The rate of growth of population is proportional to the number of inhabitants. If the population doubles in 25 years and the present population is 1,00,000, when will the city have population 4,00,000?Solution : Let ‘P’ be the population at time ‘t’.…

Q.3 (C)(ii)

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

Solution : Let ‘P’ be the population at time ‘t’. Since rate of growth of population is proportional to the no. of inhabitants : \(\dfrac{dP}{dt} \propto P\)

\(\therefore\) Differential equation can be written as \(\dfrac{dP}{dt} = kP\), where \(k\) is constant of proportionality.

\(\therefore \dfrac{dP}{P} = k\cdot dt\). On integrating we get \(\square = kt + c\) .....(i)

(i) When \(t = 0,\ P = 1{,}00{,}000\), \(\therefore\) from (i) \(\log 1{,}00{,}000 = k(0) + c\), \(\therefore c = \square\), \(\therefore \log\left(\dfrac{P}{1{,}00{,}000}\right) = kt\) .....(ii)

(ii) When \(t = 25,\ P = 2{,}00{,}000\) as population doubles in 25 years, \(\therefore\) from (ii) \(\log 2 = 25k\), \(\therefore k = \square\), \(\therefore \log\left(\dfrac{P}{1{,}00{,}000}\right) = \left(\dfrac{1}{25}\log 2\right)\cdot t\)

(iii) \(\therefore\) when \(P = 4{,}00{,}000\): \(\log\left(\dfrac{4{,}00{,}000}{1{,}00{,}000}\right) = \left(\dfrac{1}{25}\log 2\right)\cdot t\), \(\therefore \log 4 = \left(\dfrac{1}{25}\log 2\right)\cdot t\), \(\therefore t = \square\) years