Balloon Problem: Solving for Radius Change due to Air Loss at 36π cu. in.

[Brandon_R]

A punctured balloon, in the shape of a sphere, is losing air at the rate of
2 in.3/sec. At the moment that the balloon has volume 36π cubic inches,
how is the radius changing?



I got

dr/dt = -1/18π in/sec.

Is that correct?

 

[HallsofIvy]

No, it is not. Had you shown your work, I could have helped you more.

 

[Brandon_R]

Sorry about that Sir, here is what i did.

Volume = (4/3)$$\pi$$r³

dv/dt = -2 in³/sec

dv/dt = 4$$\pi$$r²dr/dt

I substituted 36$$\pi$$ into the volume equation to get the radius which is equal to 3

dr/dt = -2/(4$$\pi$$r²)

dr/dt = -1/(18$$\pi$$)

 

[Brandon_R]

Could anyone spare the time to shed some light on this problem please. Thanks :)

 

[Dick]

Could anyone spare the time to shed some light on this problem please. Thanks :)

That looks fine to me. Halls may have called it wrong since you wrote -1/18 pi instead of -1/(18 pi).

 

[HallsofIvy]

Ah, thanks, Dick!