Chapter 9, Problem 5P

To determine

To sketch: The vector field $F\left(x,y\right)=\frac{yi+xj}{\sqrt{x^2+y^2}}$.

Explanation of Solution

The vector field is a function of a space which gives a vector quantity at each point. It is a collection of arrows with a given magnitude direction.

The general form of vector field is given below,

$F\left(x,y\right)=xi+yj$ (1)

Compare the vector field $F\left(x,y\right)=\frac{yi+xj}{\sqrt{x^2+y^2}}$ with equation (1).

$x=\frac{y}{\sqrt{x^2+y^2}}$

$y=\frac{x}{\sqrt{x^2+y^2}}$

The vector field at several points shown is below Table 1.

$\left(x,y\right)$	$F\left(x,y\right)$	$\left(x,y\right)$	$F\left(x,y\right)$
$\left(1,0\right)$	$\langle 0,1\rangle$	$\left(-1,0\right)$	$\langle 0,-1\rangle$
$\left(0,1\right)$	$\langle 1,0\rangle$	$\left(0,-1\right)$	$\langle -1,0\rangle$
$\left(0,2\right)$	$\langle 1,0\rangle$	$\left(0,-2\right)$	$\langle -1,0\rangle$
$\left(2,0\right)$	$\langle 0,1\rangle$	$\left(-2,0\right)$	$\langle 0,-1\rangle$
$\left(1,1\right)$	$\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle$	$\left(-1,-1\right)$	$\langle -\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\rangle$
$\left(2,2\right)$	$\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle$	$\left(-2,-2\right)$	$\langle -\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\rangle$

Table 1

From Table 1, the graph of vector field is given below in Figure 1.

From Figure 1, it is noticed that the all vectors have different directions but same velocity or magnitude.

Thus, the Figure 1 shows the vector field $F\left(x,y\right)=\frac{yi+xj}{\sqrt{x^2+y^2}}$.