SCHAUM'S OUTLINE SERIES
THEORY AND PROBLEMS OF
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P R O S
SCHAUM'S OUTLINE SERIES
McGRAW-rilLL BOOK
COMPANY
C o p y r i g h t © 1971 by M c G r a w - H i l l , I n c . A l l R i g h t s R e s e r v e d . P r i n t e d i n the
U n i t e d States of A m e r i c a . No p a r t of this publication m a y be reproduced,
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07-060216-6
2 3 4 5 6 7 8 9 0
SHSH
7 5 4 3 2 1
Chapter
5
Vector Analysis
VECTORS A N D
There
are quantities i n physics characterized
magnitude
force
and
SCALARS
and
direction, such
acceleration.
To
as
describe such
i n t r o d u c e t h e c o n c e p t o f a vector
called
t h e terminal
point.
length
both
velocity,
quantities,
We
we
point t o a n o t h e r p o i n t
denote vectors b y
faced letters or letters w i t h an a r r o w over them.
is d e n o t e d b y A o r A
by
as a d i r e c t e d l i n e s e g m e n t
PQ f r o m one p o i n t P c a l l e d t h e initial
Q
displacement,
as i n F i g . 5 - 1 .
The
Thus
magnitude
bold
PQ
or
Fig.5-1
o f t h e v e c t o r i s t h e n d e n o t e d b y \PQ\, PQ, |A| o r \A\.
O t h e r q u a n t i t i e s i n p h y s i c s a r e c h a r a c t e r i z e d b y m a g n i t u d e o n l y , s u c h as m a s s , l e n g t h
and temperature.
S u c h q u a n t i t i e s a r e o f t e n c a l l e d scalars
to distinguish t h e m f r o m
v e c t o r s , b u t i t m u s t be e m p h a s i z e d t h a t a p a r t f r o m u n i t s s u c h as f e e t , d e g r e e s , etc.,
they are n o t h i n g more t h a n real numbers.
W e can t h u s denote t h e m b y o r d i n a r y letters
as u s u a l .
VECTOR
ALGEBRA
The operations of a d d i t i o n , s u b t r a c t i o n and m u l t i p l i c a t i o n f a m i l i a r i n the algebra of numbers are, w i t h suitable
definition, capable of extension to an algebra of vectors.
The f o l l o w i n g definitions are fundamental.
1.
T w o v e c t o r s A a n d B a r e equal i f t h e y h a v e t h e
same m a g n i t u d e a n d d i r e c t i o n regardless of t h e i r
i n i t i a l p o i n t s . T h u s A = B i n F i g . 5-1 above.
2.
A vector h a v i n g d i r e c t i o n opposite to t h a t of vector
A b u t w i t h t h e s a m e m a g n i t u d e is d e n o t e d b y — A
[see F i g . 5 - 2 ] .
3.
T h e sum o r resultant o f v e c t o r s A a n d B o f F i g . 5-3(a) b e l o w i s a v e c t o r C f o r m e d
by p l a c i n g the i n i t i a l p o i n t of B on the t e r m i n a l p o i n t of A a n d j o i n i n g the
i n i t i a l p o i n t o f A t o t h e t e r m i n a l p o i n t o f B [see F i g . 5-3(b) b e l o w ] .
The sum C
is w r i t t e n C = A + B . T h e d e f i n i t i o n h e r e is e q u i v a l e n t t o t h e parallelogram
law
f o r v e c t o r a d d i t i o n as i n d i c a t e d i n F i g . 5-3(c) b e l o w .
A + B
(b)
Fig.5-3
121
Fig.5-2
VECTOR
122
[CHAP. 5
ANALYSIS
Extensions t o sums of more t h a n t w o vectors
Fig.
5-4
below
shows
how
to
obtain
the
sum
are immediate.
or
resultant
For
E
of
example,
the
vectors
A, B, C and D.
F i g . 5-4
4.
T h e difference
of vectors A a n d B , represented b y A - B , is t h a t vector C w h i c h
added to B gives A .
Equivalently, A - B
m a y be d e n n e d as A + ( — B ) .
t h e n A - B i s d e n n e d as t h e null o r zero vector
If
A = B,
a n d i s r e p r e s e n t e d b y t h e s y m b o l 0.
T h i s has a m a g n i t u d e of zero b u t i t s d i r e c t i o n is n o t denned.
5.
M u l t i p l i c a t i o n of a vector
A by a scalar m produces
a vector
mA
with
magni-
t u d e \m\s t h e m a g n i t u d e o f A a n d d i r e c t i o n t h e s a m e as o r o p p o s i t e t o t h a t
o f A a c c o r d i n g as m i s p o s i t i v e o r n e g a t i v e .
L A W S OF VECTOR
If
m = 0,
mA = 0,
the null
vector.
ALGEBRA
I f A , B and C are vectors, and m and n are scalars, t h e n
Commutative L a w for Addition
1.
A + B
= B + A
2.
A + ( B + C) =
3.
ra(nA)
4.
(m + n)A =
m A + nA
Distributive L a w
5.
m(A + B ) =
mA + m B
Distributive L a w
Associative L a w f o r A d d i t i o n
(A + B) + C
= (mn)A
=
Associative L a w f o r Multiplication
w(raA)
N o t e t h a t i n these l a w s o n l y m u l t i p l i c a t i o n o f a v e c t o r
d e f i n e d . O n p a g e s 123 a n d 124 w e d e f i n e p r o d u c t s o f v e c t o r s .
UNIT
by
one
or
more
scalars
is
VECTORS
U n i t vectors are vectors h a v i n g u n i t length.
I f A is a n y vector w i t h length
t h e n A / A i s a u n i t v e c t o r , d e n o t e d b y a, h a v i n g t h e s a m e d i r e c t i o n as A .
RECTANGULAR UNIT
Then
VECTORS
The rectangular u n i t vectors
i, j a n d k are u n i t
vec-
t o r s h a v i n g t h e d i r e c t i o n o f t h e p o s i t i v e x, y a n d z a x e s
o f a r e c t a n g u l a r c o o r d i n a t e s y s t e m [see F i g . 5 - 5 ] . W e use
r i g h t - h a n d e d r e c t a n g u l a r coordinate systems unless o t h e r w i s e specified.
Such systems derive t h e i r name f r o m the
f a c t t h a t a r i g h t t h r e a d e d screw r o t a t e d t h r o u g h 90° f r o m
Ox t o Oy w i l l a d v a n c e i n t h e p o s i t i v e z d i r e c t i o n . I n g e n e r a l ,
p j . 5.5
g
A > 0,
A = Aa.
VECTOR
C H A P . 5]
123
ANALYSIS
three vectors A , B and C w h i c h have coincident i n i t i a l points and are not coplanar are said
t o f o r m a right-handed
system
o r dextral
i f a r i g h t threaded screw rotated t h r o u g h
system
a n a n g l e less t h a n 1 8 0 ° f r o m A t o B w i l l a d v a n c e i n t h e d i r e c t i o n C [see F i g . 5-6
below].
0
AdJ
A-,k
A:)
Fig.5-7
Fig.5-6
COMPONENTS OF A
VECTOR
A n y v e c t o r A i n 3 d i m e n s i o n s c a n be r e p r e s e n t e d w i t h
o f a r e c t a n g u l a r c o o r d i n a t e s y s t e m [see F i g . 5-7 a b o v e ] .
coordinates of the t e r m i n a l p o i n t of vector A w i t h
A z j a n d A k a r e c a l l e d t h e rectangular
component
3
o f A i n t h e x, y a n d z d i r e c t i o n s r e s p e c t i v e l y .
components,
o r s i m p l y components,
initial point at the origin
i n i t i a l p o i n t a t 0.
An a n d A
u
The vectors
o r s i m p l y component
vectors,
A
0
L e t ( A i , At, As) be t h e r e c t a n g u l a r
3
are called the
Ad,
vectors,
rectangular
o f A i n t h e x, y a n d z d i r e c t i o n s r e s p e c t i v e l y .
T h e s u m o r r e s u l t a n t o f Ad, A j a n d A k is t h e v e c t o r A , so t h a t w e c a n
2
3
A
=
=
|A| =
write
Ad + A j + A k
3
(1)
^AT+AT+A
(2)
2
T h e m a g n i t u d e o f A is
I n p a r t i c u l a r , t h e position
written
o r radius
vector
r
and has m a g n i t u d e
r = j r j = yjx
DOT OR SCALAR
PRODUCT
2
=
vector
r from 0
t o t h e p o i n t (x, y, z)
xi + yj + zk
is
(3)
+ y + z.
2
2
T h e dot or scalar p r o d u c t of t w o vectors A and B, denoted by A • B (read A dot B)
is d e f i n e d as t h e p r o d u c t o f t h e m a g n i t u d e s o f A a n d B a n d t h e c o s i n e o f t h e a n g l e
between t h e m . I n symbols,
A-B
=
AB cose,
0 g e S -
N o t e t h a t A - B is a s c a l a r a n d n o t a v e c t o r .
The f o l l o w i n g laws are valid:
1.
A-B
=
B-A
Commutative L a w for Dot Products
2.
A • ( B + C) =
3.
m ( A • B) =
4.
i'i =
A •B + A • C
(mA) • B =
i ' j - k - k
=
l ,
Distributive L a w
A • (mB) =
i'} =
(A • B)m,
j ' k =
k«i
=
w h e r e m is a s c a l a r .
0
(4)
VECTOR
124
5.
If A = A J + A j + A k
2
and
3
B = B i i + 5 j + Bak, t h e n
2
A-B
=
AiBi
A-A
=
A
B •B
6.
If
A -B = 0
[CHAP. S
ANALYSIS
+ AaBa
2
— A\ At + Al
2
=
+ AtB
B°- =
B\ Bl + B\
and A and B are not null vectors, then A
CROSS OR V E C T O R
and B are perpendicular.
PRODUCT
T h e cross or v e c t o r
p r o d u c t of
A
and B
is a v e c t o r
C = A x B
(read A
cross
B).
T h e m a g n i t u d e o f A x B is d e n n e d as t h e p r o d u c t o f t h e m a g n i t u d e s o f A a n d B a n d t h e
sine of t h e angle between t h e m .
to the plane of A
S
y
m
b
°
l
8
T h e d i r e c t i o n of the vector
a n d B a n d such
'
A x B
that A, B
=
C = A x B
AB s i n 9 u ,
In
(5)
0^9
If
w h e r e u is a u n i t vector i n d i c a t i n g the d i r e c t i o n of A X B
t o B , t h e n s i n 9 = 0 a n d w e define
is p e r p e n d i c u l a r
and C f o r m a right-handed system.
A = B
o r i f A is p a r a l l e l
A x B = 0.
The f o l l o w i n g laws are valid:
1.
A x B
2.
A x ( B + C) =
3.
m ( A x B) =
4.
i x i =
5.
If
A
=
- B x A
( C o m m u t a t i v e L a w f o r Cross P r o d u c t s Fails)
A x B + A x C
(mA) x B
j x j = k x k
=
=
=
Distributive L a w
A x (mB)
0,
A i i + A j + Ask
and
2
=
i X j =
B
(A x B)m,
k,
=
j x k
|A x B| =
If
TRIPLE
i,
k x i =
P a + B j + Bak,
2
i
A X B
w h e r e m is a s c a l a r .
=
=
j
then
k
]
Ai
A
2
Aa
Bi
B%
Bz
t h e a r e a o f a p a r a l l e l o g r a m w i t h sides A a n d B .
A x B = 0 and A and B are not null vectors, t h e n A and B are parallel.
PRODUCTS
D o t a n d cross m u l t i p l i c a t i o n o f t h r e e v e c t o r s A , B a n d C m a y p r o d u c e m e a n i n g f u l
p r o d u c t s o f t h e f o r m ( A - B ) C , A - ( B x C) a n d A x ( B x C ) . T h e f o l l o w i n g l a w s a r e v a l i d :
1.
(A-B)C + A(B-C)
2.
A-(BxC)
=
i n general
B - ( C x A)
=
C-(AxB)
=
volume of a parallelepiped h a v i n g A , B ,
a n d C as edges, o r t h e n e g a t i v e o f t h i s v o l u m e a c c o r d i n g
do n o t f o r m a r i g h t - h a n d e d s y s t e m .
and
C =
Cii + C j + Csk,
2
A x ( B x C) ¥• ( A x B ) x C
4.
A x (BxC)
(AxB)xC
=
=
(A-C)B (A-C)B-
=
as A , B
A J + A j + A k, B
2
3
=
a n d C do
B\ BA +
or
B;k
then
A-(BxC)
3.
If A
=
Ai
A2
A
Bi
B%
B3
Ci
C
C3
2
3
(6)
(Associative L a w f o r Cross P r o d u c t s Fails)
(A-B)C
(B-C)A
T h e p r o d u c t A - ( B x C ) is s o m e t i m e s c a l l e d t h e scalar triple
a n d m a y be d e n o t e d b y [ A B C ] .
T h e p r o d u c t A x ( B x C ) is
product.
product
o r box
c a l l e d t h e vector
product
triple
VECTOR
C H A P . 5]
I n A • ( B x C) p a r e n t h e s e s
parentheses
m u s t be
( A x B ) • C.
used
ANALYSIS
125
are sometimes omitted and we
in A x ( B x C )
(see
Problem
w r i t e A • B x C.
5.25).
Note
that
a function
2
+
=
5.22].
FUNCTIONS
I f corresponding
A (u)j
A-(BxC)
T h i s is o f t e n expressed b y s t a t i n g t h a t i n a scalar t r i p l e p r o d u c t t h e d o t a n d
t h e cross c a n be i n t e r c h a n g e d w i t h o u t a f f e c t i n g t h e r e s u l t [see P r o b l e m
VECTOR
However,
t o e a c h v a l u e o f a s c a l a r u w e a s s o c i a t e a v e c t o r A , t h e n A is c a l l e d
of u denoted
b y A(u).
I n three dimensions
we
can
write
=
A(u)
A,(w)i +
A (u)k.
3
T h e f u n c t i o n c o n c e p t is e a s i l y e x t e n d e d .
T h u s i f t o e a c h p o i n t (x, y, z) t h e r e c o r r e s p o n d s a v e c t o r A , t h e n A is a f u n c t i o n o f (x, y, z), i n d i c a t e d b y A(x, y, z) — Ai(x,y,z)\
A ( x , y, z)j + A»{x, y, z ) k .
a
W e s o m e t i m e s say t h a t a v e c t o r f u n c t i o n A(x, y, z) defines a vector
a t e s a v e c t o r w i t h each p o i n t o f a r e g i o n .
S i m i l a r l y 6(x,y,z)
field s i n c e i t a s s o c i -
defines a scalar
field s i n c e i t
associates a scalar w i t h each p o i n t of a r e g i o n .
LIMITS, C O N T I N U I T Y A N D D E R I V A T I V E S OF VECTOR
FUNCTIONS
L i m i t s , c o n t i n u i t y and derivatives of vector f u n c t i o n s f o l l o w rules s i m i l a r to those
f o r scalar f u n c t i o n s already considered.
The following statements show the analogy
w h i c h exists.
1. T h e v e c t o r f u n c t i o n A(u) is s a i d t o be continuous
a t uo i f g i v e n a n y p o s i t i v e n u m b e r e, w e c a n find s o m e p o s i t i v e n u m b e r 8 s u c h t h a t \A(u) — A(u )\ i
whenever
\u — uo\ 8.
T h i s is e q u i v a l e n t t o t h e s t a t e m e n t l i m A(u) =
A(u ).
0
0
2. T h e d e r i v a t i v e of A(u)
is d e f i n e d as
dA
du
provided this l i m i t exists.
A
A(u + Au) — A(u)
Mi
I n case A(u) = A (u)i
+ A (u)j
1
dA
du
_
~
dAi .
du
2
3. I f A(x, y, z) = Ai(x, y, z)i + A (x,
dA
etc., c a n be s i m i l a r l y
s
=
-^dx
+
then
3
dAs
du
J
y, z)j + A (x,
2
(7)
+ A (u)k;
2
dA-i .
du
'
1
H i g h e r d e r i v a t i v e s s u c h as cPA/du ,
is t h e differential
..
„_o
y, z)k,
then
—dy
+
defined.
—dz
(8)
of A .
4. D e r i v a t i v e s o f p r o d u c t s o b e y r u l e s s i m i l a r t o t h o s e f o r s c a l a r f u n c t i o n s .
w h e n cross p r o d u c t s a r e i n v o l v e d t h e o r d e r m a y be i m p o r t a n t .
w
»<•*>
-
* s
icj
|(AXB)
=
A x f
+
^
+
A
§
-
x
B
However,
Some examples are:
VECTOR
126
GEOMETRIC
INTERPRETATION
OF
A VECTOR
I f r is t h e v e c t o r j o i n i n g t h e o r i g i n 0 of
c o o r d i n a t e s y s t e m a n d t h e p o i n t (x,y,z),
specification
(see
point
Fig.
of
5-8)
r
a space
parametric
curve
fixed
dr
(9)
is a u n i t v e c t o r i n t h e d i r e c t i o n of the
vector.
and
is
called
the
tangent
unit
tangent
F i g . 5-8
I f u i s t h e t i m e t, t h e n
dr
=
dt
is t h e velocity
(10)
v
w i t h w h i c h t h e t e r m i n a l p o i n t of r d e s c r i b e s t h e c u r v e .
dr
dt
ds
dt
dr ds
ds dt
_
T
dh_
2
(U)
is v =
ds/dt.
Similarly,
(12)
i s t h e acceleration
w i t h w h i c h the t e r m i n a l p o i n t of r describes the
have i m p o r t a n t applications i n
mechanics.
AND
have
a
dt
DIVERGENCE
We
vT
f r o m w h i c h w e see t h a t t h e m a g n i t u d e o f v , o f t e n c a l l e d t h e speed,
GRADIENT,
r(u)
then
ds
curve
—
I f the parameter u
z = z(u).
p o i n t on the curve,
Ar = r(« + Am)
equations
is t h e a r c l e n g t h s m e a s u r e d f r o m s o m e
the
a
then
A s u changes, the
describes
having
x = x(u), y = y(u),
to
DERIVATIVE
o f t h e v e c t o r f u n c t i o n r(u) defines
x, y a n d z as f u n c t i o n s o f u.
terminal
[CHAP. 5
ANALYSIS
curve,
These concepts
CURL
Consider the vector operator V
(del) d e f i n e d
=
V
i f
dx
+
by
j f
dy
+
k f
dz
v
T h e n i f <]>(x, y, z) a n d A.(x, y, z) h a v e c o n t i n u o u s first p a r t i a l d e r i v a t i v e s i n a r e g i o n (a
t i o n w h i c h i s i n m a n y cases s t r o n g e r t h a n n e c e s s a r y ) , w e c a n d e f i n e t h e f o l l o w i n g .
1. G r a d i e n t .
T h e gradient
of
is d e f i n e d
d<f>.
—l
+
dx
=
T h e divergence
div A
=
dij> . , d<p
T^J + ^ - k
dy
dz
o f A is d e f i n e d
V •A
=
condi-
by
A n i n t e r e s t i n g i n t e r p r e t a t i o n is t h a t i f <f>(x, y,z) = c
t h e n V?> is a n o r m a l t o t h i s s u r f a c e (see P r o b l e m 5.36).
2. D i v e r g e n c e .
(IS)
'
,
of a surface,
by
( i ^ + i ^
6Ai
is t h e e q u a t i o n
dA
2
* (Aii + Aaj+Ask)
,
dAi
(15)
VECTOR
C H A P . 5]
3. C u r l .
T h e curl
curlA
of A is denned
=
V X A
=
by
i
J
k
A
A
A
dx
dy
dz
A
A
Ai
d
Note
that
i n the expansion
^ + k ^ ) X (Aii + A * j + A s k )
( i ^ + J
_
p r e c e d e Ai, A ,
127
ANALYSIS
3
2
dy
d
dz
A
A
2
a
a
a
dX
az
3a;
ay
Ai
A
Ai
A
d_
-
i
3
/aAs_aA2\
/AAi
[dy
[dz
of
dz J
the
(16)
+
k
2
8A \
a
dx J
determinant, the
/flAa
8AA.
[dx
dy J
1
operators
2
d/dx, d/dy, d/dz
must
A3.
2
FORMULAS
INVOLVING V
I f the p a r t i a l derivatives of A , B, U and V are assumed to exist, t h e n
1.
V(U
=
2.
V
3.
V x (A + B)
4.
V(UA)
5.
VX(tfA)
6.
V ( A X B )
7.
V X ( A x B )
=
8.
V(A-B)
(B-V)A
9.
v ( V ^ )
+ V)
VU
(A + B)
=
+
or
t?V
V •A +
=
=
V
or
=
(B-V)A
VU
Z
+
-
V
2
=
c u r l ( A + B)
div A + div B
=
curl A +
A - ( V x B)
B ( V A )
-
(A-V)B
+ B x
Q^TJ
d^U
d^XJ
-jjjg + — j - + - ^ - j
=
a
^5
a
+ ^2
2
A(V'B)
(V x A) + A x (V
m
a
+ ^2
+
*
c a l l e d t h e Laplacian
s
=
0.
T h e c u r l o f t h e g r a d i e n t o f U is z e r o .
11.
V ' (V^A)
=
0.
T h e divergence of t h e c u r l of A is zero.
12.
V X ( V X A )
ORTHOGONAL
x
we
operator.
A
CURVILINEAR COORDINATES.
T h e transformation
[where
2
o f £/
2
VX(VLO
V ( V - A ) - V
XB)
is c a l l e d t h e Laplacian
10.
=
curl B
xA)
(A-V)B
2
and
grad u + grad V
div (A + B)
or
C7(v
B - ( V x A) -
=
=
L 7 ( V A)
A +
(W)x
=
B
V X A + V X B
(VET)'A +
=
grad(£/ + y)
JACOBIANS
equations
=
f(Ui, Ui, Us),
a s s u m e t h a t / , g, h a r e
y
=
g(ui,U2,u ),
z
3
continuous,
have
— h(ui,u ,u )
2
continuous
(17)
3
partial
derivatives
h a v e a s i n g l e - v a l u e d i n v e r s e ] e s t a b l i s h a one t o one c o r r e s p o n d e n c e b e t w e e n
xyz a n d U\U u
2
3
and
points i n an
rectangular coordinate system.
I n vector notation the t r a n s f o r m a t i o n
(17)
r
+ g(u ,U2,u )j
(18)
c a n be w r i t t e n
=
xi + yj + zk
=
f(u ,u ,u )i
1
2
3
1
3
+ h(ui,U2,u )k
3
VECTOR
128
[CHAP. 5
ANALYSIS
A p o i n t P i n F i g . 5-9 c a n t h e n be d e f i n e d n o t o n l y
b y rectangular
2
(x, y, z) b u t b y c o o r d i -
coordinates
n a t e s (Ui,u ,u )
as w e l l .
3
curvilinear
3
2
the
3
of the point.
coordinates
I f u% a n d u
W e c a l l (ui,u ,u )
are constant,
t h e n as U\,
r d e s c r i b e s a c u r v e w h i c h w e c a l l t h e Ui
S i m i l a r l y w e define t h e u
curve.
coordinate
and u
2
coordi-
3
n a t e c u r v e s t h r o u g h P.
From
(18), w e h a v e
dr
—
-~-dui
dlli
T h e v e c t o r dr/diti
in
this
write
tively.
+
we
= he
2
+
2
~-du
d«3
(19)
3
2
F i g . 5-9
I f ei is a u n i t v e c t o r a t P
i s t a n g e n t t o t h e U\e c u r v e a t P.
direction,
dr/du
-^-du
dU
2
can w r i t e
and
2
dr/du
d r / d w i = hid
where
= he,
3
3
where
3
Similarly
hi - \dr/dUi
h% — |dr/dW2.|
and
we
can
hz = \dr/du \3
T h e n (19) c a n be w r i t t e n
dr
T h e q u a n t i t i e s hi, h , h
2
I f ei, e , e
2
=
hi du e i +
2
2
+
2
a r e s o m e t i m e s c a l l e d scale
3
are m u t u a l l y perpendicular
3
c a l l e d orthogonal.
h du e
(20)
h du e
3
3
3
factors.
a t a n y p o i n t P, t h e c u r v i l i n e a r c o o r d i n a t e s a r e
I n s u c h case t h e e l e m e n t o f a r c l e n g t h ds i s g i v e n b y
ds
=
2
dr-dr
=
h\du\
h\du\
h\dul
and corresponds to the square of the length of t h e diagonal
A l s o , i n t h e case o f o r t h o g o n a l
dV
—
(21)
i n t h e above
parallelepiped.
coordinates the volume of the parallelepiped
\(hiduiei)'(h du e )
2
2
X (h du e )\
2
3
3
is g i v e n b y
hih h duidu du
3
2
3
2
3
w h i c h c a n be w r i t t e n as
dV
=
dr
dr
dr
dlli
du
du
2
duidu du
2
d(u u ,u )
u
is c a l l e d t h e Jacobian
2
z)
d(Ui,U ,U )
2
dX
dX
dX
dUi
dU
dU
d(x, y, z)
where
B(x,y,
—
3
3
3
2
dui du du
2
3
(28)
3
3
dy
dy
dy
dUi
dU
dU
2
(U)
3
dz
dz
dz
dUi
du
du
2
3
of the t r a n s f o r m a t i o n .
I t i s clear t h a t w h e n t h e J a c o b i a n is i d e n t i c a l l y zero t h e r e is n o p a r a l l e l e p i p e d .
I n such
case t h e r e i s a f u n c t i o n a l r e l a t i o n s h i p b e t w e e n x, y a n d z, i . e . t h e r e i s a f u n c t i o n 4> s u c h t h a t
<j>(x, y, z) = 0
identically.
GRADIENT, DIVERGENCE,
CURVILINEAR
CURL
AND LAPLACIAN
IN ORTHOGONAL
COORDINATES
I f $ is a scalar f u n c t i o n a n d A = A i e i + A e + A e
a vector function
c u r v i l i n e a r c o o r d i n a t e s Ui, u , u , w e h a v e t h e f o l l o w i n g r e s u l t s .
2
2
1.
V *
=
grad $
i a*
r- 7— e i
hi dUi
2.
V - A
=
div A
2
3
3
3
2
e
du
2
2
1
gj
h
du
3
e
3
3
^(h h Ai) ±-(h hiA )+±-(hih A )[
hih h
2
+ h
2
3
3
+
3
2
2
3
of
orthogonal
CHAP.
3.
VECTOR
5]
V x A
=
curl A
v*
=
he
_3_
Bui
8
Bu
hiAi
hA
2
2
hae3
2
du-3
2
2
1
hih h
Laplacian of $
2
hid
hihohs
129
ANALYSIS
ftsAa
2
r 9 fhihi 3 $
[.dui \ Bui
3
d (hih
2
du \
+
3
//l3&i 3 *
dUi \ du>.
3$
dU3
3
T h e s e r e d u c e t o t h e u s u a l e x p r e s s i o n s i n r e c t a n g u l a r c o o r d i n a t e s i f w e r e p l a c e [ui, u , Us)
2
b y (x,y,z),
i n w h i c h case e i , e
SPECIAL CURVILINEAR
and e
2
x =
where
pCOS<p,
psin<£,
y =
- «
( p . * . *)
z = z
<
< «.
Z
hi = 1 , fo = p, fes = 1
2
of arc length:
t
T, •
Jacobian:
- ha = 1.
2
See F i g . 5 - 1 0 .
(p,<j>,z).
0 ^ <f> < 2TT,
Sccde factors:
Element
hi - h
equations:
^ 0 ,
P
and
COORDINATES
1. C y l i n d r i c a l C o o r d i n a t e s
Transformation
are replaced b y i , j a n d k
s
-)
ds
— dp +
2
2
P2
d<b + dz°2
d(x,y,z)
' = p
d( , 4>, z)
P
Element
of volume:
dV =
pdpd<p dz
Laplacian:
1 3 /
V
2
[
"
/
dU\
papK^p;
!
cPU
~ p -3</>
2
dz
2
2
N o t e t h a t c o r r e s p o n d i n g r e s u l t s c a n be o b t a i n e d f o r p o l a r c o o r d i n a t e s i n t h e
p l a n e b y o m i t t i n g z d e p e n d e n c e . I n s u c h case f o r e x a m p l e , d s = d p + p d</> , w h i l e
t h e e l e m e n t o f v o l u m e is r e p l a c e d b y t h e e l e m e n t o f a r e a , dA = pdp d<p.
2
2. S p h e r i c a l C o o r d i n a t e s (r, 6,<p).
Transformation
2
2
2
See F i g . 5 - 1 1 .
equations:
P(X.V,t)
(r, a, <()
x = r s i n 9 cos <p, y = r s i n 9 s i n 4>, z = r cos 0
where
r ^ 0, 0 S 0 S n, 0 g <p < 2 T .
Sccde factors:
Element
2
Element
Laplacian:
2
of arc
ds
Jacobian:
hi = 1, h — r, h = r s i n 9
=
3
z
V
length:
dr + r d9 + r s i n 9 d<t>
2
2
2
2
2
2
J.' \ r sin 9
o(r, v, <p)
Z
of volume:
T7 U
2
2
dV =
r
2
r s i n 9 dr d9 d<j>
2
3r I
dr
1
r s i n 0 3(
2
O t h e r types of coordinate systems are possible.
sin 9
dU
4
r
2
s i n 9 3<p
2
2
[CHAP. 5
VECTOR ANALYSIS
130
Solved Problems
VECTOR
5.1.
ALGEBRA
S h o w t h a t a d d i t i o n o f v e c t o r s i s c o m m u t a t i v e , i.e. A + B = B + A .
and
Then
5.2.
OP + PQ
=
OQ
or
A + B
=
C
OR + RQ
=
OQ
or
B + A
=
C
See F i g . (a) b e l o w .
A + B = B + A.
Show t h a t the
addition
of
vectors
is a s s o c i a t i v e ,
i.e.
A + ( B + C)
=
(A + B) +
C.
See F i g . (b) a b o v e .
OP + PQ
Since
we have
=
OQ
=
and
PQ + QR
PR =
OR
=
D,
i.e.
A + (B + C )
=
D
OQ + Q R
=
OR
=
D,
i.e.
(A + B ) + C
=
D
A + (B + C) =
(B + C)
(A + B ) + C.
Problems 5.1
and 5.2
show t h a t the order of
addition of
P r o v e t h a t t h e l i n e j o i n i n g t h e m i d p o i n t o f t w o sides
o f a t r i a n g l e is p a r a l l e l t o t h e t h i r d side a n d h a s
half its length.
F r o m F i g . 5-12,
Let
and CB.
AC + CB = AB
or
b + a = c.
D E = d be the line j o i n i n g the midpoints of sides
Then
d
=
DC + CE =
£b + ^ a
=
|(b + a) =
P r o v e t h a t the m a g n i t u d e A of the v e c t o r
is
AA + Aij + Ask
F i g . 5-13.
A
=
A
yJAt+Al+At.
2
=
(OQ)
+
2
Fig.5-12
=
See
P(A,,A,,A,)
B y the P y t h a g o r e a n theorem,
(OP)
AC
-Jc
T h u s d is p a r a l l e l to c and has h a l f its length.
5.4.
=
=
E x t e n s i o n s of the results of
number of vectors is i m m a t e r i a l .
5.3.
(A + B )
OP + PR
(QP)
2
where OP denotes the magnitude of vector O P , etc.
Similarly,
Then
A
2
=
(OQ)
2
(OP)
2
= (OR)
2
= (OR)
A\ A% + A%,
2
+
(RQ) .
2
+ (RQ)
2
i.e.
A
+ (QP)
2
=
^A\
or
F i g . 5-13
any
S A P .
131
VECTOR ANALYSIS
5]
Determine
the
i / i , Zi)
and
P(xi,
find
vector
having
initial
terminal
point
Q(x2, yi, z )
its m a g n i t u d e .
See
Fig.
point
and
2
5-14.
T h e position vector of P i s r j = * i + y j + Zjk.
x
T h e position vector of Q i s r = x i + y i + z k.
2
2
r, + PQ = r
PQ
=
r, -
2
2
2
or
( x i + y£ + x£s) -
rj =
+
2
=
(x - xji + (ij 2
+ Zjk)
+ ( -i -
2
z
z
i)k
PQ
Magnitude of P Q =
V ( x - X j ) + (y. - J/]) + ( z - * i )
=
2
z
2
2
2
2
Note t h a t this i s the distance between points P a n d Q,
THE
DOT
OR
5.6.
Prove
A*b,
of
SCALAR
that
the
where
PRODUCT
projection
b
is
a
unit
of
A
on
vector
B
in
is equal
the
to
direction
B.
T h r o u g h the i n i t i a l a n d t e r m i n a l points of A pass
planes p e r p e n d i c u l a r to B a t G a n d H respectively a s i n
the a d j a c e n t F i g . 5-15; then
Projection of A on B =
5.7.
Prove
A • ( B + C)
=
GH
=
EF
A cos J =
=
A • b
A • B + A • C.
L e t a be a u n i t vector i n the direction of A ; then [see F i g . 5-161
Projection of ( B + C) on A
(B + C) • a =
=
projection of B on A + projection of C on A
B• a + C• a
M u l t i p l y i n g by A,
(B+C)-Aa
and
=
(B + C) • A
B - A a+ C - A a
=
B• A + C• A
T h e n by the commutative l a w for dot products,
A • (B + C) =
;
A • B + A • C
j
E
and the distributive l a w i s v a l i d .
5.8.
Prove that
( A + B ) • (C + D )
B y Problem 5.7,
|
F
G
A
F i g . 5-16
=
A • C + A • D + B • C + B •D.
(A + B) • (C + D) = A • (C + D ) + B • ( C + D ) = A • C + A • D + B • C + B • D .
T h e o r d i n a r y l a w s of a l g e b r a a r e v a l i d f o r dot products.
5.9.
E v a l u a t e each of the f o l l o w i n g .
(o)
i-i
(6)
i-k =
=
!i|lkjcos90
cosO
!ki [j| cos 90°
(c)
k-j =
(d)
j • (2i - 3j + k) =
(e)
(2i - j ) • (3i + k) =
D
c
If
(1)(1)(1) =
1
(1)(1)(0)
=
0
=
(1)(1)(0)
=
0
2j • i -
2
and
3
-3
3j • i -
j•k =
B = B i i + Bj
2
6 + 0 -
+ 2?sk,
0 -
prove
(A i + A j + A k) • ( B i i + S j j + B k)
=
Aii • ( B i i + Boj + B k) +A j'(B l+
=
A ^ j i - i + A , B , i - j + A ^ i - k + A.^B^ • i + A B }
i
2
3
=
2
AB
l
1
+ AB
2
2
+ A B k3
+
= A i f l i + A2B2 + AaB*
B j + B k ) + A k • ( £ i + S j + B k)
l
2
3
3
2
i - i= j -j = k*k—1
6
3
3
3
0 =
A«B
=
+ A B,k'i
since
0 - 3 + 0 =
j • (3i + k )
6i • i + 2 i • k -
A = Aii + A j + A k
A-B
3j • j + j • k =
2 i ' ( 3 i + k) -
=
5.10.
=
=
A B
3
2
j + A B k-k
3
3
3
and a l l other dot products a r e zero.
2
x
2
• j + ABj
2
a
«k
3
5.11.
[CHAP. 5
VECTOR ANALYSIS
132
If
A = AA + A j +Aak,
show that
2
A-A
=
Also,
(A)(A) c o s O
=
0
A • A
A .
Then
2
=
A = y/A-A
= y'A\ A
2
+ A .
2
A = v A • A.
7
(Aii + A j j + A j k ) • (Aji + A j j + Ajk)
(AJ)(AJ) + (A )(A ) + (A )(A )
=
2
2
3
3
=
A\+
A\+A\
by Problem 5.10, t a k i n g B = A .
Then
A = ^A • A -
\/A
THE
CROSS OR
VECTOR
5.12.
Prove A x B = - B x A.
2
+ A
2
+ A
2
is the magnitude of A .
Sometimes
A • A is written A .
2
PRODUCT
(a)
(6)
F i g . 5-17
A X B = C has magnitude AB
system [ F i g . 5-17(a) above].
s i n s and direction such t h a t A , B and C f o r m a r i g h t - h a n d e d
B X A = D has magnitude BA
system [ F i g . 5-17(6) above].
s i n e and direction such that B , A and D f o r m a r i g h t - h a n d e d
T h e n D has the same magnitude as C but is opposite in direction, i.e.
- B X A.
C = —D
T h e commutative l a w for cross products is not v a l i d .
5.13.
P r o v e t h a t A x ( B + C) = A x B +
A x C f o r t h e case w h e r e A is p e r p e n d i c u l a r t o B a n d also t o C.
Since A is perpendicular to B , A X B is a
vector perpendicular to the plane of A and B
and h a v i n g magnitude AB sin 90° = AB
or
magnitude of A B . T h i s is equivalent to m u l t i p l y i n g vector B by A and r o t a t i n g the
r e s u l t a n t vector through 90° to the position
shown in F i g . 5-18.
S i m i l a r l y , A X C i s the vector obtained
by m u l t i p l y i n g C by A and rotating the
r e s u l t a n t vector through 90° to the position
shown.
I n like m a n n e r , A X (B + C) is the vector
obtained by m u l t i p l y i n g B + C by A and
r o t a t i n g the r e s u l t a n t vector through 90° to
the position shown.
Since A X (B + C) i s the diagonal of the
p a r a l l e l o g r a m w i t h A X B and A x C as sides,
we have A X ( B + C ) = A X B
+AXC.
Fig.5-18
or
A X B =
VECTOR ANALYSIS
Prove that
A x ( B + C) = A x B +
133
A x C
i n t h e g e n e r a l case w h e r e A , B a n d C a r e
non-coplanar.
See F i g . 5-19.
Resolve B into two component vectors, one
perpendicular to A a n d the other p a r a l l e l to A ,
and denote them b y B , a n d B . respectively.
Then B = B + B .
Bi
{
±
A
\
M
I f 0 is the angle between A and B , then
B i = B sin e. T h u s the magnitude of A X B ,
is AB sin e, the same a s the magnitude of A X B .
Also, the direction A X B is the same as the
direction
I
\
of A X B ,
Hence
A X B
B,
C„
/
AXB.
1
S i m i l a r l y i f C is resolved into two component vectors C | a n d C j _ , p a r a l l e l a n d perpendicular respectively to A , then A X C , = A X C .
(
A l s o , since B + C
+ B,| -r C
±
A X ( B
Fig.5-19
+ C„ =
(B
+ C J
i
=
+ C ) + (B,
x
±
it follows t h a t
Cm)
A X ( B + Cj
Now B , a n d C , a r e vectors perpendicular to A a n d so by Problem 5.13,
A X (B
Then
=
+ C
±
A XB
A X C1
= A X B + A X C
A X (B + C)
and the distributive l a w holds. M u l t i p l y i n g by —1, u s i n g Problem 5.12, this becomes ( B + C ) X A =
B X A + C X A . Note t h a t the order of factors in cross products is important. T h e u s u a l l a w s of
algebra apply only i f proper order is m a i n t a i n e d .
5.15.
If
A = A i i + A j + Auk
and
2
A X B
=
B = B i i + B j + S k,
2
3
prove that
A x B
( A , i + A j + A k ) X ( . ^ i + J? j + B k )
2
2
3
P
+ £ j + £ k ) + A j x (B i + Btf + B k ) + A k X ( £ , i + B.j + B k)
Aii X
2
2
3
s
3
3
A = 3i - j + 2k
3
2
A Bnk X i + A B k X j + A F k X k
3
3
(A B -A B )i
2
z
3
2
3
2
3
2
3
+ ( A s B i - A ^ j
3
+ (A , B , -
A B,)k
2
A j
A 2 A
3
B
B
3
and
B = 2i + 3j - k,
j
1
=
k
3 - 1
2
find
2
A x B .
-1
2
3
3 - 1
-5i
2
-1
+
-
3
2
j 2 -1
+
k
3
- I
2
7j + I l k
Prove t h a t the area of a parallelogram w i t h
sides A and B is |A x B j .
A r e a of p a r a l l e l o g r a m
See F i g . 5-20.
=
h |B|
=
|A| sin e |B|
= |AXB|
Note t h a t the a r e a of the t r i a n g l e w i t h sides A
B = £ | A X B|.
B
Fig.5-20
3
B
3
Bz
2
AjBji X i + AjBji X j + A , B i X k + A B]j X i + AjBjj X j + A B j X k
A X B
and
Si
k
A
2
3
1
5.17.
j
A
=
=
If
i
Ai
-
+
5.16.
=
VECTOR
134
5.18.
F i n d the area of
the triangle w i t h
PQ
PR
A r e a of t r i a n g l e
=
5.19.
at P(2,3,5),
=
( 4 - 2)i + (2 - 3)j + ( - 1 - 5 ) k =
=
(3 - 2)i + (6 - 3)j + (4 - 5 ) k
= \
TRIPLE
vertices
11 PQ X P R |
=
[CHAP. 5
ANALYSIS
=
1
j
k
2
-1
-6
1
3
Q(4,2,-l),
2i -
=
72(3,6,4).
j - 6k
i + 3j -
k
-J- [ (2i - j - 6k) X (i + 3j - k) |
-
\ 19i - 4j + 7k |
-1
| 7 ( 1 9 ) + ( - 4 ) + (7)2
2
2
\s[m>
=
PRODUCTS
Show
that
A • ( B x C)
is
in
value equal to the volume
lelepiped
Fig.
with
sides A , B
absolute
of
a
and
paral-
C.
See
5-21.
J.
_L»_
h
L e t n be a unit n o r m a l to p a r a l l e l o g r a m
I, h a v i n g the direction of B X C , a n d let h be
B
the height of the t e r m i n a l point of A above
the p a r a l l e l o g r a m / .
F i g . 5-21
V o l u m e of parallelepiped
If
5.20.
If
A , B and C do not form
A = A,i + A j + A k,
2
3
=
(height 7i)(area of p a r a l l e l o g r a m / )
=
(A-n)([BxC!)
=
A • {|B X C| n }
a right-handed
B = 5,i + B]
A • ( B X C)
=
A-
Bi
Cj
j
A • n < 0
a n d the volume — |A • ( B X C)
C = Cd + C j + Csk
+ Bsk,
2
A - ( B x C )
i
system,
A • (B X C )
=
show
2
=
Ai
A
Bx
B
Ci
C
that
A
2
3
2
2
B
3
C3
k
B
B
2
C
3
C
2
3
=
( A i + A j j + A k ) • [(B C
=
A ^ B ^ - B
t
2
3
- B C )i
3
3
2
+ (BgC, - B , C ) j + ( B , C 3
B-AJk)
2
Aj
3
C
2
) + A ^ B a d - B ^ a ) + A^B^-B C )
2
=
V
C]
5.21.
F i n d t h e v o l u m e of a p a r a l l e l e p i p e d w i t h sides
A = 3i -
j , B = j + 2k,
=
]A • ( B X C)| =
| 0
5.22.
Prove that
A • ( B x C) = ( A x B ) • C,
i.e. t h e d o t a n d
A\
B y Problem 5.20:
A • (B X C)
1—201 =
2
A
B\ B
3
2
C\
2
C
C
3
c
2
3
2
20
cross can
be
interchanged.
C,
('AXB)-C
B
4
3
,
3
2
0
1
1 5
=
A
2
C = i + 5 j + 4k
3 - 1
B y Problems 5.19 a n d 5.20, volume of parallelepiped
A
B , B
=
3
Since the two determinants a r e equal, the r e q u i r e d r e s u l t follows.
C'(AXB)
=
C
2
A\
A j
Bj
B
2
C I
3
3
2
B
3
VECTOR
: H A P . 5]
5.23.
Let r
and
= xA + yij + Z i k , r
2
r
= xi + yj
3
3
v e c t o r s o f p o i n t s P i (x
3
3
3
the
position
Z\), P (x , yz, z )
u
Find
P (x ,y ,z ).
3
be
y
u
and
= x-A +1/2] + 2 2 k
2
+ 23k
3
135
ANALYSIS
2
2
an
2
equation
the
plane passing t h r o u g h P i , P
See
Fig.
2
for
and
P.
3
5-22.
W e assume that P P a n d P do not lie i n
the same s t r a i g h t line; hence they determine
a plane.
v
2
3
L e t r = x i + y) + z k denote the position
vector of a n y point P(x, y, z) i n the plane. Consider vectors P i P = r j — r j , P 1 P 3 = r — r !
and P j P = r — r
w h i c h a l l lie i n the plane.
2
3
x
T
h
6
or
PiP'P.Pa X P P
n
t
=
3
0
(r - r , ) • ( r - r , ) X ( r - r , )
2
3
=
0
I n terms of r e c t a n g u l a r coordinates this becomes
[(x - x i ) i + (y-
F i g . 5-22
Vi)] + (z - Zi)k] • [(x - x ) i + (y 2
+ (z -
2
x
Z])k]
2
X [ ( x - Xi)i + (y - 3/1)) + ( z 3
3
x or, using Problem 5.20,
y -
Vi
y
2
- Vi
x
-
y
3
- V\ - z ,
*i
F i n d an equation f o r the plane passing
=
0
z - Z]
~ »t
3
5.24.
Xx
x
2
z^k]
3
z, -
z,
=
0
3
t h r o u g h the points P i ( 3 , 1 , - 2 ) ,
P (-l,2,4),
2
P (2,-l,l).
3
T h e position vectors of P ,P ,P
X
r,
Then
=
3i + j -
2k,
P?!
r
a n d a n y point P(x,y,z)
2
3
2
=
- i + 2j + 4k,
P P ! = r —rj,
2
required equation is
r
P 3 P 1 = r —r
2
3
3
=
on the plane a r e respectively
2i - j + k,
x i + yj + z k
r =
a l l lie i n the r e q u i r e d plane a n d so the
1 (
(r — r ^ • ( r — T ] ) X ( r — r i ) = 0, i.e.,
2
3
{(x - 3)1 + (y - l ) j + (z + 2)k} • { - 4 i + j + 6 k } X { - i - 2j + 3k}
{ ( * - 3)i + (y - l ) j + (z + 2)k} • {151 + 6j + 9 k }
15(x - 3) + 6(y - 1) + 9(z + 2) =
Another method.
x - 3
3
y -
1
z + 2
2 -
1
4 4-2
2 - 3 - 1 - 1
If
A = i + j,
(a)
A X B =
i
j
1
i
BX C =
2 - 3
0
0
C = 4j - 3k,
find
Then (A X B) X C
( A X B ) X C,
(6)
j
k
1 -1
-5
0
k
1
11
i
i - j - 5k.
1
j
5x + 2y + 3z =
(a)
k
0
2 - 3
(6)
11
1 + 2
B = 2i — 3j + k,
1
0
5x + 2y + 3z =
or
0
B y Problem 5.23, the required equation is
-1 -
5.25.
0
=
=
T h e n A X (B X C ) =
4 - 3
I t follows that, i n g e n e r a l ,
1
5
( A X B ) X C # A X (B X C ) .
23i + 3j + 4k.
4 - 3
i
5i + 6j + 8k.
A X (B x C).
j
1
k
=
0
6
8
8 i - 8 j + k.
VECTOR
136
[CHAP. 5
ANALYSIS
DERIVATIVES
5.26.
If
t =
( a )
r =
(i
0
3
+ 2t)i
and
Tt
t -
At
F r o m (a),
( c )
5¥
<3
+
2 t
)
+ ft
i
[dr/dtj =
^(^)
(~
V(2)
2
{(3
-
2
5.27.
|d r/dt | = 12
2
x = t + 2t,
time,
these
t i o n s of
*u
J
s
d
- j - (A • B)
du
'
/.
d
>
- f - (A • B )
aw
D
6e
~
2tj +
represent
lim
,
W
(d)
at
dt
6e-2tj +
2
10 cos 5t k
(
'
A
B
)
£ ( 3 , y, z) =
10cos5tk
I
A • AB +
<t>A
=
(,lA)
~
6ti
l
2
~
e
2
t
i
~
50 s i n 54 k
the
velocity,
magnitude
t = 0
of
a particle moving
where
A
and
of
the
along
velocity,
the
space
B
are
differentiable
f
unc-
A • B
AA • B 4- A A • A B
/
A«-
AB
\M
AA
—
h
A"
2
+ A
^ +
dB
=
, \
d
• AB
dA
A •
-=—h
an
y
B = B ^ 4-B j 4-B k.
3
-7— •
B
ait
Then
3
A.B,)
dB
2 - r - + A
du
2
A
AA
—
• B +
AM
2
d B \,
- ^
4dM /
dA
_ l B
3
3
3
2
\
I
1
dA
4 - ^ B
aw
3
2
2
4 - - ^ B
^M
f
A - ^ 4 - ^ - B
au
du
and
dfdz~
> =
0.
Alt
dB,
^
+
du
A
a; ?/z
— (<6A)
t =
at
/
—
^(AiBi
=
2
2y 85
AM
,.
Inn
=
=
A =
3cc ?/i 4- y z j 2
(x yz)(3x yi
2
2
2
+ yz j
-
2
4
y = -2,
If
A
x
s i n y i 4- z
=
2x s i n y i
2
dy~ ( *y
=
x = 1,
2
Sx
2 = -1,
2
cos
xzk)
2
this becomes
y j -
a;i/ k,
2
2
find
=
2
2
+ 3x 2/ z j -
H
xzk,
x*yz k)
— ( 3 x « z i 4- x i y z 3 j -
If
~
=
J
2
2
(e)
2
(3* + 2)i +
respectively
AU-O
=
=
d r
2
,
dt
0.
Let A = A , i 4-A j 4-A k,
a !
=
k
( A 4- d A ) • ( B 4- A B )
•—
—
i-
=
AU-*O
Method
* )
dB
d A.
A • - 3 — h -j- • B ,
du
du
=
_
—
Method 2.
dr
(b)
z — 2 s i n 54.
2t
,.
5.29.
5
VliO
=
2
\u - 0
If
n
u.
i
Method 1.
5.28.
i
of the acceleration at
K
M
ft&
+
+
4=
y = —3e~ ,
3
that
2+ 2)i
at
2
t represents
Prove
~ ,
_l2j.
acceleration a n d magnitude
curve
)J
3 e _ 2 (
(a)
significance.
+ (6)2 4- (10)
2
^ *
=
d r/d4
F r o m (c),
If
find
2
d r / d t = 2i + 6j + 10k.
t = 0,
At
(d)
<
0,
(b)
=
3 e " ' j + 2 s i n 5t k,
g i v e a possible physical
ft
=
-
2
~12i -
find
(^A)
3x 2/ zi
4
=
2
=
the
4- x 2/ z=j -
3x*y*i
2x*yzk)
at
2
+
x yz k
2
3x y z j
2
2
3
-
2
6 x % i 4- 6x yz j
2
12j 4- 2k.
dA.
j/ k,
2
~
=
«2 cos y i -
z
2
sin y j -
2a^k,
^
2
2x*yzk
-
2
1.
—
point
2xHk
(1,
-2,1)
VECTOR
C H A P . 5]
= ^p-dx
dA
dx
ANALYSIS
137
+ ^-dy + ^-dz
9y
dz
-
(2x s i n y i - y k) dx + {x cos y i -
=
(2x s i n y dx + x cos «
=
d(x sin j/)i + d ( z cos
=
(2x sin y dx + x cos y dy)i + (2z cos y dz - z sin y dy)j -
2
z sin y j -
2
2
2xyk) dy + (2z cos y j ) dz
+ (2z cos y dz - z s i n v <*2/)j -
2
( j / dx + 2xy dy)k
2
2
Method 2.
dA
5-30.
2
— d(x2/ )k
2
2
2
A p a r t i c l e m o v e s a l o n g a space c u r v e
some i n i t i a l
time.
( i / dx + 2xy dy)k
2
r = r(t), w h e r e
I f v = \dr/dt\ d s / d i
2
t is the t i m e measured
is t h emagnitude
of t h e velocity
p a r t i c l e (s i s t h e a r c l e n g t h a l o n g t h e s p a c e c u r v e m e a s u r e d f r o m t h e i n i t i a l
position),
a o f the particle is given b y
prove t h a t the acceleration
a
=
+
p
dt
w h e r e T a n d N a r e u n i t t a n g e n t a n d n o r m a l v e c t o r s t o t h e space c u r v e
d r ~
2
<PxV
ds
!
ds
, [&y\
2
2
U s
T
T h e velocity of the particle is given by v = vT.
dv
d , _,
dv _
dt
dt
dt
v
from
of the
'
.
Since T h a s unit magnitude, we have
/
2
and
fd^z
U s
2
T h e n the acceleration i s given by
dT
dv _
dt
dt
T • T = 1.
,
d T ds
dv _
ds dt
dt
,
, dT
ds
T h e n differentiating w i t h respect to s,
+
^ . x = 0,
2 T - ^ = 0
or
T . £ = 0
ds
ds
ds
ds
from w h i c h i t follows that d T / d s is perpendicular to T . Denoting by N the u n i t vector i n the direction of d T / d s , a n d called the principal normal to the space c u r v e , we have
T
. ^
S
where K is the magnitude of d T / d s .
dT/ds = d r/ds .
Hence
2
=
N o w since
*
«
N
T = dr/ds
(9), page 126], we have
[see equation
2
dfrl
ids 1
JAM* , (dhLV
\v / ' v^y
_
ds2
2
p = 1/ , (2) becomes
Defining
dT/ds
K
= N/p.
a
v v
U2
ds
T h u s from (J) we have, a s required,
dv _
dt"
=
,(d?z\^
+
, v „.
"p"
2
T h e components d v / d t a n d v / p i n the direction of T a n d N a r e called the tangential
a n d normal
components of the acceleration, the latter being sometimes called the centripetal
acceleration.
The
quantities p a n d K a r e respectively the radius of curvature
a n d curvature
of the space curve.
2
GRADIENT, DIVERGENCE
5.31.
I f <j> = x yz
2
(</>A),
CURL
a n d A = xzi — y j + 2x yk,
3
2
2
find
(e) c u r l (<pA).
=
(6)
AND
V - A
2xj/« i + x z j 43
=
2
3
(j^i + ^j
Zx yz k
2
+-j^k)
2
• (xzi-y j
2
+ 2x vk)
2
(a) V < P , (&) V • A ,
(c) V x A ,
(d) d i v
VECTOR
138
(c)
V XA
# - i + 4- i + T - k ) X (xzi - y*j +
dx
dy
dz
=
i
j
d/dx
div (0A)
(e)
(2x y)
"
j i +
2
- ^(2x y)
dx
(
{xz)
dy
k
V • (x^yz^i — x # z j + 2 x ? / z k )
=
+-(x yz ) + +-(-x 2/ 2 ) + -^-(2xVz )
dx
dy
dz
3
2
4
2
V X (0A)
3
s
2
3
k
d/dy
d/dz
—x y z
(4x*yz
4
2
3
=
3
5x yz*
=
+
3
3
-
2
Sx y z
2
2
+
3
Gx*y z
2
2
2x y z k)
i
2
3
3
3
- 8x y z )j
3
3
2
-
3
(2xyH
+ x z*)k
3
3
^ V ' A ) ,
V - ( 0 A i 4 - s A j + 0A k)
1
i ^
=
2
(Ax yz
2
( V ^ ' A +
V-(0A)
2
2x y z
3
+ 3x 2/ z )i +
3
V • (<pA) =
3
4
V x (x yz*i - x y z j
=
2
3
3
i
4
3
3
d/dx
3
=
( -<-»•)
dx
+ (x — 4xy)j
V • (0A)
=
)j +
2
=
c u r l (0A)
Prove
2
- ~(-y )
dz
2
v
x yz
5.32.
2x y
2
2
(d)
2
k
—y
2x i
2x yk)
d/dy d/dz
xz
dy
[CHAP. 5
ANALYSIS
(
A
l
9
)
2
3
+
Tz^
+
30 .
, 30 .
,
3A
3A
2
3
3a;
\dx
=
5.33.
Prove that
dy
(Aa + A J + A a k )
dz
( 7 0 ) . A + 0 ( 7 - A)
V<P is a v e c t o r p e r p e n d i c u l a r t o t h e s u r f a c e
w h e r e c is a
<j>(x,y,z) = c,
constant.
r = x i + j / j + zk be the position vector to a n y point P(x, y, z) on the s u r f a c e .
Let
dr = dx i + dy j 4- dz k lies i n the p l a n e t a n g e n t to the s u r f a c e a t P . B u t
Then
d0
^dx
dx
S
+
^dy
dy
+ ^dz
dz
d
=
0
or
, .^ i. + ^. j , .+ ,^ k ] - ( d x i + d y j + d z k )
\dx
dy
dz
=
0
1
7 0 • dr = 0 so t h a t V<p i s p e r p e n d i c u l a r to d r a n d therefore to the s u r f a c e .
i.e.
5.34.
=
Find
a unit
normal
to the surface
2a; + Ayz - 5z
2
= -10
2
at the
point
P(3, - 1 , 2 ) .
B y Problem 5.33, a vector n o r m a l to the s u r f a c e i s
7(2x + 4j/z-5z )
2
2
=
4 x i + 4 z j + (Ay - 10z)k
=
12i + 8j -
12i + 8j -
T h e n a unit n o r m a l to the s u r f a c e a t P i s
24k
V(12) + (8) + ( - 2 4 )
2
A n o t h e r unit n o r m a l to the s u r f a c e at P i s 5.35.
If
$ = 2x y - xz ,
2
(o)
70
(6)
7 0
2
3
=
=
find
(a) V4>
| | i + |£j + M
L a p l a c i a n of 0
and
k
=
=
7 •70
3 l
2
+
2
j
24k
_
at
(3,-1,2)
3i + 2j - 6k
2
7
— .
(6) V <£2
(4xj/~**)»
+ 2x j 2
3
6'x
(4xy-z )
3
3xz k
2
+ ^"(2x ) + |^(-3xz )
2
2
=
4y - 6xz
VECTOR
C H A P . 5]
Another
method.
V d
ax
2
=
5.36.
Prove
139
ANALYSIS
aj/
2
2
3i/
dx
2
5 x 2
Zx-y
o2_
— xz°
32'
(2x y — xz )
2
3
2
2
4y — 6xz
d i v c u r l A = 0.
i
div c u r l A
=
V'(VXA)
=
A
=
V
\dy
dz )
/dA
a
dz J
d°-A,
aA
dxdy
dxdz
2
2
A
t
dx j
s_ f3Ai
dy J
d
f 2
dA
dx J ^ dz \
a A,
aA
dy dz
dy dx
2
' \
1
_3Ag\
dy \
+
k
d/dy d/dz
' \
l
dAA
3
dx \1y
j
d/dx
V
2
dy
3 A,
d Ai
2
2
s
' dz dx
0
dz dy
a s s u m i n g that A h a s continuous second p a r t i a l derivatives so t h a t the order of differentiation i s
immaterial.
5.37.
Find
equations f o r
= 0
F(x,y,z)
(a) t h e t a n g e n t p l a n e
a t t h e p o i n t P(x ,yo,z ).
0
See
Q
and
(5)
A vector n o r m a l to the s u r f a c e a t P is N„ = V F j .
(a)
Then if r
P
respectively from
the n o r m a l line
to the
0
a n d r a r e the vectors d r a w n
0 to P(x , y , z ) a n d Q(x, y, z) on the plane, the equation
0
0
0
(r-r ).N„
0
since r — r is perpendicular to N .
0
0
F \(x-x )
x
=
=
( r - r ) - VF\
0
of the plane i s
0
I n r e c t a n g u l a r f o r m this is
+ F \(y-y )
0
surface
F i g . 5-23.
y
+
0
F,\(z-z )
=
0
0
I f r i s the vector d r a w n from 0 in F i g . 5-23 to a n y point (x, y, z) on the n o r m a l line, then
r — r i s collinear w i t h N a n d so
(6)
0
0
(r-r )XN
0
0
=
(r-r )XVF;
0
P
=
0
w h i c h i n r e c t a n g u l a r form is
Fig.5-23
5.38.
Find
* -
equations
/(«),
for
V = g(u),
(a)
Fig.5-24
the tangent line,
z = h(u)
(6)
at the point where
the
n o r m a l plane
u = u.
0
See
to
a
Fig. 5-24.
space
curve