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 cm^3/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.

To find the rate at which the depth of the water is changing, we need to use the concept of related rates.

Given:
Radius of the tank (r) = 16 cm
Height of the tank (h) = 8 cm
Rate of water drainage (dV/dt) = 4 cm^3/min
Depth of the water (x) = 6 cm (the variable we want to find)

We need to find the rate at which the depth (dx/dt) is changing when the depth is 6 cm.

Let's start by determining a relationship between the variables involved.

We know that the volume of a cone is given by:
V = (1/3) * π * r^2 * h

We can solve this formula for height (h) in terms of the volume (V) and radius (r):
h = (3V) / (πr^2)

Differentiating both sides with respect to time (t) will give us the related rates equation:

dh/dt = [(dV/dt) * (3/(πr^2))] [by applying the chain rule]

Now, let's substitute the given values into this equation:

dh/dt = (4 * 3) / (π * 16^2)

Simplifying further:

dh/dt = 12 / (π * 256)

Now, we need to find dh/dt when the depth is 6 cm. To do that, we can substitute x = 6 into dh/dt:

dh/dt = 12 / (π * 256) ≈ 0.00467 cm/min

Therefore, the depth of the water is changing at a rate of approximately 0.00467 cm/min when the water in the tank is 6 cm deep.

hint

dh/dt * surface area = 4 cm^3/min