Logistic Models

Exponential growth can’t last forever. A population of fish in a lake, a rumour in a school, or the number of people who own a new phone all grow quickly at first, then slow down as they run out of room, food, or new people to reach. The logistic model captures this. It starts out like exponential growth but levels off at a maximum called the carrying capacity. On the AP BC exam, you’re expected to read a lot of information straight from the differential equation, without solving it.

Key ideas

The logistic differential equation

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

• $k$ is the growth constant: when $P$ is small, $1 - \dfrac{P}{L} \approx 1$ and $\dfrac{dP}{dt} \approx kP$, which is exponential growth.
• $L$ is the carrying capacity. As $P$ gets close to $L$, the factor $1 - \dfrac{P}{L}$ gets close to $0$ and growth slows to a stop.

You’ll also see the form $\dfrac{dP}{dt} = kP(L - P)$. It’s the same idea with a different constant out front. In either form, read $L$ from the factor that is zero when $P = L$. For example, $\dfrac{dP}{dt} = 0.002P(800 - P)$ has $L = 800$.

Reading behaviour from the equation

The sign of $\dfrac{dP}{dt}$ tells you everything about the long run:

Starting value	Sign of $\dfrac{dP}{dt}$	What happens
$P(0) = 0$ or $P(0) = L$	$0$	$P$ stays constant (equilibrium)
$0 \lt P(0) \lt L$	positive	$P$ increases toward $L$
$P(0) \gt L$	negative	$P$ decreases toward $L$

So for any starting value $P(0) \gt 0$:

$$\lim_{t \to \infty} P(t) = L$$

Fastest growth at half the carrying capacity

Think of $\dfrac{dP}{dt} = kP - \dfrac{k}{L}P^2$ as a function of $P$. It’s a downward-opening parabola with zeros at $P = 0$ and $P = L$, so its maximum is halfway between, at $P = \dfrac{L}{2}$.

That means the population is growing fastest when $P = \dfrac{L}{2}$. On the graph of $P$ against $t$ this is the inflection point: below $\dfrac{L}{2}$ the curve is concave up, above it the curve is concave down. You can confirm it with the second derivative:

$$\frac{d^2P}{dt^2} = k\,\frac{dP}{dt}\left(1 - \frac{2P}{L}\right),$$

which changes sign at $P = \dfrac{L}{2}$ (when $0 \lt P \lt L$).

Solutions of $\frac{dP}{dt} = 0.4P\left(1 - \frac{P}{500}\right)$: every positive starting value leads to $P \to 500$, and the rising curves are steepest at $P = 250$.

The solution formula

Solving the logistic equation by separation of variables (see Example 4) gives

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

As $t \to \infty$, $e^{-kt} \to 0$, so $P \to L$, matching what you read from the equation. The AP exam focuses on reading the differential equation; you won’t usually be asked to derive this formula, but it’s good to see where it comes from.

Worked examples

Example 1: Reading a logistic equation

A deer population satisfies $\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right)$, where $t$ is in years, and $P(0) = 100$.

• (a) Find $\displaystyle\lim_{t \to \infty} P(t)$.
• (b) For what value of $P$ is the population growing fastest? How fast is it growing then?

Solution.

(a) The carrying capacity is $L = 500$, and $0 \lt 100 \lt 500$, so $P$ increases toward $500$: $\displaystyle\lim_{t \to \infty} P(t) = 500$.

(b) Growth is fastest at $P = \dfrac{L}{2} = 250$. Then

$$\frac{dP}{dt} = 0.4(250)\left(1 - \frac{250}{500}\right) = 0.4(250)(0.5) = 50 \text{ deer per year}$$

Example 2: Starting above the carrying capacity

A culture of yeast satisfies $\dfrac{dy}{dt} = 0.002y(800 - y)$ with $y(0) = 1000$.

• (a) Find $\displaystyle\lim_{t \to \infty} y(t)$.
• (b) Is the graph of $y$ concave up or concave down for $t \ge 0$?

Solution.

(a) The factor $800 - y$ is zero at $y = 800$, so $L = 800$. Since $y(0) = 1000 \gt 800$, $\dfrac{dy}{dt} \lt 0$ and $y$ decreases toward $800$: the limit is $800$.

(b) Differentiate with the product rule:

$$\frac{d^2y}{dt^2} = 0.002\frac{dy}{dt}(800 - y) + 0.002y\left(-\frac{dy}{dt}\right) = 0.002\frac{dy}{dt}(800 - 2y)$$

For $y \gt 800$, $\dfrac{dy}{dt} \lt 0$ and $800 - 2y \lt 0$, so $\dfrac{d^2y}{dt^2} \gt 0$. The graph is concave up: it falls, but more and more gently, as it levels off toward $800$.

Example 3: Using the solution formula

The deer population from Example 1 is $P(t) = \dfrac{500}{1 + 4e^{-0.4t}}$.

• (a) Check that $P(0) = 100$.
• (b) Estimate the population after $5$ years.
• (c) When is the population growing fastest?

Solution.

(a) $P(0) = \dfrac{500}{1 + 4} = 100$. ✓ (Also, $A = \dfrac{500 - 100}{100} = 4$.)

(b) $P(5) = \dfrac{500}{1 + 4e^{-2}} \approx 324.393$, so about $324$ deer.

(c) Growth is fastest when $P = 250$:

$$\frac{500}{1 + 4e^{-0.4t}} = 250 \;\Rightarrow\; 1 + 4e^{-0.4t} = 2 \;\Rightarrow\; e^{-0.4t} = \frac{1}{4} \;\Rightarrow\; t = \frac{\ln 4}{0.4} \approx 3.466 \text{ years}$$

Example 4: Deriving the solution (optional extension)

Solve $\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right)$ with $P(0) = 100$.

Solution. Rewrite the right side as $\dfrac{0.4}{500}P(500 - P)$ and separate:

$$\int \frac{500}{P(500 - P)}\,dP = \int 0.4\,dt$$

By partial fractions, $\dfrac{500}{P(500 - P)} = \dfrac{1}{P} + \dfrac{1}{500 - P}$. For $0 \lt P \lt 500$:

$$\ln P - \ln(500 - P) = 0.4t + C \quad\Rightarrow\quad \ln\frac{P}{500 - P} = 0.4t + C$$

At $t = 0$, $P = 100$: $C = \ln\dfrac{100}{400} = \ln\dfrac{1}{4}$. So

$$\frac{P}{500 - P} = \frac{1}{4}e^{0.4t}$$

Solve for $P$. Taking reciprocals, $\dfrac{500 - P}{P} = 4e^{-0.4t}$, so $\dfrac{500}{P} - 1 = 4e^{-0.4t}$ and

$$P(t) = \frac{500}{1 + 4e^{-0.4t}}$$

Common mistakes

Misreading the carrying capacity. In $\dfrac{dP}{dt} = 0.002P(800 - P)$, $L = 800$, not $0.002$ or $\dfrac{1}{0.002}$. Find the value of $P$ that makes the bracket zero.

Saying growth is fastest at the carrying capacity. At $P = L$ the growth rate is $0$. Growth is fastest at $P = \dfrac{L}{2}$, halfway up.

Giving the time instead of the population (or the reverse). “For what value of $P$” wants $\dfrac{L}{2}$. “At what time” needs the solution formula and a logarithm.

Assuming the population always increases. If $P(0) \gt L$, the population decreases toward $L$. The limit is still $L$.

Using $\dfrac{dP}{dt}$ when you need $\dfrac{d^2P}{dt^2}$. Concavity questions need the second derivative, found by differentiating the right side with respect to $t$ and substituting $\dfrac{dP}{dt}$.

Practice

1. (Warm-up) A population satisfies $\dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{2000}\right)$ with $P(0) = 150$. Find the carrying capacity and the population at which it grows fastest.

Solution

$L = 2000$. Growth is fastest at $P = \dfrac{2000}{2} = 1000$.

2. (Warm-up) For the same equation, suppose instead $P(0) = 2500$. Is the population increasing or decreasing at first? Find $\displaystyle\lim_{t \to \infty} P(t)$.

Solution

At $P = 2500$, $\dfrac{dP}{dt} = 0.3(2500)\left(1 - \dfrac{2500}{2000}\right) = 0.3(2500)(-0.25) = -187.5 \lt 0$, so it is decreasing. It decreases toward the carrying capacity: the limit is $2000$.

3. (Warm-up) A quantity satisfies $\dfrac{dy}{dt} = 0.01y(300 - y)$ with $y(0) = 20$. Find $\displaystyle\lim_{t \to \infty} y(t)$, the value of $y$ where $y$ is growing fastest, and the growth rate at that moment.

Solution

$L = 300$, so $y \to 300$. Fastest growth is at $y = 150$, where

$$\frac{dy}{dt} = 0.01(150)(300 - 150) = 0.01(150)(150) = 225$$

4. (Core) The number of trout in a lake satisfies $\dfrac{dF}{dt} = 0.5F\left(1 - \dfrac{F}{6000}\right)$, where $t$ is in years, with $F(0) = 1500$.

• (a) Find $\displaystyle\lim_{t \to \infty} F(t)$.
• (b) How many trout are there when the population is growing fastest, and how fast is it growing then?
• (c) Is the graph of $F$ concave up or concave down at $t = 0$? Justify.

Solution

(a) $L = 6000$ and $0 \lt 1500 \lt 6000$, so $F \to 6000$.

(b) At $F = 3000$: $\dfrac{dF}{dt} = 0.5(3000)(0.5) = 750$ trout per year.

(c) $\dfrac{d^2F}{dt^2} = 0.5\dfrac{dF}{dt}\left(1 - \dfrac{2F}{6000}\right)$. At $F = 1500$, $\dfrac{dF}{dt} = 0.5(1500)(0.75) = 562.5 \gt 0$ and $1 - \dfrac{3000}{6000} = 0.5 \gt 0$, so $\dfrac{d^2F}{dt^2} \gt 0$: concave up. (This makes sense: $1500$ is below half the carrying capacity.)

5. (Core) A population is modelled by $P(t) = \dfrac{900}{1 + 8e^{-0.5t}}$, with $t$ in years.

• (a) Find $P(0)$ and $\displaystyle\lim_{t \to \infty} P(t)$.
• (b) At what time is the population growing fastest? Give your answer to 3 decimal places.

Solution

(a) $P(0) = \dfrac{900}{9} = 100$. As $t \to \infty$, $8e^{-0.5t} \to 0$, so $P \to 900$.

(b) Fastest growth when $P = 450$: $1 + 8e^{-0.5t} = 2$, so $e^{-0.5t} = \dfrac{1}{8}$ and

$$t = \frac{\ln 8}{0.5} = 2\ln 8 \approx 4.159 \text{ years}$$

6. (Core) Show that $\dfrac{dP}{dt} = 2P - 0.004P^2$ is a logistic equation. Find $k$, the carrying capacity, and the value of $P$ where growth is fastest.

Solution

Factor out $2P$:

$$2P - 0.004P^2 = 2P(1 - 0.002P) = 2P\left(1 - \frac{P}{500}\right)$$

So $k = 2$, $L = 500$, and growth is fastest at $P = 250$.

7. (Core) For $\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{100}\right)$, find $\dfrac{d^2P}{dt^2}$ in terms of $P$ and show that it is zero when $P = 50$.

Solution

Write $\dfrac{dP}{dt} = 0.2P - 0.002P^2$. Differentiate with respect to $t$:

$$\frac{d^2P}{dt^2} = (0.2 - 0.004P)\frac{dP}{dt} = 0.2(1 - 0.02P) \cdot 0.2P\left(1 - \frac{P}{100}\right)$$

At $P = 50$, the factor $1 - 0.02(50) = 0$, so $\dfrac{d^2P}{dt^2} = 0$. For $0 \lt P \lt 50$ it is positive and for $50 \lt P \lt 100$ it is negative, so the graph of $P$ has an inflection point at $P = 50$.

8. (Challenge, optional extension) Solve $\dfrac{dP}{dt} = 0.5P\left(1 - \dfrac{P}{40}\right)$ with $P(0) = 10$, using separation of variables and partial fractions.

Solution

Rewrite as $\dfrac{dP}{dt} = \dfrac{0.5}{40}P(40 - P)$ and separate:

$$\int \frac{40}{P(40 - P)}\,dP = \int 0.5\,dt$$

Partial fractions: $\dfrac{40}{P(40 - P)} = \dfrac{1}{P} + \dfrac{1}{40 - P}$. For $0 \lt P \lt 40$:

$$\ln\frac{P}{40 - P} = 0.5t + C$$

At $t = 0$: $C = \ln\dfrac{10}{30} = \ln\dfrac{1}{3}$. So $\dfrac{P}{40 - P} = \dfrac{1}{3}e^{0.5t}$, which gives $\dfrac{40 - P}{P} = 3e^{-0.5t}$, so $\dfrac{40}{P} = 1 + 3e^{-0.5t}$ and

$$P(t) = \frac{40}{1 + 3e^{-0.5t}}$$

Check with the formula: $A = \dfrac{40 - 10}{10} = 3$. ✓

9. (Challenge) In a school of $1200$ students, a rumour spreads at a rate proportional to both the number of students $R$ who have heard it and the number who haven’t. At $t = 0$ days, $50$ students have heard it.

• (a) Write a differential equation for $R$.
• (b) How many students have heard the rumour when it is spreading fastest?
• (c) The solution is $R(t) = \dfrac{1200}{1 + 23e^{-0.6t}}$. When is the rumour spreading fastest? Give your answer to 3 decimal places.

Solution

(a) $\dfrac{dR}{dt} = kR(1200 - R)$ for some constant $k \gt 0$.

(b) This is logistic with $L = 1200$, so the rumour spreads fastest when $R = 600$ students.

(c) Set $R = 600$: $1 + 23e^{-0.6t} = 2$, so $e^{-0.6t} = \dfrac{1}{23}$ and

$$t = \frac{\ln 23}{0.6} \approx 5.226 \text{ days}$$

(Check the starting value: $R(0) = \dfrac{1200}{24} = 50$. ✓)