F. Population Dynamics · Population DynamicsGeometric and Exponential Growth Models
F2.2 F2.1 A1.12 A1.13About 75 minutes
Learning goals- I can tell when to use a geometric (discrete) model and when to use an exponential (continuous) model.
- I can calculate future population size with Nₜ = N₀λᵗ and Nₜ = N₀eʳᵗ.
- I can find λ or r from population data and calculate doubling time.
- I can recognize a J-shaped curve and explain why exponential growth cannot continue for long in nature.
Words to know
geometric growth
Growth in which a population is multiplied by the same factor λ each breeding season (discrete, non-overlapping generations): Nₜ = N₀λᵗ.
λ (lambda)
The geometric growth rate (finite rate of increase): λ = Nₜ₊₁ ÷ Nₜ. λ > 1 means growth; λ < 1 means decline.
exponential growth
Continuous growth in which the rate of increase is proportional to N: dN/dt = rN, so Nₜ = N₀eʳᵗ.
r (intrinsic rate of increase)
The per capita growth rate in continuous time, r = b − d (per capita births minus deaths). rₘₐₓ is its maximum under ideal conditions.
doubling time (t_d)
The time for a population to double: t_d = ln 2 ÷ r ≈ 0.693 ÷ r, or about 70 ÷ (percent growth rate).
J-shaped curve
The shape of an exponential growth graph: slow at first, then steeper and steeper.
Growth that feeds on itself
In Measuring Population Change you saw that 10% growth adds more individuals each year because 10% of a bigger population is bigger. When the per capita growth rate stays constant, the population grows faster and faster. Biologists model this in two ways, depending on how the organism reproduces.
Math connection: these are the exponential functions you study in Advanced Functions (MHF4U). You will need the ln key on your calculator.
Model 1: geometric growth (breeding seasons)
Many organisms in Canada breed once a year in a short season: most songbirds, many insects, annual plants, fish that spawn each spring. Their populations change in steps. Ecologists compare the population from one breeding season to the next:
λ = Nₜ₊₁ ÷ Nₜ (lambda, the geometric growth rate)
If λ stays constant, then after t seasons:
Nₜ = N₀ λᵗ
- λ > 1: growing (λ = 1.25 means 25% growth per season).
- λ = 1: stable.
- λ < 1: declining (λ = 0.9 means a 10% loss per season).
Worked exampleLet's model an invasive fish geometrically
A small population of 500 round goby (an invasive fish in the Great Lakes) colonizes a new bay. It grows with λ = 1.6 per breeding season. (a) Find N after 5 seasons. (b) After how many seasons will it first exceed 100 000?
- (a) N₅ = N₀λ⁵ = 500 × 1.6⁵ = 500 × 10.486 = 5243 goby (about 5200).
- (b) Solve 500 × 1.6ᵗ = 100 000, so 1.6ᵗ = 200.
- Take logarithms: t = ln 200 ÷ ln 1.6 = 5.298 ÷ 0.470 = 11.3.
- Because breeding happens in whole seasons, check: N₁₁ = 500 × 1.6¹¹ ≈ 87 961 (not yet) and N₁₂ = 500 × 1.6¹² ≈ 140 737.
- So the population first exceeds 100 000 after 12 seasons.
(a) About 5243 goby. (b) After 12 breeding seasons.
Geometric growth of the round goby example: each season's population is 1.6 times the one before (500, 800, 1280, 2048, 3277, 5243).Model 2: exponential growth (continuous breeding)
Bacteria, yeast, and humans reproduce continuously: births and deaths happen all the time, not in one season. Then growth is modelled with calculus:
dN/dt = rN
This says the growth rate (dN/dt) is proportional to the population size. The constant r is the per capita rate: r = b − d, where b and d are per capita birth and death rates. Its maximum possible value under ideal conditions, rₘₐₓ, is the species' biotic potential expressed as a number.
The solution (you can check it with derivatives in Calculus and Vectors, MCV4U) is:
Nₜ = N₀ eʳᵗ (e ≈ 2.718)
The two models are linked: λ = eʳ, or r = ln λ. A population with λ = 1.25 per year has r = ln 1.25 ≈ 0.223 per year.
Doubling time
An exponentially growing population has a constant doubling time. Set Nₜ = 2N₀:
2N₀ = N₀eʳᵗ, so 2 = eʳᵗ, so t_d = ln 2 ÷ r ≈ 0.693 ÷ r
A quick mental version is the rule of 70: doubling time ≈ 70 ÷ (growth rate in percent). A population growing 3.5% per year doubles in about 70 ÷ 3.5 = 20 years. (The exact answer, ln 2 ÷ 0.035 ≈ 19.8 years, is very close.)
On an ordinary graph, exponential growth makes a J-shaped curve. On a graph with a logarithmic vertical axis, the same data fall on a straight line, with slope related to r. Ecologists use this to test whether data really are exponential.
Worked exampleLet's model bacteria with Nₜ = N₀eʳᵗ
Under ideal lab conditions, E. coli can divide about every 20 minutes. A culture starts with 100 cells. (a) Find r per minute. (b) How many cells after 3 hours? (c) Check your answer by counting doublings.
- (a) Doubling time t_d = 20 min, so r = ln 2 ÷ 20 = 0.6931 ÷ 20 = 0.0347 per minute.
- (b) t = 3 h = 180 min. Keep r unrounded (store ln 2 ÷ 20 in your calculator): rt = 0.034657… × 180 = 6.238, so Nₜ = 100 × e^6.238 = 100 × 512 = 51 200 cells. (Rounding r first changes the answer: r = 0.0347 gives about 51 600 and r = 0.035 gives about 54 500.)
- (c) 180 ÷ 20 = 9 doublings, and 2⁹ = 512, so 100 × 512 = 51 200 ✓.
r ≈ 0.0347 per minute; after 3 hours about 51 200 cells (9 doublings).
Worked exampleLet's find r and doubling time from field data
A newly established colony of wild turkeys grows from 40 birds to 250 birds in 5 years. Assuming exponential growth, find r and the doubling time.
- Start with Nₜ = N₀eʳᵗ: 250 = 40e^(5r).
- Divide: e^(5r) = 250 ÷ 40 = 6.25.
- Take ln of both sides: 5r = ln 6.25 = 1.833, so r = 0.367 per year.
- t_d = ln 2 ÷ r = 0.693 ÷ 0.367 = 1.89 years.
- Sense check: 5 years is about 2.6 doubling times, and 2^2.6 ≈ 6.1, close to the observed 6.25 ✓.
r ≈ 0.37 per year, doubling time ≈ 1.9 years.
Why exponential growth never lasts
In On the Origin of Species (1859), Charles Darwin chose the elephant, one of the slowest-breeding animals, and calculated that a single pair, breeding at its natural rate, would have about 19 million descendants after roughly 750 years. Even slow breeders explode if nothing stops them.
Exponential growth does happen in nature, but only for short periods: when a species colonizes a new habitat (an invasive species in a new lake), recovers after a crash, or has just been freed from predators or disease. Sooner or later food, space, predators or disease slow it down. The next lesson, Logistic Growth and Carrying Capacity, adds that brake to the model.
Just for funDarwin's elephant calculation is a good reminder that even an animal pregnant for nearly two years can, in theory, out-populate a planet. In practice, the elephants ran out of peanuts first.
Hands-on activityDoubling grains of rice
You need: uncooked rice, a small bowl, a kitchen scale if you have one (or a measuring spoon), paper, a calculator.
- Put 1 grain on the table (generation 0). Each generation, double the number: 2, 4, 8, ... Count out the grains up to generation 8 (256 grains). Time how long it takes.
- Make a table of generation (t) and N. Plot N against t on ordinary graph paper (J-curve), then plot log₁₀N against t (straight line).
- Weigh your 256 grains (or measure their volume in teaspoons). Use proportion to predict the mass of rice at generation 20 (2²⁰ = 1 048 576 grains) and generation 30.
- Calculate the doubling time and r if one generation is one day.
What to look for: the first few doublings are trivial; the last ones take most of your counting time. If 256 grains weigh about 5 g, generation 20 is about 4096 × 5 g ≈ 20 kg (a large sack) and generation 30 is about 1024 times more, roughly 21 tonnes. With one doubling per day, t_d = 1 day and r = ln 2 ≈ 0.693 per day. Keep rice away from small children.
Your turn
1When is a geometric model more appropriate than an exponential model?
When the organism reproduces in distinct breeding seasons (discrete generations), such as many songbirds, insects and annual plants.
2A population of 2000 has λ = 1.15. Find N after 4 seasons.
N₄ = 2000 × 1.15⁴ = 2000 × 1.749 = 3498 (about 3500).
3A population has λ = 0.92. What does this mean?
It is declining by 8% per breeding season (λ < 1).
4Find N after 8 years if N₀ = 2000 and r = 0.12 per year (continuous growth).
N = 2000e^(0.12 × 8) = 2000e^0.96 = 2000 × 2.612 = 5223 (about 5200).
5Use the rule of 70 to estimate the doubling time of a population growing 5% per year.
70 ÷ 5 = 14 years (exact: ln 2 ÷ 0.05 = 13.9 years).
6Convert λ = 1.25 per year to r.
r = ln 1.25 = 0.223 per year.
7Why does a graph of exponential growth look like a straight line on a log scale?
log N = log N₀ + rt × log e is a linear function of t, so equal time steps give equal increases in log N.
More practice: warm-up
1Write the equation for geometric growth.
2Write the equation for exponential growth.
Nₜ = N₀eʳᵗ (from dN/dt = rN).
3What is the formula for doubling time?
t_d = ln 2 ÷ r ≈ 0.693 ÷ r.
4What shape does exponential growth make on an ordinary graph?
5If λ = 1, what is happening?
The population is stable (no change from season to season).
More practice: on your own
1A population of house sparrows grows from 120 to 150 in one breeding season. Find λ, then predict N after 3 more seasons at the same λ.
λ = 150 ÷ 120 = 1.25. N after 3 more seasons = 150 × 1.25³ = 150 × 1.953 = 293 (about 290).
2A culture of yeast grows from 5.0 × 10³ to 8.0 × 10⁴ cells in 6 hours. Find r and the doubling time.
e^(6r) = 8.0 × 10⁴ ÷ 5.0 × 10³ = 16. 6r = ln 16 = 2.773, r = 0.462 per hour. t_d = 0.693 ÷ 0.462 = 1.5 hours (16 = 2⁴, so 4 doublings in 6 h ✓).
3How long does it take a population with r = 0.05 per year to grow from 1000 to 8000?
8000 = 1000e^(0.05t), e^(0.05t) = 8, 0.05t = ln 8 = 2.079, t = 41.6 years. Check: 8 = 2³, so 3 doublings × 13.9 years = 41.6 ✓.
4Two populations each start at 100. A has λ = 1.30 per year (geometric); B has r = 0.30 per year (continuous). Which is larger after 10 years, and why?
A: 100 × 1.30¹⁰ = 100 × 13.79 = 1379. B: 100e³ = 100 × 20.09 = 2009. B is larger because r = 0.30 is equivalent to λ = e^0.30 = 1.35 per year, which is a bigger yearly factor than 1.30.
5Explain why exponential growth is often seen in invasive species shortly after they arrive.
In a new habitat they often have plentiful resources and few predators, parasites or competitors adapted to them, so their per capita growth rate stays near rₘₐₓ. Environmental resistance is low until the population becomes large.
Challenge
1A human population grew from 2.5 billion in 1950 to 5.3 billion in 1990 (approximately). (a) Assuming exponential growth, find r. (b) Find the doubling time. (c) If that r had continued, predict the population in 2030. (d) The actual population around 2030 is projected at about 8.5 billion. What does the comparison tell you?
(a) e^(40r) = 5.3 ÷ 2.5 = 2.12; 40r = ln 2.12 = 0.7514; r = 0.0188 per year (1.88%). (b) t_d = 0.693 ÷ 0.0188 = 36.9 years. (c) N = 5.3e^(0.0188 × 40) = 5.3 × 2.12 = about 11.2 billion. (d) The projected 8.5 billion is much lower, so the per capita growth rate has fallen since 1990: human growth is no longer exponential at a constant r (see Human Population Growth and the Demographic Transition).
2Show algebraically that the doubling time of Nₜ = N₀eʳᵗ does not depend on N₀, and explain what this means biologically.
Set Nₜ = 2N₀: 2N₀ = N₀eʳᵗ. Divide by N₀: 2 = eʳᵗ, so t = ln 2 ÷ r. N₀ has cancelled, so the doubling time depends only on r. Biologically, a population of 10 and a population of 10 million growing at the same per capita rate both double in the same time; the large one simply adds far more individuals per unit time.
3A population of a monocarpic plant (it flowers once and dies) has λ = 0.85 per year after a drought. Starting at 4000, how many years until it falls below 1000? What does λ < 1 mean for its R₀ from Fecundity, Mortality, Life Tables and Survivorship Curves?
4000 × 0.85ᵗ < 1000, so 0.85ᵗ < 0.25. t > ln 0.25 ÷ ln 0.85 = (−1.386) ÷ (−0.1625) = 8.53, so after 9 years (check: 4000 × 0.85⁸ ≈ 1090; 4000 × 0.85⁹ ≈ 926). For a species with one generation per year, λ < 1 corresponds to R₀ < 1: each individual is not replacing itself.
Quiz
1. A population of 300 has λ = 1.2 per season. What is N after 2 seasons?
2. What is the doubling time of a population with r = 0.10 per year?
3. Which organism is best modelled by geometric (discrete) growth?
4. On a graph with a logarithmic vertical axis, exponential growth appears as:
5. A population with λ = 0.9 is:
6. If r = ln λ, what is r when λ = 1?
7. Using the rule of 70, a population growing at 2% per year doubles in about:
Big idea (F2.2, F2.1): geometric growth (Nₜ = N₀λᵗ) fits species with breeding seasons; exponential growth (Nₜ = N₀eʳᵗ, from dN/dt = rN) fits continuous breeders. Constant per capita growth gives a constant doubling time (ln 2 ÷ r) and a J-shaped curve, which never lasts long in nature.
How to help: your teen needs a scientific calculator with ln and eˣ keys. If they are taking Advanced Functions (MHF4U), point out that this is the same algebra as their exponential and logarithmic functions unit. The rice activity is a memorable way to feel how fast doubling runs away.
Common difficulties:
- Mixing λ and r (λ = eʳ; λ = 1 and r = 0 both mean no growth).
- Rounding r too early, which causes large errors after exponentiating.
- Forgetting that geometric answers must be whole seasons when asked "after how many seasons".
Sources: standard population ecology models; Darwin's elephant calculation is in Chapter 3 of On the Origin of Species (1859); human population figures are approximate UN estimates. The E. coli 20-minute division time is the commonly quoted figure for ideal laboratory conditions.