An inverted cylindrical cone, 52 ft deep and 26 ft across at the top, is being filled with water at a rate of 11 ft3/min. At what rate is the water rising in the tank when the depth of the water is 1 foot?

Answer 1

At a given time of t minutes,
let the height of water be h, let the radius of water level be r
by simple ratios,
r/h = 13/52 = 1/4
so r = h/4

given: dV/dt = 11 ft^3/min
find : dh/dt when h = 1

V = (1/3)π r^2 h
= (1/3) π (h^2/16)h
= (1/48) π h^3
dV/dt = (1/16)π h^2 dh/dt
11 = (1/16)π(1) dh/dt
dh/dt = 176/π ft/min

check my arithmetic