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Calculus/Related Rates

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Expert: - 7/9/2010


Question
Water is leaking out of an inverted conical tank at a rate of 12500 cm3/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.

I just can't seem to figure out how to solve this problem. Any help at all would be great!
Thanks!

Answer
The tank has a 6m diameter at the top that is 4m.
The wates is rising at 20 cm/min.
The water is leaking out at 12,500 cm³/min.

The currrent height is 2m, so the which is half of the height.
This means the diameter at that point is one half of the 6m, so it is 3m.
This means the surface area is πr², so it is 9π cm².
If that water is leaking out at 12,500 cm³/min, then the loss of height would be 12,500/(9π).

The water level is rising at 20cm/min, so total change is 20 + 12,500/(9π) =
(180π+12,500)/(9π) cm/min.

Multiplying this by the surface area would give us the incoming rate.
Note that the surface area has been put in the denominator of (180π+12,500)/(9π) cm/min,
so the numerator is the incoming rate.

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