Question

The radius of a circle is increasing uniformly at the rate of $3\text{cm/s}$. Find the rate at which the area of the circle is increasing when the radius is $10\text{cm}$.

Answer 1

Let $r$ be the radius of circle
Given, Radius of a circle is increasing at the rate of $3\text{cm/s}$
Thus,
$\frac{dr}{dt}=3\text{cm/s}$
We know that
Area of circle $A=\pi r^2$
Differentiating w.r.t $t,$

$\frac{dA}{dt}=\frac{d\left(\pi r^2\right)}{dt}$

$\Rightarrow \frac{dA}{dt}=\pi \frac{d\left(r^2\right)}{dt}$

$\Rightarrow \frac{dA}{dt}=2\pi r\frac{dr}{dt}=6\pi r[\because \frac{dr}{dt}=3]$

When $r=10\text{cm}$

$\frac{dA}{dt}=60\pi {\text{cm}}^2/\text{s}$

Hence, area of the circle is increasing at the rate of $60\pi {\text{cm}}^2/\text{s},$ when radius is $10\text{cm}$.