# Calculus

A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2cm/min. At what rates are the volume and surface area of the balloon increasing when the radius is 5cm?

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1. V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
when r = 5 and dr/dt = 2
dV/dt = 4π(25)(2) = 200π cm^3/min

SA = 4πr^2
d(SA)/dt = 8πr dr/dt
= 8π(5)(2) = 80π cm^2/min

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