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 the number of bacteria at a given times 't' be N
Then,
$\frac{dN}{dt}=kN$ where $k$ is the proportionality constant.
$\frac{dN}{N}=kdt$
${\int }_{N_0}^N\frac{dN}{N}={\int }_0^{t}kdt$
$ln\left(\frac{N}{N_0}\right)=kt$
$N=N_0{e}^{kt}$ ...(i)
If $N_0={10}^5$ and $t=2hours$ and
$N=(1+\frac{10}{100})\times {10}^5$
$=1.1\times {10}^5$
Hence
$ln\frac{N}{N_0}=kt$
$ln\left(\frac{1.1\times {10}^5}{{10}^5}\right)=2k$
$ln\left(1.1\right)=2k$
$k=\frac{1}{2}ln\left(1.1\right)hour{s}^{-1}$
$=ln\left(\sqrt{1.1}\right)hour{s}^{-1}$
Now
$ln\left(\frac{N}{N_0}\right)=kt$
N is now $2\times {10}^5$
Hence
$\frac{N}{N_0}$
$=\frac{2\times {10}^5}{{10}^5}$
$=2$
Thus
$ln\left(\frac{N}{N_0}\right)=kt$ implies
$ln2=ln\sqrt{1.1}t$
$ln2=\frac{ln\left(1.1\right)}{2}t$
$2ln2=ln\left(1.1\right)t$
$t=\frac{ln4}{ln\left(1.1\right)}$
$=14.54$ hours