Problem 21 - Essential Calculus Early Transcendentals

Question

If \(\mathbf{v}\) lies in the first quadrant and makes an angle \(\pi / 3\) with the positive \(x\) -axis and \(|\mathbf{v}|=4,\) find \(\mathbf{v}\) in component form.

Step-by-Step Solution

Answer

\(\mathbf{v} = \langle 2, 2\sqrt{3} \rangle\) in component form.

Step 1: Understand the Problem

We need to express the vector \(\mathbf{v}\) in component form given that it lies in the first quadrant, makes an angle of \(\frac{\pi}{3}\) with the positive \(x\)-axis, and has a magnitude of 4.

Step 2: Use Trigonometric Functions to Find Components

The vector \(\mathbf{v}\) can be expressed as \(\mathbf{v} = \langle v_x, v_y \rangle\). Since \(\mathbf{v}\) makes an angle of \(\frac{\pi}{3}\), we use the cosine and sine functions to find the components:\[v_x = |\mathbf{v}| \cos(\theta) = 4 \cos\left(\frac{\pi}{3}\right) = 4 \times \frac{1}{2} = 2\]\[v_y = |\mathbf{v}| \sin(\theta) = 4 \sin\left(\frac{\pi}{3}\right) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\]

Step 3: Write the Vector in Component Form

Now that we have the components \(v_x\) and \(v_y\), we can write the vector \(\mathbf{v}\) in component form:\[\mathbf{v} = \langle 2, 2\sqrt{3} \rangle\]

Key Concepts

First Quadrant

When discussing vectors in a Cartesian coordinate system, understanding the quadrants is crucial. The coordinate plane is divided into four quadrants. The first quadrant is defined by both x and y values being positive. This means any vector located in this quadrant will point somewhere between the positive x-axis and the positive y-axis.
These vectors have components that move in the direction of both the positive x and y axes. Since both components are positive, the overall direction of the vector is upwards and to the right. This is a helpful initial check when determining the components' signs in any vector problem.

Angle with x-axis

The angle a vector makes with the positive x-axis is a key characteristic, especially in the context of trigonometry. In our exercise, the vector makes an angle of \(\frac{\pi}{3}\), or 60 degrees, with the x-axis. This indicates the vector is oriented quite upwards from the x-axis.
Knowing this angle allows us to use trigonometric functions to find the horizontal and vertical components of the vector. It's essential to always measure this angle from the positive x-axis, ensuring consistency and correctness when calculating vector components. Furthermore, this angle dictates how much of the vector's strength, or magnitude, is directed horizontally versus vertically.

Magnitude of Vector

The magnitude of a vector indicates its size or length, basically how much force it exhibits. In our problem, the magnitude is given as 4. Imagine a straight-line arrow extending out from the origin; this magnitude tells us how long that arrow is.
Magnitude combined with direction forms a vector's essence. To accurately find the vector's components, you apply this magnitude in conjunction with the direction angle. By multiplying the magnitude by the cosine or sine of the given angle, you translate the vector's overall "size" into precise horizontal and vertical impacts on the coordinate plane. Therefore, understanding this concept is foundational in using trigonometry to deconstruct vectors into components.

Cosine and Sine Functions

These functions are fundamental in breaking down a vector into its components. The cosine of the angle a vector makes with the x-axis helps determine the horizontal component of the vector, while the sine helps with the vertical component.
For an angle \(\frac{\pi}{3}\), we have specific values for these functions: \(\cos(\frac{\pi}{3}) = \frac{1}{2}\) and \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\). When we multiply these values by the vector's magnitude, we effectively "project" the vector's overall force onto the x and y axes.

• The term \(v_x = |\mathbf{v}| \cos(\theta)\) shows how much of the vector is in the horizontal direction.
• The term \(v_y = |\mathbf{v}| \sin(\theta)\) tells us how much is in the vertical direction.

These functions and their values are essential for anyone learning trigonometry and vector mathematics, as they create a bridge between geometry and algebra in the context of physics and engineering.