Question

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was $200000$ in $1990$ and $250000$ in $2000$, what will be the population in $2010$?

Answer 1

Let $P_i$ be the initial population and the population after time $t$ be $P$.

Now, according to the question,

$\frac{dP}{dt}\propto t$

$\Rightarrow \frac{dP}{dt}=kP$

$\Rightarrow dP=kPdt$

$\Rightarrow \frac{dP}{P}=kdt$

Integrating both ides, we have

$\int \frac{dP}{P}=\int kdt$

$\Rightarrow \log P=kt+\log c.....\left(1\right)$

Given:- $P=200000$ when $t=1990$

$\therefore \log \left|200000\right|=1990k+\log c.....\left(2\right)$

Also, $P=250000$ when $t=2000$,

$\therefore \log \left|250000\right|=2000k+\log c.....\left(3\right)$

On subtracting equation $\left(2\right)$ from $\left(3\right)$, we have

$\Rightarrow \log \left|250000\right|-\log \left|200000\right|=k(2000-1990)$

$\Rightarrow \log \frac{250000}{200000}=10k$

$\Rightarrow \log \frac{5}{4}=10k$

$\Rightarrow k=\frac{1}{10}\log \frac{5}{4}$

Substituting the value of $k$ in equation $\left(2\right)$, we have

$\log \left|200000\right|=\frac{1}{10}\log \frac{5}{4}\times 1990+\log c.....\left(4\right)$

Now, substituting the value of $k,\log c$ and $t=2010$ in equation $\left(1\right)$, we have

$\log P=\frac{1}{10}\log \frac{5}{4}\times 2010+\log \left|200000\right|-\frac{1}{10}\log \frac{5}{4}\times 1990$

$\Rightarrow \log P=\log \left|200000\right|+\frac{1}{10}\log \frac{5}{4}\times 20$

$\Rightarrow \log P=\log \left|200000\right|+2\log \frac{5}{4}$

$\Rightarrow \log P=\log \left|200000\right|+\log {\left(\frac{5}{4}\right)}^2$

$\Rightarrow \log P=\log [200000\times {\left(\frac{5}{4}\right)}^2]$

$\Rightarrow P=200000\times \frac{25}{16}$

$\Rightarrow P=312500$

Hence the population in $2010$ will be $312500$.