Question

gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute. it forms a pile in the shape of a right circular cone whose base diameter and height are always equal. how fast is the height of the pile increasing when the pile is 11 feet high? recall that the volume of a right circular cone with height h and radius of the base r is given by $v = \frac{1}{3}\pi r^2 h$

Explanation:

Step1: Relate radius and height

Given base diameter \(d = h\), so radius \(r=\frac{d}{2}=\frac{h}{2}\).

Step2: Substitute r into volume formula

Volume of cone \(V = \frac{1}{3}\pi r^{2}h\). Substitute \(r=\frac{h}{2}\), we get \(V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{1}{12}\pi h^{3}\).

Step3: Differentiate V with respect to time t

Differentiate both sides with respect to \(t\): \(\frac{dV}{dt}=\frac{1}{12}\pi\times3h^{2}\frac{dh}{dt}=\frac{\pi h^{2}}{4}\frac{dh}{dt}\).

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

We know \(\frac{dV}{dt} = 30\) cubic feet per minute and \(h = 11\) feet. Substitute into the equation: \(30=\frac{\pi(11)^{2}}{4}\frac{dh}{dt}\). Then \(\frac{dh}{dt}=\frac{30\times4}{\pi\times121}=\frac{120}{121\pi}\approx\frac{120}{380.13}\approx0.316\) (but we keep it in exact form for now). Wait, let's recalculate: \(30=\frac{\pi h^{2}}{4}\frac{dh}{dt}\), so \(\frac{dh}{dt}=\frac{4\times30}{\pi h^{2}}=\frac{120}{\pi h^{2}}\). When \(h = 11\), \(\frac{dh}{dt}=\frac{120}{\pi\times121}=\frac{120}{121\pi}\approx\frac{120}{380.1327}\approx0.3157\) ft/min. Wait, maybe I made a mistake in substitution. Wait, original volume after substitution: \(r = h/2\), so \(V=\frac{1}{3}\pi (h/2)^2 h=\frac{1}{3}\pi \frac{h^2}{4}h=\frac{\pi h^3}{12}\). Then derivative: \(dV/dt=\frac{\pi}{12}\times3h^2 dh/dt=\frac{\pi h^2}{4} dh/dt\). So solving for \(dh/dt\): \(dh/dt=\frac{4 dV/dt}{\pi h^2}\). Plugging \(dV/dt = 30\), \(h = 11\): \(dh/dt=\frac{4\times30}{\pi\times11^2}=\frac{120}{121\pi}\approx\frac{120}{380.13}\approx0.3157\) ft/min. Let's compute it as \(\frac{120}{121\pi}\) or approximately \(0.316\) ft/min. Wait, but let's check the steps again. Step 1: diameter = height, so radius is half of height, correct. Step 2: volume formula substitution, correct. Step 3: differentiation, correct (power rule: derivative of \(h^3\) is \(3h^2\), times 1/12 π gives 3/12 π h² = π h² /4, correct). Step 4: substitution, correct. So the formula for \(dh/dt\) is \(\frac{4 dV/dt}{\pi h^2}\). So with \(dV/dt = 30\), \(h = 11\), we have \(dh/dt = \frac{120}{121\pi}\) ft/min, which is approximately \(0.316\) ft/min. Wait, but let's calculate it more accurately. \(121\pi\approx121\times3.1416 = 380.1336\). Then \(120\div380.1336\approx0.3157\), so approximately \(0.316\) ft/min. But let's see, maybe I messed up the volume formula. Wait, volume of cone is \(1/3 \pi r^2 h\), correct. Radius is h/2, so \(r^2 = h²/4\), so \(V = 1/3 \pi (h²/4) h = \pi h³/12\), correct. Derivative: \(dV/dt = \pi/12 * 3 h² dh/dt = \pi h² /4 dh/dt\), correct. So solving for \(dh/dt\): \(dh/dt = (4 dV/dt) / (π h²)\). So plugging in \(dV/dt = 30\), \(h = 11\): \(dh/dt = (430)/(π121) = 120/(121π) ≈ 120/380.13 ≈ 0.316\) ft/min. So the answer is \(\frac{120}{121\pi}\) ft/min or approximately \(0.316\) ft/min. Wait, but let's check with exact calculation: \(120\div(121\times3.1415926535) = 120\div380.13271108 = 0.315685\) ft/min, which is approximately \(0.316\) ft/min.

Answer:

\(\boxed{\dfrac{120}{121\pi}}\) (or approximately \(\boxed{0.316}\) ft/min, but the exact form is \(\dfrac{120}{121\pi}\) which is approximately \(0.316\) ft/min)