Math Core

Exponential growth $\dfrac{dP}{dt} = kP$ predicts that a population grows forever, faster and faster. Real populations run out of food, space or customers, and their growth levels off. The logistic differential equation models that leveling off, and AP Calculus BC expects you to read a lot from it without ever solving it.

The logistic differential equation

In the logistic model, the growth rate is proportional both to the population $P$ and to how far $P$ is from a ceiling $L$ .

Definition

Logistic differential equation

\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right), \qquad k \gt 0,\ L \gt 0.

The constant $k$ is the relative growth rate when $P$ is small, and $L$ is the carrying capacity: the population level the environment can sustain.

The factor $\left(1 - \dfrac{P}{L}\right)$ is what makes this model different from exponential growth:

When $P$ is small compared with $L$ , the factor is close to $1$ and $\dfrac{dP}{dt} \approx kP$ . Growth looks exponential.
As $P$ approaches $L$ , the factor approaches $0$ and growth slows to a stop.
If $P$ is larger than $L$ , the factor is negative, so $P$ decreases back toward $L$ .

You will also see the equivalent form $\dfrac{dP}{dt} = cP(L - P)$ . Here $L$ is still the carrying capacity, and $c = \dfrac{k}{L}$ . To find $L$ from any logistic equation, factor out $P$ and find the value of $P$ that makes the remaining factor zero.

Equilibrium solutions and long-run behavior

Setting $\dfrac{dP}{dt} = 0$ gives $P = 0$ or $P = L$ . These constant functions are the equilibrium solutions. Every other solution with $P(0) \gt 0$ heads toward $L$ :

starting value	sign of $\dfrac{dP}{dt}$	behavior	$\displaystyle\lim_{t \to \infty} P(t)$
$P(0) = 0$	$0$	stays at $0$	$0$
$0 \lt P(0) \lt L$	positive	increases toward $L$	$L$
$P(0) = L$	$0$	stays at $L$	$L$
$P(0) \gt L$	negative	decreases toward $L$	$L$

Where the population grows fastest

The rate $\dfrac{dP}{dt} = kP - \dfrac{k}{L}P^2$ is a downward-opening quadratic in $P$ with zeros at $P = 0$ and $P = L$ . A parabola reaches its maximum halfway between its zeros, so the growth rate is largest when $P = \dfrac{L}{2}$ .

You can also see this through the second derivative. Differentiating with respect to $t$ (using the chain rule, since $P$ depends on $t$ ):

\frac{d^2P}{dt^2} = \left(k - \frac{2k}{L}P\right)\frac{dP}{dt}.

For $0 \lt P \lt L$ , the factor $\dfrac{dP}{dt}$ is positive, so the sign of $\dfrac{d^2P}{dt^2}$ matches the sign of $k - \dfrac{2k}{L}P$ : positive when $P \lt \dfrac{L}{2}$ and negative when $P \gt \dfrac{L}{2}$ .

Reading a logistic equation

For $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{L}\right)$ with $0 \lt P(0) \lt L$ :

$P$ is increasing, and $\displaystyle\lim_{t \to \infty} P(t) = L$ .
$P$ is growing fastest when $P = \dfrac{L}{2}$ . This is the inflection point of the solution curve.
The graph is concave up while $P \lt \dfrac{L}{2}$ and concave down while $P \gt \dfrac{L}{2}$ , which gives the familiar S-shape.

Worked example: Reading the equation

A fish population satisfies $\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{2000}\right)$ with $P(0) = 300$ . Find the carrying capacity, the population when it is growing fastest, and the growth rate when $P = 500$ .

Solution. The carrying capacity is $L = 2000$ fish. Growth is fastest at $P = \dfrac{2000}{2} = 1000$ . When $P = 500$ ,

\frac{dP}{dt} = 0.4(500)\left(1 - \frac{500}{2000}\right) = 200(0.75) = 150 \text{ fish per unit of time}.

Worked example: The other form

A rumor spreads through a school of 800 students according to $\dfrac{dy}{dt} = 0.003\,y\,(800 - y)$ , where $y$ is the number of students who have heard it after $t$ hours. What is the maximum rate at which the rumor spreads?

Solution. The factor $800 - y$ is zero at $y = 800$ , so $L = 800$ . The rate is greatest when $y = 400$ :

\frac{dy}{dt} = 0.003(400)(400) = 480 \text{ students per hour}.

Rewriting as $0.003(800)\,y\left(1 - \dfrac{y}{800}\right)$ shows $k = 2.4$ .

The logistic solution

The logistic equation is separable (with partial fractions, a BC technique), and its solution is

P(t) = \frac{L}{1 + Ae^{-kt}}, \qquad \text{where } A = \frac{L - P(0)}{P(0)}.

You can check the value of $A$ : at $t = 0$ , $P(0) = \dfrac{L}{1 + A}$ , so $1 + A = \dfrac{L}{P(0)}$ . As $t \to \infty$ , $e^{-kt} \to 0$ and $P(t) \to L$ , matching everything above.

P(t) = 1000/(1 + 9e^(−0.5t)) rises in an S-shape toward the carrying capacity 1000. It grows fastest at P = 500, when t = 2 ln 9 ≈ 4.39.Open in grapher →

Worked example: Using the solution

A population satisfies $\dfrac{dP}{dt} = 0.5P\left(1 - \dfrac{P}{1000}\right)$ with $P(0) = 100$ . Write $P(t)$ and find when the population is growing fastest.

Solution. Here $L = 1000$ , $k = 0.5$ , and $A = \dfrac{1000 - 100}{100} = 9$ , so

P(t) = \frac{1000}{1 + 9e^{-0.5t}}.

Growth is fastest when $P = 500$ : $1 + 9e^{-0.5t} = 2$ , so $e^{-0.5t} = \dfrac{1}{9}$ , and $t = 2\ln 9 \approx 4.394$ .

Common mistake

The fastest growth happens at $P = \dfrac{L}{2}$ , which is a population value, not a time. If a question asks when growth is fastest, you need the solution formula (or given data) to turn $P = \dfrac{L}{2}$ into a value of $t$ . And if $P(0) \gt \dfrac{L}{2}$ , the population never passes through $\dfrac{L}{2}$ , so the growth rate is largest at the start.

Tip

A quick way to spot a logistic equation: $\dfrac{dP}{dt}$ is a quadratic in $P$ with no constant term and a negative $P^2$ coefficient, like $\dfrac{dP}{dt} = 0.2P - 0.0005P^2$ . Factor out $P$ to read off the carrying capacity: $0.2P\left(1 - \dfrac{P}{400}\right)$ gives $L = 400$ .

Practice

Practice 1

A population satisfies $\dfrac{dP}{dt} = 0.05P\left(1 - \dfrac{P}{1200}\right)$ . What is the carrying capacity?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A population satisfies $\dfrac{dP}{dt} = 0.002P(900 - P)$ with $P(0) = 50$ . For what value of $P$ is the population growing fastest?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

For the population in the previous problem, $\dfrac{dP}{dt} = 0.002P(900 - P)$ , what is the maximum value of $\dfrac{dP}{dt}$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

A population satisfies $\dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{1000}\right)$ with $P(0) = 1500$ . Find $\displaystyle\lim_{t \to \infty} P(t)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Let $P(t)$ satisfy $\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{1000}\right)$ with $P(0) = 300$ . Which describes the graph of $P$ for $t \ge 0$ ?

Practice 6

Find the particular solution $P(t)$ of $\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{600}\right)$ with $P(0) = 100$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 7

For $P(t) = \dfrac{600}{1 + 5e^{-0.2t}}$ from the previous problem, at what time $t$ is the population growing fastest? Give an exact answer or a decimal to the nearest hundredth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A population satisfies $\dfrac{dP}{dt} = \dfrac{P}{5} - \dfrac{P^2}{4000}$ with $P(0) = 20$ . Find $\displaystyle\lim_{t \to \infty} P(t)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.