Problem 68 - Precalculus with Limits

Question

Finding the Component Form of a Vector In Exercises \(67-74\) , find the component form of \(v\) given its magnitude and the angle it makes with the positive \(x\) -axis. Sketch y. $$\begin{array}{ll}{\text { Magnitude }} & {\text { Angle }} \\\ {\|\mathbf{v}\|=1} & {\theta=45^{\circ}}\end{array}$$

Step-by-Step Solution

Answer

The component form of vector \( v \) with magnitude 1 and angle \( 45^{\circ} \) with the positive x-axis is \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

Step 1: Convert Given Information

Firstly, note that \( v = \|v\| \cdot (cos(\theta), sin(\theta)) \), where \(\|v\|\) is the magnitude of vector \( v \) and \(\theta\) is the angle it makes with the x-axis. Here, \(\|v\|\) is given as 1 and \(\theta\) is given as \(45^{\circ}\). You have to convert this angle to radians since here, 360 degrees is equivalent to \(2π\) radians. Hence, \(45^{\circ} = \frac{45}{360} \times 2π = \frac{π}{4}\) radians.

Step 2: Input Substitute Values

Substitute the given values of \(\|v\|\) (magnitude) and \(\theta\) (angle in radian) into the component form of the vector \(v\). That is, \(v = 1 \cdot (cos(\frac{π}{4}), sin(\frac{π}{4}))\), which simplifies into \(v = (cos(\frac{π}{4}), sin(\frac{π}{4}))\).

Step 3: Compute Component Values

Compute the values of the cosine and sine of \(π/4\). Both these functions return \(\frac{\sqrt{2}}{2}\) for an angle of \(π/4\) radians. Thus the component form of vector \(v\) is \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

Step 4: Sketch the Vector

To sketch the vector, start at the origin (0,0) and draw a line segment to the point \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\). The resultant line segment represents the vector \(v\). Remember that this is a unit vector that makes a \(45^{\circ}\) angle with the positive x-axis.

Key Concepts

Vector Magnitude

Understanding vector magnitude is crucial when dealing with vectors in mathematics and physics. The magnitude of a vector refers to its length or size. It is analogous to measuring the distance between two points. A vector is often represented with an arrow, and the magnitude is the length of that arrow from its start point to its end point.

In mathematical terms, if a vector \( \mathbf{v} \) is represented as \((x, y)\) in a two-dimensional space, its magnitude \( \|\mathbf{v}\| \) can be calculated using the Pythagorean theorem:

• \( \|\mathbf{v}\| = \sqrt{x^2 + y^2} \)

This formula illustrates how the magnitude is derived from the horizontal and vertical components of the vector.

In our specific example, the magnitude is given as \(1\). This means that regardless of the direction indicated by the angle, the length of the vector remains consistent at one unit. Having a magnitude of \(1\) simplifies calculations and often represents a unit vector, which is a vector of length one used primarily to indicate direction.

Angle with X-Axis

The angle a vector forms with the x-axis reveals its direction in a 2D plane. This angle is often denoted as \( \theta \) and plays a pivotal role in defining the vector's direction.

For the vector description, this angle can be in degrees or radians. In mathematical contexts such as calculus or trigonometry, angles are more commonly converted and used in radians. To convert degrees to radians, use the formula:

• \( \theta_{rad} = \theta_{deg} \times \frac{\pi}{180} \)

In our exercise example, \(45^{\circ}\) is converted to \( \frac{\pi}{4} \) radians. Angles measured from the positive x-axis toward the vector also help identify components of the vector on the coordinate plane.

These angles provide insight into the direction and orientation of the vector in relation to the x-axis. A 45-degree angle suggests that the vector is equally aligned with both the x and y-axes, making its horizontal and vertical components equal, evident from its components being the same \( \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \).

Trigonometric Functions

Trigonometric functions are essential in calculating components of a vector once its magnitude and direction are known. The primary functions used in this context are sine and cosine. These functions relate the angles to the components of a right triangle formed by the vector in the coordinate plane.

The component form of a vector can be expressed using cosine and sine as follows:

• \( v_{x} = \|\mathbf{v}\| \cdot \cos(\theta) \)
• \( v_{y} = \|\mathbf{v}\| \cdot \sin(\theta) \)

In our character vector example, because the magnitude \( \|\mathbf{v}\| \) is \(1\) and the angle \( \theta \) is \( \frac{\pi}{4} \), the components become:

• \( v_{x} = 1 \cdot \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
• \( v_{y} = 1 \cdot \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)

Both the sine and cosine of \( \frac{\pi}{4} \) equal \( \frac{\sqrt{2}}{2} \). Therefore, in this scenario, the vector's x and y components are equal, reflecting the angle's position between the axes. Understanding these functions allows you to break any vector into its relevant parts along the x and y axes.