Calculus and balloon inflation rates

Hey guys, I've been stuck on a problem for a while now and the more I think about it, the harder it becomes to wrap my head around it.

It goes something like this,Air is pumped into a spherical balloon at a rate of 2 cm^{3}/s (two cubic centimetres per

second). Denote by V the volume of the balloon, r the radius of the balloon, and S the

surface area of the balloon.

(a) Find $\displaystyle \frac{dr}{dt}$ when V = 30 cm^{3}.

(b) Find $\displaystyle \frac{dS}{dt}$ when V = 30 cm^{3}.

You may need these formulas: $\displaystyle V = \frac{4}{3}\pi r^3$, $\displaystyle S = 4\pi r^2$.

While I found a similar question, it dealt more with a change in radius and surface area, rather than volume. Hoping you guys can put me on the right track; I'm not sure what equation I need to find, or how to get there.

Thanks!

Re: Calculus and balloon inflation rates

Quote:

Originally Posted by

**bradbury17** Hey guys, I've been stuck on a problem for a while now and the more I think about it, the harder it becomes to wrap my head around it.

It goes something like this,[INDENT]Air is pumped into a spherical balloon at a rate of 2 cm^{3}/s (two cubic centimetres per

second). Denote by V the volume of the balloon, r the radius of the balloon, and S the

surface area of the balloon.

(a) Find $\displaystyle \frac{dr}{dt}$ when V = 30 cm^{3}.

Well

$\displaystyle \frac{dV}{dt} = 2$

and

$\displaystyle \frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$

So what does dr/dt equal?

Quote:

Originally Posted by

**bradbury17** (b) Find $\displaystyle \frac{dS}{dt}$ when V = 30 cm^{3}.

Do this the same as the other. Let us know if you run into any problems.

-Dan