Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6m and the diameter at the top is 4m. If the water level is rising at a rate of 20cm/min when the height of the water is 2m, find the rate at which water is being pumped into the tank.

Question

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First, I know we'll be using the formula for volume of a cone, V=(1/3)pi*r^2*h. I know dV/dt = -10000, h=600cm, d=400cm (or 200cm for radius), and dh/dt=20cm/min.
I also know when h=200cm, r=150cm (I did a proportion with the given heights and diameter/radius).
I believe the question is asking for another dV/dt, but I don't know how to go about solving for that with all this given information. Help please?

Thank you!

Answer 1

V=1/2 PI r^2 h
dv/dt= 1/3 PI (r^2 dh/dt + 2hr dr/dt)

now relate dr/dt to dh/dt (proportion)
then solve for dv/dt given dh/dt