Water main construction A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the east direction. Determine the angle u required in the water main for the turn from north to east.

To determine the angle \( u \) required in the water main for the turn from north to east, we need to analyze the given information in terms of vectors or slopes, using the concept of gradients and angles.

### Step 1: Represent the grades as vectors
- A 20% grade in the north direction means that for every 100 units of horizontal distance, the elevation changes by 20 units. This can be represented as a vector in the north direction \( \mathbf{N} = (0, 1, 0.2) \), where the first component is for east, the second is for north, and the third is for the vertical (elevation) change.
- A 10% grade in the east direction means that for every 100 units of horizontal distance, the elevation changes by 10 units. This can be represented as a vector in the east direction \( \mathbf{E} = (1, 0, 0.1) \).

### Step 2: Calculate the dot product
The angle between two vectors \( \mathbf{N} \) and \( \mathbf{E} \) is given by the formula:
\[
\cos(u) = \frac{\mathbf{N} \cdot \mathbf{E}}{\|\mathbf{N}\| \|\mathbf{E}\|}
\]
Where:
- \( \mathbf{N} \cdot \mathbf{E} \) is the dot product of the two vectors.
- \( \|\mathbf{N}\| \) and \( \|\mathbf{E}\| \) are the magnitudes (lengths) of the vectors.

### Step 3: Dot product calculation
The dot product of \( \mathbf{N} = (0, 1, 0.2) \) and \( \mathbf{E} = (1, 0, 0.1) \) is:
\[
\mathbf{N} \cdot \mathbf{E} = 0 \times 1 + 1 \times 0 + 0.2 \times 0.1 = 0 + 0 + 0.02 = 0.02
\]

### Step 4: Magnitude calculation
The magnitude of \( \mathbf{N} \) is:
\[
\|\mathbf{N}\| = \sqrt{0^2 + 1^2 + 0.2^2} = \sqrt{1 + 0.04} = \sqrt{1.04} \approx 1.02
\]
The magnitude of \( \mathbf{E} \) is:
\[
\|\mathbf{E}\| = \sqrt{1^2 + 0^2 + 0.1^2} = \sqrt{1 + 0.01} = \sqrt{1.01} \approx 1.005
\]

### Step 5: Calculate \( \cos(u) \)
Now we can calculate \( \cos(u) \):
\[
\cos(u) = \frac{0.02}{1.02 \times 1.005} = \frac{0.02}{1.0251} \approx 0.0195
\]

### Step 6: Determine the angle \( u \)
Now, find the angle \( u \) by taking the inverse cosine:
\[
u = \cos^{-1}(0.0195) \approx 88.88^\circ
\]

Thus, the required angle for the turn from north to east is approximately \( u \approx 88.88^\circ \).

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Would you like more details on any part of the solution or have any questions?

### Related Questions:
1. How is the dot product used to find the angle between two vectors?
2. What is the significance of vector magnitudes in this context?
3. How would the solution change if the grades were different?
4. Could this problem be solved graphically instead of algebraically?
5. How does the angle change if one of the grades is zero?

**Tip**: Always check the units of measurements (percent grade vs degrees) when dealing with real-world slope problems.

A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the east direction. Determine the angle u required in the water main for the turn from north to east.

Determine the angle required in a water main construction project with a 20% grade north and 10% grade east. This solution uses vector algebra, the dot product formula, and trigonometry to calculate the angle, making it suitable for advanced high school or undergraduate students.

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