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Population growth and doubling time

Suppose we have a population \(x\) of some organism, whether cells in a laboratory, animals in the wild, or humans in a society. Letting \(t\) denote time, a population sometimes approximately ⁶ obeys a differential equation

\[ \dfrac{dx}{dt} = kx, \]

with constant continuous growth rate \(k\). The population will then grow exponentially.

However, exponential population growth is usually unrealistic. The differential equation above expresses the idea that the rate of increase of the population is proportional to the population; and indeed, we might expect a larger population to produce more offspring. But increases or decreases in a population often depend on many other factors, including ecology, economics and culture. Obviously, on a finite planet, exponential population growth cannot continue indefinitely!

Since at present we are focusing on exponential growth, we investigate the differential equation

\[ \dfrac{dx}{dt} = kx, \]

but always remembering that the results obtained for populations may be unrealistic. We will consider a more realistic model in the section Links forward (Logistic growth).

Example

Today there are 1000 birds on an island. They breed with a constant continuous growth rate of 10% per year. To three significant figures, how many birds will be on the island after seven years?

Solution

Let \(x\) be the number of birds on the island after \(t\) years. With continuous growth rate \(k = 10\% = 0.1\), we have

\[ \dfrac{dx}{dt} = 0.1 \, x \]

and hence, solving the differential equation,

\[ x(t) = C e^{0.1 \, t}, \]

for some positive constant \(C\). Since \(x(0) = 1000\), we have \(C=1000\), so

\[ x(t) = 1000 \, e^{0.1 \, t}. \]

After seven years, the number of birds on the island is \(x(7) = 1000 \, e^{0.7} \approx 2010\), to three significant figures.

In this example, the bird population more than doubled in seven years.

Exercise 2

For the bird population in the previous example, how long does it take for the population to double?

Just as an exponentially decaying quantity has a half-life, an exponentially growing quantity has a doubling time — the time it takes for the quantity to double. Calculating doubling time is very similar to calculating half-life, as we see in the next exercise.

Exercise 3

Prove that, for a population with constant continuous growth rate \(k\), the doubling time \(T\) is given by

\[ T = \dfrac{1}{k} \log_e 2. \]

For \(k = 0.1\), the doubling time should agree with the answer to exercise 2. In fact, in the absence of ecological constraints, bird populations can grow far more rapidly!

Exercise 4

In 1937, eight pheasants were introduced to an island off the coast of the USA and proceeded to reproduce rapidly. For the next few years, the pheasant population grew with a continuous growth rate of over 100%.

Assuming a constant continuous growth rate of 100%, what is the number of pheasants on the island after 60 years? (Give the answer to three significant figures.)
Assume each pheasant weighs 1.5 kilograms. After 60 years, which has more mass: the pheasants or the earth? (The earth weighs approximately \(5.98 \times 10^{24}\) kilograms.)
Discuss the realism of these questions. (In fact, the growth rate of 100% only continued for about three years.)

Reference
These events are discussed by David Lack in The Natural Regulation of Animal Numbers, Clarendon Press, 1954, pp. 11–12.

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