Problem 6 - Calculus: Early Transcendentals

Question

Find \(|\boldsymbol{v}|, \boldsymbol{v}+\boldsymbol{w}, \boldsymbol{v}-\boldsymbol{w},|\boldsymbol{v}+\boldsymbol{w}|,|\boldsymbol{v}-\boldsymbol{w}|\) and \(-2 \boldsymbol{v}\) for \(\boldsymbol{v}=\langle 1,3\rangle\) and \(\boldsymbol{w}=\langle-1,-5\rangle .\)

Step-by-Step Solution

Answer

\(|\boldsymbol{v}|=\sqrt{10}, \boldsymbol{v}+\boldsymbol{w}=\langle 0,-2\rangle, \boldsymbol{v}-\boldsymbol{w}=\langle 2,8\rangle, |\boldsymbol{v}+\boldsymbol{w}|=2, |\boldsymbol{v}-\boldsymbol{w}|=\sqrt{68}, -2\boldsymbol{v}=\langle-2,-6\rangle\).

Step 1: Find Magnitude of Vector \( \boldsymbol{v} \)

To find the magnitude of the vector \( \boldsymbol{v} = \langle 1, 3 \rangle \), use the formula \(|\boldsymbol{v}| = \sqrt{v_1^2 + v_2^2}\). Here, \(v_1 = 1\) and \(v_2 = 3\). So, \(|\boldsymbol{v}| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}\).

Step 2: Add Vectors \( \boldsymbol{v} \) and \( \boldsymbol{w} \)

To add two vectors \( \boldsymbol{v} = \langle 1, 3 \rangle \) and \( \boldsymbol{w} = \langle -1, -5 \rangle \), use component-wise addition: \(\boldsymbol{v} + \boldsymbol{w} = \langle 1 + (-1), 3 + (-5) \rangle = \langle 0, -2 \rangle\).

Step 3: Subtract Vector \( \boldsymbol{w} \) from \( \boldsymbol{v} \)

To subtract the vector \( \boldsymbol{w} = \langle -1, -5 \rangle \) from \( \boldsymbol{v} = \langle 1, 3 \rangle \), use component-wise subtraction: \(\boldsymbol{v} - \boldsymbol{w} = \langle 1 - (-1), 3 - (-5) \rangle = \langle 2, 8 \rangle\).

Step 4: Find Magnitude of \( \boldsymbol{v} + \boldsymbol{w} \)

To find the magnitude of the vector \( \boldsymbol{v} + \boldsymbol{w} = \langle 0, -2 \rangle \), use \(|\boldsymbol{v} + \boldsymbol{w}| = \sqrt{0^2 + (-2)^2} = \sqrt{0 + 4} = \sqrt{4} = 2\).

Step 5: Find Magnitude of \( \boldsymbol{v} - \boldsymbol{w} \)

To find the magnitude of the vector \( \boldsymbol{v} - \boldsymbol{w} = \langle 2, 8 \rangle \), use \(|\boldsymbol{v} - \boldsymbol{w}| = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68}\).

Step 6: Multiply Vector \( \boldsymbol{v} \) by \(-2\)

To multiply vector \( \boldsymbol{v} = \langle 1, 3 \rangle \) by \(-2\), multiply each component by \(-2\): \(-2 \boldsymbol{v} = \langle -2 \times 1, -2 \times 3 \rangle = \langle -2, -6 \rangle\).

Key Concepts

vector addition

Vector addition is a fundamental operation where two or more vectors are combined to produce a new vector, known as the resultant. In this simple adding process, we focus on joining two vectors end-to-end.
Mathematically, vector addition occurs by adding corresponding components from each vector. For example, if you have vectors \( \mathbf{v} = \langle a, b \rangle \) and \( \mathbf{w} = \langle c, d \rangle \), their sum \( \mathbf{v} + \mathbf{w} \) is given by:

• \( \mathbf{v} + \mathbf{w} = \langle a+c, b+d \rangle \)

This means adding the first components of both vectors and the second components separately.
For instance, when adding \( \mathbf{v} = \langle 1, 3 \rangle \) and \( \mathbf{w} = \langle -1, -5 \rangle \), it results in \( \langle 0, -2 \rangle \). This vector addition implementation exemplifies how oppositely directed vectors can cancel out each other's effect.

vector subtraction

Vector subtraction is akin to vector addition but instead involves taking the difference of corresponding components.
This operation allows us to determine the vector pointing from one vector to another. When you subtract \( \mathbf{w} = \langle c, d \rangle \) from \( \mathbf{v} = \langle a, b \rangle \), the result is:

• \( \mathbf{v} - \mathbf{w} = \langle a-c, b-d \rangle \)

In the given exercise, subtracting vector \( \mathbf{w} = \langle -1, -5 \rangle \) from vector \( \mathbf{v} = \langle 1, 3 \rangle \) yields \( \langle 2, 8 \rangle \).
This resultant vector stretches across by increasing the magnitude, which visually represents moving across the plane in a particular direction from one vector's tail to the other's head.

vector magnitude

The magnitude of a vector, often referred to as its "length," provides the distance of the vector from the origin in the coordinate system.
Using the Pythagorean theorem in two dimensions, we calculate the magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \) using the formula:

• \(|\mathbf{v}| = \sqrt{a^2 + b^2}\)

In the original exercise, the magnitude of \( \mathbf{v} = \langle 1, 3 \rangle \) calculates as \( \sqrt{10} \), whereas that of \( \mathbf{v} + \mathbf{w} \) and \( \mathbf{v} - \mathbf{w} \) is \( 2 \) and \( \sqrt{68} \) respectively.
Magnitude is crucial for understanding both the size and distance of vectors in diverse fields like physics and engineering.

scalar multiplication

Scalar multiplication involves multiplying each component of a vector by a real number called a scalar. This operation scales the vector either by stretching or compressing it without altering its direction, unless multiplied by a negative scalar which also reverses its direction.
Let's say you have a vector \( \mathbf{v} = \langle a, b \rangle \) and a scalar \( k \). The result of scalar multiplication is given by:

• \( k\mathbf{v} = \langle ka, kb \rangle \)

For example, multiplying \( \mathbf{v} = \langle 1, 3 \rangle \) by \(-2\) results in \( \langle -2, -6 \rangle \).
Importantly, scalar multiplication adjusts the vector's magnitude by \(|k|\) while its direction flips when \(k\) is negative. This concept is integral in physics for altering vector quantities like velocity and force.