If the population grows at the rate of 8 % pe | vedclass

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(D) Let the initial population be $P$ and the rate of increase be $8 \%$ per year. The differential equation representing the growth is $\frac{dP}{dt}

Vedclass · Accepted Answer

$8.64$

Vedclass · Answer

(D) Let the initial population be $P$ and the rate of increase be $8 \%$ per year. The differential equation representing the growth is $\frac{dP}{dt} = \frac{8}{100} P = 0.08 P$. Separating the variables,we get $\frac{dP}{P} = 0.08 dt$. Integrating both sides,we have $\int \frac{dP}{P} = \int 0.08 dt$,which gives $\log P = 0.08 t + C$. At $t = 0$,the population is $P_0$,so $C = \log P_0$. Thus,$\log P = 0.08 t + \log P_0$,or $\log \left( \frac{P}{P_0} ight) = 0.08 t$. For the population to double,$P = 2 P_0$,so $\log 2 = 0.08 t$. Given $\log 2 = 0.6912$,we have $0.6912 = 0.08 t$. Therefore,$t = \frac{0.6912}{0.08} = 8.64 	ext{ years}$.

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled is (Given $\log 2=0.6912$ ) (in $\text{ years}$)

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