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

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