Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 2 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 2 feet high?

Answer 1

`(2)/(9\pi)ft^3/min`

Explanation:

Given

`(dV)/(dt)=2ft^3/min\\ \\(dh)/(dt)=?\\ \\V_(cone)=(1)/(3)\pi r^2h\\\\d=2r=3h\\\\(d)/(2)=r=(3)/(2)h\\\\h=2`

Calculation

`V=(1)/(3)\pi r^2h\\ \\V=(1)/(3)\pi((3)/(2)h) ^2h\\\\V=(1)/(3)\pi((9)/(4)h^2)h\\ \\V=(9)/(12)\pi h^3\\ \\V=(3\pi)/(4)h^3\\ \\(dV)/(dt)=(9\pi)/(4)h^2(dh)/(dt)\\ \\2=(9\pi)/(4)(2)^2(dh)/(dt)\\ \\ 2=9\pi(dh)/(dt)\\ \\ (2)/(9\pi) =(dh)/(dt)`

Therefore, the rate that the height of the pile is changing when the pile is 2 feet high is
`(2)/(9\pi)ft^3/min`.