Calculus and balloon inflation rates

**bradbury17** · Sep 17th 2013, 07:39 AM

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 cm³/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 $\frac{dr}{dt}$ when V = 30 cm³.

(b) Find $\frac{dS}{dt}$ when V = 30 cm³.

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

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!

**topsquark** · Sep 17th 2013, 08:12 AM

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 cm³/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 $\frac{dr}{dt}$ when V = 30 cm³.

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

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

So what does dr/dt equal?

Originally Posted by bradbury17

(b) Find $\frac{dS}{dt}$ when V = 30 cm³.

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

-Dan

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