{N-P}\right| = kNt + C\text{.} \end{equation*}

From the definition of the logarithm, replacing \(e^C\) with \(C\text{,}\) and letting \(C\) absorb the absolute value signs, we now know that

\begin{equation*} \frac{P}{N-P} = Ce^{kNt}\text{.} \end{equation*}

At this point, all that remains is to determine \(C\) and solve algebraically for \(P\text{.}\)

If the initial population is \(P(0) = P_0\text{,}\) then it follows that \(C = \frac{P_0}{N-P_0}\text{,}\) so

\begin{equation*} \frac{P}{N-P} = \frac{P_0}{N-P_0}e^{kNt}\text{.} \end{equation*}

We will solve this equation for \(P\) by multiplying both sides by \((N-P)(N-P_0)\) to obtain

\begin{align*} P(N-P_0) &= P_0(N-P)e^{kNt}\\ &= P_0Ne^{kNt} - P_0Pe^{kNt}\text{.} \end{align*}

Solving for \(P_0Ne^{kNt}\text{,}\) expanding, and factoring, it follows that

\begin{align*} P_0Ne^{kNt} &= P(N-P_0) + P_0 Pe^{kNt}\\ &= P(N-P_0 + P_0e^{kNt})\text{.} \end{align*}

Dividing to solve for \(P\text{,}\) we see that

\begin{equation*} P = \frac{P_0Ne^{kNt}}{N-P_0 + P_0e^{kNt}}\text{.} \end{equation*}

Finally, we choose to multiply the numerator and denominator by \(\frac{1}{P_0}e^{-kNt}\) to obtain

\begin{equation*} P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\text{.} \end{equation*}

While that was a lot of algebra, notice the result: we have found an explicit solution to the logistic equation.

Solution to the Logistic Equation.

The solution to the initial value problem

\begin{equation*} \frac{dP}{dt} = kP(N-P), \ P(0) = P_0\text{,} \end{equation*}

is

\begin{equation} P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\text{.}\label{gnE}\tag{7.6.2} \end{equation}

For the logistic equation describing the earth's population that we worked with earlier in this section, we have

\begin{equation*} k=0.002, N= 12.5, \ \text{and} \ P_0 = 6.084\text{.} \end{equation*}

This gives the solution

\begin{equation*} P(t) = \frac{12.5}{1.0546e^{-0.025t} + 1}\text{,} \end{equation*}

whose graph is shown in Figure 7.6.5.

Figure 7.6.5. The solution to the logistic equation modeling the earth's population.

The graph shows the population leveling off at 12.5 billion, as we expected, and that the population will be around 10 billion in the year 2050. These results, which we have found using a relatively simple mathematical model, agree fairly well with predictions made using a much more sophisticated model developed by the United Nations.

The logistic equation is good for modeling any situation in which limited growth is possible. For instance, it could model the spread of a flu virus through a population contained on a cruise ship, the rate at which a rumor spreads within a small town, or the behavior of an animal population on an island. Through our work in this section, we have completely solved the logistic equation, regardless of the values of the constants \(N\text{,}\) \(k\text{,}\) and \(P_0\text{.}\) Anytime we encounter a logistic equation, we can apply the formula we found in Equation (7.6.2).

Activity 7.6.3.

Consider the logistic equation

\begin{equation*} \frac{dP}{dt} = kP(N-P) \end{equation*}

with the graph of \(\frac{dP}{dt}\) vs. \(P\) shown in Figure 7.6.6.

Figure 7.6.6. Plot of \(\frac{dP}{dt}\) vs. \(P\text{.}\)

  1. At what value of \(P\) is the rate of change greatest?

  2. Consider the model for the earth's population that we created. At what value of \(P\) is the rate of change greatest? How does that compare to the population in recent years?

  3. According to the model we developed, what will the population be in the year 2100?

  4. According to the model we developed, when will the population reach 9 billion?

  5. Now consider the general solution to the general logistic initial value problem that we found, given by

    \begin{equation*} P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right)e^{-kNt} + 1}\text{.} \end{equation*}

    Verify algebraically that \(P(0) = P_0\) and that \(\lim_{t\to\infty} P(t) = N\text{.}\)

Hint

  1. Small hints for each of the prompts above.

Answer

  1. When \(P = \frac{N}{2}\text{.}\)

  2. When the population is 6.125 billion.

  3. \(P = \frac{12.5}{1.0546e^{-0.025t} + 1} \text{;}\) \(P(100) = 11.504\) billion.

  4. \(t = \frac{1}{-0.025} \ln \left( \frac{\left( \frac{12.5}{9} - 1 \right)}{1.0546}\right) \approx 39.9049\) (so in about year \(2040\)).

  5. \(\lim_{t \to \infty} P(t) = N\text{.}\)

Solution

Consider the logistic equation \(\frac{dP}{dt} = kP(N - P)\text{.}\)

  1. The maximum value for \(\frac{dP}{dt}\) will occur when \(P = \frac{N}{2}\) since this is the highest point on the graph.

  2. The model we created for the population of the earth was \(\frac{dP}{dt} = P(0.025 - 0.002P\text{,}\) which can be rewritten as \(\frac{dP}{dt} = 0.002P(12.5 - P)\text{.}\) So for this model, the maximum rate of change for the earth's population will occur when the population is 6.125 billion. This value is slightly less than the earth's population in recent years.

  3. So for the logistic model, we have \(k = 0.002\text{,}\) \(N = 12.5\text{.}\) Using the year 2000 as \(t = 0\text{,}\) the initial population is 6.084. So the solution for the differential equation in part (b) is

    \begin{equation*} P = \frac{12.5}{1.0546e^{-0.025t} + 1}\text{.} \end{equation*}

    For the year 2100, we use \(t = 100\) and this model predicts that in the year 2100, the earth's population will be 11.504 billion.

  4. To determine when this model predicts that the earth's population will be 9 billion, we solve the equation

    \begin{equation*} \frac{12.5}{1.0546e^{-0.025t} + 1} = 9 \end{equation*}

    for \(t\text{.}\) Multiplying both sides by \(\left( 1.0546e^{-0.025t} + 1 \right)\text{,}\) we have

    \begin{equation*} 12.5 = 9 \left( 1.0546e^{-0.025t} + 1 \right)\text{.} \end{equation*}

    Dividing by \(9\) then subtracting \(1\) from both sides, \(1.0546e^{-0.025t} = \left( \frac{12.5}{9} - 1 \right)\text{.}\) Thus, \(e^{-0.025t} = \frac{\left( \frac{12.5}{9} - 1 \right)}{1.0546}\text{.}\) Using logarithms in the usual way, \(t = \frac{1}{-0.025} \ln \left( \frac{\left( \frac{12.5}{9} - 1 \right)}{1.0546}\right) \approx 39.9049\text{.}\) The earth's population will reach 9 billion in about 2040.

  5. The solution for the general logistic initial value problem is

    \begin{equation*} P(t) = \frac{N}{\left( \frac{N - P_0}{P_0} \right) e^{-kNt} + 1}\text{.} \end{equation*}

    For this solution, we see that

    \begin{align*} P(0) \amp = \frac{N}{\left( \frac{N - P_0}{P_0} \right) + 1}\\ \amp = \frac{N}{\frac{N - P_0 + P_0}{P_0}}\\ \amp = \frac{N P_0}{N}\\ \amp = P_0 \end{align*}

    To determine \(\lim_{t \to \infty} P(t)\text{,}\) we first note that since \(-kN \lt 0\text{,}\) \(\lim_{t \to \infty} e^{-kNt} = 0\text{.}\) Hence,

    \begin{equation*} \lim_{t \to \infty}\left[ \left( \frac{N - P_0}{P_0} \right)e^{-kNt} + 1 \right] = 1\text{,} \end{equation*}

    and therefore,

    \begin{equation*} \lim_{t \to \infty} P(t) = \lim_{t\to \infty} \frac{N}{\left( \frac{N - P_0}{P_0} \right) e^{-kNt} + 1} = \frac{N}{1} = N\text{.} \end{equation*}

Subsection 7.6.3 Summary

  • If we assume that the rate of growth of a population is proportional to the population, we are led to a model in which the population grows without bound and at a rate that grows without bound.

  • By assuming that the per capita growth rate decreases as the population grows, we are led to the logistic model of population growth, which predicts that the population will eventually stabilize at the carrying capacity.

Exercises 7.6.4 Exercises

1. Analyzing a logistic equation.
2. Analyzing a logistic model.
3. Finding a logistic function for an infection model.
4. Analyzing a population growth model.
5.

The logistic equation may be used to model how a rumor spreads through a group of people. Suppose that \(p(t)\) is the fraction of people that have heard the rumor on day \(t\text{.}\) The equation

\begin{equation*} \frac{dp}{dt} = 0.2p(1-p) \end{equation*}

describes how \(p\) changes. Suppose initially that one-tenth of the people have heard the rumor; that is, \(p(0) = 0.1\text{.}\)

  1. What happens to \(p(t)\) after a very long time?

  2. Determine a formula for the function \(p(t)\text{.}\)

  3. At what time is \(p\) changing most rapidly?

  4. How long does it take before 80% of the people have heard the rumor?

Answer

  1. \(p(t) \to 1\) as \(t \to \infty\) provided \(p(0) \gt 0\text{.}\)

  2. \(p(t) = \frac{1}{9e^{-0.2t} + 1} \text{.}\)

  3. \(t = -5 \ln(1/9) \approx 10.986\) days.

  4. \(t = -5 \ln(0.2/9) \approx 19.033\) days.

Solution

  1. The equilibrium solutions for this logistic equation are \(p = 0\) and \(p = 1\text{.}\) Since \(p = 1\) is the carrying capacity and thus a stable equilibrium, \(p(t) \to 1\) as \(t \to \infty\) provided \(p(0) \gt 0\text{.}\)

  2. We know that the general solution to the logistic equation \(\frac{dP}{dt} = kP(N-P), \ P(0) = P_0\) is

    \begin{equation*} P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\text{.} \end{equation*}

    In the context of this problem, we know that \(k = 0.2\text{,}\) \(N = 1\text{,}\) and \(p(0) = 0.1\) and therefore the solution is

    \begin{equation*} p(t) = \frac{1}{\left(\frac{1-0.1}{0.1}\right) e^{-0.2t} + 1} = \frac{1}{9e^{-0.2t} + 1}\text{.} \end{equation*}

  3. We know that \(p\) is changing most rapidly when its derivative, \(p'\text{,}\) is largest. Recalling the original differential equation, we know that

    \begin{equation*} p' = 0.2p(1-p) \end{equation*}

    which shows that \(p'\) is a quadratic function of \(p\text{.}\) That quadratic function has zeros at \(p = 0\) and \(p = 1\text{,}\) so the vertex of this concave down quadratic function lies at \(p = \frac{1}{2}\text{,}\) and at this value of \(p\) the rate of change of \(p\) is largest. Thus, we solve the equation \(p(t) = 0.5\) to find the time at which the population is increasing fastest. We see that

    \begin{equation*} 0.5 = \frac{1}{9e^{-0.2t} + 1} \end{equation*}

    so

    \begin{equation*} 9e^{-0.2t} + 1 = 2 \end{equation*}

    and \(e^{-0.2t} = 1/9\text{.}\) It follows that \(t = -5 \ln(1/9) \approx 10.986\) days.

  4. Since \(p(t)\) tells us the percentage of people who've heard the rumor, we need to solve the equation \(p(t) = 0.8\text{.}\) Thus we have

    \begin{equation*} 0.8 = \frac{1}{9e^{-0.2t} + 1} \end{equation*}

    so

    \begin{equation*} 9e^{-0.2t} + 1 = 1.2 \end{equation*}

    and \(e^{-0.2t} = 0.2/9\text{.}\) It follows that \(t = -5 \ln(0.2/9) \approx 19.033\) days.

6.

Suppose that \(b(t)\) measures the number of bacteria living in a colony in a Petri dish, where \(b\) is measured in thousands and \(t\) is measured in days. One day, you measure that there are 6,000 bacteria and the per capita growth rate is 3. A few days later, you measure that there are 9,000 bacteria and the per capita growth rate is 2.

  1. Assume that the per capita growth rate \(\frac{db/dt}{b}\) is a linear function of \(b\text{.}\) Use the measurements to find this function and write a logistic equation to describe \(\frac{db}{dt}\text{.}\)

  2. What is the carrying capacity for the bacteria?

  3. At what population is the number of bacteria increasing most rapidly?

  4. If there are initially 1,000 bacteria, how long will it take to reach 80% of the carrying capacity?

Answer

  1. \(\frac{db}{dt} = \frac{1}{3000} b(15000 - b)\)

  2. \(b = 15000\text{.}\)

  3. When \(b = 7500\text{.}\)

  4. \(t = -\frac{1}{5} \ln(1/70) \approx 0.8497\) days.

Solution

  1. Let \(y\) be the per capita growth rate as a function of \(b\text{.}\) Using the given data in the problem statement, we know that this linear function passes through the points \((6000,3)\) and \((9000,2)\text{.}\) The equation of this line is thus

    \begin{equation*} y - 3 = -\frac{1}{3000}(b - 6000) \end{equation*}

    so \(y = (3 + 2) - \frac{1}{3000}b = 5 - \frac{1}{3000}b\text{.}\) Since the logistic equation comes from taking the product of the population at time \(t\) with its per capita growth rate at that time, we find that

    \begin{equation*} \frac{db}{dt} = b \left( 5 - \frac{1}{3000}b \right) = \frac{1}{3000} b(15000 - b) \end{equation*}

  2. From the form of the logistic equation, we see that the carrying capacity for the bacteria is \(b = 15000\text{,}\) which is a stable equilibrium for the population.

  3. The bacteria is increasing most rapidly when \(\frac{db}{dt}\) is greatest. Since

    \begin{equation*} \frac{db}{dt} = \frac{1}{3000} b(15000 - b) \end{equation*}

    is a quadratic function of \(b\text{,}\) \(\frac{db}{dt}\) is maximized at the vertex of this concave down parabola, which occurs halfway between its zeros: \(b = 0\) and \(b = 15000\text{.}\) Thus, when \(b = 7500\text{,}\) the population is growing fastest. To find the exact \(t\)-value when this occurs, we'd need to know an initial condition like in the next part of the problem.

  4. Using \(b(0) = 1000\text{,}\) we solve the logistic equation IVP with

    \begin{equation*} \frac{db}{dt} = \frac{1}{3000} b(15000 - b)\text{.} \end{equation*}

    Using \(N = 15000\text{,}\) \(k = \frac{1}{3000}\text{,}\) and \(b_0 = 1000\text{,}\) we know from the standard form of the solution to the logistic equation that

    \begin{equation*} b(t) = \frac{15000}{\left(\frac{15000-1000}{1000}\right) e^{-5t} + 1} = \frac{15000}{14e^{-5t} + 1}\text{.} \end{equation*}

    To find how long it takes the bacteria to reach 80% of the carrying capacity, we solve the equation

    \begin{equation*} 0.8 \cdot 15 = \frac{15000}{14e^{-5t} + 1}\text{.} \end{equation*}

    That equation implies that

    \begin{equation*} 0.8 = \frac{1}{14e^{-5t} + 1}\text{,} \end{equation*}

    so \(14e^{-5t} + 1 = 1.2\) and thus \(e^{-5t} = 0.2/14 = 1/70\text{.}\) It follows that \(t = -\frac{1}{5} \ln(1/70) \approx 0.8497\) days.

7.

Suppose that the population of a species of fish is controlled by the logistic equation

\begin{equation*} \frac{dP}{dt} = 0.1P(10 - P)\text{,} \end{equation*}

where \(P\) is measured in thousands of fish and \(t\) is measured in years.

  1. What is the carrying capacity of this population?

  2. Suppose that a long time has passed and that the fish population is stable at the carrying capacity. At this time, humans begin harvesting 20% of the fish every year. Modify the differential equation by adding a term to incorporate the harvesting of fish.

  3. What is the new carrying capacity?

  4. What will the fish population be one year after the harvesting begins?

  5. How long will it take for the population to be within 10% of the carrying capacity?

Answer

  1. 10000 fish.

  2. \(\frac{dP}{dt} = 0.1P(10 - P) - 0.2P \text{.}\)

  3. \(8000\) fish.

  4. \(P(1) \approx 8.7899\) thousand fish.

  5. \(t = -1.2 \ln(5/11) \approx 0.9461\) years.

Solution

  1. From the form of the given logistic equation, we can see the carrying capacity is 10000 fish.

  2. In this situation, the initial condition is \(P(0) = 10\text{,}\) and by removing \(20\)% of the population per year, we subtract \(0.2P\) from the rate that defines the differential equation, thus finding

    \begin{equation*} \frac{dP}{dt} = 0.1P(10 - P) - 0.2P\text{.} \end{equation*}

  3. Expanding and rewriting the differential equation in (b), we see that

    \begin{equation*} \frac{dP}{dt} = P - 0.1P^2 - 0.2P = 0.8P - 0.1P^2 = 0.1P(8 - P)\text{.} \end{equation*}

    This, too, is a logistic equation, with carrying capacity \(8000\) fish.

  4. We can solve this logistic equation in the usual way (keeping in mind the initial condition \(P(0) = 10\text{,}\) and noting that the carrying capacity is \(8\)):

    \begin{equation*} P(t) = \frac{8}{\frac{8-10}{10}e^{-0.1 \cdot 8 t}+1} = \frac{8}{-0.2e^{-0.8t}+1} \end{equation*}

    After one year, the population of fish will thus be

    \begin{equation*} P(1) \approx 8.7899 \end{equation*}

    thousand fish.

  5. To be within 10% of the carrying capacity, we need \(0.9(8) \le P(t) \le 1.1(8)\text{.}\) Since the initial condition is larger than the carrying capacity, we want to know when \(P(t) = 1.1(8) = 8.8\text{.}\) Thus we solve the equation

    \begin{equation*} 8.8 = \frac{8}{-0.2e^{-0.8t}+1}\text{,} \end{equation*}

    which implies \(-0.2e^{-0.8t}+1 = 8/8.8 = 10/11\text{.}\) Thus, \(e^{-0.8t} = 5/11\text{,}\) so \(t = -1.2 \ln(5/11) \approx 0.9461\) years.

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