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A balloon is being filled with helium at the rate of 4ft^3/min. the rate, in square feet per minute, at which the surface area is increasing when the volume is 32π/3 ft³ is

User BushyMark
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The rate at which the surface area is increasing when the volume is 32π/3 ft³ is 16π ft²/min.

Given that the balloon is being filled with helium at a rate of 4ft^3/min and the volume is 32π/3 ft³, we need to find the rate at which the surface area is increasing.

The formula for the volume of a sphere is V = (4/3)πr³, where r is the radius.

Since we are given the volume, we can solve for the radius: 32π/3 = (4/3)πr³. Simplifying, we get r³ = 8/3, and taking the cube root of both sides, we find r = 2.

The formula for the surface area of a sphere is A = 4πr².

Using the radius, we can find the surface area: A = 4π(2)² = 16π.

Therefore, when the volume is 32π/3 ft³, the surface area is increasing at a rate of 16π ft²/min.

User Rohan Fating
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