1
Homework 1 Solutions
1.
2 CHAPTER 1
1-2.
a)
x1
A
B
C
D
α
βγ
O
E
x2
x3
From this diagram, we have
cosOE OAα =
cosOE OBβ = (1)
cosOE ODγ =
Taking the square of each equation in (1) and adding, we find
2 2 2 2cos cos cos OA OB ODα β γ+ + = + +
2 2 2OE (2)
But
2 2
OA OB OC+ =2 (3)
and
2 2
OC OD OE+ =2
(4)
Therefore,
2 2 2
OA OB OD OE+ + =2 (5)
Thus,
2 2 2cos cos cos 1α β γ+ + = (6)
b)
x3
A A′x1
x2O
ED
C
Bθ
C′
B′E′D′
First, we have the following trigonometric relation:
2 2
2 cosOE OE OE OE EEθ2′ ′+ − = ′ (7)
2MATRICES, VECTORS, AND VECTOR CALCULUS 3
But,
2 2 22
2 2
2
cos cos cos cos
cos cos
EE OB OB OA OA OD OD
OE OE OE OE
OE OE
β β α
γ γ
′ ′ ′ ′= − + − + −
′ ′= − + −′ ′
′+ −′
α
(8)
or,
2 2 22 2 2 2 2 2
2 2
cos cos cos cos cos cos
2 cos cos cos cos cos cos
2 cos cos cos cos cos cos
EE OE OE
OE OE
OE OE OE OE
α β γ α β
α α β β γ γ
γ
α α β β γ
′ ′= + + + + +′ ′ ′
′− + +′ ′ ′
′= + − + +′ ′ ′ γ ′ (9)
Comparing (9) with (7), we find
cos cos cos cos cos cos cosθ α α β β γ γ= + +′ ′ ′ (10)
1-3.
x1e3′
x2
x3
O
e1e2
e3
A e2′
e1′e2
e1
e3
Denote the original axes by , , , and the corresponding unit vectors by e , , . Denote the new axes by , , and the corresponding unit vectors by
1x 2x 3x 1 2e 3e
1x′ 2x′ 3x′ 1′e , 2′e , e . The effect of the rotation is e e , , e . Therefore, the transformation matrix is written as:
3′
1 3′→ 2 1′e e→ 3 → 2′e
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
cos , cos , cos , 0 1 0cos , cos , cos , 0 0 1
1 0 0cos , cos , cos ,
′ ′ ′
λ ′ ′ ′= =
′ ′ ′
e e e e e e
e e e e e e
e e e e e e
1-4.
a) Let C = AB where A, B, and C are matrices. Then,
ij ik kjk
C A= B∑ (1)
( )tji jk ki ki jkij
k k
C C A B B A= = =∑ ∑
..
4
3.26 CHAPTER 1
1-36.
d
z
x
y
c2 = x2 + y2
The form of the integral suggests the use of the divergence theorem.
(1) S V
d⋅ = ∇ ⋅∫ ∫A a A dv
Since ∇⋅ , we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is
1=A
(2) 2
Vdv c dπ=∫
1-37.
z
y
x
R
To do the integral directly, note that A , on the surface, and that . 3rR= e
5
rd da=a e
3 3 24 4S S
d R da R R Rπ π⋅ = = × =∫ ∫A a (1)
To use the divergence theorem, we need to calculate ∇⋅A . This is best done in spherical coordinates, where A . Using Appendix F, we see that 3
rr= e
( )22
15rr
r r2r
∂∇ ⋅ = =
∂A A (2)
Therefore,
( )2 2 2
0 0 0sin 5 4
R
Vdv d d r r dr R
π π 5θ θ φ π∇ ⋅ = =∫ ∫A∫ ∫ (3)
Alternatively, one may simply set dv in this case. 24 r drπ=
4. ..
24 CHAPTER 1
1-32. Note that the integrand is a perfect differential:
( ) (2 2d d
a b a bdt dt
⋅ + ⋅ = ⋅ + ⋅r r r r r r r r� � �� � )� (1)
Clearly,
( ) 2 22 2 cona b dt ar br⋅ + ⋅ = + +∫ r r r r �� � �� st. (2)
1-33. Since
2
d r rdt r r r 2
rr
−= = −
r r r r r� �� � (1)
we have
2
r ddt dt
r r dt r− =∫ ∫
r r r�� (2)
from which
2
rdt
r r r− = +∫
r r rC
�� (3)
where C is the integration constant (a vector).
1-34. First, we note that
( )ddt
× = × + ×A A A A A A� � � �� (1)
But the first term on the right-hand side vanishes. Thus,
( ) ( )ddt dt
dt× = ×∫ ∫A A A A�� � (2)
so that
( )dt× = × +∫ A A A A C�� � (3)
where C is a constant vector.
1
5.
24 CHAPTER 1
1-32. Note that the integrand is a perfect differential:
( ) (2 2d d
a b a bdt dt
⋅ + ⋅ = ⋅ + ⋅r r r r r r r r� � �� � )� (1)
Clearly,
( ) 2 22 2 cona b dt ar br⋅ + ⋅ = + +∫ r r r r �� � �� st. (2)
1-33. Since
2
d r rdt r r r 2
rr
−= = −
r r r r r� �� � (1)
we have
2
r ddt dt
r r dt r− =∫ ∫
r r r�� (2)
from which
2
rdt
r r r− = +∫
r r rC
�� (3)
where C is the integration constant (a vector).
1-34. First, we note that
( )ddt
× = × + ×A A A A A A� � � �� (1)
But the first term on the right-hand side vanishes. Thus,
( ) ( )ddt dt
dt× = ×∫ ∫A A A A�� � (2)
so that
( )dt× = × +∫ A A A A C�� � (3)
where C is a constant vector.
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