Water is draining from an inverted conical tank that has a height of 12 metres and diameter of 8 metres.

Question

a) As the water is draining out, the depth of the water, h, in the conical tank is decreasing at a rate of

(h – 12) m/minute. At what rate is the volume of the water in the conical tank changing when the depth of

the water, h, is 3 metres? (Need to use volume of a cone)

b) The conical tank is draining into a cylindrical tank that has a base with an area of 400(pi)(m²).At what rate is the depth of the water in the cylinder increasing when the depth of the water in the conical tank, h, has reached 3 metres? (Need to use volume of a cylinder)

Answer 1

First get an equation for the volume. V = (1/3)πr²h

The height is 12m when the radius is 4m, therefore the height of the water is always going to be 3 times the radius of water due to similar triangles.

V = (1/3)π(h/3)²h = (π/27)h³

Next take the derivative of both sides with respect to time

dV/dt = (π/27)(3h²)dh/dt

dV/dt = (π/9)(h²)dh/dt

dV/dt = (π/9)(h²)(h - 12)

Lastly, solve by plugging in h = 3

b)

Use the equation for a volume of a cylinder with area of base = 400π

V = 400πh

Take the derivative of both sides with respect to time:

dV/dt = 400π(dh/dt)

The rate at which the water is flowing into the cylinder is going to be negative of the rate that the water was flowing out of the cone. So plug that into dV/dt and solve for dh/dt.