How to Calculate the Rate of Change of Surface Area for an Expanding Sphere?

[JKLM]

How do you get the volume of an expanding object?
For example:
A sphere's is increasing at 12ft/sec, the volume is 36, how fast is the surface area of a sphere

V=4*pi*r^3/3 S=4*pi*r^2

dV/dt= 12
So i took the derivative of the volume function

dV/dt=4*pi*r^2*dr/dt
12=4*pi*r^2*dr/dt
how do you solve for dr/dt

 

[FZ+]

Seems like a simple case of the chain rule.

dS/dt = dV/dt * dS/dV

= dV/dt * dS/dr * dr/dV

And no, I can't get the hang of latex either.

 

[HallsofIvy]

Doesn't it strike you that your question and example have nothing to do with one another?

Your question was "How do you get the volume of an expanding object?".

Your example had to do with "rate of change of surface area".

"a sphere's is increasing at 12ft/sec"

A sphere's what?? Okay, I know you meant to type "radius" but I had to look at the "12ft/sec" to be sure. I, myself, never mist83*3!

You know that, for any sphere, the surface area is given by
S= 4πr² so dS/dr= 8πr. You want "dS/dt" (because you are asked for rate of change of surface area) and, as FZ+ said, that means you use the chain rule: dS/dt= (dS/dr)(dr/dt)= (8πr ft)(12 ft/sec). Of course in order to get that you need to know r (at that particular time). THAT'S why you need volume. Not to "find" volume but to use it to find r. Since that at a particular time, the fact that the sphere is changing has nothing to do with it. Solve (4/3)πr³= 36 for r and plug that into your formula for dS/dt.