# Related Rates of a water leak

1. Oct 24, 2004

### catch.yossarian

Hi guys, I'm completely stuck here when doing related rates questions. Here is the question, and following the setup I have so far.
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Water is leaking out of an inverted conical tank
at a rate of 10,00cm^3/min at the same time that water is
being pumped into the tank at a constant rate. The tank has
height 6m and diameter at top is 4m. IF the water level is
rising at a rate of 20cm/min when the height of water is
2m, find teh rate at which water is being pumped into the
tank.

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d = 4m
h = 6m
2m = 20cm/min.
v = (1/3) (pi) r^2 h
dl/dt = 10,000cm^3/min (rate at which water leaves cone)
dp/dt = ? (rate at which water is being pumped into cone)

thats really all I have. I don't know where to go after that. Any help is appreciated.

2. Oct 24, 2004

### HallsofIvy

this belongs under "Calculus" not "Differential Equations".k

The volume of a cone of radius r and height h is (1/3)&pi;r2h. In this case entire tank has r= 3 and h= 4. Whatever the level of the water, the ratio of height to radius must be the same: r= (3/4)h so the volume is given by
V= (1/3)&pi;(9/16)h3= (3/16)&pi;h3 (I've written r in terms of h because we are given the rate of change of the height, h).
Now differentiate with respect to t: dV/dt= (9/16)&pi;h2 (dh/dt).

You are told that h= 2 and dh/dt= 20 cm/min= .2 m/min so dV/dt= (9/16)&pi;(.04)(20)= 0.4 &pi cubic m /min= 400000 cubic cm/min. Since we are also told that "Water is leaking out of an inverted conical tank at a rate of 10,00cm^3/min", in order to account for the water leaking out AND the increase in volume, the water must be coming in at 400010 cubic cm./min.