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Transcript of Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2

• 1

Homework 1 Solutions

1.

2 CHAPTER 1

1-2.

a)

x1 A

B

C

D

α

β γ

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 2 OE (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

E D

C

C′

B′ E′D′

First, we have the following trigonometric relation:

2 2

2 cosOE OE OE OE EEθ 2′ ′+ − = ′ (7)

• 2MATRICES, VECTORS, AND VECTOR CALCULUS 3

But,

2 2 2 2

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.

x1 e3′

x2

x3

O

e1 e2

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 0 cos , 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 kj k

C A= B∑ (1)

( )t ji jk ki ki jkij k k

C C A B B A= = =∑ ∑

..

• 3

2.

• 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 V

dv c dπ=∫

1-37.

z

y

x

R

To do the integral directly, note that A , on the surface, and that . 3 rR= e

5

rd da=a e

3 3 24 4 S 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

( )221 5rrr r 2r

∂ ∇ ⋅ = =

∂ A A (2)

Therefore,

( )2 2 2 0 0 0

sin 5 4 R

V dv 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 2 d da b a b dt 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 r dt r r r 2

r r

− = = −

r r r r r (1)

we have

2 r d

dt dt r r dt r − =∫ ∫

r r r (2)

from which

2 r

dt r r r − = +∫

r r r C (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 dtdt× = ×∫ ∫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 2 d da b a b dt 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 r dt r r r 2

r r

− = = −

r r r r r (1)

we have

2 r d

dt dt r r dt r − =∫ ∫

r r r (2)

from which

2 r

dt r r r − = +∫

r r r C (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 dtdt× = ×∫ ∫A A A A (2) so that

( )dt× = × +∫ A A A A C (3) where C is a constant vector.