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Q2:
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase ((r)) of the population?

Solution :

Step 1: Theoretical Foundation of Exponential Growth

The exponential growth of a population is mathematically represented by the differential equation $\frac{dN}{dt} = rN$. Upon integration, this yields the equation for population size at a given time $t$:

$N_t = N_0 e^{rt}$

Where:

  • $N_t$ = Population size at time $t$
  • $N_0$ = Initial population size
  • $e$ = Base of the natural logarithm ($\approx 2.71828$)
  • $r$ = Intrinsic rate of natural increase
  • $t$ = Time interval

Step 2: Assigning Given Variables

Based on the problem statement, we have the following parameters:

  • Doubling time ($t$) = $3$ years
  • Final population ($N_t$) = $2N_0$

Step 3: Algebraic Derivation for $r$

Substitute the known values into the growth equation:

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

[By dividing both sides by $N_0$]:

$2 = e^{3r}$

[By taking the natural logarithm ($\ln$) on both sides to solve for the exponent]:

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

[Using the logarithmic identity $\ln(e^x) = x$]:

$\ln(2) = 3r$

$r = \frac{\ln(2)}{3}$

Step 4: Numerical Calculation

Given that $\ln(2) \approx 0.6931$:

$r = \frac{0.6931}{3}$

$r \approx 0.2310$

Visual Representation of Exponential Growth

Time (t in years) Population (N) Nâ‚€ 2Nâ‚€ (at t=3)

Final Solution: The intrinsic rate of increase ($r$) for the population is approximately $0.231$ per year.


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