Question

In a culture,the bacteria count is 1,00,000.The number is increased by $10\%$ in 2 hours. In how many hours will the count reach 2,00,000,if the rate of growth of bacteria is proportional to the number present?

Answer 1

Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
$∴\frac{dy}{dt}∝y$
$⇒\frac{dy}{dt}=ky$(where k is a constant)
$⇒\frac{dy}{y}=kdt$
Integrating both sides,we get:
$∫\frac{dy}{y}=k∫dt$
$⇒logy=kt+C...(1)$
Let $y_0$ be the number of bacteria at t=0.
$⇒logy_0=C$
Substituting the value of C in equation(1),we get:
$logy=kt+logy_0$
$⇒logy-logy_0=kt$
$⇒log(\frac{y}{y_0})=kt$
$⇒kt=log(\frac{y}{y_0})...(2)$
Also,it is given that the number of bacteria increased by $10\%$ in 2hours.
$⇒y=\frac{110}{100y_0}$
$⇒\frac{y}{y_0}=\frac{11}{10}...(3)$
Substituting this value in equation(2),we get:
$k.2=log(\frac{11}{10})$
$⇒k=\frac{1}{2}log(\frac{11}{10})$
Therefore,equation(2) becomes:
$\frac{1}{2}log(\frac{11}{10}).t=log(\frac{y}{y_0})$
$⇒t=\frac{2log(\frac{y}{y_0})}{log(\frac{11}{10})}...(4)$
Now,let the time when the number of bacteria increases from 100000 to 200000 be $t_1$.
$⇒y=2y_0\space at\space t=t_1$
From equation(4),we get:
$t_1=\frac{2log(\frac{y}{y_0})}{log(\frac{11}{10})}=\frac{2log2}{log(\frac{11}{10})}$
Hence,in $\frac{2log2}{log(\frac{11}{10})}$ hours the number of bacteria increases from 100000 to 200000.