Engineering Mathematics Bhaskar Rao

ENGINEERING MATHEMATICS -I SECOND EDITION

P.B. Bhaskar Rao M.Sc., Ph.D. Retd. Professor, Former Chairman, Board of Studies, Department of Mathematics Osmania University Hyderabad

S.K.V.S.Sriramachary M.A., M.Phil., B.Ed. Professor & Head (Retd.) Department of Mathematics University College of Engineering (Autonomous) Osmania University Hyderabad

M. Bhujanga Rao M.Sc., Ph.D. Professor, Dept. of Mathematics University College of Engineering (Autonomous) Director of Centre for Distance Education Osmania University Hyderabad

BSP BS Publications 4-4-309, Giriraj Lane, Sultan Bazar, Hyderabad - 500 095 A.P. Phone: 040 - 23445688 e-mail: [email protected]

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Printed at Adithya Art Printers Hyderabad.

ISBN: 978-81-7800-151-7

i

Contents

CHAPTER -1 Ordinary Differential Equations of First Order and First Degree ..................................................... 1 CHAPTER -2 Linear Differential Equations with Constant Coefficients and Laplace Transforms ...................... 69 CHAPTER-3 Mean Value Theorems and Functions of Several Variables .............................................. 111 CHAPTER-4 Curvature and Curve Tracing ................................................ 213 CHAPTER-5 Application of Integration to Areas, Lengths, Volumes and Surface areas ........................ 313 CHAPTER-6 Sequences of Series .............................................................. 385 _ CHAPTER-7 Vector Differentiation ............................................................. 475 CHAPTER-8 Laplace Transforms ............................................................... 623

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1 Ordinary Differential Equations of First Order and First Degree 1.1

Introduction Differential euqtions play an important role in many applications in the field of science and engineering, such as (i) problems relating to motion of particles (ii) problems involving bending of beams (iii) stability of electric system, etc. For example, Newton's law of cooling states that the rate of change of temperature of a body varies as the excess temperature of the body to that of its surroundings. If 8(t) is the temperature of the body at time 't' and 8 0 is the temperature of the room

de

in which the body is kept, then dt gives the rate of change of temperature with time.

de dt

= K(8 - 8 0) ; K is constant

Similarly Newton's second law of motion for a particle of mass m moving in a straight line can be written as

d 2x m dt 2

=F

Where m is the mass, x is the distance of the particle at time 't' measured from a fixed origin and F the external impressed force.

Engineering Mathematics - I

2

A differential equation is an equation involving an unknown function and its derivatives. Ifthere is only one independent variable and one dependent variable the equation is called (Ill ordinary differential equation. If there are more than one independent variable the equation is called a partial differential equation as this involves partial derivatives. For example:

d3y

dy

_ y=e(

.... (a)

d3y J4 ( d2YJ8 (dY) 12 6 _ 8 (-dx;3- + -dx-2 + -dx + Y -x

.... (b)

4

dx

3

+3x

dx

.... (c)

.... (d)

.... (e)

.... (t)

.... (g)

The first four equations (a), (b), (c) and (d) are ordinary differential equations and the remaining three are partial differential equations.

Order 0/ a differential equation: The order of a differential equation is the order of the highest ordered derivative appearing in the equation.

0/ a differential equation: The degree of a differential equation is the power to which the highest ordered derivative appears in the equation after clearing the radicals if any.

Degree

Ordinary Differential Equations of First Order and First Degree

3

In the above examples:

Example: Example: Example: Example:

1.2

1.1(a) is a differential equation of order 3 and degree I. 1.1(b) is of third order and fourth degree differential equation. 1.1(c) is a second order, first degree differential equation. 1.1(d) is a second order, second degree ditferential equation.

Example Formation of an ordinary D.E : The differential equations ar~ formed by eliminating all the arbitrary constants involved in the functional relationship between the dependent and independent variables.

th~t are

For example:

y

=

cx2 + c 2 where c is an arbitrary constant.

.... (I)

To eliminate 'c': (only one constant) From(l)

dv _0

dx

=

c.2x+ 0

I (~V c= 2x dx Substitution of c in (I) gives

y

=

d ~ ( dx

_1_(dy )2

_I dy x2+ 2x d\: 4x 2

d);

)2 + 2x 3 2d

=

dx

- 4x2y

0

is the required D.E and y = cx2 + c 2 is called the solution of the D.E.

Note: Depending on the number of constants in the given equation differentiate it as many number oftimes successively. Then the elimination of the arbitrary constants from the resulting equations and the given equation gives the required differential equation whose order is equal to the number of constants.

1.3

Example Eliminate the arbitrary constants a, b from xy + x 2 = aeX + be-X and form the differential equation.


4

Solution: The given equation is xy + .x2 = ae-'" + be-x

..... ( I)

The number of arbitrary constants is two. Differentiating (I) w.r., to 'x' two times successively.

dy x dx

_

+ Y + 2x = ae-' - be-x

d2y

ely dy + - + - + 2.1 d-c dx dx

X ---2

=

aex + be-x

.....(2)

.... (3)

From (I), (2) and (3) el imination of a, b gives the D.E. from (I) and (3) we get

d l )' 2dy . . + - - + 2 = xy + x 2 IS the requIred D_E. dxdx

x

1.4

--?

Example Form the differential equation by eliminating the constants a and b from 1

a.x2 + by =

Solution: Differentiating ax2 +

by =

I w.r.t 'x'

dy 2ax+ 2byd-c

=

0

.... (I)

.... (2)

Again differentiating wr.t., 'x'

d 2y dy dy 2a+2by - ? +2b-.- =0 dxdx dx Elimination of a, b from (I), (2) and (3) gives

x-?

i

x

yYl (Y.h + yl2

-I

0 =0 0

Expanding the determinant we get

2 y x d y +x(d )2 2 dx dx

_ y(dY)=o dx

.... (3)


1.5

5

Example Form the differential equation by eliminating the constants from

y

=

a secx + b tan x

Solution Given equation is

y= asecx + btanx

.... (I)

Differentiating w.r. to 'x'

dy dx

=

asecx tanx + bsec 2x

.... (2)

=

secx[a tan x + b sec xl

.... (3)

dy dx

Further differentiation gives

d 2y -, 2

(X

= a sec x tan 2x

+ a sec3 x + h2sec 2x tan x

2

I.e.,

I.e.,

d Y

--1

dx-

d 2y

dx 2

=

2

asecx tan x

1 + bsec-xtanx + bsec 2xtanx + asec 3x

= secx t'lnx(atanx + bsecx) + sec 2x(btanx + asecx) ..... (4)

Substituting

and

asecx + btanx

=y

atanx + bsecx

= --

from (I)

(~) from (2) secx

in (3) we get

[-f) d2 dx

i.e.,

dY) ( secx tanx ~ + sec x(y) secx --

=

2

d 2y dy - -2 - tanx - - ysec2x dx

dx

=

0

6


1.6

Example Form the differential equation of all circles passing through the origin and having their centres on the x - axis.

Solution Take any tangent to the circle as y-axis, the centre lies on x-axis. Let 'a' be the radius of the circle. Then centre is (a, 0) :. Equation of the circle is (x - a)2 +

.r

=

a2

.... (1)

y

x

,

x

y' Fig. 1.1 Differentiating (I) w.r. to 'x'

dy dx

2(x - a)

+ 2y -

x- a

dy -ydx

=

=

0

dy dx

a=x+ y -

.... (2)

From (1) and (2)

[x-(x+ .r

y:lJ' y' =[x+ y:J +

(:r +.r

2XY :

+ x2 -

= x2

+.r

(:r +

2XY ( : )

.r = 0 is the required D.E

<

7


1.7

Example Form the differential equation of all central conics whose axes coincide with the axes of coordinates.

Solution The equation of all central conics whose axes coincide with the axes is ax2 +

bl =

.... (1)

1

Differentiating (1) w.r.t., x

dy 2ax+ 2by dx

=

0

.... (2)

Differentiating (2) w.r.t., 'x' again

(d

2 y )2 d y a+by - 2 +b =0 dx dx

.... (3)

Eliminating a, b from (1), (2)(3)

x2

y2 dy ydx

X 2

y~+ ~ d 2 (d dx dx

r

o =0 0

y Xyd2; +x(d )2 _y(dY)=O is the required D.E. dx· dx dx

=>

Exercise - 1(a) 1. Eliminate the arbitrary constants from the following and find the corresponding differential equation : (i) y

= mx + c (m, c arbitrary constants)

(ii) y

=

a e2x + b e-2x

(a, b arbitrary. constants)

d2

[ Ans : ~ - 4y - 0] dx 2

8


(iii)

y

=

ax + bx 2

(a, b arb. constants)

d2 [ADS: x 2

dy ----? 2x -d + 2y dx x

(iv)

(x - h)2 + (y - kf = a2, (h, k a, b arb. constants)

(v)

y

=

y

=

0]

=

OJ

(a + bx)e-X , (a, b arb. constants)

d 2y 2dy [ Ans: - 2 + +Y dx dx (vi)

=

a sin x + b cosx (a, b arb. constants)

d 2y

[ADS: - l 2 (X

+ Y = 0]

2. Find the differential equation of all circles with centre on the line y = x and having radius' I '.

3. Form the differential equation of all the circles with centre on the line y passing through the origin. [ADS. : (xl + y)

(:

-1) =

2(y -

=

-x and

x) (x + y:)]

4. Find the differentiall equation of all the parabolas with vertex at the origin and foci on the x - axis. [ADS:

y

dy -2xy dx

=

0]

5. Find the differential equation of all parabolas with the origin as focus and axis along x-axis.

dy [ADS: 2x dx + Y

?

dy -

(

dx) -Y

=

0]


9

Methods to Solve

1.8

The differential equations of the first order and of the First Degree:

1.8.1 Separation of Variables Sometimes the differential equation /

dy dx

=

q(x, y)

can be written as f(x) dx + g(y)dy = 0

.... (I)

if the variables can be separated. Integration of (1) gives the solution of the equation.

Jf(x)dx+ Jg(y)dy=c

i.e.,

where c is an arbitrary constant.

1.8.2 Example Solve et"tany dx + (1-~sec2ydy

=

0

Solution The given equation

e-'"tany dx + (1-e-'")sec 2ydy =0 can be rearranged as e( sec 2 y - - d x + - - dy=O

I-eX

tany

2

Integrating

eX Jsec y --dx+ - - dy=c

J I_eX

tany

-Iog( e-'" -I ) + log t~ll1Y = c i.e.,

tany -X-I

e -

=

c or tan y = c(e X -I)


10

1.8.3 Example dy Solve - = 1 + xl + dx

Y + xly

Solution The given differential equation can be written as dy dx

=

(1 +xl) (1 + y)

dy x3 -1- 2 = (1+xl)dx => tan-I y = x + - + C +y 3

1.8.4 Example dy Solve dx - 2xy = x, where YCO) = 1

Solution

dy

- =x(1+2y) dx dy I+2y =xdx On integration

x2

I

"2 Iog(1+2y) = 2 +c Giveny= 1 when x = 0 Substituting in (1) 1

"2 Iog(3) = 0 + C C=

~ log3 = log(v'J)

Hence the solution is 2

"21 Iog(I +2y)= 2x + Iogv'J (i.e.,) log (1+2Y) - 3 - =xl

.... (1)


11

1.8.5 Example dy Solve - =(4x+y+ If dx

Solution dy = (4x + y + 1)2 dx Substituting 4x + y + I = t in (I)

.... (I)

-

dy 4+dx i.e.,

dy dx

=

dt dx

=-

dt -4 dx

dt - -4 = t 2 dx Integrating

i.e.,

I t -tan-I-=x+c

2

2

tan-I (4X+;+ I) = 2(x + c) The solution can also be written as 4x + y + I = 2tan(2x + c) where C is an arbitrary constant.

Exercise -1(b)

1.9

Solve the Following Differential Equations 1.

:

=~Y

2. (2 - x)dy - (3 + y) dx = 0

[ADS:

~

+ e-Y

[ADS (3 + y) (2 -x)

= =

c] c]


12

[ ADs: yc

dy 6. (x-y)2 dx

= (a + x) (I - ay) ]

[ADs: (x - y) + log ( x- y-a) x-y+a

= a2

7.

dy - =(3x+4y+ 1)2 dx

8.

dy =(2x+y+ dx

9.

dy dx = tan (x + y)

= x + c]

[ADs: 2(3x+4y+ 1)= .J3tan- 1 2.J3x+c]

If

[ ADs:

1 tan-I (2X + .J2 .J2Y + I )

=

x +c]

[ADs: log[sin(x + y) + cos(x + y)] = x - y + c]

10 dy = 2 . dx (x+2y-3)

[ADs: (x + 2y - 3) - 410g (x + 2y + I)

=x + c ]

1.9.1 Homogeneous Equations The differential equation of the form

dy dx

f(x,y) g(x,y)

where f, g are homogeneous functions of same degree in x, y is called a homogenous differential equation. Such a differential equation can be written as

.... (I)

Substitutingy = vx dy dv - =v+xdx dx

Ordinary Differential Equations of First Order and First Degree The D. E (I) becomes

dv \11 (v) v + x dx = ~(v)

~(v)dv

dx

-; + v~(v)-\lf(v)

0

=

Integration yields the solution

dx

J-

X

+

J"'()~(v)dv( ) = c where v = -y v'" v - \11 v

x

\

1.9.2 Example Solve

(xl +

dy I) -dx

=xy

Solution

· . dy dv S U bstltuttngy = vx, -d = v + x x dx

dv v + xdx

xvx =

2

2

x +v x

dv v+ x dx

=--

dv xdx

v - -2- v

=

2

v 1+ v 2

1+ v

dv _v 3 x-=-dx 1+ v 2

dx J I JI-dv=c ~+ J-+ X v3 v

13

14


v-2 logx + + logv

-2

logxv- -

1

=

e

e

=

2V2

(y)

X2

logx - ~ x 2y i.e.,

2ylogy

=

=

e ~ 2ylogy - xl

2ey + x 2

1.9.3 Example Solve

dy y2 x-+-=y dx x

Solution

dy xl-+y=xy dx i.e.,

Substitutingy = vx, dy dv -=v+xdx dx

dv v+x - =v-vl dx dv x - =-vl dx dx

dv

-+=0 X v2 Integration yields V-2+1

logx+ - - =e -2+1

1

~logx--

v

=c

=

2ey


x logx - y

ylogx -x

=

=

c cy

or

ylogx

=

x + cy

1.9.4 Example Solve [x+ YSin(Yx)]dx = xSin(Yx)dy Solution The given differential equation can be written as dy

x+ ysin(X)

= __

--,---.0.-,.-----.:-

dx

Substitutingy = vx

xSin(j~)

~

dv v+x dx

dy dy - = v + x. dx dx x+ vxsinv xsinsin(v)

= ----

dv l+vsin v+x- = - - dx sinv dv l+vsinv 1 -v=-.x-= dx sin v SI11 v dx sin v dv = x Integrating -cosv = logx + c logx + cos(Yx) = c

15


16

Exercise - 1 (c)

1.10

l.

Solve the Following Equations dy y2 dx - xy-x2

2. (2 - 2xy)dx

=

[ADS : y = ceix ]

(x 2 - 2xy)dy

[ ADS : xy(y - x) = c ]

3. 2xy + ~ - r) dy = 0 dx

[ADS:

r + y = cy ]

[ADS: logx = 2tan- i (Yx) + c]

5. xdy - ydx = ~ x 2 + y2 dx y [ ADS: cos - = logcx ] x

7. xcos(Yx) (ydx + xdy) = ysin (Yx) (xdy - ydx)

8. (ry - x 3 )dy - ~

9. (r + y) dx

10.

1.10.1

X

+ xy) dx

=

0

[ ADS:

= 2xydy

[ADS: sec (Yx) = yxc]

y~ x + y = cx.e 2

2

( ADS: (r -

dy dx = y[logy - logx + 1]

tan

-I(YI) Ix

]

y) = xc ]

( ADS : y = xecx ]

Non-homogeneous Differential Equations The D.E of the form

dy dx

=

ax+by+c Ax+By+C

.... (J)

where a, b, c A, B, C are constants, is called a non-homogeneous dif.ferential equation.


17

Case (i) If

a

b

A

B

--::;:.-

Substituting x = X + h, Y = y + K where h, k are (constants) to be chosen so as to satisfy. ah + bk + c = 0, Ah + Bk + C = 0

Solving these equations, values of h, k are obtained. The given D.E then reduces to a homogeneous D.E. dY

aX +bY

dX

AX +BY

which is then solved taking Y = VX and then substitute X

=

x - hand Y

=

y - k in the solution.

Case (ii) If

a

b

A

B

Then the differential equation will be of the fonn dy _ (ax + by)+c dx - m(ax+by)+C

since Ax + By will be constant m times ax + by. Now substitute ax + by = t,

Differentiation gives a+bdy=dt dx dx

i.e.,

dt --a dy dx -=-dx b

D.E (2) then reduces to dt

--a

t+c

b

mt+C

~=---

.... (2)


18

Then the solution is obtained by using the method of separation of variables.

1.10.2

Example Solve

dy dx

x+2y-3 2x+ y-3

Solution dy x+2y-3 = dx 2x+ y-3

... - (I)

Substituting x = X + h, y = Y + k where h, k are chosen to satisfY h + 2k - 3 = 0 and 2h + k - 3 = 0 solving we get h=I,k=I

.... (2)

i.e., we take x = X + I, Y= Y + I The D.E (I) reduces to dY

dX

X-2Y 2X+Y

=---

dY

dV

Substituting Y = VX, dX = V + X dX dV X+2VX V+X-=--dX 2X+VX V+XdV =1+2V dX 2+V

~

XdV =1-V

dX

2+v dv=dX I-v X dX +(1 +_3_) dv = 0 X v-I

10gX + v + 3 log(v-I) = loge

2

2+V


v + log( v-I )3 + logX = loge v-log(v-I)3 X = loge

Y X +10 I.e.,

~-X

3

V-x

)

X=loge

10g(Y - X)3 + XY = Xloge log(y - \ - x + \)3 + (x - \) (y - I) = (x -I) loge log(y - x)3 = (x - 1) (loge - y + \)

1.10.3

Example Solve (2x + 3y + \) dx + (2y- 3x + 5)dy = 0

Solution dy dx

2x+3y+ 1 2y-3x+5

Substituting x = X + h, y = Y = k, :

= :

Choosing h, k so that 2h + 3k + 1 = 0, 3h - 3k - 5 = 0 we get h = 1, k

= -1

The given differential equation reduces to

dV dX

2X+3Y 3X-2Y

Substituting Y=YX

dV dV =Y+XdX dX dV Y + X dX

2X+3YX 2YX

= 3X -

Y+X dV =2+3Y dX 3-2Y

X dV dX

= 2(1 + y2 ) 3-2Y

19


20

3-2V)dV= 2dX ( 1+ V2 X

Integrating

f

fdX

2v 3 -dv - - --dv=2 -+c 1+ v 2 1+ v 2 X

f

3tan- 1(V) - log( I + v2)

3tan B~

1

= 210gX + c

(~ )-IO~ X';/') ~ 210g)( + c

X=x-h=x-I Y=y-k=y+1

3tan-1 (~::

)-IOg[ (x -Il~: ~ + I)'] ~

1.10.4, Example Solve

dy =x-y+l dx 2x-2y

Solution

(x- y)+1 -dy = --'------'-'-dx

2(x- y)

Substituting x - y = V dY=I_ dV dx dx l_dV=V+I dx 2V dV_ dx

V+l_V-1 2V 2V

-- I------

2V dV=dx V-I

210g(x-

I) + c

21


Integrating 2 JV-I+I dV=x+c V-I 2[V + 10g(V-l)]

=

x+c

2[x - y + log(x - y -1)1 210g(x - y - I) (x - y -1)2

or

1.10.5

=

x+c

2y - x + c

=

= e2j~x

C

Example Solve

dy

4x+6y+1 2x+3y -5

dx

Solution (~V

-=-

d'(

2(2x+3y)+ I (2x+3y)-5

Substituting 2x + 3y = V dy (N 2+3-=dx dx dy dx

=~[dV 3 £Ix

-2]

The differential equation reduces to

~(dV -2)=- -2V + I 3 dx

V-5

dV

6V+3 =2+ - dx 5-V

5-V 4V+13 dV=dx

IO-2V+6V+3 5-v

= ------

13+4V 5-V

22


Integrating

fl 1J~

3

3

4V + 13

) dv = _ 4 Idx + e

33

v- -log(4V+ 13)=-4x+e 4 33 (2x + 3y) - - log[4(2x + 3y) + 13] = - 4xm + C 4 i.e.,

4(6x+3y-e)=33Iog[4(2x+3y)+ 13]

Exercise -1(d)

1.11

Solve the Following Equations dy y-x+5 \. - + -=------dx y+x+3

[Ans:tan- J

2.

(

dy x+2y-3 dx - 2x+3y-5

~] -

[Ans: (2+Ji) IOg[y-1 x-I ,,3

3

1

y+4 1 y+4 x-I ) +"2 log[ ( x-I ) 2 +1 =log(x-I)+loge]

~]+ ,,3~

(2-Ji) log [y-I + x-I ,,3

logx= loge]

_dy = _2y'------_x_-_4 . dx y-3x+3 [Ans: (x - 2)2 - 5(x - 2) (y - 3) + (y - 3)2 = e {2(y-3)-[5+h1{X-2)]}] 2{y -3)-(5 -h1(x-2»

4. (2x+5y+ l)dx-(5x+2y- l)dy=O [ Ans : (x + y) 7 = e ( x - y -

~

r]


5.

dy dx

23

y-x+ 1 y+x+5 y-2) I [ADS : tan~l ( x _ 3 + log [(y - 2)2 + (x - 3)2]

2

6.

loge]

=

dy _ 3y+2x+4 dx 4x+6y+5

[ADS: .7'j(2x+3y)-

:9

(14x+2ly+22)=x+c]

7. (4x-6y-l)dx+(3y-2x-2)dy=O 3 [ADS: x - y + 41og(8x - 12y - 5) = c ]

8.

dy dx

x-y+3 2y + 5

= 2x -

[ ADS: (x - y) + log[2 + x - y ]

x = -

2

+ C]

1.12 Linear Differential Equations A differential equation of the from :

+ py = q where p, q are functions of 'x',

alone or constants is said to be a linear differential equation offirst order. Multiplying both sides of the equation by el pta [called the integrating factor (I.F)] we get,

el ptlr dy + el ptlr py = q .e Iptlr dx The left hand side is the differential coefficient of y. e (i) can be written as

d(y.e

1ptlr

)

=

qe 1ptlr

Integrating

which gives the desired solution.

.... (i)

l pcb:


24

Note: In some cases a differential equation can be reduced to the linear form by taking 'y' as independent variable and x as the dependent variable. The D.E is written as

p" q I are functions of y or constants Now the I.F

=

eJ

Pldy

Solution is

x.e

1.12.1

JPldy

=

fql·e

Jpl"Y

•

+c

Example Solve

(I + x 2 )

dy + 2yx - 6x2 = 0 dx

-

Solution Rearranging the given differential equation to the form

dy dx + py = q We have

dy 2x 6x 2 -+--y=-dx I +X2 1+X2 Here

6x 2

2x p=

I+X2

,q=

I+X2

~dx

I.F= e

Jpdx =e J I+x2

log(I+X2 )

I.F = e

=

(1 + x 2 )

Solution is given by

y(1. F) =

fp(I.F)dx + c


1.12.2

Example Solve

xlogx

dy - +y=2Iogx dx

Solution dy y 2 -+--=dx xlogx x

Here

I 2 p= - - andq=xlogx x

Solution is

ylogx =

J~ logxdx + c

ylogx=2

1.12.3

(IogX)2 2

+c

Example Solve

dy Y (xsmx+cosx . ) =I xcosx -+ dx

Solution xsin+ cosx dy -+y.---dx xcosx

p=

xsin x + cosx

xcosx

xcosx

I ,q=--

xcosx

25


26

J

~Slll X~.COS~ dx

I.F

=

e

XCDSX

= e(log(xsecx)

I. F

= e1og(x sec x)

xsecx

=

Solution is

y(xsec x)

xysec x xysecx

1.12.5

f_lxcosx

=

= =

x

x sec x dx + c

2

Jsec xdx E c tanx+c

Example Solve

dy + 2ytanx = sinx given that y ~

=

0 where x

=

~, 3

Solution dy + 2ytanx = sinx dx p

=

2tanx

q

=

sinx ef

2 tan xd!:

IF

=

~

= e210gsecx

ylF=

JqxIF.dx+c 2

ysec 2x = Jsecx.sec xdx + c


ysec 2x ')

Given that y

Jsec x tan xlir + c

=

ysec-x

secx + ('

=

.... (I)

;7j

0 when x =

=

1C

o=

sec - + c => 3

C =

Substituting c = -2 in (I)

ysec 2x

=

secx - 2

is the required solution

1.12.6

Example dy Solve (x + 2.v) -,

=

(X

y

Solution

+

2.v

dx dy

x y

X

=

--- =

y yd dx

2.v

is in the form of

P

-1

I

= -

Y

y

27

'

q

1

=

2,1 Y

-2


28

x-1

f2y

=

y

2

1 x-dy+c y

x

- =y+C Y

1.13.7 Example dy dx

(x + y+ I) -

Solve

=

1

Solution dy dx- X =y+l PI=-lq\,=y+1

IF =

e

f-1dy

e- Y

=

Solution is given by x(IF)

xe-Y

= fql(/F)dy+c

=

f(y + I)e-Y dy + c

or

i.e.,

x +y + 2

=

ce Y

Exercise 1(e)

1.13 Solve the Following Differential Equations I. (I + y)dx

= (t~n-Iy

- x)dx [ADS: xetan- I y = tan-Iyetan-Iy --etan- I y + c]

dy 2. cos2x - + Y

dx

= tanx [ADS: Y = ce-tanx - tanx -1]

29


dy 3. x - +2y-x2 1og=0 dx (ADS: y

4.

dy dx + ycot x = 4x cosec.x, if y

=

c =

-?

x-

I .r2 +- x210gx --I 4

16

n/ 0, when x = ~2 n2 (ADS: ysinx = 2x2 - 2

5. yeYdx

=

(y

+ 2x&)dy

6. (x + 31) dy dx

7. (xy -I )

=

dy dx+ y

3

=c-

( ADS : x

21 + cy ]

+ y3

3

=

= 0

I ADS: x dy dx

e~Y

y

I

8.

I

( ADS : xy~2

= x 3 - 2xy if y = 2 when x =

.

= ce Y

I

+ - + II Y

I

[ADS: 2y - x 2 + 1 = 4 el~~ ]

dy x+ ycosx 10. dx = I +sinx [ADS: y(l+sinx)

=

x2

c - -] 2

30


1.13.1 Non-linear Differential Equation of First Order Ber noulli j. equation: The differential equation of the form ely dx + PY

=

llyn

.... (I)

where p, q are functions ofx alone is said to be a Bernoulli's differential equation. Dividing (I) throughout by yn

dy =q dx Substitutingynt-l(coefficient ofp) "" v y-n_ + py-n+l

.... (2)

1 dy dv (1-n) - - = y" dx dx (2) reduces to I dv - - - +pv=q

(I-n)dl'

dv - + (1-n)pv = (I-n)q dx which is linear in v. The avove D.E can be solved by using the method given in 1.12.1 example.

1.13.2 Example Solve dy - ytanx = ysecx dx Solution

dy

dx - ytanx = ysecx

dy 1 y-2 -d - - tan x = sec x x Y Substituting

-y =v,

.... (I)


1 dy

dv

dx

dx

+--=/

(1) reduces to dv dx + vtan x

=

sec x

is linear in v. Here

p

tanx, q

=

IF

=

e

=

e

secx

=

fpd< flanXd<

= elogsec x =

secx

SolutIon of th [) E (I) IS therefore

v(lF) = Jq(IF}dx + c

v. secx

=

Jsecx.secxdx + C 2

vsecx.= J sec xdx + c Substituting

1

v=-Y 1

- - sec x y i.e.,

=

tan x + c

y(tan x + c) + sec x

1.13.3 Example Solve (3xy + .0) dx - 3i2dy = 0

Solution dy 3x2 - - 3xy dx

=

.0

=

0

31

32


is in Bernoulli's fonn 1 dy _ dx - xy - 3x 2

"7

Substituting

-1 =v

y

D.E reduces to

dv I 1 ---v=dx xy 3x 2 which is linear in v Here

1

1 q= 3x 2

p=~,

IF = e =

e

Jpdx

J!d\ x

=x Solution of the D.E (1) is

v(lF) = fq(lF)dx+c -..!..x= f_l_. xdx + c y 3x 2

x 1 --=-Iogx+c Y 3 i.e.,

y(110g x + c) + x = 0

1.13.4 Example dy Solve dx + (2xtan- 1y -.x3) (1 + y) = 0 Solution

dy 1+ y2 dx

dv

= dx

.... (1)


33

The given differential equation becomes dv

-

d-c

+ 2xv=x3

.... (1)

which is linear in v Here

p = 2x, q = x 3

IF

e

=

J

2xdx

=e

x2

Solution of the D.E (1) is

v(lF) = Writing

Jq(IF}dx+ c

x2 = t xdx

1

= 2dt 2

r

I

dt

v.e x = {e 2+c

1.14.5 Example dy Solve tany dx + tanx

= cosy cos3x

Solution dy tany d'C + tanx

= cosy cos 3x

Dividing by cosy throughout

dy secy tany dx + secy tanx = cos 3x Substituting secy = v, we get

dy secy tany dx

dv d'C

34


dv

dx + v . tal1.X = cos3x

.... (I)

is linear in v p = tal1.X, q = cos3x

Here

IF =

e

f pdt

=

e

flallXdy = secx

Solution of the D.E (I) is v(IF)

= fq(IF)dx + c

v. secx

=

3

fcos xxsecxdx+c

secy secx

=

secx secy

=

2

fcos xdx + c I +cos2x 2 +c

I

x sin2x secx secy = -+--+c

2

4

1.13.6 Example dy x

Solve -d

.

= (SII1.X -

. cosx smy) - cosy

Solution

dy. ." cosy dx + smycosx = SII1.XCOSX Substituting siny = v

dy dv cosy-=dx dx The gIven equation reduces to

dv . dx + vcosx = SII1.XCOSX

.. .. (I)

35


is linear in v

p

Here

=

~

IF

COSX,

= e

q = SIl1 X cos X

fpdr = e feo') '(tl\

=e

~1Il X

Solution of the D.E (I) is v.e SlIlt

=

J. xcosx.e m"'. I + c SII1

(X

sinx = t=> cosx dx = £II in the RHS

write

siny

e;lIlt

sinyeSlI1\"

= [te t =

el] + c

eSItlX [sinx-l] + c

Exercise - 4(f)

1.14 Solve the Following Differential Equations I. (ylogx -I )ydx =

x(~y

1 (Ans: y 2.

dy . - cosx + YSlnx = dx

r.::::::::

( Ans : dy tany ---£Ix l+x

=

1 + logx + ex )

...; ysecx

(Ans :2y y

4.

=

(I+x)

~ y-

= -

12

-Jsecx

=

tanx + 2c )

sin 2x -sinx -~ + ce 2SIIl\"

2

)

~secy

[Ans: siny = (1 + x) (eX + c) )

36


dy 5. x - + ylogy = xyeX dx

I

+c

J

x3

2

dy

ADs : xlogy = (x - I)e"

6. 3-+--y=-) dx x+ I y

I ADs: 7.

dy + ytanx dx

=

(x +

Iii =

dy dx

X

2

6

2x 5

+-

X4

+- +C 5 4

isecx

I ADs: 8.

x6 -

cos2x

=

y(c + 2sin x) J

1

Y +xy

1.14.1 Exact Differential Equations Let us consider the differential equation Mdx + Ndy = 0 where M, N are functions ofx,y. If this equation is to be exact, then it must have been derived by directly differentiating some function ofx,y. Hence

Mdx + Ndy = du, say

.... (I)

But from differential calculus

au

au

du = -dr:+-dy ax ay From (I) and (2) we get

au

N=

M= ax'

Now

aM

a2u

ay

ayax

-=--

au

ay

and -ax- - -ax-ay-

.... (2)


aM

-

oy

oN.IS tIle con d·· ItlOn ..lor exactness.

= -

AX

:. The differential equation Mdx +

N~v =

0

.IS exact 'f aM aN I -=-

oy

ax

Then the solution is expressed in the form

+ (integrate w.r.t y those terms that are independent of x)

(treatingyas constant integrate w.r.t x)

Note: IfN has no term independent ofx then the solution is fMdx

=

e

1.14.2 Example Solve (x + 2y - 3) dy - (2x - Y + \ )dx = 0

Solution (x+2y - 3)dy - (2x - Y + I) dx = 0 M = -(2x - Y+ \) N = (x + 2y - 3)

aM

-=\

oy

aN

-=\

ax

The given differential equation is exact The solution is - f(2x - y + \ )dx +

fe x + 2y - 3)

_2X2 2i - -2- + yx - x + 2

=>

I - x 2 + xy -

x - 3y

=

e

=

- 3y =

(ry = e

e

37

38


1.14.3 Example Solve (x 2+ y.)dr: + 2xy dy = 0

Solution

(x 2 + .V) dx + 2xy dy

=

0

M=x2+y.

N

aM av

-

=

2xy

aM ax

-=2y

=2y

The given differential equation is exact 2

Solution is f(x + y2

}h + f2xydy = c

1.14.4 Example Solve (I + e

-,:~, )dx + /Y

[I.'... ;

1

dy = 0

Solution (I + e'x Y ) dx + e x'Y

(

J -;;, ) dy = 0

x'y M = J + e:t Y , N = e,i

aM

-x

~

i

-=-

[

x)

J- Y

,y' e'Y

oM oN oy

ox

The given differential equation is exact

39


Solution is

X

+ e

x

Y

(y)

=

c

Exercise - 4(g) 1.15 Solve the Following Differential Equations 1. (el' + 1)cosxdx +

eJ'sinx~v =

0

lADs: (oY + 1)sinx = c 2. (vcosx + siny + y)tb' + (sinx + xcosy + x)dy

=

I

0

I ADs: ysinx + (siny + y)x = c I 3. (x 2

-

ay)dx = (ax - ;l)(~v

ADS: x 3 + ),3 - 3ll.\y

= C

I

c

I

4c

I

4. (ax + hy + g)d'C + (hx + hy + j)dy"= 0

I ADs·. 5. (x 2 +

1- a2 )xdx + (x 2 -.v - P)ydy =

2 ax - 2 + (liy + g)x +

hi':' .!.

+ fy

=

0

I ADS

: x4 + 2x21

- 2a2x2 - l - 2b 2;l =

1.15.1 Integrating factors If the differential equation Mdx + Ndy = 0 is not exact, it can be made exact by multiplying it with some function of x, y. Such a function is called an integrating

jactor. Rule!t' for fillt/illg the illtegrtltillg factors : 1. Integrating factors found by inspection:

Example Solve x dy- ydx

=

0

Solution

xdy- ydx

=

0

40


Dividing by x2

xdy- ydx

---'-,=-=--

2

x

=

0

On integration

Yx

=c

First method /0 find an integrating jac/or : If the differential equation Mdx + Ndy is not exact, but is homogenous and Mx + Ny

7:-

0, then the integrating factor is

1 . Multiply the differential Mx+Ny

equation by IF. The DE becomes exact.

1.15.2 Example Solve

(x 2y - 2xy2)dx - (x 3

-

3x2y)dy

=0

(x 2y - 2xy)dx - (x 3

-

3x2y)dy

=0

Solution .... (I)

The differential equation (I) is homogeneous

M

= x 2y- 2xy

aM

-

oy

N

= -

aN

=x2-4xy

-

ax

(x 3

-

3x2y)

=-3x2 + 6xy

The DE is not exact and

Mx + Ny = x 2y IF

=

7:-

0

) 1 Mx+ Ny - x 2 y2

Multiplying the DE by the integrating factor ~ • x y

(

y2 X2Y -2X 2

X

Y

2

)d _(X2 -3X3Y )dY X

2

X

Y

2

=

0

.... (2)


1

2

Y

X

-x 3 N =-+I y2 Y

write

M

then

--=-=--

= --I

aMI

aNI

ry

ax

y2

DE (2) is exact Solution is

x - - 210gx + 3logy + c y i.e., x

or

1.15.3 Second Method to Find the Integrating Factor If the differential equation Mdx + Ndy = 0 is not exact and is of the form

j(xy)ydx + g(xy)xdy

=

0

and

Mx- Ny *- 0

then

1 Mx _ Ny is an integrating factor

1.15.4 Example Solve (x 2y2 + xy + 1)ydx + (x2y2 - xy + 1)xdy

=0

Solution

(x2y2 + xy + 1)ydx + (x2y2 - xy + 1)xdy = 0

= x2j3 + xy2 + y, N = x2y2 - x2y + x Mx - Ny = 2x2y2 *- 0

M

1

Hence the

1

IF=-- Mx- Ny 2x 2y2

.... (1)

41


42

Multiplying (I) by IF

2 (X y 2 +xy+ 2x 2y2

1)+ (X

2

y 2 -xy+ 2x 2y2

I) = 0

I{ I 2} I{ II}

- y+-+dx+- x - - + -1 2 x x 2y 2 y xy-

dy=O

.... (2)

DE (2) is exact Solution of the DE (I) is

~ 'y+~+-!-tx+~ 'x_J.-+_I_ )dY = c 2

Jl

x

x- y

1[

r

2

Jl

y

1

xy

I] I

- xy + log x - - - -logy = c 2 xy 2

1.15.5 Example Solve (xysin xy + cos xy)ydx + (xysin xy - cos xy)xdy = 0

Solutiou (xysin xy + cos xy)dx + (xysinxy - cosxy)xdy = 0 M

.... (1)

= (xysin xy + cos x,v)y, N = (xysin xy - cos xy)x

Mx - Ny

=

2xycosxy

IF=---

Mx- NY

=F-

0

2xycosxy

Multiplyingthe DE by 2xycosxy The DE reduces to

~(ytanXY+~)dX+~( xtanxy- ~) dy

=

c

.... (2)


43

which is exact (verify) .. Solution is

1

1

1

-Iogsecxy + -Iogx - - log y 2 2 2

=

c

I.e.,

logxsecxy = 2c + logy. Taking 2c as log A

~

xsecxy

= Ay Exercise - 4(h)

1.16 Solve the Following Differential Equations I. (x 3;

+ x 2y2 + xy + 1)ydx + (x 3;

-

xli - xy +

1)xdy

=

0

r ADS: xy 2.

(xli + xy +

I )ydx + (xli

-

1 - - 210gy = c xy

I

xy + 1)xdy = 0

r ADS:

1 xy + logx - logy - xy

=

c]

[ ADS: logx2 - logy - _I = c ] xy

i

4. (x4y4 + x 2

+ xy)y dx + (x 4y4 - x 2i + xy)xdy

=

0

I ADS : ~ x 2i 2

- _I xy

logx - logy

=

c ]

1.16.1 Third Method to Find the Integrating Factor

aM If Mdx + Ndy = 0 is not exact and

then the IF

=e

J

j (x)dr

aN

_a-=-~_ _ax_ N

is a function of x alone say fix),


44


(x 2 + y + 6x )dx + yxdy = 0

Solution (x 2 + Y + 6x)dx + yxdy = 0

M = x2 +

Y + 6x

..... (1 )

N = yx

aM aN ay= 3y, fu = y2 aM

aN

ay-ili

3y2 - y2 2

N

y x

2 X

is a function of x alone IF = e f~

2tJx

= x2

Multiplying the given DE by x2

x2(x2 + y + 6x)dx + yi3dy = 0 aMI

-

ay

..... (2)

=3x2y

DE (2) is exact Solutionis 4

f(x + x

2

/

2 2

+ 6x 3 }tx + fy x dY = c


(x2 + y) dx - 2xydy = 0

Solution (x2 +

y)dx - 2xydy = 0

..... (1 )


M

45

= .xl + .0, N = - 2xy

aM ay

aN 2y, ox

=

-2y

=

aM aN ax

ax N

2y+2y -2xy

-2 x

is a function of x alone

IF

= e

f -~dx 1 x =x2

MuItiplyingthe DE by

IF= -

1

x2

The given DE reduces to

[ l)

2y 1+dx,.--dy=O 2 x

aM, _ 2y

-a --2' y X

..... (2)

x

oN,

2y

-=-

(2) is exact

Solution of (1) is

y2

2y

J--dy=c Jl+-dx+ x x 2

y2

x- -

x

=c

·1.16.4 Fourth Method to Find the Integrating Factor

aN aM If the differential equation Mdx + Ndy

= 0 is not exact and

ax M By

[ function of y alone, say fly)

1

is a


46

Then

IF

=

e

f

t(y)
1.16.5 Example Solve (xy + y) dx + 2(.x2.0 + x + y4)dy

=

0

Solution (xy + y)dx + 2(.x2.0 + x + y4)dy = 0 M =

.xy3 + y

oM 8y

oN

3xy + 1

--- =

oN -8; OM] -a; (

N = 2(.x2.0 + x + y4)

=

-

ox

=4xy2+2

(4xy2 + 2)-(3xy2 + I)

M

xi+y

y is a function ofy alone

F=e

f~d)' Y

=y

Multiplying the differential equation by I.F =Y

(xy4 + .0) dx +

2(.x2Y + xy + yS)dy = 0

oMI .,3 --=4xy +2y 8y

..... ( I)

oNI .,3 =4xy +2y

ox

The differential equation (I) is exact Solution is f{xl +

i }:Lx + 2 f(x 2 i

+ xy + yS) dy

=

0


47

1.16.5 Example Solve (3x 2y4 + 2xy)dx + (2x 3_Y'

x2)(~V = 0

-

Solution

2

aN _ aM _ (6x /

ax

-2x)-(12x 2y 3 +2xL 6x 2/ -4x -_3. 3x 2 oJ- 2x - y{3x 2/ + 2x ) - Y

l

ay -

is a function of y alone

IF

=

e

-J~

I

dy

=-?

Y

Y Multiplying the DE by

~, the DE reduces to y

2X) dx+ (3 x2) dy= 0 2x y-7 (3x 2y 2+-y aNI =6xy- 2x

ax :. DE (I) is exact Solution of the DE is

2

x3 2 2 x 3-y+--=c 3 y 2

i


48

Exercise - 4(i)

1.17 Solve the Following Differential Equations I. (x + y3 + I )dx + ydy = 0 ADs:

(ADs:

2 3x 2 x 3x 2 e 3x X e 3 e 3x - - - - e +--+(y + l)-=e I

3

9

933

2. (x 2 + Y + 2x)dx + 2ydy = 0 3. 2xydy - (x 2 + Y + I)dr

=

0

I ADs: y

5. (y4 + 2y)dx + (xy3 + 2y4 - 4x)dy

=

- x 2 + I = ex I

0 2x I ADs: xy + -) + Y.1 = e ]

y

1.17.1 Fifth Method for Finding Integrating Factor If the differential equation Mdx + Ntry= 0

.... (l)

is not exact and is of the form xliyb(mydx + nxdy) + Xf y(pydx + qxdy)

=

0

when a, b, r, s, m, n, p, q are constants Then the

IF

=

xlII

where h k are constants such that the equation (I) after multiplying with IF becomes exact.


49

1.17.2 Example Solve xy\ydx + 2xdy) + 3ydx + 5xdy) = 0

Solution xy3(ydx + 2xdy) + (3ydx + 5xdy)

The

IF

=

0

..... (1 )

x"1

=

Multiplying (I) by IF =

xhl it must become exact

(xh+1 1+1 + 3x" 1+1)dx + (2x ilj .? + 1+3 + 5xil f-/ I){~Y

=

0

is exact if

aM -

ay

aN

-

ax

=

=

(4 + k)xH1 .1+3 + 3x\k + 1)1 .

2(h + 2) xh+I /+ 3+5(h +

1)x"1

For DE (2) to be exact

aM

aN

ax

ax

Comparing the coefficients on both sides 2h - k

Solving

=

0 and Sh - 3k = -2

h = 2, k= 4

Substituting h = 2, k = 4 in (I) DE (I) reduces to (x 3y 8 + 3x2yS) dx + (2x4y1 + 5x3y 4)dy = 0 which is exact

.. Solution is

..... (2)


50

1.17.3 Example Solve (3x + 2y)ydx + 2x(2x + 3y)dy = 0

Solution (3x + 2y)ydx + 2x(2x + 3y)dy = 0

IF

=

.... (1)

0/,

MUltiplying (1) by IF (0/,+2 + 30+2 /,+I)dx + (2x h+3/,

oM

-

ox oN ox

-

0+ 1/,+I)dy

=

0

.... (2)

3(k + 2)0/,+1 + 2(k + 1) 0+2yK

=

= 2(h + 3) x h+2/, ---(h + 1)0/,+ 1

The DE(1) is to be exact

oM

ON

ox

ox

2(k + 1) = 2(h + 3) => h = -

%k = -/i

Comparing the coefficients

k + 2 = - (h + J) and

2(k+ 1) = 2(h + 3) => h =

-,% k= -/i

Substituting h, k values in (2) -5/

(

3/

X /2 y/2

II) ( II -II + 2x -1/ 12 yl2 dx + 2X 72 Y 12 I

(3) is an exact DE Solutionis

2

--x 3

-2/

3/

13 yl2

II

II

+ 4X 72 y72 = c

-3/

II) dy = 0

_ X /2 y72

.... (3)

51


Exercise - 4(j) I

Solve the following differential equations I. x(3ydx + 2xdy) + 8y4(vdr + 3xdy = 0

I Ans:

4

3x

3.

(y2 + 2x2y)dx + (2x 3 -

xy)dy

=

10 +1

2 3+l

4

3

y

3

X 1

---y 10 - - +I 3

+1

7

3

=c]

0 I

( Ans : 4x 2 y

I

2

3

2 __ X

3

2Y 2 =

ex ]

3

4. (2.~y - 3y4)dx + (3x 3 + 2xy3)dy

=

0

(Ans: 5x

- 36, 13

24

y 13_12x

_10 ' I]y

15 I]

=c]

1.17.4 Applications to geometry, law of natural growth and Newton's law of cooling: When some action is applied on a quantity the changes and the action affects all the parts equally. The rate of change depends on the original quantity. For example the total population 'p' of a country increases with respect to time 't', say, then its rate of change with time is

~ . Under ideal condrtions the rate of

change of the population will be proportional to the total population at any given time and is called the law of natural growth. The growth of the population satisfies the differential equation

dp =kp dt

-

where k > 0 is a constant Solving the differential equation we have

p

= cekl

dx The rate at which a quantity x is decreasing is given by dt and this is negative.

52

Engineering Mathematics - I This rate of decrease or decay is found to be proportional to x itself

dx =-kx dt

-

(where k> 0 is a constant)

is the law of decomposition

= ce-kl

Solving the DE x

Newton s law of cooling: Newton's law of cooling states that the rate of decrease of the temperature of a body is proportional to the difference of the temperature 9fthe body and that of its surrounding medium. Let 0 be the temperature of the body at time 't' and 00 the temperature of its surrounding medium. :. Difference between the temperatures = 0 - 8 0 (0 > 00) and the rate of decrease

dO

of the temperature of the body is - dt' since the body is cooling oc. Using Newton's law ofcooling

or

-dO

dt dO

-

dt

=

= k(O -

de - oc(O - 0 ) dt 0

00 ) where k> 0 is a constant of proportionality

-k (0 - 0 ) 0

f~=-k fdt+c 0-0 0

10g(0 - 00) = - kt + c

or

0=0 o + Ce-kl

1.17.4 Example The mass of crystalline deposit increases at a rate which"js proportional to its mass at that time. The deposit has started around a crystal seed of 5 grams. Find an expression of its mass at time 't'. Ifin 30 minutes the mass of the deposit increases by'l' gram, what will be the mass of the deposit after 10 hours.

Solution Let m be the mass of crystalline at time 't' then by law of growth

dm=k' dt '

dm -=dt

km


53

on integration log m = kt + c

.... (I)

initially when t = 0, m = 5 (1) =>

log5 = 0 - c c=-log5

SubstitutilllJ; c value in (1) log m = kt-log5

I.e.,

kt= 10g(

~)

.... (2)

When t = 30 minutes mass deposit increases by , I ' gram m = 5 + 1 = 6 grams Substituting t=

"21 (hours), m =6

Substituting in (2) log(1Y t= 109; Now to find the mass after 10 hours (i.e t = 10) from (3) we get

IOg(H IO~ log 7~ log (7) ~ IOg(%)" x

(

6)20

m = 5"5

grams

.... (3)

54


1.17.5 Example The rate at whi~h a certain substance decomposes in a certain solution at any instant is proportional to the amount of it present in the solution at that instant. Initially, there are 27 grams and three hours later, it is found that 8 grams are lett. How much substance will be left after one more hour.

Solution If m grams is the amount of the substance left in the solution at time 't', then the rate at which it decomposes is dm , which is proportional to m. dl By law of decay

dm

dt

= -

f~

km (k> 0)

=-k fdt+c

logm = -kt+ c

.... (I)

Initially when t = 0, m = 27 From (1) we have log27 = - k.O +c

=>

c= log27

Substitution of'c' in (I) gives log m = - kt + log27

10g(;; )= -

kt

It is given that m = 8 when t = 3 .. From (2) 8 log (2 ) = - k, 3 7

8 -k= log ( 27 2 -k=log3

)X

.... (2)


55

Then (2) becomes

10g~ =IOg(%}

... (3)

when t = 4

III

log 27

2

4

= log ( "3 )

m=27 x (%r grams

m=

16

3

grams.

1.17.6 Example The number x of bacteria in a culture grow at a rate proportional to x. The value ofx was initially 50 and increased to 150 in one hour what will be the value ofx after

1

12" hour. Solution

dx -=/0: dt

dx =kdt x

-

logx = kt + c c is the constant of integration when

t= O,x= 50

..

log 50 = k.O + c

or

c= log50

(1) =>

logx = kt + log50

.... (I)


56

x log- =kl 50 x = 150, when 1= 1 ..

150 log 50

or

k= log3

= k.I

(2) then gives log (

x

5O) = flog3

we want to find x when I

3

=

X

50

"2

3

=

(3)2

3

X

= 50 (3)2 grams

1.17.7 Example The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the air temperature is 20°C and the body cools for 20 minutes from 140°C to 80°C, find when the temperature will be 35°C. Solution If 8 is the temperature of the body at time '1' then from Newton's law of cooling

-d8 -a(8-20) dl = f~ 8-20

~

de

- =-k(8-20) dl

k rldl + c

J'

log(8 -20) = - kt + c Initially when t = 0, 8 = 140 log(8 - 20) = 0 + c,

.... (I)


57

c = log( 120), (I) reduces to log(140 - 20) = - kt + log 120 or

+kt = log( 120) - 10g(S - 20)

.... (2)

It is given that q = 80 when t = 20 minutes k. 20= logI20-log(80-20) I (120) k= 20 log 60 I k= -log2 20

Substituting in (2) 1 (2 0 IOg2)t= logI20-log(S-20) It is required to find t when

0= 35°c

t=

logI20-log(35 - 20) I -log2 20

0 20 10g( \25 ) log2

10g(8) 2010g23 60 10g2 . t= 20 - - = - - = - - =60mmutes log2 log2 log2

Exercise - 4(k) I.

In a certain reaction, the rate of conversion of a substance at time "t' is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour 60 grams while at the end offour hours 21 grams remain. How many grams of the first substance was there initially? ( ADs: 85 grams approximately I

2.

The rate of growth of a bacteria is proportional to the number present. If initially there were 100 bacteria and the amount doubles in '1' hour, how many bacteria will

1 be there after 2"2 hours. (ADs: 564

I

58


3. Under certain conditions cane sugar in water is converted into dextrose at a rate which is proportional to the amount unconverted at any time. 1f75 grams was there at time t = 0.0 and 8 grams are converted during the first 30 minutes find the amount I converted in 12 hour. [ADS: 21.5 gms]

4. The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the surrounding air is kept at 30°c and the body cools from 80°c to 60°c in 20 minutes. Find the temperature ofthe body after 40 minutes. [ADS: 48°c J

5. If the air is maintained at 30°c and the temperature of the body cools from 80°c to 60°c in 20 minutes. Find the temperature of the body after 40 minutes. [ ADS: 48°c]

6. The rate at which a heated body cools in air is proportional to the difference between the temperature of the body and that of the surrounding air. A body originally at 80° cools down to 60°c in 20 minutes the temperature of the air being 40°c what will be the temperature of the body after 40 minutes from the original temperature. [ADS: 50°c J

7. The rate at which bacteria multiply is proportional to the instantaneous number present. If the original number doubles in 2 hours in how many hours will it triple. [ ADS: 210g3 J

log2 8. Water at temperature 100°C cools in 10 minutes to 80°C in a room of temperature 25°C. Find the temperature of water after 20 minutes. [ ADS: 65.5°c]

9. A cup of coffee at temperature 100°C is placed in a room whose temperature is 15°C and it cools to 60°C in 5 minutes, find its temperature after a further interval of 5 minutes. [ADS: 38.8°c ]


59

Orthogonal Trajectories 1.18.0 Definition (1) A trajectory of a family of curves is a curve cutting all th.e members of the system according to some law. For example a curve cutting a family of curves at a constant angle is a trajectory.

Definition (2) If a curve cuts every member of a given family of curves at right angles, it is called an orthogonal Trajectory. The orthogonal trajectories of a given family of curves _ themselves form a family of Curves. If the two families of curves are such that each member of either family cuts each member of the other family at right angles then the members of one family are known as the orthogonal trajectories of the other. In two dimensional problems in the flow of heat, the curves along which the heat flow takes place and the isothermal curves or loci of points at the same temperature are orthogonal trajectories. In hydrodynamics, the flow of water from a lake into narrow channel produces a family of streamlines which are orthogonal trajectories to the curves of equal Velocity Potential. In the flow of electricity in thin conducting sheets, the paths along which the current flows are the orthogonal trajectories of the equipotential curves and vice - versa.

1.18.1 Orthogonal Trajectories: Cartesian Coordinates

f(x,y,c)=O

Let

..... (1)

be a family of curves, where c is a parameter. We can form a first order differential equation by eliminating 'c' from (1) \

F(X,y,:)=o

i.e.,

..... (2)

is the differential equation whose general solution is (1 ) If the two curves are orthogonal (curves intersecting at right angles) the product of the slopes of the tangents at their point of intersection must be equal to -I. Suppose

(x, y)

trajectory .

is the point of intersection of the curve (I) and its orthogonal

60


At this point slope of the tangent to the curve (I) is dy and -dx is the slope of

dx

dy.

the tangent to the orthogonal trajectory. Therefore on replacing dy by - dx in (2)

dx

dy

, the equation thus obtained is the differential equation of the family of orthogonal trajectories of the family(l) .

. . ¢(

xJI,

-

:)=0

...

(3)

is the differential equation of the system of orthogonal trajectories, and its solution is the fami Iy of orthogonal trajectories of (I) .

1.18.2 Orthogonal Trajectories - Polar Coordinates Suppose

f(r,O,c) = 0

..... (1)

is the family of curves where c is the arbitrary constant. We can form a differential equation

F(r,o, :~ )= 0

..... (2)

of the family (I), after elimination of the constant 'c'. Let ¢ be the angle between the radius vector and the tangent at any point

(r, 0) .

on a member of the family of curves. Then

dO tan¢=rdr

.... (3)

Let ¢I be the angle between the radius vector and the tangent at any point (lj, ~ ) on the trajectory. Then

tan;/, Y'I

dBI = r,·1 _ dr,

..... (4)

I

At a point of intersection of the given curve and the orthogonal trajectory

lj=r,

01 =0

Hence tan¢1

From (3) and (4), lj

dOl dr l

1 dr

= --; dO

1 dr i.e., -; dO

dOl

= -lj dlj

= -cot¢


61

Hence the differential equation of the orthogonal trajectory is obtained by

·. dB fi 1 dr. J dB fi dr Sll bstltutmg - r - or - - , I.e., -r- ordr

r dB

dr

Therefore the differential equation F

dB

(r, B, _r2 ~~) = 0

..... (5)

gives the differential equation of orthogonal trajectory of the family of cllrves(l) Solution of (5) is the orthogonal trajectory of the family of curves.

1.18.3 Example Find the orthogonal trajectory of the family of curves a/

= x 3 , where a is variable

parameter.

Solution: Given family of curves is

a/ = x

3

Differentiating (I) w.r.t. 'x'; 2ay dy

dx

..... ( I)

= 3x 2

..... (2)

Eliminating 'a' from (1) and (2)

. (X3) .y dy _ dx -3x

.. 2 /

2x

2

dy =3y dx

is the differential equation of the family (I). Now replacing :

..... (3)

by - : in(3),

gives the differential equation of the orthogonal trajectory to (I) as -2x dx = 3y

dy

Integrating both sides 2X2

23 2 -2~ = L + c . Therefore 2 2

+ 3/ = c is the equation of orthogonal trajectory of (I)


62

1.18.4 Example Find the orthogonal trajectory of the family of parabolas y2 parameter.

= 4ax

where 'a' is the

Solution:

y2 =4ax

..... (1)

differentiating (I) w.r.t. 'x';

y dy dx

Eliminating 'a' from (I) and (2).

.. y=2x

= 2a

..... (2)

y2 = 2X(Y:)

dy dx

..... (3)

is the differential equation of the family (1) Replacing

dy by _ dx in(3) dx dy

Y=2X( -: J

..... (4)

is the differential equation of the orthogonal trajectory to (I). Integrating (4)

fydy = -2 fxdx + c; 2X2 + y2 = c is the orthogonal trajectory of (I) 1.18.5 Example 2

Find the orthogonal trajectories of ~2 + (

a

2

a +A

= 1 whereA is the parameter.

Solution:

X2

y2

-+ =1 a2 a2 +A Differentiating (I) w.r.t. 'x';

..... (1)

2x2 + 22y dy a a +A dx

=0 ..... (2)


63

)

xxy -2- - - = 1 a a 2 -dy dx

Eliminating A from (I) and (2)

:. ( x

2

-

a 2 ) : = xy

..... (3)

is the differential equation of the family (I) ·· -dx - fOor -dy.111 (3) S ub stltutmg

dy

y

dx

dY~-( x' x a

2

;

Y (2 X -a 2)( -d- ) = xy dx

. I.e,

)dx

is the differential equation of the orthogonal trajectory to (1) 2

Integrating we get

x +l 2

= 2a

2

2

Y 2

X

2

-=a logx---

2+c

log x + C is the orthogonal trajectory of the family of curves( 1)

1.18.6 Example Find

the

orthogonal

trajectories

of

the

family

of

coaxial

circles

x + l + 2gx + C = 0 where g is the parameter. 2

Solution: Given

x 2 + y2 + 2gx + C = 0

..... (1 )

Differentiating (I) w.r.t. 'x' ;

dy 2x+2y-+2g+0=O dx

..... (2)

Eliminating 'g' from (I) and (2)

x + y2 - 2x ( x + y : ) + c = 0

2

y2 _ x 2 _ 2xy dy + c = 0 dx

..... (3)

is the differential equation of the family of (1 ) Substituting· - dx for dy in (3) we get,

dy

dx dx y2 _ x 2 + 2xy - + C = 0 dy

..... (4)

---


64

Which is the differential equation of orthogonal traject{)ry. SimplifYing

dx 1 2 c+ / 2x---x = - - dy y y

..... (5)

is a linear equation of the Bernoulli's form (non - linear differential equation of first order and first degree) Substituting x 2 = t,

2x dx = dt in (5) dy dy

dt t ---= dy

y

(c+ / ) y

..... (6)

is linear in t. I· --dy 1 I :.I.F=e Y =-

Y General solution of(6) is

t.! = y

f +yy2 .!dy+k y C

t

C

y

y

-=-y+-+k.

x 2 + Y - ky -

C

=0

( since t = x 2 )

is the equation of the orthogonal trajectory of (I)

1.18.7 Example Find the equation of the system of orthogonal trajectories of the parabolas

r

=

2a , where 'a' is the parameter. l+cosO

Solution:

2a = r (1 + cos 0) 2a = 2r cos 2 0/2

..... (1)

~ a = rcos 0/2 2

log a = logr + 21ogcosO/2 .Differentiating w.r .to '0 '

o=! dr + 2 (-sinO/2).(l/2)=O i.e, r dO cosO/2


1 dr ---tanB/2 = 0 r dB 2

65

..... (2)

£10 elr. for m (2) we get dr dB

Substituting -r -

1(

- -r-?dB) -tanB/ 2 =0 r dr

dr

-

=

r

-cotB/2dB

..... (3)

is the differential equation of the 0I1hogonai trajectory . Integrating (3)

log r

= -2 log sin B/2 + log c

log(~) = logsin c

.

2

/

(

2

B/2

1- cos B)

-=sm B 2=--'--------'r 2

:. 2c = r (1 - cos D) is the equation of orthogonal trajectory of the fami Iy of ( I ) 1.18.8 Example Find the orthogonal trajectory of

rill

= alii

cosmB where 'a' is the parameter.

Solution: Given

rill = a cosmB ln

..... (1)

mlogr = mloga+ log cos mB Differentiating w.r.to 'B' ;

~ dr r

dB

=

m -dr= -

r dB

1

cosmB

( -msmmB . )

-tan(mB)

is the differential equation of the family of curves (I) Replacing

dr

-

dB

2

dB . m(2) d,.

by -r -

..... (2)

66


~.(_r2 da) = -tan (rna) r

dr

dr = cot (rna)da

..... (3)

r

is the differential equation of the orthogonal trajectory. Integrating (3) 'log r = J.-Iogsin (rna) + loge

rn log rln = log e sin (rna) ; rill lll

= e lll sin (rna) is the orthogonal trajectory of the

family (I) Exercise 4 (s) 1.

Find the orthogonal trajectories of the following family of curves, where a

IS

a

variable parameter.

(ii) (iii)

x 2 -xy+ l =a 2

(iv)

2X2 + y2

(v) (vi)

2.

2

y=ax 2 xy=a 2

(i)

[ADS: x +21 =e] 2

?

[ADS: x - y-

= kx

x 2 _ y2 = a2 X2/3 + y2/3 = a2/3

=e ]

[ADs: (x-y)=e(x+y)2] [ ADs: x

2

= - y2 log ( ;

)]

[ADs: xy = e] [ADs: y4/3 _ X4/3 = e 4/3}

Show that the system of confocal and coaxial Parabolas

l

=

4a (x + a)

is self

orthogonal.

3.

3 Find the orthogonal trajectories of the curves 3xy = x 3 - a , a being the parameter.

4.

Prove that orthogonal trajectories of a system of circles x 2 + y2 - ay = 0 is

x 2 + y2 -bx =0

Ordinary Differential Equations of First Order and First Degree 5.

67

Find the orthogonal trajectories of the following family of curves. -

(i)

r=aB

(ii)

r =e aO

(iii)

rn sinnB = a"

2 (iv) reosB = asin B (v)

r = a(l +eosB)

r = aeosB 2 (vii) r2 = a eos2B (vi)

0' --

= ce 2 I (log)2 + B2 = c I

[ADS: r [ ADS:

[ADS: rll eosnB = e"

I

[ ADS: r = e (3 + eos 2B) ] [ADS: r = c(l-eosB) I [ADS: r [ADS: r ?

= esinO I

= C- Sin 20 ) ?


2 Linear Differential Equations with Constant Coefficients and Laplace Transforms 2.1.1 Let (i) the differential operator" ~" be donoted by 'D'

dx

(ii) PI' P2' P3' ............. Pn be either functions of x or constants (iii) R be a function of x then the general form of a linear differential equation (L.D.E) of order n is given by D nY + PI Dn- I Y + P2 Dn- 2Y + ................. PrJ! -- R

.... (I)

or simply (Dn + ppn-I + ............ +Pn)Y = 0, If PI' P2' ..................., Pn are constants then (I) is called a L D E with constant coefficients. Denoting the differential operator (Dn + PI Dn - I + .............. + Pn) by f (D), (I ) can be written as f(D)y

=

R

2.1.2 Consl·d er the D.E, dy dx + Py = 0

Separating the variables dy = - P(x) dx y

.... (2)

70


Integration yields logy= fp(x)dx => y =Ce -fp(x)dx + logc Solution isy= Ce-

fpdx

Suppose p is a constant = -m, say, then solution is y (i.e.) the solution of (D - m)y = 0 is y = Ce rnx

2.1.3

..... (3)

= Cernx .... (4)

In this chapter the attention is mostly confined to L.D.E.S with constant coefficients. If R = 0 then equation (2) becomes f(O)y

=

0

We shall take, here afterwards, f(m)

= 0 as the auxiliary equation, (A.E).

2.1.4 Consider a second order L.O.E (D2 + a\ D + ~)y

A.E. is m 2 + aim + a2

=0

..... (6)

= o.

Let m\, m 2 be the roots of this equation. Now four cases arise. (i) m\, m2 are real and distinct

(ii) m\, m 2 are real and equal (iii) m\, m2 are complex and distinct (iv) m\, m2 are complex and equal

2.1.5 Case i : (6) can be written as [0 2 - (m\ + m 2 ) D + m\m 2 ] y

Call (0 - m l ) y

=

= 0 or (D - m2 ) (0 - m\) y = 0

.... (7)

y

(7) now becomes (0 - m 2) Y = 0

From (4) it follows that Y = Ce D12X From (7) (D - m\)y = Ce D12X I.F.

=

e-rn\X

Solution is y . e-m\x = C fe(m,-m')x dx + C\

y.e-rn\x=

C m 2 -m!

e(rnrrn\)x dx + CI' Call

.... (8)

C m 2 -m!

as C 2 .

-

Linear Differential Equations with Constant Coefficients ...

(i.e.,) y

=

71

C2 e-1Il 2x + C, elll,x where C, and C 2 are arbitary constants

Similarly ifm" m2, .......... I11n are real and dinstinct roots off(l11) = 0 then y = C, e'n,~ -I- C 2 el11 2x -I- ....•....•..•..•.. + C n elllnx

.... (I)

(I) is the solution of f(O)y = 0 where C" C 2, ........... , C n are arbitrary constants.

2.1.6 (Case ii) : m,

=

m2,

From (8) it follows that the solution of(6) is given by

:. y

= (Cx

+ C,) elll,x

Similarly if m, is repeated say 'r' times then solution of(D - m,Y y

=

0 is

y = (C, + C 2x + C 3x2 + .............................. + c;-r-') elll,x

..... (II)

(II) is the solution corresponding to a root m, repeated r times

2.1.7 Case (iii) : Ill" m2 are complex, say (a ± if3) Then solution of(6) is given by

y

=

A eta ± I~)X + B e(a-

(i.e.,) y =

=

IfJJX

where A and B are arbitrary constants

Ae ax . e Vlx + Beax.e-1jJx = eax [A(cosjJx + isinjJx) + B (cosjJx' - isinjJx)]

eax [C, cos/lr + C 2 sin,lk]

..... (III)

where A + B = C, and i(A - B) = C 2 (III) gives the solution corresponding to two complex conjugate roots (a± ifJ)

2.1.8 Case (iv) : If (a ± ifJ) are repeated say's' times then the corresponding solution (follows from case (ii) and (iii) is given by y = ~[(C, + C 2x + C 3x2 + ........................ + CsXS- ' ) ] cosf3x

+ (d, + d 2x + d 3x2 + ....................... + dsxS- ' ) + sinf3x] Example 2.1.9 Solve

d2 y

dy

dx

dx

- 2 +5-+6y=O

Sol. A.E is m2 + 5m + 6 => (m + 2) (m + 3) = 0 :.m=-2,m=-3. Solution is y

=

Ae-2x -I- B e-3x

.... (IV)


72

Example 2.1.10

d2y

dy

Solve dx 2 -11 dx +30y = 0

Sol. A.E is m2 - 11 m Solution is y =

+ 30 => (m -

Ae 5x +

B

5) (m - 6) = 0

e6x

Example 2.1.11

d2y

dy

Solve2 dx 2 +5 dx -12y=0

Sol. A.E is 2m 2 + 5m - 12 => (2m - 3)(m + 4) = 0 m =3/2, m=-4 3

:. Solution is y = C 1 e2x + c 2 e--4x

Example 2.1.12 d 3y d2y Solve -+4--6y=0 dx3 dx 2

Sol. m3 + 4m2 + m - 6 = 0 (m - I) (m + 2) (m + 3) = 0 :. Solution is y = cl~ + c 2 e-2x + C 3 e-3x

Exercise - 2(a) Solve the following differences:

1.

d 2y dy -+2--3y=0 2

2.

d y _ 3y = 0 dx 2

3.

d 3y d2y dy -+2--5--6y=0 2 3

4.

d2 y dy 9 -+18--16y=0 2

5.

d2y dy -+3-+2y=O 2

dx

dx

2

dx

dx

dx

dx

Ans : y

= A~ + Be-3x

Ans: y =

dx

A~

+ Be-X

Ans : y = Ae-X + Be2x + Ce-3x

dx

dx

Ans : y

= Ae-X + Be-2x

73


Solved Examples Example 2.1.13

dZy dy Solve -z +6-+9y=0 dx dX' Sol. A.E is mZ + 6m + 9 =

°

i.e. (m + 3)z = 0, m = -3,-3 :. Solution is y = e-3x (A + Bx) Example 2.1.14 d3

dZ

dx

dx

Solve~+~=O 2 3 Sol. A.E is m + mZ = 3

°

Z

=> m (m

+ I) =

°

:. m=O,O,-1 Solution is y = (A + Bx)e oX + C e-x = (A + Bx) + Ce-X Example 2.1.15

Solve

d4y

dZy

dx

dx~

- 4 + 181

+ 81y=0

Sol. A.E. is m4 -18m 2 + 81 = :. (mz - 9) (m z - 9) =

°

°

:. m = 3, 3, -3,-3 Solution is y

I.

Solve

=

(A + Bx) e-3x + (C + xD)e 3x = (C(x + Cz)e-3x + (C 3x -t- C 4 )e 3x Exercise 2 (b) d2y

-Z

dx

dy +10- +25y=0 dx

Ans : y = e-5x (Ax + B)

2.

Ans : y

=

(Ax + B)e2x + cex

3.

Ans : y

=

(Ax + B) + (Cx +

4.

Ans : y =

(M + Bx + C) e-t

D)~

74


Solved Examples Example 2.1.16 Solve

d2 y -2

dx

dy

+4- +9y=0 dx

Sol. A.E. is m2 + 4m + 9 =0

m=-2±i/5 a

=

-2 and

f3

/5

=

:. Solution is y

=

e-2x (Acos x

/5 + Bsin /5 x)

Example 2.1.17 d 3y Solve dx 3 + y= 0

Sol. A.E. is m3 + 1 = 0 :. (m + 1)( m2 - m + I) = 0

1+J3i :.m=-I,m=-2

Example 2.1.18

d4 y

d3 y

d2 y

dy

Solve -+2-+3-+2-+y=0 dx 4 dx' dx 2 dx Sol: A.E. is m4 + 2m 3 + 3m 2 + 2m + 1 = 0 (m 2 + m + 1)2 = 0

:. m=

-1±J3i 2

-1±J3i 2

75


Exercise 2 (c) I.

2.

Ans: y

d 2y dy Solve22 +4- +3y=O dx dx

= Acos

J7x

+ Bsin

J7x

x x Ans : y = e-x (Acos.fi + Bsin .fi )

3.

4.

d4 y Solve dx4 -64y = 0 Ans : y

=

C\e2x + C 2e-2x + e-x (C 3cosfix+c 4 sin fix) + ~(C3cos.J3x+C6sinfix)

2.2.1 Consider the O.E.f{O) y

=

R

.... (I)

Operating both sides with f(IO) (called inverse operator)

1 we have f(O) [f(O)]y

=

I f(O) R

I

or y = f(O) R This solution is called the particular Integral of(l) while the solution off(O) y = 0 is called the complementary function (C.F.) Suppose y\ is C.F and Y2 is P.I for (I) Thenf{O) y\ andf{O) Y2

= 0 from (2)

= R from (1)

Hencef(O) [y\ + Y2]

=

f(0) y\ +f(O) Y2

=O+R=R

..... (2)

76


Which shows thatYI + Y2 is also a solution of(l)'Yl + Y2 (i.e.) C.F. + P.I is called the most general solution of(l).

2.2.2 Calculation of

I D-I11,

R

D _ nl, R = y, say, Operating both sides with (0-m 1), we get (0-m 1) Y = R

dy

(i.e.)

dx - m1y= R •

f-mjtU

I.F.lse =e-ml x :. Solution is given by

y.e-mf = fRe-lIl/ dx+c (i.e) y = Ce-ml X + ell/IX fRe-1Il IX dx = C.r. + P.I P.I.=e mI X fH£lIllx dx

m

If

I

=O~ '0

R= fRdx

I Thus the operator 0 stands for integration

I Find the value of - D x

Example 2.2.3

~

Sol. D 5

-5

Fe" fe".xdx ~ <' 0" [ -

+('

m} e'+ xe;" - ~;'l

x I ----5

25

Example 2.2.4

1

Sol.

I eX 02 _50+6 = (D-2){D-3) .(eX)

f

I I (eX) I 3x -3x x dx = (0-2) . (0-3) = (0-2) e e.e

=

I 3x (0-2) . e

2

f

I -I - (eX) = - -I e 2x Je r -2x .exdx = -e I 2 x e-xdx=-e I 2 x.e- t 2 (0-2) 2 2 2

• 2


77

Example 2.2.5

Solve (02+a 2) y

=

tan ax

Sol. A.E is m2 + a2 = 0 => m = ±ai C.F. is y = C I cos ax + C 2sin ax

I

I - I- - -I tan ax = (0+ ai)(O- ai) 2ai [ O-ai D+ ai

P.I =

2ai [e lax J e

r

-a/\

tan axdx - e -IllX J eat" tan axdxJ 2

alX

Je - tanaxdx

=

J. . sinax (l-cos ax) (cos ax -Ismax) - - dx = JSin ax- i dx cos ax cos ax cosax

= - -- -

a

lilly

II

J tanax

=

Je

an

tanax

dx

i isin ax -\og(sec at' + tan ax) + - a a

cosax i isinax - - - + - \og(sec ax + tan ax) - - a a a

=

(Cosax+aisinax) [-cos ax + isin ax -ilog (sec ax + tan ax]

a

[-\- icos ax log(sec ax + tan ax) + sin ax \og(sec ax + tan ax)] \

:. P.l. = - 2 ? [-2icos ax log(sec ax + tan ax)]

a-,

cos ax - - - - log(sec ax + tan ax) a :. Most general solution is y

.

= C 1 cos ax + C 2 SIl1

ax -

cos ax log(sec ax + tan ax) a

- - 2-

78


Example 2.2.6 Solve (02-50+6)y = ~

Sol. A.E. is m2-5m+6 = 0 C.F. is y

= C)e 2x + C2 e3x eX

I

P.1. isy= 0 2 -50+6 (eX)

=

2

:. Complete solution (C.S.) is y

(from 5.2.4)

= C.F. + P.I.

Exercise 2 (d) I

I I. 0 cosx

3. 0 (x 2 )

I 4. 0-2 sinx

Ans: (I)+sillx (2) e-x (3)

x3

3

(4)

(2sinx+ cos x)

Solve the following: 6.

d2y dx 2

7.

-2

8.

9.

10.

d2 y dx

-

+y=5x+3

d2 y dx 2

dy 4 dx + 3y = e2x

-

dy 3 dx + 2y = e-2x

5

79


xsin ax Ans : y = C I cos ax + C 2sin ax + C 2sin ax + - - -

II.

a

cos ax

+ 12.

(D 2 + 9) Y = tan 3x

Ans : y

=

-2-

a

log cos ax

C1cos 3x + C 2sin 3x

cos3x - - - log(sec 3x + tan 3x)

9

Methods of finding P.1. We shall now consider the methods for finding P.1. where R is of some special form

2.3.1 Particular integral Where R is of the form eax Case (i) iff(a) Since

=F

0

De ax = aeax ; D2e ar = a2 ear ...... Dn(eax) = an.ear j{D) ear

(Dn + K1D n- 1 + K 2 Dn- 2 + ...... K,)e ax

=

= (an + K1a n- 1 + .......... + Kn)e an = j{a) ear Operating with _I- on both sides

feD) - I

feD)

[f(D)e a,]

[f(a)elU]

1 f(a) eax = f(a). [ feD) I e ax] feD)

(i.e)

eax

(or)

1 a, --e feD)

=

I f D

= - (-)

=

1 at --e f(a)

1 1 1 Note: If K is a constant then feD) (K) = feD) K.eax = f(m)·K

for example -D-:02-_-2-D-+-l e

3x

I 3 I 3 9 _ 6 + I = e x = "4 e x

80


2.3.2 Case ii) If./{a) = 0 then it is possible to write./{O) as <1>(0) (O-ay where (a) i:- 0 Suppose r = 1 then

1

I

1

P.1. = - - eax = - - - f(O) O-a <1>(0)

e(lX

1 1 1 - - eax = - - - - eax O-a $(a) $(a) O-a = (la) eax

Je~a\ .eaxdx = ~a) ea\dx = ~a) ell<.x

1 1 I 1 ax ax ifr=2then f(O) e = (0-a)2 <1>(0) e = (a) (0-a)2 ea~ 1 1 - - [xeax] $(a) O-a P.1. =

1 2 1 ax- = - - e ax- -X e f(O) ( a) . L2

In general, if (O-a) is repeated 'r' times then

1 I x2 P.1. = f ( D) eax = $( a) ea~. --;:! Solved Examples Example 2.3.3 Solve

Sol:

d 2y dy + 4 - + 5y = 13e3x d~2 dx

(0 2 + 40 + 5) = 13e3x A.E. is m2 + 4m + 5 = 0, giving m = -2 ± i C.F. is y = e-2x (Acosx + Bsinx)

I PI (13e3x) .. = 0 2 +40+5 replace '0' by '3' 13e3x

13e3x

eh

9+ 12+5

26

2

.... (I)


81

h

:. Complete solution is y = C.F. + P.1. = e-2X(Acosx + Bsinx) +

e 2

Example 2.3.4

d2 y

dy

J.

dy

Solve dx 2 + 4 dx + 5y = 2e-- x , gIven that x = 0, y = I and dx =-2 Sol. Given equation isJ(O)y = 2e-2x were f(O) = 0 2 + 40 + 5

:. A.E. is m2 + 4m + 5 = 0, giving m = -2±i :. C.F. is y = e-2x (Acosx+Bsinx) I P.l. = 0 2 +40+5

e 2x 2-

2e- 2 , 2x = 4-8+5 = 2e-

:. Complete solution is y = C.F. + P.1. = e-2x(Acosx + Bsinx) + 2e-2x Now yeO) = I, (given) Substituting in (4) yeO) = I(A) + 2; A =-\. yl(x) = e-2x(-Asinx + Bcosx) -2e-2x (Acosx + B sinx) -4e-2x

yl(O) = B-2A -4, yl(O) = -2 (given)

:. -2 = B-2A-4, => B = 0 :. y(x) = e-2x (-cosx) + 2e-2x

= 2e-2x - e-2x cosx Example 2.3.5

Solve Sol.

d3 y d2 y dy - 3 + 2 -2 + - =e 2x dx dx dx A.E. is m3 + 2m2 + m = 0 ; m(m+ 1)2 = 0

m = 0, -I, -I. C.F. = C1eOx + (C 2 +C 3x)e-X I e 2x 2x P.l. = 03 + 202 + 0 e = 18

.... (4)

82


Example 2.3.6_

d2 y

dy

dx-

dx

- , -3 -

Solve

+2y=e-~

f(O) = 0 2-30+2

Sol.

A.E. is m2 _ 3m + 2 = 0 => (m - 2) (m - I) = 0; m = 1,2 here f(l) I

I

I

I

=

X

0 I

.. (0-1) (0-2) eX= (0-1) (1-2) e = (0-1) eX=-eX

~

(using 5.3.2.(3)

:. Most general solution is :. y

= CteX + C2 e2x - xe n

2.4.1 R is of the form sin ax or cos a\" 0 2 (sin ax) =:' (~a2i~ sin ax

(02f (sin ax) = (_a2)2 sin ax

(0 2)3 (sin ax)

=

(_a2)3 sin ax

(02)n (sin ax) = (_a2)n sin ax

Hence j(02) sin ax = j(-a2) sin ax, provided f( -a2)

1

2'

1 (0 ) 2

Qr sin ax =

j(O )

1(02 )

j(-a

I.

1(02 )

SIl1

_

SIl1

2

)

ax -

1

:;i:

0

1 2 . (,)j(-a )SIl1 ax 0-

sin ax sinax

ax

=

f(-a 2 )

Similarly it can be shown that

I 2 cos ax ax = I, cos ax f(-O ) 1(-a-)


83

Ifj{O) vanishes when 0 2 is replaced by _a2 then we write

I

1

j(O)

[sin ax] =

. 1ar [Imaginary part of e ]

j(O)

1 . = I.P of - - e 1ar

j(D)

1

Similarly

1

j(O)

[cosar] = R.P. of

j(O)

.

e 'ar

Solved Examples Example 2.4.2

Solve Sol.

d2 y

-

dx

2

dy

-3 -

dx

.

+ 2y = sm3x

A.E. is m2 -3m+2 = 0; m = 1,2

I. 1 . -1. (3 D - 7) . P.I., (sm3x)= 9 30 2 (sm3x)= (3 7 (sm3x)= 9D 2 49 (sm3x) 0- - 3D + 2 - + D + .) +

(3D - 7 ) . 9cos3x + 7 sin 3x = -81-49 (sm3x)=--0-2=-_-3D-+-2Hence complete solution is


84

Example 2.4.3 Solve (D3 -D) Y = sillX

Sol.

A.E. is m 3 -m

=0

m=O, m =±1

c2eX + C3e-x =

C.F. is C1e°.x +

C 1 + C2e x + C3e-x

I.Sit1X I I P.1. = D3 _ D = D2 _ 1 . D (sitlX) I -co·sx x = - - - . cosx= = --SillX D2_1 -1-1 2

The complete solution is

y = C + C eX + C e-x -~SillX I 2 3 2 Example 2.4.4 Solve (D2 - D + 1) y = cos2x

1±J3i 2

Sol. A.E. is m2 - m + 1 = 0 ~ m = ~~

fi

fi

2

2

:. C.F. = e 2 (Acos-x + B sin- x) I P.I. = D2 _ D + 1 cos2x =

(3 - D )cos2x 13

x

C.F. + P.1. = e 2

=-

I

- 4- D+1

cos2x =

1

-3 - D

9 - D2

I

1

fi BSIl1-x . fi ( Acos-x+ 2 2

Solve D2(D2+9) = sin2x + 5 A.E. is m2(m 2+9) = 0 ~ m = 0, 0, + 3i C.F.

(3-D)

13 (2sin2x + 3cos2x)

Example 2.4.5

Sol.

cos2x = -

= (C I + C2x) + C3 cos3x + C4sin3x

1 (2sin2x + 3cos2x) _13

cos2x

85


P.1.

D2 (D2 + 9») sin2x + (D2 + 9) D2

=

+ ~._I_(I) -4(-4+9z) 9 D2 sin 2x

= _ sin 2x 20

2

+ 5x 18

Example 2.4.6 Solve

Sol.

(D 4 -2D 3 + 2D2 - 2D + 1 ) y

= eX + sin2(x/2)

A.E. is f(D) = D4 - 2D 3 + 2D2 - 2D + 1 m4 -2m 3 + 2m2 - 2m + 1 = 0

= (m-l )2 (m2+1) = 0 :. m= I, I,m=i,-i C.F. is (C t +C 2x)eX + (C 3cosx + C 4sinx) 1 1 P.1. = - - eX + - - sin2(x/2) f(D) f(D)

_1_ f(D)

~~ 1 2

= -

1

=

. Sin

2

_

(x/2) -

1

[(0_1),1(0' + I) (I-COSX)] -

1 1 (cosx) - 4D D2 + 1

--.--

lID (D2) (cosx)

"2 + "4 . D2 + 1 .

= ~+~._l_.~(-sinx) 2

1

=

4 D2 + 1 (-I) 1

1

.

2

"2 + "4 . D2 + 1 . sinx

1

_

(? ) Sin (x/2) (D-It D-+l 0

?

(2

(D-It D +1

~ ~ [-I -20~ (0'

)

1)(0' + I) COS X

1

86


[)2

1 1 1 + 1 sinx = I.P of 02 + 1 ex = I.P of (D +-i)---:(-D----:-i)

I

= I.P of 2i

:. P.1.

=

ex

x

XCiX

= "2 I.P. of -i[cosx + isinx] = -~COSX

1 1 "2 - "8 x cos.\"

:. Ilence the complete solution is 2

Y =(C +C I

x)~+C

2

1 x r 1 x cosx+C sinx+ -.-e +---cosx 3' 4 2 2! 2 8 Exercise 2(e)

Solve the following (1)

(2)

(D2+D+ l)y = sin2\" Ans:

J3

J X )] e"r ( C,cos-x+ C o' S IJj / 1 - X -1 - ('J .... cos 2x+ 3' SIll '.... 2 2 13 3 x sin"2 x.sin "2

(3)

(D6 + 1) y

Ans.

(C1COSX + C 2sinx) + e- 2 (C3cos"2 +C 4 SIIl"2)+ e2- (Cscos"2 +C 6SIIl"2)

(4)

x 1 + -sinx+ -cos2x 12 126 (D3 + 4Di y = Sin2x

Ans.

y

=

jj-,

=

x

xsinx C 1 + C2 cos2x + C]sin2x --8-

I 2.5.1 P.1. is of form feD) xm

I - - xm feD)

=

[f(D)rl.x m

. x

cos2x

-16

.fh

x

. x

87


Now expand [f(0)r 1 in ascending powers of 0 and retain as far as 0 111 and then operate on xl11

Solved Examples Example 2.5.2 Solve (0 2 + 50 + 6) y = x

Sol. A.E. is m2 + 5m + 6 = 0 => m = -2,-3 C.F. is C 1e- 2x + C 2e- 3x

P.1.

=

1 0 2 + 5D + 6 x

1 (I

="6

2

0 + 50)_1 + 8 x

I 50 I 5 [1--] x= - [x--] 6 6 6 6

= -

x

= -

5 36

---

6 x

5

6

36

Solution is therefore)' = C e- 2x + C e-3x + - - I

2

Example 2.5.3 Solve

d3 v dy _ . - 3 - -2y=x2 dx' dx

Sol. A.E is m3 - 3m -2

=

0 => (m-2) (m+ 1)2

=

0

C.F. is C 1e 2x + (C 2 + C3x)e-X 2

P.1.

=

x 0- -30-2 ,

3

2

(1_(D -3D»1,.2 2 .,

0 3 -3D 02 -30 , ., [1+ +( t+ ....... ]x222 1

=--

_ -I 3D 9 2 2 _ 1 2 6x 9 _ I 2 - - [ I - - + - D ] x ---[x --+-] - - [2x -6x+9] 224 2 224

Complete Solution is y = C 1e 2x + (C 2 + C 3x)e-X

Exercise 2(f) I.

Solve (03 - 302 +2)y = x

-

~ (2x 2 4

6x

+ 9)

88


2.

Solve (D2 + 2D + 3)y = x +.~ Ans. y

3.

Solve (D2 + 2D + I) Y

=

(D2 - 4D + 4) y

=

e-x (C I cos fix + C2 sin J;:) +

14-6x

=

(y + C e-> -~(3COS2X 2

2sin 2x) + (x 2 - 2x)

x 2 + sin2x + e3t"

Ans. y

=

(C I +

, C2x)c~X

2X2 + X + 3 cos 2x 3 + - - 8-8- + e x

2.6.1 I

To find - - (e
c
=

ea\"(D+a)YI

1

Dn(elU"Y I)

=

feD) (ea\v I) Calling

ea\"(D+a)nYI =

e
f(D+a)Y I = Y, we get Y I = f(D+a) Y feD) rea,

I

f(D+a)

_I_ elU" Y

feD)

=

elU"

Y]

=

elU"Y or elU"

I

I

Y - - - elU"Y f(D+a) - feD)

I Y f(D+a)

2.6.2 Note

1.

27

cos2x + x 2

Ans. y 4.

=

While finding _1- elU" when f(a) = 0 we can employ the above method for feD)

Linear Differential Equations with Constant Coefficients

2.

Further _1_,_ [cos ax or sin ax] when f( -a2) /(0-)

=

89

000

0 we can write

_1_2- [cos ax or sin at] f(O)

1 f(02) [R.P. or I.P of eaT]

=

=

\ R.P. or I.P. of e1a\". f(O + ia)2 .\

Solved Examples 2

dy dx

dy - 4y = tfcosx dx A.E. is m2 - 2m + 4 = 0

2.6.3 Solve: -2 - 2 Sol.

m=

\

± fJi

C.F. is eX(C I cosfJx+CzSinfJx) P.1.

=

1

0 2 -20+4

tf cosx

= eX?

1

(0+\)"-2(0+\)+4

cosx

\ =

tf 02 + 20 + 1- 20 _ 2 + 4 (cosx) ~X

I ) x cosx eo' cosx (cosx = e . =--0 +3 _12 + 3 2 :. Complete solution is y = CoF. + P.1.

=

y

=

t;

- -

2

eX (C I cosfJx + C 2sin fJx) +

e'cosx 2

Example 2.6.4

Solve (02-2)y Sol.

=

tfcosx

A.E. is m2 - 2

=0 m=±

J2

C .F, is C Ie.fix + C 2 e~.fix 1

P.1.

=

(0 2

=

tf.

=

tf .

_

2) tf cosx = tf. (0 + 1/ _ 2 cosx

20-2

=

1 tf. 0 2 + 20-1 . cosx

. cosx

(20+2) eX(-2sinx+2cosx) cosx = 40 2 -4 -4-4

eX

.

= -(Sin X -

4

cosx)


90

eX +C e-- J2 • + -(sinx-cosx) Y = CeJ2x I 2 4 Example 2.6.5 3

d d ---.E. _7~ - 6y = e 2x + xe 2x 3 dx dx

Solve

A.E.ism 3 -7m-6=O:. m=-1,3,-2

Sol.

:.C.F. P.1. =

Cle-x + C2e3x + C3e-2 t"

=

(1+x) 3

o -70-6

e

2x

=e2x

1 (I+x) (0+2)3 -7(0+2)-6

= e2x

~-----3 2

0 -60 +50-12

_e2x 12

_e 2x

= -

12

50 _e 2x 5 (1+-)(1 +x)= [I+x+-] 12 12 12 17 [x+-] 12

3x 2x x Y = C I e- + C2e + C3 e- -

e 2x

-

12

17 [x + - ] 12

Example 2.6.6

Solve Sol.

d3y d2y dy - -3 3 - + 3 - - y = eX(1 +x) 2

dx

dx

dx

A.E. is m3 - 3m 2 + 3m - 1 = 0 or (m - 1)3 = 0 or m = I, I, I. C.F. - (C I

xl + C2 x + C3 )eX


P.1.

=

91

(0 _ 1)3 . (x + I) ~

e'.(x+l) 1 .1 x2 = (0+1-1)3 =~. D3 (x+ 1)=tf. 02 (T+

Complete solution is y = (C 1 x + C~ + C3) 2

X

)

x3 ~ + ~ (24 + (;) X4

Example 2.6.7

Solve

d~y

dy

- , - 4- +y = e2x sin2x de dx

A.E. is m2 - 4m + I = 0, m = 2 ±

Sol.

c.F.

= C,e(2+J3)(

+ C e(2 z

J3

,)1)(

I eh PI e2x sin2x = sin2x .. - OZ-40+1 (D+2)2-4(0+2)+1

e 2x 0 -3 Complete solution is

= - sin2x = 2

e 2x sin2x -4-3

1"

= --

7

eX sin2x

Exercise 2(g) I. 3x

Ans : y = (C 1 + C 2x) ~ + 2.

8e

(al -

4x + 3)

Solve (03 - 70 - 6) Y + e2x + x 2e3x 1 2 +-x+-) 5 169 Ans:y= C 1e-x +C 2e- 2x+C 3e 3x_e 2X_(x 12 6 72

3.

Solve (0 2 - 20 + 4)y = eXcos 2x Ans: y =

~(C, cos3Fx + C

2

sin3Fx)+le

x

-~sin 2x

92


4. Ans. y = Clcosx + C2 sinx -~ ~(2cosx - sinx) 5 5.

13 C3sm-x ' 13) - - 1 e 2x.(11cosx-3slnx) . Ans.y= C le- x +e7i (C 2COS-X+ 2 2 130 2.7.1 To find

I~O) [xv] where v is a function ofx

O(xv l) = xOv l + vI 02(xv l) = O(xDvl + vI) = x02v I + 20v I = x0 2v I + (0 2) vI Similarly 03(xv l ) = x03v I + (03) vI in general on((xv l ) = xOnv l + (on) vI (here 1 indicates differentiation w.r.t '0') ..... (1)

Thus, f(O)(xv l ) =o.x(O)vl + fl(O)vl now let f(O)vl = v

1 vI = I(O)'v From(1) j(0)

1

1

[x'/(D) v]=xv + fl(O)[/(O) v]

Operating both sides with I/O) _1_ _ _ 1_

__1_

I

_I_

x /(0) v - 1(0) (xv) + I(O)·f (0) [/(0) .v]

Linear Differential Equations with Constant Coefficients"""

93

2.7.2. Note: For finding [xlll.(sinax or cosax)] when m > 1 the above method becomes very lengthy. Hence we can take it as [Xlll. (R.P or I.P of e/a,"] and apply the method (2.6.1 ). Solved Examples Example 2.7.3 Solve (D 2 + 9) y = xsinx

Sol. A.E. is m2 + 9 = 0 =>m

=

± 3;

1 .) 1 I . -2-(XSIllX = {x--,-.2D}.-,-.sll1x D +9 D-+9 D-+9 [using the result of2.7.1. v = sitlX] P. I.

="

1 I. x. 1 2 = {x--,-2D} . - Stox= -Stox- - - - cosx D" + 9 8 8 D2 + 9 8

=

x. 2 Stox - 8"

8"

Complete solution is y

x. 1 SltlX - 32 cosx

1

. 8" . cosx = 8"

= C1cos3x + C 2sin3x + ~ sinx 8

_I cosx 32

Example 2.7.4

d2 y Find the solution of - , + 9y = xcos2x dxSol.

A.E. is m2 + 9 = 0 => m =

and

± ; Jj

1 P.1. = D2 + 9 (xcos2x)

1

= Real part of D2 +9 x(cos2x + isin2x) = R.P. of -1- xe 2"IX = R.P. ofe2"IX D2 +9

=

R.Pof -e

21x

1+ D2 + 4iD )~t x= R.P. -e 5 5

21x

(

5

1 2" 1 x = R.P. ofe IX x (D+2i)2 +9 D2 +4iD+5 (

1- D2 + 4iD) x 5


94

4iO 0 2 R.P. of -5- [I ----+----]x . 5 5 e

=

2lX

=

e 2lX 4i R.P. of -5- [x- 5]

=

5 [xcos2x + 5 sin2xJ

I

4

Most general solution is

Y

=

C ,cos3x +

c 2sIn. 3x +

. 2X ] -1 [xcos2x + -4 Sin

5

5

Example 2.7.5 Solve (0 2 - 40 + 4)y Sol.

A.E is m 2 - 4111 + 4

=

=0

8e2x x2sin2x ~ m

= 2, 2

C.F. is (C,x + C 2)e2x

·')s,·n2x - 8e2x 1 P.. I -- (0 -8 2)2 e2x . x(0 + 2 _

= 8e2x [I.P of

~2 x 2 e 2ix ] = 8e2x[I.P. of e2ix .

2i

x2sin2x = 8e2x _1_ ') sin2x 0 2 x-

1

x2 ]

(0-2i)

- 8 2x[I P f 2ix 1 x2 ] - e . 0 e -4+4iO+02

= 8e2x [I.P

4iO+02 · -1 of e 2IX (4) [1 4

= -2e2x[I.P.

of e 21x 1-

= -2e2x

.

4iO+ 0 2 4

[J.P. of e 2ix [1 + 4iD + 0

r 2

l

r

I,)

x-

x2

2 + (4iO + 0 )2]x2

4

4

=_2e2x [J.P ofe2 ix [x 2 + 8ix+2 + 16i

4

2 02

(x 2 )]

4

2.<

= -

~ [J.P. of (cos2x + isin2x) (4x 2 + 8ix - 30) 2

-e 2.<

= -2-

.J.P. of [4x2cos2x + 8ixcos2x - 30cos2x + i4x2sin2x - 8xsin2x - 30isin2x] .

= -e2x[4xcos2x + 2x2sin2x -

ISsin2x]

95


General solution is y

Y

=

=

C.F. + P.1.

(C IX + C 2)e 2\" - e2~[2x2sin2x + 4xeos2x - 15sin2x]

Exercise 2(h) 1.

Solve (0 2 -1) Y = x 2eos3x Ans. y = Cle-' + C2e- x

2.

=

Cleos2x + C 2sin2x +

=

I

X

"4 ["2

+ sin2x]

I I 3 (Clx + C 2 )eosx + (C 3x + C 4 ) sinx - 48 (x 4-9x2) cosx + x sinx

12

Solve (0 2 - 20 + 1)y = xe-'"sinx Ans. Clcos3x + C 2sin3x +

5.

26 6 50 eos3x - "5 xsin3x)

Solve (04 + 202 + 1)y = x 2eosx Ans. y

4.

10 (x 2eos3x -

Solve (0 2 +4)y = xsin 2x Ans. y

3.

1

-

I

4

"5 (xcos2x + "5 sin2x)J

Solve (0 2 - 60 + 13)y = 8e3xsin2x Ans. y = e3x[Clcos2x + C 2sin2x] - 2e3x xcos2x

6.

Solve (04

-::::

xfj

+ 0 2 + 1)y = e 2 cos (--) 2

-x

1 e2 1 1 1 + C4 sin(-xfj)] + [xcos(-xfj) + r::; xsin(-x.J3)] 2 4 2 -..13 2 7.

d2 y

Solve - , + Y = e-'"sinx dx-

Ans. Clcosx + C 2sinx -

I

"5 e-'" (2cosx -

sinx)

96


2.8.1 Cauchy's Homogenious Linear equation Most general form of the homogeneous equation is

(xnon + K1x ll- 1 on-l + ..... + Kn- 1 xO +

=X

.... (I)

Where kl' k2' ........ kn are constants and 'X' is either a constant or a function of x To solve such equations it is convenient to transform them into linear equations with constant coefficients with the substitution x = eZ i.e. z = logx.

dy d\:

= dy

dz dz' dr

=

= ~ dy

.... (2)

x dz

I 'dy 1 d 2 y -l----t----] x x dz X dz2

dy ( .: dz

= x)

__ , d2y dy - 2(d 2 - d )

z

x

..... (3 )

z

..... (4)

d

By taking '0' to stand for dz' dy (2), (3), (4) ~ x dx

=

dy dz

= Oy

..... (5)

Xll

dn dx"

-Z = 0(0-1) (D-2) ......... (O-n+l)y

By substituting (5) in the given equation (') it transforms into a linear differential equation with constant coefficients.

97


Solved Examples 2

2.8.2 Solve x 2

d y

m"2

dy -x d~" + 4y = 0

Sol. Writing x = el.

(i) reduces to

d

[D(D-I) -D + 4tv = 0, where D = -

dz

(D2_2D+4)y=00

A.E. is m2 - 2m + 4

=

:. C.F. is given by y

=

substituting

0 :::) m = 1 ± fi i

(C 1cos,,'3z+C

2 sinfiz)e

I

z = logt,

Y= Clcos (J3logx) + C~sin (J310gx) e logx y = [C ICOS (fi log x) + C 2 sin (fi logx )lx 2.8.3 Solve

d 2y

dv

x2_~ -

3x" - 4y dx

dx-

Sol. Writing x = e Z

2x2

=

..... (i)

(i) reduces to

[D(D-I) + 3D (D 2 - D

+ 3D -

4lY =

2e 2/ ,

4)y = 2e 21.

A.E. ism 2 +2m-4=0 ~m=-1 C.F. is y

=

(C1e

(I.JS)L

2

±J5

+ C 2e

")_

(h'5)L)

2

P.1. = -D-2-+-2-D - 4 (e--) = 4 + 4 _ 4 e Most general solution is

y

= (

C

e-(Il /"S)Z

1

+C

2

e (I JS)7

+ e~z

1

2_ - =

2e 2z e 2z -4- = 2

98


2

2.8.4 Solve.xl Sol.

d --?2y = x d\'

1

2

+-

X

Taking x = eZ equation (i) reduces to [D(0-1)-21v

=

A.E. is m2-m-2

:. C.F. is y

e 2x + e-Z =

(D-2)(D+l)

e2z

1 ~_ ---e-- 3(D-2) =e2z

=

3

I (D-2)(D+l)

e--Z

=

1 (D-2)(2+1)

__ e-

3(D+I)

I (I) e-3 (I) 3(D+2-2) 3(D-I+l)

3

1

+

I

.!. e2z . .!. (I) - .!.

= -

= 2,-1

C l e2x + C2e-z

=

1

P.1. =

0 ~m

D

_

e-z

3

.!. (1 ) D

I

ze2x - - ze-z 3

Complete solution C.F + P.1. I.e.

Y

=

1

C e2z + C e-= + - z( e2x I 2 3

-- e-Z )

I

Y = C1 e210gx 2 + C e-Iogx + - (e2logx

3

Y

= C

I

-

e-1ogX)logx

1 1 1 x 2 + C - + - (x 2 - - )Iogx 2 x 3 x

Sol. Taking x = ez the equation reduces to [D(O-I) (0-2) + 2D(0-1) +21v = 10ez + IOe-z

(D 3-D 2+2)y

=

IO(ez+e-Z )

e2z

+

1 (-1-2)(D+I)

e- Z


A.E. is m3-m 2+2

=

99

0

=> (m+ I) (m 2 -2m+2) = 0 (i.e.) m =-1, m = :. C.F. is y

I± i

C1e-z +e-'(C 2cosz + C 2sinz)

=

10

P.1.

=

J

,

D' - D- +2

IO z e= + D3 _ D2 + 2 e-

10e- z + - - - - :2 : - - - 1-1+2 (D + 1)(D - 2D + 2) 10e:

5e-'z +

=

1 5e-z + 2e-z D (I) = 5e-z + 2e-z . z

Complete solution =

10 e -L = 5e -L + 2e -L . - -I - - I 5( D + I) (D - I) + 1

=

= C.F. + P.1.

C1e-z + e=(C 2cosz + C sinz) + 5e-z + 2e-z . z logx 3

C1

= -

x

x

•

+ x(C 2cos(logx) + C3sm(logx)) + 5x + 2

d 2y

dy

.

2.8.6 Solve x 2 dx 2 + x dx + Y = logx sm(logx) Sol. Writing z = logx the equation reduces to [D(D-I) + D+ IlY = zsinz (D 2 + l)y = zsinz A.E. is m2 + 1 = 0 => m

C.F. is y

P.I.

= ±i

= C1cosz + C2 sinz 1

= -2-

o +1

•

zsmz 1

ze 'Z

=

I.P. of

=

I.P. of e'z ---:-,----z

-2-

o

+1

=

J.P. of e'z

1 0- -1+20i+l

1 (0+i)2 +1

z

100


=

I.P. of e=

I

Z

D2 +2Di

0)'-1

=

1 ( I P of e1z 1+ . 2Di 2i

=

I.P. of e1z

_I_(z_~) 2Di

2i

_e .i(z2 IZ

=IPof-- - - -Iz .

2

2

= Imaginary part of Z2

= - - cosz

2

Z

J

2i

[2 .(cosz + -I

I.

-- ZSInZ = 4

z. 4

- - (SInZ

. .

[22 + 2.J] Z

ISInZ)

U

+ 2zcosz)

Hence complete solution is

=

C,cos(logx) + C2sin(logx) -

lo;X

[sin(logx) +2(logx) cos(logx)]

Example 2.8.7 Solve(x2 02-xO+4)y = cos(logx) + xsin(logx)

Sol. Taking x = eZ or Z = logx the equation reduces to [0(0-1) -0 + 4lY = cosz + eZ sinz.

A.E. is m2 - 2m + 4 = 0 => m =1±J3i C.F.

=ez (C,cosJ]z+C 2sinJ]z)

P.I. =

I 2

D -2D+4

COSZ

+

2

I . Z

D -2D+4

e smz

101


1 3-20

--.cosz+

=

e

•

(D+I)--2(0+1)+4

smz

(3+20)

eZsinz

3cosz-2sinz

9+4

3

13

---cosz+ - - = The complete solution is y

=

L

1

=

eZsinz

+-3

C.F. + P.1.

~

.

x[C1cos J3 (Iogx) + C2Slll v(logx) ] +

1

.

13 [3cos(logx) -2sll1(1ogx)] +

xsin{logx) 3

Example 2.8.8 Legendre's Linear equation: An equation of the form d ll

dll-1y

(ax+b)11 dx~ + K1(ax+b)n-l dx n - I + ...... + koY

=

X

Where K's are constants and X is either a constant or a function of x is called Legendre's linear equation. To solve such an equation we transform it into linear equation with constant coeficients by the substitution. ax + b = eZ or z = log (ax +b) Now

dy dx

=

= _a_

dy dz dz'dx dy dx

(ax + b)2 -

I

Similarly (ax + b)3

dy ax+b'dz dy dz

= a-

3

d ---f dx

=

= aOy

d

where 0 = dz

a3 D(O-l) (0-2)y

by substituting these values the given equation reduces to linear equation with constant coefficients.


102

Solved Examples Example 2.8.9 Solve

d 2y dy (J+x)2 dx 2 + (I +x) dx + Y

.

= sm2[log(l +x)]

.... (I)

I +x = e => log( I +x) = Z

Sol. Write

dy

(I+x) dx

=

Dy, (I+x)2

d2y -?

dx-

= D(O-l)y where D =

Substituting these values (1) reduces to (D2_D+D+ I)y = sin2z; (i.e) (D2+1)y = sin2z A.E. is m2 + I C.F.

=

0 => m =

±i

= C ICOSZ + C2sinz 1 0 +1

P.1. = -2 - sin2z =

sin 2z

-~4+1

sin 2z

= -~3

· ·IS = C .F + PIC sin32z C omp Iete so IutlOn . = lCOSZ + C' 2Sll1Z - sin 2(1og(l + x))

.

= C1cos[log(l+x)] + C2slI1[log(l+x)]-

"

.J

Example 2.8.10 d 2y

dy

Solve (2x+3)2 dx 2 +6(2x+3) dx + 6y= log(2x+3)

Sol. Writing 2x+3 = eZ or z= log(2x+3) the given equation reduces to [2 2.D(0-1) + 6.2D + 6]y = z i.e.

(4D 2+8D+6)y = z

where D =

d dz

-

A.E. is 4m2 + 8m + 6 = 0 => 2m2 + 4m + 3 = 0 =>m= C.F. is y

~2±J2i

2

=

1

=-l+J2i

e-Z [Acos

~ z + Bsin ~ z]

-

d

dz

Linear Differential Equations with Constant Coefficients

Pol. =

I 1 z = 40- +80+6 6 1

1 6

= -

r1--40 ]z= 3

1 6

1

I

103

000

40

:m 2

40 2D2'Z = -6 [1+- +-3-rl z 3 1+-- + ---3 3 z 6

4 3

2 9

- [z- - ] = - --

Complete solution is y = C.F. + P.1. 1 B' 1 ] + z -2 =e-Z[ Acos r;::z+ SIl1-Z ..;2 ,fi 6 9 I

1

1

e x+..;2..;2

= -2 3 [Acos( r;:: log(2x+3»+ Bsin(

1

')

log(2x+3))] + -6 log (2x+3)- ::.. 9

Exercise 2(h) Solve the following differential equations I. (x 202 - 3xO + 4) Y = 2x2

2.

Ans. (C I + C2logx~ + x 2 (Iogx)2 (x 2D2 - 3xD + 5)y = sin(log(x»

3.

Ans. x 2(C Icoslogx + C2sinlogx) + (coslogx + sinlogx) (x 2D2 - xD +2)y = xlogx

4.

Ans. x(Clcoslogx + C 2sinlogx) + xlogx (x 2D2-3xD+5)y = x 2 sinlogx Ans.

5.

x2(C)coslogx +

C2sinlog~) - ~

x2 1ogxcoslogr

2 2 (_,-s_in_l--,og~x-.:.)_+_1 = -

(x D -3xD+l)y Ans.

Ans.

x

x2[C Icosh (filogx) + C.,sinh( J3log~) + _I + _1- [5sin(logx) - 6cos(\ogx)] -

6x

61x


104

8.

(x20 2+xO+I)y= (I0gx)2 + sin(logx)

Ans. 9.

(2x-I)2

Ans.

10.

C I coslogx + C2sinlogx + (Iogx? -x[2coslogx - sinlogx] d 2y

-2 +

dx

dy (2x-l) - +2y = 0

dx

C

y= C 1(2x-l)+ ~ 2x+ I

d2 dy (5+2x)2 dx; - 6 (5+2x) dx + 8y = 6x

Ans.

y

=

C (5+2x)2+.fi + C (5+2xi t fi _ 3x _ 45 I

2

2

8

2

11.

(2x+I)2 d y _ 6 (2x+l) dy + 16y = 8 (l+2x)2

dx 2

Ans. 12.

y

dx

= C 1 + C2 log(2x+ 1) + [log(2x+ 1)2](2x+ 1)2

d2y dy (2x+3) - 2 -(2x+3) --12y=6x dx dx

Ans.

C 1(2x+3)2 +

Ci2X+3f~

- 2x +

%

2.9.1 Linear Differential equations of second order d 2y dy The general form is dx 2 + P dx + Qy = R

where P, Q, R are functions of x.

Method of variation of Parameters Let the C.F. of the equation 2

d Y

dx 2

be

+ P dy + Qy = R dx

y=Au+ Bv

where A and B are constants and u, v are two independent solutions-,of d 2y

dy

dx2 + P dx +Qy = R

.... (I)


y

be

Au + Bv

=

105

.... (2)

where A and B are constants and u, v are two independent solutions of d 2y dy - 2 +P-+QllJ= 0 dx

dx

.... (3)

.J'

+ PU I + Qu = 0 and v2 + PV I + Qv = 0

such that

1I2

where

tl2 =

d 2u dx 2

ul

'

=

du dx etc.

In order to obtain the solution of the equation (I) the arbitrary constants A and Bare treated as arbitrary functions of x and are chosen in such a way that

y = A(x)u + B(x)v satisfies (1)

.... (4)

Differentiation of(2) gives = All,

+ lIAI + BVI + vB, (suffixes indicating the order of the derivative) .... (5)

Now we chose A and B such that Alu + BI V = 0 So, Again differentiating d2y dx 2 = (Au 2 + lIIAI) + (Bv2 + vIB,)

.... (6)

substituting (4), (5) and (6) in (I) we get [AlI 2 + BV2 + Alu, + Blv,] + P[Au, + Bvd + Q(Au + Bv] = R

(or)

A(u2 + Pu, + Qlt) + B(v2 + PV I + Qv) + Allt, + BI vI = R

Alu, + Biv i = R (u and v are solutions of(3)

.... (7)

Solving (5) and (7) we get dA

-

dx

=A = I

-vR

uv, --u, v

dB

and -

dx

uR

= B = --I

1/V,

-II, v

integrating we obtain A(x)=

J

-vRdx +CI;B(x)= (uv, -1I,v)

J

uR +C 2 lIV, -u,v

.... (8)

where C I and C 2 are arbitrary constants Substituting (8) in (4) we get the complete general solution of the equation (I)


106

Working Rule: First find two independent solutions 1I and v of(3). Then C.F. is given by y = Au+Bv where A and B are arbitrary constants. Treating A and B as functions of x, we have the solution of (1) as y = AI(x) 1I + BI(x) u where A(x) and B(x) are given by (8)

Solved Examples Example 2.9.2 d2

Solve by the method of variation of parameters ~- y = dx 2

2 --x

1+c

Sol. A.E. is m2 - I = 0 => m = ± I C.F. is y

= A~

+ Bc-x

Assuming A and B as functions of x such that the given equation is satisfied by

y

=

dy dx

Ac-'" + Bc-X we have =

dA

dB

dx-

dx

AcX - Be-x + ~-+e-x-

= A~

dA

- Be--x

.... (I)

dB

Choosing A and B so that ~~ + e--x dx = 0

.... (2)

.... (3)

Substituting (I) to (3) in the given equation, we get dA

~-

dx

dB

+e-x -

2

=--

.... (4)

l+c x

dx

dA

Solving (2) and (4) we get -dx dB

and -

dx

C

-x

=

_c_ eX + I

.... (5)

X

=---

eX + I

= log (~+ I) -e-x-x Integrating (5) and (6) A = _e- x + log( 1+e-') - x + C" B = -Iog( 1+~) + C 2

.... (6)


107

Hence the solution is

y=[-e-x+log(ex + y=C,ex

(or)

Remarks: If(i) I

e-x

+C 2

1)-x+C,1~+[C2-log(1 +~)]e-X

-1 +elllog(e'+I)-x]-e-'\'log(el+I)J

+P+Q=Otheny=~(ii)

I-P+Q=Otheny=e-'x

and (iii) P + Qx = 0 then y = x are solutions of D2y + POy + Qv = 0

Example 2.9.3 Solve by the method of variat ion of parameters d 3y (ix"

- \ + (I 3

Sol:

d

cOlt") ~ dx

cotx = sin 2x

.... (I)

d y dv dx 3 + (I - cotx)

dx - cotx = 0

C.F.P

=

I - cott

.... (2)

Q = -cou comparing with original eqn.

:. I - P + Q = I - I + colx - cotx = 0 showing that v = e-X is a solution or (2) To find another independent solution of(2) let

11

= eO-x . w

By this substitution, the equation (2) reduces to d 2 w dw 2 -x -+-(l-cotx--e )=0 dx 2 dr c- x

or

d2w dw - 2 = (l+cotx)dx dx

~

i(~) dw

= I

+ cou

dX dw or - =eTsinx dx

dw

log - = x + logsinx dx

:. w= JeXsinxdx=- e; (cosx-sinx) eX

.

I

.

u = e-x [--(cosx-smx)] = - - (cosx - Slnx)

2

2

The second independent solution can be taken as cosx - sinx


108

Thus solution of(2) is

y

=

A (COS X

-

silu) + Be-x

For finding the solution of (I) we treat A and B as functions of x

i.e. y = A(x) (cosx - sinx) + B(x).e-x is the solution of (I) where A(x) and B(x) are obtained by solving (cosx - sinx) A I + e-x . BI

0

=

.... (4)

and

.... (5)

dA solving (4) and (5) ~

I

-2 sinx

=

sin2x -dB = ~--ex

and

dx

4

(I-cos2x)

--'-----'4

integrating we get

f'slflxdx+C 1 =--+C cosx 1

A = - -I

2

.... (6)

2

.... (7)

and Hence the complete solution is

y

=

C 1(cosx - sinx) + C2e-x -~ (sin2x - 2cos2x) 10

Example 2.9.4 Solve by the method of variation of parameters (I -

x)

2

d y + x dy _ Y dx 2 dx

= (I

_ x)2

The given equation can be written as d 2y dx 2

x

+ I-x

dy I dx - l-x Y = I-x

x

p= _ . Q= I-x'

I

I-x

.... (I)

andX= I-x

Clearly P + Qx = 0 Hence y

.

.

d2y

= x IS a-solutIon of - 2 + dx

X

-I-

-x

dy I dx - --y = 0 I-x

.... (2)


109

To obtain the second independent solution of(2) take y = vx d 2v

x

2

I-x

x

dv

Then (2) reduces to -'} + dx (-+-.1) dx-

=

0

d . dv dv x 2 - [-]+- [-+-] = 0 dx dx dx I-x x d dv ~~

x

2

I-x-I

2

1

2

- - =- - - - =+( ) -- =+1---d,,1- x x I- x x I- x x dx

lo.g(:) =x+ log(x-I)-logx2 dv

-

dx

x-I

1

(-I)

X

x-

= -.e-t= e-t [-+-, ] 2 x

:. v

1

= e"'" .x

The second independent solution is xv = e"'" solution of equation (2) is Y = Ae"'" + Bx

.... (3)

To find the solution of (I) treat A and B as functions of x such that

dA dB e"'" - + x -

=

0

.... (4)

dA dB e"'" - + x -

=

I -x

.... (5)

dt"

and,

dx

dx

dx

dA dB Solving (4) and (5) we get d; = -xe-X and dx = I

Hence the complete solution is y

=

A e"'" + B.x

= [C1+e-x(I+x)]e"'" + (x+C 2 ) x = C1e"'" + Cr + x 2 + (I+x)


110

Exercise (i) Solve the following by the method of variation of parameters I.

d 2y

-

dx 2

+ a2y

=

sec ax

· x . I Ans. y = C,cos a:r + C'2sm ax + -SIl1 ax + -cos ax Iog(cos a,·) a a

2.

d2 y

dx 2 + Y = tanx

Ans. y 3.

=

-[Iog(secx + tanx)] cosx + C,cosx + C2sinx

d2 y

--T + 4y = cosec 2x dxAns. y = C,cos 2x + C2sin2x + xcos 2x + sin 2~ log(sin 2x)

4.

d2

dy

dx

dx

x2~-2x(l+x) -+2(1 +x)y=x3 2 2

Ans. y = C, xe 2x + C

x x r - 4-"4

3 Mean Value Theorems and Functions of Several Variables 3.1.0 This chapter deals with (i) Rolle's theorem (ii) Lagrange's mean value theorem also called as first mean value theorem. (iii) Cauchy's Mean Value theorem (iv) Higher Mean Value theorems. (v) Curvature (vi) Centre of curvature. (vii) Evolutes (viii) Envelopes 3.1.1

Rolle's Theorem Ifj{x) is (i) continuous in [a, b], (ii) differentiahle in (a, b) and (iii)j{a) = j{b), then there exists a'c'E(a,b) Proof: Suppose (i)

(ii)

3f'(C)=0

f (x) is a constant function throughout the interval [ a, b ], then f' (x) = 0 VX E ( (I, h) Hence theorem is proved ... f (x) is not a constant function in [ (I ,h ]. As f(x) is continuous in [a,b], there exists a maximum value say, at 'c' sayat'{f (a~d~h) for

f(x)

(a ~ C ~ b)

in(a,b).

and a minimum value,


112

f (c) 7:- f (d) Suppose

f( c) 7:- f( a) then

and Further

and at least one of them is different from

f(x)

and

'c + h' be a point in the neighborhood of 'c' ,.

/(c+II)- /(c) $;

h

/(c+I1)- /(c)

.... ( 1)

.... (2)

is differentiable in (a, h).

/(c+h)- /(c) h-~O h LI

1'( c) $; 0

0 when II> O.

;::: 0 when h < O.

Iz

Lt /(c + h) - /(c) ... / ,(c) == h~O h

i.e.,

f ((l) == f (b)

and

$;

0 and

As h -t 0 we get from (I) and (2) that -

/(c+h)- /(c) h~O h Lt

.

;::: 0 respectively.

1'( c) ~ 0 simultaneously => 1'( c) == 0

Similarly the theorem can be proved when

f (d) 7:- f (a)

3.1.2 Geometrical interpretation of Rolle's theorem Let P and Q be two points on the curve y=f(x). AP==BQ ordinates f(a)=f(b» and the curve is continuous from P to Q. It can be shown that there is at least one point on the curve y = f(x) between x = a and x = b at which the tangent to the curve is parallel to x-axis. . y

x=a

y=a

P

Q

.i(b)

f(a)

x'

0

A

y'

fix) is a constant function

B

x

Mean Value Theorems and Functions of Several Variables

113

x=c y

Q f(a)

o

f(b)

f(d)

x

B

A

y'

j(c) andj(d) are both different fromj(a)

=

j(b)

x=c y

x=a

x=b

f(a)

o

f(c)

f(d)

B

A

x

y'

j(c)

-:F j(a)

andj(d)

=

j(a)

=

j(b)

x=b

x=a y

x=d f(b)

f(a) f(d)

x'

A

0

B

y'

j(d)

* j(a) and j(c) = j(a) = j(b)

x

Eng~neering

114

3.1.3

Mathematics - I

(x) = log { :;a:a:) }in ( a,b )

Verify Rolle's theorem for / Solution: Consider the function

J (x) = 10g{ x(2 + ah)} in the interval ( a, b ). x a+b

J(x+h)- J(x) h~O± h

Lt

2

=

I[ (x+h)2 +llh I x +abj - log - og--+ h~O± h ( a + b) ( x + h) ( a + b) x

Lt

~rlOg (x +_~h) + 2xll+h2 -log x+ hj 2

=

Lt

+h~O±

hl

(x 2 +ah)

x

1

= Lt -1 [ log {'1+ 2xh + h

}

7

r"~O± h

x- + ab

-log {h}] 1+x

Lt [2X

=

+h~O± x 2 + ab

1 ] --+O(h) x

where O(h) indicates terms of order h and higher powers of h.

I

I(

X)

which indicates that

=

2x - ~ x +ab x 2

I (x)

is differentiable in ( a,b ) and hence continuous also.

2

I

Further

(a +ab) (a ) = log ( ) a a+b

= log 1 = 0

2

I (b) = log

(b +ab) (

b a+b

) = 0.

Thus

I (a ) = 0 = I (b)

All the conditions of Rolle's theorem are satisfied. Hence 3c( a < c < b) such that

I

I(

C)

=0 II () C =0

Clearly c

= +ab

. gIves

2

2c

c +ab

1 c

--=O~c

is the G.M of a and b and so

The theorem is thus verified.

E

2

=ab or c=±ab

(a, 1)


3 .1.4

115

j"() S1l1X.10 x = ---

. Ven'fy Rolle's t I1eorem '"lor the function

eX

(0 ,Tl )

Solution:

f(x) =~lIl
(O,Tl)

f(x+h)- f(x) h

Lt h-... O±

= Lt -1 [Sin(x+h) -Sinx] h--.O±

h

e( ail)

e'

e< [Sin ( x+ h) - e" sin x] = Ltx iI--.O±

h

ex+1I .e

2

Sin(x+h)-Sinx{l+ h + h + .. ..1 1 I! 2!

=

f

-----------X+-il~x--------~

LI h

e

}HO±

.f!

~

~

1 2 cos ( x + ). sin ( ) - h {sin x + 0 ( h )}

=h--.O± Lt h

----~--~--~-----------ext-II

. (h)

S1l1

=II--.O± Lt !h 2COS(X+!!'). ( h ) -{sinx+O(h)} 2 2

f' ( x ) = cos x ~ S1l1 x e Thus

f (x)

Further

f

is differentiable in

(0) = 0 = f

and hence continuous there.

(Tl) .AIl the conditions of Rolle's theorem are satisfied.

f'( e) = 0 cose-sine .. value) -----= 0 => e = -Tl (pnnclpal

:. 3e( 0 < e < Tl) (i.e.,)

(0, Tl)

and clearly e =

eC

such that

4

Tl E (0, Tl) . Hence the theorem is verified. 4


116

3.1.5 Example Verify Rolle's theorem for the function.f{x) = Ixl in (-1, I). Solution Here

.f{x) = - x for -I < x < 0

= 0 for x = 0

=x for 0
.f{-I)=.f{I)

/'{x}

=-1

/'{x} =

1

for

-1.:sx.:sO

for

:. f'{x} does not exist at x = 0 and hence j(x) is not differentiable in (- 1, I)

.. Rolle's theorem is not applicable to the functionj(x) = Ixl in (-1, I). y y = f(x) = Ixl

------------------~~------------------x

y'

Fig. 3.1 (The curve is not smooth at x = 0).

117


Exercise - 3(A) I. Verify Rolle's theorem for the following functions : 1. lex)

= x (x + 3)e-xI2

2. j(x) = sin x

3. j(x)

m

(-3.0)

m

(O.n)

(.'s;+6:

= (x- a)m (x- b)n in (a,

6. fix)

= xl -

3.2.1

(2,3)

m

5. fix)

[Ans : c = n/2]

(:.?~)

= e" (sin x - cos x m

4. j(x) = log

[Ans:c=-2]

b), m > 0,

[Ans : c = n]

[Ans : c =

11 > 0

5x + 7 in (2, 3)

[Ans : c =

J6]

{mb+ na} {m+ 11}

]

[Ans : c = 5/2]

Lagrange's Mean Value Theorem (First mean Value Theorem): Ifj(x) is (i) continuous in [a, b] and (ii) differentiable in (a, b) then ::3 at least one "alue 'c' in (a, b)

3

f'{c)

= f{bi- f{a) -a

Proof: Define a new function «j>x where A is a constant

3

fix) + A.\

..... ( I)

«j>(a) = «j>(b)

i.e.,

fia) + A (a)

i.e.,

A

= _

=

=

fib) + Ab

[f{b)- }'(a)] b-a

..... (2)

j(x) and Ax are continuous in [a, b]

Hence «j>(x) is continuous in [a, b]

..... (3)

j(x) and A x are differential in (a, b),

Hence «j>(x) is differential in (a, b) andj(a)

=

«j>(b)

..... (4) ..... (5)


118

(3), (4), (5) show that ~(x) satisfies all the conditions of Rolle's theorem.

3 atleast one value c in (a, b) ;) ~'

(x)

~' (c)

A

~' =

0

f' (x) + A

=

= f' (c) + A = 0 f'(c)

= -

..... (6)

From (2) and (6) it follows that

f'{c) = j'(b)- f{a) b-a 3.2.2 Geometrical Interpretation of Langrange's Mean Value Theorem P and Q are two points on the continuous curve y = j(x) corresponding to x = a and x = b respectively. .. P[a, j(a)],

Q [b,j(b)] are two points on the curve.

· ... h . PdQ· SI ope on t he Ime JOll1lng t e pomts an IS

f{b)f{a) ' R·IS a pomt . h b_a on t e

curve between P and Q corresponding to x = c, so that f'{c) is the slope of the tangent line at R [c,j(c)].

f'{c) =

f{b)- f{a) b_a

means that the tangent at R is parallel to the chord P Q.

y Q

p

f(a)

x'

0

x=a

f(c)

x=c

f(b)

x=b

x

y'

Fig. 3.2 Hence this theorem tells that there is at least one point R on the curve PQ where the tangent to the curve is parallel to the chord PQ.

119


3.2.3 Example Verity Lagrange's theorem for the functionf(x) = (x - I) (x- 2) (x- 3) in (0, 4).

Solution j(x)

=

(x - I)(x - 2)(x - 3) and a

=

0, b

=

4

j(x) = x 3 - 6 Xl + II x - 6 is an algebric polynomial and (0, 4) is a finite interval. Hencej(x) is differentiable in (0, 4) and is continuous in [0,4] showing that the conditions of Lagrange's Mean Value theorem are satisfied.

:3 atleast one value 'c' in (0,4), such that f'{c)

f{b)- f{a) b-a j(0) = - 6, f(4) = 6 =

..... ( I)

f'{x) = 3 x2 - 12 x + II f'{c) = 3c2 - 12 c + 11

3c2 _ 12 c + II

From (I)

=

3c2 - 12 c + 8 = c=

°

6±2fj 3

c = 6 ± 2fj

The point

6 - (- 6) 4-0

3

W h·ICh CIear Iy

°)

I·Ie .111 ( ,4 .

3.2.4 Example Verify Langrange's Mean Value theorem for the functionj(x)

= ~

in (0, I) :

Solution j(x) = ..

~

is differentiable in (0, I) and continuous in [0, I] :3 atleast one value 'c' in (0, 1) such that 3

3

l'{c)

'{ ) f c

=

f'{x)

=

..... ( I)

f(l) - f(O) 1-0

=

j(0) = eO

f{b)- f{a) b-a

=

..... ( 1)

I, f(l) = e

~ gives

f'{c)

=

e

C


120

eC =

From (I)

e-I ' 1-0

-

c=log(e-I), thisc EO, I)

3.2.5 Example Verify Langrange's Mean Value theorem for the functionj(x) in (3,4).

= 5x2 + 7x + 6

Solution j(x) is an algebraic polynomial and the interval (3,4) is finite, j(x) is differentiable in (3,4) and continuous in [3,4]. . . 3 atleast one value c r.(3, 4) such that j'(e) = f(ll) - f(3) 4-3 3

f'{c}

..... (1 )

f{h}- f{a} b-a

=

j(3) = 72,j(4) = 114, f'{x} = lOx + 7 114-72

IOc+7=

c = 3, 5

4-3 (3, 4)

E,

Exercise - 3(8) I. Verify Langrange's Mean Value theorem for the following functions : I. j{x)

=

x(x - I) (x - 2) in (0,

X)

2. j(x)=logxin(1,e) x 2 _ 3x _

I

in ( -~

1, I;)

3. j(x)

=

4. j(x)

= a2 - 7x + lOin (2, 5)

3.3.1

Cauchy's Mean Value Theorem If two functionsj(x) and g(x) are (i) continuous in [a, b] (ii) differential in (a, b) and (iii) g'(x) =F- 0 in (a, b) then 3 atleast one value 'c' in (a, b) 3

f'{c} g'{c}

f{b}- f{a}

= g{b}-g{a}

121


Proof: Define a new functionj(x)

= j(x) + A g (x)

.....( I)

Where A is a constant such that 4>(a) + 4>(b)

..... (2)

j(a) + A g (a) = j(h) + A g (h)

[j{b}- j{a}]

..... (3)

A=- [g{b}-g{a}] j{x), A g (x) are continuous in [a, b)

Hence 4>(x) is continuous in [a, b) j(x), A g (x) are differentiable in (a, h)

..... (4)

Hence 4>(x) is differentiable in (a, b)

..... (5)

and 4>(a) = 4>(b)

..... (6)

(4), (5), (6) show that 4>(x) satisfies all the conditions of Rolle's theorem 3 atleast one value c in (a, b) ;) 4>' (c)

But

=

0

4>'{x} = f'(x} + Ag' (x) 4>'{c} = f'{c} + Ag' (c) = 0

=>

A= -

/'{c} g'{c}

From (3), (7) we get

/'{c} g'(c}

..... (7)

/{b}- /(o) =

g(h}- /(o)

3.3.2 Example Verify Cauchy's Mean Value theorem for j(x)

=

1 and g(x) x-

-?

Solution j(x), g(x) are differentiable in (a, b) and continuous in [a, b)

1 in (0, h) x

= -

122


::3 atleast one value 'c' in (a, b) :)

f'{c)

g'{c) Here

f{b)- f{a) g{b)- f{a)

=

f'{x) = ~ X3 -I

g'{x)

=7

-2 C 3

=--IT

=

1

//c 2

a+b ab

2

c c

a

b

2ab a+

= --b which is the Harmonic Mean of , a' and 'b'

c E(a, b)

3.3.3 Example Verify Cauchy's

Mea~

Valve Theorem for fix)

=

e, g(x) =

Solution fix), g(x) are differentiable in (3, 7) and continuous in [3, 7]

::3 atleast one value 'c' in (a, b) :)

f'{c) f{b)- f{a) g'{c) = g{b)-g{a) Here

f'{x)

=

e

g'{x)

=

e-x

c = 5 E (3, 7)

e-X in (3, 7)


123

Exercise - 3(C) I. Considering the functionsj{x):::: x 2 , g (x):::: (x) in Cauchy's Mean value theorem for (a, b) prove that 'c' is the arithmetic mean between a and b. 2. Verify Cauchy's Mean Value theorem for j{x):::: sin x, g(x):::: cos x in (a, b) 3. Verify Cauchy's Mean Value theorem for j{x)::::

3.4.1

I

fx, &>(x):::: fx

in (a, b)

Higher Mean Value Theorem with Lagrange's form of remainder (Taylor's theorem with Lagrange's form of remainder): If a functionj{x) is such that

(i) j(x); f'{x}, r{x} ..... fn - I (x) are continuous in [a, a + h] (ii)

fW{x} exists in (a, a + h), then :3 at least one number '8' between '0' and' I' h hn - ' h" 3j{a + h):::: j{a) + ,f'{a} + ..... + - ( \"f(n-I)(a) + -F (ll + 9h) 1. n -I,. n!

Proof: Define a new function. j{x):::: j{x) + (a+h-x) f'{x} + (a+h-xY r(x) + ... J! 2!

+

(a + h - X ),,-1 /"-1 (a + h - x)" (-I) (x) + .A n . n.,

where A is a constant

3

..... (1 )

cp (a) :::: cp (a + h) 2

h j" (a) + 2! h fW (a)+ cp(a) :::: j{a) + 1! hn - I

hn

..... + - ( I",,r-I(a) + n- J!

n!

A

..... (2)

and

cp (a + h) :::: j{a + h)

..... (3)

but

cp (a) :::: cp (a + h)

..... (4)

Thus

j(a + h) :::: cp (a + h) :::: cp (a)


124

Hence using (2) we get

h2 I" (a) + ..... . 2!

h J!

j(a + h) = lea) + - j(a) + -

..... (5) j(x), I'(x) , I" (x ), ... .fn-I(x) and (a + h - x) (a + h - x)2 etc continllolls in [a, a + 17] and differentiable in (a, a + h)

Hence

~(x)

is continuous in

ra, a + 17] and differentiable in (a, a + 17)

..... (6)

(4) and (6) show that ~(x) satisfies all the conditions of Rolle's theorem.

:.

:3 atleast one value 'c' (a < c < 0 + 11) ;)

~' (c) =

0

Write c = a + e 17 where 0 < e < I

:.

:3

e E (0,

I) ;)

~'

(0 +

e 17) = 0

..... (7)

Differentiating (I) with respect to 'x'

.,.

+ [(a+h-x)"J

(n-I)

I" (x)- (n-IXa+h-xt-

2

(n-I)

In-J/(x)j_ n(a+h-xtF;'n!

J

A ..... (8)

From (7) we get ~' (0

+ e 17) =

(h - eh),,-J

(n-I)

[rea + eh]

=0 ..... (9)

From (5) and (9) it follows that 2

j(a + h)

=

hi' (a) + -, h I j(a) + -1' .

2.

"

(a) + ...

(0 < e < 1)


The last term

3.5.1

125

~ .!' (a + 0 h) is called Lagrange'sform of remainder. n!

Higher Mean Value Theorem with Cauchy's form of remainder (Taylor's theorem with Cauchy's form of remainder): If a functionf{x) is such that (i)f(x), f'{x) , r{x) ..... fll(X) arc continuous in

[a, a + 11] and (ii),f'1(x) exists in (a, a + h) then 3 atleast one number '0' between '0' and' 1' such that 2

f(a + h)

=

I!

h '" «(I) + ... .f{a) + h f' () a + j!f

Proof: Define a new fUllction ~(x) = .f{x) +

(a+h-x) () (a+h-x)2 () f' X + f" x + .... I! 2! '

(a + h - x)" where A is constant

From (1)

~

(a + h)

I

{n-l}!

+

..... (1)

fll-l(x)+(a+h-x)A

(a + h) = ~ (a)

3 ~

..... (2)

= f(a + 11) h .

~(a) = f(a) + -1'.1

'I

{a} +

h2

f 2!

"

h"-' , {a} + ... + -(-1)'.1 11-1 (a) + II

n- .

A

..... (4)

From (2), (3), (4) f{a + h)

=

h " j

.f{a) + -"

.

,J

n- f"

(0) + -,

2.

(a) + .......

.. ... (5) f(x), f'{x) , r(x) ..... jil- I (X) and (a + h - x), (a + h - x)2 ...... are continuous in [a, a + h] and differentiable in (a, a + 1/)

126

Engineering Mathematics - I Hence (x) is continuolls in [a, a + h]

..... (6)

(x) is differentiable in (a, a + h)

..... (7)

(2), (6) (7) show that (x) satisfies all the conditions of Rolle's theorem. :. 3 atleast one number '0' in between '0' and' I' Differentiating (I)

3 <1>'

(a

+ 0 h) = 0

..... (8)

to 'x'

W.f.

'{x) = I'{x) + [(a+h-x) IW{x) - f'{x)

[

(a+h-x f I"'{x)- 2(a+h-x) f"{x)] 2! 2 + ... + ... +

+ [(a+h-x

llu1 )

(11-1).

F(x)_{n_I){a+h-x)"2 f"'{X)] -A)

(n -I).

(a+h-x)"-' {n-I}. f"(x)-A

I.e., '{x) =

..... (9)

From (8) and (9) we get

{h-Oh)"-I <1>'

..

=

(a + 0 h)

A =

{n -I}.

fll (a + 0 h) -

A=0

h,,-I (I - 0),,-1 (n-I)' fll(a+Oh)

..... (10)

From (9) and (10) 2

flo + h)

=

The last term

hi' h j.w j(a) + -" (a) + -, (a) + .... . 2.

h"{I-OY--1 (

)

n-I!

fll (a + 0 h) is called Cauchy's form of remainder.


3.5.2

127

Alternate form of Lagrange's Mean Value theorem If f(x) is i) continuous in the closed interval [a. a + h] and ii) derivable in the open interval ( a . a + h ) then there exists at least one number B, 0

< B<1,

such that f(a+h)=f(a)+f'(a+Bh).

Proof: In ( 2.1 ) put b exists 'c' ,

=a + h

f'(c)

such that

,then from the proof of L . M . V theorem there

= f(a+h)- f(a)

a+h-a writing c = a + Bh we have 0 f( a + h) = f(a)+ hf'(a+Bh) a+h-a 3.5.3 Example: If f(x+h)=f(x)+hf'(x+Bh), O
Solution: Let

f (x) = ax 2 + bx + c , then f(x+h)=a(x+h)2 +h(x+h)+c

It is given that

= f(x)+h(2ax+b+ah)

.......(1)

= f (x + h) = f (x) + hf' (x + Bh)

....... (2) From ( I ) and (2) we get,

f'(x+ Bh) = 2ax +b +ah (i.e.,) 2a(x + Bh)+ h = 2ax +b +ah

.

(sincef'(x) = 2ax+b) => 2B = 1 or 3.5.4

B=

1;2

Alternate form of Cauchy's Mean value theorem If f (x) and g (x) are (i) continuQ,Us in the closed interval [ a, a+h] (ii) derivable in the open interval ( a . a+ h ) and (iii) g' ( x) *- 0 at any point in the open interval

( a, a+ h ) , then there exists at least one number B , 0 < B < 1, such that

f'(a+Bh)

f(a+h)- f(a)

g'(a+Bh) - g(a+h)-g(a) Note: Lagrange's Mean Value theorem can be deduced from C . M . V theorem replacing g (x) with x in the interval ( a, b )

by


128

= b , g ( a ) = a and g X ) = 1 f'(C) f(b)-l(a) -- = . which is L. M. V. theorem. 1 b-a

We get g ( b)

I(

Example: Using Cauchy's Mean Value theorem, show that sin b - sin a < b - a given that 0 < a < h <

Jr 2

Solutio,,: Taking

f (x)

3e

=

sin x and g (x) = x and applying C.M. V theorem we get that,

such that

. Furt her cose < ] 111

sinb-sina

b-a

(0 Jr)2 ;-

:.

= cos

Jr

e where 0 < a < c < b < -

sin b - sin a <

b-a

3.5.5 Example: I f f '( x) is continuous in [ a, b] and

l' .

2

' (I.e.,) Sin b -SIna < (b -a)

fll (x) exists in ( a, b ) then show that I

f(b)-f(a)=(b-a)f '(a)+(b-a)2 fll(a) 2!

where

a
Solution:

g( x) = h( x) _(b - X)2 h( a) where b-a h(x) = f(b)- f(a) -(b -X)f'(X) ........(1) g(a)=h(a)-h(a)=O and g(b)=h(b)-O=O from (I) and L.M.V

Choose a function

Now

g( x)

such that

theorem.

g(a)=O=g(b) . It is given that f'ex) is continuous in [ a, b] and differentiable inC £I, b) .

h(x) and g (x) . :. g (x) satisfies all the conditions of Rolle'stheorem.Hence 3e (a
g

I (

X)

= h' ( X ) + 2 (b - C! h ( a) (b-a)

from (I )

=-(b-X)f'1 (x)+ 2(b-X![f(b)_ f(a)-(b-a)/(a)] (b-a)

........(2)


129

Using (2) we get,

-C!

(b-c )f"( c) = 2( b [f(b) - f(a) -(b - a)f'( a)] (b-a) (b - a)2 f(b)=f(a)+(b-a)f'(a)+ 2! f"(c) ........(3)

or

Taking

(' = a + Biz where

h = b - a and

(3) gives

f(a + h) = f( a)+ Jif'(a) + ~2! f'(a +Oh)

0
3.5.6 Example:

f (x), g (x)

and

h(x)

are three functions derivable in ( a , b ), show that there

exists c in ( 'I, b ) such that

f'(c) g'(c) h'(c) f(a) g(a) h(a) =0 f(b) g(b) h(b) deduce L. M . V and C . M . V theorems.

Solution:

Consider the function

Clearly Also :.

f'(x) g'(c) h'(c) ¢ ( x) = f (a) g ( a) h(a) f(b) g(b) h(b)

=0

¢(a) = 0 = ¢(b)

f( x),g( x)

and

h( x)

are all derivable in (a, b),

All the conditions of Rolle's theorem are satisfied. Hence 3c E

that ¢' ( c ) = 0

(i.e.,)

........ ( I )

f'(c) g'(c) h'(c) f (a) g ( a) h(a) = 0 f(b) g(b) h(b)

........ (2)

(a,b)

such


130

g(x) =x

Taking

f'(e) I f(a) a f(h) b f'(e) =

(i.e.,)

h(x)

and

I we have from (2)

=

0 1

=O=>f'(c)(a-h)-[f(a)-f(h)J=O

1

f(h)- f(a) h-a

where

e E (a,b)

........ (3)

(3) shows that L . M . V theorem can be deduced from (I) Taking only h (x)

f'(e) g'(c) f (a ) g ( a ) j(b) g(b)

we have,

=I

0 1

= 0 where

c

E (

a, b)

=> f'(e)[g(a)-g(b)J-g'(c)[f(a)- f(b)J=O

f'(c)

f(b)- f(a) ()

=> -(-) = () g' e

where c

g h - g a

E

(a,b)

. .

which

IS

C. M . V theorem.

3.5.7 Example: Apply Maclaurin's theorem with Lagrange's from of remainder for the function 2

f (x) = eX

and show that 1 + x + ~

2

~ e ~ 1 + x + ~ eX for every x ~ 0 . 2 2 t

Solution: Maclaurin's theorem with L form of remainder is 2

n-I

n

f(X) = f(o)+~r'(o)+~ f"(O)+ ... +1~_1 in-I) (0)+ ~~ f" (Ox) ~

2

3

~

lJ

X + -X eX = 1+ X + T=)

~

n-I

~

n

X Ox (SInCe . Dn ( eX) = eX) + ...... + IX ~ _ 1 +- e ~ ~

........ (1) where 0 < (} < 1 Taking n = 2 in (1) we get X2

eX

= 1+ X + ~ eOx

For every X ~ 0 we have eOx ~ eX where 0 < (} < 1

........ (2)

Mean Value Theorems and Functions of Several Variables 1

1

II

~

131

x- Ox 1 x- x 1+x+-e· S +x+-e ........ (3) Ox

Further e

> 1 whenever x ~ 0 and 0 < 0 < 1 X

2

X

2

l+x+-sl+x+-e

II

fh

II

........ (4) From (2), (3) and (4) it follows that 2

2

2

II

II

II

1 x Ox x x, l+x+-s +x+-e =e sl+x+-e X

3.5.8 Example 2

. Taylor's theorem to prove that loge ( 1 + x ) < 1- x + x Apply

3

2 3

whenever x> 0 .

Solution: According to Taylor's theorem

h"- I

h2

h

I( a+h) = l(a)+11 1'( a)+ ~ f"(a) + ... + In-l f"- (a)+ R" where

Rn

=

h" (1-0),,-" ~ n- .p

RII with Lagrange's form

I" (a+Oh)

(n = p)is Rn =

I

, 0<0<1 h"

~ f" (a+Oh)

Substituting a = I and h = x in (1)

1(1 + x) = 1(1)+ ~f'(I)+ Taking

2

x

(_1)"-1 In-l n

X

3

II 3(l+Bx)

Foranyx>O, (1+0x»1

II

~ 1"(1)+ ... + 1:-1 in-I) (1) + ~ I" (1 +Ox)

I (x ) = log x, I" (x ) =

x2 .. 10g(l+x)=O+x--+

n-I

30
132

3.6


Functions of Several Variables

3.6.1 Students are quite familiar with functions of a single independent variable. Functions which depend on more than one independent variable are called Functions of Several Variables. (i) Volume V = nr2h ofa cylinder, r = radius, h = height, curved surface area (ii) Area of a triangle A =

A = 2prh.

I

"2 xy, x = base, y = altitude

(iii) Volume of a rectangular parallelopiped v = xyz, where x = length, y = bredth z = height are all examples offunctions with more than one variable. (i) and (ii) are examples offunctions oftwo variables and (iii) is a function of three variables. Let z =j(x, y) is a function of two independent variables. Here x, yare independent variables and z is the dependent variable and let the function f(x, y) be defined in a region R. Suppose (x, y) be a movingpoint and (a, b) a fixed point in the region R ofthe xyplane. The point (x,y) may approach (a, b) along different paths (see figure) PI' P2' P3 etc.

3.6.2 Concept of a Limit Let

and

(i)

f(x, y) be defined in a region R

(ii)

(x, y) tend to (a, b) along any path

(iii)

E

< 0 be given.


If ::3 0 > 0 such that Ij(x, y) -II <

133

V Ix - al < 0, Iy - bl < 0, thenj(x, y) is said to

E,

tend to I.

j(x, y)

Lt

We write

=

'I

x~a

y~b

I(x, y)

Lt

or as

=

I

(x,y)~(a,b)

3.6.3 Example

x2

x2 + y2

Lt

x-~2

Consider

---'--, it can be seen that if 4xy

y->!

+ y2

5 - -8 <

4 xy

E

whenever Ix - 21 < 0.02,

E =

lv -

0.01 (say) then

II < 0.02 thus 0

=

0.02

satisfies )

)

x- +,v-

4xy

5

-"8

5

< 0.0 I. Hence the desired limit is

"8 .

Note: 1 Let then

Lt

«,y)-->(a,b) .

(i)

f(x, y)

=

Lt

[f ± g]

Lt

[fg]

(x,y)->(a,b)

(ii)

I and

Lt

=

=

(x,y)->(a,b)

and

(iii)

Lt (x,y)->(a,h)

g("(, v)

=

(X,Y)-'(a,b)'

m

I±m

/m

1

I

g

III

Note: 2 (x,y) LI ~ (a, b) exists iff

LI

[

LI

(x-~a) y~b

1

I(x,b) ==

Lt

[

Lt

(y~b) x~a

f(x,h)

1


134

3.6.4 Example ~

f(x, y)

=

x- + V . 2 ' find Lt as (x, v) 2x+ y .

~

(2, I)

Solution

LI [Lt

y~ x~2

1 [4

2

y]

x +y LI + 5 2X+;2 . = y~ 4+ y2 =5=1 ..... (i)

Lt [Lt

x~2 y~l

2

x

Lt

+y ]

2x+ y2

=

x~2

[X2 +

I] 5

2x+1 =5=1

The two limits (i) and (ii) are equal. 'J

Lt

Henee

x' + Y ---=-;;-=1

(x,y) ~ (2,1) 2x + y2

3.6.5 Example J

f(x, v) .

=

)

x- - Y' ? J find whether the limit exits as (x, y) x- + y-

~

(0, 0)

Solution

I,I

Thus

X

L/

~ 0 [Y ~ 0

2 2]

x - Y J

')

x- + y-

lIenee the limit does not exist.

3.6.6

Example

Lt Find

xy ~

(x,y)

~

')

(0,0) y- - x-

LI

7:-

y

I[""'

I,/

x] - Y

~ 0 ly ~ 0 x- + y ?

2] )

..... (ii)

135


Solution Puty =

11/X

Lt

Lt

xy

(X,y)~(O,O)

/

__ X2

mx

2

X~O m 2 x 2 __ X2

It is clear that limits will be different for different values of m i.e., the limit depends upon the slope of the path along which (x. y) approaches (0,0). Hence the limit does not exist.

Exercise - 3(0) I. Examine whether the following limits exist. Find them if they exist. x 2 ~ y2 + 4

Lt (i)

(x.y)

~

Lt (iii)

(x.y)

3.7.1

yx~2y

J

(x, J~

-~

(0,0) y2

(iv)

X

4

~ 2X2

Y4 (x,y) ~ (0,0 ) x-J ~ y-J Lt

~(2,2) xy~2x

Lt (v)

(ii)

3xy2

(2,1)

x 2 + 4y2

Lt

~

J

x-y-

(x,y) ~ (0,0) x 2 + y2

Concept of Continuity Suppose

Lt

(i)

j(x. y) exists

(~,y) ~(a,b)

and

Lt

(ii)

f(.~, y)

=

j(a, b)

(x,y)~(a,b)

thenj(x. y) is said to be continuous at (a. b).

Note: 1 Ifj(x. y) is said to be continuous at every point of a region R, then it is said to be continuous in R.

Note: 2 Letj(x. y) and g(x, y) be continuous at (a, b) then f ± g,fg, and are all continuous at (~ b).

f

g

(g -:t 0)


136

3.7.2 Example Consider the function/ex, y) Lt

(x,y)

~(I,I)

= x 2 + Y - 2x when (x,

f(x,y) =

Lt [Lt x~

~

1 y

I

(x 2

y) 7= (0, 0) and.l( I, I)

=0

+ y-J -- 2x) ]

=/(I, I) Hence the function is continuolls at (1, 1)

3.7.3 Example 2

2

Consider the function f(x, y) = ~ y J when (x, y) 7= (0,0) and.l(O, 0) = 0 . . x + yLet (x, y)

~

(0, 0) along the path y

=

Lt (.Y,y)

~

mx Lt

(0,0)

2

2

~y 2 (x, y) ~ (0,0) x- + y

f(x,y) =

x 2 m 2 y2

Lt

(x, y) ~ (0,0) x 2 + 1112 x 2

m2x 2 (.r, y) ~ (0,0) I + m Lt

-2 =

°

=.1(0 0) ,

lienee .l(x, y) is continuous at (0, 0)

3.7.4 Example Consider the function.l(x, y) Lt

=

xy x2 _ y2 (x, y 7= 0, 0), .1(0, 0)

(x, y) ~ (0,0) x 2 _ y2 along the path y

•

Lt

xy

mx

=

°

2

(x, y) ~ (0,0) x 2 (I - m 2 )

= niX III

= - - 7=

1-JIl

2

.1(0, 0) except when

Hence the given function is discontinuous at (0, 0)

III

=

°


137

3.7.5 Example X

Consider the functionj(x, y) = =

2

Y

2

(x, y) cf. (0, 0)

~X2 _ y2

°

at (x, y)

=

(0, 0)

In this case it is convenient to introduce (polar co-ordinates). Substitutions x = rcosB, y

=

rsinO.

x 2 y2

r4cos20sin2B

l"(2 + y2 - Jr2 (cos

2

0 + sin 20)

r'

-(sin 2 20) 4

4

{x2 + y2

=

y2

4 Now

. I (x-, +y-, )3 2 ~ -E provided /x/ < E· 3

thus when

4

2

/x - 0/ <

E

I· i3

and

lY - 0/ <

I

E3

..... (i)

Hence

x 2 y2

Lt

---;===== =

(x,y)~(0,0)~X2 + y2

°

frol11 (i)

=j(O, 0) 2 2

j(x, y)

=

Jxx Y+ y 2

2-

is continuous at (0, 0)


138

Exercise - 3(E) I. Investigate the continuity of the following: (i)

j(x, y)

=

2x + ),2

(x, y) t= (2, 3)

}

=0

(x, y) = (2, 3)

at (2, 3)

Ans: Not continllous (ii)

xy j(x.y) = - - , ~Xl + y2

(x, y) t= (0, 0) at (0,0)

=0

(x, y)

=

(0, 0)

Ans : Not continuolls. x-y (iii) j(x,y) = - - , 2y+x

(x, y) t= (0, 0)

=0

3.8.1

(x, y) = (0,0)

Ans : Not continuolls

Partial Differentiation Consider z = j(x, y) where x and yare independent variables and z is the dependent variable. Keeping one of the two variables x and y as constant and allowing the other to vary we get a partial derivative of 'z' with respect to the variable that is varied. Keep 'y' constant and allow x to vary, then partial derivative of z w.r. to 'x' is obtained and is denoted by az or f (x, y) or Dj{x, y) or 8f ax x ax az So Similarly

ax az

By

= =

Lt

j(x+Jx,y)- j(x,y)

ax~o

bx

Lt

f(x,y+Jy)- j(.x:,y) by

ay~o

Note: 1

z = j(x, y) represents a surface in the cartesian co-ordinates (x, y, z) system. The section of the surface z = jex, y) with the plane x = k (parallel to yz plane) is a curve. (Similarly the sections with planes parallel to xy - plane and zx - planes also will be curves).

Mean Value Theorems and Functions of-Several Variables

139

Note: 2

az) ( Ox [

gives the slope of the tangent at (x, y, z) to that curve x=k

~z 1 ~

gives the slope of the tangent at (x. y. z) to the curve obtained as

y=k

the section of the surface z =f(x. y) with the plane y

3.8.2 Example z = x3

(i) To find

+ x2y + I

az , we keep 'y' as constant ay

partial derivative w.r. to x of x 3 = 3x2 partial derivative w.r. to x of x 2y partial derivative w.r. to x of I

az

.. (ii) To find

ax

=

=

=

2xy

0

3x2 + 2xy

az , we keep 'x' as constant

ax

partial derivative w.r., to y ofxJ = 0 partial derivative w.r., to y of x 2y

=

x2

partial derivative w.r., to y ofy2 = 2y

az

-

ay

=x2 + 2y

3.8.3 Example f= x 2 + 1- 2xy

af

- =2x-2y ax

af

-

ay

=2y-2x

3.8.4 Example f= (x - y) (x + 2y)

=

k.


140

This is a product, and the usual product rule applies

at' = (x - y) (1 + 0) + (x + 2y)( I - 0) -' ax

= x - y + x + 2y = 2x + y aj

-

ay

=

(x - y) (0 + 2) + (x + 2y) (0 - 1)

=

2x - 2y - x - 2y = x - 4y

3.8.5 Example y Iff= x-2 , Find af and aj x+y ax ay Solution

Applying the Quotient rule

aj

(x+ y)(I-O)-(x-2y)(I+O)

ax

(x + y)2

aj

(x+ y)(-2)-(x-2y)(O+I) (x + y)2

ay

3y (x+ y)2 -3x

= (x + y)2

3.8.6 Example Iff

=

sin(ax + by), sino aj and aj ax ay

aj

-

ax

a

= cos(ax + by) x -

ax

a

ay

=

cos(ax + by)

x

a

ay

(ax + by)

= acos(ax + by)

(ax + by) = bcos(ax + by)

3.8.7 Example Consider f Then

=

ax2 + hxy - by aj

ax

at

= 2ax + hy - ' = 2by + hx ' ay

The expressionix = 2ax + hy is itselfa function ofx andy. We could therefore find its partial derivative with respect to x or y.


141

If we differentiate it partially w.r. to x, we get

(i)

2

I]

~ [Of] ax ax

and this is

a a [- , = - (2ax + hy) = 2a ax-

ax

This is called the second partial derivative of 'f \V.r. to 'x'. If we differentiate Of partially w.r. to 'y' we get

ax

!L (a~) ay

ail ayax

written as - -

Thus

al ax

-

= 2ay

+ hy,

-8f ay

=

hx - 2by

Similar steps can be carried out with the expression for

a~ 2

ay

=

2

aI axay

=

~ (8f) ~ ay al ax

=

(8f) ~

-2b

=

h

8fay

ax

and this is -


142

3.8.8 Example 2

2

2

2

al 8f a 1 a f a 1 a 1 Iff= ax3 + hx2y + bl, find - ' - ' - 2 ' - 2 , - - , - ax ay ax ay ayox axay

Solution

af

-

ax

8j

=3ax2+2hxy~ -

ay

') =hx2+3by

a21 a2 f

In this example also, we see that - - = - ' -

ayax axay

3.8.9 Example Iff= log(x2 + y), prove that

a21

a21

axay ayax Solution

8j

2x

=

a2f -4xy a2f -4xy 2 2 ayax = (x + y2 axay = {x + y2

r

r

axay ayax 3.8.10 Example

)

au au ax ay

au au ax ay

Ifu(x + y) =x2 + y~ then prove that ( - - - ) 2 =4( 1 - - - -)


143

Solution Since

u=

x+y 2

x 2 +2xy- y2 . _au = _X +2xy+ y2 (X+y)2 'ay (X+y)2

au ax

(

au _ au 2 ax ay )

=[7 x- +2xy- y 2 +x-7-2xy- y-7]2 (x+ y)2 ..... (i)

..... (ii)

(i) and (ii) prove the result

3.8.11 Example If z = 2(ax + byi - (x 2 + y) and a2 + b2 = 1, then prove that

Solution

az

ax

=

a2 z

-

=

ax 2

az

ay

4(ax + by) . a - 2x

=

4a2 -')~

4(ax + by) . b - 2y

a2 z a2 z

-2 +-2 =

ax

ay

0


144

3.8.12 Example

If z

=

f ( -Xl] y

aZ az then prove that x - + 2y- = 0 8.:r Oy

Solution

x az + 2y az =f'(Xl][2X2 _ 2X2]=0 ax ay y y y

..... (i)

Differentiating (i) partially w.r., to 'x' . a 2z az a 2z x-+-+2y--=0 ax 2 ax axay

..... (ii)

Differentiating (i) w.r., to 'y' partially a 2z az a 2z x--+2-+2y-=0 ayax ax ay2

Multiplying (ii) by 'x' and (iii) by 'y' and adding 2 a'Z ::.) 2 a z az a2 z ::.2 u Z 2 u- Z x -+x-+2xy--+xy--+2y-+2y -=0 ax 2 ax axOy Oyax Oy ~l

az az x-+2y-=0 ax Oy

But

)

2

::.)

2 a-z a z 2 u-Z x -+3xy--+2y -=0 axay ax 2 Oy2 3.8.13 Example

If u = log(x3 + T + z3 - 3xyz) then prove tllat

..... (iii)


145

Solution

...... (i)

Now

au ax

-3- - - - - - ( 3 x 1 - 3 yz) x + y3 + Z3_ 3xyz

all

3

3

3

(3y2-3xz)

~y

x + y +z -3xyz

au az

1 ~ -3- - - - - - ( 3 z - -3xv) x + y3 + Z3 --3xyz .

au

au

au

- +- +ax ay OZ

3x 2 +3yz-3 y 2 -3xz+3z 2 -3xy Xl +./ +Z3 -3xyz

= -----'--,-----'-------,.-------'-

3

(.: (x + y + z)

x+y+z

(Xl

+ y2 + Z2 - xy - yz - zx)

= x 3 + y' + Z3 - 3xyz) Hence from (i) LHS

=

(!

+

=3(-

~ + ! J( x + : + z 1 1

{x + y + Z )2

-9 {x+ y+ Z)2

=--------:-

_

1

{x + Y + Z )2

-

1

{x + Y + z

f

1


146

3.8.14 Example r'

If II

=

1a (r? au en e 40 then find 'n' so that -au = -)

ao ,.2 ar

a,.

Solution

au -_ nen ae

1

e

-r' -40

+

--I' ( ? On -40-~

.e

.

4e

2

1 ..... (i)

also

-1[4lt

?]

== 2e r 20 + 3r-u

..... (ii) Equating (i) and (ii) we get

Hence

n

-3

e

2e

-3

n==-

2


147

Alternative method :

Taking logarithms on both sides )

r-

log u = n log 0 - -

..... (I)

40

Diff. (I) partially w.r.t 8

au

11

r2

u ae

8

u0 2

I

au (

r2)1l - = 11+-

--=-+-

ae

40 8

..... (2)

Diff.( I) partially w.r.to 'r' I

au - r

u

ar

au - u. r or 28

--=-=:)-=--

20

) au

-11.,.3

r--=---

ar

-

28

a( au) _-I [ u. 3r

ar

I .2 -

--

ar

2

28

+r

.1(--1/1")] 20

..... (3)

From (2) and (3) we get

r2

_3

r2

11+-=-+-

48

2

40

-3

.... 11=2

3.8.15 Example ?

If x- yY zZ

=

k then prove that at x = y

=

a2 z

z, - -

axBy

= [x log exr I


J48

Solution Applying logarithms to.r y zZ =

k,

we get xlog x + ylog)' + zlog z = log k.

Differentiating this implicit function partially w.r., to 'x'.

1

(I

x-+logx+ z.-+Iogz )az -=0 x z ax (I + logx) + (1 + logz).

/Illy

az

(I + logy)

ax

l+logz

az

=

ay

az

ax = 0

..... (i)

(I + logy) 1+ logz

Differentiating (i) partially w.r., to 'y'

{1+logz )-a2Z - +aZ[1 - - .az] - =0 ayax ax.z oy Substituting x

=

y

=

z, we get

a I =0 (I + logx)_z_+2

axay

-axay- - -

x

I I x{1 + logx) - x{loge + logx)

xlogex

or

- (xlogex)

I

Exercise - 3(F) x+4y OJ of 1. If f= 4x _ y' find -8 and 7) x

2.

If z

Y

az az

=

ax2 + bxy + by/inti ax' ay

-17y

17x (ADS: (4x- y)2' (4x-- y)2 ]

(ADS: 2q.x +

~y,

bx + 2by]


.

3. Ifz = (ax + by)(a\" - by), tind

OZ

{)z

-ax '-0y

oz oz

4. Ifz=tan(ax+ by), find sin(ax + by) 5.

Ifz=

[ADS: asec 2(a," + by), bsec 2(ax + by»)

ox'ay oz OZ

find

xy

[ADS: [

ox' OU.J'

Y2 )

a\"cos(a\" + by) - sin(a\" + by) bycos(a\" + by) - sin(ar + b x2y , x2y

6. If 11 = log(siny + ysinx), then prove that

a2 u

a211

ax~v

~vJx

-----

)

)

x- - y-

ayax 8. If u = tan

9. If u =

10. If 11

I

149

XY

.

~X2 + y2 + I

X

2

+ y-J

then prove that

--I .. x'-- prove that ct 2 e 4c 't

au ) a2u -=c--at ax 2 = .I(x + Ky) + g(x - Ky) then show that a2u = K2 a2u ay2

ax 2

I


150

V

II. If

1I

=~

z

z + - prove that x

x au + y ay + z

ax . uy

12. I f

II = log(x3 t-

y3

-f :;3 ~3xyz),

au =: 0

oz

prove that

all au

UII

3

ax

az

x+y

-+-t----=:~--

13. I f

1I

=.I(r)

14. [1' 11 =

W Ilere

x

~-

y+z

r

~)I

t- :;

I

I,,)

.jx" + y + z- prove t HIt

=

y

z

z+x

y+z

+ _.- -+ - - then prove that

lADs: -I)

16. If u = xyf (;

l'

prove that

au

all

x-- + y-- = 211 Dx . ~v 17. If x = rcosO, y

18. If 1-1 = logr, r

=

=

rsin0, then prove that

0 20 0 1 0 - , +-) =0 ox- ay 3 x + );J - x 2y - xy2, prove that 4

19. lf 11

=

x 2y + J,2z + z2x , prove that

au+ -au+ -au= (x+ V+z-)~ ax

~

az

.


20. If u = .x3 +

151

i - 3ary then prove that a2u

a211

21. If u = e ax + by j(ax - by), prove that

au

all

ax

0;

h-+a-=2abu X

22. If

2

- ?-

a- +u

y-)

z-)

+ - 2- + - 2 - - = I , prove that b +u c +u

-2( allax

+ (au)2 + (au)2 (-au)2 ax 0; az 3.9.1

au au) x-+ y-+zBy

az

Composite Functions If u is a function of two variables x and yare themselves functions of an independent variable I. then

au au ax au av at ax at ay al

-=-.-+-.-'

..... (i)

Suppose now that 'u' is a function of the variables x and y, and x and yare themselves functions of two other variables rand s. then

au au ax au ay ar ax ar ay ar all au ax au 0; -=-.-+-.'as ax as 0; as

-=-.-+-.-

..... (A)

..... (8)

II Ily if u is a function of rand s, where rand s are themselves functions of x and ythen

au au or ax as ax ar ax as ax au au or au as -=-.-+-.0; ar ay as ay -=-.-+-.-

..... (C)

The second and higher-order partial derivatives of'u' can be obtained by repeated application of the above formulae. Also, the formulae can be extended to functions of three and more variabl~s.


152

3.9.2 Example If z = fix, y), x = ell + e- v, y = e-l/ eV then show that

az az az az ---=x--yau av ax ay Solution

az GZaX azOy az u az au ax .au Oy .ax ax . av .

-=--+--=-e --e

az az ax az Oy

az

-=--+--=--e av fu·av Oy·av fu·

v

u

az Oy.

--e

v

By substraction

az - az)_( - - e (au av

v)aZ - - (" e -e v)aZ ax ay az y-az =x--ax ay

3.9.3

11

+e

Example If u

=

j(x, y), x

=

rcos(J, y

=

sinO prove that

Solution we know that

au = au. ax + all .ay = au cosO + au sin 0 ar ax ar Oy ar ax Oy au au ax au Oy au . au - = - . - + - . - = -(-rsll1 0) + -(rcosO) as ax as ay ae ax ay au lau au +2cosOsmO-.-+ . auau ()2 =cos-O.()2 (-Or )2+2 rae ax axOy 1


.

~

au 2 .

SIIl-O(- ) +SIIl 2 c3y

au -~ 0(-) ax

153

. au au ax ay

-2cosOsIIlO-.-+ COS 2

3.9.4 Example If u

=

j{x - y, y - z, z - x) prove that

au + au + al~ = 0 ax c3y az Solution Let X = x - y, y = y - z, Z = z -x then u

=

f(x, Y, Z)

all all ax au ay au az au au ax ax ax ay ax az ax ax az au au all au au az ay = ay - ax' az az ay

-=-.-+-.-+-.-=---

lilly

Adding all these gives

au + au + au =0 ax c3y az 3.9.5 Example If

z = j{u,

u x2 -y and v = 2xy az y-=2(x· az x-- y)az )ax ay au

v) Where

=<

prove that

?

(a)

(b) Solution

u = x2 - y, v = 2xy au au =-2y, av ax =2x - 'c3y ax =2y, av c3y =2x

0(-au )2 c3y


154

..... (i)

..... (ii)

From these, we get

at at 2 2 aZ x - - y - = 2(x + y ) ax ay au From (i) and (ii) we obtain

(-J

az -

ax

7

az +( By

az az az az)2 = {[ 4 x-+y- J? + [-y-+x- J? } au av all av

3.9.6 Example If x = reos Bandy = rsill B, prove the following:

ax 1. -=eose· as ' 2.

3.

ar ax

ax . e. -=-rsll1

as

x

~X2 + y2

,

(:r +(: r

=1

ay = sine· By = rsinE> ar 'as

'

ar By

y

~X2 + y2

,

as

-

ax

=-

y x- + y-? ?

as

'

By

x x2 + y2


155

Solution I. We have x we get

=

reos O. Differentiating this relation partially w.r.t., rand w.r.t., '0,

ox = cosO· -o=r . or ' ao -rsl110

-

..... (i)

Similarly, the relation y = rsinO yields

oy = sin {} oy = reosO or 'ao 2. From the relations x = reosO, y

.... (ii)

rsinO, we get

=

x 2 +1=,2

..... (iii)

y = tall 0 x

.... (iv)

or

Dr Differentiating (iii) paliially w.r.t., 'x' and w.1:l. ), 'we get -0 ;-0 x y This gives

or y or x x ox r Jx 2 + y" '

y ..... (v)

Differentiating again (iv) partially w.r.t., 'x' and 'y' we get

y( __1 ) = sec" 0 ao , x2

ox

~ = sec" 0 ao ~v

x

which gives

y = cos " 0 = --, y =, -ao = - -,

ox

x·

00

cos" 0

x

x

,.2

r-

-=--=-

oy

y , x- + y

..... (vi)

x

.... (vii)

3. From (v) we get

2 or)2 +(or)" =X +y2 =1 ( ox oy r- rr2

=< =<

(::l\(:r~1

..... (viii)


156

4. From (v) we also get ..... (ix)

..... (x) Adding (ix) and (x)

a2 ,. a2r ax ay2

I r

-+ 2 -=-

..... (xi)

5. From (ix) and (x), we get ..... (xii)

Differentiating

ar = Y

ay

partially w.r.t., 'x' we get

r

So that

..... (xiii)

From (xii) and (xiii) we get that ..... (xiv) 6. From (vi) we get

Similarly from (vii), we get ..... (xvi)

157


Adding (xv) and (xvi) we get 2

~

a 0 a-A 2xy 2xy 0 ----:? + - 2 = -4- - - . 1 = ax ~v r r

Note: I. The variables (I', 0) related to (x, y) through the relations x = rcosO, y = rsinA are the polar co-ordinates of the point whose cal1esian co-ordinates are (x. y). 2. In obtaining the resllits in (i) rand 0 are regarded as independent variables and x and yare regarded as dependent variables. 3. In obtaining the results (ii) - (vi), x andy are regarded as independent variable~ and rand 0 are taken as dependent variables. 4. After partial differentiation is carried out, the final expression for the partial derivatives are to be expressed in terms of the independent variables.

.

or ax ax ao ax ar as ax

.

ar is not the ax

5. From the expressIOn for - , - ; - , - , It may be observed that -

. I reclproca

0

fax ax.IS not the reclproca . I 0f or an d ae

ao ox'

ar = ax and ax = 1'2 as . Similarly ar = 0' ay = 1'2 as ax ar ao ax 0' ar ' as 0' 6.

I n Clact we note IJere

.

The results (i - vi) arc useful in calculations involving polar co-ordinates.

Exercise - 3(G) I. If u = x + y, v = xy and f = feu, v), show that

aj Of Of Of x-+ y-=u-+2vax 0' au av 2. If z = 1(r, y) and x = e" + e- X and y = e- II -ev show that

az az az az ---=x--yau av ax 0'

158


3. If z = ttu, v) where

II

'--=

OZ x - --

x2 - y2 and v = 2xy prove that UZ

2

()z

2

v- == 2(x + y ) -

ox . oy

4. If II

=

(!"siny. v

5. If Z

=

/og(u 2 + v).

OU

e'cosy and z =.f(1I, v) prove that

=

U =

eX' +/.

V =

!!..~ _ 2.\"(21/ 2 + I)

ox -

/1

6. If w =.f(u. v) and u = x 2

x2

-f

y, show that 2

OZ _ 4Y1l + 1 • ~v 1/ + v

-

2+V

-I, v = 2.\:v prove that

r 1 l--+-ov-

J o2 W U-W 4(.2 2) ) ) .\ +Y --+--=

olr

7. Ifu

x

~v-

J ;-,J o-W u-W 2)

OU

V and w =zandf=}(u, v, z

= -;V=-'--

z

ll~

show that

Of Of Of Of x-+y-+z-=w-

ox

8. If z = }(x, y), x

9. If z = }(u, v),

= 11

It =

oy

oz

cosh v, y =

Ix +

1I

ow

sinh v prove that

II~Y, l' =- ~v

-

I11X,

show that

a-J + z 0 2z (/ 2 0 Z+02Z - = +111 2)(2 2 ox ~2 0,,2 ov 2 10. If z =f(u, v),

U =

x2

-I,

v

=

1- x 2 show lhat

OZ oz x-+ y-=O ~v

ox

1


II. If x

ell /e/1/V.)' C~ ell sec)'

=

159

fino the vallie of

vv + }'-~l ( x~l!-ax + .}' Dy(~)' 1x (x ax'

lADS: 0

~v

(

12. If w

I Iog(y-? -. x-) ) . . II == 2 Ilmt an d v::= Sill =

I(

yx

I

II

y, )~ .. z2, 22 -- x 2) prove that

.f(x2 -

-.!. ow + _~_ ow + _~ all' == 0 x ax )' ~y z Dz 13. If w

=

j(x, y),

11

0=

e\ v

D2 w

eY show that

3 2 \1'

--:=

3x~v

14. If x::= -!vw,y::=

~

---.(111')

ouvv

j;;,z = -J;;;,

Prove that

o

ax . ~Y

oZ

all

ov

all'

where
3.10.1 Homogeneous Function

Definition A functionj(x, y,

Z, II .. .)

written in the form xl1

Ex.

I. z = sin-I

is said to be homogeneous in x, y,

g(Yx '!-,!~ ..... ) and is said to be of degree x X

(~ ) + tan 1(~)

Z=xO[sin-I(~J+tan 2•

Z

= x

2

+

y

x-y

Z, 11 .....

2

. IS

if it can be

11.

is homogeneolls and is of degree '0'. Since

I:]

IlOmogeneous an d'IS 0 f'd egree Isll1ce '

Engineering Mathematics - I .

160

3.

~ = x 3sin(~) + A)!ZIOg(;1+ XZ2 tan I(~ 1

~ x'ISin(~)<~lot H~r tan'(; llence

~

lJ

is of degree' 3' and homogeneous in x, y, z.

3.10.2 Euler's Theorem The following theorem, known as the Euler's theorem gives a very lIseful formula for a specific combination of partial derivatives ofhomogeneolls functions.

Theorem :

ffz is a homogeneous function ofx andy with degree '11' then

oz

oz

x-+ y-=nz ox

.

..... (i)

~y

Proof : Since Z is a homogeneous function of degree '11' we can write 'z' in the form.

z=

Now

xllf(~)

~: = Il - ~) + x l /(:)( ~;) I1X

=

1f

(

nxlllf(~)_xll 2y/(~)

and

xoxoz + yoz = f( xy) -x" 'y/ (y) + x" 'y/ (:Ix) x x ~y

I1X"

= nxllf(:Ix) =nz

[.:Z=Xllf(;)]


uz

oz

oX .

~V

x-+ V-=IIZ

Thus

·3.10.3 Ir z is a homogeneous functions of x and y with degree 'n' then 2 ]OZ

J

o-z uxoy

)

X - , + 2xy - - + y

ax -

20-Z

-i = uy

(

11 II -

)

1z

..... (ii)

Pr(}(l: Diflerentiating (i) pUl1iaily w.r., to x, we obtain

02 Z OZ 02 Z OZ x--+-+ y--=ll-2 ox ox oxoy ax 02 Z 02 z UZ x-) + yx--=(n-I)-

I.e.,

ox'

oxoy

ax

.... (iii)

Differentiating (i) par1ially w.r., to 'y', we obtain

02 Z 02 Z oz y-+x--=(n-1)oyJ oyox oy

..... (iv)

Multiplying (iii) by 'x' and (iv) by 'y' and adding we get

02 Z 02 Z ) 02 Z (OZ Oz) x--? +2xy--+ y-? =(n-I) x-+y- =n(n-l)z ox' oxuy oy' ox ry ?

3.10.3 General Formula The Euler's theorem may be generalized to homogeneous functions of more. than two variables. For a homogeneous function j(x, y, z) of degree 'n' in three variables, the theorem can be read as :

x oj + y OJ + Z oj ax ry oz

= nj

..... (v)

In general, for a homogeneous functionj(xl ....x,,) of degree 'n' in 'k' independent variables x\ .... x k the theorem reads

xl -

OJ

aXl

+x2 -

OJ

aX 2

+ ..... +xk -

oj

ax k =l1j

The proofs of (v) (vi) are left to the student

..... (vi)


162

3.10.4 Example

.

1/ 1I :~. cosec

-[;-;+JY] au __ ~ if;: + if;; ,thcn prove that x au ax + y ay - 6 tanu I

Solution We note that from the given cxpression for' u' we can write cosecll

;-;+JY . ;1.: f (say)

= "il

'Y.\"

+vY

=:

..... (i)

Then

:. fis a homogeneous function of degree

~,

Hence by Euler's theorem

W WI. I x-+ y-=- j =-cosecu ax

I.e.,

a

Oy

6

6

a

I x-(cosecu) + y-(cosecl~ =: -cosecli ax ay 6

..... (ii)

163


Carrying Ollt partial differentiation, we get

I

--cosecu 1 au a 6 x-+ y-=-=----=--tanu II

~v

ax

- cosec II. cot u

6

au + yau- = - -1 tanlt

x-

ry

ax

6

3.10.5 Example

x y z Ifu= j o( -,-,yzx

J prove that x-+y-+z-=O au au au ry

ax

az

Solution 11=

o(x

z)

y j -,-,y z x

all au au 0'( -----+---+x z x y y z x-+y-+z-=j ax ry az y x y z z x =0 3.10.6 Example If u = sin

1; + tan -1 ~

au

Solution

u = f(x. y) is of degree '0' By Euler's theorem

all

prove that x ax + Y ry = 0

J


164

au

ali

ax

0'

x-+ y-=o.u=o

.

,x - = v, y

Sill

Alter:

,y

tan - -

x

=w

u=v+w

au avow -::::-+au av Ow ax ax ax' ay 0' 0' x . 1 av -:::: Sill V, - = cosvy y ax y - y ? aw - = tan w, -} = sec- w.x xax -x av I ) aw --:::: cosv- - = sec"w.-

-=-+-

0"

y2

x

0'

=0

3.10.7 Example If

ZI =

X3 + 1'3)

tan' - __ ( x+y 0

(i)

prove that

au au . ax oy 2 a-u? au. . 2 a-u X - 2 +2xv--+ y-} =sm4u-sm2u ax . axay ay-

x - + y - = s1112u J

(ii)

-

?

Solution

Proof: For the given expression for 'u' we have

x3 + y3

tanu = ----'-- = f, say x+y


Ilere

~f'

165

is a homogeneous functions of degree'2'

By Euler's thcorem

at" of

.\"-' Ox +Oy I.e.,

.1' 7f' = 1l.J =~ --:

X~(lanl/) I~ y~~~(tanll)= 2tanll ax

~v

Carrying oul partial derivatives, we get

:"\ vII

? xsec-II-;-~

vx

all

?

+ ysecII

au

all

aX

(~y

x - + y~~~

2 tan 11

= --,,-

sec ~

- 2tanll

~v

.

ceo

S1l1211

..... (i)

u

Next dirterentialing bL)th sides of the resull (i) partially \Y.r., to 'x' and 'J" wc get alII

all

ax-

D.\·

Ou

alII

x ~-, + ~- + v .~-- =-- 2 cos 211 -

..... (ii)

a"11 D) II UII all x -~ + )' - , + -~ = 2 cos 211 ~~

..... (i i i)

(~VaX

. axD)'

~v'

Dx

~v

(Iy

Multiplying the result (ii) by x, the result (iii), by 'y' and adding the results we get

,a"II +

a~1I + y' x', ~-,(1.\'-

~-:,~

ay

'1

= (2cos211 =

a"II

~xy-- =

OXG)'

(2cos211 ~-l (all all) x - + y--Dx

-I ) sin211,

sin411 - sin211

3.10.8 Example

Solution Let

and

(from i)

~v


166

Then v and ware hOl1logeneous./llllclioll.\' (!ldegree '2 '. Applying 3.10.3 to 'v' and 'w' we get ..... (i)

!Illy

..... (ii)

(i) - (ii) gives

Note: Letj(z) be a homogeneous function ofx and y of degree Il ; then from Euler's theorem

a" =I/f + y_:'l ax Dy

('JI"

x_:'l

i.e.,

.' f)z , az x..f . ax + xi ay =

.

oz

OZ

Ilf

1?/ => X ax + y ay = 7

DifTerentiating (i) w.r., to y'partially

j

--JJ a'z aZ }I--=Il a'z [j") 'fj'" aZ \'-+-+ ' 2 . ax Dx . oxoy /2 -ax J

J

Differentiating 0) w.r., to 'y' partially 2

2

[/2/2-jt' 1

a z a z az x--+y-+-=n . axay ay2 ry

-az ay

..... (i)

..... (ii)

..... (iii)

(ii) x + (iii) Y gives

=

Il[j '2

~ f( ][x.az + y_~~]

j2

OX~}I

..... (iv)


167

Using (i) in (iv)

..... (v)

llere

3.10.9 Example If z = log(,3 + .vI __ x 2y __xy2), prove that (i)

a=

oz

x-+ v- =3

ax . ry

Note: (i) and (iv) can be llsed as formula in such cases

Hence

f

=

t!, f'

=

c=, f"

=

e= and

1/

az

From (i) of3.1 0.8 note

=3

u=

x-+ v--=3 Dx . ~v 1

Form (v) on. 10.8 note

3.10.10

)

,

lO-Z a·z 2a'z " x - , +2xv--+y -,=3.2--9=-_, ax' . axoy ~v'

Example Verify Euler's theorem for the function 1/(x, y, z)

=

Jx',z+y;

Solution 1I

is a homogeneolls function of degree

-=J. 2

..... (i)

So that

uu

,

, _,

2u- = -z(x- + v-) . . 2x 0;'" .


168

au

)

2u~=-z(x-

oy

).)

+ y) - .2y

-z[

au y~+z-·=au ou x--+

gives

ax

oy

oz

211

=

) I )

x- + y-

2 U I - - - -:;: ----1/

211

2

1 by (i)

Hence the result

3.10.11

Example J

If f

=

cos·

X

I

I

J

+Y -

3

Z

"X 1 -t )'4 - Z4

of Df of then show that x - + }'- + z-'- = -cotf AX . ~v oz

Solution ..... (i)

So that

coif=

11

From (i) we see that

11

is a homogeneous function of degree I.

By Euler's theorem

all

all

au

ax

~

az

x~+ y~+Z--=-ll

...... (iii)

Form (ii), we find that

au = -sin l. Of, ou = -sinf i?l, au = -sin fo.l ax . ax oy ~v oz . oz Using these and (ii) and (iii), we get

.f[

-Sill

Of) = cos j'

Of + y-'of + z-'-

x~

ox

ay

oz

Of Of Of x-- + y - + z~ = -cotf

ax

~v

oz

Mean Value Th.eorems and Functions of Several Variables

3.10.12

169

Example

, 2 2 2 . 1all au ,all ... Fmd x- - , + 2xy-- + y- - , ,If u = sm-- I ax' axoy ~~!.

' ,]1

x' + y3

[Xl + y2 ,

,

Solution

, x3~ + y31

Taking z = , , [ Xl + y2

]~

then z = sinu =.f(u) say z is a homogeneous function or

I degree - - = n 12

From Euler's theorem

au all j(u) x - + y - = 11-,ax ay f (ll) ,

2

I sin II 12 cosu

1

= - - - - = --tan II 1

12 .

1 ) ,"

a'u .... a 11 2 a-lI ( I) f 11- f -I x -+.ay--+y -,=1111- - , , a.'/ axay ay' f f 3 2

Again

=

(-~)(~ -I)tanu - _I_(-tan u).tan 12 12 144 {-:./ = COSll, (' =

=

3.10.13

2

1I

-sin 11)

tanu 1 --(13 + tan-u) 144

Example 2 1 _ Y )a u a 2II 1 a-lI If u = tan' ( -=--- ) ,find x' --+ 2xy--+ y--2 X ax axay ay

Solution

Let z =

az

2:. degree of z is 0 and z = tanu x

az

x-+ y-=o ax ay

az ax

1

au

-=sec'u-

ax

af.

,all

- =sec'uoy ~v


170

..... (i) x 02~ +

au + y

. uxoy oy

02U =

0

..... (ii)

~l

(i) x + (ii) y gives

) 0\'

OlU

(au au)

0\.

,

x--+2xv--+ y-=- x-+ y - =0 0'(2 . ax~y oy' ax oy

Exercise - 3(E) I. Verify Euler's theorem for the following functions:

xy x+y

(a) z=--

(b)

•

z = SIl1

(d)

( e)

_.

Z-SIl1

--I

(

X

2. If z = xyf

X 3

3}) Y +Y

(I)

3

(~) then show that y

Z = (x

2

3. Show the following by using Euler's theorem: (a)lfz=

x3 + / OZ OZ 5 . thenx-+y-=-z ~x+ y ax· ~Y 2

(b) Ifz= log

x 3 + y3

OZ OZ ,then x-+ y-=2 x+ y ax oy

3 3) ( x-y

x-y 87 OZ (c) Ifz=sin- I - - - - then x--=' + y - == 2 tan z

ax

~y

(

+ y-?+ z-?)

x OZ + Y OZ = 2z ax ~

Y1 ) x-y

1 X} -

1

2


l'

J

4. If z = tan-I ( x + /

lx+y 5. Ifz =

then x uz + y OZ = sin 2u

8x

sec~l[x2 + yll, then

'~y

x a::: + yUZ =cotz

x+ y

~

ox

6. If z is a homogeneous function of x and y with degree' 11' and z = f(u), prove that

au

(u)

ux . ~

/(11)

all

x-+

7.

If z = x 3

. (y) 1 SIll·-1 - ,show that x

8. If z = tan-I ~ X4 + y4

az ax

oz oy

,

y-=n-'--

a 2

a 2

z + 2xy-11 fiz x 2 --T + y2 -2- = 20z

ax-

axay

show that

.

(i) x - + y - = sm2z

,a

a

2 z 0 2z , 2z . . ii) x) + 2xy + v-) = s1I14::: - Sill 2z (

ax'

az ax

ax~v'

~v'

8z I (i) x-- + y - = -tanz

ay

2

1 10.lfz=tal1- 1 [;l+YSin- [.:l show that

II. If z = cosec-I

\

\-

x2

+ y2

then show that

~y

171

Engineering Mathematics - 1

172

2

X

a2 z -2

ax

0-" z 2 0-" Z + 2xy - - + y ---:;aX~Y

ay

ll3

tan z 1 J ] = - - -+-tan- z 12 12 12

2 2) ,then show that

12.lfz=tan- 1 x +y

(

x-y

2

13. If z = log rand ,2 = x 2 +

14. If z = sin-I

X4I

+ y4I

!

!

r

y

2

2

axay

ay

"a z a z )a z then show that x- - 2 + 2xy-- + y - " + 1 == () ax

j then show that

x5 _ y5

')

2

a-z

2

a z )a z 1 2 X - - 2 +2xy--+ y - , =--tanz(tan z-19) ax ax~ ay 400 2

~ x 2+ y2] 15. If z = log (

x+y

;\2;\ 2 ; \2UZ 2 2uZ uZ

find the value of x

GX2

+ 2xy axay + y ay2

3.11.1 Jacobian Ifu and v are functions of two independent variables x and y then the determinant

au au ax Ov ax

~ i.e.,

Ov

ux

uy

v,

vy is called the Jacobian of u, v with respect to x and y

au

and is denoted by

a(u, v) a(x,y) If u,

l~

or

.J(~) x,y

ware functions of three independent variables x, y, z then the determinant


lI(

UI'

1I,

V(

vI'

Vz

W(

Wy

lV,

is called the Jacobian of 1I.

V,

w with respect to x.

a{/I, v, w) a{x,y,=)

)~

173

z and is denoted by

)(U,\" w)

or

x,y,z

Similarly aUI

uU I

a{U p 1l2·····uJ _ aX I aX 2 a{X I'X 2.....x,,) - all" OU" oX I aX2

(J/I~

ax" au" ax"

SOME PROPERTIES OF JACOBIAN 3.11.2 Property

a(u,v) a{x'Y)_1 a(x,y)' a{u, v) where

J = a{lI,

vI and

a{x,y)

_ a{x,y) .J - 0(11, v}

Proof: Let u = u(x. y) and v = lI(x. y) so that u and v are functions of x and y. Then differentiating these partially each with respect to II and v, we get I :=

au . ax + all . 0' ax all

ay au


174

av ax av ay ax au ay au

O= - . - + - . -

1=

Now

av .ax + av .ry ax av ay av au ax av ax

a(lI, v) a(x,y) a(x,y)' a(u, v)

all oy av ay

au -au ax ay av av ax ay

ax ax all av ry ay au av

x

-

ax au ax av

-

x

ry

all ry

av

(Interchanging rows and columns in 2 nd determinant)

ali ax au oy ax au oy au av ax av ry -.-+-.-au au ry au

au ax au ay ax' av ay' au av ax av ay --+--' ax' av ~v' av

-.-+-.- --+--

=I~ ~1=1

Proved

3.11.3 Property

a{u,v) _ a{u,v) a{r,s) a{x,y) - a{r,s)' a(x,y) where u, v are functions of r, sand r, s are functions of x, y.

Proof By differentiation offunction ofa functions, partially; we get

au au ar au as ax ar ax as ax

-=-.-+-.-;

Mean Value Theorems and Functions of Several Vanables

175

au -+--'---_. or oy OS

ou

-=-

or .~ as· oy ,

~

..... (i)

oval' - + avos -_. ox or .ox os· ox ' Ov 01' or 01' as -=- -+--_. ~y or .oy os· ~y , 01'

-=~-

and

all au

o(u, v) a{r,s) (1(r,s)· a(x,y)

(lr

=?----=

uv i"lr

os elv us

x

(lr (Ix us

(Is (Iy us

ilyl

(!y.

or as or os x ax ax ov -oV (II' as or as oy oy

Oll

all

(Interchanging rows and columns of the second determinant)

or ou os or ox os ox Ov or OV os -.-+-.-or ox os ox

or Oll as or oy os oy ov or ov as -.-+-.or oy as ~

all

ou

-.-+-.- -.-+-.-

ou -ou ox ~ _ o(lI,v) ov Ov - o(x,y) ox ~

-

using (i)

-

3.11.4 Property If u, v are functions of two independent variables x and y thell u, v, are independent if

.

Otherwise they are dependent. oo((U,V))oFO. x,y

.

.

. o(u, v, w)

(functions of x, v, z) are II1dependent If and only If

(

a x,y,Z

)

oF

Similarly if

0

II,

v,

W

Englneenng Mathematics - I

176

Theorem If the functions 11 1, 1I::...... 1I11 of the variables xI' x 2 ....• xn be defined by the relations Il, ==.ll'",), 112 =flx ,. x 2), 113 = .li x ,. x 2• x 3 ) ..... 1/11 .I;,
O(lI p 1/2 .....U,,}

Then

OU, UU] OU 3

o(x"x2.....x,,) -

ox, . oX1 'a~~

OU II OXII

We know that Similarly lor

Oil,

Oil 2

ox,

ox,

Oil,

0111

0(U"l/2 .. ·..U,J _ aX 2 o(x, 'X 2 ..... X II )

Then

oX1

all,

~1I2

aU

OXII

OXII

OXII

OU,

Oll]

OU"

ox ,

ox ,

ox ,

Olll.

Ollil oX1

0

oX 2

o

OU, Olll ox , . oX 1

o

3.11.5 Example

If x

= 112

-vl . Y

_ o(x,y)

=

2uv ,. tllld --( --) 0 1l,V

Solution We have

x

=

u2 - v2

~

-ox

all

=

2lI.

ox

--=-2v OV

Ii

OU II OXII

177


cYv

0)1

v = 2uv => --- = 2 v -"-- = 211 V I I ' 01'

-

ox AX V{X,y) _ or ()I~ ~ 1211 - 21'1 O(II.~) - VI' ~F 21' 211 au VV

:=

4(u-» +) 1'-

)

3.11.6 Example

V(x, y) V r,O

a( t,

°)

If x ~ reose. y ~ rsintJ find - (- ) lllld-(--)

,

a x,y

Solution

x = reosq, (It

=>

~ =

y

rsinO

=

cos()o

(1,.

-ax = -rsiIlO,

ov = rc()sf)

---

0(1

of)

ax ax

Then

o(x,y) _ or

ao

12 = x 2

- sinO

ao

or Again

-rsinol . r Sill 0

_Icoso

~v

0(1',0) - ~v

+ y2 ; 0 =

tal1- 1 ~

x or

2r-- = 2x

ox

-- y

(l()

ax

0

r+y or

-

ax Dr ay

-y 0 r-

0

y

,.

+x

= -

I'

(1()

ilx

x2

-y

-y

+)'2

r

)

=

r(cos 2 0 + sin 20)

= r


178

x

x

lar

ar

x

a(,., 0) ~ ax ~}! aCy,y) ~ ao ao --

Y y

~

vy

lay

3.11.7 Example If x

=

uv, y

=

11 -

then show that .1./ ::.: I

v

Solution ax

We have

-=V,

all

al'

~F

ay

v

VII

But

')

lr =

aI'

=

11

--)

v·

VU )I, ax .

DII 2u-=x

2u- =

xy

,

and

at

-=11

v·

x

=-, 2v

J =

y

av =!...

aI'

ax y'

21'-= ay

y

x

211

II"

lly

2u

Vr

v)'

I

~x

2vy

21:J!

X

X

4uvy

411vy

ay

X

2uv· Y

I? =---v211v

V

2u

x --?

y.


179

- 3.11.8 Example Prove that JJ

1 for x

=

e V secu, y

=

e V tanu

=

Solution

ax = e secu tanu, all and

oy

au

=

=

- = ('

"

I

I'

)

e sec1I =

ye-

.

tanu

v

e secutanu

2v =

e secu

=

e" tan 11

II

1

Y

eVtanu

--e" secll

x

y

SlIlli = X

U

(Y) an d v '21 Iog(x

·-1 = Sill -;

Du

y

X~X2

i1x

av ax

=

x- -

X~X2 -y 2

(Jy

- /

Dv y

? '

- Y-))

all

,

x 1

2

y ?

?

x- - y-

i1y

Y J

lIx

lIy

Vt

VI'

X~X2

_ y2

X

x2 _ y2

~X2

_ y2

Y

x-1 -

1

y-

2"

-e secu = -xe

1

y, also

e2v = x 2 -

v

av

sec 2u - lan 2

x 2 e- 2v -

e V secli

0;

eV sec 2 u;

x" Yv Now

ax av

-=

I'


180

, \' I JJ =xe ,-=1 \'

xe

3.11.9 Example VIV Wlt lIV I I I O(X, y, z) If x=-,y=--,z=-- t Jen SlOW tlat - ( - '-)=4

v

II

W

01l,V,\I'

Solution

ax Oll

o(x,y,z) _ oy 0(1/, V,-H')

Oil OZ 011

ox ov oy ov oz OV

oX

VI\'

all'

2

u

lI'

v

It

1/

11'

Wit

II

V

2

oz

V V

v

ow

IV

IV

oy ow

- VlI'

Wli

VlI'

- Wli

VlI'

lIW

II

uv

ltV ltV '-2 '-2 ' 2 II v W --l/V

-I -I -I =-1(1-1)-1(-1-1)+ 1(1+1)=4

Exercise - 3(1) I. If y + x + Z = 1I, Z + Y

=

1Il: Z = lIVW, find o(x,y,z) o(u, v, w)

[ADS:

(1I

2

v) J

lSI


2.

If

II

2/ - Z2, V = x2yz, W = 2X2 -xy find

= x+

o(u, v, w)

at the point (1,-1,0)

o(x,y,z)

Ans:19

3.

If x==rcosB,}J==rsinB, find .

o(x,y)

o(r,O)

o(r,O)

o(x,y)

--'--.

~(x,Y)

and

!JS!~~~ 8(x,y)

and show that

= 1 (Hint :relcr 3.6.6) X X,

XOXI

8(r,0)

XIX)

o(YI,VnY,)

4.

I ,y, = - - - ,show that ----'----'--== 4 If YI = - --- ,Y) = --XI x2 x, 0(X I ,X2,X3)

5.

If

6.

If x=a(coshu)cosv,y=a(sinhll)sinv, then show that

YI ==1-XI'Y2

o(x, v)

a

=xl (l-x2 ),y] ==xlxl(l-x,),provcthat

O(YI' Y), y, ) 0(X I ,X2 ,X,)

= (1)3 - XI 2 x"

2

--,-'- = - (cosh 211 - cosh 2v) 2

0(11, v)

7.

If II = /(x l ), v = ¢(x l , x 2 ), W = If/(x!, X 2' Xl) then show that

Oell, v, w) 0(X I ,X2 ,X3 ) 8.

If

X

. 1')

= r S1l1 r.

Solution:

011 Ov

ox

do • 0 . do 0 I oC X, y, z) ). 0 cos 'I', Y = r Slll S1l1 or, Z = r cos - S lOW that - - - = ,.- S1I1

o(x,y, z) o(r,O,¢)

=

uw

=--' oX • ox} 2 i

o(r,a,¢)

=

ox

ox

uy

or

00

o¢

ay

oy

uy

--

or

UO tJ¢

OZ

OZ oz --

or

oe

sinOcos¢ reosOcos¢

= sin Osin ¢ cosO

r cos Osin ¢ -rsine

- r sin Osin ¢ rsinO eos¢

o

o¢

sin ecos¢[ 0 +,.2 sin 2 Oeos¢ ] + r coseeos¢.r sineeosecos¢ +

+( -r sin esin ¢)[ -r sin 2 Osin¢ - r cos 2 esin ¢ ] == r2 sin 3 Oeos 2 ¢ + r2 sin Ocos] Oeos 2 ¢ +,.2 sin 3 Bsin 2 ¢ + r2 cos 2 Osinesin 2 ¢ == r2 sin 3 0 + r1 sin () cos 2 () = ,.2 sin O.


182

3.12.0 JACOBIAN OF COMPOSITE FUNCTIONS: Let (u,v) be functions of x,y where x,y themselves are functions of r,e .Then 1I,V arc

. fi .. . e 8(1l, v) composite unctlOlIls of r, and

= 8(u, v) . 8(x,y)

.

(from 3.6.3)

8(r,B) 8(x,y) 8(r,B) Note 1: Ifu,v are functions ofx,y we may regard x,y as functions ofu,v. Taking r = ll, = v in the above result, we have

e

au

8u

8u 8v

8v

1 0

av

0

811

8v

-

8(1l, v) 8(x,y)

---

8(x,y) . 8(1l, v)

..

=

8(u,v) 8(1I, v)

8(1l, v) 8(x,y)

=

8(x,y) . 8(1l, v)

=1

1

1 => .1.1 1 = 1

I where J = 8(1l,:i,J = 8(x,y)

8(x,y)

Note 2:

8(u, v, w) 8(r,e,¢)

are functions of

8(1l, v)

=

8(u, v, w) 8(x,y,z)

.

8(x,y,z) 8(r,e,¢)

where u,V,W are functions of x,y,z and x,y,z

r, B, ¢ .

Also

J./' = 1; where J= 8(1l, v, w) , .II 8(x,y,z)

3.12.1 Example: If

II =

= 8(x,y,z)

8(u,v,w)

xyz, V = x 2 + y2 + Z2, W = x+ y+ z find .I = 8(x,y,z) . 8(1l, v, w)

Solution: we first evaluate .I

I

=

8(1l, V, w) 8(x,y,z)

,

since u,v,w

yz zx xy functions of x,y,z.

.II

= 2x 1

2y 2x

1

Hence from the relation that .1.1

1

= -2(x -

y)(y - z)(z - x)

1

= 1,

we have

J = 8(x,y,z) = -] 8(u, v, w) 2(x- y)(y-z)(z-x)

are explicitly given as

183


Exercise - 3(G) I.

)

1

C(ll, v)

•

o(r,O)

2.

If x =

;;;;,y = Jwu,z =-r;;;;

where II = r sin OcosrjJ, v =,. sin Osin rjJ, W = rcosO, show that

3.

3

Ifu=2xy,v=x--y, whcrex=rcosO,y=rsmO show that ---=-4,..

·

If x =

ltV )' ,

v(x,y,z) a(r,O,rjJ)

ll+V

_

= -1I-V'

1 4

= --,.

2

•

Slll

0

v(u,\')

(1I-V)2) (A liS : --'------'-

fll1d - - o(x,y)

411v

·

yz zx xy 8(x,y,z) V = HI = prove that = x ' y , z 8( ll, V, lV) 4

·

x +Y

4.

II II = -

5.

1I1I=--,O=tan-l(x)-tan-- (y) show that =0. 1- .xy v( x, y)

6.

If lI=x--2y,v=x+y+z,w=x-2y+3z show that ----=1Ox+4

•

D( ll, 0)

I

v(u,V,lI') 8(x,y,z)

1

3.12.2 Jacobian of implicit functions Let

1I1' 1I2' 1I3,

be implicit functions of x,,

jJu" Then

then

1I2' 113'

x 2' x3

x" x2, x3 ) = 0,

so that r= 1,2,3

oj; + a.t; .aU I + a.t; .aU 2 + a.t; .aU 3 aX I aUI oX I oU 2 aX I aU 3 oX I

= 0 and so on

a(.t; ,j~ '/1) 0(11" u), uJ -a(u"U 2 ,1I))' a(X I ,X2 ,X3 ) I

aj~

all,. =

ax!

I af2 . OUr Ollr

I

ax!

a.t:l }u au,.

(by the rule of prodllct of determinants)

oj~ . OUr I a.f! . Ollr au, oX2 all, oX3 I of2 . all,. I a.f~ . all r au,. aX 3 OUr aX 2 I af3 . au,. I 'fJj~ . OUr all, aX3 au,. oX2

. all,. I

r

ax!


184

_ aJ; ax, _ ~f2

au, 0f, ax,

W, _ ~t; aX2 ax)

_ aj~

- aj~

au) 0f, ax)

QU 1

aj; aX2

= (_ I)' a(J;,j2''/;)

a(x"X2 ,X3 )

aCt; ,j~,f,)

a(u" 112' uJ _(_1)3 a(;;~-x~~~,"J a(x, x2 ,xJ ~V;,fI,lj) , a(U"l/2,lIJ

Hence

(The result if obvious)

3.12.3 Example

x y z If u = r:--:;' v = r:--:;' W = r:--:; prove that v'1-r2 v'1-r2 v'1-r2

a(u, v, w) _ 1 a(x,y,z) - (l-r2 where

-? =

x2 +

(1-/ 2 )

=x2

r+

y;

z2

Solution we have

II

W, x, y,

=>

II(u, v,

Similarly

./; == v2 (1 - x 2 ~

13

z)

== w2 (1 - x 2 ~

= u2 (I _x2 -

r - z2) - r

r-

r

_z2) -

=

0

z2) - z2 =

0

x2 = 0

.... (i)

(see 3.13.2)


aCt; ,.I~ ,.I;) _

J

- 2u 2 x- 2x

- 211-Y

- 2112 z

-2v 2 x

-2v 2 y-2y

- 2v 2 z

8(x,y,z)

-2w 2 x

- 211'- y

)

(-2x) (-2y)(-2z)

=

- 2w 2 z -2z

)

u2 + 1 v)

w-

185

?

)

u-

11-

2

v +1

V

2

w2 +1

)

11'-

u-?

0

)

v-

-8xyz -I

w2 +1

-I

0

by C 1 -C 2

C2 -C 3 =

-8xyz(u2 + v2 + w2 + I)

=

-8xyz

[

X2

y2

?

')

')

x- + y + z- + 1-

=-8xyz

;also.

2u(l-x2 - / 8(ui' v2 , w3 )

x

o

_Z2)

o o

=

8l1vw(l- Xl

8

,.,

1'-

1-r-

=--2

8U;,.I;, .I;)

]

J

-8xyz I-r

Z2

+--- 2 +--? + 1 I-r- I-r l-r-

--J

~

.

o

2u(l_x2 -

0

y2 _Z2)

o

2u( 1- Xl

_ yl _ Z2 )

_ y2 _ Z2 )3

Y

JI-?

.

z

(l_r2)3=8~yz(l_r2)112

.J1-r 1

_. 8(11, V, w) 1)3 -8xyz ~ Hence trom (I), we have - - - - = (- - , 8 (I 2)1/2 8(x,y,z) I-rxyz -r

=

I

r:-? ',"1_1'2

X Y Z Cor: Given U=- V=- W=k' k' k

where k

= VII -

X

2

-

' f iIn d •,(x,y,z) Y, - Z-, (u, v, w)

in terms ofk. lAns: k

S

]


186

Exercise - 3(H) I.

Ifx+y+z = u ,y+z = uv, z = uvw, then prove that

8(x,y,z)

----'----=--~

=- II

2

V

8(u, v, w) ~)

8(u,v) x--y2. Ifx + l + u _v = O,uv + xy = 0, prove that 8(x,y) u 2 + v 2 then 3. If u1 =x1 +X2 +x3 + x,pU 1U 2 =x2 +X3 +X4 ,1lIU2113 =x4 8(X p X2 ,X3,X4 ) 3 2 =1I1 .il2 ·113 8(u l , 112 , u3' 114) 4. If u 3 +V3 +W3 =X+ y+z,u 2 +V2 +W2 =X3 + / +Z3, 8(1I,v,w) (y-z)(z-x)(x-y) t Ilen SlOW t Ilat = ---"'----'---'..-----'-----'-----'----"-I 8(x,y,z) (v- w)(w-u)(u - v) 2

2

2

prove

that

3.7.4 Use of Jacobians in determining functional dependence and independence of functions. Let u and v be two functions of x and y connected by the relation v = f(u), then we say that u and v are functionally dependent. We shall prove that the condition for functional dependence is

o(u, v) = 0 o(x,y) Consider

w = v - feu) = 0

Now w is a function of u and v where u, v are functions of x and y. w is a composite function of x and y

ow ow au Ow Ov -=-.-+-.-

ax ax ax ov ax ow ow au ow Ov -=-.-+-.ax au ax ov ax But w considered as a function of x, y is identically zero. i.e.,

Ow -=0

ax

'

..... (i)

..... (ii)


187

From (i) and (ii) we get

all ow ov au ax av ax

Ow

-.-+---.--=0

Dw

all ow av

au

~v

--.--+-.-=

and

Eliminating

D1'

..... (iii)

0

ay

..... (iv)

Ow aw -a-·-o from (iii) and (iv) we have 11 I'

all aI' ax ax all -av Oy

=0

~v

au au ax ay ou au =0 ax Oy

--

a(u,11=o

=>

a(x,y)

The concept of functional dependence can be extended to any number of variables. Thus if u, 1\ 11' are functions of x. y, z the 1I, v, w will be functionally dependent (i.e., there will exist a relation between ll. 1\ 11') if

Q~!_,V, w) = 0 a(x,y,z) 3.14.5 Example Are x -I- y - z, x - y + z, x 2 + y2 + z2 - 2yz functionally dependent? I f so, find a relation between thelll.

Solution Let

lI=x+y-z

..... (i)

1'=x-y-l-z

..... (ii)

11' =

Xl

-I-

I

-I-

i2 -

2yz

..... (iii)


188

o(U, v, w) = o(x,y,z)

Then

\ \

2x

-\

-\

2y-2z

2z-2y

-\

-\

2y-2z

2y-2z

= 2x

=0

(since 2nd and 3 rd columns are identical)

:. u, v, ware functionally dependent To find the relation between u, v, w we notice that, from (i) and (ii) 11

+ v = 2x,

v = 2(y-z)

U -

From (iii)

u2 + v2 = 2w, the required relation 3.14.6 Example

i

11,

If 11 = x + Y + z, v = x 2 + y2 -I- z2, W = x 3 + + z3 - 3xyz then prove that l~ ware not independent and find the relation between them.

Solution ..

ti

c:

.

CondItIOn or lunctlOna

Now

Id d . epen ence IS

o(u, v, w) 0 ( ) =

o x,y,z

o(U, v, w) 2x 2y 2z o(x,y,z) - ? ? 2 3x- -3yz 3y- -3zx 3z -3xy

=6

X

x

2

-

yz


=6

189

0

0

x-y

y-z

(x- y)x+ y+z

y - z(x + y + z) z- -xy

o =

6(x - y) (y - z)

z )

o z

I

x+ y+z

x+ y+z

z -xy

=0 11, V, IV

are functionally dependent. We shall now find the relation between them

we have

w

Now

u2 -

=

(x + y + z) (.x2 + I + z2 - xy - yz - zx) V =

..... (A)

(x + y + z)2 - (x 2 + I + z2)

= 2(xy + yz + zx) We write (A) as

W=U

-[v -U-V] 2

Thus

3.14.7 Example

x+ y x-y

Test whether u = - - , v = (

xy )2 are functionally dependent and if so, find x-y

the relation between them.

Solution Treating'lI, vas functions of x, y the condition for functional dependence is

a(u,v) =0 a(x,y) x+y x-y

11=--,

xy

V=-~-c-

(x- y)2


190

We have

ux

=

{x-y)'I-{x+y)'1 (X_y)2

II

==

--'---~---'--'---'--~'--'--'--

=

(X- y).(I).{X+ y){l) (X_y)2

y

-2y {X_y)2 2x {x-yf

= {x- y)2 y-2{x- y).l.xy

V

(x _ y)4

x

_ {(X- y)y-2xy)

(x- y)

-

xy - y2 - 2xy _ - y{y + x)

{x- y)3

- (X- YY

X(X+ y) x-y )3

lilly

Vy={

a{u, V)

lIx

ll"

a(x,y) = Vx

Vy

=UV - v v x y x y

'-2y 2 .x{x+ y} + y{X+ y!. 2x J =0 (X- y) (X- y) {X- y} (X- y)•.

U,

v are functionally dependent

We shall now find the relation between them We have

x+y x-y

u=--

x-y x+ y

u

By componendo and dividendo

x

1+u

Y

I-li

-=-

A

191


Now

Y

xy

o X

v = (x - y)' = (;

r

V

1

x

1+11

y

l-lt

'V=C~:~2 J u2(4v+ 1) = I The required relation of functional depedence between

3.14.8 Example Examine for functional dependence of u = sin-I x + sin-1y

v=x~l-y2 +y,JI-x 2 and find the relation between them, if it exists.

Solution We have

u =-===

,

~1_x2

lilly

a{u, v)

1I (

It )'

a{x,y)

Vx

Vy

lI,

v.

by (A)

192

:. ll, V


are functionally dependent we shall now determine the relation between u and v Let A=sin-1x,B=sin-1y; =>x=sinA y=sinB The given function can then be written as u = A +B 2

2

v = sin A)1-sin B +sin BJl-sin A = sin(A+B)= sinu Hence the required relation of functional dependence between u and v is v = sinu

3.14.8(a) If u

= eX sin y, v = eX cos y,

show that u,v are functionally independent

au ax

au ay

av

av

ax

ay

-

a(u, v) Solution: The Jacobian a(x,y)

X

•

X

e smy e cosy eX

cos y

_ex

sin y

=

_ex

* o.

:. x, v are functionally independent.

.

x-y x+z

x+z y+z

3.14.9 VerifY whether u = - - , v = - - are functionally dependent, and if so find the relation.

Solution: Treating u,v as functions of x,y regarding z as an absolute constant, the condition for functional dependent in

a(u, v) a(x,y)

=0

au ax

Now - =

,

y + z au 1 av 1 av = - = - - - and -=~-, (x+z/' ay x+z ax y+z ay

x +z (Y+Z)2


B(u,v) B(x,y)

=

y+z (x + Z)2

193

---

X+Z

=0

x+z (y + Z)2

1

y+z

u, v are functionally dependent.To lind the relation between u and v, we eliminate 'x' between the given relation viz: u

x-y

= ------------- (A),

x+z

v = ---------------(B)

X+Z

y+z

+Y 1-11

liZ

from(A) ux+vy = x-yo =>(I-u)x=uz+y => x = - liZ

+Y

Substituting for 'x' in 'B', v = 1-11

+z

y+z

=

(y+Z) I-l/' y+z

I-u

Hence the required functional relation between 'u' and 'v' is v( I -u) = I.

Exercise - 3(1) I.

Verify whether the following are functionally dependent and if so, lind the relation between them.

x- y x+ y x+y x x-y ( b) u = - - , v = tan -\ x + tan -\ y I-xy (a) II = - - , V = - -

(c)u=x-y,v= xy x+y (X+y)2 2.

(Ans: u( I+v) = 2) (Ans: v = tan (Ans: 4v + 1I

2

-\

II)

=1)

Prove that the following functions are not independent.

= X + 2y + Z, v = x- 2y + 3z, W = 2xy - 2z + 4yz - 2Z2 2 (b) II = x+ y- z, v = x- y+ z, w= x + l + Z2 - 2yz 3 (c) u = x + y + z, v = xy + yz + zx, w = x + l + Z3 - 3xyz

(a) II

3.

If u

= x + y "v = y + z Z

functional relation.

x

w

= y(x + y + z) , xz

show that u,v are connected by a


194

4.

x-y l-xy

Examine for functional dependence between u = - - , v = tan-I x - tan -I y.1f ( Ans: u=tan v)

dependent, find the relation. 5.

Prove that functions

II

=Y + z, v = x + 2z 2 , W = x -

dependent and find the relation between them.

4yz - 21 are functionally

(Ans: V=W+2l?)

3.15.1 Maxima and Minima The method of finding the maxima and minima of functions of two and three independent variables is discussed in this topic. The maximum or minimum values are also called Stationary or extreme or turning values. For finding these values we use Taylor's theorem for functions of two variables.

3.15.2 Taylor's theorem for a function of two variables: We know that Taylor's theorem for a function f(x) of a single variable 'x' is

f(x+h) = f(x) + hj\x) +

~~ /I(X)+ ...... .

....( I) Now, consider f(x,y), a function of two independent variables. Keeping 'y' constant and following (I) we get

. a h a j(x+h,y+k) = f(x,y+k)+h-f(x,y+k)+---J j (x,y+k)+ .........(2) 2

ax

2

.

2! ax-

Keeping 'x' constant and by applying (I), we get

a. ea f(x+h,y+k) = f(x,y)+k-j(x'Y)+---2 f(x,y)+ ...... . 2

ay

2! ay

.... (3) Substituting (3) in(2) we get

. a. ea j(x,y)+k a;f(x'Y)+2T ay2 f(x,y)+ ...... . 2

f(x+h,y+k) = [

]

2

a [ f(x,y)+k-j(x,y)+---) a. e a f(x,y)+ ...... .] + hax ay 2! ay 2 2 112 a [ aj. a ] + 2T al f(x,y)+k iY (x,y)+2T al f(x,y)+ ....... +..... .

e


195

Substituting (iii) in (ii) we get 2

2

f(x + hy + k)

=

k a f(x,y)+ ....] a [f(x,y)+k-f(x,y)---2 ay 2! ay

2 a [ f(x,y)+k-f(x'Y)+---2 a a f(x,y)+ .....] +hax ay 2! ay

e

2 2 2 h a [ f(x,y)+k-f(X'Y)+---2 a a f(x,y)+ .....] +---2 <2 ax ay < 2 ay

e

+ .....

[

aj

[aj

2 2

2

k a f + .....] + h-+hk--+ a f .....] = f(x,y)+k-+-2 ay

< 2 ay

+[~ a <2 ax

+. . .]

2

{

f(x + h. y + k)

Thus

=

ax

f(x, y) + (h! + k

axay

~ )f

(a

a)2 f+·····

1 +~hax+kay

..... (iv)

Expression (iv) if known as Taylor's expansion for functions of two variables. Writing x

=

a, y

=

b expansion about (a, b) is given by

a

a)2

[a

a

If(a+h,b+k)=f(a,b)+ ( hax +kay

ira,b)

1 h-+k- ]2 i((jb)+ ..... +1! ax ay , Now if we call h = x - a, y - b = k we get

f(x,y) =f(a, b) + ((x-a)!

+(y-b)~)i((j'b)


196

1 ( (x-a)-+{y-b)a a )2 fiab) + ..... . +-

ax

2!

ay'

..... (v) a = 0 = b, we have

Writing

f(x, y)

j(0, 0) + (x

=

~+y;

)./(0,0) + ......

.. ... (vi)

This is Maclaurin's series for two varialbes. 3.15.3

Example

Solution

Ix (x, y)

=

fyy (x, y)

=> f{I,I) = e 2

f(x, y) =- ~2 + y2

We have

=

2xe x

e

2+ 2 Y

X2 + y2

=> Ix (I, I)

2

+ 4Y eX

2

7

=

2e

2

=> fyy (I, I)

+Y-

=

6e2

Hence putting these values in second form of Taylor's theorem, we get e,2+y2

1 =e2[l +2(x-I)+2(y-I)]+ 2! [6(x-I)2+8(x-I)(y---I)+6(y---If+ .... ]

3.15.4 Example

Expand

~

siny in powers of x and y as far as terms of third degree

Solution f(x, y)

= ~siny

=> j(0, 0) = 0

Ix(x, y)

= ~siny

,=>Ix(O, 0)

=

J;,(x, y)

= ~co,';y

=>J;,(O, 0)

=

0 1


197

lxlx. y) = ~siny -=>lxx(O, 0) = 0 lx/x. y)

= ~C()sy

/y/x. y)

=

-=>fxy(O, 0) = I

~siny -=>/y/O,

0)

"=

0

lxxi'" y) = ~siny -=>j~x..l0' 0) = 0 lxx/x. y) = ~sillY -=>j~-X/O, 0) = 1 j~J'Y(x, y)

= ~siny -=>fxy/O, 0) = 0

fy;./x. y)

=

~C()sy -=>1;'JY(x, y)

= -1

Then by Taylor's theorem we have f(x. y)

j{0, 0) + [xlx(O, 0) + y1;. (0, 0)]

=

I 2 .2 + 21 [xfxx (0,0) + 2xylxy (0,0) + yjXy(O, 0) ] + ...... J

J

y-

x-

J

x-

0 + x(O) + y( 1) + - (0) + xy( 1) + - (0) + - (0) + ..... 226

=

2

i

x y =y+xy+---+ ..... 2 6

3.15.5 Example j(x. y)

=

x 2y + 3y - 2 in powers of (x-I) and (y + 2) by Taylor's theorem

Solution j(x, y)

f(a, b) + [(x-- a)lx (a, b) + (y - b)fy (a, b)]

=

I

21 (x -

+

a)2 fxx(a, b) + 2(x -a) (y - b)j~ (ab)

+ (y - b)2.fyy (ab) + .....

Here a

=

.... (i)

1, b = -2 f(x. y)

=

x 2 + 3y - 2 -=> j{ 1, -2)

=

-10

fJx. y) 2xy -=> J,x (I, -2) = -4 x /y(x. y) =x2 + 3 -=> 1;,(1, -2) lxix. y)

=

=

1+ 3 = 4

2y -=> lxx(\' -2) = 2(-2) =-4

fxy(x. y) = 2y -=> fxy(\' -2) = 2( 1) =2 j~(x.

y) = 0 -=> j~(1, -2) = 0


198

11)=0 Jrxx.\ (x'J/

' /x Y')/ j xxy\"

f'

=

--- Jrxxx (I ' -2)=0 ---,I'

2 => j'xx)'(I , -2)

=

2

=> j'XJ'}'(I , -2) = 0

/x y,)/ = 0

, AYY\"

' /x Y')/ = () => j'»)')'(I , -2) = 0 j .Y.ly\" Substituting a = -1, b = 2 and above values in (i) we have

x 2y + 3y - 2

10 -4(x -I) + 4(y + 2) - 2(x-l;2

= -

+ 2(x- -1) (y + 2) + (x -

1)2

Cv + 2)

3.15.6 Example Evaluate log[(l.03) I] + (0.98) '4 -1] ,approximately.

Solution

h)~ + (y + k)~ -

Then

f(x + h. y + k)

Assuming

x = 1, y

1, II

=

0.03, k

We get

F(x+h, y + k)

=

IOg[(l.03)~ +(0.98)~ -IJ

=

=

of ox

I -x 3

2

3

IOg( (x +

=

[

oy

x 3 + y4 _I

=

0 .02

oF

OF)

.

F(x, y) + h-+ k - approximately ..... (A) ox oy

1 --3/4 -y

of

1

= -

I)

4 I -

-

I

x 3 + y4 -1

F(x, y) + h

. ox + k [OF) oy approximately (OF)


199

~ IOg[ (1.03)~ + (0.98)~ -I] =

F(I, I) + 0.03

(aF)

ax

+

(_O.02)[aF)

(1,1)

ay

(1,1)

~O.03 ll+I-IJ r j to.oJ ~ 1~O.005 ll+I-IJ

approximately

(-: F(l,I) =

0)

Exercise - 3(J) I.

Obtain the expansionllsing Taylor's theorem of the following:

I. xy2 +

cos(xy) at

(I'~)

2. xY at (I, I) lIpto second term

I ADS: 3. sinx. siny in powers of

xY = 1

+(x -I ) +(x -I) x (y -I ) + -21 (x_I)2 I

(x -~) and (y -~) ~+~(x_~)+~(V_~)_~(X_~)2

( ADS. ·2242'

4.

444

eax sinby at (0, 0) ( ADS: hy + xyah + '" )

200


.

5. Prove that sm(x +y)

( x+

=x +Y-

y)

<3

3

+ .....

2

6. Show that eY 10g(1 + x) = x + xy - ~ approximately 2

7. Expandf(x, y)

=

sinxy in powers of (x -1) and (y [ ADs • I - -1t

•

~) upto second degree term.

2 (x - 1)2 - -1t(x -1 \

8

2

y - -1t) - -1 ( y - -1t)2 + ..... I 2 2 2

3.16.7 Maxima and Minima The function F(x, y) is maximum at (x, y) if for all small positive or negative values of hand k; we have

f(x + h, y + k) - F(x, y) < 0

.... (i)

Similarly, F(x, y) is minimum at (x, y) iffor all small positive or negative values of hand k; we have

F(x + h, y + k) - F(x, y) > 0 Thus ifF(x, y) has a minimum or maximum value then [F(x + h, Y + k) -F(x, y)] keeps the same sign for all small positive or negative values of hand k. Now using Taylor's expansion, we get

F(x + h, y + k) - F(x, y)

=

+

-f

(

aF) haF -+k ay

ox

2 2 2 -2-(h2 a F +2hk a F +k2 a F) 21 ax 2 axay 0'2 + .....

..... (iii)

Assuming h, kto be sufficiently small; the sign of the expression on the left can

aF aF be made to depend on that of h ax + k 0' Hence the necessary condition for f(x, y) to be a maximum or minimum is that


of

of

ox

oy

201

- = 0 and - = 0

of p

I.e.,

=

0 and q

then (iii) ~ f(x + h y .

-I-

0

=

where p = ox'

k) - j(x y)

= -

I[

'2!

2 2 2 ,0 F 0 F , 0 F] 11- --) + 2hk-- + K- --) + .....

or

I 2!

2

=r- [ h-,

2!

r

=-

2!

z= 0

ox8y

oy-

r rh 2 + 2shk + tk2 ]

= -

02F s=-ox8y'

of q = oy

~l

~y-

s t ,] + 2hk-+-K-

r

r

s - s ) k, )+ (2 -,-k- +1-) [( h+-k r ) rr-

II

=~[(h+~k)2 +(rt-.~.2)k:l 2! r rIt has the same sign for all h, k if and only if (rt - s2) is positive. When

(i)

(rt - s2) positive,j(x, y) is maximum for negative 'r' and minimum for positive 'r'.

(ii)

(rt - s2) is negative, we have neither a maximum nor minimum and such stationary points are called saddle points.

Solvingp = 0, q

=

0, we get the extreme points

Extreme value :

.I(a, b) is said to be an extreme value ofJ(x, y) if it is either maximum or minimum. 3.16.8 Example Find the maximum and minimum values of x 3 + 3xl- 31 + 4.

Solution We have

J(x, y)

=

x 3 + 3xl- 3x2

fx = 3x2 + 31- 6x,

fy

-

31 + 4

= 6xy - 6y,

f)y = 6x- 6,

1;)' = 6y


202

We now solve

Ix = o,f;, =

0 simultaneously

I.C.,

x2+y~2x=0

and

xy

~Y =

..... (i)

0

..... (ii)

xy ~ y = 0 => y = 0 or x = I From

x 2 + y ~ 2x

when

y = 0, .~ ~ 2x = 0

0 we get

=

x = 2 or x = 0

when

x= l,y~1

=

0 i.e.,y= I or~1

Thus the points are (0, 0), (2, 0), (I, I), (I, ~ I) (a) For

x = 0 , y = 0 , r =j'xx = ~6 ' s = 0 ') t = ~f) rt ~

s2 =

36 ~ 0 > 0

j(x, y) is stationary at x = 0, y = 0 But

r=/xx =

~6

<0

fix, y) is maximum at x = O,y = 0 and the maximum valuej{O, 0)

=4 (b) Por Again

x = 2 , .y

= 0,

r =/xx = 6 ' s =/.xy = 0 , t =j'n' = 6

rt ~s2 > 0

j(x, y) is stationary at x But

=

2, Y

=

0

r =/xx = 6 > 0 j{x, y) is minimum at x = 2, y = 0 and minimum value j(2, 0) = 0

(c) At

x

=

I, Y = I or at x = I, Y = ~ I

rt ~ s2 > 0 and we can reject these points as they are not stationary points.

3.16.9 Example Discuss the maximum and minimum of f('(,y)

=

x2 ~

Y + 6x -- 12

P = 2x + 6, q = ~2y, r = 2, s = 0, t = ~2 p = 0 gives x = ~3, q = 0

~

y =0


203

Stationary point is (-3, 0); and here r = 2 >0; and rl-s~ == 4-0 == 4 > 0 ) ) - 1,... x- -.V + 6x - .!. is minimum at (-3,0) thc minimum valuc is

j'(-3,O)==-21. Exercise -3(k)

....

Find the maximum and minimum valucs of: x' + 3x/ - 3/ + 4 jAns : (0,0) max. its valuc 4,(2,0) min. its value 'W] 3xy ) - 3.c)-3Y) + 7 x'+ [Ans :max . value 7 at (0,0) min.value '3' at (2,0))

3.

x~ + /

I. I. ')

+6x+ 12

[Ans: mill. value '3' at (-3,O)j

3

4.

x + x/ + 21x -12x2 - 2./

5.

X4

6.

X4

+x/ + / + / _ x2 _

l Ans : max.value

10 at ( 1,0) min. value -98 at (7,0) 1

jAns: minimum value '0' at (0,0)1 y2

+1

lAns: max.value I at (0,0) min.value Y2 at t(Hlr pOints( ±

7. 8.

9.

)

3

0

2

~

~ ± ~)]

1 (1 1)

3

jAns: max.valuc --at -,- ] 324 2 2 ) ) ) 3 Discuss the stationary values of II == x-y-3x- - 2y -4y + [Ans: u is a maximum at(O,-I). The maximumvalue is 5, u is neither maximulll nor minimum at(±4,3) Find the maximulll and minimum values ofthc following functions. (a) x'/(12-3x-4y) jAns:maximulllvalucat(2,1)] X

Y -x Y -x Y

(b)

Xl

+ / - 3axy( a > 0)

[Ans: minimum value is _(/' at (a,a)1 1

[Ans:maximum value is ~ at (a/3,a/3)]

(c) xy(a-x-y) (a>O) 10.

27

Determine the values of x,y for which the following functions are maximum or mllllmum .. (a) x ' y 2(1-x- y) l Ans: maximum for x = 112 , Y = I/3 j (b) (x 2

+ y2)"

-2a\x~

- /)

[Ans: minimum for x == ± a,y = OJ [Ans : maximum for x = y = 7[ /3]

(c) sinx siny sin(x+ y) (d)a[sinx+siny+ sin(x+y)] [Ans: maximum for x = y =7[ /31 (e) x/(3x+6y-2)

[Ans: minimum at x = y = 1/6J

204


Lagrange's method of undetermined multipliers: 3.17.0Theorem: Find the maximum and minimum values of

f(x"X 2'X3' ......xlI )

where x" x2 ' x 3 ' •••••• XII are connected by the following m equutions.

¢, (x, ,x2 ' •••••••• x,J= 0 ¢2(X"X2' ....... .xJ = 0

¢m(X, ,X2 ,.·· .... .xll ) = O. Proof: Let u

=

f(xl'x2, .........x,J .Since n variables are connected by the m given

relations, only n-m of the variables are independent. The maximum and minimum values ofu can be found by the method of Lagrange's method of undetermined Multipliers .For u to be max or min of u , du i.e.

=

0

au au au -dx, +-dx2 + ........ +-dxn =0 ax, aX2 aXil also from, we have

..... ( 1) multiplying the above equations by 1, ~, ~, ....Am respectively and adding, we get

au a¢, a¢2 a¢m ) ( au a¢, a¢l a¢m ) ( -+~-+~-+ ........ +AI/I- dx, + -+~-+~-+ ........ +AI/I- dx2 ~ ~ ~ ~ ~ ~ ~ ~

+.................................... . au a¢, ~ a¢2 1 a¢m) _ + ( -+~-+"'2-+ ........ +AI/I-- dxn -0. ax" aXil aXil Ox

. .... (2)

ll

The values ~,~, .... Am are at our choice .We can, therefore choose them so as to satisty 'm 'linear equations .

.. .. (3) I(is immaterial which n - m of the n variables are regarded as independent. Let these bexl/I+"x"H2' .......x".Then since the n - m quantities dxm+"dxm+2, .......dx" are all independent, their coefficient must be separately zero. This gives additional equations.

205


..... (4)

au axil

+

A, atA aXil

+

~ a¢2

+ ........ + Alii

- axil

a¢m = 0 aXil

Now we have m+n equations in all and given relations ¢,

= 0, ¢2 = 0, .......¢III = 0

The equations (3) and n-m equations (4) are sufficient to determine the m + n quantities A,,~, .... AII/ and

x"x2 ' ••••••••• x

lI

for which maximum and minimum values of u

are possible .The m multipliers are called Lagrange's Multipliers.

3.17.0 (a)Theorem: If u=
discuss the

maximum and minimum values ofu

Proof: For a maximum or minimum, we have du I.e.

=0

af" af" af" -'-' d" + -'-' dy + _0_' dz = 0 ax ay oz

Also from

j; = 0 andJ; = 0

FaJ; dx+ aJ; dy+ a.t; dz=O ax iY az

'" .. ( 1)

aJ; dx+ aJ; dy+ aj; dz = 0 ax iY az Multiplying these equations by 1, A" ~ respectively and adding

a¢ + A, aJ; ( ax ax ...

+

~ aJ;) dx + ( a¢ + A, a.t; + ~ aj ;) dy + l( a¢ + A, a.t; + ~ a.t;) dz = 0 ax

liY

iy

iy

fu

~ ... (2)

The lagrange's multipliers

A"

~ are choosen such that

fu

fu


206

o¢ +:1/11 ()I; + /l, ~ 01; oy oy - ~v

=

0

.... (4)

then equation (2) becomes

o¢ +:1/11 ~/~ + /L1) ~/; ()z oz - oz

=0

•.••

These equations (3) ,(4), (5) and given

(5)

1;<'t,y,z)=0, .I;(x,y,z)=O are

sufficient to find ~, ~ ,x,y,z which give maximum and minimum values of u ..

3.17.0(b) Theorem: If u

=

¢(x,y, z) and f(x,y,z) = 0 find the maximum or

minimum values ofu.

Proof: For a maximum or a minimum I.e.

o¢ ox

o¢ oy

o¢ oz

~.dx+~.dy+~.dz

II

,we have du

=

0

=0

.... ( I)

and

f(x,y,::) = o. of ol ~l - .dx + -.Jy + -.Jz = 0 ox oy OZ

from given

.... (2)

Multiplying (1) by 1 and (2) by A and adding

o¢ + A of)dx+( o¢ + A ol)dY+(o¢'+ A ol)dZ = 0 ( ox ox oY oy OZ oz

.... (3)

Lagrange's multiplier A is choosen such that

o¢ +A ~l =0 ox ox .... (4) and

o¢ +A ~l Oy oy

=0

.... (5)

The equation (3) gives

o¢ +A ~l =0 oz oz The equations (4),(5),(6) and

.... (6)

1 (x,y, z) = 0

which u is a maximum or a minimum.

enable us to tind the values of

A "x,y,z for

207

Mean Value Theorems and Functions of Several Vanables

Exercise - 3(1) 3.17.1 Example

Find the point upon the plane ax + by + cz

=

p at which the function

$ = x 2 +1+ z2 has a minimum value and find this minimum $. Solution

ax+by+:::=p

p-ax-hy e

~:::= ~---~

..... (i) )

Again

$=

')

. .2

J

+ .V"" + z-

x~

= r

')

J

+ ), +

(

17 - axc- ·bV)-

Then for min./max., we have

ox = 2.x -

0$

2a (p _ ax - by) = 0 e2

~$ = 2y- 2~ e-

uy

From (8)

x

(p _ ax - hy) = 0

p-ax-hy

a

) c-

y

p-lLr-hy

h

c2

~-=

~-=

x a

.

bx y=. (l Now, substituting this value ofy in (ii) we obtain y

~-=-'--~

b

2

X) =0 x- 2a ( P - l I Xb- e 1I c 2x - (lP

Then,

and

+ a2x + b2x

=

0

+

x

ap bx bp ) ) and Y = =) ) ) a +h- +c a a- +b- +c-

=

2

b2

( 2) =

x( a 2

+

(lP

..... (ii) ..... (iii)

208


Hence 3 a min, when

p -x =-y =----:---=-::----c2 2 2 Also

a

b

Z =

p-ax-by c

Z

I.e.,

a +b +c

=~[p_ c

2

a p

a2 + b2 + c2

P

=

Hence 3 a min. at

x

y Z abc

-==-

4> . =.xl + y2

+ z2 =

P 2 a +b 2 +c 2

a 2 p2 + b 2 + p2 +C2p2

(a 2 + b2 + c2 r

----'---,---...:...--.---:'--

mm

3.17.2 Example Divide 24 into the three parts such that the continued product of the first, square of the second and cube of third is maximum.

Solution Let 24 be divided into parts x, y, z then x + y + z

x 2 y2z

Given

4>(x, y, z)

then

4>(x, y) =.xl;? (24 .- x - y) from (i)

or

4>(x, y) = 24x3y2 - x4y2 - x 3y3

=

Differentiating, we get

p

= ~! = n.xly2 -

a 4> r =-?

4~y2 -

3x2y3

2

ax-

=

144xy2 - 12x2y2 - 6xy3

=

24

..... (i)

209


- - --- - - - - - - - - - - - - - - - - - - - - - - - - - and

S

a 1

=-02~ - = - (o~ oxoy oX ~Y

144x2y-? - 8x3y - 9x 2y-?

=

.

For a maximum value of ~(x, y)we have

a~ = 0 => ox

i - 3x2y3 =

72x 2y"2 _ 4x 3

0

=> x 2i (72 - 4x - 3y) = 0 => 72 - 4x - 3y = 0,

o~ = 0 => 48x3.v -

and

~

x = 0, y = 0

2x4y - 3x3 v2 .

=

..... (ii)

0

=> x 3y (48 - 2x - 3y) = 0 => 48 - 2x - 3y = 0, Solving (ii) and (iii) we get x

x = 0, y = 0

..... (iii)

= 12 and y = 8 etc

At (12, 8), we have

r= 144, 12(8)"2- 12(12)2(8)-6(12)(8)3 = 12(8)2 (144 - 144 - 48) = 12(8)2 - (-48)

r=48(12)3_2(12)4_6(12)3.8 = (12)3 (48 - 24 - 48) s = 144( 12)2 . 8 - 8( 12)3 . 8 - 9( 12)2 (8)2 ==

(12i. 8 (144 -96 -72) == (12)2.8(-24)

=12. (8)2 (-48). (12)3 (-24) == (12)4.8 2 .24 (48 - 24)

:. rt - s2

==

-l (12)"2.8.

(-24)]2

(12)4.8 2 .242> 0;

Since rl - s2 > 0 and r < 0, therefore ~ (x, y) is maximum at (12, 8). Putting x

==

12, and y

==

8 in (i), we get z == 4

The values of x, y, z are 12, 8, 4 respectively. This is the division of 24 for maximum ~(x, y, z).

3.17.3 Example A rectangular box, open at the top, is to have a volume of 32 c.c. Find the dimensions of the box requiring least material for its construction.


210

Solution Let I, band h be the length, breadth and height of the box respectively. Then, wc have 2(1 + h) h + Ih

v

=

Ihh

=

32 ; surface

S

=

2(1

I-

b)h + Ih and b

=

=

32 =

-

III

32) h + I (32) 32 s = 2 ( 1+= 2111 + -64 +III liz I h

~~

Now

=

2/-(:;)

For maximum and minimum ofS, we get

as =0= as al

ah

as = 0 ~ 2h _ 642 = 0 h = 32 0/

'

12

as = 0 ~ 2/- 32 = 0

and

and

1

ah

82~ = +2

8h-

h2

so 'S' is minimum for

1= 4,b = 4,h = 2.

S (say) .....(i)


I

211

1

1

3.17.4 Example: If u=a'x"+h'/+c3z1 where -+-+-=1 show that the x y z stationary point of u is given hy x

a+h+c

a+h+c

a

h

=- - - , y =

a+h+c

,,"" - - - c

Solution: For a stationary value of 'u', du =0 I.e.

1

a ' 2xdx+h'-2ydy+c 2:::dz

Also we have the given relation

=0

..... ( I )

~~ + l + l = 1 x

y

:::

1 1 1 -) dx+-) dy+-) dz = 0

y

x~

..

~

..... (2)

Equating the coefficients of dx , dy ,dz from (1)+ A(2) separately to zero

A

I

A y

I

1

A --

We get a x + -) = 0, h x + -) = 0, c x + -, = 0 . x-

1

or

:. a x ' =h ' /=c ' z'-=(-A) ax = by = cz = (-At] = k(say) x = k I a; y = k I h; z = k I c ..... (3)

· . j'or x, y, Z SU hstltutll1g

.

1 + -1 + ~~1 = I Y z

In X

abc k k k :. k = a+h+c

-+-+-= 1

..... (4)

Substituting in (3) we get X=

a+b+c II

,y=

a+b+c a+b+c , z = - - - for which u is stationary. c b Exercise 3(M)

I.

Find the maximum and minimum distances from the ongll1 to the curve 5x" + 6xy + 5 y2 - 8 = O. [Ans : max. 4, min .16,6,3]

2.

Fin~the dimensions of the rectangular box, with out a top of maximum capacity rAns:6",6",3" J whose surface is 108 Sq. inches.


212

3.

If

f = u 1 +j'1 +W1

where

11

+ v + w = 1, show that

f

IS stationary when

1

U=V=W=~

3 4.

Use Lagrange's method of multiplies to determine the minimum distance from the origin to the plane x + 2y + 3z = 14.

Jl4"1

[Ans:

5.

Find the maximum value of / = xyz when xy + yz + zx = k. [Ans: (k / 3)3/?]

6.

If the distance from the origin of any point

3x+2y+z-12 = 7.

°

Lagrange's method). Find the maximum

is P

=~X2 + y2 +Z1

volume

of a

p(x,y, z) on the plane

,find the minimum value of p(use the

parallelepiped

inscribed

b-

h

8a c] [Ans: 3 (:;3

c-

-V j

Given x+y+z=k,findthemaximumvalueof XII,yll,ZY [Ans: If

r2

=

x

2

+

l

+

Z2

12.

1

aa./3II.rY.k
and x+y+z =30 find the values of x,y,z for which

r

IS a

[Ans: x = y = z = 10]

mll111nUm 11.

]

Find the volume of the largest rectangular parallelepiped that can be inscribed in

a

10.

2

3J3

, , ) · ·d x- Y z- 1 t I1e e II IpSo! - 2 +-) +-) = . 9.

a sphere

8r [Ans:

' x-, + y-) +z 2 = r.

8.

III

Find the dimensions of a rectangular box without a top of maximum capacity whose surface area is 108 square inches. [Ans: 6,6,3 inches] Find the dimensions of a rectangular box with open top, so that the total surface area S of the box is a minimum, given that the volume 'V' of the box is constant. fAns: x = y = 2z = (2\,)1/3]

13. 14.

Show that the rectangular solid of maximum volume of that can be inscribed in a sphere is a cube I f the total surface area of a closed rectangular box is 108 sq.cm. Find the dimensions of the box having maximum capacity

[Ans:

.Ji8,.Ji8,.Ji8]

15.

Perimeter of a triangle is constant, prove that the area of this triangle is maximum when the triangle is equilateral.

16.

If u=¢(x,y)where/(x,y)=O, find the maximum and minimul.1 values of 11 using Lagrange's multipliers Method

4 Curvature and Curve Tracing

4.1.1

Curvature The curvature of a curve (bending of a curve) varies from point to point on the curve. Let P be a point on the curve and Q be

c.l

point nearer to P, are OQ = 0 s

Let tangents at P and Q make angles, \.11, and \.11 + 0\.11 with the X- axis.

LT R T' = d\jJ is the angle through which the tangent at P turns as a point moves along the curve form P to Q through a distance os along the curve and hence 0\.11 depends on the arc length

os.

0\.11 is defined as the total bending or total curvature of the arc P Q. 0\.11

8s

is defined as the average curvature of the arc P Q.

Now os

~

Lt

~\.11

&s~O uS

0 as Q ~ P =

dd\.l1 is defined as curvature of the curve at the point P and is denoted S

by K.

Thus

d\.l1 K=ds


214

x'

0

T

Fig. 4.1

Radius of Curvature The reciprocal of the curvature of a curve at any point' P' is called the radius (?f curvature at that point and is denoted by 'p'.

ds

Thus p = -d is called the radius of curvature of the curve at the point P. \1' .

4.1.2 Theorem If 'r' is the radius of a circle then the radius of curvature of the circle is same as its radius.

Proof: Let 'c' be the centre and 'r' be the radius ofa circle. P be any point on it and Q be a neighbouring point, arc PQ = ds. The tangents at P and Q make angles \If and \If + 8~, respectively. From the figure

LT R T' = 8 \If LP R Q'

= 180 - 8\1f

LP C Q'

= 8\jf,

8s

~ = u

~,

from the sector PCQ.

r and as Q ~ P, 8s

~

0

Curvature and Curve Tracing

215

Lt 8 \1' = ~ 8s r

01->0

1

I.e.,

P

r

p=r

y

o

x'

x

y'

Fig. 4.2 4.1.3 Cartesian form of the Radius of Curvature Let y = j{x) be the Cartesian ~quation of a curve and tan (\lJ) be the slope of the tangent to the curve at a point P.

dy tan \lJ = dx Differentiating w.r to 'x'

We know that

ds dx


216

... ,..

,

1+(ddxy )2

[I+(~~)' I+(~)' 1 p=

3/

. _I[ + y( ) '2 I.e.,pwhereYI y~

dy d 2Y = -d 'Y2=-2 X dx

is the radius of curvature of the curve (Cartesian form).

4.1.4 Parametric form of Radius of Curvature Let x = fil), Y = get) be the Parametric form of the equation of a curve.

-d'( = f\t() -dy = g ,(t ) dl

i.e.,

x'

' dl

= /'(t),

y'

= gl(t),

dy dy dl -=dx dx dt

'., dy dx

y'

- = -I X

.

-,

217


Differentiating w.r. to 'x' on both sides

" .- y·x '" •,{2y X 'Y --=

{X')3

3/

dy

d'y

Substituting -d ' - I 2 values in p = X

ex

[l+(~)T d2

.

Y dx 2

[l+(Hr p=

-

'----'..,..

x'y" - y'x"

{X')3 [(XI)2 + {y')2 y~ P=

where

x'y" - y'x"

I dx x =-

dt '

I

dy dt

y=-

is the parametric form of the radius of curvature,

4.1.5 Polar form of Radius of Curvature Let P (r, 8) be a point on the curve r =f(O). If '0' is the pole, 0 X is the initial line.

LXOP =0 The tangent at P makes an angle \I' with the initial line.

$ is the angle between the radius vector OP and tangent PT.


218

From figure

\If=e+cj> d\lf de d -=-+ds ds ds

de d de +-.ds de ds

1 p

-=-

!

p

[1

= de + d] ds

..... (1 )

de

x

Fig. 4.3 We know

ds de

and

tan

i.e.,

tan cj>-=dr de

Differentiating w.r. to

r2+

(dr)2 de

de

= r-

dr

r

'e' on both the sides

..... (2)


(1 + tan

219

2

d~

de

<1»

=

d~

de

d de

The value of

I+

~~

r-~~

t~

I + r' +

de

1+

d,h

-'I'

r2

r

(dr)2

+ de

r2 +2(_dr)2 _rtl_2_r 2

de de = --'-----'------:---

de ds Substituting the value of de'

r2

+(-:r ,

I+

d~

de

from (2), (3) in (I)

..... (3)


220

is the polar form of radius of curvature.

4.1.6 Pedal form of Radius of Curvature Let P be any point on the curve. the tangent P T and P makes an angle \jJ with the initial line OA, LP 0 X = e (see the figure in 4.1.5) \jJ=e+~

..... ( I)

Differentiating w.r. to s

d\jJ de d~ -=-+d\' ds ds -

I

p

I p

-

We have

=

I de d~ dr -.r- + - r ds dr cis d~

I r

= - (sin~)

de

+ dr I p

sin~=r-, ~

p

I [ sm~+rcos~. d~] ,cos~=dr

= -

~

r

I d(r sin ~) r dr

_ = ~dp p

cos~,

r dr

:. p

~

p = r sin ~

=

dr rdp

where p is the perpendicular distance from pole to the tangent.

4.1.7 Examples 2

Prove that the radius of curvature of the curve y = a cosh (x/a) is

Solution y

=

a cosh (x/a)

L

a


dy

-

=

dx

221

I a sinh(x/a)a

d 2y -2 =

dx

I cosh (x/a)a

dy d 2 y Substituting the values of - , - - ? in p = dx dx-

( dy)2l·~2

[ I + ~j; d2

Y

dx 2

]3 2

?

P =1[ + sinh - x / a cosh{x / a ).~ p

= aeos h2(x/a)

p

~ a( ~:),

as y

~ cosh(x/a)

p~ ( : ) 4.1.8 Example Prove that for the rectangular hyperbola xy = c2 the radius of curvature at any point r3

is given by p =

-2

2c

where 'r' is the distance of the point from the origin.

Solution Differentiating xy = e 2 w.r. to 'x'

dy

x - +y=O dx

222


..

dy y .

[

( ~b:dy)2l~~

d 1+ Substltutmg for -d ,--? m P = /2 X dx( Y 2

dx 2

p=

as ,.2 = x 2 + rand xy = c 2

4.1.9 Example Prove that the square of the radius of curvature at any point on the parabola

y = 4 ax varies as the cube of the focal distance of the point. Solution From

y=4ax

dy dx d 2y dx 2

-

=

~ f~ I -3/ lax /2 2 -va.

= -

..... (1 )

223


p=

Hence

y

x'

o

x

y'

Fig. 4.4

p=

-1 r -~~ x 2

-~"a

p

=

1a(x+a)~

..... (2)

The focal distance of the point p (x, y) sp=

~(x-aY +(y-oy

= ~X2 +a 2 -2ax+ y2 =

sp=

~ x 2 + a 2 - 2ax + 4ax

using(l)

~(x+aY

SP = focal distance = (x + a)

..... (3)

224


From (2) and (3)

-2

P = ~ [Focal distance]3/2

4

p2 = - [Focal distance]3

a

p2

00

= cube of the focal distance.

4.1.10 Example

Show that the radius of curvature at an end of the major axis of the ellipse Xl

a2

y2

+---;;: =

1 is equal to the semi

latu~-rectum

of the ellipse.

Solution

i ..e.,

..... (1 )

Differentiating (1) w.r. to 'x'

dy dx

2

2fy-xdyl dx

d y b dx 2 = - --;;

/

y2


225

I

y

x'

o

A'(-a,O)

(0,0)

A(a,O)

y'

Fig. 4.5

d 2y dx

1

b2 2

( using ax'2 + b2 y' = I

-2 ----

a

i

[1+(:)']" p=

Hence

d 2y dx 2

[ l+---~ h" 4 a i p= _b 4 2 a

r;

(a 4i+b 4x2) a 4b 4

i

The coordinates of the ends of the major axis are A (a, 0) and A' (-a, 0) Taking the end A (a, 0) we have from (2)

pl(a,o)

b2 = -;; =

Semi latus-rectum of the ellipse.

..... (2)

226


4.1.11 Example J

J'

5<

Prove that p at any point of the curve x;; 3 + y/3 =a 3 is three times the length of the perpendicular from origin to the tangent at that point.

Solution The parametric equations of the curve are

x

=

acos 3e,

y = asin3e

dy

3a sin 2 e cose dy - 3a cos 2 e sin e dx

de

dy dx

-------,,---- - - = -

dx

=

tan e

de and

d 2y dx

de -sec2edx = - sec2e / (-3acos 2 e sin

-= 2

e)

3a cos 4 e sin e

gives

Hence

p=

(I + tan 2e

f2

I 4

3a cos e sin e p

=

..... ( I)

3a cos e sin e

The perpendicular distance from the origin to the tangent

dy y-xp

= ---;===dx==

I+(:J

P = a sin e cos e

From (1)&(2) p=3p

2

a sin 2e - cos e(- tan e)

2 JI + tan e

..... (2)

227


4.1.12 Example Find p at any point for the cycloid

x = a (8 - sin 8) y = a(1- cos 8) and also find p at 8 = 90°

Solution dy dy _ de _ a(sin8) dx - dx - a(l-cos8)' de 2

d y

and

-

dx

2

=

dx

2 8 1 d8

cosec - - 2 2 dx

1 28 -2cosec 2

d 2y 2

=

-1

a(1-cos8) ,

[1+(:)'r

..

p=

d 2y dx

p

8 dy = cotdx 2

-

= - 4a sin

( at 1+cot

p=

2

2

2

-1 4a sin4 ~ 2

8

2

at 8 = 90°, P = - 4a sin 45° i.e., p = -

212 a

4.1.13 Example Find p at (r, 8) for the curve r = a (1 + cos 8) and also find at 8 =

Solution r

= a (1 + cos 8)

dr de = r) = a(- sin 8) d 2r - 2 =r =acos8 de 2

~.

228


[a 2(1 + cos e)2 + 2a 2 sin 2 e + a 2cos e(1 + cos e)] [2a 2 (1 +cose)7i] 2

2

2

a + 2a + 3a cose

2

[2a .2cos

2

~~]7i

3a .2cos ~~ 2

2

=

.iacos e/ 3 /2

e=~

at

4 P = -acos 45° 3

=

2 r;:; -,,2a 3

4.1.14 Example Find p for the curve r'" = if! cos me Solution r'"

= if! cos me

mlog r = m log a + log (cos m e)

1 dr m -. r de r)

= -r

=

1 cos me

(-

m sin m e)

tan me

..... ( I)

ar r2 = -r m sec2 me - tan m e. de r2 = - rm sec2 me + r tan 2 me From (1)

=

3 {r 2+ r2 tan 2me)2 r2 + 2r2 tan 2me + mr2 sec 2me - r2 tan 2me

..... (2)

229


r

{m + J)cos me am

r m

{m+J( m a

For the curve ,-In = d" cos m

P=

e

(m + J)r m- 1

4.1.15 Example Find the radius of curvature of the curve p2 = ar

Solution

p2

=

ar

..... ( 1)

Differentiating w.r. to 'r'

2p dp dr dp dr p

=

a

=

a

2p

r dr

= r.

dp

2p a

From(l) 2r

p=-j;;; a

4.1.16 Example 1 Find the radius ofcurvature at the point (p, r) on the ellipse

-2

P

J

J

= -2 + -b2 a

r2 ~b2

a


230

Solution

Differentiating 2 dp p3 dr

-2r a 2b 2 2

=>

dr a 2b r-=-dp

p3

4.1.17 Example

Find the radius of curvature p of the curve ,2 cos28

=

if

Solution

,2 = cos28 = a2

..... ( I)

Taking logarithms on both sides we get

2 log r + log cos 2 8 = 2 log a Differentiating w.r.to '8' 1 dr 2 -;: d8 - 2 tan 28 = 0

d8

r-=cot28 dr

d8

tan <\> = r dr tan <\> = tan

=

cot 28

(~ - 28 )

1t

<\>=2- 20 We have

P = rsin<\> => p = r sin P = r cos 28

Substituting

cos28 =

a2 -2

r

from (1)

(~-28)


231

dp = dr

~ ,.2

dp ,. p=r-=---) dr -(r

4.1.18 Curvature at Origin (i)

When y can be expanded in powers ofx by some algebraical or trigonometrical method from the equation of the curve as qxl

y = px + -

2!

dx

y

x=O

y=o

d2 q = --) dx- x=o

y=o

3

..

p (at ongll1) =

(ii)

dyl

+.... then p = -

(I + p).i

--'----~

q

Newtonian Method: If a curve passes through (0, 0) and the tangent at (0,0) IS

X - axis, then we have x = 0, y =

°and d,Y = °at origin. {X

The expansion of y by Maclaurin's theorem reduces to Xl

Y

= (yo)

+ (YI)O +

° °+ °. +

2

(Y2)O

Taking upto terms of (x 2 ) Y=

x

x2 q - +... 2!

Dividing each term by x 2 and taking limits 2y q= L t x->o x 2

232

Engineering Mathel\latics - I

.,

We have

P = (at ongln)

as

p = 0 at (0,0)

and

Lt 2 q = x~Ox

and

P (at origin) =

3

(I + p)2

= -'-------=--<-q

2y I

q

p (at origin) = Lt

x~o

(~) 2y

(iii) Similarly, if a curve passes through (0, 0) and the tangent at (0, 0) is Y-axis 2

then

p (at origin) = Lt (Y2 x~o

X

)

(iv) Curvature at pole: If the initial line is the tangent to the curve at the pole then

2 X2 ) r2 cos 8 (.: x = r c.os 8) p= Lt = Lt x~o ( 2y x~o 2rsin8 y=rsm8 r

8

2

P = Lt -·--·cos 8 e~o 28 sin8

-

Lt - r )

p - e~o ( 28

8

(as Lt - .- ~ 1.cos28 ~ I) e~osm8

4.1.19 Example Find p at the origin of the curve y4 + xl + a (xl + y) - a2y method.

= 0 by Newtonian

Solution Equating to zero the terms of the lowest degree in the equation of the curve

- a2y =?

=

0

y= 0

i.e.,

x - axis is the only tangent at origin

Then

'p' at (0,0) when X - axis is the tangent is given by

x2 p= Lte~o2y

..... (1)

233


Dividing the given equation by y x-J

I

+ x· -

x2

+ a. - + ay = y y

a2

x~O

Let So that

y~O

and

-

From (1)

o + 0.2p + a2p

x

2

=2p

y

+ a.O = a2 => 2 a p = a2 => p =

a

"2

4.1.20 Example Find p at the origin for the curve x = a (8 + sin 8), y = a (I - cos 8) by Newtonian method.

Solution

x = a (8 + sin 8),

y = a( 1 - cos8) is a cycloid

X-axis is the tangent to the curve at (0, 0)

p at (0, 0) = Lt

(~l

x---+o 2y

2 Lt a (8 + sin 8)2 0---+0 2a(1- cos 8)

a (8 + sin 8)2 L t - -'-----'-0---+02 l-cos8 ' =

(Of the form % )

Lt a (8 + sin 8)2(\ + cos 8) sin 2 8

0---+02

=

Lt

!!.(~+ 1)2 (1 + cos 8)

0---+02 sm8

a

8

= -2 (1 + 1)2 (I + 1) using Lt = I 0---+0 sin8

a [8] 2 P = 4a

p=

-


234

4.1.21 Example Find pat (0, 0) for the curve 2x4 + method.

31 + 4x2y + x Y - y2 + 2x = 0 by Newtonian

Solution Equating to zero the terms of the lowest degree in the given equation of the curve 2 x = 0 => x = 0 y-axis is the tangent to the curve at (0, 0) 2

P at (0, 0) = Lt L x->o2x ..... ( I)

So that Dividing the given equation by 'x' 2x3 + 3y2. L

2

x

Taking

y2

+ 4x y + Y - -

x

+2

=

0

x~O,y~O

2.0 + 3.0· (2p) + 4.0 + 0 - 2 P + 2 = 0; p = 1

Exercise - 4(A) I. Find p for the following curves : 1.

y

=

4 sin x - sin 2x at x

n

=

2" (Ans:

515 ..

-I 4

3J2a

[Ans: - - ] 16

235


5.

E +JY =J;

at

%,~ a

lADs: J21

= 1t

6.

x = a (0 - sin 0), y = a (I - cos 0) at 0

7.

= a sin 2 t (I + cos 2t) Y = a cos 2 t (I - cos 2 t) at any point 't'

8.

x = a (cos 0 + 0 sin 0) y = a (sin 0 - 0 sin 0)

9.

? = a 2 cos20

lADs: - 4 al x

lADs: 4 a cos 3 tl

lADs: a 01

a2

[ADS: );]

10.

r =

~

(\

+ cosO) at

(~,~) J2 [ADS·. -3a ]

an

lADs:

4.2

{n + 1)rn-1 1

Centre of Curvature, Circle of Curvature and Evolute 1. Definition: The centre of curvature at a point 'P' of a curves is the point 'C' which lies on the positive direction of the normal at 'P' and is at a distance 'p' (in magnitude) from it. 2. Definition: The circle of curvature at a point' P' of a curve is the circle whose centre is at the centre of curvature 'C' and whose radius is 'p' in magnitude.

236


y

o

x'

y'

Fig. 4.6

4.2.1

Centre of Curvature Let P (x, y) be any point on the curve and C (X, Y) be the centre of curvature at 'P' which is on the normal PC at P. Then PC =: p = radius of curvature at 'P'. The tangent P T at P to the curve makes an angle \jJ with X - axis. P B, C A are perpendiculars to X - axis, and P R is perpendicular to C A. L.PTX=L.RPT=L.PCR=\jJ I cos \jJ = - sec \jJ

I

(I + tan 2 \jJ2 )2" ..... ( I)

dy dx

..... (2)

237


..... (3)

and we have

(i.e.,)

x = 0 A = 0 B - A B = OB x = x - P C sin \jJ X

=

x - p sin

PR (From

\jl

From (2) and (3)

X=

X=x-

d~[1 +(tfl)2] dx

dx

d 2y dx 2

Y=CA=RA+RC=Y+PCcoo\jJ From (I) and (3)

~

P R C, PR = PC sin \jJ)

238

Engineering Mathematics - ,

y

o

x'

x y'

.. The coordinates of the centre of curvature of a curve

p (x, y) is C

at

2

2

2

_ dy _ d Y [ _ Yl (I + Yl ) 1+ Yl ] , . Denoting Yl - d 'Y2 - - - 2 we can WrIte C x ,Y + x dx Y2 Y2 C (X, Y) is the centre and 'p' is radius ofthe circle of curvature .. The equation of the circle of curvature is (x -

xi + (y - Yi = p2

4.2.2 Example Find the centre of curvature of the curve Y

= x 3 - 6x2 + 3x + 1 at (1, -1)

Solution

dy = 3x2 - 12x + 3 dx

-


239

dy

_

dX(I,I)

--6

=-6 dX2 (1,_1)

X=

YI (I +

X -

2

yn

Y X= I-

(-6XI+36)

-6

I+ YI2

and Y = Y + - Y2 1+36 and Y = I + - -

-6

-43 X = - 36 and Y = 6 -43) The coordinates of the centre of curvature are ( - 36, -6-

4.2.3 Example Find the centre of curvature for the cycloid x

=

ace - sine),

Y = a(1 - cos e)

Solution

dy = a sin e _ cot e/ dx a(l-cose) 12 d 2y 2 e 1 de 1 1 -=cosec -.--=x---,------,dx 2 2 dx 2· 2 e a(l cos e) Sin 2

. 4 4aSIn

e

--

2

240


• 4

4a SIO

X=

ace -

e

-

2

. 4e e 4 aslO -cot-sine) + _ _-'2"'--------=-2

. e

SIO-

2

ace - sine) + 2a sin e X = ace - sine) X=

2

1+ YI

Y=y+-Y2

(l+cot2E!) Y=a(l-cose)12 . 4 - 4 a SIO

e

-

2

Y = a(l- cose)- 4a sin 2 ~ 2 Y = a(l - cose) - 2a(1 - cose) Y = -a(l - cose) .. The coordinates of the centre of curvature are

[ace + sine),

-

a(1 -

cose)]

4.2.4 Example Find the centre ofthe curvature at any point (x, y) on the eIlipse.

Solution ..... (1 )

241


Diff. (I) w.r. to 'x'

2b 2x + 2Q2y dy dx

=

0

..... (2)

..... (3)

Using (I) (2) and (3)

X= x _ YI (1 + yO Y2

[~: (a' - h'), -;.' (a' - h' 4.2.5

ij

is the centre of curvalure,

Example Find the circle of curvature of the curve

fx + fY

=

fa

at the point

(~, ~)

Solution ..... (1)


242

Diff (1) w.r. to 'x' 1 1 -1 -I 1 --I dy -x 2 +_y2 -=0

2

2

dx

dy

JY

..... (2)

dx--.r; and

:10/

=

ai

14' /4

[

d 2y dx 2

~1

, I ~-I ----..jy-x dy C 1 ~-I] ..;x-y 2 dx 2

=

X

d 2y

dx

..... (3)

4

..... (4)

a

0/ (// /4' )14

y~ ¥= [1 + 1]% = ~

p = [1 + Y2

i

.fi

a a a is the radius curvature of the curve at ("4'"4 )

X= x _ YI (1 + YI2 ) Y2

a 1(1+1) 3a X=-+--=4 4 4 a

Y=y+

(1 + y2) 1

Y2 a

(I + 1)

Y="4 + -4- =

3a

4

a 3a 3a)

Coordinates of the centre of curvature ( 4' 4

..... (5)

243


Equation of the circle of curvature of the curve (x - X)2 + (v _ y)2

3a)2 ( 3a)2 _ ( X -4- + y -4- -

=

p2

(~)2 Ji

Exercise - 48 I.

Find the coordinates of the centre of curvature of the curve y = x J at the point

7 31 IAns: - - ) 64' 38 2

2. Show that the centre of curvature for the curve y

=

x x+ 9 at

(3,

6) is

(3, I;)

3. Show that the centre of curvature at the point 't' on the ellipse x = a cos t, .

4.

a- - b 2) . 2 , )cos-I,- () b Slll-'

a - ba

.

y=bSllltlS

[(

1

Find the centre of curvature at (

1

1

3;1, 3;1) of the folium x

3

+I

=

3a xy

21(1 21a IAns: - - --I 16 ' 16

5. Find the centre of curvature at the point '0' of the curve

+ a sin 0

x

=

(I - aO) cos 0

y

=

(I - aO) sin 0 + a cos 0 IAns : a sin 0, - a cos 01

6.

Find the coordinates of the centre of curvature of the cycloid x = a (0 + sinO), y

=

a( I - cos 0) (Ans: a (0 -sin 0), a (I + cos 0) 2

7.

For the curve y x2

+I

= 111X

= a 2 (I

+ ~ show that the circle of curvature at the origin is a

+ 1112) (v -

I1lx).


244

4.3 4.3.1

Evolutes Definition .' Corresponding to each point on a curve we can find the curvature of the curve at that point. Drawing the normal at these points we can find centre of curvature corresponding to each of these points. Since the curvature varies from point to point, centres of curvature also differ. The totality of all such centres of curvature of a given curve will define another curve and this curve is called the evolute of the curve. Thus the locus of centres of curvature of a given curve is called the evolute of that curve.

Notes: (i)

Elimination of x, y from the coordinates of the centre of curvature (X, Y) gives an equation in terms of X, Y which is called evo/ute of the given

curve y = fix). (ii)

When the equation of the curve is given in the parametric form (say t) elimination of '1' from the coordinates of the centre of curvature (X, Y) gives an equation in terms of X, Y which is called evo/ute of the curve x = fit), y = get).

4.3.2 Example Find the evolute of the parabolay = 4 a x

Solution Differentiating

y

=

4ax

2ydy dx dy dx

=

4a

Fa Fx -Fa

=

2

d y dx2

w.r. to 'x'

..... (1)

=-3-

..... (2)

2x2

X= X _

YI

(1 + yl2 ) Y2

X=x

-~(1 +~) Fx -Fa 3

2x 2

x


245

X=3x+2a

..... (3)

I-Fa

I + Cl ) 1 + y2 X Y = Y+ _ _I = Y+ ~----=c'y)

,

2x 2

..... (4\

Squaring y2

4 -.x3

=

a

X-2a

From (3) substituting x = --3- in (5)

4(X-2a)'

y2= _ a 27 ay2 27 a

r

3 =

4(X - 2 a)3

=

4(x - 2 0)3 is the evolute of the curve.

4.3.3 Example

Solution ..... ( I)

Differentiating (I) with respect to 'x' dy 2b 2x + 2a2y dx

dy dx

=

=

0

2

-b x 02y

..... (2)


246

d 2y dx2 X

=

=

_b 2 a 2b2 a 2 y2· a 2;.

=

_b 4 a 2y 3

..... (3)

(I

x _ YI + yl2 ) Y2

X

Substituting c?y2

=

=

x

x - -4 (a4y2 2

a b

+ b4;x2)

a 2b2 - b2x 2 from( I)

..... (4)

b4 x 2 1+4""2

=y+

ay

_b 4

a

2

i

247


Y

Substituting b2x 2

=

y - ~ (0 4; a 2b 4

= 02b 2 -

Y

=

Y

=

y

a 2;

from (I)

-+ a b4

a 2 _b 2

+ b4x 2 )

[a 4;

+

b

2 2

a (a 2

-;)]

y3

b4

..... (5)

Eliminating x, y from (4) (5)

(aX)32+ (bY)}2= (a 2 - b 2 )2:I is the evolute of the given curve.

4.3.4 Example Show that the evolute of the curve 2

2

2

X3 +y3 +a 3

2

2

~

= is(X+Y)3 +(x-Y)l =2a 3

Solution The parametric equation of the curve is x = acos 3S, y = a sin 3S dy dy dx

= de dx

3 a sin 2 ScosS - 3a cos 2Ssin S

de dy - =-tan S dx

d2y de - 2 =-sec 2 S dx dx

248


30 cos 4 esin e ..... ( I)

(I + tan

2

e)

1

..... (2)

From (1) and (2) (X + Y) = a(cose + sine)3 and X - Y = a(cose - sine)3 ')

I

~

'),

{X+yt3 +{X-Y)"'3 J

(x + y)~ + (x 4.3.5

=

/

2a 3

2

J

y)~ == 2a:1

is the evolute of the given curve.

Example Show that the evolute of the curve x

=

a(cose + sine), y

=

a(sin e - e cose) is

x 2 +y=a2 Solution x

=

a(cose + sine) and y

=

a(sin e - e cose)


249

Differentiating w.r.t 0

dx

..

= a(- Sll10 +

-;-0 a

dx dO

dy £10

.

Sll10 + OcosO), -'- = {{(cosO - cosO + 0 SIIIO)

dy. aO cosO, dO = aO SII10

=

dy dx

aOsinO (JOcosO

ely

d 2y dx 2

..... ( I)

tan 0

-, = (X

..... (2)

(l0 cos'O

-

y{l+y2)

X = x- .

1

. 1

=

0)

. tan{l+tan 2 a(cosO + 0 S1l10) _ ----"-------:-----"--

Y2

I

aOcos 1 e X

=

a cos 0

y

=

I+ y2 y + --'_I Y2

..... (3) =

a(sin 0 - 0 cosO) +

(I + tan 2 0 ) I

aOcos '0 Y

=

(I

sin 0

Eliminating '0' from (3) & (4)

X2 + y2 :.

= (/2

x 2 + 1= (12 is the evolute of the given curve.

4.3.6 Example Show that the evolute of the curve

y

=

a cos h

x a( cos t + log tan ~) ,y asin =

=

x

a

Solution

x

=

aCcost + log tan

'2t ) y =

1I

sin t

t is


250

r.

j

1 2 -.t 1 -dy =acost -dx =a -smt+--.sec dy t t 2 2 ' dt an2

j [.

r·

-dx = a -smt+ =a -smt+-.IJ dt 2 sm-. t1cos-t sm t

2

acos 2 t dy . and -d SIn t t

dx dt

2

= a cost

dy dy

.J!L dx dt

dx

--

acost

= tan t

acos 2 t sin t

X = a( cos t + log tan

t

t

"2) - a cos t = a log tan "2

..... ( 1)

1+ yf Y2 y

=

a sin t+

a

y=sin t From (1)

t tan 2

From (2)

SIn t=

.

= eX/a a

Y

1+ tan 2 t sin t

..... (2)


251

2

(i.e.,)

2 tan t 2

a

1+ tan 2 t . 2

y

x

2e

a

2X

I+e

y y

=

a cos II

=

([

Y

a

~( /

a

+ e - X,,)

= ((

cosh

(%)

(%) is the evolllte of the given curve. Exercise - 4(C)

I. Find the evolllte of the curve x

2.

=

c t and y

c = -

t

Prove that the equation to the evolute of the parabola x 2 27 a x 2

3. Show that the evolute of the hyperbola

x2

4ay is 4(y - 2(/)3

=

I is (ax)2/3 - (by)213

=

=

(a 2 + b2 )2/3

4. Show that the evolute of the cycloid x equal cycloid.

=

a (q - sin q) y

=

a (I - cos q) is another

4.4 4.4.1

Envelopes Family of curves: Let us consider the equation of a straight line x cosa + y sina = p. Where a is a parameter. For different values of a, x cosa + y sina = p gives different straight lines but all of them are at a constant distance 'p' from the origin. The set of all of these straight lines is said to form a family of straight lines and 'a' is called the parameter of the family. For a given p, different values of'a' give different straight lines which touch the circle x2 + .Y = p2. The circle x2 + 1 = p2 is called the

envelope of the family of straight lines.

252


A curve, which touches each member ofa given family of curves and each point is the point of contact of some member of the family is called the envelope of the family of curves.

Letj(x, y, a)

=

0 be the given family of curves and a is the parameter.

. . WIt . h respect to " . II y Now d ·f'" I Jerentlatmg a partla

a/(x,y,a) aa

=

0

· · 0 f' a' f romJ'"(x, y, a ) = 0 an daf(x,y,a) = o· . .III EI· Imlllatlon gIven an equatIOn terms of x, y which is called the envelope

aa 0/ fhe family of curves.

If the given equation of the family ofcurvesj(x,y, a) = 0 is the quadratic in 'a', say of the form Aa2 + Ba + C = 0 where 'a' is the parameter and A, B, Care functions of x, y. Then the envelope of the family of curves is B2 - 4AC = o.

4.4.2

Example Find the envelope of the family of straight lines y

= mx + ~ m

where 'm' is the

parameter.

Solution a

y=mx+ -

m

III

..... ( I)

is the parameter

Diff(l) w.r. to 'm' a

o=x--2 m

111=

~

Substituting for' m' in (I)

y

=

2.[;;;

Squaring we gety = 4

ax is the envelope of the family of curves

..... (2)


Aliter:

253

y

a

=

mx + -

III

m2x -

111 Y

+a= 0

is quadratic in parameter '/11' (A = x, B = - y, C = a)

8 2 - 4 AC = 0 then gives the required envelop. .. Envelope of the family of curves is

i

i - 4(a)(x) = 0

= 4 ax is envelope.

4.4.3 Example Find the envelope of the family of curves I J J J y=mx+ va-nr+bw here'm'ist heparameter.

Solution I

J

J

J

y=mx+ vlrm-+b(y - I1lx)2 = ([2m 2 + b 2

m 2(x 2 - a 2) + m(- 2 y x) + (V

- b2) =

0

is quadratic in parameter '111' The envelope of the family of curves (-2 x y)2 - 4(x 2 - a 2)(1- b2) = 0 b 2x 2 + a 2 = a 2b 2

i

2

2

.;. + -=;- = I is the envelope of the family of curves. a b 4.4.4 Example Find the envelope of the family of curves x cos 38 + y sin 3 8 = c when 8 is the parameter.

Solution Given x cos 3 8 + y sin 3 8 = c

(8 is parameter)

Diff. (I) partially w.r. to '8' - 3 x cos 2 8 sin 8 + 3y sin 2 8 cos 8 = 0 Hence

tan8= y

x

..... ( I)


254

Substituting these values in (]) X

[ ~ X2 Y+ y2

x 21 4.4.5

= C(X2

3 ]

[

+y

13

j=c

x ~ X2 + y2

+ I) is the envelope of the family curves.

Example 2

2

4

Find the envelope of the ellipse -;- + a b

=

I where

a, b are parameters connected by the relation a2 + b 2

=

c2

Solution Given

x2

y2

a

b

-2+ 2

=]

..... (1 )

a2 +b2 =c2

..... (2)

Assume that' a' and' b' are functions of' t' Diff. (]) and (2) w.r. to '1'

2) da + I a3 dt

2 (-

x

x 2 da

y2

(- 32)db

db

b

-3 - + -3 - =0 a dt b dt Diff.

(i.e.,)

a2 + b2 = c2 w.r.

=

0

dt

..... (3)

to '1'

da db 2a- +2b- =0 dt dt da db a- + b- = 0 dt dt

Eliminating da, db from(3),(4) dt dt

..... (4)

255


Using (I) and (2) we get

x

2

/

£=~=-2 b

a

(i.e.,)

[/4

(14

c

c2

=x2c2

(12= XC

/

-=-

Substituting (12, h2 , values in a 2 + h2

x 2c2 + Ic 2

xl + J,z = I 4.4.6

=

= ("2

c2

is the envelope of the family of curves.

Example Find the envelope of the family of curves

~ + Eh

= I where (I,

b are parameters

{/

connected by the relation ah

= ('2.

Solution Given

~+E a b ab

=

I

= ('2

..... ( I)

..... (2)

Assume that a, b are functions of 't' Diff. (1) and (2) w.r. to 'I' - I-) + da y.-( -I -) = db O x( a 2 dt h 2 dt

x da

y db

- - - + 2- - =0 (/2 £II h dl

Diff.

ab

= ('2

..... (3)

w.r. to 'I'

db da a-+b-=O dt dt

..... (4)

256


da db Eliminating dt ' dt from (3), (4)

x

y

x

y

-

-

ab

ab

x

y

-+!L.=lL~-.!L=JL=~ 2

2

b

a

2ab

Using (I) and (2) we get y

x

£=~=-2 b

2c

a

x

a = 2c

2

ab

Using (2) c2 ~= 2c 2 X

..... (5)

a=2x

a y

ab 2c 2

-=--

b

Using (2)

b=2y

..... (6)

Substituting a, b values from (5), (6) in (2)

(2 x)(2 y)

=

c2

4 xy = c2 is the envelope of the family of curves.

257


Exercise - 4 (0) 1. Find the envelope of the family of curves: (a) y

(b) ;

= mx + a ~I + m 2 when =

111

is parameter.

2a(x - (I) where a is the parameter.

2; =

IAns : x 2 (c) x cos

(d)

~ +~ a

b

=

=

p where

I when a + b = e

(a, b parameters).

(Ans : (e)

~ + ~ = 1 when a2 + b2 = e2 a

b

ff H +

=

I when a + b = e

fx

+

.JY = -!c ]

fx

+

.JY = -!cl

(a, b parameters).

IAns:

(f)

01

(a, b parameters). 2

2

(Ans : x 3 + y3 In

2

=c 3 ]

m

~ + L = I when dnb n = elll + n ani bill

(g)

Find the envelope of the family of curves

(h)

Find the envelope of the family of straight lines

~+~

=

1 where a & bare

related by the equation d l + bn = en, e being a constant. n

(Ans :

xn+1

II

+ yll+1

/J

= C"+ I ]


258

4.5

Curve Tracing Generally a curve is drawn by plotting a number of points and joining them by a smooth line. If an approximate shape of the curve is sufficient for a given purpose then it is enough to study certain important characteristics. This purpose is served by curve tracing methods. The points to be observed for tracing of plane algebraic curves are given below.

1. Whether the curve is passing through the origin, if so the equations of the tangents to the curve at the origin. Suppose F(x,y) = F(O,O) =

° =>

°

is the algebraic form of the equation of the curve.

the curve is passing through origin (i.e.,) If there

IS

no

constant term in F(x,y) then origin lies on the curve. The equations of the tangents to the curve are obtained by equating the lowest degree terms in F(x,y) to zero. If at 0(0,0) the tangents are: (i) real and coincident then '0' is called a cusp (ii) real and different then '0' is called a node (iii)imaginary then '0' is called a conjugate point 2.

Whether the curve is symmetric about an axis or about other any line. If (i) F(x, y) = F(x-y)

=> curve is symmetric about x-axis

(ii) F(-x, y) =F(x, y)

=> curve is symmetric about y-axis

(iii) F(-x,-y) =F(x, y)

=> curve is symmetric in opposite quadrants.

(iv) F(y, x) = F(x, y) curve is symmetric about y = x (v) F(-y,-x) =F(x, y) curve is symmetric about y =-x


259 y

y

----+-----------~~x

--------~~~----------.. x

y' =4ax

(i)

(ii)

y

y

------~~~----~x

----~~~----------------~x

x

(iii)

y

3

+y'

~~V

(iv)

-----..;:::.~"'-------~x

(v)

3. Weather the curve intersects the axes, if so the tangents to the curve at these points. 4.

Find the region in which the curve exists (i.e.) the curve is defined. The values of x for which y is defined gives the extent parallel to x-axis and the values of y for which x is defined gives the extent in a direction parallel to y-axis. If y is imaginary for values of x in a certain region then the curve does not exist in that region.

260


Similarly with respect 10 values ofy . 5.

Finding the asymptotes. An asymptote is a line that is at a finite distance from (0, 0) and is tangential to the curve at infinity (i.e.) the curve approaches the line at infinity. (i) Sum of the coefficients of the highest degree terms in x equated to zero gives the equations of the asymptotes parallel to x - axis. (ii) Sum of the coefficients of the highest degree terms in y equated to zero gives the equations of the asymptotes parallel to y-axis. (iii) To find the asymptotes that are neither parallel to x-axis nor parallel to yaxis (i.e.) oblique asymptotes, the following method is suggested. Substitute y =mx + c equation in 'x'as

$

n

(m)x

n +$

F(x,y) = 0 and rewrite the equation as a polynomial

111

n-

l(m)x

n-I + .... +$ =0 n

The slopes of the asymptotes are given by $I/(m) = O. Let the slopes be

The values of'c' can be obtained trom $I/-I(m)

= 0,$1/_2(m) = 0

(if necessary).

Let the corresponding values of c be c1' c2 , •••••

Note: if $1/ (m) is a constant then there are no oblique asymptotes to the curve. 6. Obtain dy from F(x,y) = 0 by differentiation.

dx

If in the interval for

and :

x, dy > 0 then the curve is increasing dx

< 0, then the curve is decreasing in that region.

. · ·· . by -dy 7. TIle po liltS 0f maxima an dminima are given

dx

=

0.


261

y

~o+-------------------~X

dy dx

-

=

o.gIves

XI,X)

-

If in an interval for x 2

(i)

~

el

~-

> 0 then the curve is concave upwards

(X), X 4 ' .... in -

the figure) and has

a minima in that interval. 2

(ii)

el

~

llx-

< 0 the curve is concave downward (XI' xw .... in the figure) and has a

-

maxima in that interval. 2

(iii)

If at a point

X)

-

(say) in the interval d

~

dx-

=

0 then that point is called a point

of inflection and at that point the curve changes it concavity to the opposite direction.

4.5.1 Example Trace the curve y

X2

+2x

.... ( 1)

=

x+ 1 Solution: Substitution of

X

=0

in (1) gives y

F(x,-y):f:- F(x,-y):f:- F( -x, y) etc 1.

=

:. The curve is not symmetric about any line

can be written as y (x + 1) - x (ie) x

2

-

0:. origin lies on the curve.

2

-

2x = 0

.... (2)

xy - (y - 2x) = 0 Lowest degree term is y - 2x = 0

Equation of the tangent at (0,0) is Y Substitutingy

=

0 in (1) we have x(x+ 2)

= 2x .

= o,x = O,x =-2


262 :. The curve crosses x -axis at (0,0) and (-2,0)

Since x =0 in (I) gives y = 0 only, there is no intercept on y-axis

x :. The curve is defined in the region R -

The

curve

is

not

defined

-I.

at

Hence

it

IS

discontinuous.

{-I}

Coefficient of highest degree term in x «i.e.) of Xl) Hence there is no asymptote parallel to x -axis.

IS

I which is a constant.

Coefficient of highest degree term in y is x + I. :. x + 1=0 is the asymptote parallel to y- axis. From

(2)

dy dx

which

is >

0 always except

at

x

Hence the curve is increasing in the regions. 00

< x < -2, -2 < x < -1, -1 < x < 0 and 0 < x < 00 \ \

\~

\..-

\j.:, \ \ \ \ \ I \

/

\ /

1

2

Fig. 4.7

4.5.2 Example 2

Trace the curve

y-

x +1

- Xl

-1

Solution: ....... ( I)

= -1


263

°

x= in (I) gives y y intercept is -I.

r (-x,y) = r (x,y) y=

°

gives x

2

=

-I :. Hence curve does not pass through (0,0), and

curve is symmetric about y-axis

°

+ 1= =>

= ± i curve does not cross

x

Coefficient of highest degree term in x is y-I

x -axis.

(from( I»

llence y = I is the asymptote parallel to x -axis Coefticient of highest degree term in y is to y-axis

dy =

Hence y

dx

I

=0

x2 -]

.

gives x

=0

x = ±] are asymptotes parallel II

and Yx=o =

-34 < 0 and =I:- 0

:. The maximum point on the curve is (0,-1) and there are no points of inflection.

y

~

~

l~Y.,

I'J

Ix

I I I

I I

--------1---_ I

----r--------I

I I I

I

I I

I

I I

0

I --------1---

X

I I __ ..JI ________ _ (0,-1) Y=-1

I I I

I

I I I I

I I I

I

Fig. 4.8 The curve is decreasing in the interval

.. an d ·· IS IIlcreaslllg 111

-00

°

<

Ixl < ] and I< x <

00

as dy is -ve there

dx

. t I" i < X < 1 as -dy.IS +ve III l1S mterva dx

4.5.3 Example: Trace the Folium x

3

+ / == 3axy

.... ( I)


264

Solution: Clearly (0,0) lies on the curve .No intercepts on y-axis.

F (y,x)=/+x'-3ayx=0=F (x,y)

:. curve is symmetric about x-axis

Coefficient of highest degree term in y is I. a constant, and hence no asymptote parallel to y-axis. Lowest degree term equated to zero gives xy = 0, x = 0, y = 0 (i.e) the axes are the tangents to the curve at the origin.

y = x in (I) gives 2x' - 3ax 2 = 0 x = 0, 3;1 (ie) the line y = x meets the curve at (O,O)and F (-x, - y)

(3~, 3~)

= x' - /

+ 3axy

= F (x, y) :. curve is not symmetric about (0,0)

To obtain oblique asymptotes: Substitute y 3

= mx + c in (I) and

rewriting in terms

2

2

of 'x' we get (I+m )x'+(3m c-3am)x + ..... =0 (ie) ¢3(m)=I+m'and

¢2 (111) = 3m 2c - 3am, ..... ¢, (m)

= 0 gives m= -1, there is only one oblique asymptote with slope-I

,

,

,

x+y=3a

,

VI

Fig. 4.9

¢2(m)=0 gives 3m(cm-a)=0 => cm-(l=o :.Theasymptoteis

y=-x-a or y+x+a=O

(e) c=-a(-:m=-l)

.... (2)


265

--------------------~----------------------------------------~

dy= From (I) dx

2

ay-x ~ :. (dY - ) =-1 y -ax dx e~/,32")

Equation of the tangent to the curve at this point is y

+ x = 3£1 which is parallel to (2)

4.5.4 Example: Trace the curve a 2y 2- x 2( a 2- x 2) = 0 ........ ( I )

Solution: x

=0

F

(-x, y)

in (I) gives y = 0,0 (ie) double point on the curve =

F

(x, -y) = F (x, y) curve is symmetric about both the axes.

Lowest degree terms, in (I), equated to zero gives a are the tangents to the curve at (0.0). y

=

°

in (I) gives x

The curve intersects the x-axis at

(a, 0)

and

Ixl > a

l

is negative (ie) the curve does not exist for

Tangents at the points

ddx2y ) at x ( 2

=

°

gives

x

2

)

= 0 ~ y = ±x

(-(1,0)

From (I) it can be seen that

Ixl > a

= 0, ± .i2' points of maxima and III inima.

a x = ± J2 to the curve are parallel to x-axis .Further

a . . = J2 IS negative.

in 0< x <.i2

x

(/ -

= ±a

for

YI

2

:. x

while decreasing in

a .IS a maxImum . = J2

. d" . POlllt an Y IS IIlcreaslIlg

.i2 < x < a .Thus after tracing the curve in

the first quadrant, the symmetry about both the areas can be utilized for tracing the complete curve. Substituting y

= mx + c

in (I )gives

X4

+ a 2 (mx + c

r -a

2 2

x

=0

266


¢4 (m) = 1 which is a constant. Hence there are no oblique asymptotes to the curve. y

~(_-a~,O~)~--------~----------+-~~X

, Fig. 4.10

4.5.5 Example: Trace the curve y2 ( x 2 - a 2) + a 2x 2 = 0

..... (1)

Solution:

x = Oin (1) gives y = 0 :. (0,0) lies on the curve. F (x,-y) =F (x,y) =F (x,-y) :. curve is symmetric about both the areas. Highest degree terms of y equated to zero gives (X2 - a 2) l

= 0 => x = ±a

:. x = ±a are asymptotes parallel to y-axis. Highest degree terms of x equated to zero gives x 2(y2 + a 2) = 0 => x = ±ia ... No asymptotes parallel to x-axis. Equating lowest degree terms to zero we get a 2(y2 - x 2) = 0 => y ± x are tangents at (0, 0) writing y

= mx + c

in (1) and rearranging in terms of I x I

m2x 4+2mcx3+(c 2 _a 2m2+a 2)x2

-2a 2mcx-c 2a 2 =0

=> c is indeterminate since m = 0 ¢2 (m ) = 0 gives c 2+ a 2 = 0 => c = ±ai oblique asymptotes do not exist. Further there are no intercepts on either of the axes.

267

Curvature and Curve Tracing 2

( 1) can be written as

l

=

a 2x x _a

-2 - which indicates that curve does not exist 2

when Ixl > a :. The curve exists only in the region -a < x < a. y

x

Fig. 4.11

dy a4 x = and y> 0 in first quadrant. Hence curve is increasing in 0 < x < a . dn y 4.5.6 Example: Trace the curve

l (2 - x) - x2 (1 + x) = 0 ....... ( 1)

Solution: x=Oin (1) gives

y=O

:.(0,0) lies on the curve

F(x,-y)

=

F(x,y) :.curve is

symmetric about x-axis . Lowest degree terms equated to zero gives

2l- x 2 = 0 => y = ±

12

tangents at (0,0) to the curve y

=

0 in (\) gives -1,0

curve passes through (-1,0) in addition to (0,0) Coefficient of highest degree terms of y equated to zero gives 2 - x = 0 (ie) x = 2 is an asymptote parallel to y-axis 2

No asymptote parallel to x-axis as coefficient of x is 1,a constant


268 Substituting Y

x

3

(

= mx + c

in (I) and rearranging

2 2 2 2 2 m + 1) + x ( 1+ 2mc - 2m ) + ( c - 4mc ) x - 2c

¢3 (m) = m 2 + 1 => ¢3 (m) = 0 gives m = ±i :.

=0

There are no oblique asymptotes.

--~----~~--------~----~x ...

~...,

I (2,0) I '" J2 I "'",. I I I -x

y=-

I I

Fig. 4.12 From (I) Y1 = y2

2

+ (2x + 3x h' h h O· 0 < X < 2 were h ) . T IS sows t at YI > 111 t he curve

2y 2-x

is increasing and x = 2 is an asymptote. YI

= 0 => x = 0, x =

5±J5j 4

.

.. In the secon d quadrant y IS maxImum at x

5-£1 =-4

where the tangent IS

parallel to x-axis It can be seen that the curve forms a loop in

[-1,0]

due to symmetry about x-axis.

4.5.7 Example Trace the curve /

-(x-2)(x-4)2 =0

Solution: y = O. (I) gives x = 2, 4 Curve intersects the x-axis at (2,0) and (4,0)

.. , .. (1)


269

F(x, -y) = F (x,y) :. curve is symmetric about x-axis For x < 2 , /

is -ve :. therefore curve does not exist for x < 2

From (I) it can be seen that x = 2 is a tangent at (2,0) Coefficient of x 3 = l,a constant :. No asymptote oarallel to x-axis Differentiating (1) w.r.t 'x'

dy _ (x-4)(3x-8)

dx At

2y

(4.0) dy

= Lt ± (x-4)(3x-8)

dx

Lt± 3x-8 = ±J2 2.Jx-2

2.J x - 2 ( x - 4)

H4

H4

Equation of tangents at (4,0) are y =

±J2(x -

4)

Further dy --}- 00 as x --}- 2 :. x = 2is a tangent at (2,0)

dx

y

--~--~----~~~~~x

,,

(2,0)

Fig. 4.13

dy = dx

°

gives x = 8/3. Thus the tangents at x = 8/3 are parallel to x-axis. For the

region. 2 < x < 8/3, dy >

dx

°

above x-axis and dy <

dx

° °

extreme values. For the region 8/3
dx

x -axis. :. the curve forms a loop in 2 ~ x ~ 4

below x- axis. At x = 8/3, y takes above x -axis and dy >

dx

°

below


270

4.5.8 Example 2

Trace the curve y ( x + 4a

2

) -

8a 3

=0

.... (1)

Solution:

y

* 0 when x = O.

:. origin does not lie on the curve.

Curve is symmetric about x-axis ,since F

(-x,y)=F (x,y)

y is +ve for all real x :. Curve completely lies above x-axis. Substituting y = mx + c in (I)

m3 + cx 2 + 4a 2 mx + 4a 2 c - 8a 3

¢3 ( m ) = 0 => m = 0 and

=0 ¢2 ( m ) = 0

gives c = 0

dy

(ie) y = 0 is an asymptote to the curve From (I) -

dx In first quadrant dy < 0 and hence y

IS

dx

maximum at

16a 3 x

t

= - ----

(X2 + 4a2

decreasing. dy

dx

o at (0, 2a). y s

(0, 2a) .

----------~--------------~x

Fig. 4.14

4.5.9 Example: Trace the curve y- c cosh x / c = 0

Solution:

x= 0 in (1) gives y = (c/2)[e O +e-D] = r

.... (1)


271

:. Curve does not pass through (0,0) and intersects y-axis at (0, c). F

(-x,y)= y -(c/2)( e- x / c +e
curve is symmetric about y-axis

It can be seen from (I) that y>o for all real x .

dy= Sill . h -

dx

( x /) C = -1 [x/c e -x -Xlc] 2

which is >0 for x >0 and <

°

for x <0. y

Y=c

(D,c) ------~~------------~x o

Fig. 4.15 Thus the curve is increasing in

°

(0,00), and decreasing in (-00,0).

When x = we have dy = 0. Further y = c for x = 0.

dx

(i.e.) The tangent to the curve at (0, c) is parallel to x-axis.

4.5.10

Example

Trace the curve

(x / a )2/3 + (y /a )2/3 -1 = 0

Solution:

y

:I;

0 when

x =

° :.

It can be observed that F

(0,0) does not lie on the curve.

(-x,y)= F (x,y)= F (x,-y)

The curve is symmetric about both the axes.

.... ( 1)

Engineering Mathematics-I

272

--~---------+----------~----~x

(0, - b)

-

Fig. 4.16

x = 0 gives / = b 2 => Y = ±b y = 0 gives x = ±a. Thus the curve meets x axis. at (-a,o),( a,o) and y axis at (o,b ),( o,-b). For

Ixl> a or \YI >b

(x/a) 2/3 +(y/b)2/3 -1> 0

Equation is not satisfied.

Ixl> a or IYI >b. :. Curve completely lies in the region Ixl ~ a and Iy ~ bl :. Curve does not exist for

y -!!..( a )I/3. In the first quadrant y is decreasing as a bx

From (I) dy

dx

dy < 0 for dx

O 0

Solution: Clearly y F

(x, - y)

= 0, =

0 when x = O. Origin is a double point on the curve.

F ( x, y). Curve is symmetric about x-axis.

.... (1)


273

dy

Differentiating (I) w.r.t, 'x' we get -

Fx (3a - 2x )

= -

dx

3jJ

2 (a-x) -

dy dx

1(0,0) =

o.

So x axis is a tangent to the curve at origin.

x = a is an asymptote parallel to y-axis. 3

A" coefficient of x is a constant, there is no asymptote parallel to x -axis. For x < 0 and x> a curve does not exist (as y2 is negative) Substituting y

x

3

(

= mx + c

in (1) and rearranging

m 2 + 1) + x 2 ( 2mc - am 2 ) + x ( c 2 - 2ame ) - ac 2

=0

fA (m) = m 2 + 1=> m = ±i . I I x=a

PI

I I I --~~--------.---------~x

(a,O)

Fig. 4.17 Hence there no oblique asymptotes.

3x2 + y2

From (I) YI

= (

)

2 a-x y

For 0 < x < a, YI is positive. Hence y is increasing in 0 < x < a .

. 4.5.12 Example Trace the curve x

2

(a

- x) = l

(b + x)

.. .,(\ )


274

Solution: Clearly curve passes through origin F ( x, - y ) = F

(x, y) .'. Curve is symmetric about x-axis.

Equating lowest degree terms to zero

by2 _ax 2 = 0 => y

= ±~.x

There are two real, distinct tangents to the curve at (0,0)

y

=0

in (1) gives

The curve intersects x-axis at

x2 ( a - x) = 0 => x = 0, a

(a, 0) 3

There is no asymptote parallel to x-axis as coefficient of x is a con y

I I I I

--~I~--------~~--~~~~----~x I I I l X=-b I

Fig. 4.18 Coefficient of highest degree term in y is b + x

:. x + b = 0 is the asymptotes parallel to y-axis Substituting y

= mx + c

x 3(1 + bm

in (I) and rearranging

2) + .... --0'I.e., V'3"'(m)-0' - gIves m -_+_i - Jb

Hence no oblique asymptote to the curve.

t!) can be written as y regions x > a or x < -b

= ±xJa - x b+x

which shows that y is imaginary in the

275


Hence the curve completely lies in the region -b < x

dy

From (I) - = ±

dx

The tangent

( - 2X2 - 3bx + ax + 2ab) 2(b+x)

3/2

1/)

~

a

which shows that dy ---) 00 as x ---) a

dx

(a-x)-

x = a to the curve at (a, 0) is parallel to y-axis.

Thus the curve forms a loop between

(0,0)

and

(a, 0) .

4.5.13 Example: Trace the curve x 2(y2 - a 2) - b2/

....(\ )

=0

Solution:

x = 0 in (1) gives only y = O. Curve passes through origin.only and does not intersect either of the axes:

F(-X,y)= F (x,y) = F (x, -y) :. Curve is symmetric about both the axes. 2

Lowest degree terms equated to zero gives a 2x 2+ b y2 showing that there are no tangents to the curve at

=0

(0,0)

Coefficient of highest degree terms in y (i.e.) of y2 is x 2- b 2 :. x

= ±b are the asymptotes parallel to y-axis

Coefficient of highest degree terms in x (i.e) of x 2 is y2 _a 2

= ±a are the asymptotes parallel to x- axis. Substitution of y = mx + c in (1) gives :. y

x 4 .m 2 + x 3 (2mc) +X2 (c 2 _a 2 _b 2m2)_ 2b 2mcx-b 2c 2 = 0

¢3 ( m ) = 2mc does not provide value of c since m = 0

276


±a which are parallel to x-axis.

The asymptotes are y =

Hence there are no oblique asymptotes. From (I) x

~ ±~

by

Similarly for

IYI < a , x

For

-a2

y2

Ixl < b,

is imaginary

y is imaginary. Hence curve does not exist for

Ixl < b,

Iyl
J

y=a

(0,0)

"x=b

x =-b y=-a

"

From (I) Y = ± /

"x

( Fig. 4.19

ax

2

-b

..... (2) 2

From (2) it is observed that y ~ 00 as x ~ ba and x ~ 00 as y ~ a For the first quadrant y

a 2b2

:. Yl = -

(x 2

~

=

+

~

ax

x2

_b 2

.Thus Yl < 0 for x> b . So y is decreasing when x> b

111

_b 2 )2

first quadrant.

4.5.14 Example Trace the curve

y2

(x - a) - x2 ( X + a) = 0

Solution: x = 0 gives y = 0 :. (0,0) lies on the curve.

.... (1)


277

.

(2 + y 2) = 0 => x- + y = 0

Lowest degree terms equated to zero gIves a x

J

J

:. There are no tangents to the curve at (0,0) coefficient of highest degree terms in y (i.e.) of

Y

J

IS x - a

:. x = a is the asymptote parallel to y-axis 3

No asymptotes parallel to x - axis since coefficient of x is constant) Substituting y

= mx + c

in (I)

x 3 (m 2 -1)+ x 2 (2mc-um 2 - a) + x(c 2 - 2amc)-ac 2

=0

¢3(m)=m 2 -1=0 gives m=±l ¢2(m) = 0 gives c =

a(m2 + 1)

2m

= a when m = I and = - a when m =-1

I I

V'.Sl

/\/y=x+a

/ I (a,2a) // I 1 I

--~.---~--------~x

(a,O)

"

I 1

" " l(a,-2a)

~~+a=o If""

Fig. 4.20 :. The oblique asymptotes are y From (I)

2d dx

2

=

and y + x + a

=0

2

x -ax-a (x -a).J x 2 - a 2

When x ~ -a, dy

dx

~ 00

-dy = o. gIves dx

=x + a

:.

Tangent at (-a, 0) is parallel to y-axis

x 2 - ax - a 2

1±f5 = 0 => x =-a

2

278


Tangents to the curve at x For

= 1 + Fs a (A and B in fig) are parallel to 2

0< x < a

y is imaginary

a
yexists

-a
y is imaginary

- 00

< x <-a

x -axis.

yexists

(i.e.) The curve does not exist in the region -a < x < a.

1-Fs

For x = - - a there is no tangent to the curve (since the curve itself is not

2

defined there).

4.5.15 Example Trace the curve /(a 2 _x 2)_x 2(a 2 +X2) = 0

.... ( 1)

Solution: x

=0

in (1) gives y = 0,0

:. origin is a double pt on the curve.

The curve is symmetric about both the axes sinceF(-x,y) = F(x,y) Lowest degree terms equated to zero gives y (i)

· can be written as y 2

2(2

= X a2 +x2

2)

a -x

Hence the curve lies only in

•

= ±x as the tangents at the origin.

Th"IS

III d'Icates

that y

= mx + c ¢4 (m ) = m 2 + 1 = 0

Ixl < a. in (l) gives

X4

(m 2 +1) + ... = 0

gives m = ±i

There are no oblique asymptotes From (1) dy

dx

=

4 2 2 4 a +2a x -x (a 2 _x 2)3/2(a 2 +X2yl2

In the first quadrant: dy > 0 for 0< x
dx

(i.e.) y is increasing in this region.

+

~ _00

x = ±a are asymptotes parallel to y-axis. Substitution of y

= F(x,-y).

as x

~

a.


279

y

x

Fig. 4.21


l (a 2 + x 2 ) -

x2 ( a

2

-

x

2

)

=0

.... ( I)

Solution: x = 0 gives y = 0,0 :. origin is a double point on the curve.

F(-x,y) = F(x,y) = F(x,-y) shows that the curve is symmetric about both the axes. there are no asymptotes parallel to x-axis, since coefficient of highest degree terms in x ((ie) of X4 )is constant. The coefficient of highest degree terms in y (ie) of

l

2

is x + a 2 •

:. there are no asymptotes parallel to y axis. From (1)

y

=

a 2 -x 2 ±x 2 2 a +x

For

Ixl > a,

y IS Imaginary

:. The curve entirely lies in the region -a s x sa.

y = mx + c in (l) gives ¢J. (m) = m 2 + 1 :. m = ±i


280

There are no oblique asymptotes :. y values are extreme at parallel to x-axes. 4

dy = 0 gives x = dx

x=±a~J2 -1 ,where 2 2

±a~ -1 ± J2 the tangents to the curve are

4

dy = + a -2a x _x dx - (2 a +x 2)3/2 (2 a -x 2)1/2

..... (2)

y

x

(-a,O)

Fig. 4.22

.'. x = ±a are tangents to the curve at (±a ,0). Further

(ddxY)

(0,0)

=

a:

= 1 from (2)

a

Hence y = ±x are tangents to the curve at (0,0) The curve thus forms a loop between (0,0) and ( a ,0) and another loop between (-a,O) and (0,0).


a/ - 4 x2 ( a - x) = 0 ..... ( t)

Solution: F (x, - y) = F

(x, y) :. curve is symmetric about x-axis

F (0,0) = 0 curve passes through origin.

Curvature and Curve Tracing y=

281

°

in (1) gives x = 0, a (i.e.) the curve cuts the x-axis at (0,0) and (a ,0) 2

Equating the terms of lowest degree to zero we get al- 4ax = O::::? Y = ±2x Thus Y ± 2x = 0 are the tangents to the curve at (0, 0)

dy = 4x(3a-x2) ::::?(dY ) dx 3ay dx (up)

----).00

(ie) the line x = a is a tangent to the curve at (a ,0)

= mx + c

substituting y

in (1) and rearranging the terms

4x 3 +ax 2 (m 2 -4)+ 2amcx-ac 2 = 0

fA (m) = 4 constant. For

Ixl > a we find

Hence there are no asymptotes to the curve. y is imaginary.

Curve does not exist for

Ixl > a.

________

~~--------~~~x

Fig. 4.23

4.1.18 Example: Trace the curve a

21

= x 2( 2a - x) ( x - a)

Solution: F

(+ x, - y)

=

F curve is symmetric about x-axis.

F (0,0) =0 origin lies on the curve (isolated point)


282

= 0,a,2a.

y = 0 in (1) gives x

The curve intersects the x-axis at (a ,0) and (2 a ,0) Terms of lowest degree are curve at (0,0) .

(y2 + 2X2

y

= 0 gives

l

+ 2X2 which shows that there are no tangents to the

= ±J2ix imaginary)

Proceeding as in 3.1.17 it can be seen that the curve has no asymptotes y is imaginary for x < a and x> 2a(i.e.) the curve exists only in the region a:::;x:::;2a.

Further dy

dx :. x

=a

~ 00 as x ~ aor x ~ 2a

and x

= 2a are tangents to the curve parallel to y-axis. y

x (0,0) 0 I

x=a

I

x = 2a

Fig. 4.24

4.1.19 Example Trace the curve x( x 2 + y2) = a( x 2 - y2 ),( a> 0) ........... (1) (2001 s) F (x, - y) = F (x, y) curve is symmetric about x-axis

Solution: y = 0 in (I) gives x

= 0,0, a . The curve meets the x-axis at (0,0) and (a ,0).

Equating lowest degree terms to zero we get

y2 _ x 2 = 0 => y

= ±x which are the tangents at (0,0)


dy dx

~ 00

as x

~a

283

. Thus x = a is a tangent to the curve at ( a ,0)

For x > a (1) shows that y is imaginary No part of the curve exists for x> a.

y

~ 00

as

x ~ -a :. x = - a is an asymptote to the curve.

Fig.

4.2~


l (a 2+ x 2) = x 2( a 2- x 2)

..... (1)

Solution: F(O,O) = 0 curve passes through origin F (-x,y) =F (x,y) =F (x,-y) curve is symmetric about both the axes. Lowest terms equated to zero gives a 2y2 - x 2a 2 = 0

=> y ± x which are tangents

to the curve at (0,0) (:)

~

00

as x

~ ±a

:. x

= ±a are the tangents to the curve at

y is imaginary when x < -a or x > a :. curve completely lies within the region -a

~

x~a

(±a, 0)

284

Engineering Mathematics - I y

x (-a,D)

Fig. 4.26

4.1.21 Example Yrace the curve

Solution: Let F( x,y)= y( a

2

+ x 2 ) - a2x

=0

F( -x,y) =I:- F(x,y) =I:- F(x,-y), Hence the curve is not symmetric about both the axes. The powers ofy are odd. Therefore the curve is symmetric In opposite quadrants. F(O,O)=O. So the curve passes through the origin. The coefficient of highest power of x i.e~ of x 2 is y So y=O is an asymptote parreJ to x-axis. The coefficient of highest degree of y is x 2 + a 2

= 0 => x = ±ai

Hence the extreme points are A(a,aJ2) B(-a,-aJ2) Thus the tangents at A and B are parallel to x-axis. The shape ofthe curve is as shown below.


285 y

x

Exercise - 4(E) Trace the following curves

3.

2 x-2 y = 2- x -1

5.

9ai

7.

y3

9.

y =2 - - -2 x +4a

4.

x 3 + / = ax 2

;

x> 0

= x(x-3a)2

=a 2 x-x 3 8a 3

11. i

= (x-a)(x-b)(x-c)

13. i

= x (3a-x)

; a,b,c>O

8.

i(a-x)=x 3

10.

i(a+x)=x 2 (b-x)

12. y=x

;a>O

3

2

x+a 15. y2 =(X-2)2(X-5)

17. xi =4a 2 (2a-x) 4.2.0 Tracing of curves (polar form) The equation of the curve in polar coordinates may be in one of the following three forms

F (r,B)=O

,r=F(B),

B=F(r)

A study of the following points will simplify the tracing I.

curve is symmetric (a)

About initial line B = 0 if F(r,-B)

= F(r,B)

286


(b)

About pole '0' if F(-r,B)

= F(r,B)

or

F(r,J[ + B) = F(r,O)

e = 1[/2

0=0 Pole

0

Fig.4.27(a)

2.

= J[

if F(r,J[-8)

= F(r,8)

(c)

About 8

(d)

About 8 = J[ if F(r, J[ -8) = F(r,8) 4 2

(e)

3J[ . 3J[ About 8 = - If F(r,--8)=F(r,8)

(a)

Values of 8 for which r

2

4

2

=0

gives the equation of the tangents to the curve

at the pole (b)

F(r,O) = 0 gives the points of intersection of the curve with the initial line

(c)

J[ F(r, -) = 0 gives the points of intersection of the curve with the line 2 8=J[ 2


3.

(a)

287

If 'i and r2 are the least and greatest values of r then the curve lies in the annulus between circles of radii Ii and r2

(b)

4.

Curve does not exist if r is imaginary for values of () or if for values of r

. .111 wIlIC ' h -e1r > 0 I r I •mcreases -dr < 0 I r I decreases .111 that regIOn . In t Ile region dB dB () =rr/2

__

y

o_ _ _ _ _ _ _ _ _

~~-..J...

o Let

p(r,B)

~ .

--.

X

0=0

Fig.4.27(c)

Fig.4.27(b)

5.

B is imaginary

be a point on the curve and P T is the tangent to the curve at P, OR

radius vector and ¢ is the angle T P R. Then

=r

dB gives equation ofPT

(i)

tan¢

(ii)

¢ = 0 gives tangent parallel to initial line

(iii)

do V'

= -Jr 2

dr

. gives tangent

to Il1Itla I I'me

1./(/r. ..

6.

If the curve meets the initial line at two distinct points and curve is symmetric about the initial line then part of the curve forms a loop between these two points.

7.

If for any value of B (sayB ,) such that It r 0-->00

the curve

= 00

then B = B1 is an asymptote to

288


8.

Sometimes it may be convenient to change the equation of the curve from polar to Cartesian form by substituting respectively

x = rcosO, y = rsinO y

9 = 1t/2 y

!

9=0

--nf~~Y===~)---7X

Few solved examples are given below

4.2.1 Example Trace the cardioid

r -a(l +cosO) = 0----(1) Solution: F (r,O)

=

r -a(l +cosO) = F(r,O) curve is symmetric about the initial live.

When 0 =

o=

1[

1[

we get r = 0

:. curve passes through pole and

is the tangent at the pole

Max IcosOI

=

I

:. Max r = 2a

:. The curve is entirely within the circle of radius 'a'. Hence no asymptote to the curve.

0

0

1[/4

1[/3

1[/2

21[/3

r

2a

a(l + 1/J2)

3a/2

a

a/2

r = 2 a when

0

=

0 and r = 0 when 0 =

1[

0

41[/3

31[/2

51[/2

21[

a/2

a

3a/2

2a

1[

.'. The curve intersects the initial Iine at (2a, 0) and (0, 1[ )


289

------------------~~--------------------------------------

. b The tangents to the curve are gIven y tan tan ¢ = a(l + cosO)

a(e-sine)

when

e=

= -cot e

2

AI 'I'

= rde dr

= tan

(1l2 + ( )...(2) 2

0 , (2) gives ¢ = ~ (ie)The tangents at (2 a ,O)is perpendicular to the

initial line ..

Fig. 4.28

4.2.2 Example Trace the curve r

= a(l- eose)

Solution: 8 = n/2

(+2a,n)

J--------.....

8=0

........

~---

Fig. 4.29

(0,0)

~

290


()

J[ 7J[

0, 2J[

4' 4

0

r

a(l-

2' 2

~

a

= 2a when

:. The curve intersects the initial line at tan ¢ =

a(l-cos()) asin()

J[ 3J[

3' 3

YJ2)

() = 0,2J[ and r

r = 0 when

J[ SJ[

2J[ 4J[

J[

--

3 ' 3

3~

2a

() = J[

(0, 0), (2a, J[)

()

= tan2

:. tan ¢ = 0 when () = J[ (ie) the tangent at (2a, J[) is perpendicular to the initial line. (work out remain ing points as in 3.2.1)

4.2.3 Example: Trace the curve r - a (1 + sin ())

=

°

..... (1)

Solution: r = a when () = 0, J[ r=Owhen ()=-J[/2 and r

= 2a

...... (2)

J[ when () = -

2

:. The curve intersects the initial line at and intersects the line () = J[/2 at F

(r, J[ -

(a, 0), (a, J[)

(2a"~-i)

and

(0, - J[/2)

()) = r - a ( 1+ sin ( J[ - ())) = r - a (1 + sin ()) = F

(r, ())

:. curve is symmetric about () = ~ The equation of the tangent at the pole is () = -~ (from (2)) Maxlsin the circle r

()I =1

:. Max r = 2a Hence curve completely !ie~ ".'!!hin the region of

= 2a and there are no asymptotes to the curve.


291

7r/2 57r/6 7r

0

0

7r/6

r

a

3(1/2 2a

3(1/2

27r/6 37r/2 117r/6 27r

II

0

a/2

(1/2

a

--------~--~~--~------~.O=O

0= -rr/2

Fig. 4.30

4.2.4 Example: Trace the curve r - a sin 0

=0

Solution:

0

0

r

0

7r/12

7r/4

a 2

a

-

7r/3

J3

-a 2

57r/12

7r/2 0

a 2

-

When 0 = 0 we have r = 0 :. pole lies on the curve F

(r,7r+O)

=

r-asin(27r+20)=r-asin20=

F

(r,O)

:. curve is symmetric about pole.

F (r, ; - 0 ) = r - a sin ( 7r - 20) = F (r, 0)

...... (1)

292


and

F

(r, 3; -8 ) = r -asin(3ff -

28) =

F(r,8)

... curve is symmetric about the lines 8 = ff14 and 8 = 3ff14

...... (2)

from (1) and (2) it follows that the curve is symmetric in all the four quadrants.

=0

r

gives 8

= O,ffI2,ff,3ff/2

these four lines are tangents, to the curve, at the

pole. Max r

=

Max la sin 281 = a

:. The curve completely lies within the region of

the circle r = a . The line 8 = ffl4 meets the curve at (a,ffI4). 0= rc/2

o =rc ----.-,;:~~rr_Illll:iiiii:::-----__1~ 0 =0

e =3rc/4 Fig. 4.31

Note: Similarly trace the curve r

= a cos 28 .

4.2.5 Example: Trace the curve r2 - a 2cos 28 = 0

...... ( I)

Solution: 8

0

ffl6

r

±a

±a/J2

ffl4 0

ff13 to 2ff13 imaginary

3ffl4 0

5ffl6

ff

±a/J2

±a


r

=0

gives eos2B

293 Jr

= 0 => f) = ±4

curve passes through the pole and 0

F(r,-fJ)

= F(r,fJ)

F(r,Jr -0)

= r2

4

are the two tangents at the pole

:. curve is symmetric about the initial line.

_a 2 eos(2Jr - 20)

= ,.2 -

= r2

_a 2 eos20 = F(r,fJ)

=~

curve is symmetric about 0

F(,., Jr + fJ)

Jr

= ±-

([2

eos(2Jr + 2fJ)

= F(r, 0)

curve is symmetric about the pole

8=0 __

~

~

________

~~

______

~~~~x

Fig. 4.32 Since leos 2fJI ~ 1 ,Max r

=a

Curve lies completely within the circle r

B = 0 gives r2

=a

= a 2 => r = ±a

curve intersects the initial line at (a,O) and (-a, 0)

r2 is -ve When Jr < 0 < 3Jr (": cos 20 < 0) 4 4 curve does not exist (above the initial line) in this region. Differentiating (1) w.r.t

I

fJ I 2

dr _a sin 2B dr 2 2 • 2B 0 => -=---2 r-+ a SIn dB dB r


294

dO r ff tan¢ = r - = r. 2 = -cot20 = tan(-+ 20) dr _a sin 20 2

tan¢ = co

Thus when O=O}

O=ff Hence the tangents at 0

=0

= ff

and 0

are

.lIar

to initial line

When (i.e.,) the radius vector 0

(': 0

=

=

%is tangential to the curve at the pole

%gives

r

= 0)

= 3ff

Similarly the radius vector 0

also is tangential to the curve at the pole.

4

4.2.6 Example: Trace the curve

r - a cos 30 = 0 - - - -(1)

Solution: F (r, -0) = r -acos30 == F (r,O)

Curve is symmetric about initial line Line 0

=0

intersects the curve at

(a, 0 )

Since Icos 301 ~ 1 , from (1), Irl ~ a .Curve completely lies within the region of the circle r

r=O when 30=0

=a

:.30=(2n+l)ff/2, n=0,1,2, .....

(i.e.) curve passes through the pole when 0

= (2n + 1)ff/6( n = O/oS)

These six lines are tangential to the curve at the pole The part of the curve from 0

= to 0 to

ff is AA, 0 .

6


295

This has a reflection AA20 in the initial line. Thus AAIO A2A is a loop. F (r,4n/3 -B) = r -acos( 47l' - 3B) = F (r,B) The curve is symmetric about B = 27l'/3

OBIB is the part of the curve

from B = 27l'/2 to 27l'/3. This has a reflection OB2B in line 0 ThusOBI BB2 0 forms a loop.

= 27l'/3

F(r, 8; -B ) = r -acos(87l' - 3B) = F(r,B)

Curve is symmetric about B = 4~ 0= 1t/2 B

.........

A2

..... .........

(a,O) () :::

.... ..... 111t/6

....

(-a,1t/3)

Fig. 4.33 OCIC is the part of the curve from B = 7l'/6 to 7l'/3

and

OC2C is its reflection in B = 4~ :.OCpC20 forms a loop. It can be seen that

OCPC20 is a reflection of OBI B820 in the line B = 7l'

tan¢ =

¢ ~ 00

acos3B -1 = -cot3B -3asin3B 3 when B ~ 0

:. The tangent at (a,O) to the curve is

..LIar

to the initial line.

296


4.2.7 Example: Trace the curve r2 - a 2 sec 2 f)

=0

Solution: F(r, 0) = r2 cos 2 f) - a 2 = 0 - - - -(1)

The given curve can be written as f)

=0

gives r

F(-r,f)

= ±a .Curve docs not pass through the pole.

= F(r,f)

:. Curve is symmetric about the pole

F(r,-f)

=,.2 cos 2 (-f)-a 2 = F(r,f)

Curve is symmetric about the initial line f) F(r,TC -0) = r2

COS

2

=0

(TC -f)-a 2 = F(r,f):. Symmetric about f) = TC 2

2

Since cos 0 S; 1 (i.e.,) IcosOI S; 1 we get r2 2:: a Curve does not exist in the region l' --) 00

,.2 < a

TC as f) --) -

TC

±a

l'

dl'

r=-

df)

J

= a-

1

sec- f) tan f)

2

TC

0

2

:.0 = - is an asymptote to the curve.

2 f)

2

TC

TC

TC

-

-

-

-

6

4

3

2

+ 2a

±J2a

±2a

00

-J} df)

-

dr

l'

= -) cos 2 f)cotf) a-

tan¢=rdf) =cotf)=tan(TC -f)) :. ¢=TC/2 when f)=O showing that dr 2

the tangent ate a, 0) is ~/ar to the initial line .

297


(-a,O)

-\---~-~I-

Fig. 4.34


,.2 - a

2

cos 2(} = 0

Solution:

() = 0

gives

r = ±a :. curve intersects initial line at (±a,O)

,. = 0

gives cos 2fJ

= 0 ~ B = ± 7r/4

pole lies on the curve

and () = ± 7r /4 are two tangents to the curve at the pole

Icos201 S 1 ~

IrI sa. Curve completely lies within the region of the circle

I'

Curve is symmetric about the initial line. F

(-1',0) =(rr _a 2 cos20 = F (1',0)

F

(r,7r-(})=r 2 -a 2 cos(27r-20)= 0

0 ±a I'

±Jr/6

±7r/4

±a/J2

Curve does not exist in the regions

0

symmetric about the pole F

(1',0)

symmetric about

±7r/3

±Jr/2

imaginary

7r/4 < B < 3Jr/4

and -

0=7r/2

±37r/4 0

3Jr/4 < 0 < -Jr/4

=a

298


The curve forms a loop bctween (a ,0) and (0,0) and has reflection in B = 1(/2 ~

Diffcrentiating (1)2r-

dO

=

0= 0 gives ¢

= -2a 2 sin20

tan¢

dO =r- = dl'

~2 2

a sip 20

tan (1(/2 + 20) =

1(/2

Thus the tangents at 0

=

0 (ie) (a ,0 ) and ( - ( l ,0) are perpendicular to the initial

line. y

x

Fig. 4.35

4.2.9 Example: Trace the curve

r - {a + beosB} = 0

Solution: F (r, -0) = r -

= -eot20

{a + beos ( -B)} = F (r, 0)

Curve is symmetric about the initial line. Since leosBI ~ I we have

IrI ~ a + b

The curve completely lies within the region of the circle. r

= a+b

o =0 gives r=a+b :.curvemeets 0=0 in A(a+b,O) o= 1( gives r = b curve meets B = 1( in B ( a - b, 0) {l -

.. Curvature and Curve Tracing

Thus when When

299

a*- b curve meets the initial line at A and B

a = b we get r = 0 curve passes through the pole.

Curve meets the line B = 1£/2 at (a, n/ 2) . Curve passes through the pole when () = cos--!

(-{l/ b) (since r = 0)

Hence the tangent to the curve at the pole is the line B = cos"! ( (i)

does not exist when a > b

(ii)

is B

-a/ b)

which

cos-!(-l)=n when a=band

=

(iii) exists when a < b

0

0

n/3

1£/2

2n/3

r

a+b

a+b/2

a

a-h/2

dB

tan ¢ = r -

dr

:. ¢

= n /2

= a+hcosB -hsinB

.

showmg that tan¢

(ie) the tangent to the curve at

= 00

(a + b, 0)

when is

n

--

a-b

0

.l/llr

=

0

to the initiallinc.

Further ¢ *- 0 for any value of 0 and hence no tangent to the curve is parallel to the initial line When a =h tan ¢ When Wht1f\

0

=

=

l+cosO . . -cotB/2=tan(n/2+0/2). smO

n we get ¢

=

n (ie) the initial line is a tangent to the curve at (0, n)

a*- h we have tan ¢

~ 00

as () ~ n (i.e.) The tangent to the curve at

(a - b, n) is also perpendicular to the initial line.

300

____E_n--:9:::...ineerin9 Mathematics - I

I (a+b,O) (2a,0)

I I a=b

a>b

Fig. 4.36 (ii)

Fig. 4.36 (i)

a
4.2.10 Example: Trace the curve r

= a ( sin f) + 1/sin f))

..... ( I)

This example illustrates that changing the coordinate system from polar to cartesian facilitates the tracing. (I)

can be written as

,.1 = a(r sin 0 + ~ 1 r SIn f)

..... (2)

Substituting r cos f) = x and r sin f) = y (2) reduces to ....(3)

No constant term in (3) :. curve passes through origin Lowest degree terms equated to zero gives The tangents at (0,0) are imaginary.

2/ + x = 0 => x ± yJii 2


301

--------------------=-----~------------------------------------

x = 0 in (3) gives y = 0,0,2 a y-axis cuts the curve at (0,2 a) and (0,0) is an isolated point Equating coefficient of the highest dt'gree terms in y to zero gives y = () (i.e.) x-axis is an asymptotes to the curve. F (-x, y)

=

F

(x, y) :. curve is symmetric about y-axis

Differentiating (3) W.r.t 'x',

dy

2x ( a - y) . (dY ) dx = 3/ +X2 -4ay .. dx (O,la) = 0

The tangent at (0, 2a) is parallel to x - axis (0,2a)

y = 2a

o

x

Fig. 4.37 From (3)

J2ay y-a

x = ±y

when y < a or when y > 2a ,clearly x is imaginary (i.e.,) The curve exists only in the region a < y ~ 2a

Exercise 4.2 Trace the following curves 1. r 2 cos20=a

l

4. r=3+2cosO 7. r=asin40

2. r=l+J2cosO 5. r=asin30 l 8. r=a(coso+----) cosO a2 cos20

)0. r = a(I-sinO) 11. r2 = - -

3. r=asin 2 0secO 6. r=a(cosO+secO) 9. r2 =asin20


302

4.3.0 Tracing of curves when the equation is given in parametric form: Suppose the equation of the curve is in the tonn x

= ./; (t)

and y

= ./; (t)

..... (1)

where t is the parameter. The study of the following points are useful for tracing the curve I.

If for some value for I , say I) ,

.I; (t) = 0 and

.t; (I) = (l

then the curve passes through the origin. 2.

If.1; (I) is an odd function and

.t; (I)

is an even function

then the curve is symmetric about y-axis. 3.

If

J; (I)

is even and

J; (I)

is odd

then the curve is symmetric about x - axis 4.

Intercepts on X(Y)a.xis are obtained by solving

.1;(1)=0 (.t;(t) =0)

respectively. 5.

Greatest and least value of

.I; (I)

and

.t; (I)

give the region in which the curve

exists.

It (t)

6.

If

or

7.

(ddxY)I~II = 0

1; (I) tends to

00

as I -) t) then t = t) is an asymptote to the curve.

indicates that t = t) is a tangent to the curve parallel

to the

x-axis. ) -) 00 as I -) ') then I

= I)

is a tangent parallel to the y-axis.

8.

(:

9.

If It(t+a)=j;(1) and .t;(t+a)=.t;(t) then the curve is periodic of period a

10. Sometimes it may be convenient to transform the equation of the curve from parametric to cartesian form for tracing the curve

4.3.1 Example: Trace the curve x

= at 2 ,y = 2at


303

Solution: Eliminating t from x and Y

l

= 4a 2t 2 = 4a.at 2 = 4ax y

~~+-

(0,0)

__

~

______-+x

l = 4ax ---...;..Fig. 4.38

This is a parabola with vertex at origin y aXIs is a tangent to the curve at the origin.

4.3.2 Example: Trace the curve x = acos/,Y = bsint

Solution: Eliminating t from both x and Y

(= J+( ; J~

1 wh ich is an ellipse (standard fonn) y (O,b)

---------+----~~----~--------~.x (-a,O) (a,O)

(O,-b)

Fig. 4.39

4.3.3 Example: 3

Trace the curve x

1

= (2 ,Y = t-3

304


--------------------------------------

Solution: Eliminating't'

2 y2 =t (I-t/{f

=x(l-73'f =i(3-X)2

9i = X(3-X)2 -----(1)

(i.e.,)

y

=0 gives x =O,x =3.

The curve passes through origin and again intersects x - axis

F(x,-y)

at (3,0).

= 9y2 -x(3 _X)2 = F(x,y)

The curve is symmetric about x - Q.'Cis and hence it forms a loop between (0,0) and (3,0) y

=±

(3-x)Fx 3

which shows that the curve does not exist for x < 0

(i.e.,) to the left of Y-axis Lowest degree terms equated to zero gives x curve at origin.

=0

(i.e.,) y-axis is a tangent to the

Coefficient of x 3 is I (a constant) and coefficient of

i

is a constant.

Hence, there are no asymptotes parallel to either x or y-axis Substituting, y

= mx + c 3

in (I) and rearranging 2

_x + x 2 (9m +6) + ........ = 0

fA (m) =-1

a constant,

fA (m) =0 gives

:. No oblique asymptotes exist

Fig. 4.40

m

=±J{i


305

--------=-----------------------------------------dy =±![l-X] £Ix 2 ~

Y (d )

=

dx ,~I

elY) ( dx

0:. The curve is extreme at x = 1

1 J3

=+ ___ (3,0)

-

:. J3Y = ±(x - 3) are tangents to the curve at (3,0) 4.3.4 Example: Trace the curve x = a(t + sin I), Y = a(l + cos t)

Solution: For no value of f both x and Y simultaneously vanish :. The curvc does not pass through the origin. x is an odd function while y is even :. The curve is symmetric about y -llxis

x(t) = 0 gives t

=0

.Correspondingly x = 0 and y = 2a

dy =dYidt =_ llsint =-tany,;=tan(TC-t/) dx dx/dt a(i+cost) .2 12 When 1=0 slope tan¢ = tanTC ~ ¢ = TC :. The tangent at (0,2a) is parallel to x - axis

t

-TC

X

llTC

y

_5% -~ -a(~+ 5;) -a(; +1)

° a(l+ ~)

0

0

0

2a

5% a(I+;) a(~+ 5:) ~

a

(/(1+~)

TC

(lTC

°


306

y(t) = 0 gives cosl = -1 ~ I = ±7l",±37l", ..... . (i.e.,) 1= ±(2n + 1)7l"

corresponding x is

n=1,2,3, ....

±a7l",±3a7l", ....

Further x ~ 00 as I ~ 00 and

IYI ~ 2a

:. The curve completely lies within the region 0 ~ Y ~ 2a and is period of period 27l" y

Fig. 4.41

4.3.5

Example:

Trace the curve x

= a(t -sinl);y = a(l-cost)

Solution:

X(I)

=0

only when 1=0

y(t) = 0 gives 1= ±2n7l" x(O)

n=0,1,2, ....

= 0 = y(O):. curve passes through origin.

Since Icos'l ~ I we have y 2 0 The curve is completely above x - axis and lies within the region 0 ~ y ~ 2a

(y = 2a when' The curve meets x -

= ±(2n + 1)7l",n = 0,1,2, .. ) axis at x = ±2an7l", n = 0,1,2, ....

y is an even function and x is odd

307


:. curve is symmetric about y-axis

LI x(t) t~oo

= 00

but LI y(t) is finite l~cIJ

:. There are no asymptotes to the curve

= cot ~ = tan ( ;

tan ¢ = :

Jr

-

~)

I

~¢=---

2 2

The tangents to the curve at t

= ±2nJr, n = 0,1,2, .......

are parallel to y-axis

The curve is periodic of period 21f y

~------~------~--~--~--------~----~x

-4a1t

-2a1t

2a1t

o

4a1t

Fig. 4.42

4.3.6 Example: Trace the curve x

= aeos 3 e,y = bsin 3 e

Solution: The functions cos 3 e and sin 3 e are periodic of period 2Jr . Hence it is sufficient to trace the curve for one period. For no value of

e both

x and

y vanish.

x( e)

=0

gives

e=±~ ,

yeO)

=0

gives

e = O,±Jr

(0,0) does not lie on the curve


308

:. The curve meets x - axis at x = ±a and y-axis at y = ±b Since

Icos

01 s:; 1 and Isin '01 s:; 1 we have Ixl s:; a, Iyl s:; b

3

e

0

Jr/6

x

a

3J3 --a

y

h

8 dy

tan ¢ = -

dx

tan¢

a/8

0

3J3 b

8

0

Jr/2

=0

2Jr/3

5Jr/6

Jr

-2J3a

8

1 -b

Jr/3

b

= - - tan

when

a

-a/8

8

~J3 b

-a

b

-

8

8

0

e

e = ±(2n+ l)Jr

(i.e.,) The tangent to the curve at these points is parallel to x - axis tan ¢ ~ 00 when

e

~±

2n+l) Jr

(-2-

:. The tangent to the curve at these points is parallel to y-axis It can be observed that for corresponding to x for some and corresponding to x for some

e there is

-

y

for - e .

:. The curve is symmetric about both the axes (O,b)

o =...nE-_ _-+ __~3!Jii-(a,o) (-a,O)

0=0

0= -n12 (O,-b)

Fig. 4.43

e there is -x

for Jr -

e


309

4.3.7 Example: Trace the curve x = a[ cose +

~ log tan ~ J,:I = asin e 2

Solution:

yeO) = 0 but x(O):f:- 0 . (0,0) docs not lie on the curve yeO) = 0 gives

e= 0

=a

and lyllll'Lx

x is an even function and y is odd. So the curve is symmetric about the x - axis Corresponding to each x( e) there is -x for 1C -

e

:. Curve is also symmetric about y-axis Since Isin el -(I

S;

S; 1

we have

IYI S; a.:.

The

curve entirely lies in the region

Y S; a .

x( Similarly,

~) = a [ cos ;

+ log tan : ]

= a [0 + log 1] = 0

x[ -;] = 0

Corresponding y values are y(; )

=a

and y( -; )

=-£1

Thus the curve intersects the y -axis at (0, a) and (0, -(I)

tan~ =
2 Hence,

¢ =0

1

l

a -sine+ - .~ [ = sinti = cos' 0 acosO

J

= cot 0 = tan (1C _

sinOcosO

e

when

e = ~ and ¢ = 1C

when

e = -~

2

e)


310

(i .e.,) The tangents to the curve at Further .". X -

e~

00

or

-00

e =:= ± 7r/2

according as

are parallel to x - axis

e ~ 0 or 7r

axis is an asymptote to the curve.

(O,-a)

Fig. 4.44

4.3.8 Example:

3al 2 l+t'

3at l+r'

Trace the curve x = - - , ,y = - - , a > 0

Solution: X

3

3 3 = (3a)3 - , .I,y = (3a)3 -3 .1 6

l+r

.". x 3 + y3

1+1

3 3 3a )3 27a t 3 ) = = ( __ 1\1 +/ 3 3

?

1+1

(l+t

t

3at 3at 2 l+t l+t

=3a'--3 ' - - 3 = 3axy 3

Thus the Cartesian form of the curve is x + For tracing the curve see example 1.3

4.3.9

Example:

Trace the curve

x

=

a(l-t 2 ) y 1+ t 2 '

2bl

= - -2 1+ t

i

=

3axy


311

Solution:

(1_12]2 i

2

x a2 Thus

= 1+t 2

'

2 2 ~ + L -_ I a

b

b2 ( see examp Ie 3.3 .L-'"')

2

Exercise 40

= aCt +sin t),y = a(l + cost) x = a sin 21(l + cos 2t), Y = a cos 2t(l- cos 2t)

I. x

2.

3. x = a[ cosO -log(l +cosO) ],y

= asin 0

4. x = aU -sint),y = a(l-cost) Exercise 3.1 I . An asymptote to the curve (i) x-a=O

i (a + x) = x (3a - x) is

(ii) x+a

2

=0

(iii) x-3a=O

(iv) none of these

I ADS: (ii») 2. The curve a 2y2 (i) x-axis

=x2 ( a2 -

Xl) is symmetric about

(iii) both the axes

(ii) y-axis

(iv) none of these

I ADS: (iii) I 3

3. The curve x +

i = 3axy

is symmetric about the line .............. .

[ADS: y = x) 4. If the tangents to the curve

i

= (x - a)( x - b)2 ,( b > a,)

at the point x = bare

inclined at angle 0, tan 0= .............. .

[ADS: 5. The two tangents to the curve

i

±.jb-a )

= x 3 at the origin are real and distinct. I ADS: false)


312

6. For the curve r

= a sec B tan B the origin is a cusp. [Ans: true)

7. The curve x

= a (t + sin t), Y = a (1 -

cos t) is symmetric about x-axis. [Ans: false)

8. x - 2a = 0 is an asymptote to the curve

l

(2a - x) = x 3 [Ans:true)

5 Application of Integration to Areas, Lengths, Volumes and Surface areas 5.1

LAGRANGE'S METHOD OF UNDETERMINED MULTIPLIERS The definite integrals are useful in formulating important physical applications like (i) lengths (ii) areas and (iii) volumes and surgace areas of sol ids otrevolution. 5.1.1 (a) The area bounded by a curve y = f(x} , the axis of x and the two ordinates x = a and x = b is given by the definite integral. S:ydx= J:f(x)dx y

= f(x)

. x'-~----!---

o

"'---=--x

y'

Fig.5.1.1(a)

Similarly the area bounded by a curve x = fry) the axis of y and the two abscissae y = c and y = d is given by the definite integral.

S:xdy = S:j(y)dy


314

x' ---:,-+----------------, o x y'

(b)

Fig.5.1.1(b) If the equation of the curve is in prametric form, then the area =

f

ll

I,

(c)

dx y - dt or dt

f

l

II

2

dy x - dt dt

where tl' t2 are the limits of integration depending on the boundary of the curve. The area enclosed by the polar form of the curve r = f(B) and two radii vectors (} = a, (} = f3,, is

= fPCd(}=~ a

2

2

fP[r«(})Jd(}

(area of sector OPQ

a

o~~----------------

Fig.5.1.1(c)

5.1.2 Example Find the total area within the curve c?y

"

=

c?x2 -

x4

,,

Y

/ , / 'y

Y = - x" " " "", ,,

,

=x

...........8

A

x ' ---~---~~~~~~~-~x

,,

,,

,

y'

Fig. 5.1.2 --

,,

,

,,

=

1 2

2

- r d(})

Application of Integration to Areas, Lengths, Volumes and Surface Areas

315

Solution: The curve has two equal loops between the lines x = a and x = -a and is symmetric about x axis. The total area within the curve = 4 (the area of the half of the loop i.e., 'OABO')

= 4 Laydx =4

ax

1~2--2

fo -va a

-x dx

=~x_l fa~2_x2j2(_2x}lx -2

a

0 3

( 2 _ -2\a -x 2 )-2 4( 2~=--\O-a J2 3 3 a 2 o

=

4

3a

2

Sq.units

5.1.3 Example: Find the complete area of the curve a 2y2

=

:x3(2a - x)

y x = 2a B

x'

x

0 (0, 0) A(2a,0)

y'

Fig. 5.1.3 Solution: The curve is symmetrical about x-axis it cuts the x-axis at the points 0(0,0), A(2a,0). The curve consists of a loop lying between x = 0, x = 2a The required area =20ABO 2

= 2 =

Substituting x

fo ;dx

2f =

2a X 3/ 2.J2a - x

o a 2a sin2 ()

dx


316

We have

dx = 4as in e cose x=O=>O=O x=2a=>O=Jr/2

Area

=~ a

=

Jll 0

32a 2

2

(2a)12 sin 3eEa cose.4asine cose de

J:

2 sin

4

e cos 2 e de

/2 2 = 32 a 23.1.1 --Jr = 1111 Sq. units

6.4.2 5.1.4 Example: Find the area of the segment cutoff from the parabolay y = 4x -1. Solution: The given curves are y = 2x Y = 4x - 1

=

x'-----Ib-~_1_-=--------

2x, by the straight line

..... (i) ..... (ii)

x

y'

Fig. 5.1.4 Solving (i) and (ii) we get the points of intersection as

A(.!2'I}'Lll8' J.! -.!} 2 The required area

=

area AOBA (shaded region)

= area CBEAOOC - area BOADOCB =

J ! {y + 1)dy- J 1 -124

1 -12

y2 dy (from equations (i) and (ii) respectively), 2

~ ~[~2 +yL -[Y:L 9

= 32 Sq. units


5.1.5 Example: Find the area between the curve.xlY

=

a2(y

y

- x 2)

and its asymptotes.

B

x' ---j===::;>+-:;::::=::::f-..!..--- x

'y

Fig. 5.1.5 Solution: The given equation can be written as

~2X2 2

y2{a 2 _X2)= a 2x 2 => y2 =

a- -x The asymptotes to the curve are given by

a 2 _x 2 = O=> x::;:±a The curve is symmetric about both the axes and passes through the origin. The shaded portion is the required area. The complete area = 4(areaOAB in the first quadrant)

= 4 Laydx =

4J" 12ax 0

= -

va -x

2a X

~2

2

dX=-2J" -2ax ,~dx oi2 2)1/~

2 - X )

1

12 a

10

-x

= -4a[:2 -J a -

x

2

Jl0I

2 =

--4a[O - a]

=

4a2 Sq.units.

5.1.6 Example: Find the area bounded by the curves

y

= 9x, x 2 = 9y,

Solution: Solving

y

=

9x and.xl

=

9y

The points of intersection of the two curves are 0(0,0) A(9,9)

317

318


x'----+----~=F=-.:=--~-----x

Fig. 5.1.6 The required area = area OBACD =

area ODACO - area OBACO

S:=oydX - S:=oydx (parabola y = 9x) (parabola x 2 = 9y)

=

X2 = f 93 J";dx- f9 dx = 27 sq. units 009

5.1.7 Example:

Find the area bound by the cissoid x

=

asin 2t, y

=

asin 3 t . and Its asymptote. cost

Solution: The equation of the curve is

x

=

asin 2t, y

asin 3 t

= ---

cost y

x=a x,-----l..~;;;;;;;~~~--x

o

y'

Fig. 5.1.7

y

=

a 2 sin 6 t cos 2 t

a-x


319

The equation of the given curve is transformed into cartesian form. The curve is symmetric about x-axis and x = a is the asymptote. The required area = 2(shaded area)

=2Jao ydx =2J: _x_ a-x 0 (

Writing

x

1/2 )

' dx

=

asin 2 B, dx=2asinBcosBdB

=

4a

2fl! 2Sin. 4ede = 4a 2.-.-.3 I n 042 2

4 5.1.8 Example: Find the area included between the cycloid x = a(e - sinO), y = a(1 - cosO) and its base. Solution: The equation ofthe curve is x = a(B- sinO) y = a(1 - cosO) y

x ' - - - - - - - - - : : ' I ' -.......~.........'-'-'~.........'---'-:::-...".----- x

e =0

:

A(2an ,0)

'y

Fig. 5.1.8 The required area = area OABO =

f02;, dx

=

J2l! Y dx .dO o

dO

=

f ;(1 -

=

4a

2 0

cose )a(I - cose)de

2f 2l! Sin. 4-de e = 8a- fl!·Sin 4-de e ?

o

2

e = 2n

0

2


320

f

",2

? 8a-.2

=

371a2 Sq.units

0

()

sin 4 tdt

=

(taking -=t) 2

5.1.9 Example: Find the area between the curve x

=

a(O + sinO), y

=

a(1 - cosO) and x-axis.

Solution: The required area =

2 area OAB

=

2f"ydx .dO =2f"aQ-cosO)aQ+cosO)dO o dO 0

=

2a .2

2

f

"/2

0

sin 2 0dO y .-------f------. A

x,---=--~~~~~~l--- x o =-1t 0 0 =0 8 'y

Fig. 5.1.9 =

2

I

1t

4a .-.-

2 2

= mJ Sq. units 5.1.10 Example: Find the complete area of the curve given by the equations.

x

=

acos3 fJ, y

=

bs in 3 ()

Application of Integration to Areas, Lengths, Volumes and Surface Areas Y 8(0, b)

x'---~?-+=~----x

A(a, 0)

Y' Fig. 5.1.10

Solution: The required area

= 4 area OABO

=4Jaydx =4Jo °

=

y dx de de

4 I:'2bsin3B~3Bcos2BsinB)

= 12ab =

1t 2

1t/2

4

2

J° sin e cos e 3 1 1

de

1f

12abx-.-.-.-

642 2

3

= 81fQb Sq. units. 5.1.11 Example: Find the area of loop of the curve

-? =

a 2 cos2f)

e =1[/4

e =0

x'----~~--------~~------~~~~

8

c Y'

Fig. 5.1.11

x

321

322


Solution:

The required area =

4 areaOABO

=

2f"4~9 = o

=

"!4

fo

f" 4r2d9

2

0

2

a cos 29 d9

a2

,,'4

= -~in29]o 2

a2 2

5.1.12 Example:

Find the area of the curve r

=

a(l + cosO)

Solution:

B

c

() = 1t

A

Fig. 5.1.12

Required area 2 . (area OABCO) r2 =2 -d9 =

f

"

2

o

= f: a Q+cos9 )2d9 2

=

J:

= 4a 2 = 4a 2 =

8a

4(cos %Jd~ f"o cos ~9 2 f cos t .2dt

a

2

2

2

4

"/2

0

3 I 1r 422

4

3Jlll 2

---=--

2

(where t = 0/2) •

Sq. umts


323

5.1.13 Example: Find the area of the portion included between the carbo ids , a(l - cosB').

=

a (l + cosB') and

, =

Solution: Required area =

4(OCBDO)

[I

1t 0

=4

2,2 d8

2

1

4-~-~r::---_~ r = aQ

A

+ cas8 ) 8 =0

Fig. 5.1.13

I

a-V -cas8 )2d8

=

2a 2

L1t/2a 2 Q- c~s8 )2d8

=

2a

= 2

1t / 2 0

1 {,

~!2 +~.~}= a2 (3n -8) Sq.units 2

2

{

[8 - 2sin8]

5.1.14 Example: Find the area afthe curve

r2 =

a2sin28.

8=~ 4

8

=n/4 x

Fig. 5.1.14

324


Solution: The total area of the curve = 2 area of one loop of the curve 1 = 2 . -1 fll .!2r 2de = f1l 2? a- sin 2ede

2

=

0

0

2

2

2

2

~ [- cos28] ~ 2 =~ n+ 1]=a2 Sq . units Exercise 5 (A)

i

2

1.

x Find the whole area of the ellipse ~ + [;2 = 1

2.

Find the whole area of the curve x 2(x 2 + .I)

3.

Find the area of the curve a2x 2

=

a 2(x 2 -

(Ans:

.I)

1t

ab]

[Ans: cl(Jr - 2)J =

y(2a - y)

5.

(Ans: Jra2 ] Find the area bounded by the curve xy = 4el(2a - x) and its asymptote. (Ans: 4Jra 2 ) Find the area included between .I = 4ax and y = mx

6.

8a 2 (Ans: - 33 ) m 4a(x + a) and

4.

Find the area included between the parabolas .I 4b (b - x)

=

.I =

(Ans:

7.

~(a + b}Fab )

3 Find the area common to the circle x 2 + y2 = 9 and parabola x2 = 8y

. [ADS: "31 {r;:; 2,,2 + 27 Sin 8.

Show that the area of the loop of the curve ely

9.

Find the area of the loop of the curve r

=

-1(2J2i} - 3- ) I 3JTa

=

2

r(2a - x)(x - a) is -8-

asin2B. JTa2

[Ans: 10.

Find the area common to the circles r

=

8

]

aJ2. - 2acosB [Ans: el(Jr - 1)]


325

5.2.1 Lengths of plane curves:

(a)

The length of the arc of the curve y = fix) included between the points whose abscissae are 'a' and 'b' is

(d)2 dx I+~

s=J h (b)

or

dx

The length of the arc of the curve x = f(y) included between the points whose ordinates are 'c' and 'd' is 2

dx

d

s=J ('

1+dy ( dy )

(c)

The length of the arc of the curve x = fit), y = g(t) included between two points whose parametric values are 'a' and '/3' is

(d)

The length of the arc of the curve r = j(B) included between two points whose vectorial angles are 8/ and 82 is

s=

J-

9??

r-

9,

(e)

2

dr

+( r -

de

)

dt

The length of the arc of the curve 8 = j(r) from r = r/ to r = r2 is given by s=

J,,-,. ,

(de )2 dr

1+ r dr

5.2.2 Example:

Find the complete length of the curve x2(a2

-

x 2)

=

y

8a2


326

Solution: y

8

x'------4--------lIf-------+------ x A(a, 0)

A'(a,O)

'y

Fig. 5.2.2 The curve is symmetrical about both the axis, one loop is formed in between x and x = a and another loop is formed in between x = 0 and x = - a. The total length of the curve Total length = s = 4

f:

=

I+( :

=

0

4(1ength OABO)

J

.... ( I)

dx

The equation of the curve is

:x2(a2 - x 2 )

=

8a2y2

Differentiating with respect to 'x'

16a 2y dy =2xa2 _ 4x 3 dx dy _ x(a 2 _2X2) dx 8a 2 y

r

x2(a2 _2X2 dy )2 1+ ( dx =1+ 64a 4/

Substituting for y2 from the equation of the curve 8a2y2 = x 2(a 2 - x 2 )

Application of Integration to Area!, Lengths, Volutnes and Surface Areas

327

9a 4 -12a 2 x 2 +4X4 8a 2{a 2 -x2)

.... (2)

= s=

.J2 2a2 sin- I (I}=2ha. Jr a

.

2

.J27«1

5.2.3 Example:

Find the length of the loop of the curve

3ay2 = x(x - a)2 Solution: y

8 x'-------*J.U.jtJ.LL~~+_-- X

o

(0,0) 'y

Fit. S.l.3

328


The curve is symmetrical about x - axis, the loop is formed between lines x and x = a, differentiating 3ay = x(x - a)2 w.r., to x.

=

0

= x.2{x-a)+ {x-a)2

6ay dy dx

dy = (x-aX3x-a) dx 6ay dy )2 1+ (dx

=

(x-a)2(3x-a)2 1 + -'---------''------'::--::-----"-36a 2y2

... (1)

Substituting 3ay2 = x(x - a)2 in (I) we get

dy )2 {x-aY{3x-aY = I + -'---------''--'--------'-dx 12ax{x-a)2

1+ (-

12ax+9x 2 +a 2 -6ax 12ax

1+

(:r

=

(3~2+;)2

.... (2)

The length of the loop of the curve

=2

s=

r

(3x+a) dx o 2Fa-Fx

~Ja3-Fx+ax-12dx

2,,3a =

1 ,,3a

r:

r::;- L2x

0

32

+ 2ax

12] a 0

= ~ [2a 3/ 2 + 2aFa]= 4~ ,,3a

,,3

5.2.4 Example: Find the perimeter of the loop of the curve 3ay

=

r(a - x)


329

Solution:

x'-------;~t..J.J,.t..J.J,."'f__:__--

x

Fig. 5.2.4 The curve is symmetrical about x-axis and the loop of the curve lies between x and x

=

a.

Perimeter of the curve

s= 2

=

20ABO

(4a-3x) dx o 2J3a.Ja-x

f

a

s = _1-

afj

fa a+3{a -X~lx 0

.Ja-x

=

-1-[-2a{a-x) 2 -2{a-x)'2] ~

=

Ir:;- [0 + 2afa + 2a 3!2]= 4a/ fj

afj

a-v3

5.2.5 Example: Find the length of the arc of the parabola y2 = 4ax cut off by the line 3y = 8x.

=

0

330


Solution:

...------x

x'-----~!OII"I'!J

Y'

Fit. ,.2.6 Solving the equations of the J>araboLa anct the fine gives the points of intersections

0(0,0) and A

(~: ' 3; )

y=4ax 2ydy =4a dx

2a y

dy dx

-=-

1+

(dy)2 =1+L dx 40 2

The length of the arc OBA of the

s= .

=

_1

[y ~y2 +4a2+ 40 log~~~y24a2 }T2

2a 2 =

a

2

a[~~ + IOg2]

2..

Jo


5.2.6 Example:

Find the length of an arc of the cycloid x = a(t - sin t), y = a( I - cos t). Solution: y

------~o~--~a-n--~~a~n----A~------x

(2a1t,0 )

Fig. 5.2.6

dx dt

= a(l- cost), dy = asint dt

(~~J +(~J =~2Q-costY+a2sin2t] 1/2 =

2asint/2

The length of an arc of the cycloid

= 2

Io" 2asin 7i dt

=

4a[- 2cost/2] ~

=

8a

5.2.7 Example:

Find the total length of the curve X2/3

y2/3

a 2/ 3 + b 2/ 3 =I

331

332


Solution: y B(O,b) A(a,O) x'-----E:-~-~------x

B'(O,-b)

y'

Fig. 5.2.7 The parametric form of the curve is x = acos 3 (J, y = bsin3 0

The total length of the curve = 4. length OABO

s=4

f1t2[(-3asinecos2e)+~bsin2ecoseJ 2

0

:\2J12

de

Substituting and

clcos 2o + b 2sin 2 0 = z2 (b 2 - a 2) 2sinO cosO dO

s = 12

=

=

2z dz

z 12 Z3/ h z. ( 2 2) dz = 2 2 a ,b -a b -a 3 a

f

h

~ 4 ~ (b b--a-

3 _

a3

)= 4(a

2

2

+ b + ab) a+b

5.2.8 Example: Find the length of the arc of the curve given by x = asin21(J + cos21), y = acos2t(J-cos2t) measured from the origin to any point.


333

Solution: x

asin2t(1 +cos2t)

=

dx dt = 2acos2t(1 + cos2t) + asill2t( -2sin2t)

and

dx dt

= 2a(cos2t + cos4t)

dx dt

=

y

.... ( I)

4acos3tcost

acos2t(1 - cos2t)

=

dy dl

= -

2asin2t(1 - cos2t) + acos2t(2sin2t)

dy dl = 2a(sin4t - sin2t) = 4acos3tsint The length of the curve from origin to any point

=

L (~;

r r +(;

dt

Using (I) and (2)

f~ J(4a cos3/COS/)2 + (4acos 31 sin 1Ydl s = f'4a cos 3tdt = sin 3/11 s=

lIa

0 3 0

s=

4a . 3 3

-Sin I

5.2.9 Example: Find the perimeter of the cordioid r

=

a(1 + cosO).

Solution:

B

e = 1t

0=0

------~~------~-----------x

Fig. 5.2.9

.... (2)


334

r

=

a( I + cos 0)

dr

:=)

dB

=

-a sin 0

The perimeter of the cordiod =20ABO

+(~

J

=

2 fox

=

2f" Ja 2 Q+ COSO)2 + (- asinO )2dO

=

2

=

1'2

dO

o

L" J2a

2

Q+cosO )dO

f"o 4a COS2~2 dO 0 2 f 2a cos-dO 2

2

x

=

o

,..,

L

X

=

4a.2.sinB/

=

8a

20

5.2.10 Example: Find the perimeter of the curve r

=

2acosB.

Solution: The equation of the curve is r

=

2acosB.

It is a circle passing through pole whose centre is on the initial line at a distance 8/2 from pole.

0=0

Fig. 5.2.10


335

The perimeter of the curve

=20ABO =2J

n'2

r2+(dr)2dB dB

o

= 2

f: ~(211~-;;~OJ ~-(12

2asinB)1 dO

J(

=

4a2

=

2aJ(

5.2.11 Example Find the length of the arc of the parabola Ijr

Solution: LSL' = 21 is the latus rectum of the parabola.

Fig. 5.2.11 The length of the arc L' AL = 2AL

=2

I

n

o

;2

(dr)2 dB

1 r+ -

(IB

Equations of the curve is Ijr logr =

1 + cosOc logl + log(l + cosO) =

~ dr = 0 _ - sin 0 rdO l+cosO

=

1 -t cosOcut off by its latus rectum.


336

dr -=rtanB/2 dB The length of the arc = 2 =

f

1t/2 ~ 2

?

?

r + r- tan-9/2d9

0

1t/2 0

f rsec9j2d9 I 2 fo 1+ cos9 .sec9j2d9 2 fo 2cos 9 2 .sec9j2d9 9 I sec-.sec 9j2d9 fo 2 2

1t2

=

1t 2

=

/

2

2

1t!2

=

= / L1t/2 ~I + tan 2 9j2 sec 2 9j2d9 writing

tan Bj2

t

=

I

sec 2 B/2.-dB = dt 2

= 2{~F2 +~log(1 +F2)] =

l[ F2 + log(1 +

F2)]

Exercise - 5(8)

1.

Find the length of the arc of the parabola y

=

4ax cut off by its latus rectum. [ADS: 2a[F2+log(l+F2)]}

2.

Find the perimeter of the loop of the curve 9ay

=

(x - 2a}(x - 5ai

(ADS: 4.fja J


3.

337

Find the length of the arc of the parabola x2 = 4ay from vertex to one extremity of the latus rectum. [Ans: a[ 12 +Iog(l +

4.

Find the length of an arc of the parabola y = x2 measured from the vertex. [Ans:

5.

12)]J

Find that the length of the arc of the curve y

=

~Jl+4x2

+-.!.-sin- I {2x)] 4

2

log tanh (x/2) from x

=

1 to x 2

=

2

I)

e + [Ans: log ( -e- ] 6.

Find the length of the arc of the curve y

=

x(2 - x) as x varies from 0 to 2. [Ans: -.!.-log(2 + 2

7.

Find the length of the curve x

8.

Find the length of the arc of the curve x

B

9.

=

=

a(B + sinO) y =

=

J5)+ J5 ]

a(l- cosO)

eOsinB, y

=

[Ans: 8a) eOcosB from B = 0 to

1[/2

Prove that the loop of the curve (3

X = (2,

10. 11.

Y

= (- -

3

is of length 4fj

Find the perimeter of the cardioid r

=

a(l- cosO)

Show that the arc of the upper half of the curve r

e=

=

[Ans: 8a) a(l - cosO) is bisected by

21[/3

5.3.1 Volume and Surface of solids of revolution: (a)

Volume of the solid generated by the revolution of the area bounded by the curve y = f(x), the x-axis, the ordinates x == a and x == b about the x-axis is

(b)

Volume of the solid obtained by the revolution of the curve x abscissae y = c and y = d about the y-axis

Ld

1tX

2

dx.

=

J:

1t)12 dx.

fry), the y-axis the

338

(c)


Volume of the solid obtained by the revolution of the curve x x-axis is f

'2 ny2

dx dl dt

'] (d)

= f '2 n ~ Cr)r ']

=

}(t), y

=

$(t) about

=

j{t), y

=

¢i...t) about

dfCr ).dl dt

Volume of the solid obtained by the revolution of the curve x y-axis is f'21t\"2 dy dl = f'2n[f'Cr)Y d$Cr).dl

'] (e)

dt"

dl

Volume ofthe solid obtained by the revolution ofthe cure r = j{ fJ) about the initial line IS

=

02

f.

.

f\)2

S n\"sIno

d{rcose )d0 .

de

OJ

(f)

Volume of the solid obtained by the revolutuon of the curve r perpendicular to initial line =

f.

h?

81

=

f

82

f.

0)2

n \" COSo

0]

d{rsine),lO

de

.U'

The volume of the solid generated by the revolution about the initial line (x-axis) of the area bounded by the curve r = j{ fJ) with the radii vectors B = a, B = {3 is =

I

O=~

O=a

(h)

j{fJ) about the line

dv f rrx- dy = 1°2 n \" cose )2 -"-.de dO a

(g)

=

2nr 3 SIlIO '_0 d{\ 0 3

The volume of the solid generated by the revolution of the area about the line B = ni2 (y-axis) of the area bounded by the curve r = f(e) with the radii vectors

B=a, B={3is =

o=~ 2n -r3 cos OdO 8=a 3

I

5.3.2 Example: Find the volume of the solid formed by the revolution of the loop of the curve y(a + x) = x 2(a - x) about x-axis.


339

Solution: y

= 0 gives x = 0, a

:. The loop is formed between x = 0 and x = a x = - a is the asymptote to the curve. y

x=-a

Fig. 5.3.2 :. The volume formed by the revolution of the loop about x-axis

=

(Jr x2 (a-x)l 1(

a+x

o

ex

~ 1{- ~ + ax' -2a'x+2a'log(a+x)I = 21(a 3 [ log 2a - log a -1 ] 3

= 2Jra [IOg2-1] cubic units. 5.3.3 Example: Find the volume of the solid generated by revolving the curve xy y-axis.

=

4(2 - x) about


340

Solution: y

x' -------,o..-+----f.,-,,;-,~O)r- x

Fig. 5.3.3

The given curve can be written as

xcY+4)=8 The volume of the solid obtained by revolving the given curve about y-axis

Substitutingy

64

J y+4 )2 OC!

= 2n

o

(~

= tanO ~

= 8n

f

1t /

0

2

dy dO

dy

= 2sec 2 OdO

I +cos2e de

1

+2 sin28

=

8n [ e

=

4n2 cubic units.

]

1t

2

0

5.3.4 Example:

The part of the parabola y2 = 4ax cut off by the latus rectum revolves about the tangent at the vertex. Find the volume of the reel thus generated.


Solution: y

L(a,2a)

S

x'------------+-M+--+-------------x J?--+--4.I

L'

y'

Fig. 5.3.4 Volume of the reel generated

=

[ '}Y y-

211

=2fo

1[-

4a

Jf

Y-1

]2
8a 2 [

5

0

4Jfa 3

= --

5

cubic units

5.3.5 Example: Find the volume ofa sphere of radius 'a'.

Solution: Equation of the circle of radius 'a' is x 2 +

Y

=

a2 .

y

x'------------~~~~~~~~---------------x

B

A

(-a,O)

(-a,D)

y'

Fig 5.3.5

341

342

Engineering Mathematics - I By revolving this circle about x-axis we get a sphere of radius 'a'.

=

Volume of the sphere = rr

=

f

ll (2 ~11 \q -

f

a

1

~"rry-dx

x-' ) dx

• "34 JlU 3 cubic. Ul1lts.

5.3.6 Example: Find the volume of the solid generated by the revolution of the curve x 3 = bsin () about x-axis.

=

(l

em;3 (),

y

Solution: y

B(D,b)

x'

(a,D)

x

B'(O,-b) 'y

Fig. 5.3.6 The volume of the solid generated by the revolution of the curve x

y

=

hsin3 () ((i.e.,) x: ',33 + <~: a-

.h-·

=

1)

IS

dx =2fO 21tJ' de .dB = 2fO rr(bsin e y~3acos2esine )de 2

7t

3

11/2

32JlUh 2 - - - cubic units.

105

=

aeos 3 (),


343

5.3.7 Example: Find the volume of the solid fonned by the revolution of x

y

=

=

a(

e- Sillf/),

a(l - cosf/) about its base.

Solution: y

o = n/2

Fig. 5.3.7 Volume obtained by the revolution of the area OABO about the base (x-axis).

=

n

f

O=21t

dx(10. y2dO

()=o

= na

3f8=21t1J 8=0 ~

3

= na"

.8

J

21t

0=0

-

cosB

)~

dB

2sin 6 O/2dO (e12 = t => dB = 2dl)

sin6 tdt = 32u3n(6 -I 6 - 36 - 5 .n I 2) 66-26-4

=

16na3.2 J1t o

=

5,(2a3 cubic units.

2

5.3.8 Example: Find the volume generated by revolving the curve r = a( 1 - cosq) about the initial line.


344

Solution: O=~ 2

o = It A

Fig. 5.3.8 The volume generated by the curve =

=

2n 3

f1t r3 sin8 d8

2n 3

f1t a Q-cas8)1 sin8 d8

0 3

0

3

=

2na J1t Q-cas8 -3 0

ysin8 £18

1- cosO = t sin OdfJ = dt [

and():O~

1

2

8Jfa 3

= - - cubic units. 3

5.3.9 Exam'pre:

Find the volume orthe solid formed by the revolution of the curve r (a
Volume

=2n -

3

')n 3

= ~

f1t r3 sin8d8 0

f1t {,a+hcos8 )3 sin8d8 0

(a+bcOS()=t

~ sint1f() = - ~ ,t :a+b ~ a-b)

=

a + hcosO


=

J

h

U

21t 3

-

u+h

345

p(_ dt) b

SU-h t3d!

= - 21t 3b

u+b

~~[,;[

==

~[(a+bt -(a-b)4]

=

41l'Q(2 2) -3-\a + b cubic units.

6b

5.3.10 Example: Find the volume of the solid generated by revolving the lemniscate? about the line B=

tr

2

Solution:

Fig. 5.3.10 The curve is symmetric about q = tr/2 Volume

=

23. 1t I1t 4,3 case de 3

0

4

= =

3

1t

~ S4(a 2COS2e)2 cosede 3

i3 1ta

0

3

I1t4 (cos2e l2 case de 0

=

a2 c:os2B


346

(taking

-

.J2 sinO =

sint =>

.J2 cos(}dO =

cos/dt)

2.J27lZJ3 [4--I 4 --3 7C] -3

4 4-22

.J27C 2a3

- - - cubic units. 8

Exercise - 5(C) 1.

Find the volume of the solid generated by revolving the clips :: x (a)

About the major axis

(b)

About the minor axis

~: = I

[ADS: 2.

yea

The curve + x) = by the curve.

4

2

37lZJ b J

r(3a - x) revolves about the x-axis. Find the volume generated [ADS: ~(8Iog2 - 3)J

3.

Show that the volume of the solid generated by the revolution of the curve (a -x)y

7C 2a3

= a2x, a~out its asymptote is -2-. 4.

Find the volume of the solid generated by the revolution ofthe curve y(a2 + x2) about the asymptote

=

a3


5.

347

Find the volume of the solid obtained by revolving the loop of the curve a 2 == x 2(2a - x)(2a - x)(x - a) about x-axis.

1

(Ans:

6.

Find the vloume generated by the portion of the arc y ==

x

~

0 and x

=

JI + x"

60

lying between

4 as it revolves about the x-axis. (Ans: 767[/3)

7.

Show that the volume of the solid generated by the revolution of the cycloid

. . 3 , 1 87[' IS - 7[" £l - 2 3 Find the volume of the solid generated y revolving the cycloid x y = aU + cosO) about its base.

x

8.

==

. a( 8 + SinO), Y ==

aU -- cosO) about the y-axIs

=

a(8 + sinO), IAns: 5~£l31

9.

Find the volume of the solid generated by revolving

r

=

(Ans:

10.

Find the volume of the solid generated by revolving r

=

a2cos28about initial line. 7[(1'

Ii rl3log"\/2 ( h + 1)-"\/2h] I

12

acos38 between 8 == -7[/6

and e == 7[/6 about the initial line. (Ans:

197[(1' 968 )

5.4.1 Area of the Surface of revolution (a)

Surface area of the solid generated by the revolution about the x-axis of the area bounded by the curves y = fix), the x-axis the ordinates x = a, x = b is

h

I (b)

X=Q

21ty £Is ==

Ii> X=U

ds 21ty - £Ix

dx

Surface area of the solid generated by the revolution about y-axis of the area bounded by the curve y = fix), the y-axis and the abscissae y = c, y == £I is

I

Y=J 21tX cis == IJ

X=Q

Y={

cis (~V dy

21tX -

[ d~== dy

1+(dX)2] dy


348

(c)

The parametric form is

f

'=/2

/=/1

(d)

[ d~

ds .21ty-dt dt

dt

==

(dX)2 + ({~y)2l dt

dt

The polar form is

I - 21trsine (dS-de )de O?

91

5.4.2 Example: ? '0 Find the surface area of solid generated by revolving the curve x-i-' + y'? '3 == a-I?',

about x-axis. Solution:

The parametric form of the curve is x

=

acos3 e, y

=

asin 3 B.

Y B(O,a)

x

x'

A'(-a,O) 8'(0,-a)

Y' Fig. 5.4.2 The surface area of the solid due to revolution about x-axis == 2(surface area of the solid generated by revolving an arc in the first quadrant of the astroid is) =

ds " 21ty-.de f° de dx d 41t f Y - ) +(2) .de ° de' de 2

2

?

2

lt/2

=

=

(

41t

I

It/2

0

asin 3 e ~(\.3acos 20 sine )2 +( 3asin 2 e cose )2de


=

12na 2

"n

J

0' -

349

sin-le cosO dO ffn

sinSe - 12JlU 2 12JlU [ - 5 - j0 = - 5 - Sq. units. 2

=

5.4.3 Example:

Find the area of the surface formed by the revolution of the ellipse x2 + 4y about its: (a) major axis

(b) minor axis

Solution:

Equation of the ellipse is

ely dx

2

2

16

4

~ +L

=

I

x 4y

16y2

+X2

16/ (a)

Surface area formed by the revolution about major axis -l ds = 2 2ny-dx o dx

J

=

16

350


.Jl[~(~ r Jl] =

-x' +

=

8"[1 + 4"9

(b)

*

~: Sin-{ )I

Sq. units.

Surface area formed by the revolution about minor axis

=

41t

= 41t

=

L2 J

Xl

J:

+ 16yl dy 2

J16-4 y 2 +16y dy

8Jl. s: (1 r

+ y' dy

o


351

5.4.4 Example: Find the surface area formed by revolving cycloid x = a(B+ sinO),y = a = (/ - cosO) about the tangent at the vertex.

Solution:

e

y

=-1[

e

a1[

=-1[

a1[

0=0

x' - - -.........-.....;~"""'"--..I---- X y'

Fig. 5.4.4 Surface area required

= 2(surface area generated by revolving the arc =

2

I

x

o

= 4 IX II

in the first quadrant about OX)

ds 21ty- dB

de

y)2 de (dX)2 +(d de de

1t)'

. -cos e -e de =161t a-. IX sm?

ry

o

2

2

(%=t => dB = 2dt) = 321tary

32JlU 2

I

= --

3

X/2 0 sin 2

t cost dt

Sq. units.


352

5.4.5 Example: Find the area of the surface of reel thus generated by revolution of the part of the parabola = rax bounded by the latus rectum about the tangent at the vertex.

y

Solution: y

x'------+-~-+_------ x

Fig. 5.4.5

y

Surface area

=

4ax

dy

2a

dx

y

=

2

ds 21tX-dx o dx

f

a

d =2fa 21tX 1+.1:'. ( o dx

=

4n

)2 dx

fo" x.J x 2 + ax dx a

= 4n

J:[(x+~r -(~rl2dx


=

353

4lT

o = JZa

2

[3J2 -log(J2 + 1)] Sq.units.

5.4.6 Example: Find the area of the surface of revolution formed by revolving the curve r about the initialline~

Solution: Equation of the circle is

r

=

2acosB

dr= -2 ' B asm dB

y

....--x

x'-------~I----o

y'

Fig. 5.4.6

The surface area

=

L

lIi2

=

J" 21ty -dO dO 2

ds

o

dr 21trsinO r2 + ( dO

2 )

dO

J:/ 21trsinOJ4a 2cas 0 + 4a sin 0 dO = 81ta J sinO cosO dO =

2

2

=

2

"/2

0

4mr Sq. units

2

2

=

2acosB


354

5.4.7 Example:

Find the surface area of the solid formed by the revolution of the cardioid = a(J + cosO) about the initial line.

r

Solution:

Equation of the curve r

=

a(J + cosO)

dr . B -=-asm dB

0=0

The surface area =

Io"2ny-de de

Fig. 5.4.7

ds

= 2n

L" rsine

= 2n

I:

r2 +(*

J

2 rSineJa (i +cose

de

y +a 2sin 2e de

L" a(i + cose)sine .Ja 2(i + cose)2 + a = 16na I" COS3~ sin ~ cos ~ de o 2 2 2 e sm"2 . e d8 = 16na I" cos "2 =

2n

2

2

=

0

4

[25 2eJ"

16a 2n - - cos 5 -

0

32mi

= --

5

Sq. units

2

sin 2e de


355

5.4.8 Example:

Find the surface of the solid generated by revolving the lemniscate,.2 about the initial line. Solution:

Fig. 5.4.8

Equation of the curve is ,.2 Surface area

=

a 2cos2B

" ds fo 21ty-dO dO 4

=

2

41t

In

11/4

dr rsinO r2 + ( dO

1[/4 0 rsinO

f

=

41t

=

41ta 2 [_ cosO ] ~

2 )

dO

~ a4 • ~ r- +-~ sm-20 dO r-

4

=

a 2cos2B

356


Exercise - 5(0) I.

Find the total area of the surface obtained by revolving the ellipse

?

?

a-

b-

x~ + y~ = 1 about

its major axis.

2.

Find the surface ofa sphere of radius 'a'. (Ans: 4Jlu 2 ]

3.

Find the surface of the solid generated by revolving the arc of the parabolay bounded by its latus rectum about x-axis. (Ans:

4.

3 Find the area of the surface generated by the revolution of the cycloid x y = a(l - cost) about x-axis.

=

a(t - sint), 64

3

2

7m )

Find the area of the surface of the solid formed by the revolution of the cardiod r = a(l - cosO) about the initial line. (Ans:

6.

4ax

~7lU2(2J2 -I)J

[Ans: 5.

=

32

2

SJl"a )

The lemniscate? = a2cos2B revolves about a tangent at the pole. Show that the surface area generated is 47lU2 •

5.5.1 Double and triple integrals: Letj(x,y) be a continuous and single valued function of x and y within a region R bounded by a closed curve 'c' and upon the boundary c. Let the region R be subdi vi doo in rrty mrrtne- into n subregi ons of areas 8R\, 8R2 .... 8Rn •

Let (x" y,) be any point in the subregion of area 8Ri . Consider the sum

The limit of this sum as n ~ 00 (i == 1,2, .... ) is defined as the double integral ofj(x,y) over the region R and is written as

fff(x,y}iA


Hf(x,y}ixdy (a)

I f(x" yJ) R,

Lt

=

n~oo ,=\

II

Suppose the region R is described by the inequalities c ::::; y ::::; d and g(y) ::::; x ::::; heY) y y=d ...--_ _.,....::--....fX = h(y)

x = g(y)

y-c

x'---------+--------x o Y' Fig. 5.5.1 (a) Then t

="

ff f(x,y )dydx=Jff(x,y )dxdy = f fX=II(Y)f(r,y)dXdy x=g(y) II

(b)

Y~

II

Jfthe region R is described by inequalities a ::::; x ::::; band g, (x) ::::; y ::::; h,(x) y

x ' - - - - -....... oof----I----""----x y'

Fig.5.5.1(b) x=h

Then

II

(c)

y=lI\ (.)

ff r(x,y}ixdy = fff(x,y}iydx = f f y=g\ (.) f(x,y)dx.dy II

x=a

If the region R is bounded by the lines x = a, x = b, y = c, y = d (rectangle) Then

Hf(x,y}dxdy II

=

h

"

" h

a

c

( a

f ff(x,y}dydx= f ff(x, y}ix dy

Note: The order of integration is immaterial for constant limits

357


358 5.5.2 Example: Evaluate

I

2

o

I

2 I I{x + y2 }Ix dy

Solution:

I

=

8 I 2dy -+2y 2 ---y 03 3

I

il

o =

3

Y 7 I Y +- dy=-+-yl 3 3 3 0 7

2

J

1 7 8 333

-+-=-

5.5.3 Example: 4

Evaluate

K-~

I

Ixydydx o

Solution: 4

4

J4-x

I'=J4-x

I

fxydydx = f·

f [xydy] dx

I

0

0

4

x=l

y=K-~

f

fxydy.dx

x=l

=

0

2

4f~(4 -X}tx = 4x _~14

9

I

2

x=l

2

4

6

Application of Integration to Areas, Lengths, Volumes and Surface Are."'''

359

5.5.4 Example: 1 2- r

J Jx dXdY 2

Evaluate

o J~'

Solution:

v=1

J1(2- Y )' -l(J.i~)\~v

y~O

J8 - -

)'=1

= -1

3

3

V1

-12)1 + 6v2 --') V- l IV ~

\'~O

=

1[S-1-6+2-lJ 67 60

5.5.5 Example: Find the value of Hxy(x+ y}/xdy taken over the region enclosed by the curves

y

=

x and y

=

x2.

Solution:

y

y=x

x' _ _ _ _ _

~~:-:.....-----

y'

Fig. 5.5.5

x

360


R is the rogion bouuded by the curves y = x and J' =

1=

7

x~

ff'"y(x + y')dxcly II

=

x5 x5 x7 x8 I 10+]5-14- 24 10

=

I I I 1 -+-----

10

15

14

24

3

=-

56

5.5.6 Example: Evaluate

IfR 1- Y

dx{1y where A is the area in the positive quadrant of the circle

A

x2 + y2

=

1.

Solution: y

x'----------~r-~~~~~~~------x

8(1,0)

y'

Fig. 5.5.6


~ 1- (2

_I

±ofx[~ -/'2-~--] ()

= -

I

£Ix

I

fX{I[-Q-X 2)]2 -I}dx

= --

o I

f{x x}It 2

= -

-

o

-x 3 X-J ]1 - --+-

-

[

3

2 ()

-I

1

1

3

2

6

= -+-=-

5.5.7 Example: Evaluate ffe

tlHhY

dt dy over the triangle x

=

0, y = 0, ax + by = I

Solution:

x:: : Ol----t---" x'------------~----~~----~~----x

~

AU,a)

Fig. 5.5.7

361

362


1 = ffea<+hY dx dy

1= I I'" ,=0

I = _

lY: '-S-f:/\ e

j

+hI' I I -(Y(X

y=o

-;-R~ e ,(I-a<) _e llX+)

-----

h

lLt

1dx

+O

bo

=

~[.:. -.:. - 0 + ~l b a a a ab

5.5.8 Example: Evaluate ffydxdy where R is the region bounded by the parabolas y R

xl

=

4y.

=

4x and


Solution: y

x'--------~~--~------x

y'

Fig. 5.5.8 Solving the curves i' = 4x, x 2 = -/y

X2 )2 ( - 4 =4x x= O,x= 4 (0,0), (4,4)

1= ffydxdy 11

48 5

363


364 5.5.9 Example: Evaluate

ff(f; - y2 }iydx

where R is a triangle with vertices at (0, 0), (10, )),

II

(I, I).

Solution:

=1 ,....-.,------""!:II ... A(10.1)

8(1.1) Y

x· ____________ ________________---------x ~

0(0.0) y'

Fig. 5.5.9 The triangle OAB is within the limits y I

II

y=o x=y

JelY ~~~y- /y

I

=

o

3 ]IOY

[

y

x

~

lOy and 0 ~ y

x~IOy

J J(FxY -

ff(J~ - y dydx) = 2

~

Y

y2)dxdy

~)


365

Exercise -: 5(E) I. 1.

Evaluate the following double integrals: 2

2

0

0

f f(x 2 + y2 ~xdy 32 3

IAns: - I 3

x+2

f f~vdx

2.

-I

x2

IAns:

} Jdxdy J

3.

1

8

-I 3

4

3

J

(x + y)25 IAns: log-I 24

II

4.

~1I2_x2

f

fidydx

o

0

25(/5 (Ans: -15- 1

5.

II

h

o

0

f f(x 2+ y2 )ixdy

J)

ah (J IAns: 3\U- + h- ]

[Ans:

3Jru 4

--I 4

366


a

8.

2a-x

f o

fxy{O;dx );2

a

3a 4 [Ans: - ] 8 J

9.

Find the value of ffxydxdy taken over the positive quadrant of the ellipse

a Ja 2 -y2 10.

f

f

o

0

~a2 _x 2 -

J

~: + ;~ = I

2

y dxdy 7fa

3

(Ans: - ] 6 x=1 y=x

II.

f

y

f eX dxdy

x=o y=O

e-\ [Ans: - ) 2

12.

[Ans:

13.

7f

4

log(.J2 + 1)1

j fSin(x+ y)dxdy x

0

o

[Ans: I]


14.

Evaluate H-,ydxcZV where R is the quadrant of the circle x 2 +

367

y2 = {/2

when

x~O,y~O. {/4

[Ans:

15.

8

Evaluate Hxydx((v whose R is the region bounded by x-axis, ordinatex ~ 2a and the ?

curve x-

.Jay.

=

[Ans:

16.

-I

Evaluate

H(x + )')d'({~}' whose R is the triangular region bounded by y

=

2x, y

a4

3 =

1

~

/I

and y

17.

=

3 ~ x.

Evaluate and C(9,

Hxydx(~Y

where R is the triangular region with vertices A(~6, 2), B(~ 1,3)

~7). ~1025

[Ans: 18.

----:n- 1

Evaluate Hxydn'y over the region in the positive quadrant for which x + y

~ 1

I [Ans: 241

5.5.10 Change of variables: The variables x, y in Hf(x, y ~/xdy are changed to It, v with the help of the relations II

x = j(u, v), y = filt, v) then the double integral is transferred into

fIt!; (11, v)f2 (It, v)]!JI}illdv 1/'

dx dx du elv where J = dy cZY and 'R' is the region in the uv plane corresponding to the region du dv R in the xy plane.

368


5.5.11 Changing from Cartesion to Polar co-ordinates: x = reosO, y = rsinB dx (be J = dr dO =ICOSB -rsinBI =r dy dy sinO rcosB ---

dr dB Jff(x,y)dxdy =

Jff~·cos9,rsinO] /Jld9dr

/I

/I'

= Jff[rcosO,rsinOp'dltlr /I'

Note: In polar form dxdy is replaced by rdBdr


Jf~ a 2 - x 2 - y2 dxdy

over the semi circle

xl + Y

=

quadrant.

Solution: Substituting x

=

reose,

y = rsinB, in xl + Y = ax ,.2eos2 0 + ,.2sin2 0

=

an'osO

r

=

aeosO

which is the equation ofthe given circle in polar co-ordinates.

y

x' --------~I'-.-lo._~-_+-_o A(a,O)

Y' Fig. 5.5.12 From the figure it can be sedn that

r

=

OP

=

aeos8

X

ax in the positive


for the upper halfofthe circle, Ovaries form 0 to value of 0, r varies from 0 to

369

"27T where as for any intermediatary

or i.e., () to acm·O. 1[

2 acosO,--_ _ _ __

JJ~a2-x2-idx(ZV= f II

I) ~O

fJa2-r2cos20-r2sin20rdrdO r~O

where R is the shaded region.

1t

3 2

=

~ f~ 3

-sin 0)dO 2

0

5.5.13 Example: {/

I

2 J \ a - y-

Evaluate f o

f{x 2 + y2 }Ix~v by changing into polar coordinates. 0

Solution: From the given limit x = ~ a 2

-

i

it is clear that one of the boundaries is the circle x 2

+ y = (r, taking x = rcasO, y = rsillO. The given rigion is the first quadrant of the circle. Here r varies from '0' to

'(I'

and 0 varies from '0' to 1t

Ja 2 -y2 2 a f f~2 + y2 )dxdy = f fr2.rdrdO

£l

o

a

O~~O 1t

7T "2.

370


(14

1l

=-x-

4

2

8 5.5.14 Example: 00

(2 2)

W

Transform the integral I Ie - '+Y dxdy to polar coordinates and evaluate it. o r=d Solution: The limits of x and yare both from 0 and' 00 '. Therefore the region is in the first quadrant where r varies from '0' to ' 00 ' and B varies from '0' to Substituting x

=

1l

2

rcosB, y = rsinB, and dxdy = rdrdB. 1t

1t

=

1t

e-r 2 [2jW

I --2

o

1t

de

1 ?=-

2

0

1

2

rde =-[e] J' 2 0

0

4

Changing of order of inte8,ration: The double integration can be integrated with respect to y first and then with respect to c or it can be integrated with respect to 'x' first and then with respect to y'. In the former case, the limits of integration are determined for the given region by drawing strips parallel to y-axis while in the second case by drawing strips parallel to x-axis. 5.5.15 Example: ,

a

2a-x

Evaluate by changing the order of integration I

Ixydx«:v

o " a


371

Solution: 2

The given limits are

x:O~a,y:~~2a-x a

2

i.e., the region is bounded by x = 0, x = a and y =

~ y = 2a - x i.e., x + y = 2a. a

x' _ _ _ _ _ _.-.."""'....._ _ _ _ _---::a...-_ _ X

Y' Fig. 5.5.15 a

2a-x

o

x2/a

J Jxydxdy

Hxydxdy + Hxydxdy

=

OA(,O

y=a x=[:ry

=

J

a

=

('AIl(,

y=2ax=2a-y

Jxydxdy +

J

Jxydxdy

1 2(/ JY(2a- yydy

~ Ji dY+2" o

(/

5.5.16 Example: a

Change the order of integration and evaluate

H

J J(x 2 + y2 }txdy

o

x a


372

Solution:

The given limits are x = 0 x = 0 and y =

~a

and y = .[; i.e., a

ay = x.

y

x'------------~~-------------x

Y'

Fig. 5.5.16

f

2J~2 y2)1xdy =:tEX2 y~dx}y +

+

a

=

f ~3 i I (

3

o

- a 3 l + ai - ay4 } y 3

a3 a =-+28 20 5.5.17 Example: a

Evaluate

2&

f fx 2dxdy by changing the order of integration. o

0

Solution:

The given limits are x = 0, x = a and y = 0, y = 2.[;;; i.e.,

y

= 4ax.


y

x=a

x' -----...",..jl--""-::-I------ x

y'

Fig. 5.5.17 I

=

3"

f

2a [

3

a -

[') Y 4a

]3] dy 3"[a y - 64a I

Y7]2(1

I

3

=

3

x

7

0

5.5.18 Example:

f yfx

4

Change the order of integration and evaluate

o

4

2 X

2 dxdy

+y

Solution:

y

x=4

x'------..,t"f'---~~---

y'

Fig. 5.5.18

x

.

373

374


The given limits are y =

°y

= 4 and x = y, x = 4.

4

4

o

0

r -I dx= I7r-dx=-7r [x ]4 =7r = ian 0

4

4

5.5.19 Example: a

'22

vaL _yL

Evaluate by changing the ord¢r of integration f o

fxydxdy 0

Solution: y

x'------------+---~+_~--~-----------x

y'

Fig. 5.5.19 The given limit are y = 0, y = a and x = 0, x = ~a2

2l~ dx

xL

X o

2

0

_ /

i.e., x 2 + y2 =a2

Application of Integration to Areas, Lengths, Vo'lumes and Surface Areas

375

8

Triple integration: Let u = j(x,y,z) be a single valued function of the independent variables x, y, z defined through the region V. Divide the region V into 11 subregions t5V I' t5V 2 ..•. t5V nand P(x l , YI' z) be any point inside or on the boundary of the subregion t5V /' then the sum II

If(x"ypzJdv, is defi·ned as triple integral.

under the limit Lt 11

~oo ,~I II

If(xpy"z, )dv,

Lt 11

~

00

=

,~I

HJf(x,y,z}1v /'


\

z

x+z

-I

0

x-z

JJ J(x + y + z }ixdyd::.

Solution: \

I

z

x+z

Jdz Jdx J(x + y + z }ixdydz

=

-I

\

0

x-z

z

Jaz fi2z(x + z)+ 4xz}ix -\

0


376

1

JZ 3 + 2z 3 + 2z 3dz -I

1 [ 4]1 =SJ.::\lz=S 24 =0 -I

-I

5.5.21 Example: 2

Evaluate

r

x+y

J J JeX+Y+ZdXdydz

000

Solution:

HJe'[J'dZH'LT+[[e'.(e' r t"dy

fe<[ feY{e-HY -\ }IY}b: o

2

0

4<

e - 3 7< +e-rdx ---e-o 2 2

J


5.5.22 Example:

Evaluate 1

1

f f[xz t~dnly

I=

\'=0

1

r:::

1,2

)

1 \=1

f .~ - X3

dy t:..::y2

\'=-0

f (I-6 1

y=()

y6l

y4 _0_+dy

2

3

~ [~< +~;I 3~ 5.5.23 Example:

Evaluate Solution:

4[-~+-S'11 x 42

f

z=O

2

2

2

4In::dZ [z22rJo = 1f

=81f

°

~1

X {-}

J4;

]2.[; dz <=0

377

378


5.5.24 Example: Evaluate HJryzdzdydx over the volume enclosed by three coordinate planes and the I'

plane x + y + z = I.

Solution: The plane x + y + z = I meets the coordinate axes in A( I ,0,0),8(0, I ,0) and C(O,O, I).

z C(O,O,1)

x Fig. 5.5.24

WX;adZdyd!~ ,1, :I('JpIz }!YdX

1(2 I

=

XY[

~

r' }n dy

I-x

J JX;[(I-X- y )2_ 0}/YdX x=o y=o

=

~

I

I-x

J JXy(I+X2+y2_2X-2y+2Xy~ydx x=O ),=0


379


ff~X2 + i

+

z2 }Izdydx: where V is the volume of the cube bounded by the

I'

coordinate planes and the planes x

=

y

= z = {[,

Solution: z

y

Fig. 5.5.25

ff~x2 + y2 + z2 )lzdydx = I'

'f 'f x~o

\'=0

2

Z~O

Jra-))

(/

x=o

)Iz(zvch

'f(x + y2 + z2

a

4

a

4

+-+-dx 3 4

I

= [ ~' a' + ';' x+ ~' x

=d


380

5.5.26 Example:

Evaluate by tripple integration the volume of a hemisphere of radius o. Solution:

Fig. 5.5.26

The volume of the hemisphere

a

V=4

r;;C;'1-

X~O

V

2 3

,----

J J [z]ga -r

= -1rG

y~O

3

2

2

-1'2

dyd"(


381

Exercise - 5(F) 1.

Evaluate the following integrals by changing the order of integration: [4-:':;'

3

I.

J

J(x+ y}lxdy

0

241 IAns: 60 5

2.

I

x2

J fx{x + i }Ixdy 2

0

0

29 56 IAns: 24· I

" 2 ,2,,1,,2_x

f

3.

fy 2dx((V 0

0

({~

lW4

IAns: - + - 1 16 32 4.

I

2-x

0

x

I

j;

J J!...dxdy v .

II

IAns: 2 log2 5.

f

0

fxy(X + y}lxdy I'

IAns:

,.

3 56

1

4£1 2\'at

6.

f

o

Jdydx x2/4a

IAns: 2

7.

2

3

4

Jfx I

16a

-I

2

2

+ y dxdy

3

IAns: 10)

382

II.


Evaluate the following tripple integrals: u

I.

x

•

y

f f fxyzdxdydz 000

a6 [Ans: 48]

2.

f 'J 1-](> + i o

0

+ Z1 }lwlydz

0

I [Ans: 20]

3.

, , J-XIY~ dydz

f II' ()

0

I

[Ans:

3"]

122

4.

2

f f fx yzdxdydz (J

U

I

[Ans: I]

III 5.

f f f(x + y + z)dxdydz

()

()

()

[Ans: I

6.

f o

3

2]

JI-x 2 ~~2_y2 f

fxyz dtdydz

0

0

I [Ans: 48]


7.

3

I

I

I x

Fv

JJ

Jxyz dxdydz 0

(Ans:] ( I

8.

13 1 1og3 ) 9-6 ]

I-x x+y

JJ

o

383

0

z

Je dxdydz 0

1 (Ans: -] 2

9.

4

x

!BY

o

0

0

JJ

Jz dxdydz (Ans: 16]

o

10.

J 02_x 2

J J

o _J02_x2

niX

2

Jz dxdydz o


6 Sequences of Series 6.0 Sequence A function fN ~ S, where S is any nonempty set is called a Sequence i.e., for each nE N, ::l a unique elementj{n) E S. The sequence is written asj{I),j{2), j(3), ..... j{n).... ,and is denoted by (j(n)}, or or ( all ). Here j{ n) or all are the

d h terms of the Sequence. 6.1.1 Example: 1 ,4,9, 16, ......... n 2

, ••••• (or)

< n2 >

s


386

6.1.2 Example:

~3 ,~,-;-, ~ (or)(~) I 2- J n n •••••

••••

s

N

6.1.3 Example: 1, 1, 1. ..... I.. ... 0r

6.1.4 Example: 1 ,-1, 1, -1, ......... or (( - I

Note

1. 2.

r-

I )

If S c R then the sequence is called a real sequence. The range of a sequence is almost a countable set.

387

Sequences of Series

6.1.5 Kinds of Sequences 1. Finite Sequence :A sequence < all > in which an

= 0 Vn > mEN

is said to

be a finite Sequence. i.e., A finite Sequence has a finite number of terms.

2. Infinite Sequence: A sequence ,which is not finite is an infinite sequence. 6.1.6 Bounds of a Sequence and Bounded Sequence 1. If :3 a number 'M' ~ an:::; M, Vn E N, the Sequence < an > is said to be bounded above or bounded on the right.

Ex: 2. If

I,.!.), ,....... here all :::; 1Vn EN 2 3

:3 a number 'm' ~ a" ~ m, Vn E N, the sequence < a" >

IS

said to be

bounded below or bounded on the left. Ex : 1 , 2 , 3 ,..... here a" ~ 1 Vn EN

3. A sequence which is bounded above and below is said to be bounded. Ex:

Let

an = (-1

n

r (1 +;) 2

3

4

3/2

-4/3

5/4

2-1 __

o

2

3

4

5

n

-1 -

From the above figure (see also table) it can be seen that m :. The sequence is bounded.

=

-2 and M

=

l. 2


388

6.1.7 Limits of a Sequence A Sequence < (Ill > is said to tend to limit 'I' when, given any + ve number' E however small, we can always find an integer 'm' such that and we write Lt

=I

all

n->co

If an

Ex:

(an

or

=

n2 + 1 2

2n +3

"

Ian -II
~ I)

then
1

>~-.

2

n

6.1.8 Convergent, Divergent and Oscillatory Sequences ). Convergent Sequence: A sequence which tends to a finite limit, say' I' is called a Convergent Sequence. We say that the sequence converges to 'I' 2. Divergent Sequence: A sequence which tends to ±oo is said to be Divergent (or is said to diverge). 3. Oscillatory Sequence: A sequence which neither converges nor diverges ,is called an Oscillatory Sequence. Examples

3 -4 ,-,..... 5 here 1. Const'der the sequence 2, -,

234

1 n

an =1 +-

The sequence < all > is convergent and has the limit 1

1 1 1 1 an -1 = 1+ - -1 = - and - < E whenever n > n

nnE 1 Suppose we choose E= .001, we have - < .001 when n> 1000. n 2. If all

1 =3+(-lr-- converges to 3. 'n'

3. If an =n 4. If an 6.2

2

+(-lr .n, diverges.

= ..!. + 2 (-1 n

r '< an > oscillates between -2 and 2.

Infinite Series

6.2.1 If < UII > is a sequence, then the expression u1+ u2 + u3 + ........ + un co

infinite series. It is denoted by

I

Un

or simply

I

+ ..... is called an

Un

11=1

The sum of the first n terms of the series is denoted by sn i.e.,

sn = U, +u2 +u3 + ...... +un;sps2's3' ....sn are called partial sums.

Sequences of Series

389

6.2.2 Convergent, Divergent and Oscillatory Series Let be an infinite series. As ~ 00, there are three possibilities.

Iu"

n

COl1vergent series: As n

(a)

~

oo,s"

~

a finite limit, say's' in which case

the series is said to he convergent and's' is called its sum to infinity. Thus

Lt

8"

=S

=8

(or) simply LIs"

11----)00 00

This is also written as

+ 1I} + II] + ..... + 11" + .. .1000 = s. (or)

III

I

UII = S

11=1

(or) simply

IlIlI = s. 8 11 ~ 00

(b)

Divergent series: If

(c)

Oscillatory Series: If

or

-00,

the series said to be divergent.

does not tend to a unique limit either finite or

8 11

infinite it is said to be an Oscillatory Series. Note: Divergent or Oscillatory series are sometimes called non convergent series.

6.2.3 Geometric Series . 1 1 II-I TIle senes, +x+x- + ..... x + ... (i) Convergent when

•

IS

Ixl < 1, and its sum is

(ii) Divergent when x ~ 1. (iii) Oscillates finitely when x

=

_I_ I-x

-I and oscillates infinitely when x < -I.

Proof: The given series is a geometric series with common ratio 'x'

I-x"

sn

(i)

=- -

When

I-x

when x *1 [By actual division - verify]

Ixl < 1: LI s n;->oo n

=

1

( 1)

Lt - - - Lt (XII] - - =-1- x n->oo 1- x 1- x

n->oo

[ since xn ~ 0 as n ~ 00 ] . 1 :. T he senes converges to - -

I-x

(ii)

When x ~ 1: sn

x" -1

=- x-I

and

:. The series is divergent. (iii) When x = -1 : when n is even,

sn ~ 00

sn ~

:. The series oscillates finitely.

as n ~

00

0 and when n is odd,

sn ~

1


390

(iv) When x < -1, s"

~ 00

or

-00

according as n is odd or even.

:. The series oscillates infinitely.

6.2.4 Some Elementary Properties of Infinite Series 1. The convergence or divergence of an infinites series is unaltered by an addition or deletion of a finite number of terms from it. 2. If some or all the terms of a convergent series of positive terms change their signs, the series will still be convergent. 3.

Let

Iu"

converge to's'

Let 'k' be a non - zero fixed number. Then Ikll" converges to h.

I

Also, if 4.

Let (i) (ii)

I

II"

u" diverges or oscillates, so does converge to 'I' and

I

I

v" converge to

ku" '111 '.

Then

I

(u" + v,,) converges to ( 1+ 111 ) and I(u" -v,,) converges to (/-m)

6.2.5 Series of Positive Terms Consider the series in which all terms beginning from a particular term are +ve . Let the first term from which all terms are +ve be ul • Let

I

Un . be such a convergent series of +ve terms. Then, we observe that the

convergence is unaltered by any rearrangement of the terms of the series.

6.2.6 Theorem If Iu" is convergent, then It un n->oo

= o.

Proof:

Sn=u l +U2 +······+U" sn_1 = ul + u2 + ...... + U,,_I,' so that, un = sn - sn_1 Suppose

IU n =/ then It sn =/ and Lt sn_1 =/ n---+oo

It un = It (Sn-Sn_I); n~oo

n---+oo

n~oo

It s,,- It n~oo

S,,_I

=/-/=0

n~oo

Note: The converse of the above theorem need not be always true. This can be Observed from the following examples.

Sequences of Series

(i)

1 2

.

1 3

1 n

1+ - + - + ....... + - + .... ;

Consider the series,

But from p-series test ( (ii)

.

2.7) it is clear that

lin

1 n

= -, It

n~oo

lin

=0

I!n is divergent.

1 1 1 1 + -) + -) + ..... + -) + ..... . 1 2- 3n-

ConsIder the sertes,

-2

= ~, It

lin = 0, by p series test, clearly Ir n~OCJ Note: If Lt lin ::t:- 0 the series is divergent; lIl/

391

,

I

~ n-

converges,

IJ-too

Ex:

2n -1 1'" = - 2 ,here Lt n

lin

Il---Ht)

I

lin

=1

is divergent.

Tests for the Convergence of an Infinite Series In order to study the nature of any given infinite series of +ve terms regarding convergence or otherwise, a few tests are given below.

6.2.7 P-Series Test i ' sertes, . ~1 Th e '111f illite ~= -1+ 1 - +1 - + ...... , n~1

1"

n"

2"

IS

3"

(i) Convergent when p > I, and (ii) Divergent when p ~ 1. [JNTU Dec 2002, A 2003] Pro(~r:

Case (i) Let p> I . J) > 1 3" > 2'" => _1 < _1_

,

,

'3"

1

1

1

1

2

2"

3"

2"

2"

2"

2"

-+-<-+-=Similarly,

111111114 -+-+-+-<-+-+-+-=4" 5" 6" 7" 4" 4" 4" 4" 4" 1

1

1

8

8"

9"

16"

8"

-+-+ .... +-<-,

and soon.

Adding we get

1 2 4 8 ~ n" 2" 4" 8" 1 1 1 1 n" < 1+ 2(,,-1) + 2 2(,,-1) + 2 3(,,-1) + ..... .

"-<1+-+-+-+ .... I.e.,

I


392

The RHS of the above inequality is an infinite geometric series with common ratio

1

2,,-1 < 1(since p > 1) The sum of this geometric series is finite. Hence

f ~n

is also finite.

11=1

:. The given series is convergent.

Case (ii) : Let p = I; We have,

1 1 1 1 I - = 1+ - + - + - + ...... n" 2 3 4 1 1 1 1 1 -+- >-+- =3 4 4 4 2 111111111 -+-+-+->-+-+-+-=567888882 11 111 11 -+-+ ....... - >-+-+ ..... - = - and so on 9 10 16 16 16 16 2 I_I = 1+(!+!)+(!+!+!+!)+ ..... n" 2 3 4 5 6 7 1 1 1 ~1+-+-+-+ ..... 2

The sum of RHS series is

00

2

2

1 (Since sn = 1 + n ~ = n; 1and /1~'" SII = 00 )

:. The slim of the given series is also

00;:.

f_l-n 11=1

P

(p

= I ) diverges.

Case(iii):Let p
1 I7=1+

1

+ 1 +..... 2P Y

Since

1 1 1 1 p < 1,- > - - > - ...... and so on 21' 2' JP 3'

1 1 1 1 I->I+-+-+-+ ....... P n 2 3 4 From the Case (ii), it follows that the series on the RHS of above inequality is divergent.

I Jl is divergent, when n P

P< I

Note: This theorem is often helpful in discussing the nature of a given infinite series.

Sequences of Series

393

6.2.8 Comparison Tests 1. Let Un and

I

I

I

Then

be two series of +ve terms and let

Vn

I

VII

be convergent.

converges,

Un

1. If un ~ Vn' \:;f n EN

2. or

U

-.!!.. ~

k\:;fn EN where k is> 0 and finite.

vn

3. or

Proof: 1. Let

un vn

~ a finite limit> 0

I

Vn

Then, u l

=I

(finite)

+u2 + ....• +lln

+ ...... ~VI

Since I is finite it follows that

2.

un vn

ll

,

Ikvll is convergent:. IU

3. Since Lt

Then

and

UII

I

Un

VII

is convergent and k (>0)

IS

is convergent.

~ < k\:;fn EN vn

:. from (2), it follows that

I

n

I

is finite, we can find a +ve constant k,3

un n-'>oo vn

>0

is convergent

UII

~ k ~ UII ~ kv \:;fn EN, since

finite,

2. Let

I

+v2 + ..... VII + ..... ~l

I

IU

II

is convergent

be two series of +ve terms and let

Vn

I

VII

be divergent.

diverges,

* 1.

If

un ~ VII' \:;fn

or

* 2.

If

un vn

or

* 3.

If Lt ull is finite and non-zero.

EN

~ k, \:;fn EN where k is finite and"* 0

n-'>OO Vn

Proof: 1.

Let M be a +ve integer however large it may be . Science a number m can be found such that VI +V2 + ..... +vn > M, \:;fn > m

u +u2 + ...•. +un > M, \:;fn > m(un ~ vJ l

I

VII

is divergent,


394

I 2

11,

I

1I"

2:: kv,,\:;In

v" is divergent ~

I 3.

is divergent

I

kv" is divergent

u" is divergent

Since Lt

~ is finite, a + ve constant k can be found such that

un

> k, \:;In

(probably except for a finite number oftenns ) :. From (2), it follows that

I

is divergent.

Un

Note: I. In (I) and (2), it is sufficient that the conditions with * hold \:;In> mEN Alternate form of comparison tests : The above two types of comparison tests 2.8.( I) and 2.8.(2) can be culbed together and stated as follows: If

Iu"

I

and

Vn

are two series of + ve terms such that Lt "->00

where k is non- zero and finite, then

I

u" and

I

Vn

un =

vn

k,

both converge or both

diverge.

Note: I. The above fonn of comparison tests is mostly used in solving problems. 2. In order to apply the test in problems, we require a certain series Vn whose

I

nature is already known i.e., we must know whether divergent. For this reason, we call

I

Vn

I

Vn

is convergent are

as an 'auxiliary series'.

3. In problems, the geometric series (2.3.) and the p-series (2.7) can be conveniently used as 'auxiliary series'.

Solved Examples 6.2.9 Example Test the convergence of the following series:

3456 1 8 27 64

(a)

-+-+-+-+.....

(c)

00 [ ~ (n

4

+1) 1/4 -n ]

(b)

4567 1 4 9 16

-+-+-+-+ .....

Sequences of Series

395

Solution (a)

Step 1:

d"

To find "ul/" the

term of the given series. The numerators 3, 4, 5,

6 ...... ofthe terms, are in AP.

n'" term tl/

3 + (11- 1).1

=

=

n+2 n+2

3 2' 33 43 · Denomlllators are 1 ,-, , ..... n ,,, term = n \ ;

Step 2: To choose the auxiliary series

I

VI/ .

111/

= --3n

In 1I11 , the highest degree or 11 in the

numerator is I and that of denominator is 3. :. we take,

= ----:l=I = 2

VII

n

11

Step 3 :

~=

It 1/-->00

vl/

2 It n+2xn = Lt n+2 = It 3

I/~OO

n

I/~OO

11

I/~OO

(1+~}=1 11

'

which is non- zero and finite. Step 4: Conclusioll:

It ~=1 n-----}oo

and

:. IUI/

I

VI/

VII

I

,,-?1

= ~n-

v" both converge or diverge (by comparison test). But

is convergent by p-series test (p

=

2 > 1); :.

Ill"

is

convergent. (b)

4

5

6

7

1

4'-9

16

-+-+-+-+ ..... Step 1 : 4 , 5, 6, 7, ..... in AP , tn Step 2 : Let "~ v"

= -1

= 4 + (n -1) 1 = n + 3

. be tIle auxi'1'lary series

11

Step 3: Lt I/~'"

finite.

~ = Lt (11~3)xn = It (1 v" n,,~oo

n+3

I/~OO

+l) = 1, n

which is non-zero and


396

Step 4 : :. By comparison test, both

I

utI and

I

v" converge are diverge

together. But

Iv" =I-n1

is divergent, by p-series test (p

= 1); :.

Iu"

is

divergent.

=n

1+-'4n-.+ ~G-l)~+ . . _1 2! n

= 4~3 -

3;n 7 + .... =

~3 [~- 3;n

Here it will be convenient if we take v"

(! -

!, and I

4

=n[-'-.-~+ . . .J

+ ....

4n

32n

.J

=~ n

Lt utI = Lt -1-4 + ..... ) = which is non-zero and finite vn n~oo 4 32n 4

n~oo

:. By comparison test, by p-series test

IV

I

Un

1 n =-3

n

VII

both converge or both diverge. But

is convergent. (p

convergent.

6.2.10 Example

=

If u n

{j'--3n-2-+-1

~hn3 +3n+5

show that

I

Un-

is divergent

Solution As n increases, utI approximates to r:::-?

~ 3n 2 l

II

Jl3

2/

n73

II

3/ 3

1

~2n 3 = 2Y.; x nYt = 2}~ . nX2

=

3 > 1); :.

Iu"

is

Sequences of Series

397 Ii

= - 1-/

If we take vn

U

,

nl12

[(or) Hint: Take vn

Lt -.-!!.. n-->""VII

1

= ~ , where n' ,

)13

=-I

which is finite.

274

II and 12 are indices of 'n' of the

largest terms in denominator and nominator respectively of ull • Here

1

VII

1

=-3-2 =-1 ]

n4

3

nl2

I

By comparison test,

"L..,. VII

VII

and

I

U II

converge or diverge together. But

, 1 .IS d'Ivergent by p =, L..,.-I,-'

. test (smce . senes p

= -1 < 1)

Wl2

L un

12

is divergent.

6.2.11 Example Test for the convergence of the series.

Jf ~ J! ~ +

+

+

+ ......

Solution

=~

n:

Here,

un

Take

1 1 vn =-1-I =-0=1, Lt u. ----

n2

I

VII

1;

n

n-->"" V

~

Lt ) 11-->""

n

2

11

= 1 (finite)

1+n

is divergent by p - series test. (p = 0 < 1)

:. By comparison test,

I

Un

is divergent, (Students are advised to follow the

procedure given in ex. 1.2.9(a) and (b) to find" un" of the given series.}

6.2.12 Example 1 1 1 Show that 1 + IT + l2 + ....... + l!:! +..... is convergent. Solution

un

=

l!:!1

(neglecting 1sl term)

1 1 1 - - - - < -----==,--- = 1.2.3 ......n 1.2.2.2 .....n -ltimes (2n-l)


398

1

1

1

"1I" < 1+-+-? +-, + ..... . ~ 2 2- 2' which is an infinite geometric series with common ratio

" 2,,-1 1 IS . convergent. (1.2.3(a)). Hence :. ~

I "/I

~ <1 2

is convergent.

6.2.13 Example

_1_ + _1_ + _1_ + ...... . 1.2.3 2.3.4 3.4.5

Test for the convergence of the series,

Solution

"

Take

1 . n(n+1)(n+2)'

=

u

1 V"=-3

n

3

Lt

U"

,,~ao V

n)(1+-2 )= 1 (finite)

= n~ao Lt

( 3 1 n 1+-n

n

By comparison test, series test,

I

v" =

I

I

n

U" '

~

and

I

v" converge or diverge together. But by p-

is convergent (p = 3> 1 ); :. Iu"

n

is convergent.

6.2.14 Example If u" =

-J n4 + 1 - -Jn4 -1, show that I

Un

is convergent. [JNTU, 200s]

Solution 1

u"

= n2 ( 1+ n1)2 4

I

1)2

n 2 ( 1- n 4

~n2[(1+ 2~4 - 8:' + 16~12 - ....)+- 2:4 - 8:' -16~12 - .....)] =n Take

vn

2 [-;-

n

1

= -2 n

'

+

~ + .... J= ~ [1 + ~ + .... J 8n n 8n

hence Lt ,,~ao

u' -E...

v"

=1

Sequences of Series

399

I


I

Vn

=~ n

and

lin

I

v" converge or diverge together. But

I

is convergent by p -series test (p = 2 > 1):.

Un

is convergent.

6. 2.15 Example

1 1 1 I-' · Test the senes - - + - - + - - + ..... lor convergence.

l+x

2+x

3+x

Solution

Take

1

un

=--;

VII

=-,

n+x

n

UII

then

_

n n+x

=

1+~ n

Lt

(_1_] 1; I =

n~""

VII

=

X

1+n


I

I!n

is divergent by p-series test (p =1 )

is divergent.

Un

6.2.16 Example Show that

fSin(!)n

is divergent.

n=1

Solution Take

. (1) sint =

Yn

SIn U Lt --E... n~"" V n

n = n~"" Lt ( ) 1 _

Lt -

I~O

t

(where t

= 1n) = 1

n

Iu I ,

n

VII

.both converge or diverge. But

(p -series test, p

= 1);

:.

I

Un

is divergent.

I

VII

=

I

~

is divergent


400

6.2.17 Example Test the series

L sin

-I ( : )

for convergence.

Solution • _I 1 u =sm _. n n' 1 vn =n

Take

-{~)

u sin ( ()) ( . . _ 1 ) Lt 2..= Lt ( ) ;= Lt - . - =1 Takmgsm '-=() v tHoo 1 8---*0 sm () n

/1---*00

/I

_

n But

L v"

is divergent .Hence

L u"

is divergent.

6.2.18 Example

1

33

22

Show that the series 1 + - 2 + - 3 + - 3 + ..... is divergent.

234

Solution

1

22

33

Neglecting the first term, the series is 1 + - 2 + - 3 + - 4 + .....

n" U

-

-

234 nn n-

.Therefore nn

"~(n+l)"" ~ (n+l)(n+l) ~ n(l+~).n"(I+~J 1

1

Take vn =-

n

1 e

which is finite and

L vn = L!n is divergent by p -series test (p L un is divergent.

=

1)

Sequences of Series

401

6.2.19 Example . 1 3 5 . SIlOW t Ilat t Ile series - - + - - + - - + .......00 IS convergent.

1.2.3

2.3.4

1

3

5

1.2.3

2.3.4

3.4.5

3.4.5

Solution

--+--+--+ .......00

Iii

n

2n-1

term~ u" ~ n(n+l)(n+2) ~ n' ·(I+~)(I+~)

Take

1 n

Vn = - 2

it un

n~oo VII it HOO

Un Vn

= it HOO

=(

LVII =

L n1

-2

_I (2-~ ) (_1 ) n

2

(1 + ~'~)(1+ 2n)

2)(

0

1 +0 1+0

:. By comparison test But

1 ( 2- n )

)

Llln and

=2

n

2

which is finite and non-zero

LVII

is convergent. :.

+

converge or diverge together

LUll

is also convergent.

6.2.20 Example 1

LII~I J;;n + ,J;;+l n+ I 00

Test whether the series

Solution 1 L-==---== J;; +.j;;+1 00

The given series is

11=1

1

is convergent


402

Take

which is finite and non-zero

I

Using comparison test But

I I

Vn =

l

I

Un

and

I

Vn

converge or diverge together.

is divergent (since p

= Ii)

is also divergent.

Un

6.2.21 Example Test fo~ convergence

f[ ~n + 1 - nJ [JNTU 1996,2003, 2003s] 3

n=1

nih term

Un

= n[(1 =

f~L

Then

3:2- 9:5

I

Vn

+ ...... = :2

(1- 9:3 + ......) ;

X=n~Jlj - X

n 3 + ....

:. By comparison test, But

+~))~ -1) .~+ n -1] = n[1 +~+ 3n Jj U~ 1.2 n ..... -1]

I

Un

and

I

Vn

Let

1 n

Vn =-2

)= lj *0

both converge or diverge.

is convergent by p -series test (since p = 2 > 1)

convergent.

6.2.22 Example Show that the series, for p 5, 2

~ + ~P + ~P + ....... JP

2

3

is convergent for p > 2 and divergent

Sequences of Series

403

Solution III

••

n term of the given senes =

U

I

n +1

n ( 1+ In)

n"

n"

= --~ =

=

( 1+ 1~) 11

,,-1

Lt u = l:;t: O· I = _1_. n,,-I '11-"') /v ... L 1111 and L VII both converge or diverge by comparison test. BlIt LVII = L }~>-I converges when p -I> 1 ; i.e., p >2 and diverges when

Let

liS

take v

n/

'

lI

p -1 ~ 1 i.e p 6.2.23 Example

~

2; Hence the result.

+ L (?II 7-3J 3 +1 OCJ

Test for convergence

2 1

11=1

Solution

"" J 2"(1+ 3~,,)r' l3

11

1 (

+

'~3 1 ) J

" Jf. VII U

V

Take

n -

_.

3"'

I )1,3

1+ 3 2 - [ 1 1,' + n

--'!...-

Lt !!A = 1 :;t: 0; :. By comparison test, v

1l-7r:iJ

L un

and

L

VII

behave the same way.

ll

"

BlIt L... vn

=

~(2)"2 = ~- + -2+ (2)3 -

L... 11=1 3

common ratio

3

3

I. I. a geometflc···1 sefles Wit

2

+ ..... , W lIC 1

3

IS

1

)34, «I) :. LVII is convergent. Hence LUll

is convergent.

6.2.24 Example

. Test for convergence of the sefles,

1

4.7.10

+

4

7.10.13

+

9

10.13.16

+ ..... .

Solution

and

4,7, 10, .............. is an A. P; 7, 10, 13, ............ is an A. P; 10, 13, 16 , ............. is an A. P;

til

= 4 + (n -1) 3 = 3n + 1

'II = 7 +(n-l)3 = 3n+4 'II = 10 + ( n - 1) 3 = 311 + 7


404

n2

ll=

n

=---:-----:------:-----;----;----;-

(3n+1)(3n+4)(3n+7)

II

3n(1+

V

" 3n

).3n(1+4

3n

).3n(1+ 7/ ) ·3n

1

27n(1+

=!

Taking vn

Lt IHOO

n

,we get

~ = _1_:;t: 0; v

27

II

1, )(1+ 413n ;' )(1+ 7/ ) /3n

/3n

LUn and L vn

:. By comparison test, both

same manner. But by p -series test,

L

VII

is divergent, since p

=

behave in the I. :.

Lilli

is

L

is

divergent.

6.2.25 Example Test for convergence

",hn2 -Sn+1 ~ 3 2 4n -7n +2

Solution Iii

••

n term of the gIVen senes = un Let

vn

=-2

L

lin

=

.J2n2 - Sn + 1 3 2 4n -7n +2

n

L

l l

nloo ~ = IH~· L

nloo

~/ ~/?

n 2 - / n + / n- n 2 n3 /~ +~<3) x-I

(4-

~/ V J

2 - / n + / n2

(4 _ 7/ +


/n

L un

convergent. [p series test - p

=

2/) 'n

and

3

J

r;; '112

= 4 :;t:

L vn

2 > I] :.

0

both converge or diverge. But

L un

is convergent.

6.2.26 Example

·" Test tI1e senes ~ un ,w hose n Iii term

• ( IS

I .) 4n 2 - I

VII

Sequences of Series

405

Solution

1 Ull

Let

VII

:. I

Ull

and

I

Vn

= (4n 2 -/.);

=-\, n-

Lt 11->00

_llil VII

=Lt lrn Il->ao

2 2

1

n /'n

(

4-

)1=-4 *0 " o

both converge or diverge by comparison test. But

convergent by p -series test (p

= 2 > 1) ; :.

I

UII

is convergent.

6.2.27 Example If un

= (; ).Sin (;), show that

IU

II

is convergent.

Solution Let

VII

=

1 -?

n-

,so that

'" VII is convergent by lJ -series test. ~

J

(1)n - It (-sin I ) (1 n ) - 1l~
ll Lt _" - LI sin

11->00 ( V

- n->ao n

(~J =1 * 0 :. By comparison test, I is convergent.

where t = lin, Thus Lt

Jl~
VII

Un

6.2.28 Example Test for convergence Take vn

I

1tan(l;>

= II ~/; Lt [ulIl ] = 1 * 0

In

2

n--->oo

I VII

Hence by comparison test,

IIIIl converges as I

6.2.29 Example Show that fsin 11=1

(~s in above example)

2(~) is convergent. n

VII

converges.

I

VII

IS


406

Solution

. ?(I) ; Take vn =-?1

Let un =sm- n

Lt

I/~Cf)

( l n-

(~)= Lt

n~oo

V

n

sin 1/

. ')12 =Lt(smt)
2

I~O

,'n

Lt (lin/v/ ) = 12 = 1 0 :. By comparison test, I and I behave the same way. But I is convergent by p- series test, since p 2 > 1; :. I t = II ;

where

In

:;t:

n~oo

n

UI/

Vn

=

Vn

UI/

is convergent.

6.2.30 Example: Show that

f

Vlog (n1/) is divergent.

n=2/

Solution U II

Similarly

= Inlogn' II .

> II . 12log2 72'

log 2 < 1 => 2log 2 < 2 => 1/

}jIOg3> X,.····Xlogn> Yn,nE N

L In 1/ Iogn > L In II ; But L II In

is divergent by p-series test.

By comparison test, given series is divergent. [If

u/I ~ vn"i/n then

IU

n

I

Vn

is divergent and

is divergent.]

(Note: This problem can also be done using Cauchy's integral Test. See ex 1.6.2) 6.2.31 Example 00

I

Test the convergence of the series

(c + n

rr (d + nr' ,

where c, d, r, s are all

n=1

+ve. Solution · Th e n Ih term 0 ft h e senes

Let

1

= un =

vII = -r+.\- Then

n

r

1

(c+n) (d+n)

U/' _

,

1

--------=------

Vn

n

r (

1+

~

J +~- J (1+~J (1+ ~J .n" (

1

Sequences of Series

Lf 11-)00

_

1111 -_

vJl

407

1 --," -;-- 0 :. L)III and " ~ v" both converge are diverge, by comparison

test.

L

But by p-series test,

:. L

II"

converges if (r + .\) > 1 and diverges if (r + s) ~

VII

converges if ( r + s ) > I and diverges if (/"1 s) ~ I.

6.2.32 Example Show that

~

~n

-(1+ 1 )

"is divergent.

1

Solution 11 "

Take

VII

=n

"=--

n.n 1n

1I 1 1 =- ; Lt -" = Lf -

n

11---)00

Il-)ct)

y

11-+00

"

n

In

= 17:- 0

II

VII

11-)00

«(:)

= eO = 1

using L Hospitals rule)

L

1I"

and

diverges (since p

=

I); Hence

By comparison test both

L

V

1 1 1 Lt -v = y say; log y = Lt - - .log n = - Lt -.-!1 = 0 n ,,-too n 1

For let

test,

1

-(1+ 1 )

LV" L

converge or diverge. But p-series lIlI

diverges.

6.2.33 Example

~

(n+ar II~I (n + b)" ( n+ c )"

Test for convergence the series ~

Solution

, a, h,

C

,p, q, r, being +ve.


408

v

Take

II

1 =- . It u = 1"* 0 .' I1P+lJ-r' II~OO V _II

II

Applying comparison test both Rut by p-series test,

I. Hence ~

I

lIlI

I

VII

I

and

lin

I

VII

converge or diverge.

converges if (p+ q -r) > I and diverges if (p + q -r)

converges if (p + q - r) > 1 and diverges if (p + q - r) ~ I.

6.2.34 Example Test the convergence of the following series whose nl" terms are:

(311+4)

(a)

( 211 + 1)( 211 + 3)( 211 + 5)

(c)

1

tan-;

(b)

11

(d)

11

(3 +5 (e)

11 )'

11. 311

Solution (a)

Hillt:Take

VII

=~;Iv" 11-

is convergent;

It

,,~oo

(u,,)=~"*o (Verify) v 8 "

Apply comparison test:

I

lIlI

is convergent [the student is advised to work out this problem fully]

(b)

Proceed as in 1. 2.16;

(c)

Hint: Take v"

1 11

=-2

Iu" ;

is convergent.

(u) = It ((l+J~lr = 3e = 1 "* 0

It -"

,,~oo

VII

II~OO

1 + 3/~

)11

-?

e

e-

v" = -\- is convergent (work out completely for yourself) 11

1

(d) U"

1

~ 3" +5" ~ 5"

II

1

+(~J]

1

;Take v" ~~ ; "If.

(Un)

~ ~l ,,0

Sequences of Series

LUll

409

L

and

VII

behave the same way. But

a geometric series with common ratio

:. LUll (e)

is convergent since it is

~<1 5

is convergent by comparison test.

1 1 - - ~ - \;fn EN, n.3"

L v"

3" '

n.3";:::: 3n ;

since

1

1

L -3;;n.
.....( I)

_1

The series on the R. H .S of (1) is convergent since it is geometric series with 1

r=-<1. 3

.

:. By comparison test

I -I-. n. 3"

IS

convergent.

6.2.35 Example Test the convergence of the following series. (a) (b)

21+~+3 J + J 1+~+32+4 2 +.............. . 1 +2-+3- 1-+2 +3 +4 12+22+3 2 12+22+3 2 +42 1+ 3 + 3 + +............. . 3 3 1 +2 3 1 +2 +3 C+2 3 +3 3 +4 3 1+

~+2J

1 +212+22

+

Solution

(a)

U _

Take

But

(b)

_

3

12 + 22 + 3 2 + .....wJ - n ( n + 1) _on (2 -t-_ 1) - (2 n + 1) 6

n -

I

1+2+3+ .... +n _

(n+ 1) n-2

Un

I

Vn

and Vn

U = Lt ( 3n- ) =-*0 3 Lt _" n ..... '" 2n + 1 2

=-

n

I

n .....'" VII

VII

behave alike by comparison test.

is diverges by p-series test. Hence

I

Un

is divergent.

(2n+ 1) 12+22+ .... +n2 n(n+1)~6-- 2(2n+1) un = 13 + 23- + .....n 3 = ( )2 = 3n (n + 1) 2 n +1 n ---4


410

Hillt: Take vn

=! n

and proceed as in (a) and show that

IU

II

is divergent.

Exercise 6(a) 1. Test for convergence the infinite series whose nih term are: (a)

(b)

1 n-Fn

[Ans: divergent)

.Jn~I-Fn

[Ans : convergent)

n

(c)

.Jn2 + I-n

[Ans : divergent)

(d)

Fn n 2 -1

[Ans : convergent)

(e) (f)

(g) (h)

.Jn3 +1-N 1

[Ans : divergent] [Ans : divergent]

~n(~+I) Fn n 2 +1 2n3 +5 4n 5 +1

[Ans : convergent] [Ans : convergent]

2. Determine whether the following series are convergent or divergent.

1

2

·3

(a)

1 + 3- 1 + 1 + T2 + 1 + T3 + ............

[Ans : divergent]

(b)

12 22 32 2+ IOn r+23+]T+ ........ + n3 +..... .......

[Ans: convergent]

(c)

J1 +.J2 + .J2 +J3 + J3 + J4 +...... ........

[Ans : divergent]

(d)

234 2+-2 +2+ ............... 345

[Ans : divergent]

(e)

2+-3 +4+ .......... 123

1

1

1

1

1

1

[Ans : convergent]

Seq uences of Series

411

~n2 + 1

00

~ ~4n2 +2n+3 f(8 Yn -1) ....................

(f) (g)

(Ans : divergent] (Ans : divergent]

J

3n L 5n

3

00

(h)

+8

5

(Ans : convergent]

...................... ..

+9 123 - + - + - + .......... . 1.3 3.5 5.7 J

(i)

(Ans : divergent]

6.3 6.3.1 0' Alembert's Ratio Test Let (i)

LU

n

beaseriesof+vetermsand(ii) Lt un +J

Then the series

II~OO

L un

un

=k(~O)

is (i) convergent if k < 1 and (ii) divergent if k > I.

Proof:Case(i): Lt un + J =k«I) n~oo

un

From the definition of a limit, it follows that U

3m > 0 and 1(0 < I < 1) 3 ~ < IV n ~ m

un

i.e.,

U m+ J

Um

Um

U m+ 2

.

< I ,..........

U m+ J

+ Um+J + Um+ 2 + .............. =

Um

[1 + uu

m

+

J

+ um+ 2 + ..... ]

m

um

< um(1 +1 +/2 + ...) = um.-I-(I < 1) I-I 1 But

Um . - -

I-I

L un 00

is a finite quantity:.

is convergent

By adding a finite number of terms u J + u2 + ...... + um- J ' the convergence of the 00

series is unaltered.

L un n=m

is convergent.


412

C(l!t'e (ii)

U

Lt ---.!!.±l. = k > 1 n-->oo

11n

There may be some finite number of terms in the beginning which do not satisfy the condition ull +' 21. In such a case we can find a number 'm'

3

Ull

ll

UII

+ 112 + llj + ......... , we have

Omitting the first 'm' terms, if we write the series as u,

!!2. 2 1, u3 u,

U,

u +' 21, 'lin 2 m

112

21, u4 2 1 .......... and so on

u3

+ 112 + ...... + till = u,

(1 + u + 2

ll,

113 tl2

.!!2. + ..... J

(to 11 terms)

u,

2 ull (1 + I + 1.1 + ....... to n terms) II

Lt n~oo

Iu" 2 Lt n.u, which ~ 00; :. IU 11=1

Il

Il~CO

is divergent

6.3.2 Note: 1 The ratio test fails when k = 1. As an example, consider the series,

I ~n n='

p

u

n P n + 1)

Lt ---.!!.±l. = Lt - -

Here

n-->oo

U

n-->oo (

n

1 --

= Lt

1+ 1/ )

n-->oo (

=1

,In

k = I for all val ues of p,

i.e.,

But the series is convergent if p > 1 and divergent if p::; 1, which shows that when

k = 1, the series may converge or diverge and hence the test fails.

Note: 2 Ratio lest can also be stated as follows: If

I

Ull

is series of +ve terms and if Lt 11-->00

~ = k ,then u +'

If k > 1 and divergent if k < 1 (the test tilib

ll

'.I.

hen k

=

I).

I

Un

is convergent

Sequences of Series -~---~

~-----

----~----

413

-----

Solved Examples Tests for convergence of Series

6.3.3 Example X

X

X

\

-+--+---+ ................ .

(a)

1.2

2.3

3.4

Solution x" 111/

Xl/II

= n (11 + 1) ;111/+1 = (11 + 1)( n + 2)

~/!.!..'.l=

Xl/+I

(n+

1/"

Therefore

II

LI

_1/_,1

11----)00

1111

:. By ratio test

Ixl>

r

1

n(n+1)

1)(n + 2

,

~ (I + ~

x"

r.

=X

IUI/

is convergent When

Ixl

<

and divergent when

I;

When x = I, ul/

=

1, 1 II ) ; fake vn = - 2 ; Lt _1/ = 1 n- 1+ lin n v1/ . J (

11->00

:. By comparison test Hence

I

IUI/

is convergent.

Ixl ~ 1

Un is convergent when

and divergent when

Ixl > 1

3

1+ 3x + 5x 2 + 7 x + ...... .

(b) Solution

Un

= (2n-l)xn-l;ul/+

1

=(2n+l)x"

ul/+ I L t (2n 1) x=x -+ L1-= 1/-->00

:. By ratio test

lin

I

1/-->00

2n-l

Un is convergent when

Ixl > 1 When x

= 1: ul/ = 211 -1; Lt un = 00; 11----)00

Hence IlIl/ is convergent when 00

(c)

1/

I-+-··········· n- + 1 1/;1

:.

I

UI/

Ixl < 1

and divergent when

is divergent.

Ixl < 1 and divergent when Ixl ~ 1


414

Solution U n

x" n +1' (

-;;: =

(n+l)2 +1

n+1

n +1 ) n 2+ 2n + 2 x 2

Lt un +1 = Lt

n~'"

n

(

1 +

>;:2)

II~'" n2(1+~+2) 2

Un

n

:. By ratio test,

When

I

=----

U

2

un +1

Hence

X"+

=2 -- .

L un

is convergent when

1 x=I'u = - - . Take • n n2 + 1 '

LU


n

(x)=x

n

Ixl < 1

and divergent when

Ixl > 1

1

v =n n2 is convergent when

Ixl ~ 1

and divergent when

Ixl >1 6.3.4 Example Test the series

f (n:n -1}n, x> +1

n~'"

0 for convergence.

Solution

U" =(:::: )X";U"" = Lt un+1 n~'" un

[~:: :i: :: ]X'HI 2

2

= Lt [( 2n + 2n ) ( n2 + 1)] x n~'" n + 2n + 2 (n - 1) . 4 n (1+2/n)(1 +l/n) =nIf", [ n4(l+2/n+2/n2)(1-l/n)

:. By ratio test,

LU

n

is convergent when

n

n2 -1 n +1

= 2- -

=x

x< 1 and divergent when x> 1 when

x = 1, U

1

Take vn

1 n

=-0

Sequences of Series

415

Applying p-series and comparison test, it can be seen that

L un is divergent when

x= 1.

.'. L un

is convergent when x < 1 and divergent x ~ 1

6.3.5 Example

2P 3P 4P + l] +

Show that the series 1+ ~

l1 + ..... , is convergent for all values ofp.

Solution

Lt .un +1

n~oo

=

un

l.fl]=

Lt [(n+IY x In + 1 nP

n~oo

= Lt ( n~oo

L

lll/

1

Lt {

n~oo

(I

) x Lt 1+ - )P n + 1 n~oo n

1

(n + I)

(n+I)P} n

= 0 < 1;

is convergent for all 'p'.

6.3.6 Example Test the convergence ofthe following series

1 F

1 3

1 5

1 7

-+-+-+-+ P P P ........... . Solution

u = 1/

ul/+ 1

un

1

(2n-IY'

=

1

·u

=---

(2n+ly (2n-Iy _ 2P.n P(I-l/2nY. (2n+ly - 2PnP(I+l/2ny , 1/+1

Lt un+1 =1 HOO

un

.'. Ratio test fails.

1 = _. n nP'

Take v

U

-E... vn

=

nP

(2n -I)P

=

which is non - zero and finite 7 .'. By comparison test,

LU

n

ar.d

1

U . Lt -E... = -1P ( I)P , I/~OO V 2 , 2 P 1-n 2n

L

vn

both converge or both diverge.

EngineerinG M~th~ni1!!l'S~

·H6

I

But by p - series test,

-=

VII

I -I

'~onverges when p > I and di\ c,

11"

when p ~ I

:. I

iI"

I and divergent if p ~ I .

is convergent if /'

6.3.7 Example Test the com crgence of till'

" ' (n -1- l).\""

"cric~

L.. - 1/,

-,--.X

I

>0

11

Solution

( n + I) x" lin =

( n + 2) X"-l I

1

;1I1Itl - - - - , - , -

(11+1)

n

; I

By

i'".1i

j()

illi rl

LI

--

11->''0

111/

test,

I

=

II

..: ' ! ,,- _.. ,

L,

" ,.w

UII

~

;

1I

1

i

: , "" - - - ,

.x = X

+ 11 ( 1+ 1 )' n, \.. n

converges when x < 1 and diverges when x > 1 .

n+ I

When ..x= I ' IIII = n3Take v"

.'. I

= -~ ; By comparison test n

UII

I

UII

.is convergent ( give proof)

is convergent if x ~ 1 and divergent if x > I.

6.3.8 Example Test the convergence of the series

. ~(n2

1) (.. 1 2.5.8 2.5.8.11 ... ) 1 1.2 1.2.3 II) + - - + + ... (III - + - + - - + n 1.5.9 1.5.9.13 3 3.5 3.5.7

(I) L.. -+-2 II_I

2"

Solution

2 1 (n2 1) n (i) I ---;+-2 I-II + I-2 2 n 2 n 00

00

00

11=1

11=1

=

II-I

Let

:'t_ '

Sequences of Series

417

= (n+1)2

U

.1l11+1

2"rl'

11+1

2"tl·

U"

11

2

1(1 +-1)2 --< _1 1

_ L Lt1111+1 -t-. 2

I1~OO lin

convergent. :. The given series

2

n

11-+00

By ratio test

(ii)

= (n+1)2 ~

I

Un

(I

Un

is convergent .By p -series test,

+

I

v,,)

I

v"

IS

term

=

is convergent.

Neglecting the first term, the series can be taken as,

2.5.8 2.5.8.11 --+ + 1.5.9 1.5.9.13 Here, 1sl term has 3 fractions ,2 nd term has 4 fractions and so on . :. nih term contains (11 + 2) fractions 2.5. 8....... are in A. P.

t ( n + 2t (n + 2

term = 2 + ( 11 + 1 ) 3 = 3n + 5 ;

1. 5. 9, ....... are in A. P.

U

term = 1 + ( n + I ) 4 = 4n + 5

2.5.8 ..... (3n+5) 1.5.9 ..... ( 4n + 5)

=----'--~

n

un+1

2.5.8 ..... (3n+ 5)(3n+ 8) = 1.5.9 ..... (4n+5 )( 4n+9 )

un +1 _ (3n+8) . -;,:- - (4n + 9) ,

n(3+~)

Lt u +1 = Lt n HOO un HOO n ( 4 + By ratio test, (iii)

I

Un

<1

4

is convergent.

1,2,3, ........ are in A. P

2n+ 1

=l

!)

l1

nih

tenn =

n ;

3.5. 7 .......... are in A.P.

d"


418

1.2.3 .....n [ un = 3.5.7 ..... (2n+l)

1.2.3 .....n(n+l) = [ 3.5.7 ..... (2n+l)(2n+3)

n 1

U + Il

U +

U

1

1

=(~)

I

l

2n+3

n.(l+!) Lt

UIl +I

n~oo

= Lt

Un

/HOO

L un

:. By ratio test,

n =

n(2+~)

! <1 2

is convergent.

6.3.9 Exercise

L

Test for the convergence of the series

1.2.3.(....n ) 3.5.7 ...... 2n+ 1

Solution

1.2.3 ...... n 1.2.3 ...... (n+l) un = 3.5.7 ...... ( ) ; U n+1 = ); 2n+l 3.5.7 .....: ( 2n+3 U n +1

un :. By ratio test,

= n+l . 2n+3'

LUll

Lt /HOO

= Lt n(l+ un HOO n( 2+

U,,+I

/~) =!
is convergent.

6.3.10 Example

~ 1.3.5 .... (2n-l)

Test for convergence L.. 11=1

n

I(

.x - x> 0

)

2.4.6 .... 2n

Solution

1.3.5 .... (2n-l)

The given series of +ve terms has u"

=

1.3.5 .... (2n+l)

n

and

U,,+I

= 2.4.6 .... (2n+2 ) x

2.4.6 .... 2n

nl

.x -

Sequences of Series

419

2 1 x = Lt ( ~ //-->00 2n + 2 )

Lt 1111+1 U//

IHOO

I

By ratio test,

x

= ),

u//

IHOO

)

2n .x=x 2n( 1+ 22n)

is converges when x < 1 and diverges when x > I when

the test fails.

Then

=

U

1.3.5 .... (2n -1)

2.4.6 ..... 2n

II

:. I

2n(1 + 1

Lt

=

U/I

is divergent. Hence

<1

Lt u//

and

;f::.

0

11-700

I

is convergent when x < ), and divergent

UII

when x ~ 1

6.3.11 Example

2// -2 ) x /I-I + ..... ( x > 0) Test for the convergence of 1+ -2 x + -6 x 2 + ........ + ( 5

2/1+1

9

Solution

II,

Om itting 1" tcnn,

=( ~: :~ ) x'-' ,(n " 2)

'II,' arc all eve.

and

= (2"+1 - 2) x/l. Lt (~) = Lt (211+1 - 2) x ( 2// + 1) X

U IHI

(

2//+1 + 1)

'/1-->00

u

11-->00· 2"+1 + 1

ll

2n+1 (1- 12/1) 2// ( 1+ 12/1) =

//~~ [ 2//+1 (I +

Hence, by ratio test, When x Hence

I

Un

Un

1

l''2/1+1 )·2/1 (1- 2, 2/1 ).x = x ;

converges if x < 1 and diverges if x> 1.

= ), the test fails. Then

I

u/I

2/1-2 2 +1

= -n - ; Lt

un

= 1 ;f::. 0;

IHOO

is convergent when x < ) and divergent x > 1

6.3.12 Example

I

00

Using ratio test show that the series

/1=0

(3 -4i)" n!

converges

Solution

_ (3-4i)" un -

2/1 - 2 .

I .

_(3 -4i),,+1 /

In! ,un+1 -

. /(n+l)! '

:.

I

Un

diverges

Engmeerlng ---------------------------------=--

420

MC3tht::rnatl~~

1II1tl)_ 1 I (3-4i)_0 ---<1 1 t ( --llil ":'h n + 1 -

II ',,,,

II

Hence, by ratio test,

l:

l:onverges.

1111

6.3.13 Example

. . - 2 x + - 3 x 2 + - 4 x 3 + ........00 (x0 DlsclIss t Ile nature 0 f t he series, >)

3.4

4.5

5.6

Solution Since x > 0 , the series is of +ve terms; llil

(n + 1) = (n+2)(n+3)x

II

> Zln+1

[(n + 2)2.X

Lt IIn+ I II~OO - - ; ; : -

(

HOO

r 2 (1 + 7;;)2 n

l

I

2

(

n 1+

Un

.

n

=!. n

'

.. I :. I

Un Un

]

Yn + j~2 )

= X.

,

=(

(n+l) )( ) ;

n+2 n+3

Lt = un = 1 :;t: 0 vn

n~oo

:. By comparison test by p-series test. (p

.x

converges if x < 1 and diverges if x> 1

When x = 1, the t(.st falls; Then un Taking v

]

n + 1)( n + 4)

= Lt

Therefore by ratio test,

(n + 2) X/HI = (n+3)(n+4)

IU

n

and

I

Vn

behave same way. But

= 1);

is diverges when x

=1

is convergent when x < 1 and divergent when x ~ 1

6.3.14 Example

.

.

~

DIscuss the nature of the senes ~

3.6.9 .... .3n.5n ( )( ) 4.7.10 ..... 3n+l 3n+2

I

Vn

is divergent

Sequences of Series

421

Solution Here,

=

U n

3.6.9 ..... 3n 5" . 4.7.10 ..... (3n+l)(3n+2)' 3.6.9 .... .3n(3n + 3)5"+ 1

U II + I

= 4.7.1O ..... ( )( ) (3n+5 ); 3n+l 3n+4

Lt

11 I ~

11->00

llil

= Lt (3n+2)(3n+3).5

--'-----_-'--c-'---,-----_'---:-

5.9n2 (1 + 2 3" )(1 + 3/3' Inn 2 9n + 4 3n )( + 5, 3n)

Lt

=

11->00

I

:. By ratio test,

(3n+4)(3n+5)

11->00

(1

f

1

)1 = 5 > 1

is divergent.

lin

6.3.15 Example 00

I

Test for convergence the series

n l - II

Solution

= n ; u +1 = (1)-11 n+ ; I-II

1I11

lI

':;,:' = = (n;I,!"

Lt un+1 = Lt

n~'" U

I

Un ,

n

1+ 1,'

,'n

is convergent

6.3.16 Example

I

00

Test the series

11=1

2

3

~ ,for convergence. ~

Solution

2n 3 Un

= ~ ;

U II

=~C:J

!.[_l_,]" =O}=O<1

11->00

II

:. By ratio test

n(n:l)"

+I

2(n+ 1)3 = In + 1

e


422

lln+1

= 2(n+l)3 x

~

un

In + 1

2n

u

Lt

= (n+l)2 = (1+;~r

n3

3

n

= 0 < 1;

--.!!..±l

n-+oo Un

L u/l

:. By ratio test,

is convergent.

6.3.17 Example

2/1 ,

Test convergence of the series

L~ nn

Solution

n 1

+ (n + I)! /1 ( ~ )11 ~=2 (n + 1)"+1 . 211 n! n+ 1

un +1 = 2

1111

U

Lt

--.!!..±l

IHOO

ull

:. By ratio test,

= 2 Lt

LUll

(e)

3n + 7n ( 5n 9 +

2

11

Lt n-+oo

L un (b)

1.2.3 ....n )2 ( 4.7.10 .... .3n+3

(a)

< 1 (since 2 < e < 3)

e

where u/I is

x n- I

(c)

Solution

=-

is convergent.

n2 + 1 3/1 +1

3

2

Hoo(l+ )-~r

6.3.20 Example Test the convergence of the series (a)

1

I)

(d)

--,(a>O) (2n+lf ~ ~1+3n

)xn

u = Lt [(n+l)2 +1 x3/1-+1] (~ ull n-+oo 3n+1 + 1 n2 + 1

423

Seq uences of Series

1

-<1 3

=

LUll

:. By ratio test,

Lt

(b)

is convergent.

(u )= Lt [ (2n + 3r x~(2n+lf] x n+ I

11-->00

xn

un

n I -

n-->oo

aa

= Lt IHOO

By ratio test,

LUll

[ 2a na (l+ }Snf

convergence if x < 1 and diverges if x> 1.

When x = 1, the test fails; Then,

=

1I /I

Lt n-->OO

(Un) = Vn

(_n_)a =

Lt n-->oo

l

2 n (I+!' )a . 2n .x = X

2n + 1

un

If

; Taking v

1 = _1 a 2 + ;!~r 2

Lt n-->oo (

L


1 (2n+

and

L

VII

= _1a

II

* 0 and finite

have same property

But p -series test, we have

and

(i)

LVII convergent when a> 1

(ii)

divergent when a

(ii)

LU x > 1, L

(iii)

x

:. To sum up, (i)

(iv)

x < 1,

n

un

~

1

is convergent Va. is divergent Va.

L x = 1, a ~ 1, L =

1, a> 1,

un

un

is convergent, and is divergent.

awe have,

n

(since a > 0).


424

Lt

U

I

~= IHoc> ltll

(c)

IHoc>

Lt [

=

IHoo

:. By ratio test,

[

Lt

I.2.3 ....n(n+I) 4.7.1O .... (3n+3)]2 x-----'--------<4.7.IO .... (311+3)(3n+6) 1.2.3 ....n

(n + I) X]2 3 ( n + 2)

IlIlI

=! lIlI

LI

(e)

U

= Lt

--"..±.!..

n->oc>

LI

lr 3 ( n + 1)3 + 7 ( n + I

t

S( n + 1 + II

(I

{(I + .1/)3 +2_(1 + 1/)2} n 3n n

n->oc>

Sn :. By ratio test,

9

I

{(

Un

Sn 9 + II ] xx 3 3n + 7

x

~ )3 +7n2(1+ .1.1)2 Sn9(I+l!'s/ 9) ] .n n x n xX 3 9 Sn (I + V)9 + II 3n + 7;1 .n / 3n i

Lt

r

f3n3(1+

n->oc>

3

is convergent.

Hoc>

= Lt

U"

3n =

U"

I+ }~r + SI~_}

(1 I/VS

Sn 9 +

n

x

3n

3

(1 + .%n

U

= 3n3(1+ jjn)

" Taking

Vn =

Sn

9 (

1+ IYsn9)

=~ 6

Sn

~, we observe that n

3)

) xx

=X

3)

converges when x < 1 and diverges when x > 1.

When x = 1, the test fails, Then

9

1

(1+ l~n)

(I + IYsn

Lt

U" =

"->OC>~

9)

l;f: 0 S

Sequences of Series

425

:. By comparison test and p series test, we conclude that

:. L u"

L un

is convergent.

is convergent when x::::; 1 and divergent when x > 1.

Exercise - 1(b) 1. Test the convergency or divergency of the series whose general term is :

x"

(b)

nx n-I .......................... .

Ixl < legl, Ix ~Ildgt ] I Ans : Ixl < legt, Ix ~Ildgt I

(c)

(~:.

I Ans :

Ixl < legl, Ix ~Ildgt

I

(d)

(:::: )X" ......... .

[ Ans :

Ixl < legl, Ix ~Ildgt

I

(e)

lfl

[ Ans: cgt.]

(f)

4"·lfl

[ Ans: dgt.]

(a)

(g)

I Ans:

n

:n

X

"'

•••••

nil (

n 3 +1 )"

[ Ans: cgt.]

(3 + 1) n

2. Determine whether the following series are convergent or divergent :

(a) (b)

x3

x2

x

- + - + - + .............. 1.2 3.4 5.6 X x2 x3 1+-2+-2 +-2 + ............. . 234 1

x2

X

(Ans:

Ixl::::; legl,lxl > ldgt

]

I Ans :

Ixl : : ; legt, Ixl > Idgt

]

( Ans :

Ixl::::; legt, Ixl > Idgl

] J

(c)

- - + - - + - - + ..... . 1.2.3 4.5.6 7.8.9

(d)

1+-+-+-+ ..... - 2- + ....

(Ans:

Ixl::::; legt,lxl > ldgt

(e)

1.2 2.3 3.4 -+-2 +-3 + .............. .

[Ans:

Ixl > legt,lxl::::; Idgt]

X

X

2

x

2

5

x

X

3

10

x

x

n

n +1


426

6.4

Raabe's Test

Let

I u" be series of +ve terms and let E!. {n(:.:, - I)} ~ k

Then (i) If k > 1, ~>n is convergent. (ii) If k < 1,

L

un

is divergent. (The test fails if k = 1)

Proof: Consider the series

Case (i): In this case,

Lt n{~-l} =k

n-+oo

U + n 1

>1

We choose a number 'p' 3 k > p > 1 ; Comparing the series which is convergent, we get that

L

un

L

U"

with

L

will converge if after some fixed number

of terms un

Vn

U + n 1

i.e. If.,

> Vn+1

Un)

i.e., If

(n+ l)P

= -;;-

n ( - - 1 > p+ Un+1

p(p-l) 1 .-+ ......... from (I) L2 n

Lt n(~-I» P

n-+oo

U + n 1

Vn

Sequences of Series

427

Llln

i.e., If k > p, which is true. Hence

is convergent .The second case also

can be proved similarly.

Solved Examples 6.4.1 Example Test fe>r convergence the series

1

x 3

1.3

x

x 7

1.3.5

x+-.-+-.-+--.-+ ..... 2 3 2.4 5 2.4.6 7 Solution Neglecting the first tern ,the series can be taken as ,

1

x

3

1.3

x

5

1.3.5

x

7

-.-+-.-+--.-+ ..... 2 3

2.4 5

2.4.6 7 1.3.5 .... are in A.P. nih term = 1+(n-1)2=2n-1 2.4.6 ... are in A.p. n'h term

2 + (n

=

-1) 2 = 2n

3.5.7 ..... are in A.P nih term = 3+(n-1)2 = 2n+ I uti ( nih term of the series)

=

1.3.5 .... (2n-1)

X

21l 1

+

=----'----;------::--"--

2.4.6 .... (2n) 1l1l+1

=

1.3.5 .... (2n-1)(2n+ 1)

() 2.4.6 .... ( 2n)2n+2

X

2n

+

2n+1

3

. 2n+3

1.3.5 .... (2n+1) X 2n +3 2.4.6 ... .2n (2n+1) -;;:- 2.4.6 .... (2n+2)"(2n+3)"1.3.5 .... (2n-1)" X 2n + 1 lIn+1 _

(2n + 1)2 x 2 ( 2n + 2) ( 2n + 3)

1)2

4n-1 ( 1+-2n

II

LI ~ = LI IHOO

:. By ratio test, If

LU"

Ixl = 1 the test fails.

Un

n~oo 4n

converges if

2

(

1+ 22~ )(1+ ~~ )

x

2

1

= X-

lxl < 1 and diverges if Ixl > 1


428

Then

=1

x2

~ = (2n + 2)( 2n + 3 )

and

(2n+

1111+1

1r

~ -1 = (211 + 2)( 2n + 3) _ 1 = 6n + 5 (2n+1)2

111/+1

ul/ n -

LI tHOO

{ (

ul/+

1J}

(2n+l)2

1

2 6n + 5n I/~OO 4n 2 + 4n + 1

= l I (

1

n Lt

=

11-->00

By Raabe's test,

Ixl ~ 1

I

UI/

n

2

2 (

~)

6+

3 =- >

(4 + ~n + _n-\ ) 2

1

converges. Hence the given series is convergent when

an divergent when

Ixl > 1 .

6.4.2 Example Test for the convergence of the series

3 7

3.6 7.10

2

1+-x+--x +

3.6.9 3 x + ..... ;x> 0 7.10.13

Solution Neglecting the first term,

=

ul/

3.6.9 .... 3n 1/ .x 7.10.13 .... 3n+4 3.6.9 .... 3n(3n+3)

= 7.10.13 .... ( ) ( 3n+7 ).x 3n+4

lIl/+1

= 3n + 3.x

:. By ratio test,

; Lt

3n + 7

lIl/

When x

1/+1

ul/+ 1

I

UII

lIll+1

is convergent when x I.

= I The ratio test fails. Then u 3n + 7. lin 1 _ ll

_

1I1/+1 -

=X

lIlI

1/-->00

3n+3' lIn+1

-

-

4 3n+3

LI {n(~-l)}= Lt (~)=i>l

11-'>00

lln+1

11-->00

3n+3

3

Sequences of Series

429

--~-----------------------------------------------------------

LUll

:. By Raabe's test,

is convergent .Hence the given series converges if x ~ 1

and diverges if x> 1.

6.4.3 Example Examine the convergence of the series

f 12.5 2.92.... (4n-3)2 42.g2.122 .... (4n)2

11=1

Solution

3)2 ~ ~ 4-.8-.12-.... 4n )2

12.5 2.92•••• ( 4n -

= ---------'-------'-

U II

7

(

12.5 2.9 2.... ( 4n -

litH

3)2 (4n + 1)2 1= -~--~---2--'---(-----')2'---(--'--------'-)24-.8-.12 .... 4n

Lt u

lI

+

1

=

un

n->oo

:. The ratio test fails. Hence by Raabe's test,

4n+4

( 4n + 1)2 ~ 11->00 ( 4n + 4 Lt

LU

t

n

= 1 (verify)

is convergent.

6.4.4 Example Find the nature of the series

L (l!:!)2 x

n

~

,

(x > 0)

Solution ull

=

ulI +1

(l!:!)2 ~

ull

n +1

:. By ratio test,

(l~±l)2 11+1

(n+l)2 X' (2n+l)(2n+2) '

_

Lt u 11->00 U

II

.x ; un + 1 = 12n+2 .x

n

L un

=

n2 ( 1+ 1/ ')2 ;' n

Lt n->004n2(1+}/ , 2n

)(1+2/) 2n

.x = ~

4

converges when:! < 1, i. e ; x < 4; and diverges when x >4;

When x = 4, the test fails.


430

x = 4 ~ = (2n + 1)( 2n + 2 ) '1l11+1 4(n+l)2

When

-2n-2

U

r

_" -1=

LI

Hence

I

Un

[n(~-11]= -12 <1 li

I

:. By ratio test,

. 2 ( n + 1) ,

4 (n + 1

lln+1

II-->if.>

-1

=

ntl

is divergent

lin

is convergent when x < 4 and divergent when x> 4

6.4.5 Example Test for convergence of the series

,,4.7 .... (3n+ 1)

~

1.2.3 .... n

n

x-

Solution

_ 4.7 .... (3n+1) lin -

n.

X

1.2.3 ....n

_

4.7 .... (3n+l)(3n+4)

'U n + 1 -

II Lt ~ = Lt [(3n+4) .x] n-->oo ( n + 1)

()

1.2.3 ....n n + 1 .

n+1

X

= 3x

n-->oo Un

:. By ratio test

I

Un

converges if 3x

x=

When

x ~:3' n u.:, -1

" lin ~

II

]

~n

[(n+ 1)3

3n + 1 -1

]

~

[-1 ]

n 3n + 4

n[~-I] = -!3 < 1

:. By Raabe's test, :.

3

3

1 [

Lt

3

!, the test fails

If

n-->
<1 Le., X ! ;

II

11+1

IU

n

is divergent.

. convergent wl IS len I x
3

1 ~-

3

~-

1

(3+ ~

1

Sequences of Series

431

6.4.6 Example

,

1

4x 2 5x ++ ~ + ........... (x > 0 ) 2 3 4

3x

1 est for convergence 2 + -

Solution The

nih

=(n+l)x

II

term

H

I •

tl -

'

11

_ (n+2) (11+1) X

.tl.

II

Hel

-

11

Lt

= LI

is.convergent if x I

II

Ifx= I, the test fails.

Lt

11

[

II

_tl -

1] -1 [ n(n+2) 1 =LtI1 =0<1 [11 ( 111] +

1 = Lt

IIl/el

II--'>UJ

(11 + 1)-

11

,I/---W'

2)

11->00

:. By Raabe's test

:. I

Zlil

X

2

11

IU

Then

1(1+ 2) 11 .x = (1 1 11 + )1

H~-'UJ

11 tl

:. By ratio test,

_11(11+2) (11+1)2 .x

lr

I

_H_'

H-N,

II'HI

'--;;,:-

III//

is divergent

is convergent when x < I and divergent when

x 21

6.4.7 Example

.

.

3 4

3.6 3.6.9 + + ......00 4.7 4.7.10

FlI1d the nature of the series - + -

Solution

3.6.9 ..... 311(311 + 3)

3.6.9 ..... 3n 1I

11

= 4.7.10 ..... ();1I1/+ = 4.7.10 ..... ( )( ) 3n+l 311+1 311+4

1111+1

ZlII

1

3 3 =~. LI llll+1 3n + 4' /1--'>00 llll

Ratio test fails.

LI 11 { - [ ,lll/+1 Zltl

tl--'>oo

I}]

) 3n = 1 11->00 3n( 1+ 4 ) 3n

= LI

3n(I+3

- LI [ n (3n+4 --/1--'>00 3n + 3

I)]


432

= Lt

11

,,->003(n+1)

:. By Raabe's test

L

U"

=

n Lt 11->003n(1+,]n)

1

=- <1

3

is divergent.

6.4.8 Example If p, q > () and the series

1 p 1.3.p(p+1) 1.3.5 p(p+1)(p+2) 1+"2 q + 2.4.q(q + 1) + 2.4.6 q(q + l)(q +2) + .... is convergent, find the relation to be satisfied by p and q. Solution

un

=

1.3.5 ..... (2n-1) p(p+l) ..... (p+n-1) . I [negle"tmg l' term 1 2.4.6 .... .2n q(q + 1) ..... (q + n-1)

lI"tl =

1.3.5 ..... (2n-1)(2n + 1) p(p+ l) ..... (p + n -l)(p+ n) 2.4.6 .....2n(2n+ 2) q( q + 1) ..... ( q + n-1)( q + n)

U,,+l _

--;z- Lt 11->00

(2n+1) (p+n). (2n + 2) (q + n) ,

~ = Lt [2n (1+ 12n ) n (1+ ~'~ ) J= 1 lI"

11->00

2n( 1+ 1"2n)" n( 1+ ~~)

:. ratio test fails. Let us apply Raabe's test

,~.H::':, ~1)]~ ,~.H ~::~)~~::~~ ~1}] 1 Lt [n {2q ( n + 1) - ~ (2n + ) + nIl n 2 ( 1+ ~/~ ) ( 2 + },~)

11->00

r

Lt 2q(1+ )~)-p(2+ II->ool 2

J~)+11=2q-2p+l 2

433

Sequences of Series

Since

~

.

L..,.llil IS

convergent, by Raabe's test,

Yi,

=> q - p >

2q-2p+l >1 2

is the required relation.

Exercise 1 (c) 00

1. Test whether the series

I

Un

is convergent or divergent where

I

II II

=

2 2.4 2.6 2..... ( 2n - 2 )2 3.4.5 ..... (2n-I)(2n)

2n

X

[ADS:

Ixl:::; legl, Ixl > Idgl ]

2. Test for the convergence the series

f 4 .7 . 10 ..... (3n+l) x" I

[ADS:

lfl

1 I Ixl < -egl,lxl ~ -dg/] 3 3

[ Ans : divergent] (ii)

3.4 5.6 3 + ..... ( x> 0) x +4.5 - x 2 +-x 1.2 2.3 3.4 [ ADS: cgt if x :::; 1 dgt if, x > 1 ]

(iii)

~ 1.3.5 ..... (2n -1) xn ( ) L.. . ( ) x>O 2.4.6 ..... 2n 2n + 2 [ ADS: cgt if x :::; 1 dgt if, x > 1

(iv)

X + (1l)2 x 1+ (U)2 II l1

2

+

(lJ)2 x

l2

3

+

(x > 0)

..... .

[ ADS: cgt if x < 4 and dgt if, x ~ 4

6.5

Cauchy's Root Test Let

L un

I'

be a series of +ve terms and let LI u/n = I . Then

when I < 1 and divergent when I> 1

n-->co

L un

is convergent


434

Proof: I,'

II

Lt u/" =1<1=>3a +venumber 'A'(!rn

(i)

I1~OO

Since A < 1,

(or) u ll rn

I

A" is a geometric series with common ratio 00

UII

/11 = I> 1

:. By the definition of a limit we can find a number r 3

u/ > 1Vn > r 'II

i.e., ull > Vn > r i.e., after the 1sl 'r ' terms, each term is> 1.

Lt 11-->00

Note: When

Lt{ In) Un

=

2>11 =

00

LUll

:.

is divergent.

1, the root test can't decide the nature of

LUll' The

fact

n-->oo

of this statement can be observed by the following two examples.

1. Consider the series

I Yn3n :-LfU'/" =Lt 1/

11-->00

2. Consider the series

L Yn,

(

11-->00

in which

1 ) J~

-3

n

= Lt ( V1 11-->00

nl II

LfUlly" = Lt ~ = 1 11-->00

11-->00

n11l

1/

In both the examples 'given above,

LfU/" = 1. But series (1) is convergent

(p-series test) And series (2) is divergent. Hence when the limit= 1, the test fails.

Solved Examples 6.5.1 Example Test for convergence the infinite series whose (i)

1

n'"

(ii)

1

Oogn)"

)3 =1

nIh

(iii)

terms are: 1

[1 +~r

Sequences of Series

435

Solution

. Lt u ,

By root test

(ii)

u

=

n

I

Un

n~co

,

1n

II

1

= lI~oo Lt -n2 = 0 < 1· '

is convergent.

1 ~. 1 ·u n = - - . (log n)'" n log n '

By root test,

I

Un

,.

Lt u 'n

=

n

n->oo

1 n->oo log n

Lt --=0<1·

is convergent.

(iii)

6.5.2 Example Find whether the following series are convergent or ,) i vergent. (i)

'" 1 I-n-1 n=13

Solution 1/ In

1

1/

7n

Lt un Yn

n~ct)

By root test, (ii)

1

= n----+oo Lt

I

Un

is convergent.

111 1+-2 +3+-4 + .....

234

1 3

=-<1·

'

'


436

Solution

1 ' Lt U in 1/ == Il~OO Lt ( - 1 )}~ = 0 < 1 nil 'Il~OO n nil

1I 11

By root test,

I

lin

is convergent.

(iii) Solution =

U n

Lt

[(n+1)x n"+1

u};, =

n---j.c:/J

r

-'=--'----':--=----

n

r

II

[{(n+1)x ],'n

Lt

n"+ 1

Il~OO

~

1]), _ Lt [{ ( n + ] ) X}" .n n

n-->oo

n + 1) Lt ( - x '1I-'

n-->oo

n

nl n

(l+!)X.+, = Lt x.+,=x n n'II n-->oo n/ n Un is convergent if Ixl < 1 and divergent if Ixl > 1 and when Ixi = 1 the test Lt

n-->oo

:. I fails. Then

Take

r

1 n

vn =-

r

Un = (n + 1 .n = (n + 1 = (1 + !)n ; vn n"+ 1 nn n :. By comparison test,

(I

Uti

11-->00 VII

is divergent.

diverges by p -series test)

Vn

Hence

I

U

I

Un

is convergent if

Ixl < 1 and divergent Ixl ~ 1

6.5.3 Example ,/

If un =

n

(n+l)

,,2'

show that

I

Ull

is convergent

Lt-"=e>l

437

Sequences of Series Ill

l

2

I n n II n /1 Lt u" " = Lt 2 ; = Lt = " = Lt (-)-" "~a) II~OO n + 1)" ,,~oo n + 1) ,,~oo n + 1

(

J

(

~

Lt [_1_]" = < 1 ;:. "u" converges by root test. ,,~OO l I e L. +n

=

r

6.5.4 Example

J

~ + ( % + ( % + .' ......

Establish the convergence of the series Solution U

=

11

(_n_)" . . . . . 2n+l

Lt u };,

,,~oo n

By root test,

I

UII

(verify);

= Lt (_n_)=!<1 2n + 1

II~OO

2

is convergt:nt.

6.5.5 Example Test for the convergence of

Ioog --.x

/I

n+ 1

11=1

Solution

U/I

=

[_11 ]~ .x"; 1+n

Lt u"·v,, ,,~oo

= Lt

Il---+oo

[_11 JI 1+n

x

=x

IU is convergent if Ixl < 1 and divergent if Ixl > 1. When Ixl = 1 : un = J n ,taking vn = ~ and applying comparison test, it can n+I n :. By root test,

n

be seen that is divergent

I

U/I

is convergent if

Ixl < 1

and divergent if

Ixl ~ 1 .


438

6.5.6 Example

Show that

~ ( n~ _ I) n converges.

Solution

=( nYn -If

Un

Lt n~oo

:. L

Un

unYn

= Lt

n . . . . . ct)

(n~-I)=I-I=O
Lt 1l~Cf)

n'~:' =1);

is convergent by root test.

6.5.7 Example

. the convergence 0 f the series whose nI h Examine term·IS

(

n + 2 )n .Xn n+3

--

Solution

Un = ( n+ 2 )n .xn; n+3

:. By root test,

When

L un

unYn =

Lt

n~

converges when

Ixl = 1 : un = (n +

2) ; Lt Un n~
n~
2 (nn+3 + )

X

=

X

Ixl < 1 and diverges when Ixl > 1

n

n+3

Lt

= Lt

n~
2)n (1+-

nn

( 1+-3 ) n

e

2

I

- 3 = - '" 0

:. L un

e

e

is divergent. Hence

and the terms are all +ve .

L un

is convergent if

Ixl < 1

and divergent if

Ixl~l. 6.5.8 Example Show that the series,

-HI +[~: -H' +[:: -H +. . . u" ~ [(n :":t -n: r~ ( n: T[( n: r r [~:

3

isconvergent

1

1

-1

Sequences of Series

Lt un

~ II

IHOC)

:. By root test,

439

1 1 1 e-l

1 e-l

=-.--=--<1

LUll

is convergent.

6.5.9 Example 00

Test

Llllll for convergence when u

m =

111=1

e

-Ill

2

(1+2 mfll

Solution

LU

Hence Cauchy's root tells us that

m

is divergent.

6.5.11 Example: Test the convergence of the series '" ~ ~ 112

e

Solution II II

Lt 11-)-00

U /n n

= n--.r:1J Lt - = 0 < 1 en

:. By root test,

6.5.12

n/n

LU

n

is convergent.

Example

r l

( n + 1 .x 2 32 2 Test the convergence of the senes, -2 x + - 3 X + ... I + ...... , x > 0 1 2 n'H

.


440

Solution

Lt unYn I

n~oo

= Lt

[(

II~OO

= Lt n~oo

L un

:. By root test, When x

n+I ) 11 .xn ]);, n"+ 1

[( + -1). 1] 1

n

1

I/.x n l II

II~OO

=

n)'n

[.

I.I.x = x SInce Lt n/Vn = 1] II~OO

converges if x < 1 and diverges when x> 1 .

= 1 , the test fails.

Then

II11

I)n _. 1 1 = 1 +. , Take vn =(

Lt un n~oo

Vn

n

= Lt II~OO

n

n

(1 + !)n =e 0 n :;t:

By comparison test and p-series test, Hence

n+I n

= Lt [ (- - ) .-/.x]

L un

L un

is divergent.

is convergent when x < 1 and divergent when x ~ 1.

Exercise 1 (d) 1. Test for convergence the infinite series whose (a) (b)

(c) (d)

1 2" -1 1

..............................

---'2::--n • (

(log)

d h terms are: [ Ans : convergent] [ Ans : convergent]

n :;t: 1)

3n+ 1 )n ................................ . ( 4n+3· x

[Ans:

4 4 Ixl < -cgt,lxl ~ -dgt ] 3 3

[ Ans: cgt for all x

~

0)

(e)

[ Ans: convergent)

(t)

[ Ans : convergent]

Sequences of Series

441

(2n2 -I (g)

(h)

r

[ Ans: convergent]

(2n )211

(n}~ -I

rn

[ Ans: convergent]

-n) . . . . . . . . . . . . . . . . . . _112

(i)

(j)

n-I

(

[ Ans : divergent]

(n:l)"'(x>o) ........................ .

[ Ans : x < 1 cgt, x

2. Examine the following series for convergence : x x2 x3 (a)

1+2+)2+43 + ..... ,x>O

(b)

~+(H

................. .

+UoJ +. . . . . . . . . . . . . ..

[ Ans: x

~

~

1 dgt ]

1cgt,x > Idgt ]

[ Ans : convergent]

6.6 6.6.1 Integral Test +ve term series,

¢(1)+¢(2)+ ..... +¢(n )+ ..... where ¢ (n) decreases as integral

Proof: Let Sn

r

n increases is convergent or divergent according as the

¢ (x) dx is finite or infinite.

= ¢(l) +¢(2)+ ..... +¢(n)

From the above figure, it can be seen that the area under the curve y = ¢ (x) between any two ordinates lies between the set of exterior and interior rectangles formed by the ordinates at n = I, 2, 3, .....n, n + I , ..... . Hence the total area under the curve lies between the sum of areas of all interior rectangles and sum of the areas of all the exterior rectangles. Hence {¢(l) + ¢(2) + ..... + ¢( n)}

.. Sn ~

r

~

¢(x)dx~Sn+I-¢(l)

r+ ¢(x)dx ~ 1

{¢(2)+ ¢(3) + ..... + ¢( n+ I)}


442

v= O(n)

P, P2

0

1

Cn +1

Q

Bntl

°3 0

2 A2

1

AI

n ~ 00,

As

°Il

°11~1

3

n

A)

An

n

+1

~

Lt SI1 is finite or infinite according as [

¢ (x)dx is finite or infinite.

Hence the theorem.

Solved Examples 6.6.2 Example

1 I --n logn 00

Test for convergence the series

n=2

Solution

r

1

- xlogx

dx - n~oo It [ r' 1 dx] 1 xlogx

=

It [log log

11~00

By integral test, the given series is divergent. 6.6.3 Example Test for convergence of the series

Solution

[ _1 xP

1 dx - It [fdx] x 11~00

=

P

p 1

It [ x- + ]11 n~oo - p + 1 I

_1_ Lt [ n l - p -1 ] 1- P IHOO

x]~ =

00

Sequences of Series

443

ClIse (i)

,,1 . L..J IS convergent. n"

If P > I, this limit is finite; ClIse (ii)

,,1 . . L..J- IS dIvergent. n"

If P < 1, the limit is in finite; Ca.\·e

(iii)

Ifp= 1, the limit Lt

logxl;' = Lt (logn)=oo;:. I_l-

11...--.)00

I_ln"

Hence

n"

11-)00

IS divergent.

is convergent if p > I and divergent if p :S

6.6.4 Example

n "-2 ~e" 00

Test the series

for convergence.

Solution 1111

¢ (n)

=

n = ¢(n){say); e"

-2

n increases. So let liS apply the integral test.

is +ve and decreases as

I e-,2

1¢(x){lx= x I

=! le- dt{t=x 1

IlIl1

-"21e

2

,dt=2xdx}

2I

I

=

By integral test,

dx 00

-I 1

I

1' W IlIC. h·IS fiillIte. . = -"21(0 --;;1) = 2e

is convergent.

6.6.5 Example

(7f)

1 Apply integral test to test the convergence of the series "-sin 00

~n2

11

Solution Let ¢ (n)

= ~2 sin ( :

); ¢ (n) decreases as

J-21

00

J¢(x)dx =

00

2

2 X

sin

(}x 7f X

11

increases and is +ve .


444

Letix =t

J

-! sintdt = !costl~/=! TC

~

TC

finite,

TC

/2

_TC/ / x

2dx = dt

Yx2dx=- ~dt :. By integral test,

I

converges x

Un

= 2 => t = ~ x = 00 => t = 0

6.6.6 Example Apply integral test and determine the convergence ofthe following series. (a)

(c)

(b)

Solution (a)

¢ (n) -

3n is +ve and decreases as n increases - 4n 2 + 1 2 j¢(x)dx= ~x dx (4X +1=t=>Xdx= Ysdt ) I 14x +1 x=l=>t=5,x=00=>t=00

j

= Lt [3-log t -log 5]= I¢ ( x ) dx = Lt [3-8 f-dt] t 8 I

00

n~oo

I

:. By integral test, (b)

¢ ( n) =

~n3

3n +2

j¢(x)dx =

I

Un

diverges.

decreases as n increases and is +ve

1~X3

dx 3x +2 1 OOfdt 1 = (; = (;[logt]5 = 00

I

00

n-->oo

5

1

t

00

4

[where t= 3x + 2]

5

By integral test, (c)

I

Un

is divergent.

¢ (n) = _1_ is +ve, and decreases as n increases. 3n+l

1 1 dt 1 f¢(x)dx= f-dx= f--[t=3x+l]=-logtl~=00 I I 3x+ 1 4 3 t 3

00

00

00

445

Sequences of Series

:. By integral test,

I

Un

is divergent.

6.7 6.7.1 Alternating Series A series, u l - u2 + u3 - u4 + - + ..... + ( -1 ),,-1 un + ..... , where un are all +ve, is an alternating series.

6.7.2 Leibneitz Test If in an alternating series

I (-1 ),,-1 .u

n '

where

un

are alI +ve,

> un+I ' Vn, and Lt ull = 0, then the series is convergent.

(i)

un

(ii)

n ........ oo

Proof: Let u l

-

u2 + u3 - u4 + ...... be an alternating series (' un 'are all +ve)

Let ul > u2 > u3 > u4 •••••• , Then the series may be written in each of the following two forms: ......... (A) (UI -U 2)+(U3-U4)+(U5-u6)+····· ul

(A) (B)

-(u 2 -U3 )-(U4 -u5 )-·····

............ (B)

Shows that the sum of any number of terms is +ve and Shows that the sum of any number of terms is < ul . Hence the sum of the series is finite. :. The series is convergent.

Note: If Lt ull n--.
-:t:-

0, then the series is oscillatory. Solved Examples

6.7.3 Example . 1--+--1 1 1 ..... Cons!·derthe series

234

In this series, each term is numerically less than its preceding term and ~O as n~oo. :. By Leibneitz's test, the series is convergent. (Note the sum of the above series is Loge2 )

6.7.4 Example Test for convergence

L (-I ),,-1 2n-1

nih

term


446

Solution The given series is an alternating series We observe that (i)

> 0, Vn (ii)

un

un

I (-1 ),,-1

un

,where

= _I_

U n

> Un+ I ' Vn (iii) Lt

Un

2n-I

=0

/1---700

:. By Leibneitz's test, the given series is convergent.

6.7.5 Example 111 Show that the series S - 1- - + - - -

3 9

27

+ ...... converges.

Solution The given series is series in which 1.

~ ~

un

( ),,-1 -1 1 3n-

= "'(_1)/1-1 u" L.J

u, = _1_1 is an alternating

where

I

> 0, Vn 2.

un

3/1-

I

> U/I+1' Vn and 3. Lt

un

= 0;

ll---7oo

Hence by Leibneitz's test, it is convergent.

6.7.6 Example X

x3

2

Test for convergence of the series, - - - ~ + - - 3 - +..... , 0 < x < 1 1+x 1+x- 1+x Solution n-1 n-1 ( -1 ) .xn The given series is of the form'" = "'(-1) u,

L.J

x"

where

UII=--II

. I+x'

Further,

U II

-u

L.J

n

Since OO,Vn;

x"

n+1

1+ XII

X"+1 I+XIl+1

=-n ----

l+x

x" _X"+1

x" (I-x)

=--:-----:--:------;-

( 1+ x" ) ( I + x"+ 1 ) 0< x < 1 +ve.

Again,

~

all tenns in numerator and denominator of the above expression are

x" ~ 0 as x" ~ 00 since 0 < x < 1; .'. Lt

:. By Leibneitz's test, the given series is convergent.

II~OO

o =0 1+0

U = --. II

447

Sequences of Series

6.7.7 Example

(-1 ),,-1

L ----r======== 00

Test for convergence

1I~2 ~11(11+1)(n+2)

Solution The given series is an alternating series "(-1)"-1 L..J ull where Again,

1

1111

~11 ( 11 + 1)( 11 + 2)

=

; u > 0, \1'11 ; ll

~(11+1)(n+2)(n+3) >~n(I1+1)(n+2) 1

1

~(11+1)(n+2)(11+3) < JI1(I1+1)(11+2) ,\1'11 I.e.,

Further,

Lt u IHOO

= Lt

II

By Leibnitz's test,

11-->00

1

J 11 ( 11 + 1)( 11 + 2 )

f( -1)"-1 un

=0

is convergent

2

6.7.8 Example Test for the convergence of the following series,

1

2

3

4

5

---+---+--+ 6 11 16 21 26 .... Solution

L (-1) 00

Given series,

n=1

n-I

11

- - = L (-1)

II-I

is an alternating series

Un

511 + 1

11 n 511 + 1 11 11+1

u =-->0\1'11 . -----

511+1

Again,

511+6

'

-1 (

511+1

11 IHOO 511 + 1

)(

511+6

)

~

Un

< Un+ I' \1'11

1 5

Lt u = Lt --=-:;t:0 n-->oo

n

Thus conditions (ii) or (iii) of Leibnitz's test are not satisfied. The given series is not convergent. It is oscillatory.


448

6.7.9 Example Test the nature of the following series.

I

00

(a)

1

(

r-

-1

I

I --'--::--'r--(

(b)

f;z +J;+i

-1

I

(c)

2

n +1

Solution (a)

u,,=

1

~>OVn

I

vn+vn+l

1

U" -

U,,+I

=

1

f;z + J;+i - J;+i + J n + 2

:. By Leibnitz's test the series converges. 1

1

1

u" =-2->0,Vn; - 2 - > 2 =>U" >un+I,Vn; n +1 n +1 (n+l) +1

(b)

Lt

UI/

n-->oo

(c)

un

=0 1

= I ~.l

1

~

:. By Leibnitz's test, given series converges.

> 0, Vn ;

1 UI/ > lln+P V n ~~

In + 2 > In + 1 => I ~

1 I

')

By Leibnitz's test, given series converges. 6.7.10 Example Test the convergence of the series

111 sJ2 - sJ3 + sJ4 - +..... .

Solution

I

00

The series can be written as

n=1

(_1)"-1 -'-;=====

sJ;+i

1

=---;==

U /I

SJn+l

(i)

(ii) (iii)

Lt un

n-->oo

=0

.

'

By Leibnitz's test, the given series converges.

Sequences of Series

6.7.11

449

Example Test for convergence the series, 1-

h + ~ - ~ + .....

Solution The given series can be written as

(-If (omitting \'1 term) I-2n

1 1 1 1 ->OVn; ->--=>u >lll/+I'Vn; Lt -=0 2n 2n 2n + 2 2n lI

IHOO

:. By Leibnitz's test,

6.7.12

(-If I-2n

is convergent.

Example

1 3!

1 5!

1 7!

Test for convergence the series, 1- - + - - - + ..... .

Solution

General tenn of the series is /-1)"-;

2n-I !

The series is an alternating series; (

1 ) > OV n

2n-I !

1 1 1 >( ) => Un> Un+ I' Vn EN; Lt ( ) =0 2n - 1 ! IHOO 2n - 1 ! ( 2n - I! ) By Leibnitz's test, given series is convergent.

6.8

Absolute convergence A series UII is said to be absolutely convergent if the series

I

convergent

6.8.1 Consider the series

1 1 1 IU =1--3 +3"--3 +-+ ...... 234 1 1 1 1 Ilun l=1+-3 +3"+-3 + ...... = I-3 2 3 4 I n By p - series test, IUn is convergent (p = 3 > 1 ) II

00

I

I

I

IU I IS II


450

I

Hence

Note: 1. If

I

Un

Un

is absolutely convergent. is a series of +ve terms, then

I

lIlI =

I

IUI/ I·

For such a series, there is no difference between convergence and ahsolute convergence. Thus a series of +ve terms is convergent as well as absolutely convergent. 2. An absolutely convergent series is convergent. But the converse need not he true.

I (_1)"-1.-!-n =1-1..+1.. -1..+ ..... . 2 3 4

Consider

1

This series is convergent (1.7.3)

I I(-1 ),,-1 )~I =1+ ~ + ~ + ~ + ....... is divergent (p-series test). Thus I is convergent need not imply that II I is convergent (i.e., I is not absolutely convergent). But

Un

UII

Uti

6.9 6.9.1

Conditional Convergence [fthe series is divergent and

Ilu,,1

I

Un

is convergent, then

I

Un

is said

to be conditionally convergent.

6.9.2

Consider the Series

1-1..+.!-1.. ....... 234 But

:. I

I

IU

II

is convergent by Leibnitz's test. (Ex. 1.7.3)

IUn I= 1+ ~ + ~ + ~ ..... Uti

is divergent by p - series test.

is conditionally convergent.

6.10 6.10.1

Power Series and Interval of Convergence A series, ao + a1x + a2 x 2 + ..... + anx n + ..... where 'an' are all constants is a power series in x. It may converge for some values of x .

. Sequences of Series

451

Lt n~
UII +I till

= Lt

a/HI •x (I st term is omitted.)

n~
all

Lt

=kx (say) where

n~'"

Then, by ratio test, the series converges when i.e., it converges '\Ix

E (

,)

(~1 ~

The interval

a ll +1 = all

k

Ikxl < 1 .

~1 , ~ )(k *- 0)

is known as the interval of convergence of the given

power series.

Solved Examples 6.10.2

Example

I \ '"

Find the interval of convergence of the series

11;1

II

n

Solution

"Lf.( U~:, )="Lf.(n: l)'·x =H.[ ~.~ 1

By ratio test, the given series converges when When x

= 1, L un = L

:. L un

in

3'

I

J:

Ixl < 1, i.e., x

= E

x

(-1,1)

which, is convergent by p series test.

is convergent when x

=1

Hence, the interval of convergence of the given series is (-1, I]

6.10.3

Example Test for the convergence of the following series. (a)

1-

YJ2+ YJ3- YJ4+··········

(b)

1+

X2- }i2- ~2+ Ys2+ ~2- (;2- h2+ . . . .


452

(c)

I-X~+X~-~+ ......... .

(d)

I(-I)"(n + l)x

n

,

with x

<-.!..

o

2

Solution (a) The series is of the form

L (-I),,-I un

where

un =

y.;;;

It is an alternating series where (i) un > OV n (ii) Un > 1l,l+IVn and

(iii) Lt n~oo

Again

un = 0;:.

the

By Leibnitz test, the series is convergent. 1+

series

YJ2 YJ3 YJ4 +

+

+ ......

is

divergent,

p - series test. Hence the given series is conditionally convergent. 00

(b)

L lu,,1 = L;;:2

which is convergent by p - series test.

p=1

The given series is absolutely convergent. It is convergent.

(c) The given series is 2n-2

"(_I)n-l. x . ="(-1 ),,-I U · . Iu ~ (2n-2)! ~ n' •• n

-U n +1 = u"

By ratio test, the series

2n-2

I=_x __ (2n-2)!

1 1 21 ; Lt -lln+1 .x (2n -1 )(2n) tHoo ll"

Llu,,1 converges Vx;

=0 < 1

L u"

i.e.,

is absolutely

convergent Vx; (d) Here,

Iunl =(n + I)xn;1 un+1 1= (n + 2)x n+1 (neglect 1

sl

lu I

Lt ~ = Lt IHOO

Iunl

:. Llunl

IHOO

(n + 2)

(n + 1)

Ixl = Lt IHOO

term)

7n)~ Ixl =Ixl < 1 (1 + Yn)

(1 +

}i

(": x < 12 )

is convergent Vx, i.e., given series is absolutely convergent

and hence convergent.

by

453

Sequences of Series

6.10.4 Show that the series 1 + x + x/6 + x/6 + ..... converges absolutely \;Ix Solution

Lt '1->00

IU'HII =13 =0 < 1 when lUl I

n

Ilunl

:. By ratio test, When x

x*-O [since .

111" 1= (n-I)!' Ix"-II ·Iu 1= Ix"l ] n! IItl

is convergent \;Ix *- O.

= 0, the series is (1 + 0 + 0 + ...... ) and is convergent

:. Ilu,,1 converges ~ Ill" is absolutely convergent 6.10.5

\;Ix.

Example Show that the series, 1 -

~ + ~ - ~4 + ........ 3

3-

3

is absolutely convergent.

Solution , which is a geometric series with common ratio

~ <1 3

:. It is convergent. Hence given series is absolutely convergent.

6.10.6

Example Test for convergence, absolute convergence and conditional convergence of the series, 1 ] 1 1--+---+ ......... .

5

9

13

Solution The given alternating series is of the form u" > O\;ln EN;

Hence,

u -u n

I

(-1

r-

1

I

Un'

where,

= _1_ . 4n-3

1

U,,+I = 4 (n+ 1) -3 = 4n+ 1

1 1 4n - 3 4n + 1 4n+I-4n+3

=----,,+1

=

4

- - - - - - > 0, \;In E N

(4n - 3)( 4n + 1) 1 I.e., un > un+I ' \;In EN Lt II = Lt - - = 0; n-+oo 11 fl~a) 4n - 3 All conditions of Leibnitz's test are satisfied. I Hence (-1 un is convergent.

I

un

r-

(4n - 3)( 411 + 1)


454

lu 1=-1-. 4n - 3 /l

Take v

I series test, I

But by p -

6.10.7

lunl = Lt

Lt //--+00

v"

111// I and

I

n

By comparison test,

Ilu// I

=! .

11'

'

,,--+00

n I n( 4 - 3/) n

=! *- 0

and finite.

4

v" behave alike.

v" is divergent (since p = 1) .

is divergent and :. The given series is conditionally convergent.

Example

I (_1)"-1.

Test the series

//=1

I, ,for absolute / conditional convergence. }\1 n

Solution The given series is an alternating series of the form

I (-1 r-Iu// .

Here

1

un = 3vn"

(i)

Vn EN

(ii) 3(n+l»3n=>3-Fn+l >3,Jn,Vn.

1 1. :. 3.Jn+l <3,Jn ,I.e., And Lt

n~oo

un =

Lt n---)-oo

\-I

ll'/+I

N

1

, =0 3" n

:. By Leibnitz's test, the given series is convergent. But

I

1

(_1)//-1.__ = 3,Jn

I_I3,Jn

is divergent by p- series test (since

1

p= -<1)

2

:. The given series is conditionally convergent.

6.10.8

Example Test the following series for absolute / conditional convergence. (a)

(c)

I(-If-I , sin(~a) //=1 n

I

,,=0

(-1)" (2n)!

(b)

I(-I)"-I.~ n=1

n +1

" (_1)"-1.

nJr

n

(d)

L..J

e3n+1

Sequences of Series

455

Solution

1 [ ] lu,,1 = Isinnal 112 < n 2 since Isin nal < 1

(a)

considering

comparison and p - series tests, we get that

VII

II = 1112

and using

I lUl I is convergent I

lin

is

absolutely convergent. By Leibnitz's test, the series converges.

(b)

. Tak1l1g v"

1

=-

n

.

. ,by companson and p- series tests,

"n. n- +1 2

~ - ,- , IS

seen to

be divergent. Hence given series is conditionally convergent. Take

(c)

lu,,1 = _1_; 2n!

Lt ,,->0')

II I I= 0 < 1; By ratio test, I lUl I is convergent; III + lIlI

Hence given series is absolutely convergent. Jt'''

lu,,1 = e~)"+I

(d)

;

given senes is absolutely

By root test, is convergent,

convergent. [In problems (a) to (d) above, hints only are given. Students are advised to do the complete problem themselves]

6.10.9

Example Find the interval of convergence of the following series. (a)

:t (_1)"-1 xX ,,~I

(c) Solution (a)

(b):t (_1)"-1 n(x:3 2)"

3

II~I

log(l+x)

Let the given series be

I

lIlI ;

Then

\ "\ \"+1\ Iunl = ~ ; lun+,1 = X 3 n

(n+ 1)

/3 [_11J' ·lxHx: 00

:. By ratio test,

Ilu,,1

is convergent if

Ixl < 1

I Un is absolutely convergent if Ixl < 1 ; :. I u" is convergent if Ixl < 1

i.e.,

1+-n


456

1 1 1 Ifx= I,thegivenseriesbecomes 1-23"+ 33 - 4 3 + ....... .

. W h·IC I1 IS

." 1 IS . convergent. convergent, sIllce ~ -3 n

Similarly, if x

=

convergent. Hence the interval of convergence of (b)

I

-1, the series becomes

I

UII

-~ = -I ~ n

n

which is also

(-1 ~ x ~ 1)

is

Proceeding as in (a),

lulI+11

Ix+21

Lt-=--

Iu'"

IHOO

3

:. I U is convergent if Ix + 21 < 3, i.e., if -3 < x+ 2 < 3, i.e., if II

-5 < x < 1.

I

If x = -5, Ifx= I,

lIlI

=I (-llll-I .n

,and is divergent ( in both these cases

IU = I( -1)"-1 .n, and is divergent II

Lt

UII:I;

0 )

ll--i.rJJ

Hence the interval of convergence of the series is (-5 < x < I) log (I +x)=

(c)

x2

x3

I( -1)"-1 ~ = IU 00

=

II

n=1

Ixnl

X4

x--+---+ ...... . 234 'n

II

(say)

111+11

IUnl=-;-; lun+1= :+1 lun+11 1 I x I= I I Lt - = Lt Iunl (1 +~) 1

X

/1-+00

/1-+00

By ratio test, i.e.,

I Un

I lUl I is convergent when Ixl < 1 x

is absolutely convergent and hence convergent when -1 < < 1.

When x = -I,

"~ u = "( ~ -1 )11-1 . -1 = 1- -1 + -1 - -1 + ....... , ll

n

2

3

4

which is convergent by Leibnitz's test. (give the proof)

457

Sequences of Series

:L>n

Whenx= 1,

L ~n Hence I since

=

-(1+~+!+~+ ...... ) 234

which is divergent,

is divergent by p -series test (prove).

Un

is convergent when -1 < x ::; 1

Interval of convergence is ( -1 < x ::; 1 ).

Exercise -1 (e)

1. Use integral test and determine the convergence or divergence of the following series: 1

Ln

[ Ans : convergent]

2

1

00

2.

~ n{logn )2

2. Test for convergence of the following series: 1 1 1 l-~

'"

2.

+l:!: -l2 + ......................................

{-If-l

~(2n-I)(2n)

[ Ans : convergent]

[ Ans : convergent]

[ Ans : convergent] [ Ans : convergent]

I

3. Classify the following series into absolutely convergent and conditionally convergentseries: [ Ans : abs.cgt ] [ Ans : abs.cgt ]

3.

I

{-If n{logn)2

4. Find the interval of convergence of the following series 2n x n 1.

L

~

[ Ans : abs.cgt ]

[ Ans :

-00

< x < 00 ]


458

[ Ans : - 1 ::;; x ::;; 1 ]

r Ans : -1 < x ::;; 1 ] 111 - 2 + .... is absolutely convergent. 2 3- 4 111

(a) Show that 1- - 2 + - ? -

5.

(b) Show that 1 -

J2 + j j - J4 + ....


Tests of Convergence - A Summary 00

I. The geometric series

Lx

n I -

converges if 1x 1< 1 , diverges if x ~ 1, and oscilates

n=1

when x::;;-1 2. If

L u"

is convergent, Lt ull n->oo

f ~n

3. P - series test:-

=0

[The convergent need not be necessary ]

is convergent if p > 1 and divergent if p::;; 1

11=1

4.-C-omparison lest :- The series

LUll

and

L vn

are both convergent or both

~ is finite and non - zero.

divergent if Lt

n->oo VII

L un

5. D 'A/ember! 'sRatio test:-

Lt n->oo

(

U

2:':l.

u

< 1 or > 1

ll

Alternately, if Lt n->oo

6. Raabe's test:

converges or diverges according as

L un

~ >1 u ll

+

or < 1) . If the limit = 1, the test fails

1


.~HU::1 -I}]> lor
LU

n

converges or diverges according as Lt

n->oo

(Iflimit = 1, the test fails.)

(U}) < lor>

1

459

Sequences of Series

8. Integral test: A series

L ¢ (n) of +ve terms where ¢ (n) decreases as n increases co

is convergent or divergent according as the integral

J¢ (x) dx is finite or infinite.

1 co

9. Alternating series - Leibnitz 's test: An alternating series

L (-1 ),,-1 un

convergent

n=1

if(i) un

= ull +1' \In

and (ii)

Lt un

=0

Il---+CO

10. Absolute / cO':lditional convergence: (a) un is absolutely convergent if

L

L Iun I is convergent.

(b)

LUll is conditionally convergent if LUn is convergent and Llun I is

(c)

divergent. An absolutely convergent series is convergent, but converse need not be true . i.e., a convergent series need not be convergent.

Miscellaneous Exercise - 1 (e) 1. Examine the convergence of the following series:

111 1.2· 3.4 5.6 22 32 2. -3-+-3-+-3-+ ..... . 1 +1 2 +1 3 +1 2 22 23 3. -+-+-+ ...... . 1 2 3 1 2 3 4. 1+2 + 1+22 + 1+23 + .....

1. - + - + - + ........................................... .

( cgt.]

e

( dgt.] ( cgt.]

3

x2 x 5. --+--2 +--3 + ..... (x>O) ........ l+x l+x l+x 2 3 3x 4x 6. 2x+-+-+ ..... (x > 0) .......................... X

8

27

7. 1+!+ 1.3 71.3.5 +....

................................. 2.4.6 32 32.5 2 32.5 2.7 2 8. 2" + -2-2 + 2 2 2 + ... ... ............................... 6 6 .8 6 .8 .10 2

2~4-

9. 3.4 + 4.5 + 5.6 +....

1.2

( dgt.]

2.3

3.4

................................

( cgt. if

x~l dgt. ifx> I]

( cgt. if x ~ 1 dgt. if x > 1) (dgt.]

[ cgt ] (dig.]


460

10.

(l!)2 (tJ)2 2 (lJr ~ tJ .X+ l1 x + 12 x + .... (X>O) .............

X x2 x3 II. 1+ 22 +j2+4T+ ..... (x>O)

(4

5

........................ .

12.

3x 5)2 x 2+(6)3 +x3+ ..... (x>o) 4+

13.

L(l+~r

............ .

[ege ifx<4,dgt. if x24 ] [ egt.if x::::; 1, dgt. if x > I] [ egt if x < 1, dgt. if x 2 1 ]

[ dgt. ] [ egt. ]

15.

a" 2:--2,a<1

[egt.]

l+n

111 16. 1 - - + - - - + - + -..... 2.2 3.3 4.4

[ Abs. egt.]

2. Examine for absolute and conditional convergence of the following series:

n "(_I)n 33 n 3

I. L.,;

2.

2

.......... ......... ............. ....

(-l .n 2: 2f

[ Abs. egt.]

[ Abs. egt. ]

n

[Cond. cgt]

[ Cond.egt.]

3. Determine the interval of convergence of the following series:

x 2 x 3 X4 x+-+-+-+ ..... 234 " (x+lf

2. L.,;

n.2"

.............................. . [-3
Sequences of Series

461

Exercise - 1 (g) (Objective type questions) 1. The infinite series (i) (iii)

convergent oscillatory . 1+n 2. T he senes - - 2 1+n (i) convergent (iii) oscillatory

.

1

(ii) (iv)

divergent ndne of these

I ADS :(i) ]

(ii) (iv)

divergent none of these

I ADS :(ii) ]

IS

1

1

1

3. The senes 1.J}- 2J2 + 3J3 - 414 + ..... (i) (iii)

4.

oscillatory conditionally convergent

The series (i) (iii)

1-.!. + ..!. -.!. ......

(ii) (iv)


IS

oscillatory convergent

-oo
1.2

3.4

X

X

3

X

4

I ADS :(iii) ] .

X--+---+ ...... ,IS (ii) (iv)

6. The series _1_+~+~+ .........00 (i) (iii)

absolutely convergent none of these [ADs :(ii)]

234

5. The interval of convergence of the series (i) (iii)

(ii) (iv)

234 -1 < x < 2

none of these

I ADS :(iii) ]

(ii) (iv)


I ADS :(ii) ]

(ii) (iv)

convergent none of these

I ADs :(iii) ]

IS

5.6

convergent oscillatory

· 123 . 7. T he senes - + - + - + .........00 , IS 1.3 3.5 5.7 (i) (iii)

conditionally convergent divergent

.2345

8. The senes - + - P + - + - P + ........00 V 2 JP 4

is convergent if

(i) p < 2 (ii) p= 2 (iii) p> 2 (iv) none of these 9. The series 6 - 10 + 4 + 6 - 10 + 4 + 6 - 10 + 4 + 6............. 00 is (i) convergent (ii) oscillatory (iii) divergent (iv) none of these

I ADs :(iii) I [ADS :(ii)]


462

111

10. The series - + - + - + .........

2.4

(i) (iii)

4.6

IS

6.8

convergent oscillatory

(ii) (iv)


2. Indicate whether the following statements are true or false: 1 I. The series " is convergent. LJ 1+ Til

[Ans :(i) ]

[ False]

2

n +5 2n2 + 7 is not convergent.

2. The series

I

3. The series

1 1 1 U + 12 + lJ + ........ x

3

.. ................... .

. d' IS Ivergent... ................ .

x

4. The series x - - + - - + - ....... , converges when -1:s x :S 1.. 3 5

[ True] [ False]

[True]

(_1)"-1 5. The series " LJ n.5/l

is absolutely convergent. ........................

6. The series x + 2X2 + 3x 3 + 4X4 + ....00 is convergent if x> I........ x2

[ True]

[ False]

x3

+ - + ........00 is divergent if x ~ 1 ............... [True] 3 x 2! 3 3! 3 4! 4 8. The series 1+-+2"x +-3 x +-4 x + ...... +00 is convergent 2 3 4 5 7. The series x + -

2

if x> e 9. The series

10. The series

[ True]

I

-"J;; ( 1 + .);; )

is divergent ............................ .

( 4 )3 x 3+ ....... IS. convergent "21 +"32 x + ( 4"3)2 x 2+"5

ifx< I .

II. The series 1- 2x + 3x 2 - 4x 3 + .....00 ( x < 1) is divergent... ........ . x . 12. The senes - - .

[ False]

l+x sin x

2

3

[ True] [ False]

4

x +-x 3 - -x . T] 4 + ......00 IS convergent ...... [ rue l+x l+x l+x sin 2x sin 3x

--2

13. The senes -3- - --3- + --3- + ......00 converges absolutely ...... [True]

1

2

3

Sequences of Series

14. The series

463

I (~1-1


[ True]

2

•

3n + 5

Ih.

15. The sertes whose n

term

IS

(n+2)"

is convergent.

[ False]

3. Fill in the Blanks :

L ar l 00

1. The geometric series

I

[ Ans: 1r

converges if_ _ _ __

1< 1]

11=1

2. If a series of +ve terms "" L.. 3.

I {{!n +I-n} 3

un

is convergent, n->oo Lt

un =- - -

[Ans: 0 ] [ ADs: convergent. ]

IS _ _ __

11=1

L 3n3-4 00

4. If

1/=1

(n+5Y

is divergent ,value ofp is _ _ __

5. The interval of convergence of

6.

L un

I

u" where u"

~ ( :: :~

r

[ Ans: ::;; 4]

x, is _ __ [ Ans: -1 < x < 1]

is convergent series of +ve series. Then Lt

11-»00

(un

11 1( )

is

----

( ADs: < 1] ( ADs: Oscillatory]

7. The series 8 - 12 + 4 + 8 - 12 - 4 + ....... is

8. If

un >

0,

V nand

LUll

is convergent, then Lt 11->00

[n {~-I}] is _ __ u +

n 1

[ADS: > 1]

9. If the series

I(-Ifan,(all

>OVn) is convergent, then for all values of

n,!!JL

~

aMI

( ADS: > 1]

is- - 10. If

un =(I+!)-n2 , Lt u l/ n= _____ n

n->oo

lI

[ ADS:

1/e]


464

Tests of Convergence - A Summary 00

1. The geometric series

I

x~ 1

converges if Ixl< I, diverges if

X"-I

and oscilates

n~1

when

x~

1.

2. IfIll" is convergent, Lt u"

=0

[The converse need not be necessary ]

Il-~OO

3. p - series test:-

f ~n

~1

is convergent if p > I and divergent if p

,,~I

4. Comparison test: The series

L u"

and I

Vn

are both convergent or both

divergent if LI u" is finite and non - zero. n~oo

v"

5. D'ALEMBERT'S Ratio test :

IU

n


< l o r > 1 (Alternately, L t1l,,+1 =

>1). If the limit

I, the test fai Is.

6. Raabe's test:

I

U"

converges or diverges according as Lt

[n{~ Unt-I

n-)oo

7. Cauchy's root test: I

U"

converges or diverges according as Lt n~oo

-I}] I I > or <

(u~ J< 1 or> 1 ;

(If the Limit = I, The test fails).

8. Integral test: A series

L rj> ( n) of +ve terms where rj> ( n) decreases as n increases

frj> ( x ) dx is finite or infinite.

00

is convergent or divergent according as integral

1 00

9. Alternating series - Leibnitz's test: An alternating series

I

(-1 ),,-1 un (where

n~1

Un>

OVn ) is convergent if(i)

un

>

lin+1 '

Vn (ii) Lt

Un

=0

n~oo

10. Absolute/ conditional convergence: (a) LUn is absolutely convergent if II Un (b)

IU

n

is conditionally convergent if

I is convergent.

Iu"

is convergent and

II U" I

divergent. (c) An absolutely convergent series is convergent but converse need not be true. i.e., a convergent series need not be convergent.

is

465

Sequences of Series

Solved University Questions

1. Test the convergence of the series:

123 1.2.3 2.3.4 3.4.5

- - + - - + - - + ........ . Solution Let ull be the

d h term of the series;

Then,

U

II -

1 n(n+1)(n+2) - (n+1)(n+2)

n

1

Let

VII

=

-

U

then, Lt

--'!.. 11->00 V

-2 ;

n

Lt (

=

tHoo

II

11->00 (

LUll

But

L

VII

and

L

n +1 n +2

VII

)

1

Lt

=

Which is non-zero and finite. :. By comparison test, both

n2 )(

=1

1) (

2)

1+;; 1+;;

,


is convergent by p-series test (p > 1):.

LUll

is convergent.

2. Show the every convergent sequence is bounded Solution Let

Lt

(all) all

be a sequence which converges to a limit '/ 'say.

= I =>

given any +ve number

E,

however small ,

1l~C()

we can always find an integer'

III ' , ) ,

Ian -II
'ifn ~ m

jaIl -II < 1; I.e., (I - 1) < all < (i + 1), 'if n ~ m Let A = min {a"a 2 , •••••••• am _,,(1 -1)} ,and fl = max {a"a 2 , •••••••• am-,,(I + 1)} Then obviously, A ~ an ~ fl, 'ifn EN; Hence (an) is bounded.

Taking

E

=

1, we have,

3. Show that the series,

1 1 1 S =1--+---+ .......... converges. 3! 5! 7!


466

Solution

2)-1 ),,-1 u" '

The given series is an alternating series where u" (i) (ii)

ll"

Lt

=(

1

)

211-1 !

We observe that,

> 0 and u" > 1l,,+1' '1111 and II

,,--+00"

1 =0 IHoo ( 211 - 1) !

= Lt

:. By Leibneitz's test [7 .2 ] the given series converges. 00

4.

Show that the geometric series

I

qlll = 1+ q + q2 + ....... converges to the slim

111=0

1 - - when /q/ < 1 and diverges when /q/ ~ 1 l-q Solution See theorm 2.3 (replace' x ' by 'q') . 5. Define the convergence of a series. Explain the absolute convergence and conditional convergence of a series. Test the convergence of series

I[l+ J,;r' Solution For theory part, refer 2.1,2.2, 8.1 ,9.1, and 9.2

,

Problem:

Let

u" = (1 + _l_)-n ; Lt (u~{,) = Lt (1 + _1_)-n J;; J;; 11--+00

~ H-[l+ By Cauchy's root test,

6.

I

Un

1

J,;J

,,--+00

1

=7<1

is convergent.

1 2

1.3 2.4

2

1.3.5 2.4.6

3

Test the convergence of the series, 1+-x+-x +--x + .....

Given that x> O. Solution

1.3.5 ..... (211-1)

Omitting the first term of the series, we have, ------'--------'- x

2.4.6 .... .211

n

and

Sequences of Series

467

1.3.5.(2n-l) 1/

UI/ =

2.4.6 .... 2n

1.3.5 ..... (2n+l) ; UI/+ I =

X

(

2.4.6 ..... 2n + 2

n+1

).x;

L (2n + 1 )

L u n+ 1

1/100 ---;;: = 1/100 2n + 2 .x = x By ratio test,

LUI/

is convergent when x < 1, and divergent when x > 1

The ratio test fails when x = 1 Whenx= 1,

~-1 = 2n+2 -1 =_1_ 2n + 1

U n +1

2n + 1

Lt [n(~-I)]= Lt (_n 2n +

n~oo

:. By Raabe's test,

U Il +1

LUI/

IHOO

1

)=!<1

2

diverges.

:. The given series converges when x < 1 and diverges when x 21 . 7.

( 3 )2 x 2 +"5 (4 )) x 3 +.........x > 0 "3 +"4

Test the convergence of the series, 1 + ( 2 ) x

2"

Solution Neglecting the 1st term,

u"

=[(=:~H'

1/ u 1n II

Lt uny.n =

=

[1+ 1/]

(n+l) - - X= _ _ n x n+2 1+2/ n

x; By Cauchy's root test, "~u// is cgt. when x 

ll----too

1; when x = 1 , the test fails. When

:. LU

x

n

=

1, un

(1 + l/~r

=(

, )n;

1+ 3/~

Lt Un = ee = -e1* 0 -2

IHOO

is divergent.

:. is cgt. when x < 1 and dgt. when x 2 1 .

468

8.


nih

Test the series whose

term is

(3n -1) / 2n

for convergence.

Solution

= n

U

(3n-l)

U,,+I _

un :. By ratio test,

2"

. '

{3(n+l)-I}

U

=-'--------'-

(3n + 2) 2(3n-l)

L un

2"+1

IHI

- - -- 1 <1 L (Un+1

2

n->oo Un

is convergent.

L

1

9.

Show by Cauchy's integral test that the series

n;2

n(logn

r

converges if p > 1

and diverges if 0 < p ~ 1 Solution 1 Let ¢ ( x ) = ; x ~ 2 ; Then ¢ ( x) decreases as x increases in x(logxy
00

ax

f¢(x)dx=f x(logx)

2

2

p

=

d

I --;;; U

[2,00]

I-p

II

=~ P

log2

. I [Takmg log x = u, - ax = du x = 2 => u = log 2 and x x Case (i) : p > 1 => 1- P < 0 => Integral is finite, and

1

00

log2

;

= 00 => U = 00 ]

Case (ii) : 0 Integral is infinite. Hence, by integral test, the given series converges if p > I and diverges when O
. JI

10.

Test the convergence ofthe series

Solution

I (I + J,;)"""

Sequences of Series

469

1 <1 e

1

L1

= -

U;,II

n~oo

I

By Cauchy's root test,

[2
u" is convergent.

" I,,~2 -(1)" .x)' 0 < x < I n-l 00

11.

Test the convergence of the se'ries,

(

11

Solution The given series is of the form

This is an alternating series in which (i) further L1

ll"

xn u" = ( ) n n-l

I (-1)" 11" ,where lin

> 0 and

> 1l1l+IVn EN.

1I"

= o. Hence, by Leibnitz test, the series is convergent.

11-»00

12.

. DISCLISS

. the convergence of the series,

1 J.

2'1'1

+

x2

x6

X4

r;; + r:; + t. + ......... . 3'1'2 4'1'3 5'1'4

Solution 2n

n

Iii

· term 0 f th e series

=

x r--:-:; u = (n+2)'I'n+l

"1I+1,J;;+2 rn+i

21 X llt

~= I/~OO lrVn.M. (+ 3.~) 2l=x LI

2

X

"II

ll

2

;

1

It"

= 1, u =

2

.x

LI

I converges if x > 1 , i.e., if Ixl > 1 ;

When x

(n + 3)

1=

By ratio test, x

1+2

(n + 3).J n + 2 ; ---;;: =

U

/HOO

2

( . . I sl term ) omlttmg

ll

2

< 1 , i.e., if

rn+i ;taking

1

(n+2) n+l

vn

Ixl < 1, and diverges if

= -3-'

,

n i2

]I

Lt

U _II

n~oo vn

=

n/ 2

Lt

n~oo n%

By comparison test,

I

Un

(1 + ,'n 2/) \fIT?n hl and

I

VII

=1

both converge or diverge together;


470

L vn is convergent by p-series test. :. L un is convergent if Ixl ~ 1 and divergent if Ixl > 1

But

211

00

13.

<::1 Test the convergence of the series

" " x L..--~=

n~1 (n+ 1)J;;

Solution X 211

=

U

x2n+2

. U

(n+l)J;;'

n

=----==

~

,r-:-;

u n+1

vnvn+l. x2 = \jIT /~ .x 2 ; Lt n+2 (l+;ln)

=

un By ratio test, When

11-+00

3/ (

n

test, we observe that

L

2

1 .) taking 1 + 1/

=

VII

LUll

~

Ixl ~ 1

and diverges when

Find the interval of convergence of the series,

3

4

234

11+1

x = --; un +1 n+1

n+2

II)

1+ Lt un +1 = Lt ~ n~'" U n~oo ( 1+ 2/ By ratio test,

L un

When x

U

Taking

=

I

,

n

X=X

/n

n

converges when

Ixl < 1 i.e., -1 < x < 1

1 =--

n+1

un

;-=--., Vn

1+;11

n

Ixl > 1

.

~+~+~+ .........oo

Solution · . For tIle gIven serIes, un

.

and applying the comparision

is convergent.

2

14.

Un

n/ 2

.n

converges when

lin

u n +1 =x2;

L un converges when Ixl < 1 and diverges for Ixl > 1

Ixl = 1 , un =

Hence

(n+2)rn+i

11+1

Sequences of Series

471

Lt un

and,

= 1 *- 0

and finite.

II~'" VII

IU

:. Both But

I

When x

and

II

Vn

=

I

diverges:.


VII

IU

also diverges when x ~ I.

n

-1, the given series is I . . -1 --I + -1 --1 + wI·h· lIC IS a ternatll1g senes \\ 1111 2 3 4 5 Ut •

> un+IVn and un

By Leibneitz's test

IU

~ 0 as n ~

converges when x =-1

II

.. Interval of convergence is [(-1, 1) i.e., -1 15.

00

~

x <1

~ 1.3.5 .... (211+1) Test the convergence of the series L. 11=1

2.5.8 .... (311+2)

Solution

1. 3.5.... ( 2n + 1)

un

Un +1 _

Un

By ratio test, 16.

1.3.5 .... {2n+3) un +1 = 2.5.8 .... (3n+5 )

= 2.5.8 .... (3n+2 );

I

Prove that the series

2n+3. 3n+5' UII

Lt

n~'"

~ = Lt r2 + ( n~'"

Un

3

l3 + ( 5

11 )] ) 11

is convergent.

(-If I--n(logn)3

converges absolutely.

Solution

I Iu 1- n(1ognr

;

(where t = log x)

= - : -2 = - -

n -

By integral test

" L un

LlulIl

"'f

dx

= "'f d!

2

X(logxf

log2

-11 t

1

g

is convergent.

converg..:s absolutely.

log 2

r'

,which is finite.

~

= <1

3

472

17.


'" (2n3 + I) X ,x> 0 n +1 II

Test the convergence of the series L.J

Solution . . n/I! term 0 ft l le given series,

U n

I = -2n+ 3-X

n +1

_[2(11+1)+1] n+1 un + 1 1 X (n+1) +1 II

1

Lt

~I/~t

IH"O

1111

=

1,1

2n+3 n+1 ,x (n+1} +1

_

-

(2n+3).x"H

11->00 { (

II

(113+1)

r

n + 1 + I}

x --'-----'---

xn ( 2n + I)

x=.x

l1y ratio test, IlIlI converges if x I. If x

Whenx

I,

=

2n+ I . un = - , - ; Takmg

It

LI _"

II~'OO

I But I Thus,

lin Vn

I

n +1

?

11->00

I

Vn

n +1

1 2 n

I the test fails.

'

•

and filllte


converges:.

I

Un

also converges.

lIlI converges when x ~ 1 and diverges when x > 1.

I

00

18.

=

I = LI -2n3-+X n- = 2"* 0

VII

and

VII

=

Test the series

II~I

(-I)" (log n ) 2

'

for absolute/conditional convergence

n

Solution

Un = (-1)" (;ogn) ; IUIII=(lo~n) n

n

Sequences of Series

473

"'flO~, x dx -- ' 'f le-'dl 2

x-

I/ [taking logx=l, x=e ' , /xdx=log/]

log2

~ -Ie -, + e-, I:.2 ~ 0 - [I-log 2].e -'"" ~ i(!Og 2 -I), which is finite.

I

I

:. By integral test ~]un is convergent => (Note that 19.

I

UII

1I11 converges absolutely.

is cgt. by Leibnitz's test).

Test the convergence of the series

I

I

(log log n )"

Solution

=

Given that u, I

LI u'1/n n-->oo n

= LI

n-->oo

[I] = log log n

I

By Cauchy's root test, 20.

I

(log log n)"

Un

0<1

is convergcnt.

Find the interval of convergence of the series, 7 I x 3 1 3 x 5 1.3.5 x x+-.-+-.-.-+--.-+ ..... . 2 3 2 4 5 2.4.6 7

Solution Term of the series, ull

=

1.3.5 ..... (2n-l)

lI

+1

=

2.4.6 .... .2n ( 2n +) 2

Xlll'l

)

2n + 1 X211 +1

2.4.6 ..... 2n

1.3.5 ..... (2n-I)(2n+l)

u

.(

(neglecting I st term)

. (2n + 3) 2

Lt u

lI +1

11-->00

un

By ratio test, 2

I

= Lt [ n-->oo

Un

(2n + 1)2 X2] (2n + 2)( 2n + 3

r

2

converges when x < 1 , i.e.,

When x = 1 ,the test fails;

~/n

4 Lt [ n ( + + I n2) X2] = x2 I/-'>OO n 2( 4 + ~ + 6 n2

19

Ixl < 1 => -1 < x < 1

r


474

Then

~_1=(4112?+1011+6 -1)= ~11+5 u

ll

411- + 211 + 1

+1

Lt IHOO

411- + 211 + 1

[11(~-1)]= Lt n2(8+1~) lIn+1

By Raabe's test,

IHOO

LUll

n ( 4 + 7~ + /~~2 )

converges when x

:. Interval of convergence of

=2>1

2

L

lin

is

2

= 1, i.e., x = ±1.

(-1:::; X:::; 1)

7 Vector Differentiation

7.1.1

Vector Point function and vector field Let P be any point in a region 'D' of space. Let r be the position vector ofP. Ifthere exists a vector function F corresponding to each P, then such a function F is called a vector point function and the region D is called a vector field.

Note: In what follows i,j, k are unit vectors along X, Y, Z axes respectively For example, consider the vector function

F = (x - y) i + xyj + yzk

..... (1 )

Let P be a point whose position vector is r = 2i

+ j + 3k in the region D of space.

At P, the value of F is obtained by putting x = 2, Y = I, z = 3 in F.

I.e.

At P,

F = i + 2j + 3 k

Thus, to each point P of the region D, there corresponds a vector F given by the vector function (I). Hence F is a vecotr point function (of scalar variables x, y, z) and the region D

is a vector field.


476

Scalar point junctioll and scalar fielit Ifthere exists a scalar 'f' given by a scalar function 'f' corresponding to each point P (with position vector r) in a region 0 of space, of' is called a scalar point function and 0 is called a scalar field. As an example, let P be a point whose position vector is r = 2i + j + 3k. Consider f= xyz + xy + z Then the value of f at P is obtained by putting x = 2, y = I, z = 3

f= 2.1.3 + 2.1 + 3 = II

i.e., At P,

Hence the scalar' II ' is attached to the point P. The function 'f' is a scalar point function (of scalar variables x, y, z), and 0 is a scalar field.

Note : There can be vector and scalar function of one or more scalar variables.

7.1.2 Differentiation of a vector Ifr (u) = r,(u)i + r2(u}j + riu)k, (where rl' r2, r3 , are scalar functions of 'u') be a vector function of' u' , then,

Lt

dr du

Or

~Oou

ou

~ dr,.

dlj. du

Lt

r{ u + ou) - r{ u)

ou ~O

ou

dr2 . du

dr3 k du

= L.,-l=-l+-}+-

du

Example If r(u) = (3u 2 + 5u + 6) i + 3U7 - 4uk, Find dr , when u = 1 du

dr du

= {~(3U2 +5U+6)li+{~(3U2)}j+{~(-4U)}k du

1

du

du

Note: We can apply the above rule of derivative to the case of partial derivatives also ex : If A

=

a

(ryz)i + (xyz}j - (3;Jyz2)k

2 A find - - at the point (I, -I, 2)

axOv

Vector Differentiation

aA ay

-

477

)t.

=

{ - a (2 x yz ay

=

(x 2z)i + (2xyz)j - (6x1yz2) k

=

a {xy 2Z }. j -a- {3x·yz3 ') ')} k 1+Oy

Oy

2xz i + 2yzj - 18x2yz2 k

At the point (I, -1,2)

a2 A

ax~v

=

4i - 4j + 72k

7.1.3 Application to space curves Let ret) = x(t)i + y(t)j + z(t) k represent the position vector of a point (x. y. z) on a space curve whose equations are given by x time.

=

x(t), y = yet), z

= z(t), where 't' is (K + M, Y + ~Y, z + 6Z)

dr dx. dy. dz Then -=-I+-J+-k dt dt dt dt and x

(i) dr represents the velocity vector v (or tangent vector) of the point (x, y, z) dt (ii) d 2 r represents the acceleration vector a at the point (x, y, z) dt 2 Ex:

If a particle moves along a curve x

=

e-t, y

=

2 cos 2t, z = 2 sin 2t, where 't' is time.

I) find velocity and acceleration at time t = 0, and 2) find also their magnitudes Sol:

r = xi =

+ yj + zk

(e-t)i + (2 cos 2t)j + (2sin 2t)k


478

dr d v= - = dt dt =

(e-~i

d d + - (2cos 2t}j + - (2sin 2t)k dt dt

(-e-t)i -(4sin 2t}j + (4 cos 2t)k

d 2 r dv d d. d a= -=-=-(-e-t)i-(4sm2t}j+ - (4cos2t)k 2 dt dt dt dt dt = (e-t ) i-(8cos 2t)j-(8sin 2t) k

.... (2)

Putting t = 0 in (I), velocity as t = 0 is v = -i + 4k Magnitude =

10

putting t = 0 in (2) acceleration at t = 0 is a = i-8j Magnitude =

J65

7.2 Gradient of Scalar Function 7.2.1

The Vector differential operator 'DEL' or 'NABLA', denoted as 'V' ' is defined by V'

=i

a .a ax ay

k

a az

-+ }-+ -

(i,j, k are unit vectors in x, y, z directions)

This operator' V' ' is used in defining the gradient, divergence and curl. Properties of' V' ' are similar to those of vectors. The operator is appled to both vector and scalar functions.

7.2.2 Gradient If ~ (x, y, z) is a scalar function, defined at each point (x, y, z) in a certain region of space and is differentiable, the gradent of as,

.a .a ( ax ay

a) az

grad ~ = V' ~ = 1-+ }-+k-- ~

(which is a vector function)

~

(shortly written as grad

~)

is defined


479

:. If ~ defines a scalar field, 'grad' ~ or V ~ defines a vector field.

7.2.3 Physical significance of 'grad

~'

:

If ~ (x, y, z) = c (c being a constant) represents a surface, then 'grad ~'represents the normal vector to the surface at the point (x, y, z) For, if r = xi + yj + zk, is the position vector of the point (x, y, z) on the surface, we have, dr = (dx) i + (dy) j + (dz) k which is in the tangent plane to the surface of (x, y, z)

(.,'

~

=c)

:. The vector' V ~' which is 1.- r to the tangent plane is the normal vector to ~ =

c at(x, y, z)

7.2.4 Directional Derivative

V~.a

If a be any vector,

lal

which represents the component of V

.'~

~

in the direction of

a is known as the directional derivative of' ~ , in the direction of a, (1) Physically the directional derivative is the rate of change of' ~ , in the direction

ofa. (2) The directional derivative will be maximum in the direction of V a=

V~) and the maximum value of the directional derivate = It~~

=

~

(i.e.,

IV~I·

7.2.5 Some basic properties of the gradient If

~

and \jJ are two scalar functions,

'J) grad (~+\jJ)

Proof: grad

=

(~+ \1')

grad ~ + grad \jJ(or) V(~+\jJ)= V~+V\I'

=

~

V( + \jJ)

=

{! (~+

\jJ)} i + { ;

(~ + \jJ)} j + {! (~+ \I')} k


480

8ljI. 8ljI )

o~. ~. ~ ) + (Oljl. = -I+-j+-k -I+-j+-k ox oy OZ ox ay OZ

(

= Y'~+Y'ljI

(2) grad (~ljI)= ~ (grad\fl) + ljI(grad~) (or) Y'(~ljI) = ~ (Y'ljI)+(Y'~)\jJ Proof: grad

=

{! (~ljI)}i

(~ljI)

Y'(~ljI)

=

+{;

(~ljI)} j + {! (~ljI)}k <.

oljl a~).1+ = ( ~-+ljIox ox

(8ljI

O~).j+ (8ljI ~)k ~-+ljI-

~-+-

ay

oy

=

~(Oljl i + oljl j + oljl k)

+ ljI

=

~(grad \fI) + ljI(grad ~)

ox

ay

oz

~) (3) If ljI :;t: 0, grad ( ljI

=

OZ

OZ

(o~; + o~ j + o~ k) = ~(Y'ljI) + ljI(Y'~) ox

oy

OZ

ljI (grad~ ) - ~(gradljl ) (ljI)2

(Proof is left to the reader.) Solved Examples

Ex. 7.2.6

Iff= x2yz, find gradfat the point (I, -2, 1).

Sol:

f= x2yz; ": grad ~ = Y'~ = 0/ + ay } + OZ k,

o~.

o~.

o~

o 0 0 gradf= -8 (x2yz); + - (x2yz}j + -8 (x2yz)k x ay 'Z = (2xyz)i +(x2z}j + (x 2y)k

:. At the point (1, -2, ]), grad

~

= - 4; + j - 2k

Ex. 7.2.7

Find the unit normal to the surface xy +yz + zx = 3 at the point (I, 1, I).

Sol:

If ~ = c is a surface,

Y'~

is the normal to it.


481

Here f= xy +yz + zx

= (y + z)i + (x + z)j + (y + x)k :. normal at(l, 1,1) = 2i + 2j + 2k 2;+ 2j + 2k . :. Umt normal = ~ 22 + 22 + 22

Ex.7.2.8

(a)

J3

Find the directional derivative ofl= towards the point (2, I, 3).

1 = 2e2x-ytz;

Sol :

;+ j+k

2e2x-ytz

at (1,3, I) in a direction

Vf = 2e 2x- ytz (2i - j + k )

Vll(1,3,1) = 2.e 2- 3+1(2i - j + k) = 4i - 2j + 2k Let A = (1,3, I) and B = (2, 1,3), AB = (2-1)i + (l-3)j + (3-1)k = i - 2j + 2k = a (say)

¥

. . I denvatlve . . JI1 . the d'IrectlOn . 0 f a = Vf. a DlrectlOna

= (b) Ans:

(4i-2j+2k).(i-2j+2k) ,Jl +4+4

=

(4+4+4) 3

=4

In the 'Problem (a)' find the maximum value of the directional derivative

Maximum value of directional derivative =

IV11

= I 4i - 2j + 2k I = ,J16 + 4 + 4 = 2 J6

Ex.7.2.9

Find the acute angle between the surface point (2, 1, 1).

Sol :

Let 1 = xy2z = 4 be the surface (1)

xyz = 2 and:x2 + y + z3 = 6 at the

-


482

Normal vector to (1) at (2, 1, I) = Vfl(2

1,1)

= 1(Vz); + (2xyz}j + (xy)k 1(2, I,

I)

= ; + 4j + 2k = a (say). Let g = (.xl y + .?) = 6 be the surface (2) Normal vector to (2) at (2,1,1) = Vg 1(2,1,1)= (2x; + 2yj + 2zk)i(Z,I,I)

= 4; + 2j + 2k = b (say) :. Angle between the surfaces

= Angle between the normals to them = Angle between a and b - cos -

I

=

cos

_1(a,b)1 =cos -II

-1 (

ab

4+8+4 .Jl+16+4.J16+4+4

J2t16) J24 = cos

-I (

16)

6.J14 = cos

-I (

I

8)

3.J14

7.2.10

Find the constants p and q such that the surfaces px2 4x2y + z3 = 4 are orthogonal at the point (1, -1, 2)

Sol:

Letf= px2 - qyz - (p + 2)x = 0 be surface (1), and

-

qyz

(p + 2)x and

=

Let g = 4.xly + z3 = 4 be surface (2) Normal to (1) at (1, -1, 2) = Vf

(1-1

"

2)=(81 ; + 81 j + af k 11(1_12) ax ay az

y"

= [(2px- P - 2); -{qz)j - (qy}k]I(I,-1,2)= (p - 2); - (2q)j -(q)k Normal to (2) at (1, - 1, 2) = Vg 1(1,-1,2)

= [(8xy); + (4x 2)j + (3z 2)k]1 (1,-1,2) = -8; + 4i + 12k = b (say) Since the surfaces (1) and (2) are orthogonal, a.b = 0

:. -8(p - 2) + 4(-2q) + 12(q) = 0

=> -8p + 16 - 8q + 12q = 0

=

a (say)

483


=> -8p + 4q + 16 = 0 => 2p - q

=

..... (i)

4

Since the point (1, -1,2) lies on (1), we have,

p + 2q - p - 2 = 0 => q

=

I :. p = 512, q = 1

from (i) we get,p = 5/2

7.2.11

If r = xi + yj + zk and r = Irl

Sol:

Let ~ = r3 a~

Then, ax

=

(x 2 +

=

2"3

a~

similarly 8y

grad =

~=

=

~ x 2 + y2 + Z2 show that grad (r3) = 3r r

T + z3)3/2

(x 2 + T + z2)3/2-1 . 2x = 3x r a~

=

a~.

3yr; and az 3zr ~.

~

- l + - ) +- k ax 8y az

(3xr) i + (3yr) j + (3zr) k

= 3r (xi + yj + zk) = 3r r Aliter:

Ifr2

=

xl + T + z2, we have, 2r ar = 2x => ar _~

ax

.. ar Slmtiarly 8y

ax

r

yarz =r'az r

=- .-

grad ~ = Vr 3

a 3. a 3 . a 3 ..1 ar 2 ar. 2 ar = -(r )l+-(r )}+-(r )k =3r-i+3r -}+3r - k ax 8y az ax 8y az

zk] =3rr

[x.

..1 y. =3r -l+-)+r r r

7.2.12

Evaluate grad r"

Sol:

Let ~= rn

= (xl + T + z2)nl2

~ =!!{x 2 + y2 + Z2 ax

2

r/

t

2 -

.2x =nx{x 2 + y2 z2

r/z-

t


484

..

sImilarly -o~ = ny ox grad rn = grad

~

(X 2 + y2 + Z2 )'?z-/,I and -o~ oz

(222)";-1

= nz x + y + z

o~. ~. ~k = - I + - j +ox 0' oz

= n, {x 2 + y2 + Z2 )'fi- I • (xi + yj + zk) = n rn-2.r Aliter:

g

o~ or. o~ or. o~ or 1+ (o~l. - j+ (~) - k=-.-I+-.-j+-.-k (-o~). ox oy OZ orox or0' oroz

~

rad

= nrn- 1 . !.. i + nxn- I . y j + nxn- I . !... k = nrn-2.r r

r

r

Ex.7.2.13

If A is a constant vector prove that grad (r . A) = A

Sol:

Let A = A I i + Aj + A3k r

= xi + yj + zk

r. A

=

A1x + A2Y + A3z = f(say)

of ox = AI'

81

Ex.7.2.14 Prove V Let f=

81

-

ox

=

~

",I

81

0'

:. grad (r.A)

Sol:

(A I' A 2 , A3 being constant functions)

= A2, OZ = A3

= gradf= 81 i+ 81 j+ 81 k ox 0' oz

=A i+Aj+A k=A I

3

~I(r).r

(~(r»=--

r

(r)

ar

(r) -

ox

'Y

=

~

or ox

I'X

(r).r

81 =~I(r)ar =~I(r).y 0' 0' r 81 = ~I(r) or = ~I(r).!... OZ

.. V

OZ

or

81. of. 81 0/ + 0' ) +

oz'''

1(r)

=

W

r

or

y

0'

r

-az =-r (see Aliter of ex. 7.2.11)

r

(~ (r» = V f=

z

x

1(r)r

(xi+yj+zk) =

W

485


Ex.7.2.15 Find the equations for the tangent plane and normal line to the surface z = x 2 + at the point (2, I, 5).

.r

Sol:

Let r = xi + y) + zk be the position vector of any point P(x, y, z) on the surface. Let r l = xli + yJ + zl k be the position vector offixed point A(xI'YI' z,) on the surface. Then AP = (x - xI) i + (y - YI») + (z - z\) k

= r - r,

Let n be the normal to the surface at A. Then, since AP is perpendicular to n, we have, ..... (I)

(r-r\)n=O which is the equation to the tangent plane at A. Here, in the given problem r-r, =(x-2)i+(y-I»)+(z-5)k and 11 = V (xl + =

.r -

z) at A (2, I, 5)

(2xi + 2y) - k)I(2.).5)

= 4i + 2) - k

:. The tangent plane at (2, I, 5) is, (from (I», 4(x - 2) + 2(y - I) - I(z - 5)

=0

~4x

+ 2y - z = 5

.... (2)

From (2), the direction ratios of the normal Iine at A are 4, 2, -I :. Equation 10 tpt( pormal line at A are,

y- Y) z-z) he \ w re abc

x-x)

-- =

~,=--

(xl'YI' zl)

= (2,_1,5) and (a, b, c) = (4; 2, -I)

:. The equations of normal line are x - 2 4,

= Y -I = z 2

5

-I

From the above example. we have to remember the following : Let (I)

+

=

c be any given surface and (xl' YI' zl) be a point on it; then

Equation to the tangent plane to

+=cat(xI'Y"z,) is

486


Equations to Normal line at (xl,y\, zl) are

(2)

= Y- y, = z-z, 8<1>/8x 8<1>/ay 8<1>/8z x-x,

Exercise 7(a) I.

If
=

2xz3 - 3xlyz, find V and IVI at the point (2, 2, -I) [Ans: (i) 22i+ 12}-12k(ii)2M3]

2.

3.

If V = 2x i - 3y) + z 3k, and and Vx V at the point (1,2,3) [Ans . (I) - 426 (2) 6i + 352} t 156k] Iff=2xyzandg=xly+z,fing V (f+g) and V (fg) at the point(I,-I,0) [Ans : -2i + ) - k; 2k ] 2

4.,Jr + 1 ~)

7

[AI:1S : (6 - 2~/2 - 2r -3) r ]

4.

Evaluate V (3r

5. 6.

If = r2 e- , show that grad = (2-r) e-r r Find a unit normal vector to the surface z = xl + Y at the point (I, -2, 5)

-

~r

r

I

[Ans: 7.

51 (2i -

4j - k)

Finq the equations to the tangent plane and the normal line to the surface xz2 + xly = z - I at the point (I, -3,2)

°

"

. .. x-I y+3 z-2 [Ans: (I) 2x- y- 3z + 1 = (n) -2 =-1- =-3-

8.

Find ~uations to the Tangent plane and normal line to the surface y at the point (1, 5, -2)

= xl + z2

x-I y-5 z+2 [Ans: (i) 2x-y-4z= 5 (ii) -2-=-=1=--=-4 9.

Find the directional derivative ofU = 4xz3 - 3x2yz at (2, 1, -2) in the direction of (3i - 2) + 6k). [Ans:

10.

Find the directional derivative of

384

7]

= 4e2x- y+z at the point (1, I, -1) in a direction

towards the point (2, 3, 1) [Ans : 8/3]

487


II.

axl

Find the values of the constants a, b, c so that the directional derivative off= + byz + cz2x 3 at (1,2, -I) has a maximum magnitude 64 in a direction paraIlel to z-axis. [Hint:

/0:./ at (I, 2, -I) is IlleI to z-axis.

:. Equate coefficients of i and j to zero and IV ~ = 64. Thus get 3 equations in a, b, c and solve them] [Ans: a =6, b = 24, c = - 8] 12.

Find the acute angle between the surfaces xlz = 3x + z2 and 3x2 - l + 2z = I at thepoint(I,-2, I)

J3

[Ans : cos- I - - ]

7J2

13.

Find grad ~ if r = xi + yj + zk, r = Irl and

(i)

~

= Log r (ii)

~

I = r

(iii)

~

=r [Ans: (i) r/~ (ii) -r/~ (iii) r/r]

14.

Find the directional derivative of g = x 2l + lz2 + z 2x 2 at the point (I, I, -2) in the direction of the tangent to the curve x = e-t, y = 2 sint + I, z = t - cos t at t = o. [Hint: Tangent vector to the given curve is

dx. dy. dz k -l+-}+dt dt dt 2

[Ans: 15.

Find the acute angle between the normals to the surface xy = z2 at the points (1,9,3) and (3, 3, -3)

16.

J6]

If r

= xi + yj + zk and

4> = x 3 +

Y + z3 -

3xyz, show that r, grad 4> = 34>.


488

7.3

The Divergence of a Vector Function

7.3.1

If A = A)i + A2k is a vector function, defined and differentiable at each point (x, y, z) in a certain region of space [i.e., A defines a vector field], then the divergence of A (abbreviated as 'Div A') is defined as, DivA= V.A

. a }-+ . a k a) = (1-+ ax ay -az' (A )1'+A21'+A3k) =

(aAax i + aAay + aA3) az 2

I

(since i.i = j j = k.k = 1)

Note: (1) Div A is a scalar field (2) V.A

7:-

A.V

7.3.2 Physical significance of the divergence If A represents the velocity of fluid in a fluid flow, Div A represents the rate offluid flow through unit volume. (or) Div A gives the rate at which fluid is originating at a point per unit volume. Similarly if A represents the Electric flux or heat flux, Div A represents the amount of electric flux or heat flux that diverges per unit volume in unit time.

7.3.3 Some properties of Divergence If A, B , are vector functions and '/' is a scalar function, then, prove that (1)

Div (A + B) = Div A + Div B (i.e) V. (A + B) = -V . A + V .8

Proof Let A = A)i + Aj + A3k B = 8 1i + B~ + B3k A + B = (A) + 8) i + (A2 + 8 2 )j + (A3 + 8 3) k

a ax

a

a az

Div A = -(AI +B l )+-(A 2 + 8 2 )+-(A 3 + 8 3 )

ay

= (aA1 + aA2 + aA3)+(aBI + aB2 + aB3)

ax

ay

~

= Div A + Div B.

az

ax

ay

az

489


(2)

Prove that, Div (fA) = (grad.f) .A + .f{Div A) i.e. V .(fA) = (V .f). A + .f{ V A) Proof: Let A = Ali + A'll + A3k then fA = fAli + fA'll + fA3k V .(fA)

a

a

a

= ax (fA.)+ ay (fA2)+ az (fA3) = fQ~+A. af + f aA2 +A2 af + f aA3 +A3 8j -Ox

ay

ax

ay

az

az

..... (1)

..... (2)

..... (3) (1), (2), (3) => V .(fA) = (v.f).A + f(v .A)

7.3.4 Solenoidal vectors: A vector A is said to be solenoidal if Div A = 0 Solved Examples

Ex. 7.3.5

If A = (.x2y) i + (xy2z)j + (xyz)k, find div A at the point (1, -1,2).

Sol:

A = (.x2y) i + (xyz)j + (xyz)k

. aA. aA 2 aA 3 DIVA=-+-+ax ay az =

a 2 a 2 a -(x y) + -(xy z) + -(xyz) ax ay az

=

2xy + 2xyz + xy

= xy (2z ,+ 3) :. At (1, -1, 2), Div A = (1) (-1) [4 + 3] =-7


490

Ex. 7.3.6

If V = 2xyi + 3x2yj - 3pyz k is solenoidal at (1, 1, 1), find 'p'.

Sol:

. Dlv V

=

a a 2 a -(2xy)+-(3x y)+-(-3pyz) ax oy oz

= 2y + 3xl - 3py At (1. 1, 1). Div V = 5 - 3p Since V is solenoidal, Div V = 0 :. p =

5

3"

Ex.7.3.7

Ifr = xi + yz + zk, and r = Irl, show that Div (r3 r) = 6 r3

Sol:

r= ~X2+y2+Z2

:. r3 r = (xl+y +z2)3/2 (xi+yj+zk)

= Ali + Aj + A3k (say) then, A I = x(xl + y + z2)312

A2

=

y(x2 + Y + z2)3/2

A3 = z(xl + y + z2)312

oA 3 2 .. l =3z r+r

oA2 .1. .. l similarly ~ =3y r+r~

OZ

. oA I oA2 oA3 DlvA= - + -+ax oy oz = 3r (xl + Y + z2) + 3r3 = 3r3 + 3r3 = 6r3 Aliter:

r3 r = r3 xi + r3 yi + r3zk.

a Or x -8 (r3 x) = r3.1 + x . 3r2 -8 = r3 + x . 3r2 . x'

=

r3 + 3xlr

x

r


Similarly

491

o

ay

0

(~y) = ~ + 3yr and oz (r3z) = ~ + 3z2r

DivA=6~

Ex. 7.3.8

l

Evaluate Div [ r grad (r-3) ] or V. {rvC 3 )}

(x.

_ 4 or . ar . --4 -or k--3r _ --4 -i+-j+y. Z --3r -1-3r4 -j-3r ox c:y oz r r r

=

-3r-5 (xi + yj + zk)

r grad (~) = -3r--4 (xi + yj + zk)

= Ali + Aj + A3k

(say)

where AI = -3r--4x, A2 = -3r--4y , A3 = -3r--4z

aA I = ~ [-3r--4 x]

ax

ax

= -3r--4 . 1 + x. -3. -4 r- 5

. ::

= -3r--4 + 12 x r- 5

= -3r--4 + 12 x 2 r-6 Similarly,

oA c:y2

=-3r-4 + 12yr-6

:. Div (r grad r-3) =

oA ax

oA

oA

Oy

az

_I + __ 2 +_3

= -9r-4 + 12 r-6 (x 2 +

y + z2) = -9r-4 + 12 r--6 . r2 = 3r-4

(Alternate method is left to the reader).

.

~

k)


492

Ex.7.3.9

Show that V .(x" r)

Sol:

r" r

= (n + 3) r". Hence show that r/f3 is solenoidal.

= r" (xi + yj + zk)

V .(r" r)

a

= -a (xr")+ x

a

a az

+-

~, (yr") v)'

(zr")

z)

y + z.= 3r" + n ['1-1 ( x.-X + y.r

=

r

r

3r" + n r"-2 . r2 = (n + 3) r"

Ifn = -3, (r-3 r) = (-3 + 3) r-3

=0

r

:. 3 is solenoidal r

Ex.7.3.10 Prove that Div (CIA + C 2 B) = C I Div A + C 2 Div B, where CI' C 2 are constants.

Sol:

Let A

= A I i + Aj + A3k

B = Bli+ Bj+ Bl CIA + C2B = (CIA I + C 2B I) i + (C IA2 + C 2B2U + (C IA3 + C 2 B 3 )k Div (CIA + C2B) =

a ax

a

(CIA I + C2B I) + Oy (C IA2 + C 2B2)

=C (MI+M2+M3)+C(aBI+aB2+ I

ax

Oy

az

2ax

ay

aB

3)

az


493

Ex.7.3.11 If A = 2xi + 3yj + 5zk and f= 2xyz, find div (fA) at (1,2,3). Sol:

2xyz (2xi + 3yj + 5zk) = (4x 2yz)i + (6xyz)j + (10xyz2)k

fA

=

a a a Div (fA) = ax (4x2yz) + ay (6xyz) + az (10xyz2)

= 40xyz :. At (\,2,3), div (fA) = 240 Aliter:

div (fA) = D.(fA) = (V f). A + f( V A) Vf

= 2yzi + 2xzj + 2xyk

V fA

= 4xyz + 6xyz + \ Oxyz = 20xyz

f(v .A) = 2xyz (2 + 3 + 5) = 20xyz Hence V .(fA) = 40xyz and at (\,2,3) V .(fA) = 240

Ex.7.3.12 Iff, g are scalar fields show that V fx V g is solenoidal

vfx V g

=

a.fjax agjax

j

k

aljay agjay

atjoz agjaz (Suffixes denote partial derivatives)

a

Div (V fx V g) = L ax Vygz - fzgy) =

L

(!y gxz + gzfxy - fz gxy -

=0 :. V fx V g is solenoidal

~fxz)


494

Exercise 6(b) I.

If V

=

(xlz)i - (2y3z1)j + (vz)k, find div A at the point (I, -I, I) [Ans. - 3]

2.

Ifr = xi + yj + zk, find div r

3.

If F = (3,xyz2)i + (2xy3)j - (xlyz)k, and ¢ = 3xl - yz,

[Ans.3]

find (i) Div F (ii) Div (¢F) and (iii) Div (grad¢); at the point (I, -I, I) [Ans. (i) 4 (ii) I (iii) 6 ] 4.

If V = (3xly - z)i + (xz 3 + y4)j - 2x3z 2k, find grad (Div V) at the point (2, -I, 0) [Ans. - 6i + 24j - 32k]

5.

Evaluate: (I) Div (r2r) (2) Div (r r) (3) grad Div (r/r) (4) div (r/r3) [ Ans. (1) 5r2 (2) 4r (3) -2r/r3 (4) 0 ] 3y4z2i + 4x3z7 + 6x2y3k is solenoidal

6.

Show that V

7.

Show that the vector F = (2xl + 8xyz)i + (3~y - 3xy)j - (4yz2 + 2x3z)k is not solenoidal, but G = xyz2 F is solenoidal.

8.

If a is a constant vector and V solenoidal vector.

9.

Determine the constant 'b' such that the vector, V = (2x + 3y)i + (by - 3z)j + (6x- 12z)kis solenoidal

10.

Ifr, and r2 are vectors joining fixed points A (x"YI' z,) and B(x2'Y2' z2) to a variable point P(x, y, z) prove that r, x r2 is solenoidal.

11.

If r

12.

If g = r-2n , find div (grad g) amd find 'n' such that 'g' is solenoidal.

=

= xi + yj + zk and r = Irl

=

a x r, where r

=

xi + yj + zk, show that V is a

show that, Div (grad rn)

= n( n+ I )rn- 2 .

[A ns.

7.4

2n(2n -I) I ·n-?n+2 ,- 2 ] r-

Curl of a vector function

7.4.1 If A is a differential vector function, then curl A is defined as, curl A = V' x A

If A

= Ali + Aj + A3 k , then

Curl A = a/ax AI

j

k

a/ay A2

a/az A3

495


=

(aAJ _

0'

aA 2 )i + (aA I _ aA3)j + (aA 2 _ aA I )k az az ax ax 0'

Note: The curl of a vector is also a vector

7.4.2 Physical significance of curl Let r = xi + yj + zk be the position vector cf a point P(x, y, z) of a rigid body rotating about a fixed axis about the origin 0 with an angular velocity ffi = ffili + ffi.j + ffi)k. Then the velocity V of the particle P is given by,

V

= ffi

x r

=

j

k

ffil

ffi2

ffi3

X

Y

z

a/ax

Curl V=

=

i

i (ffi 1+ ffi I) + j

j

k

8/ay

a/8z

(ffi2 + ffi 2) +

k ( ffi 3 + ffi3) = 2ffi

Thus the curl of velocity vector is twice the angular velocity of rotation.

7.4.3 Irrotational Vector: A vector V whose curl is zero is said to be an irrotational vector. 7.4.4 Properties :(1) V x (A + B) = V x A + V x B (or) curl (A + B) = curl A + curl B Proof:

Let A = Ali + A.j + A)k and B

= Bli + B.j + B)k so that


496

i

Curl (A+B) =

j

k

ajax ajay AJ+BJ A2 +B2 j

ajaz A3 +B3 j

k

k

= ajax ajay ajaz + ajax ajay ajoz = curl A + B (2)

If ¢ is a scalar function and A is a vector function Curl (¢A) = ((curl A) + (grad¢)

x

A

(or) V x (¢A) = ((V x A) + (V {P) x A Proof: If A = Ali + Aj + A3k, then cj>A = cj>Ali + cj>Ai + cj> A3k

j L\ x (cj>A)

k

= ajax ajay ajaz cj>AI

cj>A2

cj> A 3

= i[~(q,A3)-~(cj>A2)]+ j[~(cj>AJ)-~(cj>A3)J+ k[~(cj>A2)-~(cj>AJ)] ay az az ax ax ay =

"i[.I."'ayaA3 ~

+ A aq, _.I. OA2 - A aq,] 3ay "'az 2az

= .I. L( aA 3 _ aA2)i + L(A aq, _ A acj»i '" ay az 3ay 2az j k cj>ajax ajay ajaz i

=

AI

A2

j k aq,jax aq,jay acj>jaz i

-?>

A3

=cj>( V x A) + (V cj» x A

7.4.5 Conservative vector field: A vector field F, which can be derived from a scalar field q, such that F = Vcj>, is called a conservative vector field and q, is called the scalar potential ofF.

497


Solved Examples Ex.7.4.6 If A = (xy)i + (yz}j + (zx) Ii; find (a) curl A and (b) curl curl A at (1,2, -3)

Sol :

A

= xyi + yzj + zxk j

(a)

curl A = V x A =

k

a/ax a/ay a/az xy

yz

zx

a a].[ a a ] +k[-(yz)--(xy) a a] = I.[-(zx)--(yz) + } -(xy)--(zx) ay

= i(O =

&

&

&

&

ay

y) + j(O - z) + k (0 - x)

-yi - zj - xk

:. curl A at (1, 2, -3) = -2i + 3j - k (b)

curl A= V x (V xA)

=V

(-yi-zj-xk)

x

j

i

=

k

a/ax a/ay a/az -y

-z

-x

a].[ a a ] +k[-(-z)--(-y) a a ] = i -(-x)--(-z) + } -(-y)--(-x) [aya & fu fu fu ay = i(O - (-I) + j(O - (-1» + k(0 - (-I» =i+j+k

:. curIAat(I,2,-3)=i+j+k Ex.7.4.7 Show that V = xi + Ij + z3k is irrotational -

Sol:

Curl V = V x V

a 3)---(y a 2]) + }.[-(x)--(z a a 3]) +k[-(y a 2)--(x) a ] = I.[-(z ay

=0

az

:. V is irrotational

oz

&

fu

ay


498

Ex.7.4.8 If F = (4x + 3y + az)i + (bx - y + z)j + (2x + cy + z)k is irrotational, find the constants a, b, c i

j

k

ajax

ajay

ajaz

(4x+3y+az)

(bx- y+z)

(2x+cy+z)

Curl F =

Sol:

=

i[~(2X+CY+Z)-~(bXy+Z)]+ j[~(4x+3y+aZ)-~(2X+CY+Z)J ay az az ax k[~(bXy+Z)-~(4x+3y+az)] ax ay

+

= (c -

l)i + (a - 2)j + (b - 3)k

Since F is irrotational, curl F = 0 :. c - I = 0, a - 2 = 0, b - 3 = 0

i.e., a = 2, b = 3, c = 1 Ex.7.4.9 If r = xi + yj + zk, and r = Irl, find curl (rn r) Sol :

r = xi + yj + zk, r = Irl = ~ x 2 + y2 + Z2 rn r

=

(.xl + Y + :?)n/2 (xi + yj + zk) k

j

curl (rn r) =

ajay

ajax III

x(x 2 + y2 + Z2Y2

Y(X2 +

l

ajaz 1/ 1

+ Z2Y2

(? + y-?+z-?hJLY- Y."2n (2 x + Y 2 +z 2) 2z J

n x= I.[ z."2

n 2 + y 2 +z 2 ).2z-z."2(x n 2 + y 2 +z 2 ).2x ] + ).[ x."2(X

= i(O) + j(O) + k(O) =0

Z(X2 +

l

+ Z2 )"2

499

Vector Differentiation Aliter:

rn r

=

rn (xi + yj + zk)

curl (rn r) =

= "'.[ L..,l - a

ay _

-

'" .[

L..,l

j

k

ajax

ajay

ajaz

xrn

yr"

zr"

( zr n) - a ( yr n)] = "'.[ ar - y.nr n-I -ar] L..,l, z.nr n-I az ay az

z]

Y - y.n.r II_I -; f/zr n-I -;

= i( 0) + j( 0) + k (0) = 0 Ex. 7.4.10 Prove that, ifF

Sol:

Curl F =

= (x + Y +

ajax

j

k

a/ay

ajaz -x-y

x+ y+1 = i

I)i + j - (x + y) k, F. curl F = 0

[-I -0] + j[O + I] + k[O -1]

= -;

+j - k

:. F curl F = -I (x + y + 1) + 1.1 + 1(x + y)

=0

Ex. 7.4.11 P(x, y, z) is a variable point and Q(xl' YJ, zJ)' R(x2, Y2' z2) are fixed points. If U = QP and V = RP; Prove that curl (U x V) is equal to 2(U - V).

Sol:

P(x, y, z), Q = (xl' Yl' zJ)' R = (x 2, Y2' z2) :. QP = (x -xJ); + (y - YJN + (z - zJ)k = U RP

= (x -

x 2); + (y - Y2 N + (z - z2)k j

k

UxV= (X-XI)

(Y-YI)

(Z-Zl)

(x-x 2)

(Y-Y2)

(Z-Z2)

= ~{(y =~

=V

YJ) (z - z2) - (z - z\) (y - Y2)} ;

{y(z\ - z2) - z(yJ - Y2) + (yJ z2 - Y2 z J)}i


500

Curl (U

x

V)

=

I[ ~{x(y, -!

- Y2)- Y{X, -X2)+{X'Y2 -x2 y,)}

{Z{X, -X2)-x{Z, -Z2)+{Z,X2 -Z2XJ}}

= L(-(XI - X2) - (xI - X2); = L -2(x l - X2 ); = 2L(X2 - xI); = 2(U - V) Ex. 7.4.12 If F is a conservative vector field show that curl F = 0 Sol.

F is a conservative vecor field.

:. There exists a scalar field '~' such that F = j

k

ajay

ajaz 8$jaz

; :. Curl F = ajax a~jax

a~j~

i( a ~ a ~ 2

== L

V'"

'I'

8$. 8$. 8$ k = ax 1 + ~ J + az

2

~az

_

az~

)

== 0

Ex. 7.4.13 Show that F == (6xy + iJ)i + (3r -z)j + (3xz 2 - y)k is irrotational. Find ~ such that F == V ~ i

Sol:

Curl F =

ajax 6XY+Z3

j k aj8y ajaz 3X2 -z 3xz2-

J

== i (-1 + 1) + j (3z2 - 3z2) + k(6x - 6x) ==0 :. F is irrotational L tF== e

:. :

V~==

8$. 8$. a~k

-I+-J+ax ay az

== 6xy + iJ

..... (I)


501

a~ -

Oy a~

-

az

=

3x2 -z ?

= 3xz~

..... (2)

- y

..... (3)

Integrating equation (1) with respect to x we get ~ = 3.x2y + xz 3 + fly, z)

... :. (4)

Differentiating (4) partially w.r.t 'y' a~

af(y,z)

ay = 3.x2 +

ay

..... (5)

From (2) and (5) we have

af(y,z)

ay

..... (6)

=-z

Integrating (6) w.r.t 'y' we get fey, z) = -yz + h(z) ~ = 3x2 + zx 3 - yz + h(z)

Hence

..... (7)

I

difffrentiating (7) w.r.t 'z' we get ..... (8)

Comparing (3) and (8) we have h l(z) = 0, :. h(z) = constant, 'c' say Hence ~ Aliter:

=

3.x2y + xz 3 - yz + c -~ =6xy+z3

ax

a~

ay

=

3.x2 - z

(from 7) .... (I)

.... (2)

..... (3)


502

Integrating the above equations respectively w.r.t x, y, and z, we get

= 3x2y + xz3 + f(y, z) ~ = 3x2y - yz + g(z, x)

.... (4) .... (5) ..... (6)

~

~= ~

xz3 - yz + h(x. y)

should satisfy all the above three equations simultaneously

(i.e.) (4), (5) & (6)

3x2y + xz3 - yz + c, where c is a numerical constant. [Herej= -yz, g = xz3, h = 3x2y will satisfy (4), (5) & (6) :. ~ =

Note:

Ex.6.4.14 Iff(r) is differentiable, show that f(r)r is irrotational Sol:

fir) r = fir) (xi + yj + zk)

Curl (f(r) r)

j o/Oy

%x

=

k

%z

j

xj{r) y j{r) zj{r =

..

I{; {zj{r)}- ! {y j{r)})

=

I{z./{r)~ - y./{r):)

=

I{z./{r)~ -y./{r);) =0

f{r)r is irrotational

Ex. 7.4.1SlfU and V are irrotational, prove that U Sol:

Let U = Uti + U2i +

x

V is solenoidal

Ui

V=V t i+V2i+ Vi U is irrotational

=> Li[OU 3

oy

..

sImilarly,

_

:. curl U

OU 2 ]

oz

=

=

0

'" .[OV 3 oV 2] = 0 Oy oz L-l -

- -

0

..... (1)

..... (2)


503

}

k

VxV= VI

V2

V3 =1:;(U 2V 3 -U 3 V 2)

VI

V2

V3

..... (3) au 3 _ au 2 • au I _ au 3 . au 2 _ au I From (I) ,we have, - - - - - , - - - , - - - 0' az az ax ax 0' av3 _ av2 • aV _ av3 . av2

aV

an d from (2) we have - - - -I- - - - -I , '0' az' az ax' ax 0' _

..... (4)

..... (5)

Substituting the six equations of(4) & (5) in (3), we observe that all the 12 terms of (3) will get cancelled. Hence Div (U x V) = 0 ~

U x V is solenoidal Exercise - 7(c)

I. If V = (2xz2);_ (yz)j + (3xz3)k, andf= xlyz, find the following at the point (I, I, I) (a) curl V (b) curl (jV) (c) curl (curl V) [Ans. (a) ; + } (b) 5; - 3} - 4k (c) 5; + 3k] 2. If 'g' is a scalar function, show that curl (g grad g) = 0 3. Find the value of the constant 'p' for which the vector V = (pxy - z3) + (p - 2) xl} + (I - p) xz2k is irrotational. [Ans: 4] 4. Find the constants a, b, c so that the curl of the vector A = (x + 2y + az); + (bx - 3y - z)j + (4x + cy + 2z)k is identically equal to zero [Ans. 4, 2, -1] 5. If r =

xi + y) + zk, r = Irl, show that curl (;Z ) = 0 and find a scalar function of' such r

that 2"= -v J,.f(a) = 0 where a> r

o. [Ans :f= log

(~)]


504

6. If r = xi + yj + zk, and p, q are constant vectors, show that (I) curl [(r x p) x q] = ( p x q) and (2) curl [(p.q)r)] = o.

7.5

Laplacian Operator: V2

7.5.1

V2 =

=

V.V

=

(i~+ j~+k~) ax j~+k~).(i~+ ry az ax ry az

a2 + -a2' + -a22 ) .IS called the Laplacian . Operator ( ax 2 ay2 az -

'V2' can be applied to both scalar and vector functions as shown below. a2~

2

a·2~

a2~

V~=-+-+-

ax 2 ry2 az2

where '~' is scalar function If A = Ali + Aj + A3k, is a vector function, then V2 A = =

a2 a2 a2 ) ( ax2+ ay2 + az2 (Ali + Aj + A3k)

(V2AJ+(V2A2)j+(V2A3~

7.5.2 Vector Identities: we shall give below some vector identities with proofs. \. If '¢' is a scalar function curl grad

Proof: Curl grad cjI= V x

~

= 0, (or) V x V ~ = 0

acjl acjl. acjl ) (-i+-}+-k ax ry az k

j

ajax ajry ajaz acjljax acjljay acjljaz = L>[~(acjl)-~(a~)l ay az az ry

= 0, assuming that

'~'

partial derivatives.

possesses continuous second order

505


(2)

IfY is a vector function, Div (Curl A) = 0 (or) V.(Vx

A) = 0

Proof: Let A = Ali + A-j + A3k CurIA=

3 -aA-, ) " .(aAay az ~I

"~a {aA3 aA)} ax -ay- - -aza~A3 a2 A2 a2A4 a~A3 a~A~ a~AI = -----+-----+----axay axaz ayaz ayax azax azay

Div (curl A) =

=0 (3)

Proof:

If A is a vector function, curl (curl A)

curlA=

grad (Div A) -

3 "~l.(aAaA, ) ay -az-

j

Curl (curIA)=

=

k

a/ax a/ay a/az aA3_ aA2) aA I _ aA3) aA2_ aAI) ( ay az ( az ax ( ax ay

I[~(aA2 _ aA I)_~(aAI _ aA 3)]i ay ax ay az az ax

V2 A (or)


506

=

(4)

-'1 2A + '1('1.A)

'1.(AxB)=B.('1xA)-A.('1xB), (A, B are vector functions) (or) Div(AxB)=B. curl A-A. curl B. Proof: Let A = A) i + A?J + A3k and B = B) i + B?J + B3k

j

k

B = AI

A2

A)

BI

B2

B)

i

A

x

=

I

i(A2B) - A)B2)

a ox

Div (A x B) = "-(A2B) -A3B 2) LJ = "

L:.

'1xA

=

(A OB3 + B OA2 + A OB2 + B OA3) 2 AX 3 AX 3 AX 2

ax

..... (1)

a/ax AI

..... (2)

507


Similarly, A. (V'

x

" (BB3 0' - BB?) Bz-

B) = L.J AI

..... (3)

Expanding the summations of (I), (2) & (3), we observe that, Div (A x B) = B. (V' x A) - A. (V' x B)

7.5.3 Operation of V' on product of two functions Suppose ~ and 'P, are two scalar or vector point functions. When V' is operated on the product of ~ and \j1, the following rule is useful. V' (~'P) = V' (~o \{J) + V' (~'P 0) wherein the suffix '0' indicates that the function is not to be varied, that is, V' is not to be operated on that function. After the completion of operation of V' the suffixes are dropped. While proving the identities the following are useful. (a)

a.b = b.a

(b)

a x b = -b x a

(c)

axa=O

(d)

a.(bxc)=(axb).c

(e)

a x (b x c) = (a. c)b - (a. b) c

(f)

(a. c) b = a x (b x c) + (a . .b}C

(g)

a. a = a2

(h)

x operation is always between vectors only

vector identities using V' operation: (i) V' .(FG) = V' . (FG o) + V' . (FoG) = V' F.G o + Fo V' .G :. div (fG) = G. grad F + F div G (ii) V' x (FG) = V' x (FG o) + V' x (FoG) V' x (FG o) = V' F x Go V' x (FoG) = (V' x G)Fo=Fo(V' xG) :. V' x (FG) = V' F x G + F( V' x G) (i.e.) curl (FG) = (grad F) x G + F curl G

due to (h)


508

(iii) V'. (F x G) = V' .(F x Go) + V'. (Fo x G) V' . (F x Go) = (V' x F) .G o

due to (d)

Go .( V' x F)

due to (a)

=

V' . (Fox G) = - V' . (G x F0)

due to (b)

(G. F0)

due to (d)

= -

V'

x

=-(V' x G). Fo

due to (a)

=-Fo(V' x G) :. V'. (F x G) = G. (V' x F) -F. (V' x G)

Thus div (F x G) = G. curl F - F. curl G (iv) V' x (F x G) = V' x (F x Go) + V' x (Fo x G) V' x (F

x Go) = (V' . Go) F -

due to (e)

(V' . F) Go

due to (a)

= (Go· V') F - Go (V' . F)

due to (e)

V' x(FoxG)=(V'.G)Fo-(V'·Fo)G

due to (a)

= Fo (V' . 0) - (F o · V') G :. V' x (F x G) = (G . V') F - G( V' .F) + F (V' .G) - (F. V' ) G

Thus curl (F x G) = (G. V' ) F - (F. V' ) G + F (V' .G) - G (V' .F) "I-

(v) V' (F . G) = V' (Fo . G) + V' (F .G o) { V'(Fo.G)=Fox(V' xG)+(Fo. V')G

due to (t)

V' (F . Go)

due to (t)

=

V' (Go. F)

=

Go x (V' x F) + (Go. V') F

:. V' (F . G) = F x (V' x G) + G x (V' x F) + (F . V') G + (G . V') F

Thus grad (F . G) = F x curl G + G x curl F + (F . V') G + (G . V') F (vi) curl (grad F) = V' x (V' F) = (V' x V') F = 0 (vii) div (curl F) = V' .( V' x F) = V' x (V' . F) = ( V' x V' ). F = 0

due to (c) due to (d) due to (c)

(viii) curl (curl F) = V' x (V' x F) = (V' . F) V' - (V' . V' ) F =

V' ( V' . F) - V' 2 F

=

grad (div F) - V' 2 F

due to (e) (since V'. must not appear at the end)

509


Solved examples

Ex.7.5.4 If/= x 21z2, find V 2/ at (1,2, 1) Sol:

Y

V

=

=

0

2

2

/

ox- 0/

0

2

/

oz-

21z2 + 6x2yz2 + 2x21

:. at (1, 2, 1),

Ex.7.5.5 Show that, if r Sol:

0

/

--J +--+-J-

v2j= 16+ 12+ 16=44. =

xi + yj + zk, r =

\rI, then

~ = (x2 + Y + z2)n12, (·:r = ~x2 + y2

V 2r ll = n(n + 1) rll

2

+Z2 )

... (1) 2

Similarly -02 (r n) = n ( n- 2)y 2r n-4 + nr n-2

... (2)

, oy

2

( n) J n-4 and -a2 I! =n ( n- 2)z-r +nr n- 2

... (3)

oz

=

3n,.n-2 + n(n-2) ~-4 (x 2 +y + z2)

= 3n~-2 + l1(n - 2) r"-2

= n(n +

1) ~-2

adding (I), (2) & (3)

(": x 2 +

y + z2 = r2)


510

ar 11-1 .x -a (r") =nr 11-1 .-=nr ax ax r

Aliter:

=

n,.rr2 . x

~~II)= n(r 2 az

ll

-

2.1 + x.{n _ 2)r"-3. ar) ax

~ n(,"-' + (n-2}r"-'. x:) =

nr"-2 + n(n-2) r"--4. xl

... (I)

2

Similarly -a2 (r ") =nr 11-2 +n (n- 2)r n-4 .y 2

... (2)

, ay

and

~2 ~n ) = nr n- 2 + n{n - 2 )r n-4.Z2

... (3)

az

Adding (I), (2) & (3) we get,

=> V2rn = 3nr"- 2 + n{n - 2)r n- 4.r2 = ,.rr2[3n + n2 - 2n] = n(n + I) r"-2 Note: Ifn =-1, we have

Ex.7.S.6 If V .U =0,

vt~ )= 0, which means that (.; ) satisfies the Laplace's equation

av

the wave equation V 2U Sol:

au

v .v =0, v xU=-at' v xV = at' show that Uand V satisfy a2u =

at 2

vx(vxU)= v x (_av) = -a{vxv)

at

at

2 -a(au)_ -a u

= at

at - ----ai2

... (1 )


But

511

V x (V x U) = V (V .U) - V 2U

= - V 2U from ()) & (2),

(": V U

= 0)

... (2)

a,2 = V2U , which shows that U satisties the wave equation.

i'iu

Similarly; Vx (Vx and

au a at at

-a~v

v)= Vx- =-(Vx u)=-2

... (3)

at

vx(vxY)=V(V.Y)-V 2 Y=-V 1 Y(": V.Y=O) ... (4)

(3) and (4)

=> Y satisfies the wave equation.

Ex.7.5.7 If V.Y = 0, show that V x[V x {V x (V x V)}] = V 4y (or) curl [curl {curl (curl Y)} ] = V 4y

Sol:

We know that [from 7.5.2 (3)]

V x (V x Y) = V (V .Y) - V 2y = - V 2y (": V.Y = 0) = -

U (say)

Then the given expression = - V x( V x U) = V 2U - V CV .U) = V 2( V 2y) - V (V . U)

[from 7.5.2 (3)]

(": V 2y = U)

= V 4y - V (V . U) V . U = V .( V 2y) =

{~) ~}{V2V}

=

I {i ~ }.(V2(~i + V2i + V k))

=

V2(a~ + aV2 + aV3)

=

V2(DivV)= V2(0) =

3

ax

Hence the problem

0'

az

0

(": Div Y

= V . Y = 0)

Engineering Mathematics - 1

512

Exercise 7(d) (I)

(2)

ry( v. r"r) =-;-:t 2

(3)

Show that V-

(4)

Show that V- r

ry () = ~ry d" 2 df +-dr-

r dr

~'ry

1

show also that, if V- = 0, then = C] + -- where CI' C 2 are constants

r

xi + )1 + zk,

7.6

VECTOR INTEGRATION

7.6.1

Ordinary integration of vectors

(I)

If A(u) = A](u)i + A2(u}j + Aiu)k be a vector function of a scalar variable 'u' (A,(u), Aiu), A 3(u) assumed to be continuous in any given interval), the indefinite integral of A(u) is given by.

r

(k.

I grad -) = 0

If r =

prove that, curl

(k x grad

I -) + grad

(5)

r

fA(u)du = i fAl(ll)du+} fA" (u)du +k fA3(1l)du (2)

If there exists a vector B(u) such that A(u) =

d

du

(B(u)), we can write

fA(U )du = f:U (B(ll ))dll = B(u) + c (3)

where c is a constant of integration independent of u. The definite integral of A(u) between the limits 'a' and 'b' in the above, is written as, h

b

d

b

fA(u)du = f- (B(u))du = B(u)+cl du a a a

= B(b) - B(a)

7.6.2 Line Integrals: Let A(x,y,z) be a continuous vector function defined in the entire region of space. Let c be any curve in the region. Divide c into n intervals by taking points A = Bo, BI' B2 ... Bn(= B). Let Pi be any point in the interval B i_, Bi 0"

' •.;.;.


513

A= 8 0

Let ro' rJ ...... rn be the position vectors of points Bo' B)' B2 ...... Bn respectively. Let us consider the sum,

The limit of this sum as n ~ 00 and

1&,:1 ~ 0

is defined as the line integral of A

along the curve c and is denoted symbolically by

f A.dr or

dr

f A.-dt; dt

If c is a closed curve, the integral is written as

which is a scalar.

fA.dr

Cartesian form of line integral: If A=AJi+A-j+Ai dr = (dx)i + ({Mi + (dz)k,

f A .dr = fA1dx + A2 dy + A3 dz c

Note:

fdr, and fAX dr are also examples of line integrals.

7.6.3

Physical appliations:

(1)

Work done by a force (I) If A represents a force and dr is an element of the path of the particle along a curve c, then the line integral {J

fA .dr

(P, Q are 2 points on c)

J'

represents the work done by force A in moving the particle from P to Q.

514

(2)


flow or circulation: (2) If A is the electric field strength, the line integral given above i.e., JA.dr, is called the flow of A along c. If c is a closed curve it is often referred to as circulation of A around c. {J

In general, the line integral JA.dr, will depend on the path from P to Q /'

7.6.4 T"eorem: Prove that the necessary and sufficient condition for the integral fA.dr, c

to be independent of the path c joining any two points is that A is a conservative vector field, (or) there exists a scalar field I/> such that A = VI/> (or) curl A = 0, [i.e., the work done by the force A in moving the particle from one point to another is independent of the path if A = VI/>]

Proof Let P = (x I ,y I ,Z I)' Q = (x2'Y2,z2) be any two given points on the curve c. Let A = VI/> where I/> is single-valued and has continuous derivatives. (J

(1)

Work done

=

JA.dr

I'

Q

fVI/>.dr=

p

G~ i-+j-+k01/> 01/> 01/» p ox ry oz

(dxi+dyj+dzk)

Qol/> 01/> 01/> (J J-dx+-dy+-dz= fdl/> = I/>(Q)-I/>(p) /' ox ry oz I' = ~(x2'Y2,z2) -1/>(xI'Yl'zl) i.e., the integral depends upon the two points only but not on the path joining them. (This is true only if I/> is single-valued at all points P and Q). (2)

The integral is independent of the path. Then, (x,y,z)

I/>(x,y,z)

=

f A.dr = (Xl ,YI ,Zl)

(x,y,z)

f

dr A. ds ds

(Xl ,)'1 ,zl )

Differentiation with respect to's' gives

dl/> =A. dr ds ds

...... (I)


515

d =Vdr ds ds

But

...... (2)

dr s

dr ds

(I) - (2) => (A - V
:. A= V

Solved Examples 4

Ex. 7.6.5:

F(t) = (3t 2-t)j + (2-6t}j -

4tk, find

(a) JF{t}dt

(b) JF{t}dt 2

Sol:

(a) JF{t}dI=; J{3/

2

-/~t+ j J{2-61}dI-k J41dl

~ (I' - I~} +(21-31'); -2t'k+c (b) fF{t}dt

=(/3 -~lj + (21 -3t 2)j - 2t 2k + cj = 50; - 32j - 24k 2

2

2

x/2

Ex.7.6.6 Evaluate: n{3sin e)i + {2cose)j}ie

Sol:

o Given integral xl2

xl2

x/2

ri/2

=; J(3sine}de+ j J{2cose}de=-3cosel;

+2sinejJ o

0 0 0

Ex.7.6.7 The acceleration a ofa particle at any time 't' a

~

=3;+ 2j

0 is given by,

= e-tj - 6(t + 1}j + (3sint)k

Find the velocity V and displacement r at any time 't' given that V = 0 when t = 0 and r = 0 when t = O.

Sol:

a = e-t ;

:. V

-

6(t + l}j + (3sint)k =

d 2r -2

dt

~ : ~ Jadl ~ - e-t i - 6(1~ +I

lj -

(3cost)k + C,

(C 1 is constant of integration) But V

= 0 when t = 0

:.-i-3k+C 1 =0=> C 1 =;+3k Substituting in (I), we get Velocity V

= (I - e-t)i - (3t2 + 6t}j + (3 - 3cost)k

..... (1 )

516


Integrating, r

=

(I + e-t)i - (t3 + 3t2)j + (3t - 3sint)k + C 2

(e 2 being constant of integration)

But r

= 0 when t = O.

:. (2) => + i + C 2

=

0 :. C 2 =

-

i

From (2), r = (t - 1 + e-t)i - (t3 + 3t2)j + (3t - 3sint)k

Ex.7.6.8 Show that d2F dF Fx--=Fx-+c 2 dt dt We know that

f

Sol:

~(FxdF)=FX~(dF)+dFxdF dt

dt

dt

dt

dl

dl

Hence the result.

Ex.7.6.9 If A = 2ti + 3t7 - (4t + 1)k, and B = ti + 2j + t 2k, find 2

2

f(A x B}dt . o (i) A.B = (2t)t + (3t2) (2) - t2 (4t + 1) = 2t2 + 6t2 - 4t3 - t2 = 7t2 - 4t3

(i)

f(A.B}dt

(ii)

o

Sol:

i (ii) A x B = 21 t

j 2

3/ 2

=

(3t4 + 8t +2)i + (--4t2 - t - 2t3)j + (4t - 3t3 )k

=

-1--j-4k 5 3

196.

62.

..... (2)


517

Ex.7.6.10 If F(2) = ; + 2j - 2k and F(3) = 6; - 2j + 3k,

dF

3

evaluate IF.-du

du

2

Sol:

From vector differentiation,

d dF dF (dF) We get, -(F.F)=F.-+-.F=2 F.- so that du

F. dF 'du

du

du

du '

=~~(F.F)=~~IFI2 2 du

2 du

But F(2) = ; + 2j - 2k ~

IFI2 = I + 4 + 4 = 9, when u = 2

and F(3) = 6; +2j - 3k ~ IFI2

= 36 + 4 + 9 = 49, when u = 3

3 dF I . IF.-du=-[49-9] =20 ., 2 du 2

Ex. 7.6.11 If F = (.xl

- 2y); - 6yzj + 8xz2k, evaluate IF.dr

from the point (0,0,0) to the

c

point (I, I, I) along the following paths (1) x

= t, Y = t2 , z = t3 .

(2) the straight line from (0,0,0) to (I ,0,0), then to (1, I ,0) and then to (1,1, I) and (3) the straight line joining (0,0,0) to (1, I, I).

Sol.

(1) x = t;

dx = dt;

Y = t 2; dy = 2t dt;

z = t 3,.

dz = 3t2 dt·,

when t = 0, the point is (0,0,0), when t = 1, the point is (1,1, I)

IF.dr= I(x 2 -2y}tx-6yzdy+8xz 2 dz c

c /;1

J~2 -2t 2 }it -6.t 2 .t 3(2t}dt + 8t~3 J{3t 2 }it

/;0

1

=

2

6

9

J-t dt -12t dt + 24t dt o


518

=

Irf

J~24t

9

o

6 2 \, 24t lO 12t 7 (3 I -12t -t p t = - - - - - - ! 10 7 30

12 12 1 252-180-35 37 = 5 7 3 105 105 Along C, F = (t2 - 2t2 )j - 6. t 2 t 3j + 8.t.t6 k = - t 2 j - 6tJ + 8t7k dr = (dx)i + (dy}j + (dz)k = dt i + (2t dt}j + (3t2 dt)k -----=

Aliter:

I

JF.dr c

=

9 6 2 J(- t -12t + 24t ~t

(Taking dot product of F and dr)

1=0

37 (2)

105 Let 0 = (0,0,0), P = (1,0,0), Q = (I, 1,0), R = (I, 1,1 ). Then along OP, y = 0, z = 0, dy = 0, dz = 0 and x varies from 0 to 1. I

I

2 2 :. JF.dr= J(x -2.0}tx-6(OXOXO)+8x(0)2(0)= Jx dx=± ~

x~

.... (I)

0

Along PQ, x = I, z = 0, dx = 0, dz = 0 and y varies from 0 to 1. I

:.

JF.dr=

J(12-2y~-6y(0)dy+8.1.(0)2(0)= Jo=O

.... (2)

y=O

I'Q

Along QR, x = I, Y = 1, dx = 0, dy = 0 and z varies from 0 to I. I

I

2 :. JF.dr= J(12-2.IXO)-6.1.z(0)+8.1.z2.dz= J8z dz=% QR

z=O

Adding (I), (2), (3), fF.dr

(3)

.... (3)

0

=~ + 0 +! = 3 3

3

The equation of the straight line joining (0,0,0) to (1,1, I) is

f

=

f

~ = = t (say),

so that x = t,y = t, z = t, dx = dy = dz = dt 't' takes values from 0 to I. I

I

.. JF.dr = n~2 2t)- 611 + 8tJ }it = J(8t c o o 2

3

-

2 5t - 21

'Vt

519


Ex. 7.6.12 Find the total work done by a force F = 2xyi - 4zj + 5xk along the curve x = t2, Y = 2t + 1, z = t3, from the points t = 1 to t = 2. Sol: x = t2 , dx = 2t dt; y = 2t + 1, dy = 2dt; z = t\ dz = 3t2 dt Total work done = fF.dr c

= f(2xyi-4zj + 5xk}.{(dx}i + (dy}j + (dz}k} = f2xydx-4zdy+ 5xdz c 2

2

'~I

1

= fl2J 2(21 + I}J2Idl _{4./ 3 )2dt + {5/2 )3/2 dl = f{8/ 4 + 4/ 3 -81 3 + 15/ 4)dl

{ 713 638 23/ 4 -4/ }11= [1 23--/4 ]2 =23 - (32-1)-(16-1)= - - 1 5 = f 5 5 5 5 5

2

3

1

1

Ex.7.6.13 If c is the curve y = 3x2 in the xy - plane and F = (x + 2y)i - xyj, evaluate fF.dr, from the point (0,0) to (I ,3). S'ol:

Since c is a curve in xy - plane, we take r = xi + yj, so that F.dr= {(x+2y)i-xyj}. {(dx)i+(dy}j} =(x+2y)dx-xydy. 1st Method: By taking the curve in parametric coordinates as, x = t, Y = 3t2, dx = dt, dy = 6t dt, so that t varies from

°

to 1 to get the points

(0,0) & (1,3): We have (1,3)

fF.dr = (0,0)

1

1

'~O

0

fv + 6t 2) dt -/{3t 2) (6tdt) = fv + 6t 2 -181 4)dt

t2 2

6/ 3 3

18t 5 1 5 0

1

18

5

18

11

2

5

2

5

10

= -+---1 = -+2--=---=-2nd Method:

>: =3x2, dy = 6x dx and x varies from

1

.. fF.dr c

=

f(x + 6x 2}Ix - x.3x 2.6x.dx

0

1

~( 2 , 11 = Jx+6x -18x 4 \px =-o 10

°

to 1.


520

Ex. 7.S.14 Find the work done by the force F in moving a particle once around the circle 'C' in xy - plane, ifthe centre of the circle is origin and radius is '2' and F = (x +y + z)i + (2x + y}j + (2x - y +z)k. Sol:

xy - plane is z = 0, :. F = (x + y)i + (2x +y}j + (2x - y)k and r = xi +yj F. dr = (x

+ y)dx + (2x + y)dy.

~

dr = (dr)i + (dy}j y

The equation of the circle is x = 2cosS, Y = 2sinS :. dx =

-

2sinS dS, dy = 2cosS dS

S varies from 0 to 2n Work done = fF.dr

=

2lt J(2COSS + 2sin sX- 2sinS)de + (2.2cosS + 2sin SX2cosS)dS o

-~

2lt 2lt 2 2 2 2 J{- 2sin2S -4sin S + Scos S + 2sin 2S}ie = J{scos S - 4sin S ~S o 0

2lt 2lt = n4(1 + cos2S)- 2(1- cos2S)}iS = J(2 + cos2S)dS = [2S + 3sin 2S~lt o 0 = 4n Ex.7.S.1S Show that the necessary and sufficient condition for a vector field V to be conservative is curl V = 0 Sol: a) Necessary condition: If V is conservative, :3 a 'cp' 3 V = V cpo curl V·= curl (V cp) = 0 (see 7.5.2 (1)) b) Sufficient condition: Let V = V I i + V'li + V 3k i

Curl V = V x V

~

0

=

~

j

k

8/ Ox 8/ By 8/ 8z = 0

I(8VBy 8V8z )i = 0 3 _

2

.... (1)

521


The work done by the force field V in moving a particle from (x\,y\,z\) to (x,y,z) is IV.dr c

= Iv; (x,y,z)dx + V2 (x,y,z)ay + V3 (x,y,z)dz c

where V is a path joining (x\,y\,z\) to (x,y,z) Let us choose a particular path consisting straight line segments from (x\,y\,z\) to (x,y\,z\) to (x,y,z\) to (x,y,z) and denote the work done along this path by a scalar function ~(x,y,z); x

:.

z

y

~(x,y,z) = IV;(X'Yl,ZI)dx+ IV2 (x,y,zJly+ IV3 (x,y,z)dz XI

..... (2)

Yl

From (2), it can be seen that,

8cI>

.... (3)

8z = Vix,y,z)

.z

= V2 (X,y,ZI)+ V2(X,y,Z~

= V2(X,y,ZI)- V2 (x,y,z)- V2(X,y,ZI) .... (4)

from y

= V$x,y\,z$ + V;(X,y,ZI~

Yl

= V $x,y\,z$ + V $x,y,z$ = V $x,y,z)

(1)

z

+ V;(x,y,z~

Zl

V \(x,y\,z$ + V $x,y,z) - V \(x,y\,z$

.... (5)

(3),(4),(5) => :i+ : j + : k=V;(x,y,z)i+V2 (x,y,Z)j+V3 (x,y,z)k=V

=> V =

V~

Hence the proof.


522

Ex. 7.6.16a) Show that F = yi + (2xy +z2)j +2yzk is a conservative force field. b) Find its scalar potential. c) Find the work done in moving an object in this field from (1,2,1) to (3, 1,4)

Sol:

a) V x F= Max

i

j

k

a/fJy 2xy+Z2

a/az

2yz

= i(2z - 2z) + j(O - 0) + k(2y -2y) = 0 :. F is a conservative force field. b) Let 'cjl' be the scalar potential of F. 1st method:

2

:. acjl ax = y2

acjl_2 z az - y .... (3)

.... (1) :=2Xy +z .... (2)

Integrating (1) w.r.t x, (2) w.r.t. y, & (3) w.r.t. z, respectively, we get,

cjl =

xY + j(y,z),

cjl =

xY +yz2 + g(z,x), and

cjl

=

yz2 + h(x,y).

These equations will be consistent iff, g, h are taken as

j(y,z) Hence cjl

=

yz2,

g(z,x) = 0, h(x,y)

=

xy.

= xy + yz2 + constant

2nd Method: Since F is conservative, JF.dr is independent of path joining c

(Xl,yl,zl) to (x,y,z)

using method of problem 7.6.15(b),

cjl =

x

Y

Z

XI

YI

zl

J&12}ix + J(2Xy + z/ }ry + J2yzdz = xy~ ( +(xy 2 + Z12 y) r+yz2 f Xl

.Y]

= xYI - XlYI + xy +z 12y - ry I - ZI 2y I + yz2 - yz I = xy2 + yr - x1yl2 - Zl2Yl = xy2 + yz2 + constant.

2

3rt/ Method: Since F. dr = Vcjl.dr = acjl dx + acjl dy + acjl dz =dcjl ,

ax

fJy

az

zl

523

Vector Differentiation d~ == cY)dx + (2xy + z2)dy + (2yz)dz == cYdx + 2xydy) + (z 2dy +2yz)dz

== d(xy) +d(yz2) == d(xy2 + yz2)

=>

~ == xy + yz2 +constant

(0,1)8

1'2=(3,1,4)

fd~

f F.dr ==

c) work done ==

IHI,2,1)

c

(3,1,4)

~(P2) - ~(PI) = xy + xz2

==

'=:-:::----i----"'I\( 1,0)

I

(1,2,1)

=(3.1 2 +3.4 2)-(1j2+ 1.1 2)=51-5=46.

Ex. 7.6.17 Evaluate f(x 3dy + y 2dx) where c is the boundary of the triangle whose vertices c

are (0,0), (1,0), (0, I). Sol:

f(x 3dy + y2 dx)

Let I ==

c

I

== II +12 +1 3, where II ==

f, 12 == f, 1 f 3=

OA

(i)

Along OA, y == 0, => dy == . ••

(ii)

AH

Jx 3 •0 + 02 •dx

11 ==

AlongAB, o

HO

°

== 0

x+y=1 ,y=l-x=> dy=-dx,xvaries from

I toO.

:.1 2 = fx 3 {-dx)+{I-x)2 dx I

= J(x 2_x 3 +1_2X)dx=_x

3

o

(iii)

Along BO,

:. 1

3

x

_~o =0_(_~_~)= __1 4

I

3

4

=°=> dx =° :. 13 =fo.dy+ i.o =° 1

= II + 12 + 13 = - 12

Ex.7.6.18: If 1= xy2z2 ,evaluate fldr where the curve 'c' is given by

x = (, Y = (2,

Z

=(3

from t

=

°

to I .

12

524


f=xy2 Z2 =/(/2y .(/3

Sol:

Y=1

11

dr=dx i+dyj+dz k 2 2 dr =(dt)i + (2tdl )j + (3/ dl )k =(i + 2tj + 3/ k )dl 1

Jfdr

=

c

Jill

(i + 2tj + 3/ 2k )dl

1

=i

1=0

1

Jill + dl+ j J2/

0

12

1

dl+ k J3/ 13 dl

0

0

./12 I • 2/ 1 3/ 14 I 1. 2. 3 =1- J+l-J+k- J=-I+-l+- k 12 0 13 0 14 0 12 13 14 Ex. 7.6.18: If A = 3zi - 2xj + yk , and c is the curve given by 13

x = cos I, Y = sin 1 , Z = 2cos I, evaluate

JA x dr from I =0to I = 2

1t .

c

i j k A x dr = 3z - 2x y dx dy dz

Sol:

= (-2x dz- y dY)i+(y dx-3z dZ)j +(3z dy+2x dx)k x

= cos I, Y =sin I, z = 2 cos I ~ dx = (-sin/)dl,dy = (cos I) dl, dz = (-2sin/)dl

:. (1)

~

Ax dr = i[(-2cos/)(-2sin/) - sinl . cost]dt +j[(sin/)(-sint) - 3(2cost)

(-2sin/)]dl + k[3(2cos/)( cost) + (2cos/)(-sin/)]d/.

=i(3sint cost) dt + j[(12 sinl COS/) -

sin2/)] dl + k(6cos2/- 2sint cost) d/.

J

J

1C/23 ft/2[ 1 Axdr=i J-sin2Idt+ j 6sin2t--(1-cos2/) dl .. c o 2 0 2

!

1C12

+ k ~3(1 + cos2/)-sin 2/}it o

.3 (_COS2/)1C12.[ 1 sin 21 Jft/2 [ 3. COS2/]ft/2 = 1+ J -3cos2/--+-+k 3/+-sI02/+--

24

220

= {

0

3 -4 )(-1-1)+ j[-3(-1-1)- : + 0]+k[3;

= %;+(6-

:}+e;

-1)k

2

+O+~(-l- ])]

20

525


Exercise - 7(e) 4

I)

IfU(t)=(2t2 -I)i+3tj+(2-t)k. Find (a)

JU(t)dt,(b) Ju(t)dt 2

l

12

3 2 t2 2/ \06 [Ans:(a) ( ~-I i+ 3t j + ( 2t2 k,(b) 3i+18j-2k] 11/2

2) Evaluate: J(6sillu)i -(3cosu)j +uk o 1t

2

[Ans: 6i-3j+-k] 8 2

3) If A(s) = si - s2j + (s + l)k, B(s)

=

2

2si + 6sk find (a) JA.Bds, (b) JA x Bds . o

[Ans: (a) 4) If A = ti - j + 2tk, B = t2 ; 1

-

0

100

3' (b) -

48i - 12j + 16k]

tj +2k, C = ; - 2j + 2k, evaluate 1

(a) J{A.(BxC)}dt,(b) J{Ax(BxC)}dt o 0

I 3'

5. 37. 8 k +-} +- ] . 2 6 3 5) The acceleration of a particle a at any time t ~ 0 is given by a = eli + (2cos2t)j + (2sin2t)k If the velocity V and displacement r are both zero at t = 0, find V and r at [Ans' (a) -- (b)

--I

any time t. [Ans: V = (el

r

-

I)i + (sin2t)j + (1 - cos2t)k,

=(i -t-I)+~(l-COS2t)j +(t -~Sin2t}]

3

6) Evaluate

fA. dA

2

du, given that, A(2) = 4i - 2j + 3k and A(3) = 2i + j + 2k

du

[Ans: - 10]


526

Exercise 7(f) I. If ~

= xyz, valuate I~dr, where c is the curve x = t3, y = t2, z = t, from t = 0 to I c

[Ans: 2. If F

= xi - yzj + z2k,

and c is the curve given by x

(i) IF.dl" and (ii) fF.dr, from t = 0 to t e

I. I k JI i +4J+ 7 1

= t, Y = t3, z = t 2, evaluate

= I.

C

.) [A ns: ( I

5. 7. 1\ k .. -71-I5J +1"2 ,(II)

23] 24

3. If A = (2x + 3)i + xyj + (zx - y)k, evaluate fF.dr along c where c is (a) the curve x

= t3,y = 2t2, z = t from t = 0 to I.

(b) The straight lines from (0,0,0) to (1,0,0), then to (1,0, I) and then to (1,2, I) (c) The straight Iinejoining (0,0,0) and (I ,2, I). 491 13 14 [Ans: (a) 105' (b) (c)

2'

4. If A = (3xy - 2r)i + (x - y)j, evaluate

"3]

fA.dr along the curve c in xy - plane given c

by y

= x 3 from the point (0,0) to (2,8). 1308

[Ans:

3.5]

5. IfF = (2x- y)i + (x- 2y)j, evaluate fA.dl" where c is the closed curve shown in the c

figure below. y

x

527


6. If A = (3x + 2y)i + (x + y)j, and c is the boundary of the tringle whose vertices are (0,0), (1,0), (0, I), evaluate fA.dr. c

3

[Ans:

2" J

7. If A = (2x + y)i + (3x - 2y)j, compute the circulation of A about the circle C: xl + 4, traversed in the positive direction. [Ans: 81t] 8. Find the work done in moving a particle in the force field. F = 2x2i + (2yz - x)j + yk, along (a) the straight line from (0,0,0) to (3, I ,2) (b) the space curve x = 3t2, Y = t, z = 3t2 - t from t = 0 to t = I

1=

113

[Ans: (a)

58

6' (b) 3]

9. (a) Prove that V = (2x siny - 3)i + (x 2 cosy + z2)j + 2(yz + I)k is a conservative force field. (b) Find the scalar potential ofY. (c) Find the work done in moving an object in this field from (1,0,-1) to (2,

2"1t ,I).

[Ans: (b) x 2 siny + yz2 - 3x + 2z + constant, (c) 10. If A

= (9xly -

1t

2"

+ 5]

2xz3 )i + 3x3j - 3x2z2k, (a) prove that fA.dr is independent of the c

curve 'c' joining two given points. (b) show that there is a differentiable function
7.7

SURFACE INTEGRALS

7.7.1

Let S be a two-sided surface. Let one side be taken as the positive side. If S is a closed surface, the outer side is considered as the positive side. Let A be a vector function. Consider an element of area 'ds' in the surface. Let n be the unit normal vector to ds in the positive direction. It can be seen that A.n = A cose. (where '8' is the angle between A alld n and A = IAI) is the normal component of A. Let ds be a vector whose magnitude is ds and whose direction is that ofn. :. ds

=

n ds.


528

z

Then the integral,

ffA.ds = ffA.Il .ds

..... (1)

s

is an example of a surface integral which is also called as flux of A over s. If'f' is a scalar function,

ff4>ds

..... (2)

..... (3)

ffAx ds = ffA x nds

Note

..... (4)

are some other examples of surface integrals. (1) The surface integrals can also be defined in terms oflimits of sums. (see 7.7.2) (2) The notation

If is also used to denote a surface integral over the closed

surface s. (3) Sometimes the notation

f may also be used for surface integrals.

(4) Surface integrals can be conveniently evaluated by expressing them as double integrals over the projected area of s on one of the coordinate planes {see 7.7.3)

529

Vector Differentiation /

7.7.2 Definition of surface integral as the limit of a sum:

.,J------...... y

The area S is divided into 'L' elements of area Mill' m = 1,2, .... L, let

P'II =

(xm'YIII,zm) be any point in

&\)111.

Let A(xm,Ym,zl/.)= AHI

Let nlll be the positive unit normal to I'1S111 at Pm. Then «(Am.nm) is the normal component of Am at Pm. Consider the sum, /,

I

Am.nIllMm

..... (I)

1/1=1

The limit of the sum (l)as L ~oo such thatthe largestdimens~on of each I'1S", ~ 0 (if the limit exists) is known as the surface integral of the normal component of A

IIA.n.d'}

over S an d denote d by ,

7.7.3 Evaluation of a surface integral To evaluate surface integrals, it is convenient to express them as double integrals taken over the projected area of the surface S on one of the coordinate planes (xy,yz, or zx planes). If R is the projection of S on the xy - plane, it can be shown that

ffA.nds ,

= ffA.n d\XdY, /I

n.k

From 7.7.2, the surface integral is the limit ofdle sum /,

I

111=1

Am .nmMIII

..... (1)


530

z

;---~----~~~--.Y

The projection of ASmon thexy- plane is

InmA.S'II/.kl (or) Illm.kIMi'm

which is equal

:. The sum (I) becomes

By the fundamental theorem of integral calculus the limit of this sum as L

~ 00

in

such a manner that the largest Llx m and AYm approach zero is

dxdy

IfA.n In.kl

which is the required result.

R

Note: Similarly ifR is the projection ofS on yz and zx planes respectively, it can be seen as

ff A .nds = ffA.n dl'Y~ZI

S

and

R

nJ

ffA.nds = ffA.n dz~xl S

R

ln.]

7.7.4 Physical interprettion of surface integrals: Let A denote the velocity of a moving fluid. Let S be a fixed surface in the fluid. Let ds be an element of surface. Then A.n ds = A.dS represents the amount offluid that passes normally through dS in unit time at any point. If the direction ofn is outward or positive, the amount offluid flow is positive. Similarly if dS' is another element for which n is in the negative direction, the fluid flow is negative at that point.

531


to fA.nds and it is known as the total flux of A through the entire surface S.

s A can be a vector denoting physical quantities such as electric force, magnetic A

n....---...---~C"

force, flux of heat or gravitational force etc. In all these cases,

fA.nds

denotes

s total flux of A through S.

Sioved Examples Ex.7.7.5 Evaluate

fJA.nd\" where A = {x + y2} - 2xj + 2yzk and S is the surface of the

plane 2x + y + 2z = 6 in the first octant.

Sol:

A = (x + y)i - 2.>.] + 2yzk Let ~ = 2x + Y + 2z - 6

z

V ~ = 8q>i + aqy + 8q>k = 2i + j + 2k

ax

~

az

V~

2i+j+2k Unit normal n to S = IV~I = ~22 + 12 + 22 2i+ j + 2k 3

x


532

= .![2x+ 2y2 - 2x + 2y(6 - 2x - y)] 3

[Substituting 2z = 6 - 2x - y ( ... 2x + y + 2z = 6)]

= .!(12y-4xy) 3

If R be the projection of S on the xy - plane.

dxdy

2

3

In.kl = -3 => ds = -l/l.kl- = -dxdy 2 .. IfA.nds = If(A.n) dXdY /I /I ln.k I If .!(12y - 4);y )~dxdy /I 3 2 =

If(6 y - 2Xy}dXlry

..... (I)

/I

To evaluate this double integral over R, (i) Keep x fixed and integrate w.r.t. y from y w.r.t. x from x = 0 to x = 3. :. Given integral

=

0 to (6 - 2x), ii) and then integrate

3 6-2x

J J(6y-2xy)1ydx

x=O y=O

3 3 3

=

TI3i _xy2 r:~x dx =

J(3 -xX6- 2xY dx =

2 3 J(J08-108x+36x -4x }Ix

= [\08x-54x2+12x3-x4]~ =324-486+324-81 =81 Ex.7.7.6 If F = 4xzi -

yJ + yzk, evaluate

IfF.nds where S is the surface of the cube s

bounded by x =

O~

x = I, Y = 0, Y = 1, z = 0 and z = 1.

533


Sol.

The surface S can be divided into 6 faces (see figu --,

z

o

(i) S\: Face EPFA (ii) S2: Face OBOC

E f---+----(

(iii) S3: Face PFBO y

(iv) S4: Face OCEA (v) Ss: Face POCE

x (vi) S6: Face OBFA

IfF.nd5 = IfF.nds + IfF.nds + IfF.nds + IfF.nds + IfF.nds + IfF.nd.,. S ~ ~ ~ ~ ~ % On S \: n = i x = 1 1

1

IfF.nds =

JJ(4Zi - y2 j + yzk }idydz JJ4zdydz

SI

0

0

On S2: n =

-

1

0

0

1

1

o

0

=

i, x

1

JJ(- y2 j + yzk X- i)dydz JJ(O}lydz =

o

=

1

1

0

0

J2Z21~

0

1

1

1

1

1

J J(-I)1xdz= J-zi

.'13

0

0

0

0

0

0

OnS 4 :n=-j,y=O

:. HF.n.ds

1

1

o

0

=

1

1

o

0

J J(4xzi).(- j)dxdz J J(O)1ydz =

= 0

On Ss: n = k, z = 1 1

1

:. HF.n.ds=J J(4Xi-/j+yk}kdxdy= 0

8S

On S6: n = - k,

Z 1

=

0

=0 1

J J{- y2 j }(- k )dxdy =

0

0

0

. JJF.n.ds =2 + 0 - 1 + 0 + ~ + 0 = ~ .. 2 2

s

J2dY = 2 0

=I 1

.'16

=

= 0

HF.n.ds=J J(4xzi-j+ zk).jdxdz =

HF.n.ds

1

0

=

1

HF.n.d.,. = On S3: n = j, y

1

1

1

1

21

0

0

0

0

J Jydxdy= J~ I

1

dx= J-Idx=-l 0


534

Ex.7.7.7 If A = z 2i + xlj - yZk, and S is the surface ofthe cylinder xl + Y = 16 included

Sol:

in the first octant between z = 0 and z = - 5, evaluate jfA.n.ds s z

f

d~- ~~'

n

,. 5 R Q

r=-----'JL..-- y

,....

'<.

4

- ' z=Q

x Project s on xz plane and let the projection be R. (See figure)

HA.n.ds = HA.n dx,d~, S

The norma I to x 2 +

..... (I)

n.)

R

Y = 16 is

V(X2y2 )=~(X2 + y2) +~(X2 + y2)j

ax

ay

= 2xi+ 2yj . Ul1Itnormaln=

Z

A.n

2(xi + yj)

xi + yj =-4-

(.: xl + Y

2X+X 2 y 4

= ------'--

.

y

n.J= :. From (I)

2xi + 2yj

~(2xY+(2yy = 2JX2+y2

"4

J

JA.nds

=

S

J

V2X+X2y dxdz 4 y/4

R

[.: J

x dx=4, o ~16 _x 2

3

= 4z + 64zf = 820 3

3

0

3

16)


Ex.7.7.8 Evaluate

535

I Ind~

where S is the surface of problem 7.7.7 above and

=

x~z

s

Sol:

We have

dxdz

·1 IfnI

Ifs

nds =

xi+ yj using n = - 4 - ' n.j

1

-

R

80

64 Z. 3

64Z. 3

-l+-}

I 64(z2. 83 2

n.)

II

"4' the integral on R.H.S. becomes,

f

Z

.]1

5

Z2 2

y

=

-.- -1+-)

0

8 25(. 3 ~

.)

100(. 3

.)

=-.--;;-1+) =-1+.1

Ex.7.7.9 If A = yi -+ (x - 2xz)j - xyk, evaluate

I I{curIA).nd~· s

the sphere x2 +.0 + z2

=

4 above the xy plane

Sol: Z

n

I--~-+--Y

x

CurlA=

j

k

a/ax

a/fJy

a/az

y

x-2xz

-xy

where S is the surface of


536

= i{~(-xy)-~(x-2xz)} + j{~(y)-~(-xy)} +k{~(x-2xz)-~(y)}

&

~

&

fu

fu

~

= xi + yj - 2zk The normal to the surface is V (xl + .0 + z2) = 2xi + 2yj +2zk

2xi + 2yj + 2zk Unit normal n = I ,,(2x)2 + (2y)2(2z)2 xi+ yj+zk 2 x 2y2 _2z2

(Curl A). n =

2

The projection of S on xy plane is the circle xl

+.0 = 4, Z = 0 (see figure)

.. f.f(curl A.n}ds = f [(CUr! A.n) I:'~f =

fpX2 +i

- 2Z2 ) dxdy 2

/I

z 12

2

( 2 ) 3 x + y2 - 8dxd I 2 2 OJ x=-2Y=_~4_x2 ,,4-x y 2

4x

f f

Changing into polar coordinates by taking x = r cose, y = r sine, dx dy =r dr de, the integral becomes

2

de= f(8-8}ie=o 8=0

537


Ex.7.7.10 Evaluate J JA.nds where A = yzi + zx} + xyk and S is the part of the sphere s x 2 + y2 + z2 = 9 which lies in the first octant.

Sol:

V x 2 + y2 + Z2

.

Untt normal n to S =

IV x

2

'

,

+ y +z-

2xi+2y}+2zk

xi+ y}+zk

~4X2 + 4y2 + 4Z2

3

A.n

=

(.: x2 +y2 +z2 =

9)

z

nk= -3 .

3xyz;

If R is the projection of S on xy - plane, we have,

Jsf A.nds

=

f fA.n d1xdY /I n.k I

= f f 3xyz dxdy /I z/3

=9 f fxy dxdy II

The region R is bounded by x-axis, y-axis and the circle x 2 + y2 = 9; z = 0 chang,iog to polar coordinates, the last integral becomes ",/2

9 J

3

rr/23

f(rcos9Xrsin9)rdrd9=9 J[Jr 3 cos9sin9dr}19

8;0 r;O

=

9

0 0

fo ~4)31 cos9sin9d9=9.-811[/2fcos9sined9

1[/2(

4

4

0

0

1[/2

(

I)

': !cos9Sin9d9=2"

Exercise - 7(g)

I. If F = 18zi - 12x} + 3yk, evaluate J f F.nds where S is that part of the plane s 2x + 3y + 6z = 12 which is located in the first octant (Hint: Take projection ofS on the xy plane and ds =

dxdy -1-1 ) n.k [Ans: 24]


538

2. Evaluate f f V.nd~ , where V = yi + 2xj - zk and s is the surface of the plane s

2x + Y = 6 in the first octant cut off by the plane z = 4 [Ans: 108] 3. If r = xi + yj + zk find the value of the integral f fr.nds over s

a) the surface S of the unit cube bounded by the coordinate planes and the planes = y = z = I and b) the surface of the sphere of radius 'a' with centre at the origin [Ans: a) 3, b) 41ta3]

x

4. Evaluate f fF.d~ over the entire surface ~fthe region above the xy plane bounded s by the cone :l-

= x2 + Y and the plane z = 4, if F = 4xzi + xyzj + 3zk

5. IfS is the surface of the parabolic cylinder

y

[Ans: 3201t] = 8x in the first octant bounded by the

planes y = 4 and z = 6, evaluate f fF.nds, where F = 2yi - zj + x 2k s [Hint: f f F .nds = f fF.n S

R

dIY~IZ , where R is the projection ofS on the yz plane] nJ

[Ans: 132] 6. Evaluate f fV.nds over the entire surface S of region bounded by the cylinder S

x2 + :l- = 9, x = 0, y = 0, z = 0 and y = 8 if V = 6zi + (2x + y}j -

xk [Ans: I81t]

7. Evaluate f f A .nds where A = zi + xj - 3yzk and S is the surface of the cylinder s

x2 + Y = 16 included in the first octant between z = 0 and z = 5 [Ans: 90]

8. If V

= yzi + zxj + xyk and S is that part of the sphere x2 + Y +:l- = 1 which lies in

the first octant, find the value of f fV.nds . s [Ans:

3

"8]

539


9. Evaluate

JJ{(x

3

-

yz2)i _(2x2y)j + 2k}.ds over a cube with edges of length 'r'

s parallel to the coordinate axes

I

[Ans:

2"]

10. If V = xi + yj + zk, and S is the 'triangle with vertices at (1,0,0), (0, 1,0) and (0,0, I), find the value of I IV.ds . s

I

[Ans:

2"]

II. If A = xi - yj + (z2 - I )k, find the value of I IA.nds, where S is the closed surface s bounded by the planes z = 0, z = I and the cylinder x 2 +

y=I [Ans: 1t]

7.8

VOLUME INTEGRALS

7.8.1

Consider a closed surface in space enclosing a volume V. Then, integrals of the form I I IAdv and I I I~dv, [A is a vector function, f is a scalar function] are

v

"

examples of volume integrals.

7.8.2 Expression of volume integral as the limit of a sum: Let A be a continuous vector function. Let S be a surface enclosing the region D. Divide this region D into a finite number of subregions D, ... ,= 1,2, ..... n. Let L1v, be the volume of the subregion D, enclusing any point whose position vector is rio Consider the sum II

V =

LA~JL1v, ,;1

The limit of this sum as n ~oo such that L1V, ~ 0, is called the volume integral of A over D and denoted by I I I Adv J)

If A = A I (x,y,z)i + Aix,y,z}j + A 3(x,y,z)k,


540 so that dv = dx dy dz

f f fAdv = iff fA,(x,y,z}lxdydz + j f f f A2(X,y,z}!xaydz + D

D

D

k f f f A3(X,y,z}lxdydz f)

Solved Examples

Ex.7.8.3 If F

=

(2x 2 - 3z)i - 2xyj - 4xk, evaluate f f fV.Fdv where V

IS the closed

v

region bounded by the planes x = 0, y = 0, z = 0 and 2x +2y + z = 4.

Sol:

VF

.

=

~(2X2 -3z)-~(2xy)-~(4x)=4x-2x=2x ry

ax

az

2 2-x4-2x-2y 2 2-x 4-2x-2y .. f f fV.Fdv= f f f 2xdzdydx= f f [ f 2xdz ]dydx v x=Oy=o z=O x=o y=o z=O 2 2-A 4-2x-2y 2 2-x 2 2-x 2 f f 2xz l dydx = f f 2x(4-2x-2y}!ytlx= f[ f(8x-4x -4xy)t(y]dx A=O y=O Z=O x=O y=O x=O y=O 2 2-x = f8xy-4x2y-2xy2fdx x=O y=O

2

f[8x(2 - x)- 4x 2(2 - x)- 2x(2 - X)2}1x x=O 64 3

8 3

=16--+8=-

Ex.7.8.4 Evaluate

ff j(V.A}tv over the region bounded by x + Y = 4, z = 0 and z = 3, 2

v

where A = 4xi - 2yj + z2k.

Sol:

a(4x ) --(2y a 2 a 2 )=4-4y+2z V.A=)+-(z ax ry az x=2 y=J4-x 2 3 .. fff(v.A}!v= fff(4-4y+2z}lv= f [J(4-4y+2z)dzf/ytlx v v x=-2 y=-v4-x~ 12 z=O

f


541

2

J[

dydx==

x~-2

)'~f4-x2

JC 2l - 12 y)dy]dx ;--- ?

y=-Y4-r-

x r.-? x 2 n ==84[--v4-x 2 +2sin- ' C-)] ==84[O+2C-)-O]==84n 220 2

Ex.7.8.5 Evaluate

JJJ
os:; z < c Sol:

and ~ == 2{x + Y + z)

fJf¥v ~ ff f2{x+ Y+ z}Jv ~ ,U[}{x + a

b

x=O

0

== J2exy+ cy2 +e 2YI

y + z)

}YdY

a

a

x=o

0

dx == JC2bex+eb2 +e 2b)dx==bex2 +(b 2e+be 2)xl

== a 2 be + (J(b 2 e + be 2) == abe {a +b + e) Ex. 7.8.6 If ~ == 4y + 2xz, evaluate x2

Sol:

JJJ~dv over the region in the first octant bounded by J'

+1 = 9, z = 0, z = 2.

JJJ~dv== JvJJ{4y+2xz}lv I'

3

x=O

0

J J

=

~2 ? 2 [J(2xz+4y}lz]dydx== JC4yz+xz )f dydx

J.=3y=J9-x 2 z=2

y=o

3

~x2

o

0

J[

3

J

==0

J(8y + 4xX/y]dx

3

=

0

~2

JC4y2 +4xy)1 0

= J[4(9-x2)+4x~9-x2]dx==108 o

0

0

dx


542

Ex.7.8.7 Evaluate f f fFdv where F = xzi - 2xj + 2y.k and V is the region bounded by v

the surfaces x

= 0, y = 0, y = 6, z =.x2, and z = 4

Sol: z

y=

s

Q

\:

.Q...~ 2

z=4

z=x

\

,I '\

~y

=6

L _________ J//

y

x -i

The region V is covered by (see the figure) (a) keeping x andy fixed and integrating from z = .x2 to z = 4 (base top of column PQ) (b) then by keeping x fixed and integrating from y = 0 to y = 6 (R to S in the slab) and (c) finally integrating from x = to x = 2 (where z =.x2 meets z = 4).

°

:. The required integral is 2

6

f f[

4

f(xzi-2xj+2y 2k)dz}1ydx

x=Oy=O z=x2

=i

264

264

264

f f[

f f[

f f[

fxzdz]dydx+j

x=O y=O z=x2

26

= i f f x; x=O y=O

f- 2xdz ]dydx+k

J

26

4

dydx+ j f f- 2xz

z=x2

x=O y=O

2

x=O y=O z=x 2

x=O y=O z=x2

24

f2y dz}1ydx

J

z=x2

26

4

dydx+k f f2 y 2 z x=O y=O

J

dydx

z=x2

<


543

26

5

26

J[ J(8X-~)dy}lx+j J[ J(2x

=i

x=() y=O

J[ J(8i- 2x2 y)dy]dx

-8x)dy]dx+k

~=o y~o

x=o y=O

2 8 y3 2X2 3 dx+k J-'--_Y)'~O 3 3 x=()

X5 J8xy-~ I dx+ j J2x'y-8xyl 2

2

=

26 3

i

6

6

)'~()

x=o

2

2

J( 48x - 3x

=i

5

£Ix

2

0

1 = i(24x 2 -:;- x 6 ) ~

4~i

I

y=O

)dx + j J(l2x 3 - 48x)dx + k J(576 - 144x2 )dx

o

= 64i -

6

2

I

0

2

2

~=O

x=o

+ j(3X4 - 24x2) I + k(576x - 48x 3 ) I

x=o

+ 768k

Ex.7.8.8 Find the volume of the region common to the intersecting cylinders x 2 + and

x2

+z2

r =b

2

= b2 .

Sol: z

2

2

x +z ::: b

/ I

/

",

I

I-

II

V

2

/'"

-

2

2

_''x +y ::: b

2

V

~o

y

x

Required volume is equal to 8 times the volume of the region shown in the figure (as the axes cut the volume into 8 equal parts one in each octant)


544

Exercise - 7(h) 1. Evaluate

Iff( 2x + y) dv where V is the closed region bounded by the cylinder 80 3

z = 4 - x 2 and the planes x = 0, y = 2 and z = 0 . 2.

I ADS: - I

If A = (2X2 - 3z)i - 2xyj - 4xk , and V is the closed region bounded by the planes

x=O,y=O,z=Oand 2x + 2y + z = 4,findthevalueof

Iff( curlA )dv [ADS:

3.

Evaluate

Ifff dv where f

1u

-k) )

= 45x 2 y and v is the closed region bounded by the

planes

[ADS: 128) 4x+2y+z = 8,x = O,y = 0 apd z = 0 4. If A = 2xzi - xj + y2 k and v is the region bounded by the surfaces x = O,y = O,y= 6,

z= x

2

5.

and z = 4, find the value of

Evaluate

ffI(Div A)dv

parallelepiped,

o~ z ~ 3 , where A = (X2 - yz) i + (i - zx) j + ( Z2 - xY) k

[ADS:3 6 1

6. Evaluate

0~y

~

over

ADS: 128i-24j+384k rectangular

o~ x ~ 1,

taken

Iff Adv the

2,

Iff(Div F)dv for the volume of a cube with edges of length unity

parallel to the coordinate axes where F

= (x 3 -

yz2 ) i - ( 2X2Y ) j + 2k [ADS: 1/3)

545


7.9

CURVILINEAR COORDINATES

7.9.0 The students are already familiar with the coordinate systems such as i) cartesian ii) Polar and iii) parametric. A study of yet another coordinate system called "curvilinear coordinate system" will be now taken up.

7.9.1

Transformation of coordinates Suppose the rectangular coordinates (x,y,z) of any point be expressed as functions of variables vI' v2' v3 as ...... ( I)

Let the above system of equations (i) be solved for vi' v2' v3 and another system of equations ..... (2)

Ifthe functions in (I) and (2) are assumed to be single valued and to have continuous derivatives, the correspondence between the variablesx,y, z and vI' v 2' v3 is unique. Thus, to a given point P with cartesian coordinates (x,y,z), we can associate from (2) a unIque set of coordinates (vl'v 2,v3 ) which are called curvilinear coordinates. The systems of equations (I) and (2) give the "transformation of coordinates".

7.9.2 Orthogonal Curvilinear coordinates V3

z

........

;

curve

"

>-----------~-------------y

x

The surfaces vI = ci' v2 = c 2 ' v3 'coordinate surfaces'. (see figure).

= c 3. (cl'c2,

c 3 beIng constants) are called

Each pair of the above surfaces intersect in curves called 'coordinate curves or lines'. i.e., vI = c I and v 2 = c2 intersect in v3 curve, v 2 = c2' v3 = c 3' vI = c I in v 2 curve.


546

If the coordinate surfaces taken in pairs intersect at right angles the coordinate system is called 'an orthogonal curvilinear coordinate system'.

Note: The coordinate surfaces described above are similiarto the coordinate planes of the rectangular system and the coordinate curves are similar to the coordinate axes. Infact, the rectangular coordinate system is also an example of curvilinear system with VI = x, v2 = y, v3 = z; c I = c2 = c 3 = 0 7.9.3 Unit vectors of a curvilinear system Consider the position vector r = xi + yj + zk, of a point P. using (I), it can be written as r = r(v l ,v2,v3) (i)

A tangent vector to the VI curve at P(for which v2' v3 are constants) is

av,

I I'

.

the Ul1lt tangent vector we have e I

~. If e l is

ar/av ar

=

av,

lilly if

ar , ar are tangent vectors to the v2 and v3 curves at P respectively and e2,

av aV3 2

e3 are unit vectors in these directions,

ar

we can write , -av = A2 e2 , 2

ar

av3 = A3e3

-

where The unit vectors e I' e2 , e3 are in the directions of increasing v\' v2' v3 respectively. The quantities A\, A2 , A3 are called 'scale factors'. (ii)

'Vv, ' is a vector at P normal to the surface

VI

= c 1. If EI is a unit vector in this

direction,

VV 2

lilly the unit vectors, E2 = IVv v2

=

c 2'

and

v3

= cy

2

1'

E3 =

Vv

IVv: I at P are respectively normal the surfaces

547


V3

curve

Thus, in general, at a given point P there exist two sets of unit vectors, (i) el' e2, e 3 tangential to the coordinate curves and

(ii) E], E2, E3 are normal to the coordinate surfaces (see figure)

7.9.4 Representations of a vector (a) A vector A can be represented in terms of the unit base vectors e], e2 , e3 or EI' E2, E3 in the following forms: (i) A (or)

= ale]

+ a 2e2 +a3e3

(ii) A = A]E] + A2E2 + A3E3'

Here a\, a2 , a3 or AI' A 2, A3 are the components of A in the two systems

.

8r

8r

8r

respectively (b) The vectors - , - , - and Vv" Vv J , Vv3 , are called

av, av2 av3

-

'Unitary base vectors (need not be unit vectors in general). We can also represent A in one of the following forms; as

Here CI' c2' c3 are called the 'contravariant components of A and CI' C 2, C 3 are called the 'Covariant components of A


548

7.9.5 Arc length, area (surface) and volume elements (i) Let r

= r(vl'v2,v3)

:. dr=

~dvJ + 8v J

ar dV 2 + ar dV3 =AJ(dvJe J+A 2 (dv 2 )e2 +A 3(dv 3)e3 ..... (3) aV2 aV3

Since ds 2 = dr. dr, the differential ds of arc length is given by,

ds 2 = AI2dvl2 +A}dv} +A/dv/ for an orthogonal coordinate system (.,' e l .e2 = e2·e3 = e3.e l = 0) (ii) Let dvl' dv 2, be vectors specifying arc-elements along VI and v2 curves respectively at point P. i.e., dV I = dr along VI curve and dV 2 = dr along v2 curve. Since v2 ' v3 are constants along VI curve and vI' v3 are constants along v2 curve, we have, from (3),

..... (4) The area of the parallelogram formed with the vectors of( 4) as adjacent sides is the area element or surface element 'ds 3 ' on the surface v3 = c3 (see fig) and it is given by

ds 3 = Idv i Oily ds I dS 2

=

x

dv21 = IAIA2dvldv2e31 = AIA2dv ldv2

V2

curve

A2A3dv2dv3

= A3Aldv3dvi

are surface elements on surfaces VI = c i and v2 = c2 respectively. (iii) The volume element 'dv' for an orthogonal curvilinear coordinate system is equal to the volume of the parallelopiped whose coterminous edges are dv I' dv 2 , dv 3 i.e., Aldvle l, ~dv2e2' A3dv3e3 (see figure) :. dv

= I(Aldv1e(). (A2dv 2e2) x (A3dv3e3~ = AIA2A3 dv(dv2dv3 le(.e 2 x e31 = A(A2A3

dv 1dv2dv3

(.,' le 1.e2 x e31'; 1)


549

7.9.6 EXPRESSIONS FOR GRADIENT, DIVERGENCE AND CURL IN AN ORTHOGONAL CURVILINEAR COORDINATE SYSTEM: 7.9.6(1)

~:

Gradient of a scalar function

Prove that V'~ (i.e. grad~) =

[ Ia~c7v ~ I

Proof: Let V'~

= gle l

+ g2e2

I~

I~

el +~ c7v e2 +~ c7v e3

I

2

2

3

1

3

+g3e3' where gl' g2' g3 are to be determined

ar

ar

ar

c7v1

c7v 2

c7v3

dr= -dvI +-dv2 +-dv3 =

d~

=

(A1el)dv 1 + (A 2e2)dv2 + (A 3e3)dv3 V'~.dr,

d~ = A1g1dv 1 + A2g 2dv2

Again,

d~ = ~ dVI + ~

av

l

av

+

dv 2 +

2

A3~3dv3

~

av

d,V3

..... (8)

3

Note: The above expression indicates that V'

7.9.6(2)

..... (A)

el

~

I

I

= [ ~a;

e2

~

e3

~

2

2

3

3

+ ~ c7v + ~ c7v

Ifvl' v2' v3 are orthogon~1 coordinates, prove that (a) lV'vII = A,I,(I= 1,2,3)

(b) e l = E1(I= 1,2,3) We know that, (from 7.9.6 (I))

V~=~ a~ +!2~ +~ a~ Al c7v1

A2 c7v2

A3 c7v3

1


550

Hence

lilly

e e = ~ IVV3 I= -:;: = ~-] I V I= -:;: 2

n

v 2

3

-] 1\,2,

1\,3 ,

2

which proves (a)

3

(by definition, see 7.9.3(ii))

[from (a)]

/= 1,2,3.

y, 7.9.S. (3) Prove that, in orthogonal curvilinear coordinates, V.(A,eJ=

1

a (A,AzA3) and hence show that,

A]A2A3 Ov]

e, VV 2 xVV3 = -1- (e2 xe3 ) = A2 A3 AzA3

e]

A2A3 (VV2 X Vv 3) lilly e2 = A3A, (VV3 x Vv,) and e3 = A,A 2(Vv,

..... (C)

=

X

r

VV 2

V.(A]e,)= VAA 2 A3 (Vv 2 x Vv 3) =

V(A,A 2A3}(Vv2 x Vv 3)+ A,A zA3 {V.(Vvz x Vv 3 )}

=

V(A,A2A3).~ + A'A2A3{V.~} A2 A3 A2 A3

..... (D)

(Using (C)) Now,


551

(Writing the expression for V ) =

_I ~(_I_)+O+O AI Ovl A2 A3

=0

=

1

a (AIA2 A3)

..... (E)

a (A2A3AI)

..... (F)

AIA2A3 Ovl

(Operating V)

lilly, we get,

V.A 2 e2 ==

1

AIA2 A3 Ov2

and

..... (G)

Hence, if A == AIel + A2e2 + A3e3,

Div A = V.(Ale l + A2e2 + A3e3) =

7.9.6 (4)

V.(Ale l )+ V.(A 2e2 )+ V.(AJ e3)

Prove that in orthogonal curvilinear coordinates,

Vx(AleJ==~~(AIAJ-~~(AIAI) and hence, A3 AI Ov3

AIA2 Ov2


552

e .: VVI = -.L, we have,.

AI

V x (Ale l )= V x (A,A,Vv,) =V(A,AI)xVv, +(A,A,)(VxVv,) (from 7.5.2 (i) curl grad

=

=

vI

=

0)

z [!l~(AIAI)+ e ~(AIAJ+!2.~(AIAJlx!l AI oV A Ov 1..,3 oV AI

z

I

3

2

(substituting for V)

..... (i)

lilly, it can be shown that V x (A 2 e2)= ~~(A2AJ--e'-~(A2AJ 1..,11..,2 Ov l AzA3 OV3

..... (ii)

el - -0- (A3A3 ) - -ez- -0- (A3A3 ) Vx (A3e3 ) = 1..,21..,3 Ov z 1..,31..,1 Ov l

..... (iii)

and

:. curl A = V x A = V x (Aiel + A2 e2 + A3e3) = Vx

=

(Ale,)+ V(A 2e2)+ V(A 3 e3)

1 _e_ A2 A3

{~(A3A3)-~(A2AJ1+~{~(A'AJ-~(A3A3)1 oV2

f

Ov3

1..,31..,1

Ov3

Ovl

f

(Adding R.H.S. of (i), (ii) and (iii))

Ale,

A2e2

A3e3

--olov, olOv2 olOv3 1..,11..,21..,3

"I

A,/\',

A2A2

A3 A3

553


Ex. 7.9.6. (5) Derive an expression for V2~ in orthogonal curvilinear coordinates. Sol:

(from 7.9.6(1»

A=_I~ I

AI Ov I '

.. V2~

= V.V~ =

V.A

Note: In the expressions for V ,

grad~, Div A, curl A, and V2~ derived in 7.9.6(1) to

7.9.6(5), if we put vI = X, v 2 = y, v3 = z; AI =A2 = A3 = I, and e l = i, e2 = j, e3 = k; we obtain the corresponding expressions in (rectangular) Cartesian coordinate system. [The reader is advised to verify]

7.9.7 Spherical Polar coordinates (Spherical Coordinates) If we take vI = r, v2 = e, v3 =~, the curvilinear coordinate system becomes a "spherical polar coordinate system".

7.9.7(1}

z

x


554

The equations of transformation are,

x = r sin ecos~,} y = rsinesin~

..... (i)

z =rcose ~

where r

7.9.7(2) (i)

0, O:s ~ < 2n, and O:s e:s n .

Coordinate surfaces and coordinate curves. The coordinate surfaces (level surfaces) in a spherical coordinate system are r = c] which are spheres with centre at.origin [or origin itself if c] e

=

0]

c2 which are cones with verex at origin

=

(lines if c2 = 0 or nand xy-plane if c2 = n12) ~

(ii)

= c3 which are planes through the z-axis

The coordinates curves are: (a) Intersection ofe

= c2 and ~ = c3 is the r--curve and it is a line

(b) Intersection of r = c] and

~

(c) Intersection of r ",,-c] and e

7.9.7(3)

= c3 is the e--curve. It is semi circle if c] =

c2 is the

~urve.

It is a circle (or point).

Scale factors: r

= xi + yj +zk (rsinecos~)i + (rsinesin~)j + (rcose)k

=

~ = Or = (sinecos~)i + (sinesin~)j + (cose)k Ovl

Or

18rl=~(sinecos~)2+(sinesin~?+(cose?

A = ] 8r =1

8r = 8r Ov2 as

=r

:j:.

=(rcosecos~)i+(rcosesin~)j-(rsine)k

0

555


-

or = -or = (-rsm. 0 Sill", . .t.).1+ (rSIl1 . 0 cos",.t.) }.

Ov 3

0<1>

lorl=~(-"sinOsin

A = 3

=

r sinO

: .. The scale factors of spherical coordinate system are

A( = I,

A2 = r,

A3

=

r sinO.

7.9.7(4) Base vectors: The base vectors e (' e 2, e 3 are taken for the sake of convenience as er , eo' etjl respectively. They are,

e r

=

e

=

(

_I ~=_I or = (sin Ocos
AI Ov l

eo = e2 =_1

AI

or = ~[(rcosOcos
AI GO

=

r

(cosOcos
eel> =e3 =_1 or

and

or

A3 0<1>

=-~-(-rsinOsin
= (- sin

)j

From the above expressions, it can be observed that

e,..e s = es.eq. = er.eq. = 0 er xes =eq.; efJ xeq. =e,.; eq. xe r =eo;

and

[students are advised to verify the above equations] which show that "the spherical polar coordinate system is a right-handed orthogonal coordinate system".

7.9.7(5) Arc length, Area element, and volume element in spherical coordinates: A( = I, A2 = r, A3 = r sinO

.) dS 2 = 11.1 ~ 2d 2 ~2d 2 ~ 2d 2 VI + 11.2 V2 + 11.3 V3

I

= 12(dr)2 +r2(dO)2 + (rsinO)2(df gives the arc length 'ds'.

(from 7.9.5(1))


556

and ds r , ds s ' dStjl are the area elements on the respective surfaces r

= cl'

iii) From 7.9.5(3),

e

= c2'

dv

=

and

cj>

= c3.

1..)1..21..3 dv)dv2dv3

= I.r.{rsin e)dr.d9.dcj> i.e., dv = r2 sin edr.d9.dcj> is the volume element.

7.9.8 Cylindrical polar coordinates (cylindrical coordinates) 7.9.8(1)

In this case, we take,

Then the curvilinear coordinate system becomes a 'cylindrical coordinate system'.

The equations of trans,formation are x=pcose, y=psine, z=z. where

7.9.8(2)

Scale factors: r

= xi + yj +zk = (pcose)i + (psin e)j + zk

557


-8r = -8r = (cos 8) I.+ (. sm 8)·} av l 8p 8r I= -J cos 8 2 + sm • AI = 8p 82 = I

I

. ) i+ (pcos8 ) j -8r= -8r= (-psm8 8v2 as

A2

=1:1=~(-psin8)2 + (pcos8)2 =p A1 = I8r I= 1 8z

8r = 8r = k; 8v) 8z

:. scale factors are I, p, 1. 7.9.8(3) Coordinate surfaces and coordinate curves: (i) The coordinate surfaces (level surfaces) in a cylindrical coordinate system are.

p = cl' which are cylinders coaxal with z-axis (or z-axis itself if c 1 = 0) 8 = c2' which are planes through z-axis and z = c3' which are planes

~r

to z-axis

(ii) Coordinate curves are: (a) Intersection of 8 = c2 and z = c3 is the p-curve which is a straight line. (b) Intersection of p

= c 1 and z = c3 is the 8-curve which is a circle (or Point)

(c) Intersection of p = c 1 and 8 = c2 is the z-curve which is a straight line. 7.9.8(4) Base vectors: (ep,eo,e z ) ep = _I 8r = (cos 8 )i + (sin 8)j

AI 8p

eo =_1 8r

1<.2

as

=~[(-psine)i+(pcose)j] p

=

(-sin8)i+(cose))


558

1 8r e =--=k· ~ A3 8z '

which shows that cylindrical coordinate system is a righthanded orthogonal coordinate system.

7.9.8(5) Arc length,Area element and volume element in cylindrical coordinate system: (i) =

12 (dpy + p2(de)2 + 12 (dzY

=

(dpy + p2(d8)2 + (dZ)2

gives arc length ods'.

ds p' ds s' ds z represent the area elements on the surfaces p respectively.

=

c l ' 8 = c2 ' and z = c 3

(iii) Volume element 'dv':

=

l.p.dp.dedz = pdp.de.dz.

7.9.9: EXPRESSIONS FOR GRADIENT, DIVERGENCE, CURL, LAPLACIAN AND JACOBIAN IN SPHERICAL AND CYLINDRICAL COORDINATES: 7.9.9(1)

Gradient of a scalar function 'f

18/

18/

18/

Grad f= V j=---e l + - - e2 +---e3

AI

av

l

A2

av

2

A3

av

3

(from 7.9.6(1))

559


a) spherical coordinates: Al = 1, el

=C r

A2 = r,

A3 = rsin 0

=(sin8cos~)i+(sin8sin~)j+(cos8)k

e2 = Co = (cos8cos~)i + (cos8sin ~)j ~(sin 8)k

e, =e =(~sin~)i+(cos~)j

v f= of e +! Wc or r as r

+_I_ofe 0

rsine

as

b) cylindrical coordinates:

Al = 1,

A) = 1

A2 = p,

el =ep =(cose)i+(sine)j e2

=eo =(~sine)i+(cose)j

7.9.9(2) Divergence of a vector function

If A = AIel + A2e2 + A3 e3, the expression for Div A is given by Div A=

[~(AIA2A3)+~(A2A3AI)+~(A3AIA2)l(from 7.9.6(3» 8v 8v

1

AIA2 A3 8v1

2

3

(a) In spherical coordinates

:. Div A =

2

1.

[~~2 sin eA )+ ~(rsineA2)+ ~(rA3)l as o~ J I

r sll1e or

(b) In cylindrical coordinates Al =1,

A2 =p,A 3 =1,

:. Div A

= -

VI

=p,v 2 =e,

V3

=Z

1[0-(pAI)+-(A2)+-(pAJ 0 0 J

p op

as

oz


560

7.9.9(3) Laplacian

(V 2 ): Let f be a scalar function. Then the expression for

V2 j

m

curvilinear coordinates is given by

[from 7.9.6(5)]

a)

In spherical coordinates:

Ai =1,

1.,2 =r,

V2j =

1.,3 =rsiIl9,

v2 =9, v)

=r,

Vi

j

1 [ -a ( r2sm9. a ) +a (.sm9OJ ) +a ( -1- aj - )] r2 sin 9 ar ar 09 09 a sin 9 O

a2j ar2

aj 1 a2j r ar r2 ae

2

cot 9 aj

1 sin 9

= -- +- - +- - 2 2 + - - - + ---::---:----

r2 ae r2

a2f a2

(after ditferntiation and simplification) b)

In cylindrical coordinates:

Al = I,

..

V2j

1.,2

=

= p,

1.,3

= 1,

VI

= p,

V2

= 9,

V3

=Z

![~(p Of)+~(! Of)+~(p Of)] pap op OOpOO OZ GZ a2f ap2

1 af p ap

1 a2j p2 09

a2f az

= - - + - - + - - -2+ - 2 (after differentiation and simplification)

7.9.9(4)

Curl of a vector function A.

If A = Aiel + A2 e2 + A3 e3 '

(from 7.9.6(4»

a) In spherical coordinates:

Ai =1,

1.,2 =r,

A)

=r~';n9,

VI

=r,

v 2 =9,

v) =


561

1

..

curl A = r2 sin 8

er rea {rsin8}eq, ajar ajae aj~ AI

rA2

(r sin 8 )A3

b) In cylindrical coordinates:

A2 = p,

AI =1,

7.9.9(5)

A3 = 1, VI =P,

v2 = 8,

V3 = Z

Jacobian:

(i) Orthogonal curvilinear coordinates: Let x = x{V"V 2 ,V3); Y = Y{V"V 2 ,V3); Z = z{V"V 2 ,V3) represent an orthogonal curvilinear coordinate system

Jacobian =

since

r

=

J(

x,y,z ) = v" v2 ' V3

axjavl ayjav l azjavl axjav2 ayjav2 azjav2 axjav3 ayjav3 azjav3

xi +yj + zk

[Scalar Triple Product]


562

(ii) Spherical coordinates:

x=rsinOcos<\> y=rsinOsin<\>

z = rcosO, are the equations of transformation of spherical coordinate system.

. =.,(x,y,z) Jacobian -r,O,<\>

ax/ar ay/ar az/iJr ax/a8 ay/ao az/80 ax/a<\> ay/a<\> az/a<\> sin Ocos<\>

sinOsin<\>

cosO

rcosOcos<\>

rcosOsin <\>

- rsin 0

-rsinOsin<\>

rsinOcos<\>

0

= r2 sinO

(or) [J = AIA2A3 = I.r.r sinO =

r2 sinO]

(iii) Cylindrical coordinates:

x = pcosO,y = psin 0, z = z, are the equations of transformation.

Jacobian =

ax/ap ay/ap az/ap J[x,y,z) = ax/88 ay/88 az/88 p,O,z. ax/az ay/az az/az eosO

sinO

0

-psinO

peosO

0

o

0

[(or) J

=

AIA2A3

=

Lp.l

=

p

=

p]

The summary of important results of this chapter is given in a table in the next page for the convenience of the student.

Cartesian Coordinates (x, y, z)

Spherical Coordinates (r, 0, ¢)

Equations of transformation of coordinates

x=x y=y

x

2)

Scale factors

AI = 1,

3)

Base vectors

i, j, k

1)

=r sinO coS¢ y =r sinO sin¢ z = r cosO

z=z

Jacobian (J)

x =pcosO y =psinO z=z

A2= I,A. 3= 1 e,= (sinO coS¢)i + (sinO sin¢)j + (cosO)k ep = (cosO)i + (sinO)j ee= (cosO coS¢)i + (cosO sin¢)j-(sinO)k ee= HinO)i + (cosO)j e~=

4)

Cylindrical Coordinates (p ,0, z)

8(x, y, z) a(x,y, z)

=(dx)2 + (dy)2 + (dz)2

(-sin¢)i + (coS¢)j

ez=k

8(x, y, z) _ 2 • Ll a(r. 0, ¢) - r smu

=(dr)2 + r2(d~)2 + r1 sin 2O(d¢)2

5) (Arc Length)2

(ds)2

6) Area elements

ds , = r2 sinO dO d¢

dsp =pdO d-

on the coordinate

=dy dz ds 2 =dz dx

surfaces

ds 3

=dx dy

ds e= r si nO d¢ dr dSIj>= r dr dO

ds Q=d'"-, dp ds z pdpdO

7) Volume

dv

=dx dy dz

dv = r1 sinO dr dO d¢

dv

element (dv,\

ds,

(dS)2

~

=

= pdpdOdz

Cartesian Coordinates

Spherical Coordinates (r, 0, ¢)

(x, y, z)

8)

Grauf

9)

Div A

()A I

a~

a~

-+-+Ox ay clZ

Div A =

+

10) Laplacian

go

Cylindrical Coordinates (p ,0, z)

~ O[ ~

....2

rsm

(Jr

(r sinO A2) +

(r2sinO AI)

~~ (r A3)]

~ 2!Ji. I 2:J. cotO!Ji. - ao tJr2 -r ar +r2- ()02 + r2

!!i.!J:i.iEJ. or + iJy2 + iJ z 2

(v2j)

+ 2

r2sin O

f2:[ + I .!Ji. + I Sf Up2 Ii up 7Y a(J2

+iEJ. at

m

::J

co ::J

!Ei U¢2

(l) (l) ~.

::J

co

s::

Q)

j

11) CuriA

a/{)x AI

OliJy A2

k

alaz A3

r2sinO

er

reo

alar

alfJo

AI

rA2

(r sinO)e¢

()/af/J (r sin{})A3

I

-p

peo

e:

cl/(Jp

(J/r)O

i.JI (Jz

AI

pA2

A3

eI'

=r (l)

3Q)

oen·


565

Solved examples

Ex. 7.9.10 Obtain the equations of transformation from cylindrical to cartesian coordinates. Sol:

Equations of transformation from cartesian to cylindrical coordinates are x=pcos8

..... (1),

y=psin8 ..... (2)

z=z ..... (3)

from (I) and (2), we get, p2 = x 2 + y2

~p=~X2 + y2 ,p> 0 y from (I) and (2), we also get - = tan 8

x

~8=tan-l(~) :. The required equations are

p=~X2+y2, 8=tan-'(~}

z=z

If a point lies on z-axis, x = 0, y = 0 => 8 is indeterminate. These points on z-axis are known as singular points of the transformation. Ex. 7.9.11 Represent the vector A = 2xi - yj +z2k in cylindrical polar coordinates. Sol: The base vectors in cylindrical coordinates are

Note:

ep =(cos8)i+(sin8)j eo = (-sin8)i+(cos8)j

..... (2)

ez = k

..... (3)

Solving (\) and (2),. we get i=(cos8)ep -(sin8h j

Then

A = 2[(cosS)ep

= (sin8)ep + (cosS)eo

-(sin8)eo~-[(sinS)ep +(cos8~o}y+z2e~.

A = (2xcosS - ysinS)ep -[2xsin8+ ycosSh + z2e~

= (2pcos 2S - psin 28 ~p - [2psin8cos8 + psin8cos8to + Z2 ez (substitutingx= pcos8, =

...... (\)

y= psin8)

p(2cos2S-sin28~p -(3psinScos8)eo +z 2e:


566

Aliter: Since cylindrical coordinates form an orthogonal coordinate system, we can write A as A = a,ep + a2ee + a3ez From (1) we find,

A.ep =a"

A.eo =a2 ;

.•..•

(I)

whereat'~, a 3 are to

be determined.

A.ez =a3

..... (2)

For cylindrical coordinates, we have, x=pcos8, y=psin8, Z=Z given A becomes, A= (2pcos8)i-(psin8)j+z 2 k

..... (3)

ep = (cos8)i+(sin8)j } Also, eo = (-sin8)i+ (cos8)j

..... (4)

ez =k using (3) and (4) in (2), we get

a, = 2pcos 2 8 - psin 2 8; a 2 =-2psin8cos8-psin8cos8=-3psin8cos8; a 3

=Z2

A = a,ep + a2 ee + a3 ez = (2 cos 28 - sin 2 8) pep - (3p sin8 cos8) eo + z2 ez

Ex. 7.9.12 Sol:

Represent the vector A = xyi - zj + xzk in the spherical coordinate system. Since the base vectors er, eo' A

=

ecj>

are mutually orthogonal, we can write

a,er + a2 eO + a3eq,

..... (J)

where aI' a 2, a 3 are to be found. From (I) we get, A.er =a,. A.ee =a2 • A.eq, =a3 For spherical coordinates, we have, x = r sin8 cos, y = r sin8 sin, 2

2

z

•••••

= r cos8

2

A = {r sin 8sin cos
(2)

..... (3) ..... (4)

er = {sin 8cos
={cos8cos )j - {sin 8)k

eq,

= (-sin4»i+ {cosiJj

Using (4) and (5) in (2), we get,

a, = 1'2 sin 3 8sin 4>cos 2 4> - rsin 8cos8sin + 1'2 sin 8cos 2 8cos4> . a2 = r2 sin 2 8cos8sin cos 2 - rcos 2 8sin cp - r2 sin 2 8cos8coscp a3

= _1'2 sin 2 8sin 2 cpcos - r cos 8 cos cp

..... (5)


567

Hence A = aler + a 2ee + 01 eq, =

{r

2

sin} 8sin $cos 1 $ - rsin 8cos8sin $ + r2 sin Ocos 2 8cosc/>k + {r 2 sin 2 8cos8sin $cos 2 $ - rcos 2 8sin $ - r2 sin 2 8cos8cos$ ~o 1

2

2

+ (-r sin 8sin $cos$-rcos8cos$h

Ex. 7.9.13 Prove the following:

Sol:

(i)

e = Seo+ {sin 8 )~eq,

(ii)

eo = -Ser + {cos8)~eq,

r

(iii) eq, = (- sin 8 )~er - (cos8 )~eo where' • ' denotes differentiation w.r.t. time 't'. (i) We know that er = (sin 8cos$)i + (sin 8sin $)j + (cosS)k " cos8cos$8"} i + { " sinOcosc/>.$"} j - (sin8)8k " er ={sin 8(- sin$)$+ cos8.sin$.8+ =

e{(cos 8 cos $)i + {cos8sin$)j - (sin 8 )k} + {{- sin8sinc/»i + {sin8cos$)j }

" " = 8e e + sin 8.$eq,

(ii) ee

={cos8cos$}i + (cos8sin C/>}j - {sin 8}k

ee =( - sin8.8cos$ -cos8sin$~ } + ( - sin8.8s;". + coseco.~++ )j -(cOS8.e)k = -

e{(sin 8cos$)i + (sin 8sin$}j + (cos 8 )k} + cos8.+{(- sinc/»i + {cos$)j}

= -

" " 8e r + cos8$eq,

(iii) eo!> =(-sin$)i+{cos$}j

eo!>

=-cos$~i-sin$~j (- sin 8 }~er - cos8~ee

= ~[- sin 8{sin 8cos$i + sin 8sin $ j + cos8k} - cos 8 {cos 8cos c/> i + cos8sin c/> j - sin 8 k}]

..... (i)


568

o

0

= - coscj).cj)i - sin cj).cj) j

= e~ Ex. 7.9.14

from (i)

If ep , eo are base vectors in cylindrical coordinates, show that

(i)(ep)=ee o and

(ii)(eo)=-eep,

where'.' denotes differentiation w.r.t. 't' Sol:

ep = (cosS)i+ (sinS)j

= e{(-sinS)i + (cosS)j}

eo = -sin Si + cosS j, eo = - S(cosS)i -e(sin S)j = -Sep

Ex. 7.9.15 Express the velocity V and acceleration a ofa particle in cylindrical coordinates. Sol:

In cartesian coordiantes, the position .vector r = xi + yj + zk .

. dr dx. dy. dz VelOCIty V = -=-I+-j+-k dt dt dl dt Acceleration a=

d 2r

d 2 x.

d 2y

d 2z

dt

dt

dr

dt

-=-1+-, j+-k 2 2 2

In cylindrical coordinates, x = pcosS, y = psin S , z = z and ep=(cosS)i+(sinS)j, eo =(-sinS)i+(cosS)j, ez

=k

~'. i = cosSep - sin Seo ; j :::: (sin S)ep + (cos e)eo (Solving above equations for i andj) :. r = (pcosS XcosSe p - sin See)+ psin e(sin Sep + cosSee)+ zez = pep + ze z

dr dp de dz . V = -=-e +P-P +-e .. dt dt p dt dt Z

(0) 0

o = pep +p Sea +zez

[

.: dez dt

=~(k)=O] dt

(from Ex. 7.9.14)

569 .


dV

a = dt

d.

.

.

= dt [pep + pOeo + zeJ (since

....

..

..

....

= pep + pep + pOeo + pOeo + pOeo + ze z

ez = 0)

pBen + pep + pBeo + pBeo + pB(-Bep ) + ze z ' (using results of (ex 7.9.14»

=

[p- pe 2 ]e p +[pO + 2pB]eo + zez Ex. 7.9.16 If t

= xyz,

find 'gradf' in (a) Cylindrical coordinates (b) Spherical

coordinates Solution: In cylindrical coordinates (p,O,z), we have, x = pcosO,y = psinO,z =

Z

..... ( I)

and 2

f = xyz = p2 Z sin 0 cos 0 = p

2

Z

sin 20

af = pz sin 20· af = p2 Z cos 20. at = p2 sin 20 ap 'ao 'az 2 2

:. (I) gives, gradf = Vf = (pz sin 20)e p + (pz cos 20)eo + (~ sin 20)ez (b)

In spherical coordinates (r,O,tjJ), we have,

x = rsinOcostjJ,y = rsinOsintjJ,z = rcosO, and

Vf= af e +! af e +_1_ af e ar r r ao 0 r sin 0 atjJ ; Here

f = xyz = 3r 2sin 2 OcosOcostjJsintjJ :. af ='csin2tjJ{-sin 3 0+cos02sinOcosO} ao 2

..... (2)


579

aj

o~

=rJsin20cosO.cos2~

:. from (2) )

~l=(3r2sin20cosOsin+cos~~,. + '; sin2~{-sin30+2sinOcos20}ee + (r2 sinOcosOcos2~~~. E.7.9.17 Itf= pzcosO (in cylindrical coordinates) find Vf. Sol: f= pz cosO.

-aj = zcoso of

op

Hence Vf

,

=

aj

00

=

-pzsin 0 aj = pcosO ' oz

I aj

of

op ep +p ae eo + oz ez

= (zcosO)ep -(zsinO)eo + (pcosO}ez Ex.7.9.18 Iff=

Sol:

,.'2

sin2Gsin+ in spherical coordinates, find Vf.

aj = 2rsin20sin4>

or

-8f = 2r2 cos 20sin 4>

ae

aj =r2 sin20cos~ o~

.. ~l

8j = =

Jo/

1

Of

or e,. +-; ae ce + rsinO o~ e.p

(2rsin20sinq,)e,. +(2rcos20sin~)eij + (2rcosOcos4»e.p

Ex. 7.9.19 Show that the vectorfieldA =

z{(sin O)ep + (cosoh }-(pcosO}ez, in cylindrical

polar coord inates is solenoidal.

Sol:

IfA= Aiel' +A2 co +AJe z then, AI =zsinO; A2 =zcosO; A3 =-pcosO

571


=

~[~(pz sinO)+~(zcOsO)+~(- p2 coso)l

=

~[zsinO-zsinO+O] = 0

as

p ap

aZ

J

p

:. A is solenoidal

I

1

I

)

Ex.7.9.20 IfA= ( r-cosOer +-eo +-.-Oe+ ,in spherical coordinates, find Div A. r

Sol:

rS1l1

I I Here A =r 2 cosO, A2 =-, A3 = - . -

r

I

rS1l10

1

3

=

=

[4r -sin20+cosO+O ,.- S1l1 0 2 I

1.

cotO

4rcosO+-2r

Ex.7.9.21 Ifj= p2Z2cos20, show that V2j=2p2cos20 Sol:

In cylindrical coordinates, 2

af ap2

1 af p ap

1 p2

a2 f

a2 f

ao

az

V j = - + - - + - -2- + 2

j

= p2 Z 2 cos20 2

Qf = 2 pz 2 cos2 0 . a j = 2 z 2 cos 20 . ap , ap2 ,

-

..... (I)

at ') as = -2p- Z 2 si1120·'

_'J


572

:. From (1) we get, yo2 f = 2Z2 cos 26 + 2Z2 cos 29 - 4Z2 cos 29 + 2p2 cos 29 = 2p2 cos 29 .

Ex.7.9.22 Using spherical coordinates, show that yo2r" = n{n + l)r,,-2 when n is a Sol:

constant and r = 0 if n<2 In spherical coordinates, 2 2 yo2f=a2f+~Of +_1 a f + cot 9 Of + 1 d f ar2 r ar r2 00 2 r2 a9 r2 sin 20 d~2 Here, f= rn 2 Of = nrfl-I. a f = n{n- 1)r"- 2

..

8r

.. (I)

' Br2

~ yo2f =V2r" = n{n-l)r"- 2 +~.nr"-I

Ex.7.t.23 Iff=

r 2 = n{n -1 )r"-2 + 211r"- = n{n + I )r"- 2 .

r 2 sin29+cos 2 «p, find 2

Sol:

Of = 2rsin29. a

or

Of -=-sin2+

0+

'

yo2f.

~ = 2sin 29

ar~

a2f ' 0+

-=-2cos2~ 2

a 2 f 20f 1 a 2f cot90f a 2f yo2f - --+--+---+----+-::---::-2 2 .. - ar2 r ar r2 00 ,.2 a9 r2 sin 9 a~2 + [CO/9x2r2cos29+ r2

2

r

~

?

sm~9

X{-2COS2«P)]

= .2sin 29 +~(2rsin29)+ ~(- 41'2 sin 29)+ co~9 x -2cos2«p r r r = 2sin 29+ 4sin 29 -4sin 29 + 2cot9.cos20 -4cosec 29.cos2«p r

= 2sin 29 + 2 cot 9.cos 29 -4cosec 2 9.cos2«P r

..... (I)

573


Ex.7.9.24 If A is a vector field, find curl A where,

{r:~sino~,+(~cosO}e+ r l2eq,

(\) A=

(2) A = {zcosO)ep -{ZSiIlO)eo Sol:

If A = Ale, + A2eO + A3eq., We have, in spherical coordinates,

reo alar a/ae r-smO 1 ~.

CurlA=

e,

{rsinO)eq,

Al

(rsinO}A3

rA2

2

Here, AI =r sinO, A2

:. Curl A =

=~cosO, r

ajar

2

=

r2

3

{rsin O)eq,

a/ae

a/a$

cosO

(I/r}sine

\ [- a {( -smO 1. ) - a (cosO )}e,+ {-a (r 2smO . )-a (sinO)} - - rea r a$ a$ ar r

~.

==

=

A =-

e,

\

r sin 0 r2 sin 0

=

a/a~

r smO ae

7

1.

[~coso.e, +-;-sinO.re e _r2 coso.rsinoeq,] r

r- smO r

1 [\-cosO.e, +--ee sin 0 -r 3 smO.cosOeq, • ] smO r r

2.

r

cotO

I

-3-e, +]ee -rcosOeq, r

r

we have, Curl A =

\

-

p

ep

pea

ez

a/ Bp alae a/az AI

A2

A3


574

Here, A I =

cosO, A2 = -z sine, A3 = o.

Z

I

:. Curl A =

-

e

peo

ez

a/ap

a/ae

a/az

-pzsinO

0

P

p zcose

a a .}e + {-(zcose)--(o) a a} = -I [{ -(o)--(-pzsme) poO az Paz ap pee

=

~[psineep +pcosOeo +(-zsinO+zsine)ezl p

= (sinO)ep + (cose)eo

Ex.7.9.25 Show that the vectorfieldA= (pzsin2e)ep + (pz cos 2e)ee irrotational.

Sol:

Here, AI =pzsin2e, A) =pzcos2e, A) -

=~p2sin2e 2

ep

peo

ez

a/ap

alae

a/az

p pzsin2e

p2zcos2e

~p2sin2e 2

+(~p2sin2e)e= is

575


:. A is irrotational.

Ex.7.9.26 Show that div (curl A) = 0, where A is a vector function, in orthogonal curvilinear coord inates.

Sol:

Let A = Aiel + A2e 2 + Aj e3 Then

.

. '"

I

Dlv j = ~ A A A I

=0

2 3

Ale]

A 3 e3

%v2

Ojov)

A2A2

A)A,

[~( Ov V··A A ) I I

I

2

3

Engineering Mathematics - •

576

Exercise - 7(i) l. Show that the spherical coordinate system is an orthogonal system. 2. Show that the cylindrical coordinate system is an orthogonal system. [Hint for I and 2: Show that e,.e 2

= e2 .e3 = e3 .e, = 0

etc 1

3. Express the vector, F = 2xi - 3y.j +z,Yk, in cylindrical polar coordinates. 2 2 [Ans: F= (2pcos S-3p 2 sin S}p

-~sjn2S+3p2sin2Sh +(pzcosskJ

4. Represent the vector, A = xyi + yzj +zxk, ill spherical polar coordinates. [Ans: A = r2 sinS{sin 2 Scos 2 ~sin~+sinScosSsin2 + (sin Scos8cos

2

~sin~ + cos 2 8sin 2 ~ -

~+COS2 Scos~k

sin ScosScos~~o

+ (cosSsin~cos~-sinSsin2 ~cos~}~ 1 5. Express the vectors a) A = yi + xj and b) F = zi - 2xj + yk, in cyl indrical coordinates.

[Ans: a) A = peo b) F

= (zcosS - p sin2S)e p - (zsinS + 2pcos2S)eo + (psinS)ezJ

6. Express the vectors a) A coord inates.

= -

yi + xj, b) F

=

xi + 2y.i + yzk, in spherical polar

[Ans: a) A = r sinS eq, b) rsinS [{sinS(I + sin2~) + rcos2~ sin~}er + {cosS(l + sin2~)- rsinScosS sin~}e() + sin~cos~ eq,l 7. Find the gradients of the fol/owing function

i)f= pz sinS

ii)f=

sinS -2-

r

[Ans: i) Vf = (zsin S)ep + (z cos S)eo + (psinO)ez ii)

Vf=~[(-2sinS)er + (cosO)eoJ r

iii) Vf =(2rsin

2

~}r - (rsin 2S)eo -(~cosecscos2 ~sin~ )e~1

8. Find the divergence ofth fol/owing vector functions. i) A = (cosS + sinS) ep + (cosS - sinS)e o + ez

ii) F =

~2 sinS}r + (~COSO )eo + rl2 e~ , iii) V = (pcosO)er + (psinO)eq,


577

.. 4 . 0 1 cos 20 (II) rsm +2-.-r Sill 0

IAns: (i) 0

9.

(iii) 3cosO )

Show that the following vector fields are solenoidal. (i)

10.

A= (zcosO)c" -( z - sin O)co(ii) F = (~cot O)c r r

-~c() + (r cosO)c¢ r

Find the curl of the following vector fields: (i) A= (zsin O)C" + (zcosO)ctJ -(pcosO)c z (ii) F =(rsinO)c r

...

+(~COSB )c +(~ )e¢ r

1 0 = -tan -e¢ ,. 2

(III)V

IAns:(i) (sinO-cosO)c p +(sin¢+cos¢)Cf) (ii) (,.\ cot 0 11.

If ' f'

jer -(cosB)c¢ (iii) 1~2 c¢)

is a scalar function in orthogonal curvil inear coordinates v" v2 , V3 prove

that 'Vf' is irrotational.

7.10

GREEN'S THEOREM IN THE PLANE

7.10.1 Green's Theorem: Let (i) (ii)

R be a closed region of the xy plane bounded by a simple closed curve C

P(x,y,) and Q(x,y) be continuolls functions of x and y having continolls aQ derivatives in R. Then r Pdx + Qdy = c - ap )d'Cdy .l 11 ax ay

JI

where C is in the positive direction.

Proof:

Y

E B

a

___ .lI __ I

~

_ _7

A

I I I o~----~c------------~--x


578

Let the equations of the curves DAB and DEB (see the figure) be respectively y = fl (x) and y = f2 (x). Let R be the region bounded by the curve C. Then,

oP " [ 12(X) oP } f f=-dxdy = f f -dy ix II CY x=c y=fI(x) CY

fp(x,y)ix = - f f~xdy II CY

i.e.,

..... (1)

c

lilly if the equations of the curves ADE be and ABE be taken respectively as

x

=

gl(Y) and x

=

g2(Y)' we have,

a

h

h

fiQ(g2,y)-Q(g\,y)}iy= fQ(g\,y)iy+ fQ(g2,y)iy

=

a

h

a

c

Thus

f

fQ(x,y)iy = I~Qdxdy c

II

~

Adding (1) and (2),

rt

J pdx +Qdy = f' oQ _ oP xdy

'!

}lox

oy

..... (2)

579


7.10.2 Vector notation of Green's theorem Green's Theorem in the plane can be put in vector notation in the following way. Let F

=

P(x, y)i + Q(x, y)j

and r = xi + yj, so that dr

=

F. dr

(dx)i + (dy)j

P dx + Q dy

=

Again,VxF=%x

j

k

o/ay

%z

P =

(

Q

OQ OP)

ox oy

°

k, so that

(v x F).k = oQ _ oP ox oy Taking dR = dxdy, Green's theorem in the plane can be stated in the vector form as,

fF .dr

JJ(V x F)JaIR

=

II

..... (A)

7.10.3 Physical interpretation of Green's theorem I.

Let F denote the force field acting on a particle. Then

fF .dr represents the work done in movi.g the particle around a closed

curve C. :. From (A) it follows that the work done is determined by curl F = V x F. 2.

In particular, if V x F = 0 i.e., ifF is conservative (or F = V t) Then fF.dr = O. i.e., the work done is independent of the path.

3. Conversely, if the integral is independent of the path, i.e., if

fF .dr =0, then vx F=O In the plane, V

x

F = 0 is equivalent to saying that :

=

~~

where F = Pi +Qj.


580

7.10.4 Application of Green's theorem to the evaluation of area of a simple closed curve. The area bounded by a simple closed curve C

=

lfxu:y - ydx c

Proof By Green's theorem, we have,

.fPdx + Qdy fI!~{auax - apay rtXl~Y ==

I"

if we put P =

-

y, and Q = x,

.I.e., -ap =-1, ay

au _ap =2, ax ay

aQ_ == I,

ax

:. We get,

fXdY - ydx = I"

i.e., A =

ff2dxdy

=

2A , where A is the required area.

II

~fxdy -

ydx .

2 ('

Solved Examples ~

~

Ex. 7.10.5 Find the area of the ellipse ~+L = I. a 2 h2 Sol:

Parametric equations of the ellipse are, = a cosO, y = b sinO dx = - a sine de, dy = b cose dO. :. By Green's theorem,

x

Area of the ellipse

=

lfXdY - ydx L

I

2n

=-

2

f(acosehcos 0 + bsinea sine}le 0

I 2n == - fahdO = rrah

2

0


581

Ex. 7.10.6 Evaluate f(y - sinx}dx + (cosx}1y, a) directly and b) lIsing Green's theorem,

where c is the boundary of the triangle in xy-plane whose vertices are (0,0), (%,0) and (%,1) traversed in the positive direction.

Sol:

Let 1= f{y-sinx}d\" + {cosx)dy 1!

X =-

a)alongOA:

y=o,

2

dy=O 1['2

n<2

f(y - sin x}1x + {cos x }dy = f- sin xdx = cosxi 0

OA

alongAB:

=-1

0

d\"=O

.. f(y -sinx}lx + (cosx~/y = fOdy

=

°

Ali

2 along BO: Equation of 013 is y = 2x , dy = -dx , 1t

7t

f(y-sinx}dx+cosxdy=

1[(2: 11

/JO

2

2.

x 0 = -+cosx+-stnxi 1t

1t

-situ )+;cosx}x

~

2)

4

11/2

:. I = Sum of the integrals along OA, AB and 130 7t 2 7t 2 =-1 +0+1- - - - =- - - -

4

7t

2

1t =(0+ 1 +0)- (1t -+0+- =1----.

4

7t

1t

4

1t

582


(b)

By Green's theorem,

J Pdx + Qdy =

'jc

8Q ax

Ii( 8Q - ~~ldX dy, dx dy

Here

P = y-sinx, Q = cosx

I I

ap = 1

= _ sin x

'ay [ =

=

!,f( ~sinx~ l)dr try

l)dy

Jd< = ~i ~ysinx~ y'l d<

2x. 2x f 2 2 fo --smx-x = -- f xsmx d'(-- f xd'(

7r

2(

7r

1C

2.

7r

1C X~O

1C

2

1C ()

Ex.6.10.7 Evaluate f(3x+4y)dx+(2x-3y)dy where C is the circle in xy plane with

centre at origin and radius 2 units.

Sol:

By Green's theorem,

fPdx + Qdy =

8Q 8Pl dxdy, Ii(--dx dy II

Here

8~=4

P=3x+4y, Q=2x-3y ;

The given integral

=

Oy

aQ =2

'ax

ff(2-4)dxdy

= -2 ffdxdy = -2A, II

where A is the area of the circle C.

= -2 x

1C

x 22

= -81C


583

7.10.8 Verity Green's theorem in the plane for the integral 1(3x 2 - 8y2 )dx + 4( 4y - 6xy)dy where C is the boundary of the region given by (I)

(2)

x = 0, y = 0, x

+ Y= I

(I) The given region is shown in the figure below. y

._~

______ X

The points of intersection of y = x 2 and y = We have to integrate [ along (I) y = x 2 from 0 to A. along y =

(2)

fx

fx

are (0, 0) and (I, I).

from A to 0, and add the two values.

Alongy=x2, dy=2xdx x~1

:.1= f (3x 2

-

8X4)dx + (4x 2

6x')2xdx = f(3x 2

-

X~O

AU

Alongy =

-

20x 4 + 8x 3 )dx = x 3

I

-

4x 5 + 2X4 I =-1 0

fx ; x = y, dt = 2ydy

yeO II 0 5 :.1= J(3 y 4_8 y 2jdy+(4y-6x 3 )dy= J(6yS-22y3+4Y)(2Y~/y=y6_~y4+2y21=2 I 2 AO )' I

:.

5 3 1=-1+-=-

2

2

By Green's Theorem, 1Pdt + Qdy = c

Here,

P

=

Ii OQ. OP)dXd )l ox ~v Y

3x2 - 8y ; Q = 4y - 6xy

oQ _ oP =-6v-(-16v)=IOy

ox

oy

.

.

.


584

:.I=f

I[E 1 I

flOydxdy= f

flOydydx= fSy

x=O y=x 2

R

2

x=O

J~l dx = x2

I

4

3

fCSx-Sx )dx=2 x=o

Green's theorem is verified. 2.

The given region is shown in the figure below. y

(0,1)8

x=O o~--~~----------x y=0 A(1,0) y

AlongOA,

=

o

=

f~X2 -

I

I

x=o

0

3 2 f 3x dx = x 1= I

:. Given integral = y AlongAB, :. Given integral

0 ::) dy = 0

=

I - x ::) dy =

-

dx 0

8(1- x 2 )}dx + {4(1- x) - 6x(1- x)C-dx)} = fC-Ilx 2 + 26x -12) dx ' x=1

x=1

II 3 2 0 II 8 =--x +13x -12xl =--13+12=-

3

AlongBO,

I

3

3

x = 0 ::) dx = 0 y=O

Given integral =

0

f 4ydy = 2y21 =-2 y=1

:. The given integral = 1 +

I

8

S

"3 - 2 ="3

By green's theorem,

1Pdx + Qdy = f JlnaQ. ap)dXdY ax ay C

R


585

Here, P = 3x2 - 8y2,

aQ ap ax· ay

Q = 4y - 6xy

= -6y - (-16y) = lOy

= 5(1-x)3i =-~(O-I)=~ 3x-1 0 3 3 Hence the theorem is verified. (2,1)

Ex. 7.10.9

J(10x

Evaluate

4

2x/ dx - 3xl.v dy along the path x4 - 6xl - 4.0 = 0

-

(0,0)

P = IOx4 - 2xl, Q = 3x2.0

Sol:

ar

~

-=-6xy-

ay

8Q

=-

ax

The integral is independent of the path. Hence we can use any path. For example, if we use the path from points (0, 0) to (2, 0) and then from (2, 0) to (2, 1); we can evaluate the integral. y= 0, dy = 0

(i) From (0, 0) to (2, 0) ;

.

:. The tntegral =

2

J

4

lOx dx - 2x

512

= 64

0

o

x= 2, dx = 0

(ii) From (2, 0) to (2, I);

2 2 2 5 4

:. The integral = JIOx dx = 2x J= 64 o

0

2

J

12y 2 dy = -

4/1

=-4

0

:. The value of the integral = 64 - 4 = 60 Aliter: Since

ar = aQ, we know that {I Ox4 ay ax

differential of (2x 5 -

2x1)dx - 3xl.0 dy} is an exact

xlI). (2.1)

The given integral =

(2,1)

2

J d(2x 5 - x /)= 2x (0,0)

5

-

2

2

3

x y3 J= 2.25 - 2.1 = 60 (0,0)

586


Ex.7.10.10

X VerifY Green's theorem for f(e- siny)dx + (e-X cosy) dy where C is the c

boundary of the rectangle whose vertices are (0, 0) (n, 0) (n, ~ ) and (0, ~ ) 2 2 traversed in the +ve direction. Sol: y

I

O(~O~,O~)----~'~----~A~(n~,O~)-x

By Green's theorm, fPdx+ Qdy = c

Here, P = e-x siny,

aQ

ap

In)l ax

8Q - 8P)dXdY

Q

=

ay

e-x cosy

= -

e-x cosy - e-x cosy

fPdx+Qdy=

If -2e x cosydxdy

c

R

-

8x

- -

ay

x

=-2

x

y=O

=-2 fe-xdx=2e

x=()

xl: =2(e-

c

Again,fPdx+Qdy=f f+ f+ f+ f c OA

..

y = 0,

AB BD DO

dy = 0

fPdx+Qdy=O OA

Alo~gAB:

x

=

n,

x/2

f [ f e-xcosdyJdx=-2 f eX siny fdx

x

Along OA ;

-2e-x cosy

x/2

x=o

c

=

dx= 0

X

-1)

0

.....(i)

587


f Pdx+ Qdy

1tI2

Je

=

AB

11

cosydy = e- 1I sin YI~/2 = e

11

0

AlongBD; o

t

0

p dx+Qdy= fe-xdx=~ =-1 + e

J

BD

JP

°

11

x = 0, dx =

Along DO ;

o

dx + Qdy = feos ydy =

DO

11

-- 1 11

sinYI~/2 = - 1

11/2

..... (ii) c

:. From (i) and (ii) it is proved that

J

H( 8Q - 8P) dxdy 8x

p dx + Qdy =

c

8y

R

Hence the theorem is verified.

Ex. 7.10.11

Apply Green's theorem to obtain the area bounded by the curve x2/3

Sol:

+ y2/ 3 = a 2/3 , a>

°.

The parametric equations of the curve are x = a cos3e, y = sin 3e. A rough sketch of the curve is given below:

A'

x~'--~~~~~~-----x (6

=rc)

A(6

=0)

,

y

x

= acos3e,

y

= asin 3e

dx = -3 acos 2e sine de, dy = 3a sin 2ecose de

By Green's theorem, the area bounded by a simple closed curve C is given by

!

J(xdy- ydx) 2c


588

:. The area bounded by the given curve C 21<

=

!

J{acos 30.3asin 2 0cosO+asin 30.3acos 2 0sinO}dO

2 {)=o

ar 2 21t

3

3

3 3 af·} (lr n:a sm-20dO=-lJ-cos40)dO=--

2·2

2 br

2 2x

2

= - JCos Osm OdO=-

2

8

0

16 ()

0

8

Verify Green's theorem for J(2xy-x 2)dx+(x+y2)dy, where C is the c closed curve in xy-plane bounded by the curves y = x 2 and = x.

Ex.7.10.12

y

Sol:

By Green's theorem,

JPdx+Qdy= Jfc

A(1,1)

- ap)dxdy ax ay

R

C

y

aQ

2 aQ ap Here, P = 2xy - x , Q = x +,1 - - -ry Y, ax

=

1 - 2x o-=:;;~-----x

aQ of Jjr---)dxdy= J[ f (1-2x)dyJdx= JY-2xyl~:;; dx ax ry y=~

I

x=() y=x2

()

I

R

I

3

5

3

4 I

= jr-v'x -2x-v'x _x 2 +2X3 )dx = '!:x 2 -2.'l:x 2 -~+2.~ o 3 5 3 40 1

4

1

1

1

2

5

3

2

30

=-----+-=-

..... (i)

Again fPdx+Qdy= fCPdx+Qdy)+ J(Pdx+Qdy) ~

r

~

Along C I' Y = x2, dy = 2xdx, "x" varies from 0 to 1 1

I

:. fpdx+Qdy= f C2x 3 -x2)dX+CX+X4)2xdx= fC2x 3 +X2 +2X5)dx (')

x=o

0


AlongC 2,

589

'y' varies from I to 0

x=y, dx=2ydy, o

0

:. JPdx+Qdy= J(2/_y4)2ydy+(y2+ y 2)dy= J(4 y 4_2 y 5+2 y 2)dy (',

y=1

I

..... (ii) From (i) and (ii), the theorem is verified.

Exercise -7(j) 1.

Evaluate 1(x 2 + y2)dx+3xy 2dy, (a) directly (b) by Green's theorem, where cis the circle x 2 +

y = 4, traversed in the +ve direction.

(Ans: 121t) 2.

Evaluate 1(x 2 + 2xy)dx + (x 2 y + 3)dy around the boundary C of the region given by y = 8x and x = 2, (a) directly and (b) by Green's theorem. (Ans: 128)

3.

VerifY Green's theorem for the integral 1(3x 2 + 2y)dx - (x + 3cos y)dy where Cis the boundary of the parallelogram with vertices at (0,0), (2, 0), (3, I) and (1, I) (Ans: -6) (n,2)

4.

Evaluate

J(6xy - /)dx+ (3x

2

-

2xy)dy along the cycloid x

(0,0)

y

= l-cos8. VerifY the result by using Green's theorem.

(Ans : 61t 2 - 41t)

= 8 -sin8,

590

5.


Using Green's theorem find the area bounded by one arch of the cycloid

x=a(B-sinB), y=a(l-cosB),a>O, and x-axis. [Ans: 3J[a 2 ] 6.

Evaluate f{2x 2 - y2 )dx + (x 2 +

i)

by Green' theorem where C is the boundary of

c

the surface in the xy plane enclosed by x axis and the semi-circle y

= ..Jl- x 2

[Ans: 4/3] 7.

Evaluate f(cos x sin y - xy)dx + sin x cos y,using Green's theorem where c is the c

[Ans: 0] 8.

VerifY Green's Theorem in the plane for f{x 2_xy 3 )dx+ &2 -2xy)where C is the c

square with vertices at ( 0,0) , ( 2,0) and (0 ,2) [Ans: 8] (2,1)

9.

f (12 x3- 2x/ ) dx - 3x2y2 dy along the path x3- y3 + Y - 4xy

Evaluate

(0,0)

[ Hint: Proceed as in aliter of 7.1 0.9] 4

2 31(2,1) (0,0)

[Ans: 3x - x y

= 44 ]

= 0

591


7.11

Gauss Divergence Theorem

7.11.1 Gauss' Divergence Theorem Let (I) V be the volume bounded by a closed surface S (2) A be a vector function of position with continuous derivatives. Then,

fffV .A .dv ffA .nds 1A .ds =

=

where n is positive unit normal (outside drawn normal) to S.

Proof: z

t------y ..... R

,

x S is a closed surface Let any line II lel to the coordinate axes cut S in at most two points. Let z = gl(x, y) and z = gix, y) be the equations of the upper portion SI and lower portion S2 respectively. Let R be the projection of S on the xy-plane.

If A = Al i + A::J + A3 k


592

Then,

oA OA fff OZ 3 dv = fff OZ 3 dzdydx v

v

oA 0/ dz] dydt=

Iq(x,y)

ff[ S

z~g2( x,y)

II

ff

AJX,y,Z)1

z~g,

dydx

z~g2

II

f f[A 3 (x,y,g,)- A 3 (x,y,g2)]dydx

=

R

For the upper portion SI' dy dx = (cosyl)dS I = k. n l dS I since the normal n l to SI makes an acute angle YI with k. For the lower portion S2' dydx

= (-cOSY2) dS 2 = --k.n 2 dS 2,

since the normal n 2 to S2 makes an obtuse angle Y2 with k.

:. f f A3(X,y,g, )£lydx =:f f A 3k .n, dS, and R

"

So that, f f[A 3 (x,y,g,) -A3 (x, y, g2) ]dydx

= f f A 3k .n,dS, + f fA3k.n2dS2 8,

II

f

= fA3k.ndS s Similarly, by projecting S on the other coordinate planes, we get,

f f fO~, dv= f fAl nds v

S2

..... (I)

..... (2)

•

and f f f v

oA2

cry

dv = f f A 2j.n ds

..... (3)

s

(I) + (2) + (3) =>

fff(o~, + O~2 + o~3)dV=

ff(A ,i+A 2j+A3k).ndS s

v

i.e. f f fV.A dv = f fA.n dS which proves the theorem. S

593


7.11.2 (a) Express Gauss' Theorem in words and (b) obtain its Cartesian form. (a)

Gauss' Theorem states that "The surface integral of the normal component of vector A taken over a closed surface is equal to the integral of the divergence of A taken over the volume enclosed by the surface. Let A=A 1i+A 2 j+A 3k,

(b)

.

Then dlv A = V.A =

aA aA? aA 3 + -"'- + --ax ay az

~-I

Let the unit normal n to S make angles

a,P,r

with the +ve coordinate axes so

that, cosa=n. i, cosp =n. j, cosr=n.kand cosa, cosp, cosr are the direction cosines of n

n = ( cos a )i +( cos P)+( cos r

)

[ n. i =

"I = cos a

etc.]

:. A.II =Alcosa+A 2 cosP+A 3 cosr Hence the divergence theorem can be written in Cartesian form as,

JJr{ aA + aA + aA )dXdYdZ = v

Jl ax

ay

az

=

J J(A I cosa + A2 cos p + A3 cosr)dr;

s

JIA,dydz + A 2dzdx + A dzdy 3

s

Solved Examples Ex. 7.11.3:

Evaluate

using

the

divergence

theorem = J J F .n

d~

where

s

F = 2xyi + yz2 j + xzk and S is the surface of the parallelepiped bounded by x = 0, y = 0, z = 0, x = 2, Y = 1 and z =3 . Sol: By the divergence theorem; HF.n ds = JHV.Fdv s Here, V.F

=~(2Xy)+~(yz2)+~(XZ) ax ay ay =(2Y+Z2 +x)

I IF.n ds = HI(2Y+Z2 +x)dv s


594

2

I

3

2

I

3

I I Ie 2 y + Z2 + x)dxdydz = I I [ I(2y + Z2 + x)dz] dxdy X;O y;O z;O 2 I

X;O y;O =;0 3

II2Yz+~+xz

=

3

00

2

3

2 I

dxdy== Ifc 6 y+9+3x)dxdy 0

00

I

2

I[ I (6y+9+3x)try]d\" = I 3i

=

X;O y;O

3 22 +9y+3xyl dx== I(12+3x)d\"=12x+ ; I

2

0

0

0

0

=24+6= 30

Ex. 7.11.4 Verify the divergence theorem for F = (4xy)i - (j2)j + (xz )k, over the cub( bounded by x = 0, x = I, Y = 0, Y = 1, z = 0 and z = 1. Sol:

By the divergence theorem II F.n dS == I I IV.F dv \

V

Here F

= (4xy)i - (j2)j + (xz)k

a a 2 a V.F==-(4xy)--(y )+-(xz) =4y-2y+x=x+2y ax ay az I

I

I

I

I

I

I I IV.Fdv == I I Ie x + 2y)dzdydx == I I [ Ie x + 2y)dz ]a:vdx x;Oy;Oz;O

v

x;Oy;O =;0 I

== I

I

fi xz + 2 y z1 dydx

x;Oy;O I

==

I

If I(x+2y)dy]dx=

X;O y;O

To evaluate

I

Ixy+y x;O

I

2 1

0

I

dx=

2

3

IeX+I)d~==~ +xl~==2 x;o

IIF.ndS

z

The surface S contains 6 faces (see figure) S) - Face EPFA, S2 - Face OBDC

S3 - Face PFBD, S4 - Face OCEA

J---+--+---y

Ss - Face PDCE, S6 - Face OBFA x

595


The surface integral

IIF.ndS is equal to the sum of the surface integrals on the

above 6 faces.

x = I, F.n = 4y; dS = dy dz

on S" n = i, J

J

J

J

J

:. JJF.ndS= J J 4ydydz= J[ J4ydz]dy= J y;O :;0

on S2' n = -i, x = 0, F.n = :.

°

F-O

:~O

J

4yzl~dy= J 4ydY=2/1~ =2 0

y~O

IIF.n~S=O

on S3' n = i, y = I, F.n = -I, dS = dx dz. I

I

I

J

I[ I-Idz]dx= I-z dx= I-dx = --xl~ =-1

:. I IF.ndS=

on S4' n = -i, y = 0, F.n =

:. I IF.ndS=o

°

84

on S5'

11

= k, z = I, F.n = x ; dS = dx ely J

J

J

I

J

J

1

2 1

:. I I F.lldS = I I xdxdy = I [ Ix dy 1 dx = I xy dx = I xdx = ~ = 85 X~O y=o X;O y;O x;(} 0 x=O 2 0 on S6' n = k, z = 0, F.n =

:. I IF.ndS =

°

Hence :. II F.ndS

~

°

= 2 - 1 + ~ =%

8

The divergence theorem is verified

Ex.S.11.5 :

If r = xi + Yi + zk, evaluate :.

IIr.n dS where S is any closed surface. s

Sol:

By the divergence theorem

:. I Ir.n dS = I I Iv." dv ,where V is the volume enclosing S. s v


596

V.r

000

= -(x) + -(y) + -(z) = 3

ox

0'

oz

:. I Ir.n ds = I I I3 dv = 3v

Ex.7.11.6 :

If '\jI' is a scalar function, prove that I I IV\jIdv = I I'P n ds

Sol: By divergence theorem I I IV.Adv = I IA.n ds

..... (I)

Take A = \jIc, C = cli + c-j + c3k is any constant vector.

:.(l)=> JIIV.(lPc)dv= fI('Pc).nds

..... (2)

s

v

But,

= (V \jf).C

o )+-(c~)+-(c 0 0 l [ ":V.c=-(c ox oy- OZ3 )=0 ,

= c. V \jf and

(\jIc).n = c.(\jJl1)

cl' c2 ' c3' being constant] [ ": a.b = b.a]

so that (2) becomes,

IJI(c.V'P) dv = JI{c.('Pn)}ds

=> c. JI IV'Pdv = c. JI'-IJn ds => JI IVlPdv = JI'Pn ds v

(.: c is arbitrary)

Ex.7.11.7 : Ifs is a closed surface enclosing a volume V and ifF = (Ix)i + (my)j + (llz)k,

I, m, n being constants, evaluate JIF.nds


Sol:

597

By Gauss' theorem HF.nds= Hfv .Fdv

a ax

a

a az

V.F = -(Ix) + -(my) + -(nz) = I + m + n :. HF.nds

ry

=

Hfu + m + 11) dv

=(/+m+n)V

Ex.7.11.8: IfE=curlA,evaluate HE.nds wheresisanyclosedsurface Sol:

By Gauss' theorem,

HE.n ds = Hf(div E) dv where V is the volume enclosed by s .\'

V

But Div E = div (curl A) = 0

[ See 7.5.2 (2) ]

:. HE.nds=O

Ex. 7 .11.9 : If s any closed surface and n is unit +ve normal to s, show that Hn ds = 0 Sol:

Consider a constant vector a

=

ali + ~ + a3k

( i.e.

a2, a3 are constants)

al'

Then, a·[Hndv] = H(a.n)ds =

Hf(V.a)dv

(by divergence theorem)

aa aa ax ay az

a.

[ ':V.a=-) +_2 +_3 =0]

=0

Thus a. a.[ Hn dv] = 0 is true where a is any arbitrary vector.

:. Hnds=O

Ex.7.11.10 : Evaluate H(F.n)ds, where s is the region bownded byy2 = lx, x = 2, s Z = 3,

and F = 2xi + 3yj +

1

"3 z3k, using Gauss' theorelll

Z

= 0,


598

Sol: By the divergence theorem,

ff(F .n) ds

= fff(V.F)dv

s

V.F

=~(2X)+~(3y)~~(Z3) = 5+z ax ay ay 3

From (I), ff(l1~.Il)ds s

2

=

.... (1)

v

2.[i-;

f f

2

= fff(5+z 2 )dv,

3 3

5z + ~

x=O Y= 2 Ex

dydx 0

~--------r-----X

2

=

) 2

f 96J2xdx = 96J2 ~

= 192J2 0

x=o

Ex. 7.11.11

ff( 2xi - 31 j + Z2 k ).n ds over the surface bounded by

Evaluate

s 2

x + y2

= 1, z = 0,

Z

= 2, using Gauss' theorem.

Sol: By the divergence theorem,

ff(F .n) ds

= fff(V.F)dv ,

s

Here,F= 2xi-3y2j+Z2k DivF

=~(2x)-~(31)+~(z2) ax ay ay = 2-6y+2z +1

ff(F.n)ds=

M

f f

2

f(2-6y+2z)dzdydx

x=ly=~O


=

599

16[~~I-X2 +~sin-1 XJ1 2

Ex.7.11.12:

2

=8n

.1

Evaluate using the divergence theorem H(F.Il)dS where S is the surface

of the sphere x 2 + I + z2

Sol:

=

h2 in the first octant and F = yi + zj + xk

By divergence theorem,

f

:. H F.1l d~' = HV'·F dv

..... (i)

v

F = yi + zj + xk

V'.F= 0 :. H(V'.F) dv = 0

..... (ii) z

~--+B--Y

x Let us evaluate the surface integrals over the faces OAB, OBC and OCA. b

Jj;2~)

f fF.n ds = - f fx dx(ry OA B

x=O

y=O

(,,' n = -k)


600

=

If F.ndv

Similarly

DAB

If F.nds + If F.nds + If F.nds + If F.nds OAB

ABC

ABC

DCA

3

= -Jrb + If F.nds

.... (iii)

AIlC

From (i), (ii), and (iii), we get 3

0= -Jrb + If F.nds ABC

If F.nds

Jrb 3

=

ABC

VerifY divergence theorem for F=4xi-2y2j+z2k taken over the region bounded by

Ex. 7.11.13: 2

x + / = 4,

Z

= 0 and z = 3.

Sol: By the divergence theorem, we have

Iff DivFdv (1) DivF

=

IfF.~ ds

=~(4x)-~(2/)+~(Z2) =4-4y+2z Ox

L.R.S. 0[(1)

~

.... ( I )

s

,Ll"

By

~ ,t~~l,JL (4-4y+2z)dz ]dydt

[4z-4yz+z'JC

dydx~

)11

(21-12y)dy

f ~4_X2dx=84Jr

x~-2

[Do the integration w.r.t.x yourself, taking x (2)

.,.

8z

2

=42

..

Evaluate of surface integral IfF.~ ds s

= 2sin ()]

Jdx

601

Vector Differentiation Z

Z=3

y

x

The given surface of the cylinder can be divided into 3parts, namely (a)

8 1 : the circular surface z = 0

(b)

S2: the surface z == 3 (circular) and I

(c)

S3: the cylindrical portion of 8: Xl + i

= 4,

z == 0, z

=3

we now find JJF.~ ds over SI ,82 , S3 .If we add them, we get R.H.S of (I). (a)

on SI:Z=.O;

~=-k; F.~=-(4xi-2ij).k=0;:. JJF.~ds'=O. s,

(b)

- - ( ) dxdy on S2:z==3; n=k ; F.n= 4xi-2y2j+9k .k=9; d\'=-I':' -I =dxdy

n.k

:. JJF.1i (is' = JJ9dxdy = 9A , where A is the area of the circle s, x 2 + y2

(c)

s,

= 4, = 9J'l' (22) = 36J'l'

on S3: Let fjJ =x2+i-4=0;

-

V fjJ

2 ( xi + yi)

!VfjJ!

2~X2 + i

n =- - =

F.n=

=

xi + yi ( .

2

SInce x + y

2

= 4)

2

4 x2 -2y 3 22 3 =x-y; 2

To evaluate

JJF.1i ds, take x = 2cosB,y = 2sinB, S3


602

= 2dO dz

and ds

; limits of z are 0 to 3 and those of 0 are 0 to 27r . 2lf

Hence

fJF.~ ds S3

=

3

f f (8cos 0-8sin 0)2d 0 dz 3

2

(}=02=0

2"

=

16

f [( cos 0-sin O)z ]

3

2"

3

2

(}=o

2

3

(}=O

2=0

[ Do the integration w.r.t.

f (cos 0-sin O)dB =487r

dO = 48

e yourself]

:. R.H.Sof(I)= 0+367r+487r=847r;:. L.H.S= R.H.S Hence the theorem is verified.

Exercise - 7K 1.

Verify Gauss's divergence theorem for A

= (x 2 -

yz)i +

(i - zx) j + (Z2 - xy)k

taken over the rectangular parallalopiped 0 ~ x ~ 2, 0 ~ y ~ 3, 0 ~ z ~ 1.[Ans:36] 2.

Use the divergence theorem to find ff F.ndS, where --

f

s

= (3x + 2Z2)i - (Z2 - 2 y) j

+ (/ - 2z)k and S is the surface of the sphere with

centre at (2, -I, 3) and radius 2 units. [Ans:32 TC]

3.

Verify

the

divergence

theorem

for

the

unit A = (4xz)i - (i)j + (yz)k , taken over the x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1.[Ans:3/2] 4.

If r

= xi + yj + zk , and S

vector cube

function, bounded

by

is the surface of the rectangular parallelepiped bounded

by planes x = O,y = O,z = O,x = a,y = band z = c, find the value of ffr.ndS s using Gauss's theorem. Verity your answer by direct evaluation of the integral. [Ans:3abc] 5.

Use the divergence theorem to evaluate ffA.nds for A where s is the sphere given by (x _1)2

6.

If

V=(lx)i+(my)j+(nz)k

ff V.ds s

= 327r (l + m + n). where 3

(X-3)2 +(y-2i +(z-I)2

= 4.

+ i + Z2 = 1 I, m, n

and

S

= (2x)i-(2y)j + (3z)k

IS

being the

[Ans:4 7r] constants surface

show the

that

sphere

603


7.12

Stoke's Theorem

7.12.1 Stoke's Theorem Let (1) S be an open, two-sided surface bounded by a simple closed curve C. (2) A be a vector function having continuous derivatives Then, fA.dr

=

H(Y' x A).nds = H(Y' x A).d\'

where C travels in the +ve direction and n is the unit +ve (outward drawn) normal to S.

Proof:

z

Let S be the surface. Let the projections of S on the coordinate planes be regions bounded by simple closed curves. Let 'R' the projection ofS on xy plane be bounded by C l . (see the figure above). Let the equation of S be z differentiable function.

1(x, y) where <1>1 is a single valued, continuous and

=

j

Then

Y'x(Ali)=alax alay AI

0


604

{V x (A]i}}.n ds =(8A] _ 8A] ).nds = {8A] (n.}) _ 8A3 (n.k)}ds

8z

8y

8z

8y

.....

(.) 1

The position vector r of any point on S can be taken as r

=

xi + y) + zk

= xi + y} + t/>/x, y)k 8r _. 0+] and - - } + - k ay 8y 8r

But

ay

8r

being the vector tangent to S, it is .i, to n.

.. ay' n = 0

~

n.} =

3] -

8y (n.k)

. 8AI 84>1 8AI :. (I) => { V x A\,)} . n ds = (---n.k--n.k)ds 8z ay 8y ..... (ii)

on S,

A\(x y, z)

= A\(x, y, 4>\(Y» = G(x, y) (say)

8AI 8AI 8z aa -+-.-=ay 8z 8y 8y

:. (ii) becomes, { V

x

(A\i) ] .n ds

H(VxA\i)}.nds= ,I

aa

aa

= - ay (n.k) ds = - 8y dxdy

r·:

n.k. ds

=

dxdy]

Jf- aaay dxdy=fCdx-, by Green's theorem in the plane. Now (' R

at each point (x, y) ofC, the value ofG is the same as the value of AI at each point (x, y, z) of C, and since dx is same for both curves, we have

I

605


H{V x(Ali}}·n ds = fAldx

Hence,

s

..... (iii)

c

lilly by projecting S on yz and zx planes, it can be shown that,

H{V x (A2i)}·n ds = fA 2dY s

c

H{V x (A3 k )}.n ds = f A 3dz c

s

Adding (iii), (iv) and (v),

HVx A.n ds = s

..... (iv)

..... (v)

fc A. dr

(since A. dr = AI dx + A 2dy + A 3dz) Hence the theorem is proved. Ex.7.12.1 :(a) Express Stoke's theorem in words and (b) obtain its cartesian form. Sol :(a)

The I ine integral of the tangential component of a vector A taken around a simple closed curve C is equal to the surface integral of the normal component of curl A taken over a surface Shaving C as its boundary.

(b) As in 7.11.2(b) A=A l i+Aj+A3k n = (cosa)i + (cos~)j + (cosy)k

Then,

Vx A = ajax

) ajay

k ajaz

A2

A3

Al

=(aA 3 _ ay

aA2)i + (aA I az az

_

aA3)} + (aA 2 _ aA I )k ax ax ay

aA 3 aA 2 aA aA 3 aA 2 aA I (V x A)Jl =( - -)cosa + ( -I - -)cos~ +( - -)cos y ay az az ax ax ay

Engine~ring

606

Mathematics - I

Hence the cartesian form of the Stoke's theorem can be stated as,

=fA ,dx+A 2dy+A 3dz

c

Solved Examples Ex.7.12.3 : Sol:

VerifY Stoke's Theorem for A = (x - 2y)i + yilj + y2zk, where S is the upper half of the sphere xl + y2 + z2 = 1 and C is its boundary.

The boundary of the projection ofS in the xy-plane is a circle with centre at origin and unit radius. Its parametric equations are x = cosS, y = sinS, Z = 0, 0:::: S < 2n

dx = (-sinS)d9, dy = (cosS)dS. f Adr =f(x-2y)dx+ yz 2dy+ y2 zdz c c 21t

(.: z =0)

= f[cosS - 2sinS] (- sinS )dS e~()

-_ 2~[-sin2S JI + I -cos2S]dS- cos2S - - + S -sin2SI - - -_2 n o 2 4 2 0 21t

i

Y' x A =

ajax (x-2y)

k

j

ajry ajaz = 2k yz2

y2 z

H(Y' x A).nds = H2(n.k )ds = 2 Hdxdy R

s

(n.k ds

=

dxdy and R is the projection of S on the xy plane)

Hence Stoke's theorem is verified.

607


Ex.7.12.4: Prove that a necessary and sufficient condition that fF.dr for every closed ("

curve C is that '\I x F

Proof: (a)

=0 .

The condition is necessary: Let '\IF = 0; Then by Stoke's theorem, fF.dr

= If('\I x F).n ds = 0

("

(b)

S

The condition is sufficient:

Suppose that fF.dr

=0

around every closed path c.

"* 0

at some point P. then, assuming that '\IF is continuous,

('

Assume that '\I x F

there exists a region with P as its interior point where '\IF = 0 .Let S be surface contained in this region and let the normal n to S at each point has the same direction as '\IF . Then '\IF = an, (a being a +ve constant);

Let C be the boundary of S.

Then by Stokes theorem, f F .dr = If ('\IF).n ds =a Ifn.n ds > 0 c s s Which is a contradiction to the hypothesis that f F.dr

=0 ;

:. '\IF

=0

Note: It follows that '\IF = 0 is also a necessary and sufficient condition for the line 1'2

integral f F.dr to be independent of path joining the points ~ and ~ .(see 7.6.4) I;

Ex.7.12.S

If r = xi + yj + zk , show that f r .dr = 0 c

By Stoke's theorem, Jr.dr c

= If(Curl r).n ds ('

..... (1)


608

j

i

k

But curl r= a/ax a/ay a/dz =0

x

y

z

:. (1) => Jr.dr = 0 ('

Ex.7.12.6:

If'f and 'g' are scalar functions, show that Jf( grad g). dr =- Jg(grad j).dr ('

('

Sol: By stoke's theorem, J{grad (fg)·dr

= Jficurl{grad(fg)}ln ds = 0, since curl grad (fg) = 0

('

S

..... (1)

.. J{grad (fg)}·dr =0 ('

But grad(fg)

=

f(grad g) + g(grad j)

Hence the result

..... (2) /

[from (1) and (2)]

If A is any vector function, prove by stoke's theorem that div curl A = O.

Ex.7.12.7:

Sol: Let V be any volume enclosed by a closed surface S. Then by Gauss' divergence theorem. We get, JJJV.(curl A)dv = JJ(curl A).nds

Divide the surface S into two portions SI and S2 by a closed curve C. Then

JJ(curl A ).n ds

= JJ(curl A).ndst + JJ(curi A).n dS 2 = JA.dr c

JA.dr c

..... (1)

609


= 0, by Soke's theorem, since the +ve directions along the boundaries ofS, andS 2 are opposite. :. (I)

~ fffV.(clirl A)dv=O

Since this is true for all volume elements V, we have, V . curl A = 0

~

div(curl A) = 0

Ex.7.12.8: Use Stoke's theorem and prove that curl gradJ= 0, where 'f' is a scalar function.

Sol:

IfS is a surface enclosed by a simple closed curve C, we have, by Stoke's theorem,

H{curl(grad J)}.nd.. = f(grad J).dr

.... (I)

('

Now, grad! dr

=(8f i + aJ j ax 0'

+ 8f k ).( dx i + dy j + dz k) az

= aJ dx+ 8f dy+ 8f dz=dJ ax ay az A

:. f(grad J).dr = fdf ('

A

= f(A) - f(A) = 0, where A is any point on C :. (1) ~ ff{curl(grad f)}·nds

=0

Since this equation is true for all surface elements S, we have,

Ex.7.12.9:

curl (grad j) = 0

Veri1)r stoke's theorem for A = yj - 2xyj taken round the rectangle bounded by x = ±b, y = 0, y = a y F~

x=b

x=-b

--____

__~__~y=a ______~B

~----~--~----~~--L---~~X

D

A

y=O


610

i

j

Sol: curl A = ajax

k

ajay ajaz = - 4yk

y2

-2xy

0

For the given surface S, n = k :. (curl A). n

4y

= -

Hence ff(curIA).n ds a

a

h

h

U

= ff- 4 ydxdy= f[ f -4ydx]dy= f -4xyl dx= f- 8bydy y=o

s

J.=~h

0

~h

0

a

= -4by 2/ =-4a 2b o fA.dr

=

C

J

f+ f+ f+ DA AB BF

..... (I)

FD

fA.dr = y2 dx - 2xydy y

AlongDA,

0, dy = 0, => JA.dr=O

=

( .: A.dr = 0)

DA

x= b, dx

AlongAB,

=

0

a

2 f A.dr= f-2bydy=-by2/: =-a b

:.

y=O

AB

y=a,dy=O

AlongBF, -b

:. JA.dr= BF

Ja

2

dx

=

-2a2b

b

x

Along FD,

=

-b, dx

0

=

o 0 2 :. fA.dr= f 2bydy=-bi/ =-a b FD

a

a

:. fA.dr= 0 - a 2 b - 2a2 b - a 2 b

= -

4a2b

C

From (1) and (2), fA.dr = ffcurl(A).nds

c

s

Hence the theorem is verified

..... (2)


Ex.7.12.10:

611

Use stoke's theorem to evaluate the integral fA.dr where A = 2y.i +3xY('

(2x +z)k, and C is the boundary of the triangle whose vertices are (0,0,0), (2,0,0), (2,2,0). )

Sol:

k

Curl A = ajax ajay ajaz 2 2/ 3x -2x-z = 2}

+ (6x - 4y)k

Since the z-coordinate of each vertex of the triangle is zero, the triangle lies in the xy-plane.

:. n

=

k.

:. (curl A). n = 6x - 4y consider the triangle in xy-plane. Equation of the straight line OB isy = x By Stoke's theorem, f A .dr c

=

fsf(CurlA)nds

I

2

2

o

3

= f6xy - 2/ dx = f(6x - 2X2 )dx = 4 ~ x=O

Ex. 7.12.11

A(2,O)

(0,0)

2

1= 332 .

0

Use Stoke's theorem to evaluate

ff(curlA)nds, where A

=

2yi + (x- 2zx)j

+ xyk, and S is the surface of the sphere .xl + Y. + z2 = b 2 above the xy-plane. Sol:

The boundary C of the surface S is the circle.xl + The parametric equations of C are x By Stoke's theorem, we have,

fsf(CurIA).nds =cf A.dr

=

bcose, y

=

y. + z2 = b2, Z = o.

bsine, z = 0, 0 ~ e < 27t

612


= f2ydx+(x-2zx}dy+xydz = f2ydx+xdy

(0: z=O,dz=OonC)

('

('

2"

[-: x = bcos8 => dx = - bsin8 d8

f(2bsin8)(-bsin8)d8+bcos8.bcos8.d8 o

and y = bsin8 => dy = bcos8 d8]

= h 2 ](cos 2 8 -2sin 2 8~8 = b 2 2][1 + cos 28 - (l-cos28)18 o 0 2

r

J"

b2 [ 3sin28]b2 = - r(-1+3cos28)=+- -8+ =--.2n=-nb 2 2 222 o h221t

t

Ex.7.12.12

Apply Stoke's theorem to evaluate f A.dr .where A = (x- y)i + (2y + z}j +

(y-z)k and C is the boundary ofthe triangle whose vertices are

and

Sol:

(O,O,~)

LetA=

(~, 0, 0) (O,~, 0) z

(~,O,O)

B=

(O,~,O) C= (O,O,~).

The equation of the plane ABC is (by intercept form), x y z 1/6 + 1/3 + 1/2

=

..... (I)

I =>6x+3y+2z= 1

The direction ratios of the normal to (1) are 6, 3, 2 .. . 6 3 2 :. D IrectlOn cosmes are -,-,777 Ifn is the unit normal to the plane, n =

§..i +l j

A = (x - y)i + (2y + z}j + (y - z)k i

VxA= a/ax x-y

.. (Vx A}.n =3. 7

j

k

aj8y a/az =k 2y+z y-z

7

y

7

+3. k 7


613

:. By Stoke's theorem,

fA.dr = Ifs (VF).n ds = ~7sIfds = ~7

..... (2)

(Area of triangle ABC)

('

To find the area of triangle ABC:

AB=

AC=

(H+(H ~ ~ (H+(H ~~

Direction ratios of AB are -1, Direction ratios of AC are

!,

0 .

6 3 -1 2

15' 15'

0

. . ratios . 0 f AC a r _-1 0.~Direction e..

flO' , flO

cos CAB = (

sin CAB =

Ts )(~ )+

0 +0 =

)so

.:i.-EO

:. Area of triangle ABC = 1. AB.AC sin CAB = 1.. 15 2

2

6

.JlO ._7_ = 2. 6 J50 72

JA.dr=~x-2..=_1 [from(2)]

(' Ex.7.12.13Evaluate

7 72

If(curl A).n ds

36

taken over the portion s of the surface

s

X2+/+Z2_2/x+ft=0

above

the

xy

plane

z

0,

if

A = ~::CX2 + /-z2)i and verity Stoke's theorem. Let'S' denote the portion of the surface, x 2 + y2 + Z2 - 2ft + ft = 0 above the xy-plane z= O. The surface S meets the xy-plane in the circle 'C', whose equations are

Solution:

2

x + / - 2ft = 0, Z

= O.


614

:. The parametric equations of 'C' can be taken as

x == j + j cosS,

y== jsinS,

(0~S<2n;)

z=o

Let S) denote the plane region bounded by C.lfS) is the surface consisting ofS and SI' SI is a closed surface. :. From example 7.11.8 on divergence theorem, we have

=0 i.e., f f(Curl A).2 ds + f f(Curl A).n ds = 0 s

[.: S) consists of Sand S)]

I.e., f f(Curl A).n ds - f f(curIA).kds == 0 s

'"

f f(Curl A).nds = f f(curIA).kds

..... (I)

s

a/at

Now, curl A=

=

:. (curl A). k

j

k

a/ay

a/az

i(- 2y - 2z) + j(-2z - 2x) + k(- 2x - 2y)

= -

2(x + y)

:. From (I), ff{curIA).nds=-2 f f(x+ y}is

.

"

.'I'

Polar equation ofS) is r =

~cosS.

:. changing to polar coordinates, It

ff(curlA).nds=-2 f s, 9=0

2/eos9

f(rcosS+ rsinS)rdedr r~O


615

8(,1t

= -

=

16f31t

3

4

2 x -'- f(eosO + sin e )eos 8de = --3- feos OdO

3

0

0

1t

[.: feos 3 Osin Od8 = 0] o

16f3 1t,12 4 32f3 3 I n .3 - - x 2 feos Od8=---.-x-x-=-2nj 3 0 3 4 2 2

..... (2)

fA.dr

Again,

('

f(x 2 + i - Z2 }h: + (i + Z2 - x2~y + (Z2 + x2 -- y2 )1z

=

C

f(x 2 + i }Ix + (y2 - x2)1y

=

[.: on C

C

z = 0,

dz = 0]

ftU + f cos 8)2 + f2 sin e](- fsine)de

2"

2

o 21t

+

f [f2 sin 2e- (f + f cos8)21f cose)de

o

fffiki + =

cosS)' + sin' S}(- sinS) + {'in'S""

(I + cosS' )}coril]nIl

f3[2I_ 2sin ede +2I_ 2sin 2ede.+ 2I_ 2cosed8 + 2I_ cos o

0

0

21t

21t

o

0

2

0

1

- f(I+Cos2e)de+ fSin ecos8de =

f'(-2n)=-2nf3

(all other integrals vanish)

from (2) and (3), we have

f f(CurIA).nd\· = fA.dr s

('

which verifies Stoke's theorem.

3 eaU

(on simplification) ..... (3)


616

Exercise-7(1) 1.

f F=(xe )i+(3/)j-(z)dz,and X

2

C is the X +y2=9,z=2, evaluate fF.dr ('

using Stoke's theorem. [Ans: OJ

2.

Apply Stoke's theorem to obtain the value of the integral

fV.dr ,where ('

v = (3/)i + (2X2)j -(x+ 2z)k

and C is the boundary of the triangle whose veltices are (0,0,0),(1,0,0) and (I, I ,0) [Ans: I]

3.

Verify Stoke's theorem for F = (2x - y)i - (yz 2)j - (/ z)k if S is the upper halt surface of the sphere x 2 + y2

+ Z1

= I and C is the boundary.

[Ans: Jr]

4.

+ 2)i + (yz + 4)j - (xz)k and S represents the surface of the cube x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the xy plane , verify that

If F = (y - z

f

ff(Curl F).ds = F.dr ,C being the boundary of S traversed in the +ve direction. s c [Ans: Each integral =-4]

5.

Find the value of the integral f(yz )dx + (zx )dy + (xy )dz ,using Stoke's theorem c where C is the Curve x 2 + y2

= 4, z = y2 [Ans:O]

6.

Verify Stoke's theorem for the function V = (3x )i + (2xy)j , integrated along the square x = 0, y = 0, x = I,y =) in thexy-plane.. [Ans: 0]

7.

Evaluate

2

by

Stoke's

theorem

= (2sinz)i-(3cosx)j+(siny)k; o~ x ~ Jr, 0 ~ Y ~ 1, z = 0 A

[Ans:6]

the

integral

where

c where C is the boundary of the rectangle


617

Exercise - 7(m) 1.

If !(x,y,z)=x 1y lllzn-1 , find the directional derivative of direction of (i+2j+2k)

f

at (1,1,1) in the

1 3

[Ans: -(I + 2m+ 2n)] 2.

Find the acute angle between the surfaces

x 2 + i + Z2

= 6 and 3xyz + i

Z-

xy + 3 =

°

at (1, -I, 2)

IAns: cos'( ~} r = xi + yj + zk , and Div{(rx p)xq} = -2(p,q)

p, q

3.

If

are

constant

4.

If F=(x 2y)i-(iz)j+(z2x )k,findcurIFat(1,-2,3)

S.

If !=xyz(x+y+z),provethat curl gradf=O

6.

A fluid motion is given by V=(z3)i-(i)j+(3xz 2 )k. Show that it irrotational. Find its velocity potential rjJ sllch that V

vectors,

show

that

IS

= V rjJ 3

[Ans: rjJ = xz 3 _l_ + c ]

3

(2,1)

7.

Evaluate

J(4x -12x 2i)dx-(8x y)dy along the path x 3-3xi =2/ 3

3

(0,0)

[Ans: 16] 8.

2

If V=(xy)i-(yz)j+(zx )k and S is the surface of the cube bounded by

x

= 0, x = 2, y = 0, y = 2, z = 0, z = 2 , evaluate

JV.n ds

s [Ans: 32/3]

9.

If x-,

! = 4x + yz , evaluate

HJ!dv over the region in the first octant bounded by v

+ y 2 = 1, z = 0, z = 3

[Ans: 11/2]

618

10.


Express F = xi -

i

j + zk in (a) cylindrical polar coordinates (b) spherical polar

coordinates.

+(r sin OcosOcos 2 ¢

_,.2

sin 2 OcosBsin 3 ¢ - rsinBcosB)eo

+ (-rsinBsin¢cos¢-r 2 sin 2 Bsin 2 ¢cos¢)e~] II.

Iff = pz sin 2B ,find grad f in cylindrical coordinates [(zsin 20)ep + (2zcos 2B)eo + (psin 2B)ez

12.

If A

= (rcosB}er -(!sinB)eo +re~, find the

curlA in spherical coordinates

r

[Ans: (cotB)e r +2eo -(2sinB)e~] 13.

Verify Green's theorem in the plane for

f(x

3

-

y2)dx + (x 2 - 2xy)dy ,where C is

('

the boundary of the squart: bounded by O:S x:S 1, O:s y:s 1

14.

Using

Gauss

divergence

theorem,

prove

that

If F.n ds =

[Ans:1] TCr

2

P,

where

s

f = (y2 Z ); + (xz)j + (z2)k ,and

S

is

the

surface

bounded

by

x 2 + y2 = r2 , z = 0, and z = I

15.

Show that the Stoke's theorem, when restricted to the xy-p/ane, is Green's theorem in the plane. (Hint: In Stoke's theorem, take A

= Pi + Qj;

n = k; and ds = dxdy)


619

Exercise-7(n) I.

Choose the correct answer in the following questions

I.

The tangent vector at the point t=1 on the curve x = t + 1, y 2

(a) 2i-4j+3k 2.

if

(c) 18

(b) 9

(b) I

(c) 2

-, The maximum rate of change of f (a)

5.

(d) 2i+4j-3k

[b]

(d) 3

[c]

(d) 3

Cd]

1= xyz, the value of Igrad II at the point (I, 2, -I) is

(a) 0

- 4.

(c) 2i-4j-3k

The magnitude of acceleration at 0 = 0 on the curve x = 2cos30,y = 2sin 30,z = 30 is (a) 6

3.

(b) 2i+4j+3k

= 4t - 3, z = t 3 is

Jli

= x/ + yz + zx 2

at the point (I, I, I) is

(c) 3

(b) 0

(d) none

The angle between the normals to the sphere x 2 + y2 +

Z2 =

Cd]

9 at the points (1,2,

2) and (2, 1,2)is 1C

1C

(b) -

(d) -

2 6.

If a is a constant vector and r

(a) 0 7.

= xi + yj + zk , then

(b) a

(c) r

[a]

'\l(a.r) is (d) r

[b]

If (x + 3y)i + (2 - 3y)j + (x + az)k is a solinoidal vector, the value of a is (a) 0

8.

4

If r

= xi + yj + zk,

(a) 0

(b) 1

and a = (b) I

(c) 2

~ r, 3

(d) 3

[c]

(d) 2

[b]

div ii = (c) -1


620

Exercise - 7(0) I. Iff(x,yx) = X lymz"_I, find the directional derivative of fat (1,1,1) in the direction of (i+2j+2k)

[Ans:

2. Find the acute angle between the surfaces x 2 +

I

"3

(I + 2m + 2n)

y +:? = 6 and

3xyz + yz - xy + 3 = 0 at (1,-1,2)

[Ans: cos

_,(Fs] 3]

3. If r = xi + yj + zk, and p, q, are constant vectors show that Div {r x p} x q}

=-

2(p.q)

4. If F = (ry)i - (y2z)j + (:?x)k, find curl curl F at (1,-2,3) [Ans: 6i - 2j + 4k] 5. Iff= xyz(x + y+ z), prove that curl gradf= 0

6. A fluid motion is given by Y = (z3)i - (y2)j + (3xz 2)k. Show that it is irrotational. Find its velocity potential

+such that Y =

V~

[Ans: ~ = xz 3 (2,1)

7. Evaluate

3

f{4x -12x2 y2

)Ix - {8x y}Jy 3

along the path x 3 - 3xy

=

y3 - -

3

+c ]

2;

(0,0)

[Ans: 16] 8. IfY = (xy)i - (yz}j + (zr)k and S is the surface of the cube bounded by x = 0, x

=

2 , y=O " y=2 z=Oandz=2 , evaluate f V .nds s

[Ans: 32/3] 9. If f = 4x + yz, evaluate f f ffdv over the region in the first octant bounded by v x 2 + y2 = I, z = 0, z = 3

[Ans: 1112]


10. Express F = xi -

621

y.j + zk in

a) cylindrical polar coordinates b) spherical polar coordinates. [Ans: a) p(cos28 - pSin38)e p - psin8cos8(1 + psin8).ea + ze z b) (rsin 28cos2coscj> - ~sin28sin2cj>coscj»e.] 11. If/= pzsin28, find grad/in cylindrical coordinates [Ans: (zsin28)e p + (2xcos20)e a + (psin28)e z] 12. If A = (rcosO}er -(-;,sin 0

)eo + re4>, find curl A in spherical coordinates [Ans: (cot8)e r + 2eo - (sin8)e.]

13. Verify Green's theorem in the plane for J(x 3 c boundary of the square bounded by 0 ~ x

~

-l )tx + (x 2 -

2xy}ty, where C is the

I, 0 ~ Y ~ I [Ans: 1]

14. Using Gauss' divergence theorem, prove that

J

JF.nds = 1tr2[2 , where s

F = (y2 Z} + (xz)j + (Z2 ~ , and S is the surface bounded by

x2 + y2 = r2 , z =0, and

z=l 15. Show that the Stoke's theorem, when restricted to the xy plane, is Green's theorem in the plane [Hint: In Stoke's theorem, take A = Pi + Qj; n = k; and ds = dx dy]


8 Laplace Transforms 8.1

Laplace Transforms Introd uction The theory of Laplace transform is an essential part ofthe mathematical background required by engineers, physists and mathematicians. It gives an easy and effective means for solving certain types of differential and integral equations. It is the foundation of the modern form of operational calculus, which was originated in an attempt to justify certain operational methods used by an electrical engineer Oliver Heavinide, in the latter part of the 19th century for solving equations in electromagnetic theory. The Laplace transform reduces the problem of solving a differential equation to an algebraic problem. It is particularly useful for solving problems where the mechanical or electrical driving force has discontinuties, acts for a short time only or is periodic but not merely a sine or cosine function.

8.1.1 A class of linear transformations of particular importance is that of the integral transforms. The Laplace transformation (L.T) is a special case of an integral transformation and is defined h~' cO

F(s) = L [f{t)1

=

Je-stf(t) dt o

624

Def.


Piecewise continuity ofa function: A function off(t) is sectionally continuous (Piecewise continuous) in an interval o~ t ~ b iff(t) is continuous over every finite interval 0 ~ t ~ b except possibly at a finite number of points, where there are jump discontinuties and at which the function approaches different limits from left and right.

f(t)

a

Sectionally continuous function Def.

Functions of exponential Order:A function f(t) is said to be of exponential order cr, as t

3/f(t)/ < Meat, or equivalently, le--crtf(t)1 ~ M,

\;f

\;f

~

00

if 3 M > 0

t~0

.... (1)

t ~ o.

Remark: eat and sinbt are of exponential order, while e

t2

is not of exponential order since

which can be made larger than any given constant by increasing 't' indefinitely.

8.1.2

Conditions for L. T to exist By using above remarks we can prove the conditions that are sufficient for the existence of Laplace transforms.

Theorem: Iff(t) is piecewise continuous in every finite integral [0, N] for N > 0, and of exponential order cr,that is If(t)1 ~ Meat, \;f t ~ 0 and M> o. (from I) then the Laplace transform ofF(t) exists for all s> 00

00

00

o

0

0

0'

Proof: LetI = /F(s)1 = Je-stf(t)dt ~ ~ f(t)/ e-stdt ~ JM eate-stdt

=-=M

s

cr

Here, s > 0' since I is non-negative everywhere. This shows that above conditions are sufficient for the existence of L [f(t)].

625

Laplace Transforms

Properties of Laplace Transforms

8.1.3

Linearity property: Let f,(t) and fit) be two functions on [0, co] such that the Laplace transforms L[f,(t)] and Lltit)] exist. If k, and k2 are two constants, then L[k/,(t) + klit)]

=

k,L[f,(t)J + k2 L[fit)J

..... (i)

This property is valid since 00

00

00

=

f[ktf; (t) + k 2 f2(t)]e-'1 dt fk\.!; (t)e-'I elt + fk2.f; 0 0 0

(t)e -,I lit

= k\L[j; (I)] + k 2 L[f2(t)] ~

This result can be extended to the linear combinations of more than two functions.

8.1.4

Iff(t) is piecewise continuous on [0, N] for each N > 0, and ttt) is of exponential order 0", that is by 8.10.1 (1), If(t)1 ::s Meat, V t ~ 0 and M > 0 and ifL [f(t)] = F(s) then we have (1) L [eatf(t)]

,

= F(s-a) V s > 0" + a va> 0

(2) L[f (tn.

=

e-aSF(s)

where f,(t)

=

{

and (3) L (ttat»

..... (i)

f(t -a) t > a} 0 t < a (a>O)

=

±F(~)

(for any a> 0)

_a_

Proof(l)

--~------~----~~s

From definition, we have 00

L[f(t)]

= fe-Mf(t) dt = F(s) o 00

so that

L[eatf(t)]

= fe-ot[eal f(t)dt o

00

= fe-(s-a)1 f(t)dt = F(s - a) 0

626


Th is property shows sh ifting on the s-axis and is called the first shifting property. This means that replacing s by s - a in the Laplace transform corresponds to multiplying the original function by eat a

00

(2)

00

L(fl(t) = fe-Ilf; (t)dt = fe-'If; (t)dt + fe-sl.t;(t)dt o 0 a a

00

fe-.II.Odt+ fe-lIf(/-a) dt

=

o

from definition offl(t)

a

00

00

= fe-·llf(t-a)dt== fe-I(p+a)f(p)dp o

(where t = p+a)

0 00

= e -a.1 fe -P' f(p )dp == e -a.1 F(s) o

(3)

This property indicates shifting on the t-axis and is called the second shifting property d L[f(at)] = fe-'I f(at)dt == fe-·I(p/a) f(p)-'£ [taking t = pia]

=

00

00

o

o

a

~ }e-IP/af(p)dP==~F(~)

a0 a a This property is known as change of scale property.

8.1.5

Laplace Transforms of standard functions

(a)

L (e-at )

Pro()f:

we have L(e-at ) =

I

= --

(s> a)

s+a

.

rJe o

-.1/

_ .e -aidt-

rJe

-(Ha)1

d _ - 1 [ -(Ha)I]OO t--- e 0

_

1 s

Ifa=OtheL(I)= -

1 s+a

---

s+a

0

...... ( I)

(s> 0)

..... (2)

Also changing the sign of , a' we get L(eat )

(b)

Pro()f:

I

= --

..... (3)

s-a

L(sinat) =

a 2

S

+a

2

and L(cosat) =

s J

s- +a

2

1 s+ia L(cosat+isinat) = L(e lat ) = --.- ==-)-S -lU s- + a 2

(s> 0) from (a)

627

Laplace Transforms

Equating real and imaginary parts we get L(sinat) =

a s 2 ,L(cosat) = 2 2 +a s +a

2 S

(c)

Proof:

L(sinhat) =

2

S

.

a 2 and L(coshat) = -a

(e _eOI

L(smhat)=L

s 2

S

I]

from (a)

s+a

a

Thus (sinhat) =

2

S

-a

2

similarly we can show that L(coshat) = L(tn) =

I(n + 1)

----;+]

s

laD

1

-----

2 s-a

(d)

(s>

2 -(I

1 01 1 -01 =-L(e )--L(e ) 222

1[1

- -

ol

..... (4)

s s

2

2

•••.

(5)

-(I

[(n + I) > 0 and s> 0]

Proof:

OOfe-P.pnd =T(n+l) o

S

'P

n+1

S

11+1

00

Since T(n+I)= fe-xxndx

and

T(n+ I)

=

n!

ifn is a +ve integer

o

n!

:. L(tn) = ----;;-+J s From (6) we have

1 L(I)= -, s

L(t) =

..... (6)

1 -2 ,

s

Solved Examples

8.1.6

Find the Laplace transform of 1. 2t3 + cos4t + e-2t 2. e2t + 4t3 - sin2tcos3t 3. cos(wt + ~)

4. 3t2 + e- t + sin 32t 5. cos32t


628

Sol: I.

2.

6 s 1 12 s 1 + - - =-+ +-L(2t3 + cos4t + e-2t ) = 2x-+ S4 S2 + 16 S + 2 S4 S2 + 16 S + 2 L(e2t ) + 4L(t3 ) - L(sin2tcos3t)

I

6 (5

1)

S - 2 + 4x7 - S2 + 25 - n 2 + 1 [.: sin2tcos3t = 3.

1

"2 (sin5t -

sint)]

L cos(wt + 13) = L[coswt.cosj3 - sinwtsinj3]

=cosfJL(coswt)-sinfJL(sinwt)= cos fJ 4.

S

2

+w L[3t2+e-t+sin 32t] = 3L(t2 ) + L(e-t ) + L (sin 32t)

2

.

-sm

fJ

S

2

S

2 1 3 3 =3.-+--+ - -?-S3 s+1 2(S2+4) 2s-+36 2

1

3

1

3

1

=-+--+-------S3 S + 1 2 S2 + 4 2 S2 + 36 1

4" L[3sin2t -

(.: L(sin32t) = 5.

sin6t]

L(cos 2t) = L

[cos6t + 3cos2t] 4

1

(s 3S) S2 + 36 + S2 + 4

3

=

4"

Exercise 8(A) Find the Laplace transform of 1.

4t2 + sin3t + e-2t

8 3 1 Ans. - + - - + - S3 S2 +9 s+2

2.

(sin2t - cos2tf

Ans. S

3.

4. 5.

1

4

S2 + 16

Ans. !(!+_S_) 2 S S2 -64

w +w

2

629

Laplace Transforms

6.

Ans. - - + ? ? 2 s s- + 16b-

162 Ans. -----::,---------:--(S2 -81)(.~·2 -9)

7.

8.

sinhat- sinbt

An s.

8.1.7

b

a -?::-------::-)

s- -(r

S(S2 - 28) Ans. -----:----::--2 (S2 - 36)(5 - 4)

9.

10.

s)

1(1

cos 2 (2bt)

coshat - cosbt

Ans.

s 2 S

s 2

-(I

Examples on properties 8.1.4 Find the Laplace transform of

2. t 3 e-4t

1. e-btcosat 4. e-3t sin3t

5. e-2t[2cos3t-3sin3t]

Sol. S

1.

L( cosat)

=)

s- +0

2

(from 8.1.4( I) first shifting theorem)

2.

6 L(t3 ) = 4

s

By shifting theorem 8.1.4(1) 6 L(e-4t. t 3) = - - (s

3.

+ 4)4

. 3 L(sm3t) = -2-S +9 by 8.1.4(1), 2t

L( e sin3t)

=

(5 _

3 2)2 + 9.

Engil~eering

630

4.

Mathematics - I

e-3t .sin3t

[~(3Sin 1 - Sin3/)] = ~ L(sin I) - ~ L(sin 3/) 4 4 4

L(sin3t) =- L 3

1

1

3

By 8.1.4(1) -3t .

L(e

5.

-

~

sm3t) - 4·

(s+

I 3)2

+I

e-2 t(2cos3t - 3sin3t) s L(2cos3t) = 2.-2s +9

9

~9

L(3sin3t) =

8

+

By shifting theorem 8.1.4( I)

2L(e-2tcos3t) - 3L(e-2t sin3t) 2(s + 2) = (s

8.1.8 SoL

9 + 2)2 + 9 - (8 + 2)2 + 9 3s

If L[f(t)] = 'L[f(t)]

~9 . 8

+

find L[f(3t)]

38 +9

= S2

s

I By 8.1.4 (3), L[f(3t)]

=

3")

38

3(8)2 - +9 3

8

2

+27

Exercise 8(8) Find the Laplace Transform of 1.

Ans. 4cosa

2

8-3 +sin-----,:----(s - 3)2 + 16 (8 - 3)2 + 16

2. 3.

cosh2t.cos2t

Ans

~[ 8-2 + __8_+_2_] . 2 (8 - 2)2 + 4 (8 + 2)2 + 4

Laplace Transforms

4.

631

coshat sinat

Ans.

a(s2 + 2( 2) S4

+ 2a 4

S(S2 -

5.

cos3t.cosh4t

Ans. S4

6.

e- 21(2cos3t-3sin3t)

7)

+ 625 -14s 2

2s-5

Ans. S2

+4s+ 13 S3

7.

Ans.

coshat cosat

S4

8.

elsin4t.cos2t

Ans.

Ans.

Ans.

8.1.9

+ 4a 4

3 (s _1)2 +36

5s 2 -3s+2 S3

6 n+ 1 (8 + a)"+1 + (.\'2 + 1)(s2 + a)

Property (4) : Effect of multiplication by t If L[f(t)l = F(s) then L[tf(t)] = -FI(s) 00

Proof:

We know that F(s) = Je-"f(t)dt. o differentiating with respect to s 00

a

00

F(s)= J-[e-"f(t)]dt= Je-"'f(t)dt o 0

as

00

:. -FI(s)

=

Je-" [if(t)]dl = L[if(t)] o

8.1.10 Examples I.

d 1 _ 1 - -3- L(Ie -2')_ --------_o_ L( e-2')_ s+2' ds s + 2 (s + 2)2

2,

d d ( 2 2S2 L(lcosat)=(-I)-[L(cosal)]=-2 = 2 2 2 ds ds s + a (s + a )

3.

L(t sin I) = __d d..

s)

(_I_) = S2

+1

---,-2_s---,(S2 + 1)2

632

4.


a

L(atsinat-cosat)=aL(tsinat)-L(COsat)=a[-dd ( 2S S +a 2as]

s = a [ (S2 +a2f~ - (S2 +a 2)

5.

2)]- s 2+a 2 S

2 2a s s = (S2 +a 2 )2 - (S2 +l?)

L(1+tet)3 = L[I+3et+3t2e2t+t3e3t] = L(I) + 3L(tet)+ 3L(t2e 2t) + L(t3e3t) 2 3 d 1 d 1 d ( 1 ) '1 3 6 6 =~-3 ds (s-I) +3 ds 2 (s-2)ds 3 s-2 =2+ (s-1)2 + (S-2)3 + (s-3)4 1

6.

7.

d

d [

=-

41'

d [ 2 ds S2 -85+20

J

] by shifting property

45 -16

= (S2 -8s+20)2

2

8.

2

L(te4t sin2t) = -L[e sm 2/1 = - 2 ds ds (s-4) +4

2

l d d [ s -1 ] L(t2etcos4t) = ds2 L(e cos4/) = d~2 (s _1)2 + 16 by shifting theorem

=_~[

S2-2S-15]= ds (S2 - 2s + 17)2

3 2 (2s -6s -90s+94) (S2 - 2s + 17)3

Exercise 8(C) Find the Laplace transform of the following:

I

H~

Ie-"'sjn2.

Ans.

2.

3t te- sin3t

6s-18 Ans. (S2 _ 6s + 18)2

3.

Ans,

(s

(S2

3)' -

(X~~::15;),]

1.

2s+4 + 4s + 5)2

633

Laplace Transforms

4.

2tsin3t-3tcos3t

Ans.

5.

tetsin2tcost

Ans.

8.1.11

Property (5) If L[f(t)] = f(s), then L

9+ 12s -3s 2 2 2 (s +9) 35-3 2

?

(s - 2s + 10)-

U.f(t)]

=

J/(5) ds ,

00

Proof:

f(s)= fe-'I/(t)dt Integrating bothsides w.r.t 's' between the limits s, integration on right hand side (R.H.S),

'1Je-.\II(t)dS]dt = J-[e~\1 l(t)] "tL

J/(S)ds

=

0

8.1.12

dt

0

and changing the order of

= Je-\I

I~t) dr

0

Examples Find the Laplace Transforms of the following

1.

~(1-cosal)

2.

I

Sol.1.

00

00

~(e-I sinal) I

1

L(I-cosat) = L(l) - L(cosat) = - S

(

)K

s 2 S

+a

2

] [

]

[-1

-1

l I s 1 2 L -(I-cosat) = - - 2 2 ds= logs--log(s2+a) t 2 , s s +a

2.

L(e-tsinat) =

00

0

(2? 1

1 s + a=-Iog 2 2 s

a 2

(s+l) +a

2

I ISlOat) . ] a2 (S+I)]OO 1t s+1 _I(S+I) L -(e= oof 2ds= tan - =--tan --=cot -[t s (s + 1) + a a ,2 a a


634

3.

L(1-e2 t)

=

L(I) - L(e2t)

1 1 s s-2

= ---

X

21] I I)ds= [logs-log(s-2)Js100 =Iog-(s - 2) I L[-(1-e ) = ---t s s-2 s s 4.

L(e-3tsin2t) =

L(!et

3/

Sin2t)= } s

1t =-=tan 2

5.

L(eal )

2 (S+3)2 +4

---::---

_1(S+3) - - =cot _1(S+3) -2

[

2

1 s-a

I L(e ht )= - -

= --,

s-b

1 al -e bl] L [ -(e ) = t

=

22 ds=tan- I (S+3)00 (s + 3) + 4 2 s

s-a log( - )] s-b

00

s

XI s

100 - - - -1) - ds= [log(s-a)-Iog(s-bh,. s-a s-b

s-a s-b ) ( =-IOg(-)=IOgs-b s-a

Exercise 8(0) Find the Laplace transfonn ofthe following:

1 (hI]

1.

1 -(1-cost) t

Ans. 2"log - s -

2.

I _I • -e SlOt t

Ans. coc 1 (s+l)

3.

1 - (-at e - e-hI) t

Ans.log C+b) -s+a

4.

!

(e-3tsin2t)

S-3) Ans. coC I ( -2-

5.

!

(sin2t)

Ans.

t t

1 (.'+4]

"4 log

~

635

Laplace Transforms

6.

(S2+b 2)

I - (coat - cosbt)

I Ans. -log

2

t

2

S

+(1

2

(S2 +9) --2-

I - (I - cos3t) t

I Ans. -log 2

8.

!

Ans.

8.2.1

Theorem: Iff(t) is continous ¥ t ~ 0 and of exponential order, say 0" and has a derivative ~(t) which is piecewise continuous on every finite interval [0, N] for each N > 0, then the Lapalce transform of the derivative ~(t) exists for s > 0" and

7.

(sin3t)

t

S

_) 4"I [3cot

S-

cot

_)

"3s]

L{~(t)} = sLf(t) - f(O) 00

L(f(t)]

Proof:

=

je-'I f(t)dt o 00

L(~(t)]

=

je-'I fl(t)dt

o

Integration by parts gives

L(~(t)] =

00

[e-.'I f(1)1 + s je-'I f(t)dt

o Since f(t) is of an exponential order, the integrand in first integral on the R.H.S. is zero at the upper limit when s> 0' and iff(O) at the lower limit. Thus, we have 00

L( ~(t)]

= 0 - f(O) + S

je

-;t

f(t)dt =' sLf(t) - f(O)

o

If Lt f(t)

= f(O+)

exists, but is not equal to f(O), which mayor may not exist then

1-)0

L(~(t)]

Note:

= sLf(t) - f(O+).

Iff(t) and its derivatives ~(t), ~I(t) .............. fI- 1 (t) are continous ¥ t ~ 0 and are functions of exponential order for some M > 0 and 0', and the derivative fI(t) is Piecewise continous on every finite interval in the range t ~ 0, then the Laplace transform offl(t) exists when s> 0' and is given by L[fI(t)] = s"[Lf(t) - s" - If(O) - s" - 2 ~(O) ............... _tn - 1(0)] If

n = 2, L [~I(t)]

=

s2[L(f(t»] - sf(O) -~(O) and so on.

636

8.2.2


Example Prove that L(cosat) =

Sol.

S2

+ a2

We can show this with the help of above theorem Let f(t) = cosat => f(O) = 1 then [I(t) [l1(t)

= -asint => fl(O) = 0

= -a2cosat

Hence, we have L[fll(t)]

= s2Lf(t) - sf(O) - fl(O)

:. L[-a 2cosat]

= s2L(cosat) - s - 0 2 s2L(cosat) + a L(cosat) = s

:. L( cosat)

=

s2L(cosat) - s

s

=

2

2

+a Theorem: Iff is Piecewise continous on [0, N] for each N > 0 and is of exponential S

8.2.3

I I order cr, then L ff(u)du = -L[/(/)] v s > CT, o s I

Proof.

Let «p(t)

= ff(u)du. Then we have «p1(t) = f(t) and f(O) = 0; o

L[~(t)l =s.L[~(t)l-~(O) =s[L(~(t»l =SL[!f(U)du1 :. L Iff(u)du [ o 8.2.4

III

= -L[¢I(/)] = -[L(f(/»] s

s

Examples

Find the Laplace transforms of I

I. fsin 2pdp o

Sol.

I

2. fp cosh p dp o

2 1. L(sin2t) = Z--4 = f(s) , say s + 112 fsin 2pdp =- f(s) = --=-2-o S s(s +4)

r

I

3.

I

•

mp

o p

dp

rp

.

sm P

4. je . - - d'P o P

637

Laplace Transforms

2. 2

I

)

I (s + (1-) then L pcosh pdp = - I(s) =- ) ) o 8 (8- -a-)s

J

3.

. L(smt)=

I ~I ;

s +

L

(I .

-SlOt

)

t

J-2s +1

00

=

I

ds

0\

p dOp=-Ijo() I -I L 'JSin -s =-cot s o P s s

4.

I L(sint)= -2-1 s +

I ... L( etsint) = --.,,---(s _1)2 + I

ds2 I L [ -ee' sint) ] = ooJ t (s -I) +1 ,

'f

sin p 1 1 1 :. L e P --dp = - I(s) = -cot- (s -1) o P s s

Exercise 8(E) Find the Laplace Transform of I

I

I. Jcos2pdp

3 2. Jpe- sin 4 pdp

0

Ans. I.

0

s S(S2

+ 4)

2.

0

8(s + 3) S(S2

I

Jp sin 3pdp

I

3.

+ 6s + 25)2

I _I 3. -cot s s


638

Laplace Transforms of Periodic Functions

8.2.5

A function f(t) in said to be periodic with period p>O iff(t+p) = f(t) for all t>O Since periodic appear in many practical problems and in most cases are more complicated than single sine or cosine functions, we find it useful to establish the following result: I I' L([f(t)] = I-e- Ps fe-Slf(t)dt o

Proof:

Let f(t) be a periodic function of period 'p' so that f(t) = f(t+p) = f(t+2p) = ...... . I'

00

L[f(t)] = fe -si f(t)dt o

21'

= fe -s/ f(t)dt + o

fe -si f(t)dt + ............... . I'

writing t = u+p 21'

I'

now fe-ollf(t)dt= fe-o,(u+p)f(u+ p)du 0

I'

I'

= e- Ps fe-OVlf(t + p)dt

(change u to t)

o I'

f -vlfCt )dt = e -1" Je

[.: f(t+p)=f(t)]

0

o 31'

I'

similarly, fe-oIl f(t)dt

= e- 21'S fe-ollf(t)dt

and so on

0

21'

I'

L[f(t)] = (1+e-ps+e-2ps + .................... 00) fe-ollf(t) o The expression within parantheses is a G.P. with c.r. = e-PS :. L[f(t)] = 1_:-1'$ Je-Slf(t)dt o

639

Laplace Transforms

Solved Problems

8.2.6

Find the Laplace Transform of

2.

t - for 0 < t < P and f(t) = f(t+p) P f(t) = 1 for 0 < t < a and f(t) = -1, a < t < 2a and f(t) is periodic with period 2a

Sol.

Since f(t) is periodic with period p.

l.

f(t)

=k

L[f(l)] =

2.

P

P

1. Je-s1f(l)dt=

l-e- sp

o

1

l-e-"P

k

Je-"I.~dt

0

p

f(t) is peirodic with peirod 2a

L[f(t)] =

1 2a si 1 [a 2a. Je- f(t)dt = Je-sl.l.dt + Je-·'I (-I)dt 1 _ e- 2as 1- e-2as 0 0 0

_ 1 [{_e-SI}a -1_e-2a, - s - 0

-

1

SI {_e- }2aj_l 1 (l -a,)2 - s - 0 - s l_e-2as ' -e

1-

]_ 1[/f - e-'T --tanh 1 [-as ] - 1s [11+- ee -2as --. -2as s 2

--.

S

8.2.7

~

e 2 +e

-~ 2

Find the Laplace transform of the rectangular wave shown in the figure: f(t)

1s

o

b 2b

3b

Rectangular wave

640


Sol:

The period of the given function is '2b'. I

Hence L[f(t)]=

2h. -2h, Je-Mj(/)dt=

l-e

hs

-

e 2 -e

0

I

[ h .S1 2b -2h.\ J1.e- j(t)dt+ J(-I)e-S1dt

l-e

0

1

b

-bs

-

2

I bs =-tanhs 2

8.2.8

What is the Laplace transform of the staircase function (fig(a)) given by f( t) = n+ I, np < t « n+ 1)p for n = 0, 1, 2 ............ .

Sol.

The easient way of getting the Laplace transform of the staircase function is to consider it as the difference between the two functions shown in Fig(b) the transform ofthe linear function (t + p) can easily be found by using first shifting, P second shifting and change of scale properties. Thus, we have

f(t)

o

f(t)

P 2p

3p

(a) Staircse function (b) solution of the stair case function (c) Saw-tooth function.

Laplace Transforms

641

t The function fl(t) = - is called a saw-tooth function (fig(c» and its transform is p obtained as shown below.

L[ft(t)] =

1 P 1 1 [ -,I ]P _, fe-'I~d'= _.- _e_(_st_1) 1- e Ip p 1- e "P p S2 o

0

1- (1 + ps)e - P' ps2(1-e-"P) This can be simplified to

L[ I' (I)] JI

-

~[1 + ps P S2

1

p s(1-e- PS )

Thus from (1) and (2), we get

L[f(t)]=~(~+P)_~(I+PS_ p- 1 p s- s P S2 s(1-e P") 1

Exercise 8(F) 1.

If Lf(t) = (s), Prove that L[~II(t)] = s3<1>(s) - s2 (s) - sl(s) - 1I(s)

L[~I(t)] = Tan-I (+), f(O) =,2 and ~(O) = -1 find L[f(t)]

2.

If

3.

A periodic function f(t) of period 2a is defind by f(t) = t, = 2a - t,

for 0 .'S t .'S a, for a < t < 2a

7

show that L[f(t)] = 1 tanh

(as) 2

..... (2)

642


Iff(t) = t2,

4.

°

< t < 2 and f(t+2)

=

f(t)

find L[f(t)].

5.

Iff(s) = L(f(t»), Prove that

F~) and hence generalise the result.

L[Sdl SJ'(U)duj == o 0 s

8.2.9

Laplace transform of the unit step function The unit step function orthe Heaviside function is defined as H(t-a) =

°

H(t-a)

lift < a

=

ift
where a is a non-negative constant. In particular, when a = 0, we have

°

O,i/ 1 < H(t)= [l,ifl>O

The unit step function is the basic building block of certain functions whose knowledge greately increases the usefulness of the Laplace transformation. It is easy to show that the Laplace transform of H(t-a) is given by e- U \ L(H(t-a»==s a

00

00

L[H(t - a) == fe-vI H(t - a)dt = fe-VI O.dl + fe-·\/l.d' o 0 a 00

-e -sl s o

e-as s

8.2.10

. 1 In Particular, when a=O, we get L(H(t» = s Heaviside shift theorem: If Lf(t) = f(s), then L[f(t-a)H(t-a) = e-as f(s).

Proof:

By defnition, we have a

00

00

L[f(t - a)H(t - a») == fe- si f(t - a)H(t - a)dt == fe- si f(t - a)O.dt + fe- si f(t - a)l.dt o 0 0 00

00

== fe -s(p+a) f(p )dp == e -sa fe -sp (p )dp o 0

(consider p = t-a)

643

Laplace Transforms

8.2.1

Examples

Find the L.T. of the function f(t)

1.

=1 =0

} if 0 < t < 1t if1t
ift>21t

=sint SoL

Let us represent f(t) interms of unit step functions as flt)=H(t)-H(t-1t)+H(t-21t) sint -2 It' 1 -It' L[f(t)] = L[H(t)] - L[H(t-1t)] + L[H(t-21t)sint] = __ _e _ + _e_ s s s

2.

Find L.T. oft2H(t-3) Let us express (t) = t 2 as a function of(t-3) by using Talor's series, which states that

_

fI

f(x) - f(a) + (x-a) (a) +

¢(t)=94-(t - 3).6+

(t

3)2

2!

(x-a)2 II 2! f (a) + ................. + ............. .

.2

Now, by the above result L[f(t).H(t-a)] = L[9+6(t-3) + (t-3fH(t-3)] = L[f(t-3).H(t-3)] = e-3S Lf(t) L[t2H(t-a)]

=

where f(t)

=

9+6t+t2

e-3s [ -9 + -6? + 32 + ......... .] s s- s

3.

Express F(t) = (t-2)2 when t>2 and f(t) = 0 when 0
Sol.

It is easy to see that F(t) = (t-2)2 = (t-2f.H(t-2) (since for t<2, H(t-2)=0 and for ~2, H(t-2)=1) where f(t) = t2 L[f(t).H(t-2)] = L[(t-2)2H(t-2)] = e-2s .Lf(t) =e

-2,\

2! .3 s


644

Exercise 8(G) I.

Find the Laplace tranform off(t) using the unit step function, where ifO
I f(t) =

2 ifl
4

-2 if 4
2.

Determine the Laplace transform of the following function: (i) f(t)

=

6-2H(t-1t) - 4H(t-21t)

6-2e-7t'i -4e-2"" Ans.-----s

(ii) f(t)

=

e2I H(t-log5)

Ans. - 5-s

(iii) f(t) = (t-a)H(t-a)

(iv) f(t)

= t2 H(t-2)

5 2-,

e

-(1\'

Ans.

-2

Ans.

e-2"(2 4s 4s 2 ) --'-----.:3--...:...

s

+

+

S

(v) f(t)

=

H(t-e t )

1 Ans. s

(vi) f(t) = tH(t-2) (vii) f(t) = (l +2t-3t2 +4t3)H(t-2)

25 38 42 24] Ans. e-2s - + - + - + [ s S2 S3 S4

(viii) f(t) = sint.H(t-1t)

Ans. e-ltS

1 --

S2

+1

J

645

Laplace Transforms

8.3.1

Inverse Laplace Transforms If L[f(t)] = F(s) then f(t) is called the inverse Laplace transform of F(s) and is denoted by L-I[F(s)] = f(t) if fl and f2 are continuous functios on [0, 00] such that L[fl(t)]

L[f2(t)] or FI(s) = F2(s) (v- s>O) then fl(t) = fit) (v- t::::O) Some of the important properties of the inverse laplace transforms are analogous to the corresponding properties of Laplace transforms.

1.

2.

=

Linearity Property: If kl' k2 are any constants and F I(s) and Fis) are the Laplace transforms of fl(t) and fit) respectives, then L-I[kIFI(s) + k2Fis)] = klfl(t) + k/i t). First shifiing property: If L- I [F(s)] = f(t), then

jor t > a for t < a 3.

Change of scale property: if L-' [F(s)] = f(t), then

4.

Inverse Laplace transforms of derivatives: If L-'[F(s)] = f(t) then

5.

Inverse Laplace transfonn of integrals: If L-'[F(s)] = f(t) then

6.

Multiplication by's': If L-'[F(s)] L-[sF(s)]

7.

=

=

f(t) and f(O)

fI(t)

Division by's': If L-I [F(s)]

=

f(t) then

=

0, then


646

F(S»)

. result can be extended to L- I ( - This s" I I

=

I

ff..·················ff(u)du". 00

0

(the proof is left to the student)

8.3.2

Table of inverse transforms L- I [F(s)]

L[f(t)] L(1)

I

s

L(e-at)

L(tn )

L- I

= -

1

L- I

= --

s+a

n!

1-1

= S"+I

(~)

=

1

(_I ) s+a ( I) + I)! =

-

S'1+1

-

e-at

(n

t"

if 'a' is a positive integer L(sinat) =

L(cosat)

rl (

s =

L(sinhat)

L(coshat) 8.3.3

S2

a +a 2

=

=

I

2) =

~a sinat

L-I ( 2 S

2)

a _a 2

L_ I ( 21

2)=~sinhat a

s _a 2

L- I (

S2

+a 2

S2

S2

2

s +a

s +a

s -a

S2

=

cosat

s 2 ) coshat _a

Using the above results inverse Laplace transforms of some simple functions can b~ obtained as below. Find the inverse Laplace transform of:

2s+3

I.

3.

3 6.s+ 4

2s+ I 7. (s + 1)

• S2

I

5. 4s-5

3s+5 2 9s -25

2 --

+9

1_ 4._ 4s+5 s+3

8·~4 s +

647

Laplace Transforms

Sol.

From the above table, we get

I.

2. =

3.

I L- (

3~+5

9s- - 25

2cos3t + sin3t

)=L-I(

3s )+L-I( 2 5 9s 2 - 25 9s - 25

)=~COSh(~)+~sinh(~) 3

3

3

3

4.

5.

6.

L- (_3_) = 3rl_1_ = 3es+4 s+4

7.

-2L-I(_s ) + L-I(_l )_" . L-I 2s+1 1 1 2 - .... cost+smt (s- + 1) s- + 1 s +1

8.

L

I

-I( S+3) -I( -1--

s- + 4

=L

41

-I(

s ) +L s- + 4

-1--

3 ) =cos2t+-sm21 3 . s- + 4 2

-1--'

Exercise 8(H) Find the invere Laplace transform of:

2

1

S

S3

1

1. -+-+-S

+4

4s+ 15 5. 16s 2 _ 25


648

[2

Ans.

1.2+ - + e-4t 2

2. 3cos4t + sin4t

3. 2cosh3t+sinh3t

,4

4.1 + t 2 + 24

Examples 8.3.4 Find the inverse Laplace transform of: 1.

6s-4 --"-2- - -

s -4s+ 20

2. (s _ 3)3

3. (s _ 2)4

4.

s-2 + -=-2- - (s-l) s -4s+5 I

2

Sol.

4 L- I

(

.

1 ) + L-I (s - 2) (S-2)4 (s2-4s+5)

-I(

=L

-l(

1 ') ) + L (s-2t

= L-I(

S-2) (S-2)2 + 12

=

I ) + L- I ( (s - 2) ) (s-2i s2-4s+4+1

') et.t + e~t.cost

Exercise 8(1) Find the inverse Laplace transform of:

s+4 S2 -4s+ 13

(i)----

(ii)

s +6 S2 -Ss +25

(iii)

4s+ 12 S2 +Ss + 16

-=----

649

Laplace Transforms

B.3.5

Method of partial fractions: Whenever possible we exress the given function F(s) into the sum of linear or quadratic partial fractions as F(s) =

A Bs+C r + ? 2 r and then use the standard results to find the inverse L.T. (s + a) (s- + a )

Examples (I) Find Sol.

4] for all s>O, s +s

r l [ S3 +

By partial fraction expansion, we have s+ 1 A Bs+C ---=-+---

S(S2 + I)

S

S2

+1

s+1 = A(s2+1) + (Bs+c)s = As2 + A + Bs2 + cs comparing coefficients of different powers of s from both sides O=A+B )=

c

:. I = A, B =-1 .

s+1 1 -s+1 =-+-+ I) S S2 + 1

.. S(S2

By the linearity property of L-I, we have

- - = L-l(I) - - L-1(-s- ) + L-1(-I- ) L-l(S+l) S3 + s ' S S2 + 1 S2 + I

. =I -cost+smt

2S2 -6s+5

2.

If L[f(t)] =

Sol.

By partial fractions we have

S3

+ 6s2 + lIs -6 find [f(t)]

J(s) = S3

2S2 -6s+5 =_A_+_B_+_C_ -6s 2 +Ils-6 (s-l) (s-2) (s-3)

:. 2S2 -6s + 5 = A(s -2)(s -3)+ B(s -l)(s -3)+C(s -l)(s -2) put s = I we get A= 1/2

I =2A put s

=

B=-1

2

650


Similarly s = 3 gives 5/2

C

=

. ..

L-I[f'( S.)]_ I L- I - I - L- I ( -I- ) +5 L- I ( -A- )_ 1 1 -e 21 +-e 5 31 ---e

.

2

(s-I)

8-2

2

8-3

2

2

3s+ I )( 2 I ' find f(t) s-I s + )

3.

If Lf(t) = (

Sol.

By partial fractions we have

j(s)=~+ Bs+C

s -1 S2 + I :. Solving for A, B C

we have A = 2, B = -2, C = -I L- I [j(s)] = 2C l _1__ 2CI _s_ + L- I _)_1_ S -1 S2 + I s- + I =

4.

If L[f(t)]

Sol.

j(s) =

2e t - 2cost + sint

s+2 (s + 3)(s + 1)3 find f(t)

=

s+2 (s+3)(s+I)3

=~_I__ ~_I_+~ I +1 1 8s+3 8s+1 4(8+1)2 2(8+1)3

(after expressing the function in partial fractions)

:.rl[f(8)]=~rl-I--~rl-I-+ 1 8

8+38

?

8+14(8+1)-

+ I 1 2(8+1)3

=~e-31 - ~e-I -t ~/e-I +t 2e-1 = ~[(2t2 + 21 -1)e-' + e-31 ] 8

8

5.

If L[f(t)]

Sol:

Let f(s) :.S2

=

=

4

(S2

8

(S2 + 2s - 42) + 28 + 5)(8 2 + 2s + 2) find f{t)

as+b 82 + 28 + 5 +

S2

c8+d + 28 + 2

+ 28 -4 =(a8 +b)(8 2 + 28 + 2)+ (cs + d)(S2 + 2s + 5)

Equating the coefficients of like powers of s we get a = 0, b = 3, c = 0, d = -2.

Laplace Transforms

651

3 :·f(s)= s2+2s+5

-I(

:.L

f(s)]=3L

-I(

2 s2+2s+2

-I[

12 ? ) -2L (s+l) +2-

If L[f(t)]

Sol.

By change of scale property we have

-I[

L

2 (s+I)2+12

12 ? (s+l) +1-

1

2s 4s2 + 16 ' find f(t)

6.

=

3 (s+I)2+22

since L- I [

2s

] =-cos 1 21 4s +16 2 2

2 S

=

]

s +16

cos4t

e- 5• 7.

If L[f(t)]

Sol.

We know

=

(s _ 2)4 find f(t)

rl[

1

(s - 2)4

1 L-I[_I] = e 21

S4

=

e2/.~3! = ~t3e21 6

Thus by second shifting property

L- I [

e -5s (S-2)4

1={Ig

-(I - 5)3e2(t - 5)

ift>5 if 1 < 5

Exercise 8(J) Determine the inverse Laplace transform of each of the following functions. (i)

(v)

3 s+2 s-5 S2 +6s+ 13

(ii)

s s2+2s+6

(iii)

(vi)

e -2 .. s(s + 1)

(vii)

2s+ 1 s(s + 1) S2 (S2 + 9)(S2 + 16)

(iv)

10g(~) s+2


652

2S2 + s -I 0

(viii)

-(ix) log (I + Sl2 )

(s - 4)(S2 + 2s + 2)

(xii)

(xi)

s(s + I)(s + 2)(s + 3) 2

8

(xiii) 8

+ lOs + 13

-I 0(8

2

-

(xiv)

5s + 6)

8+29

(xv)

(xvi)

(S+4)(S2 +9)

(xvii)

48+4

(8

4

+ 64)

s (s4+s2+1) s

2

(8 + 1)(s2 +4)

5s+3 (s -1)(s2 + 2s + 5) S2 + 1

(xviii)

(8 _1)2(8 + 2)

(x)

s3

+3s 2 +2s

. ) -1 (-2/ e -e -3/) ( IV t

Ans.

(v) e-3tcos2t - 4e-3tsin2t

(vi) 4(1 - e-2t)(t - 2)

(viii) e4t - e-t(cos2t+2sint)

2 (ix) -(I-cost)

t

. I I _/ 1 -2/ 1 -3/ -.l--e +-e --e 6 2 2 6

(XI)

(xiii) 12et - 37e2t + 25e3t

(xviii)

.! - 22

et

+

~ 2

e-2l

(x)

(xiv)

(vii)

I

"8

(~Sin4t -%Sin3t)

sin2t sinh2t

. J32(. TJ5 t.sln'ht]2"

(XII)

"3I

Sin

(cost - cos2t)

(xv) e--4t - cos3t +

"35

sin3t

Laplace Transforms

8.4

653

Convolution Theorem If L[f(t)] = F(s) and L[g(t)] = G(s) the inverse transfonn of F(s). G(s) denoted by H(t), is called the convolution of f and g is defind as given below. I

H(t) = (f*g)(t) = f(t-u)g(u)dll,vtzO o

..... ( I)

The convolution operation * has the following properties. (f*g)(t) = (g*f)*t),

8.4.1

v t

~

0 (commutative)

[f*(g*h)](t)

=

[(f*g)*h](t)

v- t

~

0 (associative)

[1'*(g+h)](t)

=

(f*g)(t) + (f*h)(t)

v- t

~

0 (dstributive)

Theorem: Let f(t) and get) be the inverse transforms of F(s) and G(s) respectively, and satify the hypothesis of the existence theorem (8.10.2) then the the inverse transform h(t) of the product F(s) G(s) is the convolution of f(t) and get). I

h(t) = (f*g)(t) = fl(t-u)g(ll)du o I

i.e.

L-'[F(s)G(s)]

=

fl(t - u)g(u)du

..... (i)

o

Proof:

It is sufficient if we show that F(s) G(s)

Now F(s)G(s)

~

L[

1f

(l- u)g(u)du

~ [fe" f( v)dv] =

1

[V·

g(u)du ]

rrllo

Ile-(U+V)S J(v)g(u)dvdu = le-S(V+U) J(V)dV]g(U)dU' 00

where the integration is extended over the first quadrant (u ~ 0, v .:=:: 0) in the uv-plane. We now introduce a new variable 't' in the inner integral of the last expression by taking v = t-u so that u + v = t and dv = dt (u being fixed during this integration). Thus

F(s)G(s) =

rrllule->t

J(t - U)dt]g(U)dU =

Ile->t J(t -u)g(u)dtdu 0

U


654

al

a

s, I

F(s)G(s)= fTIe-SI!(t-u)g(u)du]dt= fe- jlr{t-u)g(u)du}tt 00

0

0

8.4.2 Examples (I)

Evaluate

~ r' [ s2(s1+ I )2l using the convolution theorem

Sol.

we have

L-'C, )~I

and

r'((S~I)' )~/e-'

By the convolution theorem =L-

I

[

2(

2

U

U

1 y]= J(ue- -{t-u}du= J(ut-u )e- )du s s+1 0 0

= [(ut -u2)e-u - (t-2u)e-u + (-2)(-e-U) = t-e-t + 2e- 1 + t-2 (2)

~valuate L-

Sol.

Let F(s) =

I

[

2 S 2 2] using the convolution theorem.

(s +2 )

s s ' G(s) = - 2 - and Let s + s +4

~4

_I s f(t) = L-I [F(s)] = L (-2-) = cos2t s +4

1 g(t) = L-l[G(s)] = r-I (-2-_) = ! sin2t s +4 2

f(u).;: cos2u CIld g(t-u) =

2"1 sin2(t -u)

~s)

= F(s) G(s)

655

Laplace Transforms

I I C l [tp(S)] = JCOS 2u -sin 2(1 - u) du

2

o I

~ JSin 2t + sin 2(t -

=

2u )dll

40

=~[sin2t.U+~COS2(t-2U]1 ~tsin2u 4

(3)

Evaluate L-

Sol.

Let F(s) = ¢i...s)

I

[

2

~ 1using convolution theorem.

s s+4 ~

G(s) =

I

I

vs+4 F(s).G(s)

=

04

.. f(t) = L- [F(s)] =

r

(s + 4)

1

- and choose s

~

= e-4t

CI_1_ =e-41

s~

t~ = r;; r41

e-

1

2

2

1 =1 s

g(t) = L-1[G(s)] = L- 1 (-)

e --4u rand g(t-u) = 1

f(u) =

'\I1tU

By convolution theorem I

-4u

L-1[$(s)] = Jer- du o '\I 1tU

where 4u

=

.xl

=

1 r

Je-

2!i

x2

dx

'\I 1t 0

2du =xdx

J

I

(where erf(t)

= 2r e- x '\I1t 0

(erft stands for error function)

2

dx

'\I1tl

656


Exercise 8(K) Using the convolution theorem find the inverse Laplace transforms of

(v) log«s+2» S +1

(vi)

1

?

(s + 1)(s- + 1)

(ii)

(iv)

e

I o

-/I

-e u

-211

du

1

"3 (2sin2t-sint)

(iii)

1

"3 tsint

I f4 -cost-smt . ] (vi) -Le 2

8.4.3 Applications to differential equations Suppose we wish to find the particular solution of the differential equation

lay" + /3y' + yy = J(t)

..... (1)

=

which satisfiestheinitial conditionsy(O) Yo and yeO) = Yo Taking the Lapa1ce transform on bothsides of (l) and using its linearity, it follows that

aL(y") + /3L(y') + yL(y) =

L[J(t)]

but L(y") = s2L(y) -sy(O) _yl(O)

L(y') = sL(y) - yeO) a[s2L(y) - sy(O) - yl(O)] + /3[sL(y) - yeO)] + yL(y) and hence (as 2 + /3s + y)L(y) - L[f(t)] + (as + /3)Yn + ayJ

= L[f(t)]

=> L(y) = L[/(t)] + (as + /3)yo + aYb as 2 + /3s +y The function L[f(t)] is a specific function of s since f(t) is known; and since a, /3, y. }'0 and y J are constants, L(y) is completely known as a function of s. The inverse Laplace transform of L(y) will be the solution of the given diff. eqn. (Here it is assumed that the functions f( t), y, y' and y" must have Laplace transforms).

657

Laplace Transforms

Solved Examples

8.4.4 2

' d th ' 0f d2 y + -dy - 2Y = smt ' w h'IC h satls 'files t h'" , e soiutlOn e Imtla I cond'ItlOI1S y = 0, Fm dt dt Sol.

y' = 0 when t = 0 Applying the Laplace transforms to both sides of the given equation and by linearity, we have L(yW) + L(y') - 2L(y) = L(sin t)

1 As we know that L(sint) = -2-1 ;s > 0 s + [s2L(y) -sy(O) - y' (0)] + [sL(y) - yeO)] -2 L(y) Given conditions yeO) = 0, y' (0) = 0

=> s2L(y) + sL(y) -2 L(y) = L(y)=

1 ?

(s- + S - 2)(S2 + 1)

1 -2-

s +1

=-----:---

(s -IXs + 2)(S2 + 1)

we have the partial fraction expansion

1 (s -1)(s + 2)(S2 + I)

A B C . +D s -I s + 2 S2 + I

- - - - - - - : - - = - - + - - + -":--

where

A

=.! B =-=! ,C =-=! and D = - 3 6'

15

10

10

Therefore we get •

1 1 yet) = L- 1 - - - - - - 1 (s -1)(s + 2)(S2 + 1)

as the solution to the given initial value problem,

1

= -2-1 s +


658

8.4.5 Example Solve the initial value problem y"-3y'+2y=e 31 .y=l,y'=0 wheret=O Sol.

Applying Laplace transform to both sides of the differential equation, we get [s2Ly(t) - sy(O) - y' (0)] -3 [sLy(t) - yeO)] +2 L(t)

=

L(e3!) = -

1

s-3

Applying the given conditions, we get (s2 - 3s + 2) Ly(t) = s-3 + -

1

s-3

s-J

I

:. L[y(t)] = (S2 _ 3s + 2) + (s - 3)(S2 - 3s + 2)

1

s-3

:. L(y(t)] = (s _ 3)(S2 _. 3s + 2) + (8

2-

3s + 2)

1

8-3

(8-1)(s-2)(s-3)

(s-I)(8-2)

--------+------

.... (1)

To find yet), we expand each term on the right hand side of (I) in partial functions, Thus ABC ------=----+--(s-I)(s-2)(s-3) s-1 s-2 s-3 1 = A(s-2) (s-3) + B(s-l) (s-3) + C(s-I)(s-2) I

substituting s = I gives A = 2" ;similarly substituting s = 2 gets B = -1 and substituting I s = 3 gives c ="2

hence

- - -1- - - = (s-I)(s-2)(s-3)

1

2(8-1)

Similarly we can also write s-3 D E ----=-+-(s-I)(s-2) s-1 s-2

--- +---s-2

2(8-3)

659

Laplace Transforms

s-3 = D (s-2) R(s-I) Taking s =: 2, we get E =:-1 similarly s = I we get D = 2 L[f(t)] =

Hence

I

I

I

s-2

2{s-3)

--+

2(s-l)

I

I

s-I

s-2

+---

221 - - - + -,----- 2( s - I ) s - 2 2{s - 3)

-

Example 8.4.6 Solve yll + 3yl + 2y = 2t3 + 2t + 2 with yeO) = 2, yl(O)

Sol.

Applying Laplace transform to both sides of the given equation we get L(yll) + 3L(yl) + 2L(y) = 2L(t2) + 2L(t) + 2 ? I 4 2 2 [s-Ly(t) - sy(O) - yeO)] + 3 [sL[y(t)] - yeO)] + 2Ly(t) = 3 + -2 + -

_

s

s

s

using the given conditions (1) gives 2

(

3 2

()

s + s+ )Ly t

=

2(2+s+s2) 3

2

+ s+

6

3

=

2(S4 +3s +S2 +s+2)

S

By partial fractions, yet) = yet)

3L-I(~) -2r

=3 -

L\ ) 2L-ILI

1

2t + t 2 - e-2t

3

S

+

3) -

L-Ie ~ 2)

.... (1)

660


Exercise 8(1)

Using Laplace transform method solve the following differential equations with the given conditions.

Ans.l. y

I.

yll + 2yl - 3y = 0 at, y = 0,

yl = 4 when x = 0

2.

yl = 0 when t = 0

3.

yll + 4y = 4t at y = I, yll - 2yl + Y = et at y = 2,

4.

yll + k2y = coskt at y = 0, yl =

5. 6.

yll- y = 1 at y = -I, yl = I when t = 0 yll + 2yl + 3y = 3e-t sint; yeO) = yl(O) = 0

7.

xii + 4xl = -8t given x(O) = 0, xl(O) = 0

8.

yll + 9y = 18t given yeO) = 0, yl(O) = 0

9.

yll + Y = sint given yeO) = -I, yl(O) =

= eX -

-I

1

8 +"2

t - t2 +

I

"2

when t = 0

"21

1 I I 2. y= - t- - sint + cost 3. y = 2et - tet + -2 t2et 4 8

e-3x

1. k 1 . 4. Y = 2k Sill t + "2 tSll1t 7. x =

yl = -I when t = 0

I

"8

5. y = -1 + et - cosht

e-4t 8. y = 2t -

2

"3

sin3t

6. y = 3e-tsint

Engineering Mathematics Bhaskar Rao

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