Problem 36 - Precalculus with Calculus Previews

Question

Express the given vector (a) in trigonometric form and (b) as a linear combination of the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\). $$ \langle-4,-4\rangle $$

Step-by-Step Solution

Answer

(a) Trigonometric form: \( 4\sqrt{2}(\cos(\frac{5\pi}{4})\mathbf{i} + \sin(\frac{5\pi}{4})\mathbf{j}) \), (b) Linear combination: \( -4\mathbf{i} - 4\mathbf{j} \).

Step 1: Identify the components of the vector

The given vector is \( \langle -4, -4 \rangle \). It has components that can be described as \( x = -4 \) and \( y = -4 \).

Step 2: Calculate the magnitude of the vector

The magnitude \( r \) of a vector \( \langle x, y \rangle \) is calculated using the formula \( r = \sqrt{x^2 + y^2} \). Substituting the values gives: \( r = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \).

Step 3: Find the angle of the vector

The angle \( \theta \) can be found using the tangent function: \( \tan(\theta) = \frac{y}{x} \). Therefore, \( \tan(\theta) = \frac{-4}{-4} = 1 \). The angle whose tangent is \( 1 \) is \( 45^\circ \) or \( \frac{\pi}{4} \) radians. However, since the vector is in the third quadrant (negative x and y), we must add \( \pi \) radians, giving \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \).

Step 4: Express the vector in trigonometric form

The vector's trigonometric form is expressed as \( r(\cos(\theta)\mathbf{i} + \sin(\theta)\mathbf{j}) \) where \( r = 4\sqrt{2} \) and \( \theta = \frac{5\pi}{4} \). This leads to: \( 4\sqrt{2}(\cos(\frac{5\pi}{4})\mathbf{i} + \sin(\frac{5\pi}{4})\mathbf{j}) \).

Step 5: Express the vector as a linear combination of unit vectors

To express \( \langle -4, -4 \rangle \) as a linear combination of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), use the form \( x\mathbf{i} + y\mathbf{j} \). This gives \( -4\mathbf{i} - 4\mathbf{j} \).

Key Concepts

Trigonometric Form

The trigonometric form of a vector is a useful way to represent its direction and magnitude using angles. The basic notation is:

• The vector is expressed as \( r(\cos(\theta)\mathbf{i} + \sin(\theta)\mathbf{j}) \)
• Where \( r \) is the magnitude of the vector, and \( \theta \) is the angle formed with the positive x-axis.

In our example, the given vector is \( \langle -4, -4 \rangle \). First, we found the magnitude \( r = 4\sqrt{2} \). Next, we determined the angle \( \theta = \frac{5\pi}{4} \) based on the vector's position in the third quadrant.
This position influences the directional angle because the signs of the components minus 4 and minus 4 indicate that both x and y are negative.
Thus, in trigonometric form, our vector is: \( 4\sqrt{2}(\cos(\frac{5\pi}{4})\mathbf{i} + \sin(\frac{5\pi}{4})\mathbf{j}) \).

Magnitude of a Vector

The magnitude of a vector, sometimes referred to as the vector's length, provides a measure of how far the point is from the origin in the coordinates plane. Think of it as the "size" of the vector. It is always a non-negative value.
The formula used to calculate the magnitude \( r \) is:

• \( r = \sqrt{x^2 + y^2} \)

For our vector \( \langle -4, -4 \rangle \), this calculates to:\( r = \sqrt{(-4)^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2} \).
The magnitude tells us how far the vector extends in the coordinate plane, regardless of direction.

Unit Vectors

Unit vectors are the building blocks of vector representation. They are standardized vectors that point in a direction and have a magnitude of 1.
Typically, in a 2D space, we use unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), which point in the directions of the x-axis and y-axis respectively:

• \( \mathbf{i} \equiv \langle 1, 0 \rangle \)
• \( \mathbf{j} \equiv \langle 0, 1 \rangle \)

When expressing a vector \( \langle x, y \rangle \) in terms of unit vectors, it is written as \( x\mathbf{i} + y\mathbf{j} \). For our case, expressing \( \langle -4, -4 \rangle \) using unit vectors gives \( -4\mathbf{i} - 4\mathbf{j} \). This expresses the vector neatly in terms of directional components.

Quadrants

The coordinate plane is divided into four parts, called quadrants. Each quadrant is named based on the signs of the x and y coordinates:

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

In this exercise, we are dealing with the vector \( \langle -4, -4 \rangle \), both coordinates are negative, placing it in Quadrant III.
This is important because it affects the calculation of the angle \( \theta \), noting that angles in this quadrant fall between \( \pi \) and \( \frac{3\pi}{2} \).

Linear Combination

Linear combination is a method of expressing a vector as a sum of scaled vectors. The scaling factors are called coefficients. Specifically, in the context of unit vectors, any vector \( \mathbf{v} \) in the plane can be described as a linear combination of \( \mathbf{i} \) and \( \mathbf{j} \).
For a vector \( \langle x, y \rangle \), the expression is \( x\mathbf{i} + y\mathbf{j} \).
For our problem, vector \( \langle -4, -4 \rangle \) can be expressed as:

• \( -4\mathbf{i} - 4\mathbf{j} \)

Each coefficient (in this case -4) represents how much of \( \mathbf{i} \) and \( \mathbf{j} \) is needed to obtain the vector.
This form is beneficial for easily visualizing the direction and magnitude of vectors and is straightforward for computation when adding or subtracting vectors.