If a population growing exponentially double in size in $3$ years, what is the intrinsic rate of increase $(r)$ of the population?
A population grows exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation:
$N_{t}=N_{0} e^{r t}$ Where,
$N_{t}=$ Population density after time $t$
$N_{0}=$ Population density at time zero
$r=$ Intrinsic rate of natural increase
$e=$ Base of natural logarithms
$(2.71828)$
From the above equation, we can calculate the intrinsic rate of increase $(r)$ of a population.
Now, as per the question,
Present population density $=x$
Then,
Population density after two years $=2 x$
$t=3$ years
Substituting these values in the formula, we get:
$\Rightarrow 2 x=x e^{3 r}$
$\Rightarrow 2=e^{3 r}$
Applying log on both sides:
$\Rightarrow \log 2=3 \mathrm{r} \log e$
$\Rightarrow \frac{\log 2}{3 \log e}=r$
$\Rightarrow \frac{\log 2}{3 \times 0.434}=r$
$\Rightarrow \frac{0.301}{3 \times 0.434}=r$
$\Rightarrow \frac{0.301}{1.302}=r$
$\Rightarrow 0.2311=r$
Hence, the intrinsic rate of increase for the above illustrated population is $0.2311$.
The formula for exponential population growthis
The growth of population is determined by
Rapid decline in a population due to high mortality rate is