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Population Growth Models — Quantitative

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One equation family, two behaviors

Both population models describe the same quantity — dN/dt, the number of individuals added (or lost) per unit time — but they differ in whether resources run out. Exponential growth (dN/dt = rN) assumes unlimited resources: the per-capita rate r is constant, so the population accelerates without bound in a J-curve. Logistic growth (dN/dt = rN((K − N)/K)) adds a single throttle, the factor (K − N)/K, that measures the fraction of carrying capacity still unused. That factor slides from near 1 (empty habitat, growth ≈ exponential) down to 0 (full, growth stops), bending the J into an S.

Exponential growth
dN/dt = rN
Whole-population growth rate = per-capita rate r × current size N. The per-capita rate itself, (1/N)(dN/dt), is simply the constant r — it never changes no matter how large N gets.
Logistic growth
dN/dt = rN((K − N)/K)
The (K − N)/K term is the density-dependent brake. The per-capita rate here is (1/N)(dN/dt) = r((K − N)/K), which shrinks toward 0 as N → K. This is the equation form of "competition intensifies with crowding."
Worked example

A population grows logistically with r = 0.8 per year and carrying capacity K = 2,000. Compute dN/dt at N = 500, N = 1,000, and N = 1,500. At which N is growth fastest?

  1. 1.At N = 500: (K − N)/K = (2,000 − 500)/2,000 = 1,500/2,000 = 0.75. Then dN/dt = rN × 0.75 = 0.8 × 500 × 0.75 = 400 × 0.75 = 300 individuals/yr.
  2. 2.At N = 1,000: (K − N)/K = (2,000 − 1,000)/2,000 = 1,000/2,000 = 0.50. Then dN/dt = 0.8 × 1,000 × 0.50 = 800 × 0.50 = 400 individuals/yr.
  3. 3.At N = 1,500: (K − N)/K = (2,000 − 1,500)/2,000 = 500/2,000 = 0.25. Then dN/dt = 0.8 × 1,500 × 0.25 = 1,200 × 0.25 = 300 individuals/yr.
  4. 4.The rates are 300, 400, 300 — the peak (400) sits at N = 1,000, which is exactly K/2. Growth at N = 500 and N = 1,500 is identical, showing the curve is symmetric about K/2.
Answer: dN/dt = 300, 400, and 300 individuals/yr at N = 500, 1,000, and 1,500. Growth is fastest at N = 1,000 = K/2.

Why K/2 is the sweet spot

Logistic growth is a tug-of-war between two factors: N (more breeders pushes dN/dt up) and (K − N)/K (crowding pushes it down). When N is small the brake is nearly 1 but there are few breeders; when N is near K there are many breeders but the brake is near 0. The product rN((K − N)/K) is largest exactly in the middle, at N = K/2, where the maximum growth rate equals rK/4. On the S-curve this is the inflection point — the steepest part of the climb, after which growth decelerates even though the population is still rising toward K.

Maximum growth rate
peak dN/dt = rK/4, at N = K/2
Substituting N = K/2 into the logistic equation: r(K/2)((K − K/2)/K) = r(K/2)(1/2) = rK/4. This is the maximum sustainable yield point exploited in fisheries and wildlife management.
Checkpoint

For a population following the logistic model, at what population size N is the growth rate dN/dt the greatest?

Watch out

Keep dN/dt and the per-capita rate distinct. dN/dt is the whole population’s change; the per-capita rate is (1/N)(dN/dt). In the logistic model the per-capita rate = r((K − N)/K) falls steadily as N rises, but dN/dt first rises (to its peak at K/2) and only then falls. Two different curves — do not conflate them.

Checkpoint

A population grows logistically with r = 0.5 per year and K = 1,000. When N = 750, what is the per-capita growth rate, (1/N)(dN/dt)?

The life history behind r and K

The parameters are not just math — they map onto strategies. r-selected species bet on a high r: many small offspring, no parental care, early reproduction, short lives. They win in empty or unstable habitats where growth is near-exponential and the (K − N)/K brake barely applies. They show Type III survivorship (massive early death, a few long-lived survivors). K-selected species are adapted to life near K: few large offspring, heavy parental investment, long lives, and success under crowding where competition dominates. They show Type I survivorship (low early mortality, most deaths late in life). Type II — a constant death rate at every age (many birds, rodents) — sits between them.

On the exam

A quantitative FRQ often hands you r, K, and N and asks for dN/dt. Write the logistic equation, substitute, and show the (K − N)/K step explicitly — that is where points live. Then interpret: is N below, at, or above K/2? Below K/2, growth is still accelerating; above it, growth is decelerating toward the S-curve plateau at K.

Answer the 2 checkpoints as you read.

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