Question

A bacterial infection in an internal wound grows as $N^{\prime }\left(t\right)=N_{0}exp\left(t\right)$, where the time $t$ is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as $\frac{dN}{dt}=-5N^2$. What will be the plot of $\frac{N_0}{N}vst$ after $1hour$?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The expression for bacterial growth is
$N=N_0{e}^t$
$\frac{N_0}{N}=e^{-t}$
From 0 to 1 hour $N^{\prime }\left(t\right)=N_0{e}^t$
From 1 hour onwards, $\frac{dN}{dt}=-5N^2$
On integrating the above equation from $N^{\prime }$ to $N$ we get.
$\underset{N^{\prime }}{\overset{N}{\int }}{N}^{-2}dN=-5\underset{1}{\overset{t}{\int }}dt$
$\Rightarrow [\frac{1}{N}{]}_{N^{\prime }}^N=5[t{]}_1^t$

$\Rightarrow \left[\frac{1}{N}\right]-\left[\frac{1}{N^{\prime }}\right]=5[t-1]$

$\therefore At1hour,N^{\prime }=e{N}_0$

$\Rightarrow \left[\frac{1}{N}\right]-\left[\frac{1}{e{N}_O}\right]=5[t-1]$

Multiply both sides by $N_0$, we get
$\Rightarrow \frac{N_0}{N}-\frac{1}{e}=5N_0(t-1)$

$\Rightarrow \frac{N_0}{N}=5N_0(t-1)+\frac{1}{e}$

$\Rightarrow \frac{N_0}{N}=5N_0\left(t\right)+\frac{1}{e}-5N_0$

On comparing the above equation with equation of straight line,
$y=mx+c$
We get $m=5N_0,c=\frac{1}{e}-5N_0$
$\therefore$ Plot of $\frac{N_0}{N}vst$ is shown below.