The population of a village increases at a ra | vedclass

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(B) We have $\frac{dP}{dt} \propto P$,which implies $\frac{dP}{dt} = kP$. Integrating both sides,we get $\int \frac{dP}{P} = \int k dt$,so $\log P = k

Vedclass · Accepted Answer

$1,60,000$

Vedclass · Answer

(B) We have $\frac{dP}{dt} \propto P$,which implies $\frac{dP}{dt} = kP$. Integrating both sides,we get $\int \frac{dP}{P} = \int k dt$,so $\log P = kt + \log c$. At $t = 0$,$P = 20,000$,so $\log 20,000 = \log c$. At $t = 10$,$P = 40,000$,so $\log 40,000 = 10k + \log 20,000$. This gives $\log \left(\frac{40,000}{20,000}ight) = 10k$,so $10k = \log 2$,or $k = \frac{1}{10} \log 2$. The general equation is $\log P = \left(\frac{1}{10} \log 2ight) t + \log 20,000$. We need the population after another $20$ years,which means at $t = 10 + 20 = 30$ years. Substituting $t = 30$: $\log P = \frac{30}{10} \log 2 + \log 20,000 = 3 \log 2 + \log 20,000 = \log (8 	imes 20,000) = \log 1,60,000$. Therefore,$P = 1,60,000$.

The population of a village increases at a rate proportional to the population at that time. In a period of $10$ years,the population grew from $20,000$ to $40,000$. Then,the population after another $20$ years is:

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