Problem 14 - Thomas Calculus

Question

Find the component form of the vector. The unit vector that makes an angle \(\theta=-3 \pi / 4\) with the positive \(x\)-axis

Step-by-Step Solution

Answer

The component form is \( \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \).

Step 1: Understand the Unit Vector

A unit vector is a vector with a magnitude of 1 that points in a specific direction. We are tasked with finding the component form of a unit vector that makes an angle \( \theta = -\frac{3\pi}{4} \) with the positive \( x \)-axis.

Step 2: Calculate the Components

The component form of a unit vector in the direction of an angle \( \theta \) is given by \( \langle \cos(\theta), \sin(\theta) \rangle \). Here, \( \theta = -\frac{3\pi}{4} \).

Step 3: Find the Cosine Component

Calculate \( \cos\left(-\frac{3\pi}{4}\right) \). Since cosine is an even function, it means that \( \cos(-x) = \cos(x) \), so \( \cos\left(-\frac{3\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

Step 4: Find the Sine Component

Calculate \( \sin\left(-\frac{3\pi}{4}\right) \). Since sine is an odd function, it means that \( \sin(-x) = -\sin(x) \), so \( \sin\left(-\frac{3\pi}{4}\right) = -\sin\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

Step 5: Combine Components

Combine the calculated cosine and sine components to get the vector: \( \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \).

Key Concepts

Component Form

A vector is often represented in terms of its components along the coordinate axes. This representation is known as the component form of the vector. Understanding how to express vectors in component form is essential, especially when dealing with mathematical operations like vector addition, subtraction, and scalar multiplication.

To find the component form of a vector, you need to express the vector in terms of its individual horizontal and vertical components. For a 2D vector, these components correlate with the

• Horizontal Component (x-component)
• Vertical Component (y-component)

Specifically, for a unit vector that makes an angle \( \theta \) with the positive x-axis, the component form is given by \( \langle \cos(\theta), \sin(\theta) \rangle \). This formula reflects the right triangle relationship in the unit circle, where the magnitude of 1 implies that our x and y components are simply the cosine and sine of the angle, respectively.

Trigonometric Functions

Trigonometric functions like cosine and sine are fundamental in dealing with angles and vectors. When a vector is specified by an angle, it's these trigonometric functions that allow us to break down the vector into its component form.

Cosine (\( \cos \)) of an angle in a unit circle represents the x-coordinate, while Sine (\( \sin \)) represents the y-coordinate.

Using the angle \( \theta = -\frac{3\pi}{4} \) from the example, let's dive into these functions:

• Cosine is an even function: \( \cos(-x) = \cos(x) \); for \( \theta = -\frac{3\pi}{4} \), \( \cos \left( -\frac{3\pi}{4} \right) = \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} \)
• Sine is an odd function: \( \sin(-x) = -\sin(x) \); for \( \theta = -\frac{3\pi}{4} \), \( \sin\left( -\frac{3\pi}{4} \right) = -\sin\left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} \)

With these identities, you can calculate each part of the component form of a unit vector easily, understanding deeply how angles influence the trigonometric outputs.

Angle with x-axis

In trigonometry, angles are often measured with respect to the positive x-axis in a counterclockwise direction. This is crucial in coordinate geometry as it standardizes the way we interpret angles, ensuring consistency in mathematical calculations and graphical representations.

In the exercise provided, the unit vector angle \( \theta = -\frac{3\pi}{4} \) is given in radians. It indicates a rotation in the clockwise direction, as negative angles are typically measured clockwise from the positive x-axis.

To visualize, consider \( -\frac{3\pi}{4} \):

• This angle is equivalent to \(-135^\circ\), placing it directly in the third quadrant of the coordinate plane.
• Unit vectors in this quadrant will have both negative x and y components.

This direction gives insight into where the vector is facing, making it easier to predict and verify the components \( -\frac{\sqrt{2}}{2} \) for both x and y, which were calculated accordingly.