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Exercises · 11.2

Q.If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?

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✓ Free question

For exponential growth, doubling time and the intrinsic rate of increase (r) are linked by $r = \ln(2)/t_d$; with a doubling time of 3 years, $r \approx 0.231$ per year.

When a population grows exponentially, its size at any time is given by the integrated growth equation:

$$N_t = N_0 e^{rt}$$

"Doubling time" ($t_d$) is simply the value of $t$ at which $N_t$ becomes exactly $2N_0$. Substituting that condition into the equation:

$$2N_0 = N_0 e^{r t_d}$$

The $N_0$ on both sides cancels, leaving:

$$2 = e^{r t_d}$$

Taking the natural logarithm of both sides:

$$\ln(2) = r t_d$$

$$r = \frac{\ln(2)}{t_d}$$

This is the key relationship: a population's intrinsic rate of natural increase can be recovered directly from how long it takes to double, without needing to know its starting size at all.

Substituting the given doubling time of 3 years:

$$r = \frac{\ln(2)}{3} = \frac{0.693}{3} \approx 0.231 \text{ per year}$$

Note

This value is far higher than the $r$ values the NCERT textbook quotes for slow-growing species -- 0.015 for the Norway rat, 0.12 for the flour beetle -- which makes sense, since a doubling time of only 3 years is genuinely fast population growth.

✓Final answer

The intrinsic rate of increase of the population is $r = \ln(2)/3 \approx$ 0.231 per year.

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