Question

The radius of a cone is increasing at a constant rate of 8 feet per second. The volume remains a constant 56 cubic feet. At the instant when the radius of the cone is 3 feet, what is the rate of change of the height? The volume of a cone can be found with the equation V = farh. Round your answer to three decimal places.

Answer 1

`\Rightarrow (dh)/(dt)\approx -31.690`

Step-by-step explanation:

The volume of a cone, V = 1/3πr^2h

Where r is the radius, and h is the height of the cone.

Given:

Volume is constant at 56 cubic feet.

Radius of cone increases at a constant rate of 8 feet per second.

The radius of the cone is 3 feet.

When radius is 3 feet

`\Rightarrow h= (3V)/(\pi r^2)`

`\Rightarrow h= (3* 56)/(\pi * 3^2)`

`\Rightarrow h= (3* 56)/(\pi * 9)`

`\Rightarrow h= ( 56)/(3\pi)`

Where h is in feets.

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

`\Rightarrow (dV)/(dt)=((1)/(3)\pi (dr^2)/(dt)h)+((1)/(3)\pi r^2 (dh)/(dt))`

`\Rightarrow (dV)/(dt)=((1)/(3)\pi (dr^2)/(dr).(dr)/(dt)h)+((1)/(3)\pi r^2 (dh)/(dt))`

`\Rightarrow (dV)/(dt)=((1)/(3) \pi 2r(dr)/(dt)h)+((1)/(3)\pi r^2 (dh)/(dt))`

Since the volume is constant so
`(dV)/(dt)=0`

The rate of change of radius is
`(dr)/(dt)=8`

`\Rightarrow 0=((1)/(3) \pi (2* 3)(8)((56)/(3\pi)))+((1)/(3)\pi (3^2) (dh)/(dt))`

`\Rightarrow 0=((56* 16)/(3))+((1)/(3)\pi (9) (dh)/(dt))`

`\Rightarrow 0=((56* 16)/(3))+(3\pi (dh)/(dt))`

`\Rightarrow 3\pi (dh)/(dt)=-(56* 16)/(3)`

`\Rightarrow (dh)/(dt)=-(56* 16)/(3* 3 * \pi)`

`\Rightarrow (dh)/(dt)=-(56* 16)/(9\pi)`

`\Rightarrow (dh)/(dt)=-31.689518`

`\Rightarrow (dh)/(dt)\approx -31.690`

The minus indicates that the rate of change of height is decreasing.