A container in the shape of an inverted cone has radius 6 ft and height 12 ft. It is being drained at 2〖ft〗^3/min. Find the rate of change of the height of the liquid in the cone when the height is 3 feet. The ratio of the radius to the height remains constant.

Answer 1

when the contents have depth y, the radius of the surface of the liquid is y/2

So,

v = pi/3 r^2 y = pi/3 (y/2)^2 y = pi/12 y^3

dv/dt = pi/4 r^2 dy/dt

Now just plug in your numbers.