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

Let us assume that the radius of circle $= r$

And area of the circle,$A = \pi {r^2}$

Therefore,$\frac{d}{{dt}}A = \frac{d}{{dt}}\pi {r^2}$

$\Rightarrow$ $\frac{{dA}}{{dt}} = 2\pi r \cdot \frac{{dr}}{{dt}}$ …….(i)

Since, the area of a circle increases at a uniform rate, then
$\frac{{dA}}{{dt}} = k$ ……..(ii)

where, $k$ is a constant.

From Eqs. (i) and (ii), $2\pi r \cdot \frac{{dr}}{{dt}} = k$

$\Rightarrow$ $\frac{{dr}}{{dt}} = \frac{k}{{2\pi r}} = \frac{k}{{2\pi }} \cdot \left( {\frac{1}{r}} \right)$ …….(iii)

Let the perimeter, $P = 2\pi r$

Differentiating P w.r.t. t. we get :

$\frac{{dP}}{{dt}} = \frac{d}{{dt}} \cdot 2\pi r \Rightarrow \frac{{dP}}{{dt}} = 2\pi \cdot \frac{{dr}}{{dt}}$

$= 2\pi \cdot \frac{k}{{2\pi }} \cdot \frac{1}{r} = \frac{k}{r}$ [By using Eq.(iii)]

$\Rightarrow$ $\frac{{dP}}{{dt}} \propto \frac{1}{r}$ Hence proved.