BiologyISC Class 12 · Organisms and Populations

Population Growth: Exponential and Logistic Models, ISC Class 12

Updated 15 October 2026 · Fact-checked

With unlimited resources a population grows exponentially, dN/dt = rN, giving a J-shaped curve. With limited resources growth is logistic, dN/dt = rN(K − N)/K, giving an S-shaped (sigmoid) curve with lag, acceleration, deceleration and an asymptote at the carrying capacity K. Logistic growth is more realistic because resources are finite.

Understand Population Growth: Exponential and Logistic

What changes population density. Four basic processes: natality (B) and immigration (I) add individuals; mortality (D) and emigration (E) remove them.
N(t+1) = N(t) + [(B + I) − (D + E)].
Normally births and deaths matter most; immigration matters more when a new habitat is being colonised.

Natality: number of births during a period added to the initial density.
Mortality: number of deaths during a period.
Immigration / emigration: individuals of the same species coming into / leaving the habitat during the period.

Per capita rates. Birth rate b = births ÷ initial N; death rate d = deaths ÷ initial N (per individual per unit time).

1. Exponential (geometric) growth. When food and space are unlimited, each species realises its full innate potential (as Darwin noted).

dN/dt = (b − d) × N; with r = b − d, dN/dt = rN.
r is the intrinsic rate of natural increase, used to assess the effect of any biotic or abiotic factor on growth. Examples: Norway rat 0.015, flour beetle 0.12, human population of India in 1981, 0.0205.
Integral form: N_t = N₀ e^(rt), where e = 2.71828.
Plot of N against t: J-shaped curve. Even slow breeders (Darwin's elephant example) reach enormous numbers without checks; the chessboard grains (1, 2, 4, 8…) and a Paramecium doubling daily show how fast doubling escalates.

2. Logistic (Verhulst-Pearl) growth. No population has unlimited resources, so individuals compete and the 'fittest' survive and reproduce. A habitat can support a maximum number, the carrying capacity (K).

dN/dt = rN (K − N)/K.
Phases on the curve: lag phase → acceleration → deceleration → asymptote at K.
Plot of N against t: sigmoid (S-shaped) curve.
When N is small, (K − N)/K ≈ 1 and growth is almost exponential; as N approaches K, (K − N)/K → 0 and growth stops. Growth is fastest at N = K/2.
Considered more realistic because resources are finite and become limiting sooner or later.

Comparison.

	Exponential	Logistic
Resources	unlimited	limited
Equation	dN/dt = rN	dN/dt = rN(K − N)/K
Curve	J-shaped	S-shaped (sigmoid)
Upper limit	none	carrying capacity K
Environmental resistance	absent	present, rises with N

Life history variation and reproductive fitness. Populations evolve to maximise reproductive fitness, also called Darwinian fitness (high r), in their habitat. Some breed once in a lifetime (Pacific salmon, bamboo); others many times (most birds and mammals). Some produce many small offspring (oysters, pelagic fishes); others few large offspring (birds, mammals). Life-history traits evolve under the abiotic and biotic constraints of the habitat.

Diagram checklist. One graph: x-axis 'time (t)', y-axis 'population density (N)'; curve a (J) rising steeply; curve b (S) levelling off; dashed horizontal line labelled K (carrying capacity); label lag, acceleration, deceleration and asymptote on curve b; label the gap between the curves 'environmental resistance'.

Population Growth: Exponential and Logistic: key facts

Change in density: $N_{t + 1} = N_{t} + [(B + I) - (D + E)]$; B births, I immigrants, D deaths, E emigrants.
Exponential growth: $\frac{d N}{d t} = r N, r = b - d$; Integral form N_t = N_0 e^{rt}.
Logistic growth: $\frac{d N}{d t} = r N (\frac{K - N}{K})$; Do not drop the N; growth is fastest at N = K/2.
Per capita rates: $b = \frac{births}{N _{0}}, d = \frac{deaths}{N _{0}}$; Per individual per unit time.

How to answer Population Growth: Exponential and Logistic questions

ISC sets J and S curves to identify and draw, the equations with all terms defined, reproductive fitness, and sums on natality, change in density and growth rate.

1Identify the curve from its shape: steep, unchecked rise = exponential (J); levelling at K = logistic (S).
2Write the equations in full, then define N, r, K and t.
3In logistic sums, compute (K − N)/K first, then multiply by rN.
4For change in density, collect B + I and D + E before subtracting.
5For per capita rates, divide by the starting population, not the final.
6Draw the S curve with labelled axes, K line and the four phases.

Population Growth: Exponential and Logistic in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Logistic growth and sustainable harvesting

Suppose a village pond can support at most 2000 fish (K) of a species whose intrinsic rate of increase r is 0.4 per year, and the villagers want the largest catch every year without emptying the pond.

At what population should they keep the fish, and about how many fish can they take each year?

Try it first, then show the solutionSolution

Logistic growth dN/dt = rN(K − N)/K is greatest at N = K/2 = 1000.
At N = 1000: dN/dt = 0.4 × 1000 × (2000 − 1000)/2000 = 0.4 × 1000 × 0.5 = 200 fish per year.
Catching about 200 a year keeps the population steady at 1000, where it regrows fastest.
Overfishing below this level slows regrowth; leaving it near K also gives little growth because resources are nearly used up.

Answer: Keep about 1000 fish (K/2) and harvest about 200 per year.

Memory hook: Half full grows fastest.

Tricks to remember Population Growth: Exponential and Logistic

Short memory aids. Say each one aloud once, then use it on a question the same day.

J for Just keeps going, S for Slows down
J-curve: no brakes (exponential); S-curve: brakes at K (logistic).
Use it when: Naming curves.
BIDE
Births and Immigration add; Deaths and Emigration subtract.
Use it when: N(t+1) equation.
L-A-D-A
Lag, Acceleration, Deceleration, Asymptote.
Use it when: Phases of the S curve.

Population Growth: Exponential and Logistic: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two curves · 3 marks · Short answerBoard level

Distinguish between exponential and logistic growth, giving the equation and the shape of the curve for each.

Try it first, then show the solutionSolution

Exponential: unlimited resources; dN/dt = rN; J-shaped curve (1 mark).
Logistic: limited resources with carrying capacity K; dN/dt = rN(K − N)/K; S-shaped (sigmoid) curve (1 mark).
N = population density, r = intrinsic rate of natural increase, K = carrying capacity (1 mark).

Answer: J-shaped dN/dt = rN versus S-shaped dN/dt = rN(K − N)/K.

Example 2: Change in density · 3 marks · Short answerBoard level

Over twelve months a herd of 800 deer gains 120 fawns and 30 newcomers, while 70 die and 40 leave. Work out the herd size at the end of the period and the per capita birth and death rates.

Try it first, then show the solutionSolution

N(t+1) = 800 + (120 + 30) − (70 + 40) = 800 + 150 − 110 = 840 (1 mark).
Birth rate b = 120/800 = 0.15 per deer per year (1 mark).
Death rate d = 70/800 = 0.0875 per deer per year (1 mark).

Answer: 840; b = 0.15, d = 0.0875 per individual per year.

Example 3: Growth slows near K · 3 marks · Short answerHarder board level

For a population with r = 0.2 per month and K = 1000, calculate dN/dt at N = 100, 500 and 900 under logistic growth, and compare with exponential growth at the same N.

Try it first, then show the solutionSolution

N = 100: 0.2 × 100 × 900/1000 = 18 (exponential 20) (1 mark).
N = 500: 0.2 × 500 × 500/1000 = 50 (exponential 100) (1 mark).
N = 900: 0.2 × 900 × 100/1000 = 18 (exponential 180); logistic growth is greatest at K/2 and falls towards zero near K (1 mark).

Answer: 18, 50, 18 per month (exponential 20, 100, 180).

Example 4: Integral form · 2 marks · Short answerHarder board level

A culture of 150 cells grows exponentially with r = 0.25 per day. Using N_t = N₀ e^(rt), estimate the number of cells after 4 days (e = 2.718).

Try it first, then show the solutionSolution

rt = 0.25 × 4 = 1, so N_t = 150 × e¹ (1 mark).
N_t = 150 × 2.718 ≈ 408 (1 mark).

Answer: About 408.

Example 5: Reading two growth curves · 3 marks · Short answerHarder board level

On one set of axes, curve A rises ever more steeply while curve B rises and then levels off at a dashed horizontal line. (a) Name the growth model and the shape of each curve. (b) What does the dashed line represent, and what does the vertical gap between A and B represent? (c) Pacific salmon breed only once in a lifetime, whereas most mammals breed many times. Explain how both can have high reproductive fitness.

Try it first, then show the solutionSolution

(a) A: exponential growth, J-shaped curve; B: logistic growth, S-shaped (sigmoid) curve (1 mark).
(b) The dashed line is the carrying capacity K; the gap between the curves is environmental resistance, the effect of limiting resources (1 mark).
(c) Reproductive (Darwinian) fitness is the ability to leave the most surviving, fertile offspring; each life-history pattern has evolved to maximise this under the constraints of its own habitat, so both can be highly fit (1 mark).

Answer: A exponential (J), B logistic (S); dashed line K, gap = environmental resistance; both life histories maximise reproductive fitness in their habitats.

Example 6: Introduced species · 4 marks · Short answerHOTS / competency

Suppose 30 goats are left behind by sailors on an uninhabited island covered with shrubs, where nothing preys on them. (a) Predict the shape of the growth curve in the first few years and give the equation. (b) What will happen later, and which model then applies? (c) Give two factors that would set K on the island.

Try it first, then show the solutionSolution

(a) Nearly exponential (J-shaped) because resources are effectively unlimited at first: dN/dt = rN (1 mark).
(b) As numbers rise, shrubs become scarce, competition rises and growth slows; the population levels off near K: logistic, dN/dt = rN(K − N)/K (2 marks).
(c) Amount of shrubs (food), space and shelter, fresh water, disease (any two) (1 mark).

Answer: J at first, then S levelling at K set by food, water, space and disease.

Where students lose marks in Population Growth: Exponential and Logistic

The slips examiners see most often, and what to write instead.

Mistake: Dropping N from the logistic equation
In 2024 examiners reported dN/dt = r[(K − N)/K] written without N.
Do this: Write dN/dt = rN(K − N)/K and define every symbol.
Usually costs: 1 mark.
Mistake: Wrong curve drawn
In 2025 examiners found a J-shaped or species-area curve drawn for logistic growth, with unlabelled axes and phases explained wrongly.
Do this: Draw an S curve against labelled axes (N, t) with K marked and the four phases named.
Usually costs: Up to 3 marks.
Mistake: Swapping figures or giving circular reasons
Examiners (2024) noted Figures A and B swapped, and 'unlimited supply of nutrients' repeated as the reason.
Do this: Link the shape to the cause: J = no environmental resistance; S = limiting resources.
Usually costs: 1 mark.
Mistake: Explaining reproductive health for reproductive fitness
Examiners (2024) reported reproductive health explained instead of reproductive fitness.
Do this: Fitness = ability to leave the most fertile offspring (Darwinian fitness).
Usually costs: 1 mark.

Exam tips

Write r as 'intrinsic rate of natural increase' in full once.
Put a dashed line for K on any S curve.
Show substitution in every sum before the answer.
Mention 'environmental resistance' when explaining the gap between J and S curves.
Logistic growth is also called Verhulst-Pearl logistic growth.

Practice questions on Population Growth: Exponential and Logistic

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the logistic growth equation, K stands for:
(a) intrinsic rate of increase (b) carrying capacity (c) population density (d) time
Hint
Maximum supported.
Answer
(b) carrying capacity.
Q2 · 2 marks · Short answerBoard level
Name the four phases of the logistic growth curve.
Hint
Starts slowly, ends flat.
Answer
Lag phase, phase of acceleration, phase of deceleration, and asymptote (when density reaches K).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The logistic model is considered more realistic than the exponential model.
Reason (R): Resources for growth are finite for most populations and become limiting sooner or later.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Finite resources.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
A flour beetle population has r = 0.12 per day and N = 500. Find dN/dt if growth is exponential.
Hint
rN.
Answer
dN/dt = rN = 0.12 × 500 = 60 beetles per day.
Q5 · 2 marks · Short answerHarder board level
Give one example each of organisms that (i) breed only once in a lifetime and (ii) produce a large number of small offspring.
Hint
Salmon; oysters.
Answer
(i) Pacific salmon fish or bamboo. (ii) Oysters or pelagic fishes.
Q6 · 2 marks · Short answerHarder board level
Why does logistic growth stop when N = K? Show using the equation.
Hint
(K − N)/K.
Answer
At N = K, (K − N)/K = 0, so dN/dt = rN × 0 = 0: births are balanced by deaths and the population stops growing, because the habitat cannot support more individuals.
Q7 · 3 marks · Short answerHOTS / competency
Suppose a Paramecium culture starts with one cell that divides once a day with unlimited food. (a) How many cells are there after 10 days and after 20 days? (b) Why can this never continue for long in a real pond?
Hint
2ⁿ.
Answer
(a) 2¹⁰ = 1024 cells after 10 days; 2²⁰ = 1 048 576 after 20 days. (b) Food and space in a real pond are limited; competition rises, the death rate increases and growth slows to the carrying capacity (logistic growth).
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a fish farmer's pond has K = 2000 for a fish with r = 0.4 per year.
(a) At what population is the yearly increase greatest? (1)
(b) What is that increase? (1)
(c) Write the equation you used, defining each term. (1)
Hint
K/2.
Answer
(a) N = K/2 = 1000. (b) 0.4 × 1000 × (2000 − 1000)/2000 = 200 fish per year. (c) dN/dt = rN(K − N)/K, where N is population density, r the intrinsic rate of natural increase, K the carrying capacity and t time.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is r?Show answer
The intrinsic rate of natural increase, b − d.
2.Shape of the exponential growth curve?Show answer
J-shaped.
3.What is carrying capacity?Show answer
The maximum population a habitat can support.
4.At what N is logistic growth fastest?Show answer
K/2.
5.Which two processes increase population density?Show answer
Natality and immigration.
6.r of India's human population in 1981?Show answer
0.0205.

Population Growth: Exponential and Logistic: frequently asked questions

What is the difference between exponential and logistic growth?

Exponential growth occurs with unlimited resources (dN/dt = rN, J curve); logistic growth occurs with limited resources and levels off at carrying capacity (dN/dt = rN(K − N)/K, S curve).

Why is logistic growth more realistic?

Because resources are finite for most populations and become limiting sooner or later.

What does K mean in ecology?

Carrying capacity: the maximum number of individuals of a species that a habitat can support.

What is reproductive fitness?

Darwinian fitness: the ability of an individual or population to leave the maximum number of fertile offspring (high r).

How is population density after a period calculated?

N(t+1) = N(t) + [(B + I) − (D + E)], adding births and immigrants and subtracting deaths and emigrants.