A conical tank (with vertex down) is 10 feet acros the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Let h be the depth.

Then the base radius (at the top) at depth h is 5*h/12. So volume of water is
V= 1/3 PI (5h/12)^2 h
take the derivative, set it equal to 10, solve for dh/dt when h=8

To find the rate of change of the depth of water, we need to find the derivative of the depth with respect to time.

Let's denote the depth of the water as h and the rate of change of the depth as dh/dt.

The volume V of a conical tank can be calculated using the formula: V = (1/3) * π * r^2 * h, where r is the radius at any given depth h.

In this case, the radius at any given depth h can be calculated using similar triangles. Since the top of the tank is 10 feet across, the radius at the top is 5 feet. Using the proportionality of similar triangles, we can set up the relationship:

r / h = 5 / 12.

Simplifying this equation, we can express r in terms of h:
r = (5/12) * h.

Substituting this expression for r into the volume formula, we get:
V = (1/3) * π * ((5/12) * h)^2 * h.

Now, we can find the rate of change of the volume with respect to time, dV/dt, by differentiating the volume formula with respect to t and applying the chain rule:

dV/dt = (1/3) * π * (2 * (5/12) * h) * ((5/12)^2 * h^2) * (dh/dt).

Given that dV/dt is 10 cubic feet per minute, we can rearrange the equation to solve for dh/dt:

dh/dt = (3 * dV/dt) / (π * (2 * (5/12) * h) * ((5/12)^2 * h^2)).

We are interested in finding the rate of change of the depth when the water is 8 feet deep (h = 8).

Plugging the given values into the equation, we get:

dh/dt = (3 * 10) / (π * (2 * (5/12) * 8) * ((5/12)^2 * 8^2)).

Simplifying this expression will give us the solution for the rate of change of the depth of the water when it is 8 feet deep.