Problem 3 - University Calculus: Early Transcendentals

Question

Let \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle-2,5\rangle .\) Find the (a) component form and (b) magnitude (length) of the vector. $$\mathbf{u}+\mathbf{v}$$

Step-by-Step Solution

Answer

\( \mathbf{u} + \mathbf{v} = \langle 1, 3 \rangle \), \| \mathbf{u} + \mathbf{v} \| = \sqrt{10} \).

Step 1: Understand Vector Addition

Vector addition involves adding corresponding components of the vectors. If \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), then \( \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle \).

Step 2: Add the Components of \( \mathbf{u} \) and \( \mathbf{v} \)

Given \( \mathbf{u} = \langle 3, -2 \rangle \) and \( \mathbf{v} = \langle -2, 5 \rangle \), compute \( \mathbf{u} + \mathbf{v} \) by adding the components: \( \langle 3 + (-2), -2 + 5 \rangle \).

Step 3: Calculate Component Form

Perform the arithmetic to get the component form: \( \langle 3 + (-2), -2 + 5 \rangle = \langle 1, 3 \rangle \).

Step 4: Understand Magnitude Calculation

The magnitude (length) of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is given by \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).

Step 5: Calculate the Magnitude of \( \mathbf{u} + \mathbf{v} \)

Substitute the components of \( \langle 1, 3 \rangle \) into the magnitude formula: \( \| \langle 1, 3 \rangle \| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \).

Key Concepts

Component Form

Understanding the component form of a vector is essential in vector arithmetic. It involves expressing a vector in terms of its horizontal and vertical components. For a vector defined as \( \mathbf{a} = \langle a_1, a_2 \rangle \), the components \( a_1 \) and \( a_2 \) are the respective values along the x-axis and y-axis.This notation helps us quickly perform operations like vector addition and subtraction. In the exercise, we began with vectors \( \mathbf{u} = \langle 3, -2 \rangle \) and \( \mathbf{v} = \langle -2, 5 \rangle \). By adding corresponding components from these vectors, we find the resulting vector's component form:

• The x-component is \( 3 + (-2) = 1 \)
• The y-component is \( -2 + 5 = 3 \)

Thus, the component form of \( \mathbf{u} + \mathbf{v} \) is \( \langle 1, 3 \rangle \).This method allows us to visualize and calculate vectors easily, simplifying further calculations.

Magnitude of a Vector

Finding the magnitude of a vector is about measuring its length or size. Once a vector is expressed in component form, calculating its magnitude becomes straightforward using the Pythagorean theorem.Given a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), its magnitude is calculated using the formula: \[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \]For the vector \( \mathbf{u} + \mathbf{v} = \langle 1, 3 \rangle \),we calculate its magnitude as follows:

• Square the x-component: \( 1^2 = 1 \)
• Square the y-component: \( 3^2 = 9 \)
• Add them together: \( 1 + 9 = 10 \)
• Take the square root: \( \sqrt{10} \)

The magnitude of \( \mathbf{u} + \mathbf{v} \) is therefore \( \sqrt{10} \). Understanding the magnitude is crucial as it gives us the size of the vector independent of its direction.

Vector Arithmetic

Vector arithmetic involves basic operations performed on vectors, such as addition, subtraction, and scalar multiplication. In vector addition, like the problem in our exercise, we add respective components of two vectors to form a new vector.This operation is straightforward:

• If \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), their sum \( \mathbf{a} + \mathbf{b} \) is \( \langle a_1 + b_1, a_2 + b_2 \rangle \).

Another key aspect is scalar multiplication. It differs from addition because it involves multiplying a vector by a scalar (a single number), which stretches or compresses the vector.For instance:

• Suppose a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), multiplying by a scalar \( k \) results in \( \mathbf{a}' = \langle ka_1, ka_2 \rangle \).

Understanding these vector operations is crucial for deeper engagement with physics and engineering problems.They provide tools to manipulate and use vectors in practical scenarios, making them a foundation in vector calculus.