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| Math Central | Quandaries & Queries |
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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.
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