Problem 8 - Thomas Calculus

Question

In Exercises \(1-8,\) let \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle- 2,5\rangle .\) Find the (a) component form and \((\mathbf{b})\) magnitude (length) of the vector. $$ -\frac{5}{13} \mathbf{u}+\frac{12}{13} \mathbf{v} $$

Step-by-Step Solution

Answer

Component form: \(\langle -3, \frac{70}{13} \rangle\); Magnitude: \(\frac{\sqrt{6421}}{13}\).

Step 1: Calculate the Component Form

To find the component form of the vector \(-\frac{5}{13} \mathbf{u} + \frac{12}{13} \mathbf{v}\), begin by calculating each component separately. Start with \(\mathbf{u}\):\[-\frac{5}{13} \langle 3, -2 \rangle = \langle -\frac{5}{13} \times 3, -\frac{5}{13} \times (-2) \rangle = \langle -\frac{15}{13}, \frac{10}{13} \rangle.\] Next, calculate for \(\mathbf{v}\):\[\frac{12}{13} \langle -2, 5 \rangle = \langle \frac{12}{13} \times (-2), \frac{12}{13} \times 5 \rangle = \langle -\frac{24}{13}, \frac{60}{13} \rangle.\] Now, sum these two component forms:\[\langle -\frac{15}{13}, \frac{10}{13} \rangle + \langle -\frac{24}{13}, \frac{60}{13} \rangle = \langle -\frac{15}{13} - \frac{24}{13}, \frac{10}{13} + \frac{60}{13} \rangle = \langle -\frac{39}{13}, \frac{70}{13} \rangle.\] Simplify to get the final component form:\[\langle -3, \frac{70}{13} \rangle.\]

Step 2: Calculate the Magnitude of the Vector

Find the magnitude (length) of the vector \(\langle -3, \frac{70}{13} \rangle\). The formula for magnitude is \(\sqrt{x^2 + y^2}\).\[\text{Magnitude} = \sqrt{(-3)^2 + \left(\frac{70}{13}\right)^2}.\] Calculate each term:\[(-3)^2 = 9\] and \[\left(\frac{70}{13}\right)^2 = \frac{4900}{169}.\] Add them together in the square root:\[\text{Magnitude} = \sqrt{9 + \frac{4900}{169}} = \sqrt{\frac{1521}{169} + \frac{4900}{169}} = \sqrt{\frac{6421}{169}}.\] Simplify:\[\text{Magnitude} = \sqrt{\frac{6421}{169}} = \frac{\sqrt{6421}}{13}.\] The magnitude simplifies to \(\frac{\sqrt{6421}}{13}.\)

Key Concepts

Component Form

When dealing with vectors, the **component form** is essential for simplifying calculations and understanding vector operations. In component form, a vector is expressed as a combination of its horizontal and vertical components, usually denoted as \( \langle x, y \rangle \). It shows how far the vector goes in each direction.

For the vector expression \(-\frac{5}{13} \mathbf{u} + \frac{12}{13} \mathbf{v}\), we first take the vectors \(\mathbf{u} = \langle 3, -2 \rangle \) and \(\mathbf{v} = \langle -2, 5 \rangle \) and apply the scalars to each component:

• For \(\mathbf{u}\): \(-\frac{5}{13} \times 3 = -\frac{15}{13}\) and \(-\frac{5}{13} \times (-2) = \frac{10}{13}\)
• For \(\mathbf{v}\): \(\frac{12}{13} \times (-2) = -\frac{24}{13}\) and \(\frac{12}{13} \times 5 = \frac{60}{13}\)

Once individual components are determined, they are added together. This gives us the combined vector as \(\langle -\frac{15}{13} - \frac{24}{13}, \frac{10}{13} + \frac{60}{13} \rangle = \langle -3, \frac{70}{13} \rangle\).

This structured approach makes it easier to handle complex vector equations.

Vector Magnitude

The **magnitude of a vector** is its length or distance from the origin. Mathematically, it is like measuring how far a point is from the center in a coordinate system. This is crucial in understanding the vector's impact or strength.

The formula to find the magnitude of a vector \(\langle x, y \rangle \) is \(\sqrt{x^2 + y^2}\). For our given vector \( \langle -3, \frac{70}{13} \rangle \), this becomes:

• Calculate \((-3)^2 = 9\)
• Calculate \(\left(\frac{70}{13}\right)^2 = \frac{4900}{169}\)
• Sum them: \(9 + \frac{4900}{169}\)
• Find the square root: \(\sqrt{\frac{6421}{169}}\)

After simplifying further, the magnitude comes out as \(\frac{\sqrt{6421}}{13}\).

By calculating the magnitude, you gain insight into how intense a vector's effect is. This can be particularly useful in physics, engineering, and various fields involving spatial analyses.

Vector Addition

**Vector addition** is a fundamental operation that combines two or more vectors to produce a new vector. This operation is vital for many applications, from calculating forces in physics to creating movements in computer graphics.

In our exercise, vectors \(\mathbf{u} = \langle 3, -2 \rangle \) and \(\mathbf{v} = \langle -2, 5 \rangle \) are adjusted by scalars to result in reduced components of \( \langle -\frac{15}{13}, \frac{10}{13} \rangle \) and \( \langle -\frac{24}{13}, \frac{60}{13} \rangle \).

The steps of combining these vectors involve:

• Adding the horizontal components: \(-\frac{15}{13} + (-\frac{24}{13}) = -\frac{39}{13}\)
• Adding the vertical components: \(\frac{10}{13} + \frac{60}{13} = \frac{70}{13}\)
• Resulting in the combined vector: \(\langle -3, \frac{70}{13} \rangle\)

Through vector addition, diverse sets of vector quantities are united into a singular expression, making it easier to understand the overall effect or outcome.

Scalar Multiplication

**Scalar multiplication** is the process of multiplying a vector by a scalar (a simple number). This operation scales the vector either by stretching or compressing it, while also potentially reversing its direction if the scalar is negative.

For the vectors given in the exercise, identifying the impact of scalars is the first step:

• Multiplying \(\mathbf{u} = \langle 3, -2 \rangle \) by \(-\frac{5}{13}\) results in \(\langle -\frac{15}{13}, \frac{10}{13} \rangle\)
• Multiplying \(\mathbf{v} = \langle -2, 5 \rangle \) by \(\frac{12}{13}\) results in \(\langle -\frac{24}{13}, \frac{60}{13} \rangle\)

This multiplication adjusts the size and, at times, the direction of the vector components, which is crucial for properly balancing equations and understanding vector dynamics.

The alteration in length and direction possible through scalar multiplication has profound implications in diverse fields such as physics, where it can represent changes in speed or force magnitude.