## Cones, volume, and related rates problem help

This is my first year of calculus, so be gentle please. The (downwards facing) cone has a height of 15 cm and a radius of 5 cm, and water is trickling into the cone at a constant rate of 10cm^3 per hour, but a small hole at the bottom of the cone is allowing water to leak out at a rate proportional to the height of the water in the cone. When the volume is π cm^3, the instantaneous rate of change of the total amount of water in the tank is -2cm^3 per hour.

A) Determine the constant of proportionality for the rate at which water is escaping.

B) Determine the rate at which the height is changing when the volume of water in the cone is changing at a rate of 1cm^3 per hour.

C) At some position, the water will cease to decrease, and the amount of water in the cone will stabilize. Find the height of the water in the cone when that happens.

My attempt:

A) This is probably wrong, but -> dV/dt = 10t-(he^kt) -> When V = π, -2t = 10t-(he^kt)

1/3π(h/3)^2 * h = π -> (h^3) /9 = 3 -> h=3, which I then plugged into the previous equation:

-2t = 10t-(3e^kt) -> 12t = 3e^kt -> k=(ln(4)+ln(t))/t, but I cant find k without t!

B) V = 1/3π(r^2) * h -> r/h = 5/15, so r=h/3 -> 1/3π((h^2) /9)*h

Taking the derivative: dV/dt = 1/9(πh^2) * (dh/dt) -> Given dV/dt = 1 -> 9/(πh^2) = dh/dt, but I can't find h without finding k and t!

C) dV/dt = 0 -> 0=10t-he^kt -> 10t = he^kt, but again, I can't find this without finding t and k!