Asked by Peggy
A spherical balloon is inflated with helium at the rate of 100pi ft ^3/min. How fast is the ballon's radius increasing at the instant the radius is 5ft.?
Answers
Answered by
Steve
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
100 = 4 pi * 25 dr/dt
dr/dt = 1/pi
dv/dt = 4 pi r^2 dr/dt
100 = 4 pi * 25 dr/dt
dr/dt = 1/pi
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