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.