Calculus
posted by Jay .
A water tank has a shape of an inverted circular cone with base radius 3 m and height of 5 m. If the water is being pumped into the tank at a rate of
2 m^3 /min, find the rate at which the water level is rising when the water is 3 m deep.

let the height of the water be h m
let the radius of the water be r m
by ratios : h/r = 5/3
5r = 3h
r = 3h/5
Volume = (1/3)π r^2 h
= (1/3)π(9h^2/25)h
= (3π/25)h^3
d(Volume)/dt = (9π/25)h^2 dh/dt
2 = (9π/25)(9)dh/dt
dh/dt = 50/(81π) m/s
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