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 Question from Ray, a student: Water is passing through a conical filter 24 cm deep and 16 cm across the top into a cylindrical container of radius 6 cm. At what rate is the level of water in the cylinder rising when the depth of the water in the filter is 12 cm its level and is falling at the rate of 1 cm/min?

Ray,

You called this problem a related rates problem and you are correct. The fundamental question as I see it is how to you relate what is going on inside the filter to what is going on inside the cylinder. The water is flowing from the filter into the cylinder so the volume of the water in the filter is decreasing at the same rate as the volume of the water in the cylinder is increasing. To say this mathematically I need some variables.

Let t be the time measured in minutes, Vf be the volume of the filter measured in cubic centimeters and Vc be the volume of the cylinder measured in cubic centimeters. The fact I know then is

dVf/dt = - dVc/dt.            (*)

Suppose at some time t the height of the water in the filter is hf, the radius of the surface of the water in the filter is rf and the height of the water in the cylinder id hc, all measured in centimeters. Thus you know that

Vc = π 62 hc

and

Vf = 1/3 π rf2 hf

Use similar triangles to express Vf in terms of hf alone. Differentiate Vf and Vc with respect to t. Relate the expressions using equation (*). Substitute the values you know when hf = 12 cm, nd solve for dhc/dt.

Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.