The population of a town increases at a rate | vedclass

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(C) Let $P(t)$ be the population at time $t$. The rate of change is given by $\frac{dP}{dt} = kP$. Integrating this,we get $\ln P = kt + C$,or $P(t) =

Vedclass · Accepted Answer

$320$

Vedclass · Answer

(C) Let $P(t)$ be the population at time $t$. The rate of change is given by $\frac{dP}{dt} = kP$. Integrating this,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$. At $t = 0$,$P(0) = 40,000$. So,$P_0 = 40,000$. At $t = 20$,$P(20) = 80,000$. Thus,$80,000 = 40,000 e^{20k}$,which implies $e^{20k} = 2$. We want to find the population after another $40$ years,which is at $t = 20 + 40 = 60$ years. $P(60) = 40,000 e^{60k} = 40,000 (e^{20k})^3$. Substituting $e^{20k} = 2$,we get $P(60) = 40,000 	imes (2)^3 = 40,000 	imes 8 = 320,000$.

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 $20$ years,then the population in another $40$ years will be (in $,000$)

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