To sketch this vector field, choose a sample of points from each quadrant and compute the corresponding vector. The following table gives a representative sample of points in a plane and the corresponding vectors.

Figure 2(a) shows the vector field. To see that each vector is perpendicular to the corresponding circle, Figure 2(b) shows circles overlain on the vector field.

Figure 2. (a) A visual representation of the radial vector field [latex]\small{{\bf{F}}(x,y)=\frac{x}2{\bf{i}}+\frac{y}2{\bf{j}}}[/latex]. (b) The radial vector field [latex]\small{{\bf{F}}(x,y)=\frac{x}2{\bf{i}}+\frac{y}2{\bf{j}}}[/latex] with overlaid circles. Notice that each vector is perpendicular to the circle on which it is located.

Solution

Create a table (see the one that follows) using a representative sample of points in a plane and their corresponding vectors. Figure 5 shows the resulting vector field.

[latex](x, y)[/latex]	[latex]F(x, y)[/latex]	[latex](x, y)[/latex]	[latex]F(x, y)[/latex]	[latex](x, y)[/latex]	[latex]F(x, y)[/latex]
[latex](1, 0)[/latex]	[latex]\langle0,-1\rangle[/latex]	[latex](2, 0)[/latex]	[latex]\langle0,-2\rangle[/latex]	[latex](1, 1)[/latex]	[latex]\langle1,-1\rangle[/latex]
[latex](0, 1)[/latex]	[latex]\langle1,0\rangle[/latex]	[latex](0, 2)[/latex]	[latex]\langle2,0\rangle[/latex]	[latex](-1, 1)[/latex]	[latex]\langle1,1\rangle[/latex]
[latex](-1, 0)[/latex]	[latex]\langle0,1\rangle[/latex]	[latex](-2, 0)[/latex]	[latex]\langle0,2\rangle[/latex]	[latex](-1, -1)[/latex]	[latex]\langle-1,1\rangle[/latex]
[latex](0, -1)[/latex]	[latex]\langle-1,0\rangle[/latex]	[latex](0, -2)[/latex]	[latex]\langle-2,0\rangle[/latex]	[latex](1, -1)[/latex]	[latex]\langle-1,-1\rangle[/latex]

Figure 5. (a) A visual representation of vector field [latex]{\bf{F}}(x,y)=\langle{y},-x\rangle[/latex]. (b) Vector field [latex]{\bf{F}}(x,y)=\langle{y},-x\rangle[/latex] with circles centered at the origin. (c) [latex]{\bf{F}}(a,b)[/latex] is perpendicular to radial vector [latex]\langle{a},b\rangle[/latex] at point [latex](a, b)[/latex].

Analysis

Note that vector [latex]{\bf{F}}(a,b)=\langle{b},-a\rangle[/latex] points clockwise and is perpendicular to radial vector [latex]\langle{a},b\rangle[/latex]. (We can verify this assertion by computing the dot product of the two vectors:[latex]\langle{a},b\rangle\cdot\langle{-b},a\rangle=-ab+ab=0[/latex].) Furthermore, vector [latex]\langle{b},-a\rangle[/latex] has length [latex]r=\sqrt{a^2+b^2}[/latex]. Thus, we have a complete description of this rotational vector field: the vector associated with point [latex](a, b)[/latex] is the vector with length [latex]r[/latex] tangent to the circle with radius [latex]r[/latex], and it points in the clockwise direction.

Sketches such as that in Figure 6 are often used to analyze major storm systems, including hurricanes and cyclones. In the northern hemisphere, storms rotate counterclockwise; in the southern hemisphere, storms rotate clockwise. (This is an effect caused by Earth’s rotation about its axis and is called the Coriolis Effect.)

To visualize this vector field, first note that the dot product [latex]{\bf{F}}(a,b)\cdot(a{\bf{i}}+b{\bf{j}})[/latex] is zero for any point [latex](a,b)[/latex]. Therefore, each vector is tangent to the circle on which it is located. Also, as [latex](a,b)\to(0,0)[/latex], the magnitude of [latex]{\bf{F}}(a,b)[/latex] goes to infinity. To see this, note that

[latex]\large{||{\bf{F}}(a,b)||=\sqrt{\dfrac{a^2+b^2}{(a^2+b^2)^2}}=\sqrt{\dfrac{1}{a^2+b^2}}}[/latex].

Since [latex]\frac1{a^2+b^2}\to\infty[/latex] as [latex](a,b)\to(0,0)[/latex] then [latex]||{\bf{F}}(a,b)||\to\infty[/latex] as [latex](a,b)\to(0,0)[/latex]. This vector field looks similar to the vector field in Example “Chapter Opener: Drawing a Rotational Vector Field”, but in this case the magnitudes of the vectors close to the origin are large. The table below shows a sample of points and the corresponding vectors, and Figure 6 shows the vector field. Note that this vector field models the whirlpool motion of the river in Figure 1(b). The domain of this vector field is all of [latex]\mathbb{R}^2[/latex] except for point [latex](0, 0)[/latex].

[latex](x, y)[/latex]	[latex]F(x, y)[/latex]	[latex](x, y)[/latex]	[latex]F(x, y)[/latex]	[latex](x, y)[/latex]	[latex]F(x, y)[/latex]
[latex](1, 0)[/latex]	[latex]\langle0,-1\rangle[/latex]	[latex](2, 0)[/latex]	[latex]\langle0,-\frac12\rangle[/latex]	[latex](1, 1)[/latex]	[latex]\langle\frac12,-\frac12\rangle[/latex]
[latex](0, 1)[/latex]	[latex]\langle1,0\rangle[/latex]	[latex](0, 2)[/latex]	[latex]\langle\frac12,0\rangle[/latex]	[latex](-1, 1)[/latex]	[latex]\langle\frac12,\frac12\rangle[/latex]
[latex](-1, 0)[/latex]	[latex]\langle0,1\rangle[/latex]	[latex](-2, 0)[/latex]	[latex]\langle0,\frac12\rangle[/latex]	[latex](-1, -1)[/latex]	[latex]\langle-\frac12,\frac12\rangle[/latex]
[latex](0, -1)[/latex]	[latex]\langle-1,0\rangle[/latex]	[latex](0, -2)[/latex]	[latex]\langle-\frac12,0\rangle[/latex]	[latex](1, -1)[/latex]	[latex]\langle-\frac12,-\frac12\rangle[/latex]

Figure 6. A visual representation of vector field [latex]\small{{\bf{F}}(x,y)=\frac{y}{x^2+y^2}{\bf{i}}-\frac{x}{x^2+y^2}{\bf{j}}}[/latex]. This vector field could be used to model whirlpool motion of a fluid.

To find the velocity of the fluid at point [latex](1, -1)[/latex], substitute the point into [latex]\bf{v}[/latex]:

[latex]\large{{\bf{v}}(1,-1)=-\frac{2(-1)}{1+1}{\bf{i}}+\frac{2(1)}{1+1}{\bf{j}}={\bf{i}}+{\bf{j}}}[/latex].

The speed of the fluid at [latex](1, -1)[/latex] is the magnitude of this vector. Therefore, the speed is [latex]||{\bf{i}}+{\bf{j}}||=\sqrt2[/latex] m/sec.

To show that [latex]\bf{F}[/latex] is a unit field, we must show that the magnitude of each vector is [latex]1[/latex]. Note that

[latex]\hspace{4cm}\large{\begin{align} \sqrt{\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2+\left(-\frac{x}{\sqrt{x^2+y^2}}\right)^2}&=\sqrt{\frac{y^2}{\sqrt{x^2+y^2}}+\frac{x^2}{\sqrt{x^2+y^2}}} \\ &=\sqrt{\frac{x^2+y^2}{x^2+y^2}} \\ &=1. \end{align}}[/latex]

Therefore, [latex]\bf{F}[/latex] is a unit vector field.