Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...

Engineering Mathematics - II - Vector Calculus - 2007
Syllabus
Vector Differentiation
- Velocity and Acceleration
- Gradient
- Divergence
- Curl
- Laplacian
- Solenoidal and Irrotational Vectors
Recap
Scalar product
: a .
b
a1 b1
a2
b2
a3
b3
|
a | | b | cos
,
is
the
angle
between
a
and
b
Pr ojection :
projection
of
a
on
b
is
a . b.
Work done : If a force
F displaces are provide from A to B then
the work done is
F . AB
Vector
product :
a
b
i
a1
b1
j
a2
b2
k
a3
b3
| a | | b | sin
n
where
n
is
unit
vector
perpendicular
to
both
a
and
b
Moment: If the vecto r moment of a force
F about a po int A
through a point
P
is
M
AP
F
Scalar
triple
product : [ a ,
b ,
c ]
a .( b
c )
a1
b1
c1
a2
c2
c3
a3
b3
c3
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Engineering Mathematics - II - Vector Calculus - 2007
Equation of line :
Equations of a line pas sin g through (x1, y1, z1)
and
direction
ratios
a, b, c are
x x1 y y1
z z
1
a b c
Equation of a plane : is of the form
ax + by + cz + d = 0 where a,b,c are the DR's of the normal.
Introduction
In this chapter the basic concepts of Differential Calculus of scalar functions are
extended to vector functions.
Here we study vector differentiation, gradient, divergence, curl, solenoidal and
irrational fields.
Also we study about vector integration and verification of the integral theorems.
Vector Function of a Scalar Variable
Let f f1
i f2
j f3
k where f1, f2
and f3
are
functions
of
a
var iable
t
is called a vector function.
If f1, f2 and f3 are differentiable then we define
d f f(t t) f(t)
df
1 df
2 df
lim i j
3
k
dt t0
t dt dt dt
Similarly higher order derivatives can also be defined.
Note:
d f
0
dt
iff
f
is
a
cons tan t
vector .
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Engineering Mathematics - II - Vector Calculus - 2007
Rules of differentiation
1. If g(t) is a scalar function of t and f (t) is a vector function of
d d f d
scalar t then ( f) f
dt dt dt
2. If
and
are
scalar
cons tan ts,
f
and
g
are vector
functions of
t
then
d
dt
( f
g)
d f
dt
d g
dt
3.
d
dt
Proof:
4.
d
dt
Proof:
( f . g)
d g
f .
dt
d f
. g
dt
Let f f1
i f2
j f3
k
f . g
g g1
i g2
j g3
k
f1 g1
f2
g2
f3
g3
d dg1
df1
dg2
df2
dg3
df3
( f . g) f1 g1
f2
g2
f3
g3
dt
dt dt dt dt dt dt
dg
1 dg
2 dg
3 df
1 df
2 df
f
3
1 f2
f3
g1
g2
g3
dt dt dt dt dt dt
d g
f .
dt
( f
df
.
dt
g
d g d f
g) f
dt dt
Let f f1
i f2
j f3
k
g g1
i g2
j g3
k
g
f g
i
f1
g1
j
f2
g2
k
f3
g3
Page 3 of 72

Engineering Mathematics - II - Vector Calculus - 2007
f g i (f2
g3
g2
f3)
d
dt
( f g)
d
dt
i (f2
g3
g2
f3)
d f
i
2
dt
g3
f2
dg3
dt
dg2
dt
f3
g2
df3
dt
d f
i
2
dt
dg
g
2
3
dt
f3
i
f2
dg3
dt
g2
df3
dt
d f
dt
g
f
d g
dt
5.
d
dt
f , g, h
d
f
,
dt
g, h
f
,
d g
d h
,h f , g ,
dt
dt
Proof:
d
dt
f ,
g,
h
d
dt
f
g .
h
d
dt
f
g .
h
f
g .
d h
dt
d f
dt
g
.
h
f
d g
.
dt
h
f g .
d h
dt
d
f
,
dt
g , h
d g
f , , h
dt
f ,
g,
d h
dt
d f
6. If | f | cons tan t then f . 0
dt
Proof:
Let | f |
C
| f | a
2
where a is constant
f .
f
cons tan t
d
dt
f .
f
0
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Engineering Mathematics - II - Vector Calculus - 2007
d f d f
f . . f 0
dt dt
d f
f .
dt
0
7. If
f
has
a
fixed
direction
then
f
d f
0
dt
Proof:
Let
f
f e
Now
f
d f
dt
f e
d
dt
(f e)
df d e
f e e 0
dt dt
f
df
dt
(e e)
df
f (0)
dt
0
8.If
f
d f
dt
0
then
f
has
a
fixed
direction.
Equation of a Space Curve
Let P (x,y,z) be any point in space then its position vector is
OP
r
x i
y j
z k
If x = x(t), y = y(t) and z = z(t) are functions of a scalar variable
t then
r r (t)
describes
a
curve
in
space
called
the
space
curve.
Page 5 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Note 1:
d r
dt
represents
the
velocity
v
(rate
of
change
of
position
with
which
the
ter min al po int
of
r
describes the
curve )
Note 2:
d r
dt
is
along
the
direction
of
the
tan gent
to
the
space
curve
at
P (x, y, z)
Note 3:
The
unit
tan gent
vector
to
the
space
curve
is
denoted by
t
and
is given by
t
d r
dt
d r
| |
dt
Note 4:
d v
dt
d
2
r
is called the acceleration a
dt
2
of
the
curve
at time
t.
Unit Normal to a Space Curve
Let A be a fixed point on the curve and s be the length of the arc AP where
P(x,y,z) is a point on the space curve.
Page 6 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Let
t
be
the
unit
tan gent
vector
at
P.
Consider
d.w.r.t s
t . t
1
t
.
d t
ds
d t
. t
ds
0
d t
2 t .
ds
0
t
d t
.
ds
0
d t
ds
is
a normal
to
the
space
curve
and
n
d t
ds
d t
| |
ds
is
called
the
unit
normal to
the
curve.
Scalar and Vector Fields
Scalar Valued Function
Let P(x,y,z) which is a scalar is called a scalar point function.
Note: A scalar point function is also called a scalar fixed . = (x,y,z)
Ex.: (x,y,z) = x 2 + y 2 + z 2
Vector Valued Function
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Engineering Mathematics - II - Vector Calculus - 2007
Let P(x,y,z) be a point. A vector point function is a vector whose
components are real valued functions of x,y,z. A vector point function is called a
vector
field
F .
Ex:
f f (x, y, z) f1
i f2
j f3
k
F xyz i x
2
y j yz k
Differential
operator
:
denotes
the
operation
i
x
y
j
k
z
i
x
does
not
represent
a
vector
if
only
defines
the
differential
operator
Gradient of a Scalar Field
Let(x, y, z) be a scalar field, then i k i
x z x
is called the gradient of denoted by .
The
following
are
some
results
obtained
by
the
definition
of
.
1. ( )
,
is a scalar cons tan t.
2. ( )
,
and are scalar fields.
3. ( )
,
and are scalar cons tan ts
4.()
5.
2
6. d dx dy dz
x y z
i
x
j
y
k
z
i
dx
j dy
k dz
d
. d r
Page 8 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Geometrical Interpretation
Q(x, y, z)
c represents a surface in space and
to the surface at any point.
represents a normal
Note 1: Since = c on the surface.
d . d r
0
which
shows
that
is
perpendicular
to
the
tan gent
plane
at
any
po int .
Note
2 : Unit
vector
along
is
denoted by
n
is
| |
called
the
unit
normal
to the surface (x,y,z) = c.
Directional Derivative
Let
a
be
a
vector
inclined at
an
angle
with
then
.
a is
called
the
directional
derivative
along
a .
Note : Maximum value
of directional
derivative
is
n | |
Page 9 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Divergence of a Vector Field
Let f f1
i f2
j f3
k be a vector field then divergence of f
denoted by
div
f
or
.
f
is
defined
as
f
f
1 f2
f
3 i
div. f . f
x y z x
Physical Interpretation
If
v the
velocity
of
a moving
fluid
at
a po int
P
at
time
t,
then
div v
represents
the rate at which the fluid moves out of a unit volume enclosing the point P.
Solenoidal Vector
A
vector
field
f
is
said
to
be
solenoidal
if
div. f
0.
Properties
1. div ( f g) div f div g , where f and g are vector fields.
2. div ( f ) div f where is a scalar cons tan t.
3.
div ( f g)
cons tan ts.
div f
div g
where
and
are
scalar
4.
If
is a scalar field and f is a vector field then div ( f ) div f .
Proof:
Let f f1
i f2
j f3
k
then f f1
i f2
j f3
k
div ( f ) ( f1 ) ( f2
) ( f3)
x y z
f1
x
f1
x
f
2
y
f2
y
f
3
y
f3
y
f1
x
i
x
. f1
i
div f
.
f
Page 10 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Laplacian of a scalar field
Let
then
= (x,y,z) be a scalar field
is a vector field
and
div
( )
.
x
2
x
2
x
y
2
y
2
z
2
y
z
z
denoted by
2
is
called
the
Laplacian
of
a
scalar
field.
Note 1:
2
2
x
2
2
y
2
2
z
2
Note 2:
A
scalar
field
is
called
a harmonic
function
if
2
0
Note 3:
2
()
2
, where
2
and are cons tan ts,
and
are scalar fields.
Curl of a Vector field
Let f f1
i f2
j f3
k
then
curl
of
f
denoted by
curl
f
or
f
is
defined
as
f
i j k
/ x / y
/ z
f1 f2
f3
f
3 f
2
i
y z
Physical Interpretation
If v is the velocity of a particle in rigid body rotating about a fixed
axis
then
the
angular
velocity
at
any
po int
is
given by
1
2
curl v .
Page 11 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Irrational Vector
f is said to irrational if curl f 0
Properties
(i)
If
f
and
g
are
vector
fields then
curl ( f
g)
curl
f
curl
g
(ii) Curl ( f ) Curl f , where is a scalar cons tan t.
(iii)
If
is
a
scalar
field
and
f
is
vector
field
then
curl ( f )
curl
f
f
Proof:
Curl ( f ) ( f3)
( f2
) i
y z
f
3
y
f
2
i
z
f3
y
f2
i
z
f f
Curl ( f )
3
2
i f3
f2
i ...(1)
y z y z
Consider
f
i j k
/ x / y / z
f1 f2
f3
f3
y
f2
z
i
Curl ( f )
curl
f
f
(iv)
Curl ( )
0
Proof:
Curl ( )
i
y z z y
0
(v). If f is a vector field then div (Curl f ) 0
Proof:
Page 12 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Page 13 of 72
k
y
f
x
f
i
x
f
z
f
i
z
f
y
f
f
Curl
1
2
3
1
2
3
y
f
x
f
x
x
f
z
f
z
z
f
y
f
x
f)
(curl
div
1
2
3
1
2
3
y
z
f
x
z
f
x
y
f
z
y
f
z
x
f
y
x
f 1
2
2
2
3
2
1
2
2
2
3
2
= 0
Some important Identities
(i) Divergence of a vector product:
then
fields
vector
are
g
and
f
If
g
curl
.
f
f
g .curl
g)
( f
div
Proof:
k
f
j
f
i
f
f
Let 3
2
1
k
g
j
g
i
g
g 3
2
1
k
)
g
f
g
(f
j
)
g
f
g
(f
i
)
f
g
g
(f
g
f 1
2
2
1
3
1
1
3
3
2
3
2
2
3
3
2 g
f
g
f
x
g)
div (f
g)
( f
x
.
i
x
g
f
g
x
f
.
i
f
x
g
.
i
g
x
f
.
i
f
.
x
g
i
g
.
x
f
i
f
.
g)
(
g
.
)
f
(
g
Curl
.
f
f
Curl
.
g
(ii)
Curl of Curl of a vector:

Engineering Mathematics - II - Vector Calculus - 2007
Curl (Curl f ) grad (div f )
2
f
Proof:
Let f f1
i f2
j f3
k
f3 f2 f1 f2 f2 f
f i j
1
k
y z z1
x x y
Curl (Curl f )
i
f
2
y x
f1
y
z
f1
z
f
3
x
2
f
2
f
2
2
1
1 f
1 f
3
i
y
x
y
2
z
2 z x
2
f
2
f
2
f
2
3
2
2 f
1
j
z
y
z
2
x
2 x y
2
2
f
2
f f
2
1 3 3 f2
k
x
z
x
2
y
2 y z
adding
and
subtracting
2
f1
,
y
2
2
f3
z
2
to
each
of
the
terms.
i
x
f1
x
f
2
y
f
2
3 f
1
z
x
2
2
f
2
y
2
2
f
1
z
2
i (div f )
x
2
f
Note 1:
(div f )
2
f
The operator f . (f i f j f k) .
1 2 3
i
x
f1
f2
f3
x
y
z
j
y
k
z
is called the Linear differential operator.
Note 2: If is a scalar field
Note 3:
f .
f1
f2
f
3
x y
z
Page 14 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Page 15 of 72
field
a vector
is
g
If
z
g
f
y
g
f
x
g
f
g
.
f 3
2
1
(iii)
Gradient of a scalar field:
then
fields
vector
are
g
and
f
If
f
)
( g .
g
)
.
( f
f
Curl
g
g
Curl
f
g)
.
( f
Proof:
)
g
.
( f
x
i
g)
.
( f
Consider
x
g
.
f
g
.
x
f
i
...(1)
x
g
.
f
i
x
f
g .
i
x
f
i
g
f
Curl
g
Now
x
f
i
g
x
f
i
g .
i
x
f
g .
f
g .
x
f
g .
i
...(2)
x
f
g .
i
g
)
( g .
f
Curl
g
Similarly
...(3)
x
g
.
f
i
g
)
.
( f
g
curl
f

Engineering Mathematics - II - Vector Calculus - 2007
Page 16 of 72
From (1), (2) and (3)
f
)
( g .
g
)
.
( f
f
Curl
g
g
Curl
f
g)
.
( f
Note:
f
)
.
2( f
f
Curl
f
2
)
f
.
( f
then
g
f
If
f
)
.
( f
f
Curl
f
)
f
.
( f
grad
2
1
or
(iv) Curl of a vector product
then
fields
vector
are
g
and
f
If
g
)
.
( f
f
)
( g .
f
g div
g
div
f
g)
( f
Curl
Proof:
g)
( f
x
i
g)
( f
Curl
x
g
f
g
x
f
i
x
g
f
g
x
f
i
x
g
f
i
g
x
f
i
x
g
)
f
.
( i
f
x
g
i
g
x
f
.
i
x
f
g)
.
( i
x
g
f
x
g
.
i
f
x
f
.
i
g
x
f
g 1
1
g
)
.
f
(
g
div
f
f
g div
f
)
( g .
g
)
.
( f
f
)
( g .
f
div
g
g
div
f
Integration of Vector functions
a
be
C
and
space
in
R
region
a
over
defined
field
vector
a
be
f
Let

Engineering Mathematics - II - Vector Calculus - 2007
curve in R whose equation is
r x(t) i y (t) j z(t) k
Let A and B be two points on Curve C corresponding to t = a and t = b,
We
have
t
d r
dt
d r
| |
dt
AB
f . t ds
Note 1:
Note 2:
Note 3:
Note 4:
The
is
Since t
called the line int egral
above int egral
d r
ds
of f
is also represented as
C
f . t ds
C
along curve C between
f .d r
f .d r
f1 dx
f2
dy
f3
dz
C C C C
Also
b
dx dy dz
1 dt
dt dt dt
a
f .d r
f
f2
f3
C
C
f . t ds
A
and B.
Physical Meaning:
Page 17 of 72

C
Engineering Mathematics - II - Vector Calculus - 2007
f .d r represents the work done in moving a particle of unit
along the Curve C from A to B.
mass
Surface Integral of a vector function
The
surface int egral
of
f
over a
surface
S
is
defined
as
S
f .n
dS
where
n
is the unit normal to the surface S and dS = dx dy
Physical Meaning:
The
surface int egral
of
f
gives
the
total normal
flux
through
a
surface.
Volume integral of a vector function
The
volume int egral
of f
over
a
volume V
is
defined
as
V
f dV
f1d V
i
f2
dV
j
d V
k
V V V
Integral Theorems
Green's Theorem (Statement only)
Let M(x,y) and N(x,y) be two functions defined in a region A in the xy plane with
a simple closed curve C as its boundary then
C
(M dx
Mdy)
N
x
A
M
dx dy
y
Strokes Theorem (Statement only)
Let S be an open surface bounded by a simple closed curve C for a field
Page 18 of 72

f
defined
Engineering Mathematics - II - Vector Calculus - 2007
over a region containing S and C.
C
f .d r
S
(Curl
f ) .n
dS
Gauss Divergence theorem (Statement only)
Let S be the closed boundary surface of a region of Volume V. Then for a
vector field f defined in V and on S.
S
f .n
dS
S
i.e., in Cartesian form
div f dV
f
f
f
1 dx dy dz
x
y
z
V
f dy dz f
1
2
3
2 dz dx f3
dx dy
S
Note:
Page 19 of 72

For
Engineering Mathematics - II - Vector Calculus - 2007
any vector field f , and any closed surface S
S
Curl f . n
dS
0
Page 20 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Summary
Vector function: A vector function of a scalar variable t is of the form
f1(t)
i f2
(t) j
f3(t)
k
Differentiation of a Vector:
d f df
1 df df
i
2
j
3
k
dt dt dt dt
Equation of a space curve: Equation of a space curve is represented by
r (t)
x(t) i y(t) j z (t) k
d r
Velocity :
dt
space curve.
is
the
velocity v
and is
along
the
tan gent
to
the
Unit
tan gent vector
: t
d r
dt
d r
| |
dt
Unit normal : n
|
d t
ds
d t
|
ds
is called the unit normal to the curve.
Scalar field: A scalar corresponding to each (x,y,z) in a region R.
Vector field: A vector corresponding to each (x,y,z) in a region R
r f1
(x, y, z) i f2
(x, y, z) j f3(x, y, z) k
Gradient: If
is
a scalar field
then
i
x
j
y
k
z
is
called
the
gradient of .
Unit Normal to the surface : If = c is a surface then
| |
n
is
the unit
normal
to the surface.
Page 21 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Directional derivative: . a is called the directional derivative in
the direction of a .
Divergence of a vector field :If f f1
i f2
j f3
k
is
a vector
field
then
div
f .
f
f
1
x
f
2
y
f3
z
f1
x
Solenoidal:
f
is
solenoida.l of
div f
0
Laplacian of a scalar field: Let be a scalar field then
2
2
x
2
2
y
2
2
z
2
Curl
of
a
vector
field: Let
f
be
a
vector
field
then
curl
of
f
Curl
f
f
i j k
x
y
z
f1 f2
f3
f
3 f
2
i
y z
Irrational: f
is
irrational if
Curl
f 0
Introduction
In this chapter the basic concepts of Differential Calculus of scalar functions are
extended to vector functions.
Here we study vector differentiation, gradient, divergence, curl, solenoidal and
irrational fields.
Also we study about vector integration and verification of the integral theorems.
Vector Function of a Scalar Variable
Let f f1
i f2
j f3
k where f1, f2
and f3
are
functions
of
a
var iable
t
is called a vector function.
Page 22 of 72

Engineering Mathematics - II - Vector Calculus - 2007
If f1, f2 and f3 are differentiable then we define
d f f(t t) f(t)
df
1 df
2 df
lim i j
3
k
dt t0
t dt dt dt
Similarly higher order derivatives can also be defined.
Note:
d f
0
dt
iff
f
is
a
cons tan t
vector .
Rules of differentiation
1. If
g(t) is
a
scalar
function
of
t
and
f (t) is a
vector
function
of
d d f d
scalar t then ( f) f
dt dt dt
2. If
and
are
scalar
cons tan ts,
f
and
g
are vector
functions of
t
then
d
dt
( f
g)
d f
dt
d g
dt
3.
d
dt
Proof:
( f . g)
d g
f .
dt
d f
. g
dt
Let f f1
i f2
j f3
k
f . g
g g1
i g2
j g3
k
f1 g1
f2
g2
f3
g3
d dg1
df1
dg2
df2
dg3
df3
( f . g) f1 g1
f2
g2
f3
g3
dt
dt dt dt dt dt dt
dg
1 dg
2 dg
3 df
1 df
2 df
f
3
1 f2
f3
g1
g2
g3
dt dt dt dt dt dt
d g
f .
dt
df
. g
dt
Page 23 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Page 24 of 72
g
dt
f
d
dt
d g
f
g)
( f
dt
d
4.
Proof:
k
f
j
f
i
f
f
Let 3
2
1
k
g
j
g
i
g
g 3
2
1
3
2
1
3
2
1
g
g
g
f
f
f
k
j
i
g
f
)
f
g
g
(f
i
g
f 3
2
3
2
)
f
g
g
(f
i
dt
d
g)
( f
dt
d
3
2
3
2
dt
df
g
f
dt
dg
dt
dg
f
g
dt
d f
i
3
2
3
2
3
2
3
2
dt
df
g
dt
dg
f
i
f
dt
dg
g
dt
d f
i
3
2
3
2
3
2
3
2
dt
d g
f
g
dt
f
d
dt
d h
g ,
,
f
,h
dt
d g
,
f
h
g,
,
dt
f
d
h
g,
,
f
dt
d
5.
Proof:
h
.
g
f
dt
d
h
g,
,
f
dt
d
dt
d h
.
g
f
h
.
g
f
dt
d
dt
d h
.
g
f
h
.
dt
d g
f
h
.
g
dt
f
d
dt
d h
g,
,
f
h
,
dt
d g
,
f
h
g ,
,
dt
f
d

Engineering Mathematics - II - Vector Calculus - 2007
d f
6. If | f | cons tan t then f . 0
dt
Proof:
Let | f |
C
| f | a
2
where a is constant
f .
f
cons tan t
d
dt
f .
f
0
f .
d f d f
. f
dt dt
0
d f
f .
dt
0
7. If
f
has
a
fixed
direction
then
f
d f
0
dt
Proof:
Let
f
f e
Now
f
d f
dt
f e
d
dt
(f e)
df d e
f e e 0
dt dt
f
df
dt
(e e)
df
f (0)
dt
0
8.If
d f
f
dt
0
then
f
has
a
fixed
direction.
Page 25 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Equation of a Space Curve
Let P (x,y,z) be any point in space then its position vector is
OP
r
x i
y j
z k
If x = x(t), y = y(t) and z = z(t) are functions of a scalar variable
t then
r r (t)
describes
a
curve
in
space
called
the
space
curve.
Note 1:
d r
dt
represents
the
velocity
v
(rate
of
change
of
position
with
which
the
ter min al po int
of
r
describes the
curve )
Note 2:
d r
dt
is
along
the
direction
of
the
tan gent
to
the
space
curve
at
P (x, y, z)
Note 3:
The
unit
tan gent
vector
to
the
space
curve
is
denoted by
t
and
is given by
d r
t
dt
d r
| |
dt
Page 26 of 72

Note 4:
d v
dt
Engineering Mathematics - II - Vector Calculus - 2007
d
2
r
is called the acceleration a of
dt
2
the
curve
at time
t.
Unit Normal to a Space Curve
Let A be a fixed point on the curve and s be the length of the arc AP where
P(x,y,z) is a point on the space curve.
Let
t
be
the
unit
tan gent
vector
at
P.
Consider
d.w.r.t s
t . t
1
t
.
d t
ds
d t
. t
ds
0
d t
2 t .
ds
0
t
d t
.
ds
0
d t
ds
is
a normal
to
the
space
curve
and
n
d t
ds
d t
| |
ds
is called
the
unit
normal to
the
curve.
Page 27 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Scalar and Vector Fields
Scalar Valued Function
Let P(x,y,z) which is a scalar is called a scalar point function.
Note: A scalar point function is also called a scalar fixed . = (x,y,z)
Ex.: (x,y,z) = x 2 + y 2 + z 2
Vector Valued Function
Let P(x,y,z) be a point. A vector point function is a vector whose
components are real valued functions of x,y,z. A vector point function is called a
vector field F .
Ex:
f f (x, y, z) f1
i f2
j f3
k
F
xyz i
x
2
y j
yz k
Differential
operator
:
denotes
the
operation
i
x
y
j
k
z
i
x
does
not
represent
a
vector
if
only
defines
the
differential
operator
Gradient of a Scalar Field
Let(x, y, z) be a scalar field, then i k i
x z x
is called the gradient of denoted by .
The
following
are
some
results
obtained
by
the
definition
of
.
1. ( )
,
is a scalar cons tan t.
2. ( )
,
and are scalar fields.
3. ( )
,
and are scalar cons tan ts
4.()
5.
2
Page 28 of 72

Engineering Mathematics - II - Vector Calculus - 2007
6. d dx dy dz
x y z
i
x
j
y
k
z
i
dx
j dy
k dz
d
. d r
Geometrical Interpretation
Q (x, y, z)
c
represents
a
surface
in space
and
represents
a normal
to the surface at any point.
Note 1: Since = c on the surface.
d . d r
0
which
shows
that
is
perpendicular
to
the
tan gent
plane
at
any
po int .
Note
2 : Unit
vector
along
is
denoted by
n
is
| |
called
the
unit
normal
to the surface (x,y,z) = c.
Directional Derivative
Let
a
be
a
vector
inclined at
an
angle
with
then
.
a is
called
the
directional
derivative
along
a .
Page 29 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Note : Maximum value
of directional
derivative
is
n
| |
Divergence of a Vector Field
Let f f1
i f2
j f3
k
be
a
vector
field
then
divergence
of
f
denoted by
div
f
or
.
f
is
defined
as
f
f
1 f2
f
3 i
div. f . f
x y z x
Physical Interpretation
If
v the
velocity
of
a moving
fluid
at
a po int
P
at
time
t,
then
div v
represents
the rate at which the fluid moves out of a unit volume enclosing the point P.
Solenoidal Vector
A
vector
field
f
is
said
to
be
solenoidal
if
div. f
0.
Properties
1. div ( f g) div f div g , where f and g are vector fields.
2. div ( f ) div f where is a scalar cons tan t.
3.
div ( f g)
cons tan ts.
div f
div g
where
and
are
scalar
4.
If
is a scalar field and f is a vector field then div ( f ) div f .
Proof:
Page 30 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Let f f1
i f2
j f3
k
then f f1
i f2
j f3
k
div ( f ) ( f1 ) ( f2
) ( f3)
x y z
f1
x
f1
x
f
2
y
f2
y
f
3
y
f3
y
f1
x
i
x
. f1
i
div
f
.
f
Laplacian of a scalar field
Let
then
= (x,y,z) be a scalar field
is a vector field
and
div
( )
.
x
x
y
y
z
z
2
x
2
2
y
2
z
2
denoted by
2
is
called
the
Laplacian
of
a
scalar
field.
Note 1:
2
2
x
2
2
y
2
2
z
2
Note 2:
A
scalar
field
is
called
a harmonic
function if
2
0
Note 3:
2
()
2
, where
2
and are cons tan ts,
and
are scalar fields.
Curl of a Vector field
Let f f1
i f2
j f3
k
then
curl
of
f
denoted by
curl
f
or
Page 31 of 72

Engineering Mathematics - II - Vector Calculus - 2007
i j k
f is defined as f / x / y
/ z
f1 f2
f3
f
3 f
2
i
y z
Physical Interpretation
If v is the velocity of a particle in rigid body rotating about a fixed
axis
then
the
angular
velocity
at
any
po int
is
given by
1
2
curl
v .
Irrational Vector
f
is
said
to
irrational if
curl
f
0
Properties
(i)
If
f
and
g
are
vector
fields then
curl ( f
g)
curl
f
curl
g
(ii) Curl ( f ) Curl f , where is a scalar cons tan t.
(iii)
If
is
a
scalar
field
and
f
is
vector
field
then
curl ( f )
curl
f
f
Proof:
Curl ( f ) ( f3)
( f2
) i
y z
f
3
y
f
2
i
z
f3
y
f2
i
z
f f
Curl ( f )
3
2
i f3
f2
i ...(1)
y z y z
Consider
f
i j k
/ x / y / z
f1 f2
f3
Page 32 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Page 33 of 72
i
f
z
f
y
2
3
f
f
curl
)
f
(
Curl
(iv)
0
)
(
Curl
Proof:
i
y
z
z
y
)
(
Curl
0
(v). 0
)
f
(Curl
div
then
field
vector
a
is
f
If
Proof:
k
y
f
x
f
i
x
f
z
f
i
z
f
y
f
f
Curl
1
2
3
1
2
3
y
f
x
f
x
x
f
z
f
z
z
f
y
f
x
f )
(curl
div
1
2
3
1
2
3
y
z
f
x
z
f
x
y
f
z
y
f
z
x
f
y
x
f 1
2
2
2
3
2
1
2
2
2
3
2
= 0
Some important Identities
(i) Divergence of a vector product:
then
fields
vector
are
g
and
f
If
g
curl
.
f
f
g .curl
g)
( f
div
Proof:
k
f
j
f
i
f
f
Let 3
2
1
k
g
j
g
i
g
g 3
2
1
k
)
g
f
g
(f
j
)
g
f
g
(f
i
)
f
g
g
(f
g
f 1
2
2
1
3
1
1
3
3
2
3
2
2
3
3
2 g
f
g
f
x
g)
div (f

Engineering Mathematics - II - Vector Calculus - 2007
i . ( f g)
x
i
.
f
x
g
f
g
x
f
i .
x
g
g
i .
x
f
i
f
.
x
g
i
g
.
x
f
( f ) .
g
(
g) .
f
g . Curl
f
f . Curl
g
(ii)
Curl of Curl of a vector:
Curl (Curl f )
grad (div
f )
2
f
Proof:
Let f f1
i f2
j f3
k
f3 f2 f1 f2 f2 f
f i j
1
k
y z z1
x x y
Curl (Curl f )
i
f
2
y x
f1
y
z
f1
z
f
3
x
2
f
2
f
2
2
1
1 f
1 f
3
i
y
x
y
2
z
2 z x
2
f
2
f
2
f
2
3
2
2 f
1
j
z
y
z
2
x
2 x y
2
2
f
2
f f
2
1 3 3 f2
k
x
z
x
2
y
2 y z
adding
and
subtracting
2
f1
,
y
2
2
f3
z
2
to
each
of
the
terms.
i
x
f1
x
f
2
y
f
2
3 f
1
z
x
2
2
f
2
y
2
2
f
1
z
2
Page 34 of 72

Engineering Mathematics - II - Vector Calculus - 2007
i (div f )
2
f
x
Note 1:
(div f )
2
f
The operator f . (f i f j f k) .
1 2 3
i
x
f1
f2
f3
x
y
z
j
y
k
z
is called the Linear differential operator.
Note 2: If is a scalar field
Note 3:
If
f .
f1
f2
f
3
x y
g is a vector field
z
g g
f . g f1
f2
f3
x
y
g
z
(iii)
Gradient of a scalar field:
If f
and g are vector fields
then
( f . g)
f
Curl
g
g
Curl
f
( f . )
g
( g . )
f
Proof:
Consider
( f . g) i ( f . g)
x
f g
i . g f .
x x
f
i g .
x
g
i f .
x
...(1)
Now g Curl
f
g
i
f
x
Page 35 of 72

Engineering Mathematics - II - Vector Calculus - 2007
f
g i
x
f
g . i
x
g . i
f
x
i
f
g .
x
g .
f
g
Curl
f
( g . )
g
i
f
g .
x
...(2)
Similarly
f
curl g
From (1), (2) and (3)
Note:
( f . g)
( f . )
g
f Curl
g
g
i f .
x
g Curl
f
...(3)
( f . )
g ( g . )
f
If
f
g
then
( f . f )
2 f
Curl
f 2( f . )
f
or
1
grad ( f . f )
2
f
Curl
f
( f . )
f
(iv) Curl of a vector product
If f
and
g
are
vector
fields
then
Curl ( f
g)
f div
g g div f ( g . )
f ( f . )
g
Proof:
Curl ( f g)
i
x
( f g)
i
f
x
g
f
g
x
Page 36 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Page 37 of 72
x
g
f
g
x
f
i
x
g
f
i
g
x
f
i
x
g
)
f
.
( i
f
x
g
i
g
x
f
.
i
x
f
g)
.
( i
x
g
f
x
g
.
i
f
x
f
.
i
g
x
f
g 1
1
g
)
.
( f
g
div
f
f
g div
f
)
( g .
g
)
.
f
(
f
)
( g .
f
div
g
g
div
f
Question and answers
01. If
h
g,
,
f
are vectors functions of a scalar variable t. Prove that
dt
d h
x
g
x
f
h
x
dt
d g
x
f
h
x
g
x
dt
f
d
h
x
g
x
f
dt
d
Suggested answer:
h
x
g
dt
d
x
f
h
x
g
x
dt
f
d
h
x
g
x
f
dt
d
dt
d h
x
g
h
x
dt
d g
x
f
h
x
g
x
dt
f
d
dt
d h
x
g
fx
h
x
dt
d g
x
f
h
x
g
x
dt
f
d
02. If
k
t
j
sin t
i
cos t
g
2t k,
j
e
i
e
f
,
e
t
t
2t

d
find i)
dt
Engineering Mathematics - II - Vector Calculus - 2007
d
d
f , ii) f g iii) f x g
dt
dt
Suggested answer:
i) f e
t
i e
3t
j
2te
2t
k
d
dt
f e
t
i 3e
3t
j
2
2te
2t
e
2t
k
ii)
d
dt
f g (e
t
sin t) i ( e
t
cos t) j
(
2 1) k
iii)
f x g
i j
e
t
e
t
cos t sin t
k
2t
t
i( te
t
2t sin t)
j(te
t
e
t
cos t) k(e
t
sin t e
t
)
d
( f x g) i te
t
e
t
2t cos t 2 sin t
dt
j e
t
te
t
e
t cos t e
t
sin t
k e
t
cos t e
t
sin t e
t cos t e
t
sin t .
03. A particle moves along the curve whose parametric coordinates are x = e -t ,
y = 2 cos 3t, z = 2 sin 3t. Here t is time i) Determine its velocity and
acceleration at time t ii) find the magnitude of velocity and acceleration at t =
0.
Suggested answer:
Page 38 of 72

Engineering Mathematics - II - Vector Calculus - 2007
i) Equation to the curve is
r e
t
i
2 cos 3t j
2 sin3t k
v
d r
dt
e
t
i 6 sin3t j
6 cos 3t k
a
d
2
r
dt
2
e
t
i 18 cos 3t j
18 sin3t k
ii) At t = 0,
v i 6 k,
a i 18 j
and
| v | 37, | a | 325
04. Find the unit tangent vectors at any point on the curve x = t 2 +1, y = 4t - 3,
z = 2t 2 - 6t. Determine the unit tangent at the point t = 2.
Suggested answer:
Equation to the space curve is
r (t
2
1) i
(4t 3) j
(2t
2
6t) k
d r
dt
2t i 4 j
(4t 6) k
d r
dt
t2
4 i 4 j
2 k
unit tangent
t
d r
dt
d r
dt
Page 39 of 72

Engineering Mathematics - II - Vector Calculus - 2007
4 i 4 j
2k
36
2
3
i
2
3
1
j
k
3
05. A particle moves so that its position vector is given by
r cos t
i
sin t
j,
is
a
cons tan t
show
that
i) velocity v
is perpendicular to
r
ii) acceleration
to the distance from the origin.
a is directed towards the origin and has magnitude proportional
iii) r x v is a cons tan t vector.
Suggested answer:
i)
r cos t
i sin t
j
d r
dt
sin t
i cos t
j v
Now
v . r 0
v r
ii)
a
d
2
r
dt
2
2
cos t
i sin t
j
a
2
r
| a | 2
| r |
iii)
r x v
Page 40 of 72

Engineering Mathematics - II - Vector Calculus - 2007
(cos t
i sin t
j)x( sin t
i cos t
j)
2 cos t( i x j) sin
2
t( j x i)
cos
2
t k
sin
2
t
k
k
= constant vector.
d f
06. Two vector f and g such that x f ,
dt
d
for some , prove that ( f x g) x( f x g)
.
dt
d g
dt
x g
Suggested answer:
d
dt
d g d f
( f x g) f x x g
dt dt
f x( x g) ( x f )x g
( f . g) ( f . )
g ( g . )
f ( g . f )
( g . )
f ( f . )
g
x( f x g)
07. If x = t 2 +1, y = 4t - 3, z = 2t 2 - 6t represents the parametric equations of a
space curve. Find the angle between the unit tangents at t = 1 and t = 2.
Page 41 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Suggested answer:
r (t
2
1) i (4t 3) j
(2t
2
6t) k
d r
dt
2t i 4 j
(4t 6) k
t
2t i 4 j
(4t 6)k
4t
2
16 16t
2
36 48t
2t i 4 j
(4t 6)k
20t
2
48t 52
t t1
2 i 4 j
2 k
24
a
(say)
t t2
4 i 4 j
2 k
36
2 2 2
i j
k b (say)
3 3 3
Let be the angle between
a
and b
a . b
cos 1
|
a || b |
5
cos 1
3
6
08. If
f cosnt
sinnt
where n is a constant scalar and
and
are
constant vectors prove that
Page 42 of 72

Engineering Mathematics - II - Vector Calculus - 2007
d f
i) f x n( a x b)
dt
ii)
d
2
f
dt
2
n
2
f
iii)
f ,
d f d
2
f
, 0
dt dt
2
Suggested answer:
Given
f cos nt a sinnt b
d f
i) f x
dt
d f
n sinnt a n cos nt b
dt
(cos nt a sinnt b )x( n sinnt a n cos nt b )
n cos
2
nt( a x b ) n sin
2
t(b x a)
n( a x b )
d
2
f
ii)
dt
2
n
2
cos nt a n
2
sinnt b
f
iii) f ,
d f
,
dt
n 2 0
d
2
f
d f
f , , n
2
f
dt
2
dt
09. A particle moves along the curve x = t 3 + 1, y = t 2 , z = t + 5 where t is the
time. Find the components of velocity and acceleration along the vector
Page 43 of 72

Engineering Mathematics - II - Vector Calculus - 2007
c i 3 j
2 k at t = 1.
Suggested answer:
r (t
3
1) i t
2
j
(t 5) k
d r
3t
2
i 2t j
k
dt
d
2
r
dt
2
6t i 2 j
d r
dt
t1
v 3 i 2 j
k
Now velocity along
c is
v c
11
14
12
Acceleration along c is a . c .
14
d d A d
2
A d A d
3
A
10. Show that A, , A . x .
dt
dt dt
2
dt dt
3
Suggested answer:
d
dt
A,
d A
,
dt
d
2
A
2
dt
d A
,
dt
d A
,
dt
d
2
A
A,
dt
2
d
2
A
,
dt
2
d
2
A
A,
dt
2
d A
,
dt
d
3
A
3
dt
Page 44 of 72

Engineering Mathematics - II - Vector Calculus - 2007
d A d
3
A
0 0 A,
,
dt 3
dt
11. If R A e
nt
B e
nt
, where A and B are cons tan t vectors
d
2
R
then show that n
2
R .
dt
2
Suggested answer:
d R
dt
ne
nt
A ne
nt
B
d
2
R
dt
2
n
2
e
nt
A n
2
e
nt
B
n 2 R
12. Find unit vector normal to the curve 4 sin t i 4 cos t j
3t k .
Suggested answer:
We have
r 4 sin t i 4 cos t j
3t k
d r
dt
4 cos t i 4 sin t j
3k
d r
dt
5
Page 45 of 72

t
d r
dt
d r
dt
Engineering Mathematics - II - Vector Calculus - 2007
4 cos t i 4 sin t j
3k
5
d t
ds
d t dt
dt ds
d t/ dt
ds / dt
also
ds d r
dt dt
5
d t
ds
1
25
( 4 sin t i 4 cos t j)
and
N
4
(sin t
25
i cos t j)
16 /(25)
2
(sin t i cos j)
13. Find the unit normal vector to the curve x = 3 cost, y = 3sint, x = 4t.
Suggested answer:
We have
r 3 cos t i 3 sin t j
4t k
d r
v
dt
| v | 5
3 sin t i 3 cos t j
4 k
(tan gent vector)
Page 46 of 72

Engineering Mathematics - II - Vector Calculus - 2007
t
d r / dt
ds / dt
d t/ dt
| d r / dt |
1
( 3 cos t i 3 sin t j)
25
unit normal vector
d t/ ds
N
| d t/ ds |
3
(cos t i sin t j)
25
3 / 25
(cos t i sin t j)
14. The position vectors of a moving particle at time t is r t
2
i t
3
j
t
4
k .
Find the tangential and normal components of its acceleration at t = 1.
Suggested answer:
r t
2
i
t
3
j
t
4
k
v
d r
dt
2t i 3t
2
j
4t
4
k
v t1
2 i 3 j
4 k
v
1
(2 i 3 j
4 k)
29
a t1
2 i 6 j
12 k
Page 47 of 72

Engineering Mathematics - II - Vector Calculus - 2007
tangential component acceleration is
a .v
70
29
Normal component of acceleration |
a ( a .v) v |
82 i 36 j
68 k
29
1
29
82
2
36
2
68
2
= 3.8774
15. If is a function of r only, then prove that
1 d
d
r r .
r dr dr
Suggested answer:
We have
r
2
r . r x
2
y
2
z
2
r
x
x
r
,
r
y
y
r
,
r
z
z
r
i
x
j
y
k
z
d
i
dr
x
r
d
j
dr
y
r
d
k
dr
z
r
1 d
(x i y j
z k)
r dr
Page 48 of 72

Engineering Mathematics - II - Vector Calculus - 2007
1 d
d
r r d
r r
r dr dr r dr
16. Find
in the following
i)
ii)
x
2
y
3
z
4
at (1, 0, -2)
1
r
2
Suggested answer:
i)
2xy
3
z
4
i 3x
2
y
2
z
4
j
4x
2
y
3
z
3
k
( )
(1,0, 2)
2 i 3 j
4 k
ii)
1
r
2
we have
1 d
r
r dr
1 2
r
r r
3
2 r
r
4
17. Find the equation of the tangent plane to the surface xyz = 6 at the point (1,
2, 3).
Suggested answer:
xyz
yz i xz j
xy k
(1,2,3)
6 i 3 j
2 k
Page 49 of 72

Engineering Mathematics - II - Vector Calculus - 2007
unit normal to the surface
n
|
|
6 i 3 j
2 k
7
Equation of the tangent plane in vector form is ( r a).n 0
Where
r x i y j
z k,
a i 2 j
3k
ie
6(x 1) 3(y 2) 2(z 3) 0
i.e., 6x + 3y + 2z = 18
18. Find the angle between the surfaces xlogz = y 2 - 1 and x 2 y = 2 - z at the
point (1, 1, 1).
Suggested answer:
Let
2
1 x log z y
2 x 2 y z
1
log z i 2y j
x
z
k
2
2xy i x
2
j
k
( 1)
(1,1,1 )
2
j
k
( 2)
(1, 1, 1)
2 i j
k
The angle between the surfaces is the angle between 1
and 2.
Page 50 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Let be the angle between 1 and
2
.
cos
1 1
2
| 1
|| 2
|
cos 1 1
30
cos 1
1
30
19. Find the constants a and b such that the surfaces ax 2 - byz = (a + 2)x and
4x 2 y+z 3 are orthogonal at (1, -1, 2).
Suggested answer:
Let ax
2
1 byz (a 2)x
...(1)
4x
2
y z
3
2 4
...(2)
1
[2ax (a 2)] i bz j
by k
2
8xy i 4x
2
j
3z
2
k
( 1)
(1, 1,2)
(a 2) i 2b j
b k, ( 2)
(1, 1,2)
8
i 4 j
12 k
given 1 and 2
are orthogonal
1. 2
0
i.e., -8(a - 2) + 4(-2b) + 12b = 0
2a
b 4
...(3)
Page 51 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Also 1 2
at (1, -1, 2)
a 2b (a 2) 4
8 4
i.e., 2b - 2 = 0
b 1
substituting in (3), we get
a = 5/2.
20. Find the directional derivative of xyz
along the direction
of the normal to the surface x
2
z y
2
x yz
2
3, at( 1,1,1).
Suggested answer:
Let
xyz,
x
2
z y
2
x yz
2
yz i
xz j
xy k
at
(1, 1, 1) i j
k
(2xz y
2
) i (2yx z
2
) j
(x
2
2zy) k
at
(1, 1, 1) 3 i 3 j
3k
n
|
|
i j
k
3
Directional directive of along
n is
Page 52 of 72

Engineering Mathematics - II - Vector Calculus - 2007
1(1) 1(1) (1)
. n
3
3
21. If F (x
3
y
3
z
3
3xyz) find div F and curl F .
Suggested answer:
F (3x
2
3yz) i
(3y
2
3xz) j
(3z
2
3xy) k
div F 6x 6y 6z
i j k
Curl F
x y z
3x
2
3yz 3y
2
3xz 3z
2
3xy
i(
3x
3x) j( 3y
3y) k( 3z
3z)
0
22. If
is a constant vector and
v x r
then show that
div v 0.
Suggested answer:
Let
r x i y j
z k
v x r
Page 53 of 72

Engineering Mathematics - II - Vector Calculus - 2007
i j k
1 2
3
x y z
v i[ 2z
3y]
j[ 1z
2x]
k[ 1y
2x]
div v ( 2z
3y)
( 1y
g3x)
( 1y
2x)
x
y
z
= 0
23. If f (x 2y z) i (x y z) j
( x
y 2z)k find div f .
Suggested answer:
div f
(x 2y z) (x y z) ( x
y 2z)
x
y
z
1 1
2
2
24. If
f x
2
i
y
2
j
z k and g yz i zx j
xy k
show that
f x g
is a solenoidal vector.
Suggested answer:
f x g
i
x
2
yz
j
y
2
zx
k
z
2
xy
f x g i{xy
3
xz
3
} j{x
3
y yz
3
} k{zx
3
zy
3
}
Page 54 of 72

div
Engineering Mathematics - II - Vector Calculus - 2007
( f x g) (y
3
z
3
) (z
3
x
3
) (x
3
y
3
)
0
f x g is solenoidal.
25. Prove that
div (r
n
r ) (n 3)r
n
.
Suggested answer:
div (r
n
r ) r
n
. r r
n
div r
nr
n1
x
r
i nr
n1
y
r
z
nr
n 1
k.
r r
n
r
x
x
y
y
z
nr
n1
{x i y j
z k}. r r
n
.(3)
r
nr
n2
.r
2
r
n
(3)
( n 3)r
n
26. If
a is a constant vector then prove that
i) div( a x r ) 0
ii) div ( r x( r x a)) 2( r . a)
iii) div{r
n
( a x r )} 0, n being
a
cons tan t.
Suggested answer:
Let a a1
i a2
j
a3
k
Page 55 of 72

Engineering Mathematics - II - Vector Calculus - 2007
r x i y j
z k
i j k
i)
a x r a1
a2
a3
x y z
( a2z
a3y)
i (a3x
a1z)
j
(a1y
a2x)
k
div( a x r ) (a2z
a3y)
(a3x
a1z)
(a1y
a2x)
0
x
y
z
ii) r x( r x a) ( r . a) r ( r . r ) a
div( r x( r x a)) ( r . a). r ( r . a)div r div(r
2
a)
a . r ( r . a)3 a . r
2
r
2
div a
a . r 3( a . r ) a{2r
x
r
i
y z
2r j
2r k}
r r
a . r 3( a . r ) 2( a . r )
2( a . r )
iii) div{r
n
( a x r )} r
n
.( a x r ) r
n
div( a x r )
r
nr
n1
.( a x r ) r
n
div( a x r )
r
Page 56 of 72

Engineering Mathematics - II - Vector Calculus - 2007
0 r
n
div( a x r )
i j k
a x r a1
a2
a3
x y z
i(a2z a3y) j(a1z a3x) k(a1y a2x)
div( a x r ) (a2z
a3y)
(a3x
a1z)
(a1y
a2x)
x
y
z
= 0 + 0 + 0
= 0
div(r
n
( a x r )) 0
27. If
a and
b
are constant vectors, prove that
i) div{( a x r ) b} a . b
ii) div { a x( r x b )} 2( a . b )
iii) div {( a x r )x(b x r )} [a,
b , r ]
Suggested answer:
Let a a1
i a2
j
a3
k
b b1
i b2
j
b3
k
Page 57 of 72

Engineering Mathematics - II - Vector Calculus - 2007
c x i y j
z k
i)
div[( a . r ) b ] ( a . r ). b a . r div b
a . b ( a . r )(0)
a . b
ii) we have a x( r x b ) ( a . b ) r ( a . r ) b
div{ a x( r x b )} div{( a . b ) r } div{( a . r ) b}
( a . b )div r ( a x r ). b ( a . r )div b
( a . b )3 a . b ( a r )0
2( a . b )
iii)Let
a x r p
( a x r )x(b x r ) p x(b x r )
(p . r ) b (p . b ) r
{( a x r ). r } b {( a x r ). b} r
2( a . b )
Page 58 of 72

Engineering Mathematics - II - Vector Calculus - 2007
iii)Let a x r p
( a x r )x(b x r ) p x(b x r )
(p . r ) b (p . b ) r
{( a x r ). r } b {( a x r ). b} r
0 [ a, r , b ] r
( a x r )x(b x r ) [a,
b ,
r ] r
div{( a x r )x(b x r )} [a,
b , r ]. r [a, b , r ]div r
r . [a,
b ,
r ] [a, b , r ](3)
...(1)
a1
a2
a3
[ a, b , r ] b1
b2
b3
x y z
a1(b
2z
b3y)
a2(b1
z b3x)
a3(b1
y b2x)
[ a, b , r ] (a2b3
a3b2
) i ( a1b
3 a3b1
) j
(a1b
2 a2b1
) k
a x b
Now r . [a,
b , r ] r .( a x b )
( a2b3
a3b2
)x (a3b1
a1b
3)y
(a1b
2 a2b1)z
Page 59 of 72

Engineering Mathematics - II - Vector Calculus - 2007
[a, b , r ]
In (1)
div{( a x r )x(b x r )} [a, b , r ] 3[a, b , r ]
[a, b , r ]
28.
If a
and
b
are vectors, prove that
1
1
a . {b . } {3( a, r )(b . r ) r
2
( a . b )}.
r r
5
Suggested answer:
1
1
r
b .
b .
r
r
2
r
1
(b . r )
r
3
1
b .
r
1 1
(b . r ) (b . r )
3
3
r
r
1 3 r
[b ] (b . r ).
r
3
r
4
r
1
b .
r
b 3(b . r )
r
r
3
r
5
1
b 3(b . r )
a . b .
a . . a . r
r
r
3 r
5
Page 60 of 72

Engineering Mathematics - II - Vector Calculus - 2007
1
r
5
3( a . r )(b . r ) ( a . b )r
2
29. If F (x y 1) i j
(x y)k show that F .Curl F 0.
Suggested answer:
Curl F
i
x
x y 1
j
y
1
k
z
(x y)
i(
1)
j( 1)
k( 1)
Curl F i j
k
F .Curl F (x y 1)( 1)
1(1) (x y)( 1)
= -x - y - 1 + 1 + x + y
= 0
30. If F x
2
y i 2xz j
2yz k find x(
x
F ).
Suggested answer:
x
F
i
x
x
2
y
j
y
2xz
k
z
2yz
i(2z 2x) j(0) k( 2z
x
2
)
Page 61 of 72

Engineering Mathematics - II - Vector Calculus - 2007
i j k
x(
x
F )
x
y
z
2z 2x 0 2z x
2
(2x 2) j
31. Find the directional derivative of
x
2
yz 4xz
2
at (1, -2, -1)
in the direction 2 i j
2 k .
Suggested answer:
(2xyz 4z
2
) i x
2
z j
(x
2
y 8xz) k
at
(1,
2,
1)is 8 i j
10 k
Let
c 2 i j
2 k
Directional derivative in the direction of
c
is
(2 i j
2 k)
. c (8 i j
10 k).
3
1
(16
3
1 20)
37
3
32. Find the directional derivative of 4xz
3 3x
2
y
2
z at (2, -1, 2)
in the direction of 2 i 3 j
6 k .
Page 62 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Suggested answer:
(4x
3
6xy
2
z) i ( 6x
2
yz) j
(12xz
2
3x
2
y
2
) k
at
(2, 1,2)
8 i 48 j
84 k
c 2 i 3 j
6 k
1
. c [8(2) 48( 3)
(84)(6)]
7
376
7
33. Find the directional derivative of
x
2
y
2
z
2
at (1, 1, -1)
in the direction of the tangent to the curve x = e t , y = 1 + 2sint,
z = t - cost, 1 t 1.
Suggested answer:
x
2
y
2
z
2
2xy
2
z
2
i 2x
2
yz
2
j
2x
2
y
2
z k
at
(1, 1,
1) 2 i 2 j
2 k
r e
t
i (1 2 sin t) j
(t cos t) k
d r
dt
at t
0
is
i 2 j
k
Page 63 of 72

t
d r
dt
d r
dt
Engineering Mathematics - II - Vector Calculus - 2007
i 2 j
k
6
directional derivative of in the direction of
c
is
( i 2 j
k)
(2 i 2 j
2 k).
6
1
[2 4 2]
6
4
6
34. Find the equation of the tangent plane and the normal line to the surface z =
x 2 + y 2 at (2, -1, 5).
Suggested answer:
x
2
y
2
z
2x i 2y j
k
Equation to the normal line at (2, -1, 5) is
x 2
4
y 1
2
z 5
1
Equation the tangent plane is 4(x - 2) + (-1)(y + 2)(-1)(z - 5) = 0
i.e., 4x - 2y - z - 5 = 0
Page 64 of 72

Engineering Mathematics - II - Vector Calculus - 2007
35. Find the angle between the surface x 2 + y 2 + z 2 = 9 and z = x 2 + y 2 -3 at
the point (2, -1, 2).
Suggested answer:
Let
x
2
y
2
z
2
1 9
x
2
y
2
2 z 3
1
2x i 2y j
2z k
1
at
(2, 1,2)
4 i 2 j
4 k
2
2x i 2y j
k
2
at (2, 1,2) 4 i 2 j
k
Angle between the surfaces is the angle between 1 and 2.
.
cos
1 1 2
|
1
|| 2
|
4(4) ( 2)( 2) 4( 1)
cos 1
36 21
8
cos 1
3
21
36. Find the values of a and b such that the surfaces ax 2 - byz = (a + 2)x and
4x 2 y + z 3 = 4 are orthogonal at (1, -1, 2).
Suggested answer:
Let : ax
2
1 byz (a 2)x 0
...(1)
Page 65 of 72

Engineering Mathematics - II - Vector Calculus - 2007
: 4x
2
y z
3
2 4 0
...(2)
Given (1, -1, 2) is a point on (1) and (2)
1 2
at (1, 1,2)
i.e., a + 2b - (a + 2) -4 + 8 - 4
2b - 2 = 0
b 1
1
(2ax (a 2)) i
bz j
by k
1 at (1, 1,2) (a 2) i 2 j
by k
2
8xy i 4x
2
j
3z
2
k
2
at (1, 1,2) 8
i 4 j
12 k
since 1 and 2
are orthogonal
1. 2
0
-8(a - 2) - 8 + 12 = 0
i.e., a = 5/2
a 5 / 2
and b 1
37. Find the constant a, b, c so that the vector
f (x 2y az) i (bx 3y z) j
(4x cy 2z)k
is irrotational.
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Engineering Mathematics - II - Vector Calculus - 2007
Suggested answer:
F is irrotational if
Curl F 0
i j k
CurlF
0
x y z
x 2y az bx 3y z 4x cy 2z
c 1 0, 4 a 0, b 2 0
a 4,b 2, c 1
38. If and
are scalar fields prove that
i) div{ }
2
.
ii)
2
( )
2
2
2.
Suggested answer:
i) div( )
.
div( )
.
2
...(1)
ii) similarly
div( ) .
2
...(2)
adding (1) and (2)
div( )
div( )
2
2
2.
...(3)
Also div(
) div( ) div( ( ))
2
( )
2
( )
2
2
2.
39. Evaluate
i)
2 r
2
Page 67 of 72

ii)
Engineering Mathematics - II - Vector Calculus - 2007
2 r
div
r
2
Suggested answer:
i)
2
r
2
2
2
2
(r
2
) (r
2
) (r
2
)
x
2
y
2
z
2
x y
2r
2r
x
r y
r
2r
z
z
r
= 2 + 2 + 2
= 6
ii)
r 1 1
div
. r
r
2
r
2
r
2
div r
2 r 1
.
r (3)
r
3 r r
2
2 3
r
2
r
2
1
r
2
40.
3 Find the value of a if f(axy x) i(a 2)x j(1
2 a )xz
2
k
is irrotational.
Suggested answer:
Since
f is irrotational
Page 68 of 72

Engineering Mathematics - II - Vector Calculus - 2007
Curl f 0
i
i.e.,
x
axy z
3
j
y
(a 2)x
2
k
z
(1 a)xz
2
0
(1
a)z
2
3z
2
k (a
2)2x ax
i{0 0} j
0
z 2
1
a 3 0 and x (a
2)2 a 0
a 4
41. Show that
r
r
3
is both solenoidal and irrotational.
Suggested answer:
r 1
div
r
3
r
3
1
div r
. r
r
3
1
(3)
r
3
3 r
.
r
r
4 r
3 3
r
3
r
3
= 0
r 1 1
Curl
Curl r
x r
r
3 r
3
r
3
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Engineering Mathematics - II - Vector Calculus - 2007
1 3 r
(0) x
r
r
3 r
4 r
0 0
0
42. If
a is a constant vector prove the following
i)
Curl( a . r ) a 0
ii)Curl{ r x( a x r )} 3( r x a)
iii)Curl{r
n
( a x r )} (n 2)r
n
a nr
n2
( r . a) r
Suggested answer:
i)Curl{( a . r ) a} ( a . r )Curl a ( a . r )x a
( a . r ) 0 a x a ( a . r ) a
0 0
0
ii) Curl{ r x( a x r )} Curl{( r . r ) a ( r . a) r }
Curl{( r . r ) a} Curl{( r . a) r }
( r . r )Curl a ( r . r )x a ( r . a)x r
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Engineering Mathematics - II - Vector Calculus - 2007
r
0 2r x a ( r . a) 0 a x r
r
2( r x a) r x a
3( r x a)
iii) Curl{r
n
( a x r )} r
n
Curl( a x r ) r
n
x( a x r )
r
r
n
Curl( a x r ) nr
n1
x( a x r )
r
r
n
Curl( a x r ) nr
n2
{( r . r ) a ( r . a) r }
r
n
Curl( a x r ) nr
n
a nr
n2
( r . a) r
...(1)
i j k
Consider ( a x r ) a1 a2
a3
i(a2z
a3y)
j(a3x
a1z)
k(a1y
a2x)
x y z
Curl( a x r )
i
x
a2z
a3y
j
y
a3x
a1z
k
z
a1y
a2x
2 a
in (1)
Curl{r
n
( a x r )} 2r
n
a nr
n
a nr
n2
( r . a) r
(n 2)r
n
a nr
n2
( r . a) r
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Engineering Mathematics - II - Vector Calculus - 2007
Page 72 of 72