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5.
Vector Calculus
5.1. Differentiation of a Vector Function
If a(t )  a1 (t )i  a2 (t ) j  a3 (t )k is a vector function of a scalar variable t then the derivative of a(t)
with respect to t is
d
 a (t   t )  a (t ) 
a (t )  lim 


t

0
dt
t

a (t   t )  a3 (t )
a (t   t )  a1 (t )
a (t   t )  a 2 (t )
 lim 1
i  lim 2
j  lim 3
k
t 0
t 0
t 0
t
t
t
da
da
da
 1i 2 j 3 k
dt
dt
dt
Example
If r(t) = x(t)i+y(t)j+z(t)k is the position vector of a moving point, where t represents time,
dr
r
 lim
dt  t 0  t
represents the velocity of the point and is seen
to be a vector tangential
to the path of the point.
r (t   t )
d 2r
Similarly 2 represents
dt
the acceleration of the point.
O
r (t   t )  r (t )   r
r (t )
Example
A particle moves such that its position vector at time t is r  e t i  2 cos 3tj  3 sin 3tk
Determine its velocity and acceleration at time t = 0.
Sol:
velocity = v 
dr
 e t i  6 sin 3tj  9 cos 3tk
dt
v(0)  i  9k , magnitude
acceleration =
1
325
82 , direction
1
82
(i  9k )
d 2r
a  2  e t i  18 cos 3tj  27 sin 3tk , a(0)  i  18 j , magnitude
dt
(i  18 j )
The usual rules of differentiation apply
1
325 , direction
d
d
d
( a (t )  b(t ))  a (t )  b(t )
dt
dt
dt
d
d
( c a (t ))  c a (t )
c  constant
dt
dt
and also
d
at   bt   d at   bt   at   d bt 
dt
dt
dt
d
at   bt   d at   bt   at   d bt .
dt
dt
dt
If a  a1i  a2 j  a3 k where a1 , a 2 , a3 are functions of several variables t1 , t 2 ,..., t n , then partial
derivatives
a
a
a a1

i 2 j 3 k
t i t i
t i
t i
may be defined in the obvious way. Also vector integration may be defined.
Example
If a particle moves with velocity v  2 sin ti  (cos t  1) j  6tk and r = 0 at t = 0, determine its
position vector at time t.
Sol:
dr
dt
dt
  2 sin ti  cos t  1 j  6tk dt
r  2 cos ti  sin t  t  j  3t 2 k  c
r
At t = 0, 0  2i  c , hence c  2i and r  (2  2 cos t )i  (sin t  t ) j  3t 2 k .
Example
Let r in polar form be the position vector of a moving particle in R 2 . Find the velocity and the
acceleration of r in polar form.
Sol:
 cos  
 a unit vector and r  r
Represent r in polar form, then r  rR , where R  
 sin  
The velocity v of the moving particle in polar form is:
d r dr
dR dr
dR d

Rr

Rr
dt dt
dt
dt
d dt
dR   sin  
  P . Notice that R is perpendicular to P.

Let
d  cos 
Then
v
2
v
dr
d
Rr
P.
dt
dt
And v is the linear combination of two perpendicular unit vectors.
The acceleration of the moving particle in polar form is:
dv d 2 r
dr dR d dr d
d 2
d dP d
 2 R

Pr 2 Pr
dt dt
dt d dt dt dt
dt d dt
dt
2
2
d r
dr d
d 
d dP d
 2 R2
Pr 2 Pr
dt dt
dt d dt
dt
dt
2
 d 2r
1 d  2 d 
 d  
  2  r
 R 
r
P
r dt  dt 
 dt  
 dt
a
where
dP   cos 
  R

d   sin  
5.2. Vector Fields
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
then v ( x, y , z ) is a vector function and we say that we have a vector field.
Examples
(i) A magnetic field B in a region of space, B  B1i  B2 j  B2 k
N
S
(ii) The velocity field of water flowing in a pipe, v ( x, y , z ) .
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(i) The earth’s gravitational field.
5.3. The Vector Operator 
Recall that for any scalar field  ( x, y, z ) we may define its gradient
grad 



i
j
k
x
y
z
This is also an example of a vector field which gives the magnitude and direction of the greatest rate
of change of  at a point.
If we define the vector operator



i  j k
x
y
z
(analogous to D 
d
), known as del (or nabla), then
dx
grad  
We may also apply this operator to vector functions.
5.4. 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 is a vector field, then its divergence is defined
as
div v 
v1 v 2 v3


x
y
z
In terms of the operator  we may write
 


  v   i  j  k   v1i  v 2 j  v3 k 
y
z 
 x
v
v v
 1 2  3
x
y
z
  v  divv
Note   v is a scalar function.
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Example
If
v  x 2 zi  2 y 3 z 2 j  xyz 2 k
divv    v  2 xz  6 y 2 z 2  2 xyz
Physical Interpretation
The idea of divergence is probably most easily illustrated by considering the motion of an
incompressible fluid (e.g. Water) where we take the vector filed to be the velocity
v  v1i  v2 j  v3k of the fluid.
Consider a small imaginary rectangular box of dimensions x, y and z centred at ( x , y , z ) in the
fluid.
z
D
D
A
A
C
C
B
B
y
x
We consider the flow of fluid through this box. The volume of fluid crossing a surface in unit time is
an example of flux.
Taking the midpoint of a face as a representative point, the flux through the face ABCD out of the
box is
x
, y, z ) y z
2
( v2 and v3 are parallel to ABCD and make no contribution)
v1 ( x 
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
5
x
x
 



v1  x  2 , y , z   v1  x  2 , y , z yz



 
2
3


1 x  v1  x, y , z  1  x   2 v1  x, y , z  1  x   3 v1  x, y , z 


  v1  x, y , z  
  





2
3


1
!
2

x
2
!
2

x
3
!
2

x






2
3 3
2


 v1  x, y , z   1   x   v1  x, y , z   1   x   v1  x,2 y , z   1   x   v1  x,3 y , z   yz


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
3
x
3!  2 
x
1! 2

  v1  x, y , z  2 x 2  3 v1  x, y , z 





xyz
x
3! 2 3
x 3


Similarly by considering the two other pairs of faces, we have that the total flux out of the box is
  v1  x, y , z   v 2  x, y , z   v3  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 v3  x, y , z 




 xyz
3
3
3
3
3
3
x
3! 2
y
3! 2
z
 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

 

Take the limit as the volume of the box shrinks to zero we obtain
   v1  x, y , z   v 2  x, y , z   v3  x, y , z 
divv  lim




x 0 
x
y
z



y  0
z  0
 2 x 2  3 v1  x, y , z  2 y 2  3 v 2  x, y , z  2 z 2  3 v3  x, y , z 
 



  
3
3
3
3
3
3
x
3! 2
y
3! 2
z
 3! 2
 
 v1  x, y , z   v 2  x, y , z   v3  x, y , z 



x
y
z
As a measure of the net outward flux per unit volume at a point (x, y, z), positive divergence
indicates a source of fluid at the point whilst negative divergence indicates a sink.
If E is an electric field vector, div E measures the charge density at a point.
6
If for a vector field v, div v  0 then v is said to be solenoidal i.e. no net outflow or inflow of fluid/
no electric charge.
Example
The electric field at position r due to a point charge at the origin is
E
r
Q

3
40 r
40
xi  yj  zk
Q
x
 y 2  z 2 2
3
2


 
x
  similar te rms
 divE    E 
3

40 x  2
2
2
 x  y  z 2 
Q

Q 1
 3 2x 
Q
x
 similar te rms 
3
5 

40  r
2 r 
40
 3 3x 2  y 2  z 2 
 r3 
0
r5


divE  0 except at the origin.
Hence E is solenoidal ( r  0 ) which is consistent with the above interpretation.
5.5. Curl of a Vector Field
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
 v
v 
 v v 
j  1  3   k  2  1 
x 
y 
 z
 x
Note   v is a vector function.
Example
If v  x 2 zi  2 y 3 z 2 j  xyz 2 k
i
j


curlv    v 
x
y
2
x z  2 y3z2
k

 xz 2  4 y 3 z i  x 2  yz 2  j
z
xyz 2
Physical Interpretation
Again we shall illustrate the idea of curl using the motion of an incompressible fluid with velocity
7
v1i  v2 j . Consider just the xy-plane and a small imaginary rectangle of sides  x and  y centered
at (x, y).
y
D
C
(x,y)
B
A
x
Flow along AB = length of AB  velocity component along AB  v1 ( x, y 
CD  v1 ( x, y 
 y
2
 y
2
) x and flow along
)x
 y
 y 

sum  v1 ( x, y 
)  v1 ( x, y 
) x
2
2 



v
v
1
1
 v1 ( x, y )   y 1 ( x, y )  ....  v1 ( x, y )   y 1 ( x, y )  ... x
2
y
2
y


v
  1  x y  ...
y
Similarly the sum of flows along BC + DA is
v2
 x y  ....
x
The total circulation (net flow around a closed path) around the rectangle is
 v2 v1 

 x y  ....

 x y 
If we consider circulation per unit area and take the limit as the area of the rectangle shrinks to zero
we have
v 2 v1

x
y
If this quantity is positive/negative/zero a small paddle wheel placed in the fluid at the point P would
rotate anticlockwise/clockwise/remain stationary.
By analogy with a rigid body rotating about an axis with angular velocity  for which the angular
velocity vector is  n , where n is a unit vector along the axis of rotation , we may associate with the
rotational effect of the fluid a vector
 v 2 v1 

k

 x y 
This is the third component of curl v. Similarly by considering flow in 3-dimensions we may derive
8
the full expression for curl v which gives a measure of the rotational effect of the fluid flow at a
point.
If H is a magnetic field vector, curl H measures the current density at a point.
Example: (An Interpretation of Curl)
Angular Velocity  as the Curl of the Tangential Velocity T
Suppose a body rotates with uniform angular speed  about a line L. See the figure below.
The angular velocity vector  has magnitude  and is directed along L as a right-handed screw
would progress if given the same sense of rotation as the object. Let L go through the origin and let
R  xi  yj  zk for any point (x, y, z) on the rotating object. Let T ( x, y , z ) be the tangential velocity.
Then T   R sin     R , where  is the angle between R and L. Since T and   R are in the
same direction, then T    R .
Now, write   ai  bj  ck .
Then
i j k
T    R  a b c  (bz  cy )i  (cx  az ) j  (ay  bx )k .
x y z
But
9
i
j
k



curlT    T 
x
y
z
bz  cy cx  az ay  bx
  ( ay  bx )  ( cx  az )    ( ay  bx )  (bz  cy )    ( cx  az )  (bz  cy ) 
 



i  

 j  
y
z
x
z
x
y
 

 

 2ai  2bj  2ck  2
Therefore

1
1
CurlT    T
2
2
The angular velocity of a uniformly rotating body is a scalar multiple of the curl of the tangential
velocity. Because of this interpretation, curl was once written rot (for rotation), particularly in British
books on mechanics. This is also the motivation for the term irrotational for a vector field whose
curl is zero.
Example
Consider a current-carrying conductor with circular cross-section of radius R. Then the magnetic
field H is
  yi  xj
 2 r 2
H 
  yi  xj
 2 R 2
y
outside conductor
inside conductor
R
x
1
2 2
where r  ( x 2  y )
H field
Outside the conductor,
i
j
k



curlH    H 
x
y
z
y
x
0
2 x 2  y 2  2 x 2  y 2 
 1
2x2
1
2y2 
 0i  0 j  k 



 crulH  0
2
4
2
4
2

r
2

r
2

r
2

r


Inside the conductor,
10
i
j
k



crul H    H 
x
y
z
y
x
0
2 R 2 2 R 2
1 
 1
 k

2
2 R 2 
 2 R
k

 R2
J
- the current density in the conductor.
If for a vector field v, curl v  0 , then v is said to be irrotational
Examples
(i) The magnetic field above, outside the conductor, is irrotational.
(ii) The electric field E due to a point charge is irrotational (exercise).
5.6 Vector Identities
Some useful results involving the vector operator  are:
f and g are scalar fields, v and w are vector fields.
(i)  f  g   f  g
(ii )  fg   fg  gf
(iii )   v  w     v    w
(iv)
( v)
( vi)
( vii)
( viii)
(ix)
( x)
   fv   f  v  f  v
  v  w     v    w
   fv   f  v  f  v
  v  w   w    v  v    w
  v  w   w   v  v   w    w v    v w
  f  0
    v   0
( xi)
  f   2 f 
2 f 2 f 2 f


x 2 y 2 z 2
Example
Prove (vi)   ( fv)  f  v  f  v
Proof:
11
i

   fv 
x
fv1
j

y
fv2
k
 
 







 i   fv3    fv2   j   fv1    fv3   k   fv2    fv1 
z
z
x
y

 y
  z
 x

fv3
v
v
 f
v   v f
f
f 
 i  v3  f 3 
v2  f 2   j  f 1 
v1  f 3 
v3 
y z
z   z z
x x 
 y
 v
v
f
f 
 k  f 2  v2
 f 1  v1 
x
y
y 
 x
 f
 v
v   f
f 
f 
 v v 
 i  v3 
v2   fi 3  2   j  v1 
v3   fj 1  3 
z 
z   z
x 
x 
 z
 y
 y
 f
 v
v 
f 
 k  v2
 v1   fk  2  1   f  v  f  v
y 
y 
 x
 x
Example
(ix) curl ( grad f )    (f )  0
i

  f  
x
f
x
j

y
f
y
k

z
f
z
 2 f 2 f 
 
 i

 yz zy 
0
 2 f 2 f   2 f
2 f 
  k 

j 


 zx xz   xy yx 
The converse of this result is also true i.e. that every irrotational vector field is the gradient of some
scalar field, and has important applications in electromagnetism and fluid dynamics.
Similarly the converse of result (x), which says that every solenoidal vector field is the curl of some
vector field, arises in applications.
-End-
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