If the area of a circle increases at a uniform rate,then prove that the rate of change of its perimeter varies inversely as the radius.

(N/A) Let the radius of the circle be $r$ and the area of the circle be $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \quad \dots(i)$
Given that the area of the circle increases at a uniform rate,let $\frac{dA}{dt} = k$,where $k$ is a constant.
From equation $(i)$,we have:
$k = 2\pi r \cdot \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{k}{2\pi r} \quad \dots(ii)$
Let the perimeter of the circle be $P = 2\pi r$.
Differentiating the perimeter with respect to time $t$:
$\frac{dP}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \cdot \frac{dr}{dt}$
Substituting the value of $\frac{dr}{dt}$ from equation $(ii)$:
$\frac{dP}{dt} = 2\pi \cdot \left( \frac{k}{2\pi r} \right) = \frac{k}{r}$
Since $k$ is a constant,we have $\frac{dP}{dt} \propto \frac{1}{r}$.
Thus,the rate of change of the perimeter varies inversely as the radius.