# Calculus

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Air is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm?
V= 4/3*pi*r^3
S= 4 pi r^2

• Calculus -

from V= 4/3*pi*r^3
dV/dt = 4pi(r^2)dr/dt
so when dV/dt=30 and r = 19
30 = 4pi(361)dr/dt
dr/dt = 30/[4pi(361)]

now in
A = 4pir^2
dA/dt = 8pi(r)dr/dt
= 8pi(19)*30/[4pi(361)]
= you finish it.

• Calculus -

Got it down to 3.157 which is correct. Thanks!

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