Question

the base of a cone-shaped tank is a circle of radius 5 feet, and vertex of the cone is 12 feet above the base. the tank is filled to a depth of 7feet, and water is flowing out of the tank at the rate of 3 cubic feet per minute. how would I find the rate of change of the depth of water in the tank?

Answer 1

Let:
H be the height of the tank,
R be the radius of the base,
v be the volume of water in the tank at time t,
h be the depth of water at time t,
r be the radius of the tank at height h.

From similar triangles:
r = R(H - h) / H

The volume of a slice of the cone of thickness dh at height h is:
dv = pi r^2
= pi R^2(H - h)^2 dh / H^2

dv / dh = pi R^2 (H - h)^2 / H^2

dh / dt = { H^2 / [ pi R^2 (H - h)^2 ] } (dv / dt)

When h = 7ft and dv / dt = - 3 ft^3 / min:
dh / dt = - [ 3 * 144 / (625 pi) ] ft / min
= - 0.220 ft / min

Answer 2

I'm sorry I meant −0.11225 ft / min

Answer 3

thank you

Answer 4

how would one draw this out?