Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 16cm and a height of 8cm, at the rate of 4 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 6 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer.

Let h=height of water
Let r=radius of water surface
r/h=16/8 =2, so r=2h.
The volume of water is:
v=(1/3)×π×r²×h
=(1/3)×π×(2h)²×h
=4/3πh³
dv/dh=4πh^2
By chain rule:
dv/dt=dv/dh×dh/dt
but
dv/dt=4 
thus:
4=(4πh)×dh/dt
dh/dt=4/(4πh²)
when h=6cm we have:
dh/dt=4/(4π6²)
=0.00884 cm³/min