# Math Help - Radius of Speherical Ballon increasing at time

1. ## Radius of Speherical Ballon increasing at time

2. Originally Posted by CAIR

This is a related rate problem, which means that you need to differentiate a standard equation for the volume of a sphere with respect to time. Do you know how to do that?

3. Hello. This is a related rates problem.

You are given the following:

DV/dt = 40cm^3, when r = 10cm

You want: dr/dt:

Dv/dt = DV/dr . dr/dt

-> 40 = 4. pi. r^2 . (dr/dt)

-> 40 = 4 . pi . (100) . (dr/dt)

-> dr/dt = 40 / ((400) . (pi))

-> dr/dt = 1 / ((10) . (pi) )

-> dr/dt = 0.031830988

To test:

DV/dt = DV/dr . (dr/dt)

-> DV/dt = 400 pi . (0.031830988)
-DV/dt = 40

Therefore dr/dt does, in fact = 0.031830988

Best of luck.