Air is being pumped into a shperical balloon at the rate of 20 cubic ft. per sec. At what rate is the radius of the balloon increasing at the instant when radius equals 15 ft.?

Question

allank · Answer

Nice question. You need to use the formula for the volume of the balloon, as a function of the radius.      V(r)=4/3*pi*r^3
Now, you have the rate of volume increase, meaning that V'(r) = 20

anonymous · Answer

oh thanks.

anonymous · Answer

V=4/3 * pi * r^3
V' = 4 pi * r^2 * r'

replace the given then solve for r'

anonymous · Answer

so is the answer 2/(90pi) ft. per sec

anonymous · Answer

yep... looks good  but simplify that....

anonymous · Answer

so it would be 1/(45pi)

anonymous · Answer

thanks! :)