Problem 5
Question
In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y, z)=x \mathbf{i}+0 \mathbf{i}+\mathbf{k} $$
Step-by-Step Solution
Verified Answer
The vectors in \( \mathbf{F} = x \mathbf{i} + \mathbf{k} \) spread out along the x-axis and point consistently upward in the z-direction.
1Step 1: Analyze the Vector Field
The given vector field is \( \mathbf{F}(x, y, z) = x \mathbf{i} + 0 \mathbf{j} + \mathbf{k} \). This means that in this field, the x-component of the vectors changes with \( x \), the y-component is always zero, and the z-component is constant at 1.
2Step 2: Determine Vector Components
For any point \( (x, y, z) \) in the field, the vector can be written as \( \langle x, 0, 1 \rangle \). This shows that horizontally, the vectors will extend in the x-direction and have no movement in the y-direction, while vertically they maintain a constant size of 1 in the z-direction.
3Step 3: Choose Sample Points
Select sample points along the x-axis, such as \( x = -2, -1, 0, 1, 2 \). For each of these points, y can be any value since it does not affect the vector. Choose a few z-values for visualization, like \( z = 0, 1, 2 \).
4Step 4: Plot Vectors at Sample Points
For each chosen sample point, plot the corresponding vector components. For example, at the point \( (2, 0, 0) \), plot the vector \( \langle 2, 0, 1 \rangle \). Continue this for other sample points like \( (-2, 0, 0) \to \langle -2, 0, 1 \rangle \), emphasizing how the vectors spread out along the x-axis and point upward along the z-axis.
Key Concepts
Vector ComponentsVector VisualizationSample Point Selection
Vector Components
Understanding vector components is crucial in vector field analysis. The vector field provided is \( \mathbf{F}(x, y, z) = x \mathbf{i} + 0 \mathbf{j} + \mathbf{k} \). This equation represents a vector for each point \((x, y, z)\) in space. It breaks down into three parts, each associated with one of the three coordinate axes.
- **X-component**: This varies with \( x \), so the vector's strength or length in the x-direction depends on the x-coordinate. - **Y-component**: Here, it is always zero \(0 \mathbf{j}\), meaning there's no change or movement along the y-axis. - **Z-component**: Is constant at \( \mathbf{k} \), which means every vector maintains a steady upward direction regardless of where it is on the x-y plane.
These components collectively determine the full direction and magnitude of any given vector in the field. Having this break down helps in visualizing how the vectors behave at any point.
- **X-component**: This varies with \( x \), so the vector's strength or length in the x-direction depends on the x-coordinate. - **Y-component**: Here, it is always zero \(0 \mathbf{j}\), meaning there's no change or movement along the y-axis. - **Z-component**: Is constant at \( \mathbf{k} \), which means every vector maintains a steady upward direction regardless of where it is on the x-y plane.
These components collectively determine the full direction and magnitude of any given vector in the field. Having this break down helps in visualizing how the vectors behave at any point.
Vector Visualization
Visualizing vectors in a field relies on plotting them based on their components. In this task, you can think of each vector as an arrow emerging from a specific point in 3D space. The direction and length are determined by its components \( (x, 0, 1) \).
- **Horizontal Projection**: Since the y-component is zero, all vectors lie flat on the x-z plane and have no height change in the y-direction.- **Vertical Projection**: The constant z-component \( \mathbf{k} \) gives each vector a uniform upward segment, defining a 'lifting' effect along the z-axis.
To better grasp this visualization:- Extend the vector in the x-direction proportionately to its x-value.- Note that regardless of x, vectors always point in the z-direction by one unit.
Plotting these vectors at made-up views of x and z will help realize how they sort themselves along these axes.
- **Horizontal Projection**: Since the y-component is zero, all vectors lie flat on the x-z plane and have no height change in the y-direction.- **Vertical Projection**: The constant z-component \( \mathbf{k} \) gives each vector a uniform upward segment, defining a 'lifting' effect along the z-axis.
To better grasp this visualization:- Extend the vector in the x-direction proportionately to its x-value.- Note that regardless of x, vectors always point in the z-direction by one unit.
Plotting these vectors at made-up views of x and z will help realize how they sort themselves along these axes.
Sample Point Selection
Proper selection of sample points plays a key role in efficiently representing a vector field. In this example, we chose positions mainly along the x-axis, since it influences the x-component of the vector:
- **X-values**: Practical choices as \( x = -2, -1, 0, 1, 2 \) illustrate how vector lengths vary with different x-values. They portray the spread of vectors horizontally.- **Z-values**: Adopting sample heights like \( z = 0, 1, 2 \) can diversify the vertical positioning of vectors. Even if z does not alter vector magnitude, having different z-heights can demonstrate their constant nature.- **Y-value**: As the y-component is zero, any selection along this axis won't affect the vectors.
Picking these points allows a spectrum of how vectors appear when distributed through the given frame, providing a comprehensive depiction of the vector field.
- **X-values**: Practical choices as \( x = -2, -1, 0, 1, 2 \) illustrate how vector lengths vary with different x-values. They portray the spread of vectors horizontally.- **Z-values**: Adopting sample heights like \( z = 0, 1, 2 \) can diversify the vertical positioning of vectors. Even if z does not alter vector magnitude, having different z-heights can demonstrate their constant nature.- **Y-value**: As the y-component is zero, any selection along this axis won't affect the vectors.
Picking these points allows a spectrum of how vectors appear when distributed through the given frame, providing a comprehensive depiction of the vector field.
Other exercises in this chapter
Problem 4
Evaluate \(\iint g(x, y, z) d S\). \(g(x, y, z)=2 y^{2}+z ; G: z=x^{2}-y^{2}, 0 \leq x^{2}+y^{2} \leq 1\)
View solution Problem 4
Evaluate each line integral. \(\int_{C} x e^{y} d s ; C\) is the line segment from \((-1,2)\) to \((1,1)\).
View solution Problem 5
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=x^{2} y z \mathbf{i
View solution Problem 5
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C}\left(x^{2}+4 x y\right) d x+\left(2 x^{2}+3 y\right) d
View solution