Your formula for Surface Area is incorrect, anyhow ....
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when r = 9, and dV/dt = 90
90 = 4π (81) dr/dt
dr/dt = 90/(324π) = (5/18π)
SA = 4πr^2 <---- the correct formula
dSA/dt = 8π r dr/dt
for the given data:
dSA/dt = 8π(9)(5/18π) = 20 cm^2/s
Air is being pumped into a spherical balloon so that its volume increases at a rate of 90cm^3/s. How fast is the surface area of the balloon increasing when its radius is 9cm? Recall that a ball of radius r has volume V=4/3 pi r^3 and surface area S=4/3 pi r^2
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