To find how fast the water level is rising in the inverted conical vessel, we can use related rates and the concept of similar triangles.
Here's a step-by-step approach to solving this problem:
1. **Identify the Given Information**:
- Water flow rate = 24 m³/min
- Depth of water = 4 m
- Height of the vessel = 8 m
- Radius at the top = 2 m
2. **Determine the Volume of the Water in the Vessel**:
- The volume of a cone is V = (1/3) * π * r² * h, where r is the radius and h is the height.
- Substituting the values, V = (1/3) * π * (2)² * 4 = 16π m³.
3. **Apply Related Rates**:
- The rate of change of volume with respect to time is dV/dt = 24 m³/min.
- We want to find the rate at which the water level is rising, dh/dt.
4. **Use Similar Triangles**:
- By considering similar triangles, we can set up a proportion between the large and small cones formed by the changing water level.
- The ratio of the radius of the small cone to the large cone is 2:4 or 1:2.
5. **Relate Volumes and Heights**:
- Since the ratio of volumes is the cube of the ratio of heights, we have (1/8) = (1/2)³.
- Now differentiate with respect to time to get (1/8) dV/dt = (1/2)³ dh/dt.
6. **Solve for dh/dt**:
- Substituting dV/dt = 24 m³/min into the equation, we get (1/8) * 24 = (1/8) dh/dt.
- Solving, dh/dt = 3 m/min.
Therefore, the water level is rising at a rate of 3 meters per minute in the inverted conical vessel when the depth of water is 4 meters.
