posted by Helga .
A container in the form of a right circular cone of height 16 cm and base radius 4 cm is held vertex downward and filled with liquid. If the liquid leaks out from the vertex at a rate of 4 cm^3/s, find the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out.
When the water is x cm deep, and the surface is a circle of radius r
r/x = 4/16
r = x/4
v = pi/3 r^2 x = pi/3 x^2/16 x = pi/48 x^3
Now, what is x when half the liquid is gone?
pi/3 r^2 x = pi/3 * 16 * 16 / 2
x^3/4 = 128
x^3 = 2^11 = 2^9 * 4
x = 8 cbrt(4)
dv/dt = pi/16 x^2 dx/dt
4 = pi/16 * 64 cbrt(16) dx/dt
dx/dt = 64/pi / (128cbrt(2)) = .126 cm/s
Better check the details...