SECTION
17.1
Vector Fields
(ET Section 16.1)
1061
19. F
= h
x
,
0
,
z
i
SOLUTION
This vector feld is shown in (A) (by process oF elimination).
20. F
=
*
x
p
x
2
+
y
2
+
z
2
,
y
p
x
2
+
y
2
+
z
2
,
z
p
x
2
+
y
2
+
z
2
+
The unit radial vector feld is shown in plot (D), as these vectors are radial and oF uniForm length.
21. F
= h
1
,
1
,
1
i
The constant vector feld
h
1
,
1
,
1
i
is shown in plot (C).
In Exercises 22–25, fnd a potential Function For the vector feld
F
by inspection.
22. F
= h
x
,
y
i
We must fnd a Function
ϕ
(
x
,
y
)
such that
∂
∂
x
=
x
and
∂
∂
y
=
y
. We choose the Following Function:
(
x
,
y
)
=
1
2
x
2
+
1
2
y
2
.
23. F
=

ye
xy
,
xe
®
The Function
(
x
,
y
)
=
e
satisfes
∂
∂
x
=
and
∂
∂
y
=
, hence
is a potential Function For the
given vector feld.
24. F
=

yz
2
,
xz
2
,
2
xyz
®
We choose a Function
(
x
,
y
,
z
)
such that
∂
∂
x
=
2
,
∂
∂
y
=
2
,
∂
∂
z
=
The Function
(
x
,
y
,
z
)
=
2
is a potential Function For the given feld.
25. F
=

2
xze
x
2
,
0
,
e
x
2
®
The Function
(
x
,
y
,
z
)
=
ze
x
2
satisfes
∂
∂
y
=
0,
∂
∂
x
=
2
x
2
and
∂
∂
z
=
e
x
2
, hence
is a potential
Function For the given vector feld.
26.
±ind potential Functions For the vector felds
F
=
e
r
r
3
and
G
=
e
r
r
4
in
R
3
.
We use the gradient oF
r
,
∇
r
=
e
r
, and the Chain Rule For Gradients to write
∇
µ
−
1
2
r
−
2
¶
=
r
−
3
∇
r
=
r
−
3
e
r
=
e
r
r
3
=
F
∇
µ
−
1
3
r
−
3
¶
=
r
−
4
∇
r
=
r
−
4
e
r
=
e
r
r
4
=
G
ThereFore
1
(
r
)
=−
1
2
r
2
and
2
(
r
)
1
3
r
3
are potential Functions For
F
and
G
, respectively.