12 min readConcept sheet1-mark MCQ or assertion-reason on the curves, a 3-mark comparison of the two models, and a 5-mark answer with graphs; also a NEET favourite

Growth models: exponential and logistic growth, and life history variation

With unlimited resources a population grows exponentially (J-shaped curve). With limited resources it grows logistically (S-shaped curve), levelling off at the carrying capacity K.

Exponential growthdN/dt = rN; integral form Nₜ = N₀ eʳᵗ (e = 2.71828); J-shaped curve

Logistic growthdN/dt = rN (K − N)/K; sigmoid (S-shaped) curve

Intrinsic rate of natural increase (r)r = b − d, the per capita birth rate minus the per capita death rate

Carrying capacity (K)The largest population that the resources of a habitat can support

Exponential growth: building the equation

1
Assume unlimited food and space, so each species can realise its full potential to grow in number, as Darwin noted.
2
In a population of size N, let b be the per capita birth rate and d the per capita death rate.
3
The change in N per unit time is dN/dt = (b − d) × N.
4★
Write (b − d) = r, the intrinsic rate of natural increase. So dN/dt = rN.
Textbook r values: Norway rat 0.015, flour beetle 0.12, India's human population in 1981 0.0205
5
The integral form is Nₜ = N₀ eʳᵗ: Nₜ is the density after time t, N₀ the density at time zero, and e = 2.71828, the base of natural logarithms.
6
A plot of N against time rises slowly and then shoots up: a J-shaped curve. So, with unlimited resources, a population grows in a geometric (exponential) way. Even a slow breeder like the elephant could reach huge numbers this way, as Darwin showed.

★ key step

Logistic growth: when resources run short

No real population has unlimited resources. As numbers rise, individuals compete for food and space, and the fittest survive and reproduce. Every habitat can support only a maximum number of a species. This limit is its carrying capacity (K). A population in a habitat with limited resources shows a lag phase, then acceleration, then deceleration, and finally an asymptote, where it levels off at K. The plot of N against time is a sigmoid (S-shaped) curve. This is Verhulst-Pearl logistic growth: dN/dt = rN (K − N)/K. When N is small, (K − N)/K is close to 1, so growth is almost exponential. As N nears K, (K − N)/K nears 0, so growth stops. Resources become limiting sooner or later, so the logistic model is more realistic.

Exponential vs logistic growth

Exponential growthResources unlimited. dN/dt = rN. J-shaped curve with no upper limit; growth keeps speeding up. Seen only for short spells, such as bacteria in fresh culture medium or a weed spreading in new land.

Logistic growthResources limited. dN/dt = rN (K − N)/K. S-shaped (sigmoid) curve: lag, acceleration, deceleration and asymptote. Levels off at the carrying capacity K. More realistic for most populations.

Both curves start the same way. They separate once resources begin to run short.

How to draw the two growth curves

Draw two axes: time (t) on the x-axis and population density (N) on the y-axis.
Draw a curve that starts low and bends sharply upward. Label it (a) Exponential growth (J-shaped) and write dN/dt = rN beside it.
On the same axes, draw a curve that rises slowly, then steeply, then flattens. Label it (b) Logistic growth (S-shaped) and write dN/dt = rN (K − N)/K beside it.
Draw a dashed horizontal line where curve (b) flattens. Label it Carrying capacity (K).
Write the title: Population growth curves.

Worked examples: using the growth equations

Problem

(a) A bacterial culture starts with N₀ = 1,000 cells and grows exponentially with r = 0.5 per hour. Find N after 2 hours, taking e = 2.718. (b) A deer population in a sanctuary has N = 300, K = 1,200 and r = 0.2 per year. Find dN/dt under logistic growth.

Working

(a) Nₜ = N₀ eʳᵗ, and rt = 0.5 × 2 = 1, so eʳᵗ = e¹ = 2.718.
Nₜ = 1,000 × 2.718 = 2,718 cells.
(b) dN/dt = rN (K − N)/K.
rN = 0.2 × 300 = 60, and (K − N)/K = (1,200 − 300)/1,200 = 900/1,200 = 0.75.
dN/dt = 60 × 0.75 = 45 deer per year.
Check: with unlimited resources, dN/dt would be rN = 60. Limited resources slow growth to 45.

Answer

(a) About 2,718 cells. (b) 45 deer per year.

Life history variation: different ways to maximise fitness

Breed once or many timesSome organisms breed only once in their lifetime, e.g. Pacific salmon fish and bamboo. Others breed many times, e.g. most birds and mammals.

Many small or few large offspringSome produce a large number of small offspring, e.g. oysters and pelagic (open-sea) fishes. Others produce a small number of large offspring, e.g. birds and mammals.

Populations evolve to maximise their reproductive fitness, also called Darwinian fitness (a high r value). Ecologists suggest these life history traits evolved under the limits set by the abiotic and biotic components of each habitat.

In real life

Extra reading (not in your textbook): in Mizoram, a bamboo species flowers all together about once in 50 years and then dies. The flood of seeds lets rat numbers grow almost exponentially, and the rats then destroy crops, a famine known as mautam.

Exam angle

2 marks: “Give one example of an organism that breeds only once in its lifetime and one that produces many small offspring.” Breeds once: Pacific salmon fish or bamboo. Many small offspring: oysters or pelagic fishes.

3 marks: “Distinguish between exponential and logistic growth. Which is more realistic, and why?” Exponential: unlimited resources, dN/dt = rN, J-shaped curve. Logistic: limited resources, dN/dt = rN (K − N)/K, S-shaped curve levelling at K. Logistic is more realistic because resources become limiting sooner or later.

5 marks: “Explain the logistic growth model with its equation and a labelled curve. What is carrying capacity?” Define K; limited resources and competition; phases: lag, acceleration, deceleration, asymptote; equation with each symbol; labelled S-curve with a dashed K line; contrast with the J-curve; conclude that it is realistic.

Remember it

J for Jumping (no limits); S for Stopping (levels off at K).