Chapter 14
Vector Analysis
14.1 Introduction and Vector Fields
Thinking Back
Work as an integral of force and distance
Since work is the application of force through a distance, the work is (using integration by parts with
u = x and dv = sin x dx)
Z π/2 Z π/2
x sin x dx = [− x cos x ]0π/2 + cos x dx = 0 + [sin x ]0π/2 = 1.
0 0
Vector geometry
√
. Since F( x, y, z) = z( x − y)2 i + y j + k, we have
√ √ √ D √ E
F(3, 3, 13) = 13(3 − 3)2 i + 3j + k = 0, 3, 1 .
√ D √ E
The norm of this vector is 0 + 3 + 1 = 2, so a unit vector in this direction is 0, 23 , 1
2 .
D √ E
. A vector equation for the line is h0, 0, 0i + t 0, 23 , 1
2 .
Calculus of vector-valued functions
. To differentiate, we differentiate each component:
1 − t2
d d d d t
(r(t)) = (3 cos2 t)i + (5t)j + k = −6 sin t cos ti + 5j + k.
dt dt dt dt t2 + 1 (1 + t2 )2
. To integrate, we integrate each component:
Z Z Z
1
Z
r(t) dt = et dt i + t3 dt j + −4 dt k = et i + t4 j − 4tk + (C1 i + C2 j + C3 k).
4
,704 Vector Analysis
Concepts
1. (a) False. A vector field is a function whose outputs are vectors. See Definition 14.1.
(b) True. See Definition 14.1.
(c) False. A vector field in R2 has as inputs points of R2 , while a vector field in R3 has as inputs
points of R3 . See Definition 14.1.
(d) True. See part (c), and Definition 14.1.
(e) True. If f is a potential function for F, then so is f + α for any real number α, since ∇( f + α) =
∇ f + ∇ α = ∇ f = F.
(f) False. A vector field F is a gradient if and only if it is conservative (Definition 14.2), which is
true if and only if (in the case of a vector field on R2 ) ∂F 1 ∂F2
∂y = ∂x . The analogous condition for
vector fields on R3 is more complicated.
(g) False. f and f + α for α ∈ R always have the same gradient, since ∇( f + α) = ∇ f + ∇ α =
∇ f.
(h) True. This is the definition of work; see Definition 6.12 and Example 1 in Section 6.4.
2. (a) For example, in R2 , the vector field F( x, y) = h x, 0i:
3
2
1
0
-1
-2
-3
-3 -2 -1 0 1 2 3
For example, in R3 , the vector field F( x, y, z) = h1, 1, 1i:
,14.1 Introduction and Vector Fields 705
2
y
0
-2
2
z
0
-2
-2
0
x
2
∂F1 ∂F2 ∂y
(b) The vector field h x, yi is conservative, since ∂y = ∂x
∂y =0= ∂x = ∂x . A potential function
1 2 1 2
for this vector field is 2x + 2y .
∂x2
(c) The vector field y, x2 is not conservative since ∂F1 ∂y ∂F2
∂y = ∂y = 1 while ∂x = ∂x = 2x.
3. By Definition 14.1, a vector field in the Cartesian plane, which is R2 , is a function from points in
R2 to vectors in R2 . So its inputs are points in the Cartesian plane (or R2 ).
4. By Definition 14.1, a vector field in R3 is a function from points in R3 to vectors in R3 . So its inputs
are points in R3 .
5. By Definition 14.1, a vector field in the Cartesian plane, which is R2 , is a function from points in
R2 to vectors in R2 . So its outputs are vectors in R2 .
6. By Definition 14.1, a vector field in R3 is a function from points in R3 to vectors in R3 . So its
outputs are vectors in R3 .
7. A vector field is conservative if it is the gradient of some function. So, for example, a vector field
∂f
h F1 ( x, y), F2 ( x, y)i in R2 is conservative if there is some function f ( x, y) such that ∂x = F1 and
∂f
∂y = F2 .
8. They point away from the origin. For example at ( x, y) = (2, 2), the vector is h2, 2i, which points
up and to the right, which is away from the origin at (2, 2). As another example, at ( x, y) = (2, −2),
the vector is h2, −2i, which points down and to the right, which is away from the origin at (2, −2).
9. The vectors F( x, y) and G( x, y) at any point ( x, y) point in the same direction, since h2x, 2yi =
2 h x, yi. But the vectors in F( x, y) are twice as long at each point as the vectors of G( x, y).
10. Since G( x, y, z) = −F( x, y, z), the vectors of G are the same length at each point as those in F, but
point in the opposite direction.
11. Since F( x, y) = −G( x, y) for any point ( x, y), the vectors of F are the same length at each point as
those in G, but point in the opposite direction.
12. Since G( x, y, z) = −F( x, y, z), the vectors of G are the same length at each point as those in F, but
point in the opposite direction.
, 706 Vector Analysis
13. We have
2F( x, y, z) = 2 hyz, xz, xyi = h2yz, 2xz, 2xyi ,
so let G( x, y, z) = h2yz, 2xz, 2xyi.
14. F is a vector field in R2 , while G is a vector field in R3 . At any point ( x, y, z) ∈ R3 , G( x, y, z) is
parallel to G( x, y, 0), since they are both equal to x2 , y2 , 0 . Regarding the xy plane as R2 , we see
that this vector is just F( x, y). Thus G( x, y, z) consists of an infinite number of copies of F( x, y)
stacked on top of each other, one for each z ∈ R.
∂F1 ∂F2
15. If F = h F1 ( x, y), F2 ( x, y)i, then if ∂y 6= ∂x , the vector field is not conservative. This follows from
∂f
the fact that if it were conservative, then there would be some function f ( x, y) with ∂x = F1 and
∂f ∂2 f ∂F1 ∂2 f ∂F2
∂y = F2 . Since mixed partials are equal, we would have ∂x ∂y = ∂y = ∂y ∂x = ∂x .
16. If F = h F1 ( x, y, z), F2 ( x, y, z), F3 ( x, y, z)i, then F is not conservative if any of the following condi-
tions fail to hold:
∂F1 ∂F ∂F1 ∂F ∂F2 ∂F
= 2, = 3, = 3.
∂y ∂x ∂z ∂x ∂z ∂y
This follows from the fact that if it were conservative, then there would be some function f ( x, y, z)
∂f ∂f
with ∇ f = F. Thus for example ∂x = F1 and ∂y = F2 ; equality of mixed partials would then imply
the truth of the first equality above.
Skills
17. Using the procedure in the text, integrate the first component with respect to x:
Z
3x2 cos y dx = x3 cos y + α.
Since the only term the second component also depends on x, no further work is required, so
x3 cos y + α is a potential function for any α ∈ R. As a check, ∂y x3 cos y + α = − x3 sin y, which
∂
is the second component of the vector field.
18. Using the procedure in the text, integrate the first component with respect to x:
Z
ey sec2 x dx = ey tan x + α.
Since the only term in the second component also depends on x, no further work is required, so
ey tan x + α is a potential function for any α ∈ R. As a check, ∂y
∂
(ey tan x + α) = ey tan x, which is
the second component of the vector field.
19. Using the procedure in the text, integrate the first component with respect to x:
Z
(5x4 + y) dx = x5 + xy + α.
The term −12y3 in the second component does not depend on x, so we must integrate it as well,
with respect to y: Z
(−12y3 ) dy = −3y4 + β.
Add these results and set α + β = γ, to give x5 + xy − 3y4 + γ as a potential function for any
γ ∈ R. As a check,
∂ 5 ∂ 5
( x + xy − 3y4 + γ) = 5x4 + y, ( x + xy − 3y4 + γ) = x − 12y3 .
∂x ∂y
Vector Analysis
14.1 Introduction and Vector Fields
Thinking Back
Work as an integral of force and distance
Since work is the application of force through a distance, the work is (using integration by parts with
u = x and dv = sin x dx)
Z π/2 Z π/2
x sin x dx = [− x cos x ]0π/2 + cos x dx = 0 + [sin x ]0π/2 = 1.
0 0
Vector geometry
√
. Since F( x, y, z) = z( x − y)2 i + y j + k, we have
√ √ √ D √ E
F(3, 3, 13) = 13(3 − 3)2 i + 3j + k = 0, 3, 1 .
√ D √ E
The norm of this vector is 0 + 3 + 1 = 2, so a unit vector in this direction is 0, 23 , 1
2 .
D √ E
. A vector equation for the line is h0, 0, 0i + t 0, 23 , 1
2 .
Calculus of vector-valued functions
. To differentiate, we differentiate each component:
1 − t2
d d d d t
(r(t)) = (3 cos2 t)i + (5t)j + k = −6 sin t cos ti + 5j + k.
dt dt dt dt t2 + 1 (1 + t2 )2
. To integrate, we integrate each component:
Z Z Z
1
Z
r(t) dt = et dt i + t3 dt j + −4 dt k = et i + t4 j − 4tk + (C1 i + C2 j + C3 k).
4
,704 Vector Analysis
Concepts
1. (a) False. A vector field is a function whose outputs are vectors. See Definition 14.1.
(b) True. See Definition 14.1.
(c) False. A vector field in R2 has as inputs points of R2 , while a vector field in R3 has as inputs
points of R3 . See Definition 14.1.
(d) True. See part (c), and Definition 14.1.
(e) True. If f is a potential function for F, then so is f + α for any real number α, since ∇( f + α) =
∇ f + ∇ α = ∇ f = F.
(f) False. A vector field F is a gradient if and only if it is conservative (Definition 14.2), which is
true if and only if (in the case of a vector field on R2 ) ∂F 1 ∂F2
∂y = ∂x . The analogous condition for
vector fields on R3 is more complicated.
(g) False. f and f + α for α ∈ R always have the same gradient, since ∇( f + α) = ∇ f + ∇ α =
∇ f.
(h) True. This is the definition of work; see Definition 6.12 and Example 1 in Section 6.4.
2. (a) For example, in R2 , the vector field F( x, y) = h x, 0i:
3
2
1
0
-1
-2
-3
-3 -2 -1 0 1 2 3
For example, in R3 , the vector field F( x, y, z) = h1, 1, 1i:
,14.1 Introduction and Vector Fields 705
2
y
0
-2
2
z
0
-2
-2
0
x
2
∂F1 ∂F2 ∂y
(b) The vector field h x, yi is conservative, since ∂y = ∂x
∂y =0= ∂x = ∂x . A potential function
1 2 1 2
for this vector field is 2x + 2y .
∂x2
(c) The vector field y, x2 is not conservative since ∂F1 ∂y ∂F2
∂y = ∂y = 1 while ∂x = ∂x = 2x.
3. By Definition 14.1, a vector field in the Cartesian plane, which is R2 , is a function from points in
R2 to vectors in R2 . So its inputs are points in the Cartesian plane (or R2 ).
4. By Definition 14.1, a vector field in R3 is a function from points in R3 to vectors in R3 . So its inputs
are points in R3 .
5. By Definition 14.1, a vector field in the Cartesian plane, which is R2 , is a function from points in
R2 to vectors in R2 . So its outputs are vectors in R2 .
6. By Definition 14.1, a vector field in R3 is a function from points in R3 to vectors in R3 . So its
outputs are vectors in R3 .
7. A vector field is conservative if it is the gradient of some function. So, for example, a vector field
∂f
h F1 ( x, y), F2 ( x, y)i in R2 is conservative if there is some function f ( x, y) such that ∂x = F1 and
∂f
∂y = F2 .
8. They point away from the origin. For example at ( x, y) = (2, 2), the vector is h2, 2i, which points
up and to the right, which is away from the origin at (2, 2). As another example, at ( x, y) = (2, −2),
the vector is h2, −2i, which points down and to the right, which is away from the origin at (2, −2).
9. The vectors F( x, y) and G( x, y) at any point ( x, y) point in the same direction, since h2x, 2yi =
2 h x, yi. But the vectors in F( x, y) are twice as long at each point as the vectors of G( x, y).
10. Since G( x, y, z) = −F( x, y, z), the vectors of G are the same length at each point as those in F, but
point in the opposite direction.
11. Since F( x, y) = −G( x, y) for any point ( x, y), the vectors of F are the same length at each point as
those in G, but point in the opposite direction.
12. Since G( x, y, z) = −F( x, y, z), the vectors of G are the same length at each point as those in F, but
point in the opposite direction.
, 706 Vector Analysis
13. We have
2F( x, y, z) = 2 hyz, xz, xyi = h2yz, 2xz, 2xyi ,
so let G( x, y, z) = h2yz, 2xz, 2xyi.
14. F is a vector field in R2 , while G is a vector field in R3 . At any point ( x, y, z) ∈ R3 , G( x, y, z) is
parallel to G( x, y, 0), since they are both equal to x2 , y2 , 0 . Regarding the xy plane as R2 , we see
that this vector is just F( x, y). Thus G( x, y, z) consists of an infinite number of copies of F( x, y)
stacked on top of each other, one for each z ∈ R.
∂F1 ∂F2
15. If F = h F1 ( x, y), F2 ( x, y)i, then if ∂y 6= ∂x , the vector field is not conservative. This follows from
∂f
the fact that if it were conservative, then there would be some function f ( x, y) with ∂x = F1 and
∂f ∂2 f ∂F1 ∂2 f ∂F2
∂y = F2 . Since mixed partials are equal, we would have ∂x ∂y = ∂y = ∂y ∂x = ∂x .
16. If F = h F1 ( x, y, z), F2 ( x, y, z), F3 ( x, y, z)i, then F is not conservative if any of the following condi-
tions fail to hold:
∂F1 ∂F ∂F1 ∂F ∂F2 ∂F
= 2, = 3, = 3.
∂y ∂x ∂z ∂x ∂z ∂y
This follows from the fact that if it were conservative, then there would be some function f ( x, y, z)
∂f ∂f
with ∇ f = F. Thus for example ∂x = F1 and ∂y = F2 ; equality of mixed partials would then imply
the truth of the first equality above.
Skills
17. Using the procedure in the text, integrate the first component with respect to x:
Z
3x2 cos y dx = x3 cos y + α.
Since the only term the second component also depends on x, no further work is required, so
x3 cos y + α is a potential function for any α ∈ R. As a check, ∂y x3 cos y + α = − x3 sin y, which
∂
is the second component of the vector field.
18. Using the procedure in the text, integrate the first component with respect to x:
Z
ey sec2 x dx = ey tan x + α.
Since the only term in the second component also depends on x, no further work is required, so
ey tan x + α is a potential function for any α ∈ R. As a check, ∂y
∂
(ey tan x + α) = ey tan x, which is
the second component of the vector field.
19. Using the procedure in the text, integrate the first component with respect to x:
Z
(5x4 + y) dx = x5 + xy + α.
The term −12y3 in the second component does not depend on x, so we must integrate it as well,
with respect to y: Z
(−12y3 ) dy = −3y4 + β.
Add these results and set α + β = γ, to give x5 + xy − 3y4 + γ as a potential function for any
γ ∈ R. As a check,
∂ 5 ∂ 5
( x + xy − 3y4 + γ) = 5x4 + y, ( x + xy − 3y4 + γ) = x − 12y3 .
∂x ∂y
