# The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.

Question:

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Solution:

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\therefore \frac{d y}{d t} \propto y$

$\Rightarrow \frac{d y}{d t}=k y$                              ( $k$ is a constant)

$\Rightarrow \frac{d y}{y}=k d t$

Integrating both sides, we get:

$\log y=k t+C \ldots(1)$

In the year $1999, t=0$ and $y=20000$.

Therefore, we get:

$\log 20000=C \ldots$ (2)

In the year $2004, t=5$ and $y=25000$.

Therefore, we get:

$\log 25000=k \cdot 5+\mathrm{C}$

$\Rightarrow \log 25000=5 k+\log 20000$

$\Rightarrow 5 k=\log \left(\frac{25000}{20000}\right)=\log \left(\frac{5}{4}\right)$

$\Rightarrow k=\frac{1}{5} \log \left(\frac{5}{4}\right)$                     ...(3)

In the year 2009, t = 10 years.

Now, on substituting the values of tk, and C in equation (1), we get:

$\log y=10 \times \frac{1}{5} \log \left(\frac{5}{4}\right)+\log (20000)$

$\Rightarrow \log y=\log \left[20000 \times\left(\frac{5}{4}\right)^{2}\right]$

$\Rightarrow y=20000 \times \frac{5}{4} \times \frac{5}{4}$

$\Rightarrow y=31250$

Hence, the population of the village in 2009 will be 31250.