Calculus I - Ballon Variation when perforated by a needle

Question

A spherical balance is pierced by a needle so that its volume comes to decrease a rate of $50\, \mathrm{cm}^3 / \mathrm{min}$. Determine an index of variation of your surface area on the moment when the balloon radius is $15$ cm. I have tried: $$ V = \frac{4}{3}\pi r^3.$$ Then I did the derivative of it: $$ V' = 4\pi r^2.$$ But I don't know what to do with the $50\, \mathrm{cm}^3 / \mathrm{min}$ to find the aswer

Answer 1

$\frac {dV}{dt} = 4\pi r^2 \frac {dr}{dt}$

$\frac {dV}{dt}$ is the rate that the volume is changing. i.e. $50 \frac {\text{cm}^3}{\text{sec}}$

$\frac {dr}{dt}$ is the rate at which the radius is changing, and what you need to solve for.

$50 \frac {\text{cm}^3}{\text{sec}} = 4\pi(15 \text{ cm})^2\frac {dr}{dt}$