Problem 57 - A Graphical Approach to Precalculus with Limits

Question

Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle in \(\left[0,360^{\circ}\right)\). $$\langle 15,-8\rangle$$

Step-by-Step Solution

Answer

Magnitude: 17; Direction angle: 331.9°.

Step 1: Formula for Magnitude

The magnitude of a vector \( \langle a, b \rangle \) is calculated using the formula: \[ \sqrt{a^2 + b^2} \]. Here, \( a = 15 \) and \( b = -8 \).

Step 2: Calculate Magnitude

Substitute values into the magnitude formula: \[ \sqrt{15^2 + (-8)^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \]. The magnitude of the vector is 17.

Step 3: Formula for Direction Angle

The direction angle \( \theta \) of the vector is found using the tangent function: \( \tan \theta = \frac{b}{a} \). Use the inverse tangent to find \( \theta \): \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

Step 4: Calculate the Initial Direction Angle

Using the inverse tangent, calculate \( \theta = \tan^{-1}\left(\frac{-8}{15}\right) \approx \tan^{-1}(-0.5333) \). This gives \( \theta \approx -28.1^{\circ} \).

Step 5: Adjust Direction Angle

Since direction angles are given in the range \([0, 360^{\circ})\), and the initial calculation gives \( \theta \approx -28.1^{\circ} \), add \( 360^{\circ} \) to place it in the correct range: \( -28.1^{\circ} + 360^{\circ} = 331.9^{\circ} \).

Step 6: Conclusion of Calculation

For the vector \( \langle 15, -8 \rangle \), the magnitude is 17 and the direction angle is \( 331.9^{\circ} \).

Key Concepts

Magnitude of a Vector

When working with vectors, one of the first things you'll often need to do is find the magnitude. Think of the magnitude as the length or size of the vector. If you imagine the vector as a straight line, then the magnitude is simply how long that line is.
To calculate the magnitude of a vector given by \( \langle a, b \rangle \), you use the formula:

• \[ \sqrt{a^2 + b^2} \]

In this exercise, we have a vector \( \langle 15, -8 \rangle \). By substituting these values into the formula, you get:

• \[ \sqrt{15^2 + (-8)^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \]

Therefore, the magnitude of the vector \( \langle 15, -8 \rangle \) is 17. It's important to remember that magnitude is always a non-negative number as it represents distance, which isn't directional.

Direction Angle

The direction angle of a vector provides information about its orientation in the plane. It's the angle formed between the vector and the positive x-axis. While it sounds complicated, finding this angle involves straightforward trigonometry.
To find the direction angle \( \theta \) of our vector \( \langle a, b \rangle \), you use the formula:

• \( \tan \theta = \frac{b}{a} \)

This formula tells us that tangent of the angle is the ratio of the y-component to the x-component. Next, we use the inverse tangent function to determine \( \theta \):

• \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

For our vector \( \langle 15, -8 \rangle \), substituting \( a = 15 \) and \( b = -8 \) gives us:

• \( \theta = \tan^{-1}\left(\frac{-8}{15}\right) \approx \tan^{-1}(-0.5333) \)

Initially, this results in approximately \( -28.1^{\circ} \). Since direction angles must lie between \( 0^{\circ} \) and \( 360^{\circ} \), we adjust by adding \( 360^{\circ} \), providing the final direction angle as \( 331.9^{\circ} \). Adjusting for a positive angle is standard when the calculated angle is negative.

Inverse Tangent Function

The inverse tangent function, often written as \( \tan^{-1} \) or \( \arctan \), is a key tool in calculating the direction angle of a vector. This function helps us retrieve the angle from a known tangent value. Essentially, it reverses what the tangent function does.
In trigonometry, if \( \tan \theta = x \), then \( \theta = \tan^{-1}(x) \).
Using the vector \( \langle 15, -8 \rangle \), we calculated \( \tan \theta = \frac{-8}{15} \). To find \( \theta \), we solve \( \theta = \tan^{-1}(-0.5333) \), resulting in an angle of approximately \( -28.1^{\circ} \).
The inverse tangent function is particularly useful in converting the ratio from a vector's components into an angle that shows the vector's direction. Remember, the inverse tangent can provide angles in various ranges. However, we often adjust the angle to ensure it fits into the desired interval of \([0, 360^{\circ})\) when dealing with direction angles in this context.