Question

The radius of a balloon is increasing at the rate of $10$ cm/sec. At what rate is the surface area of the balloon increasing when the radius is $15$ cm?

Answer 1

Let r be the radius and S be the surface area of the balloon at any time t. Then,

$S=4\pi r^2$ and $\frac{dr}{dt}=10cm/sec.$

Now, $S=4\pi r^2$

$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

$\frac{dS}{dt}=80\pi r$

$(\frac{dS}{dt}{)}_{r=15}=80\pi (15)=1200\pi c{m}^2\sec .$