Question

In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{at}\mathrm{any}\mathrm{time}\mathrm{the}\mathrm{bacteria}\mathrm{count}\mathrm{be}N. \\ \mathrm{Given}: \\ \frac{dN}{dt}\alpha N \\ \Rightarrow \frac{dN}{dt}=\lambda N \\ \Rightarrow \frac{1}{N}dN=\lambda dt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{N}dN=\int \lambda dt \\ \Rightarrow \ln N=\lambda t+\ln \mathrm{C}...\left(1\right) \\ \mathrm{Given}: \\ att=0,N=100000 \\ therefore,\ln C=\ln 100000 \\ Puttingthevaluein\left(1\right)weget, \\ \ln N=\lambda t+\ln 100000 \\ Also,att=2 \\ N=110000 \\ \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}t\mathrm{and}N\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \ln 110000=2\lambda +\ln 100000 \\ \Rightarrow \frac{1}{2}\ln \frac{11}{10}=\lambda \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\ln C\mathrm{and}\lambda \mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \ln N=\frac{1}{2}\ln \left(\frac{11}{10}\right)t+\ln 100000....\left(2\right) \\ \mathrm{When}N=200000,\mathrm{let}t=T. \\ \mathrm{Substituting}\mathrm{these}\mathrm{values}\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \ln 200000=\frac{T}{2}\ln \left(\frac{11}{10}\right)+\ln 100000 \\ \Rightarrow \ln 2=\frac{T}{2}\ln \frac{11}{10} \\ \Rightarrow T=2\frac{\ln 2}{\ln {\displaystyle \frac{11}{10}}} \\ \mathrm{Therefore},\mathrm{in}2\frac{\ln 2}{\ln {\displaystyle \frac{11}{10}}}\mathrm{hours},\mathrm{the}\mathrm{count}\mathrm{will}\mathrm{reach}200000. \end{gathered}$$