Question

Gravel is being dumped from a conveyor belt at a rate of 35 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 8 ft high? (Round your answer to two decimal places.)

Answer 1

0.70 ft

Explanation:

The volume V of a cone is given by

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

where r is the radius of the base and h is the height of the cone. Since the radius is half the diameter D, we can rewrite V as

`V = (\pi)/(3)\left((D)/(2)\right)^2h = (\pi)/(12)D^2h`

But since the base diameter D is always equal to the cone's height, we can write the volume in terms of the height h as

`V = (\pi)/(12)h^3`

Now gravel is being added to the conical pile at a rate of 35 cubic ft per minute so to find the rate at which the height of the pile is increasing, we are going to take the derivative of V with respect to time:

`(dV)/(dt) = (d)/(dt)\left((\pi)/(12)h^3\right)`

`\:\:\:\:\:\:\:= (\pi)/(12)\cdot 3h^2(dh)/(dt) = (\pi)/(4)h^2(dh)/(dt)`

Rearranging the variables and solving for dh/dt, we get

`(dh)/(dt) = (4)/(\pi h^2)(dV)/(dt)`

`\:\:\:\:\:\:\:\:= \frac{4}{\pi(8\:\text{ft})^2}(35\:\text{ft}^3\text{min})`

`\:\:\:\:\:\:\:=0.70\:\text{ft/min}`