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.

Answer 1

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

Answer 2

dv/dt = 3pi/27 x^2 dx/dt
-4 = pi/9 (25) dx/dt

dx/dt = -36/25pi = -.46 in/s