Oil is being poured into an inverted cone at a rate of 28π cubic meters per second. At the very instant, the height of the oil is 4 meters, the volume of the oil is 12π cubic meters. At what rate is the height of the oil changing?

Answer 1

v = 1/3 πr^2 h
Since v=12π when h=4, r = 3
So, at all times, r = 3/4 h

So, v = 1/3 π (3/4 h)^2 h = 3/16 π h^3
dv/dt = 9/16 πh^2 dh/dt

so, when h=4,
28π = 9/16 π * 16 dh/dt
dh/dt = 28/9 m/s