James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1070 Chapter 16 Vector Calculus
It appears from Figure 5 that each arrow is tangent to a circle with center the origin.
To confirm this, we take the dot product of the position vector x − x i 1 y j with the
vector Fsxd − Fsx, yd:
x Fsxd − sx i 1 y jd s2y i 1 x jd − 2xy 1 yx − 0
This shows that Fsx, yd is perpendicular to the position vector kx, yl and is therefore
tangent to a circle with center the origin and radius | x | − sx 2 1 y 2 . Notice also that
| Fsx, yd | − ss2yd2 1 x 2 − sx 2 1 y 2 − | x |
so the magnitude of the vector Fsx, yd is equal to the radius of the circle.
Some computer algebra systems are capable of plotting vector fields in two or three
dimensions. They give a better impression of the vector field than is possible by hand
because the computer can plot a large number of representative vectors. Figure 6 shows a
computer plot of the vector field in Example 1; Figures 7 and 8 show two other vector
fields. Notice that the computer scales the lengths of the vectors so they are not too long
and yet are proportional to their true lengths.
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5
6
5
_5 5
_6 6
_5 5
_5
FIGURE 6
F(x, y)=k_y, xl
_6
FIGURE 7
F(x, y)=ky, sin xl
_5
FIGURE 8
F(x, y)=kln(1+¥), ln(1+≈)l
Example 2 Sketch the vector field on R 3 given by Fsx, y, zd − z k.
SOLUtion The sketch is shown in Figure 9. Notice that all vectors are vertical and
point upward above the xy-plane or downward below it. The magnitude increases with
the distance from the xy-plane.
z
0
y
FIGURE 9
Fsx, y, zd − z k
x
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We were able to draw the vector field in Example 2 by hand because of its particularly
simple formula. Most three-dimensional vector fields, however, are virtually impossible
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