CONTINUOUS ASSESSMENT correction
Exercise 1
Part I:
Let E be the vector field defined by E  ( y 3  6 xy 2 ,3xy 2  6 x 2 y )
1) Show that this field derives from a scalar potential
 E E   E E   E E 
rot E   Z  y  i   Z  x  j   y  x  k
z   x
z   x
y 
 y
Ex  y 3  6 xy 2 , E y  3xy 2  6 x 2 y , E z  0
E y
x
 3 y 2  12 xy ,
Ex
 3 y 2  12 xy
y
rot E  0 . Hence, the vector field E derive from a scalar potential
2) Determine this scalar potential
E  gradV

E i  E
x
y
 V
V
V
j  Ez k  
i
j
y
z
 x


k

 V
3
2
 x  Ex  y  6 xy
 V

 E y  3 xy 2  6 x 2 y
 y
V
 Ex  y 3  6 xy 2
x
 V ( x, y)    y 3  6 xy 2 dx  xy 3  3x 2 y 2  c( y)
V 
c( y )
  xy 3  3x 2 y 2  c( y )   3xy 2  6 x 2 y 
 3xy 2  6 x 2 y
y y
y
 c( y )  K
Hence V ( x, y )  xy 3  3x 2 y 2  K
3) Assuming the constant is zero, determine the divergence of this scalar field
PS: we don’t calculate the divergence of a scalar field, but only for a vector field.
For example: divE 
Ex E y
; and the result is a scalar field

x
y
Part II:
Let f be the function defined by f ( x, y )  xy ( x  y  1)
1) Determine all the critical points of this function
 f
 x  y ( x  y  1)  xy
 f
  x( x  y  1)  xy
 y
Applying
f
f
 0 and
 0 lead to the system below:
x
y
 y (2 x  y  1)  0

 x( x  2 y  1)  0
The first equation lead to y  0 or (2 x  y  1)  0
 Considering y  0 and replacing into the second equation lead to the critical points
below: A(0,0) and B(1,0).
The second equation lead to x  0 or ( x  2 y  1)  0
 Considering x  0 and replacing into the first equation lead to the critical points below:
A(0,0) and C(0,1).
2 x  y  1  0
 Considering now the system 
. The resolution lead to the critical point
x  2 y 1  0
D(1/3, 1/3)
Hence, we have four critical points which are: A(0,0) and B(1,0), C(0,1) and D(1/3,1/3)
2) Determine for all (x,y) 
  2 f ( x, y )

x 2
Hf ( x, y )   2
  f ( x, y )

 yx
2y

Hf ( x, y )  
2
x

2 y 1

2
the Hess matrix H defined as
 2 f ( x, y ) 

xy 
 2 f ( x, y ) 

y 2

2 x  2 y  1

2y

3) Say for each critical points whether it is a saddle point, a local maximum or a local
minimum.
 For A(0,0),
det Hf (0, 0)  1
 For B(1,0),
det Hf (1, 0)  1
 For C(0,1),
det Hf (0,1)  1
 0
Hf (0, 0)  
 1
0
; thus the critical point A is a saddle point
0
Hf (1, 0)  
1
0
1
3
1

0
; thus the critical point C is a saddle point
 For D(1/3,1/3),
det Hf (1 / 3,1 / 3) 
1

2
; thus the critical point B is a saddle point
2
Hf (0,1)  
1
0
-1

0
 1 1   2/3
Hf  ,   
 3 3  1 / 3
0
1/3 

2/3 
; Tr (H) = 4/3 0 ; thus the critical point D is a local minimum