Start from the integral form of the exponential equation.

\( N_t = N_0 e^{rt} \)

Substitute what is given. The population doubles, so N_t = 2N₀, and t = 3 years.

\( 2N_0 = N_0 e^{3r} \)

\( 2 = e^{3r} \)

Take natural logarithms of both sides.

\( \ln 2 = 3r \)

\( r = \dfrac{\ln 2}{3} = \dfrac{0.693}{3} = 0.231 \)

Answer: r = 0.231 per year (r is per capita per unit time, so the unit is 'per year' here).

A useful thing to notice. The doubling time depends only on r, never on the starting size: t_double = 0.693/r. So r = 0.231 means a doubling every three years whether the population is ten individuals or ten million. Compare it with the r values in the chapter — 0.015 for the Norway rat, 0.12 for the flour beetle, and 0.0205 for the human population of India in 1981 — and you can see that 0.231 is a very fast rate of increase.

The curve. A population growing in a habitat with limited resources shows initially a lag phase, followed by phases of acceleration and deceleration, and finally an asymptote, when the population density reaches the carrying capacity K. A plot of N against time gives a sigmoid or S-shaped curve. This is Verhulst-Pearl Logistic Growth, described by dN/dt = rN(K−N)/K. (Refer to the growth-curve diagram above, curve b, with the dashed line marking K.)

Phase by phase.

Lag phase. N is very small, so even though the per capita rate is high, rN is small in absolute terms and the curve rises almost imperceptibly. Individuals are also establishing themselves in the habitat.

Acceleration. N has grown large enough that rN is substantial, while N is still far below K, so (K−N)/K is still close to 1. Growth is nearly exponential and the curve climbs steeply — this is where the logistic and exponential curves are hardest to tell apart.

Deceleration. N is now a large fraction of K, so (K−N)/K becomes small and drags dN/dt down. Competition between individuals for limited resources intensifies; the population is still growing, but more and more slowly.

Asymptote. N = K, so (K−N)/K = 0 and dN/dt = 0. The population has reached the maximum possible number the habitat's resources can support, beyond which no further growth is possible.

Why this model rather than the exponential one. Since resources for growth for most animal populations are finite and become limiting sooner or later, the logistic growth model is considered the more realistic one.

It states what a population would do if nothing stopped it. Ideally, when resources in the habitat are unlimited, each species has the ability to realise fully its innate potential to grow in number — as Darwin observed while developing his theory of natural selection. The exponential model measures that innate potential, and r, the intrinsic rate of natural increase, is the number that expresses it.

That number is what makes comparison possible. r is a very important parameter chosen for assessing the impact of any biotic or abiotic factor on population growth. Measure r before and after a change — a pesticide, a predator, a drought — and you have quantified the effect. Without a model of unimpeded growth there is no baseline against which to measure any restraint.

It describes real situations, briefly. Early in the logistic curve, when N is far below K, the factor (K−N)/K is close to 1 and growth is essentially exponential. A newly inoculated culture, an invasive species newly arrived in a land without its natural predators, a population recolonising after a catastrophe — all grow exponentially for a time.

And it carries the argument that matters most. Darwin showed how even a slow-growing animal like the elephant could reach enormous numbers in the absence of checks. Since no such numbers are observed, checks must exist — and the search for those checks is most of ecology. The exponential model is valuable precisely because reality departs from it.

The common goal. Populations evolve to maximise their reproductive fitness, also called Darwinian fitness — a high r value — in the habitat in which they live. Under a particular set of selection pressures, organisms evolve towards the most efficient reproductive strategy. Both species are solving the same problem, under different conditions.

Why many small offspring works for the oyster. Oysters and pelagic fishes produce a large number of small-sized offspring. A pelagic larva faces mortality that is enormous, unpredictable and almost entirely beyond any parent's influence — currents, predators, chance. No amount of parental investment in an individual larva would appreciably improve its odds. The winning strategy is therefore to buy as many lottery tickets as possible, and a few will survive by chance alone.

Why few large offspring works for the bird. Birds and mammals produce a small number of large-sized offspring. Here survival does depend on things a parent can supply — food, warmth, protection, and in many species learning. Investment in a single offspring raises its chance of surviving substantially, so concentrating the same total resource into two well-provisioned young yields more surviving descendants than scattering it among thousands.

The general conclusion. Ecologists suggest that life history traits have evolved in relation to the constraints imposed by the abiotic and biotic components of the habitat in which organisms live. There is no strategy that is best in the abstract. The same reasoning explains the other axis of variation: breeding once in a lifetime, as in Pacific salmon and bamboo, against breeding many times, as in most birds and mammals.

What the observation means. A population that has stopped growing and settled at a steady level has reached the asymptote of the logistic curve: N = K, the carrying capacity, which is the maximum possible number the habitat's resources can support, beyond which no further growth is possible. In the equation, (K−N)/K has fallen to zero, so dN/dt = 0.

Why importing deer will not work. Immigration is one of the four processes that increase density, so N will rise to 700 immediately. But at N > K the factor (K−N)/K becomes negative, so dN/dt is negative: the population declines. It does so through the other two processes — mortality rises as competition between individuals for limited resources intensifies, and natality falls as underfed animals breed less successfully — until N returns to about 500. The manager will have bought 200 deer in order to kill 200 deer, mostly by starvation.

What she should do instead. Raise K, not N. Carrying capacity is a property of the habitat, not of the herd: more forage, more water, more space, or removal of whatever resource is limiting. If the reserve is fenced, enlarging it or opening a corridor also raises K.

One caveat in the herd's favour. If the reserve population were very small rather than at K, importing animals would be sound — not to add numbers but to add genetic variation, since it is at the population level that natural selection operates to evolve the desired traits. At N = K, however, that is a genetic argument and not a numerical one, and it needs far fewer than 200 animals.