Water is running into an open conical tank at the rate of 9 cubic feet per minute. The tank is standing inverted, and has a height of 10 feet and a base diameter of 10 feet. At what rate is the exposed surface area of the water changing when the radius is 2 feet?

To find the rate at which the exposed surface area of the water is changing, we can use the concept of related rates. We'll start by finding an equation that relates the variables involved.

Let's denote the radius of the water in the tank as "r" and the height of the water as "h."

We know that the tank is conical with a height of 10 feet and a base diameter of 10 feet. Since the base diameter is 10 feet, the radius of the base is 10/2 = 5 feet. This means that when the radius of the water is 2 feet, the height of the water can be determined using similar triangles:

(r/h) = (5/10)
r = (h/2)

Now, let's express the volume of the water in terms of "h" since we are given the rate of change of volume with respect to time:

V = (1/3) * π * r^2 * h

The rate at which the volume is changing is given as 9 cubic feet per minute, so we can write:

dV/dt = 9

To find the rate at which the exposed surface area of the water is changing, we need to find dA/dt (the derivative of the surface area A with respect to time t).
The surface area A can be expressed in terms of "r" and "h" as follows:

A = π * r * √(r^2 + h^2)

Now, we can differentiate A with respect to t using the chain rule:

dA/dt = (dA/dr) * (dr/dt) + (dA/dh) * (dh/dt)

Next, we need to find expressions for (dA/dr), (dA/dh), (dr/dt), and (dh/dt).

To find (dA/dr), we differentiate A with respect to r, treating h as a constant:

(dA/dr) = π * (√(r^2 + h^2)) + r * (1/2) * (2r / √(r^2 + h^2))

To find (dA/dh), we differentiate A with respect to h, treating r as a constant:

(dA/dh) = π * r * (1/2) * (2h / √(r^2 + h^2))

Now, to find (dr/dt), we go back to the relationship between r and h:
r = h/2, so dr/dt = (1/2) * (dh/dt)

We are given that the rate of change of volume (dV/dt) is 9 cubic feet per minute. This relates to (dh/dt), the rate at which the height of the water is changing. We can substitute the equation for the volume:

dV/dt = (1/3) * π * r^2 * (dh/dt)

Since r is given as 2 feet, we can solve for (dh/dt):

9 = (1/3) * π * 2^2 * (dh/dt)
dh/dt = 9 /[(1/3) * π * 4]

Finally, we can substitute the values we have found into the expression for (dA/dt):

dA/dt = (dA/dr) * (dr/dt) + (dA/dh) * (dh/dt)
dA/dt = [π * (√(r^2 + h^2)) + r * (1/2) * (2r / √(r^2 + h^2))] * [(1/2) * (dh/dt)] + [π * r * (1/2) * (2h / √(r^2 + h^2))] * [9 /((1/3) * π * 4)]

Simplifying this equation will give us the rate at which the exposed surface area of the water is changing when the radius is 2 feet.