1.
page 9, Eigenvalues and Eigenvectors (21/9/2001)
Original version:
Let P  x1 x 2  x n  be the matrix whose columns are the n eigenvectors of A. Then
AP  A x1 A x 2  A x n   1 x1 2 x 2  n x n 
 1 x1  02    0n 01  2 x 2  03    0n  01    0n 1  n x n 
Revised version:
Let P  x1 x 2  x n  be the matrix whose columns are the n eigenvectors of A. Then
AP  A x1 A x 2  A x n   1 x 1 2 x 2  n x n 
 1 x 1  0 x 2    0 x n 0 x1  2 x 2  0 x 3    0 x n  0 x 1    0 x n 1  n x n 
2.
The following example of solving a system of linear equations has been added to page 13 of the
note of Matrix Algebra.
Example
Suppose the following matrix in reduced row echelon form is the augmented matrix of some
inhomogeneous system. Solve the system and express the solutions in vector form.
0

0
0

0
1 2 0 0 1
0 0 1 0 0

0 0 0 1 2

0 0 0 0 0
Sol:
As there are no pivots in columns 1 & 3 , x1 , x3 are therefore free i.e. you can assign parameters to them.
However, there are pivots in columns 2 , 4 , 5, x2 , x4 , x5 are therefore restricted i.e. they must be
expressed in terms of the free variables or the constants.
Let x1  s, x3  t; where s, t are any real numbers.
From row three we have x5  2 .
From row two we have x4  0 .
From row one we have x2  1  2 x3  1  2t .
 x1   s   0   s   0   0   1   0 
  
     
      
 x 2   1  2 t   1   0    2t   1   0    2 
 x    t    0    0    t    0   s  0   t  1  , where s, t are any real numbers.
 3 
     
      
 x4   0   0   0   0   0   0   0 
 x   2   2 0  0   2 0  0 
     
      
 5 
1
3.
Page 6, Real-valued Functions of Several Variables
Original Version:
Limit, Continuity and Differentiability
The concepts of limit, continuity and differentiability naturally generalize from the one variable situation.
Two variables will be considered as the extension to three or more variables will be clear.
Revised Version:
Limit, Continuity and Partial Derivative
The concepts of limit, continuity and derivative naturally generalize from the one variable situation. Two
variables will be considered as the extension to three or more variables will be clear.
4.
Page 7, Real-valued Functions of Several Variables (24/9/2001)
Original Version:
Examples
(ii)
Show that f ( x, y ) 
x2
has no limit at the origin.
x2  y2
Proof:
x2
As ( x, y )  (0,0) along the line y = 0, f ( x,0)  2  1 and the limiting value is 1.
x
As ( x, y )  x0 , y0  along the line x  0, f (0, y ) and the limiting value is 0.
Revised Version:
Examples
(ii)
x2
Show that f ( x, y )  2
has no limit at the origin.
x  y2
Proof:
x2
As ( x, y )  (0,0) along the line y = 0, f ( x,0)  2  1 and the limiting value is 1.
x
As ( x, y)  0,0 along the line x  0, f (0, y ) and the limiting value is 0.
5.
Page 10, Real-valued Functions of Several Variables
Original Version:
Recall that for a function f(x) of one variable
f x0  h   f x0 , y 0 
df
x0   lim
h 0 
dx
h
Revised Version:
Recall that for a function f(x) of one variable
f ( x0  h )  f ( x0 )
df
x0   lim
h 0
dx
h
2
6.
Page 10, Real-valued Functions of Several Variables
Original Version:
made counterclockwisely from the positive direction of x-axis to the line x  t cos  , y  t sin  and t is
any real number.
Then if cos  0
lim f ( x, y )  lim f (t cos  , t sin  )  lim
( x , y ) ( 0 , 0 )
t 0
t 0
x t cos
2t 3 cos  sin 2 
t 2 cos 2   t 4 sin 4 
y t sin
2t 3 cos  sin 2 
3
2 cos  sin 2 
t
 lim 2
 lim
 0  f (0,0)
t 0 t cos 2   t 4 sin 4 
t 0 cos 2 
4
 t sin 
t3
t
Revised Version:
made counterclockwisely from the positive direction of x-axis to the line x  t cos  , y  t sin  and t is
any real number.
Then if cos  0
lim f ( x, y )  lim f (t cos  , t sin  )  lim
( x , y )( 0 , 0 )
t 0
x t cos
t 0
2t 3 cos  sin 2 
t 2 cos 2   t 4 sin 4 
y t sin
2t 3 cos  sin 2 
2
2t cos  sin 2 
0
t
 lim 2
 lim

 0  f (0,0)
t 0 t cos 2   t 4 sin 4 
t 0 cos 2   t 2 sin 4 
cos 2 
t2
7.
Page 13, Real-valued Functions of Several Variables (27/9/2001)
Original Version:
f (0, y )  f (0,0)
00
f y (0,0)  lim
 lim
0
x 0
x 0
y
y
Revised Version:
f (0, y )  f (0,0)
00
f y (0,0)  lim
 lim
0
y 0
y

0
y
y
8.
Page 34,35, Real-valued Functions of Several Variables (11/10/2001)
Original Version:
Examples
(i)
f  x, y   x 2  2 y 2
f
f
 2x
 4y
x
y
f
f
 0 and
 0  ( x, y )  (0,0) is the only stationary point.
x
y
3
f (0,0)  0 and clearly f ( x, y )  0 hence this is a local minimum.
f ( x, y)  x 2  2 y 2
y
x
(ii)
f  x, y   2 x  x 2  2 y 2
f
f
 2  2x
 4 y
x
y
f
f
 0 and y  0   x, y   1,0 
x
is the only stationary point.
f  x, y   f 1,0   2 x  x 2  2 y 2  1
  x  1  2 y 2
2
 0   x, y   1,0 
Thus f ( x, y )  f (1,0) and (1,0) is a local maximum point.
f ( x, y)  2 x  x 2  2 y 2
y
x
(iii)
4
f  x, y   x 2  2 y 2
f
f
 2x
 4y
x
y
f
f
 0 and
 0   x, y   0,0 
x
y
is the only stationary point and f(0,0) = 0.
However f(x,y) may be either positive or negative depending on the relative magnitudes of x and y .
If y  0
f ( x,0)  x 2  0 always
If x  0
f (0, y)  2 y 2  0 always
Thus (0,0) is a saddle point
f ( x, y)  x 2  2 y 2
y
y
f<0
f 0
f>0
x
f  0 if x  2 y
x
f  0 if x  2 y
f<0
Revised Version:
Examples
(i)
f ( x, y )  x 2  2 y 2 , f x ( x , y )  2 x , f y ( x , y )  4 y
f x ( x, y )  2 x  0 & f y ( x, y )  4 y  0  ( x, y )  (0,0) is the only stationary point.
f xx x, y   2, f yy x, y   4, f xy ( x, y )  0 and f xx 0,0  2  0
f xx 0,0 f yy 0,0  f xy2 (0,0)  8  0 .
f (0,0)  0 is a local minimum and clearly f ( x, y )  0 hence this is a global minimum.
5
f ( x, y)  x 2  2 y 2
y
x
(ii)
f ( x, y )  2 x  x 2  2 y 2 , f x ( x, y )  2  2 x, f y ( x, y )  4 y
f x ( x, y )  2  2 x  0 & f y ( x, y )  4 y  0  ( x, y )  (1,0) is the only stationary point.
f xx  x, y   2, f yy  x, y   4, f xy ( x, y )  0 and f xx 1,0  2  0
f xx 1,0 f yy 1,0  f xy2 (1,0)  8  0 .
f (1,0)  1 is a local maximum. In addition,
f ( x, y )  f (1,0)  2 x  x 2  2 y 2  1  x  1  2 y 2  0
Thus f ( x, y )  f (1,0) and (1,0) is a global maximum point.
2
f ( x, y)  2 x  x 2  2 y 2
y
x
(iii)
f ( x, y )  x 2  2 y 2 , f x ( x, y )  2 x, f y ( x, y )  4 y
f x ( x, y )  2 x  0 & f y ( x, y )  4 y  0  ( x, y )  (0,0) is the only stationary point.
f xx x, y   2, f yy x, y   4, f xy ( x, y )  0
Since f xx 0,0 f yy 0,0  f xy2 (0,0)  8 , (0,0) is a saddle point.
(0,0) being a saddle point can also be observed by noting that f ( x, y ) may be either positive or negative
depending on the relative magnitudes of x and y .
6
If y  0 , x  0, f ( x,0)  f (0,0)  x 2  0 . If x  0, y  0
Thus (0,0) is a saddle point
f (0, y )  f (0,0)  2 y 2  0 .
f ( x, y)  x 2  2 y 2
y
y
f<0
f 0
f>0
x
f<0
9.
Page 43, Real-valued Functions of Several Variables (12/10/2001)
Original Version:
Q
s
x  x cos 
y   sin 
y

x
P
From Taylor’s Theorem
f 

f
x
f
 c o  s s s i
y
n
f
f
x  y
x
y
f f
f

cos   sin 
x x
y
And as s  0
df f
f

cos   sin 
ds x
y
(*)
Revised Version:
Q
s

P
f  0 if x  2 y
x
y
x
x  s cos
y  s sin 
From Taylor’s Theorem
7
f  0 if x  2 y
f
f
x  y
x
y
 f f
f

cos  
sin 
s x
y
f 
f 
f
x
x 
f
y
y
df f
f

cos   sin   (*)
ds x
y
10.
Page 3, Vector Calculus (17/10/2001)
Original Version:
If at each point (x,y,z) there is an associated vector
v( x, y, z)  v1 ( x, y, z)i  v2 ( x, y, z) j  v2 ( x, y, z)k
And as s  0
Revised Version:
If at each point (x,y,z) there is an associated vector
v( x, y, z )  v1 ( x, y, z )i  v2 ( x, y, z ) j  v3 ( x, y, z )k
11.
Page 4, Vector Calculus
Original Version:
5.4.1. Divergence of a Vector Field
If v( x, y, z )  vi ( x, y, z )i  v2 ( x, y, z ) j  v3 ( x, y, z )k
Revised Version:
5.4.1. Divergence of a Vector Field
If v( x, y, z )  v1 ( x, y, z )i  v2 ( x, y, z ) j  v3 ( x, y, z )k
12.
Page 6, Vector Calculus
Original Version:
Example
The electric field at position r due to a point charge at the origin is
Q r
Q
xi  yj  zk
E

3
3
4 0 r
4 0
4 0 x 2  y 2  z 2 2


Q  
x
  similar terms
divE    E 
3
4 0 x  2
2
2 2 
 x  y  z  

Q 1
 3 2 x  similar terms
x

3

4 0  r
2 r 5 


Q  3 3 x2  y2  z2 
 

4 0  r 3
r5

divE  0
except at the origin.

Revised Version:
The electric field at position r due to a point charge at the origin is
8
E
r
Q

3
40 r
40
xi  yj  zk
Q
 divE    E 
x
 y 2  z 2 2
3
2


 
x
  similar te rms
3

40 x  2
2
2
 x  y  z 2 
Q
Q 1
 3 2x 
Q  3 3x 2  y 2  z 2 

x

similar
te
rms


0
40  r 3
2 r 5 
40  r 3
r5

divE  0 except at the origin.

13.
Page 7, Vector Calculus
Original Version:
We define
 


curl v    v   i
 j
 k   ( v1i  v2 j  v3k )
y
z 
 x
i


z
v1
j

y
v2
k
 v
 v
v   v v 
v 

 i  3  2   j  1  3   k  2  1 
z
z   z
x 
y 
 y
 x
v3
Revised Version:
We define
 


curl v    v   i
 j
 k   ( v1i  v2 j  v3k )
y
z 
 x
i


x
v1
j

y
v2
k
 v
v 

 i  3  2  
z
z 
 y
v3
14.
Page 5, page 6, Vector Calculus (19/10/2001)
 v
v 
 v v 
j  1  3   k  2  1 
x 
y 
 z
 x
Taking the midpoint of a face as a representative point, the flux through the face ABCD out of the box is
x
v1 ( x 
, y, z ) y z
2
( v2 and v3 are parallel to ABCD and make no contribution)
Similarly the flux through face A' B' C' D' into the box is
v1 ( x 
x
2
, y, z ) y z
and the net flux out of the box is
9
 
x
x



v1  x  2 , y, z   v1  x  2 , y, z yz



 
x v1

x, y, z     v1 x, y, z   x v1 x, y, z   yz
 v1  x, y, z  
2 x
2 x


v
 1 xyz  
x
Similarly by considering the two other pairs of faces, we have that the total flux out of the box is
 v1 v 2 v3 
 x  y  z xyz  


If we consider the flux per unit volume and take the limit as the volume of the box shrinks to zero we
obtain
v
v v
divv  1  2  3
x
y
z
Revised Version:
Taking the midpoint of a face as a representative point, the flux through the face ABCD out of the box is
x
v1 ( x 
, y, z ) y z
2
( v2 and v3 are parallel to ABCD and make no contribution)
Similarly the flux through face A' B' C' D' into the box is
v1 ( x 
x
2
, y, z ) y z
and the net flux out of the box is
x
x
 



v1  x  2 , y , z   v1  x  2 , y , z yz



 
2
3
2











v
x
,
y
,
z

v
x
,
y
,
z
 3 v1  x, y , z 
1

x
1

x
1

x




1
1


  v1  x, y , z  
  
v
x
,
y
,
z

  


1
2
3


1! 2
x
2!  2 
x
3!  2 
x


2
3
2











v
x
,
y
,
z

v
x
,
y
,
z
 3 v1  x, y , z 
1

x
1

x
1

x






1
1
 v1  x, y , z     
yz


  
v
x
,
y
,
z






1
2
3


1
!
2

x
2
!
2

x
3
!
2

x








 2 x  v1  x, y , z  2  x  3  3 v1  x, y , z 


  


yz
x
3!  2 
x 3
1! 2

  v  x, y , z  2 x 2  3 v1 x, y , z 

 1

 xyz
3
3
x
3! 2
x


Similarly by considering the two other pairs of faces, we have that the total flux out of the box is
10
  v1  x, y , z   v 2  x, y , z   v 3  x, y , z 



xyz
x
y
z


 2 x 2  3 v1  x, y , z  2 y 2  3 v 2  x, y , z  2 z 2  3 v 3  x, y , z  



xyz
3
3! 2 3
3! 2 3
x 3
y 3
z 3
 3! 2

If we consider the flux per unit volume , we have
2
2
2
  v1  x, y, z   v 2  x, y , z   v3  x, y , z   2 x   3 v1  x, y , z  2 y   3 v 2  x, y , z  2 z   3 v3  x, y , z  







  3! 2 3
x
y
z
x 3
3! 2 3
y 3
3! 2 3
z 3

 

15.
Page 2, Multiple Integrals
Original Version:
(1)
(2)
y
y
d
d
Sy
Sy
c
c
x
a
Sx
x
a
b
Sx
b
Revised Version:
(1)
y
(2)
y
d
d
y
y
c
c
x
a
x
b
a
x
x
b
x
16.
Page 2, Page3 Multiple Integrals
Original Version:
Example
Evaluate
  (2 xy  y
2
)dxdy where S is the rectangle 1  x  2, 0  y  1 .
S
Sol:
11
  2 xy  y ds dy   x y  xy  dy
  4 y  2 y  y  y dy
1
2
y 0
x 1
2
1
y 0
1
2
2 2
1
2
2
y 0
1
 y2 y3 
 3
 
3 0
 2
 11 6
or
1
3
 1 2 xy  y 2 dy dy dx  2  xy 2  y  dx

x1 y 0
x1 

3 0
2 
1
   x  dx
x 1
3

2


2
 x2 1 
   x
 2 3 1
4 2 1 1
   
2 3 2 3
11

6
Revised Version:
Example
Evaluate
  (2 xy  y
2
)dxdy where S is the rectangle 1  x  2, 0  y  1 .
S
Sol:
1 2
1
1


2
2
2 2
2
2
0 1 2 xy  y dxdy  0 x y  xy 1dy  0 4 y  2 y  y  y dy
 3 y 2 y 3  1 11
 
  
3 0 6
 2
or
2 1
2
2


 2 y3  1
 x 2 x  2 4 2 1 1 11
1

2
       


2
xy

y
dy
dx

xy

dx

x

dx





0
1 0


3
3



 2 31 2 3 2 3 6
1 
1



17.
Page 3, Multiple Integrals
Original Version:
Example (Non-rectangular regions)
12
y
y
h(x)
y
h(x)
d
S
h( y )
c
g (x )
a
S
g ( y)
S
g (x )
x
x
b
a
b
a  x  b, g ( x)  y  h( x)
c  y  d , g ( y )  x  h( y )
a  x  b, g ( x)  y  h( x)
 
 
 
b
h( x)
xa
y g ( x)
f ( x, y )dy dx

d
h( y )
x c
yg ( y)
f ( x, y )dx dy

b
h( x)
xa
y g ( x)
x
f ( x, y )dy dx

Variable Limits unless a
rectangular region
Constant
Limits
Revised Version:
Example (Non-rectangular regions)
y
h(x)
y
y
h(x)
d
S
h( y )
c
g (x)
a
g (x)
x
x
b
a  x  b, g ( x )  y  h ( x )
c  y  d , g ( y )  x  h( y )
 

b
h( x)
xa
y g ( x)
S
g ( y)
S
f ( x, y ) dy dx

 g ( y ) f ( x, y )dx dy

y c 
 x h ( y )
d
a
b
a  x  b, g ( x )  y  h ( x )
 
b
h( x)
xa
y g ( x)
f ( x, y )dy dx

Variable Limits unless a
rectangular region
Constant
Limits
18.
Page 1, Multiple Integrals (22/10/2001)
Original Version:
For a function f (x ) of one variable, we define the definite integral of f from a to b as

N
f  x dx  lim  f  xi xi
N 
i 1
Revised Version:
For a function f (x ) of one variable, we define the definite integral of f from a to b as
13
x
b

f ( x )dx  lim
N 
a
N
 f ( x )x
i
i
i 1
19.
Page 18, Multiple Integrals (24/10/2001)
Original Version:
x x x
u v w
  x, y, z  y y y
J

 u , v, w u v w
x x x
u v w
Revised Version:
 x

 u
 ( x, y, z )  y
J

 (u, v, w)  u
 z

 u
x
v
y
v
z
v
x 

w 
y 
w 
z 

w 
20.
Page 17, Real-valued Functions of Several Variables (24/10/2001)
Original Version:
2
2
f
d 2x
2 f
2 f
dy
dx
 dx 
d z

(
1
,
1
)

(
1
,
1
)

(1,1)


t 1
t 1
t 1
2 t 1
2
2 t 1
x
dt
x
yx
dt
dt
 dt 
dt
f
d2y
2 f
dx
dy
2 f
 dy 
(1,1) 2 t 1 
(1,1)

(1,1)  t 1
t 1
t 1
2
y
dt
xy
dt
dt
y
 dt 
 e  2  e  4  2e  1  2  e  0  2e  2  1  e  0  14e
Revised Version
2
2
f
d 2x
2 f
2 f
dy
dx
 dx 
d z

(
1
,
1
)

(
1
,
1
)

(1,1)


t 1
t 1
2 t 1
2
2 t 1
x
dt
x
yx
dt
dt
 dt 
dt
2

f
d2y
2 f
dx
dy
2 f
 dy 
(1,1) 2 t 1 
(1,1)

(1,1) 
t 1
t 1
2
y
dt
xy
dt
dt
y
 dt 
 e  2  e  4  2e  1  2  e  0  2e  2  1  e  1  15e

21.
Page 4, Multiple Integrals (24/10/2001)
Original Version:
y
1
y  2x
y  x2
S
0
1
y0
2
x
14
2
t 1
t 1
Revised Version:
y
1
y  x2
y  -x+2
S
0
1
y0
2
x
22.
Page 12, Vector Integration (8/11/2001)
Original Version:
Example
Evaluate
  F dS
where F  x 2 i  xyj  z 2 k and S is the surface of the cube bounded by x = 0, x = 2,
S
y = 0, y = 2, z = 0 and z = 2.
Sol:
We integrate over each face in turn and sum the results.
For the face ABCD:
dS  dydz , the outward normal n  i , x  2 .
 F  d S 
ABCD

2
z
2 G
 F  idS
F
C
D
ABCD

2
x 0 y 0
x 2 dydz  
2

2
x 0 y 0
4dydz  16
2
O
2
x
For the face BEFC:
dS  dxdz , the outward normal n  j , y  2 .
A
y
y
E
B
y
 F  d S   F  idS
BEFC

2
BEFC

2
x 2 dxdz  
x 0 z 0
2
2
 2 xdxdz  8
x 0 y 0
 F  d S  16,  F  d S   F  d S   F  d S  0
Similarly,
CFGD
Hence
OEFG
OADG
OABE
  F  d S  40
S
Revised Version:
Example
Evaluate
  F dS
S
where F  x 2 i  xyj  z 2 k and S is the surface of the cube bounded by x = 0, x = 2,
y = 0, y = 2, z = 0 and z = 2.
Sol:
We integrate over each face in turn and sum the results.
For the face ABCD:
dS  dydz , the outward normal n  i , x  2 .
z
2 G
F
C
D
2
O
15
2
x
y
A
E
B
y
y
 F  d S   F  idS
ABCD

2
ABCD

2
z 0 y 0
x dydz  
2
2

2
z 0 y 0
4dydz  16
For the face BEFC:
dS  dxdz , the outward normal n  j , y  2 .
 F  d S   F  jdS
BEFC

2
BEFC

2
x 0 z 0
xydxdz  
2
2
 2 xdxdz  8
x 0 z 0
 F  d S  16,  F  d S   F  d S   F  d S  0
Similarly,
CFGD
OEFG
OADG
OABE
Hence
  F  d S  40
23.
A remark (in red) on surface integral has been added in page 18, 19. See below:
S
Now we are going to make a few concluding remarks on the ways of computing the surface integral of
second kind, that is, compute  F ( x, y, z )  d S , where F ( x, y, z ) is a vector field defined on S.
S
Suppose the surface is given implicitly by  ( x, y, z )  0 and we are going to project S onto xy-plane with
the projection, the region  xy . Assume we have chosen the normal n to S at the point (x, y, z) by
considering the cosine of the included angle between the normal and the positive direction of z-axis.
Then
 
  
S F  d S  S F  ndS  S F   dS   F    Z dxdy
xy
  F    
 xy
1
z
dxdy
Whether the positive sign or the negative sign is chosen is depending on the problem raised.
In addition, if the surface is given explicitly by z   ( x, y )  0 and still we are going to project S onto xyplane with the projection, the region  xy , then
 F  d S   F  ndS   F 
S
S
   X i  Y j  k 
 X  Y  1
2
S
2
dS 
   X i  Y j  k   X  Y  1
F
dxdy   F     X i  Y j  k dxdy

1
 2  2 1
 xy
 xy
2
X
2
Y
16