Vector Calculus

Vector Differentiation

dF d 2F dF d 2F
If F(t)=ti+t2j+t3k find dt , dt 2 , dt x dt 2

dF d 2F
dt =i+2tj+3t2k dt 2 =2j+6tk

dF d 2F i j k 6t 2i  6tj  2k
dt dt 2 1 2t 3t 2
x = 0 2 6t
=
Product formula
d ( F .G ) dG dF
 F.  G.
dt dt dt
d ( FxG ) dG dF
 Fx  xG
dt dt dt
Problem
d ( PxP)
 PxP
Show that dt
d ( PxP)
 PxP  PxP  PxP
dt

Problem of partial differentiation

e st i  (2 s  t ) j  t sin sk  2r  2r r r
x
If r= find s 2 , t 2 , s t

r  2r
 te st i  2 j  t cos sk  t 2e st i  t sin sk
s s 2

r  2r
 se st i  j  sin sk  s 2e st i
t t 2

r r i j k (2 sin s  t cos s )i  j ( se st t cos s  e st sin s )  k (te st  2 se st )

x
s t te st 2 t cos s
= se
st
 1 sin s =
Problem
f f f
i  j k
Find a unit vector in the direction of x y z at (1, 0, 2)
where f ( x, y, z )  4( x 2  y 2 )  z 2
f f f
 8x  8y  2 z
x y z
f f f
i  j k
x y z =8xi+8yj-2zk=8i-4k
f f f 8i  4k 2i  k
i  j k
Unit vector along x y z is 80 = 5

Gradient, Divergence and Curl

   
i  j k
The vector operator (del or nabla) is defined by x y z
Definitions:
1. Gradient- If    ( x, y, z ) is scalar point function then gradient
of  written as grad  or   is defined as
   
   i  j  k  
 x y z 
  
i  j k
x y z (vector quantity)

2. Divergence- If v= v (x,y,z)=v1(x,y,z)i+v2(x,y,z)j+v3(x,y,z)k is
a vector point function then divergence of v written as divv or
. v is defined as
.    
  i  j  k 
v  x y z  .( v1i+v2j+v3k)
v1 v2 v3 .  
 
= x y z (scalar quantity) v v.

3. Curl- If v= v (x,y,z)=v1(x,y,z)i+v2(x,y,z)j+v3(x,y,z)k is a
vector point function then curl of v written as curlv or  x v is
defined as
    
  i  j  k 
xv  x y z  x( v1i+v2j+v3k)

2
 i j k  v v   v v   v v 
i 3  2   j  1  3   k  2  1 
  
 y z   z x   x y 
x y z
= v1 v2 v3 = (vector quantity)

Problem
1. If f ( x, y, z ) =3x2y-y3z2 find f at the point P(1,-2,-1)
f  (3x y  y z )
2 3 2
 (3x y  y z )
2
 (3x 2 y  y 3 z 2 )
3 2
i j k
= x y z
=6xyi+(3x2-3y2z2)+2y3zk
f (1,-2,-1) =-12i-9j-16k

1 r  x2  y 2  z 2 

2. Ex. If r and find
 1  1  1  1
 i   j  k  
= r = x  r  y  r  z  r 
  1  r   1  r   1  r r 2  x 2  y 2  z 2 2r r  2 x
i    j   k  
= r  r  x r  r  y r  r  z , x
1 1 1 r x r y r z
 3 xi  3 yj  3 zk   
= r r r x r y r z r
1 1
 3 ( xi  yj  zk )  3
= r = r r

3. Ex. If A=x2z2i-2y2z2j+xy2zk find . A at P(1,-1,1)

. A1 A2 A3
 
A= x y z
  
= x (x2z2)+ y (-2y2z2)+ z (xy2z)
=2xz2-4yz2+xy2
. A(1,-1,1) =2+4+1=7

4. Ex. If A=x2z2i-2y2z2j+xy2zk find  xA at P(1,-1,1)

 i j k
  
x y z
xA= x2 z 2  2 y2z2 xy 2 z

3
   ( xy 2 z )  (2 y 2 z 2 )    ( x 2 z 2 )  ( xy 2 z )    (2 y 2 z 2 )  ( x 2 z 2 ) 
i    j     k  
=  y z   z x   x y 
=(2xyz+4yz2)i+(y2z-2x2z)j
hence at P,  xA=2i+j

1. Prove that  .(  A) = (  ).A+  (  .A)

hence find  .( r 3 r)
Let A=A1i+A2j+A3k
 .(  A)=  .(  A1i+  A2j+  A3k)
     
= x ( A1)+ y ( A2)+ z ( A3)
 A1  A  A 
 2  3
= x +A1 x + y +A2 y + z +A3 z
    A1 A A
 2  3
= x A1+ y A2+ z A3+ x + y + z
      A1 A2 A3
 i  j k  z
=  x y z  . (A1i+A2j+A3k)+ ( x + y + )
=(  ).A+  (  .A)
 .( r 3 r) = ( r 3 ).r+ r 3 (  .r)
.    
  i  j  k 
r  x y z  .(xi+yj+zk)
x y z
 
= x y z =3
 3  3  3
r 3
i  
r  j  
r k r  
= x y z
 3 r  3 r  3 r r
i  
r j r  
k r  
r 2  x2  y 2  z 2 2r  2x
= r x r y r z , x
x y z r x r y r z
3r 2 i  3r 2 j  3r 2 k   
= r r r x r y r z r
= 3r ( xi  yj  zk )
= 3r r
 .( r 3 r) =(3rr).r+ r 3 (3)= 6 r 3

 2  2  2  2
 
= x 2 y 2 z 2

4
  2                  
2 2 2
 i  j  k . i j k   2 2
= . =  x y z   x y z  = x y z
2

2  2 2 2
 
= x 2 y 2 z 2 Laplacian

2. Prove that  x(  A) =  (  xA)+  xA

Let A=A1i+A2j+A3k

 x(  A) =  x(  A1i+  A2j+  A3k)

i j k
  
x y z
= A1 A 2 A 3
  (A 3 )  (A 2 )    ( A1 )  (A 3 )    (A 2 )  ( A1 ) 
i    j    k  
=  y z   z x   x y 
=
 A 3  A 2  
i   A3    A 2 
 y y z z 
 A  A 3    A 2  A  
 j 1  A1    A 3   k    A2   1  A1 
 z z x x   x x y y 
 A A   A A   A A  
  3  2 i   1  3  j   2  1 k 
=  y z   z x   x y  
+
            
 A3  A 2 i   A1  A 3  j   A2  A k 
 y z   z x   x y  
  i j k
  
x y z
= ( xA)+ A1 A 2 A3
=  (  xA)+  xA

Alternative method
   
  i
x
j
y
k
z
i
= x
 
=  i x

5
A  A1i  A2 j  A3k
A A A
. A  1  2  3
x y z
A  A A A  A
i.  i.i 1  j 2  k 3   1
x  x x x  x
A  A A A  A
j.  j.i 1  j 2  k 3   2
y  y y y  y
A A3
k. 
z z
A A A A
. A  i.  j.  k.   i.
x y z x

. 
=  x
i.

v=v1i+v2j+v3k
    
  i  j  k 
xv  x y z  x( v1i+v2j+v3k)
i j k  v v   v v   v v 
i 3  2   j  1  3   k  2  1 
  
 y z   z x   x y 
x y z
= v1 v2 v3 =

v  v v
iX  i X (v1i + v 2 j + v3k)  k 2  j 3
x x x x
v  v v
jX  j X (v1i + v 2 j + v3k)  k 1  i 3
y y y y
v  v v
kX  k X (v1i + v 2 j + v3k)  j 1  i 2
z z z z
Xv v v v v
iX
= x
 jX
y
kX
z =
 i X x
X 
=  iX
x

Prove that
3. . (FxG) = G.curlF-F.curlG

6
. 
 i. x ( FXG )  i. Fx XG  FX Gx 
(FxG) = =  
 F   G 
 i. XG   i. FX
=  x x


 F   G 
=
 G. i X    F .
x   x
X i 

F G
G . i X  F . i X
= x x
=G.curlF-F.curlG

4. X (FxG)=FdivG-GdivF+(G.  )F-(F.  )G

5.  (F.G)=FxcurlG+GxcurlF+(F.  )G+(G.  )F

Vector analysis by Raisinghania

Ex. 1. Prove that  x  =0 curlgrad  =0

    2  2    2  2    2  2 
  i     j    k
i j k  yz z y   z x x z   xy y x 
  
x y z
  
x = x y z = =0
Remarks: If  xv=0 then v must be 

Ex. 2. Prove that  . (  xA)=0 divcurlA=0

    
  i  j  k 
xA  x y z  x( A1i+A2j+A3k)
i j k  A A   A A   A A 
i 3  2   j  1  3   k  2  1 
    y z   z x   x y 
x y z
= A1 A2 A3 = (vector quantity)

7
         A A  A A  A A 
 i  j  k  i 3  2   j  1  3   k  2  1 
. ( xA)=  x y z  .   y z   z x   x y 
  2 A3  2 A2    2 A1  2 A3    2 A2  2 A1 
          
= x y xz   yz yx   z x z y 
=0
Remarks: If  .v =0 then v must be  xA

Definition:
Level surface: The family of surfaces f(x,y,z)=k is called iso-
surface or level surface. For different values of c the surface
such as x 2  y 2  z 2  c 2 represents a family of concentric spheres
with center at the origin and varying radius c and they constitute
a level surface. The surfaces (i) of constant temperature known
as isothermal surface (ii) of constant gravitational or electric
potential known as equipotential surface are examples of level
surfaces.


x y z c
2 2 2 2
o

x2  y2  z 2  c2 o

Level curve: In two dimensions the family of curves f(x,y)=k is

called level curve. For different values of c the curve such as
x 2  y 2  c 2 represents a family of concentric circles with centre at
the origin and varying radius c and they constitute a level curve.
The curve of constant temperature known as isothermal curve is
an example of level curve.

8
  normal
x2  y2  c2 o

Geometrical meaning of gradient of scalar 

Theorem
Prove that f is a vector perpendicular to the level surface
f ( x, y, z )  c where c is a constant.

Let r=xi+yj+zk be the position vector to any point P(x, y, z) on

the surface. Then dr=dxi+dyj+dzk lies in the tangent plane to
the surface at P.
f ( x, y , z )  c
df  0
f f f
dx  dy  dz  0
x y z
f f f
i  j k
( x y z ). (dxi+dyj+dzk)=0
f .dr =0
So that f is perpendicular to dr i.e. perpendicular to the tangent
plane to the surface at (x,y,z).
Hence f is a vector perpendicular to the surface f ( x, y, z )  c at
any point (x, y, z).

 ( x, y , z )  c
r
r+  r 
p
r

9
 o

Formula
The unit normal to the surface f ( x, y , z )  c at the point
f
(x,y,z) is n= f
Problem
Find the unit normal to the surface x2y+2xz=4 at the point
(2,-2, 3).

  ( x 2 y  2 xz )  ( x 2 y  2 xz )  ( x 2 y  2 xz )
i j k
(x2y+2xz)= x y z
=(2xy+2z)i+x2j+2xk
=-2i+4j+4k
The unit normal to the surface is
f  2i  4 j  4k 1 2 2
 i j k
n= f = 36 = 3 3 3
Problem
Find the angle between the surfaces xy 2 z  3x  z 2 and
3 x  y  2 z  1 at (1, -2, 1)
2 2

Let f(x, y, z)= xy 2 z  3x  z 2

g(x, y, z)= 3x 2  y 2  2 z  1
 ( xy 2 z  3x  z 2 )  ( xy 2 z  3x  z 2 )  ( xy 2 z  3x  z 2 )  ( xy 2 z  3x  z 2 )
i j k
f= = x y z
=i(y2z-3)+j2xyz+k(xy2-2z)
=i-4j+2k
 g=6i+4j+2k
 f.  g= f . g cos 
 f .g (i  4 j  2k ).(6i  4 j  2k ) 6
cos = f g = 21 2 14 = 14
6
  cos 1
14
=============================================

10
Slope/gradient/rate of change in different dimensions in
differential caculus
One dimensional body T=f(x)
yor T

T=x2+2
o x
T=x +2
2

x
o
A piece of heated wire
Geometrically y or T = f(x) is the equation of a curve
dy

dx rate of change of y along x-axis/directional derivative of y
along x-axis/ gradient of y with respect to x,
dy
dx =Slope of the tangent at P to the curve y = f(x)
dT
Ex. T=f(x), dx

Two dimensional body T=f(x,y) y

11
 y

(x,y)
(2,1) x
o
o x

T=x2+y2
z or T A sheet of heated tin

Geometrically z = f(x,y) is the equation of a surface

z z
x , y are
the rate of changes of z along x and y-axes/directional
derivative of z along x and y-axes/ gradient of z with respect to
x and y,

T T
Ex. T=f(x,y) x , y

Three dimensional body T=f(x,y,z)

u u u
u=f(x,y,z), (solid) x , y
and are the rate of changes along
z
x, y and z-axes/directional derivative along x, y and z-axes
T T T
Ex. T=f(x,y,z) , and
x y z
What is the relation between Gradient of a scalar in vector
calculus and Slope/gradient/rate of change in differential
calculus
One dimension
    (x) y  f (x) T  T (x)
  i
x

12
  
 .i 
x = Slope/gradient/rate of change of with respect to x
dy dT
or
in differential caculus = dx dx
Two dimension
     ( x, y ) z  f ( x, y ) T  T ( x, y )
  i  j
x y , ,
 
 .i 
x = Slope/gradient/rate of change of with respect to x
z T
or
in differential caculus= x x
 
 . j 
y = Slope/gradient/rate of change of with respect to y
z T
or
in differential caculus = y y

Three dimension
      ( x, y , z ) u  f ( x, y , z ) T  T ( x, y , z )
  i  j k
x y z
 
 .i 
x = Slope/gradient/rate of change of with respect to x
u T
or
in differential caculus = x x
 
 . j 
y = Slope/gradient/rate of change of with respect to y
u T
or
in differential caculus = y y
 
 .k 
z = Slope/gradient/rate of change of with respect to z
u T
or
in differential caculus = z z

Directional derivative
  
x , y , z are called directional derivative in vector calculus

 .i 
x

13
The directional derivative of  along x-direction is obtained by
taking dot product of  and unit vector along x-direction

The directional derivative of  along any direction is

obtained by taking dot product of  and unit vector along
that direction.
The directional derivative/rate of change of  along any vector
a is  .â where â is unit vector along a.

Maximum rate of change

Geometrical meaning of 
 .aˆ   1cos   cos where â is unit vector along a
This rate of change is maximum when   0 .
The rate of change is maximum along the normal to the surface
 ( x, y, z )  c and it is equal to 
  2 2 2
        
       
 x   y   z 

a
normal
tangent



x2  y2  c2 o

14
 tangent normal

x2  y2  z 2  c2 o

Problem
Find the directional derivative of  =x2yz+4xz2 at P(1,-2,1) in
the direction of the vector a=2i-j-k. Also find the maximum
rate of change of  .
 =x2yz+4xz2
  
  i j k
x y z
 ( x 2 yz  4 xz 2 )  ( x 2 yz  4 xz 2 )  ( x 2 yz  4 xz 2 )
i j k
= x y z
= i(2xyz+4z2)+j(x2z)+k(x2y+8xz)
=j+6k
a 2i  j  k
 aˆ 
The unit vector along a is a 6
 .â  2i  j  k 7
the directional derivative is ( j+6k). 6 = 6
The maximum rate of change of  is
 2 2 2
02  12  62 37
        
       
=  x   y   z  = =

Physical Interpretation of Curl of vector function

  1 
Ex If v= xr, Prove that where = 2 curlv
is a constant
vector.
Let  = 1 i+ 2 j+ 3 k be the angular velocity and r=xi+yj+zk

15
 i j k 2 z  1 y 3 x  1 z 1 y  2 x
1 2 3
xr= x y z =( )i+( )j+( )k
  i j k
  
x y z
curlv= x( xr) = 2 z  1 y 3 x  1 z 1 y  2 x
=2( 1 i+ 2 j+ 3 k)
=2 
 1
= 2 curlv
Thus physically interpreted, curl of linear velocity of any
particle is twice the angular velocity of the particle.
Curl has the effect of rotation curlA=rotA

Condition of irrotatioal motion

curlv= 2  , If curlv=0 the motion is irrotational
Also we know that  x  =0,
If the motion is irrotational i. e. curlv=0 then v must be  ,
 is called scalar potential function.
Ex. Determine the constant a, b, c so that vector
v=(-4x-3y+az)i+(bx+3y+5z)j+(4x+cy+3z)k is irrotational.
Find a scalar function  so that v=  .

v is irrotational if  xv=0
 i j k
  
x y z
xv= - 4x - 3y + az bx + 3y + 5z 4x + cy + 3z

16
 =(c-5)i-(4-a)j+(b+3)k =0 =0i+0j+0k

a=4, b=-3, c=5

As  xv=0, v must be 
v= 
  
i j k
(-4x-3y+4z)i+(-3x+3y+5z)j+(4x+5y+3z)k = x y z

x = -4x-3y+4z ……. (1)

y = -3x+3y+5z ……….(2)

z =4x+5y+3z ………….(3)
Integrating (1) with respect to x partially
 =-2x2-3xy+4xz+f(y,z) …..(4)
 f ( y, z )
y =-3x+ y ……….(5)
 f ( y, z )
z = 4x+ z …………(6)
f ( y, z )
Comparing (2) and (5) y =3y+5z ….. (7)
f ( y, z )
Comparing (3) and (6) z =5y+3z ….. (8)
3
Integrating (7) with respect to y, f(y, z)= 2 y +5zy+g(z)
2

f ( y, z )
 5 y  g ( z )
z ….. (9)
g ( z )  3z 3 2
g ( z)  z c
Comparing (8) and (9) 2
3 3
therefore f(y, z)= 2 y +5yz+ 2 z2+c
2

 3 3
hence =-2x -3xy+4xz+ 2 y +5yz+ 2 z2+c
2 2

Alternative method
v= 
  
i j k
(-4x-3y+4z)i+(-3x+3y+5z)j+(4x+5y+3z)k = x y z

17

x = -4x-3y+4z ……. (1)

y = -3x+3y+5z ……….(2)

z =4x+5y+3z ………….(3)
 =-2x2-3xy+4xz+f(y,z)
 3
= -3xy+ 2 y2+5yz+g(z,x)
 3
= 4xz + 5yz + 2 z2 +h(x,y)

 =-2x2-3xy+4xz+ + + +f(y,z)
 3
= -3xy+ + 2y
2
+ 5yz + +g(z,x)
 3
= + +4xz + + 5yz + 2z
2
+h(x,y)
3 3
f(y,z)= 2 y +5yz+ 2 z2
2

3
g(z,x)= -2x +4xz+ 2 z2
2

3
h(x,y)= -2x -3xy+ 2 y2
2

 3 3
hence =-2x -3xy+4xz+ 2 y +5yz+ 2 z2+c
2 2

Physical interpretation of divergence of the vector function

Consider fluid motion in space. Let P(x, y, z) be any point of
fluid at time t. Let v=ui+vj+wk be the fluid velocity at P.
Construct a small rectangular box with edges of length x, y, z
parallel to the respective coordinate axes, having P at one of the
angular points as shown in the figure.

Q( x, y  y, z )
y
Q
v
R R
u y u x
w

18
 P( x  x, y, z )
zP( x, y, z )
S ( x, , y, z  z ) S 
x
o

z
Fig18.

Then we have volume of the fluid that passes through the face
PQRS = ( y z )u per unit time
= f ( x, y, z ) per unit time (say)
Hence the fluid that passes out through the opposite face PQRS 
= f ( x  x, y, z ) per unit time
f ( x, y, z )
f ( x, y, z )  x
= x + terms containing higher powers
of x [Taylore’s theorem]
f (x) h 2
f ( x)
f(x+h)=f(x)+h + 2 +………..
The net outward flow from the rectangular box along x-axis per
f ( x, y, z )
f ( x, y, z )  x  f ( x, y , z )
unit time= x
f ( x, y, z )  u
x x uy z xy z
= x = x = x
Similarly the net outward flow from the rectangular box along
u u
xy z xy z
y-axis and z-axis per unit time will be y and z
respectively.
Hence total net outward flow per unit time (rate of flow)
through the rectangular box of volume xy z is
 u v w 
   xy z
 x y z 

The total rate of outward flow of the fluid through a unit volume

19
  u v w  u v w .
   xy z  
 x y z  x y z
= xy z = = v

Thus physically interpreted divv represents the net outward

flow of the fluid per unit volume per unit time.

Divergence=Net outward flow

Remark: The name divergence originated in the above

mentioned interpretation of divv

Equation of continuity of an incompressible fluid:

If the fluid is incompressible then fluid inflow =fluid outflow
and there will be no net outward flow of fluid. Hence divv=0
This is known as condition of incompressibility of fluid or
equation of continuity of incompressible fluid.

Solenoidal force field

Solen (Greek word) -pipe
Solenoid-shape of a pipe, strong magnet

20
 Figure of solenoid

In a solenoid the magnetic force F is uniform, So divF=0

If divF=0 then F is a solenoidal
Or, if F is solenoidal then . F =0

Problem
Ex. Determine the constant a so that vector
F =(-4x-6y+3z)i+(-2x+y-5z)j+(5x+6y+az)k is solenoidal.

F is solenoidal if . F=0
.   
F= x (-4x-6y+3z)+ y (-2x+y-5z)+ z (5x+6y+az)=0
Or, -4+1-a=0 a=-3

21