Question

The radius of a cone is increasing at a constant rate of 9 feet per minute, and the volume is increasing at a rate of 694 cubic feet per minute. At the instant when the radius of the cone is 55 feet and the volume is 239 cubic feet, what is the rate of change of the height? The volume of a cone can be found with the equation V=\frac{1}{3}\pi r^2 h.V= 3 1 ​ πr 2 h. Round your answer to three decimal places.

Answer 1

0.194 ft/min

Explanation:

Differentiating the volume equation with respect to time, we get ...

V' = (1/3)π(2r·r'·h +r²·h')

We need to find the height when the volume is 293 ft³:

239 ft³ = (π/3)(55 ft)²h

h = 3·239 ft³/(3025π ft²) = 717/(3025π) ft ≈ .07545 ft

Using volumes in ft³, areas in ft², and rates in ft/min, we can fill in the given information to get ...

694 = (π/3)(2·55·9·717/(3025π) + 3025h')

694 = 4302/55 +(π/3)3025h' . . . . simplify a bit

33868/55 = (π/3)3025h' . . . . . . . . subtract 4302/55

Now, divide by the coefficient of h.

101604/(166375π) = h' ≈ 0.194 . . . . ft/min

The rate of change of height is about 0.194 ft/min.

Answer 2

Rate of change of height is -0.108 ft per minute.

Explanation:

We have volume of cone is

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

We need to find the rate of change of the height when the radius of the cone is 55 feet and the volume is 239 cubic feet.

Substituting

`239=(1)/(3)\pi * 55^2* h\\\\h=0.275 feet`

Given

`(dV)/(dt)=694ft^3/min\\\\(dr)/(dt)=9ft/min`

Finding rate of change of volume

`V=(1)/(3)\pi r^2h\\\\(dV)/(dt)=(1)/(3)\pi (2r(dr)/(dt)h+r^2(dh)/(dt))\\\\694=(1)/(3)\pi (2* 55* 9* 0.275+55^2* (dh)/(dt))\\\\(dh)/(dt)=-0.108ft/min`

Rate of change of height is -0.108 ft per minute.