A water tank has the shape of a horizontal cylinder with radius 1 and length 2. If water is being
pumped into the tank at a rate of 1/6 m^3 per minute, find the rate at which the water level is rising when the water is 1/2 m deep.

To solve this problem, we need to apply principles of related rates and the geometry of a horizontal cylindrical tank.

### Step 1: Understanding the Problem

- The tank is a horizontal cylinder with a radius \( r = 1 \) meter and a length \( L = 2 \) meters.
- Water is being pumped into the tank at a rate of \( \frac{1}{6} \) cubic meters per minute. This is the rate of change of the volume of water, \( \frac{dV}{dt} = \frac{1}{6} \) m\(^3\)/min.
- We need to find the rate at which the water level \( h \) is rising when the water is \( \frac{1}{2} \) meters deep. This is \( \frac{dh}{dt} \).

### Step 2: Volume of Water in the Tank

When the tank is partially filled, the cross-sectional area of the water forms a segment of a circle. The volume of water \( V \) in the tank can be expressed as:

\[
V = A \cdot L
\]

where \( A \) is the area of the circular segment formed by the water, and \( L = 2 \) is the length of the tank.

### Step 3: Area of the Circular Segment

The area \( A \) of the circular segment can be found using the formula:

\[
A = r^2 \cos^{-1}\left(\frac{r-h}{r}\right) - (r-h)\sqrt{2rh - h^2}
\]

Given the radius \( r = 1 \), this becomes:

\[
A = \cos^{-1}(1-h) - (1-h)\sqrt{2h - h^2}
\]

### Step 4: Expressing the Volume and Relating it to Height

Now, using the volume \( V = A \cdot L \):

\[
V = 2\left[\cos^{-1}(1-h) - (1-h)\sqrt{2h - h^2}\right]
\]

### Step 5: Differentiating with Respect to Time

We need to find \( \frac{dh}{dt} \) when \( h = \frac{1}{2} \) meters. Differentiating both sides with respect to time \( t \):

\[
\frac{dV}{dt} = 2\left[\frac{d}{dh}\left(\cos^{-1}(1-h)\right) - \frac{d}{dh}\left((1-h)\sqrt{2h - h^2}\right)\right] \cdot \frac{dh}{dt}
\]

Given \( \frac{dV}{dt} = \frac{1}{6} \):

\[
\frac{1}{6} = 2\left[\frac{1}{\sqrt{1-(1-h)^2}} - \left(\frac{-1}{2\sqrt{2h-h^2}}(2h-2) + \sqrt{2h-h^2}\right)\right] \cdot \frac{dh}{dt}
\]

### Step 6: Substituting \( h = \frac{1}{2} \)

Substituting \( h = \frac{1}{2} \) into the above expression:

1. Calculate the values for \( \cos^{-1}(1-h) \), \( \sqrt{2h-h^2} \), and their derivatives.

2. Solve for \( \frac{dh}{dt} \).

Using these calculations:

\[
\frac{dh}{dt} = \frac{\frac{1}{6}}{2\left[\frac{1}{\sqrt{1-(1-\frac{1}{2})^2}} - \left(\frac{-1}{2\sqrt{2(\frac{1}{2})-(\frac{1}{2})^2}}(2(\frac{1}{2})-2) + \sqrt{2(\frac{1}{2})-(\frac{1}{2})^2}\right)\right]}
\]

After simplifying, the answer will give us \( \frac{dh}{dt} \).

### Conclusion

Let's simplify the calculation:

\[
\frac{1}{6} = 2\left[\frac{1}{\sqrt{1 - \left(\frac{1}{2}\right)^2}}\right] \cdot \frac{dh}{dt}
\]

\[
\frac{1}{6} = 2\left[\frac{1}{\sqrt{\frac{3}{4}}}\right] \cdot \frac{dh}{dt}
\]

\[
\frac{1}{6} = 2 \left[\frac{2}{\sqrt{3}}\right] \cdot \frac{dh}{dt}
\]

\[
\frac{1}{6} = \frac{4}{\sqrt{3}} \cdot \frac{dh}{dt}
\]

\[
\frac{dh}{dt} = \frac{\sqrt{3}}{24} \approx 0.072
\]

The rate at which the water level is rising when the water is \( \frac{1}{2} \) m deep is approximately \( 0.072 \) m/min.

### Additional Questions

1. What is the volume of the tank when it is full?
2. How does the rate of water level rise change if the pumping rate is doubled?
3. What would be the water level after 10 minutes if the tank started empty?
4. How would the calculations change if the tank's radius were different?
5. Can this method be applied to tanks of other shapes?

### Tip:

Always keep track of units when solving real-world problems involving rates and volumes to ensure consistency and correctness.

Learn how to solve a related rates problem involving a horizontal cylindrical water tank. Calculate the rate at which the water level rises when the water depth is 1/2 meter. Includes step-by-step solution using cylinder geometry and trigonometric derivatives.

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