HÌNH HỌC VI PHÂN ( Differential geometry )

MINISTRY OF EDUCATION AND TRAINING
Ho Chi Minh City University Of Pedagogy
faculty of mathematics and natural sciences


Differential Geometry
Academic supervisor: DR. Nguyễn Hà Thanh

12/2015

MINISTRY OF EDUCATION AND TRAINING


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

Ho Chi Minh City University Of Pedagogy
Faculty of mathematics and natural sciences


Differential Geometry
Academic supervisor: DR. Nguyễn Hà Thanh
Team member
1. Lê Thu Hiền
2. Thông Minh Quang
3. Thân Thị Ngọc Chi
4. Nguyễn Thị Phương
5. Trần Thị Thanh Tâm
6. Nguyễn Thị Lan Anh
7. Lê Cao Phương Trúc

8. Nguyễn Thị Kim Yến
9. Nguyễn Thị Ngọc Nhung
10. Nguyễn Thị Thùy Trang
12/2015


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

Introduction
In this essay, we give some fundamental issues about curvess and surfaces theory with
each individual part:
Part 1: Curves
Chapter 1. Vector function
Chapter 2: Parametric curve
Chapter 3. Tangent, Normal.
Chapter 4.Arc length of a curve
Chapter 5. Curvature, Switchsion
Chapter 6. Frenet trihedron, the osculating plane
Chapter 7. The application of curves
Part 2: Surfaces
Chapter 1. Surface
Chapter 2.The osculating plane and normal, binormal
Chapter 3. First fundamental form
Chapter 4. First fundamental formand Arc Length of a Curve
Chapter 5. First fundamental formand angle
Chapter 6. First fundamental formand area
Chapter 7. Second fundamental form
Chapter 8. Normal curvature

Chapter 9. Principle curvature
Chapter 10. Gaussian curvature, Mean curvature
Chapter 11. Asymptotic
In each topic, beside theories, we also give some examples. At the end of each topic is the
system of various exercises in order to consolidate knowledge and train your math skills.
Hopefully this will be a good source of practice materials and other students who concern
with differential geometry.

Acknowledgement
This research was supported by DR. Nguyen Ha Thanh. We thank our team mates for
sharing their pearls of wisdom, their comments and collect data during doing this essay.
We would like to express our special appreciation and thanks to our advisor DR. Nguyen
Ha Thanh lecturer in Mathematics - Natural Science of HCM University of Pedagogy, you have
been a tremendous menswitch for us. Who has provided valuable knowlege to us in order to
complete this essay. In the process of teaching, you alway motivated and inspired us to lead us
explore the beauty of maths in general and especially in geometry.


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

CONTENTS
Part 1:Curves........................................................................................................................5
Chaequationer 1. Vector function............................................................................6
I. Theory.......................................................................................................6
II. Exercises..................................................................................................8
Chaequationer 2. Parametric curve........................................................................22
I. Theory.....................................................................................................22
II. Exercises................................................................................................27

Chaequationer 3. Tangent, Normal, Binormal.......................................................51
I. Theory.....................................................................................................51
II. Exercises................................................................................................56
Chaequationer 4..Arc Length.................................................................................67
I. Theory.....................................................................................................67
II. Exercise.................................................................................................71
Chaequationer 5. Curvature, Torison.....................................................................75
I. Theory.....................................................................................................75
II. Exercise.................................................................................................80
Chaequationer 6. Frenet Trihedron, The osculating plane ....................................86
I. Theory.....................................................................................................86
II. Exercise.................................................................................................94
Chaequationer 7.The Application of curves...........................................................98

Part 2: Surfaces................................................................................................................103
Chaequationer 1. Surface.....................................................................................104
I. Theory...................................................................................................104
II. Exercise...............................................................................................112
Chaequationer 2. The osculating plane and normal.............................................128
I. Theory...................................................................................................128
II. Exercise...............................................................................................130
Chaequationer 3. First fundamental form............................................................135
I. Theory...................................................................................................135
II. Exercise...............................................................................................137
Chaequationer 4. The first fundamental form and arc Length.............................146
I. Theory...................................................................................................146
II. Exercise...............................................................................................146
Chaequationer 5. The first fundamental form and and angle...............................153
I. Theory...................................................................................................153
II. Exercise...............................................................................................154

Chaequationer 6. Fist fundamental formand and area.........................................163
I. Theory...................................................................................................163
II. Exercise...............................................................................................164
Chaequationer 7. The second fundamental form.................................................172
I. Theory...................................................................................................172


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

II. Exercise...............................................................................................174
Chaequationer 8. Normal curvature.....................................................................184
I. Theory...................................................................................................184
II. Exercise...............................................................................................185
Chaequationer 9. Principle curvature...................................................................192
I. Theory...................................................................................................192
II. Exercise...............................................................................................194
Chaequationer 10. Gauss curvature, Mean curvature..........................................198
I. Theory...................................................................................................198
II. Exercise...............................................................................................199
Chaequationer 11. Asymptotic.............................................................................208
I. Theory...................................................................................................208
II. Exercise...............................................................................................209
References........................................................................................................................215


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

Part 1:
C

urves
Part 1: Curves

5


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

CHAPTER 1: VECTOR FUNCTION
I. THEORY
1. Definition:
n

Let I is the subset of Euclide space R . A map:
r
r : I ̮̮ R R n
r
t a r (t )  (ri (t ))
is called vector function on I .
Where:
i)
ii)

r

I : is the domain of r .
r
r
r
r
I
Image of through : support of .

Example:
r
3
a) r : R � R
r
t a r (t )  (a0  t , b0  t  , c0  t ) ( ,  ,  ) �(0, 0, 0)
�x  a0  t
r
pass  0 (a0 , b0 , c0 )
�
�
r
 �r ( R ), t : �y  b0  t  �  � �
directio
n
of

 ( ,  ,  )
�
�z  c  t
� 0
r

3
r
b) : R � R
r
t a r (t )  (a cos t , a sin t , bt ) (a, b  0)
�x  a cos t
r
�
 ( x, y, z ) �r ( R), t : �y  a sin t � x 2  y 2  a 2
�z  bt
�
We can also let:

I   a, b  ,  a, b  ,  a, b  ,  a, � ,  a, � ,...

Remark :
r
M 1  r (t1  t )  (a cos t , a sin t , bt )
r
M 2  r (t2  t  2 )  (a cos(t  2 ), a sin(t  2 ), b(t  2 ))  ( a cos t , a sin t , bt  2 b)
M 1M 2  2 b  const

2. Operations :

Let

u
r ur
r1 , r2 : I � R n

Part 1: Curves

6


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

u
r ur
t a r1 (t ), r2 (t )

:I �R
t a  (t )
We define these operations
u
r ur
r1  r2 : I � R n
●“Add”
u
r ur
u
r
ur
t a (r1  r2 )(t )  r1 (t )  r2 (t )
●“Multiply vector function with scalar function”

u
r

( r1 ) : I � R n

u
r
u
r
t a ( r1 )(t )   (t )r1 (t )
u
r ur
r1.r2 : I � R

●“ The dot product of 2 vector function ”

n
u
r ur
t a ( r1.r2 )(t )  �r1i r2i

u
r
r1 (t )  (r11 , r12 ,..., r1n )
ur
1
2
n
r
With 2 (t )  ( r2 , r2 ,..., r2 )
●“ The cross product of 2 vector function”

i 1

u
r ur
r1 �r2 : I � R 3
u
r ur
u
r
ur
t a ( r1 �r2 )(t )  r1 (t ) �r2 (t )

3. Derivative formulas:
u
r ur
u
r ' ur '
( r1  r2 )'  r1  r2
(1)
u
r
u
r
u
r'
( r1 ) '   ' r1   r1
(2)
u
r ur
u
r ' ur u

r ur '
(r1.r2 )'  r1 .r2  r1.r2
(3)
u
r ur
u
r ' ur u
r ur '
( r1 �r2 )'  r1 �r2  r1 �r2
(4)
4. Differentiable vector function:
r
n
Let r : I ̮̮ R R
r
t a r (t )  (ri (t ))
r
r (t ) differentiable in I � ri (t ) differentiable in I .
Part 1: Curves

7


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

Remarks:
r
r (t ) �C k ( I ) � ri (t ) �C k ( I ), i  1, n

r
r (t ) �C �( I ) � ri (t ) �C �( I ), i  1, n
II.EXERCISES
u
r u
r

3

Ex1: Let differentiable vector function in R
differentiable function. Prove that:
u
r ur
u
r ur
r1  r2 '  r1 ' r2 '
a.
.
u
r
u
r
u
r
f r1 '  f ' r1  f r1 '
b.
.
u
r ur
u

r ur u
r ur
r1.r2 '  r1 '.r2  r1.r2 '
c.
.
u
r ur
u
r ur u
r ur
r1 �r2 '  r1 '�r2  r1 �r2 '
d.
.
u
r ur ur
u
r ur ur
u
r ur ur
u
r ur ur
r1 , r2 , r3 '  r1 ', r2 , r3  r1 , r2 ', r3  r1 , r2 , r3 '
e.
.



ur

 I , r  r (t )  ,  I , r

with
1

1

2

ur
ur ur
 r2 (t ) , I , r3  r3 (t )



 and



 

 






 

 

 



Solution:
Without loss of generality we suppose:
u
r
ur
ur
r1 (t )   x1 (t ), y1 (t ), z1 (t )  ; r2 (t )   x2 (t ), y2 (t ), z 2 (t )  ; r3 (t )   x3 (t ), y3 (t ), z3 (t ) 
a. We have:
u
r ur
r1  r2 '   x1  x2 , y1  y2 , z  z2  '   ( x1  x2 ) ',( y1  y2 ) ', ( z1  z2 ) '   x1 ' x2 ', y1 ' y2 ', z1 ' z2 ' 
u
r ur
  x1 ', y1 ', z1 '   x2 ', y2 ', z2 ,   r1 ' r2 '
b. We have :
u
r
f r1 '   f .x1 , f . y1 , fz1  '   ( f .x1 ) ', ( f . y1 ) ', ( f .z1 ) ' 





 

  f '.x1  f .x1 ', f '. y1  f . y1 ', f '.z1  f .z1 ' 

  f '.x1 , f '. y1 , f '.z1    f .x1 ', f . y1 ', f .z1 ' 
u
r
u
r
 f '.r1  f .r1 '
c. We have :

Part 1: Curves

8


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh
urur

 r .r  '   x .x
1 2

1

Essay Differential geometry

 y1 . y2  z1.z2  '

2

  x1 '.x2  x1.x2 ' y1 '. y2  y1. y2 ' z1 '.z2  z1.z2 ' 

  x1 '.x2  y1 '. y2  z1 '.z2    x1 .x2 ' y1. y2 ' z1 .z2 ' 
u
r ur u
r ur
 r1 '.r2  r1.r2 '
d. We have:



u
r ur
�y
r1 �r2 '  � 1
�y2



z1 z1
,
z 2 z2



  y .z
'
1

2

'

y1 �
�
y2 �

x1 x1
,
x2 x2

 



 



 

 y1.z2 '  y2' .z1  y2 .z1' , z1' .x2  z1.x2'  z2' .x1  z2 x1' , x1' . y2  x1 y2'  x2' . y1  x2 y1'



 

 y1' .z2  y2 z1' , z1' x2  z2 x1' , x1' y2  x2. y1'  y1 z2'  y2' z1 , z1 x2'  z 2' x1 , x1 y2'  x2' y1





�y ' z ' z ' x1' x1' y1' � �y1 z1 z1 x1 x1 y1 �
�1 1 , 1
,

,
,
�y z z x x y �
� �y2' z2' z2' x2' x2' y2' �
2
2
2
2
2 � �
�
�2
u
r ur u
r ur
 r1 '�r2  r1 �r2 '
e) We have:
u
v uv uv
u
v uv uv u
v uv uv
u
v uv u
v uv uv u
v uv uv

( r1 , r2 , r3 ) '  (r1 �r2 ) '.r3  ( r1 �r2 ).r3 '  ( r1 '�r2  r1 �r2 ').r3  ( r1 �r2 ).r3 '
u
v uv uv u
v uv uv u
v uv uv u
v uv uv u
v uv uv u
v uv uv
 (r1 '�r2 ).r3  (r1 �r2 ').r3  (r1 �r2 ).r3 '  (r1 ', r2 , r3 )  (r1 , r2 ', r3 )  (r1 , r2 , r3 ')
4
2
4
4
In polar coordinates: r  2b.r .cos2 =a  b (Oval Cassini).

Ex2: Find the derivative of these funtions:
r 2 ur 2 u
r ur r r r
r r
r
r2
r , r ' , r1 �r2 , r ', r '', r ''' , r '�r '' �r ''', r

 







.

Solution:
r2
r2
r  a 2  b 2  c 2 � r '  2aa ' 2bb ' 2cc '
*) r
.
r '   a ', b ', c ' 
*)
r 2
2
2
2
� r '   a '   b '    c ' 

 

 
r
��
r '
�
�

2 '

� 2a '' a ' 2b '' b ' 2c '' c '
�

�
u
r ur �y z1 z1 x1 x1 y1 �
r1 �r2  � 1
,
,
�
y
z
z
x
x2 y2 �  y1.z2  y2 .z1 , z1.x2  z2 .x1 , x1. y2  x2 . y1 
2
2
2
2
�
*)
.
r r r
� r ', r '', r '''  a '''b 'c''  a '''b''c ' , a ''b'''c '  a 'b'''c '' , a 'b''c '''  a ''b'c ''' .



 
r r r
��
r
 a  b c  a b c
 ', r '', r ''' �

�
� 
'

Part 1: Curves

4

' ''

''' ' '''



 a   b''c'  a '''b'''c ' , b  a ''c '  b''' a '''c '  b   a 'c ''  b''' a 'c ''' , c   a 'b ''  c '''a 'b ''  c   a '''b '
4

4

9

4

4

4



ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

  y1' .z2  y1.z2'  y2' .z1  y2 .z1' , z1' .x2  z1.x2'  z2' .x1  z2 .x1' , x1' . y2  x1. y2'  x2' . y1  x2 . y1'  .

  y1' .z2  y2 .z1' , z1' .x2  z2 .x1' , x1' . y2  x2 . y1'    y1.z2'  y2' .z1 , z1.x2'  z2' .x1 , x1. y2'  x2' . y1  .
u
r ur u
r ur
 r1 �r2 ' r1 '�r2 .
r
r
r ''   a '' , b'' , c ''  , r '''   a ''' , b ''' , c ''' 
*)
.
r r r
''' ' ''
''' '' '
� r ', r '', r '''  a b c  a b c , a ''b'''c '  a 'b'''c '' , a 'b''c '''  a ''b' c''' .




 
r r r
��
 a  b c  a b c  a   b c  a b c , b  a c  b a c  b   a c  b a c , c   a b  c a b  c   a b 
�r ', r '', r ''' �
� 

r r
r '�r ''    b c  b c , c a  c a , a b  a b 

*)
.
r r
 r '�r ''   b c  b c , c a  c a , a b  a b  .
'

' ''

'

4

'' '

' '''

''' '

' ''

'

''

'

''

'''

4

''' ' '''

'

' ''

''' '

'' '

''' ''' '

4

'' '

''' ''' '

4

' ''





� r2 �
� r �
� �
'





' '''





 

*)
Ex3: These following statements are true or false?
r
r
r'  r '
a)
r r
r r
r�
r '  r �r '
b)
Solution:

r
r
r   a, b, c  � r '   a ', b ', c '
Called
r
r '  (a ') 2  (b ') 2  (c ') 2
We have :
r
2aa ' 2bb ' 2cc ' aa ' bb ' cc '
r '  a2  b2  c2 ' 

2 a 2  b2  c 2
a2  b2  c2
r
r
r ' 1
r  t ,1,1
a. With
then
r
t
r '
t2  2
t
1�
, t ��
2
t 2
We have

r
r
r'
r '

Part 1: Curves

''' ' ''

''' '

r2 '
r
2aa '  2bb '  2cc ' aa '  bb '  cc '


r2
2
2
2
2
a

b

c
a 2  b2  c 2
2 r
.



' ''

'' '

r r
r ' r r ' r
r r
r  4
�
��
r ��
' r ''  �
r '''� r ' r '' r ''' r ' r '' r
�
�
6
4
4
6
6
4
4
6
6
4
 4  '''
 6 '
 c a  c a  a 'b   a '''b  , a   b '''  a   b '  b 'c    b '''c   , b   c '''  b   c '  c 'a    c '''a  



4

''' ' '''



10



.

4

''' '


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

r
r  2t ,3,1

b.
r rWith
r�

r '   2t ,3,1 �
 2, 0, 0   4t
r r
r �r '  4t 2  10 �4  2 4t 2  10

r r
r
2
r�
r ' r
4
t
�
2
4
t

10,

t
��
We have :
so

r
r'

Ex4: Prove this following property of ellipse:
Tangent to ellipse at point M is the external bisecswitch of an angle created by two radius
through the focal point.

Solution:

x2 y 2
( E ): 2  2  1
a b
Let Ellipse be
�x  a cos t
�
Parametric equations: �y  b sin t
M �E so coordinates of the point M is
M (a cos t , b sin t )
2
2
�
�F1  ( a  b , 0)
�
�F  ( a 2  b 2 , 0)
Coordinates of the focus � 2

r
Direction vector of tangent line to ( E ) at M is u  ( a sin t , b cos t )
uuuur
uuuur
F1M  (a cos t  a 2  b 2 , b sin t ) F2 M  (a cos t  a 2  b 2 , b sin t )
,

r uuuur
 a 2 sin t cos t  a sin t a 2  b 2  b 2 sin t cos t
u.F1M
r uuuur

cos(u, F1M )  r uuuur  r
u . F1M
u . (a 2 cos 2 t  a 2  b 2  2a a 2  b 2 cos t  b 2 sin 2 t

a 2  b 2 sin t ( a 2  b 2 cos t  a )

 r
u . (a 2  b 2 ) cos 2 t  2a a 2  b 2 cos t  a 2

Part 1: Curves

11


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh



Essay Differential geometry

a 2  b 2 sin t ( a 2  b 2 cos t  a)
r

u . a 2  b 2 cos t  a

a 2  b 2 sin t
r
u

(1)

r uuuur
a 2 sin t cos t  a sin t a 2  b 2  b2 sin t cos t
u.F2 M
r uuuur
cos(u, F2 M )  r uuuur  r
u . F2 M
u . (a 2 cos 2 t  a 2  b 2  2a a 2  b 2 cos t  b 2 sin 2 t

 a 2  b2 sin t ( a 2  b 2 cos t  a)

 r
u . (a 2  b 2 ) cos 2 t  2a a 2  b 2 cos t  a 2
 a 2  b 2 sin t ( a 2  b 2 cos t  a )

r

u . a 2  b 2 cos t  a

We conclude by combining (1) and (2)

r uuuur
r uuuur
cos(u, F1M )  cos(u, F2 M )

a 2  b 2 sin t
r
u

(2)

. Or tangent to ellipse at M is

external bisecswitch of an angle created by 2 radius passes through focal point.



r r
I,r  r t



Ex5: Let vector function
. Prove that:
r
r
r
uuuur
r '(t )  0, t �I � r (t )  cont , t �I
a.
r
ur
r
r (t )  r '(t ), t � r (t )  const , t �I
b.
r
r
r
r
(

t
)
�
r
(
t
)
//
r
'(t ), t �I .
c.
have const direction
r
r
r
r
r
r
r
�
�
r
(
t
)
�
r
'(
t
)

.
r
� ''(t )  0 and r (t ) �r '(t ) �0 so r ( I ) lies on a plane.
d. If �
r
r
r
r
r
r
r
�
� 0, t �I  ( a, b)
r
'(
t
)
�
r
''(
t
)
r
'(
t
),
r
''(
t
)

�
0
r
�
�
e. If
and
so ( I ) lies in a line

Solution:
a. We have:

Part 1: Curves

r
r  t   const , t �I

12


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh
r
� r (t )  (a, b, c )
r
� r '  t    0, 0, 0  (dpcm)

Essay Differential geometry

, With a, b, c is const

b. We have:
r
r (t )  const , t �I
r
� r t

r 2
r
r
r
r
r
2 �
 r (t )  const � �r  t  � 0 � 2.r  t  .r '  t   0 � r  t   r '  t 
�
�
r
r
 � We have: r (t ) have const direction � exist a constant vector v and function   t 
c.
r
r r
r
r
r
r r r
r  t     t  .v � r '  t    �
t  .v    t  .v ' � r '  t    �
t  .v as v '  0



suchr that : r
r
� r  t  and r '  t  are same direction (because same direction with v ).
r
r
r
r
r
r (t )
r (t )  r (t ) . r
 r (t ) .e
r
r (t )
e 1
 � We have
With
deduce:
r
r
r r
r
r '(t )  r  t  '.e  r  t  .e '
. other surface:
r
r
k ��\  0
r  t  and r '  t 
same direction that means exist
such that

r
r
r
r
r
r
r r
r
r
r
r
r
r
r  t   k .r '  t  � r  t   r  t  .e  k . r  t  '.e  r  t  .e ' � r  t   k . r  t  ' .e  k . r  t  .e '







2





 

r

r
e 1
e
since
multiply 2 sides with we have:
r
r
r2
r
u
rr
r
r
u
rr
r  t   k . r  t  ' .e  k . r  t  .e.e ' � r  t   k . r  t  '  0 do e.e '  0
r r
r
r uuuuur
r
e
'

0
do
k
.
r
t
�

0
�
e
 const


r
�
. So (t ) have constant direction.
r
r
r
r
r
r
�
r (t ) �r '(t ) �
.r ''(t )  0
r
(
t
)
�
r
'(
t
)
�
0
�

�
d. Since
and











r
r
ur
r  t  �r '  t 
  t  r
r
u
r
r  t  �r '  t  �   t   1
Let r
u
r
r
ur
r
ur

r  t  .  t   r '  t  .  t   r ''  t  .  t   0
So:
r
ur
r
ur
r
ur
r
ur
�
r
t
.

t

0
�
r
'
t
.

t

r
t
.


'
t

0
�
r
t
.

' t   0














�
ur
r
ur
r
ur

r
ur
�r
r
'
t
.

t

0
�
r
''
t
.

t

r
'
t
.

'
t

0
�
r

'
t
.

' t   0














�
u
r
u
r
u
r
ur
ur
  t   1 �   t    ' t 
  t

 ' t 
deduce
same direction with
; Other surface :
.
r
u
r
uuuuur
u
r
r
u
r
uuuuur
r  t     t   const
 '  t   0 �   t   const
So
deduce
.



Part 1: Curves

13



ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh
where M is the image of

r
r t

Essay Differential geometry

r
uuuur uuuuur
r
 t  have an image lies on a
so OM  const . Deduce
u
r
uuuuur
  t   const





plane passes through O and recieve
is normal vector.
r
r
r
r
r
�

r '(t ) �r ''(t ) �
� 0, t �I  (a, b) deduce r '(t ) and r ''(t ) have the same direction,
e. We have: �
r
r
r
r
r '(t )    t  .a
r
'(
t
)
as c above
have constant direction, that means :
where a is a const
vector .

r
r
r
r t  �
r '(t )dt  �
  t  .adt 

r

r

r
  t  dt  .a  b

�
b
With is const

Integral both sides, we have:
r
r
r
(
I
)
b
vector. Deduce image of
lies in a line passes through point and have direction vector is
r
a
ur u
r ur uuuuur
r0 , r1 , r2  const

Ex6: Let
. Find the support of these following vector :
r
ur u
r 2 ur
r  t   r0  t.r1  t .r2 , t ��
a.
.
r
ur

u
r
ur
r  t   r0  cost .r1  sin t .r2 ; t � 0, 2 
b.
.
r
ur
u
r
ur
r  t   r0  cht.r1  s ht.r2 ; t ��
c.
.
u
r
ur
r1
r2
Consider the case
same direction with
.
Solution:
2
O, M , M 0
On plane E choose
such that:

uuuu
r r

OM  r  t 
uuuuu
r ur
OM 0  r0
In this floating point M ,we have

O, M 0

const

a. We have :
r
ur u
r
ur
r
ur
u
r
ur
uuuuuu
r
u
r
ur
r  t   r0  t.r1  t 2 .r2 � r  t   r0  t .r1  t 2 .r2 � M 0 M  t .r1  t 2 .r2 , t ��
ur
ur
r1
r2


is not the same direction with :

Part 1: Curves

14

(1)


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

�pass M 0

Deduce: M belong to plane

In

 P  choose the basic 

 P : �
�

u
r
ur
direction vector are r1 and r2
�

.

u
r ur
M 0 , r1 , r2



so

�x  t
M  x, y  : � 2 � y  x 2
�y  t

, deduce: in

 P

M move

2
along parabol y  x .



ur
r1

same direction with

uuuuuu
r
u
r
ur
ur
M 0 M  t.r1  t 2 .r2   t 2  k .t  .r2

Let :

f  t   t 2  k .t �k0

ur
r2

. Deduce:

ur
ur
k ��: r1  k .r2

, from(1) deduce:

�passes M 0

. Deduce: M belong to

  : �
�

ur
direction vector r2
�

  .
. Deduce point M only move along half of

b. We have:
r
ur
u
r
ur
r
ur
u
r
ur
uuuuuu
r
u
r
ur
r  t   r0  cost.r1  sin t .r2 � r  t   r0  cost .r1  sin t.r2 � M 0 M  cost.r1  sin t.r2 , t ��(2)



ur
r1

is not same direction with

ur
r2

:

qua M 0
�

Deduce: M belong to plane

In

 P  choose the basis 

 P : �
�

u
r
ur
�vtcp là r1 và r2 .

u
r ur
M 0 , r1 , r2



�x  cos t
M  x, y  : �
� x2  y 2  1
 P  point
�y  sin t
so
. Deduce: in

2
2
M move along Ellipse x  y  1 .
ur
ur
ur
u
r
r1
r2
k ��: r2  k .r1

same direction with . Deduce:
, from (2) deduce:

uuuuuu
r
u
r
ur
u
r

M 0 M  cost .r1  sin t .r2   cost  k .sin t  .r1

Part 1: Curves

. Deduce: M belong to a line

15

qua M 0
�
ur
�vtcp r1

  : �
�


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Let

f  t   cost  k .sin t �  1  k 2 �f  t  � 1  k 2

segment belong to

Essay Differential geometry

. Deduce point M only move along a

  .

c. We have:
r
ur
u
r
ur
r
ur
u
r
ur
uuuuuu
r
u
r
ur
r  t   r0  cht.r1  s ht.r2 � r  t   r0  cht.r1  s ht.r2 � M 0 M  cht.r1  s ht .r2 , t ��(3)



ur
r1

ur
r2

is not same direction with

:

qua M 0
�

Deduce: M belong to plane

 P : �
�

u
r ur
u
r
ur
M
,
r
�vtcp là r1 và r2 . In  P  choose the basis 0 1 , r2



�x  c ht
M  x, y  : �
� x 2  y 2  c h 2 t  s h 2t  1
�y  s ht
so
.



 P  point M move along hybebol x 2  y 2  1 .
Deduce:
in
ur
ur
ur
u
r
r1
r2
k ��: r2  k .r1
same direction with . Deduce:
, from (3) deduce:

uuuuuu
r
u
r
ur
u
r
M 0 M  cht.r1  s ht.r2   cht  k .sht  .r1

. Deduce: M belong to a line

qua M 0
�
ur
�vtcp r1 .

  : �
�

ur r
r
�
ru '  r (u , v )
�
r (u , v)  const � �ur r
r r
2
rv '  r (u , v )
�
(
U
,
r

r
(
u
,
v
))
Ex7: Let
, U �R . Prove that:
Solution:

r
r

2
r (u , v)  const � r (u , v )  const
ur r
r2
�
2ru '.r (u , v)  0
�
� r (u , v)  const � �ur r
2rv '.r (u, v)  0
�
ur r
ur r
�
�
r
'.
r
(
u
,
v
)

0
r
�u
�u '  r
� �ur r
� �ur r
rv '.r (u, v)  0

rv '  r
�
�

Part 1: Curves

16




ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

r r
r r
2
ru ', rv '
(
U
,
r

r
(
u
,
v
))

U
�
R
Ex8:
,
and
are not the same direction. Prove that: the support
r Let
r
(
u
,
v
)
of
is a plane.

Solution:

r
r
�
ru '  r  u , v 
�
r
�r
rv '  r  u, v 
�
We have:
uuuuur

r r
r
r
ru ', rv '
is not the same direction, so r (u, v)  const .Deduce support of r (u, v) is a plane.

Ex9:Let

ur ur ur ur
r0 , r1 , r2 , r3

ur ur ur
r1 , r2 , r3

With

a.

r ur
ur
ur
ur
r  r0  u.r1  u 2 .r2  v.r3

b.

r ur
ur
ur
ur

r  r0  cos u.r1  sin u.r2  v.r3

linearly independent . Find the support of vector functions :

r ur � 1 �u
r � 1 �ur
ur
r  r0  �
u �
.r1  �
u �
.r2  v.r3
� u� � u�
c.

d.

r ur
ur
ur
ur
r  r0  u cos v.r1  u sin v.r2  u 2 .r3

Solution:

With

O, M 0

uuuuu

r ur
uuuur r
OM 0  r0
OM

r
fixed point , M floating point , Let :
,

r ur
ur
ur
ur
r  r0  u.r1  u 2 .r2  v.r3
a.
uuuuuu
r
u
r 2 u(1)
r
ur
� M 0 M  u.r1  u .r2  v.r3
(1)
ur ur ur
r1 , r2 , r3

linearly independent sou
to three-dimensional passes M 0 and have
rMur belong
ur

r ,r ,r
system of equations vector direction are 1 2 3 .
u
r ur ur
M 0 , r1 , r2 , r3
In P choose
is the basis so:
Because



Part 1: Curves



17


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh
�x  u
�y  x 2
�
2
M  x, y, z  : �y  u � �
�z  v
�z  v
�

Essay Differential geometry

.

b.

r ur
ur
ur
ur
r  r0  cos u.r1  sin u.r2  v.r3

(2)

uuuuuu
r
ur
ur
ur
� M 0 M  cos u.r1  sin u.r2  v.r3

(2)

ur ur ur
r1 , r2 , r3

linearly independent sou
to three-dimensional P passes M 0 have
rMurbelong
ur
r ,r ,r
system of equations vector direction are 1 2 3 .

u
r ur ur
M 0 , r1 , r2 , r3
In P choose
is the basis so:
Because





�x  cos u
�x 2  y 2  1
�
M  x, y, z  : �y  sin u � �
�z  v
�z  v
�

.

r
2
2
Deduce: Support of r is the intersection line between surface cylinders ellipse x  y  1 and
Plane z  v .
r ur � 1 �u
r � 1 �ur
ur
r  r0  �

u �
.r1  �
u �
.r2  v.r3
� u� � u�
c.
(3)
uuuuuu
r � 1 �u
r � 1 �ur
ur
(3) � M 0 M  �
u �
.r1  �
u �
.r2  v.r3
� u� � u�

ur ur ur
r1 , r2 , r3

linearly independent so u
to three-dimensional P passes M 0 have
rMurbelongs
ur
r ,r ,r
System of equations vector direction is 1 2 3 .
u
r ur ur
M 0 , r1 , r2 , r3

In P choose
is the basis so:
Because





1
�
�x  u  u
�
�x 2 y 2
2
2
�
x y 4
1
1
�
� 
M  x, y, z  : �y  u  � �
� �4
4
u
�z  v
�
�z  v
�
�z  v

�
�
.

Part 1: Curves

18


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

x2 y 2
r

1
4
Deduce: Support of r is hypebol 4
lies on a plane z  v .

r ur
ur
ur
ur
r  r0  u cos v.r1  u sin v.r2  u 2 .r3
d.
uuuuuu
r
ur

ur(4) 2 ur
� M 0 M  u cos v.r1  u sin v.r2  u .r3
(4)
ur ur ur
r1 , r2 , r3
M0
Because
linearly independent so u
M
have
r ubelong
r ur to three-dimensional P passes
r ,r ,r
system of equations vector direction is 1 2 3 .
u
r ur ur
M 0 , r1 , r2 , r3
In P choose
is the basis so:





�x  u cos v
�
M  x, y, z  : �y  u sin v � x 2  y 2  z � x 2  y 2  z  0
�z  u 2
�

.

r
2
2
Deduce: Support of r is x  y  z  0 .

Ex10:Write a parametric equations of
a. Sphere
b. Ellipsoid
c. Elliptic paraboloid, paraboloid hyperbolic
d. Hyperboloid of one sheet
e. Elliptic cylindrical surface, hyperboloid
Solution:
2
2
2
a. General equation of sphere : x  y  z  1

2
2
�
�x � �y �
�
�
� �
� 1
�x 2  y 2  cos 2 u � �
cos u � �cos u �
�

u �R : �
�z  sin u
�z  sin u
�

Part 1: Curves

19


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

�x
�cos u  cos v
�
�y
v �R : �
 sin v
�x  cos u cos v
cos
u
�
�
� �y  cos u sin v
�z  sin u
�z  sin u
�
�

�
2

2

2

x
y
z
 2  2 1
2
b
c
b. General equation of ellipsoid is: a

�x 2 y 2
 2  cos 2 u
�
�a 2
b
u �R : �
�z  sin u
�
�c
� x
�a cos u  cos v
�
� y
v �R : �

 sin v
�b cos u
�z  c sin u
�
�

2
2
�
� x � � y �
�
�
� �
� 1
��
�a cos u � �b cos u �
�z  c sin u
�

�x  a cos u cos v
�
� �y  b cos u sin v
�z  c sin u
�
2

2

y
z

 2  2x
2
b
c. General equation of elliptic paraboloid is: a

(1)

�y 2 z 2
�
2
2
2
2
y
z
2
2 � ( x  1)  �


x
� 1
2
2
(1) � 2  2  x  1  ( x  1)
a
b
�
�
a
b

�x  1  chu
�x  chu  1
� 2
� 2
2
2
u �R : �y
�
�y
z
z
2
2
� 2  2  x  sh u
� 2  2  2(chu  1)
b
b
�a
�a

�x  chu  1
�
2
2
��
�
� �
�
y
z


1
�
�
�
�a 2(chu  1) �
� �
�
�
�
� �b 2(chu  1) �
�

Part 1: Curves

20


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh
�
�
�x  chu  1
�
y
�
� v �R : �
 cos v
a
2(
chu


1)
�
�
z
�
 sin v
b
2(
chu

1)
�
�

Essay Differential geometry

�x  chu  1
�
� �y  a 2(chu  1) cos v
�
�z  b 2(chu  1) sin v
2

2

y
z
 2  2x
2

b
 General equation of paraboloid hyperbolic is: a
(1)
2

2

y
z
(1) � 2  2  x 2  1  ( x  1)2
a
b

�y 2 z 2
�
� ( x  1) 2  � 2  2  x 2 � 1
b
�a
�

�x  1  chu
�x  chu  1
� 2
�
2
u �R : �y
� �y 2 z 2
z
2
2

� 2  2  x  sh u
� 2  2  2(chu  1)
b
b
�a
�a
�x  chu  1
�
2
2
��
�
� �
�
y
z

1
�
�
�a 2(chu  1) �
� �
�b 2(chu  1) �
�
�
� �
�
�
�
�

�x  chu  1
�
y
�
� v �R : �
 c hv
�a 2(chu  1)
�
z
�
 s hv
�
�b 2(chu  1)

�x  chu  1
�
� �y  a 2(chu  1) c hv
�
�z  b 2(chu  1) s hv
2

2

x
y
z2
 2  2 1
2
b
c

d. General equation of hyperboloid of one sheet is: a
(1)

�x 2 z 2 � y 2
(1) � � 2  2 � 2  1
c �b
�a

Part 1: Curves

21


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh
�x 2 z 2
 2  ch 2u
�
�a 2
c
u �R : �
�y  shu
�
�b
�x
�achu  cos v
�
�z
� v �R : �
 sin v
cchu

�
�y  bshu
�
�

Essay Differential geometry

2
2
�
�x � �z �
�
�
� �
� 1
��
�achu � �cchu �
�y  bshu
�

�x  achu cos v
�
� �y  bshu
�z  cchu sin v
�
2

x
z2


1
2
c2
e. General equation of elliptic cylindrical surface: a

�x
 cos u
�
�a
u �R : �
�z  sin u
�c

�x  a cos u
��
�z  c sin u
2

x
z2

1
2
c2
 General equation of hyperbolic is: a

�x
 c hu
�
�a

u �R : �
�z  s hu
�c

�x  a c hu
��
�z  c s hu

2
 General equation of parabolic is: y  2 px (1)
2

�x  p � y 2  x 2
�
1
�
�
p2
(1) � y 2  x 2  p 2  ( x  p ) 2
� p �
�x  p
� p  chu
�
� u �R : � 2
2
�x  p(chu  1)
�y  x  sh 2u � �
�
2
�

�y  �p 2(chu  1)
� p

Part 1: Curves

22


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

Essay Differential geometry

CHAPTER 2: PARAMETRIC CURVE
I.THEORY
* Definition of parametric curve:

r
3
r
Given : I � R

r
t a r (t )  ( x (t ), y (t ), z (t ))

r
a, b  ;  a, b  ;  a, b  ;  a, b  ;  a, � ;  a, � ;  �a  ;  �, a  ;...

I
r
Where :

a map is
parametric curve.
r
r r
(
I
,
r
)
or
(
I
,
r
 r (t ))
Denote:
Where :
r
r
- ( I ) is the support of parametric curve.
3
Example: parametric curve of circle center O, radius r  0 in R is
r
r (t )  (cos t ,sin t , 0).

* Type of parametric curve:
1. Differentiable parametric curve
Differential parametric curve degree k or parametric curve class
r
k

k
r  t  �C
� x  t  , y  t  , z  t  �C
I
I
2. Compact parametric curve
r
( I , r ) is parametric curve compact if I compactin R or I   a, b  .
3. Closed rparametric curve
( I , r ) is Closed parametric curve if:
r
(
I
,
r
) is compact parametric curve
i)
r
r
r
(
a
)

r
(b) .
ii)
4. Principal parametric curve
r
(

I
,
r
) is parametric curve principal at t0 if
r
(
I
,
r
) is parametric curve principalon I if
Part 1: Curves

23

r
r '(t0 ) �0

.
r
r '(t0 ) �0, t �I .

Ck

I That means :


ACADEMIC SUPERVISOR: DR. Nguyễn Hà Thanh

-

r
r '(t0 )  0

we say

t0

Essay Differential geometry

is not principal point

Example:

r
3
r
a. Given parametric curve : R � R
r
t a r  t    1  t , t 2  2, 0 

We have:

ur
r
r�
 t    1, 2t , 0  �0, t �R

b. Given parametric curve

r

r
so is principal parametric curve.

r
r :  0, 2  � R 3

r
t a r      cos  2cos  1 , sin   2cos  1 , 0 

We have:

ur
r�
     4sin  cos  sin  , 2 cos 2   2sin 2   cos , 0 

ur
r�
    5  4 cos  �0,  � 0, 2 

Deduce
so
r
So r is principal parametric curve.
r
3
r
c. Given parametric curve : R � R
r
t a r  t    x  t  , y  t  , 0

ur
r
r�
   �0,  � 0, 2 

.

0
, t �0
�
�
y  t   �2
1
��
t sin ��
,t  0
2
�
x t  t
t
��
�
Where
and
ur
r
x�
0  y�
0  0 � r�
0   0.




We have:
r
So r is not principal at t  0 , deduce It’s not principal on R .
5. Equivalent parametric curve
r
ur
ur
r
(
I
,
r
(
t
)),
(
J
,

(
s
)).
r
Given 2 parametric curve
and  are called 2 equivalent parametric
t , s are called 2 corresponding point). That means
curves

r . if exist a transformation parameter (

r : I � R3

and

r
t a r (t )
ur
 : I � R3

ur
s a  ( s)
are 2 equivalent parametric curve if exist a diffeomorphis:
Part 1: Curves

24