F. Population Dynamics · Population Dynamics

Logistic Growth and Carrying Capacity

F2.2 F2.1 F3.3 A1.10About 75 minutes

Learning goals

I can explain carrying capacity (K) and why the logistic model includes the term (K − N) ÷ K.
I can calculate dN/dt = rN(K − N) ÷ K and identify where growth is fastest.
I can use the solution N(t) = K ÷ (1 + Ae⁻ʳᵗ) to predict population size and the time to reach a target.
I can describe the phases of an S-shaped curve and explain overshoot, time lags and the limits of the model.

Words to know

carrying capacity (K)
The largest population of a species that an environment can support indefinitely with its available resources.

logistic growth
Growth that slows as the population approaches K: dN/dt = rₘₐₓN(K − N) ÷ K.

S-shaped (sigmoid) curve
The shape of logistic growth: slow start, rapid middle, levelling off at K.

inflection point
The point on the S-curve (at N = K/2) where growth is fastest and the curve changes from bending upward to bending downward.

overshoot
When a population temporarily rises above K, often because of a time lag between using up resources and the population responding.

time lag
A delay between a change in conditions (such as food running out) and the population's response (fewer births, more deaths).

Adding the brakes

In the last lesson, exponential growth ran away because r stayed constant. Real environments push back: as a population grows, each individual gets less food, less space and more competition, and predators and disease find it easier. The per capita growth rate falls. Eventually births and deaths balance at the carrying capacity, K.

Remember from Grade 9: you learned that populations rise and fall around the carrying capacity of their habitat in a dynamic equilibrium. Now you will describe that idea with an equation.

The logistic equation

dN/dt = rₘₐₓ N (K − N) ÷ K

Read it in three parts:

rₘₐₓN is the exponential growth from the last lesson.
(K − N) ÷ K is the fraction of the carrying capacity still unused. It acts like a brake.
When N is tiny, (K − N) ÷ K ≈ 1, so growth is nearly exponential.
When N = K/2, the brake is 1/2, and growth is fastest.
When N = K, (K − N) = 0, so growth stops.
If N > K, (K − N) is negative, so the population declines back toward K.

The per capita growth rate is rₘₐₓ(K − N) ÷ K, which falls in a straight line from rₘₐₓ (when N = 0) to 0 (when N = K).

Logistic growth makes an S-shaped curve. Growth is slow at first (few parents), fastest at the inflection point (N = K/2), then slows and levels off at the carrying capacity K.

Worked example

Let's calculate logistic growth rates

A population of painted turtles in a wetland has rₘₐₓ = 0.40 per year and K = 500. Find dN/dt when N = 100, N = 250 and N = 450.

N = 100: dN/dt = 0.40 × 100 × (500 − 100) ÷ 500 = 40 × 0.80 = 32 turtles/year.
N = 250: dN/dt = 0.40 × 250 × (250 ÷ 500) = 100 × 0.50 = 50 turtles/year.
N = 450: dN/dt = 0.40 × 450 × (50 ÷ 500) = 180 × 0.10 = 18 turtles/year.
Pattern: growth is fastest at N = K/2 = 250. Check with the formula for the maximum: rK/4 = 0.40 × 500 ÷ 4 = 50 ✓.

32, 50 and 18 turtles per year. The population grows fastest at half the carrying capacity.

Graph of dN/dt against N for rₘₐₓ = 0.40 and K = 1000. The growth rate is zero at N = 0 and at N = K and peaks at N = K/2.

Predicting N with the logistic solution

Solving the logistic equation (a calculus technique you may meet in first-year university) gives:

N(t) = K ÷ (1 + A e^(−rt)), where A = (K − N₀) ÷ N₀

At t = 0, e⁰ = 1, so N = K ÷ (1 + A) = N₀ ✓.
As t grows, e^(−rt) shrinks toward 0, so N approaches K ✓.

To find the time to reach a target N, rearrange: t = ln[A ÷ (K/N − 1)] ÷ r.

The inflection point (N = K/2) is reached at t = ln A ÷ r.

Worked example

Let's predict a reintroduced population

Fifty elk are released into a large protected area that can support K = 1000 elk and r = 0.50 per year. (a) Predict N after 6 years. (b) When will the herd reach 900? (c) When is growth fastest?

A = (K − N₀) ÷ N₀ = (1000 − 50) ÷ 50 = 19.
(a) N(6) = 1000 ÷ (1 + 19e^(−0.5 × 6)) = 1000 ÷ (1 + 19 × 0.04979) = 1000 ÷ 1.946 = 514 elk.
(b) K/N − 1 = 1000/900 − 1 = 0.1111. t = ln(19 ÷ 0.1111) ÷ 0.5 = ln(171) ÷ 0.5 = 5.142 ÷ 0.5 = 10.3 years.
(c) Fastest growth at N = K/2 = 500, when t = ln A ÷ r = ln 19 ÷ 0.5 = 2.944 ÷ 0.5 = 5.9 years. That fits (a): at 6 years N ≈ 514, just past K/2.
Compare with exponential: with no brake, N(6) = 50e³ ≈ 1004, double the logistic prediction.

(a) About 514 elk; (b) about 10.3 years; (c) at about 5.9 years, when N = 500.

When reality departs from the model

The logistic model assumes that K is constant, that every individual is identical, and that the population responds instantly to crowding. Real populations often break these assumptions:

Time lags: animals born when food was plentiful still need food when it is scarce. Populations then overshoot K and die back, sometimes oscillating around K.
Changing K: a mild winter, a forest fire or new farmland can raise or lower K.
Age structure: many juveniles respond differently from adults.

A dramatic example: in 1944, 29 reindeer were released on St. Matthew Island in the Bering Sea, where there were no wolves. With rich lichen to eat, they increased to about 6000 by 1963. They had grazed the slow-growing lichen far below what could regrow; after the severe winter of 1963 to 1964 the population crashed to about 42 animals. The reindeer overshot a carrying capacity that their own grazing was lowering, an extreme case of overshoot and dieback.

Lab populations fit best. In the 1930s the Soviet ecologist G. F. Gause grew Paramecium in test tubes with a constant food supply, and his cultures followed S-shaped curves closely. You will meet Gause's experiments again in Competition, Niches and Resource Partitioning.

Exponential or logistic?

Use exponential for short periods with abundant resources (a new colony, an outbreak's early weeks). Use logistic when the population is a large fraction of what the habitat can support, or when you see growth slowing as N rises. A falling per capita growth rate is the signature of logistic growth.

Just for fun

Carrying capacity explained by a household with one bathroom: the first person up in the morning grows exponentially happy; by the fifth person, growth has levelled off at K.

Hands-on activity

Run the logistic model on a calculator or spreadsheet

You need: a calculator or a free spreadsheet (for example Google Sheets or LibreOffice), and graph paper.

Use the discrete version of the logistic model, one step per year:

Nₜ₊₁ = Nₜ + r Nₜ (1 − Nₜ ÷ K)

Set K = 1000, N₀ = 50 and r = 0.5. Calculate N for 15 years (in a spreadsheet, put 50 in cell B2 and =B2+0.5B2(1-B2/1000) in B3, then fill down). Graph N against year.
Repeat with r = 1.0. How does the curve change?
Repeat with r = 2.2. Describe the pattern after year 3.
Explain your results using the idea of a time lag: in this model, each year's growth is based on last year's population.

What to look for: with r = 0.5 you get a smooth S-curve: 50, 74, 108, 156, 222, 308, 415, 536, 661, 773, 860, 921, 957, 978, 989, 994. With r = 1.0 it rises faster and reaches 1000 by about year 8. With r = 2.2 the population jumps to 984 in year 3, then overshoots and oscillates: 1018, 978, 1026, 968, 1037 and so on, bouncing above and below K. Fast breeders with a one-year lag overshoot, just like the reindeer.

Your turn

Define carrying capacity.

For rₘₐₓ = 0.2 per year and K = 2000, find dN/dt when N = 500.

At what population size is logistic growth fastest, and what is that maximum rate for r = 0.2 and K = 2000?

What happens to dN/dt when N > K? Why?

For K = 800, N₀ = 100 and r = 0.3, find N after 5 years.

Name the three phases of an S-shaped curve.

Give two reasons a real population may overshoot its carrying capacity.

More practice: warm-up

What letter represents carrying capacity?

What shape is a logistic growth curve?

What is dN/dt when N = K?

Write the logistic growth equation.

When N is very small compared with K, what kind of growth does the logistic model approximate?

More practice: on your own

For the population in Your turn question 5 (K = 800, N₀ = 100, r = 0.3), when does growth peak?

A population has rₘₐₓ = 0.6 and K = 300. Find the per capita growth rate at N = 60, N = 150 and N = 270.

For K = 1000, N₀ = 50 and r = 0.5 (the elk example), how long until N reaches 750?

Explain, in terms of births and deaths, why a population levels off at K.

The St. Matthew Island reindeer rose from 29 to about 6000, then crashed to about 42. Use the terms carrying capacity, overshoot and time lag to explain.

Challenge

Show that the logistic growth rate rN(K − N)/K has its maximum at N = K/2. (Hint: it is a quadratic in N, or use derivatives.)

A fish population is modelled with K = 50 000 and r = 0.4 per year. Managers want to harvest the most fish possible each year while keeping the population steady. (a) At what population should they keep the stock? (b) What is the maximum sustainable harvest? (c) Why is this risky in practice?

A biologist's yeast culture grows: 0 h: 10; 2 h: 40; 4 h: 140; 6 h: 360; 8 h: 560; 10 h: 640; 12 h: 660 (units of 10⁶ cells/mL). (a) Estimate K. (b) Estimate the time of the inflection point. (c) Calculate the per capita growth rate from 0 to 2 h and from 8 to 10 h (per hour, using (N₂ − N₁) ÷ (N₁ × Δt)) and explain the difference.

Check your understanding

Quiz

In the logistic equation, what does the term (K − N) ÷ K represent?

For rₘₐₓ = 0.5 and K = 400, what is dN/dt at N = 200?

A logistic population grows fastest when N equals:

A population is above its carrying capacity. According to the logistic model, it will:

What caused the St. Matthew Island reindeer crash?

Which graph shape matches logistic growth?

For parents and teachers

Big idea (F2.2, F2.1, F3.3): the logistic model dN/dt = rN(K − N)/K adds a brake to exponential growth. Populations grow fastest at K/2 and level off at the carrying capacity K. Time lags and changing K cause overshoot and oscillation.

How to help: the spreadsheet activity is the heart of this lesson; a free online spreadsheet works fine. Ask your teen to explain what each part of the logistic equation means in words.

Common difficulties:

Thinking growth is fastest at K (it is zero there).
Sign errors in A = (K − N₀) ÷ N₀ and in e^(−rt).
Calculator order of operations: compute the denominator 1 + Ae^(−rt) first, then divide K by it.

Sources: logistic model from standard ecology texts; St. Matthew Island reindeer from D. R. Klein (1968), Journal of Wildlife Management; Gause's experiments from The Struggle for Existence (1934). Example data are invented for teaching.