Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 16cm and a height of 8cm, at the rate of 4 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 6 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer.

Question

Moises Hickman

Mathematics

2 August, 04:53

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 16cm and a height of 8cm, at the rate of 4 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 6 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer.

+3

Answers (1)

Know the Answer?

Skunk · Answer 1

Let h=height of water

Let r=radius of water surface

r/h=16/8 = 2, so r=2h.

The volume of water is:

v = (1/3) * π*r²*h

= (1/3) * π * (2h) ²*h

=4/3πh³

dv/dh=4πh^2

By chain rule:

dv/dt=dv/dh*dh/dt

but

dv/dt=4

thus:

4 = (4πh) * dh/dt

dh/dt=4 / (4πh²)

when h=6cm we have:

dh/dt=4 / (4π6²)

=0.00884 cm³/min