Chapter 1

Gradient, Divergence
& Curl
Scalar and vector fields
 Imagine a cooling system of a reactor
which is using fluid as the cooler medium
vb
va

Fluid

Tc
Td

2
FIELD is a description of how a physical
quantity varies from one point to another
in the region of the field (and with time).
(a) Scalar fields
Ex: Depth of a lake, d(x, y)
Temperature in a room, T(x, y, z)

Depicted graphically by constant

magnitude ycontours or surfaces.
d1
d3
d2

x
3
 At any point P, we can measure the
temperature T.
 The temperature will depend upon
whereabouts in the reactor we take the
measurement. Of course, the temperature
will be higher close to the radiator than the
opening valve.
 Clearly the temperature T is a function of
the position of the point. If we label the
point by its Cartesian coordinates ( x, y, z ) ,
then T will be a function of x, y and z, i.e.
T = T ( x, y , z )
.
 This is an example of a scalar field since 4
• A field is a quantity which can be specified everywhere in
space as a function of position.

• The quantity that is specified may be a scalar or a vector.

• For instance, we can specify the temperature at every point in

a room.

• The room may, therefore, be said to be a region of

“temperature field” which is a scalar field because the
temperature T (x, y, z) is a scalar function of the position.

• An example of a scalar field in electromagnetism is the electric

potential.
5
 Meanwhile, at each point, the fluid will be
moving with a certain speed in a certain
direction
 That is, each small fluid element has a
particular velocity and direction,
depending upon whereabouts in the fluid it
is.
 This is an example of a vector field since
velocity is a vector. The velocity can be
expressed as a vector function, i.e.
v = v ( x, y, z ) = v1 ( x, y, z )i + v2 ( x, y, z ) j + v3 ( x, y, z )k
v1 , v2 and v3
where will each be scalar
functions. 6
Example: Linear velocity vector field of
1-7

points on a rotating disk

7
 Physical examples of scalar fields:

Electric potential around a Temperature near a

charge heated wall

(The darker region representing higher values )

8
 Physical examples of vector fields:

+ −

Electric field surrounding a Magnetic field lines shown by iron

positive and a negative charge. filings

The flow field around an airplane 9

Hurricane
Gradient of Scalar Fields
The gradient of a scalar field is a vector
field,
 which points in the direction of the
greatest rate of increase of the scalar field,
and whose magnitude is the greatest rate
of change.

10
   x, y, z   ctt
uuu
r
dr

grad 

uuu
r
Physical meaning:grad  dr is the local variation of Φ
along dr. Particularly, grad Φ is perpendicular to the line Φ =
ctt.

11
Gradient operator

Suppose , we have a function of three variables- say,

the temperature T(x, y, z) in a room.

For the temperature distribution we see how a scalar

would vary as we moved off in an arbitrary direction.

Now a derivative is supposed to tell us how fast the

function varies, if we move a little distance.

12
 ,
If T(r) is a scalar field, its gradient is defined in Cartesian
coordinates by
∂T ∂T ∂T
grad T = i+ j+ k
∂x ∂y ∂z
∇
It is usual to define the vector operator
which is called “del”. We can write

 ∂ ∂ ∂
∇ = i + j + k  grad T ≡ ∇T
.
 ∂x ∂y ∂z 

Without thinking too hard, notice that grad T tends to

point in the direction of greatest change of the scalar
13
field T.
The significance of the gradient:

A theorem on partial derivatives states that

 ∂T   ∂T   ∂T 
dT =  dx +  
 dy +  dz
 ∂x   ∂y   ∂z 
This rule tells us how T changes when we alter all three
variables by the infinitesimal amounts dx, dy, dz. Change in T
can be written as,
 ∂T ∂T ∂T 
dT =  i+ j+ k  • ( dx i + dy j + dz k )
 ∂x ∂y ∂z 
 →   →
=  ∇T  ⋅ d l 
   
The conclusion is that, the RHS of above equation is the small
change in temperature T when we move by dl. 14
If we divide the above eq. by dl
We get
 →

dT  →   d l 
=  ∇T  ⋅ 
dl    dl 
 
 →
d l  →
but  is a unit vector in the direction of d l .
 dl 
 
So , we can conclude that, grad T has the property that the rate
of change of T w.r.t. distance in any direction â is the projection
of grad T onto that direction â.
That is
dT  ^
 → ^
 in direction of a  = ∇T ⋅ a
dl   15
the quantity dT is called a directional derivative.
dl

In general,

• a directional derivative had a different value for

each direction,

• has no meaning untill you specify the direction

16
Gradient Perpendicular to T constant surfaces

If we move a tiny amount within the surface, that

is in any tangential direction, there is no
change in T , so
dT
= 0.
dl

Surface of constant T,
These are called level surfaces. Surfaces of constant T
→ →
dl dl
So for any in the surface ∇T ⋅ =0
dl dl
Conclusion is that; grad T is normal to a surface of constant T.
17
Geometrical Interpretation of the Gradient
(1.5)

Like any vector, a gradient has magnitude and direction.

To determine its geometrical meaning, lets rewrite the
dot product In its abstract form:

dT = ∇T • dl = ∇Tdl cos θ
where θ is the angle between ∇T and dl. Now, if we fix the
magnitude dl and search around in various directions (that is, vary
θ), the maximum change in T evidently occurs when θ =0 (for then
cos θ = 1). That is for a fixed distance dl, dT is greatest when I
move in the same direction as ∇ T .

18
In the above two images, the scalar field is in black
and white, black representing higher values, and its
corresponding gradient is represented by blue arrows.
19
Example 1
If φ(x,y,z) = 3x2y– y2z2, find grad φ and ∇φ
at the point (1,2,−1).

20
 Solution
∂φ ∂φ ∂φ
grad φ = ∇φ = i + j +k
∂x ∂y ∂z
= 6 xyi + (3 x − 2 yz ) j + (−2 y z )k
2 2 2

At the point (1,2,−1),

∇φ = 6(1)(2)i + [3(1) 2 − 2(2)(−1) 2 ]j − 2(2) 2 (−1)k
= 12i − j + 8k
∴ ∇φ (1, 2, −1) = 12i − j + 8k = 12 + (−1) + 8 = 209
2 2 2

21
Example 2
 If r = x 2 +y 2 +z 2
Find out ∇r =?
→ ^ ^ ^
here r =x i +y j +z k

Q.Show that
(a) ∇( r 2 ) =2r
^
1  r
(b) ∇  =− 2
r  r
Example 3
Find φ ( x, y ) , if

∇φ = y cos x i + (sin x + e y ) j

Given φ (0,0) = 0.

23
Solution
Since ∇ φ = y cos x i + (sin x + e y
) j , we have

∂φ
∂φ = sin x + e y .....(2)
= y cos x .....(1) ∂y
∂x
Integrating (1) and (2) w.r.t. x and y
respectively, we obtain
φ = ∫ y cos xdx = y sin x + f ( y ) .....(3)
φ = ∫ (sin x + e y )dy = y sin x + e y + g ( x) .....(4)
24
Comparing (3) and (4), we can conclude
that
f ( y ) = e y + C and g ( x) = C
where C is an arbitrary constant of integration
Hence, φ ( x , y ) = y sin x + e y
+C
To find constant C, use φ (0,0) = 0.
φ (0,0) = 0 sin 0 + e 0 + C = 0
1+ C = 0
∴ C = −1
Therefore, φ ( x, y ) = y sin x + e − 1
y
♣ 25
Example 4
Find φ ( x, y, z ) if
∇φ = ( y − 2 xyz )i + (3 + 2 xy − x z ) j + (4 z − 3 x yz )k
2 3 2 3 3 2 2

and φ (0,0,0) = −2 .

26
 Solution
We have ∂φ
= y 2 − 2 xyz 3 .....(1)
∂x
∂φ ∂φ
= 3 + 2 xy − x z .....(2)
2 3 = 4 z 3
− 3 x 2
yz 2
.....(3)
∂y ∂z
Integrating (1), (2) and (3) w.r.t. x, y and z
respectively, we obtain
φ = ∫ ( y − 2 xyz )dx = xy − x yz + f ( y, z ) .....(4)
2 3 2 2 3

φ = ∫ (3 + 2 xy − x 2 z 3 )dy = 3 y + xy 2 − x 2 yz 3 + g ( x, z ) .....(5)
φ = ∫ (4 z 3 − 3 x 2 yz 2 )dz = z 4 − x 2 yz 3 + h( x, y ) .....(6) 27
Comparing (4) with (5) and (6) we get
f ( y, z ) = 3 y + z 4 + C

Therefore
φ = xy − x yz + 3 y + z + C
2 2 3 4

To find constant C, use φ (0,0,0) = −2

∴φ = xy − x yz + 3 y + z − 2
2 2 3 4

28
Problem 5
→
Show that ∇φis a vector perpendicular
to the surface φ( x, y , z ) = k where k is const.
φ =φ( x, y , z ) = k
∂φ ∂φ ∂φ
dφ = dx + dy + dz = o
∂x ∂y ∂z
 ^

 ∂φ ∂φ ∂φ 
i +j +k  •(idx + jdy +kdz )
 ∂x ∂y ∂z 
 
→ →
∇φ.d r = 0
29
Application of gradient:
Surface normal vector
 A normal, n to a flat surface is a vector
which is perpendicular to that surface.
 A normal, n to a non-flat surface at a point
P on the surface is a vector perpendicular
to the tangent plane to that surface at P.
n

z = f ( x, y )
30
 Therefore, for a non-flat surface, the
normal vector is different, depending at
the point P where the normal vector is
located. n

z = f ( x, y )

 Unit vector normal isndefined as

nˆ =
n
31
 To find the unit vector normal to the surface
z = f ( x, y )
we follow thezfollowing
= f ( x, y )steps:
(i) Rewrite the
φ ( xfunction as
, y, z ) = k , k any constant
n = ∇φ
(ii) Find the normal vector that is
∇φis
(iii) Then, the unit normalnvector
n̂ = =
n ∇φ

P ( x0 , y0the
(iv) Hence, , z0 )unit normal vector at a point
∇φ ( x0 , y0 , z0 )
nˆ =
is ∇φ ( x0 , y0 , z0 ) 32
Example 5
Find the unit normal vector of the surface
at the indicated point.

(a) z = 6 − x 2 − y 2 at (−1,3,2)
(b) xe y
+ y 3
= z 2
at (1,0,−1)

33
 Solution
(a) Rewrite z = 14 − x − y as
2
x 22
+ y 2
+ z 2
= 14

Thus, we obtain φ ( x , y , z ) = x 2
+ y 2
+ z 2

Then, ∇φ = 2 xi + 2 yj + 2 zk = 2( xi + yj + zk )
At the point (−1,3,2),
∇φ = 2(−i + 3 j + 2k ) and ∇φ = 2 (−1) 2 + 32 + 2 2 = 2 14
∇φ − i + 3 j + 2k
The unit normal vector is nˆ = =
∇φ 14
34
(b) Rewrite xe + y = z as xe + y − z = 0
y 3 2 y 3 2

Thus, we obtain φ ( x, y, z ) = xe + y − z
y 3 2

Then, ∇φ = e i + ( xe + 3 y ) j − 2 zk
y y 2

At the point (1,0,−1),

∇φ = e 0 i + [1e 0 + 3(0) 2 ]j − 2(−1)k = i + j + 2k

and ∇φ = 12 + 12 + 2 2 = 6

∇φ i + j + 2k
The unit normal vector is nˆ = =
∇φ 6
35
Divergence of Vector Fields
 The divergence is an operator that
measures the magnitude of a vector field's
source or sink at a given point
 The divergence of a vector field is a scalar
uu
r uu
r
V(x, y, z) V(x  dx, y, z)

x x+dx 36
 The divergence of a vector field
F ( x, y, z ) = F1 ( x, y, z )i + F2 ( x, y, z ) j + F3 ( x, y, z )k
is defined as
div F = ∇ ⋅ F
 ∂ ∂ ∂
=  i + j + k  ⋅ ( F1i + F2 j + F3k )
 ∂x ∂y ∂z 
∂F1 ∂F2 ∂F3
= + +
∂x ∂y ∂z

37
  Divergence
uu
r uu
r
V(x, y, z) V(x  dx, y, z) ur
v(x, y, z) is a differentiable vector field

x x+dx
ur ur v vy vz
div v =  v  x    uvu
x y z

2 – Physical meaning
ur
div v is associated to local conservation laws: for example, we
will show how that if the mass of fluid (or of charge) outcoming
from a domain is equal to the mass entering, then ur
ur div v  0
v is the fluid velocity (or the current) vectorfield 38
Geometrical Interpretation.

The name divergence is well chosen, for ∇

. F is a
measure of how much the vector F spreads out
(diverges) from the point in question.

The vector function has a large (positive) divergence

at the point P; it is spreading out. (If the arrows
pointed in, it would be a large negative divergence.)

NOTE: P=electric field due to charge (+ ve or – ve) 39

On the other hand, the function has zero divergence at
P; it is not spreading out at all.

So, for example, if the divergence is positive at a point, it

means that, overall, that the tendency is for fluid to move away
from that point (expansion); if the divergence is negative, then
the fluid is tending to move towards that point (compression).
40
 Fundamental theorem of divergence
The fundamental theorem for divergences states
that:
∫ ( ∇ • F ) d τ = ∫ F • da
volume surface
This theorem has at least three special names:
Gauss’s theorem, Green’s theorem, or, simply, the
divergence theorem.
dτ

dτ is function at the boundary element of volume (in

Cartesian coordinates, dτ = dx, dy, dz), and The
volume integration is really a triple integral.

41
 da represents an infinitesimal element of
area; it is a vector , whose magnitude is
the area of the element and whose
direction is perpendicular ( normal ) to the
surfaces, pointing outward.

On the front face of the

cube, a surface element

da1 = ( dy dz ) iˆ
is

42
 on the right face, it would be

da 2 = ( dz dx ) ˆj
whereas for the bottom it is

( )
da3 = ( dx dy ) − kˆ

43
Problem 1
 Q. Calculate the divergence of the following
vector functions?

(a ) v1 = x i + 3 xz j − 2 xzk
2 2

(b) v2 = xyi + 2 yzj + 3zxk

Problem 2
 Check the divergence theorem using the
function
∧ ∧ ∧
v = y i + (2 xy + z ) j + (2 yz ) k
2 2

And the unit cube is situated at the origin.

Z

45
X
Curl of Vector Fields
 Curl is a vector operator that shows a vector
field's rate of rotation, i.e. the direction of the
axis of rotation and the magnitude of the
rotation.
∇×v = 0

ur
curl v  0

46
47
 The curl of a vector field
F ( x, y, z ) = F1 ( x, y, z )i + F2 ( x, y, z ) j + F3 ( x, y, z )k
is defined as
i j k
∂ ∂ ∂
curl F = ∇ × F =
∂x ∂y ∂z
F1 F2 F3
 ∂F3 ∂F2   ∂F3 ∂F1   ∂F2 ∂F1 
= i −  − j −  + k  − 
 ∂y ∂z   ∂x ∂z   ∂x ∂y 

48
Problem 1
Z

 Find the curl of v3 = − yi + xj

i j k
∂ ∂ ∂
∇ × v3 = y
∂x ∂y ∂z
−y x 0

The curl of v3 points in the z-direction

X
Curl
To find a possible interpretation of the curl, let us consider a body
rotating with uniform angular speed ω about and axis l. Let us define the
vector angular velocity to be a vector of length ω extending along l in
the direction
Take the point O as the origin of coordinates we can write R = xi + yj + zk

the radius at which P rotates is |R||sinθ| Hence, the linear speed if P is

v = ω|R||sinθ| = Ω|R||sinθ|

If we take the curl of V, we therefore have

 that is

Expanding this, remembering that Ω is a vector, we find

Conclusion: The angular velocity of a uniform rotating body is thus equal

to one-half the curl of the linear velocity of any point of the body.
Example: For velocity field,

( ) ( )
^ ^
u = x + e sin ( yz ) i + x + e cos( yz ) j
x x , find the angular velocity ω.
^ ^ ^
i j k
∂ ∂ ∂
ω = ∇×u =
∂x ∂y ∂z
u v w
For the field,
u = ( x + e sin ( yz ) ) i + ( x + e cos( yz ) ) j
^ ^
x x

, we obtain:
^ ^ ^
i j k
∂ ∂ ∂
ω = ∇×u =
∂x ∂y ∂z
x + e sin ( yz )
x
x + e cos( yz )
x
0

∂
∂
( ∂
) ( ) ∂
( ) ( ) k
^ ^ ^
=− x + e x cos( yz ) i + x + e x sin ( yz ) j +  x + e x cos( yz ) − x + e x sin ( yz )
∂z ∂z  ∂x ∂y 
{ }
^ ^ ^
= e y sin ( yz ) i + e y cos( yz ) j + 1 + e cos( yz ) − ze cos( yz ) k
x x x x
 Fundamental theorem of curl
The fundamental theorem for curls, which goes by the special name of Stokes’
theorem, states that

∫ ( ∇ × v ) ⋅ da = ∫ v.dl
surface boundary
line

The integral of a curl over a region (a patch of surface) is equal to the value of the
function at the boundary (the perimeter of the patch).
54
Example 6
Find both div F and curl F at the point
(2,0,3) if

F( x, y, z ) = ze i + 2 xz cos yj + ( x + 2 y )k
2 xy

55
Solution
∂F1 ∂F2 ∂F3
div F = ∇ ⋅ F = + +
∂x ∂y ∂z
∂ ∂ ∂
= ( ze ) + (2 xz cos y ) + ( x + 2 y )
2 xy

∂x ∂y ∂z
= 2 yze 2 xy − 2 xz sin y
[Notice that div F is a scalar!]

At the point (2,0,3),

∇ ⋅ F = 2(0)(3)e 2( 2)( 0) − 2(2)(3) sin 0 = 0
56
 i j k
∂ ∂ ∂
curl F = ∇ × F =
∂x ∂y ∂z
ze 2 xy
2 xz cos y x + 2y
∂ ∂ 
= i  ( x + 2 y ) − (2 xz cos y )
 ∂y ∂z 
∂ ∂ 2 xy 
− j ( x + 2 y ) − ( ze )
 ∂x ∂z 
∂ ∂ 2 xy 
+ k  (2 xz cos y ) − ( ze )
 ∂x ∂y 
= (2 − 2 x cos y )i − (1 − e 2 xy ) j + (2 z cos y − 2 xze 2 xy )k
57
[Notice that curl F is also a vector]

At the point (2,0,3),

∇ × F = [2 − 2(2) cos 0]i − [1 − e 2 ( 0)(3) ]j
+ [2(3) cos 0 − 2(2)(3)e 2 ( 2 )( 0 ) ]k
= −2i − 6k

58
Properties of Del
If F(x,y,z) and G(x,y,z) are differentiable
vector functions φ(x,y,z) and ψ(x,y,z) are
differentiable scalar functions, then
(i) ∇(φ ±ψ ) = ∇φ ± ∇ψ
(ii) ∇(φψ ) = φ∇ψ +ψ∇φ

φ  ψ∇φ − φ∇ψ
(iii) ∇  =
ψ ψ 2

59
(iv) ∇ ⋅ (F ± G ) = ∇ ⋅ F ± ∇ ⋅ G
(v) ∇ × (F ± G ) = ∇ × F ± ∇ × G

 ∂ 2
∂ 2
∂ 2

(vi) ∇ ⋅ (∇φ ) ≡ ∇ φ ≡  2 + 2 + 2 φ
2

 ∂x ∂y ∂z 
∂φ ∂φ ∂φ
2 2 2
= 2+ 2+ 2
∂x ∂y ∂z

(vii) ∇ × (∇φ ) = 0 or curl grad φ = 0

(viii) ∇ ⋅ (∇ × F) = 0 or div curl F = 0
*Notes: In (vi), ∇ 2 is called the Laplacian operator 60