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A spherical balloon is inflated so that its volume is increasing at the rate of 20 cubic feet per minute. How fast is the surface area of the balloon increasing when the radius is 4 feet?
Hello Charlotte,
Since V=(4/3)*pi*r^3, dV/dt=4*pi*r^2*dr/dt
So that when dV/dt=20, we get 4*pi*r^2*dr/dt=20 ==> dr/dt=5/(pi*r^2)
Surface areas, A=4*pi*r^2, so that dA/dt=8*pi*r*dr/dt
Now substitute dr/dt=5/(pi*r^2) ==> dA/dt=8*pi*r*5/(pi*r^2)
==> dA/dt=40/r, when r=4 this gives dA/dt=10 square ft/min.
I hope this helps!
Abe
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Answers by Expert:
Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience
Over 15 years teaching at the college level.
Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.
Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook
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