0.70 ft
Step-by-step explanation:
The volume V of a cone is given by
[tex]V = \dfrac{\pi}{3}r^2h[/tex]
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
[tex]V = \dfrac{\pi}{3}\left(\dfrac{D}{2}\right)^2h = \dfrac{\pi}{12}D^2h[/tex]
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
[tex]V = \dfrac{\pi}{12}h^3[/tex]
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:
[tex]\dfrac{dV}{dt} = \dfrac{d}{dt}\left(\dfrac{\pi}{12}h^3\right)[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{\pi}{12}\cdot 3h^2\dfrac{dh}{dt} = \dfrac{\pi}{4}h^2\dfrac{dh}{dt}[/tex]
Rearranging the variables and solving for dh/dt, we get
[tex]\dfrac{dh}{dt} = \dfrac{4}{\pi h^2}\dfrac{dV}{dt}[/tex]
[tex]\:\:\:\:\:\:\:\:= \dfrac{4}{\pi(8\:\text{ft})^2}(35\:\text{ft}^3\text{min})[/tex]
[tex]\:\:\:\:\:\:\:=0.70\:\text{ft/min}[/tex]
