Asked by Hailey
If a spherical balloon is inflated and its volume is increasing at a rate of 6 in^3/min. What is the rate of change of the radius when the radius is 3 inches??
Answers
Answered by
Reiny
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
given: dV/dt = 6, r=3
6 = 4π (9) dr/dt
dr/dt = 6/(36π) inches/min
= 1/(6π) in/min
dV/dt = 4πr^2 dr/dt
given: dV/dt = 6, r=3
6 = 4π (9) dr/dt
dr/dt = 6/(36π) inches/min
= 1/(6π) in/min
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