Related Rate Problem?

Question

A paper cup containing water has the shape of a frustum of a right circular cone of altitute 6 inches, and lower and upper radius of 1 inch and 2 inches respectively. If water is leaking gout of the cup at a rate of 3 cubed inches per hour, at what rate is the water level decreasing when its depth is 4 inches?

if you could walk me through this i would really appreciate it.
thanks a bunch

Answer 1

The volume of the water in the cup at any moment is:

V = (1/3) pi [ (Rw)^2 + (1)Rw + (1)^2 ]
= (1/3) pi [ (Rw)^2 + Rw + 1 ] h

Rw = Radius of upper part of water
h = height of water

Rw / h = 2 / 6

6 Rw = 2 h

Rw = 1/3 h

Substitute in to the volume equation

V = (1/3) pi [ (1/9) h^2 + (1/3) h + 1] h
= (1/3) pi [ (1/9) h^3 + (1/3) h^2 + h ]

dV / dt = (1/3) pi [ (1/3) h^2 + (2/3) h + 1] (dh / dt)

Solve for dh /dt and plug in the values h=4 and dV/dt = 3