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Lesson 4 of 11

Organisms and Populations · Lesson 4 of 11

Exponential Population Growth and Intrinsic Rate of Increase

“Why populations can grow faster as they become larger; From per-capita rates to a differential equation; What positive, zero and negative r mean; Solving the continuous model; Doubling time and the chessboard illustration; When the exponential model becomes misleading”

Learning Objectives

• Derive exponential growth from per-capita birth and death rates. • Define intrinsic rate of natural increase and interpret its units and sign. • Calculate population abundance and doubling time using exponential growth. • Distinguish discrete doubling processes from continuous exponential equations.

Why populations can grow faster as they become larger

Imagine a culture of microorganisms with abundant food and room for reproduction. During a short initial interval only a few individuals divide, because the starting population is small. Once those offspring are present, there are more individuals capable of reproducing, so the absolute number of births can increase even if the average reproductive performance of each cell remains unchanged. This compounding is the heart of exponential growth. The number added per interval is related to how many organisms are already present, so a rising population can produce a progressively steeper curve.

Unlimited resources are an idealisation. Real populations encounter limits, and their birth and mortality rates may change with density, temperature, crowding, nutrients or predators. Nevertheless, exponential growth provides a baseline for understanding organisms with rapid multiplication during a short resource-rich period. Bacteria in fresh nutrient medium or a newly established population may initially approximate it. A J-shaped curve emerges when abundance N is plotted against time because the total increase per unit time accelerates as N becomes larger.

From per-capita rates to a differential equation

Let b be the average per-capita birth rate per unit time and d the average per-capita death rate over that time. If the population has N individuals, expected births per unit time are approximately bN and deaths approximately dN. Ignoring immigration and emigration for the moment, the instantaneous change is dN/dt = bN − dN = (b − d)N. Define r = b − d, the net per-capita rate of increase under the specified conditions, so dN/dt = rN. Here the symbol d in dN/dt indicates differentiation and must not be confused with d denoting the per-capita death rate.

The units are important. N is expressed in individuals, b and d in inverse time such as year⁻¹, and r also in year⁻¹. Then rN has units individuals per year, matching the left side. If b = 0.32 year⁻¹ and d = 0.12 year⁻¹, then r = 0.20 year⁻¹. For N = 150, the instantaneous increase at that size is rN = 30 individuals per year. The calculation does not mean that exactly thirty individuals appear during every later year: as N changes, rN changes too. The source calls r the intrinsic rate of natural increase; the strictly maximal value requires ideal conditions appropriate to a species.

Exponential growth: J-shaped increaseTime →NJ-shaped curveConstant per-capita increase; absolute additions accelerate
Exponential Population Growth and Intrinsic Rate of Increase — scientific schematic— Axes, labels and signs illustrate model relationships; curves are not measurements of a particular population.
Instantaneous exponential population growthLaTeX

What positive, zero and negative r mean

When r is positive, births per capita exceed deaths per capita in the closed-population approximation, so abundance increases; when r = 0 they balance and abundance is constant; when r is negative, mortality exceeds natality and abundance declines. These statements concern the simple model without migration. A population with negative intrinsic growth can still increase locally if substantial immigration occurs, while a population with positive intrinsic growth can decline if enough individuals emigrate.

The numerical value of r depends on the unit of time. A rate of 0.2 year⁻¹ is not the same numerical quantity as 0.2 day⁻¹. It also depends on environmental conditions: temperatures that favour a beetle's growth may be unfavourable to a mammal. The historical chapter gives example values for some species and the growth of India's population in an earlier year, but these are context-specific illustrations rather than universal constants. A current estimate would require contemporary counts or vital rates from a consistent dataset.

Solving the continuous model

The differential equation dN/dt = rN has the exponential solution Nₜ = N₀eʳᵗ when r is constant. Here N₀ is the abundance at t = 0, Nₜ is abundance after time t, and e is the mathematical constant approximately 2.71828. The product rt is dimensionless: if r is measured in years⁻¹, t must be in years. A rate is not added directly to a count. The equation allows us to estimate the consequences of a constant net per-capita rate over a finite interval.

For example, let N₀=100 individuals, r=0.10 year⁻¹ and t=5 years. Substituting gives N₅ = 100e^(0.10×5) = 100e^0.5 ≈ 164.87, or about 165 individuals as a model prediction. For a species counted as whole organisms, the mathematical expectation can be fractional even though actual census counts are integers. The result assumes constant r, no immigration or emigration and no density limitation; it is not guaranteed to match a real population precisely.

Continuous exponential solutionLaTeX

Doubling time and the chessboard illustration

The doubling time T is the time required for an exponentially growing population to become twice its starting size under a fixed positive r. Set 2N₀ = N₀eʳᵀ and divide both sides by N₀ to obtain 2 = eʳᵀ. Taking natural logarithms gives ln 2 = rT, hence T = ln 2/r. If r = 0.20 year⁻¹, then T ≈ 0.6931/0.20 ≈ 3.47 years. Conversely, if a population doubles every three years, r = ln 2/3 ≈ 0.231 year⁻¹. This is a continuous growth rate rather than a simple annual fraction.

The source's chessboard example illustrates multiplicative accumulation: one grain is put on the first square, two on the second, four on the third and so forth. The total after n squares is 1+2+4+...+2^(n−1) = 2ⁿ−1. After 64 squares, the total is 2⁶⁴−1 grains, an enormous number despite a one-grain start. Similarly, one Paramecium doubling in a perfectly favourable daily cycle would generate 2ⁿ descendants after n doublings, excluding mortality. This is a discrete geometric model, whereas N₀eʳᵗ represents continuous compounding. The two approaches are related but their rate parameters must be translated correctly.

Doubling time when r is constant and positiveLaTeX

When the exponential model becomes misleading

In a growing culture, nutrient depletion, metabolic waste, crowding and altered oxygen availability can gradually reduce the net per-capita rate. A predator arriving in a habitat can increase mortality, and disease transmission may intensify as hosts become numerous. Under such circumstances, extrapolating the early exponential trend far into the future can yield impossible predictions. The model is most useful over intervals during which rates are reasonably stable and resources are not yet strongly limiting.

When testing a model, use actual measurements at several times rather than assuming that any rising curve is exponential. A population that increases by ten individuals every week follows a different arithmetic pattern from one that increases by ten percent each week. A logarithmic plot of abundance against time can help reveal approximately constant exponential rates, but deviations must be interpreted biologically. The mathematical model is a tool for reasoning about conditions, not a statement that limitless growth can continue in nature.

Quick Quiz

Quiz

Quick check

What makes the number of new individuals per interval accelerate under exponential growth?

Quick check

If b = 0.30 year⁻¹ and d = 0.12 year⁻¹, what is r?

Quick check

For N₀=100, r=0.1 year⁻¹ and t=5 years, what is the closest Nₜ?

Quick check

An exponentially growing population doubles every three years. What is r per year?

Quick check

Which condition would most directly invalidate a long-term constant-r exponential projection?

Practice Problems

Practice Problems
  1. Derive dN/dt = rN from per-capita natality and mortality, defining every symbol.
  2. A population begins at 300 individuals with constant r = 0.08 year⁻¹. Estimate abundance after four years.
  3. If a population doubles in three years, calculate the intrinsic rate of increase under continuous exponential growth.
  4. Explain why the wheat-grain chessboard problem illustrates geometric growth and find the total after eight squares.
  5. Compare a constant increase of ten organisms per day with a constant increase of ten percent per day.

Expected births per unit time are bN, where b is per-capita natality, and deaths are dN, where d is per-capita mortality. In a closed population the net instantaneous change equals births minus deaths: dN/dt = bN−dN = (b−d)N. Defining r=b−d yields dN/dt=rN. If rates are measured per year, r has units year⁻¹ and rN has units individuals/year.

Key Takeaways

Key Takeaways

• Exponential growth assumes a stable per-capita net rate and no strong resource limitation. • Births and deaths contribute bN and dN, giving dN/dt=(b−d)N. • The intrinsic rate r=b−d has units of inverse time. • A positive r produces growth, zero r stability and negative r decline in the closed model. • For constant r, abundance follows Nₜ=N₀eʳᵗ. • Continuous doubling time is ln(2)/r, not simply the inverse of r. • Geometric doubling illustrates rapid accumulation but cannot continue indefinitely in real habitats.