Calculus
posted by Liz .
A container is in the shape of an inverted right circular cone has a radius of 2 in at the top and a height of 6 in. At the instant when the water in the conatiner is 5 in deep, the surface level is falling at the rate of .4 in/s. Find the rate at which the water is being drained.

when the water is at height x, by similar triangles, the radius r of the surface is given by
r/x = 2/6, so r = x/3
v = 1/3 pi (x/3)^2 * x
= pi/27 x^3
dv/dt = 2pi/27 x^2 dx/dt
4 = 2pi/27 (25) dx/dt
dx/dt = 54/25pi = .69 in/s 
dv/dt = 3pi/27 x^2 dx/dt
4 = pi/9 (25) dx/dt
dx/dt = 36/25pi = .46 in/s
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