Problem 50 - 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. $$ \vec{v}=\langle 123.4,-77.05\rangle $$

Step-by-Step Solution

Answer

The magnitude is approximately 145.46, and the angle \( \theta \) is approximately \( 327.81^\circ \).

Step 1: Calculate the Magnitude of the Vector

To find the magnitude of the vector \( \vec{v} = \langle 123.4, -77.05 \rangle \), use the formula for the magnitude of a vector:\[ \| \vec{v} \| = \sqrt{x^2 + y^2} \]Substitute the components of \( \vec{v} \):\[ \| \vec{v} \| = \sqrt{123.4^2 + (-77.05)^2} \]Calculate each term:\[ 123.4^2 = 15229.16, \quad (-77.05)^2 = 5936.0025 \]Add them:\[ 15229.16 + 5936.0025 = 21165.1625 \]Take the square root:\[ \| \vec{v} \| = \sqrt{21165.1625} \approx 145.46 \]

Step 2: Find the Direction Angle

To find the angle \( \theta \) that satisfies \( \vec{v} = \| \vec{v} \| \langle \cos(\theta), \sin(\theta) \rangle \), use the inverse trigonometric functions. The angle \( \theta \) is given by:\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]Substitute the components of \( \vec{v} \):\[\theta = \tan^{-1}\left(\frac{-77.05}{123.4}\right)\]Calculate the fraction:\[\frac{-77.05}{123.4} \approx -0.6241\]Calculate the angle using a calculator:\[\theta \approx \tan^{-1}(-0.6241) \approx -32.19^\circ\]

Step 3: Adjust the Angle to Correct Quadrant

Since the vector \( \vec{v} \) is in the fourth quadrant (positive \(x\)-component and negative \(y\)-component), adjust the angle \( \theta \) to lie within the range \([0, 360^\circ)\). An angle in the fourth quadrant can be found by adding 360 degrees:\[\theta = -32.19^\circ + 360^\circ = 327.81^\circ\]

Step 4: Verify the Angle

You can verify that \( \theta \) is correct by checking that substituting \( \theta \) back into \( \vec{v} = \| \vec{v} \| \langle \cos(\theta), \sin(\theta) \rangle \) approximately reproduces the original vector components.Use \( \cos(327.81^\circ) \approx 0.8479 \) and \( \sin(327.81^\circ) \approx -0.5302 \). Check:\[\vec{v} \approx 145.46 \langle 0.8479, -0.5302 \rangle \approx \langle 123.4, -77.05 \rangle\]

Key Concepts

Direction Angle

The direction angle of a vector tells us the orientation of the vector in its coordinate plane. It is denoted by \( \theta \), that makes the vector \( \vec{v} \) expressed as \( \|\vec{v}\| \langle\cos(\theta), \sin(\theta)\rangle \). This angle \( \theta \) indicates the direction towards which the vector points, measured from the positive x-axis.

The direction angle can be found using the tangent function which is defined for two-dimensional vectors as \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \), where \( x \) and \( y \) are the components of the vector. Once you calculate this angle, it’s important to determine in which quadrant the vector lies in order to adjust \( \theta \) appropriately:

• First Quadrant: Angle remains \( \theta \)
• Second Quadrant: \( \theta = 180^\circ + \theta \)
• Third Quadrant: \( \theta = 180^\circ + \theta \)
• Fourth Quadrant: \( \theta = 360^\circ + \theta \)

Never forget to make sure \( \theta \) lies within the range \([0, 360^\circ)\) to get the correct measure of direction.

Trigonometric Functions

Trigonometric functions are fundamental tools in mathematics, especially when dealing with vectors. The two primary functions used in finding the direction angle of a vector are cosine (\( \cos \)) and sine (\( \sin \)).

Cosine deals with the horizontal or x-component of a vector:\(\cos(\theta) = \frac{x}{\|\vec{v}\|}\), and sine handles the vertical or y-component:\(\sin(\theta) = \frac{y}{\|\vec{v}\|}\).

These functions help express a vector in terms of its magnitude and direction, as in:\(\|\vec{v}\| \langle \cos(\theta), \sin(\theta) \rangle\).• They originated from geometric circles where they relate angles to ratios of side lengths of right-angled triangles.• Trigonometric tables or calculators often provide these values to a high degree of precision by inputting an angle.• Required for determining specific angles when analyzing vector properties.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \tan^{-1} \), come into play when finding angles given certain trigonometric values. They essentially reverse the roles of input and output compared to their regular counterparts.

Take the tangent function, for instance, which is normally calculated as \( \tan(\theta) = \frac{y}{x} \). If you know \( \frac{y}{x} \), but need \( \theta \), the inverse function, \( \tan^{-1} \), provides that angle: \( \theta = \tan^{-1}\left(\frac{y}{x}\right)\).

Here are crucial aspects to note regarding inverse functions:

• Used to determine the angle when the ratio of the sides is known.
• The calculators commonly offer them and show angles typically in degrees or radians.
• Ensure understanding of their limited value ranges and keep in mind correcting them to the relevant quadrant for angles outside the standard range.

For the complete examination of vector orientations, inverse trigonometric functions are indispensable, guiding us towards correct angle values, crucial for accurate modeling and analysis.