A conical tank (with vertex down) is 16 feet across the top and 22 feet deep. If water is flowing into the tank at a rate of 18 cubic feet per minute, find the rate of change of the depth of the water

Question

A conical tank (with vertex down) is 16 feet across the top and 22 feet deep. If water is flowing into the tank at a rate of 18 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Answer 1

Vol = πr²h/3 use similar triangles to get a relationship between the dimensions of the tank and the current level and radius. This gives h/r = 22/ 8 = 11/4 or r= 4h/11 put this into the volume equation to get V=π16h³/363. The derivative of this is dV/dt= π16h²dh/121dt solve for dh/dt to get 121•18/64•16•π where h= 8 ft was used for the height, doing the math gives dh/dt= .677ft/min.

Answer 2

The volume of a cone is pi*r2*h/3 for this upside down cone with the vertex at the origin, r is a function of h. The slope is 4/11 so r = (4/11)h so, the Volume of the cone is

V = pi*(4h/11)²*(h/3) = (16*pi/363)h³

dV/dt = (16*pi/121)h² dh/dt Substituting the known information, we have

dV/dt = (16*pi/121)(8)² dh/dt = 18 ft³/min Solving for dh/dt = .677 ft/min