# Instructor's solutions manual marion, thornton classical dynamics of particles and systems, 5, Lecture notes for Classical Mechanics. Bangor University

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CHAPTER 0 Contents

Preface v

Problems Solved in Student Solutions Manual vii

1 Matrices, Vectors, and Vector Calculus 1

2 Newtonian Mechanics—Single Particle 29

3 Oscillations 79

4 Nonlinear Oscillations and Chaos 127

5 Gravitation 149

6 Some Methods in The Calculus of Variations 165

7 Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics 181

8 Central-Force Motion 233

9 Dynamics of a System of Particles 277

10 Motion in a Noninertial Reference Frame 333

11 Dynamics of Rigid Bodies 353

12 Coupled Oscillations 397

13 Continuous Systems; Waves 435

14 Special Theory of Relativity 461

iii

iv CONTENTS

CHAPTER 0 Preface

This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T. Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form. A Student Solutions Manual containing solutions to about 25% of the end-of-chapter problems is available for sale to students. The problem numbers of those solutions in the Student Solutions Manual are listed on the next page.

As a result of surveys received from users, I continue to add more worked out examples in the text and add additional problems. There are now 509 problems, a significant number over the 4th edition.

The instructor will find a large array of problems ranging in difficulty from the simple “plug and chug” to the type worthy of the Ph.D. qualifying examinations in classical mechanics. A few of the problems are quite challenging. Many of them require numerical methods. Having this solutions manual should provide a greater appreciation of what the authors intended to accomplish by the statement of the problem in those cases where the problem statement is not completely clear. Please inform me when either the problem statement or solutions can be improved. Specific help is encouraged. The instructor will also be able to pick and choose different levels of difficulty when assigning homework problems. And since students may occasionally need hints to work some problems, this manual will allow the instructor to take a quick peek to see how the students can be helped.

It is absolutely forbidden for the students to have access to this manual. Please do not give students solutions from this manual. Posting these solutions on the Internet will result in widespread distribution of the solutions and will ultimately result in the decrease of the usefulness of the text.

The author would like to acknowledge the assistance of Tran ngoc Khanh (5th edition), Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of previous versions, went over user comments, and worked out solutions for new problems. Without their help, this manual would not be possible. The author would appreciate receiving reports of suggested improvements and suspected errors. Comments can be sent by email to stt@virginia.edu, the more detailed the better.

Stephen T. Thornton Charlottesville, Virginia

v

vi PREFACE

CHAPTER 1 Matrices, Vectors,

and Vector Calculus

1-1.

x2 = x2′ x1′

45˚

x1

x3′ x3

45˚

Axes and lie in the plane. 1x′ 3x′ 1 3x x

The transformation equations are:

1 1 3cos 45 cos 45x x x= ° − °′

2 2x x=′

3 3 1cos 45 cos 45x x x= ° + °′

1 1 1 1 2 2

x x= −′ 3x

2 2x x=′

3 1 1 1 2 2

x x= −′ 3x

So the transformation matrix is:

1 1 0

2 2 0 1 0 1 1

0 2 2

  − 

        

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 Ax1

x2O

E D

C

Bθ

C

BED

First, we have the following trigonometric relation:

2 2

2 cosOE OE OE OE EEθ 2′ ′+ − = ′ (7)

MATRICES, 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= = =∑ ∑

4 CHAPTER 1

Identifying ( )tki ikB=B and ( )tjk kjA A= ,

( ) ( ) ( )t t i

t

j ik kj k

C B A=∑ (2)

or,

( )ttC AB B A= = t t (3)

b) To show that ( ) 1 1 1AB B A− − −= ,

( ) ( )1 1 1 1AB B A I B A AB− − − −= = (4) That is,

( ) 1 1 1 1AB B A AIA AA I− − − −= = = (5)

( ) ( )1 1 1 1B A AB B IB B B I− − − −= = = (6)

1-5. Take λ to be a two-dimensional matrix:

11 12 11 22 12 21 21 22

λ λ λ λ λ λ λ λ λ = = − (1)

Then,

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

2 2 2 2 2 2 2 2 2 2 2 2 2 11 22 11 22 12 21 12 21 11 21 12 22 11 21 12 22

2 2 2 2 2 2 2 2 2 2 22 11 12 21 11 12 11 21 11 22 12 21 12 22

22 2 2 2 11 12 22 21 11 21 12 22

2

2

λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ

λ λ λ λ λ λ λ λ λ λ λ λ λ λ

λ λ λ λ λ λ λ λ

= − + + + − +

= + + + − + +

= + + − + (2)

But since λ is an orthogonal transformation matrix, ij kj ik j

λ λ δ=∑ .

Thus,

2 2 2 2 11 12 21 22

11 21 12 22

1

0

λ λ λ λ

λ λ λ λ

+ = + =

+ = (3)

Therefore, (2) becomes

2 1λ = (4)

1-6. The lengths of line segments in the jx and jx′ systems are

2j j

L x= ∑ ; 2i i

L = x′ ′∑ (1)

MATRICES, VECTORS, AND VECTOR CALCULUS 5

If , then L L= ′

2 2j i j i

x x= ′∑ ∑ (2)

The transformation is

i ij j j

x λ=′ x∑ (3)

Then,

(4)

2

,

j ik k j i k

k ik i k i

x x

x x

λ λ

λ λ

    =      

  =   

∑ ∑ ∑ ∑

∑ ∑

i x

But this can be true only if

ik i k i

λ λ δ=∑ (5)

which is the desired result.

1-7.

x1

(1,0,1)

x3

x2

(1,0,0) (1,1,0)

(0,1,0)

(1,1,1)

(0,0,1) (0,1,1)

(0,0,0)

There are 4 diagonals:

1D , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D ; 1

2D , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = ; 2D

3D , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = ; and 3D

4D , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D . 4

The magnitudes of the diagonal vectors are

1 2 3 4 3= = = =D D D D

The angle between any two of these diagonal vectors is, for example,

( ) ( )1 2

1 2

1,1,1 1,1,1 1 cos

3 3 θ

⋅ −⋅ = = =

D D D D

6 CHAPTER 1

so that

1 1

cos 70.5 3

θ −  = = °  

Similarly,

1 3 2 3 3 41 4 2 4 1 3 1 4 2 3 2 4 3 4

1 3

⋅ ⋅ ⋅⋅ ⋅ = = = = =

D D D D D DD D D D D D D D D D D D D D

±

1-8. Let θ be the angle between A and r. Then, 2A⋅ =A r can be written as

2cosAr Aθ =

or,

cosr Aθ = (1)

This implies

2

QPO π = (2)

Therefore, the end point of r must be on a plane perpendicular to A and passing through P.

1-9. 2= + −A i j k 2 3= − + +B i j k

a) 3 2− = − −A B i j k

( ) ( ) 1 22 2 23 1 ( 2) − = + − + − A B

14− =A B

b)

component of B along A

B

A θ

The length of the component of B along A is B cos θ.

cosAB θ⋅ =A B

2 6 1 3 6

cos or 26 6A

θ ⋅ − + −

= = = A B

B

The direction is, of course, along A. A unit vector in the A direction is

( )1 2 6 + −i j k

MATRICES, VECTORS, AND VECTOR CALCULUS 7

So the component of B along A is

( )1 2 2 + −i j k

c) 3 3

cos 6 14 2 7AB

θ ⋅

= = = A B

; 1 3

cos 2 7

θ −=

71θ °

d) 2 1 1 1 1 2

1 2 1 3 1 2 1 2 3

2 3 1

− − × = − = − +

− − −

i j k

A B i j k

5 7× = + +A B i j k

e) 3 2− = − −A B i j k 5+ = − +A B i j

( ) ( ) 3 1 1 5 0

− × + = − − −

i j k

A B A B 2

( ) ( ) 10 2 14− × + = + +A B A B i j k

1-10. 2 sin cosb t b tω ω= +r i j

a) 2 2

2 cos sin

2 sin cos

b t b t

b t b t

ω ω ω ω

2ω ω ω ω ω

= = −

= = − − = −

v r i j

a v i j r

1 22 2 2 2 2 2

1 22 2

speed 4 cos sin

4 cos sin

b t b

b t t

ω ω ω ω

ω ω ω

t = = + 

 = + 

v

1 22speed 3 cos 1b tω ω = + 

b) At 2t π ω= , sin 1tω = , cos 0tω =

So, at this time, bω= −v j , 22bω= −a i

So, 90θ °

8 CHAPTER 1

1-11.

a) Since ( ) ijk j ki jk

A Bε× =∑A B , we have

( ) ( ) ( )

( )

,

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1

1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

( ) ijk j k i i j k

A B C

C A B A B C A B A B C A B A B

C C C A A A A A A

A A A C C C B B B

B B B B B B C C C

ε× ⋅ =

= − − − + −

= = − = = ⋅

∑∑A B C

A B C×

(1)

We can also write

( 1 2 3 1 2 3

1 2 3 1 2 3

1 2 3 1 2 3

( ) C C C B B B

B B B C C C

A A A A A A

× ⋅ = − = = ⋅ ×A B C B C A) (2)

We notice from this result that an even number of permutations leaves the determinant unchanged.

b) Consider vectors A and B in the plane defined by e , . Since the figure defined by A, B, C is a parallelepiped, area of the base, but

1

3

2e

3× = ×A B e ⋅ =e C altitude of the parallelepiped. Then,

( ) ( )3 area of the base

= altitude area of the base

= volume of the parallelepiped

⋅ × = ⋅ ×

×

C A B C e

1-12.

O

A

B

C ha

b

c

a c

c b

b a

The distance h from the origin O to the plane defined by A, B, C is

MATRICES, VECTORS, AND VECTOR CALCULUS 9

( ) ( ) ( ) ( )

( )

h ⋅ − × −

= − × −

⋅ × − × + × =

× − × + ×

⋅ × = × + × + ×

a b a c b

b a c b

a b c a c a b

b c a c a b

a b c a b b c c a

(1)

The area of the triangle ABC is:

( ) ( ) ( ) ( ) ( ) ( )1 1 1 2 2 2

× − = − × − = − × −b a c b a c b a c b a cA = − (2)

1-13. Using the Eq. (1.82) in the text, we have

( ) ( ) ( ) 2Aφ× = × × = ⋅ − ⋅ = −A A X X A A A A X A XA B

from which

( )

2A × +

= B A A

X φ

1-14.

a) 1 2 1 2 1 0 1 2 1 0 3 1 0 1 2 1 2 9 2 0 1 1 1 3 5 3 3

− −         = − =             

AB  −  

Expand by the first row.

2 9 1 9 1 2

1 2 1 3 3 5 3 5 3 − −

= + +AB

104= −AB

b) 1 2 1 2 1 9 7 0 3 1 4 3 13 9 2 0 1 1 0 5 2

−           = =                

AC

9 7

13 9 5 2

   =     

AC

10 CHAPTER 1

c) ( ) 1 2 1 8 5 0 3 1 2 3 2 0 1 9 4

−       = = − −          

ABC A BC

5 5

3 5 25 14

− −   = −    

ABC

d) ?t t− =AB B A

1 2 1 1 2 9 (from part ) 5 3 3

2 0 1 1 0 2 1 1 5 1 1 1 2 3 0 2 2 3 0 2 3 1 1 1 1 9 3

t t

−   = −    

         = − = − −         −    

AB a

B A    

0 3 4 3 0 6 4 6 0

t t

− −   − =    − 

AB B A

1-15. If A is an orthogonal matrix, then

2

2

1

1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0

1 0 0 1 0 0 0 2 0 0 1 0 0 0 2 0 0 1

t

a a a a

a a a a

a

a

=

         − =         −    

0 1

   

       =          

A A

1 2

a =

MATRICES, VECTORS, AND VECTOR CALCULUS 11

1-16.

x3 P

r θ

x2

x1

a

r

θ a

r cos θ constant⋅ =r a

cos constantra θ =

It is given that a is constant, so we know that

cos constantr θ =

But cosr θ is the magnitude of the component of r along a.

The set of vectors that satisfy all have the same component along a; however, the component perpendicular to a is arbitrary.

constant⋅ =r a

Th is

us the surface represented by constant a plane perpendicular to .

⋅ =r a a

1-17.

a

A θ

b

B

c

C

Consider the triangle a, b, c which is formed by the vectors A, B, C. Since

( ) (2

2 22

)

A B

= −

= − ⋅ −

= − ⋅ +

C A B

C A B A B

A B

(1)

or,

2 2 2 2 cosA B AB θ= + −C (2)

which is the cosine law of plane trigonometry.

1-18. Consider the triangle a, b, c which is formed by the vectors A, B, C.

A

α C B

γ β bc

a

12 CHAPTER 1

= −C A B (1)

so that

( )× = − ×C B A B B (2)

but the left-hand side and the right-hand side of (2) are written as:

3sinBC α× =C B e (3)

and

( ) 3sinAB γ− × = × − × = × =A B B A B B B A B e (4)

where e is the unit vector perpendicular to the triangle abc. Therefore, 3

sin sinBC ABα γ= (5)

or,

sin sin

C A γ α =

Similarly,

sin sin sin

C A B γ α β = = (6)

which is the sine law of plane trigonometry.

1-19.

x2

a

α x1

a2

b2

a1 b1

b

β

a) We begin by noting that

( )2 2 2 2 cosa b ab α β− = + − −a b (1)

We can also write that

( ) ( )

( ) ( )

( ) ( ) ( )

( )

2 22 1 1 2 2

2 2

2 2 2 2 2 2

2 2

cos cos sin sin

sin cos sin cos 2 cos cos sin sin

2 cos cos sin sin

a b a b

a b a b

a b ab

a b ab

α β α β

α α β β α β α

α β α β

− = − + −

= − + −

= + + + − +

= + − +

a b

β

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS 13

Thus, comparing (1) and (2), we conclude that

( )cos cos cos sin sinα β α β α− = + β (3)

b) Using (3), we can find ( )sin α β− :

( ) ( )

( ) ( )

( )

2

2 2 2 2

2 2 2 2

2 2 2 2

2

sin 1 cos

1 cos cos sin sin 2cos sin cos sin

1 cos 1 sin sin 1 cos 2cos sin cos sin

sin cos 2sin sin cos cos cos sin

sin cos cos sin

α β α β

α β α β α α β β

α β α β α α β

α β α β α β α β

α β α β

− = − −

= − − −

= − − − − −

= − +

= −

β

(4)

so that

( )sin sin cos cos sinα β α β α− = − β

j

(5)

1-20.

a) Consider the following two cases:

When i ≠ 0ijδ = but 0ijkε ≠ .

When i = j 0ijδ ≠ but 0ijkε = .

Therefore,

0ijk ij ij

ε δ =∑ (1)

b) We proceed in the following way:

When j = k, 0ijk ijjε ε= = .

Terms such as 11 11 0jε ε = . Then,

12 12 13 13 21 21 31 31 32 32 23 23ijk jk i i i i i i jk

ε ε ε ε ε ε ε ε ε ε ε ε ε ε= + + + + +∑

=

Now, suppose i , then, 1= =

123 123 132 132 1 1 2 jk

ε ε ε ε= + = +∑

14 CHAPTER 1

for , . For 2i = = 213 213 231 231 1 1 2 jk

ε ε ε ε= + = +∑ = 3i = = , 312 312 321 321 2 jk

ε ε ε ε= + =∑ . But i = 1,

gives . Likewise for i = 2, 2= 0 jk

=∑ 1= ; i = 1, 3= ; i = 3, 1= ; i = 2, ; i = 3, . Therefore,

3= 2=

,

2ijk jk i j k

ε ε δ=∑ (2)

c)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

123 123 312 312 321 321 132 132 213 213 231 231

1 1 1 1 1 1 1 1 1 1 1 1

ijk ijk ijk

ε ε ε ε ε ε ε ε ε ε ε ε ε ε= + + + + +

= ⋅ + ⋅ + − ⋅ − + − ⋅ − + − ⋅ − + ⋅

or,

6ijk ijk ijk

ε ε =∑ (3)

1-21. ( ) ijk j ki jk

A Bε× =∑A B (1)

( ) ijk j k i i jk

A B Cε× ⋅ = ∑ ∑A B C (2)

By an even permutation, we find

ijk i j k ijk

A B Cε=∑ABC (3)

1-22. To evaluate ijk mk k

ε ε∑ we consider the following cases:

a) : 0 for all , ,ijk mk iik mk k k

i j i mε ε ε ε= = =∑ ∑

b) : 1 for

0 for

ijk mk ijk imk k k

i j

j m

ε ε ε ε= = = =

= ≠

∑ ∑ and ,m k i j

i j

i

c) : 0 for

1 for and ,

ijk mk ijk ik k k

i m j

j k

ε ε ε ε= = = ≠

= − = ≠

∑ ∑

d) : 0 for

1 for and ,

ijk mk ijk jmk k k

j m

m i k i j

ε ε ε ε= = = ≠

= − = ≠

∑ ∑

MATRICES, VECTORS, AND VECTOR CALCULUS 15

e) : 0 for

1 for and ,

ijk mk ijk jk k k

j m i

i k

ε ε ε ε= = = ≠

= = ≠

∑ ∑

i j

j k

m

f) : 0 for all , ,ijk mk ijk k k k

m iε ε ε ε= = =∑ ∑

g) : This implies that i = k or i = j or m = k. or i

Then, for all 0ijk mk k

ε ε =∑ , , ,i j m

h) for all or : 0ijk mk k

j m ε ε≠ =∑ , , ,i j m

Now, consider i jm im jδ δ δ δ− and examine it under the same conditions. If this quantity behaves in the same way as the sum above, we have verified the equation

ijk mk i jm im j k

ε ε δ δ δ δ= −∑

a) : 0 for all , ,i im im ii j i mδ δ δ δ= − =

b) : 1 if , ,

0 if

ii jm im jii j

j m

m i j mδ δ δ δ= − = = ≠

= ≠

c) : 1 if , ,

0 if

i ji ii ji m j i j

j

δ δ δ δ= − = − =

= ≠

m i

d) : 1 if ,

0 if

i m imj i

i m

δ δ δ δ= − = − =

= ≠

e) : 1 if ,

0 if

i mm im mj m i m

i

δ δ δ δ= − = =

= ≠

all , ,j

f) : 0 for i j il jm iδ δ δ δ= − =

g) , : 0 for all , , ,i jm im ji m i j mδ δ δ δ≠ − =

h) , : 0 for all , , ,i jm im ij m i j mδ δ δ δ≠ − =

Therefore,

ijk mk i jm im j k

ε ε δ δ δ δ= −∑ (1)

Using this result we can prove that

( ) ( ) ( )× × = ⋅ − ⋅A B C A C B A B C

16 CHAPTER 1

First ( ) ijk j ki jk

B Cε× =∑B C . Then,

( )[ ] ( )

( )

( ) ( )

mn m mn m njk j kn mn mn jk

mn njk m j k mn jkn m j k jkmn jkmn

lmn jkn m j k jkm n

jl km k jm m j k jkm

m m m m m m m m m m m m

A B C A B C

A B C A B C

A B C

A B C

A B C A B C B A C C A B

B C

ε ε ε

ε ε ε ε

ε ε

δ δ δ δ

× × = × =

= =

  =   

= −

   = − = −    

= ⋅ − ⋅

∑ ∑ ∑

∑ ∑

∑ ∑

∑ ∑ ∑ ∑

A B C

A C A B

 

Therefore,

( ) ( ) ( )× × = ⋅ − ⋅A B C A C B A B C (2)

1-23. Write

( ) j m mj m

A Bε× =∑A B

( ) krs r sk rs

C Dε× =∑C D

Then,

MATRICES, VECTORS, AND VECTOR CALCULUS 17

( ) ( )[ ]

( )

( )

ijk j m m krs r si jk m rs

ijk j m krs m r s jk mrs

j m ijk rsk m r s j mrs k

j m ir js is jr m r s j mrs

j m m i j m i j j m

j m j m i j m j m j

A B C D

A B C D

A B C D

A B C D

A B C D A B D C

D A B C

ε ε ε

ε ε ε

ε ε ε

ε δ δ δ δ

ε

ε ε

    × × × =       

=

  =   

= −

= −

  = −   

∑ ∑ ∑

∑ ∑

A B C D

( ) ( )

j m i m

i i

C A B D

C D

     

= −

ABD ABC

Therefore,

[( ) ( )] ( ) ( )× × × = −A B C D ABD C ABC D

1-24. Expanding the triple vector product, we have

( ) ( ) ( )× × = ⋅ − ⋅e A e A e e e A e (1)

But,

( )⋅ =A e e A (2)

Thus,

( ) ( )= ⋅ + × ×A e A e e A e (3)

e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A perpendicular to e.

18 CHAPTER 1

1-25.

er

eφ

eθ

θ φ

The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by

( )

( )

( )

cos cos , cos sin , sin

sin , cos , 0

sin cos , sin sin , cosr

θ

φ

θ φ θ φ θ

φ φ

θ φ θ φ θ

= −  = −   = 

e

e

e

(1)

Thus,

( )cos sin sin cos , cos cos sin sin , cosθ φ θ φ θ θ φ φ θ φ θ θ φ θ θ= − − − −e cosr φθ φ θ+e= − (2) e

Similarly,

( )cos , sin , 0φ φ φ φ φ= − −e cos sin rθφ θ φ θ−e= − (3) e

sinr φ θφ θ θ= +e e e (4)

Now, let any position vector be x. Then,

rr=x e (5)

( )sin sin

r r

r

r r r r

r r r

φ θ

φ θ

φ θ θ

φ θ θ

= + = + +

= + +

x e e e e e

e e e

r

(6)

( ) ( )

( ) ( ) ( )

2 2 2

2

sin cos sin sin

2 sin 2 cos sin sin

2 sin cos

r r

r

r r r r r r r r

r r r r r r

r r r

φ φ θ θ

φ

θ

φ θ θφ θ φ θ φ θ θ θ θ

φ θ θφ θ φ θ φ θ θ

θ θ φ θ θ

= + + + + + + + +

= + + + − −

+ + −

x e e e

e e

e

re e e

(7)

or,

MATRICES, VECTORS, AND VECTOR CALCULUS 19

( )

( )

2 2 2 2 2

2 2

1 sin sin cos

1 sin

sin

r

d r r r r r

r dt

d r

r dt

θ

φ

θ φ θ θ φ θ θ

φ θ θ

  = = − − + −    

  +    

x a e e

e

(8)

1-26. When a particle moves along the curve

( )1 cosr k θ= + (1) we have

2

sin

cos sin

r k

r k

θ θ

θ θ θ θ

= −   = − +   

(2)

Now, the velocity vector in polar coordinates is [see Eq. (1.97)]

rr r θθ= +v e e (3)

so that

( )

22 2 2 2

2 2 2 2 2 2

2 2

sin 1 2 cos cos

2 2 cos

v r r

k k

k

θ

θ θ θ

θ θ

= = +

= + + +

= +  

v

θ θ

)

(4)

and is, by hypothesis, constant. Therefore, 2v

( 2

22 1 cos v

k θ

θ =

+ (5)

Using (1), we find

2 v kr

θ = (6)

Differentiating (5) and using the expression for r , we obtain

( )

2 2

22 2

sin sin 4 4 1 cos

v v r k

θ θ θ

θ = =

+ (7)

The acceleration vector is [see Eq. (1.98)]

( ) ( )2 2rr r r r θθ θ θ= − + +a e e (8) so that

20 CHAPTER 1

( ) ( )

( ) ( )

( )

( )

2

2 2

2 2 2 2

2 2

2

cos sin 1 cos

sin cos 1 cos

2 1 cos

1 cos 2 cos 1

2 1 cos

3 1 cos

2

r r r

k k

k

k

k

θ

θ θ θ θ θ θ

θ θ θ θ θ

θ

θ θ θ

θ

θ θ

⋅ = −

= − + − +

  = − + + + 

+  

 − = − + + 

+  

+

a e

θ

= − (9)

or,

23

4r v k

⋅ = −a e (10)

In a similar way, we find

2 sin3

4 1 cos v kθ

θ θ

⋅ = − +

a e (11)

From (10) and (11), we have

( ) ( )22r θ= ⋅ + ⋅a a e a e (12) or,

23 2

4 1 cos v k θ

= +

a (13)

1-27. Since

( ) ( ) ( )× × = ⋅ − ⋅r v r r r v r v r

we have

( )[ ] ( ) ( )[ ]

( ) ( ) ( ) ( ) ( )

( ) ( )2 2 2

d d dt dt

r v

× × = ⋅ − ⋅

= ⋅ + ⋅ − ⋅ − ⋅ − ⋅

= + ⋅ − + ⋅

r v r r r v r v r

r r a r v v r v v v v r r a r

a r v v r r a (1)

Thus,

( )[ ] ( ) ( )2r dt × × = + ⋅ − ⋅ +r v r a r v v r r a v2

d (2)

MATRICES, VECTORS, AND VECTOR CALCULUS 21

1-28. ( ) ( )ln ln i i ix

= ∂∑grad r r e (1)

where

2i i

x= ∑r (2)

Therefore,

( ) 2

2

1 ln i

i i i

i

x x x

x

∂ =

=

r

r

r (3)

so that

( ) 2 1

ln i i i

x  

=  ∑grad r er  (4)

or,

( ) 2ln r= r

1-29. Let describe the surface S and 2 9r = 1 2 1x y z+ + = describe the surface S . The angle θ

between and at the point (2,–2,1) is the angle between the normals to these surfaces at the point. The normal to is

2

1S 2S

1S

( ) ( ) ( )

( )

2 2 2 1

1 2 3 2, 2, 1

1 2 3

9 9

2 2 2

4 4 2

x y z

S r x y z

x y z = = =

= − = + +

= + +

= − +

e e e

e e e

2 −

(1)

In , the normal is: 2S

( ) ( )

( )

2 2

1 2 3 2, 2,

1 2 3

1

2

2

x 1y z

S x y z

z = =− =

= + + −

= + +

= + +