- The vector differential operator ∇ is defined as the gradient operator (∂/∂x, ∂/∂y, ∂/∂z). - The gradient of a scalar function φ is defined as gradφ = (∂φ/∂x, ∂φ/∂y, ∂φ/∂z). - For the function φ = x2 + yz + 2xy, the gradient at point P=(1,3,2) is gradφ = (2,6,4).

2. • Operator is called vector differential operator
defined as

ˆ
ˆ ˆ
i j k
x y z
 
  
   
 
  
 

3. 
( , , )
x y z

• If is a scalar function of three variables and
is differentiable, the gradient of is defined as

Where ,
is a scalar function
is a vector function
ˆ
ˆ ˆ
grad i j k
x y z
  
 
  
    
  




4. If ,determine at point P=(1,3,2).
solution
2 3 2 2
x yz xy z
   grad
     
2 3 2 2
3 2 2 2 3 2 2 2 2
3 2 2 2 3 2 2 2 2
2 , 2 , 3 2
,
ˆ
ˆ ˆ
ˆ
ˆ ˆ
2 2 3 2
(1,3,2)
ˆ
ˆ ˆ
84 32 72
x yz xy z
xyz y z x z xyz x yz xy z
x y z
Therefore
i j k
x y z
xyz y z i x z xyz j x yz xy z k
at
P
i j k

  
  


 
  
     
  
  
   
  
     

   

5. If A and B are two scalars ,then
1)
2)
( )
A B A B
     
( ) ( ) ( )
AB A B B A
    

6. Directional derivative of in the direction of a is

ˆ.
d
a grad
ds



where,
Which is a unit vector in the direction of .
dr
ˆ
dr
a
dr


7. Compute the directional derivative of at the
point (1,2,-1) in the direction of the vector A=2i+3j-4k.
Solution
2 2 2
2
x z xy yz
   
Directional derivative of in the direction of a
Where,
And ,

ˆ.
d
a grad
ds



ˆ
ˆ ˆ
grad i j k
x y z
  
 
  
    
  
ˆ
A
a
A

2 2 2
2
x z xy yz
    Hence ,
     
2 2 2 ˆ
ˆ ˆ
2 2 4 2
(1,2, 1),
ˆ
ˆ ˆ
6 9 3
xz y i xy z j x yz k
At
i j k


      

   

8. Also given ,then
ˆ
ˆ ˆ
2 3 4
A i j k
  
 
 
2
2 2
2 3 4 29
,
1 ˆ
ˆ ˆ
ˆ 2 3 4
29
,
ˆ.
51
29
A
Therefore
A
a i j k
A
then
d
a grad
ds


    
   



9. If ,the divergence of A is
defined as
ˆ
ˆ ˆ
x y z
A i j k
a a a
  
 
.
ˆ ˆ
ˆ ˆ ˆ ˆ
.
.
x y z
y
x z
divA A
i j k i j k
x y z
x y z
a a a
a
a a
 
 
  
    
 
  
 

 
  
  

10. If , determine at point (1,2,3) .
solution
2 2 ˆ
ˆ ˆ
A x yi xyzj yz k
   divA
 
.
ˆ ˆ
ˆ ˆ ˆ ˆ
.
.
x y z
y
x z
divA A
i j k i j k
x y z
x y z
a a a
a
a a
 
 
  
    
 
  
 

 
  
  
2 2
divA xy xz yz
  
(1, 2,3)
2(1)(2) (1)(3) 2(2)(3)
13
at
divA
divA
  


11. If ,the curl of A is defined by
ˆ
ˆ ˆ
x y z
A i j k
a a a
  
 
ˆ ˆ
ˆ ˆ ˆ ˆ
ˆ
ˆ ˆ
x y z
x y z
curlA A
i j k i j k
x y z
i j k
curlA
x y z
a a a
a a a
 
 
  
     
 
  
 
  
 
  

12. If ,determine curl A
at (1,3,-2).
solution
4 2 2 2 2 2 ˆ
ˆ ˆ
( ) ( )
A y x z i x y j x yzk
    
4 2 2 2 2 2
2 2 3
ˆ
ˆ ˆ
ˆ
ˆ
( 2 2 ) (2 4 )
(1,3, 2),
ˆ
ˆ ˆ
2 8 106
i j k
curlA A
x y z
y x z x y x yz
x zi xyz x z j x y k
At
curlA i j k
  
  
  
  
      

  