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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

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.

Got it down to 3.157 which is correct. Thanks!

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