Question

A bacterial infection in an internal wound grows as $N'(t) = N_0$ exp(t), 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 $\frac{N_0}{N}$ vs. $t$ after $1$ hour ?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

From $0$ to $1$ hour, $N' = N_0 \; e^t$
From 1 hour onwards $\frac{dN}{dt} = - 5 N^2$
So at $t = 1$ hour, $N'$ = $eN_0$
$\frac{dN}{dt} = - 5N^2$
$\int^{N}_{eN_0} N^{-2} dN = -5 \int^{t}_{1}dt $
$ \frac{1}{N} - \frac{1}{eN_{0}} = 5\left(t-1\right) $
$ \frac{N_{0}}{N} - \frac{1}{e} = 5N_{0} \left(t-1\right) $
$ \frac{N_{0}}{N} =5N_{0}\left(t-1\right) + \frac{1}{e} $
$ \frac{N_{0}}{N} = 5N_{0}t + \left(\frac{1}{e} -5N_{0}\right) $
which is following $ y = mx +C $