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Mathematics 19 Online
OpenStudy (anonymous):

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.?

OpenStudy (allank):

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

OpenStudy (anonymous):

oh thanks.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so it would be 1/(45pi)

OpenStudy (anonymous):

thanks! :)

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