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

Given: A circle with uniformly increasing area rate

To prove: relation between perimeter and radius is inversely proportional

Explanation: Take the radius of circle ‘r’

Let the area of the circle be A

Then $A = \pi r^2$……..(i)

After considering the give criteria of area increasing at a uniform rate,
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\mathrm{k}$
Substitute the value of equation (i) into above equation,
$\frac{\mathrm{d}\left(\pi r^{2}\right)}{d t}=k$
Differentiating with respect to t results in
$\Rightarrow \pi \times 2r \times \frac{d(r)}{dt} = k \\ \Rightarrow \frac{d(r)}{dt} = \frac{k}{2 \pi r} \quad \ldots \text{(ii)}$

Let P as the perimeter of the circle,

P = 2πr

Now differentiate the perimeter with respect to t,

$\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{d}(2 \pi \mathrm{r})}{\mathrm{dt}}$
Apply all the derivatives,
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=2 \pi \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dt}}$

Substituting of equation (ii) in the above equation,
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=2 \pi\left(\frac{\mathrm{k}}{2 \pi \mathrm{r}}\right)$

Cancelling out of the like terms results into
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\left(\frac{\mathrm{k}}{\mathrm{r}}\right)$

Now covert it into proportionality,
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}} \propto \frac{1}{\mathrm{r}}$

Hence in the given conditions, the relation between perimeter of circle and radius is inversely proportional.

Hence Proved