Question

1130 chapter 16 vector calculus
13 - 18 match the vector fields f with the plots labeled i - vi. give reasons for your choices.

1. f(x,y)=(x, - y)
2. f(x,y)=(y,x - y)
3. f(x,y)=(y,y + 2)
4. f(x,y)=(y,2x)
5. f(x,y)=(sin y,cos x)
6. f(x,y)=(cos(x + y),x)

19 - 22 match the vector fields f on ℝ³ with the plots labeled i - iv. give reasons for your choices.

1. f(x,y,z)=i + 2j + 3k
2. f(x,y,z)=i + 2j+zk
3. f(x,y,z)=xi + yj + 3k
4. f(x,y,z)

Explanation:

Step1: Analyze the vector - field $\mathbf{F}(x,y)=(x, - y)$

At the point $(x,y)$ in the $xy$ - plane, the $x$ - component of the vector is $x$ and the $y$ - component is $-y$. When $x>0$, the vectors point to the right in the $x$ - direction, and when $x < 0$, the vectors point to the left. When $y>0$, the vectors point downwards in the $y$ - direction, and when $y < 0$, the vectors point upwards. This behavior is characteristic of a vector - field that "diverges" away from the origin in a way that resembles a hyperbolic - like pattern.

Step2: Analyze the vector - field $\mathbf{F}(x,y)=(y,x - y)$

The $x$ - component is $y$ and the $y$ - component is $x - y$. Along the $x$ - axis ($y = 0$), the vectors are $(0,x)$. Along the $y$ - axis ($x = 0$), the vectors are $(y,-y)$.

Step3: Analyze the vector - field $\mathbf{F}(x,y)=(y,y + 2)$

The $x$ - component is $y$ and the $y$ - component is $y + 2$. When $y=0$, the vectors have a $y$ - component of 2 and an $x$ - component of 0. As $y$ increases, both components change in a way that the vectors have a non - zero $x$ and $y$ component depending on the value of $y$.

Step4: Analyze the vector - field $\mathbf{F}(x,y)=(y,2x)$

The $x$ - component is $y$ and the $y$ - component is $2x$. At the origin $(0,0)$, the vector is $(0,0)$. For positive $y$ values, the vectors have a positive $x$ - component, and for positive $x$ values, the vectors have a positive $y$ - component.

Step5: Analyze the vector - field $\mathbf{F}(x,y)=(\sin y,\cos x)$

The $x$ - component $\sin y$ has a periodic behavior with respect to $y$ (ranging from - 1 to 1), and the $y$ - component $\cos x$ has a periodic behavior with respect to $x$ (ranging from - 1 to 1).

Step6: Analyze the vector - field $\mathbf{F}(x,y)=(\cos(x + y),x)$

The $x$ - component $\cos(x + y)$ is periodic in $x$ and $y$, and the $y$ - component is $x$.

For example, for $\mathbf{F}(x,y)=(x,-y)$:
Let's consider some points:
At $(1,0)$, $\mathbf{F}(1,0)=(1,0)$; at $(0,1)$, $\mathbf{F}(0,1)=(0, - 1)$; at $(-1,0)$, $\mathbf{F}(-1,0)=(-1,0)$; at $(0,-1)$, $\mathbf{F}(0,-1)=(0,1)$.

We need to match each of the given vector - fields (13 - 18) with the plots I - VI by observing the behavior of the vectors at key points (such as the axes) and the overall pattern of the vector - field.

Since the problem does not provide the plots clearly in a machine - readable format for a definite match, we can only provide the general approach to solve it. But if we assume we could examine the plots visually:
1. For $\mathbf{F}(x,y)=(x,-y)$: We would look for a plot where vectors point right for $x>0$ and left for $x < 0$ on the $x$ - axis, and down for $y>0$ and up for $y < 0$ on the $y$ - axis.
2. For $\mathbf{F}(x,y)=(y,x - y)$: Analyze the vectors on the axes and in different quadrants to match the plot.
3. For $\mathbf{F}(x,y)=(y,y + 2)$: Note the behavior as $y$ varies.
4. For $\mathbf{F}(x,y)=(y,2x)$: Consider the relationship between $x$ and $y$ values for the components.
5. For $\mathbf{F}(x,y)=(\sin y,\cos x)$: Look for a periodic - like pattern in the vectors due to the trigonometric functions.
6. For $\mathbf{F}(x,y)=(\cos(x + y),x)$: Analyze the periodicity of the $x$ - component and the non - periodicity of the $y$ - component.

Since we cannot see the plots clearly to give a definite answer, we have provided the step - by - step analysis of how to match the vector fields with the plots in general.

If we assume we had the visual ability to match:
13. For $\mathbf{F}(x,y)=(x,-y)$, we would look for a plot that has vectors symmetric about the $x$ and $y$ axes in a way that they "diverge" from the origin in a particular pattern.
14. For $\mathbf{F}(x,y)=(y,x - y)$, we would analyze the vectors at different points like $(1,0),(0,1),(-1,0),(0, - 1)$ etc. to match the plot.
15. For $\mathbf{F}(x,y)=(y,y + 2)$, we note that when $y = 0$, the vector has a non - zero $y$ component of 2 and an $x$ component of 0.
16. For $\mathbf{F}(x,y)=(y,2x)$, we consider how the vectors change as $x$ and $y$ change.
17. For $\mathbf{F}(x,y)=(\sin y,\cos x)$, we expect a periodic pattern in the vectors.
18. For $\mathbf{F}(x,y)=(\cos(x + y),x)$, we analyze the combined effect of the periodic $x$ - component and the non - periodic $y$ - component.

The general approach to answer such a question is to evaluate the vector - field at key points (axes and quadrant corners) and observe the overall pattern of the vectors in the plane.

Since the plots are not clearly available for a definite match, we cannot provide a final answer in the form of "Plot X matches vector - field Y". But the above steps show how to approach the problem of matching vector fields with their plots.

Answer:

Step1: Analyze the vector - field $\mathbf{F}(x,y)=(x, - y)$

At the point $(x,y)$ in the $xy$ - plane, the $x$ - component of the vector is $x$ and the $y$ - component is $-y$. When $x>0$, the vectors point to the right in the $x$ - direction, and when $x < 0$, the vectors point to the left. When $y>0$, the vectors point downwards in the $y$ - direction, and when $y < 0$, the vectors point upwards. This behavior is characteristic of a vector - field that "diverges" away from the origin in a way that resembles a hyperbolic - like pattern.

Step2: Analyze the vector - field $\mathbf{F}(x,y)=(y,x - y)$

The $x$ - component is $y$ and the $y$ - component is $x - y$. Along the $x$ - axis ($y = 0$), the vectors are $(0,x)$. Along the $y$ - axis ($x = 0$), the vectors are $(y,-y)$.

Step3: Analyze the vector - field $\mathbf{F}(x,y)=(y,y + 2)$

The $x$ - component is $y$ and the $y$ - component is $y + 2$. When $y=0$, the vectors have a $y$ - component of 2 and an $x$ - component of 0. As $y$ increases, both components change in a way that the vectors have a non - zero $x$ and $y$ component depending on the value of $y$.

Step4: Analyze the vector - field $\mathbf{F}(x,y)=(y,2x)$

The $x$ - component is $y$ and the $y$ - component is $2x$. At the origin $(0,0)$, the vector is $(0,0)$. For positive $y$ values, the vectors have a positive $x$ - component, and for positive $x$ values, the vectors have a positive $y$ - component.

Step5: Analyze the vector - field $\mathbf{F}(x,y)=(\sin y,\cos x)$

The $x$ - component $\sin y$ has a periodic behavior with respect to $y$ (ranging from - 1 to 1), and the $y$ - component $\cos x$ has a periodic behavior with respect to $x$ (ranging from - 1 to 1).

Step6: Analyze the vector - field $\mathbf{F}(x,y)=(\cos(x + y),x)$

The $x$ - component $\cos(x + y)$ is periodic in $x$ and $y$, and the $y$ - component is $x$.

For example, for $\mathbf{F}(x,y)=(x,-y)$:
Let's consider some points:
At $(1,0)$, $\mathbf{F}(1,0)=(1,0)$; at $(0,1)$, $\mathbf{F}(0,1)=(0, - 1)$; at $(-1,0)$, $\mathbf{F}(-1,0)=(-1,0)$; at $(0,-1)$, $\mathbf{F}(0,-1)=(0,1)$.

We need to match each of the given vector - fields (13 - 18) with the plots I - VI by observing the behavior of the vectors at key points (such as the axes) and the overall pattern of the vector - field.

Since the problem does not provide the plots clearly in a machine - readable format for a definite match, we can only provide the general approach to solve it. But if we assume we could examine the plots visually:

1. For $\mathbf{F}(x,y)=(x,-y)$: We would look for a plot where vectors point right for $x>0$ and left for $x < 0$ on the $x$ - axis, and down for $y>0$ and up for $y < 0$ on the $y$ - axis.
2. For $\mathbf{F}(x,y)=(y,x - y)$: Analyze the vectors on the axes and in different quadrants to match the plot.
3. For $\mathbf{F}(x,y)=(y,y + 2)$: Note the behavior as $y$ varies.
4. For $\mathbf{F}(x,y)=(y,2x)$: Consider the relationship between $x$ and $y$ values for the components.
5. For $\mathbf{F}(x,y)=(\sin y,\cos x)$: Look for a periodic - like pattern in the vectors due to the trigonometric functions.
6. For $\mathbf{F}(x,y)=(\cos(x + y),x)$: Analyze the periodicity of the $x$ - component and the non - periodicity of the $y$ - component.

Since we cannot see the plots clearly to give a definite answer, we have provided the step - by - step analysis of how to match the vector fields with the plots in general.

If we assume we had the visual ability to match:

1. For $\mathbf{F}(x,y)=(x,-y)$, we would look for a plot that has vectors symmetric about the $x$ and $y$ axes in a way that they "diverge" from the origin in a particular pattern.
2. For $\mathbf{F}(x,y)=(y,x - y)$, we would analyze the vectors at different points like $(1,0),(0,1),(-1,0),(0, - 1)$ etc. to match the plot.
3. For $\mathbf{F}(x,y)=(y,y + 2)$, we note that when $y = 0$, the vector has a non - zero $y$ component of 2 and an $x$ component of 0.
4. For $\mathbf{F}(x,y)=(y,2x)$, we consider how the vectors change as $x$ and $y$ change.
5. For $\mathbf{F}(x,y)=(\sin y,\cos x)$, we expect a periodic pattern in the vectors.
6. For $\mathbf{F}(x,y)=(\cos(x + y),x)$, we analyze the combined effect of the periodic $x$ - component and the non - periodic $y$ - component.

The general approach to answer such a question is to evaluate the vector - field at key points (axes and quadrant corners) and observe the overall pattern of the vectors in the plane.

Since the plots are not clearly available for a definite match, we cannot provide a final answer in the form of "Plot X matches vector - field Y". But the above steps show how to approach the problem of matching vector fields with their plots.