engineering mathematics 2

7/29/2019 Engineering Mathematics 2

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LAPLACE TRANSFORMS

INTRODUCTION

Laplace transform is an integral transform employed in solving physical problems.

Many physical problems when analysed assumes the form of a differential equationsubjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable arespecified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different values of theindependent variable, the conditions are called boundary conditions.

The problem with initial conditions is referred to as the Initial value problem.

The problem with boundary conditions is referred to as the Boundary value problem.

Example 1 : The problem of solving the equation xydx

dy

dx

yd=++

2

2

with conditions y(0) =

y (0) =1 is an initial value problem


dy

dx

ydcos23

2

2

=++ with y(1)=1,

y(2)=3 is called Boundary value problem.

Laplace transform is essentially employed to solve initial value problems. This technique isof great utility in applications dealing with mechanical systems and electric circuits. Besidesthe technique may also be employed to find certain integral values also. The transform isnamed after the French Mathematician P.S. de Laplace (1749 1827).

The subject is divided into the following sub topics.

LAPLACE TRANSFORMS

Definition andProperties

Transforms ofsome functions

Convolutiontheorem

Inversetransforms

Solution ofdifferentialequations


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Definition :

Let f(t) be a real-valued function defined for all t 0 and s be a parameter, real or complex.

Suppose the integral

0

)( dttfe st exists (converges). Then this integral is called the Laplace

transform of f(t) and is denoted by Lf(t).

Thus,

Lf(t) =

0

)( dttfe st (1)

We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t)

is a function of s denoted by F(s) or )(sf .

Thus,

Lf(t) =F(s) (2)

Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denotedby L-1 [F(s)].

Thus,

L-1 [F(s)] =f(t) (3)

Suppose f(t) is defined as follows :

f1(t), 0


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NOTE : In a practical situation, the variable t represents the time and s represents frequency.

Hence the Laplace transform converts the time domain into the frequency domain.

Basic properties

The following are some basic properties of Laplace transforms :

1. Linearity property : For any two functions f(t) and(t) (whose Laplace transforms exist)

and any two constants a and b, we have

L [a f(t) +b(t)] =a L f(t) +b L(t)

Proof :- By definition, we have

L[af(t)+b(t)] = [ ]dttbtafe st

+0

)()( =

+0 0

)()( dttebdttfea stst

=a L f(t) +b L(t)This is the desired property.

In particular, for a=b=1, we have

L [ f(t) + (t)] = L f(t) + L(t)

and for a =-b =1, we have

L [ f(t) - (t)] = L f(t) - L(t)

2. Change of scale property : If L f(t) =F(s), then L[f(at)] =

asF

a1 , where a is a positive

constant.


Lf(at) =

0

)( dtatfe st (1)

Let us set at =x. Then expression (1) becomes,

L f(at) =

0

)(1

dxxfea

xa

s

=a

sF

a

1

This is the desired property.

3. Shifting property :- Let abe any real constant. Then

L [eatf(t)] =F(s-a)



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L [eatf(t)] = [ ]

0

)( dttfee atst =

0

)( )( dttfe as

=F(s-a)

This is the desired property. Here we note that the Laplace transform of eatf(t) can be

written down directly by changing s to s-a in the Laplace transform of f(t).

TRANSFORMS OF SOME FUNCTIONS

1. Let a be a constant. Then

L(eat) =

=0 0

)( dtedtee tasatst

= asas

e tas

=

1

)( 0

)(

, s >a

Thus,

L(eat) =a

1

In particular, when a=0, we get

L(1) =s

1, s >0

By inversion formula, we have

atat es

Leas

L ==

11 11

2. L(cosh at) =

+

2

atat eeL = [ ]

+

02

1dteee atatst

= [ ]

+ +0

)()(

2

1dtee tastas

Let s >|a| . Then,

+

++

=

0

)()(

)()(2

1)(cosh

as

e

as

eatL

tastas

= 22 as s

Thus,

L (cosh at) =22 as

s

, s >|a|

and so


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atas

sL cosh

22

1 =

3. L (sinh at) =22

2 as

aeeL

atat

=

, s >|a|

Thus,

L (sinh at) =22 a

a

, s >|a|

and so,

a

at

asL

sinh122

1 =

4. L (sin at) =

0

sinate st dt

Here we suppose that s >0 and then integrate by using the formula

[ ] += bxbbxabae

bxdxeax

ax cossinsin22

Thus,

L (sinh at) =22 a

a

+, s >0

and so

a

at

asL

sinh122

1 =

+

5. L (cos at) = atdte st cos0

Here we suppose that s>0 and integrate by using the formula

[ ] ++= bxbbxabae

bxdxeax

ax sincoscos22

Thus,

L (cos at) =22 a

s

+, s >0

and so

ata

sL cos

22

1 =+


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6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then

L(tn) =

0

dtte nst

Let s >0 and set st =x, then

+

=

=0 0

1

1)( dxxe

ss

dx

s

xetL nx

n

n

xn

The integral dxxe nx

0

is called gamma function of (n+1) denoted by )1( + n . Thus

1

)1()(

+

+=

n

n ntL

In particular, if n is a non-negative integer then )1( + n =n!. Hence

1

!)( += n

n ntL

and so

)1(

11

1

+=

+

n

t

sL

n

nor

!n

tnas the case may be


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TABLE OF LAPLACE TRANSFORMS

f(t) F(s)

1s

1, s >0

eatas

1, s >a

coshat 22 as

s

, s >|a|

sinhat 22 as

a

, s >|a|

sinat 22 as

a

+, s >0

cosat 22 as

s

+, s >0

tn, n=0,1,2,.. 1!+ns

n, s >0

tn, n >-1 1)1(

+

+ns

n, s >0

Application of shifting property :-

The shifting property is

If L f(t) =F(s), then L [eatf(t)] =F(s-a)

Application of this property leads to the following results :

1. [ ]ass

ass

at

bs

sbtLbteL

==22

)(cosh)cosh( =22)( bas

as

Thus,

L(eatcoshbt) =22)( bas

as

and

btebas

asL at cosh

)( 221 =


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2.22)(

)sinh(bas

abteL at

=

and

btebas

L at sinh)(1 22

1 =

3.22)(

)cos(bas

asbteL at

+

=

and

btebas

asL at cos

)( 221 =

+

4.22)(

)sin(bas

bbteL at

=

and

b

bte

basL

at sin

)(

122

1 =

5.1)(

)1()(

+

+=

n

nat

as

nteL or

1)(

!+ nas

nas the case may be

Hence

)1()(

11

1

+=

+

n

te

asL

nat

nor

1)(

!+ nas

nas the case may be

Examples :-

1. Find Lf(t) given f(t) = t, 0


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2. Find Lf(t) given f(t) = sin2t, 0

Here

Lf(t) =

+

dttfedttfe stst )()(0

=

0

2sin tdte st

= { }

0

22cos22sin

4

+

ttss

e st= [ ]se

+1

4

22

This is the desired result.

3. Evaluate : (i) L(sin3t sin4t)

(ii) L(cos2 4t)

(iii) L(sin32t)

(i) Here

L(sin3t sin4t) =L [ )]7cos(cos2

1tt

= [ ])7(cos)(cos2

1tLtL , by using linearity property

=)49)(1(

24

4912

12222 ++

=

+

+ ss

s

s

s

s

s

(ii) Here

L(cos24t) =

+

+=

+64

1

2

1)8cos1(

2

12s

s

stL

(iii) We have

( ) 3sinsin34

1sin3 =

For =2t, we get

( )ttt 6sin2sin341

2sin3

=

so that

)36)(4(

48

36

6

4

6

4

1)2(sin

2222

3

++=

+

+=

sssstL



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4. Find L(cost cos2t cos3t)

Here

cos2t cos3t = ]cos5[cos2

1tt+

so that

cost cos2t cos3t = ]coscos5[cos2

1 2ttt +

= ]2cos14cos6[cos4

1ttt +++

Thus

L(cost cos2t cos3t) =

+

+++

++ 4

1

16364

1222 s

s

ss

s

s

s

5. Find L(cosh22t)

We have

2

2cosh1cosh2

+=

For =2t, we get

2

4cosh12cosh2

tt

+=

Thus,

+=16

121)2(cosh 2

2

ss

stL

-------------------------------------------------------05.04.05-------------------------

6. Evaluate (i) L( t ) (ii)

tL

1(iii) L(t-3/2)

We have L(tn) =1

)1(+

+ns

n

(i) For n=2

1, we get

L(t1/2) =2/3

)12

1(

s

+

Since )()1( nnn =+ , we have22

1

2

11

2

1 =

=

+


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Thus,2

3

2)( tL

=

(ii) For n =-2

1, we get

stL

=

=

21

21 2

1

)(

(iii) For n =-2

3, we get

stL

222

1

)(2

12

12

3

=

=

=

7. Evaluate : (i) L(t2) (ii) L(t3)

We have,

L(tn) =1

!+ns

n

(i) For n =2, we get

L(t2) =33

2!2

ss=

(ii) For n=3, we get

L(t3) = 44 6!3 ss =

8. Find L[e-3t (2cos5t 3sin5t)]

Given =

2L(e-3t cos5t) 3L(e-3t sin5t)

=( )

25)3(

15

25)3(

32

22 ++

+++

ss

s, by using shifting property

=346

922 ++

ss

s, on simplification

9. Find L[coshat sinhat]

Here

L[coshat sinat] =( )

+ at

eeL

atat

sin2


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=

++

++ 2222 )()(2

1

aas

a

aas

a

, on simplification

10. Find L(cosht sin3 2t)

Given

+

4

6sin2sin3

2

tteeL

tt

= ( )[ ])6sin()2sin(3)6sin(2sin38

1teLteLteLteL tttt +

=

++

++

+

+

+ 36)1(

6

4)1(

6

36)1(

6

4)1(

6

8

12222

ssss

=

++

++

++

+ 36)1(

1

24)1(

1

36)1(

1

4)1(

1

4

32222 ssss

11. Find )( 25

4 teL t

We have

L(tn) =1

)1(+

+ns

nPut n=-5/2. Hence

L(t-5/2) =2/32/3 3

4)2/3( =

ss

Change s to s+4.

Therefore,2/3

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)4(3

4)(

+=

steL t

Transform of tn f(t)

Here we suppose that nis a positive integer. By definition, we have

F(s) =

0

)( dttfe st

Differentiating n times on both sides w.r.t. s, we get

=0

)()( dttfes

sFds

d stn

n

n

n

Performing differentiation under the integral sign, we get

])][()[(

)2(2222

22

aasaas

asa

++++

=


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=0

)()()( dttfetsFds

d stnn

n

Multiplying on both sides by (-1)n , we get

==

0

)]([)(()()1( tftLdtetftsFds

d nstn

n

nn , by definition

Thus,

L[tnf(t)]= )()1( sFds

dn

nn

This is the transform of tn f(t).

Also,

)()1()(1 tftsFds

dL nn

n

n

=

In particular, we have

L[t f(t)] = )(sFds

d , for n=1

L[t2 f(t)]= )(2

2

sFds

d, for n=2, etc.

Also,

)()(1 ttfsFds

dL =

and

)()( 22

21 tftsF

ds

dL =

Transform off(t)

We have , F(s) =

0

)( dttfe st

Therefore,

=

s sdsdttfstedssF

0)()( = dtdsetf

s

st

0

)(

=

0

)( dtt

etf

s

st

=

=

0

)()(

t

tfLdt

t

tfe st


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Thus,

=

s

dssFt

tfL )(

)(

This is the transform of

t

tf )(

Also,

=s

t

tfdssFL

)()(1

Examples :

1. Find L[te-t sin4t]

We have,

16)1(

4]4sin[

2

++

=

s

teL t

So that,

L[te-t sin4t] =

++

172

14

2 ssds

d

=22 )172(

)1(8

++

+

ss

s

2. Find L(t2 sin3t)

We have

L(sin3t) =9

32 +

So that,

L(t2 sin3t) =

+ 9

322

2

sds

d=

22 )9(6

+

s

s

ds

d=

32

2

)9(

)3(18

+

s

s

3. Find

t

teL

t sin

We have

1)1(1)sin( 2 ++=

steL t

Hence

t

teL

t sin= [ ]

+=

++0

1

2)1(tan

1)1(ss

s

ds

= )1(tan2

1 + s

=cot 1 (s+1)


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4. Find

t

tL

sin. Using this, evaluateL

t

atsin

We have

L(sint) =

1

12

+

So that

Lf(t) =

t

tL

sin= [ ]

=+ S

s

ss

ds 12

tan1

= )(cottan2

11 sFss ==

Consider

L

t

atsin=a L )(

sinataLf

at

at=

=

a

sF

aa

1, in view of the change of scale property

=

a

s1cot

5. Find L

t

btat coscos

We have

L [cosat cosbt] =2222 b

s

a

s

+

+

So that

L

t

btat coscos= ds

bs

s

as

s

s

+

+ 2222

=

+

+

sbs

as22

22

log2

1

=

+

+

+

+ 22

22

22

22

loglog2

1

bs

as

bs

asLt

s

=

+++

22

22

log02

1

as

bs =

++

22

22

log2

1

as

bs


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6. Prove that

=0

3

50

3sintdtte t

We have

=0 )sin(sin ttLtdttest

= )(sintLds

d

=

+ 11

2sds

d

=22 )1(

2

+ss

Putting s =3 in this result, we get

=0

3

50

3sintdtte t

This is the result as required.


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ASSIGNMENT

I Find L f(t) in each of the following cases :

1. f(t) = et , 0 3

3. f(t) =a

t, 0 t


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21.t

e t21

22.t

ee btat

23.t

t2

sin

24.t

tt 5sin2sin2

I II. Evaluate the following integrals using Laplace transforms :

25.

0

2 5sin tdtte t

26.

0

35 costdtte t

27.

0

dtt

ee btat

28.

0

sindt

t

te t


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-------------------------------------------- 28.04.05 --------------------------Transforms of the derivatives of f(t)

Consider

L )(tf =

0

)( dttfe st

= [ ]

0

0 )()()( dttfestfestst , by using integration by parts

= )()0()(( tsLfftfeLt stt

+

=0 - f (0) +s Lf(t)Thus

L )(tf =s Lf(t) f(0)Similarly,

L )(tf =s2 L f(t) s f(0) - )0(f

In general, we have)0(.......)0()0()()( 121 = nnnnn ffsfstLfstLf

Transform oft

0

f(t)dt

Let (t) = t

dttf0

)( . Then (0) =0 and (t) =f(t)

Now,

L (t) =

0

)( dtte st

=

00)()( dt

s

et

s

et

stst

=

+0

)(1

)00( dtetfs

st

Thus,

=t

tLfs

dttfL0

)(1

)(

Also,

=

t

dttftLfsL0

1

)()(1


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Examples :

1. By using the Laplace transform of sinat, find the Laplace transform of cosat.

Let

f(t) =sin at, then Lf(t) =22 as

a+

We note thatatatf cos)( =

Taking Laplace transforms, we get)(cos)cos()( ataLataLtfL ==

or L(cosat) = [ ])0()(1

)(1

ftsLfa

tfLa

=

=

+

01

22 as

sa

a

Thus

L(cosat) =22 a

s

+


2. Given2/3

12

s

tL =

, show that

stL

11=

Let f(t) =

t2 , given L[f(t)] =

2/3

1

We note that,

tt

tf

1

2

12)( ==

Taking Laplace transforms, we get

=

tLtfL

1)(

Hence

)0()()(1

ftsLftfLt

L ==

= 01

2/3

s

s

Thus

stL 11 =


3. Find

t

dtt

btatL

0

coscos

Here


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Lf(t) =

+

+=

22

22

log2

1coscos

as

bs

t

btatL

Using the result

L =t

tLfs

dttf0

)(1

)(

We get,

t

dtt

btatL

0

coscos=

++

22

22

log2

1

as

bs

s

4. Find t

t tdtteL0

4sin

Here

[ ]22 )172(

)1(84sin

+++

=ss

stteL t

Thus

t

t tdtteL0

4sin =22 )172(

)1(8

+++sss

s

Transform of a periodic function

A function f(t) is said to be a periodic function of period T >0 if f(t) =f(t +nT) wheren=1,2,3,.. The graph of the periodic function repeats itself in equal intervals.

For example, sint, cost are periodic functions of period 2 since sin(t + 2n) = sin t,cos(t +2n) =cos t.

The graph of f(t) =sin t is shown below :

Note that the graph of the function between 0 and 2 is the same as that between 2 and 4and so on.


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The graph of f(t) =cos t is shown below :

Note that the graph of the function between 0 and 2 is the same as that between 2 and 4and so on.

Formula :

Let f(t) be a periodic function of period T. Then

=T

st

STdttfe

etLf

0

)(1

1)(

Proof :

By definition, we have

L f(t) =

=0 0

)()( duufedttfe sust

= +++++ +

....)(.......)()()1(2

0

Tn

nT

suT

T

suT

su duufeduufeduufe

=

=

+

0

)1(

)(n

Tn

nT

su duufe

Let us set u =t +nT, then

L f(t) =

= =

+ +0 0

)( )(n

T

t

nTts dtnTtfe

Heref(t+nT) =f(t), by periodic property

Hence


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=

=T

st

n

nsT dttfeetLf00

)()()(

=

Tst

STdttfe

e 0)(

1

1, identifying the above series as a geometric series.

Thus

L f(t) =

Tst

sTdttfe

e 0)(

1

1


Examples:-

1. For the periodic function f(t) of period 4, defined by f(t) = 3t, 0


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T =

2. Therefore

L f(t) =

/2

0)/2(

)(1

1dttfe

e

st

s=

/

0)/2(

sin1

1tdtEe

e

st

s

= { }

/

0

22)/2( cossin1

+

ttss

e

e

E st

s

=22

/

)/2(

)1(

1

++

e

e

E s

s

=))(1)(1(

)1(22//

/

+++

see

eEss

s

=))(1( 22/ + se

Es


3. A periodic function f(t) of period 2a, a>0 is defined byE , 0 t af(t) =

-E, a


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----------------------------------------------- 29.04.05 --------------------------------------Step Function :

In many Engineering applications, we deal with an important discontinuous function H(t-a)defined as follows :

0, t a

H (t-a) =1, t >a

where a is a non-negative constant.

This function is known as the unit step function or the Heaviside function. The function isnamed after the British electrical engineer Oliver Heaviside. The function is also denoted byu(t-a). The graph of the function is shown below:

H(t-a)

1

0 a t

Note that the value of the function suddenly jumps from value zero to the value 1 as a from the left and retains the value 1 for all t>a. Hence the function H(t-a) is called the unitstep function.

In particular, when a=0, the function H(t-a) become H(t), where

0 , t 0H(t) =

1 , t >0

Transform of step functionBy definition, we have

L[H(t-a)] =

0

)( dtatHe st

=

+

a

a

stst dtedte0

)1(0

=s

e as


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In particular, we have L H(t) =s

1

Also, )(1 atHs

eL

as

=

and )(11 tHs

L =

Heaviside shift theorem

Statement :-L [f(t-a) H(t-a)] =e-as Lf(t)

Proof :- We have

L [f(t-a) H(t-a)] =

0

)()( dteatHatf st

=

a

st dtatfe )(

Setting t-a =u, we get

L[f(t-a) H(t-a)] = duufe uas )(0

)(

+

=e-as L f(t)

This is the desired shift theorem.Also,

L-1 [e-as L f(t)] =f(t-a) H(t-a)

Examples :

1. Find L[e

t-2

+sin(t-2)] H(t-2)

Letf(t-2) =[et-2+sin(t-2)]

Thenf(t) = [et +sint]

so that

L f(t) =1

1

1

12 +

+s

By Heaviside shift theorem, we have

L[f(t-2) H(t-2)] =e-2s Lf(t)

Thus,

+

+

=+ 1

1

1

1)2()]2sin([

2

2)2(

ssetHteL st

2. Find L(3t2 +2t +3) H(t-1)

Letf(t-1) =3t2 +2t +3


27/133

so thatf(t) =3(t+1)2 +2(t+1) +3 =3t2 +8t +8

Hence

sstLf

886)(

23++=

Thus

L[3t2 +2t +3] H(t-1) =L[f(t-1) H(t-1)]=e-s L f(t)

=

++sss

e s886

23

3. Find Le-t H(t-2)

Let f(t-2) =e-t , so that, f(t) =e-(t+2)Thus,

L f(t) =1

2

+

s

e

By shift theorem, we have

1)()]2()2([

)1(22

+==

+

s

etLfetHtfL

ss

Thus

[ ]1

)2()1(2

+=

+

s

etHeL

st

4. Let f(t) = f1 (t) , t af2 (t), t >a

Verify thatf(t) =f1(t) +[f2(t) f1(t)]H(t-a)

Considerf1(t) +[f2(t) f1(t)]H(t-a) =f1(t) + f2 (t) f1(t), t >a

0 , t a

= f2 (t), t >af1(t), t a =f(t), given

Thus the required result is verified.

5. Express the following functions in terms of unit step function and hence find theirLaplace transforms.

1. t2 , 1 2

2. cost, 0


28/133

f(t) =sint, t >

1. Here,f(t) =t2 +(4t-t2) H(t-2)

Hence,

L f(t) = )2()4(2 23

+ tHttL (i)

Let(t-2) =4t t2

so that(t) =4(t+2) (t+2)2 =-t2 +4

Now,

stL

42)(

3+=

Expression (i) reads as

L f(t) = [ ])2()2(23 + tHtL

= )(2 23

tLes

s

+

=

+ 3

2

3

242

sse

s

s


2. Heref(t) =cost +(sint-cost)H(t-)

Hence,

L f(t) = )()cos(sin12

++

tHttLs

(ii)

Let (t-) =sint cost

Then(t) =sin(t +) cos(t +) =-sint +cost

so that

L (t) =11

122 +

++

s

s

Expression (ii) reads as

L f(t) = [ ])()(12

++

tHtLs

= )(12

tLes s ++

=

+

++

1

1

1 22 s

se

s

s s


29/133

---------------------------------- 3.05.04 ---------------------------CONVOLUTION

The convolution of two functions f(t) and g(t) denoted by f(t) g(t) is defined as

f(t) g(t) = t

duugutf0

)()(

Property : f(t) g(t) =g(t) f(t)


f(t) g(t) = t

duugutf0

)()(

Setting t-u =x, we get

f(t) g(t) = 0

))(()(t

dxxtgxf

= =

t

tftgdxxfxtg0 )()()()(

This is the desired property. Note that the operation is commutative.

Convolution theorem :-

L[f(t) g(t)] =L f(t) . L g(t)

Proof :- Let us denote

f(t) g(t) =(t) = t

duugutf0

)()(

Consider

=0 0

)]()([)]([t

st dtugutfetL

=

0 0

)()(t

st duugutfe (1)

We note that the region for this double integral is the entire area lying between the lines u =0and u =t. On changing the order of integration, we find that t varies from u to and u variesfrom 0 to.

u

u=tt=u t =

0 u=0 tHence (1) becomes

L[(t)] =

=

=

0

)()(u ut

st dtduugutfe


30/133

=

0

)( )()( dudtutfeugeu

utssu

=

0 0

)()( dudvvfeuge svsu , where v =t-u

=

0 0

)()( dvvfeduuge svsu

=L g(t) . L f(t)Thus

L f(t) . L g(t) =L[f(t) g(t)]This is desired property.

Examples :

1. Verify Convolution theorem for the functions f(t) and g(t) in the following cases :

(i) f(t) =t, g(t) =sint (ii) f(t) =t, g(t) =et

(i) Here,

f g = t

duutguf0

)()( = t

duutu0

)sin(

Employing integration by parts, we getf g =t sint

so that

L [f g] =)1(

1

1

112222 +

=+

ssss

(1)

Next consider

L f(t) . L g(t) =)1(

11

112222 +

=+

ssss

(2)

From (1) and (2), we find thatL [f g] =L f(t) . L g(t)

Thus convolution theorem is verified.

(ii) Here

f g = t

ut duue0

Employing integration by parts, we get

f g =et

t 1so that

L[f g] =)1(

111

1

122

= sssss

(3)

Next

L f(t) . L g(t) =)1(

1

1

1122

=

ssss

(4)

From (3) and (4) we find that


31/133

L[f g] =L f(t) . L g(t)Thus convolution theorem is verified.

2. By using the Convolution theorem, prove that

=t

tLfs

dttfL

0

)(1

)(

Let us define g(t) =1, so that g(t-u) =1Then

==t t

gfLdtutgtfLdttfL0 0

][)()()(

=L f(t) . L g(t) =L f(t) .s

1

Thus

=t

tLfs

dttfL0

)(1

)(

This is the result as desired.

3. Using Convolution theorem, prove that

++=

tu

ssduuteL

02 )1)(1(

1)sin(

Let us denote, f(t) =e-t g(t) =sin t, then

=t t

u duutgufLduuteL0 0

)()()sin( =L f(t) . L g(t)

=

)1(

1

)1(

12

+

+ ss

=

)1)(1(

12

++ ss

This is the result as desired.


32/133

ASSIGNMENT

1. By using the Laplace transform of coshat, find the Laplace transform of sinhat.

2. Find (i) t

dtt

tL

0

sin(ii)

tt tdteL

0

cos (iii) t

t tdteL0

cos

(iv)

ttdte

ttL

0

sin (v) t

atdttL0

2 sin

3. If f(t) =t2, 0


33/133

10. Verify convolution theorem for the following pair of functions:(i) f(t) =cosat, g(t) =cosbt(ii) f(t) =t, g(t) =t e-t(iii) f(t) =et g(t) =sint

11. Using the convolution theorem, prove the following:

(i) ++=t u

ss

sudueutL0

22

1

)1()1(cos)(

(ii)22

0 )(

1)(

assduueutL

tau

+=


34/133

INVERSE LAPLACE TRANSFORMS

Let L f(t) =F(s). Then f(t) is defined as the inverse Laplace transform of F(s) and is denotedby

L-1 F(s). Thus L-1 F(s) =f(t).

Linearity Property

Let L-1 F(s) =f(t) and L-1 G(s) =g(t) and a and b be any two constants. Then

L-1 [a F(s) +b G(s)] = a L-1 F(s) + b L-1 G(s)

Table of Inverse Laplace Transforms

F(s) )()( 1 sFLtf =

0,1 >ss

1

asa

>

,1

ate

0,22

>+

sas

s Cos at

0,1

22>

+s

as a

atSin

asas

>

,1

22

a

athSin

asas

s> ,22 athCos

0,1

1>

+s

sn n = 0, 1, 2, 3, . . .

!n

tn

0,1

1>

+s

sn n > -1

( )1+ ntn

Examples

1. Find the inverse Laplace transforms of the following:

22)(

as

bsii

+

+

22 9

94

254

52)(

s

s

s

siii

+

+

Here

52

1)(

si


35/133

2

511

2

1

25

1

2

1

52

1)(

t

es

Ls

Li =

=

ata

bat

aLb

as

sL

as

bsLii sincos

1)(

22

1

22

1

22

1 +=+

++

=++

92

94

4

252

5

4

2

9

84

254

52)(

21

2

122

1

+

=

+

+

s

sL

s

sL

s

s

s

sLiii

= ththtt

3sin2

33cos4

2

5sin

2

5cos

2

1

Evaluation of L -1 F(s a)

We have, if L f(t) =F(s), then L[eat

f(t)] =F(s a), and so

L-1 F(s a) = eat f(t) =eat L-1 F(s)

Examples

( )41

1

13:Evaluate1.

+

+s

sL

( )( ) ( ) ( )4

13

14

1-

1

12

1

13

1

11-1s3LGiven

+

+=

+

++=

sL

sL

s

4

1

3

1 121

3s

Les

Le tt =

Using the formula

getweandntakingandn

t

sL

n

n,32

!

11

1 ==+

32

3 32 teteGiven

tt

=

52s-s

2sL:Evaluate2.

2

1-

+

+


36/133

( )

( )

( ) ( ) ( ) 41

13

41

1

41

31

41

2LGiven

2

1

2

1

2

1

2

1-

++

+

=

+

+=

+

+=

sL

s

sL

s

sL

s

s

4

13

4 21

21

++

+=

sLe

s

sLe t-t

tete tt 2sin2

32cos +=

13

12:Evaluate

21

++

+ss

sL

( )( )

( )( ) ( )

+

+

+=

+

+=

45

1

45

2

45

12LGiven

2

2

3

1

2

2

3

2

3

1

2

2

3

2

3

1-

sL

s

sL

s

s

=

45

1

45

22

123

2

123

sLe

s

sLett

=

ththe

t

2

5sin

5

2

2

5cos2 2

3

sss

ss

2

452:Evaluate4.

23

2

+

+

havewe

( ) ( )( ) 1212452

2

452

2

452 2

2

2

23

2

+

++=

++

=+

+=

+

+s

C

s

B

s

A

sss

ss

sss

ss

sss

ss

Then2s2+5s-4 =A(s+2) (s-1) +Bs (s-1) +Cs (s+2)

For s =0, we get A =2, for s =1, we get C =1 and for s =-2, we get B =-1. Using thesevalues in (1), we get

1

1

2

12

2

452

23

2

+

+=

+

+

sssss

ss

Hence

tt eess

ssL +=

+

+ 222

21 2

25

452


37/133

( ) ( )2154

:Evaluate5.2

1

+++

+ss

sL

Let us take

( ) ( ) ( ) 2112154

22 ++

++

+=

+++

+s

C

s

B

s

A

ss

s

Then

4s +5 = A(s +2) +B(s +1) (s +2) +C (s +1)2

For s =-1, we get A =1, for s =-2, we get C =-3

Comparing the coefficients of s2, we get B +C =0, so that B =3. Using these values in (1),we get

( ) ( ) ( ) ( ) 23

13

11

2154

22 +

++

+=

++++

ssssss

Hence

( ) ( ) sLe

sLe

sLe

ss

sL ttt

13

13

1

21

54 1212

12

1 +=+++

+

ttt eete 233 +=

44

3

1:Evaluate5.as

sL

Let

)1(2244

3

as

DCs

as

B

as

A

as

s

+

++

++

=

Hence

s3 =A(s +a) (s2 +a2) +B (s-a)(s2+a2)+(Cs +D) (s2 a2)

For s =a, we get A =; for s =-a, we get B =; comparing the constant terms, we getD =a(A-B) =0; comparing the coefficients of s3, we get1 =A +B +C and so C =. Using these values in (1), we get

2244

3

2

111

4

1

as

s

asasas

s

++

++

=

Taking inverse transforms, we get


38/133

[ ] ateea

sL atat cos

2

1

4

144

31 ++=

[ ]athat coscos2

1+=

1:Evaluate6.

241

++

ss

sL

Consider

( ) ( ) ( )( )

+++=

++++=

++ 11

2

2

1

111 222224 ssss

s

ssss

s

ss

s

( ) ( )

( )( ) ( ) ( )

( ) ( )

+

+=

++

++

+=

++

+=

+++

+++=

4

3

1

4

3

1

2

1

1

4

3

1

4

3

1

2

1

1

1

1

1

2

1

11

11

2

1

2

12

1

2

12

1

241

2

2

12

2

1

2222

22

s

Le

s

Less

sL

Therefore

ss

ssssssss

ssss

tt

=

2

32

3sin

2

32

3sin

2

12

1

2

1 te

te

tt

=

2sin

2

3sin

3

2 tht

Evaluation of L -1[e-as F(s)]

We have, if Lf(t) =F(s), then L[f(t-a) H(t-a) =e-as F(s), and so

L-1[e-as F(s)] =f(t-a) H(t-a)


39/133

Examples

( )41

2:)1(

s

eLEvaluate

ss

Here

( )

( )

( )

( )( )( )5

6

5

)()(2

6

1

2

1)()(

2

1)(,5

352

4

51

32

412

411

4

=

=

==

==

==

tHte

atHatfs

eL

Thus

te

sLe

sLsFLtfTherefore

s

sFa

t

s

tt

++

+

41:)2(

2

2

21

s

se

s

eLEvaluate

ss

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

22coscos

222cossin

)1(

2cos4

)(

sin1

1)(

)1(22

21

2

21

1

21

+=

+=

=+

=

=+

=

+=

tHttHt

tHttHtGiven

asreadsrelationNow

ts

sLtf

ts

LtfHere

tHtftHtfGiven


40/133

Inverse transform of logarithmic and inverse functions

( )[ ] ( ) .ttfLthenF(s),f(t)Lifhave,We HencesFds

d==

( ) )(1

ttfsFds

d

L =

Examples

++

bs

asLEvaluate log:)1( 1

( ) ( )

( )

( ) [ ]

( )

( )b

eetfThus

eetftor

eesFds

dLthatSo

bsassFds

dThen

bsasbs

assFLet

atbt

atbt

btat

=

=

=

++=

++=

++

=

1

11

logloglog)(


41/133

(2) Evaluate 1L

s

a1tan

( )

( )

( )

( )

( )

( )( )

( )( )

=

=

=

=

=

+=

=

t

t

dttfs

sFL

havewes

sFdttfSinceL

a

attf

attftor

thatsoatsFds

dLor

as

asF

ds

dThen

s

asFLet

0

1

0

1

22

1

,,

sin

sin

sin

tan)(

s

sFoftransformInverse

Examples

( )

+

221 1:)1(

assLEvaluate

( )

( )

( )( )

( )2

0

1

22

1

1

22

cos1

sin1

sin)(

1

a

at

dta

at

s

sF

LassLThen

a

atsFLtf

thatsoas

sFdenoteusLet

t

=

==+=

==+

=


42/133

( )

( )

( )

( )[ ]

( )( )[ ]

( ) ( )[ ]

[ ]

[ ] ( ) ( )

==

==

==

++=

+=+

+=

=+

=+

+

t

atat

tat-

at

t

at

at

duugutftgtfsGsFL

soandsGsFtLgtLf

eeata

dtateaass

L

atea

dtteass

LHence

teas

Lhavewe

assLEvaluate

0

1

3

0222

1

2

02

1

2

1

22

1

)()()()(

)()()()(g(t)f(t)L

thenG(s),Lg(t)andF(s)L(t)ifhave,We

1211

1111

getwethis,Using

parts.bynintegratioon,111

1

1

1:)2(

theoremnconvolutiousingbyF(s)oftransformInverse

transformLaplaceinverseforn theoremconvolutiothecalledisexpressionThis

NEXT CLASS24.05.05


43/133

Examples

Employ convolution theorem to evaluate the following :

( )( )bsasL

++ 1)1( 1

( )( )( ) ( )

( )

ba

ee

ba

ee

dueedueebsas

L

e, g(t)ef(t)

bssG

as

atbttbaat

tubaat

tbuuta-

-bt-at

=

=

==++

==

+=

+=

1

1

n theorem,convolutiobyTherefore,

getweinverse,theTaking

1)(,

1F(s)denoteusLet

00

1

( )2

22

1)2(

as

sL

+

( )

( )

( )

( )a

att

a

auatatu

duauatat

a

duauutaa

as

sL

at, g(t)at

f(t)

as

ssG

asF(s)

t

t

t-

2

sin

2

2cossin

2a

1

formulaanglecompoundusingby,2

2sinsin1

cossin1

n theorem,convolutiobyHence

cosa

sin

Then)(,1

denoteusLet

0

0

0

222

1

2222

=

=

+=

=

+

==

+=

+=


44/133

( )( )11)3(

2

1

+

ss

sL

Here

1)(,

1

1

2 +=

=

s

ssGF(s)

Therefore

( )( )( )

( ) ( )[ ] [ ]ttettee

uue

eduuess

L

t, g(t)ef(t)

ttt

tu

tut-

t

cossin2

11cossin

2

cossin2

sin11

1

havewen theorem,convolutioBy

sin

0

2

1

==

==

+

==

Assignment

By employing convolution theorem, evaluate the following :

( )( ) ( )( )

( ) ( ) ( )

( ) ( ) ( )2154

)6(

1

)3(

1

1)5(

11)2(

,)4(11

1)1(

21

222

1

22

1

221

2222

21

21

+

+

+

+++

++++

ss

s

LasL

ssL

ss

sL

babsas

sL

ssL


45/133

LAPLACE TRANSFORM METHOD FOR DIFFERENTIALEQUATIONS

As noted earlier, Laplace transform technique is employed to solve initial-value problems.The solution of such a problem is obtained by using the Laplace Transform of the derivativesof function and then the inverse Laplace Transform.

The following are the expressions for the derivatives derived earlier.

(o)f-(o)fs-f(o)s-f(t)Ls(t)f[L

(o)f-f(o)s-f(t)Ls(t)fL[

f(o)-f(t)Ls(t)]fL[

23

2

=

=

=

Examples

1) Solve by using Laplace transform method

2y(o)tyy , ==+ te


46/133

Taking the Laplace transform of the given equation, we get

( ) ( )[ ] ( )( )

( ) ( )( )

( )( )

( )( )

( ) ( )( )

( )

( )42

1

1

1

1

2

1

3114112

1

342

getwe,transformLaplaceinversetheTaking

1

342

1

121s

becomesthiscondition,giventheUsing

1

1

2

3

1

3

21

3

21

3

2

2

2

+=

++

+=

+

++++=

+

++=

+

++=

+=+

+=+

te

ssL

s

ssL

s

ssLtY

s

sstyL

thatso

styL

styLoytysL

t

This is the solution of the given equation.

Solve by using Laplace transform method :

0)()(,sin32)2( ===+ oyoytyyy

Taking the Laplace transform of the given equation, we get

[ ] [ ]1

1)(3)()(2)()()(

2

2

+=+

styLoytLysoyosytLys

Using the given conditions, we get


47/133

[ ]

( )( )( )

( )( )( )

( )

equation.giventheofsolutionrequiredtheisThis

sin2cos101

401

81

sums,partialofmethodtheusingby15

1

103

1

40

1

1

1

8

1

131

131

1)(

131

1)(

1

132)(

3

21

21

21

2

22

ttee

s

s

ssL

s

DCs

s

B

s

AL

sssLty

or

ssstyL

ors

sstyL

tt +=

+

+

+

=

+

++

++

=

++=

++=

+=+

equation.integralthesolvetomethodTransformLaplaceEmploy3)

( ) ( ) +=t

0

sinlf(t) duutuf

( ) ( )

equation.integralgiventheofsolutiontheisThis

2

11

)(1

)(

1

)(1sin)(

1

getwehere,n theorem,convolutiousingBy

sin1

getweequation,giventheoftransformLaplaceTaking

2

3

21

3

2

2

t

0

t

s

sLtfor

s

stfL

Thus

s

tfL

stLtLf

sL f(t)

duutufLs

L f(t)

+=

+=

+=

++=+=

+=


48/133

s.transformLaplaceusingtany timeatparticletheofntdisplacemethefind10,isspeedinitialtheand20at xisparticletheofpositioninitialtheIft.at timeparticletheofntdisplacemethedenotesx

where025dt

dx6

dt

xdequationthe,satisfyingpathaalongmovingisparticleA4)(

2

2

=

=++ x

getweequation,theoftransformLaplacetheTaking

10.(o)x'20,x(o)areconditionsinitialtheHere

025x(t)(t)6x'(t)'x'

asrewrittenbemayequationGiven

==

=++

[ ]

( )( )

( )

( ) ( )

problem.giventheofsolutiondesiredtheisThis2

4sin354cos20

163

170

163

320

163

70320

163

13020x(t)

256

13020Lx(t)

013020256ssLx(t)

33

21

21

21

21

2

2

tete

sL

s

sL

s

sL

s

sL

thatso

ss

s

ors

tt

+=

+++

++

+=

++

++=

++

+=

+++

=

=++

=

L

Rt

at eeaL-R

Eistany timeatcurrent

that theShowR.resistanceandLinductanceofcircuitato0at tappliedisEeA voltage(5) -at

The circuit is an LR circuit. The differential equation with respect to the circuit is

)(tERidt

di

L =+ Here L denotes the inductance, i denotes current at any time t and E(t) denotes the E.M.F.It is given that E(t) =E e-at. With this, we have

Thus, we have


49/133

at

at

EetiRtLi

orEeRidt

diL

=+

=+

)()('

[ ] [ ] orER at=+ eL(t)i'L(t)i'LLTTT

getwesides,bothon)(transformLaplaceTaking TL

[ ] [ ]a

EtiLRoitiLsL TT +=+

1)()()(

[ ]

( ) ( )

))(()(getwe,LtransforminverseTaking

)(

)(getweo,i(o)Since

11-

T RsLas

E

Lti

RsLas

EtiL

oras

ERsLtiL

T

T

T

++=

+++=

+=+=

desired.asresulttheisThis

)(

11 11

=

+

+

=

L

Rt

at

TT

eeaLR

Eti

Thus

RsLLL

asL

aLR

E

1.y(o)2,with x(o)09dt

dy

,04dtdxgiventofin termsyandfor xequationsussimultaneotheSolve)6(

===

=+

x

y

Taking Laplace transforms of the given equations, we get[ ]

[ ] 0)()()(9

0)(4x(o)-Lx(t)s

=+

=+

oytLystxL

tLy

getwe,conditionsinitialgiventheUsing

1)(5)(9

2)(4)(

=+

=+

tyLtxL

tyLtxLs

getweLy(t),forequationstheseSolving

36

182 ++

=s

sL y(t)

thatso


50/133

tt

ss

sLy(t)

6sin36cos

36

18

36 221

+=

++

+=

(1)

getwe,09

d

dyinthisUsing = x

[ ]tt 6cos186sin69

1x(t) +=

or

[ ]tt 6sin6cos33

2x(t) =

(2)(1)and (2) together represents the solution of the given equations.

Assignment

Employ Laplace transform method to solve the following initial value problems

( )

( )

( )

=+=

+=

===+

===++

=

==+

==++=++

===++

===++

====+

t

tu

t

t

fduuutfttf

dueutftf

yytHyy

yytyyy

yytyy

yyttyyy

yyeyyy

yyeyyy

yyygivenyyyy

0

0

2

2

2

4)0(,cos)(')9

21)()8

1)0(,0)0(,1)7

1)1(,3)0(,2)6

12

,1)0(,2cos9)5

0)0(,2)0(,22223)4

)0(1)0(,34)3

1)0(,2)0(,65)2

6)0(,0)0()0(022)1

any time.atcurrentthefindinitially,zeroiscurrenttheIf0.at tappliedis

sinvoltageacircuit,R-LanIn11)

t.any timeatntdisplacemethe

findrest,atinitiallyisparticletheIf.5sin805dt

dx4

dt

xd

equationby thegovernedistany timeatpointfixeda

fromnt xdisplacemeitsthatsolineaalongmovesparticleA)10

2

2

=

=++

tE

tx


51/133

3)0(,8)0(,2dt

dy

,23dt

dx

)15

3)0(,8)0(,2dt

dy,32

dt

dx)14

0)0(,1)0(,sindt

dy,

dt

dx)13

0)0(,2)0(,cosdtdy,sin

dtdx12)

:

===+=+

====

===+=

===+=+

yxyxxy

yxxyyx

yxtxey

yxtxty

t

toftermsinyandx

forequationsaldifferentiussimultaneofollowingtheSolve

LAPLACE TRANSFORMS

INTRODUCTION

Laplace transform is an integral transform employed in solving physical problems.

Many physical problems when analysed assumes the form of a differential equationsubjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable arespecified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different values of theindependent variable, the conditions are called boundary conditions.

The problem with initial conditions is referred to as the Initial value problem.

The problem with boundary conditions is referred to as the Boundary value problem.


dy

d

yd=++

2

2

with conditions y(0) =

y (0) =1 is an initial value problem


52/133


dy

d

ydcos23

2

2

=++ with y(1)=1,

y(2)=3 is called Boundary value problem.

Laplace transform is essentially employed to solve initial value problems. This technique is

of great utility in applications dealing with mechanical systems and electric circuits. Besidesthe technique may also be employed to find certain integral values also. The transform isnamed after the French Mathematician P.S. de Laplace (1749 1827).

The subject is divided into the following sub topics.

Definition :

Let f(t) be a real-valued function defined for all t 0 and s be a parameter, real or complex.

Suppose the integral

0

)( dttfe st exists (converges). Then this integral is called the Laplace

transform of f(t) and is denoted by Lf(t).

Thus,

Lf(t) =

0

)( dttfe st (1)

We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t)

is a function of s denoted by F(s) or )(sf .

LAPLACE TRANSFORMS

Definition and

Properties

Transforms of

some functions

Convolution

theorem

Inverse

transforms

Solution of

differentialequations


53/133

Thus,

Lf(t) =F(s) (2)

Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denotedby L-1 [F(s)].

Thus,

L-1 [F(s)] =f(t) (3)

Suppose f(t) is defined as follows :

f1(t), 0


54/133

=a L f(t) +b L(t)


In particular, for a=b=1, we have

L [ f(t) + (t)] = L f(t) + L(t)and for a =-b =1, we have

L [ f(t) - (t)] = L f(t) - L(t)

2. Change of scale property : If L f(t) =F(s), then L[f(at)] =

a

sF

a

1, where a is a positive

constant.


Lf(at) =

0

)( dtatfe st (1)

Let us set at =x. Then expression (1) becomes,

L f(at) =

0

)(1

dxxfea

xa

s

=

a

sF

a

1


3. Shifting property :- Let abe any real constant. Then

L [eatf(t)] =F(s-a)


L [eatf(t)] = [ ]

0

)( dttfee atst =

0

)( )( dttfe as

=F(s-a)

This is the desired property. Here we note that the Laplace transform of eatf(t) can be

written down directly by changing s to s-a in the Laplace transform of f(t).

TRANSFORMS OF SOME FUNCTIONS

3. Let a be a constant. Then

L(eat) =

=0 0

)( dtedtee tasatst


55/133

=asas

e tas

=

1

)(0

)(

, s >a

Thus,

L(eat) =a

1

In particular, when a=0, we get

L(1) =1

, s >0

By inversion formula, we have

atat eLeas

L ==

11 11

4. L(cosh at) =

+

2

atat eeL = [ ]

+

0

2

1dteee atatst

= [ ]

+ +0

)()(

2

1dtee tastas

Let s >|a| . Then,

+

++

=

0

)()(

)()(2

1)(cosh

as

e

as

eatL

tastas

=22 as

s

Thus,

L (cosh at) =22 as

s

, s >|a|

and so

atas

sL cosh

22

1 =

3. L (sinh at) =222 as

aeeL

atat

=

, s >|a|

Thus,

L (sinh at) =22 a

a

, s >|a|

and so,

a

at

asL

sinh122

1 =


56/133

4. L (sin at) =

0

sinate st dt

Here we suppose that s >0 and then integrate by using the formula

[ ] += bxbbxabaebxdxe

axax cossinsin

22

Thus,

L (sinh at) =22 a

a

+, s >0

and so

a

at

asL

sinh122

1 =

+

5. L (cos at) = atdte st cos0

Here we suppose that s>0 and integrate by using the formula

[ ] ++= bxbbxabae

bxdxeax

ax sincoscos22

Thus,

L (cos at) =22 a

s

+, s >0

and so

ata

sL cos

22

1 =+

6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then

L(tn) =

0

dtte nst

Let s >0 and set st =x, then

+

=

=0 0

1

1)( dxxe

ss

dx

s

xetL nx

n

n

xn

The integral dxxe nx

0

is called gamma function of (n+1) denoted by )1( + n . Thus


57/133

1

)1()(

+

+=

n

n ntL

In particular, if n is a non-negative integer then )1( + n =n!. Hence

1

!)( += n

n ntL

and so

)1(

11

1

+=

+

n

t

sL

n

nor

!n

tnas the case may be


58/133

TABLE OF LAPLACE TRANSFORMS

f(t) F(s)

1s

1, s >0

eatas

1, s >a

coshat 22 as

s

, s >|a|

sinhat 22 as

a

, s >|a|

sinat 22 as

a

+, s >0

cosat 22 as

s

+, s >0

tn, n=0,1,2,.. 1!+ns

n, s >0

tn, n >-1 1)1(

+

+ns

n, s >0

Application of shifting property :-

The shifting property is

If L f(t) =F(s), then L [eatf(t)] =F(s-a)

Application of this property leads to the following results :

1. [ ]ass

ass

at

bs

sbtLbteL

==22

)(cosh)cosh( =22)( bas

as

Thus,

L(eatcoshbt) =22)( bas

as

and

btebas

asL at cosh

)( 221 =


59/133

2.22)(

)sinh(bas

abteL at

=

and

btebas

L at sinh)(1 22

1 =

3.22)(

)cos(bas

asbteL at

+

=

and

btebas

asL at cos

)( 221 =

+

4.22)(

)sin(bas

bbteL at

=

and

b

bte

basL

at sin

)(

122

1 =

5.1)(

)1()(

++

=n

nat

as

nteL or

1)(

!+ nas

nas the case may be

Hence

)1()(

11

1

+=

+

n

te

asL

nat

nor

1)(

!+ nas

nas the case may be

Examples :-

1. Find Lf(t) given f(t) = t, 0


60/133


2. Find Lf(t) given f(t) = sin2t, 0

Here

Lf(t) = +

dttfedttfe stst )()(

0

=

0

2sin tdte st

= { }

0

22cos22sin

4

+

ttss

e st= [ ]se

+1

4

22


3. Evaluate : (i) L(sin3t sin4t)

(iv) L(cos2 4t)

(v) L(sin32t)

(i) Here

L(sin3t sin4t) =L [ )]7cos(cos2

1tt

= [ ])7(cos)(cos2

1tLtL , by using linearity property

=

)49)(1(

24

4912

12222

++

=

+

+ ss

s

s

s

s

s

(ii) Here

L(cos24t) =

+

+=

+64

1

2

1)8cos1(

2

12s

s

stL

(iii) We have

( ) 3sinsin34

1sin3 =

For =2t, we get

( )ttt 6sin2sin34

12sin3 =

so that

)36)(4(

48

36

6

4

6

4

1)2(sin

2222

3

++=

+

+

=ssss

tL



61/133

4. Find L(cost cos2t cos3t)

Here

cos2t cos3t = ]cos5[cos

2

1tt+

so that

cost cos2t cos3t = ]coscos5[cos2

1 2ttt +

= ]2cos14cos6[cos4

1ttt +++

Thus

L(cost cos2t cos3t) =

+

+++

++ 4

1

16364

1222 s

s

ss

s

s

s

5. Find L(cosh22t)

We have

2

2cosh1cosh2

+=

For =2t, we get

24cosh12cosh2 tt +=

Thus,

+=16

1

2

1)2(cosh

2

2

s

s

stL

6. Evaluate (i) L( t ) (ii)

tL

1(iii) L(t-3/2)

We have L(tn) =1

)1(+

+ns

n

(i) For n=2

1, we get


62/133

L(t1/2) =2/3

)12

1(

s

+

Since )()1( nnn =+ , we have22

1

2

11

2

1 =

=

+

Thus,2

3

2)( tL

=

(ii) For n =-2

1, we get

stL

=

=

21

21 2

1

)(

(iii) For n =-2

3, we get

stL

222

1

)(2

12

12

3

=

=

=

7. Evaluate : (i) L(t2) (ii) L(t3)

We have,

L(tn) =1

!+ns

n

(i) For n =2, we get

L(t2) =33

2!2

ss=

(ii) For n=3, we get

L(t3) =44

6!3

ss=

8. Find L[e-3t (2cos5t 3sin5t)]

Given =

2L(e-3t cos5t) 3L(e-3t sin5t)

=( )

25)3(

15

25)3(

32

22 ++

+++

ss

s, by using shifting property

=346

922 ++

ss

s, on simplification


63/133

9. Find L[coshat sinhat]

Here

L[coshat sinat] =( )

+ at

eeL

atat

sin2

=

+++

+ 2222 )()(21

aasa

aasa

, on simplification

10. Find L(cosht sin3 2t)

Given

+

4

6sin2sin3

2

tteeL

tt

= ( )[ ])6sin()2sin(3)6sin(2sin38

1teLteLteLteL tttt +

=

++

++

++

+ 36)1(

6

4)1(

6

36)1(

6

4)1(

6

8

12222 ssss

=

++

++

++

+ 36)1(

1

24)1(

1

36)1(

1

4)1(

1

4

32222 ssss

11. Find )( 254 teL t

We have

L(tn) =1

)1(+

+ns

nPut n=-5/2. Hence

L(t-5/2) =2/32/3 3

4)2/3( =

ss

Change s to s+4.

Therefore,2/3

2/54

)4(3

4)(

+=

steL t

Transform of tn f(t)

Here we suppose that nis a positive integer. By definition, we have

F(s) =

0

)( dttfe st

Differentiating n times on both sides w.r.t. s, we get

])][()[(

)2(2222

22

aasaas

asa

++++

=


64/133

=0

)()( dttfes

sFds

d stn

n

n

n

Performing differentiation under the integral sign, we get

=

0

)()()( dttfetsFds

d stn

n

n

Multiplying on both sides by (-1)n , we get

==0

)]([)(()()1( tftLdtetftsFds

d nstnn

nn , by definition

Thus,

L[tnf(t)]= )()1( sFds

dn

nn

This is the transform of tn f(t).

Also,

)()1()(1 tftsFds

dL nn

n

n

=

In particular, we have

L[t f(t)] = )(sFds

d , for n=1

L[t2 f(t)]= )(2

2

sFds

d, for n=2, etc.

Also,

)()(1 ttfsFds

dL =

and

)()( 22

21 tftsF

ds

dL =

Transform off(t)

We have , F(s) =

0

)( dttfe st

Therefore,

=

s sdsdttfstedssF

0)()( = dtdsetf

s

st

0

)(


65/133

=

0

)( dtt

etf

s

st

=

=

0

)()(

t

tfLdt

t

tfe st

Thus,

=

s

dssFt

tfL )(

)(

This is the transform oft

tf )(

Also,

=s

t

tfdssFL

)()(1

Examples :

1. Find L[te-t sin4t]

We have,

16)1(

4]4sin[

2 ++=

steL t

So that,

L[te-t sin4t] =

++

172

14

2 ssds

d

= 22 )172(

)1(8

++

+

ss

s

2. Find L(t2 sin3t)

We have

L(sin3t) =9

32 +

So that,

L(t2 sin3t) =

+ 9

322

2

sds

d=

22 )9(6

+

s

s

ds

d=

32

2

)9(

)3(18

+

s

s

3. Find

t

teL

t sin

We have

1)1(

1)sin(

2 ++=

steL t


66/133

Hence

t

teL

t sin= [ ]

+=

++0

1

2)1(tan

1)1(ss

s

ds

= )1(tan2

1 + s

=cot 1 (s+1)

4. Find

ttL sin . Using this, evaluateL

tatsin

We have

L(sint) =1

12 +

So that

Lf(t) =

t

tL

sin= [ ]

=+ Ss

ss

ds 12

tan1

= )(cottan2

11 sFss ==

Consider

L

t

atsin=a L )(

sinataLf

at

at=

=

a

sF

aa

1, in view of the change of scale property

=

a

s1cot

5. Find L

t

btat coscos

We have

L [cosat cosbt] =2222 b

s

a

s

+

+

So that

L t

btat coscos= dsbs

s

as

s

s

++ 2222 =

++

sbs

as22

22

log2

1

=

++

++

22

22

22

22

loglog2

1

bs

as

bs

asLt

s

=

++

+22

22

log02

1

as

bs=

+

+22

22

log2

1

as

bs


67/133

6. Prove that

=0

3

50

3sintdtte t

We have

=0 )sin(sin ttLtdttest

= )(sintLds

d

=

+ 11

2sds

d

=22 )1(

2

+ss

Putting s =3 in this result, we get

=0

3

50

3sintdtte t



68/133

ASSIGNMENT

I Find L f(t) in each of the following cases :

1. f(t) = et , 0 3

3. f(t) =a

t, 0 t


69/133

21.t

e t21

22.t

ee btat

23.t

t2

sin

24.t

tt 5sin2sin2

Dr.G.N.Shekar

DERIVATIVES OF ARC LENGTH

Consider a curve C in the XY plane. Let A be a fixed point on it. Let P and Q be two

neighboring positions of a variable point on the curve C. If s is the distance of P from A

measured along the curve then s is called the arc length of P. Let the tangent to C at P make

an angle with X-axis. Then (s,) are called the intrinsic co-ordinates of the point P. Let

the arc length AQ be s +s. Then the distance between P and Q measured along the curve C

iss. If the actual distance between P and Q isC. Thens=C in the limit Q P along C.

. . 1Q P

si e Lt

C

=

0

y

x

P(s, )

Q

A

s

s


70/133

Cartesian Form:

Let ( )y f x= be the Cartesian equation of the curve C and let ( , ) ( , )P x y andQ x x y y + +

be any two neighboring points on it as in fig.

Let the arc length PQ s= and the chord lengthPQ C= . Using distance between two

points formula we have 2 2 2 2PQ =( C) =( x) +( y)

222

11

+=

+=

x

y

x

Cor

x

y

x

C

2

1s s C s y

x C x C x

= = +

We note thatx 0 as Q P along C, also that when , 1s

Q PC

=

When Q P i.e. whenx 0, from (1) we get

2

1 (1)ds dy

dx dx

= +

Similarly we may also write

2

1

+==

y

x

C

s

y

C

C

s

y

s

and hence when Q P this leads to

2

1 (2)ds dx

dy dy

= +

Parametric Form: Suppose ( ) ( )x x t and y y t= = is the parametric form of the curve C.

Then from (1)

( , )Q x x y y + +

C

( , )P x y

s


71/133

222

11

+

=

+=

dt

dy

dt

dx

dtdx

dtdx

dtdy

dx

ds

2 2

(3)ds ds dx dx dy

dt dx dt dt dt

= = +

Note: Since is the angle between the tangent at P and the X-axis,

we have tandy

dx=

( )2 21 1 tan

dsy Sec

dx = + = + =

Similarly

( )

22 2

1 11 1 1 cot sec

tan

dsCo

dy y

= + = + = + =

i.e. Cosdx dy

and Sinds ds

= =

( ) ( ) ( )2 2

2 2 21

dx dyds dx dy

ds ds

+ = = +


72/133

We can use the following figure to observe the above geometrical connections

among , , .dx dy dsand

Polar Curves :

Suppose ( )r f= is the polar equation of the curve C and ( , ) ( , )P r and Q r r + + be

two neighboring points on it as in figure:

Consider PN OQ.

In the right-angled triangle OPN, We have

PN PN

Sin PN r Sin rOP r = = = =

sinceSin = when .is very small

From the figure we see that, (1)ON ON

Cos ON r Cos r rOP r

= = = = =

cos 1 0when = Q

( )NQ OQ ON r r r r = = + =

dx

ds

dy

C

N

O

r

P(r,)

s

x

( , )Q r r + +


73/133

2 2 2 2 2 2, . , ( ) ( ) ( )From PNQ PQ PN NQ i e C r r = + = +

2

2C rr

= +

2

2S S C S rrC C

= = +

We note that , 0 1

S

whenQ P alongthecurve also C

=

2

2, (4)dS dr

whenQ P rd d

= +

2

2 2 2 2, ( ) ( ) ( ) 1C

Similarly C r r rr r

= + = +

2

21S S C S

and rr C r C r

= = +

2

2, 1 (5)dS d

whenQ P we get rdr dr

= +

Note:

dWe know that Tan r

dr

=

2

2 2 2 2 21 secds dr

r r r Cot r Cot rCod d

= + = + = + =

Similarly2

2 21 1ds d

r Tan Secdr dr

= + = + =

1dr dCos and Sin

ds ds r

= =

The following figure shows the geometrical connections among ds, dr, d and


74/133

Thus we have :

22 2 2

1 , 1 ,ds dy ds dx ds dx dy

dx dx dy dy dt dt dt

= + = + = +

2 2

2 21ds d ds dr

r and rdr dr d d

= + = +

Example 1: 2/3 2/3 2/3ds ds

and for the curve xdx dy

y a+ =

2/3 2/3 2/3 -1/3 -1/32 2x x ' 0

3 3

y a y y+ = + =

-1 13 3

-1

3

x'

y

yy

x

= =

2 1/32/3 2/3 2/3 2/3

2/3 2/3 2/3

dsHence 1 1

dx

dy y x y a a

dx x x x x

+ = + = + = = =

Similarly

2 1/32/3 2/3 2/3 2/3

2/3 2/3 2/3

ds1 1

dy

dx x x y a a

dy y y y y

+= + = + = = =

Example 2:2

2 2

ds afor the curve y alog

dx a -xFind

=

( )2 2 2 2 2 2 22 2

log logdy x ax

y a a a a x adx a x a x

= = =

dr

ds r d


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( )

( )

( )

( )

( )

22 2 2 2 2 22 2 2 2 2

2 2 2 22 2 2 2 2 2

441 1

a x a x a xds dy a x a x

dx dx a xa x a x a x

+ + + = + = + = = =

Example 3: t tIf x aeSint, y ae Cost, find dsdt= =

t t tx aeSint aeSint aeCostdx

dt= = +

t t ty aeCost aeCost aeSintdy

dt= =

( ) ( )2 2

2 22 2 2 2t tds dx dy a e Cost Sint a e Cost Sintdt dt dt

= + = + +

( )2 2

2 2

t t

ae Cos t Sin t a e= + = ( ) ( ) ( )2 2 2 2

2a b a b a b+ + = +Q

Example 4:t

If x a Cos t log Tan ,y a Sin t, find2

ds

dt

= + =

212

2 222 2

tSecdxa Sint a Sint

t t tdt Tan Sin Cos

= + = +

( )2 211 Sin t aCos ta Sint a aCost Cot t

Sint Sint Sint

= + = = =

dyaCost

dt=

2 2

2 2 2 2 2ds dx dy a Cos tCot t a Cos tdt dt dt

= + = +

( )2 2 2 1a Cos t Cot t= +

2 2 2 2 2seca Cos t Co t a Cot t aCot t= = =

Example 5: 3 3If x a Cos t, y Sin t, findds

dt= =

2 23 , 3dx dy

aCos tSint aSin tCostdt dt

= =

2 2ds dx dy

dt dt dt

= +

2 4 2 2 4 29 9a Cos t Sin t a Sin tCos t= +


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( )2 2 2 2 29 3a Cos t Sin t Cos t Sin t aSintCost= + =

Exercise:

Findds

dtfor the following curves:

(i) x=a(t +Sint), y=a(1-Cost)

(ii) x=a(Cost +t Sint), y=aSint

(iii) x=a log(Sect +tant), y=a Sect


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Example 6: 2 2If r a Cos 2 , Show that r is constantds

d

=

22 2 22 2 2 2 2

dr dr ar a Cos r a Sin Sin

d d r

= = =

2 42 2 2 4 4 2

2

1

2 2

ds dr a

r r Sin r a Sind d r r

= + = + = +

4 4 2 4 2 4 22 2 2ds

r r a Sin a Cos a Sind

= + = +

2 2 2 22 Constanta Cos Sin a = + = = 2 2r is constant for r a Cos2ds

d

=

Example 7:2 2

-1 r dsFor the curve Cos , Show that r is constant.k dr

k r

r

=

2 2

2 2

22

2

2(1)

1 1 2

1

rr k r

d k r

dr k rr

k

=

( )2 2 2

2 2 2 2 2

1 r k r

k r r k r

+ = +

2 2 2

2 2 2 2 2 2 2 2

1 k r k

k r r k r r k r

+= + =

2 2k r

r

=

( )2 2 2 2 22 22

1 2 1k rds d r k r k

r rdr dr r r r

+ = + = + = =

dsHence r k (constant)

dr=

Example 8: ( )2

2 2

ds dsFor a polar curve r f show that ,

dr d

r r

pr p

= = =

We know thatdr d 1

ds ds rCos and Sin

= =

2 222

2

dr

1 1ds

r pp

Cos Sin p rSinr r

= = = = =Q

2 2

ds

dr

r

r p =

2ds

d

r r rAlso

pSin pr

= = =


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Exercise:

Findds ds

anddr d

for the following curves:

m

2

(i) r=a(1+Cos )

(ii) =a Cosm

(iii) aSec ( )2

mr

r

=

Mean Value Theorems

Recapitulation

Closed interval: An interval of the form a x b , that includes every point between a and band also the end points, is called a closed interval and is denoted by [ ]ba, .

Open Interval: An interval of the form a x b< < , that includes every point between a and bbut not the end points, is called an open interval and is denoted by ( )ba,

Continuity: A real valued function )(xf is said to be continuous at a point 0x if

0

0lim ( ) ( )x x

f x f x

=

The function )(xf is said to be continuous in an interval if it is continuous at every point inthe interval.

Roughly speaking, if we can draw a curve without lifting the pen, then it is a continuouscurve otherwise it is discontinuous, having discontinuities at those points at which the curvewill have breaks or jumps.

We note that all elementary functions such as algebraic, exponential, trigonometric,logarithmic, hyperbolic functions are continuous functions. Also the sum, difference, productof continuous functions is continuous. The quotient of continuous functions is continuous atall those points at which the denominator does not become zero.

Differentiability: A real valued function )(xf is said to be differentiable at point 0x if

0

0

0

( ) ( )lim

x x

f x f x

x x

exists uniquely and it is denoted by )(' 0xf .


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A real valued function f(x) is said to be differentiable in an interval if it is differentiable at

every in the interval or if0

( ) ( )limh

f x h f x

h

+ exists uniquely. This is denoted by )(' xf .

We say that either )(' xf exists or f(x) is differentiable.

Geometrically, it means that the curve is a smooth curve. In other words a curve is said to besmooth if there exists a unique tangent to the curve at every point on it. For example a circleis a smooth curve. Triangle, rectangle, square etc are not smooth, since we can draw morenumber of tangents at every corner point.

We note that if a function is differentiable in an interval then it is necessarily continuous inthat interval. The converse of this need not be true. That means a function is continuous neednot imply that it is differentiable.Rolles Theorem: (French Mathematician Michelle Rolle 1652-1679)

Suppose a function )(xf satisfies the following three conditions:

(i) )(xf is continuous in a closed interval [ ]ba,

(ii) )(xf is differentiable in the open interval ( )ba, (iii) )()( bfaf =

Then there exists at least one point c in the open interval ( )ba, such that 0)(' =cf

Geometrical Meaning of Rolles Theorem: Consider a curve )(xf that satisfies the

conditions of the Rolles Theorem as shown in figure:

As we see the curve )(xf is continuous in the closed interval [ ]ba, , the curve is smooth i.e.there can be a unique tangent to the curve at any point in the open interval ( )ba, andalso )()( bfaf = . Hence by Rolles Theorem there exist at least one point c belonging to( )ba, such that 0)(' =cf . In other words there exists at least one point at which the tangentdrawn to the curve will have its slope zero or lies parallel to x-axis.

Notation:

Often we use the statement: there exists a point c belonging to ( )ba, such that .We present the same in mathematical symbols as: c ( )ba, : 0)(' =cf From now onwards let us start using the symbolic notation.

[b,f(b)][a,f(a)]

x

y

a c b


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Example 1: Verify Rolles Theorem for 2)( xxf = in [ ]1,1

First we check whether the conditions of Rolles theorem hold good for the given function:

(i) 2)( xxf = is an elementary algebraic function, hence it is continuous everywhere and so also in[ ]1,1 .

(ii) xxf 2)(' = exists in the interval (-1, 1) i.e. the function is differentiable in (-1,1).

(iii) Also we see that 1)1()1( 2 ==f and 11)1( 2 ==f i.e., )1()1( ff =

Hence the three conditions of the Rolles Theorem hold good.By Rolles Theorem c ( )1,1 : 0)(' =cf that means 2c =0 c=0 ( )1,1

HenceRolles Theorem is verified.

Example 2: Verify Rolles Theorem for2/

)3()(x

exxxf

+=

in [-3, 0]

)(xf is a product of elementary algebraic and exponential functions which are continuous

and hence it is continuous in [ 3,0]

2/22/ )3(2

1)32()( xx exxexxf ++=

2/2 )]3(2

1)32[( xexxx ++=

2/2/2 )2)(3(2

1]6[

2

1 xx exxexx +== exists in (-3, 0)

0)3( =f and also 0)0( =f )0()3( ff =

That is, the three conditions of the Rolles Theorem hold good.

c ( )0,3 : 0)(' =cf

==+ ,2,30)2)(3(2

1 2/ cecc c

Out of these values of c, since 2 ( 3,0) , the Rolles Theorem is verified.


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Example 3: Verify Rolles Theorem for nm bxaxxf )()()( = in [a, b] where a 0.

)(xf is a product of elementary algebraic functions which are continuous and hence it is

continuous in [a, b]

11 )()()()()( += nmnm bxaxnbxaxmxf 11 )())](()([ += nm bxaxaxnbxm

11 )())](()([ ++= nm bxaxnambnmx exists in ( )ba,

0)( =af and also 0)( =bf )()( bfaf =

Hence the three conditions of the Rolles Theorem hold good.

( ),c a b : 0)(' =cf

banmnambcbcacnambnmc nm ,,0)())](()([ 11

++==++

Out of these values of c, sincenm

nambc

++

= ),( ba , the Rolles Theorem is verified.

Example 4: Verify Rolles Theorem for)(

log)(2

bax

abxxf

++

= in [a, b]

)log(log)log()( 2 baxabxxf ++= is the sum of elementary logarithmic functionswhich are continuous and hence it is continuous in [a, b]

xabx

xxf 12)(2

+

= exists in ( )ba,

01loglog)(

log)(2

22

==+

+=

++

=aba

aba

baa

abaaf

and similarly 01loglog)(

log)(2

22

==+

+=

++

=abb

abb

bab

abbbf

Hence the three conditions of the Rolles Theorem hold good. c ( )ba, : 0)(' =cf

abcabcabcc

abcc

cabc

ccf ===

+

=

+= 00

)(

20

12)( 2

2

22

2

Since abc= ),( ba , the Rolles Theorem is verified.Exercise: Verify Rolles Theorem for

(i) )cos(sin)( xxexf x = in

4

5,

4

,


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(ii) 2/)2()( xexxxf = in [ ]2,0 ,

(iii)2

sin2( )

x

xf x

e= in

4

5,

4

.


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Lagranges Mean Value Theorem (LMVT):(Also known as the First Mean Value Theorem)(Italian-French Mathematician J . L. Lagrange 1736-1813)

Suppose a function )(xf satisfies the following two conditions:

(i) )(xf is continuous in a closed interval [ ]ba,

(ii) )(xf is differentiable in the open interval ( )ba,

Then there exists at least one point c in the open interval ( )ba, such thatab

afbfcf

=)()(

)('

Proof: Consider the function )(x defined by kxxfx = )()( where k is a constant to befound such that ( ) ( )a b = .

Since )(xf is continuous in the closed interval[ ]

ba, , )(x which is a sum of continuous

functions is also continuous in the closed interval [ ]ba,

' '( ) ( ) (1)x f x kx = exists in the interval ( )ba, as )(xf is differentiable in( )ba, .

We have ( ) ( )a f a ka = and ( ) ( )b f b kb =

( ) ( )( ) ( ) ( ) ( ) (2)

f b f aa b f a ka f b kb k

b a

= = =

This means that when k is chosen as in (2) we will have ( ) ( )a b =

Hence the conditions of Rolles Theorem hold good for )(x in [ ]ba, By Rolles Theorem c ( ),a b : '( ) 0c =

''( ) 0 ( ) 0 ( ) (3)c f c k k f c = = =

From (2) and (3), we infer that c ( ),a b :ab

afbfcf

=)()(

)('

Thus the Lagranges Mean Value Theorem (LMVT) is proved.


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Geometrical Meaning of L agranges Mean Value Theorem: Consider a curve )(xf that

satisfies the conditions of the LMVT as shown in figure:

From the figure, we observe that the curve )(xf is continuous in the closed interval [ ]ba, ; thecurve is smooth i.e. there can be a unique tangent to the curve at any point in the openinterval ( )ba, . Hence by LMVT there exist at least one point c belonging to ( )ba, such

thatab

afbfcf

=)()(

)(' . In other words there exists at least one point at which the tangent

drawn to the curve lies parallel to the chord joining the points [ ], ( )a f a and[ ], ( )b f b .

Other form of LMVT:The Lagranges Mean Value Theorem can also be stated as follows:

Suppose )(xf is continuous in the closed interval [ ],a a h+ and is differentiable in the open

interval ( ), )a a h+ then (0,1): ( ) ( ) ( )f a h f a hf a h + = + +

When (0,1) we see that ( , )a h a a h+ + i.e., here c a h= +

Using the earlier form of LMVT we may write that( ) ( )

( )( )

f a h f af a h

a h a

+ + =

+

On simplification this becomes ( ) ( ) ( )f a h f a hf a h+ = + + .

[b,f(b)]

[a,f(a)]

x

y

a c b


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Example 5: Verify LMVT for ( ) ( 1)( 2)( 3)f x x x x= in [0,4]

3 2( ) ( 1)( 2)( 3) 6 11 6f x x x x x x x= = + is an algebraic function hence it is continuousin [0,4]

2( ) 3 12 11f x x x = + exists in (0,4) i.e., ( )f x is differentiable in (0,4).i.e.,both the conditions of LMVT hold good for ( )f x in[0,4].

Hence(4) (0)

(0,4): ( )4 0

f fc f c

=

i.e., 2(3)(2)(1) ( 1)( 2)( 3)

3 1 114 0

c c

+ =

i.e, 2 22

3 12 11 3 3 12 8 0 2 (0,4)3

c c c c c + = + = =

Hence the LMVT is verified.

Example 6: Verify LMVT for ( ) logef x x= in [1, ]e

( ) logef x x= is an elementary logarithmic function hence continuous in [1, ]e .

1( )f x

x = exists in ( )1,e or ( )f x is differentiable in( )1,e .

i.e., both the conditions of LMVT hold good for ( )f x in [1, ]e

Hence( ) (1) 1 log log1

(1, ): ( )1 1

f e f ec e f c

e c e

= =

1 11 (1, )

1c e e

c e = =



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Example 7: Verify LMVT for 1( ) sinf x x= in [0,1]

We find2

1( )

1f x

x

=

exists in ( )0,1 , and hence ( )f x is continuous in [0,1].

i.e., both the conditions of LMVT hold good for ( )f x in [0,1]

Hence1 1

2

(1) (0) 1 sin 1 sin 0(0,1) : ( )

1 0 11

f fc f c

c

= =

22 2

22

1 2 41

21c c

c

= = =

2 40.7712 (0,1)c

= =


Example 8: Find of LMVT for 2( )f x ax bx c= + + orVerify LMVT by finding for 2( )f x ax bx c= + +

2( )f x ax bx c= + + is an algebraic function hence continuous in [ , ]a a h+ .

( ) 2f x ax b = + exists in ( ),a a h+

i.e., both the conditions of LMVT hold good for ( )f x in [ , ]a a h+

Then the second form of LMVT (0,1): ( ) ( ) ( ) (1)f a h f a hf a h + = + +

Here ; 2 3 2 2( ) ( ) ( ) 2 (2)f a h a a h b a h c a ah a h ab bh c+ = + + + + = + + + + + 3( ) (3)f a a ab c= + + ; 2( ) 2 ( ) 2 2 (4)f a h a a h b a a h b + = + + = + +

Using (2), (3) and (4) in (1) we have:

3 2 2 3 2 12 (2 2 ) (0,1)2

a ah a h ab bh c a ab c h a a h b + + + + + = + + + + + =



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Example 9: Find of LMVT for ( ) logef x x= in 2,e e

( ) logef x x= is an elementary logarithmic function hence continuous in2,e e

1

( )f x x = exists in ( )2

, .e e

Hence by the second form of LMVT

( )

[ ] [ ]

( )

2 2 2

22

2

0,1 : ( ) ( ) ( ) ( )

( )log log

( )

( 1) ( 1). ., 2log log 2 1

1 ( 1) 1 ( 1)

21 ( 1) 1 0,11

e e

f e f e e e f e e e

e ee e

e e e

e e ei e e e

e e e

ee ee

= + +

= +

+

= + =

+ +

+ = =


Example 10: If 0x> , Apply Mean Value Theorem to show that

( ) log (1 )1

ex

i x xx

< + > < < < < + + + + + +

From (1) and (2) we get log (1 )1

ex

x xx

< + > >

( )2 2 2 2 2 2

1 1 1 1 1 12

1 1 1 1 1 1or

a c b a c b < < < <

From (1) and (2), we have1 1

1 1

2 2 2 2

1 sin sin 1sin sin

1 1 1 1

b a b a b ab a

b aa b a b

< <

engineering mathematics 2

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