A conical tank (with vertex down) is 12 feet across the top and 18 feet deep. If water is flowing into the tank at a rate of 18 cubic feet per minute, find the rate of change of the depth of the water when the water is 10 feet deep.

I got 81/200pi ft/min.

using similar triangles, we see that when the water has depth h, the radius of the water's surface is r=h/3, so the volume

v = 1/3 pi r^2 h = 1/27 pi h^3
dv/dt = 1/9 pi h^2 dh/dt
18 = 100pi/9 dh/dt
18*9/(100pi) = dh/dt
(2*81)/(100pi) = dh/dt

If you got 81/(2*100pi), then one of us has moved a factor of 2 to the wrong side of the fraction.

Well, well, well, it looks like we have a conical tank situation! Time to put on my math clown shoes and tackle this problem.

To find the rate of change of the depth of the water, we can use a little thing called related rates. Let's derive an equation that relates the depth of the water and its rate of change.

The volume of a conical tank is given by the formula V = (1/3)πr^2h, where r is the radius of the top of the tank and h is the depth of the water.

We know that the tank is 12 feet across the top, so the radius is 6 feet. Now, let's differentiate this equation with respect to time t:

dV/dt = (1/3)π(2r)(dh/dt)

The left side of the equation represents the rate at which water is flowing into the tank, which is 18 cubic feet per minute. We can plug in the values we know:

18 = (1/3)π(2 * 6)(dh/dt)

Now, we need to solve for dh/dt, the rate at which the depth of the water is changing. Let's simplify:

18 = (1/3)π(12)(dh/dt)
dh/dt = 9/(2π)

And there you have it! The rate of change of the depth of the water is 9/(2π) feet per minute. I hope I didn't drown you in math!

To find the rate of change of the depth of the water, we can use related rates.

Let's denote the radius of the top of the conical tank as R, and the depth of the water as h.

At a given time, the volume V of the water in the tank can be expressed as a function of h. We know that the volume of a cone can be calculated using the formula V = (1/3)πR^2h, where π is pi.

Given that the tank is 12 feet across the top, we can determine the radius R as half of the diameter, which is 12/2 = 6 feet. Therefore, we have R = 6 feet.

Now, we need to differentiate the volume equation with respect to time t to find the rate of change of the volume with respect to time. However, dV/dt is given as 18 cubic feet per minute, so we can substitute that directly into our equation:

18 = (1/3)π(6^2)(dh/dt)

Simplifying, we get:

18 = 36π(dh/dt)

Now, we need to find dh/dt, which represents the rate of change of the depth of the water with respect to time. Rearranging the equation, we have:

dh/dt = 18 / (36π) = 1 / (2π)

Now, we already have dh/dt in terms of the rate of change of the depth of the water. However, we need to find the specific rate of change when the depth of the water is 10 feet.

Substituting h = 10 into the equation, we get:

dh/dt = 1 / (2π)

So, the rate of change of the depth of the water when it is 10 feet deep is 1 / (2π) feet per minute.

Therefore, 1 / (2π) ft/min is the correct rate of change of the depth of the water when it is 10 feet deep.

To find the rate of change of the depth of the water, we need to use related rates and apply the concept of similar triangles.

Let's denote the radius of the top of the conical tank as R and the depth of the water as h. We are given that R = 12 feet, and the water is flowing into the tank at a rate of 18 cubic feet per minute.

First, we need to determine a relationship between R and h. Since the tank is conical with the vertex down, we can use similar triangles to relate R and h.

The ratio of corresponding sides of similar triangles is the same. In this case, if we consider the larger triangle formed by the tank and a smaller similar triangle within it, the ratio of the corresponding sides is R / h.

Therefore, we can write R / h = 12 / 18. Simplifying this, we get R = (2/3)h.

Now, let's differentiate both sides of this equation with respect to time (t) as the water level changes:

dR/dt = (2/3) * dh/dt

The dR/dt represents the rate of change of the radius with respect to time, and dh/dt represents the rate of change of the depth with respect to time. We want to find dh/dt.

Now, we have the rate at which water is flowing into the tank, which is given as 18 cubic feet per minute. The rate of change of the volume of the water in the tank is equal to the rate at which water is flowing in:

dV/dt = 18

The volume of a cone can be given by the formula:

V = (1/3) * π * R^2 * h

Substituting R = (2/3)h, we get:

V = (1/3) * π * [(2/3)h]^2 * h
V = (4/27) * π * h^3

Now let's differentiate this volume equation with respect to time:

dV/dt = (4/27) * π * 3 * h^2 * dh/dt
18 = (4/9) * π * h^2 * dh/dt
dh/dt = (9/4π) * (18 / h^2)
dh/dt = (81/2πh^2) ft/min

Now, let's substitute h = 10 into the equation:

dh/dt = (81/2π * 10^2) ft/min
dh/dt = (81/200π) ft/min

So, the rate of change of the depth of the water when the water is 10 feet deep is 81/200π ft/min.

Therefore, your answer is correct: 81/200π ft/min.