Related rate calculus problem, please help!?

Question

The base of a conical machine part is being milled such that the height is decreasing at the rate of 0.050cm/min. If the part originally had a radius of 1.0 cm and a height of 3.0 cm, how fast is the volume changing when the height is 2.8 cm ?

Answer 1

Step 1: Define radius in terms of height.
r = 1 cm
h = 3 cm
r = 1/3 h

Step 2: Define volume in terms of height, since this assumes it's a right circular cone:
V = 1/3 πr^2h (volume of right circular cone)
V = 1/3 π(1/3h)^2h
V = 1/3 π * 1/9 h^2 * h
V = 1/27 πh^3

Step 3: Take the derivative to find the rate of change of volume.
V = 1/27 πh^3
dV/dh = 1/27 * 3 * π * h^2
dV/dh = πh^2/9

Solve for dV given dh:
dV = πh^2/9 dh
dV = π(2.8)^2/9 * .05
dV = 0.137 cm^3/min (solution!)