# Rate of change of the radius of a balloon

"Helium is pumped into a spherical balloon at a rate of $4$ cubic feet per second. How fast is the radius increasing after $3$ minutes?"

So this is what I did:

$$V = \frac{4}3\pi r^3$$

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

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

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

I did not know how to proceed after this. How am I supposed to find the value of $r$? Do I have to somehow relate $3$ minutes ($180$ seconds) in somehow?

Any help?

• The volume should be $180\text{sec}\frac{4\text{ft}^3}{\text{sec}}$ at $3$ minutes right. you can get $r$ from this.
– JEM
Commented Nov 3, 2017 at 1:14
• If you call the volume $V,$ then the rate of change of volume should be $\dfrac{dV}{dt},$ not $\dfrac{dv}{dt}. \qquad$ Commented Nov 3, 2017 at 4:47

$$\frac{dV}{dt}=4$$
and also $V(0)=0$.
Hence, at time $t=3$, you can compute the volume at that time and you can use the formula
$$V=\frac{4}{3}\pi r^3$$
to solve for $r$ at that time.