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)

Explanation:

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

At the origin $(0,0)$, the vector is $\mathbf{F}(0,0)=(0,0)$. For positive $x$ - values and positive $y$ - values, the $x$ - component of the vector is positive and the $y$ - component is negative. For positive $x$ - values and negative $y$ - values, both components are positive. By observing the behavior of the vectors in different quadrants, we can match it with a plot.

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

When $y = 0$, the vector is $\mathbf{F}(x,0)=(0,x)$. When $x = 0$, the vector is $\mathbf{F}(0,y)=(y,-y)$. Analyze the direction and magnitude of vectors at key points and in different regions of the $xy$ - plane.

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

The $x$ - component of the vector depends only on $y$, and the $y$ - component is a linear function of $y$. When $y=-2$, the $y$ - component of the vector is $0$. Analyze the behavior of vectors for different $y$ - values.

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

If $x = 0$, the vector is $\mathbf{F}(0,y)=(y,0)$. If $y = 0$, the vector is $\mathbf{F}(x,0)=(0,2x)$. The direction of the vectors changes depending on the signs of $x$ and $y$.

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

The $x$ - component $\sin y$ has a period of $2\pi$ with respect to $y$, and the $y$ - component $\cos x$ has a period of $2\pi$ with respect to $x$. Analyze the values of the components at $y = 0,\pm\pi,\pm2\pi$ and $x = 0,\pm\pi,\pm2\pi$.

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

The $x$ - component $\cos(x + y)$ has a period of $2\pi$ with respect to $x + y$, and the $y$ - component is $x$. Analyze the behavior of the vectors for different values of $x$ and $y$ such as when $x + y=0,\pm\pi,\pm2\pi$.

Since we don't have the actual matching results provided in the question - asking for which plot matches which vector field, we assume we are just showing the general process of analysis. If we were to actually match:

1. $\mathbf{F}(x,y)=(x, - y)$: The vectors point away from the origin in a way that in the first quadrant $(x>0,y>0)$ they point down - right, in the second quadrant $(x<0,y>0)$ they point down - left, in the third quadrant $(x<0,y<0)$ they point up - left and in the fourth quadrant $(x>0,y<0)$ they point up - right.
2. $\mathbf{F}(x,y)=(y,x - y)$: Analyze the vectors at points like $(x = 0,y = 0),(x = 1,y = 0),(x = 0,y = 1)$ etc.
3. $\mathbf{F}(x,y)=(y,y + 2)$: When $y=-2$, the $y$ - component of the vector is zero.
4. $\mathbf{F}(x,y)=(y,2x)$: Vectors have different directions based on the signs of $x$ and $y$.
5. $\mathbf{F}(x,y)=(\sin y,\cos x)$: Due to the periodic nature of $\sin y$ and $\cos x$, the vectors have a repeating pattern in the $y$ and $x$ directions respectively.
6. $\mathbf{F}(x,y)=(\cos(x + y),x)$: The $x$ - component is periodic in $x + y$ and the $y$ - component is linear in $x$.

Answer:

Without the actual matching of plots to vector - fields given in the problem statement, we cannot provide a specific final answer for which plot matches which vector field. The above steps show the general method for making such a match.