The rate of increase of bacteria in a certain | vedclass

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(C) Let $P_0$ be the initial population and $P$ be the population at time $t$.According to the problem,$\frac{dP}{dt} = kP$,where $k$ is a constant.Se

Vedclass · Accepted Answer

$32$

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(C) Let $P_0$ be the initial population and $P$ be the population at time $t$.According to the problem,$\frac{dP}{dt} = kP$,where $k$ is a constant.Separating the variables,we get $\frac{dP}{P} = k dt$.Integrating both sides,we have $\ln P = kt + C$.At $t = 0$,$P = P_0$,so $\ln P_0 = C$.Thus,$\ln P = kt + \ln P_0$,which simplifies to $\ln \left( \frac{P}{P_0} ight) = kt$.Given that the population doubles in $5$ hours,at $t = 5$,$P = 2P_0$.Substituting these values,$\ln \left( \frac{2P_0}{P_0} ight) = 5k$,so $k = \frac{\ln 2}{5}$.Now,we need to find the population at $t = 25$ hours.$\ln \left( \frac{P}{P_0} ight) = \left( \frac{\ln 2}{5} ight) 	imes 25 = 5 \ln 2 = \ln(2^5) = \ln 32$.Therefore,$\frac{P}{P_0} = 32$,which means $P = 32 P_0$.The number of bacteria will be $32$ times the original.

The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in $5$ hours,then in $25$ hours,its number would be ......... times the original.

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