Asked by Jay
A balloon is being filled with helium at the rate of 4 ft^3/min. The rate, in feet per minute, at which the radius is increasing when the radius is 2 feet is (V= 4/3πr^3)
Answers
Answered by
Damon
you mean cubic feet per minute
surface area of sphere = 4 pi r^2
Well to avoid calculus just say the if you increase the radius by x, the volume will change by the outside surface area times that increase in radius x
so
increase in volume = 4 pi r^2 * increase in radius (that is the volume of the added skin of thickness x)
so
rate of change of volume = 4 pi r^2 * rate of increase of radius
surface area of sphere = 4 pi r^2
Well to avoid calculus just say the if you increase the radius by x, the volume will change by the outside surface area times that increase in radius x
so
increase in volume = 4 pi r^2 * increase in radius (that is the volume of the added skin of thickness x)
so
rate of change of volume = 4 pi r^2 * rate of increase of radius
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