As a ball in the shape of a sphere is being blown up, the volume is increasing at the rate of 4 cubic inches per second. At what rate is the radius increasing when the radius is 1.5 inches?
Can you also please explain "why" you do each step and not just "how" to solve the problem? Thank you!

Question

As a ball in the shape of a sphere is being blown up, the volume is increasing at the rate of 4 cubic inches per second. At what rate is the radius increasing when the radius is 1.5 inches? 
Can you also please explain "why" you do each step and not just "how" to solve the problem? Thank you!

anonymous · Answer

So we know that the volume of a sphere is $$(4/3)*\pi*r^2$$

And the rate at which the volume is increasing can be written as dv/dt

So if we have

$$v = (4/3)*\pi*r^2$$

Then we derive with respect to t to get the following:

$$dv/dt = (4*\pi/3)* 2r * dr/dt$$

Isolate dr/dt

$$(dv/dt)/(4*\pi*2r/3) = dr/dt$$

We're given dv/dt = 4 and r = 1.5.  Plug in these values.

$$4/(4*\pi*2*1.5/3)$$

You can cancel out the threes to get 4/8pi which simplifies down to 1/2pi

anonymous · Answer

Thank you!

anonymous · Answer

Actually, in the last step when you are simplifying, don't you end up with 4/4pi which = pi?

anonymous · Answer

Oh yeah, sorry about that.  It's been a bit of a long day.

anonymous · Answer

Actually, it equals 1/pi

anonymous · Answer

oh wow, right, great call. Thanks again!