Vector Calculas

www.rejin

paul.com

www.rejin

paul.com

UNIT-II

VECTOR CALCULUS

Directional derivative

The derivative of a point function (scalar or vector) in a particular

direction is called its directional derivative along the direction.

The directional derivative of a scalar point function in a given direction is the rate of change of in the direction. It is given by the component of grad in that direction. The directional derivative of a scalar point function

(x,y,z) in the direction of a is given by

a

a..

Directional derivative of is maximum in the direction of . Hence the maximum directional derivative is grador Unit normal vector to the surface

If (x, y, z) be a scalar function, then (x, y, z) = c represents A surface and the unit normal vector to the surface is given by

Equation of the tangent plane and normal to the surface

Suppose a is the position vector of the point ),,( 000 zyx

On the surface (x, y, z) = c. If

++= kzjyixr is the position vector of

any point (x,y,z) on the tangent plane to the surface ata , then the

equation of the tangent plane to the surface at a given point a on it is

given by 0. =

gradar

If r is the position vector of any point on the normal to the surface

at the point a on it. The vector equation of the normal at a given point

a on the surface is 0=

gradar

The Cartesian form of the normal at ),,( 000 zyx on the surface

(x,y,z) = c is

z

zz

y

yy

x

xx o

=

=

00

Divergence of a vector

If ),,( zyxF

is a continuously differentiable vector point function in

a given region of space, then the divergences of F is defined by

z

Fk

y

Fj

x

FiFdivF

+

+

==

.

www.rejinpaul.comwww.rejinpaul.com

www.rejin

paul.com

www.rejin

paul.com

=x

Fi

If

++= kFjFiFF 321 ,then ).( 321

++= kFjFiFFdiv

i.e., z

F

y

F

x

FFdiv

+

+

=

321

Solenoidal Vector

A vector F is said to be solenoidal if 0=

Fdiv (ie) 0. =

F

Curl of vector function

If ),,( zyxF

is a differentiable vector point function defined at each

point (x, y, z), then the curl of F is defined by

= FFcurl

= z

Fk

y

Fj

x

Fi

+

+

= x

Fi

If

++= kFjFiFF 321 ,then )( 321

++= kFjFiFFcurl

321 FFF

zyx

kji

Fcurl

=

=

+

y

F

x

Fk

z

F

x

Fj

z

F

y

Fi 121323

CurlF is also said to be rotation

F

Irrotational Vector

A vector F is called irrotational if Curl 0=

F

(ie) if 0=F

Scalar Potential

If F is an irrotational vector, then there exists a scalar function

Such that =F . Such a scalar function is called scalar potential of

F

Properties of Gradient

1. If f and g are two scalar point function that ( ) gfgf = (or) ( ) gradggradfgfgrad =

Solution: ( ) ( )gfz

ky

jx

igf

+

+

=


www.rejin

paul.com

www.rejin

paul.com

= ( ) ( ) ( )

+

+

gfz

kgfy

jgfx

i

= z

gk

z

fk

y

gj

y

fj

x

gi

x

fi

+

+

=

+

+

+

+

z

gk

y

gj

x

gi

z

fk

y

fj

x

fi

= gf

2. If f and g are two scalar point functions then ( ) fggffg += (or) ggradffgradgfggrad +=)(

Solution: ( ) = fg ( )fgz

ky

jx

i

+

+

= ( ) ( ) ( )

+

+

fgz

kfgy

jfgx

i

=

+

+

+

+

+

z

fg

z

gfk

y

fg

y

gfj

x

fg

x

gfi

=

+

+

+

+

+

z

fk

y

fj

x

fig

z

gk

y

gj

x

gif

= fggf +

3. If f and g are two scalar point function then 2g

gffg

g

f =

where

0g

Solution: =

g

f

+

+

g

f

zk

yj

xi

=

g

f

xi

=

2g

x

gf

x

fg

i

=

x

gif

x

fig

g 21

= [ ]gffgg

2

1

4. If

++= kzjyixr such that rr =

,prove that

= rnrr nn 2

Solution: nn r

zk

yj

xir

+

+

=

=

+

+

z

rk

y

rj

x

ri

nnn


www.rejin

paul.com

www.rejin

paul.com

= z

rnrk

y

rnrj

x

rnri nnn

+

+

111

=

++

r

zk

r

yj

r

xinr n 1

=

++

kzjyixr

nr n 1

=

rr

nr n 1

5. Find a unit normal to the surface 422 =+ xzyx at (2,-2, 3) Solution: Given that xzyx 22 +=

)2( 2 xzyxz

ky

jx

i +

+

+

=

= ( ) ( ) ( )xkxjzxyi 222 2 +++ At (2,-2, 3)

( ) )4()4(68

+++= kji

=

++ kji 442

63616164 ==++= Unit normal to the given surface at (2,-2,3)

=6

442

++ kji

=

++kji 22

3

1

6. Find the directional derivative of xyzxzyzx ++= 22 4 at (1,2,3) in the

direction of

+ kji2 Solution: Given xyzxzyzx ++= 22 4

)4( 22 xyzxzyzxz

ky

jx

i ++

+

+

=

= ( ) ( ) ( ) +++++++ kxyxzyxjxzzxiyzzxyz 842 222 At (1, 2, 3)

++= kji 28654

Given:

+= kjia 2

6114 =++=a


www.rejin

paul.com

www.rejin

paul.com

=a

aDD ..

= 6

22.28654

+

++ kjikji

= [ ] [ ]866

1286108

6

1 =+

7. Find the angle between the surface 5222 =++ zyx and 52222 =++ xzyx at (0,1,2)

Solution: Let 222

1 zyx ++= and xzyx 2222

2 ++=

zz

yy

xx

2,2,2 111 ==

=

zz

yy

xx

2,2,22 222 ==

=

++= kzjyix 2221

++= kzjyix 22)22(2 At (o,1,2)

+= kj 421

++= kji 4222

Cos644416

422.42.

21

21

+++

++

+=

=

kjikj

2420

20

2420

164cos =+=

= 2420

20cos 1

=

24

20cos 1

8. Find the angle between the surfaces 1log 2 = yzx and zyx = 22 at the point (1,1,1)

Solution: let zxy log21 = and zyx +=2

2

z

x

zy

yz

x=

=

=

111 ,2,log

1,,2 2222 ==

=

zx

yxy

x

+= kz

kjyiz 2)log(1

= kj22


www.rejin

paul.com

www.rejin

paul.com

Cos65

1

11414

12.

21

21 =+++

==

= 65

1cos 1

9. Find ( )nr2 Solution: ( )nr2 = ( )nr. = ( ) ( ) ( )nnn r

zkr

yjr

xi

+

+

= z

rnrk

y

rnrj

x

rnri nnn

+

+

111

++= kzjyixr

222 zyxrr ++==

2222 zyxr ++=

r

x

x

rx

x

rr =

=

22

r

y

y

ry

y

rr =

=

22

r

z

z

rz

z

rr =

=

22

( )= nr2

++

r

zk

r

yj

r

xinr n 1

=

++

kzjyixnr n 2

=

rnr

n 2

Since

+=

udivuu .

( )

=

rnrr nn 22

= ( ) +

rnrrnr nn .. 22

++

+

+

=

kzjyix

zk

yj

xir.

=1+1+1 = 3

( ) ( ) += rrnnrr nnn .3 222 = ( )( ) 242 .23 rrnnnr nn + = ( )( )22 23 + nn rnnnr

( ) [ ] ( ) 2222 1 +=+= nnn rnnnnrr


www.rejin

paul.com

www.rejin

paul.com

10. If

++= kzjyixr and rr =

.Prove that rr n is solenoidal if 3=n and

rr n is irrotational for all vectors of n.

Solution: rr n

++= krjyrixr nnn

div ( ) ( ) ( )zrz

yry

xrx

rr nnnn

+

+

=

(1)

Now 2222 zyxr ++=

Differentiating partially w.r.to x,

r

x

x

rx

x

rr =

=

22

Similarly, r

y

y

ry

y

rr =

=

22

r

z

z

rz

z

rr =

=

22

Now ( ) ( ) nnn rx

rr

rxxr

x+

=

.

= x.nnn r

r

xr +1

( ) nnn rynryry

+= 22

( ) nnn rznrzrz

+= 22

From (1) we have

( ) nnn rzyxnrrrdiv 32222 +++=

= nn rnr 3+

= ( ) nrn 3+ The vector

rr n is solenoidal if

rrdiv n = 0

( ) 03 =+ nrn 03 =+ n 3= n

rr n is solenoidal only if n = -3

Now

zryrxr

zyx

kji

rrcurl

nnn

n

=

= ( ) ( )

yrz

zry

i nn

=

z

rynrz

y

rnri nn 11


www.rejin

paul.com

www.rejin

paul.com

=

r

zynrz

r

ynri nn 11

= ( )

yznryznri nn 22 = 0

Curl (rr n ) =

++ kji 000 =0

Curl (rr n ) = 0 for all values of n

Hence rr n is irrotational for all values of n.

11. Prove that ( ) ( ) +++= kxzjxyizxyF 232 34sin2cos is irrotational and find its scalar potential

Solution:

232 34sin2cos xzxyzxy

zyx

kji

Fcurl

+

=

= [ ] [ ] [ ] 0cos2cos23300 22 =+ xyxykzzji

F is irrotational.

To Find such that gradF =

( ) ( )z

ky

jx

ikxzjxyizxy+

+

=+++

222 34sin2cos

Integrating the equation partially w.r.to x,y,z respectively

),(sin 132 zyfxzxy ++=

),(4sin 22 zxfyxy +=

),(33 yxfxz +=

,4sin 32 Cyxzxy ++= is scalar potential

12. Prove that ).().(

=

BcurlAAcurlBBAdiv

Proof : ).(

=

BABAdiv

=

BAx

i

=

+

B

x

Ai

x

BAi

=

+

B

x

AiA

x

Bi


www.rejin

paul.com

www.rejin

paul.com

=

+

B

x

AiA

x

Bi ..

=

+

BAcurlABcurl ..

13.Prove that

=

FFFcurlcurl 2

Solution:

=

FFcurlcurl

By using

=

cbabcacba ..

= ( )

FF ..

=

FF 2.

VECTOR INTEGRATION

Line, surface and Volume Integrals

Problems based on line Integral

Example 1:

If ( ) ++= kxzjyziyxF 22 201463 Evaluate

C

drF . from (0,0,0) to

(1,1,1) along the curve 32 ,, tztytx ===

Solution: The end points are (0, 0, 0) and (1, 1, 1)

These points correspond to t = 0 and t = 1

23,2, tdztdydtdx ===

C

drF . = ( ) ++C

dzxzyzdydxyx 22 201463

= ( ) ( ) ( ) ++1

0

27522 32021463 dttttdttdttt

= ( ) +1

0

962 60289 dtttt

= ( )101073 643 ttt + = ( )[ ] 50643 =+

Example 2:

Show that

++= kzjyixF 222 is a conservative vector field.


www.rejin

paul.com

www.rejin

paul.com

Solution: If F is conservative then 0=

F

Now 0000

222

=++=

=

kji

zyx

zyx

kji

F

F is a conservative vector field.

Surface Integrals

Definition: Consider a surface S. Let n denote the unit outward normal to the

surface S. Let R be the projection of the surface x on the XY plane. Letf be

a vector valued defined in some region containing the surface S. Then the

surface integral of f is defined to be dydx

kn

nfdsnf

RS

.

.

..

=

Example 1;

Evaluate dsnfS

. where

+= kzyjxizF 2 and S is the surface of

the cylinder 122 =+ yx included in the first octant between the planes z = 0 and z = 2.

Solution: Given

+= kzyjxizF 2

122 += yx

+= jyix 22

22 44 yx +=

=222 yx +

=2

The unit normal n to the surface =

= yjxiyjxi +=+

2

22

xyxzjyixkzyjxiznF +=

+

+=

.. 2


www.rejin

paul.com

www.rejin

paul.com


www.rejin

paul.com

www.rejin

paul.com

INTEGRAL THEOREMS

(i) Gausss divergence theorem

(ii) Stokes theorem

(iii) Greens theorem in the plane

Greens Theorem

Statement:

If M(x,y) and N(x,y) are continuous functions with continuous

partial derivatives in a region R of the xy plane bounded by a simple closed

curve C, then

dxdyy

M

x

NndyMdx

Rc

=+ , where C is the curve described in the

positive direction.


www.rejin

paul.com

www.rejin

paul.com

Verify Greens theorem in a plane for the integral ( ) xdydxyxc

+ 2

taken around the circle 422 =+ yx Solution: Greens theorem gives

dxdyy

M

x

NNdyMdx

Rc

=+

Consider ( ) xdydxyxc

+ 2

M = x 2y N = x

1,2 ==

x

N

y

M

dxdyy

M

x

N

R

( ) =+RR

dxdydxdy 321

= 3[Area of the circle]

= 32r

=3. 4. = 12 (1) Now + NdyMdx We know that the parametric equation of the circle 422 =+ yx x = 2 cos y = 2 sin

ddx sin2= , ddy cos2= ( ) xdydxxxNdyMdx +=+ 2

= ( )( ) ( ) dd cos2cos2sin2sin4cos2 + = d22 cos4sin8sincos2 ++ Where various from 0 to 2

( ) ++=+

2

0

2 4sin4sincos2 dNdyMdxC

=

+

+

2

0

42

2cos142sin d

= ( ) +

2

0

2cos262sin d

=

2

02

2sin26

2

2cos

+

= 122

112

2

1 =+ .(2)

From (1) and (2)


www.rejin

paul.com

www.rejin

paul.com

dxdyy

M

x

NNdyMdx

Rc

=+

Hence Greens Theorem is verified.

Example 2

Using Greens theorems find the area of a circle of radius r.

Solution: By Greens theorem we know that

Area enclosed by C = C

ydxxdy2

1

The parametric equation of a circle of radius r is x = sin,cos ryr = Where 20

Area of the circle =

2

0

)sin(sin)cos(cos2

1drrrr

= ( )

drr +2

0

2222 sincos2

1

=

2

0

2

2

1dr

= [ ] 22022

1rr =

Example 3:

Evaluate ( )[ ] c

xdydxyx cossin where c is the triangle with

vertices (0,0) ,( )0,2

and )1,

2(

Solution: Equation of OB is

02

0

01

0

=

xy

x

y2=


www.rejin

paul.com

www.rejin

paul.com

By Greens theorem dxdyy

M

x

NNdyMdx

Rc

=+

Here 1,sin =

=y

MyxM

N xx

Nx sin,cos =

=

( )[ ] ( )dxdyxxdydxyxRC

+= 1sincossin

In the region R, x varies from x = 22

toy

and y varies from y = 0 to y = 1

( ) = xdydxyxC

cossin ( ) +1

0

2

2

1sin dxdyxy

= [ ] +1

0

2

2

cos

yxx dy

= dyyy

+1

0222

cos

=

1

0

2

422sin2

+ yyy

= 2

2

42

2

+=+

Example 4

Verify Greens theorem in the plane for

( ) ( ) +C

dyxyydxyx 6483 22 where C is the boundary of the region defined

by

X = 0 , y= 0, x + y =1

Solution: We have to prove that


www.rejin

paul.com

www.rejin

paul.com

dxdyy

M

x

NNdyMdx

Rc

=+

M = xyyNyx 64,83 22 =

yx

Ny

y

M6,16 =

=

By Greens theorem in the plane

dxdyy

M

x

NNdyMdx

Rc

=+

= ( ) 1

0

1

0

10 dydxy

x

=

1

0

1

0

2

210

x

y

= ( ) 1

0

215 dxx

= ( )

3

5

3

15

1

0

3

=

x

Consider ++=+BOABOAc

NdyMdx

Along OA, y=0 , x varies from 0 to 1

[ ] 13 1031

0

2 ===+ xdxxNdyMdxOA

Along AB, y = 1 - x dxdy = and x varies from 1 to 0 .

( ) ( ) ( )[ ]dxxxxxxNdyMdxAB

+=+0

1

22 1614183

= ( ) ( ) 0

1

32

232

232

14

3

18

3

3

+

xxxxx

= 3

82312

3

8 =++

STOKES THEOREM


www.rejin

paul.com

www.rejin

paul.com

If S is an open surface bounded by a simple closed curve C and if a vector

function F is continuous and has continuous partial derivatives in S and on

C, then

=c

rdFdsnFcurl .. where n is the unit vector normal to the

surface (ie) The surface integral of the normal component of Fcurl is equal

to the integral of the tangential component of F taken around C.

Example 1

Verify Stokes theorem for ( )

= kzyjyziyxF 222 where S is the upper half of the sphere 1222 =++ zyx and C is the circular boundary on z = 0 plane.

Solution: By Stokes theorem

=sc

dsnFcurlrdF ..

( )

= kzyjyziyxF 222

zyyzyx

zyx

kji

Fcurl

222

=

= [ ] ( ) ( )

=+++ kkjyzyzi 100022

Here

= kn since C is the circular boundary on z = 0 plane

= S

area of the circle =

S

dxdydsnFcurl .

= =2)1( .(1)

ON z = 0,

=sc

dsnFcurlrdF ..

On C, x = cos sin, =y ddyddx cos,sin ==

varies from 0 to 2


www.rejin

paul.com

www.rejin

paul.com

( )( )

drdFc

sinsincos2.

2

0

=

= ( ) +

2

0

2

2

0

sinsincos2 dd

= ( )

+

2

0

2

02

2cos12sin dd

=

2

0

2

0 2

2sin

2

1

2

2cos

+

= =++2

1

2

1 (2)

From (1) and (2)

=sc

dsnFcurlrdF ..

Hence stokes theorem is verified

Example 2

Verify stokes theorem for ( ) ( )

+++= kxzjyzizyF 42 where s is the surface of the cube x = 0, x = 2, y = 0, y = 2, z = 0 and z = 2 above the xy

plane.

Solution:

By Stokes theorem

=sc

dsnFcurlrdF ..

Given ( ) ( )

+++= kxzjyzizyF 42

xzyzzy

zyx

kji

Fcurl

++

=

42

= [ ] [ ] [ ]1010 ++kzjyi

= [ ]

+ kzjiy 1


www.rejin

paul.com

www.rejin

paul.com


www.rejin

paul.com

www.rejin

paul.com

Hence Stokes theorem is verified.

Example 3:

Verify Stokes theorem for

++= kxjziyF where S is the upper half surface of the sphere 1222 =++ zyx and C is its boundary. Solution: By stokes theorem


www.rejin

paul.com

www.rejin

paul.com


www.rejin

paul.com

www.rejin

paul.com

=sc

dsnFcurlrdF ..

Gauss Divergence theorem

Statement:

The surface integral of the normal component of a vector

function F over a closed surface S enclosing volume V is equal to the volume

integral of the divergence of F taken throughout the volume V ,

dvFdsnFVS

= ..

Evaluate zdxdyxydzdxxdydzx 223 ++ over the surface bounded by z = 0 ,z = h,

222 ayx =+ Solution:


www.rejin

paul.com

www.rejin

paul.com

16

3

22

1

4

3cos

2

0

4

== d

2

3.

4adsnF

S

=


www.rejin

paul.com

www.rejin

paul.com


Vector Calculas

Documents

Transcript of Vector Calculas