Question

Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.

Answer 1

Let P be the population of the city at time t.

Then ${\displaystyle \frac{\text{dP}}{\text{dt}}}$, the rate of increase of population, is proportional to P.

∴ ${\displaystyle \frac{\text{dP}}{\text{dt}}\propto \text{P}}$

∴ ${\displaystyle \frac{\text{dP}}{\text{dt}}}$ = kP, where k is a constant.

∴ ${\displaystyle \frac{\text{dP}}{\text{P}}}$ = k dt

On integrating, we get

${\displaystyle \int \frac{1}{\text{P}}\text{dP}=\text{k}\int \text{dt}+\text{c}}$

∴ log P = kt + c

Initially, i.e. when t = 0, P = 30000

∴ log 30000 = k × 0 + c       ∴ c = log 30000

∴ log P = kt + log 30000

∴ log P - log 30000 = kt

∴ ${\displaystyle \log \left(\frac{\text{P}}{30000}\right)}$ = kt          .....(1)     

Now, when t = 40, P = 40000

∴ ${\displaystyle \log \left(\frac{40000}{30000}\right)=\text{k}\times 40}$

∴ k = ${\displaystyle \frac{1}{40}\log \left(\frac{4}{3}\right)}$

∴ (1) becomes, ${\displaystyle \log \left(\frac{\text{P}}{30000}\right)=\frac{\text{t}}{40}\log \left(\frac{4}{3}\right)={\log \left(\frac{4}{3}\right)}^{\frac{\text{t}}{40}}}$

∴ ${\displaystyle \frac{\text{P}}{30000}={\left(\frac{4}{3}\right)}^{\frac{\text{t}}{40}}}$

∴ P = 30000 ${\displaystyle {\left(\frac{4}{3}\right)}^{\frac{\text{t}}{40}}}$

∴ the population of the city at time t = 30000 ${\displaystyle {\left(\frac{4}{3}\right)}^{\frac{\text{t}}{40}}}$