   SEARCH HOME Math Central Quandaries & Queries  Question from Kimberly, a student: Water is leaking out of an inverted conical tank at a rate of 12000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. Hi Kimberly.

The two "related rates" are the rate of change of volume and the rate of change of water level. To solve the problem, you are dealing with the volume of a cone, so you should know that V = (1/3) π r2 h.

There are three variables in this, but we can reduce that to two, because the radius and the height are related (think of the similar triangles involved). Thus r = 2 m when h = 6 m. So r/h = 1/3, thus r = h/3. Now we substitute this into the formula for V. V = (1/3) π (h/3)2 h, which simplifies to just V = π h3 / 27.

We are interested in rates, so now we take the derivative of this with respect to time t.

dV / dt = ( π / 27 ) [ d(h3) / dt ] = ( π / 27) (3h2) ( dh / dt ) = ( π h2 ) / 9 ( dh / dt).

The change in volume is dV / dt, which is the amount flowing in (call this I) minus the amount flowing out (12000 cm3/min). Now we really need to pay attention to units. We are mixing meters and centimeters and can easily get confused. Let's stick with centimeters.

So we have this (using centimeters):
I - 12000 = ( π h2 / 9 ) (dh / dt).

h is given as 2 m, which is 200 cm. dh / dt is given as 20 cm/min, so we substitute those in:

I - 12000 = ( π 2002 / 9 ) (20)

And solve for I to finish the problem.

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.