Posted by **Z32** on Thursday, September 24, 2009 at 10:32pm.

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 -
**Reiny**, Thursday, September 24, 2009 at 11:02pm
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 -
**Z32**, Friday, September 25, 2009 at 12:27am
Got it down to 3.157 which is correct. Thanks!

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