Problem 13 - University Calculus: Early Transcendentals

Question

Find the component form of the vector. The unit vector that makes an angle \(\theta=2 \pi / 3\) with the positive \(x\) -axis.

Step-by-Step Solution

Answer

The vector is \(\langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \rangle\).

Step 1: Understand the Definition of a Unit Vector

A unit vector is a vector with a magnitude (length) of 1. It defines direction but not magnitude. For example, any vector that is expressed as \(\langle x, y \rangle\) and has a length of 1 is a unit vector.

Step 2: Apply the Angle to the Unit Circle

The angle \(\theta = \frac{2\pi}{3}\) radians is measured from the positive x-axis, counter-clockwise around the unit circle. This places the vector in the second quadrant of the coordinate system.

Step 3: Calculate Component Form Using Trigonometry

The x-component of the vector is found using \(\cos(\theta)\), and the y-component is found using \(\sin(\theta)\). Therefore, the vector can be expressed as:\[v = \langle \cos\left(\frac{2\pi}{3}\right), \sin\left(\frac{2\pi}{3}\right) \rangle\]

Step 4: Evaluate the Trigonometric Functions

Using trigonometric identities for common angles, we know that:\[\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\]\[\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\]Therefore, the vector is:\[v = \langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \rangle\]

Step 5: Verify Unit Vector Conditions

Ensure that the vector has a magnitude of 1 by calculating:\[||v|| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\]This confirms it's a unit vector, maintaining a magnitude of 1.

Key Concepts

Vector Components

Every vector in the Cartesian plane is composed of two essential parts: its vector components. These components are usually portrayed as coordinates \( (x, y) \) or \( \langle x, y \rangle \).
They provide a way to convey both the direction and magnitude of a vector.

• The **x-component** of a vector corresponds to the horizontal movement on the plane.
• The **y-component** relates to how much the vector moves vertically.

The length or magnitude of a vector can be calculated using its components with the Pythagorean theorem. Specifically, if you know the components \( x \) and \( y \), you can find the magnitude \(||v||\) as:\[||v|| = \sqrt{x^2 + y^2}\]
When dealing with unit vectors, both components are important for ensuring the vector has a magnitude of 1. Since if one component changes, it affects both the direction and the magnitude.

Trigonometry

Trigonometry is key when finding vector components, especially on the unit circle. It's all about using angles! Every angle on the unit circle can provide direct information about the vector.

• The **cosine function** gives us the x-component of the vector.
• The **sine function** gives us the y-component.

These functions are derived from the definitions of sine and cosine in right triangles. When applied: \[ x = \cos(\theta) \] \[ y = \sin(\theta) \]For the specific problem at hand, given \(\theta = \frac{2\pi}{3}\), you can use these trigonometric functions to determine the components directly. Angles in radian form sometimes connect more easily to these concepts, aiding in quick calculations.

Unit Circle

The unit circle is a tool in mathematics that simplifies the understanding of how angles relate to coordinates. It's a circle with a radius of 1, centered at the origin \((0,0)\). It plays a fundamental role in trigonometry and vector analysis.

• On the unit circle, any given angle \(\theta\) from the positive x-axis corresponds to a point on the circle.
• The x-coordinate of this point is \(\cos(\theta)\).
• The y-coordinate is \(\sin(\theta)\).

For the angle \(\theta = \frac{2\pi}{3}\), since it is in the second quadrant, the x-component will be negative and the y-component will be positive. Hence, the coordinates on the unit circle are \(-\frac{1}{2}\) and \(\frac{\sqrt{3}}{2}\). Utilizing the unit circle not only aids in understanding trigonometric functions but also in deducing vector directions and magnitudes effortlessly.