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 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)
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
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
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
+
−=
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
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
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
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
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−=
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
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:
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)
NCET – Class Notes 10MAT11
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
NCET – Class Notes 10MAT11
( ) 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
NCET – Class Notes 10MAT11
• 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 ++=
NCET – Class Notes 10MAT11
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 ++
−=+=
NCET – Class Notes 10MAT11
( ) ( )[ ] ( ) ( ) ( )[ ]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 +
NCET – Class Notes 10MAT11
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
−
−=
−
−==
NCET – Class Notes 10MAT11
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
NCET – Class Notes 10MAT11
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)
NCET – Class Notes 10MAT11
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 +=
NCET – Class Notes 10MAT11
( ) ( ) 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.
NCET – Class Notes 10MAT11
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
3π
ө = π
ө = 2
3π
φ
φ
P(r, ө)
r ψ
L
Y
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
=θ
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
NCET – Class Notes 10MAT11
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
NCET – Class Notes 10MAT11
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 −=−=
NCET – Class Notes 10MAT11
( )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 −=
NCET – Class Notes 10MAT11
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:-
NCET – Class Notes 10MAT11
• 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
=
NCET – Class Notes 10MAT11
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
NCET – Class Notes 10MAT11
+=
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=