A conical tank has height 3 m and radius 2 m at the base. Water flows in at a rate of 2 m^3/min. How fast is the water level rising when when the level is 1 m and when the level is 1.6 m? Use decimal notation. Round to the nearest ten thousandth.

To solve this problem, we start by defining the variables and relationships involved in the geometry of the cone and the volume of water.

### Given:
- Height of the cone, \( H = 3 \) m
- Radius of the cone, \( R = 2 \) m
- Rate of water flow in, \( \frac{dV}{dt} = 2 \) m³/min

### Volume of the Cone
The volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the water surface at height \( h \).

### Relationship between \( r \) and \( h \)
Since the cone is similar for any level of water, we can set up the ratio:
\[
\frac{r}{R} = \frac{h}{H} \Rightarrow \frac{r}{2} = \frac{h}{3} \Rightarrow r = \frac{2}{3}h
\]

### Substituting \( r \) in the Volume Formula
Now we substitute \( r \) in terms of \( h \) into the volume formula:
\[
V = \frac{1}{3} \pi \left(\frac{2}{3}h\right)^2 h = \frac{1}{3} \pi \cdot \frac{4}{9} h^2 h = \frac{4\pi}{27} h^3
\]

### Differentiate Volume with respect to Time
Next, we differentiate the volume with respect to time \( t \):
\[
\frac{dV}{dt} = \frac{4\pi}{27} \cdot 3h^2 \frac{dh}{dt} = \frac{4\pi}{9} h^2 \frac{dh}{dt}
\]

### Finding \(\frac{dh}{dt}\)
We can now express \( \frac{dh}{dt} \):
\[
\frac{dh}{dt} = \frac{27}{4\pi h^2} \cdot \frac{dV}{dt}
\]
Substituting \( \frac{dV}{dt} = 2 \):
\[
\frac{dh}{dt} = \frac{27}{4\pi h^2} \cdot 2 = \frac{54}{4\pi h^2} = \frac{27}{2\pi h^2}
\]

### Calculating \(\frac{dh}{dt}\) for \( h = 1 \) m and \( h = 1.6 \) m

1. **When \( h = 1 \) m:**
   \[
   \frac{dh}{dt} = \frac{27}{2\pi (1^2)} = \frac{27}{2\pi} \approx \frac{27}{6.2832} \approx 4.2957 \text{ m/min}
   \]

2. **When \( h = 1.6 \) m:**
   \[
   \frac{dh}{dt} = \frac{27}{2\pi (1.6^2)} = \frac{27}{2\pi (2.56)} = \frac{27}{5.12\pi} \approx \frac{27}{16.1132} \approx 1.6742 \text{ m/min}
   \]

### Final Answers
- When \( h = 1 \) m, \( \frac{dh}{dt} \approx 4.2957 \) m/min
- When \( h = 1.6 \) m, \( \frac{dh}{dt} \approx 1.6742 \) m/min

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. How does the flow rate affect the rise in water level?
2. What would happen if the tank had a different shape?
3. How would the calculations change if the height of the tank was greater?
4. Can you derive the volume formula for a different conical tank?
5. What are real-world applications of this type of calculation? 

**Tip:** Always double-check the units to ensure they are consistent when performing calculations.

A conical tank has height 3 m and radius 2 m at the base. Water flows in at a rate of 2 m^3/min. How fast is the water level rising when the level is 1 m and when the level is 1.6 m?

Explore how to calculate the rate at which water rises in a conical tank with a height of 3 m and a radius of 2 m when water flows in at 2 m³/min. Discover the calculations for water levels of 1 m and 1.6 m.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation