Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the rate of 2 cm^3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep?

Answer 1

The trick is that the rate of change of volume is the surface area times the rate of change of depth

dh/dt(pi r^2) = dV/dt = 2 cm^3/min
and r = (4/16)8 = 2 cm

Answer 2

Thank you!

Answer 3

You are welcome.