Problem 71 - 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. $$\begin{array}{ll}{\text { Magnitude }} & {\text { Angle }} \\\ {\|\mathbf{v}\|=2 \sqrt{3}} & {\theta=45^{\circ}}\end{array}$$

Step-by-Step Solution

Answer

The vector \(v\) in component form is \([2\sqrt{2}, 2\sqrt{2}]\).

Step 1: Identify the magnitude and angle of the vector

For this problem, it's given that the magnitude of the vector \(v\), denoted as \(\|v\|\), is \(2\sqrt{3}\). The angle \(\theta\) it makes with the positive x-axis is \(45^{\circ}\).

Step 2: Recall the formula for vector components

If the vector \(v\) in 2D space has magnitude \(\|v\|\) and makes an angle \(\theta\) with the positive x-axis, its x-component is \(\|v\|\cos(\theta)\) and its y-component is \(\|v\|\sin(\theta)\). This formula is derived from representating the vector in polar coordinates.

Step 3: Calculate the x-component of the vector

Apply the formula to get the x-component. This involves taking the cosine of \(\theta\) (which is 45° or \(\pi/4\) in radian) and multiplying that by the \(v\) magnitude \(2\sqrt{3}\). The x-component therefore is \((2\sqrt{3})* \cos(45 \degree) = \sqrt{2}*2\).

Step 3: Calculate the y-component of the vector

Apply the formula to get the y-component. This involves taking the sine of \(\theta\) (which is 45° or \(\pi/4\) in radian) and multiplying that by the \(v\) magnitude \(2\sqrt{3}\). The y-component therefore is \((2\sqrt{3})* \sin(45 \degree) = \sqrt{2}*2\).

Step 4: Write the vector in component form

The component form of a vector in 2D is \([x, y]\), where \(x\) is the x-component and \(y\) is the y-component. In this case, both \(x\) and \(y\) are \(\sqrt{2}*2\), so the component form of \(v\) is \([\sqrt{2}*2, \sqrt{2}*2]\).

Key Concepts

Magnitude

Understanding the concept of vector magnitude is crucial in physics and mathematics. The magnitude of a vector is essentially its length, and it is denoted by \(|\mathbf{v}|\). In simpler terms, magnitude shows a vector's size without considering its direction. In our example, we have a vector with a magnitude of \(2\sqrt{3}\).

To find a vector's magnitude when its components are known, you can use the Pythagorean theorem. If a vector \(v\) has components \(x\) and \(y\), its magnitude is given by:

• \(|\mathbf{v}| = \sqrt{x^2 + y^2}\)

However, when the vector's magnitude and direction (angle) are given, the components can be found directly, as we see in calculating vector \(v\) given a magnitude of \(2\sqrt{3}\). This allows us to compute the magnitude precisely without needing further calculations, assuming the angle and the vector's components align correctly.

Angle with Positive X-axis

The direction of a vector in a two-dimensional space can be defined by the angle it makes with the positive \(x\)-axis. This angle gives essential information about the vector's orientation.

In the given example, the angle \(\theta\) is provided as \(45^\circ\). When given in degrees, it often helps to convert this into radians for calculations, knowing that \(180^\circ = \pi \, \text{radians}\). Thus, \(45^\circ\) corresponds to \(\frac{\pi}{4}\) radians.

Angles can help determine the direction and symmetry of the vector. For instance:

• An angle of \(0^\circ\) points directly along the positive \(x\)-axis.
• An angle of \(90^\circ\) signifies pointing upwards along the \(y\)-axis.
• An angle of \(180^\circ\) indicates pointing along the opposite of the positive \(x\)-axis.

Understanding an angle's role relative to the \(x\)-axis is critical as it affects both the calculation and the graphical representation of the vector.

Cosine and Sine Functions

Cosine and sine functions play integral parts in breaking down a vector into its components. These fundamental trigonometric functions allow us to convert the polar representation of vectors into Cartesian, or component form.

The cosine of an angle relates to the \(x\)-component, while the sine connects with the \(y\)-component of the vector. The formulas are:

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

In our example, the calculations were performed using \(\theta = 45^\circ\), which means both sine and cosine for this angle are equal to \(\frac{\sqrt{2}}{2}\).

Hence, the \(x\)-component and \(y\)-component both equate to:

• \(2 \sqrt{3} \cdot \cos(45^\circ) = \sqrt{2} \cdot 2\)
• \(2 \sqrt{3} \cdot \sin(45^\circ) = \sqrt{2} \cdot 2\)

Grasping how cosine and sine functions aid in decomposing vectors is essential for understanding vector algebra and its broader applications in physics and engineering.