Question

The rate at which the population of a city increases at any time is proportional to the population at that time. If there are $1,30,000$ people in the city in $1960$ and $1,60,000$ in $1990$, what approximate population may be anticipated in $2020$?
$[{\log }_e\frac{16}{13}=0.2070,e^{0.42}=1.52]$

Answer 1

Given, $\frac{dx}{dt}\propto x\Rightarrow \frac{dx}{dt}=kx$
$\therefore \int \frac{dx}{x}=\int kdt$
$\log x=kt+C\Rightarrow x=C{e}^{kt}$
At $t=1960,x=1,30,000$
$\Rightarrow 1,30,000=C{e}^{1960k}$
$\therefore C=\frac{1,30,000}{e^{1960k}}$ ....$\left(1\right)$
At $t=1990,x=1,60,000$
$\Rightarrow 1,60,000=C{e}^{1960k}$
$=\frac{1,30,000}{e^{1960k}}{e}^{1990k}$ From $\left(1\right)$
$=1,30,000e^{30k}$
$\therefore e^{30k}=\frac{1,60,000}{1,30,000}=\frac{16}{13}$ ...$\left(2\right)$
$\therefore 30k={\log }_e\left(\frac{16}{13}\right)=0.2070;k=0.007$
$\therefore$ $p=1,30,000$; $e^{0.007}$ is the population at any time $t$
In the year $2020$, (i.e.,) when $t=60$
$x=1,30,000\times e^{0.42}=1,30,000\times 1.52=1,97,600$
So, the anticipated population is $2020$ will be $1,97,600$.