Logistic growth, explained with examples

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

Exponential growth goes on for ever, but real populations run out of food and space. The logistic model captures that: growth starts fast and then levels off at a maximum called the carrying capacity. It is a favourite AP Calculus BC topic.

From exponential to logistic

In exponential growth, dP/dt = kP: the bigger the population, the faster it grows, without limit. That is realistic for a short time, such as bacteria in a fresh dish, but eventually resources run short. The logistic model adds a brake that gets stronger as the population approaches a limit.

The logistic differential equation

dP/dt = kP(1 − P/L)

• P is the population at time t;
• k > 0 is the growth constant;
• L is the carrying capacity, the largest population the environment can sustain.

When P is small, the factor (1 − P/L) is close to 1, so growth is nearly exponential. As P approaches L, the factor approaches 0, so growth slows to a stop. If P starts above L, the factor is negative and the population falls towards L. The same equation is sometimes written dP/dt = (k/L)P(L − P).

Key facts you can read from the equation

• Carrying capacity: lim_t→∞ P(t) = L, for any starting value P(0) > 0.
• Equilibrium solutions: P = 0 and P = L, where dP/dt = 0.
• Fastest growth: at P = L/2. This is where the graph of P against t has its point of inflection.
• Shape: starting below L/2, the graph is an S-shaped curve: concave up below L/2 and concave down above it.

Why fastest growth is at L/2

The growth rate dP/dt = kP − (k/L)P² is a quadratic in P, an upside-down parabola with roots at P = 0 and P = L. Its maximum is halfway between the roots, at P = L/2. You can also show this by differentiating: d²P/dt² = (dP/dt)(k − 2kP/L), which is zero when P = L/2. On AP questions, “when is the population growing fastest?” almost always has the answer “when P = L/2”.

Worked examples

Example 1: reading the equation

A population satisfies dP/dt = 0.4P(1 − P/500). Find the carrying capacity and the population when it grows fastest.

• Compare with dP/dt = kP(1 − P/L): k = 0.4 and L = 500.
• Fastest growth at P = 250.

Example 2: a different form

dP/dt = 0.002P(800 − P). Find the carrying capacity and the maximum growth rate.

• Rewrite: 0.002 × 800 × P(1 − P/800) = 1.6P(1 − P/800). So L = 800.
• At P = 400: dP/dt = 0.002 × 400 × 400 = 320 per unit time.

Example 3: limits

With dP/dt = 0.4P(1 − P/500) and P(0) = 600, find lim_t→∞ P(t). Is P increasing or decreasing at t = 0?

• The limit is the carrying capacity: 500.
• At P = 600, the factor (1 − 600/500) is negative, so dP/dt < 0: decreasing, falling towards 500.

Example 4: concavity

For dP/dt = 0.4P(1 − P/500) with P(0) = 100, is the graph of P concave up or down at t = 0?

• P = 100 is below L/2 = 250, so the population is still speeding up its growth.
• The graph is concave up at t = 0.

The general solution

Separating variables and using partial fractions gives the solution

P(t) = L / (1 + Ae^−kt), where A = (L − P₀)/P₀.

AP Calculus BC rarely asks you to derive this in full, but you should recognise it and be able to use it.

Example 5: using the solution

A rumour spreads through a school of 1000 students according to the logistic model with k = 0.9 per day. Initially 10 students know. How many know after 5 days?

• L = 1000 and A = (1000 − 10)/10 = 99.
• P(5) = 1000/(1 + 99e^−4.5) = 1000/(1 + 99 × 0.0111) ≈ 1000/2.10.
• About 476 students, close to half the school, which is when the rumour spreads fastest.

Example 6: when is half reached?

In Example 5, when does the rumour spread fastest?

• Fastest when P = 500: 500 = 1000/(1 + 99e^−0.9t), so 99e^−0.9t = 1.
• t = ln 99/0.9 ≈ 5.1 days.

Exponential versus logistic

• Exponential: dP/dt = kP; the rate grows with P forever; no limit.
• Logistic: dP/dt = kP(1 − P/L); the rate peaks at L/2 and falls to zero at L; P approaches L.

For small P, the two models are almost identical, which is why early data from a logistic process can look exponential.

Logistic questions on the free-response section

A typical BC question gives a logistic differential equation in context, such as fish in a lake, and asks for the limit of the population, the population when it grows fastest, whether the graph is concave up or down at a given time, and sometimes an Euler's method estimate. Answer each part with a short reason drawn from the equation, such as “because P(0) < L/2, the population is increasing and the graph is concave up”.

Where logistic models appear

Logistic growth describes animal populations limited by food and space, the spread of a disease or a rumour through a fixed population, the adoption of new technology such as smartphones, and chemical reactions that slow as the reactants are used up. Each case has a natural maximum, which plays the role of L.

Common mistakes

• Saying fastest growth happens at P = L; it is at P = L/2.
• Misreading L when the equation is written as kP(L − P).
• Thinking a population above L keeps growing; it decreases towards L.
• Confusing the time of fastest growth with the population at that time.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is logistic growth on AP Calculus AB?

No. Logistic models are part of Unit 7 for AP Calculus BC only. AB covers exponential growth and decay.

Do I need to solve the logistic equation by hand?

AP BC focuses on interpreting the equation: carrying capacity, limits, and where growth is fastest. Recognising the solution form is useful, but full derivations are uncommon.

Key takeaways

• dP/dt = kP(1 − P/L), with carrying capacity L.
• P approaches L as t → ∞, from above or below.
• Growth is fastest at P = L/2, the point of inflection.
• Below L/2 the curve is concave up; above it, concave down.
• Solution: P = L/(1 + Ae^−kt).

Related guides

• Differential equations
• Euler's method
• Partial fractions