Calculus
posted by Anonymous .
A spherical balloon is being inflated at a rate of 10 cubic inches per second. How fast is the radius of the balloon increasing when the surface area of the balloon is 2pi square inches? Enter your answer correct to 3 decimal places.

V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt (#1)
when 4πr^2 = 2π
r^2 = 1/2
r = 1/√2
in #1:
10 = 4π(1/√2)^2 dr/dt
10= 2π dr/dt
dr/dt = 5/π or 1.592 inches/sec
check my arithmetic
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