Question

The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At what rate is the volume of the cone changing when the radius is 134 in. and the height is 136 in.

Answer 1

Volume of the cone is increasing at the rate
`9916\pi (in^3)/(s)`.

Explanation:

Given: The radius of a right circular cone is increasing at a rate of
`1.9` in/s while its height is decreasing at a rate of
`2.2` in/s.

To find: The rate at which volume of the cone changing when the radius is
`134` in. and the height is
`136` in.

Solution:

We have,

`(dr)/(dt) =1.9 \:\text{in/s}`,
`(dh)/(dt)=-2.2\:\text{in/s}`,
`r=134 \:\text{in}`,
`h=136\:\text{in}`

Now, let
`V` be the volume of the cone.

So,
`V=(1)/(3)\pi r^(2)h`

Differentiate with respect to
`t`.

`(dv)/(dt) =(1)/(3)\pi \left [ r^2(dh)/(dt)+h\left ( 2r \right )(dr)/(dt) \right ]`

Now, on substituting the values, we get

`(dv)/(dt) =(1)/(3)\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left ( 136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]`

`(dv)/(dt) =(1)/(3)\pi\left [ -39503.2+69251.2 \right ]`

`(dv)/(dt) =(1)/(3)\pi\left [ 29748 \right ]`

`(dv)/(dt) =9916\pi (in^3)/(s)`

Hence, the volume of the cone is increasing at the rate
`9916\pi (in^3)/(s)`.