Problem 22 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=2 \sqrt{5} ;\) when drawn in standard position \(\vec{v}\) lies in Quadrant I and makes an angle measuring arctan(2) with the positive \(x\) -axis

Step-by-Step Solution

Answer

The component form of the vector is (2, 4).

Step 1: Understand the Problem

We are asked to find the component form of vector \(\vec{v}\) given its magnitude and direction. The vector has a magnitude of \(2\sqrt{5}\) and, when drawn in standard position, it makes an angle with the positive \(x\)-axis described by \(\text{arctan}(2)\). This angle is located in Quadrant I.

Step 2: Determine the Angle

The angle \(\theta\) that \(\vec{v}\) makes with the positive \(x\)-axis can be calculated using the trigonometric function \(\text{tan}(\theta) = 2\). This means \(\theta = \text{arctan}(2)\).

Step 3: Calculate the Components Using Trigonometry

To find the component form \((v_x, v_y)\) of the vector, use the formulas for the components: \(v_x = \|\vec{v}\|\cdot \cos(\theta)\) and \(v_y = \|\vec{v}\|\cdot \sin(\theta)\).

Step 4: Use Trigonometric Identities

Since \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = 2\), we can use the identity \(1 + \tan^2(\theta) = \frac{1}{\cos^2(\theta)}\) to find \(\sin(\theta)\) and \(\cos(\theta)\). Hence, \(1 + 4 = \frac{1}{\cos^2(\theta)}\), giving \(\cos(\theta) = \frac{1}{\sqrt{5}}\) and \(\sin(\theta) = \frac{2}{\sqrt{5}}\) because \(\sin / \cos = 2\).

Step 5: Substitute to Find Components

Now substitute the values of \(\cos(\theta)\) and \(\sin(\theta)\) into the component form formulas: \(v_x = 2\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 2\) and \(v_y = 2\sqrt{5} \cdot \frac{2}{\sqrt{5}} = 4\).

Step 6: State the Final Component Form

Thus, the component form of the vector \(\vec{v}\) is \((2, 4)\).

Key Concepts

Trigonometric Functions

Trigonometric functions are essential tools in mathematics that help us relate the angles of a triangle to the lengths of its sides. They allow us to connect angles and distances, particularly in the context of right triangles. The primary trigonometric functions are sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)). For a right triangle, given an angle \( \theta \), these functions can be defined as follows:

• \( \sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} \)
• \( \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} \)
• \( \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \)

Trigonometric functions are highly useful in vector mathematics. When calculating vector components from magnitude and direction, they allow us to decompose the vector into horizontal and vertical components. In essence, the cosine function helps us find the horizontal component, while the sine function is used for the vertical component. Understanding these functions is key to solving problems involving angles and vectors.

Magnitude and Direction

Magnitude and direction are fundamental aspects of vector representation. A vector, unlike a scalar, is defined not just by its size, but also by where it points. The magnitude of a vector is a measure of its length, representing how much force or speed it carries. Magnitude is denoted by the absolute value, for example, \( \|\vec{v}\| \) for vector \( \vec{v} \).

The direction of a vector is an angle that describes the vector's orientation with respect to a reference axis, usually the positive x-axis. It can be measured in degrees or radians and is an integral part of vector specification. Together, the magnitude and direction provide a complete description of the vector's attributes. To determine the vector's components accurately, one must know both these elements.

In this exercise, the vector has a magnitude of \(2\sqrt{5}\), and its direction is determined by an angle described as \( \text{arctan}(2) \), lying in Quadrant I. These parameters are vital for breaking down the vector into its component form, which breaks down the vector's movement in each axis.

Arctangent Function

The arctangent function, denoted \( \text{arctan} \), is the inverse of the tangent function. It helps us find the angle, \( \theta \), whose tangent is a given value. For example, if \( \tan(\theta) = 2 \), \( \theta = \text{arctan}(2) \). This function is crucial in determining angles when dealing with vectors and vectors' direction angles.

The arctangent function is particularly useful when you know the ratio of the opposite side to the adjacent side of a right triangle but need to find the angle itself. In the context of our problem, knowing that \( \tan(\theta) = 2 \) allows us to pinpoint \( \theta \) using \( \text{arctan}(2) \). Once we have \( \theta \), we can use trigonometric functions to resolve the vector into its horizontal and vertical components accurately.

In general, understanding inverse trigonometric functions is indispensable for applications across physics and engineering, wherever angle determination from spatial relationships is required.

Quadrant Analysis

Quadrant analysis is a critical process in determining the precise location of an angle in the coordinate system. The coordinate plane is divided into four quadrants:

• Quadrant I: Both x and y are positive.
• Quadrant II: x is negative, y is positive.
• Quadrant III: Both x and y are negative.
• Quadrant IV: x is positive, y is negative.

Where the vector lies in these quadrants affects the sign of its components. In our exercise, since the vector lies in Quadrant I, both x and y components are positive. This directly influences the calculation of the vector's components, ensuring consistency with trigonometric values.

Understanding quadrant analysis helps avoid errors when calculating vectors' direction and components. Misplacing an angle in the wrong quadrant could lead to incorrect signs or magnitudes, impairing the accuracy of solutions. Recognizing the proper quadrant allows for correct application of trigonometric identities and signs, ensuring accurate vector representation.