Water is flowing at the rate of 50m^3/min from a concrete conical reservoir of base radius 45m and height 6m. How fast is is the water level falling when the water is 5m deep?

To solve this problem, we need to use related rates and the formula for the volume of a cone.

The volume of a cone is given by the formula V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height.

Given:
- The water is flowing at a rate of 50 m^3/min.
- The base radius of the cone is 45 m.
- The height of the cone is 6 m.

Let's denote the changing variables:
- V: Volume of the water in the cone.
- r: Radius of the water surface.
- h: Height of the water in the cone.

We are given that the water is flowing at a rate of 50 m^3/min, which means dV/dt = 50 m^3/min.

We need to find dh/dt, the rate at which the water level is falling when the water is 5 m deep.

To find dh/dt, we'll differentiate the volume equation implicitly with respect to time:

dV/dt = (1/3)π(2rh dr/dt + r^2 dh/dt)

Now, we can substitute the given values:
- dV/dt = 50 m^3/min
- r = 45 m
- h = 5 m

We also need to find dr/dt, the rate at which the radius of the water surface is changing. Since the water is flowing out radially from the cone, we can assume that dh/dt = -dr/dt.

Now, we can substitute the values in the derivative equation and solve for dh/dt:

50 = (1/3)π(2 * 45 * 5 * (-dr/dt) + 45^2 dh/dt)

Simplifying the equation:

50 = (1/3)π(-450dr/dt + 45^2 dh/dt)

Rearranging the equation:

45^2 dh/dt = 150π(dr/dt) - 50(3)π(dr/dt)

45^2 dh/dt + 150(π)dh/dt = - 150(π)dr/dt

Factoring dh/dt:

(45^2 + 150π)dh/dt = - 150πdr/dt

Dividing by (45^2 + 150π):

dh/dt = (- 150πdr/dt) / (45^2 + 150π)

We know that dh/dt = - dr/dt, so we can substitute:

- dr/dt = (- 150πdr/dt) / (45^2 + 150π)

Simplifying:

- 1 = - 150π / (45^2 + 150π)

Now, we can solve for dr/dt:

dr/dt = (150π) / (45^2 + 150π)

Let's calculate this:

dr/dt = (150π) / (45^2 + 150π) ≈ 0.0621 m/min

Therefore, the water level is falling at a rate of approximately 0.0621 m/min when the water is 5 m deep.

To find the rate at which the water level is falling, we need to use related rates. We can start by setting up a relationship between the variables in the problem.

Let's assume that the height of the water in the conical reservoir is represented by the variable "h" (in meters) and the radius of the water surface at any given time is represented by the variable "r" (in meters).

Given:
- The base radius of the conical reservoir, r = 45m
- The height of the conical reservoir, H = 6m
- The rate of water flow, dV/dt = 50m³/min (the volume of water flowing out per minute)

We want to determine:
- The rate at which the water level is falling, dh/dt (the change in height over time) when the water level is 5m deep.

To solve the problem, we need to relate the variables h and r using the volume of a cone formula. The volume of a cone (V) is given by the formula:

V = (1/3) * π * r² * h

Next, we need to differentiate both sides of the equation with respect to time (t). This will allow us to solve for dh/dt in terms of the given variables and their rates of change.

dV/dt = (1/3) * π * (2r * dr/dt * h + r² * dh/dt)

Since we want to find dh/dt when h = 5, we only need to find the value of dr/dt at that instant. This can be done using similar triangles.

When the water level is at h = 5, we can use similar triangles to relate the radius r and the height h at that specific instant:

r/h = R/H

Substituting the given values, we have:

45/5 = R/6

R = 54 (the radius of the water surface when the height is 5m)

Now we have enough information to find dr/dt. We can differentiate the equation r/h = R/H with respect to t:

dr/dt * (1/h) - r/(h²) * dh/dt = 0

Substituting the known values:

dr/dt * (1/5) - 54/(5²) * dh/dt = 0

dr/dt - (54/25) * dh/dt = 0

dr/dt = (54/25) * dh/dt

Finally, we can substitute this expression for dr/dt back into the derivative of the volume equation:

50 = (1/3) * π * (2 * 54/25 * dh/dt * 5 + (54/25)² * dh/dt)

Simplifying:

50 = (1/3) * π * (10.8 * dh/dt + 2.33216 * dh/dt)

50 = (5.4 + 2.33216) * π * dh/dt

50 = 7.73216 * π * dh/dt

dh/dt = 50 / (7.73216 * π)

Evaluating this expression, we find:

dh/dt ≈ 2.027 m/min

Therefore, the water level is falling at a rate of approximately 2.027 meters per minute when the water is 5 meters deep.

this is just like the balloon one.

what do you get?