The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 feet above the base. The tank is being filled at a rate of 3 cubic feet per minute. Find the rate of change of the depth of water in the tank when then depth is 7 feet.

Answer 1

when the water is at depth x, the radius of the surface of the water is 5/12 (12-x)

so, the volume of the air space is 1/3 pi r^2 h
= 1/3 pi (5/12 (12-x))^2 (12-x)
= 25pi/432 (12-x)^3

the volume of water is thus the tank volume less the air space:

v = pi/3 * 25^2 * 12 - 25pi/432 (12-x)^3

= 100 pi - 25pi/432 (12-x)^3

dv/dt = 25/144 pi (12-x)^2 dx/dt
3 = 25/144pi * 25 dx/dt
dx/dt = 432/(625pi) = 0.22 ft/min