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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.
Example 1 : The problem of solving the equation xydxdy
dxyd
=++2
2
with conditions y(0) =
y′ (0) = 1 is an initial value problem
Example 2 : The problem of solving the equation xydxdy
dxyd cos23 2
2
=++ with y(1)=1,
y(2)=3 is called Boundary value problem.
Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. 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
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Definition :
Let f(t) be a real-valued function defined for all t ≥ 0 and s be a parameter, real or complex.
Suppose the integral ∫∞
−
0
)( dttfe st exists (converges). Then this integral is called the Laplace
transform of f(t) and is denoted by Lf(t).
Thus,
Lf(t) = ∫∞
−
0
)( dttfe st (1)
We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t) is a function of s denoted by F(s) or )(sf .
Thus,
Lf(t) = F(s) (2)
Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denoted by L-1 [F(s)].
Thus,
L-1 [F(s)] = f(t) (3)
Suppose f(t) is defined as follows :
f1(t), 0 < t < a
f(t) = 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) = ∫∞
−
0
)(tfe st
= ∫ ∫ ∫∞
−−− ++a b
a b
ststst dttfedttfedttfe0
321 )()()(
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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 φ(t) (whose Laplace transforms exist)
and any two constants a and b, we have
L [a f(t) + b φ(t)] = a L f(t) + b Lφ(t)
Proof :- By definition, we have
L[af(t)+bφ(t)] = [ ]dttbtafe st∫∞
− +0
)()( φ = ∫ ∫∞ ∞
−− +0 0
)()( dttebdttfea stst φ
= a L f(t) + b Lφ(t)
This is the desired property.
In particular, for a=b=1, we have
L [ f(t) + φ(t)] = L f(t) + Lφ(t)
and for a = -b = 1, we have
L [ f(t) - φ(t)] = L f(t) - Lφ(t)
2. Change of scale property : If L f(t) = F(s), then L[f(at)] =
asF
a1 , where a is a positive
constant.
Proof :- By definition, we have
Lf(at) = ∫∞
−
0
)( dtatfe st (1)
Let us set at = x. Then expression (1) becomes,
L f(at) = ∫∞
−
0
)(1 dxxfea
xas
=
asF
a1
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)] = [ ]∫∞
−
0
)( dttfee atst = ∫∞
−−
0
)( )( dttfe as
= 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(eat) = ∫ ∫∞ ∞
−−− =0 0
)( dtedtee tasatst
= asas
e tas
−=
−−
∞−− 1)( 0
)(
, s > a
Thus,
L(eat) = as −
1
In particular, when a=0, we get
L(1) = s1 , s > 0
By inversion formula, we have
atat es
Leas
L ==−
−− 11 11
2. L(cosh at) =
+ −
2
atat eeL = [ ]∫∞
−− +02
1 dteee atatst
= [ ]∫∞
+−−− +0
)()(
21 dtee tastas
Let s > |a| . Then,
∞+−−−
+−
+−−
=0
)()(
)()(21)(cosh
ase
aseatL
tastas
= 22 ass−
Thus,
L (cosh at) = 22 ass−
, s > |a|
and so
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atas
sL cosh221 =
−−
3. L (sinh at) = 222 asaeeL
atat
−=
− −
, s > |a|
Thus,
L (sinh at) = 22 asa−
, s > |a|
and so,
a
atas
L sinh122
1 =
−−
4. L (sin at) = ∫∞
−
0
sin ate st dt
Here we suppose that s > 0 and then integrate by using the formula
[ ]∫ −+
= bxbbxaba
ebxdxeax
ax cossinsin 22
Thus,
L (sinh at) = 22 asa+
, s > 0
and so
a
atas
L sinh122
1 =
+−
5. L (cos at) = atdte st cos0∫∞
−
Here we suppose that s>0 and integrate by using the formula
[ ]∫ ++
= bxbbxaba
ebxdxeax
ax sincoscos 22
Thus,
L (cos at) = 22 ass+
, s > 0
and so
atas
sL cos221 =
+−
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6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then
L(tn) = ∫∞
−
0
dtte nst
Let s > 0 and set st = x, then
∫ ∫∞ ∞
−+
− =
=
0 01
1)( dxxess
dxsxetL nx
n
nxn
The integral dxxe nx∫∞
−
0
is called gamma function of (n+1) denoted by )1( +Γ n . Thus
1
)1()( +
+Γ= n
n
sntL
In particular, if n is a non-negative integer then )1( +Γ n =n!. Hence
1
!)( += nn
sntL
and so
)1(
11
1
+Γ=+
−
nt
sL
n
n or !n
t n
as the case may be
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TABLE OF LAPLACE TRANSFORMS
f(t) F(s)
1 s1 , s > 0
eat as −
1 , s > a
coshat 22 ass−
, s > |a|
sinhat 22 asa−
, s > |a|
sinat 22 asa+
, s > 0
cosat 22 ass+
, s > 0
tn, n=0,1,2,….. 1!+ns
n , s > 0
tn, n > -1 1)1(
+
+Γns
n , s > 0
Application of shifting property :-
The shifting property is
If L f(t) = F(s), then L [eatf(t)] = F(s-a)
Application of this property leads to the following results :
1. [ ]ass
assat
bssbtLbteL
−→−→
−== 22)(cosh)cosh( = 22)( bas
as−−
−
Thus,
L(eatcoshbt) = 22)( basas−−
−
and
btebas
asL at cosh)( 22
1 =−−
−−
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2. 22)()sinh(
basabteL at
−−=
and
btebas
L at sinh)(1
221 =
−−−
3. 22)()cos(
basasbteL at
+−−
=
and
btebas
asL at cos)( 22
1 =+−
−−
4. 22)()sin(
basbbteL at
−−=
and
b
btebas
Lat sin
)(1
221 =
−−−
5. 1)()1()( +−
+Γ= n
nat
asnteL or 1)(
!+− nas
n as the case may be
Hence
)1()(
11
1
+Γ=
− +−
nte
asL
nat
n or 1)(!
+− nasn as the case may be
Examples :-
1. Find Lf(t) given f(t) = t, 0 < t < 3
4, t > 3
Here
Lf(t) = ∫ ∫ ∫∞ ∞
−−− +=0
3
0 3
4)( dtetdtedttfe ststst
Integrating the terms on the RHS, we get
Lf(t) = )1(11 32
3 ss es
es
−− −+
This is the desired result.
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2. Find Lf(t) given f(t) = sin2t, 0 < t ≤ π
0, t > π
Here
Lf(t) = ∫∫∞
−− +π
π
dttfedttfe stst )()(0
= ∫ −π
0
2sin tdte st
= { }π
02 2cos22sin
4
−−
+
−
ttsse st
= [ ]ses
π−−+
14
22
This is the desired result.
3. Evaluate : (i) L(sin3t sin4t)
(ii) L(cos2 4t)
(iii) L(sin32t)
(i) Here
L(sin3t sin4t) = L [ )]7cos(cos21 tt −
= [ ])7(cos)(cos21 tLtL − , by using linearity property
= )49)(1(
244912
12222 ++
=
+−
+ sss
ss
ss
(ii) Here
L(cos24t) =
++=
+
641
21)8cos1(
21
2ss
stL
(iii) We have
( )θθθ 3sinsin341sin3 −=
For θ=2t, we get
( )ttt 6sin2sin3412sin3 −=
so that
)36)(4(
4836
64
641)2(sin 2222
3
++=
+−
+=
sssstL
This is the desired result.
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4. Find L(cost cos2t cos3t)
Here
cos2t cos3t = ]cos5[cos21 tt +
so that
cost cos2t cos3t = ]coscos5[cos21 2 ttt +
= ]2cos14cos6[cos41 ttt +++
Thus
L(cost cos2t cos3t) =
+++
++
+ 41
163641
222 ss
sss
ss
5. Find L(cosh22t)
We have
2
2cosh1cosh2 θθ +=
For θ = 2t, we get
2
4cosh12cosh2 tt +=
Thus,
−+=
161
21)2(cosh 2
2
ss
stL
-------------------------------------------------------05.04.05-------------------------
6. Evaluate (i) L( t ) (ii)
t
L 1 (iii) L(t-3/2)
We have L(tn) = 1)1(
+
+Γns
n
(i) For n= 21 , we get
L(t1/2) = 2/3
)121(
s
+Γ
Since )()1( nnn Γ=+Γ , we have 22
1211
21 π
=
Γ=
+Γ
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Thus, 2
32
)(s
tL π=
(ii) For n = -21 , we get
ss
tL π=
Γ
=−
21
21 2
1
)(
(iii) For n = -23 , we get
sss
tL ππ 2221
)(2
12
12
3−=
−=
−Γ
=−−
−
7. Evaluate : (i) L(t2) (ii) L(t3)
We have,
L(tn) = 1!+ns
n
(i) For n = 2, we get
L(t2) = 332!2ss
=
(ii) For n=3, we get
L(t3) = 446!3ss
=
8. Find L[e-3t (2cos5t – 3sin5t)]
Given =
2L(e-3t cos5t) – 3L(e-3t sin5t)
= ( )25)3(
1525)3(
32 22 ++−
+++
sss , by using shifting property
= 346
922 ++
−ss
s , on simplification
9. Find L[coshat sinhat]
Here
L[coshat sinat] = ( )
+ −
ateeLatat
sin2
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=
++
++− 2222 )()(2
1aas
aaas
a
, on simplification
10. Find L(cosht sin3 2t)
Given
−
+ −
46sin2sin3
2tteeL
tt
= ( )[ ])6sin()2sin(3)6sin(2sin381 teLteLteLteL tttt −− −+−⋅
=
++
−++
++−
−+− 36)1(
64)1(
636)1(
64)1(
681
2222 ssss
=
++
−++
++−
−+− 36)1(
124)1(
136)1(
14)1(
143
2222 ssss
11. Find )( 254 −− teL t
We have
L(tn) = 1)1(
+
+Γns
n Put n= -5/2. Hence
L(t-5/2) = 2/32/3 34)2/3(
−− =−Γ
ssπ Change s to s+4.
Therefore, 2/32/54
)4(34)( −
−−
+=
steL t π
Transform of tn f(t) Here we suppose that n is a positive integer. By definition, we have
F(s) = ∫∞
−
0
)( dttfe st
Differentiating ‘n’ times on both sides w.r.t. s, we get
∫∞
−
∂∂
=0
)()( dttfes
sFdsd st
n
n
n
n
Performing differentiation under the integral sign, we get
])][()[()2(
2222
22
aasaasasa
+++−+
=
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∫∞
−−=0
)()()( dttfetsFdsd stn
n
n
Multiplying on both sides by (-1)n , we get
∫∞
− ==−0
)]([)(()()1( tftLdtetftsFdsd nstn
n
nn , by definition
Thus,
L[tnf(t)]= )()1( sFdsd
n
nn−
This is the transform of tn f(t).
Also,
)()1()(1 tftsFdsdL nn
n
n
−=
−
In particular, we have
L[t f(t)] = )(sFdsd
− , for n=1
L[t2 f(t)]= )(2
2
sFdsd , for n=2, etc.
Also,
)()(1 ttfsFdsdL −=
− and
)()( 22
21 tftsF
dsdL =
−
Transform of t
f(t)
We have , F(s) = ∫∞
−
0
)( dttfe st
Therefore,
∫∞
∫∞
∫∞ −=
s sdsdttfstedssF
0)()( = dtdsetf
s
st∫ ∫∞ ∞
−
0
)(
= ∫∞ ∞−
−0
)( dtt
etfs
st
= ∫∞
−
=
0
)()(ttfLdt
ttfe st
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Thus,
∫∞
=
s
dssFttfL )()(
This is the transform of ttf )(
Also,
∫∞
− =s t
tfdssFL )()(1
Examples : 1. Find L[te-t sin4t]
We have,
16)1(
4]4sin[ 2 ++=−
steL t
So that,
L[te-t sin4t] =
++−
17214 2 ssds
d
= 22 )172()1(8
+++ss
s
2. Find L(t2 sin3t)
We have
L(sin3t) = 9
32 +s
So that,
L(t2 sin3t) =
+ 93
22
2
sdsd = 22 )9(
6+
−s
sdsd = 32
2
)9()3(18
+−
ss
3. Find
−
tteL
t sin
We have
1)1(
1)sin( 2 ++=−
steL t
Hence
−
tteL
t sin = [ ]∫∞
∞− +=++0
12 )1(tan
1)1( sss
ds
= )1(tan2
1 +− − sπ = cot –1 (s+1)
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4. Find
ttL sin . Using this, evaluateL
tatsin
We have
L(sint) = 1
12 +s
So that
Lf(t) =
ttL sin = [ ]∞−
∞
=+∫ S
s
ss
ds 12 tan
1
= )(cottan2
11 sFss ==− −−π
Consider
L
tatsin = a L )(sin ataLf
atat
=
=
asF
aa 1 , in view of the change of scale property
=
−
as1cot
5. Find L
−
tbtat coscos
We have
L [cosat – cosbt] = 2222 bss
ass
+−
+
So that
L
−
tbtat coscos = ds
bss
ass
s∫∞
+−
+ 2222 = ∞
++
sbsas
22
22
log21
=
++
−
++
∞→ 22
22
22
22
loglog21
bsas
bsasLt
s
=
++
+ 22
22
log021
asbs =
++
22
22
log21
asbs
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6. Prove that ∫∞
− =0
3
503sin tdtte t
We have
∫∞
− =0
)sin(sin ttLtdtte st = )(sin tLdsd
− =
+−
11
2sdsd
= 22 )1(2+ss
Putting s = 3 in this result, we get
∫∞
− =0
3
503sin tdtte t
This is the result as required.
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ASSIGNMENT I Find L f(t) in each of the following cases :
1. f(t) = et , 0 < t < 2
0 , t > 2
2. f(t) = 1 , 0 ≤ t ≤ 3
t , t > 3
3. f(t) = at , 0 ≤ t < a
1 , t ≥ a
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
14. 3
1
+
tt
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. te t21 −−
22. t
ee btat −− −
23. t
t2sin
24. t
tt 5sin2sin2
III. Evaluate the following integrals using Laplace transforms :
25. ∫∞
−
0
2 5sin tdtte t
26. ∫∞
−
0
35 cos tdtte t
27. ∫∞ −−
−
0
dtt
ee btat
28. ∫∞ −
0
sin dtt
te t
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-------------------------------------------- 28.04.05 -------------------------- Transforms of the derivatives of f(t) Consider
L )(tf ′ = ∫∞
− ′0
)( dttfe st
= [ ] ∫∞
−∞− −−0
0 )()()( dttfestfe stst , by using integration by parts
= [ ] )()0()(( tsLfftfeLt st
t+−−
∞→
= 0 - f (0) + s Lf(t) Thus L )(tf ′ = s Lf(t) – f(0) Similarly, L )(tf ′′ = s2 L f(t) – s f(0) - )0(f ′ In general, we have )0(.......)0()0()()( 121 −−− −−′−−= nnnnn ffsfstLfstLf
Transform of ∫t
0
f(t)dt
Let φ (t) = ∫t
dttf0
)( . Then φ(0) = 0 and φ′ (t) = f(t)
Now,
L φ(t) = ∫∞
−
0
)( dtte stφ
= ∫∞ −∞−
−′−
− 00
)()( dts
ets
etstst
φφ
= ∫∞
−+−0
)(1)00( dtetfs
st
Thus,
∫ =t
tLfs
dttfL0
)(1)(
Also,
∫=
−
t
dttftLfs
L0
1 )()(1
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Examples : 1. By using the Laplace transform of sinat, find the Laplace transform of cosat. Let
f(t) = sin at, then Lf(t) = 22 asa+
We note that atatf cos)( =′ Taking Laplace transforms, we get )(cos)cos()( ataLataLtfL ==′
or L(cosat) = [ ])0()(1)(1 ftsLfa
tfLa
−=′
=
−
+01
22 assa
a
Thus
L(cosat) = 22 ass+
This is the desired result.
2. Given 2/3
12s
tL =
π, show that
stL 11
=
π
Let f(t) = πt2 , given L[f(t)] = 2/3
1s
We note that, tt
tfππ1
212)( ==′
Taking Laplace transforms, we get
=′
tLtfL
π1)(
Hence
)0()()(1 ftsLftfLt
L −=′=
π
= 012/3 −
ss
Thus
st
L 11=
π
This is the result as required.
3. Find ∫
−t
dtt
btatL0
coscos
Here
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Lf(t) =
++
=
−
22
22
log21coscos
asbs
tbtatL
Using the result
L ∫ =t
tLfs
dttf0
)(1)(
We get,
∫
−t
dtt
btatL0
coscos =
++
22
22
log21
asbs
s
4. Find ∫ −t
t tdtteL0
4sin
Here
[ ] 22 )172()1(84sin
+++
=−
ssstteL t
Thus
∫ −t
t tdtteL0
4sin = 22 )172()1(8
+++sss
s
Transform of a periodic function A function f(t) is said to be a periodic function of period T > 0 if f(t) = f(t + nT) 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 2π since sin(t + 2nπ) = sin t, cos(t + 2nπ) = cos t.
The graph of f(t) = sin t is shown below :
Note that the graph of the function between 0 and 2π is the same as that between 2π and 4π 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 2π is the same as that between 2π and 4π and so on. Formula : Let f(t) be a periodic function of period T. Then
∫ −−−
=T
stST dttfe
etLf
0
)(1
1)(
Proof : By definition, we have
L f(t) = ∫ ∫∞ ∞
−− =0 0
)()( duufedttfe sust
= ∞+++++ ∫∫∫+
−−− ....)(.......)()()1(2
0
Tn
nT
suT
T
suT
su duufeduufeduufe
= ∑ ∫∞
=
+−
0
)1(
)(n
Tn
nT
su duufe
Let us set u = t + nT, then
L f(t) = ∑ ∫∞
= =
+− +0 0
)( )(n
T
t
nTts dtnTtfe
Here f(t+nT) = f(t), by periodic property Hence
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∫∑ −∞
=
−=T
st
n
nsT dttfeetLf00
)()()(
= ∫ −−
−
Tst
ST dttfee 0
)(1
1 , identifying the above series as a geometric series.
Thus
L f(t) = ∫ −−
−
Tst
sT dttfee 0
)(1
1
This is the desired result. 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,
L f(t) = ∫ −−
−
Tst
sT dttfee 0
)(1
1 = ∫ −−
−
4
04 )(
11 dttfee
sts
=
+
− ∫∫ −−−
4
2
2
04 63
11 dtedttee
ststs
=
−
+
−−
−−
−−−
− ∫4
2
2
0
2
04 6.13
11
sedt
se
set
e
ststst
s
= ( )
−−−
−−
− 2
42
4
2131
1s
seee
ss
s
Thus,
Lf(t) = )1(
)21(342
42
s
ss
essee
−
−−
−−−
2. A periodic function of period ωπ2 is defined by
Esinωt, 0 ≤ t < ωπ
f(t) =
0 , ωπ ≤ t ≤
ωπ2
where E and ω are positive constants. Show that
L f(t) = )1)(( /22 wsews
Eπ
ω−−+
Here
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T = ωπ2 . Therefore
L f(t) = ∫ −−−
ωπ
ωπ
/2
0)/2( )(
11 dttfe
est
s = ∫ −−−
ωπ
ωπ ω/
0)/2( sin
11 tdtEe
est
s
= { }ωπ
ωπ ωωωω
/
022)/2( cossin
1
−−
+−
−
− ttss
ee
E st
s
= 22
/
)/2(
)1(1 ω
ω ωπ
ωπ ++
−
−
− se
eE s
s
= ))(1)(1(
)1(22//
/
ωω
ωπωπ
ωπ
++−+
−−
−
seeeE
ss
s
= ))(1( 22/ ω
ωωπ +− − seE
s
This is the desired result. 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
L f(t) =
2tanh as
sE
Here T = 2a. Therefore
L f(t) = ∫ −−−
ast
as dttfee
2
02 )(
11
=
−+
− ∫ ∫ −−−
a a
a
ststas dtEedtEe
e 0
2
211
= ( )[ ])(1)1(
22
asassaas eee
esE −−−
− −+−−
= ( )[ ])1)(1(
)1(1)1(
22
2 asas
asas
as eeseEe
esE
−−
−−
− +−−
=−−
=
=
+−
−
−
2tanh2/2/
2/2/ assE
eeee
sE
asas
asas
This is the result as desired.
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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 0 a t Note that the value of the function suddenly jumps from value zero to the value 1 as at → 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)] = ∫∞
− −0
)( dtatHe st
= ∫ ∫∞
−− +a
a
stst dtedte0
)1(0
= s
e as−
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In particular, we have L H(t) = s1
Also, )(1 atHs
eLas
−=
−− and )(11 tH
sL =
−
Heaviside shift theorem Statement :- L [f(t-a) H(t-a)] = e-as Lf(t) Proof :- We have
L [f(t-a) H(t-a)] = ∫∞
−−−0
)()( dteatHatf st
= ∫∞
− −a
st dtatfe )(
Setting t-a = u, we get
L[f(t-a) H(t-a)] = duufe uas )(0
)(∫∞
+−
= e-as L f(t) This is the desired shift theorem. Also, L-1 [e-as L f(t)] = f(t-a) H(t-a) Examples : 1. Find L[et-2 + sin(t-2)] H(t-2) Let f(t-2) = [et-2 + sin(t-2)] Then f(t) = [et + sint] so that
L f(t) = 1
11
12 +
+− ss
By Heaviside shift theorem, we have L[f(t-2) H(t-2)] = e-2s Lf(t) Thus,
++
−=−−+ −−
11
11)2()]2sin([ 2
2)2(
ssetHteL st
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
sss
tLf 886)( 23 ++=
Thus L[3t2 +2t +3] H(t-1) = L[f(t-1) H(t-1)] = e-s L f(t)
=
++−
ssse s 886
23
3. Find Le-t H(t-2) Let f(t-2) = e-t , so that, f(t) = e-(t+2) Thus,
L f(t) = 1
2
+
−
se
By shift theorem, we have
1
)()]2()2([)1(2
2
+==−−
+−−
setLfetHtfL
ss
Thus
[ ]1
)2()1(2
+=−
+−−
setHeL
st
4. Let f(t) = 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. 1. t2 , 1 < t ≤ 2 f(t) = 4t , t > 2 2. cost, 0 < t < π
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f(t) = sint, t > π 1. Here, f(t) = t2 + (4t-t2) H(t-2) Hence,
L f(t) = )2()4(2 23 −−+ tHttL
s (i)
Let φ(t-2) = 4t – t2 so that φ(t) = 4(t+2) – (t+2)2 = -t2 + 4 Now,
ss
tL 42)( 3 +−=φ
Expression (i) reads as
L f(t) = [ ])2()2(23 −−+ tHtL
sφ
= )(2 23 tLe
ss φ−+
=
−+ −
32
3
242ss
es
s
This is the desired result. 2. Here f(t) = cost + (sint-cost)H(t-π) Hence,
L f(t) = )()cos(sin12 π−−+
+tHttL
ss (ii)
Let φ (t-π) = sint – cost Then φ(t) = sin(t + π) – cos(t + π) = -sint + cost so that
L φ(t) = 11
122 +
++
−s
ss
Expression (ii) reads as
L f(t) = [ ])()(12 ππφ −−+
+tHtL
ss
= )(12 tLe
ss s φπ−++
=
+−
++
−
11
1 22 sse
ss sπ
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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
f(t) ∗ g(t) = ∫ −t
duugutf0
)()(
Property : f(t) ∗ g(t) = g(t) ∗ f(t) Proof :- By definition, we have
f(t) ∗ g(t) = ∫ −t
duugutf0
)()(
Setting t-u = x, we get
f(t) ∗ g(t) = ∫ −−0
))(()(t
dxxtgxf
= ∫ ∗=−t
tftgdxxfxtg0
)()()()(
This is the desired property. Note that the operation ∗ is commutative. Convolution theorem :- L[f(t) ∗ g(t)] = L f(t) . L g(t) Proof :- Let us denote
f(t) ∗ g(t) = φ(t) = ∫ −t
duugutf0
)()(
Consider
∫ ∫∞
− −=0 0
)]()([)]([t
st dtugutfetL φ
= ∫ ∫∞
− −0 0
)()(t
st duugutfe (1)
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=u t = ∞ 0 u=0 t Hence (1) becomes
L[φ(t)] = ∫ ∫∞
=
∞
=
− −0
)()(u ut
st dtduugutfe
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= ∫ ∫∞ ∞
−−−
−0
)( )()( dudtutfeugeu
utssu
= ∫ ∫∞ ∞
−−
0 0
)()( dudvvfeuge svsu , where v = t-u
= ∫ ∫∞ ∞
−−
0 0
)()( dvvfeduuge svsu
= L g(t) . L f(t) Thus L f(t) . L g(t) = L[f(t) ∗ g(t)] This is desired property. Examples : 1. Verify Convolution theorem for the functions f(t) and g(t) in the following cases : (i) f(t) = t, g(t) = sint (ii) f(t) =t, g(t) = et (i) Here,
f ∗ g = ∫ −t
duutguf0
)()( = ∫ −t
duutu0
)sin(
Employing integration by parts, we get f ∗ g = t – sint so that
L [f ∗ g] = )1(
11
112222 +
=+
−ssss
(1)
Next consider
L f(t) . L g(t) = )1(
11
112222 +
=+
⋅ssss
(2)
From (1) and (2), we find that L [f ∗ g] = L f(t) . L g(t) Thus convolution theorem is verified. (ii) Here
f ∗ g = ∫ −t
ut duue0
Employing integration by parts, we get f ∗ g = et – t – 1 so that
L[f ∗ g] = )1(
1111
122 −
=−−− sssss
(3)
Next
L f(t) . L g(t) = )1(
11
1122 −
=−
⋅ssss
(4)
From (3) and (4) we find that
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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
tLfs
dttfL0
)(1)(
Let us define g(t) = 1, so that g(t-u) = 1 Then
∫ ∫ ∗=−=t t
gfLdtutgtfLdttfL0 0
][)()()(
= L f(t) . L g(t) = L f(t) . s1
Thus
∫ =t
tLfs
dttfL0
)(1)(
This is the result as desired. 3. Using Convolution theorem, prove that
∫ ++=−−
tu
ssduuteL
02 )1)(1(
1)sin(
Let us denote, f(t) = e-t g(t) = sin t, then
∫ ∫ −=−−t t
u duutgufLduuteL0 0
)()()sin( = L f(t) . L g(t)
= )1(
1)1(
12 +
⋅+ ss
= )1)(1(
12 ++ ss
This is the result as desired.
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ASSIGNMENT 1. By using the Laplace transform of coshat, find the Laplace transform of sinhat.
2. Find (i) ∫t
dtt
tL0
sin (ii) ∫ −t
t tdteL0
cos (iii) ∫ −t
t tdteL0
cos
(iv) ∫
t
t dtet
tL0
sin (v) ∫t
atdttL0
2 sin
3. If f(t) = t2, 0 < t < 2 and f(t+2) = f(t) for t > 2, find L f(t) 4. Find L f(t) given f(t) = t , 0 ≤ t ≤ a f(2a+t) = f(t) 2a – t, a < t ≤ 2a
5. Find L f(t) given f(t) = 1, 0 < t < 2a f(a+t) = f(t)
-1, 2a < t < a
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 - π) 7. Express the following functions in terms of unit step function and hence find their Laplace transforms : (i) 2t, 0 < t ≤ π (ii) t2 , 2 < t ≤ 3 f(t) = f(t) = 1 , t > π 3t , t > 3 (iii) sin2t, 0 < t ≤ π (iv) sint, 0 < t ≤ π/2 f(t) = f(t) = 0 , t > π cost, t > π/2 8. Let f1 (t) , t ≤ a f(t) = 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. Sint, 0 < t ≤ π f(t) = Sin2t, π < t ≤ 2π Sin3t, t > 2π
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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:
(i) ∫ ++=− −
tu
sssudueutL
022
1
)1()1(cos)(
(ii) 220 )(
1)(ass
duueutLt
au
+=−∫ −
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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(s) )()( 1 sFLtf −=
0,1>s
s
1
asas
>−
,1 ate
0,22 >+
sas
s
Cos at
0,122 >
+s
as aatSin
asas
>−
,122
aath Sin
as
ass
>−
,22
ath Cos
0,11 >+ s
sn
n = 0, 1, 2, 3, . . . !n
tn
0,11 >+ s
sn
n > -1 ( )1+Γ ntn
Examples 1. Find the inverse Laplace transforms of the following:
22)(
asbsii
++
22 994
25452)(
ss
ssiii
−−
++−
Here
521)(−s
i
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25
11
21
25
121
521)(
t
es
Ls
Li =−
=−
−−
at
abat
asLb
assL
asbsLii sincos1)( 22
122
122
1 +=+
++
=++ −−−
92
94
4252
5
42
984
25452)( 2
1
2
122
1
−
−−
+
−=
−−
++− −−−
s
sL
s
sL
ss
ssLiii
−−
−= ththtt 3sin
233cos4
25sin
25cos
21
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
( )41
113: Evaluate 1.
++−
ssL
( )( ) ( ) ( )4
13
14
1-
112
113
111-1s3 L Given
+−
+=
+++
= −−
sL
sL
s
41
31 1213
sLe
sLe tt −−−− −=
Using the formula
getweandntakingandnt
sL
n
n ,32!
11
1 ==+−
323 32 teteGiven
tt −−
−=
52s-s2sL : Evaluate 2. 2
1-
++
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( )( )
( ) ( ) ( ) 4113
411
4131
412 L Given 2
12
12
12
1-
+−+
+−
−=
+−
+−=
+−
+= −−−
sL
ssL
ssL
ss
413
4 21
21
++
+= −
sLe
ss Le t-t
tet e tt 2sin
232cos +=
1312: Evaluate 2
1
+++−
sssL
( )
( )( )
( ) ( )
−+−
−+
+=
−+
−+= −−
45
1
45
24
51
2L Given 2
23
12
23
23
12
23
23
1-
sL
s
sL
s
s
−−
−= −
−−
−
45
1
452
212
3
212
3
sLe
ssLe
tt
=
−
ththet
25sin
52
25cos2 2
3
sssss
2452 : Evaluate 4. 23
2
−+−+
havewe
( ) ( )( ) 1212452
2452
2452 2
2
2
23
2
−+
++=
−+−+
=−+−+
=−+−+
sC
sB
sA
sssss
sssss
sssss
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
11
212
2452
23
2
−+
+−=
−+−+
ssssssss
Hence
tt ee
ssssL +−=−+−+ −− 2
22
21 2
25452
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( ) ( )2154: Evaluate 5. 2
1
++++−
sssL
Let us take
( ) ( ) ( ) 2112154
22 ++
++
+=
++++
sC
sB
sA
sss
Then
4s + 5 = A(s + 2) + B(s + 1) (s + 2) + C (s + 1)2
For s = -1, we get A = 1, for s = -2, we get C = -3 Comparing the coefficients of s2, we get B + C = 0, so that B = 3. Using these values in (1), we get
( ) ( ) ( ) ( ) 23
13
11
2154
22 +−
++
+=
++++
ssssss
Hence
( ) ( ) sLe
sLe
sLe
sssL ttt 13131
2154 121
21
21 −−−−−−− −+=
++++
ttt eete 233 −−− −+=
44
31: Evaluate 5.
assL−
−
Let
)1(2244
3
asDCs
asB
asA
ass
++
++
+−
=−
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
2244
3
2111
41
ass
asasass
++
++
−=
− Taking inverse transforms, we get
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[ ] ateeas
sL atat cos21
41
44
31 ++=
−−−
[ ]athat coscos21
+=
1: Evaluate 6. 24
1
++−
sssL
Consider
( ) ( ) ( )( )
+−++=
+−+++=
++ 112
21
111 222224 sssss
sssss
sss
( ) ( )( )( ) ( ) ( )
( ) ( )
+−
+=
++
++−
+−=
++−
+−=
+−+++−−++
=
−−−−
43
1
43
121
1
43
1
43
121
11
11
21
1111
21
2
121
2
121
241
2
212
21
2222
22
sLe
sLe
sssL
Therefore
ss
ssssssssssss
tt
−=−
2323sin
2323sin
21 2
121 t
et
ett
=
2sin
23sin
32 tht
Evaluation of L-1[e-as F(s)] We have, if Lf(t) = F(s), then L[f(t-a) H(t-a) = e-as F(s), and so
L-1[e-as F(s)] = f(t-a) H(t-a)
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Examples
( )41
2:)1(
−
−−
seLEvaluate
ss
Here
( )
( )
( )( )( ) ( )5
65
)()(2
61
21)()(
21)(,5
352
4
51
32
412
411
4
−−
=
−−=−
==−
==
−==
−
−−
−−−
tHte
atHatfseL
Thus
tes
Les
LsFLtfTherefore
ssFa
t
s
tt
++
+
−−−
41:)2( 2
2
21
sse
seLEvaluate
ss ππ
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )ππ
ππππ
ππππ
22coscos
222cossin
)1(
2cos4
)(
sin1
1)(
)1(22
21
2
21
1
21
−+−−=
−−+−−=
=+
=
=+
=
−−+−−=
−
−
tHttHt
tHttHtGiven
asreadsrelationNow
ts
sLtf
ts
LtfHere
tHtftHtfGiven
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Inverse transform of logarithmic and inverse functions
( )[ ] ( ) .ttfL then F(s), f(t) L if have, We HencesFdsd
−==
( ) )(1 ttfsFdsdL =
−−
Examples
++−
bsasLEvaluate log:)1( 1
( ) ( )
( )
( ) [ ]
( )
( )b
eetfThus
eetftor
eesFdsdLthatSo
bsassF
dsdThen
bsasbsassFLet
atbt
atbt
btat
−−
−−
−−−
−=
−=
−−=
−
+−
+−=−
+−+=
++
=
1
11
logloglog)(
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(2) Evaluate 1−L
−
sa1tan
( )
( )
( )
( )
( )
( ) ( )
( ) ( )∫
∫
=
=
=
=
=
−
+=−
=
−
−
−
t
t
dttfssFL
havewessFdttfSinceL
aattf
attftor
thatsoatsFdsdLor
asasF
dsdThen
sasFLet
0
1
0
1
22
1
,,
sin
sin
sin
tan)(
ssFoftransformInverse
Examples
( )
+
−22
1 1:)1(ass
LEvaluate
( )
( )
( )( )
( )2
0
122
1
1
22
cos1
sin1
sin)(
1
aat
dta
atssFL
assLThen
aatsFLtf
thatsoas
sFdenoteusLet
t
−=
==+
=
==+
=
∫−−
−
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( )
( )
( )
( )[ ]
( )( )[ ]
( ) ( )[ ]
[ ]
[ ] ( ) ( )∫
∫
∫
−=∗=
=⋅=∗
==
−++=
+−=+
+−=
=+
=+
+
−
−−
−
−
−−
−−
−
t
atat
tat-
at
tat
at
duugutftgtfsGsFL
soandsGsFtLgtLf
eeata
dtateaass
L
atea
dtteass
LHence
teas
Lhavewe
assLEvaluate
0
1
3
0222
1
2
02
1
21
221
)()()()(
)()()()(g(t) f(t)L
thenG(s), Lg(t) and F(s) L(t) if have, We
1211
1111
get we this,Using
parts.by n integratioon,111
1
1
1:)2(
theorem nconvolutio using by F(s) of transform Inverse
transformLaplace inversefor n theoremconvolutio thecalled is expression This
NEXT CLASS 24.05.05
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Examples
Employ convolution theorem to evaluate the following :
( )( )bsasL
++− 1)1( 1
( )( )( ) ( )
( )
baee
baee
dueedueebsas
L
e, g(t) ef(t)
bssG
as
atbttbaat
tubaat
tbuuta-
-bt-at
−−
=
−
−=
==++
==
+=
+=
−−−−
−−−−− ∫∫
1
1
n theorem,convolutioby Therefore,
get weinverse, theTaking
1)(,1 F(s) denote usLet
00
1
( )2221)2(
as
sL+
−
( ) ( )
( )
( )a
atta
auatatu
duauatata
duauutaaas
sL
at , g(t) at f(t)
asssG
asF(s)
t
t
t-
2sin
22cossin
2a1
formula angle compound usingby ,2
2sinsin1
cossin1
n theorem,convolutioby Hence
cosa
sin
Then)(,1 denote usLet
0
0
0222
1
2222
=
−−
−=
−+=
−=+
==
+=
+=
∫
∫
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( )( )11)3( 2
1
+−−
sssL
Here
1)(,
11 2 +
=−
=s
ssGs
F(s)
Therefore
( )( ) ( )
( ) ( )[ ] [ ]ttettee
uueeduuess
L
t , g(t) ef(t)
ttt
tutut-
t
cossin211cossin
2
cossin2
sin11
1
have wen theorem,convolutioBy
sin
02
1
−−=−−−−=
−−==
+−
==
−
−−∫
Assignment
By employing convolution theorem, evaluate the following :
( )( ) ( )( )
( ) ( ) ( )
( ) ( ) ( )2154)6(1)3(
11)5(
11)2(
,)4(11
1)1(
21
2221
221
221
2222
21
21
+−+
+
+++
≠++++
−−
−−
−−
sssL
asL
ssL
sssL
babsas
sLss
L
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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.
(o)f - (o)f s - f(o) s - f(t) L s (t)f[ L
(o)f - f(o) s - f(t) L s (t)fL[
f(o) - f(t) L s (t)]fL[
23
2
′′′=′′′
′=′′
=′
Examples
1) Solve by using Laplace transform method 2 y(o) t y y , ==+′ −te
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Taking the Laplace transform of the given equation, we get
( ) ( )[ ] ( )( )
( ) ( )( )
( )( )
( )( )
( ) ( )( )
( )
( )421
11
12
13114112
1342
get we, transformLaplace inverse theTaking
1342
1121s
becomes thiscondition,given theUsing
11
2
31
3
21
3
21
3
2
2
2
+=
++
+=
++−++−+
=
+++
=
+++
=
+=−+
+=+−
−
−
−
−
te
ssL
sssL
sssLtY
ssstyL
thatso
styL
styLoytysL
t
This is the solution of the given equation. Solve by using Laplace transform method :
0)()(,sin32)2( =′==−′+′′ oyoytyyy Taking the Laplace transform of the given equation, we get
[ ] [ ]1
1)(3)()(2)()()( 22
+=−−+′−−
styLoytLysoyosytLys
Using the given conditions, we get
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[ ]
( )( )( )
( )( )( )
( )
equation.given theofsolution required theis This
sin2cos101
401
81
sums, partial of method theusingby 151
103
1401
11
81
131
1311)(
1311)(
1132)(
3
21
21
21
2
22
ttee
s
s
ssL
sDCs
sB
sAL
sssLty
orsss
tyLor
ssstyL
tt +−−=
+
−−+
+−
−=
++
++
+−
=
++−
=
++−=
+=−+
−
−
−
−
equation. integral thesolve tomethod Transform LaplaceEmploy 3)
( ) ( )∫ −+=t
0
sinlf(t) duutuf
( ) ( )
equation. integralgiven theofsolution theis This
211)(1)(
1)(1sin)(1
get wehere, n theorem,convolutio usingBy
sin1
get weequation,given theof transformLaplace Taking
2
3
21
3
2
2
t
0
ts
sLtfors
stfL
Thus
stfL
stLtLf
sL f(t)
duutufLs
L f(t)
+=
+=
+=
++=⋅+=
−+=
−
∫
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s. transformLaplace using t any timeat particle theofnt displaceme thefind 10, is speed initial theand 20at x is particle theofposition initial theIf t.at time particle theofnt displaceme thedenotesx
where025dtdx
6dt
xdequation the,satisfyingpath a along moving is particleA 4)(
2
2
=
=++ x
get weequation, theof transformLaplace theTaking10. (o) x'20, x(o)are conditions initial theHere
025x(t)(t)6x'(t)'x'
asrewritten bemay equation Given
==
=++
[ ]
( )( )
( )
( ) ( )
problem.given theofsolution desired theis This
24sin354cos20
163170
163320
16370320
16313020x(t)
25613020Lx(t)
013020256ssLx(t)
33
21
21
21
21
2
2
tete
sL
ssL
ssL
ssL
thatso
sss
ors
tt
−−
−−
−−
+=
+++
+++
=
++++
=
+++
=
+++
=
=−−++
−
=
−− LRt
at eeaL-R
E is t any timeat current
that theShow R. resistance and L inductance ofcircuit a to0at t applied is Ee A voltage (5) -at
The circuit is an LR circuit. The differential equation with respect to the circuit is
)(tERidtdiL =+
Here L denotes the inductance, i denotes current at any time t and E(t) denotes the E.M.F. It is given that E(t) = E e-at. With this, we have Thus, we have
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at
at
EetiRtLi
orEeRidtdiL
−
−
=+
=+
)()('
[ ] [ ] ( ) orER at−=+ eL(t)i'L(t)i'LL TTT
get wesides,both on )( transformLaplace Taking TL
[ ] [ ]as
EtiLRoitiLsL TT +=+−
1)()()(
[ ]
( ) ( )
))(()(get we,L transforminverse Taking
)(
)(get weo, i(o) Since
11-T RsLas
ELti
RsLasEtiL
oras
ERsLtiL
T
T
T
++=
+++=
+=+=
−
desired. asresult theis This
)(
11 11
−
−=
+−
+−=
−−
−−
LRt
at
TT
eeaLR
Eti
Thus
RsLLL
asL
aLRE
1. y(o) 2, with x(o)09dtdy
,04dtdxgiven t of in termsy andfor x equations ussimultaneo theSolve )6(
===−
=+
x
y
Taking Laplace transforms of the given equations, we get
[ ][ ] 0)()()(9
0)(4 x(o)- Lx(t) s
=−+−
=+
oytLystxL
tLy
get we,conditions initialgiven theUsing
1)(5)(9
2)(4)(=+−
=+tyLtxL
tyLtxLs
get weLy(t),for equations theseSolving
3618
2 ++
=ss L y(t)
thatso
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tt
sss Ly(t)
6sin36cos
3618
36 221
+=
++
+= −
(1)
get we,09dtdyin thisUsing =− x
[ ]tt 6cos186sin6
91x(t) +−=
or
[ ]tt 6sin6cos3
32x(t) −=
(2) (1) and (2) together represents the solution of the given equations.
Assignment
Employ Laplace transform method to solve the following initial – value problems
( )
( )
( )∫
∫
=−+=
−+=
=′=−=+′′
−=−==+′+′′
−=
==+′′
=′=++=+′+′′
′===+′+′′
=′==+′+′′
=′′=′==−′−′′+′′′
−
−
t
tu
t
t
fduuutfttf
dueutftf
yytHyy
yytyyy
yytyy
yyttyyy
yyeyyy
yyeyyy
yyygivenyyyy
0
0
2
2
2
4)0(,cos)(')9
21)()8
1)0(,0)0(,1)7
1)1(,3)0(,2)6
12
,1)0(,2cos9)5
0)0(,2)0(,22223)4
)0(1)0(,34)3
1)0(,2)0(,65)2
6)0(,0)0()0(022)1
π
any time.at current thefind initially, zero iscurrent theIf 0. at t applied issin voltagea circuit, R-Lan In 11)
t.any timeat nt displaceme the
find rest,at initially is particle theIf .5sin805dtdx4
dtxd
equation by the governed is t any timeat point fixed a fromnt x displaceme its that so line a along moves particleA )10
2
2
=
=++
t E
tx
ω
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3)0(,8)0(,2dtdy,23
dtdx)15
3)0(,8)0(,2dtdy,32
dtdx)14
0)0(,1)0(,sindtdy,
dtdx)13
0)0(,2)0(,cosdtdy,sin
dtdx 12)
:
===+=+
==−=−=
===+=−
===+=+
yxyxxy
yxxyyx
yxtxey
yxtxty
t
t of terms in y andx for equations aldifferenti ussimultaneo following the Solve
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 xydxdy
dxyd
=++2
2
with conditions y(0) =
y′ (0) = 1 is an initial value problem
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Example 2 : The problem of solving the equation xydxdy
dxyd cos23 2
2
=++ with y(1)=1,
y(2)=3 is called Boundary value problem.
Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. 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.
Definition :
Let f(t) be a real-valued function defined for all t ≥ 0 and s be a parameter, real or complex.
Suppose the integral ∫∞
−
0
)( dttfe st exists (converges). Then this integral is called the Laplace
transform of f(t) and is denoted by Lf(t).
Thus,
Lf(t) = ∫∞
−
0
)( dttfe st (1)
We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t) is a function of s denoted by F(s) or )(sf .
LAPLACE TRANSFORMS
Definition and Properties
Transforms of some functions
Convolution theorem
Inverse transforms
Solution of differential equations
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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 :
f1(t), 0 < t < a
f(t) = 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) = ∫∞
−
0
)(tfe st
= ∫ ∫ ∫∞
−−− ++a b
a b
ststst dttfedttfedttfe0
321 )()()(
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 φ(t) (whose Laplace transforms exist)
and any two constants a and b, we have
L [a f(t) + b φ(t)] = a L f(t) + b Lφ(t)
Proof :- By definition, we have
L[af(t)+bφ(t)] = [ ]dttbtafe st∫∞
− +0
)()( φ = ∫ ∫∞ ∞
−− +0 0
)()( dttebdttfea stst φ
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= a L f(t) + b Lφ(t)
This is the desired property.
In particular, for a=b=1, we have
L [ f(t) + φ(t)] = L f(t) + Lφ(t)
and for a = -b = 1, we have
L [ f(t) - φ(t)] = L f(t) - Lφ(t)
2. Change of scale property : If L f(t) = F(s), then L[f(at)] =
asF
a1 , where a is a positive
constant.
Proof :- By definition, we have
Lf(at) = ∫∞
−
0
)( dtatfe st (1)
Let us set at = x. Then expression (1) becomes,
L f(at) = ∫∞
−
0
)(1 dxxfea
xas
=
asF
a1
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)] = [ ]∫∞
−
0
)( dttfee atst = ∫∞
−−
0
)( )( dttfe as
= 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
L(eat) = ∫ ∫∞ ∞
−−− =0 0
)( dtedtee tasatst
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= asas
e tas
−=
−−
∞−− 1)( 0
)(
, s > a
Thus,
L(eat) = as −
1
In particular, when a=0, we get
L(1) = s1 , s > 0
By inversion formula, we have
atat es
Leas
L ==−
−− 11 11
4. L(cosh at) =
+ −
2
atat eeL = [ ]∫∞
−− +02
1 dteee atatst
= [ ]∫∞
+−−− +0
)()(
21 dtee tastas
Let s > |a| . Then,
∞+−−−
+−
+−−
=0
)()(
)()(21)(cosh
ase
aseatL
tastas
= 22 ass−
Thus,
L (cosh at) = 22 ass−
, s > |a|
and so
atas
sL cosh221 =
−−
3. L (sinh at) = 222 asaeeL
atat
−=
− −
, s > |a|
Thus,
L (sinh at) = 22 asa−
, s > |a|
and so,
a
atas
L sinh122
1 =
−−
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4. L (sin at) = ∫∞
−
0
sin ate st dt
Here we suppose that s > 0 and then integrate by using the formula
[ ]∫ −+
= bxbbxaba
ebxdxeax
ax cossinsin 22
Thus,
L (sinh at) = 22 asa+
, s > 0
and so
a
atas
L sinh122
1 =
+−
5. L (cos at) = atdte st cos0∫∞
−
Here we suppose that s>0 and integrate by using the formula
[ ]∫ ++
= bxbbxaba
ebxdxeax
ax sincoscos 22
Thus,
L (cos at) = 22 ass+
, s > 0
and so
atas
sL cos221 =
+−
6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then
L(tn) = ∫∞
−
0
dtte nst
Let s > 0 and set st = x, then
∫ ∫∞ ∞
−+
− =
=
0 01
1)( dxxess
dxsxetL nx
n
nxn
The integral dxxe nx∫∞
−
0
is called gamma function of (n+1) denoted by )1( +Γ n . Thus
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1
)1()( +
+Γ= n
n
sntL
In particular, if n is a non-negative integer then )1( +Γ n =n!. Hence
1
!)( += nn
sntL
and so
)1(
11
1
+Γ=+
−
nt
sL
n
n or !n
t n
as the case may be
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TABLE OF LAPLACE TRANSFORMS
f(t) F(s)
1 s1 , s > 0
eat as −
1 , s > a
coshat 22 ass−
, s > |a|
sinhat 22 asa−
, s > |a|
sinat 22 asa+
, s > 0
cosat 22 ass+
, s > 0
tn, n=0,1,2,….. 1!+ns
n , s > 0
tn, n > -1 1)1(
+
+Γns
n , s > 0
Application of shifting property :-
The shifting property is
If L f(t) = F(s), then L [eatf(t)] = F(s-a)
Application of this property leads to the following results :
1. [ ]ass
assat
bssbtLbteL
−→−→
−== 22)(cosh)cosh( = 22)( bas
as−−
−
Thus,
L(eatcoshbt) = 22)( basas−−
−
and
btebas
asL at cosh)( 22
1 =−−
−−
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2. 22)()sinh(
basabteL at
−−=
and
btebas
L at sinh)(1
221 =
−−−
3. 22)()cos(
basasbteL at
+−−
=
and
btebas
asL at cos)( 22
1 =+−
−−
4. 22)()sin(
basbbteL at
−−=
and
b
btebas
Lat sin
)(1
221 =
−−−
5. 1)()1()( +−
+Γ= n
nat
asnteL or 1)(
!+− nas
n as the case may be
Hence
)1()(
11
1
+Γ=
− +−
nte
asL
nat
n or 1)(!
+− nasn as the case may be
Examples :-
1. Find Lf(t) given f(t) = t, 0 < t < 3
4, t > 3
Here
Lf(t) = ∫ ∫ ∫∞ ∞
−−− +=0
3
0 3
4)( dtetdtedttfe ststst
Integrating the terms on the RHS, we get
Lf(t) = )1(11 32
3 ss es
es
−− −+
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This is the desired result.
2. Find Lf(t) given f(t) = sin2t, 0 < t ≤ π
0, t > π
Here
Lf(t) = ∫∫∞
−− +π
π
dttfedttfe stst )()(0
= ∫ −π
0
2sin tdte st
= { }π
02 2cos22sin
4
−−
+
−
ttsse st
= [ ]ses
π−−+
14
22
This is the desired result.
3. Evaluate : (i) L(sin3t sin4t)
(iv) L(cos2 4t)
(v) L(sin32t)
(i) Here
L(sin3t sin4t) = L [ )]7cos(cos21 tt −
= [ ])7(cos)(cos21 tLtL − , by using linearity property
= )49)(1(
244912
12222 ++
=
+−
+ sss
ss
ss
(ii) Here
L(cos24t) =
++=
+
641
21)8cos1(
21
2ss
stL
(iii) We have
( )θθθ 3sinsin341sin3 −=
For θ=2t, we get
( )ttt 6sin2sin3412sin3 −=
so that
)36)(4(
4836
64
641)2(sin 2222
3
++=
+−
+=
sssstL
This is the desired result.
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4. Find L(cost cos2t cos3t)
Here
cos2t cos3t = ]cos5[cos21 tt +
so that
cost cos2t cos3t = ]coscos5[cos21 2 ttt +
= ]2cos14cos6[cos41 ttt +++
Thus
L(cost cos2t cos3t) =
+++
++
+ 41
163641
222 ss
sss
ss
5. Find L(cosh22t)
We have
2
2cosh1cosh2 θθ +=
For θ = 2t, we get
2
4cosh12cosh2 tt +=
Thus,
−+=
161
21)2(cosh 2
2
ss
stL
6. Evaluate (i) L( t ) (ii)
t
L 1 (iii) L(t-3/2)
We have L(tn) = 1)1(
+
+Γns
n
(i) For n= 21 , we get
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L(t1/2) = 2/3
)121(
s
+Γ
Since )()1( nnn Γ=+Γ , we have 22
1211
21 π
=
Γ=
+Γ
Thus, 2
32
)(s
tL π=
(ii) For n = -21 , we get
ss
tL π=
Γ
=−
21
21 2
1
)(
(iii) For n = -23 , we get
sss
tL ππ 2221
)(2
12
12
3−=
−=
−Γ
=−−
−
7. Evaluate : (i) L(t2) (ii) L(t3)
We have,
L(tn) = 1!+ns
n
(i) For n = 2, we get
L(t2) = 332!2ss
=
(ii) For n=3, we get
L(t3) = 446!3ss
=
8. Find L[e-3t (2cos5t – 3sin5t)]
Given =
2L(e-3t cos5t) – 3L(e-3t sin5t)
= ( )25)3(
1525)3(
32 22 ++−
+++
sss , by using shifting property
= 346
922 ++
−ss
s , on simplification
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9. Find L[coshat sinhat]
Here
L[coshat sinat] = ( )
+ −
ateeLatat
sin2
=
++
++− 2222 )()(2
1aas
aaas
a
, on simplification
10. Find L(cosht sin3 2t)
Given
−
+ −
46sin2sin3
2tteeL
tt
= ( )[ ])6sin()2sin(3)6sin(2sin381 teLteLteLteL tttt −− −+−⋅
=
++
−++
++−
−+− 36)1(
64)1(
636)1(
64)1(
681
2222 ssss
=
++
−++
++−
−+− 36)1(
124)1(
136)1(
14)1(
143
2222 ssss
11. Find )( 254 −− teL t
We have
L(tn) = 1)1(
+
+Γns
n Put n= -5/2. Hence
L(t-5/2) = 2/32/3 34)2/3(
−− =−Γ
ssπ Change s to s+4.
Therefore, 2/32/54
)4(34)( −
−−
+=
steL t π
Transform of tn f(t) Here we suppose that n is a positive integer. By definition, we have
F(s) = ∫∞
−
0
)( dttfe st
Differentiating ‘n’ times on both sides w.r.t. s, we get
])][()[()2(
2222
22
aasaasasa
+++−+
=
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∫∞
−
∂∂
=0
)()( dttfes
sFdsd st
n
n
n
n
Performing differentiation under the integral sign, we get
∫∞
−−=0
)()()( dttfetsFdsd stn
n
n
Multiplying on both sides by (-1)n , we get
∫∞
− ==−0
)]([)(()()1( tftLdtetftsFdsd nstn
n
nn , by definition
Thus,
L[tnf(t)]= )()1( sFdsd
n
nn−
This is the transform of tn f(t).
Also,
)()1()(1 tftsFdsdL nn
n
n
−=
−
In particular, we have
L[t f(t)] = )(sFdsd
− , for n=1
L[t2 f(t)]= )(2
2
sFdsd , for n=2, etc.
Also,
)()(1 ttfsFdsdL −=
− and
)()( 22
21 tftsF
dsdL =
−
Transform of t
f(t)
We have , F(s) = ∫∞
−
0
)( dttfe st
Therefore,
∫∞
∫∞
∫∞ −=
s sdsdttfstedssF
0)()( = dtdsetf
s
st∫ ∫∞ ∞
−
0
)(
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= ∫∞ ∞−
−0
)( dtt
etfs
st
= ∫∞
−
=
0
)()(ttfLdt
ttfe st
Thus,
∫∞
=
s
dssFttfL )()(
This is the transform of ttf )(
Also,
∫∞
− =s t
tfdssFL )()(1
Examples : 1. Find L[te-t sin4t]
We have,
16)1(
4]4sin[ 2 ++=−
steL t
So that,
L[te-t sin4t] =
++−
17214 2 ssds
d
= 22 )172()1(8
+++ss
s
2. Find L(t2 sin3t)
We have
L(sin3t) = 9
32 +s
So that,
L(t2 sin3t) =
+ 93
22
2
sdsd = 22 )9(
6+
−s
sdsd = 32
2
)9()3(18
+−
ss
3. Find
−
tteL
t sin
We have
1)1(
1)sin( 2 ++=−
steL t
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Hence
−
tteL
t sin = [ ]∫∞
∞− +=++0
12 )1(tan
1)1( sss
ds
= )1(tan2
1 +− − sπ = cot –1 (s+1)
4. Find
ttL sin . Using this, evaluateL
tatsin
We have
L(sint) = 1
12 +s
So that
Lf(t) =
ttL sin = [ ]∞−
∞
=+∫ S
s
ss
ds 12 tan
1
= )(cottan2
11 sFss ==− −−π
Consider
L
tatsin = a L )(sin ataLf
atat
=
=
asF
aa 1 , in view of the change of scale property
=
−
as1cot
5. Find L
−
tbtat coscos
We have
L [cosat – cosbt] = 2222 bss
ass
+−
+
So that
L
−
tbtat coscos = ds
bss
ass
s∫∞
+−
+ 2222 = ∞
++
sbsas
22
22
log21
=
++
−
++
∞→ 22
22
22
22
loglog21
bsas
bsasLt
s
=
++
+ 22
22
log021
asbs =
++
22
22
log21
asbs
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6. Prove that ∫∞
− =0
3
503sin tdtte t
We have
∫∞
− =0
)sin(sin ttLtdtte st = )(sin tLdsd
− =
+−
11
2sdsd
= 22 )1(2+ss
Putting s = 3 in this result, we get
∫∞
− =0
3
503sin tdtte t
This is the result as required.
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ASSIGNMENT I Find L f(t) in each of the following cases :
1. f(t) = et , 0 < t < 2
0 , t > 2
2. f(t) = 1 , 0 ≤ t ≤ 3
t , t > 3
3. f(t) = at , 0 ≤ t < a
2 , t ≥ a
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
14. 3
1
+
tt
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. te t21 −−
22. t
ee btat −− −
23. t
t2sin
24. t
tt 5sin2sin2
Dr.G.N.Shekar
DERIVATIVES OF ARC LENGTH Consider a curve C in the XY plane. Let A be a fixed point on it. Let P and Q be two
neighboring positions of a variable point on the curve C. If ‘s’ is the distance of P from A
measured along the curve then ‘s’ is called the arc length of P. Let the tangent to C at P make
an angle ψ with X-axis. Then (s,ψ) are called the intrinsic co-ordinates of the point P. Let
the arc length AQ be s + δs. Then the distance between P and Q measured along the curve C
is δs. If the actual distance between P and Q is δC. Then δs=δC in the limit Q → P along C.
. . 1Q P
si e LtC
δδ→
=
ψ
0
y
x
P(s,ψ )
Q
A
s δs
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Cartesian Form:
Let ( )y f x= be the Cartesian equation of the curve C and let ( , ) ( , )P x y and Q x x y yδ δ+ +
be any two neighboring points on it as in fig.
Let the arc length PQ sδ= and the chord length PQ Cδ= . Using distance between two
points formula we have 2 2 2 2PQ = ( C) =( x) +( y)δ δ δ
222
11
+=
+=
∴
xy
xCor
xy
xC
δδ
δδ
δδ
δδ
2
1s s C s yx C x C x
δ δ δ δ δδ δ δ δ δ
⇒ = ⋅ = +
We note that δx → 0 as Q → P along C, also that when , 1sQ PC
δδ
→ =
∴ When Q → P i.e. when δx → 0, from (1) we get
2
1 (1)ds dydx dx
= + →
Similarly we may also write
2
1
+=⋅=
yx
Cs
yC
Cs
ys
δδ
δδ
δδ
δδ
δδ
and hence when Q → P this leads to
2
1 (2)ds dxdy dy
= + →
Parametric Form: Suppose ( ) ( )x x t and y y t= = is the parametric form of the curve C.
Then from (1)
( , )Q x x y yδ δ+ +
δC
( , )P x y
δs
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222
11
+
=
+=
dtdy
dtdx
dtdx
dtdx
dtdy
dxds
2 2
(3)ds ds dx dx dydt dx dt dt dt
∴ = ⋅ = + →
Note: Since ψ is the angle between the tangent at P and the X-axis,
we have tandydx
ψ=
( )2 21 1 tands y Secdx
ψ ψ′⇒ = + = + =
Similarly
( )
22 2
1 11 1 1 cot sectan
ds Cody y
ψ ψψ
= + = + = + =′
i.e. Cos dx dyand Sinds ds
ψ ψ= =
( ) ( ) ( )2 2
2 2 21dx dy ds dx dyds ds
∴ + = ⇒ = +
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We can use the following figure to observe the above geometrical connections
among , , .dx dy ds and ψ
Polar Curves :
Suppose ( )r f θ= is the polar equation of the curve C and ( , ) ( , )P r and Q r rθ δ θ δθ+ + be
two neighboring points on it as in figure:
Consider PN ⊥ OQ.
In the right-angled triangle OPN, We have PN PNSin PN r Sin rOP r
δθ δθ δθ= = ⇒ = =
since Sin = when . is very smallδθ δθ δθ
From the figure we see that, (1)ON ONCos ON r Cos r rOP r
δθ δθ= = ⇒ = = =
cos 1 0whenδθ δθ= →
( )NQ OQ ON r r r rδ δ∴ = − = + − =
dx
ds dy
ψ
C
N
O θ
δθ r
P(r,θ)
δs
x
( , )Q r rδ θ δθ+ +
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2 2 2 2 2 2, . , ( ) ( ) ( )From PNQ PQ PN NQ i e C r rδ δθ δ= + = +
22C rrδ δ
δθ δθ ⇒ = +
2
2S S C S rrC C
δ δ δ δ δδθ δ δθ δ δθ
∴ = = +
We note that , 0 1Swhen Q P along the curve alsoC
δδθδ
→ → =
22, (4)dS drwhen Q P r
d dθ θ ∴ → = + →
22 2 2 2, ( ) ( ) ( ) 1CSimilarly C r r r
r rδ δθδ δθ δδ δ
= + ⇒ = +
221S S C Sand r
r C r C rδ δ δ δ δθδ δ δ δ δ
= = +
22, 1 (5)dS dwhen Q P we get r
dr drθ ∴ → = + →
Note:
dWe know that Tan rdrθφ =
2
2 2 2 2 21 secds drr r r Cot r Cot rCod d
φ φ φθ θ
∴ = + = + = + =
Similarly
2
2 21 1ds dr Tan Secdr dr
θ φ φ = + = + =
1dr dCos and Sinds ds r
θφ φ∴ = =
The following figure shows the geometrical connections among ds, dr, dθ and φ
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Thus we have :
22 2 2
1 , 1 ,ds dy ds dx ds dx dydx dx dy dy dt dt dt
= + = + = +
2 22 21ds d ds drr and r
dr dr d dθ
θ θ = + = +
Example 1: 2/3 2/3 2/3ds ds and for the curve xdx dy
y a+ =
2/3 2/3 2/3 -1/3 -1/32 2x x ' 03 3
y a y y+ = ⇒ + =
-1 13 3
-13
x'y
yyx
⇒ = = −
2 1/32 /3 2 /3 2 /3 2 /3
2 /3 2 /3 2 /3
dsHence 1 1dx
dy y x y a adx x x x x
+ = + = + = = =
Similarly
2 1/32 /3 2 /3 2 /3 2 /3
2 /3 2 /3 2 /3
ds 1 1dy
dx x x y a ady y y y y
+= + = + = = =
Example 2: 2
2 2
ds a for the curve y a log dx a -x
Find
=
( )2 2 22 2 2 2
2 2log log dy x axy a a a a x adx a x a x
− = − − ⇒ = − = − −
dr
ds r dθ
φ
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( )
( )( )
( )( )
22 2 2 2 2 22 2 2 2 2
2 2 2 22 2 2 2 2 2
441 1a x a x a xds dy a x a x
dx dx a xa x a x a x
− + + + ∴ = + = + = = = −− − −
Example 3: t tIf x ae Sint, y ae Cost, find dsdt
= =
t t tx ae Sint ae Sint ae Costdxdt
= ⇒ = +
t t ty ae Cost ae Cost ae Sintdydt
= ⇒ = −
( ) ( )2 2
2 22 2 2 2t tds dx dy a e Cos t Sin t a e Cos t Sin tdt dt dt
∴ = + = + + −
( )2 22 2t tae Cos t Sin t a e= + = ( ) ( ) ( )2 2 2 22a b a b a b+ + − = +
Example 4: tIf x a Cos t log Tan , y a Sin t, find 2
dsdt
= + =
2
122 22 2 2
tSecdx a Sin t a Sin tt t tdt Tan Sin Cos
= − + = − + ⋅
( )2 211 Sin t a Cos ta Sin t a a Cost Cot t
Sin t Sin t Sin t−
= − + = = = ⋅
dy a Cos tdt
=
2 2
2 2 2 2 2ds dx dy a Cos t Cot t a Cos tdt dt dt
∴ = + = +
( )2 2 2 1a Cos t Cot t= +
2 2 2 2 2seca Cos t Co t a Cot t a Cot t= ⋅ = =
Example 5: 3 3If x a Cos t, y Sin t, find dsdt
= =
2 23 , 3dx dya Cos t Sin t a Sin t Cos tdt dt
= − =
2 2ds dx dy
dt dt dt ∴ = +
2 4 2 2 4 29 9a Cos t Sin t a Sin t Cos t= +
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( )2 2 2 2 29 3a Cos t Sin t Cos t Sin t a Sin t Cos t= + =
Exercise:
Find dsdt
for the following curves:
(i) x=a(t + Sint), y=a(1-Cost) (ii) x=a(Cost + t Sint), y= aSint(iii) x= a log(Sect + tant), y= a Sect
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Example 6: 2 2If r a Cos 2 , Show that r is constantdsd
θθ
=
22 2 22 2 2 2 2dr dr ar a Cos r a Sin Sin
d d rθ θ θ
θ θ−
= ⇒ = − ⇒ =
2 42 2 2 4 4 2
2
12 2ds dr ar r Sin r a Sind d r r
θ θθ θ
= + = + = +
4 4 2 4 2 4 22 2 2dsr r a Sin a Cos a Sind
θ θ θθ
∴ = + = +
2 2 2 22 Constanta Cos Sin aθ θ= + = = 2 2r is constant for r a Cos2dsd
θθ
∴ =
Example 7: 2 2
-1 r dsFor the curve Cos , Show that r is constant.k dr
k rr
θ − = −
2 2
2 2
22
2
2 (1)1 1 2
1
rr k rd k rdr k rr
k
θ −
− − − − = ⋅ −
−
( )2 2 2
2 2 2 2 2
1 r k r
k r r k r
+ −−= +
− −
2 2 2
2 2 2 2 2 2 2 2
1 k r kk r r k r r k r
− − += + =
− − −
2 2k rr−
=
( )2 2 2 2 22 2
21 2 1k rds d r k r kr r
dr dr r r rθ − + − ∴ = + = + = =
dsHence r k (constant)dr
=
Example 8: ( )2
2 2
ds dsFor a polar curve r f show that ,dr d
r rpr p
θθ
= = =−
We know that dr d 1ds ds r
Cos and Sinθφ φ= =
2 22
22
dr 1 1ds
r ppCos Sin p rSinr r
ϕ φ ϕ−
∴ = = − = − = =
2 2
dsdr
rr p
∴ =−
2ds
dr r rAlso pSin p
rθ φ
= = =
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Exercise:
Find ds dsanddr dθ
for the following curves:
m
2
(i) r=a(1+ Cos ) (ii) =a Cosm
(iii) aSec ( )2
mr
r
θ
θθ
=
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 [ ]ba, . 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 ( )ba, Continuity: A real valued function )(xf is said to be continuous at a point 0x if
0
0lim ( ) ( )x x
f x f x→
=
The function )(xf is said to be continuous in an interval if it is continuous at every point 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 )(xf is said to be differentiable at point 0x if
0
0
0
( ) ( )limx x
f x f xx x→
−−
exists uniquely and it is denoted by )(' 0xf .
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A real valued function f(x) is said to be differentiable in an interval if it is differentiable at
every in the interval or if 0
( ) ( )limh
f x h f xh→
+ − exists uniquely. This is denoted by )(' xf .
We say that either )(' xf exists or f(x) is differentiable. Geometrically, it means that the curve is a smooth curve. In other words a curve is said to 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 )(xf satisfies the following three conditions: (i) )(xf is continuous in a closed interval [ ]ba, (ii) )(xf is differentiable in the open interval ( )ba, (iii) )()( bfaf =
Then there exists at least one point c in the open interval ( )ba, such that 0)(' =cf Geometrical Meaning of Rolle’s Theorem: Consider a curve )(xf that satisfies the conditions of the Rolle’s Theorem as shown in figure:
As we see the curve )(xf is continuous in the closed interval [ ]ba, , the curve is smooth i.e. there can be a unique tangent to the curve at any point in the open interval ( )ba, and also )()( bfaf = . Hence by Rolle’s Theorem there exist at least one point c belonging to ( )ba, such that 0)(' =cf . In other words there exists at least one point at which the 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 ( )ba, such that ……. We present the same in mathematical symbols as: ∃ c ( )ba,∈ : 0)(' =cf From now onwards let us start using the symbolic notation.
[b,f(b)] [a,f(a)]
x
y
a c b
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Example 1: Verify Rolle’s Theorem for 2)( xxf = in [ ]1,1− First we check whether the conditions of Rolle’s theorem hold good for the given function:
(i) 2)( xxf = is an elementary algebraic function, hence it is continuous every where and so also in [ ]1,1− .
(ii) xxf 2)(' = exists in the interval (-1, 1) i.e. the function is differentiable in (-1, 1).
(iii) Also we see that 1)1()1( 2 =−=−f and 11)1( 2 ==f i.e., )1()1( ff =− Hence the three conditions of the Rolle’s Theorem hold good. ∴By Rolle’s Theorem ∃ c ( )1,1−∈ : 0)(' =cf that means 2c = 0 ⇒ c= 0 ( )1,1−∈
Hence Rolle’s Theorem is verified.
Example 2: Verify Rolle’s Theorem for 2/)3()( xexxxf −+= in [-3, 0]
)(xf is a product of elementary algebraic and exponential functions which are continuous and hence it is continuous in [ 3,0]−
2/22/ )3(21)32()( xx exxexxf −− +−+=′
2/2 )]3(21)32[( xexxx −+−+=
2/2/2 )2)(3(21]6[
21 xx exxexx −− +−−=−−−= exists in (-3, 0)
0)3( =−f and also 0)0( =f )0()3( ff =−∴ That is, the three conditions of the Rolle’s Theorem hold good. ∴ ∃ c ( )0,3−∈ : 0)(' =cf
∞−=⇒=+−−∴ − ,2,30)2)(3(21 2/ cecc c
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 nm bxaxxf )()()( −−= in [a, b] where a <b and a, b>0.
)(xf is a product of elementary algebraic functions which are continuous and hence it is continuous in [a, b]
11 )()()()()( −− −−+−−=′ nmnm bxaxnbxaxmxf
11 )())](()([ −− −−−+−= nm bxaxaxnbxm
11 )())](()([ −− −−+−+= nm bxaxnambnmx exists in ( )ba,
0)( =af and also 0)( =bf )()( bfaf =∴ Hence the three conditions of the Rolle’s Theorem hold good. ∴ ∃ ( ),c a b∈ : 0)(' =cf
banmnambcbcacnambnmc nm ,,0)())](()([ 11
++
=⇒=−−+−+∴ −−
Out of these values of c, sincenmnambc
++
= ),( ba∈ , the Rolle’s Theorem is verified.
Example 4: Verify Rolle’s Theorem for )(
log)(2
baxabxxf
++
= in [a, b]
)log(log)log()( 2 baxabxxf +−−+= is the sum of elementary logarithmic functions which are continuous and hence it is continuous in [a, b]
xabxxxf 12)( 2 −
+=′ exists in ( )ba,
01loglog)(
log)( 2
22==
+
+=
++
=abaaba
baaabaaf
and similarly 01loglog)(
log)( 2
22==
+
+=
++
=abbabb
bababbbf
Hence the three conditions of the Rolle’s Theorem hold good. ∴ ∃ c ( )ba,∈ : 0)(' =cf
abcabcabcc
abcccabc
ccf =⇒=−⇒=+
−−⇒=−
+=′ 00
)(2012)( 2
2
22
2
Since abc = ),( ba∈ , the Rolle’s Theorem is verified. Exercise: Verify Rolle’s Theorem for
(i) )cos(sin)( xxexf x −= in
45,
4ππ ,
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(ii) 2/)2()( xexxxf −= in [ ]2,0 ,
(iii) 2sin 2( ) x
xf xe
= in
45,
4ππ .
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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 )(xf satisfies the following two conditions: (i) )(xf is continuous in a closed interval [ ]ba,
(ii) )(xf is differentiable in the open interval ( )ba,
Then there exists at least one point c in the open interval ( )ba, such that ab
afbfcf−−
=)()()('
Proof: Consider the function )(xφ defined by kxxfx −= )()(φ where k is a constant to be found such that ( ) ( )a bϕ φ= . Since )(xf is continuous in the closed interval [ ]ba, , )(xφ which is a sum of continuous functions is also continuous in the closed interval [ ]ba,
' '( ) ( ) (1)x f x kxφ = − → exists in the interval ( )ba, as )(xf is differentiable in ( )ba, . We have ( ) ( )a f a kaφ = − and ( ) ( )b f b kbφ = −
( ) ( )( ) ( ) ( ) ( ) (2)f b f aa b f a ka f b kb kb a
φ φ −∴ = ⇒ − = − ⇒ = →
−
This means that when k is chosen as in (2) we will have ( ) ( )a bφ φ= Hence the conditions of Rolle’s Theorem hold good for )(xφ in [ ]ba, ∴By Rolle’s Theorem ∃ c ( ),a b∈ : '( ) 0cφ =
''( ) 0 ( ) 0 ( ) (3)c f c k k f cφ ′= ⇒ − = ⇒ = →
From (2) and (3), we infer that ∃ c ( ),a b∈ : ab
afbfcf−−
=)()()('
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 )(xf that satisfies the conditions of the LMVT as shown in figure:
From the figure, we observe that the curve )(xf is continuous in the closed interval [ ]ba, ; the curve is smooth i.e. there can be a unique tangent to the curve at any point in the open interval ( )ba, . Hence by LMVT there exist at least one point c belonging to ( )ba, such
that ab
afbfcf−−
=)()()(' . In other words there exists at least one point at which the tangent
drawn to the curve lies parallel to the chord joining the points [ ], ( )a f a and[ ], ( )b f b . Other form of LMVT: The Lagrange’s Mean Value Theorem can also be stated as follows: Suppose )(xf is continuous in the closed interval [ ],a a h+ and is differentiable in the open
interval ( ), )a a h+ then (0,1) : ( ) ( ) ( )f a h f a hf a hθ θ′∃ ∈ + = + + When (0,1)θ ∈ we see that ( , )a h a a hθ+ ∈ + i.e., here c a hθ= +
Using the earlier form of LMVT we may write that ( ) ( )( )( )
f a h f af a ha h a
θ + −′ + =+ −
On simplification this becomes ( ) ( ) ( )f a h f a hf a hθ′+ = + + .
[b,f(b)]
[a,f(a)]
x
y
a c b
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Example 5: Verify LMVT for ( ) ( 1)( 2)( 3)f x x x x= − − − in [0, 4]
3 2( ) ( 1)( 2)( 3) 6 11 6f x x x x x x x= − − − = − + − is an algebraic function hence it is continuous in [0, 4]
2( ) 3 12 11f x x x′ = − + exists in (0, 4) i.e., ( )f x is differentiable in (0, 4) . i.e., both the conditions of LMVT hold good for ( )f x in[0,4] .
Hence (4) (0)(0,4) : ( )4 0
f fc f c −′∃ ∈ =−
i.e., 2 (3)(2)(1) ( 1)( 2)( 3)3 1 114 0
c c − − − −− + =
−
i.e, 2 2 23 12 11 3 3 12 8 0 2 (0,4)3
c c c c c− + = ⇒ − + = ⇒ = ± ∈
Hence the LMVT is verified.
Example 6: Verify LMVT for ( ) logef x x= in [1, ]e
( ) logef x x= is an elementary logarithmic function hence continuous in [1, ]e .
1( )f xx
′ = exists in ( )1,e or ( )f x is differentiable in ( )1,e .
i.e., both the conditions of LMVT hold good for ( )f x in [1, ]e
Hence ( ) (1) 1 log log1(1, ) : ( )1 1
f e f ec e f ce c e
− −′∃ ∈ = ⇒ =− −
1 1 1 (1, )
1c e e
c e⇒ = ⇒ = − ∈
−
Hence the LMVT is verified.
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Example 7: Verify LMVT for 1( ) sinf x x−= in [0,1]
We find 2
1( )1
f xx
′ =−
exists in ( )0,1 , and hence ( )f x is continuous in [0,1] .
i.e., both the conditions of LMVT hold good for ( )f x in [0,1]
Hence 1 1
2(1) (0) 1 sin 1 sin 0(0,1) : ( )
1 0 11
f fc f cc
− −− −′∃ ∈ = ⇒ =− −
2
2 222
1 2 4121
c cc
π ππ π
−⇒ = ⇒ − = ⇒ =
−
2 4 0.7712 (0,1)c ππ
−∴ = = ∈
Hence the LMVT is verified.
Example 8: Find θ of LMVT for 2( )f x ax bx c= + + or
Verify LMVT by finding θ for 2( )f x ax bx c= + + 2( )f x ax bx c= + + is an algebraic function hence continuous in [ , ]a a h+ .
( ) 2f x ax b′ = + exists in ( ),a a h+
i.e., both the conditions of LMVT hold good for ( )f x in [ , ]a a h+ Then the second form of LMVT (0,1) : ( ) ( ) ( ) (1)f a h f a hf a hθ θ′∃ ∈ + = + + →
Here ; 2 3 2 2( ) ( ) ( ) 2 (2)f a h a a h b a h c a ah a h ab bh c+ = + + + + = + + + + + → 3( ) (3)f a a ab c= + + → ; 2( ) 2 ( ) 2 2 (4)f a h a a h b a a h bθ θ θ′ + = + + = + + →
Using (2), (3) and (4) in (1) we have:
3 2 2 3 2 12 (2 2 ) (0,1)2
a ah a h ab bh c a ab c h a a h bθ θ+ + + + + = + + + + + ⇒ = ∈
Hence the LMVT is verified.
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Example 9: Find θ of LMVT for ( ) logef x x= in 2,e e
( ) logef x x= is an elementary logarithmic function hence continuous in 2,e e
1( )f xx
′ = exists in ( )2, .e e
Hence by the second form of LMVT
( )
[ ] [ ]
( )
2 2 2
22
2
0,1 : ( ) ( ) ( ) ( )
( )log log( )
( 1) ( 1). ., 2 log log 2 11 ( 1) 1 ( 1)
21 ( 1) 1 0,11
e e
f e f e e e f e e e
e ee ee e e
e e ei e e ee e e
ee ee
θ θ
θ
θ θ
θ θ
′∃ ∈ = + − + −
−⇒ = +
+ −− −
= + ⇒ − =+ − + −
−⇒ + − = − ⇒ = ∈
−
Hence the LMVT is verified.
Example 10: If 0x > , Apply Mean Value Theorem to show that
( ) log (1 )1 e
xi x xx
< + <+
and 1 10 1log (1 )e x x
< − <+
Consider ( ) log (1 )ef x x= + in [0, ]x . We see that ( )f x is an elementary logarithmic function
hence continuous in [0, ]x . Also 1( )1
f xx
′ =+
exists in ( )0, .x
∴ by LMVT (0,1) : ( ) (0) (0 )f x f xf xθ θ′∃ ∈ = + +
. ., log(1 ) log1 log(1 ) (1)1 1
x xi e x xx xθ θ
+ = + ⇒ + = →+ +
Since 0x > and 0 1θ< < , we have 0 1 1 1x x x xθ θ< < ⇒ < + < +
1 1 1 11 1 (2)1 1 1 1 1 1
x x xx x x x x xθ θ θ
⇒ > > ⇒ < < ⇒ < < →+ + + + + +
From (1) and (2) we get log (1 )1 e
x x xx
< + <+
Also from (1) we have, 1 1 1 1log(1 ) log(1 )
xx x x x
θ θ+= ⇒ − =
+ +
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Since 0 1θ< < , we can write 1 10 1.log(1 )x x
< − <+
Example 11: Apply Mean Value Theorem to show that
1 12 2
sin sin , 01 1
b a b ab a where a ba b
− −− −< − < < <
− −
Consider 1( ) sinf x x−= in [ , ]a b
We see that 2
1( )1
f xx
′ =−
exists in ( ),a b
Hence ( )f x is differentiable in ( ),a b and also it is continuous in [ , ]a b .
Therefore, by LMVT ( )1 1
2( ) ( ) 1 sin sin( , ) : ( ) 1
1
f b f a b ac a b f cb a b ac
− −− −′∃ ∈ = ⇒ = →− −−
Since 2 2 2 2 2 2 2 2 21 1 1a c b a c b a c b a c b< < ⇒ < < ⇒ − > − > − ⇒ − > − > −
( )2 2 2 2 2 21 1 1 1 1 1 2
1 1 1 1 1 1or
a c b a c b∴ < < < < →
− − − − − −
From (1) and (2), we have
1 11 1
2 2 2 21 sin sin 1 sin sin
1 1 1 1
b a b a b ab ab aa b a b
− −− −− − −
< < ⇒ < − <−− − − −
Exercise: Verify the Lagrange’s Mean Value Theorem for
(i) ( ) ( 1)( 2)f x x x x= − − in 10,2
(ii) 1( )f x Tan x−= 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 )(xf and ( )g x satisfies the following conditions:
(i) )(xf and ( )g x are continuous in a closed interval [ ]ba, (ii) )(xf and ( )g x are differentiable in the open interval ( )ba,
(iii) ( ) 0 .[ ]g x for all x for all can be denoted by the symbol′ ≠ ∀
Then there exists at least one point c in the open interval ( )ba,
such that '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
Proof: Consider the function )(xφ defined by ( ) ( ) ( )x f x k g xφ = − where k is a constant to be found such that ( ) ( )a bϕ φ= . Since )(xf and ( )g x are continuous in the closed interval [ ]ba, , )(xφ which is a sum of continuous functions is also continuous in the closed interval [ ]ba,
' '( ) ( ) ( ) (1)x f x k g xφ ′= − → exists in the interval ( )ba, as )(xf and ( )g x are differentiable in ( )ba, . We have ( ) ( ) ( )a f a kg aφ = − and ( ) ( ) ( )b f b k g bφ = −
( ) ( )( ) ( ) ( ) ( ) ( ) ( ) (2)( ) ( )
f b f aa b f a kg a f b k g b kg b g a
φ φ −∴ = ⇒ − = − ⇒ = →
−
This means that when k is chosen as in (2) we will have ( ) ( )a bφ φ= Hence the conditions of Rolle’s Theorem hold good for )(xφ in [ ]ba, ∴By Rolle’s Theorem ∃ c ( ),a b∈ : '( ) 0cφ =
' ( )'( ) 0 ( ) ( ) 0 (3)( )
f cc f c k g c kg c
φ′
′= ⇒ − = ⇒ = →′
From (2) and (3), we infer that ∃ c ( ),a b∈ : '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag 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 2( )f x x= and 4( )g x x= in [ , ]a b
2( )f x x= and 4( )g x x= are algebraic polynomials hence continuous in [ , ]a b
( ) 2f x x′ = and 3( ) 4g x x′ = exist in ( ),a b also we see that ( ) 0 ( , )g x 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∈ : '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
i.e., 2 2 2 2
3 4 4 2 2 22 1 1 ( , )
24 2c b a b ac a bc b a c b a
− += ⇒ = ⇒ = ∈
− +
Hence the CMVT is verified.
Example 13: Verify the Cauchy’s MVT for ( ) logf x x= and 1( )g xx
= in [1, ]e
( ) logf x x= and 1( )g xx
= are elementary logarithmic and rational algebraic functions that
are continuous in [1, ]e
1( )f xx
′ = and 21( )g x
x−′ = exist in ( )1,e
also we see that ( ) 0 (1, )g x for all x e′ ≠ ∈ i.e., the conditions of CMVT hold good for ( )f x and ( )g x in ( )1,e .
Hence ( )1,c e∃ ∈ : '( ) ( ) (1)'( ) ( ) (1)
f c f e fg c g e g
−=
−
i.e., 2
1 log log1 1 (1, )1 1 1 11 1
e ec c c ee
c e e
−= ⇒ − = ⇒ = ∈
− −− −
Hence the CMVT is verified. Example 14: Verify the Cauchy’s MVT for ( ) xf x e= and ( ) xg x e−= in [ , ]a b
( ) xf x e= and ( ) xg x e−= are elementary exponential functions that are continuous in [ , ]a b
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( ) xf x e′ = and ( ) xg x e−′ = − exist in ( ),a b also we see that ( ) 0 ( , )g x 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∃ ∈ : '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
i.e., 2 21 1
c b a b a b ac c
c b a a bb a a b
e e e e e e ee ee e e e e
e e e
− − −
+
− − −= ⇒ − = ⇒ − =
− − −−
2 ( , )
2c a b a be e c a b+ +
= ⇒ = ∈
Hence the CMVT is verified.
Example 15: Verify the Cauchy’s MVT for ( )f x Sin x= and ( )g x Cos x= in 0,2π
( )f x Sin x= and ( )g x Cos x= are elementary trignometric functions
that are continuous in 0,2π
.
( )f x Cos x′ = and ( )g x Sin x′ = − exist in 0,2π
also we see that ( ) 0 0,2
g x for all x π ′ ≠ ∈
.
i.e., the conditions of CMVT hold good for ( )f x and ( )g x in 0,2π
.
Hence 0,2
c π ∃ ∈
:( ) (0)'( ) 2
'( ) ( ) (0)2
f ff cg c g g
π
π
−=
−
( ) (0)2 1 0,
4 2( ) (0)2
Sin SinCos c Cot c cSin c Cos Cos
ππ π
π
− ⇒ = ⇒ = ⇒ = ∈ − −
Hence the CMVT is verified.
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Exercise: Verify the Cauchy’s Mean Value Theorem for
(i) ( )f x x= and 1( )g xx
= in 1 ,14
(ii) 21( )f xx
= and 1( )g xx
= in [ ],a b
(iii) ( )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 )(xf satisfies the following two conditions:
(i) )(xf and it’s first (n-1) derivatives are continuous in a closed interval [ ]ba,
(ii) ( 1) ( )nf x− is differentiable in the open interval ( )ba,
Then there exists at least one point c in the open interval ( )ba, such that
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2 3( ) ( )( ) ( ) ( ) ( ) ( ) ( )2 3
b a b af b f a b a f a f a f a− −′ ′′ ′′′= + − + + + …..
1
( 1) ( )( ) ( )..... ( ) ( ) (1)1
n nn nb a b af a f c
n n
−−− −
+ + →−
Takingb a h= + and for 0 1θ< < , the above expression (1) can be rewritten as
2 3 1( 1) ( )( ) ( ) ( ) ( ) ( ) .... ( ) ( ) (2)
2 3 1
n nn nh h h hf a h f a hf a f a f a f a f a h
n nθ
−−′ ′′ ′′′+ = + + + + + + + →
− Taking b=x in (1) we may write
2 3 1( 1)( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ... ( ) (3)
2 3 1
nn
nx a x a x af x f a x a f a f a f a f a R
n
−−− − −′ ′′ ′′′= + − + + + + + →
−
( )( ) ( ) Ren
nn
x aWhere R f c mainder term after n termsn
−= →
When ,n → ∞ we can show that 0nR → , thus we can write the Taylor’s series as
2 1( 1)
( )
1
( ) ( )( ) ( ) ( ) ( ) ( ) ... ( ) ....2 1
( )( ) ( ) (4)
nn
nn
n
x a x af x f a x a f a f a f an
x af a f an
−−
∞
=
− −′ ′′= + − + + + +−
−= + →∑
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
2 1( 1)
( )
1
( ) (0) (0) (0) ... (0) ....2 1
(0) (0) (5)
nn
nn
n
x xf x f xf f fn
xf fn
−−
∞
=
′ ′′= + + + + +−
= + →∑
Example 16: Obtain a Taylor’s expansion for ( )f x Sin x= in the ascending powers of
4x π −
up to the fourth degree term.
The Taylor’s expansion for )(xf about 4π is
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2 3 4(4)
( ) ( ) ( )4 4 4( ) ( ) ( ) ( ) ( ) ( ) ( ) .... (1)
4 4 4 2 4 3 4 4 4
x x xf x f x f f f f
π π ππ π π π π π− − −
′ ′′ ′′′= + − + + + →
1( )4 4 2
f x Sin x f Sinπ π = ⇒ = =
; 1( )4 4 2
f x Cos x f Cosπ π ′ ′= ⇒ = =
1( )4 4 2
f x Sin x f Sinπ π ′′ ′′= − ⇒ = − = −
1( )4 4 2
f x Cos x f Cosπ π ′′′ ′′′= − ⇒ = − = −
(4) (4) 1( )4 4 2
f x Sin x f Sinπ π = ⇒ = =
Substituting these in (1) we obtain the required Taylor’s series in the form
2 3 4( ) ( ) ( )1 1 1 1 14 4 4( ) ( )( ) ( ) ( ) ( ) ....4 2 3 42 2 2 2 2
x x xf x x
π π ππ − − −
= + − + − + − +
2 3 4( ) ( ) ( )1 4 4 4( ) 1 ( ) ...
4 2 3 42
x x xf x x
π π ππ
− − − = + − − + − +
Example 17: Obtain a Taylor’s expansion for ( ) logef x x= up to the term containing
( )41x − and hence find loge(1.1). The Taylor’s series for )(xf about the point 1 is
2 3 4(4)( 1) ( 1) ( 1)( ) (1) ( 1) (1) (1) (1) (1) .... (1)
2 3 4x x xf x f x f f f f− − −′ ′′ ′′′= + − + + + →
Here ( ) log (1) log 1 0ef x x f= ⇒ = = ; 1( ) (1) 1f x fx
′ ′= ⇒ =
21( ) (1) 1f x fx
′′ ′′= − ⇒ = − ; 32( ) (1) 2f x fx
′′′ ′′′= ⇒ =
(4) (4)
46( ) (1) 6f x fx
= − ⇒ = − etc.,
Using all these values in (1) we get
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2 3 4( 1) ( 1) ( 1)( ) log 0 ( 1)(1) ( 1) (2) ( 6) ....
2 3 4ex x xf x x x − − −
= = + − + − + + −
2 3 4( 1) ( 1) ( 1)log ( 1) ....
2 3 4ex x xx x − − −
⇒ = − − + −
Taking x=1.1 in the above expansion we get
2 3 4(0.1) (0.1) (0.1)log (1.1) (0.1) .... 0.09532 3 4e⇒ = − + − =
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Example 18: Using Taylor’s theorem Show that
2 3
log (1 ) 0 1, 02 3ex xx x for xθ+ < − + < < >
Taking n=3 in the statement of Taylor’s theorem, we can write
2 3( ) ( ) ( ) ( ) ( ) (1)
2 3x xf a x f a xf a f a f a xθ′ ′′ ′′′+ = + + + + →
Consider ( ) logef x x= 2 31 1 2( ) ; ( ) ( )f x f x and f xx x x
′ ′′ ′′′⇒ = = − =
Using these in (1), we can write,
2 3
2 31 1 2log( ) log (2)
2 3 ( )x xa x a x
a a a xθ
+ = + + − + → +
For a=1 in (2) we write,
2 3 2 3
3 31 1log(1 ) log1
2 3 2 3(1 ) (1 )x x x xx x x
x xθ θ+ = + − + = − +
+ +
Since 33
10 0, (1 ) 1 1(1 )
x and x and thereforex
θ θθ
> > + > <+
2 3log(1 )
2 3x xx x∴ + < − +
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Example 19: Obtain a Maclaurin’s series for ( )f x Sin x= up to the term containing 5.x The Maclaurin’s series for f(x) is
2 3 4 5(4) (5)( ) (0) (0) (0) (0) (0) (0) .... (1)
2 3 4 5x x x xf x f x f f f f f′ ′′ ′′′= + + + + + →
Here ( ) (0) 0 0f x Sin x f Sin= ⇒ = = ( ) (0) 0 1f x Cos x f Cos′ ′= ⇒ = =
( ) (0) 0 0f x Sin x f Sin′′ ′′= − ⇒ = − = ( ) (0) 0 1f x Cos x f Cos′′′ ′′′= − ⇒ = − = −
(4) (4)( ) (0) 0 0f x Sin x f Sin= ⇒ = = (5) (5)( ) (0) 0 1f x Cos x f Cos= ⇒ = = Substituting these values in (1), we get the Maclaurin’s series for ( )f x Sin x= as
2 3 4 5 3 5( ) 0 (1) (0) ( 1) (0) (1) .... ....
2 3 4 5 3 5x x x x x xf x Sin x x Sin x x= = + + + − + + ⇒ = − +
Note: As done in the above example, we can find the Maclaurin’s series for various functions, for ex:
2 4 6( ) 1 ......
2 4 6x x xi Cosx = − + −
2 3 4 5 6( ) 1 ......
2 3 4 5 6x x x x x xii e x= + + + + + +
2 3 4( 1) ( 1)( 2) ( 1)( 2)( 3)( ) (1 ) 1 ......
2 3 4m m m m m m m m m miii x mx x x x− − − − − −
+ = + + + +
Taking 1 ( )m in iii= − we can get 1 2 3 4 5 6(1 ) 1 ......x x x x x x x−+ = + + + + + +
Replacing x by (-x) in this we get 1 2 3 4 5 6(1 ) 1 ......x x x x x x x−− = − + − + − +
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 00 , , 0 , , 0 , 10
and ∞∞×∞ ∞ − ∞ ∞
∞which do not represent any value. Such expressions are
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 ( ) 0 lim ( ) 0
x a x af x and g x
→ →= =
(ii) ( ) ( ) ( ) 0f x andg x exist and g a′ ′ ′ ≠
Then ( ) ( )lim lim( ) ( )x a x a
f x f xg x g x→ →
′=
′
The above rule can be extended, i.e, if ( ) ( ) ( )( ) 0 ( ) 0 lim lim lim .....( ) ( ) ( )x ax a x a
f x f x f xf a and g a theng x g x g x→→ →
′ ′′′ ′= = = = =
′ ′′
Note:
1. We apply L’Hospital’s Rule only to evaluate the limits that in 0 ,
0∞∞
forms. Here we
differentiate the numerator and denominator separately to write ( )( )
f xg x
′′
and apply the
limit to see whether it is a finite value. If it is still in 00
or ∞∞
form we continue to
differentiate the numerator and denominator and write further ( )( )
f xg x
′′′′
and apply the
limit to see whether it is a finite value. We can continue the above procedure till we get a definite value of the limit.
2. To evaluate the indeterminate forms of the form 0 , ,×∞ ∞ − ∞ we rewrite the
functions involved or take L.C.M. to arrange the expression in either 00
or ∞∞
and then
apply L’Hospital’s Rule.
3. To evaluate the limits of the form 0 00 , 1and ∞∞ 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 00
or ∞∞
form and then apply the L’Hospital’s
Rule. 4. We can use the values of the standard limits like
0 0 0 0 0
sin tanlim 1; lim 1; lim 1; lim 1; lim cos 1;sin tanx x x x x
x x x x x etcx x x x→ → → → →
= = = = =
Evaluate the following limits:
Example 1: Evaluate 30
sinlimtanx
x xx→
−
3 2 2 4 3 20 0 0
sin 0 cos 1 0 sin 0lim lim limtan 0 3tan sec 0 6 tan sec 6 tan sec 0x x x
x x x xx x x x x x x→ → →
− − − = = +
6 2 4 2 4 4 20
cos 1lim6sec 24 tan sec 18 tan sec 12 tan sec 6x
xx x x x x x x→
−= = −
+ + +
Method 2:
3
33 30 0 0 0
sinsin 0 0 sin 0 tanlim lim lim lim 1
tan 0 0 0tanx x x x
x xx x x x xx
x x xxx
→ → → →
−− − = = =
20 0 0
cos 1 0 sin 0 cos 1lim lim lim3 0 6 0 6 6x x x
x x xx x→ → →
− − − = = = = −
Example 2: Evaluate 0
limx x
x
a bx→
−
0 0
0 log loglim lim log log log0 1
x x x x
x x
a b a a b b aa bx b→ →
− − = = − =
Example 3: Evaluate ( )20
sinlim1x x
x x
e→ −
( ) ( )20 0 0
sin 0 sin cos 0 cos cos sin 1 1 0 2lim lim lim 10 0 2[1 0] 22 . ( 1)2 11 x x x xx xx x xx
x x x x x x x x xe e e ee ee→ → →
+ + − + − = = = = = + + −− −
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Example 4: Evaluate 20
log(1 )limx
x
x e xx→
− +
( )2
20 0 0
111log(1 ) 0 0 1 1 0 1 31lim lim lim
0 2 0 2 2 2
x x xx x
x
x x x
e e x ee x e xx e x xx x→ → →
+ + ++ − +− + + + + += = = =
Example 5: Evaluate 0
cosh coslimsinx
x xx x→
−
0 0 0
cosh cos 0 sinh sin 0 cosh cos 1 1 2lim lim lim 1sin 0 sin cos 0 cos cos sin 1 1 0 2x x x
x x x x x xx x x x x x x x x→ → →
− + + + = = = = = + + − + −
Example 6: Evaluate 20
cos log(1 ) 1limsinx
x x xx→
− + − +
2
20 0 0
11 cossin 1cos log(1 ) 1 0 0 1 1(1 )1lim lim lim 0sin 0 sin 2 0 2 cos 2 2x x x
xxx x x xxx x x→ → →
− +− − +− + − + − ++ += = = =
Example 7: Evaluate 1
lim1 log
x
x
x xx x→
−− −
2 1
1 1 1
2
0 (1 log ) 1 0 (1 log ) 1 1lim lim lim 21 11 log 0 0 11
x x x x
x x x
x x x x x x xx x
x x
−
→ → →
− + − + + + = = = = − − −
1log log 1 log (1 log )
( ) (1 log )
x
x x
Since y x y x x y x y y xy
dand then x x xdx
′ ′= ⇒ = ⇒ = + ⇒ = +
= +
Example 8: Evaluate 2
4
sec 2 tanlim1 cos 4x
x xxπ
→
−+
2 2 2 4 2 2 2
4 4 4
sec 2 tan 0 2sec tan 2sec 0 2sec 4sec tan 4sec tanlim lim lim1 cos 4 0 4 sin 4 0 16 cos 4x x x
x x x x x x x x x xx x xπ π π
→ → →
− − + − = = + − −
( ) ( ) ( )4 2 2
22 2 4 2 (1) 4 2 8 116 16 2
+ −= = =
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Example 9: Evaluate log(sin . cos )limlog(cos .sec )x a
x ec aa x→
2
cos cossin . coslog(sin . cos ) 0 cotlim lim lim cot
sec tan .coslog(cos .sec ) 0 tancos .sec
x a x a x a
x ec ax ec ax ec a x ax x aa x x
a x→ → →
= = =
Example 10: Evaluate 0
2coslimsin
x x
x
e e xx x
−
→
+ −
0 0 0
2cos 0 2sin 0 2cos 1 1 2lim lim lim 2sin 0 sin cos 0 cos cos sin 1 1 0
x x x x x x
x x x
e e x e e x e e xx x x x x x x x x
− − −
→ → →
+ − − + + + + + = = = = + + − + −
Example 11: Evaluate 20
cos log(1 )limx
x x xx→
− +
20 0
1cos sincos log(1 ) 0 01lim lim0 2 0x x
x x xx x x xx x→ →
− −− + +=
2
0
1sin sin cos0 0 0 1 1(1 )lim
2 2 2x
x x x xx
→
− − − +− − − ++= = =
Example 12: Evaluate 2
0
log(1 )limlog cosx
xx→
−
2 2
2 20 0 0 0
2log(1 ) 0 2 cos 0 2 cos 2 sin 2 01lim lim lim lim 2
sinlog cos 0 (1 )sin 0 (1 )cos 2 sin 1 0cos
x x x x
xx x x x x xx
xx x x x x x xx
→ → → →
− − − −− = = = = = − − − − −
Example 13: Evaluate 30
tan sinlimsinx
x xx→
−
2 2
3 2 2 30 0 0
tan sin 0 sec cos 0 2sec tan sin 0lim lim limsin 0 3sin cos 0 6sin cos 3sin 0x x x
x x x x x x xx x x x x x→ → →
− − + = = −
2 2 4
3 2 20
4sec tan 2sec cos 0 2 1 3 1lim6cos 12sin cos 9 sin cos 6 0 0 6 2x
x x x xx x x x x→
+ + + += = = =
− − − −
Method 2:
3
33 30 0 0 0
tan sintan sin tan sin 0 sinlim lim lim lim 1
sin 0sinx x x x
x xx x x x xx
x x xxx
→ → → →
−− − = = =
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2 2
20 0
sec cos 0 2sec tan sin 0lim lim3 0 6 0x x
x x x x xx x→ →
− + = =
2 2 4
0
4sec tan 2sec cos 0 2 1 3 1lim6 6 6 2x
x x x x→
+ + + += = = =
Example 14: Evaluate 20
tanlimtanx
x xx x→
−
3
2 30 0 0 0
tantan tan 0 tanlim lim lim lim 1
tantan 0x x x x
x xx x x x xx
xx x x xx
→ → → →
−− − = = =
2 2 2 2 4
20 0 0
sec 1 0 2sec tan 0 4sec tan 2sec 0 2 1lim lim lim3 0 6 0 6 6 3x x x
x x x x x xx x→ → →
− + + = = = = =
Example 15: Evaluate 0
limlog(1 )
ax ax
x
e ebx
−
→
−+
0 0
0 2lim limlog(1 ) 0 /(1 )
ax ax ax ax
x x
e e ae ae a a abx b bx b b
− −
→ →
− + + = = = + +
Example 16: Evaluate 20
1 loglimx
x
a x ax→
− −
22
20 0 0
1 log 0 log log 0 (log ) 1lim lim lim (log )0 2 0 2 2
x x x
x x x
a x a a a a a a ax x→ → →
− − − = = =
Example 17: Evaluate 20
log( )limx
x
e e exx→
− +
2 2 20 0 0
log( ) log (1 ) log log(1 ) 0lim lim lim0
x x x
x x x
e e ex e e x e e xx x x→ → →
− + − + − − + = =
2
0 0
110 1 1(1 )1lim lim 1
2 0 2 2
xx
x x
ee xxx→ →
+− ++ += = = =
Exercise: Evaluate the following limits.
3
30 0
2 log(1 ) log(1 )( ) lim ( ) limsin sin
x x
x x
e e x xi iix x x
−
→ →
− − + +
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20 22
1 sin cos log(1 ) logsin( ) lim ( ) limtan ( )
2x x
x x x xiii ivx x x
π π→ →
+ − + −
−
1
2 202
cosh log(1 ) 1 sin sin( ) lim ( ) limx x
x x x x xv vix xπ
−
→ →
+ − − +
2 2
0
(1 )( ) limlog(1 )
x
x
e xviix x→
− ++
Limits of the form ∞ ∞
:
Example 18: Evaluate 0
log(sin 2 )limlog(sin )x
xx→
2
20 0 0 0 0
log(sin 2 ) (2cos 2 / sin 2 ) 2cot 2 2 tan 0 2sec 2lim lim lim lim lim 1log(sin ) (cos / sin ) cot tan 2 0 2sec 2 2x x x x x
x x x x x xx x x x x x→ → → → →
∞ = = = = = = ∞
Example 19: Evaluate 0
loglimcosx
xecx→
2
0 0 0 0
log 1\ sin 0 2sin 2 0lim lim lim lim 0cos cos .cot cos 0 cos sin 1 0x x x x
x x x xecx ec x x x x x x x→ → → →
∞ − − = = = = = ∞ − − −
Example 20: Evaluate 2
log coslimtanx
xxπ
→
2 2
2 2 2 2
log cos tan tan sin cos 0lim lim lim lim 0tan sec sec 1 1x x x x
x x x x xx x xπ π π π
→ → → →
∞ − − − − = = − = = = ∞
Example 21: Evaluate 1
log(1 )limcotx
xxπ→
−
2
21 1 1 1
log(1 ) 1/(1 ) sin 0 2 sin cos 0lim lim lim lim 0cot cos (1 ) 0x x x x
x x x x xx ec x x
π π π ππ π π π π π→ → → →
− ∞ − − = = = = = ∞ − − − −
Example 22: Evaluate tan 2
0lim log tan 3xx
x→
tan 2 00
loglog tan 3lim log tan 3 lim loglog tan 2 log
ex bxx e
axx ax b→→
∞ = = ∞
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2
20 0 0
3sec 3 / tan 3 3/ sin 3 .cos3 3/ sin 3 .cos3lim lim lim2sec 2 / tan 2 2 / sin 2 .cos 2 2 / sin 2 .cos 2x x x
x x x x x xx x x x x x→ → →
= = =
0 0 0
6 / sin 6 6sin 4 0 24cos 4 24lim lim lim 14 / sin 4 4sin 6 0 24cos 6 24x x x
x x xx x x→ → →
= = = = =
Example 23: Evaluate log( )limlog( )x ax a
x ae e→
−−
log( ) 1/( ) ( ) 0lim lim lim lim 1
log( ) /( ) ( ) 0 ( )
x a x a
x a x x a x x x ax a x a x a x a
x a x a e e e ee e e e e e x a e x a e e→ → → →
− ∞ − − = = = = = − ∞ − − − +
Exercise: Evaluate the following limits.
0 0
log tan logsin( ) lim ( ) limlog cotx x
x xi iix x→ →
sin0 02
cot 2 sec( ) lim ( ) lim ( ) lim log sin 2cot 3 tan 3 xx xx
x xiii iv v xx xπ→ →→
Limits of the form ( )0×∞ : To evaluate the limits of the form ( )0×∞ , we rewrite the
given expression to obtain either 00
or ∞ ∞
form and then apply the L’Hospital’s Rule.
Example 24: Evaluate 1
lim( 1)xx
a x→∞
−
( )
11
1 2
2
1(log )( 1) 0lim( 1) 0 lim lim
1 10
xx
xx x x
a aa xa x form
x x→∞ →∞ →∞
− − − ×∞ = = −
1
0lim (log ) log logxx
a a a a a→∞
= = =
Example 25: Evaluate
2
lim(1 sin ) tanx
x xπ
→−
( ) 2
2 2 2
(1 sin ) 0 cos 0lim(1 sin ) tan 0 lim lim 0cot 0 cos 1x x x
x xx x formx ec xπ π π
→ → →
− − − ×∞ = = = = −
Example 26: Evaluate 1
limsec .log2x
xx
π→
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( )1 1
log 0limsec .log 0 lim2 0cos
2x x
xx formx
x
ππ→ →
∞× =
1
2 2
1/ 2 2lim lim1 sinsin
22 2xx
x x
xx xπ ππ π ππ
→→= = =
− −
Example 27: Evaluate
0lim log tanx
x x→
( )0 0
log tanlim log tan 0 lim(1/ )x x
xx x formx→ →
∞ ×∞ = ∞
2 2
0 0
2
sec / tanlim lim1 sin .cosx x
x x xx x
x→ →
−= =
−
2
0 0
2 0 4 0lim lim 0sin 2 0 2cos 2 2x x
x xx x→ →
− − = = = =
Example 28: Evaluate 2
1lim (1 ) tan
2x
xx π→
−
( )2
2
1 1 1 2
1 0 2 2 4lim (1 ) tan 0 lim lim2 0cot cos
2 2 2 2x x x
x x xx form x xec
ππ π π π π→ → →
− − − ×∞ = = = = −
Example 29: Evaluate
0lim tan .logx
x x→
( )0 0
loglim tan .log 0 limcotx x
xx x formx→ →
∞ ×∞ = ∞
2
20 0 0
1/ sin 0 sin 2 0lim lim lim 0cos 0 1 1x x x
x x xec x x→ → →
− − = = = = = −
Limits of the form ( )∞ − ∞ : To evaluate the limits of the form ( )∞ − ∞ , we take L.C.M.
and rewrite the given expression to obtain either 00
or ∞ ∞
form and then apply the
L’Hospital’s Rule.
Example 30: Evaluate 0
1lim cotx
xx→
−
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( )0 0
1 1 coslim cot limsinx x
xx formx x x→ →
− = − ∞ − ∞
0 0
sin cos 0 cos cos sinlim limsin 0 sin cosx x
x x x x x x xx x x x x→ →
− − + = = +
0 0
sin 0 sin cos 0 0lim lim 0sin cos 0 cos cos sin 1 1 0x x
x x x x xx x x x x x x→ →
+ + = = = = + + − + −
Example 31: Evaluate [ ]
2
lim sec tanx
x xπ
→−
[ ] ( )2 2
1 sinlim sec tan limcos cosx x
xx x formx xπ π
→ →
− = − ∞ − ∞
2 2
1 sin 0 cos 0lim lim 0cos 0 sin 1x x
x xx xπ π
→ →
− − = = = = −
Example 32: Evaluate 1
1limlog 1x
xx x→
− −
( )1 1
1 ( 1) log 0lim limlog 1 ( 1) log 0x x
x x x xformx x x x→ →
− − − ∞ − ∞ = − −
1 1 1
2
1 1 log log 0 l/ 1 1lim lim lim1 1 1 10 1 1 2log 1 logx x x
x x xx x x
x x x x→ → →
− − − − − − = = = = = − + + − + +
Example 33: Evaluate 0
1 1lim1xx x e→
− −
( )0 0
1 1 ( 1) 0lim lim1 ( 1) 0
x
x xx x
e xformx e x e→ →
− − − ∞ − ∞ = − −
0 0
1 0 1 1lim lim( 1) 0 1 1 0 2
x x
x x x x xx x
e ee xe e e xe→ →
− = = = = − + + + + +
Example 34: Evaluate 0
1 1limsinx x x→
−
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( )0 0
1 1 sin 0lim limsin sin 0x x
x xformx x x x→ →
− − ∞ − ∞ =
0 0
1 cos 0 sin 0lim lim 0sin cos 0 cos cos sin 1 1x x
x xx x x x x x x→ →
− = == = = + + − +
Example 35: Evaluate 20
1 log(1 )limx
xx x→
+ −
2
2 20 0 0 0
1111 log(1 ) log(1 ) 0 0 1(1 )1lim lim lim lim0 2 0 2 2x x x x
x x x xxx x x x→ → → →
− + − + + +− = = = =
Example 36: Evaluate 0
lim cotx
a xx a→
−
( )0 0 0
cos cos 0lim cot lim lim0sin sin
x x x
x xa x a aa aformx xx a x x
a a→ → →
− = − ∞ − ∞ = −
0 0
1sin cos . cos cos sin0lim lim0sin sin cos
x x
x x x x x xa x aa a a a a a a
x x x xxa a a a
→ →
− − + = = +
0 0
2
1 1sin sin . .cos0 0 0lim lim 01 1 1 10sin cos cos cos sin 0x x
x x x x xa a a a a a ax x x x x x xa a a a a a a a a a a
→ →
+ + = = = = + + − + −
Exercise: Evaluate the following limits.
( )0
( ) lim 2 cot( ) ( ) lim cos cotx a x
xi x a ii ecx xa→ →
− − −
2
202
1( ) lim tan sec ( ) lim cot2 xx
iii x x x iv xxπ
π→→
− −
[ ]202
1 1( ) lim ( ) lim 2 tan sectanx x
v vi x x xx x x π
π→ →
− −
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Example 37: Find the value of ‘a’ such that 30
sin 2 sinlimx
x a xx→
+ is finite. Also find the
value of the limit.
Let A= 3 20 0
sin 2 sin 0 2cos 2 cos 2lim lim0 3 0x x
x a x x a x a finitex x→ →
+ + + = = ≠
We can continue to apply the L’Hospital’s Rule, if 2+a=0 i.e., a= -2 .
2For a = − ,
20 0
2cos 2 2cos 0 4sin 2 2sin 0lim lim3 0 6 0x x
x x x xAx x→ →
− − + = =
0
8cos 2 2cos 8 2lim 16 6x
x x→
− + − += = = −
lim 2 1.The given it will have a finite value when a and it is∴ = − −
Example 38: Find the values of ‘a’ and ‘b’ such that 30
(1 cos ) sin 1lim3x
x a x b xx→
− += .
3 20 0
(1 cos ) sin 0 (1 cos ) sin cos 1lim lim0 3 0x x
x a x b x a x ax x b x a bLet A finitex x→ →
− + − + + − + = = = ≠
We can continue to apply the L’Hospital’s Rule, if 1-a+b=0 i.e., a-b = 1 .
1For a b− = ,
20 0
(1 cos ) sin cos 0 2 sin cos sin 0lim lim3 0 6 0x x
a x ax x b x a x ax x b xAx x→ →
− + + + − = =
0
3 cos sin cos 3lim6 6x
a x ax x b x a b finite→
− − −= = =
1 3 1. . ., 3 23 6 3
a bThis finite value is given as i e a b−= ⇒ − =
1 11 3 2 .2 2
Solving the equations a b and a b we obtain a and b− = − = = = −
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Example 39: Find the values of ‘a’ and ‘b’ such that 20
cosh coslim 1x
a x b xx→
−= .
20
cosh coslim0x
a x b x a bLet A finitex→
− −= = ≠
We can continue to apply the L’Hospital’s Rule, if a-b=0, since the denominator=0 .
0For a b− = ,
20 0 0
cosh cos 0 sinh sin 0 cosh coslim lim lim0 2 0 2 2x x x
a x b x a x b x a x b x a bAx x→ → →
− + + + = = = =
1. 2But thisis given as a b∴ + =
0 2 1 1.Solving the equations a b and a b we obtain a and b− = + = = = Limits of the form 0 00 , 1and ∞∞ : 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
sides and rewrite the expressions to get 00
or ∞∞
form and then apply the L’Hospital’s Rule.
Example 40: Evaluate
0lim x
xx
→
0
0lim (0 )x
xLet A x form
→=
Take log on both sides to write
0 0 0
loglog lim log lim .log (0 ) lim1/
xe x x x
xA x x x formx→ → →
∞ = = ×∞ = ∞
20 0
1/ 0lim lim 0( 1/ ) 1 1x x
x xx→ →
−= = = =
−
0
0log 0 1 lim 1x
e xA A e x
→= ⇒ = = ∴ =
Example 41: Evaluate 21
0lim(cos ) xx
x→
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21
0lim(cos ) (1 )xx
Let A x form∞
→=
Take log on both sides to write
21
2 20 00
1 log cos 0log lim log(cos ) lim log cos ( 0 ) lim0
xe x xx
xA x x formx x→ →→
= = ∞× =
2
0 0
tan 0 sec 1lim lim2 0 2 2x x
x xx→ →
− − − = = =
211
20
1 1 1log lim(cos )2
xe x
A A e xe e
−
→= − ⇒ = = ∴ =
Example 42: Evaluate cos
2
lim(tan ) x
xx
π→
cos 0
2
lim(tan ) ( )x
xLet A x form
π→
= ∞
Take log on both sides to write cos
2 2
log lim log(tan ) lim cos log(tan ) (0 )xe
x xA x x x form
π π→ →
= = ×∞
2
2
2 2 2
log tan sec / tan cos 0lim lim lim 0sec sec . tan sin 1x x x
x x x xx x x xπ π π
→ → →
∞ = = = = = ∞
0 cos
2
log 0 1 lim(tan ) 1xe
xA A e x
π→
= ⇒ = = ∴ =
Example 43: Evaluate 21
0
tanlim( ) xx
xx→
21
0
tanlim( ) (1 )xx
xLet A formx
∞
→=
Take log on both sides to write
21
20 0
tan 1 tanlog lim log( ) lim log( ) ( 0 )xe x x
x xA formx x x→ →
= = ∞ ×
2
20 0
tan sec 1log( ) 0 tanlim lim0 2x x
x xx x x
x x→ →
− = =
20 0 0
1 1 2 12 sin 2 0sin .cos sin 2lim lim lim
2 2 2 sin 2 0x x x
x xx x x x xx x x x→ → →
− − − = = =
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2 20 0
2 2cos 2 0 4sin 2 0lim lim4 sin 2 4 cos 2 0 4sin 2 16 cos 2 8 sin 2 0x x
x xx x x x x x x x x→ →
− − = == + + −
20
8cos 2 8 1lim24cos 2 48 sin 2 16 cos 2 24 3x
xx x x x x→
− − −= = =
− −
211 1
3 30
1 tanlog lim( )3
xe x
xA A e ex
− −
→= − ⇒ = ∴ =
Example 44: Evaluate 1
0lim( )x xx
a x→
+
1
0lim( ) (1 )x xx
Let A a x form∞
→= +
Take log on both sides to write
1
0 0
1log lim log( ) lim log( ) ( 0 )x xxe x x
A a x a x formx→ →
= + = + ∞ ×
0 0
log( ) 0 ( log 1) /( )lim lim0 1
x x x
x x
a x a a a xx→ →
+ + + = =
log 1 log log loga a e ae= + = + =
1
0log log lim( ) .x x
e xA ea A ea Hence a x ea
→∴ = ⇒ = + =
Example 45: Evaluate tan
2lim(2 )xa
x a
xa
π
→−
tan
2lim(2 ) (1 )xa
x a
xLet A forma
π∞
→= −
Take log on both sides to write
( )tan
2log lim log(2 ) lim tan .log(2 ) 02
xa
e x a x a
x x xA forma a a
π π→ →
= − = − ∞×
2
2
( 1/ )
log(2 ) 2 sin0 2 22lim lim lim .0cot cos 2
2 2 2x a x a x a
ax x xa a a
x x xeca a a a
π
π π π π π→ → →
−
− − = = = = − −
2 2tan
22log lim(2 ) .xa
e x a
xA A e Hence ea
ππ π
π →∴ = ⇒ = − =
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Exercise: Evaluate the following limits.
22
1
0 0
1 cos( ) lim(cos ) ( ) lim2
b xx
x x
xi ax ii→ →
+
2
11
2 log(1 )
1 0
sin( ) lim(1 ) ( ) limx
x
x x
xiii x ivx
−
→ →
−
( )cottan
0 0( ) lim(sin ) ( ) lim 1 sin xx
x xv x iv x
→ →+
( )2 tan 2cos
04
( ) lim(cos ) ( ) lim tan xec x
x xvii x viii x
π→ →
1
0
1( ) lim( ) ( ) lim1 3
x x x xx
x x
ax a b cix xax→∞ →
+ + + −
CURVATURE
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=δs. Let the tangents drawn to the curve at P, Q respectively make angles ψ
and ψ+δψ with X-axis i.e., the angle between the tangents at P and Q is δψ. While moving
from P to Q through a distance‘δs’, the tangent has turned through the angle ‘δψ’. This is
called the bending of the arc PQ. Geometrically, a change in ψ represents the bending of the
x
y
C
P
ψ ψ+δψ
δs Q
δψ
o
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curve C and the ratio s
δψδ
represents the ratio of bending of C between the point P & Q and
the arc length between them.
∴Rate of bending of Curve at P is
Q P
d Ltds sψ δψ
δ→=
This rate of bending is called the curvature of the curve C at the point P and is denoted by κ
(kappa). Thus ddsψκ =
We note that the curvature of a straight line is zero since there exist no bending i.e. κ=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.
ψκρ
ρκ
κ
dds1
letter).Greek - (rhoby denoted is and curvature of radius thecalled is 1 then 0, If
==∴
≠
Radius of curvature in Cartesian form :
Suppose y = f(x) is the Cartesian equation of the curve considered in figure.
we have ( )2
2 22
dy d y d d ds1y Tan y Sec Tandx dx dx ds dx
ψ ψψ ψ ψ′ ′′= = ⇒ = = ⋅ = + ⋅
2dsBut we know that 1dx
dydx
= +
ψ
0
y
x
c
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32 2
2 22
22
2
1ds1 1d
dydxd y dy d dy
d ydx dx ds dxdx
ψψ
+ ∴ = + ⋅ ⋅ + ⇒ =
( )
32 21 yds
d yρ
ψ
′+ ∴ = =′′
This is the expression for radius of curvature in Cartesian form.
NOTE: We note that when y’=∞, we find ρ using the formula
32 2
2
2
1 dxdy
d xdy
ρ
+
=
Example 9: Find the radius of curvature of the curve x3+y3 = 2a3 at the point (a, a).
3 3 3 2 22 3 3 0x y a x y y′+ = ⇒ + ⋅ = ( )2
2 , , 1xy hence at a a yy
′ ′⇒ = − = −
( ) ( ) ( )2 2 3 3
4 4
2 2 2 2 4, , ,y x x y y a ay hence at a a y
y a a′ − +′′ ′′∴ = − = − = −
( ) ( )
3 32 22 21 1 1
. ., 2 24 4 2
y a ai ey a
ρ ρ ′+ + − ∴ = = = ⋅ =
′′ −
Example 10: Find the radius of curvature for x y a+ = at the point where it meets
the line y=x.
On the line y x, x x a . 2 x a4ai e or x= + = = =
i.e., We need to find at ,4 4a aρ
1 1x y a 0 . , , , 14 42 x 2 y
y a ay i e y hence at yx
′ ′ ′+ = ⇒ + = = − = −
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1 12 2
,x y y
y xAlso y
x
′⋅ − ′′ = −
1 1( 1)4 4 1 12 2 ( ) ( 1) 44 4 2 2, ,
4 44 4 4
a aa a
a aat y a a a a
⋅ − −
− − − ′′∴ = − = − = − =
( ) ( )3 3
2 22 21 1 12 24 4 2
y a ay a
ρ ′+ + − ∴ = = = =
′′
Example 11: Show that the radius of curvature for the curve y 4 Sin x - Sin 2x =
5 5at x is 2 4π=
y 4 Sin x - Sin 2x 4 2 2y Cos x Cos x′= ⇒ = −
when x , y 4 2 0 2( 1) 22 2Cos Cosπ π π′∴ = = − = − − =
Also, y -4 Sin x 4 Sin 2x and when x , y -4 Sin 4 Sin 42 2π π π′′ ′′= + = = + = −
( )
3 32 2 2 21 y 1 2 5 54 4y
ρ ρ ′+ + ∴ = = ⇒ =
′′ −
Example 12: 2 3 3Find the radius of curvature for xy = a - x at (a, 0).
2 3 3 2 22 3xy a x y xy y x′= − ⇒ + = −
2 23 ( ,0),
2x yy and at a y
xy− −′ ′∴ = = ∞
2 2
dx 2 dxIn such cases we write ( , 0), 0dy 3 dy
xy and at ax y
= =− −
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( )
( )
2 22
22 2 2 2 2
dx dx3 2 2 2 6 2dy dydx 2 d xAlso
dy 3 dy 3x y
x y y x xy x yxy
x y
+ + − + − = ⇒ =
+ +
( ) ( )( )( )
22 3
22 42
3 0 0 2 0d x 6 2, 0 ,dy 9 33 0
a a aAt aa aa
+ + − − − ∴ = = = +
32 2
32 2
22
11 3
2 23
dxdy o aor
d x ady
ρ ρ
+ + ∴ = = =
−
Exercise: Find the radius of curvature for the following curves:
3 3
22
2 3 3
2 2
2 2
2 3 3
2 2
3 3(i) x 3 ,2 2
( )(ii) ( ,0)
(iii) int
( ) 1 .
( ) int ( ,0).( ) 2 (3 ) int
a ay axy at
a a xy at ax
a y x a at the po where it crosses x axisx yiv at the ends of the major axisa b
v a y x a at the po avi y x x at the po where t
+ =
−=
= − −
+ =
= −
= −2 2 2
2 2 23
tan .( ) ( ) ( 2 , 2 ).
2( )
he gents are parallel to x axisvii x y a x y at a a
ax x yviii For the curve y show thata x a y x
ρ
−
= + −
= = + +
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M.G.Geetha
INTEGRAL CALCULUS Gamma and Beta Functions
Definition:
The improper integral ( ) dxxen 1n
0
x −∞
−∫=Γ is defined as the gamma function.
Here n is a real number called the parameter of the function. Γ(n) exists for all real values of
n except 0, -1, -2., ………..the graph of which is shown below :
Recurrence formula
( ) 0 n ,1
0
>=Γ −∞
−∫ dxxen nx
∫∞
−∞
− +
=0
xn
0
xn
,dxen
xen
x integrating by parts.
Applying the definition of Gamma function and using ∞→→ xas 0x n
xe
( ) ( ) )2.......(..........1 nnnor Γ=+Γ
-4 -3 -2 -1 0 n
r(n)
( ) ( ) )1.....(....................1n
nn +Γ=Γ
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Note:
(1) Γ(n) is not convergent when n = 0, -1, -2, …….
(2) If Γ(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 Γ(1) = 1, we get
Γ(n) = (n - 1) Γ(n - 1)
= (n - 1) (n - 2) Γ(n - 2) and so on
= (n - 1) (n - 2) (n - 3) …….. 1
= (n - 1) !
Problems:
(1) Prove that π=
Γ
21
By definition, ( ) ∫∞
−−=Γ0
1nx dxxen
t xsin 212
0
2
== −∞
−∫ gudtte nt
( ) ∫∞
−−=Γ∴0
1n22x dxxe2n
∫∞
−−=0
1n22y dyye2
∫∞
−=
Γ⇒
0
2x dxe21 ∫
∞−=
0
2y dye
∫ ∫∞ ∞
+−
=
Γ∴
0 0
2y2x2dxdye 4
21
Converting Cartesian (x, y) to polar (r, θ) system, we get
∫ ∫=
∞
=
−=
Γ
2
0 0
2
)2( 221 2
π
θ
θr
r drdre
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[ ]∫ ∫ −∞− ==2
0
r-0
222
2re- ,π
θ rr edrde
∞→→θ= ∫π
r as 0e ,d22r-
2
0
= π
Taking square roots on both sides, we get π=
Γ
21
(2) Show that
Γ=∫
∞−
41
41dxe
0
4x
dyy41dxdydxx4y xLet 4
334 −=⇒=⇒=
∫∫∞
−−∞
− =∴0
43
0 414
dyyedxe yx .definition theusing ,41
41
Γ=
(3) Find ( )∫1
0
5dx xlnx
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
y65
0
5 ,dyeydxln x x interchanging the lower and the upper limits
,6
dtte0
6
5t∫
∞−= choosing 6y = t
( ) 66 6!56
61
−=Γ−=
(4) Prove that πa
xa
dxa
=
∫0 ln
using tewetxa
==
xaget ,ln
dx = -ae-t dt and t = ∞ to 0 when x = 0 to a
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∴ ∫∫∞
−−=
0
t21a
0dte.at
xaln
dx πaa =
Γ=
21
(5) Show that ( ) ( )( )∫ ++
−=
1
01n
nnm
1m!n1dxx logx .
Hence show that 4log1
0
−=∫ dxxx when 'n' is a positive integer and m > -1.
Let logx = - t ⇒ dx = -e-t dt and t = ∞ to 0 when x = 0 to 1.
( ) ( ) ( ) ( )∫ ∫ ∫∞ ∞
+−−− −=−=∴1
0 0 0
11 xlog dtetdtetedxx tmnntnmtnm
using ( ) ( ) dxytm n xlogx ,11
0
m∫=+ ( )( )
1- m ,1m
dye1m
y1 y
0n
nn >
++−= −
∞
∫
( ) ( )( ) ∫∫
∞−
++
−=∴
0
ny1n
nn
1
0
m dyye1m1dx xlogx
( )( )
( ) !n1n ,!n1m1
1n
n=+Γ
+
−=
+
choosing get we,21 m and 1 −==n the required result.
Beta function Definition: Beta function, denoted by B(m,n) is defined by
( ) ( ) dxx1xn,mB 1n1
0
1m −− −= ∫ , where m and n are positive real numbers.
By a property of definite integral, ( ) ( ) dxxxdxxx nmnm 11
0
111
0
1 11 −−−− ∫∫ −=−
Therefore we have B(m,n) = B(n,m)
Alternate Expressions for B(m,n)
( ) ( )∫ −− −=1
0
11 )1.......(....................1, dxxxnmB nm
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( ) t1
tx-1 ,1
1t1
1xLet )( 2 +=
+−=⇒
+= dt
tdxi
and t = ∞ to 0 when x = 0 to 1.
( )( )
dtttt
nm
∫∞ −−
+
+
+=∴
02
11
11
11
11 n)B(m, toreduces 1
( )
( ) ( )mnBdtt
tnm
m
,nm,B using 1
also is n)B(m, 0
1
=+
= ∫∞
+
−
( ) θθθθ ddxii cossin2sin Let x 2 =⇒= 1 to0 when x 2
to0 and =π
=θ
∫π
− θθθ2
0
1m21-2n dcossin2 toreduces (1) Then
∫ −=∴2
0
121-2m cossin2 n)B(m, π
θθθ dn
Relation between Gamma and Beta functions.
Prove that ( ) ( ) ( )( )nm
nmn,mB+ΓΓΓ
=
Proof:
We know that ( ) dxxe2n 1n2
0
2x −∞
−∫=Γ ( ) dyxeand my 12
0
2
2m −∞
−∫=Γ
( ) ( ) ( ) dxdyyxen mnyx 12
0
12
0
22
4mThen −∞
−∞
+−∫ ∫=ΓΓ
( ) θθθ= ∫∫π
=θ
−−−+∞
− dcossin drre42
0
1n21m21nm2
0
2r in polar coordinates
= Γ(m+n) B(m,n) by definitions.
Note:
1 n 0 ,nsin
dxx1
x gsinU0
1-n<<
ππ
=+∫
∞
We get Γ(n) Γ(1-n) = Γ(1) B (1-n,n) using the above relation
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( ) dxx1
xnn,-1BBut 0
1n
∫∞ −
+= using definition of Beta function.
Therefore Γ(n) Γ(1-n) = π /sin nπ, 0 < n < 1
Problems
(1) Show that
=
−∫ 2
1,52Bdx
x1
x51
05
dt54t
51dx,tget x we,t xLetting 5
15 −=== and t = 0 to 1 when x = 0 to 1
( ) dtt51t1t5dx
x1
x5 54
211
0
511
05
−−−=−
∴ ∫∫
( ) dtt1t 211
0
53 −−
−= ∫
.definitionby 21,
52
= B
(2) Evaluate ( )( ) ( )
dxx1
xdxx1
x1x
030
18
030
810
∫∫∞∞
+−
+
−
= B (11,19) - B(19,11) = 0, using B(m,n) = B(n,m).
(3) Evaluate ∫π
θθ2
0dtan
∫∫−=
2
0
21
212
0
cossintanππ
θθθθθ dd
definitionby ,41,
43
21
= B
(4) Show that ( ) 1 n 0, t ,1n,tBndxxnx1 t1t
n
0
n>>+=
− −∫
n to0 when x 1 to0 y and ndydxynx Put ===⇒=
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( ) ndyynydxxnx ttnt
n n11
1
0
1
0
11 −−− ∫∫ −=
−∴
( ) ( ) .definitionby 1,1 11
0
+=−= −∫ ntBndyyyn ttnt
(5) Prove that ( )
= − 2
1,2
1, 12 nnnB n B
∫ −=2
0
121-2n cossin2 n)B(n, that know Weπ
θθθ dn
θθ=θθθ
= ∫π
−
−cos2sin sin2 ,d
22sin2
2
01n2
1n2
∫π
−
−φ=θφ
φ=
01n2
1n22 putting ,d
2sin
( ) φφππφφφπ
sin-sin as )(0,in even is sin ,2
sin22
012
12
== ∫ −
−
dn
n
( ) 21,
21nn,B ,function Beta of definition theApplying 12
= − ngetwe n B
(6) Prove the Duplication formula
( ) ( )nnn n 222
112 Γ=
+ΓΓ −
π
Using the previous result,
+=
++
21,
21nB
21
21n,
21nB n2 …………(1)
( )
( ))2..(....................
21
2112
121
21
21,
21
21,
21
Also
Γ
+Γ+Γ
+Γ
+Γ
+Γ
=
+
++
nn
nnn
nB
nnB
( )
( ) nnn
nnFrom 22
12221n
get we(2) and (1) =Γ
Γ
+Γ
π
∴ ( ) ( )nnn n 222
1. 12 Γ=
+ΓΓ −
π
124 of 133
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(7) Prove that ( )( )
dxx1
xxn,mB1
0nm
1n1m
∫ +
−−
+
+=
By one of the definitions of Beta function, we have
( )( )∫
∞
+
−
+=
0nm
1ndx
x1xn,mB
( ) ( )
dxx1
xdxx1
x
1nm
1n1
0nm
1n
∫∫∞
+
−
+
−
++
+=
= I1 + I2 (say)
Substituting get we,Iin y1x 2=
( ) ( )
dyy1
ydyy1
y1yyI
1
0nm
1m0
12nm1n
nm2 ∫∫ +
−
+−
+
+=
−
+=
( )( )
dxx1
xxnm,B Hence1
0nm
1n1m
∫ +
−−
+
+=
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:
1. The value of ∫ ∫1
0
2
1
2 is dydxxy
(a) 1/2
(b) 7/3
(c) 7/6
(d) 7/2
2. The value of ∫ ∫1
0
2x
xisdy xdx
(a) -1/12
(b) -1/4
(c) 0
(d) -1/15
125 of 133
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3. The value of ( )∫ ∫ ∫−
+
−
++1
1
z
0
zx
zxis dxdydzzyx
(a) 1
(b) 0
(c) 2
(d) 8
4. The value of Γ(-10) is
(a) 10!
(b) 9!
(c) -9!
(d) Not defined
5. The value of Γ(-5/2) is
(a) π− 2
(b) π34
(c) π−158
(d) Not defined
6. ( )( )
tosimplifies x1
dxx1x
026
89
∫∞
+
−
(a) 0
(b) ( )!28
18! !10
(c) ( )!27
17! !9
(d) ( )!2717! !9 ,2
7. ∫π
θθ2
0 toequal is dcot
126 of 133
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(a) 2π
(b) 2π
(c) 2 π
(d) π / 2
8. The value of B(1/2, 1/2) is
(a) π
(b) π
(c) π2
(d) 1
Questions & Answers
(1) Evaluate ( )∫ ∫ ∫
− −
+++
1
0
x1
0
yx
03zyx1
dxdydz
Ans: Let this integral be I, then
( )∫ ∫∫−
−−
−
−
−
+++=
1
0x
yx1
0z3
x1
0ydxdy
zyx1dzI
( )∫ ∫−
−
−
−−
−
+++
−=
1
0x
x1
0y
yx1
0z3 dydx
zyx121
( )∫ ∫−
−
−
+++
−=
1
0x
x1
0y2 dx
zyx121
−=
852ln
21
(2) Change the order of integration and hence evaluate the integral
∫ ∫−1
0
x2
xdxdy
yx
Ans: The region of integration is the shaded portion shown in the figure below
Y y = x
x = 0
B
127 of 133
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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
∫ ∫ ∫ ∫− − −
−
−
+=1
0y
y
0x
2
1y
y2
0xdxdy
yxdxdy
yx
dy22y
y2dy
2y 2
1y
1
0y∫∫−−
−++=
= 2 ln 2 -1.
(3) Change to polar coordinates and hence evaluate
∫ ∫+
a
0
a
y22
dxdyyx
x
Ans: The region of integration is as shown below
X
Y
0
x = y
x = a
y = a
y = 0
128 of 133
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We see that θ ranges from 0 to π/4 in the shaded region. R ranges from 0 to x = a i.e., a sec θ.
The given integral
∫ ∫θ
π
θθ=seca4
0 0
22 drdcosr
( )[ ] 40
34
0
3tansecln
3adsec
3a π
π
θ+θ=θθ= ∫
( )12ln3
a3+=
129 of 133
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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 ∫∫r
dxdy
( )∫∫ ∫−−
−
−
−==3
0x
23
0x
2xx4
xydxxx3dxdy
29
3xx
23 3
23
0
=
−=
Practice Questions 1. Evaluate the following integrals
( ) ( )∫ ∫ ++3
0
2
1dxdyyx1x i
( ) ∫ ∫π
θθ2
0
a
0
2 drdsinr ii
( ) ( )∫ ∫ +1
0
x
x
22 dydxyx iii
( ) ∫ ∫a
1
b
1dydx
xy1 iv
( ) ∫ ∫ ∫ ++1
0
1
0
1
0
zyx dxdydze v
(2,4) y = x
(3,3)
130 of 133
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( ) ∫ ∫ ∫3
1
1
x1
xy
0dxdy dz xyz vi
( ) ∫ ∫ ∫π
θ −θ
2a
0
cos
0
2r2a
0ddr rdz vii
2. Change the order of integration and hence evaluate
( ) ∫ ∫1
0
y-2
y
xydxdy i
( ) ∫ ∫ 4a
0
ax2
a42x
dydx ii
( ) ∫ ∫a
0
x-2a
a2x
xydydx iii
( )( )( )∫ ∫ −−
a
0
x
0dydx
yaxaycos iv
3. Change to respective coordinate systems and hence evaluate
( ) ∫ ∫+
2
0
2x-2x
022
systempolar todydxyx
x i
( ) ∫ ∫ +a
0
2x-2a
0
222 systempolar todydxyxy ii
( ) ∫ ∫ +1
0
2x-2x
x
22 systempolar todydx yx iii
( ) polar spherical tozyx1
dzdydx iv1
0
2x-1
0
2y2x1
0222∫ ∫ ∫
−−
−−−
131 of 133
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( ) polar spherical tozyx
dzdydx v1
0
2x-1
0
1
2y2x222∫ ∫ ∫
−++
( ) polar lcylindrica toxyzdxdydz vi1
0x
1
0y
2
2y2xz
∫ ∫ ∫= = +=
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 cos2θ is a2/2.
7. Find the area of one petal of the rose r = a sin3θ.
8. Find the area of the circle r = a sinθ outside the cardioide r = a (1 - cosθ)
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. dxex Evaluate0
4x2∫∞
−
14. ( )∫1
0
3x
1log Find dx
15. ( ) 313
2
0
4x8
163x Evaluate
−−∫ dx
16. ∫ <<2
0
p 1 p 0 ,tan π
θθ dFind
17. ( )ndtet2a that Show0
2at1n2n Γ=∫∞
−−
18. a log2
1 is 0 a ,dxe that Show0
alog2x- π>∫
∞
132 of 133
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19. If m and n are real constants > -1, prove that
( )( ) 1n
n1
0
m
1m1ndx
x1logx
++
+Γ=
∫
20. Show that ( )2
cos1cos0
1 πnna
dxaxx nn Γ=∫
∞−
Hint: Write cos ax = Real part of eiax
21. Show that
=
−∫ 2
1,n1B
n1
x1
dx1
0n
22. Show that ( ) ( ) ( ) dxx1x12
1n,mB 1n1
1
1m1nm
−
−
−−+
−+= ∫
Hence deduce that ∫− −
+1
1 11
xx dx = π
23. Show that 216
dxexdxxe0
4x2
0
8x π=∫∫
∞−
∞−
24. Show that ( ) ( ) ( ) ( )n,mBabdxxbaxb
a
1nm1n1m∫ −+−− −=−−
133 of 133
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