Question

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 ). )

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 \(\frac{dV}{dt}=\frac{9}{4}\pi h^{2}\frac{dh}{dt}\):

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Answer:

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