Problem 16 - Thomas Calculus

Question

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 \(\theta\) required in the water main for the turn from north to east.

Step-by-Step Solution

Answer

The angle \( \theta \) is approximately 88.87°.

Step 1: Understanding the Problem

The problem involves finding the angle \( \theta \) between two vectors, one representing a 20\( \% \) grade heading north and another representing a 10\( \% \) grade heading east. This is a vector problem where the grades can be thought of as the slopes of vectors pointing in two different directions.

Step 2: Defining Vectors for Grades

We can define the slope vectors based on the given grades.- The north vector, \( \mathbf{N} \), can be written as \( \mathbf{N} = (0, 1, 0.2) \) because it has no component in the east direction and a vertical component of 0.2 for the 20\( \% \) grade.- The east vector, \( \mathbf{E} \), can be expressed as \( \mathbf{E} = (1, 0, 0.1) \) since it has a 10\( \% \) grade vertically.

Step 3: Calculating Dot Product

To find the angle between the vectors \( \mathbf{N} \) and \( \mathbf{E} \), we first find their dot product. The dot product is calculated as:\[ \mathbf{N} \cdot \mathbf{E} = (0 \times 1) + (1 \times 0) + (0.2 \times 0.1) = 0.02 \]

Step 4: Calculating Magnitudes of Vectors

Next, we calculate the magnitudes of both vectors.- The magnitude of \( \mathbf{N} \) is \( \|\mathbf{N}\| = \sqrt{0^2 + 1^2 + 0.2^2} = \sqrt{1.04} = 1.02 \).- The magnitude of \( \mathbf{E} \) is \( \|\mathbf{E}\| = \sqrt{1^2 + 0^2 + 0.1^2} = \sqrt{1.01} = 1.00498 \).

Step 5: Using Dot Product to Find Angle

The angle \( \theta \) between the vectors can be found using the dot product formula: \[ \cos(\theta) = \frac{\mathbf{N} \cdot \mathbf{E}}{\|\mathbf{N}\| \|\mathbf{E}\|} = \frac{0.02}{1.02 \times 1.00498} \approx 0.0198 \]By taking the inverse cosine, we find \( \theta = \cos^{-1}(0.0198) = 88.87^\circ \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way of multiplying two vectors that results in a single number. In simpler terms, it measures how much one vector extends in the direction of another. It is a key concept in vector calculus and is essential in calculating the angle between vectors. For two vectors \(\mathbf{A} = (a_1, a_2, a_3)\) and \(\mathbf{B} = (b_1, b_2, b_3)\), the dot product \(\mathbf{A} \cdot \mathbf{B}\) is given by:

• \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \)

Understanding the dot product is crucial because it allows us to determine how closely aligned two vectors are. If the dot product is zero, it means that the vectors are perpendicular. In the context of this exercise, the dot product \( \mathbf{N} \cdot \mathbf{E} = 0.02 \) tells us about the slight angle between the north and east direction vectors. With such a setup, determining angles, like \( \theta \), becomes manageable.

Vector Magnitudes

Vector magnitude is a measure of the length or size of a vector. It is a crucial element in understanding vectors in three-dimensional space. To calculate the magnitude of a vector, we use the formula:

• For a vector \(\mathbf{V} = (x, y, z)\), the magnitude \(\|\mathbf{V}\|\) is \( \sqrt{x^2 + y^2 + z^2} \)

Magnitude helps in scaling vectors and comparing their sizes. In the problem above, we identified the magnitude of the north vector \( \mathbf{N} \) as \( 1.02 \) and the east vector \( \mathbf{E} \) as \( 1.00498 \). These magnitudes are crucial for calculating the angle between the two vectors using the dot product. Think of the magnitudes as the lengths of two paths. Knowing them allows us to calculate how these paths diverge from or run parallel to each other.

Inverse Cosine

Inverse cosine, or \(\cos^{-1}\), helps us find the angle between two vectors when their dot product and magnitudes are known. This trigonometric function is highly valuable in vector calculus for translating vector relationships into practical angle measurements.To find an angle \(\theta\) between two vectors \(\mathbf{A}\) and \(\mathbf{B}\), we use:

• \( \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\| \|\mathbf{B}\|} \)
• Then, \( \theta = \cos^{-1}(\cos(\theta)) \)

In this specific problem, we calculated the cosine of the angle between vectors \(\mathbf{N}\) and \(\mathbf{E}\), and found \(\theta = 88.87^\circ\). The inverse cosine helped turn the dot product result and vector magnitudes into this specific angle, giving insight into how the water main should be constructed across the given grades.