The population of a town increases at a rate | vedclass

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(C) Let $p$ be the population at time $t$ years. Given that the rate of change of population is proportional to the population: $\frac{dp}{dt} = kp$ $

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$160$

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(C) Let $p$ be the population at time $t$ years. Given that the rate of change of population is proportional to the population: $\frac{dp}{dt} = kp$ $\Rightarrow \frac{dp}{p} = k dt$ Integrating both sides,we get: $\log p = kt + c$ At $t = 0$,$p = 40,000$: $\log 40,000 = k(0) + c \Rightarrow c = \log 40,000$ So,$\log p = kt + \log 40,000 \Rightarrow \log \left(\frac{p}{40,000}ight) = kt$ At $t = 40$ years,$p = 80,000$: $\log \left(\frac{80,000}{40,000}ight) = 40k \Rightarrow \log 2 = 40k \Rightarrow k = \frac{\log 2}{40}$ We need to find the population after another $40$ years,i.e.,at $t = 80$ years: $\log \left(\frac{p}{40,000}ight) = \left(\frac{\log 2}{40}ight) 	imes 80$ $\log \left(\frac{p}{40,000}ight) = 2 \log 2 = \log 2^2 = \log 4$ $\frac{p}{40,000} = 4$ $p = 4 	imes 40,000 = 160,000$

The population of a town increases at a rate proportional to the population at that time. If the population increases from $40,000$ to $80,000$ in $40$ years,then the population in another $40$ years will be (in $,000$)

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