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

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Question

(C) Let $P_{0}$ be the initial population and $P$ be the population after $t$ years. The rate of growth is given by $\frac{dP}{dt} = \frac{8P}{100} =

Vedclass · Accepted Answer

$8 \cdot 64$ years

Vedclass · Answer

(C) Let $P_{0}$ be the initial population and $P$ be the population after $t$ years. The rate of growth is given by $\frac{dP}{dt} = \frac{8P}{100} = 0 \cdot 08P$. Integrating the differential equation $\frac{dP}{P} = 0 \cdot 08 dt$,we get $\ln P = 0 \cdot 08t + C$. At $t = 0$,$P = P_{0}$,so $C = \ln P_{0}$. Thus,$\ln \left( \frac{P}{P_{0}} ight) = 0 \cdot 08t$. For the population to double,$P = 2P_{0}$,so $\ln 2 = 0 \cdot 08t$. Given $\log_{10} 2 = 0 \cdot 6912$,we convert to natural log: $\ln 2 = \log_{10} 2 	imes \ln 10 \approx 0 \cdot 6912 	imes 2 \cdot 3026 \approx 1 \cdot 5915$. Alternatively,using the provided $\log 2 = 0 \cdot 6912$ as $\ln 2$: $t = \frac{0 \cdot 6912}{0 \cdot 08} = \frac{69 \cdot 12}{8} = 8 \cdot 64$ years. Therefore,the correct option is $C$.

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled,is (Given $\log 2 = 0 \cdot 6912$)

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