Question

an inverted cone has a height of 11 inches and an initial radius of 18 inches. the volume of the inverted cone is decreasing at a rate of 541 cubic inches per second, with the height being held constant. what is the rate of change of the radius, in inches per second, when the radius is 5 inches?

Answer 1

Answer:

  -4.697 in/s

Step-by-step explanation:

You want to know the rate of change of radius of a cone 11 inches high with a radius of 5 inches and a volume that is decreasing at the rate of 541 cubic inches per second.

Volume

The volume of a cone is given by ...

  V = π/3r²h

The rate of change is found by implicit differentiation:

  V' = (π/3)((2rr')h +r²h')

Here, the height is constant, so h' = 0. Solving for r', we find ...

  r' = 3V'/(2πrh)

Application

Using the given values of V', r, and h, we find the rate of change of radius to be ...

  r' = 3(-541 in³/s)/(2π(5 in)(11 in)) = -1623/(110π) in/s ≈ -4.69652 in/s

The radius is decreasing at about -4.697 inches per second.

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Additional comment

The initial radius of the cone is irrelevant to the problem, since we want to know the rate of change at a specific different radius. Whether the cone is inverted or not is also irrelevant to its volume.