Question

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

Answer 1

Given that $\frac{dr}{dt}=10$ $cm/s$.
and $r=15$ $cm$
Find out $\frac{ds}{dt}=?$
We know that surface area of sphere $S=4\pi r^2$
$\frac{ds}{dt}=4\pi \left(2r\right)\frac{dr}{dt}$
$=8\pi \left(15\right)\left(10\right)$
$=1200\pi$ $c{m}^2/s$.