Class notes Maths

18MAB102T Advanced Calculus and Complex Analysis Vector Calculus


Module – 2 Vector Calculus

Review of vectors in 2, 3 dimensions – Gradient, divergence, curl – Solenoidal, Irrotational fields –
Vector identities (without proof) – Directional derivatives – Line integrals, Surface integrals,
Volume integrals – Green’s theorem (without proof) – Gauss divergence theorem (without proof),
Verification, Applications to Cubes, parallelopiped only – Stoke’s theorem (without proof) –
Verification, Applications to Cubes, parallelopiped only – Applications of Line and Volume
integrals in Engineering.

Basic Formulae
     
1.   i  j k
x y z
     
2.   grad   i j k
x y z

a
3. Directional derivative =   
a

4. Normal derivative = 


5. Unit normal vector n̂ 


1   2
6. Angle between the surfaces cos  
1  2
   
7. Let r  x i  y j  z k

r  x2  y2  z 2

r 2  x2  y2  z2

Differentiate partially w.r.t. x
r
2r  2x
x
r x

x r
r y
Differentiate partially w.r.t. y 
y r

r z
Differentiate partially w.r.t. z 
z r




SRM IST, Ramapuram. 1 Department of Mathematics

, 18MAB102T Advanced Calculus and Complex Analysis Vector Calculus


1. 
Find  if   log x 2  y 2  z 2 . 
Solution:

  
  i j k
x y z

i

x
   
log( x 2  y 2  z 2 )  j log( x 2  y 2  z 2 )  k log( x 2  y 2  z 2 )
y z
2x 2y 2z
i 2 j 2 k 2
(x  y  z )
2 2
(x  y  z )
2 2
(x  y2  z2 )


x y z
2
2
2 2  xi  y j  zk   2r 2r (r  xi  y j  zk & r 2  x 2  y 2  z 2 )


2. Find  if   x y z at the point (1, 2, 3).

Solution:
     
  grad   i j k
x y z
  
  i y z  j x z  k x y

  
 at (1, 2, 3)  6 i  3 j  2 k

3. Find  r .

Solution:
 r  r  r
r  i  j k
x y z

 x  y  z r
r  i  j  k 
r r r r

4. Find the unit normal vector to the surface x2 + xy + z2 = 4 at the point (1, –1, 2).
Solution:
Let  = x2 + xy + z2 – 4
     
  grad   i j k
x y z

  
 2 x  y,  x,  2z
x y z




SRM IST, Ramapuram. 2 Department of Mathematics

, 18MAB102T Advanced Calculus and Complex Analysis Vector Calculus



 (1,1,2)    
   
= (2 x  y )i  xj  2 zk (1, 1, 2) = i  j  4k

The unit normal vector is

 i  j  4k i  j  4k
 nˆ    .
 2 2
1 1  4 2 18

5. Find the unit normal vector to the surface x2 + y2 + z2 = 1 at the point (1, 1, 1).
i jk
Ans nˆ 
3
6. Find the directional derivative of  = 3x2 + 2y – 3z at (1, 1, 1) in the direction 2i  2 j  k .

  
Solution: The gradient of  is   i  j k
x y z

  
 6 x,  2,  3
x y z
  6 x i  2 j  3k

Directional derivative of  is

a  2i  2 j  k

a  22  22  (1)2  9  3

a   2i  2 j  k  19
 .  (6 xi  2 j  3k )    
a   3   (1,1,1) 3

7. Find the directional derivative of  = 2xy + z2 at (1, – 1 , 3) in the direction i  2 j  2k .

14
Ans
3
8. Find the directional derivative of  = x2+ y2 + 4 x y z at (1, – 2 , 2) in the direction 2 i  2 j  k .

44
Ans 
3
9. Find the directional derivative of  = x2 – y2 + 2 z2 at P (1, 2 , 3) in the direction of line PQ
where Q is (5, 0, 4).
Solution:
     
  grad   i  j k
x y z



SRM IST, Ramapuram. 3 Department of Mathematics

, 18MAB102T Advanced Calculus and Complex Analysis Vector Calculus


  
  grad   i 2 x  j (2 y )  k 4 z
  
 at (1, 2, 3)  2 i  4 j  12 k
         
a  OQ  OP  (5 i  0 j  4 k )  ( i  2 j  3 k )  4 i  2 j  k

a
Directional derivative =   
a
  
   4i  2 j  k 28
 (2 i  4 j  12 k )  
21 21
10. In what direction from (3, 1, –2) is the directional derivative of  = x2y2z4 a maximum? Find
the magnitude of this maximum.
Solution: Given  = x2 y2z4

  
 2 xy 2 z 4 ,  2 x 2 yz 4 ,  4 x2 y 2 z3
x y z
2 4 2 4 2 2 3
  (2 x y z )i  (2x yz ) j  (4x y z )k

 3, 1, 2 = 96i  288 j  288k  96(i  3 j  3k )
 The maximum directional derivative occurs in the direction of  = 96(i  3 j  3k )

The magnitude of this maximum directional derivative is
  96 12  32  ( 3) 2  96 1 9  9  96 19.

11. In what direction from (1, 1, –2) is the directional derivative of  = x2 – 2y2 + 4 z2
a maximum? Find the magnitude of this maximum.
  
Ans Directional derivative is maximum in the direction of 2 i  4 j  16 k

Maximum directional derivative = 276

12. Find the angle between the surfaces x logz = y2 – 1 and x2y = 2 – z at the point (1, 1, 1).
Solution: Let 1 = y2 – x logz – 1
   x
  log z ,  2 y, 
x y z z
  x  
 1 = – log z i  2 yj  k  1 (1,1,1)
, (  ) = 2 j  k and |  1|  22  (1) 2  5
z




SRM IST, Ramapuram. 4 Department of Mathematics