at a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet…

at a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet per minute. the diameter of the base of the cone is approximately three times the altitude. at what rate (in ft/min) is the height of the pile changing when the pile is 2 feet high? (hint: the formula for the volume of a cone is ( v=\frac{1}{3}pi r^{2}h ). )

at a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet per minute. the diameter of the base of the cone is approximately three times the altitude. at what rate (in ft/min) is the height of the pile changing when the pile is 2 feet high? (hint: the formula for the volume of a cone is ( v=\frac{1}{3}pi r^{2}h ). )

Answer

Explanation:

Step1: Express radius in terms of height

Given (d = 3h), and since (d = 2r), then (2r=3h), so (r=\frac{3}{2}h).

Step2: Substitute (r) into volume formula

The volume of a cone (V=\frac{1}{3}\pi r^{2}h). Substitute (r = \frac{3}{2}h) into it: (V=\frac{1}{3}\pi(\frac{3}{2}h)^{2}h=\frac{3}{4}\pi h^{3}).

Step3: Differentiate (V) with respect to (t)

Differentiate (V=\frac{3}{4}\pi h^{3}) with respect to (t) using the chain - rule. (\frac{dV}{dt}=\frac{9}{4}\pi h^{2}\frac{dh}{dt}).

Step4: Solve for (\frac{dh}{dt})

We know that (\frac{dV}{dt} = 4) and (h = 2). Substitute these values into (\frac{dV}{dt}=\frac{9}{4}\pi h^{2}\frac{dh}{dt}): [ \begin{align*} 4&=\frac{9}{4}\pi(2)^{2}\frac{dh}{dt}\ 4&=\frac{9}{4}\pi\times4\frac{dh}{dt}\ 4&=9\pi\frac{dh}{dt}\ \frac{dh}{dt}&=\frac{4}{9\pi} \end{align*} ]

Answer:

(\frac{4}{9\pi})