Math CentralQuandaries & Queries


Question from Katherine, a student:

sand is falling off a conveyer onto a pile at the rate of 1.5 cubic feet per minute. The diameter of the base is approximately twice the altitude. At what rate is the height of the pile changing when it is 10 feet high?

Hi Katherine,

The pile of sand will be approximately in the shape of a circular cone and the volume of a cone is given by

V = 1/3 π r2 h cubic feet

where r is the radius of the base and h is the height, both measured in feet. You are told that the diameter of the base is twice the altitude or height and the diameter is twice the radius and hence 2r = 2h or

r = h.

Substituting r = h into the volume expression you get

V = 1/3 π h2 h = 1/3 π h3.

Both the volume and height are changing so they are both functions of time, t, measured in minutes so I should really write V(t) and h(t) to emphasize the fact that they are functions of time. Hence

V(t) = 1/3 π h(t)3.

You are also told that the "sand is falling off a conveyer onto a pile at the rate of 1.5 cubic feet per minute" so

V'(t) = 1.5 cubic feet per minute.

Use implicit differentiation to differentiate both sides of V(t) = 1/3 π h(t)3 with respect to t. This will result in an equation with V'(t) on the left and an expression involving h(t) and h'(t) on the right. Substitute h(t) = 10 feet and solve for h'(t).


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