when the water is y ft deep, the radius of the surface is y/3.
v = 1/3 pi r^2 h = 1/2 pi (y/3)^2 * y = pi/18 y^3
so, y = (18v/pi)^(1/3)
dy/dt = 1/3 * (18v/pi)^(-2/3) * 18/pi
= 1/3 * 18/pi * (pi/18*8)^(2/3)
= 4∛(2/(3pi))
Water is draining at a rate of 2 cubic feet per minute from the bottom of a conically shaped storage tank of overall height 6 feet and radius 2 feet . How fast is the height of water in the tank changing when 8 cubic feet of water remain the the tank? Include appropraite units in your answer. (Note: The volume of a cone is given by V=(1/3)(pi)(r^2)(h)) Your answer may be expressed in terms of pi.
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