Problem 64 - Trigonometry

Question

Approximate the magnitude of each vector and the angle \(\theta, 0^{\circ} \leq \theta<360^{\circ}\), that the vector makes with the positive \(x\)-axis. Round your answers to the nearest tenth. $$V=-i+4 j$$

Step-by-Step Solution

Answer

Magnitude: approx. 4.1, Angle: approx. 104.0°.

Step 1: Identify the Components

The vector is given as \( \mathbf{V} = -i + 4j \). This means the vector has an \( x \)-component of \( -1 \) (for the \( i \) term) and a \( y \)-component of \( 4 \) (for the \( j \) term).

Step 2: Calculate Magnitude of the Vector

The magnitude of a vector \( \mathbf{V} = ai + bj \) is calculated using the formula \( |\mathbf{V}| = \sqrt{a^2 + b^2} \). Here, \( a = -1 \) and \( b = 4 \). Thus, the magnitude is \[ |\mathbf{V}| = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.1. \]

Step 3: Calculate the Angle with the Positive x-Axis

The angle \( \theta \) that a vector makes with the positive \( x \)-axis is calculated by \( \theta = \text{atan2}(b, a) \) (in degrees). For this vector: \[ \theta = \text{atan2}(4, -1). \]Converting \(-1\) and \(4\) into an angle,\[ \theta = \text{atan2}(4, -1) \approx 104.0^{\circ}. \] This angle is already in the required range \(0^\circ \leq \theta < 360^\circ\).

Key Concepts

Understanding Vector Components

Vectors are a fundamental concept in physics and engineering, representing quantities having both magnitude and direction. To better understand vectors, we break them down into components. Each vector can be expressed as a combination of horizontal (x-component) and vertical (y-component) axes, typically using the unit vectors \( i \) and \( j \) respectively.
In the given exercise, the vector \( \mathbf{V} = -i + 4j \) implies that it has an \( x \)-component of \(-1\) and a \( y \)-component of \(4\). This means the vector is pointing leftwards with a force of 1 unit, and upwards with a force of 4 units.
Understanding vector components is crucial for many applications, such as determining direction and other vector operations. By breaking down a vector into its components, you can analyze and manipulate the vector's behavior more easily. It also lays the foundation for further calculations, such as magnitudes and angles.

Calculating the Angle of a Vector

To find the angle a vector makes with the positive x-axis, we use a function known as the arctangent function, specifically in the form \( \text{atan2}(y, x) \). This takes the y-component and x-component of the vector and returns the angle in degrees (or radians) from the positive x-axis.
In the problem, we calculated the angle \( \theta \) for vector \( \mathbf{V} = -i + 4j \) using \( \theta = \text{atan2}(4, -1) \), resulting in an angle of \( 104.0^{\circ} \).
The \( \text{atan2} \) function is extremely useful because it considers the sign of both components, which helps determine the exact quadrant in which the angle lies, providing an accurate direction of the vector. This is especially important for cases with negative components, giving you a comprehensive angle measurement without manual quadrant adjustments.

Role of Trigonometric Functions in Vectors

Trigonometry is essential to vector analysis, allowing us to connect vector components, magnitudes, and angles. The main functions used in these calculations include sine, cosine, and tangent, with the latter being key for angle calculations.
Tan of an angle in a triangle is defined as the ratio of the opposite side to the adjacent side. In vectors, this translates to the ratio between the y-component and the x-component, giving \( \tan(\theta) = \frac{b}{a} \).
However, using \( \text{atan2}(y, x) \) as we did allows direct calculation of \( \theta \) without manual adjustment because \( \text{atan2} \) handles the signs of \( x \) and \( y \) effectively. This simplifies determining the vector's direction.
The understanding of these trigonometric concepts can significantly enhance your ability to solve vector-related problems, making it easier to visualize and calculate crucial vector properties.