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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Transcribed Image Text:The image contains four Cartesian coordinate systems, each depicting an angle of 150 degrees from the positive x-axis, illustrated by black vectors. Here are the details of each graph:
1. **Top-left graph:**
- **Axes:**
- x-axis: Horizontal axis, ranging from -1.5 to 1.5.
- y-axis: Vertical axis, ranging from -1.5 to 1.5.
- **Vector:** An arrowhead vector is drawn starting from the origin (0,0) towards (-0.5, 0.866), representing the 150-degree angle measured counterclockwise from the positive x-axis.
- **Angle:** A red arc denotes the angle 150°, measured counterclockwise from the positive x-axis.
2. **Top-right graph:**
- **Axes:**
- x-axis: Horizontal axis, ranging from -1.5 to 1.5.
- y-axis: Vertical axis, ranging from -1.5 to 1.5.
- **Vector:** An arrowhead vector starts from the origin towards (0.5, -0.866), indicating the 150-degree angle measured clockwise from the positive x-axis.
- **Angle:** A red arc denotes the angle 150°, measured clockwise from the positive x-axis.
3. **Bottom-left graph:**
- **Axes:**
- x-axis: Horizontal axis ranging from -1.5 to 1.5.
- y-axis: Vertical axis ranging from -1.5 to 1.5.
- **Vector:** An arrowhead vector starts from the origin towards (-0.5, -0.866), representing the 150-degree angle measured counterclockwise from the negative x-axis.
- **Angle:** A red arc denotes the angle 150°, measured counterclockwise from the negative x-axis.
4. **Bottom-right graph:**
- **Axes:**
- x-axis: Horizontal axis ranging from -1.5 to 1.5.
- y-axis: Vertical axis ranging from -1.5 to 1.5.
- **Vector:** An arrowhead vector starts from the origin towards (0.5, 0.866), representing a 150-degree angle measured clockwise from the negative x-axis.
- **Angle:** A red arc denotes the angle 150°, measured clockwise from the negative
![### Finding the Component Form of a Vector
Given:
- **Magnitude of the vector, \(\mathbf{v}\):**
\[
\|\mathbf{v}\| = \frac{3}{2}
\]
- **Angle \(\theta\) it makes with the positive x-axis:**
\[
\theta = 150^\circ
\]
To find the component form of \(\mathbf{v}\), we use the following formulas based on trigonometry in the context of vectors:
\[
v_x = \|\mathbf{v}\| \cos(\theta)
\]
\[
v_y = \|\mathbf{v}\| \sin(\theta)
\]
Where:
- \(v_x\) is the x-component of the vector,
- \(v_y\) is the y-component of the vector,
- \(\|\mathbf{v}\|\) is the magnitude of the vector \(\mathbf{v}\),
- \(\theta\) is the angle the vector makes with the positive x-axis.
### Step-by-Step Solution
1. **Calculate \(v_x\):**
\[
v_x = \left(\frac{3}{2}\right) \cos(150^\circ)
\]
Using the fact that \(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\):
\[
v_x = \left(\frac{3}{2}\right) \left(-\frac{\sqrt{3}}{2}\right)
\]
\[
v_x = -\frac{3\sqrt{3}}{4}
\]
2. **Calculate \(v_y\):**
\[
v_y = \left(\frac{3}{2}\right) \sin(150^\circ)
\]
Using the fact that \(\sin(150^\circ) = \frac{1}{2}\):
\[
v_y = \left(\frac{3}{2}\right) \left(\frac{1}{2}\right)
\]
\[
v_y = \frac{3}{4}
\]
### Final Component Form
Therefore, the component form of the vector \(\mathbf{v}\) is:
\[
\left(-\frac{3\sqrt{3}}{4}, \frac{3}{4}\right)
\]
This expresses the vector \(\mathbf{v}\) in terms](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fab864d-7419-4b8c-bc50-25971d273c3c%2F6adf1f75-ecc3-4a3c-b516-309c04e14e87%2Fg6n6tdh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Component Form of a Vector
Given:
- **Magnitude of the vector, \(\mathbf{v}\):**
\[
\|\mathbf{v}\| = \frac{3}{2}
\]
- **Angle \(\theta\) it makes with the positive x-axis:**
\[
\theta = 150^\circ
\]
To find the component form of \(\mathbf{v}\), we use the following formulas based on trigonometry in the context of vectors:
\[
v_x = \|\mathbf{v}\| \cos(\theta)
\]
\[
v_y = \|\mathbf{v}\| \sin(\theta)
\]
Where:
- \(v_x\) is the x-component of the vector,
- \(v_y\) is the y-component of the vector,
- \(\|\mathbf{v}\|\) is the magnitude of the vector \(\mathbf{v}\),
- \(\theta\) is the angle the vector makes with the positive x-axis.
### Step-by-Step Solution
1. **Calculate \(v_x\):**
\[
v_x = \left(\frac{3}{2}\right) \cos(150^\circ)
\]
Using the fact that \(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\):
\[
v_x = \left(\frac{3}{2}\right) \left(-\frac{\sqrt{3}}{2}\right)
\]
\[
v_x = -\frac{3\sqrt{3}}{4}
\]
2. **Calculate \(v_y\):**
\[
v_y = \left(\frac{3}{2}\right) \sin(150^\circ)
\]
Using the fact that \(\sin(150^\circ) = \frac{1}{2}\):
\[
v_y = \left(\frac{3}{2}\right) \left(\frac{1}{2}\right)
\]
\[
v_y = \frac{3}{4}
\]
### Final Component Form
Therefore, the component form of the vector \(\mathbf{v}\) is:
\[
\left(-\frac{3\sqrt{3}}{4}, \frac{3}{4}\right)
\]
This expresses the vector \(\mathbf{v}\) in terms
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