Problem 3 - Active Calculus Multivariable

Question

Perform the following operations on the vectors \(\vec{u}=\langle 0,5,-4\rangle, \vec{v}=\) \(\langle-2,0,3\rangle,\) and \(\vec{w}=\langle-3,0,1\rangle\). \(\vec{u} \cdot \vec{w}=\) _________ \((\vec{u} \cdot \vec{v}) \vec{u}=\)_________ \(((\vec{w} \cdot \vec{w}) \vec{u}) \cdot \vec{u}=\)_________ \(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}=\)_________

Step-by-Step Solution

Answer

1. \(\vec{u} \cdot \vec{w} = -4\) 2. \((\vec{u} \cdot \vec{v}) \vec{u} = \langle 0, -60, 48 \rangle\) 3. \(((\vec{w} \cdot \vec{w})\vec{u}) \cdot \vec{u} = 410\) 4. \(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} = -3\)

Step 1: Calculate the dot product of \(\vec{u}\) and \(\vec{w}\)

To calculate the dot product of two vectors, we need to multiply corresponding components and then add them all up: \(\vec{u} \cdot \vec{w} = \langle0,5,-4\rangle \cdot \langle-3,0,1\rangle = (0 \times (-3)) + (5 \times 0) + ((-4) \times 1) = 0 + 0 - 4 = -4\)

Step 2: Calculate the dot product of \(\vec{u}\) and \(\vec{v}\), and then multiply it with the vector \(\vec{u}\)

First, let's calculate the dot product of \(\vec{u}\) and \(\vec{v}\): \(\vec{u} \cdot \vec{v} = \langle0,5,-4\rangle \cdot \langle-2,0,3\rangle = (0 \times (-2)) + (5 \times 0) + ((-4) \times 3) = 0 + 0 - 12 = -12\) Now, we need to multiply the result (-12) with the vector \(\vec{u}\): \((\vec{u} \cdot \vec{v}) \vec{u} = -12 \times \langle 0, 5, -4 \rangle = \langle 0, -60, 48 \rangle\)

Step 3: Calculate the dot product of \((\vec{w} \cdot \vec{w})\vec{u}\) and \(\vec{u}\)

First, let's calculate the dot product of \(\vec{w}\) and itself: \(\vec{w} \cdot \vec{w} = \langle-3,0,1\rangle \cdot \langle-3,0,1\rangle = ((-3) \times (-3)) + (0 \times 0) + (1 \times 1) = 9 + 0 + 1 = 10\) Now, multiply this result by vector \(\vec{u}\): \((\vec{w} \cdot \vec{w})\vec{u} = 10 \times \langle 0, 5, -4 \rangle = \langle 0, 50, -40 \rangle\) Finally, calculate the dot product of \((\vec{w} \cdot \vec{w})\vec{u}\) and \(\vec{u}\): \(((\vec{w} \cdot \vec{w}) \vec{u}) \cdot \vec{u} = \langle 0, 50, -40 \rangle \cdot \langle 0, 5, -4 \rangle = (0 \times 0) + (50 \times 5) + ((-40) \times (-4)) = 0+ 250 + 160 = 410\)

Step 4: Calculate the sum of the dot products of \(\vec{u}\) and \(\vec{v}\), and of \(\vec{v}\) and \(\vec{w}\)

We already calculated the dot product of \(\vec{u}\) and \(\vec{v}\) in step 2, which gave us -12. Now, we need to calculate the dot product of \(\vec{v}\) and \(\vec{w}\): \(\vec{v} \cdot \vec{w} = \langle-2,0,3\rangle \cdot \langle-3,0,1\rangle = ((-2) \times (-3)) + (0 \times 0) + (3 \times 1) = 6 + 0 + 3 = 9\) Finally, let's sum the dot products: \(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} = -12 + 9 = -3\) All the answers are: 1. \(\vec{u} \cdot \vec{w} = -4\) 2. \((\vec{u} \cdot \vec{v}) \vec{u} = \langle 0, -60, 48 \rangle\) 3. \(((\vec{w} \cdot \vec{w})\vec{u}) \cdot \vec{u} = 410\) 4. \(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} = -3\)

Key Concepts

Dot Product

Understanding the dot product of vectors is an essential part of vector calculus. The dot product, also known as the scalar product, involves two vectors and results in a single scalar quantity. This value encapsulates how much of one vector goes in the direction of another. To find the dot product of two vectors, simply multiply their corresponding components and sum these products together. For example, with vectors \( \vec{u} = \langle 0, 5, -4 \rangle \) and \( \vec{w} = \langle -3, 0, 1 \rangle \), their dot product is calculated as follows:

\[ \vec{u} \cdot \vec{w} = (0 \times -3) + (5 \times 0) + (-4 \times 1) = -4 \]

Notice that each product results from multiplying components that occupy the same position in the vectors they're derived from.
The dot product is especially used to determine:

• Orthogonality: Two vectors are orthogonal (perpendicular) if their dot product is zero.
• Magnitude of projection: Provides a measure of one vector's magnitude along the direction of another vector.

Calculating the dot product helps solve numerous problems in physics and engineering, such as work done by a force along a path.

Vector Operations

Vector operations form the basis of vector analysis, serving as tools to manage and manipulate vectors in calculations. Common vector operations include addition, scalar multiplication, and the dot product, among others. When you multiply a vector by a scalar, you scale each component of the vector by that scalar. In the exercise, after finding the dot product of \( \vec{u} \) and \( \vec{v} \), the result was multiplied by the vector \( \vec{u} \), exemplifying scalar multiplication:

\[ (\vec{u} \cdot \vec{v}) \vec{u} = -12 \times \langle 0, 5, -4 \rangle = \langle 0, -60, 48 \rangle \]

Each component of \( \vec{u} \) is multiplied by the scalar \(-12\). Such operations are useful for changing a vector's length while keeping the same direction. Vector operations often appear across various fields:

• Physics: Calculating forces, velocities, and other physical quantities.
• Engineering: Solutions involving multiple concurrent influences.
• Computer Graphics: Scaling images or objects.

Comprehending vector operations is vital for efficiently solving multi-dimensional problems.

Multivariable Calculus

Multivariable calculus extends the fields of calculus into higher dimensions, handling functions with more than one variable. In the context of vectors, it importantly deals with functions that output vectors resulting from various inputs. Imagine calculating force fields, or analyzing pressure distribution, both of which are classic applications.

Within multivariable calculus, vector calculus plays a critical role—facilitating calculations across two or more dimensional spaces, such as calculating gradients, divergences, and curls. Referring back to our exercise, multivariable calculus helps combine several vector dot products:

We computed \( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} = -3 \). This shows how multivariable calculus handles simultaneous influences.

Important concepts include:

• Vector Fields: Functions assigning a vector to each point in space.
• Gradient: Measures the rate and direction of change.

These operations enable deeper analysis and understanding of complex systems, useful everywhere from electromagnetism to dynamics and more.