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Rate of rise of grain height in a cylindrical silo

A farmer is filling a cylindrical grain silo with corn. The silo has a fixed radius of 66 feet, and grain is poured in at a constant rate of 3636 cubic feet per minute.

Let hh be the height (in feet) of the grain in the silo and tt be time in minutes. Since the radius r=6r = 6 feet is fixed, the volume of grain is V=πr2h=36πhV = \pi r^2 h = 36\pi h. Given that dVdt=36\dfrac{dV}{dt} = 36 cubic feet per minute, find dhdt\dfrac{dh}{dt}, the rate at which the height of the grain is rising.

Hint

Since the radius doesn't change, treat πr2\pi r^2 as a constant coefficient. Differentiate V=πr2hV = \pi r^2 h with respect to tt, then solve for dhdt\frac{dh}{dt} using the given dVdt\frac{dV}{dt}.

Step-by-step solution

Start with the volume formula for a cylinder, V=πr2hV = \pi r^2 h. Since r=6r = 6 is constant over time, substitute to get V=36πhV = 36\pi h. Differentiate both sides with respect to tt: dVdt=36πdhdt\dfrac{dV}{dt} = 36\pi \dfrac{dh}{dt}.

Substitute the known rate dVdt=36\dfrac{dV}{dt} = 36 cubic feet per minute into the equation: 36=36πdhdt36 = 36\pi \dfrac{dh}{dt}. Solving gives dhdt=1π\dfrac{dh}{dt} = \dfrac{1}{\pi} feet per minute, which is approximately 0.3180.318 ft/min.

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