ncet – class notes 10mat11 chapter – 1 lesson -1...

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NCET – Class Notes 10MAT11 Chapter – 1 LESSON -1 : Successive Differentiation In this lesson, the idea of differential coefficient of a function and its successive derivatives will be discussed. Also, the computation of n th derivatives of some standard functions is presented through typical worked examples. 1.0 Introduction:- Differential calculus (DC) deals with problem of calculating rates of change.When we have a formula for the distance that a moving body covers as a function of time, DC gives us the formulas for calculating the body’s velocity and acceleration at any instant. Definition of derivative of a function y = f(x):- Fig.1. Slope of the line PQ is x x f x x f Δ Δ + ) ( ) ( The derivative of a function y = f(x) is the function ) ( x f whose value at each x is defined as dx dy = ) ( x f = Slope of the line PQ (See Fig.1) = 0 lim Δx x x f x x f Δ Δ + ) ( ) ( -------- (1) = 0 lim Δx (Average rate change) = Instantaneous rate of change of f at x provided the limit exists. The instantaneous velocity and acceleration of a body (moving along a line) at any instant x is the derivative of its position co-ordinate y = f(x) w.r.t x, i.e., Velocity = dx dy = ) ( x f --------- (2) And the corresponding acceleration is given by Acceleration ) ( 2 2 x f dx y d = = ---------- (3)

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Page 1: NCET – Class Notes 10MAT11 Chapter – 1 LESSON -1 ...msbaboo.weebly.com/.../2/2/4/4/22446366/succesive_differentiation.pdf · NCET – Class Notes 10MAT11 Chapter – 1 LESSON

NCET – Class Notes 10MAT11

Chapter – 1

LESSON -1 : Successive Differentiation

• In this lesson, the idea of differential coefficient of a function and its successive

derivatives will be discussed. Also, the computation of nth

derivatives of some standard

functions is presented through typical worked examples.

1.0 Introduction:- Differential calculus (DC) deals with problem of calculating rates of

change.When we have a formula for the distance that a moving body covers as a

function of

time, DC gives us the formulas for calculating the body’s velocity and acceleration at

any

instant.

• Definition of derivative of a function y = f(x):-

Fig.1. Slope of the line PQ is x

xfxxf

∆−∆+ )()(

The derivative of a function y = f(x) is the function )(xf ′ whose value at each x is defined as

dx

dy = )(xf ′ = Slope of the line PQ (See Fig.1)

= 0

lim→∆x x

xfxxf

∆−∆+ )()(

-------- (1)

= 0

lim→∆x

(Average rate change)

= Instantaneous rate of change of f at x provided the limit exists.

The instantaneous velocity and acceleration of a body (moving along a line) at any instant x is

the derivative of its position co-ordinate y = f(x) w.r.t x, i.e.,

Velocity = dx

dy= )(xf ′ --------- (2)

And the corresponding acceleration is given by

Acceleration )(2

2

xfdx

yd ′′== ---------- (3)

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NCET – Class Notes 10MAT11

• Session - 1

1.1 Successive Differentiation:-

The process of differentiating a given function again and again is called as

Successive differentiation and the results of such differentiation are called

successive derivatives.

• The higher order differential coefficients will occur more frequently in spreading

a function all fields of scientific and engineering applications.

• Notations:

i. dx

dy,

2

2

dx

yd,

3

3

dx

yd,…….., n

th order derivative:

n

n

dx

yd

ii )(xf ′ , )(xf ′′ , )(xf ′′′ ,…..., nth

order derivative: )(xf n

iii Dy, yD 2 , yD3 ,………..., nth

order derivative: yD n

iv y ′ , y ′′ , y ′′′ ,……, nth

order derivative: )(ny

v. 1y , 2y , 3y …, nth

order derivative: ny

• Successive differentiation – A flow diagram

Input function: )(xfy = Output function )(xfdx

dfy ′==′ (first order

derivative)

Input function )(xfy ′=′ Output function )(2

2

xfdx

fdy ′′==′′ (second order

derivative)

Input function )(xfy ′′=′′ Output function )(3

3

xfdx

fdy ′′′==′′′ (third order

derivative)

-----------------------------------------------------------------------------------------------------------------

Input function )(11 xfy nn −− = Output function )(xfdx

fdy n

n

nn == (nth

order

derivative)

Animation Instruction (Successive Differention-A flow diagram)

Output functions are to appear after operating

on Input functions, successively.

• 1.1 Solved Examples :

1. If )sin(sin xy = , prove that 0costan 2

2

2

=++ xydx

dyx

dx

yd

Operation

dxd

Operation

dxd

Operation

dxd

Operation

dxd

Operation

dxd

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NCET – Class Notes 10MAT11

Solution: Differentiating )sin(sin xy = --------- (1) w.r.t.x, we get

xxdx

dyy cos).cos(sin1 == -------------- (2)

Again differentiating dx

dyy =1 w.r.t.x gives

[ ]xxxxxdx

ydy cos)sin(sin(cos)sin)(cos(sin

2

2

2 −+−== Using product rule

[ ])sin(sincos)cos(sinsin 2

2

2

2 xxxxdx

ydy +−==

i.e.

+−= )sin(sincos)cos(sincoscos

sin 2

2 xxxxx

xy

[ ]xyxyy 2

12 costan +−= , using Eqs. (1) and (2)

or 0costan 2

12 =++ xyxyy

or 0costan 2

2

2

=++ xydx

dyx

dx

yd

2. If)(

)(

dcx

baxy

++

= , show that 2

231 32 yyy =

Solution: We rewrite)(

)(

dcx

baxy

++

= , by actual division of ax+b by c x+d, as

( ) 11 −++=+

−+= dcxkc

a

dcxc

adb

cay ---------- (1)where

−=c

adbk

Differentiating (1) successively thrice, we get

( ) 2

1

−+−== dcxkcydx

dy ---------- (2)

( ) 32

22

2

2−+−== dcxkcy

dx

yd ---------- (3)

( ) 43

33

3

6−+−== dcxkcy

dx

yd ---------- (4) From (2), (3) and (4) we get

{ }{ }[ ]432

31 )(6)(22 −− +−+−= dcxkcdcxkcyy

642

31 )(122 −+= dcxckyy

[ ]232

31 )(232−+−= dcxkcyy

Therefore 2

231 32 yyy = , as desired.

3. If tx sin= , pty sin= , Prove that 0)1( 2

2

22 =+−− yp

dx

dyx

dx

ydx

Solution: Note that the function is given in terms a parameter t. So we find,

tdt

dycos= and ptp

dt

dycos= , so that

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NCET – Class Notes 10MAT11

t

ptp

dx

dyy

dtdx

dtdy

cos

cos1 === . Squaring on both sides

( )2

22

2

22

2

222

11

)1(

sin1

)sin1(

cos

cos

x

yp

t

ptp

t

ptpy

−=

−== (by data)

∴ ( )( ) ( )222

1

2 11 ypyx −=− .

Differentiating this equation w.r.t x, we get

( ) )2()2()(21 1

22

121

2 yypxyyyx −=−+− .

Canceling 12y throughout, this becomes

( ) ypxyyx 2

12

21 −=−−

or ( ) 01 2

12

2 =+−− ypxyyx

i.e. 0)1( 2

2

22 =+−− yp

dx

dyx

dx

ydx

4. If )sin(cos tttax += , )cos(sin tttay −= , find 2

2

dx

yd

Solution: ( ) tatttttadt

dycossincossin =++−=

( ) tatttttadt

dysincossincos =−+=

∴ ttat

tat

dx

dy

dtdx

dtdy

tancos

sin===

Hence, tattat

tdx

dtt

dx

yd3

22

2

2

cos

1

cos

1secsec =

=

=

5. If ( )a

xay cosh= , prove that 2

1

2

2

2 1 yya +=

Solution: ( )( ) ( ),sinh1sinh1 ax

aaxa

dx

dyy === and

( )( ),1cosh2

2

2 aax

dx

ydy ==

∴ ( )a

xay cosh2 = , so that ( )a

xya 22

2

2 cosh=

i.e. ( ) 2

1

22

2

2 1sinh1 ya

xya +=+= , as desired.

• Problem Set No. 1.1 for practice.

1. If bxey ax sin= , prove that ( ) 02 22

12 =++− ybaayy

2. If 12 22 =++ byhxyax , prove that ( )3

2

2

2

byhx

abh

dx

yd

+

−=

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NCET – Class Notes 10MAT11

3. If )cos( ctlAey kt += − , show that ( ) 02 22

2

2

=+++ ylkdx

dyk

dx

yd

4. If ( )21log xxy ++= , prove that ( ) 012

22 =++

dx

dyx

dx

ydx

5. If ( )xy sinhtan 1−= , prove that 0tan

2

2

2

=

+dx

dyy

dx

yd

6. If ( )2tanlogcos ttax += , tay sin= , find 2

2

dx

yd Ans:

tat

4cossin

7. Find2

2

dx

yd, when θ3cosax = , θ3sinby = Ans:

2

4

3

seccos

a

ecb θθ

8. Find3

3

dx

yd, where

−= −

2

1

1tan

x

xy Ans:

( ) 25

2

2

1

21

x

x

+

9. If ttx 2coscos2 −= , ttx 2sinsin2 −= , Find 22

2 π=

x

dx

yd Ans: -3/2

10. If xx beexy −+= , prove that 022

2

=−+ xydx

dy

dx

ydx

• Session -2 1.2 Calculation of n

th derivatives of some standard functions

• Below, we present a table of n

th order derivatives of some standard functions for ready

reference.

Table : 1

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NCET – Class Notes 10MAT11

• We proceed to illustrate the proof of some of the above results, as only the above

functions are able to produce a sequential change from one derivative to the other.

Hence, in general we cannot obtain readymade formula for nth derivative of functions

other than the above.

1. Consider mxe . Let mxey = . Differentiating w.r.t x, we get

=1ymxme . Again differentiating w.r.t x, we get

( )mxmemy =2 = mxem2

Similarly, we get

3y = mxem 3

4y = mxem4

…………….

Sl.

No

y = f(x) yD

dx

ydy n

n

n

n ==

1 mxe mxnem

2 mxa ( ) mxnn aam log

3 ( )mbax + i. ( )( ) ( ) ( ) nmn baxanmmmm−++−−− 1....21 for all m .

ii. 0 if nm <

iii. ( )!n a n if nm =

iv. ( )

nmxnm

m −

− !

! if nm <

4

( )bax +1

n

n

n

abax

n1)(

!)1(++

5.

( )mbax +

1 n

nm

n

abaxm

nm++−

−+−

)(!)1(

!)1()1(

6. )log( bax + n

n

n

abax

n

)(

!)1()1( 1

+

−− −

7. )sin( bax + )2

sin( πnbaxa n ++

8. )cos( bax + )2

cos( πnbaxa n ++

9. )sin( cbxeax + )sin( θncbxer axn ++ , )(tan 122a

bbar −=+= θ

10. )cos( cbxeax + )cos( θncbxer axn ++ , )(tan 122a

bbar −=+= θ

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NCET – Class Notes 10MAT11

And hence we get

ny = mxnem ∴ [ ]=mx

n

n

edx

d mxnem .

2. ( )mbax + (See Sl. No-3 of Table-1 )

let =y ( )mbax + Differentiating w.r.t x,

1y =m ( ) abaxm 1−+ . Again differentiating w.r.t x, we get

2y = m ( )1−m ( ) 22abax

m−+

Similarly, we get

3y = m ( )1−m ( )2−m ( ) 33abax

m−+

………………………………….

And hence we get

ny = m ( )1−m ( )2−m ………. ( )1+− nm ( ) nnmabax

−+ for all m.

Case (i) If nm = (m-positive integer),then the above expression becomes

ny =n ( )1−n ( )2−n ……….3.2.1 ( ) nnnabax

−+

i.e. ( ) n

n any !=

Case (ii) If m<n,(i.e. if n>m) which means if we further differentiate the above expression,

the

right hand site yields zero. Thus ( )[ ] ( )nmifbaxDmn <=+ 0

Case (iii) If m>n, then ( )( ) ( )( ) nnm

n abaxnmmmmy−++−−−= 1......21 becomes

( )( ) ( )( )

( )( ) nnm

abaxnm

nmnmmmm −+−

−+−−−=

!

!1......21

. i.e ( )

( ) nnm

n abaxnm

my

−+−

=!

!

3. ( )mbax +

1(See Sl. No-5 of Table-1 )

( )

( ) m

mbax

baxyLet

−+=+

=1

Differentiating w.r.t x

( ) ( ) ( ) ( )abaxmabaxmy

mm 11

1 1+−−− +−=+−=

( )( ) ( )( ) ( )[ ] ( ) ( )( ) ( ) 22211

2 1111 abaxmmabaxmmymm +−−+− ++−=++−−=

Similarly, we get ( ) ( )( )( ) ( ) 333

3 211 abaxmmmym+−+++−=

( ) ( )( )( )( ) ( ) 444

4 3211 abaxmmmmym+−++++−=

……………………………

( ) ( )( ) ( )( ) ( ) nnmn

n abaxnmmmmy+−+−+++−= 1.....211

This may be rewritten as

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NCET – Class Notes 10MAT11

( ) ( )( ) ( ) ( )

( )( ) ( ) nnm

n

n abaxm

mmmnmnmy

+−+−

−+−+−+−=

!1

!11.....211

or ( ) ( )( ) ( )

n

nm

n

n abaxm

nmy

++−

−+−=

!1

!11

4.( )bax +

1 (See Sl. No-4 of Table-1 )

Putting ,1=m in the result

n

nm

n

m

n abaxm

nm

baxD

++−

−+−=

+ )(!)1(

!)1()1(

)(

1

we get n

n

nn a

bax

n

baxD

++−

−+−=

+ 1)(!)11(

!)11()1(

)(

1

or n

n

nn a

bax

n

baxD

++

−=

+ 1)(

!)1(

)(

1

• 1.2.1. Worked Examples:-

In each of the following Questions find the nth

derivative after reducing them into standard

functions given in the table 1.2.1

1. (a) )19log( 2 −x (b) [ ]75)34(log ++ xex (c) 6

2

10)1(

)32()53(log

+

−+

x

xx

Solution: (a) Let { })13)(13(log)19log( 2 −+=−= xxxy

)13log()13log( −++= xxy (Q BAAB loglog)log( += )

∴ { } { })13log()13log( −++= xdx

dnx

dx

dny

nnn

i.e n

n

nn

n

n

nx

n

x

ny )3(

)13(

!)1()1()3(

)13(

!)1()1( 11

−−+

+

−−=

−−

(b) Let [ ] 7575 log)34log()34(log ++ ++=+= xx exexy

exx elog)75()34log( +++= (Q ABAB loglog = )

∴ )75()34log( +++= xxy (Q 1log =ee )

∴ 0)4()34(

!)1()1( 1

++

−−=

−n

n

n

nx

ny Q 5)65( =+xD

0)65(2 =+xD

)1(0)15( >=+ nxD n

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NCET – Class Notes 10MAT11

(c) Let 6

2

10)1(

)32()53(log

+

−+=

x

xxy

+

−+=

6

2

)1(

)32()53(

10log

1

x

xx

e

Q 10log

loglog10

e

e XX =

+

−+=

6

2

)1(

)32()53(log

2

1

10log

1

x

xx

e

Q ABAB loglog =

Q BAB

Alogloglog −=

{ }62 )1log()32log()53log(10log2

1+−−++= xxx

e

∴ { })1log(6)32log()53log(210log2

1+−−++= xxxy

e

Hence,

+

−−−−

−−+

+

−−=

−−−n

n

nn

n

nn

n

n

e

nx

n

x

n

x

ny )1(

)1(

!)1()1(.6)3(

)32(

!)1()1()3(

)53(

!)1()1(.2

10log2

1111

2. (a) 4242 6 ++ + xxe (b) xx 4cosh4cosh 2+

(c) xxe x 2cosh3sinh− (d) 5

4)86(

)45(

1

)54(

1++

++

+x

xx

Solution: (a) Let 4242 6 ++ += xxey

4242 66 xxee +=

∴ )6(1296)( 224 xxeey +=

hence )6(1296)( 224 x

n

x

nndx

dne

dx

dney +=

{ } { }xnnxn ee 224 6)6(log212962 +=

(b) Let xxy 4cosh4cosh 2+=

24444

22

++

+=

−− xxxx eeee

( ) { }))((2)()(4

1

2

1 44242444 xxxxxx eeeeee −−− ++++=

( ) { }24

1

2

1 8844 ++++= −− xxxx eeeey

hence, [ ] [ ]0)8(84

1)4(4

2

1 8844 +−++−+= −− nnnnxnxn

n eeeey

(c) Let xxey x 2cosh3sinh−=

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NCET – Class Notes 10MAT11

+

=−−

22

2233 xxxxx eeee

e

{ }))((4

2233 xxxxx

eeeee −−−

+−=

{ }xxxxx

eeeee 55

4

−−−

−+−=

{ }xxx eee 624 14

1 −− −+−=

{ }xxx eeey 62414

1 −− −−+=

Hence, { }xnxnxn

n eeey 624 )6()2()4(04

1 −− −−−−+=

(d) Let 5

4)86(

)45(

1

)54(

1++

++

+= x

xxy

Hence, ( ){ }5

486

)45(

1

)54(

1++

++

+= x

dx

dn

xdx

dn

xdx

dny

nnnn

0)5()45(!)14(

!)14()1()4(

)54(

!)1(41

++−

−+−+

+

−=

++n

n

nn

n

n

x

n

x

n

i.e n

n

nn

n

n

nx

n

x

ny )5(

)45(!3

!)3()1()4(

)54(

!)1(41 ++ +

+−+

+

−=

• Session - 3 • 1.2.2 Worked examples:-

1. (i) 86

12 +− xx

(ii) 321

1

xxx +−−(iii)

672 2

2

++ xx

x

(iv) 9124

1

1

22 ++

+

++

xxx

x (v) ( )

ax1tan − (vi) x1tan − (vii)

−+−

x

x

1

1tan 1

• In all the above problems, we use the method of partial fractions to reduce

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NCET – Class Notes 10MAT11

them into standard forms.

Solutions: (i) Let 86

12 +−

=xx

y . The function can be rewritten as )2)(4(

1

−−=

xxy

• This is proper fraction containing two distinct linear factors in the denominator.

So, it can be split into partial fractions as

)2()4()2)(4(

1

−+

−=

−−=

x

B

x

A

xxy Where the constant A and B are found

as given below.

)2)(4(

)4()2(

)2)(4(

1

−−−+−

=−− xx

xBxA

xx

∴ )4()2(1 −+−= xBxA -------------(*)

Putting x = 2 in (*), we get the value of B as 2

1−=B

Similarly putting x = 4 in(*), we get the value of A as 2

1=A

∴2

)2/1(

4

)2/1(

)2)(4(

1

−−

+−

=−−

=xxxx

y Hence

=2

1

2

1

4

1

2

1

xdx

d

xdx

dy

n

n

n

n

n

−−

−=

++n

n

nn

n

n

x

n

x

n)1(

)2(

!)1(

2

1)1(

)4(

!)1(

2

111

−−

−−=

++ 11 )2(

1

)4(

1!)1(

2

1nn

n

xxn

(ii) Let )1)(1(

1

)1()1(

1

1

12232 xxxxxxxx

y−−

=−−−

=+−−

=

ie )1()1(

1

)1)(1)(1(

12 xxxxx

y+−

=+−−

=

Though y is a proper fraction, it contains a repeated linear factor 2)1( x− in its

denominator. Hence, we write the function as

x

C

x

B

x

Ay

++

−+

−=

1)1()1( 2 in terms of partial fractions. The constants A, B, C

are found as follows:

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NCET – Class Notes 10MAT11

x

C

x

B

x

A

xxy

++

−+

−=

+−=

1)1()1()1()1(

122

ie 2)1()1()1)(1(1 xCxBxxA −++++−= -------------(**)

Putting x = 1 in (**), we get B as 2

1=B

Putting x = -1 in (**), we get C as 4

1=C

Putting x = 0 in (**), we get CBA ++=1

∴4

14

12

111 =−−=−−= CBA

∴4

1=A

Hence, )1(

)4/1(

)1(

)2/1(

)1(

)4/1(2 xxx

y+

+−

+−

=

+

−+

−−

−+−+

−=

+++n

n

nn

n

nn

n

n

nx

n

x

n

x

ny )1(

)1(

!)1(

4

1)1(

)1(!)12(

!)12()1(

2

1)1(

)1(

!)1(

4

1121

+−+−

+

++

−−=

++ 2)1(

!)1()1(

2

1

)1(

1

)1(

1!)1(

4

111 nx

n

xxn

n

nn

n

(iii) Let 672 2

2

++=

xx

xy (VTU July-05)

This is an improper function. We make it proper fraction by actual division and

later

spilt that into partial fractions.

i.e 672

)3(

2

1)672(

2

27

22

+−

−−+=++÷

xx

xxxx

∴)2)(32(

3

2

1 27

++−

+=−

xx

xy Resolving this proper fraction into partial fractions, we get

++

++=

)2()32(2

1

x

B

x

Ay . Following the above examples for finding A & B, we

get

+−

++

+=2

)4(

322

1 29

xxy

Hence,

+

−−

+

−+=

++n

n

nn

n

n

nx

n

x

ny )1(

)2(

!)1(4)2(

)32(

!)1(

2

90

11

i.e

+−

+−=

++ 11

29

)2(

4

)32(

)2(!)1(

nn

nn

nxx

ny

(iv) Let 9124)1(

)2(2 ++

+++

=xx

x

x

xy

(i) (ii)

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Here (i) is improper & (ii) is proper function. So, by actual division (i) becomes

+

+=

++

1

11

1

2

xx

x. Hence, y is given by

2

)32(

1

1

11

++

+

+=xx

y [Q 9124)32( 22 ++=+ xxx ]

Resolving the last proper fraction into partial fractions, we get

22 )32()32()32( +

++

=+ x

B

x

A

x

x. Solving we get

2

1=A and 2

3−=B

+

−+

++

+

+=2

23

21

)32()32(1

11

xxxy

+++−

+

−+

+

−+=

+n

nn

n

nn

n

n

nnx

n

x

n

x

ny )2(

2)32(

!)1()1(

2

3)2(

)32(

!)1(

2

1)1(

)1(

!)1(0

1

(v) ( )a

x1tan −

Let ( )a

xy 1tan −=

∴( ) 2221

1

1

1

ax

a

aa

xy

+=

+=

+

=== −−22

1

1

1 )(ax

aDyDyDy nnn

n

Consider ))((22 aixaix

a

ax

a

−+=

+

)()( aix

B

aix

A

−+

+= , on resolving into partial fractions.

( ) ( )

)(

21

)(

21

aix

i

aix

i

−+

+

−= , on solving for A & B.

+

+−

=

+

−−−

aixD

aixD

ax

aD ininn 2

112

11

22

1

−−

+

+

−−

−=−−

n

n

n

n

aix

n

iaix

n

i )(

!)1()1(

2

1

)(

!)1()1(

2

111

-----------(*)

• Since above answer containing complex quantity i we rewrite the answer in terms of real

quantity, We take transformation θθ sincos rarx == where 22 axr += ,

= −

x

a1tanθ

( ) θθθ ireiraix =+=+ sincos

( ) θθθ ireiraix −=−=− sincos

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( ) n

in

innn r

e

eraix

θ

θ ==− −

11, ( ) n

in

n r

e

aix

θ−

=+

1

now(*) is ( ) ( ) [ ]θθ inin

n

n

n eeri

ny −

−−−

=2

!111

( ) ( ) ( ) ( )

θθ nr

nni

riy

n

n

n

n

n sin!11

sin22

111 −−

⇒−

=−−

(vi) Let xy 1tan −= .Putting a = 1 in Ex.(v) we get

ny which is same as above with ( )x

xr 1tan1 12 −=+= θ

( ) θθ cotcot 1 == − xorx

θθ

θθ n

nn ecrecr sin

cos

11cos1cot 2 ==⇒=+=∴

( ) ( ) ( ) xwherennxD nnn 111 cotsinsin!11tan −−− =−−= θθθ

(vii) Let

−+

= −

x

xy

1

1tan 1

put θtan=x x1tan −=θ

−+

= −

θθ

tan1

tan1tan 1y

[ ])tan(tan 41 θπ += − Q ( )

−+

=+θθ

θπtan1

tan1

4tan

)(tan44

1 x−+=+=π

θπ

)(tan4

1 xy −+=π

)(tan0 1 xDy n

n

−+=

−−

+

+

−−

−=−−

n

n

n

n

aix

n

iaix

n

i )(

!)1()1(

2

1

)(

!)1()1(

2

111

Problem set No. 1.2.1 for practice

Find the nth

derivative of the following functions:

1. )4)(1(

62 −− xx

x 2.

)12)(2( 2 +−+ xxx

x 3.

22

1423

2

−−+

++

xxx

xx 4.

34 2 −− xx

x

5. 232

3

+− xx

x 6.

−2

1

1

2tan

x

x 7.

−+−

x

x 11tan

21

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• Session 4 1. )sin( bax + .(See Sl. No-7 of Table-1 )

Let )sin( baxy += . Differentiating w.r.t x,

abaxy ).cos(1 += As XX cos)2

sin( =+π

We can write

).2/sin(1 π++= baxay

Again Differentiating w.r.t x,

abaxay ).2/cos(2 π++= Again using XX cos)2

sin( =+π

,we get 2y as

abaxay ).2/2/sin(2 ππ +++=

i.e. ).2/2sin(2

2 π++= baxay

Similarly, we get

).2/3sin(3

3 π++= baxay

).2/4sin(4

4 π++= baxay

……………………………

).2/sin( πnbaxay n

n ++=

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2. ( ).sin cbxeax + (See Sl. No-9 of Table-1 )

( ) )1.....(sin cbxeyLet ax +=

Differentiating using product rule ,we get

=1y ( ) ( ) axax aecbxbcbxe +++ sincos

i.e. =1y ( ) ( )[ ]cbxbcbxaeax +++ cossin . For computation of higher order derivatives

it is convenient to express the constants ‘a’ and ‘b’ in terms of the constants r and

θ defined by θcosra = & θsinrb = ,so t22 bar += and ( )

ab1tan −=θ .thus,

1y can be rewritten as

( ) ( ) ( ) ( )[ ]cbxrcbxreyax +++= cossinsincos1 θθ

or ( ) ( )[ ]}coscoscos{sin1 θθ cbxcbxreyax +++=

i.e. ( )[ ] )2.(.......... sin1 θ++= cbxrey ax

Comparing expressions (1) and (2), we write 2y as

( )θ2sin2

2 ++= cbxery ax

( )θ3sin3

3 ++= cbxery ax

Continuing in this way, we get

( )θ4sin4

4 ++= cbxery ax

( )θ5sin5

5 ++= cbxery ax

…………………………….

( )θncbxery axn

n ++= sin

∴ ( )[ ] ( ),sinsin θncbxercbxeD axnaxn ++=+ where 22 bar += & ( )

ab1tan −=θ

• 1.2.3 Worked examples

1. (i) xx 32 cossin + (ii) x33 cossin (iii) xxx 3cos2coscos

(iv) xxx 3sin2sinsin (v) xe x 2cos3 (vi) ( )xxe x 322 cossin +

• The following formulae are useful in solving some of the above problems.

(i) 2

2cos1cos)(

2

2cos1sin 22 x

xiix

x+

=−

=

(iii) xxxivxxx cos3cos43cos)(sin4sin33sin 33 −=−=

(v) ( ) ( )BABABA −++= sinsincossin2

(vi) ( ) ( )BABABA −−+= sinsinsincos2

(vii) ( ) ( )BABABA −++= coscoscoscos2

(viii) ( ) ( )BABABA +−−= coscossinsin2

Solutions: (i) Let ( )xxx

xxy cos33cos4

1

2

2cos1cossin 32 ++

−=+=

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( ) ( )[ ] ( ) ( ) ( )[ ]2

cos32

3cos34

1

22cos20

2

1 πππ nxnxnxynn

n +++++−=∴

(ii)Let y =

+−==

=4

2sin36sin

8

1

8

2sin

2

2sincossin

33

33 xxxxxx

[ ]xx 6sin2sin332

1−=

+−

+=2

6sin62

2sin2.332

1 ππ nx

nxy nn

n

(iii) )Let y = xxx 2coscos3cos

= ( ) [ ]xxxxxx 2cos2cos4cos2

12cos2cos4cos

2

1 2+=+

= ( )

−++

2

4cos12cos6cos

2

1

2

1 xxx

( )xxx 4cos1

4

1

4

2cos6cos

4

1−++=

4

24cos4

4

22cos2

26cos6

4

1

+−

++

+=∴

πππ

nx

nx

nxy

nn

n

n

(iv) )Let y = xsnxx 2sin3sin

( )[ ] xxx 2sin4sin2sin2

1−=

[ ]xxx 2sin4sin2sin2

1 2 −=

= ( )

−−−

xxx

6sin2sin2

1

2

4cos1

2

1

( )

−−

−= xx

x6sin2sin

4

1

4

4cos1

++

+−

+=2

6sin62

2sin22

4cos44

1 πππ nx

nx

nxy nnn

n

(v) Let xey x 2cos3= (Refer Sl.No. 10 of Table 1)

( )θnxrey x

n +=∴ 2cos3 where

22 23 +=r 13= &

= −

3

2tan 1θ

(vi) Let y = ( )xxe x 322 cossin +

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We know that [ ] [ ]xxx

xx cos33cos4

1

2

2cos1cossin 32 ++

−=+

∴y = [ ] [ ]xxex

exxex

xx cos33cos42

2cos1cossin

22322 ++

−=+

[ ] [ ]xexexeey xxxx cos33cos4

12cos

2

1 2222 ++−=∴

Hence, ( )[ ] ( ) ( )[ ]3

2

32

2

21

2

1

2 cos33cos4

12cos2

2

1θθθ nxernxernxerey xnxnxnxn

n +++++−=

where 22

1 22 +=r 8= ; 22

2 32 +=r 13= ; 22

3 12 +=r 5=

;2

1tan;

2

3tan;

2

2tan 1

3

1

2

1

1

=

=

= −−− θθθ

Problem set No. 1.2.2 for practice

Find nth

derivative of the following functions:

1. )cos(sin 23 xx + 2. xx 3cos2sin 3. xx 3sin.2cos 4. xx 2coscos

5. xx 2sinsin 6. )cos(sin 233 xxe x + 7. xxe x 4cos2cos 8. xxe x 2cossin 2−

.9. xe x 33 cos− (VTU Jan-04)

LESSON -2 : Leibnitz’s Theorem

• Session - 1 • Leibnitz’s theorem is useful in the calculation of n

th derivatives of product of two functions.

• Statement of the theorem: If u and v are functions of x, then

vuDvuDDCvuDDCuDvDCuvDuvD nrrn

r

nnnnnnn .......)( 22

2

1

1 +++++= −−− ,

wheredx

dD = , nCn =1 ,

( )( )!!

!,........,

2

12

rnr

nC

nnC r

nn

−=

−=

• Worked Examples

1. If thatproveptytx sin,sin ==

( ) ( ) ( ) 0121 22

12

2 =−++−− ++ nnn ynpxynyx (VTU July-05)

Solution: Note that the function )(xfy = is given in the parametric form with a parameter t.

So, we consider

t

ptp

dx

dy

dtdx

dtdy

cos

cos== (p – constant)

or 2

22

2

22

2

222

1

)1(

sin1

)sin1(

cos

cos

x

yp

t

ptp

t

ptp

dx

dy

−=

−==

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or ( ) ( )222

1

2 11 ypyx −=−

So that ( ) ( )222

1

2 11 ypyx −−− Differentiating w.r.t. x,

( )( ) ( )[ ] ( ) 02221 1

22

121

2 =−−−+− yypxyyyx

( ) 01 2

12

2 =+−− ypxyyx --------------- (1) [ 12y÷ , throughout]

Equation (1) has second order derivative 2y in it. We differentiate (1), n times, term

wise,

using Leibnitz’s theorem as follows.

( )[ ] 01 2

12

2 =−−− ypxyyxD n

i.e { } { } ( ) 0)1( 2

12

2 =−−− ypDxyDyxD nnn ---------- (2)

(a) (b) (c)

Consider the term (a):

( )[ ]2

21 yxD n − . Taking 2yu = and )1( 2xv −= and applying Leibnitz’s theorem we get

[ ] ...33

3

22

2

1

1 ++++= −−− vuDDCvDDCuDvDCuvDuvD nnnnnnnn

i.e

[ ] ...)1()()1()()1().()1).(()1( 23

2

3

3

22

2

2

2

2

2

1

1

2

2

2

2 +−+−+−+−=− −−− xDyDCxDyDCxDyDCxyDxyD nnnnnnnn

...)0.(.!3

)2)(1()2.(

!2

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

2

2)( +−−

+−−

+−+−= +−+−+−+ nnnn ynnn

ynn

xnyxy

( )[ ] ( ) nnn

n ynnnxyyxyxD )1(211 12

2

2

2 −−−−=− ++ ----------- (3)

Consider the term (b):

[ ]1xyD n . Taking 1yu = and xv = and applying Leibnitz’s theorem, we get

[ ] ....)().()(.)).(()( 2

1

2

21

1

111 +++= −− xDyDCxDyDCxyDxyD nnnnnn

....)0(!2

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

−++= +−+−+ nnn y

nnnyxy

[ ] nn

n nyxyxyD += +11 ---------- (4)

Consider the term (c):

n

nn ypyDpypD 222 )()( == --------- (5)

Substituting these values (3), (4) and (5) in Eq (2) we get

( ){ } { } { } 0)1(21 2

112

2 =++−−−−− +++ nnnnnn ypnyxyynnnxyyx

ie ( ) 0)12(1 22

12

2 =+−+−+−− ++ nnnnnn ypnynyynxynyx

∴ ( ) ( ) 0)12(1 22

12

2 =−++−− ++ nnn ynpxynyx as desired.

2. If )1log(2sin 1 +=− xy or

[ ])1log(2sin += xy or [ ]2)1log(sin += xy

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or )12log(sin 2 ++= xxy , show that

( ) ( )( ) ( ) 041121 2

12

2 =++++++ ++ nnn ynyxnyx (VTU Jan-03)

Out of the above four versions, we consider the function as

)1log(2)(sin 1 +=− xy

Differentiating w.r.t x, we get

+

=− 1

2)(

1

11

2 xy

y ie 2

1 12)1( yyx −=+

Squaring on both sides

( ) )1(41 22

1

2yyx −=+

Again differentiating w.r.t x,

( ) ( ) ( ) )2(4)1(221 1

2

121

2yyxyyyx −=+++

or ( ) )2(4)1(1 112

2yyyxyx ÷−=+++

or ( ) 04)1(1 12

2 =++++ yyxyx -----------*

Differentiating * w.r.t x, n-times, using Leibnitz’s theorem,

{ } 04)1()1)(()2)((!2

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

1

12

2

2

12

2 =++++

++++ −−− yDynDxgDyDnn

xynDxyD nnnnnn

On simplification, we get

( ) ( )( ) ( ) 041121 2

12

2 =++++++ ++ nnn ynyxnyx

3. If )tan(log yx = , then find the value of

( ) ( ) 11

2 )1(121 −+ −+−++ nnn ynnynxyx (VTU July-04)

Consider

)tan(log yx =

i.e. yx logtan 1 =− or xey

1tan−

=

Differentiating w.r.t x,

22

tan

111

1.

1

x

y

xey x

+=

+=

∴ ( ) ( ) 011 1

2

1

2 =−+=+ yyxieyyx -----------*

We differentiate * n-times using Leibnitz’s theorem,

We get

( )[ ] 0)(1 1

2 =−+ yDyxD nn

ie. { } { } 0....)1()()1()()1)(( 22

1

2

2

2

1

1

1

2

1 =−++++++ −− yDxDyDCxDyDCxyD nnnnnn

ie 0....0)2(!2

)1()2()1( 1

2

1 =−

++−

+++ −+ nnnn yynn

xnyxy

( ) ( ) 0)1(121 11

2 =−+−++ −+ nnn ynnynxyx

4. If xyy mm 211

=+ −, or [ ]mxxy 12 −+= or [ ]mxxy 12 −−=

Show that ( ) ( ) 0)12(1 22

12

2 =−+++− ++ nnn ymnxynyx (VTU Feb-02)

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Consider

xy

yxyym

mmm 21

21

111

=+⇒=+ −

( ) ( ) 01211 2

=+−⇒ mm yxy Which is quadratic equation in my1

∴2

442

)1(2

)1)(1(4)2()2( 221 −±

=−−±−−

=xxxx

y m

( ) ( )112

122 222

1

−±=⇒−±=−±

= xxyxxxx

m

∴ ( )mxxy 12 −±= . So, we can consider

[ ]mxxy 12 −+= or [ ]mxxy 12 −−=

Let us take [ ]mxxy 12 −+=

∴ ( )

−+−+=

)2(12

111

2

12

1 xx

xxmym

( )

+−−+=

1

11

2

212

1

x

xxxxmy

m

or

( ) myyx =− 1

2 1 . On squaring

( ) 222

1

2 1 ymyx =− . Again differentiating w.r.t x,

( ) )2()2(21 1

22

121

2 yymxyyyx =+−

or

( ) )2(1 1

2

12

2 yymxyyx ÷=+−

or

( ) 01 2

12

2 =−+− ymxyyx ------------(*)

Differentiating (*) n- times using Leibnitz’s theorem and simplifying, we get

( ) ( ) 0)12(1 22

12

2 =−+++− ++ nnn ymnxynyx

Problem set 1.3.1

In each of the following, apply Leibnitz’s theorem to get the results.

1. show that

−−−−−

=

+ n

xx

n

x

x

dx

dn

n

n

n 1.........

3

1

2

11log

!)1(log1

Hint: Take x

uxv1

;log ==

2. If ,)1( 2 nxy −= Show that ny satisfies the equation

( ) 0)1(212

22 =++−− ynn

dx

dyx

dx

ydx Hint : It is required to show that

( ) ( ) 0121 12

2 =++−− ++ nnn ynnxyyx

3. If ),sin(log)cos(log xbxay +=

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( ) ( ) 0112 2

12

2 =++++ ++ nnn ynxynyx

4. If ,1sin xmey

= Prove That

( ) ( ) ( ) 0121 22

12

2 =+−+−− ++ nnn ymnxynyx

5. If ,logcos 1

n

n

x

b

y

=

− Show that

( ) 0212 2

12

2 =+++ ++ nnn ynxynyx

6. ( ) Thatovexmy Pr,sinsin 1−=

( ) ( ) ( ) 0121 22

12

2 =+−+−− ++ nnn ymnxynyx

7. If ( ),log xxDy nn

n = Prove That

(i) )!1(1 −+= − nnyy nn (ii) !nyn =

+++++n

x1

.........3

1

2

11log

8. If ,log xxy n= Show that x

nyn

!1 =+

• Summary:- The idea of successive differentiation was presented. The computation of nth

derivatives of a few standard functions and relevant problems were discussed. Also, the

concept of successive differention was extended for special type of functions using

Liebnitz’s theorem.

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T

r = f(ө) A

P(r, ө)

O

r

Ө

Chapter – 2 : POLAR CURVES

LESSON -1 : Angle between Polar Curves

• In this chapter we introduce a new coordinate system, where we can understand the idea

of polar curves and their properties.

• Session-1

2.1.0 Introduction:- We are familiar with Cartesian coordinate system for specifying a point

in the xy – plane. Another useful system for similar purpose is Polar coordinate system, and the

curves specified by these coordinates are referred to as polar curves.

• A polar curve by name “three-leaved rose” is displayed below:

• Any point P can be located on a plane with co-ordinates ( )θ,r

called polar co-ordinates of P where r = radius vector

OP,(with pole ‘O’) ;θ = projection of OP on the

initial axis OA.(See Fig.1)

• The equation ( )θfr = is known as a polar curve.

• Polar coordinates ( )θ,r can be related with Cartesian

coordinates ( )yx, through the relations Fig.1. Polar coordinate

θcosrx = & θsinry = . system

2.1.1 Important results

ө = 0

ө = 2

π

ө = 4

π ө =

4

ө = π

ө = 2

φ

φ

P(r, ө)

r ψ

L

Y

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NCET – Class Notes 10MAT11

• Theorem 1: Angle between the radius vector and the tangent.:

i.e. With usual notation prove that dr

dr

θφ =tan

• Proof:- Let “φ ” be the angle between the radius vector OPL

and the tangent 1TPT at the point `P` on the polar

curve ( )θfr = . (See fig.2)

From Fig.2, Fig.2. Angle between radius

φθψ += vector and the tangent

tan ( )φθφθ

φθψtantan1

tantantan

−+

=+=

i.e. .................tantan1

tantan

φθφθ

−+

=dx

dy(1)

On the other hand, we have θcosrx = ; θsinry = differentiating these, w.r.tθ ,

( )

+−=θ

θθθ d

drr

d

dxcossin & ( )

+=θ

θθθ d

drr

d

dysincos //NOTE//

( )

( )

+−

+==

θθθ

θθθ

θ

θ

d

drr

d

drr

ddx

ddy

dx

dy

cossin

sincos

θθ

cos&d

drbyDrNrthedividing

( )( ) 1tan

tan

+−

+=

θθ

θθ

drrd

drdr

dx

dy

i.e. ( )( )

drrd

drdr

dx

dy

θθ

θθ

tan1

tan

+= ………………….(2)

Comparing equations (1) and (2)

we get dr

dr θφ =tan

• Note that

φd

dr

r

1cot

• A Note on Angle of intersection of two polar curves:-

If 1φ and 2φ are the angles between the common radius vector and the tangents

at the point of intersection of two curves ( )θ1fr = and ( )θ2fr = then the

angle intersection of the curves is given by 21 φφ −

• Theorem 2: The length “p” of perpendicular from pole to the tangent in a polar

curve i.e.

(i) φsinrp = or (ii)

==θddr

rrp422

111 2

• Proof:- In the Fig.3, note that ON = p, the length of the perpendicular from the pole to

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φ

O

P(r, ө)

P

N

r = f (ө)

φ

r

Ө Ψ

the tangent at p on ( )θfr = .from the right angled triangle OPN,

( ) φφ sinsin OPON

OP

ON=⇒=

i.e. )....(..........sin irp φ=

Consider φφ

ecrrp

cos1

sin

11==

( )φφ 2cot11

cos11

2

2

22+==∴

rec

rp

Fig.3 Length of the perpendicular

from the pole to the tangent

+=2

22

11

11

θddr

rrp

)..(..........111

2

422ii

d

dr

rrp

+=∴θ

• Note:-If ,1

ru = we get

2

2

2

1

+=θddu

up

• Session-2

• In this session, we solve few problems on angle of intersection of polar curves and pedal

equations. 2.1.2 Worked examples:-

• Find the acute angle between the following polar curves

1. ( )θcos1+= ar and ( )θcos1−= br (VTU-July-2003)

2 ( )θθ cossin +=r and θsin2=r (VTU-July-2004)

3. ( )2

sec16 2 θ=r and ( )2

cos25 2 θecr =

4. θlogar = and θlogar = (VTU-July-2005)

5. θθ+

=1

ar and

21 θ+=

ar

Solutions:

1. Consider Consider

( )θcos1+= ar ( )θcos1−= br

Diff w.r.t θ Diff w.r.t θ

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θθ

sinad

dr−= θ

θsinb

d

dr=

( )θθθ

sin

cos1

a

a

dr

dr

−+

= ( )

θθθ

sin

cos1

b

b

dr

dr

−=

( )( ) ( )

2cos

2sin2

2cos2

tan

2

1 θθ

θφ −=

( )( ) ( )

2cos

2sin2

2sin2

tan

2

1 θθ

θφ −=

2

cotθ−= 2

tanθ=

i.e ( ) ( )2222

tantan 11θπφθπφ +=⇒+= 211 2

tantan φφθφ =⇒=

Angle between the curves

( )222221

πθθπφφ =−+=−

Hence ,the given curves intersect orthogonally

2. Consider Consider

( )θθ cossin +=r θsin2=r

Diff w.r.t θ Diff w.r.t θ

θθθ

sincos −=d

dr θ

θcos2=

d

dr

θθθθθ

sincos

cossin

−+

=dr

dr

θθθ

cos2

sin2=

dr

dr

θ

θφ

tan1

1tantan 1 −

+= (÷ Nr & Dr θcos ) θφ tantan 2 =

i.e ( )θπθ

θφ +=

−+

=4

tantan1

1tantan 1 θφ =⇒ 2

θπφ +=⇒41

∴Angle between the curves = ( )4421

πθθπφφ =−+=−

3. Consider Consider

( )2

sec16 2 θ=r ( )2

cos25 2 θecr =

Diff w.r.t θ Diff w.r.t θ

( ) ( )2

1.2

tan2

sec32 2 θθθ

=d

dr ( ) ( )

21.

2cot

2cos50 2 θθ

θec

d

dr−=

( ) ( )2

tan2

sec16 θθ= ( ) ( )2

cot2

cos25 2 θθec−=

( )

( ) ( )2

tan2

sec16

2sec16

2

2

θθ

θθ=

dr

dr

( )( ) ( )

2cot

2cos25

2cos25

2

2

θθ

θθ

ec

ec

dr

dr

−=

( )

22tan

2cottan 1

θπθφ −==

( )2

tan2

tantan 2θθφ −=−=

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( )221

θπφ −=⇒ 22

θφ −=⇒

Angle of intersection of the curves = ( ) ( )22221

θθπφφ −−=−

2

π=

4. Consider Consider

θlogar = θlogar =

Diff w.r.t θ Diff w.r.t θ

θθa

d

dr= ( ) θθ

θ1.log

2a

d

dr−=

( )a

adr

dr θθ

θlog=

( )

−=

aa

dr

dr

θθθ

θ 2log

log

)(..........logtan 1 iθθφ = )(..........logtan 2 iiθθφ −=

We know that

( )21

21

21tantan1

tantantan

φφφφ

φφ+

−=−

( )

( )( )θθθϑθθθθ

loglog1

loglog

−+

−−=

i.e ( )( )

)..(..........log1

log2tan

221 iiiθθ

θθφφ

−=−

From the data: ( ) 1log1loglog

log2 ±==⇒== θθθθ orara

As θ is acute, we take by θ =1 NOTEe=⇒θ

Substituting e=θ in (iii), we get

( )( )

=−

=−2221

1

2

log1

log2tan

e

e

ee

eeφφ ( )1log =e

eQ

=−∴ −2

1

211

2tan

e

eφφ

5. Consider Consider

θθ+

=1

ar as

21 θθ

+=

ar

( )11111+=

+= θθ

θaar

( )r

a=+∴ 21 θ

Diff w.r.t θ Diff w.r.t θ

( )22111

θθ−=−

ad

dr

r

θθ

d

dr

ra

22 −=

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2

1

θθ a

r

d

dr

r=

θθ

d

dr

ra

r 12=

r

a

dr

dr

2θθ= i.e

θθ

r

a

dr

dr

2

−=

( )θθ

θφ

+

=

1

tan2

1 a

a

+−=

a

a 2

2

1

2tan

θθ

φ

( )θθφ +=∴ 1tan 1 ( )2

2 12

1tan θ

θφ +−=

Now, we have

( ) ( )θθθθθ

θ+=+⇒

+==

+11

11

2

2aa

ar

a

or 11 33 =⇒+=+ θθθθ or 1=θ

2tan 1 =∴ φ & ( )1tan 2 −=φ

Consider ( )( )( )21

2121

tantan1

tantantan

φφφφ

φφ+

−=−

( )

( )( )33

121

12=−=

−+−−

=

( )3tan 1

21

−=−∴ φφ

Problem Set No. 2.1.1 for practice.

• Find the acute angle between the curves

1. ( )θθ nnar nn sincos += and θnar nn sin= (ans:4

π )

2. nn anr =θcos and nn bnr =θsin (ans:2

π )

3. θar = and θar = (ans:

2π )

4. θcosar = and 2

ar = (ans:56

π )

5. θmar mm cos= and θmbr mm sin= (ans:2

π )

LESSON -2 : Pedal Equations

• Session - 1 • 2.2.0 Pedal equations (p-r equations):- Any equation containing only p & r is

known as pedal equation of a polar curve.

• Working rules to find pedal equations:-

(i) Eliminate r and φ from the Eqs.: (i) ( )θfr = & φsinrp =

(ii) Eliminate only θ from the Eqs.: (i) ( )θfr = &

2

422

111

+=∴θddr

rrp

• 2.2.1 Worked Examples on pedal equations:-

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• Find the pedal equations for the polar curves:-

1. θcos12

−=r

a

2. αθ cotcer =

3. θθ mbmar mmm cossin += (VTU-Jan-2005)

4. θcos1 er

l +=

Solutions:

1. Consider θcos12

−=r

a……….(i)

Diff. w.r.t θ

( ) θθ

sin12 2 =−d

dr

ra

a

r

d

dr

r 2

sin1 θθ

−=

θθ

sin

12

r

a

dr

dr −=

( ) ( )2

tan

2cos

2sin2

2sin2

sin

cos1tan

2

θθθ

θ

θθ

φ −=−=−

−=

( )22

tantan θφθφ −=⇒−=

Using the value of φ is ,sinφrp = we get

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

sin2

sin iirrp θθ −=−=

Eliminating “θ ” between (i) and (ii)

=

−==

r

arrrp

2

22

cos1

2sin

22222 θθ [See eg: - (i)]

.2 arp =

This eqn. is only in terms of p and r and hence it is the pedal equation of the polar curve.

2. Consider αθ coter =

Diff. w.r.t θ

( ) ( )αθαθ ααθ

cotcot cotcot erred

dr=== Q

We use the equation 2

422

111

+=θddr

rrp

( )242

cot11

αrrr

+=

( ) ( ) ααα 2

2

2

2

2

42cos

1cot1

1cot

11ec

rrrr=+=+=

α222

cos11

ecrp

=

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α2

22

cosecrp = or α222 cosecpr = is the required pedal equation

3.Consider θθ mbmar mmm cossin +=

Diff. w.r.t θ

( ) )sin(cos1 θθθ

mmbmmad

drmr mmm −+=−

θθθ

mbmad

dr

r

r mmm

sincos −=

θθθθ

θ mbma

mbma

d

dr

r mm

mm

cossin

sincos1

+

−=

θθθθ

φmbma

mbmamm

mm

cossin

sincoscot

+−

=

Consider φsinrp = , φecrp

cos11

=

φ222

cos11

ecrp

=

( )φ2

2cot1

1+=

r

+

−+=

2

2 cossin

sincos1

1

θθθθ

mbma

mbma

r mm

mm

( ) ( )( )

+

−++=

2

2

2

2cossin

sincoscossin1

θθ

θθθθ

mbma

mbmambma

r mm

mmmm

Noter

ba

rp m

mm

+=

2

22

22

11

( )

mm

m

ba

rp

22

122

+=⇒

+

is the required p-r equation

4. Consider ( )θcos1+=r

l

Diff w.r.t θ

θθ

θθ

sin1sin1

2e

d

dr

rrle

d

dr

rl =

⇒−=

( ) θφ sincot er

l =

( ) θφ sincot el

r=∴

We have ( )φ2

22cot1

11+=

rp (see eg: 3 above) Now

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+=

2

2222

22

sin11

l

rel

rp

θ

( )θ22

22

2sin1

1

lre

r+=

rle =+ θcos1

r

rle

−=θcos

−−+

=2

2

222

22

111

l

re

rlrel

rp

−=

re

rlθcos

θθ 22 cos1sin −=

On simplification lre

e

p

2112

2

2+

−=

2

1

−−=

re

rl

• Problem Set No. 2.2.1 for practice.

Find the pedal equations of the following polar curves

1. nn anr =θcos and nn bnr =θsin

2. θar = and θar =

3. θcosar = and 2

ar =

4 . θmar mm cos= and θmbr mm sin=