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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 equation
subjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable are
specified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different values of the
independent 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.
d 2 y dy
Example 1 : The problem of solving the equation

 y  x with conditions y(0) =
dx 2 dx
y  (0) = 1 is an initial value problem
Example 2 : The problem of solving the equation 3
d2y
dy
 2  y  cos x with y(1)=1,
2
dx
dx
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. Besides
the technique may also be employed to find certain integral values also. The transform is
named after the French Mathematician P.S. de’ Laplace (1749 – 1827).
The subject is divided into the following sub topics.
LAPLACE TRANSFORMS
Definition and
Properties
Transforms of
some functions
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Convolution
theorem
Inverse
transforms
Solution of
differential
equations
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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
e
 st
f (t )dt exists (converges). Then this integral is called the Laplace
0
transform of f(t) and is denoted by Lf(t).
Thus,

Lf(t) =  e  st f (t )dt
(1)
0
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 f (s ) .
Thus,
Lf(t) = F(s)
(2)
Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denoted
by L-1 [F(s)].
Thus,
L-1 [F(s)] = f(t)
(3)
Suppose f(t) is defined as follows :
f(t) =
f1(t),
0<t<a
f2(t),
a<t<b
f3(t),
t>b
Note that f(t) is piecewise continuous. The Laplace transform of f(t) is defined as

Lf(t) =  e  st f (t )
0
a
b

0
a
b
=  e  st f1 (t )dt   e st f 2 (t )dt   e  st f 3 (t )dt
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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)] =  e  st af (t )  b (t )dt


0
0
= a  e  st f (t )dt  b  e  st (t )dt
0
= 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)] =
1 s
F   , where a is a positive
a a
constant.
Proof :- By definition, we have

Lf(at) =  e  st f (at )dt
(1)
0
Let us set at = x. Then expression (1) becomes,

s
1   x
L f(at) =  e  a  f ( x)dx
a0

1 s
F 
a a
This is the desired property.
3. Shifting property :- Let a be any real constant. Then
L [eatf(t)] = F(s-a)
Proof :- By definition, we have
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



L [eatf(t)] =  e  st e at f (t ) dt
=  e ( s a ) f (t )dt
0
0
= F(s-a)
This is the desired property. Here we note that the Laplace transform of eat f(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(e ) =  e e dt   e ( s a )t dt
 st
at
at
0
0

e  ( s a )t
1

=
,
 ( s  a) 0 s  a
s>a
Thus,
1
sa
L(eat) =
In particular, when a=0, we get
L(1) =
1
,
s
s>0
By inversion formula, we have
L1
2.
1
1
 e at L1  e at
sa
s
 e at  e  at
L(cosh at) = L
2









1
=  e  st e at  e at dt
20
1
=  e ( s a )t  e ( s  a )t dt
20

Let s > |a| . Then,

1  e  ( s a )t
e ( s  a )t 
L(cosh at )  


2   ( s  a)  ( s  a)  0
Thus,
L (cosh at) =
and so
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s
,
s  a2
2
s > |a|
=
s
s  a2
2
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 s 
L1  2
 cosh at
2 
s a 
 e at  e  at
3. L (sinh at) = L
2


a
  2
,
2
 s a
s > |a|
Thus,
L (sinh at) =
a
,
s  a2
2
s > |a|
and so,
 1  sinh at
L1  2

2 
a
s a 

4. L (sin at) =  e  st sin at dt
0
Here we suppose that s > 0 and then integrate by using the formula
e ax
a sin bx  b cos bx
a2  b2
ax
 e sin bxdx 
Thus,
L (sinh at) =
a
, s>0
s  a2
2
and so
 1  sinh at
L1  2

2 
a
s a 

5. L (cos at) =  e  st cos atdt
0
Here we suppose that s>0 and integrate by using the formula
e ax
 e cos bxdx  a 2  b 2 a cos bx  b sin bx
ax
Thus,
L (cos at) =
s
,
s  a2
2
and so
L1
s
 cos at
s  a2
2
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s>0
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6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then

L(t ) =  e  st t n dt
n
0
Let s > 0 and set st = x,

then
n

1
 x  dx
L(t )   e  
 n 1  e  x x n dx
s 0
s s
0
n
x

The integral  e  x x n dx is called gamma function of (n+1) denoted by  ( n  1) . Thus
0
L(t n ) 
(n  1)
s n 1
In particular, if n is a non-negative integer then  ( n  1) =n!. Hence
L(t n ) 
n!
s n1
and so
L1
1
s n1

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tn
tn
or
as the case may be
n!
(n  1)
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TABLE OF LAPLACE TRANSFORMS
f(t)
F(s)
1
1
, s>0
s
eat
1
, s>a
sa
coshat
s
, s > |a|
s  a2
sinhat
a
, s > |a|
s  a2
sinat
a
, s>0
s  a2
cosat
s
, s>0
s  a2
2
tn, n=0,1,2,…..
tn, n > -1
2
2
2
n!
s n 1
, s>0
(n  1)
, s>0
s n1
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 :
 s 
1. L(e at cosh bt )  L(cosh bt )ss a   2
2 
 s  b  s s  a
Thus,
L(eatcoshbt)
=
sa
(s  a) 2  b 2
and
L1
sa
 e at cosh bt
2
2
( s  a)  b
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=
sa
(s  a) 2  b 2
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2. L(e at sinh bt ) 
a
( s  a) 2  b 2
and
L1
1
 e at sinh bt
2
2
( s  a)  b
3. L(e at cos bt ) 
sa
( s  a) 2  b 2
and
sa
 e at cos bt
( s  a) 2  b 2
L1
4. L(e at sin bt ) 
b
( s  a) 2  b 2
and
L1
1
e at sin bt

b
( s  a) 2  b 2
5. L(e at t n ) 
(n  1)
( s  a ) n 1
or
n!
( s  a ) n 1
as the case may be
Hence
L1
1
e at t n

( s  a) n 1 (n  1)
or
n!
as the case may be
( s  a ) n 1
Examples :1. Find Lf(t) given f(t) =
t,
0<t<3
4,
t>3
Here

3

0
0
3
Lf(t) =  e  st f (t )dt   e  st tdt   4e  st dt
Integrating the terms on the RHS, we get
Lf(t) =
1 3 s 1
e  2 (1  e 3 s )
s
s
This is the desired result.
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2. Find Lf(t) given f(t) =
sin2t, 0 < t  
0,
t>
Here


0


=  e  st sin 2tdt
Lf(t) =  e  st f (t )dt   e  st f (t )dt
0

 e  st

 s sin 2t  2 cos 2t
=  2
s  4
0
=

2
1  e s
s 4
2

This is the desired result.
3. Evaluate : (i)
L(sin3t sin4t)
(ii)
L(cos2 4t)
(iii)
L(sin32t)
(i) Here
1
L(sin3t sin4t) = L [ (cos t  cos 7t )]
2
=
1
L(cos t )  L(cos 7t ) , by using linearity property
2
=
1 s
s 
24s
 2
 2
2


2  s  1 s  49  (s  1)( s 2  49)
(ii) Here
s 
1
 1 1
L(cos24t) = L  (1  cos 8t )    2
2
 2  s s  64 
(iii) We have
sin 3  
For =2t,
sin 3 2t 
1
3sin   sin 3 
4
we get
1
3 sin 2t  sin 6t 
4
so that
L(sin 3 2t ) 
1 6
6 
48
 2
 2
2


4  s  4 s  36  ( s  4)( s 2  36)
This is the desired result.
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4. Find
L(cost cos2t cos3t)
Here
cos2t cos3t =
1
[cos 5t  cos t ]
2
so that
cost cos2t cos3t =
=
1
[cos 5t cos t  cos 2 t ]
2
1
[cos 6t  cos 4t  1  cos 2t ]
4
Thus
L(cost cos2t cos3t) =
1 s
s
1
s 
 2
  2
2

4  s  36 s  16 s s  4 
L(cosh22t)
5. Find
We have
cosh 2  
1  cosh 2
2
For  = 2t, we get
cosh 2 2t 
1  cosh 4t
2
Thus,
L(cosh 2 2t ) 
1 1
s 
 2

2  s s  16 
-------------------------------------------------------05.04.05------------------------6. Evaluate (i) L( t )
We have
(i) For n=
L(tn) =
 1 
(ii) L  (iii) L(t-3/2)
 t
(n  1)
s n 1
1
, we get
2
1
(  1)
2
L(t1/2) =
s3 / 2

1  1 1
Since (n  1)  n(n) , we have   1    
2
 2  2  2
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Thus,

L( t ) 
2s
(ii) For n = -
3
2
1
, we get
2
1
 

2
L(t 2 )  1  
s
s 2
1
(iii) For n = -
3
, we get
2
 1
  
2 
3
2
L(t 2 )    1    1  2 s
s 2
s 2
7. Evaluate : (i) L(t2)
(ii) L(t3)
We have,
L(tn) =
n!
s n 1
(i) For n = 2, we get
L(t2) =
2! 2

s3 s3
(ii) For n=3, we get
L(t3) =
3! 6

s4 s4
8. Find L[e-3t (2cos5t – 3sin5t)]
Given
=
2L(e-3t cos5t) – 3L(e-3t sin5t)
=2
=
s  3
( s  3)  25
2

15
, by using shifting property
( s  3) 2  25
2s  9
, on simplification
s  6 s  34
2
9. Find L[coshat sinhat]
Here


 eat  e at

L[coshat sinat] = L 
sin at 
2


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
1
a
a


2
2
2
2
2  ( s  a)  a
( s  a)  a 
=

a ( s 2  2a 2 )
, on simplification
[( s  a) 2  a 2 ][( s  a) 2  a 2 ]
10. Find L(cosht sin3 2t)
Given
 et  e t  3 sin 2t  sin 6t 

L 

2
4



 


=
1
3  L et sin 2t  L(et sin 6t )  3L(e  t sin 2t )  L(e  t sin 6t )
8
=

1
6
6
6
6




2
2
2
2
8  ( s  1)  4 ( s  1)  36 ( s  1)  4 ( s  1)  36 
=

3
1
1
1
1




2
2
2
2
4  ( s  1)  4 ( s  1)  36 ( s  1)  24 ( s  1)  36 
11. Find L(e  4t t
5
2
)
We have
L(tn) =
(n  1)
s n 1
L(t-5/2) =
Put n= -5/2. Hence
(3 / 2) 4 
  3 / 2 Change s to s+4.
s 3 / 2
3s
Therefore,
L (e  4 t t  5 / 2 ) 
4 
3( s  4)  3 / 2
Transform of tn f(t)
Here we suppose that n is a positive integer. By definition, we have

F(s) =  e st f (t )dt
0
Differentiating ‘n’ times on both sides w.r.t. s, we get

dn
n
F
(
s
)

e st f (t )dt
n
n 
ds
s 0
Performing differentiation under the integral sign, we get
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
dn
F ( s)   (t ) n e st f (t )dt
ds n
0
Multiplying on both sides by (-1)n , we get

(1) n
dn
F ( s)   (t n f (t )e st dt  L[t n f (t )] , by definition
ds n
0
Thus,
dn
L[t f(t)]= (1)
F ( s)
ds n
n
n
This is the transform of tn f(t).
Also,
 dn

L  n F ( s)  (1) n t n f (t )
 ds

1
In particular, we have
L[t f(t)] = 
d
F (s ) , for n=1
ds
d2
L[t f(t)]= 2 F ( s ) , for n=2, etc.
ds
2
Also,
d

L1 
F ( s)  tf (t ) and
 ds

 d2

L1  2 F ( s)  t 2 f (t )
 ds

Transform of
f(t)
t

We have , F(s) =  e st f (t )dt
0
Therefore,

  st

F
(
s
)
ds

f (t )dt  ds

  e
s
s 0


=

0

 e  st 
f (t )   dt
  t s
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
=

0



f (t )   e  st ds dt
s

 f (t ) 
 f (t ) 
=  e st 
dt

L


 t 
 t 
0
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Thus,

 f (t ) 
L
   F (s)ds
 t  s
This is the transform of
f (t )
t
Also,

L1  F ( s)ds 
s
f (t )
t
Examples :
1. Find L[te-t sin4t]
We have,
4
( s  1) 2  16
L[e t sin 4t ] 
So that,
 d 
1

L[te-t sin4t] = 4  2

 ds  s  2s  17 
8( s  1)
( s  2s  17) 2
=
2
2. Find L(t2 sin3t)
We have
L(sin3t) =
3
s 9
2
So that,
d2  3 


ds 2  s 2  9 
L(t2 sin3t) =
= 6
d
s
2
ds ( s  9) 2
 et sin t 

3. Find L
 t 
We have
L(e t sin t ) 
1
( s  1) 2  1
 et sin t 
 =
Hence L
 t 
=
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
2

ds
 (s  1)
0
2
1



 tan 1 ( s  1) s
 tan 1 ( s  1) = cot –1 (s+1)
=
18( s 2  3)
( s 2  9) 3
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 sin t 
 sin at 
4. Find L
 . Using this, evaluateL 

 t 
 t 
We have
L(sint) =
1
s 1
2
So that
 sin t 
Lf(t) = L
 =
 t 
=

s
s

2



ds
 tan 1 s S
1
2
 tan 1 s  cot 1 s  F ( s )
Consider
 sin at 
 sin at 
L
 = a L
  aLf (at )
 t 
 at 
 1  s 
= a  F   , in view of the change of scale property
 a  a 
s
= cot 1  
a
 cos at  cos bt 
5. Find L 

t

We have
L [cosat – cosbt] =
s
s
 2
2
s a
s  b2
2
So that

s 
 s
 cos at  cos bt 
L
= 2
 2
ds
2

t
s  b 2 

 s s  a
=
 s2  a2 
 s 2  a 2 
1


 2

Lt
log

log

2 
2  s   s 2  b 2 
 s  b 
 s 2  b 2 
1

= 0  log  2
2 
2
 s  a 
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
1   s 2  a 2 

= log  2
2   s  b 2  s
1  s2  b2 

= log  2
2  s  a 2 
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
6. Prove that
e
3t
t sin tdt 
0
3
50
We have

e
 st
t sin tdt  L(t sin t )
=
0
=
2s
( s  1) 2
Putting s = 3 in this result, we get

e
3t
t sin tdt 
0
3
50
This is the result as required.
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d  1 
d
L(sin t ) =   2
ds
ds  s  1
2
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ASSIGNMENT
I
Find L f(t) in each of the following cases :
1. f(t) =
et , 0 < t < 2
0, t>2
2. f(t) = 1 , 0  t  3
t, t>3
t
, 0t<a
a
3. f(t) =
, ta
1
II. Find the Laplace transforms of the following functions :
4. cos(3t + 4)
5. sin3t sin5t
6. cos4t cos7t
7. sin5t cos2t
8. sint sin2t sin3t
9. sin2 5t
10. sin2(3t+5)
11. cos3 2t
12. sinh2 5t
13. t5/2
1 

14.  t  
t

3
15. 3t
16. 5-t
17. e-2t cos2 2t
18. e2t sin3t sin5t
19. e-t sin4t + t cos2t
20. t2 e-3t cos2t
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21.
1  e 2t
t
e  at  e bt
22.
t
23.
sin 2 t
t
24.
2 sin 2t sin 5t
t
III. Evaluate the following integrals using Laplace transforms :

25.  te2t sin 5tdt
0

26.  e 5t t 3 cos tdt
0

 e  at  e bt
27.  
t
0

dt


28.
e t sin t
0 t dt
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-------------------------------------------- 28.04.05 -------------------------Transforms of the derivatives of f(t)
Consider

L f (t ) =  e  st f (t )dt
0




= e  st f (t ) 0   (s)e  st f (t )dt , by using integration by parts

0
= Lt (e
 st
t 

f (t )  f (0)  sLf (t )
= 0 - f (0) + s Lf(t)
Thus
L f (t ) = s Lf(t) – f(0)
Similarly,
L f (t ) = s2 L f(t) – s f(0) - f (0)
In general, we have
Lf n (t )  s n Lf (t )  s n1 f (0)  s n2 f (0)  .......  f n1 (0)
t
Transform of  f(t)dt
0
t
 f (t )dt .
Let  (t) =
Then
(0) = 0 and   (t) = f(t)
0
Now,

L (t) =  e  st (t )dt
0



e  st 
e  st



(
t
)
dt
=  (t )

 s  0 0
s


1
= (0  0)   f (t )e  st dt
s0
Thus,
t
L  f (t )dt 
0
1
Lf (t )
s
Also,
1

L  Lf (t )   f (t )dt
s
 0
1
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t
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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) =
a
s  a2
2
We note that
f (t )  a cos at
Taking Laplace transforms, we get
Lf (t )  L(a cos at )  aL(cos at )
1
1
or L(cosat) = Lf (t )  sLf (t )  f (0)
a
a
1  sa

 0
=  2
2
a s  a

Thus
s
L(cosat) = 2
s  a2
This is the desired result.
 t 
 1 
1
1

2. Given L 2
  3 / 2 , show that L 

s
 t 
  s
Let f(t) = 2
t

, given L[f(t)] =
We note that, f (t ) 
2
1
 2 t

1
s
3/ 2
1
t
Taking Laplace transforms, we get
 1 
Lf (t )  L 

 t 
Hence
 1 
L
  Lf (t )  sLf (t )  f (0)
 t 
 1 
= s 3 / 2   0
s 
Thus
 1 
1
L

s
 t 
This is the result as required.
 cos at  cos bt 
3. Find L  
dt
t


0
Here
t
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2
2
 cos at  cos bt  1  s  b 

Lf(t) = L
  log  2
2 
t

 2 s a 
Using the result
t
1
L  f (t )dt  Lf (t )
s
0
We get,
t
 s2  b2 
1
 cos at  cos bt 

L 
log  2
dt =
t
2s  s  a 2 

0
t
4. Find L  te t sin 4tdt
0
Here


L te t sin 4t 
8( s  1)
( s  2 s  17) 2
2
Thus
t
L  te t sin 4tdt =
0
8( s  1)
s ( s  2s  17) 2
2
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) where
n=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 4
and 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 4
and so on.
Formula :
Let f(t) be a periodic function of period T. Then
T
1
Lf (t ) 
e  st f (t )dt
 ST 
1 e
0
Proof :
By definition, we have


0
0
L f(t) =  e  st f (t )dt   e  su f (u )du
T
=
e
 su
2T
f (u )du   e
0
 e
n 0
f (u)du  ....... 
T
 ( n 1)T
=
( n 1)T
 su
 su
f (u )du
nT
Let us set u = t + nT, then

L f(t) =
T
 e
 s ( t  nT )
f (t  nT )dt
n 0 t 0
Here
f(t+nT) = f(t), by periodic property
Hence
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e
nT
 su
f (u )du  ....  
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
T
n 0
0
Lf (t )   (e  sT ) n  e  st f (t )dt
 1   st
= 
e f (t )dt , identifying the above series as a geometric series.
 ST  
1  e  0
T
Thus
 1   st
L f(t) = 
e f (t )dt
 sT  
1  e  0
This is the desired result.
T
Examples:1. For the periodic function f(t) of period 4, defined by f(t) =
3t, 0 < t < 2
6, 2<t<4
find L f(t)
Here, period of f(t) = T = 4
We have,
T
 1   st
L f(t) = 
e f (t )dt
 sT  
1  e  0
 1   st
= 
e f (t )dt
4 s  
1  e  0
4
2
4

1 
 st
3
te
dt

6e  st dt 
4 s 

1 e 0
2

    st  2 2
 st
1  
e  st 
 e
 e



=
3
t

1
.
dt

6



1  e  4 s     s  0 0  s    s



1  3 1  e 2 s  2se 4 s 
=


1  e 4 s 
s2

=


Thus,
3(1  e 2 s  2se 4 s )
Lf(t) =
s 2 (1  e 4 s )
2. A periodic function of period
Esint, 0  t <
f(t) =


2

is defined by

2
t


where E and  are positive constants. Show that
E
L f(t) = 2
2
( s  w )(1  e s / w )
Here
0,
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4
 
 
 2 
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2
T=

L f(t) =
. Therefore
2 / 
1
1 e
 s ( 2 /  )
e
 st
f (t )dt
0
=
1
1 e
 /
 Ee
 s ( 2 /  )
0
 /
 e  st

 s sin t   cos t
 s ( 2 /  )  2
2
1 e
s 
0
 s / 
E
 (e
 1)
=
 s ( 2 /  )
2
2
1 e
s 
E (1  e  s /  )
=
(1  e  s /  )(1  e  s /  )( s 2   2 )
E
=
 s / 
(1  e
)( s 2   2 )
This is the desired result.
E
=
3. A periodic function f(t) of period 2a, a>0 is defined by
E, 0ta
f(t) =
-E, a < t  2a
show that
E
 as 
tanh  
s
 2
Here T = 2a. Therefore
2a
1
L f(t) =
e  st f (t )dt
1  e 2 as 0
L f(t) =
2a
 a  st

1
 st
Ee
dt


Ee
dt
=


a
1  e  2 as  0

E
1  e  sa  (e  2 as  e  as )
=
 2 as
s (1  e )
=



 
E
1  e as
 2 as
s(1  e )
2
E (1  e  as ) 2
s(1  e as )(1  e as )
E  e as / 2  e  as / 2  E
 as 
 tanh  
 as / 2
 as / 2 
s e
e
2
 s
This is the result as desired.
=
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
 st
sin tdt
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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 is
named after the British electrical engineer Oliver Heaviside. The function is also denoted by
u(t-a). The graph of the function is shown below:
H(t-a)
1
a
0
t
Note that the value of the function suddenly jumps from value zero to the value 1 as t  a
from the left and retains the value 1 for all t>a. Hence the function H(t-a) is called the unit
step function.
In particular, when a=0, the function H(t-a) become H(t), where
0, t0
H(t) =
1, t>0
Transform of step function
By definition, we have

L[H(t-a)] =  e  st H (t  a)dt
0
a

=  e 0dt   e  st (1)dt
 st
0
=
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e
a
 as
s
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In particular, we have L H(t) =
1
s
 e  as 
L1 
  H (t  a)
 s 
Also,
1
and L1    H (t )
s
Heaviside shift theorem
Statement :L [f(t-a) H(t-a)] = e-as Lf(t)
Proof :- We have

 f (t  a) H (t  a)e
L [f(t-a) H(t-a)] =
 st
dt
0

=  e  st f (t  a)dt
a
Setting t-a = u, we get

L[f(t-a) H(t-a)] =  e  s ( a u ) f (u )du
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[et-2 + sin(t-2)] H(t-2)
Let
f(t-2) = [et-2 + sin(t-2)]
Then
f(t) = [et + sint]
so that
1
1
 2
L f(t) =
s 1 s 1
By Heaviside shift theorem, we have
L[f(t-2) H(t-2)] = e-2s Lf(t)
Thus,
1 
 1
L[e (t  2)  sin( t  2)]H (t  2)  e  2 s 
 2
 s  1 s  1
2. Find L(3t2 +2t +3) H(t-1)
Let
f(t-1) = 3t2 +2t +3
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so that
f(t) = 3(t+1)2 +2(t+1) +3 = 3t2 +8t +8
Hence
6
8 8
Lf (t )  3  2 
s
s
s
Thus
L[3t2 +2t +3] H(t-1) = L[f(t-1) H(t-1)]
= e-s L f(t)
8 8
6
= e s  3  2  
s
s
s
3. Find Le-t H(t-2)
Let f(t-2) = e-t , so that, f(t) = e-(t+2)
Thus,
e 2
L f(t) =
s 1
By shift theorem, we have
L[ f (t  2) H (t  2)]  e
2 s
e 2( s 1)
Lf (t ) 
s 1
Thus


L e t H (t  2) 
4. Let f(t) =
e 2( s 1)
s 1
f1 (t) , t  a
f2 (t), t > a
Verify that
f(t) = f1(t) + [f2(t) – f1(t)]H(t-a)
Consider
f1(t) + [f2(t) – f1(t)]H(t-a) = f1(t) + f2 (t) – f1(t), t > a
0,
ta
=
f2 (t), t > a
f1(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 their
Laplace transforms.
t2 , 1 < t  2
1.
f(t) =
4t ,
2.
t>2
cost, 0 < t < 
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f(t) =
sint, t > 
1. Here,
f(t) = t2 + (4t-t2) H(t-2)
Hence,
2
L f(t) = 3  L(4t  t 2 ) H (t  2)
s
Let
(t-2) = 4t – t2
so that
(t) = 4(t+2) – (t+2)2 = -t2 + 4
Now,
2 4
L (t )   3 
s
s
Expression (i) reads as
2
L f(t) = 3  L (t  2) H (t  2)
s
(i)
2
 e  2 s L (t )
3
s
2
4 2 
= 3  e 2 s   3 
s
s s 
This is the desired result.
=
2. Here
f(t) = cost + (sint-cost)H(t-)
Hence,
s
 L(sin t  cos t ) H (t   )
L f(t) = 2
s 1
Let
 (t-) = sint – cost
Then
(t) = sin(t + ) – cos(t + ) = -sint + cost
so that
1
s
 2
L (t) =  2
s 1 s 1
Expression (ii) reads as
s
 L (t   ) H (t   )
L f(t) = 2
s 1
s
 e s L (t )
= 2
s 1
s
 s 1 
 e s  2
= 2
s 1
 s  1
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(ii)
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---------------------------------- 3.05.04 --------------------------CONVOLUTION
The convolution of two functions f(t) and g(t) denoted by f(t)  g(t) is defined as
t
 f (t  u) g (u)du
f(t)  g(t) =
0
Property : f(t)  g(t) = g(t)  f(t)
Proof :- By definition, we have
t
 f (t  u) g (u)du
f(t)  g(t) =
0
Setting t-u = x, we get
0
 f ( x) g (t  x)(dx)
f(t)  g(t) =
t
t
=
 g (t  x) f ( x)dx  g (t )  f (t )
0
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
t
f(t)  g(t) = (t) =
 f (t  u) g (u)du
0
Consider

t
L[ (t )]   e [  f (t  u ) g (u )]dt
 st
0
 t
=
0
e
 st
f (t  u ) g (u )du
(1)
0 0
We note that the region for this double integral is the entire area lying between the lines u =0
and u = t. On changing the order of integration, we find that t varies from u to  and u varies
from 0 to .
u
u=t
t=
t=u
0
Hence (1) becomes
u=0

L[(t)] =

 e
u  0 t u
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 st
t
f (t  u ) g (u )dtdu
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


=  e  su g (u ) e  s (t u ) f (t  u )dt du
0
u





=  e  su g (u ) e  sv f (v)dv du , where v = t-u
0
0



0
0
=  e  su g (u )du  e  sv f (v)dv
= 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,
t
f g =
 f (u) g (t  u)du
0
t
=  u sin( t  u )du
0
Employing integration by parts, we get
f  g = t – sint
so that
1
1
1
 2 2
L [f  g] = 2  2
s
s  1 s ( s  1)
Next consider
1
1
1
 2 2
L f(t) . L g(t) = 2  2
s s  1 s ( s  1)
From (1) and (2), we find that
L [f  g] = L f(t) . L g(t)
Thus convolution theorem is verified.
(1)
(2)
(ii) Here
t
f  g =  ue t u du
0
Employing integration by parts, we get
f  g = et – t – 1
so that
1
1 1
1
 2   2
L[f  g] =
s 1 s
s s ( s  1)
Next
1
1
1
 2
L f(t) . L g(t) = 2 
s s  1 s ( s  1)
From (3) and (4) we find that
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(3)
(4)
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L[f  g] = L f(t) . L g(t)
Thus convolution theorem is verified.
2. By using the Convolution theorem, prove that
t
1
L  f (t )dt  Lf (t )
s
0
Let us define g(t) = 1, so that g(t-u) = 1
Then
t
t
0
0
L  f (t )dt  L  f (t ) g (t  u )dt  L[ f  g ]
= L f(t) . L g(t) = L f(t) .
1
s
Thus
t
1
Lf (t )
s
0
This is the result as desired.
L  f (t )dt 
3. Using Convolution theorem, prove that
t
1
L  e u sin( t  u )du 
( s  1)( s 2  1)
0
Let us denote, f(t) = e-t g(t) = sin t, then
t
t
0
0
L  e u sin( t  u )du  L  f (u ) g (t  u )du = L f(t) . L g(t)
=
This is the result as desired.
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1
1
 2
( s  1) ( s  1)
=
1
( s  1)( s 2  1)
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ASSIGNMENT
1. By using the Laplace transform of coshat, find the Laplace transform of sinhat.
t
t
t
sin t
2. Find (i) L 
(ii) L  e t cos tdt
(iii) L  e t cos tdt
dt
t
0
0
0
 sin t  t
(iv) L  
e dt
t 
0
t
t
(v) L  t 2 sin atdt
0
3. If f(t) = t2, 0 < t < 2 and f(t+2) = f(t) for t > 2, find L f(t)
t, 0ta
4. Find L f(t) given f(t) =
f(2a+t) = f(t)
2a – t, a < t  2a
5. Find L f(t) given f(t) =
1, 0 < t <
-1,
a
2
f(a+t) = f(t)
a
<t<a
2
6. Find the Laplace transform of the following functions :
(i) et-1 H(t-2)
(ii) t2 H(t-2)
(iii) (t2 + t + 1) H(t +2)
(iv) (e-t sint) H(t - )
7. Express the following functions in terms of unit step function and hence find their
Laplace transforms :
2t, 0 < t  
(i)
t2 , 2 < t  3
(ii)
f(t) =
f(t) =
1, t>
3t , t > 3
sin2t, 0 < t  
(iii)
sint, 0 < t  /2
(iv)
f(t) =
f(t) =
0, t>
8. Let
f(t) =
cost, t > /2
f1 (t) , t  a
f2 (t) , a<t  b
f3 (t) , t > b
Verify that
f(t) = f1(t) + [f2(t) – f1(t)]H(t-a) + [f3(t) - f2(t)] H(t-b)
9.
Express the following function in terms of unit step
function and hence find its Laplace transform.
f(t) =
Sint, 0 < t  
Sin2t,  < t  2
Sin3t, t > 2
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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:
t
s
(i) L  (t  u )e u 1 cos udu 
2
( s  1) ( s 2  1)
0
t
(ii) L  (t  u )ue au du 
0
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1
s ( s  a) 2
2
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INVERSE LAPLACE TRANSFORMS
Let L f(t) = F(s). Then f(t) is defined as the inverse Laplace transform of F(s) and is denoted
by
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 (t )  L1 F ( s)
1
F(s)
1
,s0
s
1
,sa
sa
s
,s0
2
s  a2
1
,s0
2
s  a2
1
,s a
2
s  a2
s
,s a
2
s  a2
1
,s0
n 1
s
n = 0, 1, 2, 3, . . .
1
,s0
n 1
s
n > -1
eat
Cos at
Sin at
a
Sin h at
a
Cos h at
tn
n!
tn
n  1
Examples
1. Find the inverse Laplace transforms of the following:
(i )
1
2s  5
Here
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(ii )
sb
s  a2
2
(iii )
2s  5
4s  9

2
4s  25 9  s 2
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5t
1
1
1
1
(i) L
 L1
 e2
2s  5 2
2
s5
2
sb
s
1
b
(ii ) L1 2
 L1 2
 b L1 2
 cos at  sin at
2
2
2
a
s a
s a
s a
1
5
s9
2s  5
4s  8  2 1 s  2
1
(iii ) L  2

 L
 4L 2 2
2
25
2
s 9
 4s  25 9  s  4
s 
4
1 

1
5t
5t  
3

cos  sin   4 cos h3t  sin h3t 

2
2
2 
2

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) = e at L-1 F(s)
Examples
1. Evaluate : L1
3s  1
s  14
Given  L-1
3s  1 - 1  1
s  1
4
 3et L1
 3 L1
1
s  1
3
 2 L1
1
s  14
1
1
 2et L1 4
3
s
s
Using the formula
1
L
1
s n 1
tn

n!
Given 
2. Evaluate : L-1
s2
s - 2s  5
2
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and taking n  2 and 3, we get
3e t t 2 e t t 3

2
3
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Given  L-1
 s  1  3 
s 1
1
1
 L1 
 3L1
L
2
2
s  1  4
s  1  4
s  12  4
 s  1  4 
s2
2
 et L-1
s
1
 3 et L1 2
s 4
s 4
2
 et cos 2t 
Evaluate : L1
3 t
e sin 2t
2
2s  1
s 2  3s  1
Given  2L
-1

s    1  2L s    L
1


5
5
5
s    4  s    4 s    4 
3
2
3 2
2
3
2
1
1
3 2
2
3 2
2
3t
 3t
s
1 

 2e 2 L1 2
 e 2 L1 2

s 5
s 5 
4
4


4. Evaluate :
3t
2e 2

5 2
5 
t
sin h
t
cos h
2
2 
5

2s 2  5s  4
s 3  s 2  2s
we have
2 s 2  5s  4 2 s 2  5s  4 2 s 2  5s  4
A
B
C





3
2
2
ss  2s  1
s s  2 s 1
s  s  2s s s  s  2


Then
2s2+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 these
values in (1), we get
2 s 2  5s  4 2
1
1
 

3
2
s  s  2s s s  2 s  1
Hence
L1
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2 s 2  5s  4
 2  e  2t  e t
2
2
s  s  25
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5. Evaluate : L1
4s  5
s  12  s  2
Let us take
4s  5
A
B
C



2
2
s  1  s  2 s  1 s  1 s  2
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
4s  5
1
3
3



2
2
s  1  s  2 s  1 s  1 s  2
Hence
L1
4s  5
1
1
1
 e t L1 2  3e t L1  3e 2t L1
2
s
s
s
s  1  s  2
 tet  3et  3e2t
s3
5. Evaluate : L 4
s  a4
1
Let
s3
A
B
Cs  D


 2
4
4
s  a s  a s  a2
s a
(1)
Hence
s3 = A(s + a) (s2 + a2) + B (s-a)(s2+a2)+(Cs + D) (s2 – a2)
For s = a, we get A = ¼; for s = -a, we get B = ¼; comparing the constant terms, we get
D = a(A-B) = 0; comparing the coefficients of s3, we get
1 = A + B + C and so C = ½. Using these values in (1), we get
s3
1 1
1  1 s
 


4
4
4  s  a s  a  2 s 2  a 2
s a
Taking inverse transforms, we get
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L1


6. Evaluate : L1

s3
1
1
 e at  e at  cos at
4
4
4
2
s a
1
cos hat  cos at 
2
s
s  s2 1
4
Consider

s
s
1
2s
 2
  2

2
2
2
s  s 1 s  s 1  s  s 1 2  s  s 1 s  s 1 

4


 

 






1  s2  s  1  s2  s  1  1 
1
1


 2



2  s  s  1 s2  s  1  2  s2  s  1
s2  s  1 



1
1
1
 

2
2 s 1  3 s 1 2 
2
2
4

 
 

 


3

4
Therefore
 1
1
 t
 2 t 1 1
s
1
1
2 L1
L1 4

e
L

e

2
3
s  s 1 2 
s2 
s2 
4



3

4

3
3 
1 sin
t  1 t sin
t

t
1
2 e 2
2 
 e 2
2
3
3 

2
2 

2
sin
3
 3 
t

 sin h 
t
 2 
2


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)
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Examples
e ss
s  24
Here
(1) Evaluate : L1
a  5, F ( s ) 
1
s  24
Therefore f (t )  L1F ( s )  L1
Thus
L1
1
1
 e 2t L1 4
4
s
s  2 

e 2t t 3
6
e5s
 f (t  a ) H (t  a )
s  24

e 2t  5  t  5
H t  5
6
3
 es
se 2s 
(2) Evaluate : L1  2
 2

 s 1 s  4
Given  f1 t   H t     f 2 t  2 H t  2 
1
 sin t
s 1
s
f 2 (t )  L1 2
 cos 2t
s 4
Here f1 (t )  L1
2
Now relation (1) reads as
Given  sin t    H t     cos 2t  2  H t  2 
  cos t H t     cos 2t  H t  2 
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(1)
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Inverse transform of logarithmic and inverse functions
We have, if L f(t)  F(s), then Ltf t   
d
F s 
ds
Hence.
 d

L1   F s   tf (t )
 ds

Examples
sa
(1) Evaluate : L1 log 

sb
sa
Let F ( s )  log 
  log s  a   log s  b 
 sb
Then 
d
1 
 1
F s    


ds
s  a s  b

 d

So that L1  F s    e  at  e bt
 ds

or
Thus
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t f t   e bt  e  at
e bt  e  at
f t  
b

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a
(2) Evaluate L1 tan 1  
s
a
Let F ( s )  tan 1  
s
Then 
or
d
 a 
F s    2
2
ds
 s  a 
 d

L1  F s   sin at
 ds

so that
t f t   sin at
or
f t  
sin at
a
 Fs  
Inverse transform of 

 s 
t
SinceL  f t dt 
0
F s 
, we have,
s
 F s  
L1 
   f t dt
 s  0
t
Examples


1
(1) Evaluate : L1  2
2 
s s  a 


Let us denote F s  
1
so that
s  a2
sin at
f (t )  L1F s  
a
2
1
F s 
sin at
Then L  2
 L1

dt
2
s
a
s s a
0
t
1


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
1  cos at 
a2
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

1
(2) Evaluate : L1  2
2 
 s s  a  
we have L1
1
Hence L

1
s  a 
2
 e  at t
t
1
s s  a 
  e  at t dt
2
0


1
1  e  at 1  at  , on integratio n by parts.
2
a
Using this, we get
-1
L
1
s s  a 
2
2
1
 2
a

 1  e 1  at dt
t
 at
0


1
at 1  e  at   2e  at  1
a3
NEXT CLASS 24.05.05
Inverse transform of F(s) by using convolutio n theorem
We have, if L(t)  F(s) and Lg(t)  G(s), then
Lf(t)  g(t)   Lf (t )  Lg (t )  F ( s ) G ( s ) and so
t
L1 F ( s ) G ( s )  f (t )  g (t )   f t  u g u du
0
This expression is called the convolutio n theorem for inverse Laplace transform
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Examples
Employ convolution theorem to evaluate the following :
1
(1) L1
s  a s  b
1
1
Let us denote F(s) 
, G (s) 
sa
sb
f(t)  e -at , g(t)  e -bt
Taking the inverse, we get
Therefore, by convolutio n theorem,
t
1
-1
L
s  a s  b 
 e
 a t u  bu
e
du
e
0
s
s
2
 a2
e
 a b u
du

e bt  e  at
ab

2
Let us denote F(s) 
f(t) 
t
0
 e a b t  1
 e  at 

 ab 
(2) L1
 at
1
s
, G (s)  2
2
s a
s  a2
Then
2
sin at
, g(t)  cos at
a
Hence by convolutio n theorem,
L-1
s
s
2
 a2
t
1
  sin at  u cos au du
a
0

2
1 sin at  sin at  2au 
du ,
a 0
2
t

by using compound angle formula
1 
cosat  2au 
t sin at

u sin at 



2a 
 2a
2a
0
t
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(3) L1
s
s  1 s 2  1
Here


F(s) 
1
s
, G ( s)  2
s 1
s 1
Therefore
f(t)  e t , g(t)  sin t
By convolutio n theorem, we have
t
 e u

 sin u  cos u 
e 
 2
0
1
L
 e t u sin u du
s  1 s 2  1 
-1

t




e t t
1
e  sin t  cos t    1  e t  sin t  cos t
2
2

Assignment
By employing convolution theorem, evaluate the following :
(1) L1
(2) L1
(3) L1
1
s  1 s 2  1

s  1

s
2
s
1
2
a
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s  1
2

2 2
(4) L1
(5) L1
(6) L1
s2
,a  b
s 2  a 2 s2  b2


1
s 2 s  1
2
4s  5
s  12 s  2


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LAPLACE TRANSFORM METHOD FOR DIFFERENTIAL
EQUATIONS
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 derivatives
of function and then the inverse Laplace Transform.
The following are the expressions for the derivatives derived earlier.
L[f (t)]  s L f(t) - f(o)
L[f (t)  s 2 L f(t) - s f(o) - f (o)
L [f (t)  s 3 L f(t) - s 2 f(o) - s f (o) - f (o)
Examples
1) Solve by using Laplace transform method
y  y  t e t , y(o)  2
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Taking the Laplace transform of the given equation, we get
sL yt   yo   L yt   1 2
s  1
Using the given condition, this becomes
s  1L yt   2 
1
s  12
so that
L y t  
2s 2  4s  3
s  13
Taking the inverse Laplace transform , we get
Y t   L1
2s 2  4s  3
s  13
 2s  1  12  4s  1  1  3 
L 

s  13


1
 2
1 
 L1 

3
 s  1 s  1 

1 t 2
e t  4 
2
This is the solution of the given equation.
Solve by using Laplace transform method :
(2) y  2 y  3 y  sin t , y (o)  y(o)  0
Taking the Laplace transform of the given equation, we get
s
2

Ly (t )  sy (o)  y (o)  2s Ly (t )  y (o)  3 L y (t ) 
Using the given conditions, we get
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1
s 1
2
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

L y (t ) s 2  2 s  3 
or
L y (t ) 
1
s 1
2
1
s  1s  3 s 2  1

or



1
y (t )  L1 

2
 s  1s  3 s  1 


B
Cs  D 
 A
 L1 

 2
 s  1 s  3 s  1 
s 1


 
1
1
1
1  1
10
L 

 2 5
8
s

1
40
s

3
s 1 



by using the method of partial sums,
1
1
1
 et  e  3t  cos t  2 sin t 
8
40
10
This is the required solution of the given equation.
3) Employ Laplace Transform method to solve the integral equation.
t
f(t)  l   f u sin t  u du
0
Taking Laplace transform of the given equation, we get
t
L f(t) 
1
 L  f u sin t  u du
s
0
By using convolutio n theorem, here, we get
L f(t) 
1
1 L f (t )
 Lf (t )  L sin t   2
s
s s 1
Thus
L f (t ) 
s2 1
s3
 s2 1
t2
or f (t )  L1  3   1 
2
 s 
This is the solution of the given integral equation.
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d2x
6
dx
 25 x  0 where
dt
dt 2
x denotes the displaceme nt of the particle at time t. If the initial position of the particle is at x  20
and the initial speed is 10, find the displaceme nt of the particle at any time t using Laplace transform s.
(4) A particle is moving along a path satisfying , the equation
Given equation may be rewritten as
x' ' (t)  6x' (t)  25x(t)  0
Here the initial conditions are x(o)  20, x' (o)  10.
Taking the Laplace transform of the equation, we get


Lx(t) s 2  6s  25  20 s  130  0
Lx(t) 
or
20 s  130
s 2  6 s  25
so that
 20 s  130 
x(t)  L1 

2
 s  3  16 
 20 L1
 20 e
 3t
 20s  3  70 
 L1 

2
 s  3  16 
s3
1
 70 L1
2
s  3  16
s  32  16
e 3t sin 4t
cos 4t  35
2
This is the desired solution of the given problem.
(5) A voltage Ee -at is applied at t  0 to a circuit of inductance L and resistance R. Show that the
Rt


E  at
L
e  e 
R - aL 

current at any time t is
The circuit is an LR circuit. The differential equation with respect to the circuit is
di
 Ri  E (t )
dt
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
L
Thus, we have
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L
di
 Ri  Ee at
dt
or
Li ' (t )  R i (t )  Ee at
 
LLT i' (t)   RLT i' (t)   E LT e  at
or
Taking Laplace transform ( LT ) on both sides, we get
Ls LT i (t )  i (o)  RLT i (t )  E
LT i (t )sL  R  
Since i(o)  o, we get
LT i (t ) 
1
sa
or
E
s  a   sL  R 
Taking inverse transform L-T1 , we get

E
sa
i (t )  LT1
E
( s  a )( sL  R )
E  1 1
1 
LT
 L LT1

R  aL  s  a
sL  R 
Thus
i (t ) 

E   at
e  e L
R  aL 
Rt 


This is the result as desired.
(6) Solve the simultaneo us equations for x and y in terms of t given
dy
 9 x  0 with x(o)  2, y(o)  1.
dt
Taking Laplace transforms of the given equations, we get
s Lx(t) - x(o)   4Ly(t )  0
 9 L x(t )  s Ly(t )  y(o)  0
Using the given initial conditions , we get
s L x(t )  4 L y (t )  2
 9 L x(t )  5 L y (t )  1
Solving these equations for Ly(t), we get
s  18
L y(t)  2
s  36
so that
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dx
 4 y  0,
dt
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18 
 s
y(t)  L1  2
 2
 s  36 s  36 
 cos 6t  3 sin 6t
(1)
dy
 9 x  0, we get
dt
1
x(t)   6 sin 6t  18 cos 6t 
9
Using this in
or
2
3 cos 6t  sin 6t 
3
(2)
(1) and (2) together represents the solution of the given equations.
x(t) 
Assignment
Employ Laplace transform method to solve the following initial – value problems
1) y  2 y  y  2 y  0 given y (0)  y(0)  0, y(0)  6
2) y  5 y  6 y  e 2t , y (0)  2, y(0)  1
3) y  4 y  3 y  e  t , y (0)  1  y(0)
4) y  3 y  2 y  2t 2  2t  2, y (0)  2, y(0)  0
 
5) y  9 y  cos 2t , y (0)  1, y   1
2
6) y  2 y  y  t , y (0)  3, y (1)  1
7) y  y  H t  1, y (0)  0, y(0)  1
t
8) f (t )  1  2  f t  u e  2u du
0
t
9) f ' (t )  t   f t  u cos u du , f (0)  4
0
10) A particle moves along a line so that its displaceme nt x from
a fixed point at any time t is governed by the equation
d2x
dx
4
 5 x  80 sin 5t. If the particle is initially at rest, find
2
dt
dt
the displaceme nt at any time t.
11) In an L - R circuit, a voltage E sin t
is applied at t  0. If the current is zero initially, find the current at any time.
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Solve the following simultaneo us differenti al equations for
x and y in terms of t :
12)
dx
 y  sin t ,
dt
13)
dx
 y  et ,
dt
14)
15)
dy
 x  cos t ,
dt
x(0)  2, y (0)  0
dy
 x  sin t ,
dt
x(0)  1, y (0)  0
dx
 2 x  3 y,
dt
dy
 y  2 x,
dt
x(0)  8, y (0)  3
dx
 3 y  2 x,
dt
dy
 2 x  y,
dt
x(0)  8, y (0)  3
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 equation
subjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable are
specified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different values of the
independent 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
y  (0) = 1 is an initial value problem
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d 2 y dy

 y  x with conditions y(0) =
dx 2 dx
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Example 2 : The problem of solving the equation 3
d2y
dy
 2  y  cos x with y(1)=1,
2
dx
dx
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. Besides
the technique may also be employed to find certain integral values also. The transform is
named after the French Mathematician P.S. de’ Laplace (1749 – 1827).
The subject is divided into the following sub topics.
LAPLACE TRANSFORMS
Definition and
Properties
Transforms of
some functions
Convolution
theorem
Inverse
transforms
Solution of
differential
equations
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
e
 st
f (t )dt exists (converges). Then this integral is called the Laplace
0
transform of f(t) and is denoted by Lf(t).
Thus,

Lf(t) =  e  st f (t )dt
(1)
0
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 f (s ) .
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Thus,
Lf(t) = F(s)
(2)
Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denoted
by L-1 [F(s)].
Thus,
L-1 [F(s)] = f(t)
(3)
Suppose f(t) is defined as follows :
f(t) =
f1(t),
0<t<a
f2(t),
a<t<b
f3(t),
t>b
Note that f(t) is piecewise continuous. The Laplace transform of f(t) is defined as

Lf(t) =  e  st f (t )
0
a
= e
 st
b
f1 (t )dt   e
0
a
 st

f 2 (t )dt   e  st f 3 (t )dt
b
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)] =  e
0
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 st
af (t )  b (t )dt

= a e
0
 st

f (t )dt  b  e  st (t )dt
0
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= 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)] =
1 s
F   , where a is a positive
a a
constant.
Proof :- By definition, we have

Lf(at) =  e  st f (at )dt
(1)
0
Let us set at = x. Then expression (1) becomes,

s
1   x
L f(at) =  e  a  f ( x)dx
a0

1 s
F 
a a
This is the desired property.
3. Shifting property :- Let a be any real constant. Then
L [eatf(t)] = F(s-a)
Proof :- By definition, we have



L [eatf(t)] =  e  st e at f (t ) dt
0

=  e ( s a ) f (t )dt
0
= F(s-a)
This is the desired property. Here we note that the Laplace transform of eat f(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


0
0
L(eat) =  e  st e at dt   e ( s a )t dt
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
e  ( s a )t
1

=
,
 ( s  a) 0 s  a
s>a
Thus,
1
sa
L(eat) =
In particular, when a=0, we get
L(1) =
1
,
s
s>0
By inversion formula, we have
L1
4.
1
1
 e at L1  e at
sa
s
 e at  e  at
L(cosh at) = L
2


=





1
e ( s a )t  e ( s  a )t dt
2 0


1  st at
e e  e at dt

20
=

Let s > |a| . Then,

1  e  ( s a )t
e ( s  a )t 
L(cosh at )  


2   ( s  a)  ( s  a)  0
Thus,
L (cosh at) =
s
,
s  a2
2
s > |a|
and so
 s 
L1  2
 cosh at
2 
s a 
 e at  e  at
3. L (sinh at) = L
2


a
  2
,
2
 s a
Thus,
L (sinh at) =
a
,
s  a2
2
and so,
 1  sinh at
L1  2

2 
a
s a 
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s > |a|
s > |a|
=
s
s  a2
2
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
4. L (sin at) =  e  st sin at dt
0
Here we suppose that s > 0 and then integrate by using the formula
e ax
 e sin bxdx  a 2  b 2 a sin bx  b cos bx
ax
Thus,
L (sinh at) =
a
, s>0
s  a2
2
and so
 1  sinh at
L1  2

2 
a
s a 

5. L (cos at) =  e  st cos atdt
0
Here we suppose that s>0 and integrate by using the formula
ax
 e cos bxdx 
e ax
a cos bx  b sin bx
a2  b2
Thus,
L (cos at) =
s
,
s  a2
2
s>0
and so
L1
s
 cos at
s  a2
2
6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then

L(t ) =  e  st t n dt
n
0
Let s > 0 and set st = x,

then
n

1
 x  dx
L(t )   e  
 n 1  e  x x n dx
s 0
s s
0
n

x
The integral  e  x x n dx is called gamma function of (n+1) denoted by  ( n  1) . Thus
0
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L(t n ) 
(n  1)
s n 1
In particular, if n is a non-negative integer then  ( n  1) =n!. Hence
L(t n ) 
n!
s n1
and so
L1
1
s n1

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tn
tn
or
as the case may be
n!
(n  1)
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TABLE OF LAPLACE TRANSFORMS
f(t)
F(s)
1
1
, s>0
s
eat
1
, s>a
sa
coshat
s
, s > |a|
s  a2
sinhat
a
, s > |a|
s  a2
sinat
a
, s>0
s  a2
cosat
s
, s>0
s  a2
2
tn, n=0,1,2,…..
tn, n > -1
2
2
2
n!
s n 1
, s>0
(n  1)
, s>0
s n1
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 :
 s 
1. L(e at cosh bt )  L(cosh bt )ss a   2
2 
 s  b  s s  a
Thus,
L(eatcoshbt)
=
sa
(s  a) 2  b 2
and
L1
sa
 e at cosh bt
2
2
( s  a)  b
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=
sa
(s  a) 2  b 2
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2. L(e at sinh bt ) 
a
( s  a) 2  b 2
and
L1
1
 e at sinh bt
2
2
( s  a)  b
3. L(e at cos bt ) 
sa
( s  a) 2  b 2
and
sa
 e at cos bt
( s  a) 2  b 2
L1
4. L(e at sin bt ) 
b
( s  a) 2  b 2
and
L1
1
e at sin bt

b
( s  a) 2  b 2
5. L(e at t n ) 
(n  1)
( s  a ) n 1
or
n!
( s  a ) n 1
as the case may be
Hence
L1
1
e at t n

( s  a) n 1 (n  1)
n!
as the case may be
( s  a ) n 1
or
Examples :1. Find Lf(t) given f(t) =
t,
0<t<3
4,
t>3
Here

Lf(t) =  e
 st
0
3

f (t )dt   e tdt   4e  st dt
 st
0
3
Integrating the terms on the RHS, we get
Lf(t) =
1 3 s 1
e  2 (1  e 3 s )
s
s
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This is the desired result.
2. Find Lf(t) given f(t) =
sin2t, 0 < t  
0,
t>
Here


0


=  e  st sin 2tdt
Lf(t) =  e  st f (t )dt   e  st f (t )dt
0

 e  st

 s sin 2t  2 cos 2t
=  2
s  4
0
=

2
1  e s
s 4
2

This is the desired result.
3. Evaluate : (i)
L(sin3t sin4t)
(iv)
L(cos2 4t)
(v)
L(sin32t)
(i) Here
1
L(sin3t sin4t) = L [ (cos t  cos 7t )]
2
=
1
L(cos t )  L(cos 7t ) , by using linearity property
2
=
1 s
s 
24s
 2
 2
2


2  s  1 s  49  (s  1)( s 2  49)
(ii) Here
s 
1
 1 1
L(cos24t) = L  (1  cos 8t )    2
2
 2  s s  64 
(iii) We have
sin 3  
For =2t,
sin 3 2t 
1
3sin   sin 3 
4
we get
1
3 sin 2t  sin 6t 
4
so that
L(sin 3 2t ) 
1 6
6 
48
 2
 2
2


4  s  4 s  36  ( s  4)( s 2  36)
This is the desired result.
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4. Find
L(cost cos2t cos3t)
Here
cos2t cos3t =
1
[cos 5t  cos t ]
2
so that
cost cos2t cos3t =
=
1
[cos 5t cos t  cos 2 t ]
2
1
[cos 6t  cos 4t  1  cos 2t ]
4
Thus
L(cost cos2t cos3t) =
1 s
s
1
s 
 2
  2
2

4  s  36 s  16 s s  4 
L(cosh22t)
5. Find
We have
cosh 2  
1  cosh 2
2
For  = 2t, we get
cosh 2 2t 
1  cosh 4t
2
Thus,
L(cosh 2 2t ) 
1 1
s 

2
2  s s  16 
6. Evaluate (i) L( t )
We have
(i) For n=
L(tn) =
1
, we get
2
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 1 
(ii) L  (iii) L(t-3/2)
 t
(n  1)
s n 1
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1
(  1)
2
L(t1/2) =
s3 / 2

1  1 1
Since (n  1)  n(n) , we have   1    
2
 2  2  2
Thus,

L( t ) 
2s
(ii) For n = -
3
2
1
, we get
2
1
 

1
2
L(t 2 )  1  
s
s 2
(iii) For n = -
3
, we get
2
 1
  
2 
3
2
L(t 2 )    1    1  2 s
s 2
s 2
7. Evaluate : (i) L(t2)
(ii) L(t3)
We have,
L(tn) =
n!
s n 1
(i) For n = 2, we get
L(t2) =
2! 2

s3 s3
(ii) For n=3, we get
L(t3) =
3! 6

s4 s4
8. Find L[e-3t (2cos5t – 3sin5t)]
Given
=
2L(e-3t cos5t) – 3L(e-3t sin5t)
=2
=
s  3
( s  3)  25
2

15
, by using shifting property
( s  3) 2  25
2s  9
, on simplification
s  6 s  34
2
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9. Find L[coshat sinhat]
Here


 eat  e at

L[coshat sinat] = L 
sin at 
2


=

1
a
a


2
2
2
2
2  ( s  a)  a
( s  a)  a 

a ( s 2  2a 2 )
, on simplification
[( s  a) 2  a 2 ][( s  a) 2  a 2 ]
10. Find L(cosht sin3 2t)
Given
 et  e t  3 sin 2t  sin 6t 

L 

2
4




 


=
1
3  L et sin 2t  L(et sin 6t )  3L(e  t sin 2t )  L(e  t sin 6t )
8
=

1
6
6
6
6




2
2
2
2
8  ( s  1)  4 ( s  1)  36 ( s  1)  4 ( s  1)  36 
=

3
1
1
1
1




2
2
2
2
4  ( s  1)  4 ( s  1)  36 ( s  1)  24 ( s  1)  36 
11. Find L(e  4t t
5
2
)
We have
L(tn) =
(n  1)
s n 1
L(t-5/2) =
Put n= -5/2. Hence
(3 / 2) 4 
  3 / 2 Change s to s+4.
s 3 / 2
3s
Therefore,
L (e  4 t t  5 / 2 ) 
4 
3( s  4)  3 / 2
Transform of tn f(t)
Here we suppose that n is a positive integer. By definition, we have

F(s) =  e st f (t )dt
0
Differentiating ‘n’ times on both sides w.r.t. s, we get
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
dn
n
F
(
s
)

e st f (t )dt
ds n
s n 0
Performing differentiation under the integral sign, we get

dn
F ( s)   (t ) n e st f (t )dt
ds n
0
Multiplying on both sides by (-1)n , we get

dn
(1)
F ( s)   (t n f (t )e st dt  L[t n f (t )] , by definition
ds n
0
n
Thus,
dn
L[t f(t)]= (1)
F ( s)
ds n
n
n
This is the transform of tn f(t).
Also,
 dn

L  n F ( s)  (1) n t n f (t )
 ds

1
In particular, we have
L[t f(t)] = 
L[t2 f(t)]=
d
F (s ) , for n=1
ds
d2
F ( s ) , for n=2, etc.
ds 2
Also,
d

L1 
F ( s)  tf (t ) and
 ds

 d2

L1  2 F ( s)  t 2 f (t )
 ds

Transform of
f(t)
t

We have , F(s) =  e st f (t )dt
0
Therefore,

  st

F
(
s
)
ds

f (t )dt  ds

  e
s
s 0

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
=

0


f (t )   e  st ds dt
s

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
 e  st 
f (t )   dt
  t s

=

0

 f (t ) 
 f (t ) 
=  e st 
dt  L


 t 
 t 
0
Thus,

 f (t ) 
L
   F (s)ds
 t  s
This is the transform of
f (t )
t
Also,

L1  F ( s)ds 
s
f (t )
t
Examples :
1. Find L[te-t sin4t]
We have,
L[e t sin 4t ] 
4
( s  1) 2  16
So that,
 d 
1

L[te-t sin4t] = 4  2

 ds  s  2s  17 
=
8( s  1)
( s  2s  17) 2
2
2. Find L(t2 sin3t)
We have
L(sin3t) =
3
s 9
2
So that,
L(t2 sin3t) =
d2  3 


ds 2  s 2  9 
 et sin t 

3. Find L
 t 
We have
L(e t sin t ) 
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1
( s  1) 2  1
= 6
d
s
2
ds ( s  9) 2
=
18( s 2  3)
( s 2  9) 3
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 et sin t 
 =
Hence L
 t 
=

2

ds
 (s  1)
0
2
1



 tan 1 ( s  1) s
 tan 1 ( s  1) = cot –1 (s+1)
 sin t 
 sin at 
4. Find L
 . Using this, evaluateL 

 t 
 t 
We have
L(sint) =
1
s 1
2
So that
 sin t 
Lf(t) = L
 =
 t 
=




ds
1

tan
s
S
s s 2  1

2
 tan 1 s  cot 1 s  F ( s )
Consider
 sin at 
 sin at 
L
 = a L
  aLf (at )
 t 
 at 
 1  s 
= a  F   , in view of the change of scale property
 a  a 
s
= cot 1  
a
 cos at  cos bt 
5. Find L 

t

We have
L [cosat – cosbt] =
s
s
 2
2
s a
s  b2
2
So that

1   s 2  a 2 

= log  2
2   s  b 2  s

s 
 s
 cos at  cos bt 
L
= 2
 2
ds
2

t
s  b 2 

 s s  a
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=
 s2  a2 
 s 2  a 2 
1


 2

Lt
log

log

2 
2  s   s 2  b 2 
 s  b 
=
 s 2  b 2 
1

0  log  2
2 
2
 s  a 
=
1  s2  b2 

log 
2  s 2  a 2 
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
6. Prove that
e
3t
t sin tdt 
0
3
50
We have

e
 st
t sin tdt  L(t sin t )
=
0
=
2s
( s  1) 2
Putting s = 3 in this result, we get

e
3t
t sin tdt 
0
3
50
This is the result as required.
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d  1 
d
L(sin t ) =   2
ds
ds  s  1
2
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ASSIGNMENT
I
Find L f(t) in each of the following cases :
1. f(t) =
et , 0 < t < 2
0, t>2
2. f(t) = 1 , 0  t  3
t, t>3
t
, 0t<a
a
3. f(t) =
, ta
2
II. Find the Laplace transforms of the following functions :
4. cos(3t + 4)
5. sin3t sin5t
6. cos4t cos7t
7. sin5t cos2t
8. sint sin2t sin3t
9. sin2 5t
10. sin2(3t+5)
11. cos3 2t
12. sinh2 5t
13. t5/2
1 

14.  t  
t

3
15. 3t
16. 5-t
17. e-2t cos2 2t
18. e2t sin3t sin5t
19. e-t sin4t + t cos2t
20. t2 e-3t cos2t
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21.
1  e 2t
t
e  at  e bt
22.
t
23.
sin 2 t
t
24.
2 sin 2t sin 5t
t
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
is s. If the actual distance between P and Q is C. Then s=C in the limit Q  P along C.
y
Q
s
s
P(s, )
A

x
0
i.e.
s
1
QP  C
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Cartesian Form:
Q ( x   x, y   y )
C s
P ( x, y )
Let y  f ( x) be the Cartesian equation of the curve C and let P( x, y ) and Q( x   x, y   y )
be any two neighboring points on it as in fig.
Let the arc length PQ   s and the chord length PQ   C . Using distance between two
points formula we have PQ2 = ( C)2 =( x)2 +( y)2
 C 
 y 

  1  
 x 
 x 
2

2
C
 y 
or
 1  
x
 x 
 s  s C  s
 y 



1 

 x C  x C
x 
2
2
We note that x  0 as Q  P along C, also that when Q  P,
s
1
C
 When Q  P i.e. when x  0, from (1) we get
ds
 dy 
 1  
dx
 dx 
2
 (1)
Similarly we may also write
 x 
 s  s  C s



1   
y  C y C
 y 
2
and hence when Q  P this leads to
 dx 
ds
 1  
dy
 dy 
2
 (2)
Parametric Form: Suppose x  x(t ) and y  y (t ) is the parametric form of the curve C.
Then from (1)
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2
 dy 
2
2


ds
1
 dx   dy 
 1   dt  

   
dx
dx
 dt   dt 
 dx dt 
dt


2
ds ds dx
 dx   dy 
       
dt dx dt
 dt   dt 
2
 (3)
Note: Since  is the angle between the tangent at P and the X-axis,
we have

dy
 tan
dx
ds
2
 1   y   1  tan 2   Sec
dx
Similarly
ds
1
1
 1
 1
 1  cot 2   Co sec
2
2
dy
tan 
 y 
i.e. Cos  
2
2
dx
dy
and Sin  
ds
ds
 dx   dy 
2
2
2
       1   ds    dx    dy 
 ds   ds 
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We can use the following figure to observe the above geometrical connections
among dx, dy, ds and  .
dy
ds

dx
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:
C
Q(r   r ,    )
N
s
P(r,)

r

O
Consider PN
x
 OQ.
In the right-angled triangle OPN, We have Sin  
PN PN

 PN  r Sin   r
OP
r
since Sin = when  is very small.
From the figure we see that, Cos  
ON ON

 ON  r Cos   r (1)  r
OP
r
cos   1 when   0
 NQ  OQ  ON  (r   r )  r   r
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From
PNQ, PQ2  PN 2  NQ2 i.e, ( C )2  (r )2  ( r )2
C
 S  S C  S 2  r 
r 

 r2  


r 
 


  C   C
  
  
2
We note that when Q  P along the curve,   0 also
2
S
1
C
2
dS
 dr 
 when Q  P,
 r2  
  (4)
d
 d 
C
  
Similarly, ( C )  (r )  ( r ) 
 1 r2 

r
r 
2
2
2
2
 S  S C  S
  
and


1 r2 

r C r C
r 
2
dS
 d 
 1 r2 
  (5)
dr
 dr 
2
 when Q  P, we get
Note:
We know that Tan   r
d
dr
2
ds
 dr 
2
2
2
2

 r2  
  r  r Cot   r 1  Cot   rCo sec 
d
 d 
Similarly
ds
 d 
2
 1 r2 
  1  Tan   Sec
dr
 dr 
2

dr
 Cos
ds
and
d 1
 Sin
ds r
The following figure shows the geometrical connections among ds, dr, d and 
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ds
r d

dr
Thus we have :
2
 dx 
ds
ds
 dx   dy 
 1   ,
    
dy
dt
 dt   dt 
 dy 
2
ds
 dy 
 1   ,
dx
 dx 
ds
 d 
 1 r2 

dr
 dr 
Example 1:
x
2/3
y
2/3
a
2
2
ds
 dr 
and
 r2  

d
 d 
2
2
ds
ds
and
for the curve x 2/3  y 2/3  a 2/3
dx
dy
2/3
-1
3
1
2
2
x
 y 3
 x -1/3  y -1/3 y '  0  y '  -1    
3
3
x
y3
2
x2 / 3  y 2 / 3
a2 / 3  a 

 
x2 / 3
x2 / 3  x 
2
x2 / 3  y 2 / 3

y2/3
ds
y2/3
 dy 
Hence
 1    1 2/3 
dx
x
 dx 
1/ 3
Similarly
 dx 
ds
x2 / 3
 1    1 2/3 
dy
y
 dy 
Example 2: Find
 a2 
ds
for the curve y  a log  2 2 
dx
 a -x 
y  a log a 2  a log  a 2  x 2  
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1/ 3
a2 / 3  a 
 
y2/3  y 
dy
2ax
 2 x 
 a  2
 2
2 
2
dx
a x  a x
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2
ds
4a 2 x 2
 dy 
  1    1 2

dx
 dx 
 a  x2 
a
2
 x 2   4a 2 x 2
a
 x2 
2
2

a
a
2
2
 x2 
 x2 
2
2

a2  x2
a2  x2
Example 3: If x  ae t Sint, y  ae t Cost, find ds
dt
x  ae tSint 
dx
 ae tSint  ae t Cost
dt
y  ae t Cost 
dy
 ae t Cost  ae tSint
dt
2
2
ds
2
2
 dx   dy 
        a 2 e2t  Cos t  Sin t   a 2e 2t  Cos t  Sin t 
dt
 dt   dt 
 aet 2  Cos 2t  Sin2t   a 2 et
 a  b   a  b
2
2
 2  a 2  b2 
t
ds

Example 4: If x  a Cos t  log Tan  , y  a Sin t, find
2
dt




Sec 2 t 


dx
1
2   a  Sin t 
 a   Sin t 

t
t
dt

2  Tan t 


2
Sin
Cos

2

2
2
1  Sin 2t  a Cos 2t


1 
 a   Sin t 
a

 a Cost  Cot t
Sin t 
Sin t
Sin t

dy
 a Cos t
dt
2
2
ds
 dx   dy 
        a 2Cos 2t Cot 2t  a 2Cos 2t 
dt
 dt   dt 
a 2Cos 2t  Cot 2t  1
 a 2Cos 2t  Co sec2 t  a 2Cot 2t  a Cot t
Example 5: If x  a Cos3 t, y  Sin 3 t, find
ds
dt
dx
dy
 3 a Cos 2t Sin t ,
 3 a Sin 2t Cos t
dt
dt
2

ds
 dx   dy 
    
dt
 dt   dt 
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2
 9a 2Cos 4t Sin 2t  9a 2 Sin 4t Cos 2t
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 9a 2Cos 2t Sin2t  Cos 2t  Sin2t   3 a Sin t Cos t
Exercise:
(i)
ds
for the following curves:
dt
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
Find
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ds
is constant
d
dr
dr a 2
2
2
2
r  a Cos 2  2r
 2a Sin2 

Sin2
d
d
r
Example 6: If r 2  a 2 Cos 2 , Show that r
2
ds
a4
1 4
 dr 
2
2
2
 r 
r  a 4 Sin 2 2
  r  2 Sin 2 
d
r
r
 d 
r
ds
 r 4  a 4 Sin 2 2  a 4Cos 2 2  a 4 Sin 2 2
d
 a2 Cos 2  Sin2 2  a2  Constant
r
ds
is constant for r 2  a 2Cos2
d
k2  r2
ds
r
, Show that r is constant.
Example 7: For the curve   Cos -1   
r
dr
k
 2r 
2
2
r
  k  r (1)
2
2
r2  k 2  r2 
1
d
1
1
2 k r 




 
dr
r2
k2  r2
r2 k2  r2
r2 k
1 2
k

1
k2  r2
ds
 d
  1 r2 
dr
 dr
Hence r

k2
r2 k2  r2

r 2  k 2
r2 k2  r2
ds
 k (constant)
dr
We know that Cos  
ds

dr
Also
r
r  p2
2
ds
r
r
r2



d Sin  p
p
r
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ds
r
ds r 2

,

dr
p
r 2  p 2 d
dr
d 1
and
 Sin 
ds
ds r
dr
p2
 Cos  1  Sin 2  1  2 
ds
r

k2  r2
r
k 2  r2 
r2  k2  r2 k

2 
2

1

r



r2
r
r

Example 8: For a polar curve r  f   show that


r 2  p2
r
p  rSin
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Exercise:
Find
ds
ds
and
dr
d
(i)
r=a(1+ Cos )
(ii)
r m =a m Cosm
(iii)
r  aSec 2 ( )
2
for the following curves:

Mean Value Theorems
Recapitulation
Closed interval: An interval of the form a  x  b , that includes every point between a and b
and also the end points, is called a closed interval and is denoted by a, b .
Open Interval: An interval of the form a  x  b , that includes every point between a and b
but not the end points, is called an open interval and is denoted by a, b
Continuity: A real valued function f (x) is said to be continuous at a point x0 if
lim f ( x)  f ( x0 )
x  x0
The function f (x) is said to be continuous in an interval if it is continuous at every point in
the interval.
Roughly speaking, if we can draw a curve without lifting the pen, then it is a continuous
curve otherwise it is discontinuous, having discontinuities at those points at which the curve
will 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, product
of continuous functions is continuous. The quotient of continuous functions is continuous at
all those points at which the denominator does not become zero.
Differentiability: A real valued function f (x) is said to be differentiable at point x0 if
f ( x)  f ( x0 )
exists uniquely and it is denoted by f ' ( x0 ) .
x  x0
x  x0
lim
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A real valued function f(x) is said to be differentiable in an interval if it is differentiable at
f ( x  h)  f ( x)
every in the interval or if lim
exists uniquely. This is denoted by f ' ( x) .
h
h 0
We say that either f ' ( x) exists or f(x) is differentiable.
Geometrically, it means that the curve is a smooth curve. In other words a curve is said to be
smooth if there exists a unique tangent to the curve at every point on it. For example a circle
is a smooth curve. Triangle, rectangle, square etc are not smooth, since we can draw more
number of tangents at every corner point.
We note that if a function is differentiable in an interval then it is necessarily continuous in
that interval. The converse of this need not be true. That means a function is continuous need
not imply that it is differentiable.
Rolle’s Theorem: (French Mathematician Michelle Rolle 1652-1679)
Suppose a function f (x) satisfies the following three conditions:
f (x) is continuous in a closed interval a, b
(i)
f (x) is differentiable in the open interval a, b
(ii)
f (a )  f (b)
(iii)
Then there exists at least one point c in the open interval a, b such that f ' (c)  0
Geometrical Meaning of Rolle’s Theorem: Consider a curve f (x) that satisfies the
conditions of the Rolle’s Theorem as shown in figure:
y
[b,f(b)]
[a,f(a)]
x
a
c
b
As we see the curve f (x) is continuous in the closed interval a, b , the curve is smooth i.e.
there can be a unique tangent to the curve at any point in the open interval a, b and
also f (a )  f (b) . Hence by Rolle’s Theorem there exist at least one point c belonging to
a, b such that f ' (c)  0 . In other words there exists at least one point at which the tangent
drawn 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 a, b such that …….
We present the same in mathematical symbols as:  c  a, b : f ' (c)  0
From now onwards let us start using the symbolic notation.
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Example 1: Verify Rolle’s Theorem for f ( x)  x 2 in  1,1
First we check whether the conditions of Rolle’s theorem hold good for the given function:
(i)
(ii)
(iii)
f ( x)  x 2 is an elementary algebraic function, hence it is continuous every
where and so also in  1,1.
f ' ( x)  2 x exists in the interval (-1, 1) i.e. the function is differentiable in (-1,
1).
Also we see that f (1)  (1) 2  1 and f (1)  12  1 i.e., f (1)  f (1)
Hence the three conditions of the Rolle’s Theorem hold good.
 By Rolle’s Theorem  c   1,1 : f ' (c)  0 that means 2c = 0  c= 0   1,1
Hence Rolle’s Theorem is verified.
Example 2: Verify Rolle’s Theorem for f ( x)  x( x  3)e  x / 2 in [-3, 0]
f (x) is a product of elementary algebraic and exponential functions which are continuous
and hence it is continuous in [3, 0]
1
f ( x)  (2 x  3)e  x / 2  ( x 2  3x)e  x / 2
2
1
 [( 2 x  3)  ( x 2  3 x)]e  x / 2
2
1 2
1
  [ x  x  6]e  x / 2   ( x  3)( x  2)e  x / 2 exists in (-3, 0)
2
2
f (3)  0 and also f (0)  0  f (3)  f (0)
That is, the three conditions of the Rolle’s Theorem hold good.
  c   3,0 : f ' (c)  0
1
  (c  3)(c  2)e  c / 2  0  c  3,2, 
2
Out of these values of c, since 2  (3, 0) , the Rolle’s Theorem is verified.
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Example 3: Verify Rolle’s Theorem for f ( x)  ( x  a) m ( x  b) n in [a, b] where a <b and a,
b>0.
f (x) is a product of elementary algebraic functions which are continuous and hence it is
continuous in [a, b]
f ( x)  m( x  a) m 1 ( x  b) n  n( x  a) m ( x  b) n 1
 [m( x  b)  n( x  a)]( x  a) m 1 ( x  b) n 1
 [ x(m  n)  (mb  na)]( x  a) m 1 ( x  b) n 1 exists in a, b
f (a)  0 and also f (b)  0  f (a)  f (b)
Hence the three conditions of the Rolle’s Theorem hold good.
  c   a, b  : f ' (c)  0
 [c(m  n)  (mb  na)]( c  a) m 1 (c  b) n 1  0  c 
Out of these values of c, since c 
mb  na
, a, b
mn
mb  na
 ( a, b) , the Rolle’s Theorem is verified.
mn
Example 4: Verify Rolle’s Theorem for f ( x)  log
x 2  ab
in [a, b]
x ( a  b)
f ( x)  log( x 2  ab)  log x  log( a  b) is the sum of elementary logarithmic functions
which are continuous and hence it is continuous in [a, b]
f ( x) 
2x
x 2  ab
f (a)  log

1
exists in a, b
x
a 2  ab
 log
a ( a  b)
and similarly f (b)  log
a 2  ab
a 2  ab
 log 1  0
b 2  ab
b 2  ab
 log
 log 1  0
b(a  b)
b 2  ab
Hence the three conditions of the Rolle’s Theorem hold good.   c  a, b : f ' (c)  0
f (c) 
2c
c 2  ab

1
2c 2  c 2  ab
0
 0  c 2  ab  0  c  ab
2
c
c(c  ab)
Since c  ab  ( a, b) , the Rolle’s Theorem is verified.
Exercise: Verify Rolle’s Theorem for
(i)
  5 
f ( x)  e x (sin x  cos x) in  ,  ,
4 4 
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(ii)
,
(iii)
f ( x)  x( x  2)e x / 2 in 0,2
f ( x) 
sin 2 x
e
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2x
  5 
in  ,  .
4 4 
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Lagrange’s Mean Value Theorem (LMVT):
(Also known as the First Mean Value Theorem)
(Italian-French Mathematician J. L. Lagrange 1736-1813)
Suppose a function f (x) satisfies the following two conditions:
f (x) is continuous in a closed interval a, b
(i)
(ii)
f (x) is differentiable in the open interval a, b
Then there exists at least one point c in the open interval a, b such that f ' (c) 
f (b)  f (a )
ba
Proof: Consider the function  (x) defined by  ( x)  f ( x)  kx where k is a constant to be
found such that  (a)   (b) .
Since f (x) is continuous in the closed interval a, b ,  (x) which is a sum of continuous
functions is also continuous in the closed interval a, b
 ' ( x)  f ' ( x)  kx  (1) exists in the interval a, b as f (x) is differentiable in a, b .
We have  (a)  f (a)  ka and  (b)  f (b)  kb
 (a)   (b)  f (a)  ka  f (b)  kb  k 
f (b)  f (a)
 (2)
ba
This means that when k is chosen as in (2) we will have  (a)   (b)
Hence the conditions of Rolle’s Theorem hold good for  (x) in a, b
 By Rolle’s Theorem  c   a, b  :  '(c)  0
 '(c)  0  f ' (c)  k  0  k  f (c)  (3)
From (2) and (3), we infer that  c   a, b  : f ' (c) 
f (b)  f (a )
ba
Thus the Lagrange’s Mean Value Theorem (LMVT) is proved.
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Geometrical Meaning of Lagrange’s Mean Value Theorem: Consider a curve f (x) that
satisfies the conditions of the LMVT as shown in figure:
y
[b,f(b)]
[a,f(a)]
x
a
c
b
From the figure, we observe that the curve f (x) is continuous in the closed interval a, b ; the
curve is smooth i.e. there can be a unique tangent to the curve at any point in the open
interval a, b . Hence by LMVT there exist at least one point c belonging to a, b such
f (b)  f (a )
that f ' (c) 
. In other words there exists at least one point at which the tangent
ba
drawn to the curve lies parallel to the chord joining the points  a, f (a) and b, f (b) .
Other form of LMVT: The Lagrange’s Mean Value Theorem can also be stated as follows:
Suppose f (x) 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) 
On simplification this becomes f (a  h)  f (a)  hf (a   h) .
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f ( a  h)  f ( a )
( a  h)  a
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Example 5: Verify LMVT for f ( x)  ( x  1)( x  2)( x  3) in [0, 4]
f ( x)  ( x  1)( x  2)( x  3)  x3  6 x 2  11x  6 is an algebraic function hence it is continuous
in [0, 4]
f ( x)  3x 2  12 x  11exists 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] .
f (4)  f (0)
Hence c  (0, 4) : f (c) 
40
i.e., 3c 2  1c  11 
(3)(2)(1)  (1)(2)(3)
40
i.e, 3c 2  12c  11  3  3c 2  12c  8  0  c  2 
2
 (0, 4)
3
Hence the LMVT is verified.
Example 6: Verify LMVT for f ( x)  loge x in [1, e]
f ( x)  loge x is an elementary logarithmic function hence continuous in [1, e] .
f ( x ) 
1
exists in 1, e  or f ( x) is differentiable in 1, e  .
x
i.e., both the conditions of LMVT hold good for f ( x) in [1, e]
f (e)  f (1)
1 log e  log1
 
Hence c  (1, e) : f (c) 
e 1
c
e 1

1
1

 c  e  1 (1, e)
c e 1
Hence the LMVT is verified.
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Example 7: Verify LMVT for f ( x)  sin 1 x in [0,1]
1
exists in  0,1 , and hence f ( x) is continuous in [0,1] .
1  x2
i.e., both the conditions of LMVT hold good for f ( x) in [0,1]
We find f ( x) 
Hence c  (0,1) : f (c) 

1
1  c2
c 


2
f (1)  f (0)
1
sin 1 1  sin 1 0


1 0
1
1  c2
2
 1 c 
2

2
c 
2 4
2
2 4
 0.7712  (0,1)

Hence the LMVT is verified.
Example 8: Find  of LMVT for f ( x)  ax 2  bx  c or
Verify LMVT by finding  for f ( x)  ax 2  bx  c
f ( x)  ax 2  bx  c is an algebraic function hence continuous in [a, a  h] .
f ( x)  2ax  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) : f (a  h)  f (a)  hf (a   h)  (1)
Here ; f (a  h)  a(a  h)2  b(a  h)  c  a3  ah2  2a 2h  ab  bh  c  (2)
f (a)  a3  ab  c  (3) ; f (a   h)  2a(a   h)  b  2a2  2a h  b  (4)
Using (2), (3) and (4) in (1) we have:
a3  ah2  2a 2 h  ab  bh  c  a3  ab  c  h(2a 2  2a h  b)   
Hence the LMVT is verified.
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1
 (0,1)
2
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Example 9: Find  of LMVT for f ( x)  loge x in  e, e 2 


f ( x)  loge x is an elementary logarithmic function hence continuous in  e, e 2 


1
f ( x )  exists in e, e2 .
x
 
Hence by the second form of LMVT
   0,1 : f (e2 )  f (e)  (e2  e) f  e   (e 2  e) 


 log e2  log e 
( e 2  e)
e   (e 2  e)
e(e  1)
(e  1)
i.e., 2 log e e  log e e 
 2 1 
e 1   (e  1) 
1   (e  1)
 1   (e  1)  e  1   
e2
  0,1
e 1
Hence the LMVT is verified.
Example 10: If x  0 , Apply Mean Value Theorem to show that
1
1
x
(i )
 log e (1  x)  x and 0 
 1
1 x
log e (1  x) x
Consider f ( x)  loge (1  x) in [0, x] . We see that f ( x) is an elementary logarithmic function
1
hence continuous in [0, x] . Also f ( x) 
exists in  0, x  .
1 x
 by LMVT   (0,1) : f ( x)  f (0)  xf (0   x)
i.e., log(1  x)  log1 
x
x
 log(1  x) 
 (1)
1 x
1 x
Since x  0 and 0    1 , we have 0   x  x  1  1   x  1  x
1
1
1
1
1
x
x



1

 x  (2)
1 x 1 x 1 x 1 x
1 x 1 x
x
 log e (1  x)  x
1 x
1
1 x
1
1
Also from (1) we have,


 
log(1  x)
x
log(1  x) x
From (1) and (2) we get
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Since 0    1 , we can write 0 
1
1
  1.
log(1  x) x
Example 11: Apply Mean Value Theorem to show that
ba
1 a
2
 sin 1 b  sin 1 a 
ba
1 b
2
, where 0  a  b
Consider f ( x)  sin 1 x in [a, b]
1
exists in  a, b 
1  x2
Hence f ( x) is differentiable in  a, b  and also it is continuous in [a, b] .
We see that f ( x) 
Therefore, by LMVT c  (a, b) : f (c) 
f (b)  f (a)
1
sin 1 b  sin 1 a


 1
2
ba
b

a
1 c
Since a  c  b  a 2  c 2  b2  a 2  c 2  b 2  1  a 2  1  c 2  1  b 2

1
1 a
2

1
1 c
2

1
1 b
2
or
1
1  a2

1
1  c2

1
1  b2
  2
From (1) and (2), we have
1
1  a2

sin 1 b  sin 1 a
1
ba
ba


 sin 1 b  sin 1 a 
ba
1  b2
1  a2
1  b2
Exercise: Verify the Lagrange’s Mean Value Theorem for
(i)
 1
f ( x)  x( x  1)( x  2) in  0, 
 2
(ii)
f ( x)  Tan1x in 0,1
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Mean Value Theorems
Cauchy’s Mean Value Theorem (CMVT):
(French Mathematician A. L. Cauchy 1789-1857)
Suppose two functions f (x) and g ( x ) satisfies the following conditions:
(i)
f (x) and g ( x ) are continuous in a closed interval a, b
(ii)
f (x) and g ( x ) are differentiable in the open interval a, b
(iii)
g ( x)  0 for all x.[ for all can be denoted by the symbol  ]
Then there exists at least one point c in the open interval a, b
f '(c) f (b)  f (a)
such that

g '(c) g (b)  g (a)
Proof: Consider the function  (x) defined by  ( x)  f ( x)  k g ( x) where k is a constant to
be found such that  (a)   (b) .
Since f (x) and g ( x ) are continuous in the closed interval a, b ,  (x) which is a sum of
continuous functions is also continuous in the closed interval a, b
 ' ( x)  f ' ( x)  k g ( x)  (1) exists in the interval a, b as f (x) and g ( x ) are differentiable
in a, b .
We have  (a)  f (a)  kg (a) and  (b)  f (b)  k g (b)
 (a)   (b)  f (a)  kg (a)  f (b)  k g (b)  k 
f (b)  f (a)
 (2)
g (b)  g (a)
This means that when k is chosen as in (2) we will have  (a)   (b)
Hence the conditions of Rolle’s Theorem hold good for  (x) in a, b
 By Rolle’s Theorem  c   a, b  :  '(c)  0
 '(c)  0  f ' (c)  k g (c)  0  k 
f (c)
 (3)
g (c)
From (2) and (3), we infer that  c   a, b  :
f '(c) f (b)  f (a)

g '(c) g (b)  g (a)
Thus the Cauchy’s Mean Value Theorem (CMVT) is proved.
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Example 12: Verify the Cauchy’s MVT for f ( x)  x 2 and g ( x)  x 4 in [a, b]
f ( x)  x 2 and g ( x)  x 4 are algebraic polynomials hence continuous in [a, b]
f ( x)  2 x and g ( x)  4 x3 exist in  a, b 
also we see that g ( x)  0 for all x  (a, b) since 0  a  b
i.e., the conditions of CMVT hold good for f ( x) and g ( x ) in [a, b] .
f '(c) f (b)  f (a)
Hence  c   a, b  :

g '(c) g (b)  g (a)
i.e.,
2c
4c3

b2  a 2
b4  a 4

1
2c 2

1
b2  a 2
c
b2  a 2
 ( a, b)
2
Hence the CMVT is verified.
Example 13: Verify the Cauchy’s MVT for f ( x)  log x and g ( x) 
f ( x)  log x and g ( x) 
1
in [1, e]
x
1
are elementary logarithmic and rational algebraic functions that
x
are continuous in [1, e]
f ( x ) 
1
1
and g ( x) 
exist in 1, e 
x
x2
also we see that g ( x)  0 for all x  (1, e)
i.e., the conditions of CMVT hold good for f ( x) and g ( x ) in 1, e  .
Hence  c  1, e  :
f '(c) f (e)  f (1)

g '(c) g (e)  g (1)
1
i.e.,
c  log e  log1  c  1  c  e  (1, e)
1
1
1
e 1
1
1
c2
e
e
Hence the CMVT is verified.
Example 14: Verify the Cauchy’s MVT for f ( x)  e x and g ( x)  e x in [a, b]
f ( x)  e x and g ( x)  e x are elementary exponential functions that are continuous in [a, b]
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f ( x)  e x and g ( x)  e x exist in  a, b 
also we see that g ( x)  0 for all x  (a, b) , since 0  a  b .
i.e., the conditions of CMVT hold good for f ( x) and g ( x ) in  a, b  .
Hence  c   a, b  :
i.e.,
ec
e  c

eb  ea
e b  e  a
f '(c) f (b)  f (a)

g '(c) g (b)  g (a)
 e
e 2c  e a  b  c 
2c
eb  ea
eb  ea
2c

 e 
1
 ea  eb 
1

 b

 a b 
ea 
e
 e

ab
 ( a, b)
2
Hence the CMVT is verified.
 
Example 15: Verify the Cauchy’s MVT for f ( x)  Sin x and g ( x)  Cos x in 0, 
 2
f ( x)  Sin x and g ( x)  Cos x are elementary trignometric functions
 
that are continuous in 0,  .
 2
 
f ( x)  Cos x and g ( x)   Sin x exist in  0, 
 2
 
also we see that g ( x)  0 for all x   0,  .
 2
 
i.e., the conditions of CMVT hold good for f ( x) and g ( x ) in  0,  .
 2

f ( )  f (0)
f '(c)
2
Hence  c   0,  :


g '(c) g ( )  g (0)
 2
2
 

Sin( )  Sin(0)
Cos c
  
2


 Cot c  1  c    0, 
 Sin c Cos(  )  Cos(0)
4  2
2
Hence the CMVT is verified.
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Exercise: Verify the Cauchy’s Mean Value Theorem for
(i)
f ( x)  x and g ( x) 
(ii)
f ( x) 
1
x
(iii)
2
and g ( x) 
1
1 
in  ,1
x
4 
1
in  a, b
x
f ( x)  Sin x and g ( x)  Cos x in  a, b
Taylor’s Mean Value Theorem:
(Generalised Mean Value Theorem):
(English Mathematician Brook Taylor 1685-1731)
Suppose a function f (x) satisfies the following two conditions:
(i)
(ii)
f (x) and it’s first (n-1) derivatives are continuous in a closed interval
a, b
f ( n 1) ( x) is differentiable in the open interval a, b
Then there exists at least one point c in the open interval a, b such that
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f (b)  f (a)  (b  a) f (a) 
..... 
(b  a)2
(b  a)3
f (a) 
f (a)  …..
2
3
(b  a)n 1 (n 1)
(b  a)n (n)
f
(a ) 
f (c)  (1)
n 1
n
Taking b  a  h and for 0    1 , the above expression (1) can be rewritten as
f (a  h)  f (a)  hf (a) 
h2
h3
hn 1 (n 1)
h n ( n)
f (a) 
f (a)  .... 
f
(a ) 
f (a   h)  (2)
2
3
n 1
n
Taking b=x in (1) we may write
f ( x)  f (a)  ( x  a) f (a) 
Where Rn 
( x  a)2
( x  a)3
( x  a)n 1 (n 1)
f (a) 
f (a)  ... 
f
(a)  Rn  (3)
2
3
n 1
( x  a) n ( n)
f (c)  Re mainder term after n terms
n
When n  , we can show that Rn  0 , thus we can write the Taylor’s series as
f ( x)  f (a)  ( x  a) f (a) 
( x  a)2
( x  a )n 1 ( n 1)
f (a)  ... 
f
( a)  ....
2
n 1

( x  a)n ( n)
f (a)  (4)
n
n 1
 f (a)  
Using (4) we can write a Taylor’s series expansion for the given function f(x) in powers of
(x-a) or about the point ‘a’.
Maclaurin’s series:
(Scottish Mathematician Colin Maclaurin 1698-1746)
When a=0, expression (4) reduces to a Maclaurin’s expansion given by
f ( x)  f (0)  xf (0) 
x2
x n 1 ( n 1)
f (0)  ... 
f
(0)  ....
2
n 1

x n ( n)
f (0)  (5)
n
n 1
 f (0)  
Example 16: Obtain a Taylor’s expansion for f ( x)  Sin x in the ascending powers of


 x   up to the fourth degree term.
4


The Taylor’s expansion for f (x) about
is
4
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


3
( x  )2
( x  )4
 (x  4 )

4
4 f (4) (  ) ....  (1)
f ( x)  f ( )  ( x  ) f ( ) 
f ( ) 
f ( ) 
4
4
4
2
4
3
4
4
4




1
 
f ( x)  Sin x  f    Sin 
;
4
2
4

1
 
f ( x)  Cos x  f     Cos 
4
2
4

1
 
f ( x)   Sin x  f      Sin  
4
2
4

1
 
f ( x)   Cos x  f     Cos  
4
2
4

1
 
f (4) ( x)  Sin x  f (4)    Sin 
4
2
4
Substituting these in (1) we obtain the required Taylor’s series in the form
1
 1
f ( x) 
 ( x  )( ) 
4
2
2



( x  )2
( x  )3
( x  )4
1
1
4 (
4 (
4 ( 1 ) ....
)
)
2
3
4
2
2
2





( x  ) 2 ( x  )3 ( x  ) 4


1

4
4
4
f ( x) 


 ...
1  ( x  ) 
4
2
3
4
2



Example 17: Obtain a Taylor’s expansion for f ( x)  loge x up to the term containing
 x  14
and hence find loge(1.1).
The Taylor’s series for f (x) about the point 1 is
f ( x)  f (1)  ( x  1) f (1) 
( x  1)2
( x  1)3
( x  1)4 (4)
f (1) 
f (1) 
f (1) ....  (1)
2
3
4
f ( x) 
Here f ( x)  loge x  f (1)  log 1  0 ;
f ( x)  
1
x
f (4) ( x)  
 f (1)  1;
2
6
x
4
f ( x) 
 f (4) (1)  6 etc.,
Using all these values in (1) we get
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2
x3
1
 f (1)  1
x
 f (1)  2
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f ( x)  loge x  0  ( x  1)(1) 
 loge x  ( x  1) 
( x  1)2
( x  1)3
( x  1)4
(1) 
(2) 
(6) ....
2
3
4
( x  1)2 ( x  1)3 ( x  1)4


....
2
3
4
Taking x=1.1 in the above expansion we get
 log e (1.1)  (0.1) 
(0.1)2 (0.1)3 (0.1)4


....  0.0953
2
3
4
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Example 18: Using Taylor’s theorem Show that
loge (1  x)  x 
x 2 x3

for 0    1, x  0
2
3
Taking n=3 in the statement of Taylor’s theorem, we can write
f (a  x)  f (a)  xf (a) 
x2
x3
f (a) 
f (a   x)  (1)
2
3
1
1
2
Consider f ( x)  loge x  f ( x)  ; f ( x)  
and f ( x) 
2
x
x
x3
Using these in (1), we can write,
2
3

2
1 x  1  x 
log(a  x)  log a  x        
  (2)
 a  2  a 2  3  (a   x)3 
For a=1 in (2) we write,
log(1  x)  log1  x 
x 2 x3
1
x 2 x3
1

 x

3
2
3 (1   x)
2
3 (1   x)3
Since x  0 and   0, (1   x)3  1 and therefore
 log(1  x)  x 
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x 2 x3

2
3
1
(1   x)3
1
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Example 19: Obtain a Maclaurin’s series for f ( x)  Sin x up to the term containing x5 .
The Maclaurin’s series for f(x) is
f ( x)  f (0)  x f (0) 
x2
x3
x 4 (4)
x5 (5)
f (0) 
f (0) 
f (0) 
f (0) ....  (1)
2
3
4
5
Here f ( x)  Sin x  f (0)  Sin 0  0
f ( x)  Cos x  f (0)  Cos 0  1
f ( x)   Sin x  f (0)   Sin 0  0
f ( x)   Cos x  f (0)  Cos 0  1
f (4) ( x)  Sin x  f (4) (0)  Sin 0  0
f (5) ( x)  Cos x  f (5) (0)  Cos 0  1
Substituting these values in (1), we get the Maclaurin’s series for f ( x)  Sin x as
f ( x)  Sin x  0  x (1) 
x2
x3
x4
x5
x3 x5
(0)  (1)  (0)  (1) ....  Sin x  x   ....
2
3
4
5
3
5
Note:
As done in the above example, we can find the Maclaurin’s series for various
functions, for ex:
(i) Cosx  1 
x 2 x 4 x6


......
2
4
6
(iii ) (1  x) m  1  mx 
(ii) e x  1  x 
x 2 x3 x 4 x5 x 6
 


......
2
3
4
5
6
m(m  1) 2 m(m  1)(m  2) 3 m(m  1)(m  2)(m  3) 4
x 
x 
x ......
2
3
4
Taking m   1 in (iii ) we can get
Replacing x by (-x) in this we get
(1  x)1  1  x  x2  x3  x 4  x5  x6 ......
(1  x)1  1  x  x 2  x3  x 4  x5  x6 ......
We use these expansions in the study of various topics.
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Indeterminate Forms:
While evaluating certain limits, we come across expressions of the form
0 
, , 0  ,   , 00 ,  0 and 1 which do not represent any value. Such expressions are
0 
called Indeterminate Forms.
We can evaluate such limits that lead to indeterminate forms by using L’Hospital’s Rule
(French Mathematician 1661-1704).
L’Hospital’s Rule:
If f(x) and g(x) are two functions such that
(i) lim f ( x)  0 and lim g ( x)  0
x a
x a
(ii) f ( x) andg ( x) exist and g (a)  0
f ( x)
f ( x)
 lim
x a g ( x)
x a g ( x)
The above rule can be extended, i.e, if
f ( x)
f ( x)
f ( x)
f (a)  0 and g (a)  0 then lim
 lim
 lim
 .....
x  a g ( x )
x a g ( x)
x  a g ( x )
Then lim
Note:
0 
,
forms. Here we
0 
f ( x)
differentiate the numerator and denominator separately to write
and apply the
g ( x)
0

limit to see whether it is a finite value. If it is still in or
form we continue to
0

f ( x)
differentiate the numerator and denominator and write further
and apply the
g ( x)
limit to see whether it is a finite value. We can continue the above procedure till we
get a definite value of the limit.
1. We apply L’Hospital’s Rule only to evaluate the limits that in
2. To evaluate the indeterminate forms of the form 0  ,   , we rewrite the
0

functions involved or take L.C.M. to arrange the expression in either or and then
0

apply L’Hospital’s Rule.
3. To evaluate the limits of the form 00 , 0 and 1 i.e, where function to the power of
function exists, call such an expression as some constant, then take logarithm on both
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sides and rewrite the expressions to get
0

or
form and then apply the L’Hospital’s
0

Rule.
4. We can use the values of the standard limits like
lim
x 0
sin x
tan x
x
x
 1; lim
 1; lim
 1; lim
 1; lim cos x  1; etc
x

0
x

0
x

0
x 0
x
x
sin x
tan x
Evaluate the following limits:
sin x  x
x  0 tan 3 x
Example 1: Evaluate lim
sin x  x  0 
cos x  1  0 
 sin x
0
 lim
 lim




 
3
2
2
4
3
2
x 0 tan x
 0  x0 3tan x sec x 0  x0 6 tan x sec x  6 tan x sec x 0 
lim
 lim
x 0
 cos x
1

2
4
4
2
6sec x  24 tan x sec x  18 tan x sec x  12 tan x sec x
6
6
2
4
Method 2:
sin x  x
sin x  x  0 
tan x
x3  0   lim sin x  x 0 
lim
 lim
lim
1






3
3
3
x 0 tan x
x 0
x
 0  x0  tan x   0  x0 x
0


 x 
cos x  1  0 
 sin x  0 
 cos x
1
 lim
 lim
lim





2
x 0
3x
6
 0  x 0 6 x  0  x 0 6
ax  bx
x 0
x
Example 2: Evaluate lim
ax  bx  0 
a x log a  b x log b
a

lim
 log a  log b log


x 0
x  0  x0
1
b
lim
Example 3: Evaluate lim
x 0
x sin x
e
x
 1
2
x sin x  0 
sin x  x cos x  0 
cos x  cos x  x sin x 1  1  0 2
 lim

lim

 1




2
x 0
x 0 2 e x  1 e x
x 0 2 e x .e x  (e x  1)e x 
x
0
0
2[1

0]
2






e

1
 


lim
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Example 4: Evaluate lim
x 0
x e x  log(1  x)
x2
x e  log(1  x)  0 
   lim
x2
 0  x 0
x
lim
x 0
1
1
ex  ex  x ex 
2
1  x   1  1  0  1  3
1  x  0   lim
  x 0
2x
2
2
2
0
ex  x ex 
cosh x  cos x
x 0
x sin x
Example 5: Evaluate lim
cosh x  cos x  0 
sinh x  sin x  0 
cosh x  cos x
1 1
2

 1
   lim
   lim
x

0
x

0
x sin x
sin x  x cos x  0 
cos x  cos x  x sin x 1  1  0 2
0
cos x  log(1  x)  1  x
Example 6: Evaluate lim
x 0
sin 2 x
lim
x 0
cos x  log(1  x)  1  x  0 
   lim
x 0
sin 2 x
 0  x 0
1
1
 cos x 
1
(1  x)2 1  1
1  x  0   lim

0
  x 0
sin 2 x
2 cos 2 x
2
0
 sin x 
lim
Example 7: Evaluate lim
x 1
xx  x
x  1  log x
xx  x  0 
x x (1  log x)  1  0 
x x (1  log x) 2  x x 1 1  1
lim

2
   lim
   lim
x 1 x  1  log x 0
1
1
1
  x1
 0  x1
1
x
x2
Since y  x x  log y  x log x 
and then
1
y  1  log x  y  y (1  log x)
y
d x
( x )  x x (1  log x)
dx
Example 8: Evaluate lim
x

4
sec2 x  2 tan x
1  cos 4 x
sec x  2 tan x  0 
2sec2 x tan x  2sec2 x  0 
2sec4 x  4sec2 x tan 2 x  4sec2 x tan x

lim

lim
 
 
1  cos 4 x  0  x
 4 sin 4 x
 16 cos 4 x
 0  x
2
lim
x

4
4

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2
 2
4
4
4
 2  (1)
2
16
2
4
 2
2

8 1

16 2
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log(sin x. cos ec a)
xa
log(cos a.sec x)
Example 9: Evaluate lim
 cos x cos ec a 
 sin x. cos ec a 
log(sin x. cos ec a)  0 
  lim cot x  cot 2 a
lim
 lim 


x a
log(cos a.sec x)  0  x a  sec x tan x.cos a  x a tan x
 cos a.sec x 
e x  e x  2 cos x
x 0
x sin x
Example 10: Evaluate lim
e x  e x  2cos x  0 
e x  e x  2sin x  0 
e x  e x  2cos x
11 2

lim

lim

2




x 0
x sin x
 0  x0 sin x  x cos x  0  x0 cos x  cos x  x sin x 1  1  0
x cos x  log(1  x)
Example 11: Evaluate lim
x 0
x2
1
cos x  x sin x 
x cos x  log(1  x)  0 
1 x  0 
lim
   lim
 
2
x 0
x

0
x
2x
0
0
lim
 sin x  sin x  x cos x 
 lim
x 0
1
(1  x)2
2

0  0  0  1 1

2
2
log(1  x 2 )
x 0 log cos x
 2 x 
2
1  x 2 
log(1  x )  0 
2 x cos x  0 
2 cos x  2 x sin x
20
lim

lim
 lim
 lim

2




2
2
x 0 log cos x
 0  x0   sin x  x0 (1  x )sin x  0  x0 (1  x ) cos x  2 x sin x 1  0


 cos x 
Example 12: Evaluate lim
Example 13: Evaluate lim
x 0
tan x  sin x
sin 3 x
tan x  sin x  0 
sec2 x  cos x  0 
2sec 2 x tan x  sin x  0 

lim

lim
 
 
 
x 0
sin 3 x  0  x0 3sin 2 x cos x  0  x0 6sin x cos 2 x  3sin 3 x  0 
lim
4sec2 x tan 2 x  2sec4 x  cos x
0  2 1 3 1

 
3
2
2
x 0 6cos x  12sin x cos x  9 sin x cos x
600 6 2
 lim
Method 2:
tan x  sin x
tan x  sin x
tan x  sin x 0 
x3
lim
 lim
 lim
 
3
3
x 0
x 0
x 0
sin x
x3
0
 sin x 


 x 
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sin x
1
x 0
x
lim
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sec2 x  cos x 0 
2sec2 x tan x  sin x 0 

lim
  x0
 
x 0
3x 2
6x
0
0
 lim
4sec2 x tan 2 x  2sec4 x  cos x 0  2  1 3 1

 
x 0
6
6
6 2
 lim
tan x  x
x  0 x 2 tan x
Example 14: Evaluate lim
tan x  x
tan x  x
tan x  x 0 
tan x
x3
lim 2
 lim
 lim
lim
1


3
x 0 x tan x
x 0  tan x 
x 0
x 0
x
x
0


 x 
sec2 x  1 0 
2sec2 x tan x 0 
4sec2 x tan 2 x  2sec4 x 0  2 1
 lim


   lim
   lim
x 0
3x 2  0  x0
6x
6
6
3
 0  x0
eax  e ax
x 0 log(1  bx)
ax
 ax
e e
aeax  ae ax a  a 2a
0
lim


   lim
x 0 log(1  bx) 0
b
b
  x0 b /(1  bx)
Example 15: Evaluate lim
a x  1  x log a
x 0
x2
a x  1  x log a  0 
a x log a  log a  0 
a x (log a)2 1
lim

lim

lim
 (log a)2
  x0
  x0
x 0
x2
0
2
x
0
2
2
 
 
Example 16: Evaluate lim
e x  log(e  ex)
x 0
x2
Example 17: Evaluate lim
e x  log(e  ex)
e x  log e(1  x)
e x  log e  log(1  x)  0 

lim

lim
 
x 0
x 0
x 0
x2
x2
x2
0
1
1
ex 
ex 
(1  x) 2 1  1
1  x  0   lim
 lim

1
 
x 0
2 x  0  x 0
2
2
lim
Exercise: Evaluate the following limits.
e x  e x  2log(1  x)
x 0
x sin x
(i) lim
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log(1  x3 )
x 0
sin 3 x
(ii) lim
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1  sin x  cos x  log(1  x)
x 0
x tan 2 x
(iii ) lim
cosh x  log(1  x)  1  x
(v) lim
x 0
x2
(iv) lim
x

2
log sin x

( x  )2
2
sin x sin 1 x
(vi) lim

x2
x
2
e2 x  (1  x)2
x 0 x log(1  x)
(vii) lim

Limits of the form   :

log(sin 2 x)
x 0 log(sin x)
Example 18: Evaluate lim
log(sin 2 x)   
(2cos 2 x / sin 2 x)
2cot 2 x
2 tan x  0 
2sec2 x 2

lim

lim

lim

lim
 1
  x0
  x0
x 0 log(sin x)
x 0 cot x
x 0 tan 2 x 0
(cos x / sin x)
2sec2 2 x 2

 
lim
Example 19: Evaluate lim
x 0
log x
cos ecx
log x   
1\ x
 sin 2 x  0 
2sin 2 x
0

lim

lim

0
  x 0
   lim
x 0 cos ecx 
x

0
x

0
 cos ec x.cot x
x cos x  0 
cos x  x sin x 1  0
 
lim
Example 20: Evaluate lim
x
lim
x

2
2
log cos x
tan x
log cos x   
 tan x
 tan x
 sin x cos x 0
  lim
 lim

0
   lim
2
2


tan x    x sec x
1
1
x  sec x
x
2
2
2
log(1  x)
x 1 cot  x
log(1  x)   
1/(1  x)
sin 2  x  0 
2 sin  x cos  x 0
lim

lim

lim
 lim

0




2
x 1 cot  x


   x1  cos ec  x x1  (1  x)  0  x1
Example 21: Evaluate lim
Example 22: Evaluate lim log tan 2 x tan 3x
x 0
 log tan 3x    
lim log tan 2 x tan 3x  lim 
 
x 0 log tan 2 x
x 0

  
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log b a 
log e a
log e b
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 3sec2 3x / tan 3x 
 3/ sin 3x.cos3x 
 3/ sin 3x.cos3 x 
 lim 
 lim 
 lim 



2
x 0 2sec 2 x / tan 2 x

 x0  2 / sin 2 x.cos 2 x  x0  2 / sin 2 x.cos 2 x 
 6 / sin 6 x 
 6sin 4 x  0 
 24 cos 4 x  24
 lim 
1
  lim

   lim


x 0 4 / sin 4 x
x 0 4sin 6 x
x

0



 0 
 24 cos 6 x  24
log( x  a)
x a log(e x  e a )
Example 23: Evaluate lim
log( x  a)   
1/( x  a)
(e x  ea )  0 
ex
ea

lim

lim

lim

1
  x a x x a
 
x a log(e x  ea ) 
e /(e  e ) xa e x ( x  a)  0  xa e x ( x  a)  e x ea
 
lim
Exercise: Evaluate the following limits.
log tan x
x 0 log x
(ii) lim
log sin x
x 0
cot x
cot 2 x
cot 3 x
(iv) lim
(i ) lim
(iii ) lim
x 0
x

2
sec x
tan 3 x
(v) lim log sin x sin 2 x
x 0
Limits of the form  0 : To evaluate the limits of the form  0 , we rewrite the
0 
given expression to obtain either   or   form and then apply the L’Hospital’s Rule.
0 
1
x
Example 24: Evaluate lim( a  1) x
x 
 1 
a (log a)  2 
(a  1)  0 
x 
lim(a  1) x  0   form   lim
   lim
x 
x   1 
x

 1 
0
 
 2
 x
x 
1
x
1
x
1
x
1
 lim a x (log a )  a 0 log a  log a
x 
 sin x) tan x
Example 25: Evaluate lim(1

x
2
lim(1  sin x) tan x  0   form   lim
x

x
2
Example 26: Evaluate lim sec
x 1
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
2x

.log x
2
(1  sin x)  0 
 cos x
0
 0
   lim
2
cot x  0  x  cos ec x 1
2
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limsec
x 1

2x
log x  0 
  
cos  0 
2x
1/ x
2x
2
 lim
 lim


 

  1  x1
x
 sin
2 
 sin  2 
2x
2
2 x  x 
.log x   0 form   lim
x 1
Example 27: Evaluate lim x log tan x
x 0
lim x log tan x  0   form   lim
x 0
x 0
log tan x   
 
(1/ x)   
sec2 x / tan x
 x2
 lim
x 0
x 0 sin x.cos x
 1 
 2
x 
 lim
2 x 2  0 
4 x
0
 lim
 0
   lim
x 0 sin 2 x 0
x 0 2cos 2 x
2
 
Example 28: Evaluate lim (1  x 2 ) tan
x 1
lim (1  x 2 ) tan
x 1
x
2
0  
form   lim
x 1
x
2
1  x2  0 
2 x
2
4
 lim




 x  0  x1 
x   
cot
 cos ec 2
 
2
2
2
2
Example 29: Evaluate lim tan x.log x
x 0
lim tan x.log x  0   form   lim
x 0
x 0
 lim
x 0
log x   
 
cot x   
1/ x
 sin 2 x  0 
 sin 2 x 0

lim
 lim
 0


2
 cos ec x x0
x  0  x0
1
1
Limits of the form      : To evaluate the limits of the form      , we take L.C.M.
0 
and rewrite the given expression to obtain either   or   form and then apply the
0 
L’Hospital’s Rule.
1

Example 30: Evaluate lim   cot x 
x 0 x


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1

 1 cos x 
lim   cot x   lim  
    form 
x 0 x
x 0 x
sin x 



 sin x  x cos x   0 
 cos x  cos x  x sin x 
 lim 
 lim 



x 0
x sin x   0  x0  sin x  x cos x


x sin x
sin x  x cos x
00

 0 


 lim 

0
   lim
x 0 sin x  x cos x  0
x 0  cos x  cos x  x sin x 

 

 11 0
Example 31: Evaluate lim sec x  tan x 
x
2
sin x 
 1
lim sec x  tan x   lim 

    form 


cos x 
x
x   cos x
2
2
1  sin x   0 
  cos x  0
 lim 
 lim 



 1 0


x   cos x   0 
x    sin x 
2
2
 1
x 
Example 32: Evaluate lim 

x 1 log x
x  1 

 1
 ( x  1)  x log x   0 
x 
lim 

    form   lim

 ( x  1) log x   0 
x 1 log x
x

1
x  1


 






 1  1  log x 
  log x   0 
  l/ x  1 1
 lim 

  lim
 1
    lim
 1 1
x 1 x  1
x 1
x 1
0
1

1
2



1   log x 
 2 
 log x 
x
x
x
x






1 
1
Example 33: Evaluate lim   x 
x 0 x
e  1

 (e x  1)  x   0 
1 
1
lim   x      form   lim 
 
x
x 0 x
x 0
e  1

 x(e  1)   0 



ex 1   0 
ex
1
1
 lim  x

lim







x
x
x
x
x 0 (e  1)  xe

  0  x0  e  e  xe  1  1  0 2
1
 1
 
Example 34: Evaluate lim 
x  0 sin x
x

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1
 1
 x  sin x   0 
lim 
      form   lim 
 
x 0 sin x
x

0
x

 x sin x   0 
sin x
0
 1  cos x   0 


 lim 

0
   lim
x 0 sin x  x cos x  0
x 0  cos x  cos x  x sin x 

 

 11
 1 log(1  x) 
Example 35: Evaluate lim  

x 0 x
x2

 1
1 

1

 (1  x) 2
 1 x   0 
 1 log(1  x) 
 x  log(1  x)   0 
lim  

lim

lim

lim
 x0 
  0  x0  2 x   0  x0  2
x 0 x
x2
x2








 1

 2

x
a
Example 36: Evaluate lim   cot 
x 0 x
a

x
x


cos 
cos 


a
x
a
a


a     form   lim 
a 0
lim   cot   lim  


x 0 x
x 0
x 
a  x0  x sin x 

 x sin   0 
a
a


x
x
x
x x
x

 1
 a sin a  x cos a   0 
 a. a cos a  cos a  a sin a 
 lim 
   xlim


x 0
0
x
x x
x

0


x sin
sin  cos
a
a a
a




x
x
1
x x 1
x




sin
sin  . .cos




0
00


a
a
a
a a a
a
 lim 
 lim 

0




x 0
x x
x  0  x 0 1
x 1
x x
x
1 1
 sin  cos 
 cos  cos  2 sin 
 0
a
a a
a a
a a a
 a a
a
Exercise: Evaluate the following limits.
x

(i ) lim  2   cot( x  a)
xa
a

(ii ) lim  cos ecx  cot x 
x 0



(iii) lim  x tan x  sec x 

2
x 

2
1

(iv) lim  cot 2 x  2 
x 0
x 

1 
1
(v) lim  2 
x 0 x
x tan x 

(vi) lim  2 x tan x   sec x 
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x

2
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Example 37: Find the value of ‘a’ such that lim
x 0
sin 2 x  a sin x
is finite. Also find the
x3
value of the limit.
sin 2 x  a sin x  0 
2 cos 2 x  a cos x 2  a
 lim

 finite


3
x 0
x
3x 2
0
 0  x 0
Let A= lim
We can continue to apply the L’Hospital’s Rule, if 2+a=0 i.e., a= -2 .
For a  2 ,
A  lim
x 0
2 cos 2 x  2 cos x  0 
4sin 2 x  2sin x  0 
   lim
 
2
x

0
3x
6x
0
0
 lim
x 0
8cos 2 x  2 cos x 8  2

 1
6
6
The given lim it will have a finite value when a  2 and it is  1.
Example 38: Find the values of ‘a’ and ‘b’ such that lim
x 0
x(1  a cos x)  b sin x 1
 .
x3
3
x(1  a cos x)  b sin x  0 
(1  a cos x)  ax sin x  b cos x 1  a  b
 lim

 finite


3
x 0
x
3x 2
0
 0  x 0
We can continue to apply the L’Hospital’s Rule, if 1-a+b=0 i.e., a-b = 1 .
Let A  lim
For a  b  1 ,
(1  a cos x)  ax sin x  b cos x  0 
2a sin x  ax cos x  b sin x  0 
 lim


 
2
x 0
3x
6x
 0  x 0
0
A  lim
 lim
x 0
3a cos x  ax sin x  b cos x 3a  b

 finite
6
6
1
3a  b 1
This finite value is given as . i.e.,
  3a  b  2
3
6
3
Solving the equations a  b  1 and 3a  b  2we obtain a 
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1
1
and b   .
2
2
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a cosh x  b cos x
 1.
x 0
x2
Example 39: Find the values of ‘a’ and ‘b’ such that lim
a cosh x  b cos x a  b

 finite
x 0
x2
0
Let A  lim
We can continue to apply the L’Hospital’s Rule, if a-b=0, since the denominator=0 .
For a  b  0 ,
a cosh x  b cos x  0 
a sinh x  b sin x  0 
a cosh x  b cos x a  b
 lim
 lim





2
x 0
x
2x
2
2
 0  x 0
 0  x 0
A  lim
But thisis given as 1.  a  b  2
Solving the equations a  b  0 and a  b  2 we obtain a  1 and b  1.
Limits of the form 00 , 0 and 1 : To evaluate such limits, where function to the power
of function exists, we call such an expression as some constant, then take logarithm on both
0

sides and rewrite the expressions to get or
form and then apply the L’Hospital’s Rule.
0

Example 40: Evaluate lim x x
x 0
Let A  lim x x (00 form)
x 0
Take log on both sides to write
log e A  lim log x x  lim x.log x (0   form)  lim
x 0
x 0
x 0
1/ x
x 0
 lim
 0
2
x 0 ( 1/ x )
x 0 1
1
 lim
loge A  0  A  e0  1 lim x x  1
x 0
1
Example 41: Evaluate lim(cos x)
x 0
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x2
log x   
 
1/ x   
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1
Let A  lim(cos x)
x2
x 0
(1 form)
Take log on both sides to write
1
log e A  lim log(cos x)
 lim
x2
x 0
x 0
1
log cos x  0 
log cos x (  0 form)  lim
 
2
x 0
x
x2  0 
 tan x  0 
 sec2 x 1

lim

 
x 0
2 x  0  x0 2
2
 lim
1
1
1
log e A    A  e 2 
2
e
1
 lim(cos x)
x2
x 0

1
e
x) cos x
Example 42: Evaluate lim(tan

x
2
Let A  lim(tan x)cos x ( 0 form)
x

2
Take log on both sides to write
log e A  lim log(tan x)cos x  lim cos x log(tan x) (0   form)
x

x
2
 lim
x

2

2
log tan x   
sec2 x / tan x
cos x 0

lim
 lim 2   0
 


sec x    x sec x.tan x x sin x 1
2
2
log e A  0  A  e0  1  lim(tan x) cos x  1
x

2
1
tan x x2
)
x 0
x
1
tan x x2 
Let A  lim(
) (1 form)
x 0
x
Example 43: Evaluate lim(
Take log on both sides to write
1
tan x x2
1
tan x
log e A  lim log(
)  lim 2 log(
) (  0 form)
x 0
x 0 x
x
x
tan x
sec2 x 1
log(
)

0
x
tan
x
x
 lim
   lim
x 0
x 0
x2
0
2
x
 
1
1
2
1


2 x  sin 2 x  0 
 lim sin x.cos x x  lim sin 2 x x  lim 2
 
x 0
x

0
x

0
2x
2x
2 x sin 2 x  0 
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 lim
x 0
2  2 cos 2 x
4sin 2 x
0
0
   lim
 
2
2
x

0
4 x sin 2 x  4 x cos 2 x  0 
4sin 2 x  16 x cos 2 x  8 x sin 2 x  0 
8cos 2 x
8 1


2
x 0 24 cos 2 x  48 x sin 2 x  16 x cos 2 x
24 3
 lim
1
1
1


1
tan x x2
log e A    A  e 3  lim(
) e 3
x 0
3
x
1
Example 44: Evaluate lim( a x  x) x
x 0
1
Let A  lim(a x  x) x (1 form)
x 0
Take log on both sides to write
1
1
log e A  lim log(a x  x) x  lim log(a x  x) (  0 form)
x 0
x 0 x
x
log(a  x)  0 
(a x log a  1) /(a x  x)
 lim

lim
  x0
x 0
x
1
0
 log a  1  log a  log e  log ae
 log e A  log ea  A  ea
1
x
Hence lim(a  x)  ea.
x
x 0
x tan  x
Example 45: Evaluate lim(2  ) 2 a
x a
a
x tan  x
Let A  lim(2  ) 2 a (1 form)
x a
a
Take log on both sides to write
x tan 2 ax
x
x
log e A  lim log(2  )
 lim tan
.log(2  )   0 form 
x a
x a
a
2a
a
(1/ a)




x
x
x
log(2  )
2
sin 2


2
a  0   lim 
a
2a  2
  lim .
 lim
xa
x

a
 x  0  x a  

x
 2 x


cot
 cos ec 2
 2a
2a
2a 
a




 log e A 
2

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2
 A  e
2
x tan  x
Hence lim(2  ) 2 a  e .
x a
a
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Exercise: Evaluate the following limits.
1
 1  cos x  x2
(ii) lim 

x 0
2 

b
(i) lim(cos ax) x
2
x 0
1
(iii) lim(1  x )
 sin x  x2
(iv) lim 

x 0
 x 
(v) lim(sin x) tan x
(iv) lim 1  sin x 
1
2 log(1 x )
x 1
x 0
cot x
x 0
(viii) lim  tan x 
2
(vii) lim(cos x)cos ec x
x 0
x
tan 2 x

4
1
 a x  bx  cx  x
( x) lim 

x 0
3


ax  1 x
(ix) lim(
)
x  ax  1
CURVATURE

y
Q
s
P
C

o
+
x
Consider a curve C in XY-plane and let P, Q be any two neighboring points on it. Let arc
AP=s and arc PQ=s. Let the tangents drawn to the curve at P, Q respectively make angles 
and + with X-axis i.e., the angle between the tangents at P and Q is . While moving
from P to Q through a distance‘s’, the tangent has turned through the angle ‘’. This is
called the bending of the arc PQ. Geometrically, a change in  represents the bending of the
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curve C and the ratio

represents the ratio of bending of C between the point P & Q and
s
the arc length between them.
Rate of bending of Curve at P is
d

 Lt
ds Q  P  s
This rate of bending is called the curvature of the curve C at the point P and is denoted by 
(kappa). Thus  
d
ds
We note that the curvature of a straight line is zero since there exist no bending i.e. =0, and
that the curvature of a circle is a constant and it is not equal to zero since a circle bends
uniformly at every point on it.
If   0, then
 
1


1

is called the radius of curvature and is denoted by  (rho - Greek letter).
ds
d
Radius of curvature in Cartesian form :
Suppose y = f(x) is the Cartesian equation of the curve considered in figure.
y
c

x
0
we have
y 
dy
d2 y
d
d ds
 Tan   y  2  Sec 2  
 1  Tan 2 

dx
dx
dx
ds dx
But we know that
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ds
 dy 
 1  
dx
 dx 
2
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  dy 2 
1    
2
2
d 2 y   dy   d
dy
ds
  dx  
 
 2  1     
 1   
 2
d y
dx
d
 dx 
  dx   ds
dx 2
2


ds 1   y  
 

d
y
3
3
2
2
This is the expression for radius of curvature in Cartesian form.
NOTE:
  dx  2 
1    
  dy  
We note that when y’=, we find  using the formula   
 d 2x 
 2
 dy 
3
2
Example 9: Find the radius of curvature of the curve x3+y3 = 2a3 at the point (a, a).
x3  y3  2a3  3x 2  3 y 2  y  0  y  
x2
hence at  a, a  , y  1
y2
 y 2  2 x   x 2  2 y  y 
 2a 3  2a 3 
4
 y   

 , hence at  a, a  , y   

4
4
y
a
a




1   y 2 

 

y
3
2
1   12 


4

a
3
2
i.e.,  
Example 10: Find the radius of curvature for
a
2 2
4

a
2
x  y  a at the point where it meets
the line y=x.
On the line y  x,
x  x  a i.e 2 x  a
or
x
a
4
a a
i.e., We need to find  at  , 
4 4
x y a
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1
2 x

1
2 y
y  0 i.e y  
y
a a
, hence at  ,  , y  1
x
4 4
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1
1 

 x 2 y  y  y 2 x 

Also, y   
x






 a 1
a 1 
 (1) 


4
a
1 1
 42 a
2
(  )

a a
4
4    2 2   (1)  4
 at  ,  , y   

a
a
a
a
4 4


4
4
4






1   y 2 

  

y
3
2
1   12 


4
a
3
2

a
a
2 2
4
2
Example 11: Show that the radius of curvature for the curve y  4 Sin x - Sin 2x
at x  
2
is 5 5
4
y  4 Sin x - Sin 2x  y  4 Cos x  2 Cos 2 x
 when x  
2
, y  4Cos   2Cos  0  2(1)  2
2
Also, y  -4 Sin x  4 Sin 2x and when x  
1   y 2 

  
y
3
2
1  22 


4
3
2
  
2
, y  -4 Sin   4 Sin   4
2
5 5
4
Example 12: Find the radius of curvature for xy2 = a 3 - x 3 at (a, 0).
xy 2  a3  x3  y 2  2 xy y  3x 2
 y 
3x 2  y 2
and at (a,0), y  
2 xy
In such cases we write
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dx
2 xy
dx

and at (a, 0),
0
2
2
dy 3x  y
dy
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 2

 dx

2  dx
y  2 x   2 xy  6 x  2 y  
  3x  y   2
dx
2 xy
d x
 dy

 dy

Also
 2
 2 
2
2
2
2


dy 3x  y
dy
 3x  y 




2
 2

d 2 x   3a  0   0  2a   0  6a3 2
 At  a, 0  ,



2
2
 9a 4
dy2 
3a
3
a

0




  dx  2 
1    
  dy  
   2
d x 2
dy
3
2
1  o 2 


2
3a
3
2
or
 
3a
2
Exercise:
Find the radius of curvature for the following curves:
(i)
(ii)
(iii)
 3a 3a 
x 3  y 3  3axy at  , 
 2 2 
2
a (a  x)
y2 
at (a, 0)
x
a 2 y  x 3  a 3 at the po int where it crosses x  axis
(v )
x2 y 2

 1 at the ends of the major axis.
a 2 b2
a 2 y  x 3  a 3 at the po int (a, 0).
(vi )
y 2  2 x(3  x 2 ) at the po int where the tan gents are parallel to x  axis.
(vii )
x 2 y  a( x 2  y 2 ) at (2a, 2a ).
(iv)
ax
 2 
(viii ) For the curve y 
show that 

ax
 a 
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2
3
2
 x  y
   
 y  x
2
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M.G.Geetha
INTEGRAL CALCULUS
Gamma and Beta Functions
Definition:

The improper integral n    e  x x n 1dx is defined as the gamma function.
0
Here n is a real number called the parameter of the function. (n) exists for all real values of
n except 0, -1, -2., ………..the graph of which is shown below :
r(n)
-4
-3
-2
-1
n
0
Recurrence formula

n    e  x x n 1 dx, n  0
0
  n


x x
xn

  e x   
e dx, integrating by parts.
n
n



0 0
Applying the definition of Gamma function and using
n  
xn
 0 as x  
ex
n  1
.........................(1)
n
or n  1  nn.................(2)
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Note:
(1) n is not convergent when n = 0, -1, -2, …….
(2) If n is known for 0 < n < 1, then its value for 1 < n < 2 can be found using equation (2).
Also, its values for -1 < n < 0 can be got using equation (1)
(3) If n is a +ve integer, using the recurrence relation and 1 = 1, we get
n = (n - 1) n - 1
= (n - 1) (n - 2) (n - 2) and so on
= (n - 1) (n - 2) (n - 3) …….. 1
= (n - 1) !
Problems:
1
(1) Prove that    
2

By definition, n    e  x x n 1dx
0

  e t t 2 n 1 dt u sin g x  t 2
2
0

2
 n   2  e  x x 2n 1dx
0

2
 2  e  y y 2n 1dy
0


2
2
1
     e  x dx   e  y dy
 2 0
0
2
    x 2  y 2 
  1 
 dxdy
    4   e 
  2 
0 0
Converting Cartesian (x, y) to polar (r, ) system, we get
2

2
  1 
 2   2 
  
 0
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
e
r 0
r 2
(2r )drd
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
 e

2
r 2
 d ,

 -  2re -r dr  e  r
2
0
2
0

2
2
 2  d,
 e - r  0 as r  
0
=
1
Taking square roots on both sides, we get    
2

(2) Show that
e
x4
0
1 1
dx   
4  4
Let x 4  y  4x 3dx  dy  dx 

 e
x4
0

1
dx 
4
e
y
y
3
4
dy 
0
1  34
y
dy
4
1 1
 , using the definition .
4 4
1
(3) Find
 x ln x 
5
dx
0
using ln x = -y, we get
x = e - y , dx = -e - y dy
and y ranges from  to 0 when x = 0 to 1


0
0
  x ln x 5 dx    y5e  6 y dy, interchanging the lower and the upper limits

t 5dt
  e t
66
0

a
(4) Prove that

0
using
1
66
, choosing 6y = t
6  
dx
a
ln  
 x
5!
66
a 
a
a
ln    t , we get  e t
x
 x
dx = -ae-t dt
and t =  to 0 when x = 0 to a
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a


dx

a
ln  
x
0
 t
1
0
1
n
m
 x log x  dx 
(5) Show that
0
1
log x

Hence show that
 1n n! .
m  1n 1
dx   4
when 'n' is a positive integer and m > -1.
x
0
Let
1
 a    a 
2
2 a.e  t dt
logx = - t  dx = -e-t dt and t =  to 0 when x = 0 to 1.

1
  x m log x  dx   e  mt  t  e t dt   1
n
n
0
0
t
x
m
log x 
n
dx   1
n
0
1
 x
0
m
n
e m 1t dt
0
1
using m  1 t  y,

n

yn
 m  1n e
 y dy
m 1
0
,  m  -1


 1n
log x  dx 
e  y y n dy
n 1 
m  1 0
n

 1n n!,
m  1n 1
choosing n  1 and m  
 n  1  n!
1
, we get the required result.
2
Beta function
Definition: Beta function, denoted by B(m,n) is defined by
1
Bm, n    x m 1 1  x n 1 dx , where m and n are positive real numbers.
0
1
By a property of definite integral,
 x 1  x
m 1
0
Therefore we have B(m,n) = B(n,m)
Alternate Expressions for B(m,n)
1
Bm, n    x m 1 1  x 
0
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n 1
dx...........................( 1)
n 1
1
dx   1  x 
0
m 1
x n 1 dx
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(i) Let x 
1
1
t
 dx  
dt , 1 - x 
2
1 t
1 t
1  t 
and t =  to 0 when x = 0 to 1.
m 1

 1 reduces to
 1 
0  1  t 
B(m, n) 

B(m, n) is also 
t m 1
 1  t 
m n
 1 


1 t 
n 1
 1

 1  t 2


dt


dt using Bm, n   Bn, m
0
ii  Let
x  sin 2  dx  2 sin  cos d and   0 to


when x  0 to 1
2
2
Then (1) reduces to 2  sin 2n -1 cos 2m 1 d
0

2
 B(m, n)  2  sin 2m-1 cos 2 n 1  d
0
Relation between Gamma and Beta functions.
Prove that
Bm, n  
m n 
m  n 
Proof:

2
We know that n   2  e  x x 2n 1dx

and m  2 e  y x 2 m 1dy
2
0
0

Then m n   4  e x
2
 y2
 x 2 n 1 y 2 m1 dxdy
0 0

 4e

 r 2 2m  n 1
r
dr
2
 sin
2m 1
 cos 2n 1d in polar coordinates
0
0
= (m+n) B(m,n) by definitions.
Note:
 n -1
U sin g
x

 1  x dx  sin n ,
0  n 1
0
We get (n) (1-n) = (1) B (1-n,n) using the above relation
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 n 1
B1 - n, n   
But
0
x
dx using definition of Beta function.
1 x
Therefore (n) (1-n) =  /sin n, 0 < n < 1
Problems
1
(1) Show that
2 1
dx  B , 
5
 5 2
0 1 x
5x

1
1
Letting x 5  t, we get x  t 5 , dx  t  4 dt and t = 0 to 1 when x = 0 to 1
5
5
1
1
1 1 
dx   5t 5 1  t  2 t 5 dt
5
1 x5
5x

0
1
4
0
1
 t
3
5 1  t  12 dt
0
2 1
 B ,  by definition .
5 2
 10
x
1  x8 dx  

30
0 1  x 
(2) Evaluate
x18
30
0 1  x 
dx
= B (11,19) - B(19,11) = 0, using B(m,n) = B(n,m).

(3) Evaluate
2

tan d
0

2


2
tan  d   sin
0
2
 cos
1
2
d
0

1 3 1
B , ,
2 4 4
n
(4) Show that
Put
1
by definition
n
 x  t 1
t
 1  n  x dx  n Bt, n  1, t  0, n  1
0
x
 y  dx  ndy and y  0 to 1 when x  0 to n
n
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n
 x
n
  1   x t 1dx   1  y  n t 1 y t 1ndy
n
0
0
n
1
1
n
 1  y 
t
n
y t 1 dy  n t Bt , n  1 by definition .
0
(5) Prove that
Bn, n  
1
2
2 n 1

 1
B  n, 
 2
2
We know that B(n, n)  2  sin 2n-1 cos 2 n1  d
0

sin 2n 1 2
2 
d, sin2   2sin cos
2n 1
2
0


2
sin 2n 1 
2n 1
0 2

d, putting 2  
sin 2 n1 
d , sin  is even in (0,  ) as sin  -    sin 
2 n 1
2
0
2
2
Applying the definition of Beta function , we get Bn, n  
1
2
2 n 1
 1
B n, 
 2
(6) Prove the Duplication formula
1


n  n    2 n1 2n 
2 2

Using the previous result,
1
1
1 
1 1

B n  , n   
B n  ,  …………(1)
2
2  2 2n 
2 2

1
1
1 
1


B n  , n  
 n   n  n  1
2
2
2 
2

Also
 
......................(2)
1 1
1 1


B n  , 
2n  1 n   
2 2
2 2


1

 n   n n 
1
2

From (1) and (2) we get
 2n
2
2n 2n  
1


 n . n    2 n 1 2n 
2 2

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1 m 1
Bm, n   
(7) Prove that
x
 x n 1
1  x m  n
0
dx
By one of the definitions of Beta function, we have
Bm, n  

x n 1
 1  x m  n dx
0

x n 1
1

mn
0 1  x 
dx  
x n 1
mn
1 1  x 
dx
= I1 + I2 (say)
Substituting x 
1
in I 2 , we get
y
1
 1 
y m 1

dy 
 1  ym  n dy
mn 
2
n 1
y


y
1

y


1
0
0
I2  
ym n
1 m 1
Hence Bm, n   
0
x
 x n 1
1  x m  n
dx
Conclusion: We evaluated double integrals using change of order of integration. We used
change of variables for double and triple integrals to simplify the integrals in certain cases.
We saw how Gamma and Beta functions were useful in evaluating complicated integrals.
Multiple Choice:
12
1. The value of
 xy
2
dydx is
01
(a) 1/2
(b) 7/3
(c) 7/6
(d) 7/2
1 x2
2. The value of
  xdx dy
0 x
(a) -1/12
(b) -1/4
(c) 0
(d) -1/15
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1 z xz
3. The value of
   x  y  zdxdydz
1 0 x  z
(a) 1
(b) 0
(c) 2
(d) 8
4. The value of (-10) is
(a) 10!
(b) 9!
(c) -9!
(d) Not defined
5. The value of (-5/2) is
(a)  2 
4

3
(b)
(c) 
8

15
(d) Not defined
 9


x 1  x 8 dx

26
0 1  x 
6.
simplifies to
(a) 0
10!18!
(b)
28!
9!17!
(c)
27!
2, 9!17!
(d)
27!

7.
2

cot d is equal to
0
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(a)

2
(b) 2
2
(c)
(d)  / 2
8. The value of B(1/2, 1/2) is

(a)
(b) 
(c) 2 
(d) 1
Questions & Answers
1 1 x x  y
(1) Evaluate
dxdydz
 
3
0 0 0 1  x  y  z 
Ans: Let this integral be I, then
 1 x 1 x  y
 
dz


dy dx
I   


3  

x  0  y  0  z  0 1  x  y  z   
1
1 1 x 
1 x  y

   

3


2
1

x

y

z

 z  0
x 0 y 0 

1
dydx
1 1 x 

1

 dx
2


2
1

x

y

z


x 0 y 0 
 
1
5
  ln 2  
2
8
(2) Change the order of integration and hence evaluate the integral
1 2 x
 
0 x
x
dxdy
y
Ans: The region of integration is the shaded portion shown in the figure below
Y
x=0
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B
y=x
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To get the limits for x first, we need to divide the shaded area into two parts AMB and AMO
where the x values are 0 to 2-y and 0 to y respectively. The respective y values are 1 to 2 and
0 to 1.
The given integral is now
y
1
 

y 0 x 0
1


y0
x
dxdy 
y
y
dy 
2
2 2 y
 
y 1 x  0
x
dxdy
y
2
2 y

   2 dy
y 2

y 1

= 2 ln 2 -1.
(3) Change to polar coordinates and hence evaluate
aa

0y
x
2
x y
2
dxdy
Ans: The region of integration is as shown below
Y
x=y
y=a
X
0
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y=0
x=a
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We see that  ranges from 0 to /4 in the shaded region. R ranges from 0 to x = a i.e., a sec .
The given integral

4a sec 
 r

0


cos 2 drd
0

a
a3
sec d  ln sec   tan 0 4
3
3
4 3

0

2
a3
ln
3


2 1
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(4) Find the area between the parabola y = 4x - x2 and the line y = x, using double integration.
Ans: The required area is
 dxdy
r
(2,4)
y=x
(3,3)
3 4x  x 2


 dxdy 
x 0 y  x
 3x  x
3
2
dx
x 0
3
3
x3  9
  x2   
2
3  2
0 
Practice Questions
1. Evaluate the following integrals
32
i   x1  x  ydxdy
01

2a
ii    r 2 sin drd
0 0
iii    x 2  y 2 dydx
1 x
0 x
ab
iv   
11
1
dydx
xy
111
v  e x  y  z dxdydz
000
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3 1
xy
11
0
vi     xyz dz dy dx

x
2 2
2a cos a  r
vii   
0
 rdz dr d
0
0
2. Change the order of integration and hence evaluate
1 2- y
i    xydxdy
0 y
4a 2 ax
ii    dydx
0 x2
4a
a 2a - x
iii    xydydx
0 x2
a
ax
iv   
cos y
dydx



a

x
a

y
00
3. Change to respective coordinate systems and hence evaluate
2 2x - x 2
i  

0
0
x
x 2  y2
dydx to polar system
a a 2 -x 2
ii  

0
0
1 2x - x 2
2
iii  
0
x
y 2 x 2  y 2 dydx to polar system
 y 2 dydx to polar system
x
2 2
1 1- x 2 1 x  y
iv   

0
0
0
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dzdydx
2
2
1 x  y  z
2
to spherical polar
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1 1- x 2
1
v   
0
0
1
dzdydx

2
x y z
x 2 y2
1
2
2
to spherical polar
2
vi   
x 0 y 0
 xyzdxdydz
to cylindrica l polar
z x 2  y2
4. Find the area between y = 2 - x and circle x2 + y2 = 4
5. Find the area bounded between the parabola y2 = 4ax and x2 = 4ay
6. Show that the area of one loop of the lemniscates r2 = a2 cos2 is a2/2.
7. Find the area of one petal of the rose r = a sin3.
8. Find the area of the circle r = a sin outside the cardioide r = a (1 - cos)
9. Find the volume of the paraboloid of revolution x2 + y2 = 4z cut off by the plane z =4.
10. Find the volume of the region bounded by the paraboloid az = x2 + y2 and the cylinder
x2+ y2 = R2
11. Find the volume of the portion of the sphere x2 + y2 + z2 = a2 lying inside the cylinder
x2 + y2 = ax.
12. Find the volume cut off the sphere x2 + y2 + z2 = a2 by the cone x2 + y2 = z2

13. Evaluate
4
2 x
 x e dx
0
 x
1
14. Find

3
log 1
0

2
15. Evaluate
dx

1
3x 4
3  3
dx
8

x
 16
0

16. Find
2
 tan
p
 d , 0  p  1
0

17. Show that 2a
n
2
2n 1  at
 t e dt  n 
0

2
1

18. Show that  e - x log a dx, a  0 is
2 log a
0
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19. If m and n are real constants > -1, prove that
n
n  1
 1 
m
 x log  x  dx  m  1n 1
0
1

20. Show that
x
n 1
cos ax dx 
0
1
n
n  cos
n
2
a
Hint: Write cos ax = Real part of eiax
1
21. Show that
22. Show that

dx
0 1 x
n

Bm, n  
1
2
1

Hence deduce that
1

23. Show that
1
 1  x 
b
a
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1  x n 1 dx
1 x
dx = 
1 x
0
 x  a 
m 1
1
4
x
2 x
 xe dx  x e dx 
0
24. Show that
m  n 1

8
1  1 1
B , 
n  n 2
m 1

16 2
b  x n 1 dx  b  a m  n 1Bm, n 