Calculus problem: Help needed please

Question

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 cm³/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer. 

Answer 1

Volume V of water we evaluate as volume of cone: V = 1/3 πr²h, where r is radius of base of cone, h is hite.

WE have from similyarity: r/h = 4/16, from here h = 4r or r = h/4, then V = 1/3π(h/4)²·h = 1/48πh³,

V' = 1/48π·3h²·h'; V' = 1/16πh²·h'; h' = 16V'/(πh²);

h' = (16·2 cm³/min)/(π·8²cm²) = 1/(2π) cm/min;

rate of grow depth h' = 1/(2π) cm/min