math
posted by Ravinder on .
Suppose we pump water into an inverted rightcircular cone tank at the rate of 6 cubic feet per minute. The tank has the height 9 ft and radius on the top is 8 ft. What is the rate at which the water level is rising when the water is 3 ft deep? Leave the answer approximate. You may use the formula for the volume of the rightcircular cone of radius r and height h: V=1/3pi*r^2*h

At a time of t minutes, let the
radius of the water level be r
let the height of the water be h
by ratio:
r/h = 8/9
r = 8h/9
V = (1/3)π r^2 h
= (1/3)π(64h^2/81)h
= (64π/243) h^3
dV/dt = (64π/81) h^2 dh/dt
6 = (64π/81) (9) dh/dt
I will let you solve for dh/dt 
thanks so much!