Problem 49 - 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. $$ \vec{v}=\hat{\imath}-4 \hat{\jmath} $$

Step-by-Step Solution

Answer

The vector's magnitude is approximately 4.12, and the angle θ is approximately 284.04°.

Step 1: Identify the given vector components

The vector \( \vec{v} \) is given as \( \vec{v} = \hat{\imath} - 4 \hat{\jmath} \). This means the vector components are \( x = 1 \) and \( y = -4 \).

Step 2: Find the magnitude of the vector

The magnitude \( \|\vec{v}\| \) is calculated using the formula for the magnitude of a vector: \[ \|\vec{v}\| = \sqrt{x^2 + y^2} \]Substitute \( x = 1 \) and \( y = -4 \):\[ \|\vec{v}\| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12 \]

Step 3: Calculate the angle using trigonometric functions

The angle \( \theta \) with respect to the positive x-axis can be found using the arctangent function:\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-4}{1}\right) \]Calculating this gives:\[ \theta \approx \tan^{-1}(-4) \approx -75.96^{\circ} \]Since angles are usually measured positive counterclockwise from the positive x-axis, but the result is negative, we adjust by adding 360°:\[ \theta = -75.96^{\circ} + 360^{\circ} \approx 284.04^{\circ} \]

Step 4: Verify the representation in the unit vector form

Verify that \( \vec{v} = \|\vec{v}\|\langle\cos(\theta), \sin(\theta)\rangle \):\[ \cos(284.04^{\circ}) \approx 0.24 \ , \ \sin(284.04^{\circ}) \approx -0.97 \]Thus, the unit vector in terms of \( \theta \) and \( \|\vec{v}\| \) is:\[ 4.12\langle0.24, -0.97\rangle \approx \langle 1, -4 \rangle \] Which is consistent with the original vector \( \hat{\imath} - 4 \hat{\jmath} \).

Key Concepts

Vector Components

When dealing with vectors, understanding their individual components is crucial. A vector can be thought of as an arrow pointing from one point to another in space. This arrow is made up of components along the coordinate axes. For a two-dimensional vector, these components are typically along the x-axis and y-axis.

Let's look at the vector \( \vec{v} = \hat{\imath} - 4 \hat{\jmath} \):

• \( x \)-component: 1 (represented by \( \hat{\imath} \))
• \( y \)-component: -4 (represented by \( -4 \hat{\jmath} \))

By knowing the components, we can assess the vector's direction and length. The components essentially provide a detailed view of where the vector lies in its respective space.

These fundamental components allow us to perform further calculations, such as finding the vector's magnitude and direction.

Trigonometric Functions

Trigonometric functions are essential tools in understanding vectors. These functions—such as sine, cosine, and tangent—help us relate the angles and sides of a triangle to the direction and components of vectors.

For vectors like \( \vec{v} = \hat{\imath} - 4 \hat{\jmath} \), trigonometric functions allow us to express the vector's direction in terms of angles.

• **Cosine** measures the adjacent side over the hypotenuse in a right triangle.
• **Sine** measures the opposite side over the hypotenuse.
• **Tangent** gives the ratio between the opposite and adjacent sides (\(\tan(\theta) = \frac{y}{x}\)).

These relationships are critical for converting a vector into its angular form, allowing us to find specific orientation details relative to an axis.

Angle Calculation

Calculating the angle of a vector with respect to the positive x-axis involves trigonometric functions, specifically tangent.

For the vector \( \vec{v} = \hat{\imath} - 4 \hat{\jmath} \), you find the angle \( \theta \) using the formula:\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]Here, \( x = 1 \) and \( y = -4 \), so the calculation becomes:\[\theta = \tan^{-1}\left(\frac{-4}{1}\right)\]This results in an angle approximately \( -75.96^{\circ} \). However, because standard angles are measured from the positive x-axis and counterclockwise, we adjust by adding \( 360^{\circ} \), resulting in \( 284.04^{\circ} \).

This adjusted angle helps us understand the vector's true direction in a coordinate plane.

Unit Vector

A unit vector is a vector that has a magnitude of 1, and it is often used to indicate direction. For any vector \( \vec{v} \), its unit vector is found by dividing each component by the magnitude of \( \vec{v} \).

Let's take our previous vector \( \vec{v} = \hat{\imath} - 4 \hat{\jmath} \) with magnitude \( \|\vec{v}\| \approx 4.12 \). To express \( \vec{v} \) as \( \|\vec{v}\|\langle\cos(\theta), \sin(\theta)\rangle \), we use the angle calculated:

• The cosine of the angle, representing the x-component's direction through \( \theta = 284.04^{\circ} \).
  \( \cos(284.04^{\circ}) \approx 0.24 \)
• The sine of the angle, representing the y-component's direction:
  \( \sin(284.04^{\circ}) \approx -0.97 \)

This allows us to write:\[4.12\langle0.24, -0.97\rangle \approx \langle1, -4\rangle\]This confirms the original vector matches its intended direction and magnitude through its unit vector representation.