Asked by Joan

How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dv/dt , the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/dt , the change in the radius with respect to time. Remember that the volume of a sphere is v=4/3Pir^3.

Answers

Answered by Steve
hmm. They gave you the formula, so just use it:

v = 4/3 pi r^3

dv/dt = 4pi r^2 dr/dt

So, just solve for dr/dt
Answered by Joan
= 4pi r^2(15)??
Answered by Damon
dV/dr = area of surface = 4 pi r^2
so
dV/dt = 4 pi r^2 dr/dt
so
dr/dt = (dV/dt) / (4 pi r^2)
Answered by Joan
Oh okay, thanks
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