A conical water tank with vertex down has a radius of 11 feet at the top and is 21 feet high. If water flows into the tank at a rate of 20ft^3/min,

Question

A conical water tank with vertex down has a radius of 11 feet at the top and is 21 feet high. If water flows into the tank at a rate of 20ft^3/min, how fast is the depth of the water increasing when the water is 17 feet deep?

I've tried so many different methods, but just can't seem to get the right answer. I'm using V=1/3πr^2h but I don't know where I'm going wrong.

Answer 1

h = height of water surface above the vertex

r = radius of water surface at height h

By similar triangles: r / h = 11 / 21 -> r = (11/21) h

The volume of the water with surface at height h above the vertex:

V = (1/3) π r² h = (1/3) π (11/21)² h³ (I substituted r in terms of h)

Take a derivative with respect to time:

dV/dt = (1/3) π (11/21)² 3h² (dh/dt)

You are given dV/dt = 20 ft³/min, h = 17 ft, solve for dh/dt (it will be in units ft/min).