The rate of increase of the population of a c | vedclass

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(B) Let $P(t)$ be the population at time $t$. According to the problem,$\frac{dP}{dt} \propto P$,which implies $\frac{dP}{dt} = kP$. Solving this diff

Vedclass · Accepted Answer

$30,000, \frac{4}{3}$

Vedclass · Answer

(B) Let $P(t)$ be the population at time $t$. According to the problem,$\frac{dP}{dt} \propto P$,which implies $\frac{dP}{dt} = kP$. Solving this differential equation,we get $P(t) = Ce^{kt}$. At $t = 0$,the population is $30,000$,so $P(0) = C = 30,000$. Thus,$P(t) = 30,000 e^{kt}$. At $t = 40$,the population is $40,000$,so $40,000 = 30,000 e^{40k}$. This simplifies to $e^{40k} = \frac{40,000}{30,000} = \frac{4}{3}$. Substituting this into the equation for $P(t)$,we get $P(t) = 30,000 (e^{40k})^{\frac{t}{40}} = 30,000 (\frac{4}{3})^{\frac{t}{40}}$. Comparing this with the given form $P(t) = (a)(b)^{\frac{t}{40}}$,we find $a = 30,000$ and $b = \frac{4}{3}$.

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