Calculus help!?

Question

A conical water tank with vertex down has a radius of 10 feet at the top and is 27 feet high.

If water flows into the tank at a rate of 10 cubic feet per minute, how fast is the depth of the water increasing when the water is 17 feet deep?

Answer 1

Take a differential height dh at height h from the vertex. The differential volume is dV = π*r^2*dh, where r is the radius of the tank at level h. r@27ft is 10ft, r at 0ft is 0ft, so r = (10/27)*h. Thus

dV = π*2.7^2*h^2*dh. Divide both sides by dt to get

dV/dt = π*2.7^2*h^2*dh/dt.

You are given dV/dt, that is the volumetric flow rate into the tank = 10cfm dh/dt is the rate of depth increase:

dh/dt = dV/dt / (π*2.7^2*h^2)

Answer 2

The radius 27ft from the vertex is 10 ft. The radius at x ft from the vertex will be 10x/27.

The volume of water at this (The water will be x ft deep):
V = (1/3)pi * radius^2 * height
>>
V = (1/3)pi * (10x/27)^2 * x
=(1/3)*(10/27)^2*x^3

The rate of change of volume:
dV/dt = (1/3)*(10/27)^2*3x^2 *(dx/dt)
dV/dt is given = 10 cft/min, therefore

dx/dt = 10 * (27/10)^2/(x^2)

at x = 17
dx/dt = 10 * (27/10)^2/(17^2) ~ 0.252 ft/min

Answer 3

get the radius of the cone at 17 feet and find the are of base. Then depth of the water will increase with every minute by dividing the area of base by 10.