A tumor is assumed to be roughly spherical. If the radius R is found to be increasing at the rate of 0.5 mm/day when the radius is R=2 mm, how fast is the volume changing at this time?

Answer 1

V = (4/3)πr^3
given dr/dt = .5 when r = 2
dV/dt = 4πr^2 dr/dt
dV/dt = 4π(,25)(2)
= 2π mm^3/day

Answer 2

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