Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Vector Operations
#### Problem Statement:
**Compute** \( \mathbf{u} + \mathbf{v} \) **and** \( \mathbf{u} - 2\mathbf{v} \).
#### Given Vectors:
\[
\mathbf{u} = \begin{bmatrix}
-5 \\
1
\end{bmatrix}, \ \mathbf{v} = \begin{bmatrix}
-2 \\
4
\end{bmatrix}
\]
#### Calculation:
**Step 1:** Perform the addition \( \mathbf{u} + \mathbf{v} \).
\[
\mathbf{u} + \mathbf{v} = \begin{bmatrix}
-5 \\
1
\end{bmatrix} + \begin{bmatrix}
-2 \\
4
\end{bmatrix}
\]
You add each corresponding component:
\[
= \begin{bmatrix}
-5 + (-2) \\
1 + 4
\end{bmatrix}
= \begin{bmatrix}
-7 \\
5
\end{bmatrix}
\]
**Step 2:** Perform the subtraction \( \mathbf{u} - 2\mathbf{v} \).
First, calculate \( 2\mathbf{v} \):
\[
2\mathbf{v} = 2 \begin{bmatrix}
-2 \\
4
\end{bmatrix}
= \begin{bmatrix}
2 \cdot (-2) \\
2 \cdot 4
\end{bmatrix}
= \begin{bmatrix}
-4 \\
8
\end{bmatrix}
\]
Next, subtract \( 2\mathbf{v} \) from \( \mathbf{u} \):
\[
\mathbf{u} - 2\mathbf{v} = \begin{bmatrix}
-5 \\
1
\end{bmatrix} - \begin{bmatrix}
-4 \\
8
\end{bmatrix}
\]
Subtract each corresponding component:
\[
= \begin{bmatrix}
-5 - (-4) \\
1 - 8
\end{bmatrix}
= \begin{bmatrix}
-5 + 4 \\
1 - 8
\end{bmatrix}
= \begin{bmatrix}
-1 \\
-7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F474861d3-aaf1-4b4c-afe2-e250091335e4%2Fc9b4f0c9-0bb2-4a08-9c72-52613ec5524c%2Feojsmw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Vector Operations
#### Problem Statement:
**Compute** \( \mathbf{u} + \mathbf{v} \) **and** \( \mathbf{u} - 2\mathbf{v} \).
#### Given Vectors:
\[
\mathbf{u} = \begin{bmatrix}
-5 \\
1
\end{bmatrix}, \ \mathbf{v} = \begin{bmatrix}
-2 \\
4
\end{bmatrix}
\]
#### Calculation:
**Step 1:** Perform the addition \( \mathbf{u} + \mathbf{v} \).
\[
\mathbf{u} + \mathbf{v} = \begin{bmatrix}
-5 \\
1
\end{bmatrix} + \begin{bmatrix}
-2 \\
4
\end{bmatrix}
\]
You add each corresponding component:
\[
= \begin{bmatrix}
-5 + (-2) \\
1 + 4
\end{bmatrix}
= \begin{bmatrix}
-7 \\
5
\end{bmatrix}
\]
**Step 2:** Perform the subtraction \( \mathbf{u} - 2\mathbf{v} \).
First, calculate \( 2\mathbf{v} \):
\[
2\mathbf{v} = 2 \begin{bmatrix}
-2 \\
4
\end{bmatrix}
= \begin{bmatrix}
2 \cdot (-2) \\
2 \cdot 4
\end{bmatrix}
= \begin{bmatrix}
-4 \\
8
\end{bmatrix}
\]
Next, subtract \( 2\mathbf{v} \) from \( \mathbf{u} \):
\[
\mathbf{u} - 2\mathbf{v} = \begin{bmatrix}
-5 \\
1
\end{bmatrix} - \begin{bmatrix}
-4 \\
8
\end{bmatrix}
\]
Subtract each corresponding component:
\[
= \begin{bmatrix}
-5 - (-4) \\
1 - 8
\end{bmatrix}
= \begin{bmatrix}
-5 + 4 \\
1 - 8
\end{bmatrix}
= \begin{bmatrix}
-1 \\
-7
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