Browse Questions

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

Toolbox:
• If $y = f(x)$, then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Area of the circle is
$A= \pi r^2$
Differentiating w.r.t $t$ on both sides we get,
$\large\frac{dA}{dt}= \pi.2r.\large\frac{dr}{dt}$
It is given that the area is increasing at a uniform rate.
$\therefore \large\frac{dA}{dt}=k\: \: \Rightarrow 2\pi r.\large\frac{dr}{dt}=k$
Step 2
where $k$ is a constant $\therefore \large\frac{dr}{dt}=\large\frac{k}{2\pi r}$
Perimeter of the circle is
$p=2\pi r$
Differentiatily w.r.t $t$ we get
$\large\frac{dp}{dt}=2\pi.\large\frac{dr}{dt}$
Substituting for $\large\frac{dr}{dt}$ we get
$\large\frac{dp}{dt}=2\pi.\large\frac{k}{2\pi r}$
$\large\frac{dp}{dt}=\large\frac{k}{r}$
$\Rightarrow \large\frac{dp}{dt}\: \alpha\: \large\frac{1}{r}$
Hence this proves that the perimeter varies inversely as the radius.