Problem 25 - Precalculus Mathematics for Calculus

Question

Find the component of \(\mathbf{u}\) along \(\mathbf{v}\). $$\mathbf{u}=\langle 4,6\rangle, \quad \mathbf{v}=\langle 3,-4\rangle$$

Step-by-Step Solution

Answer

The component of \( \mathbf{u} \) along \( \mathbf{v} \) is \( \langle -1.44, 1.92 \rangle \).

Step 1: Calculate the Dot Product

The first step to find the component of vector \( \mathbf{u} \) along vector \( \mathbf{v} \) is to calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \). The dot product for vectors \( \mathbf{u} = \langle 4, 6 \rangle \) and \( \mathbf{v} = \langle 3, -4 \rangle \) is calculated as follows:\[ \mathbf{u} \cdot \mathbf{v} = (4)(3) + (6)(-4) = 12 - 24 = -12 \]

Step 2: Calculate the Magnitude of \( \mathbf{v} \) Squared

Next, calculate the magnitude of vector \( \mathbf{v} \) squared. This is required to project \( \mathbf{u} \) onto \( \mathbf{v} \). The magnitude squared is calculated as:\[ ||\mathbf{v}||^2 = 3^2 + (-4)^2 = 9 + 16 = 25 \]

Step 3: Compute the Projection of \( \mathbf{u} \) onto \( \mathbf{v} \)

The component of \( \mathbf{u} \) along \( \mathbf{v} \) is the projection of \( \mathbf{u} \) onto \( \mathbf{v} \). This is given by the formula:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \right) \mathbf{v} \]Substitute the values from Steps 1 and 2:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-12}{25} \right) \langle 3, -4 \rangle \]

Step 4: Calculate the Resultant Vector

Multiply \( \left( \frac{-12}{25} \right) \) by vector \( \mathbf{v} = \langle 3, -4 \rangle \):\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-12}{25} \right) \langle 3, -4 \rangle = \langle \frac{-36}{25}, \frac{48}{25} \rangle \]

Step 5: Simplify if Necessary

The simplified form of the vector calculated in Step 4 is \(\langle -1.44, 1.92 \rangle\). Therefore, the component of \( \mathbf{u} \) along \( \mathbf{v} \) is \(\langle -1.44, 1.92 \rangle\).

Key Concepts

Dot Product

The dot product is a fundamental operation used in vector math, particularly in precalculus and linear algebra. It's a way of multiplying two vectors, resulting in a scalar. To compute the dot product of two vectors like \(\mathbf{u} = \langle 4, 6 \rangle\) and \(\mathbf{v} = \langle 3, -4 \rangle\), you multiply corresponding components of the vectors and then add up these products. This can be broken down as:

• \(4 \times 3 = 12\)
• \(6 \times -4 = -24\)

Adding these results gives us: \(12 - 24 = -12\). This value, \(-12\), is a scalar – meaning it has magnitude but no direction – and it describes a form of generalized angle between the vectors.

Magnitude of a Vector

The magnitude of a vector is akin to its length. It's a measure of how long a vector is in the coordinate space. When working with vector \(\mathbf{v} = \langle 3, -4 \rangle\), we calculate its magnitude through the formula:

• \(||\mathbf{v}|| = \sqrt{3^2 + (-4)^2}\)

Breaking it down:

• \(3^2 = 9\)
• \((-4)^2 = 16\)

Summing these gives \(9 + 16 = 25\), and the magnitude squared is simply \(25\). For projection calculations, the magnitude squared is necessary. The magnitude provides crucial insights into the physical size or intensity of the vector, retaining direction effectively.

Component of a Vector

Understanding components of a vector is essential when dealing with projections. A vector can be broken down into parts, showing how much of it lies in the direction of another vector. This is the idea behind the component of \(\mathbf{u}\) along \(\mathbf{v}\). The mathematical process involves the projection formula:

• \(\text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \right) \mathbf{v}\)

Applying this to our vectors:

• \(\text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-12}{25} \right) \langle 3, -4 \rangle\)
• Carrying out the multiplication gives \(\langle \frac{-36}{25}, \frac{48}{25} \rangle\)

Components are useful as they help decompose complex vector relations into simpler, understandable parts. This work is crucial in many fields, like physics and engineering.

Precalculus

Precalculus serves as a critical stepping stone before diving into calculus. It brings together principles from algebra, geometry, and trigonometry, forming a cohesive entry point to advanced mathematics. In the context of vectors, precalculus introduces you to operations such as the dot product and vector components. Understanding these concepts prepares you for deeper topics, such as vector calculus. You learn to handle quantities with both magnitude and direction, crucial for solving real-world problems. These skills form a foundation for studying motion, forces, and fields. Precalculus builds not only computational skills but also critical thinking needed in higher math levels.