Question

The rate of growth of population is proportional to the number present. If the population doubled in the last $25$ years and the present population is $1$ lac, when will the city have population $4,00,000$ ?

Answer 1

$\frac{dp}{dt}\propto p$

$⟹\frac{dp}{dt}=kp$ (where $k$ is constant)

$⟹\int \frac{dp}{dt}=\int kdt$

$⟹\ln p=kt+c$ (on integrating both sides)

At $t=0,p=\left(1/2\right)lac=50,000.$

At $t=25,p=\left(1/2\right)lac=1,00,000.$

$\therefore$ we get, $c=\ln (50,000)$ and $k=\ln \left(2\right)/25$

$\therefore$ For $P=4,00,000$

$\ln p=\frac{\ln \left(2\right)}{25}t+\ln (50,000)$

$\ln (4,00,000)=\frac{\ln \left(2\right)}{25}t+\ln (50,000)$

Solving for $t$, we get $t=75$.

Now as we have assumed $t=0$ after 25 years. So, city have population of $4,00,000$ before $25$ years.

Hence time $=75-25=50$years