Question

The radius of the base of a cone is decreasing at a rate of 2 22 centimeters per minute. The height of the cone is fixed at 9 99 centimeters. At a certain instant, the radius is 13 1313 centimeters. What is the rate of change of the volume of the cone at that instant (in cubic centimeters per minute)?

Answer 1

-156π

Explanation:

Answer 2

The Volume is decreasing at a rate of
`156\pi`
`cm^3` per minute.

Explanation:

Given a cone of radius r and perpendicular height h

Volume of the cone,
`V=(1)/(3)\pi r^2 h`

Since the height of the cone is fixed, the rate of change of the volume of the Cone,
`(dV)/(dt)` is given as:

`V=(1)/(3)\pi r^2 h \\(dV)/(dt)=(h\pi)/(3) (d)/(dt) r^2\\(dV)/(dt)=(2rh\pi)/(3) (dr)/(dt)`

We are to determine the rate of change of the Volume, V when:

The radius is decreasing at a rate of 2 cm per minute,
`(dr)/(dt) = -2 cm/min`

Height, h=9 cm

Radius = 13cm

`(dV)/(dt)=(2rh\pi)/(3) (dr)/(dt)\\=(2*13*9\pi)/(3)* ( -2)\\(dV)/(dt)=-156\pi \:cm^3/min`

The Volume is decreasing at a rate of
`156\pi`
`cm^3` per minute.