The growth of population is proportional to t | vedclass

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(B) Let $P_{0}$ be the initial population. Given that the rate of growth is proportional to the population: $\frac{dP}{dt} = \lambda P$. Integrating b

Vedclass · Accepted Answer

$50\left(\frac{\log 3}{\log 2}ight) 	ext{ yrs}$

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(B) Let $P_{0}$ be the initial population. Given that the rate of growth is proportional to the population: $\frac{dP}{dt} = \lambda P$. Integrating both sides: $\int \frac{dP}{P} = \int \lambda dt \Rightarrow \ln P = \lambda t + C$. At $t = 0$,$P = P_{0}$,so $C = \ln P_{0}$. Thus,$\ln P = \lambda t + \ln P_{0} \Rightarrow \ln \left(\frac{P}{P_{0}}ight) = \lambda t$. Given that the population doubles in $50$ years: $\ln \left(\frac{2P_{0}}{P_{0}}ight) = 50\lambda \Rightarrow \ln 2 = 50\lambda \Rightarrow \lambda = \frac{\ln 2}{50}$. Now,we need to find $t$ when the population triples $(P = 3P_{0})$: $\ln \left(\frac{3P_{0}}{P_{0}}ight) = \lambda t \Rightarrow \ln 3 = \left(\frac{\ln 2}{50}ight) t$. Solving for $t$: $t = 50 \left(\frac{\ln 3}{\ln 2}ight) 	ext{ years}$.

The growth of population is proportional to the number present. If the population of a colony doubles in $50$ years,then the population will become triple in . . . . . . years.

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