# How fast is the radius of the balloon changing at the instant the radius is 1ft and 2ft

Working on a problem, and I understand it mostly except for one part, which I will mark with a ? symbol.

VOLUME: A spherical balloon is inflated with gas at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is a.) 1 foot; and b.) 2 feet?

$V=\dfrac43\pi r^3$

$\dfrac{dV}{dt} = \dfrac{10}1 = 10$

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

??? HERE HOW DO THEY COME UP WITH THIS 1 OVER PART? $$\dfrac{1}{4\pi r^2}\left(\dfrac{dv}{dt}\right)$$???

I see I am differentiating with respect to volume, but this 1 numerator is confusing me.

here are the answers if they help.

when $r=1$; $\dfrac{5}{2}\pi$ ft/min

when $r=2$; $\dfrac{5}{8}\pi$ ft/min

Thanks in advance.

## 1 Answer

If you take $$\dfrac{dV}{dt} = 4 \pi r^2\dfrac{dr}{dt}$$ and make $\dfrac{dr}{dt}$ the subject (i.e. divide both sides by $4\pi r^2$), you get $$\dfrac{1}{4 \pi r^2}\dfrac{dV}{dt} = \dfrac{dr}{dt}$$

• ahh, forgot about the implied 1. Thank you! Feb 25, 2016 at 23:55