Problem 48 - College Trigonometry

Question

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\hat{\imath}+\hat{\jmath} $$

Step-by-Step Solution

Answer

The magnitude is approximately 1.41, and the angle is \(45^\circ\).

Step 1: Identify Components

The vector given is \( \vec{v} = \hat{\imath} + \hat{\jmath} \). This means the component values are \( x = 1 \) and \( y = 1 \).

Step 2: Calculate Magnitude

The magnitude \( \|\vec{v}\| \) is calculated using the formula: \[ \|\vec{v}\| = \sqrt{x^2 + y^2} \]. Substituting the component values, we have:\[ \|\vec{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.41 \]

Step 3: Determine the Angle

To find the angle \( \theta \), use the formula \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). Since both components are equal, you get:\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = 45^\circ \]

Step 4: Verify Components Using Trigonometry

Verify that \( \vec{v} = \|\vec{v}\|\langle\cos(\theta), \sin(\theta)\rangle \). With \( \|\vec{v}\| = \sqrt{2} \approx 1.41 \) and \( \theta = 45^\circ \), we compute:\[ \langle \cos(45^\circ), \sin(45^\circ) \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \]\[ \vec{v} = \sqrt{2} \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle = \langle 1, 1 \rangle \] which matches the original vector.

Key Concepts

Vector Components

Understanding vector components is essential for breaking down any vector into its basic elements along the axes of a coordinate system. For a 2D vector, these axes are typically the x-axis and y-axis. The vector \( \vec{v} = \hat{\imath} + \hat{\jmath} \) has components \( x = 1 \) and \( y = 1 \). Thus, the vector can be considered as the sum of its horizontal and vertical movements.
To find these components, always look at the coefficients of the unit vectors \( \hat{\imath} \) and \( \hat{\jmath} \). These coefficients tell you how much the vector steps forward in each direction.
Here's a summary of what you need to do:

• Identify each component in terms of \( \hat{\imath} \) and \( \hat{\jmath} \).
• Write the vector in terms of its components like \( \langle x, y \rangle \).
• Use these components to perform calculations related to the vector.

Knowing the components is the first step to solve many vector-related problems.

Angle Determination

Once you know the components of a vector, finding the angle is the next step. To find the angle \( \theta \) that the vector makes with the positive x-axis, you'll use the inverse tangent function, often denoted as \( \tan^{-1} \) or arctan.
The formula \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \) leverages the ratio of the components. For equal components, such as \( y = 1 \) and \( x = 1 \), the ratio is 1. Applying the inverse tangent function gives \( \theta = 45^\circ \), indicating the vector points diagonally at a 45-degree angle relative to the positive x-axis.

• Use the ratios of components to understand the direction and orientation.
• Remember to convert your angle to degrees or radians as needed by your context.

Angle determination is crucial in describing the vector's direction completely.

Trigonometric Functions

Trigonometric functions are powerful tools for understanding vectors in terms of angles and sides of a right triangle. For our vector example, since \( \theta = 45^\circ \), we use trigonometric identities to define its direction precisely.
The key trigonometric functions involved are \( \cos(\theta) \) and \( \sin(\theta) \). At \( 45^\circ \), we know:

• \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)
• \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)

Using these values, we confirm the components of the unit vector that align with the vector's direction:
\( \langle \cos(\theta), \sin(\theta) \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \). Finally, multiplying these by the magnitude \( \|vec{v}\| = \sqrt{2} \) results in the original vector \( \langle 1, 1 \rangle \).
This showcases how trigonometry bridges vector components and directional angles.