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8/11/2019 MATH ZC161 Lecture Notes
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Chapter 1
Functions, Limits and Continuity
Definition:
Let A and B be two non-empty sets. A function from the set A to the set B is arule which associates to each element x in A,a unique element y in B which wewrite as y = f(x). The set A is called domain of the function f, the set B is called
the codo main of f. y = f(x) is called the value of f at x and the set ( ){ } A x x f : iscalled the range of f. The graph of the function f is the set ( )( ){ } D x x f x / , where D is the domain of f. We draw the curve through these points.
Functions whose domain and codomain are both subsets of R (The set of realnumbers), are called real valued functions of a real variable. If domain is notstated then we take largest subset of R for which f(x) is defined as the domain ofthe function f.
The set { }b xa R x : is denoted by [a,b] or by b xa and called closedinterval [a,b] and the set { }b xa R x
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Then we draw the graph through these (x,y) points.
The graph is symmetric about y-axis since f(-x) = f(x)The graph is shown in the adjoining figure.
Example 3:
The function F: R R
Defined by ( ) 0== xif x x xF 0
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Left hand Limit
Let f(x) be defined on an open interval (c-h, c). A real number l is called left handlimit of f(x) at c if for all x sufficiently close to c, but xc)sufficiently close to c, f(x) can be made as close as we want to the number l .
We write ( ) l x f Limc x
=+
Limit of f(x) exists if ll = and limit does not exist if ll
Some Properties of Limits
1. C x LimK K LimC xC x
==
,
Theorem: Let f(x) and g(x) be two functions such that ( ) 1 L x f Lt C x
=
and
( ) 2 L xg LimC x
=
. Then
(a) ( ) ( )[ ] 21 L L xg x f LimC x
+=+
(b) ( ) ( )[ ] 21 L L xg x f LimC x=
(c) ( )[ ] 1 L x f LimC x
=
(d) ( ) ( )[ ] 21 L L xg x f LimC x
=
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(e)( )( )
0, 22
1 =
L L L
xg x f
LimC x
Example 1:Evaluate
(i) ( )4223
++
x x Lim x
(ii)482
2 +
+
x x
Lim x
(iii)16
1222
4
x x x Lim
x
(iv) x
x x Lim
x
+
110
Solution
(i) ( ) 19432342 223
=++=++
x x x Lim x
(ii) 24284
482
2=
+
+=
+
+
x x
Lim x
(iii)16
12lim 2
2
4
x x x
x. It is of the form
00 .
In this case we factorize the given function and cancel the common factor thenwe evaluate the limit.
( ) ( )( ) ( ) 8
74434
43
lim4434
1612
442
2
4=
+
+=
+
+=
+
+=
x x
x x x x
Lim x
x x Lim
x x x
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(iv)
+
formisit
x x x
Lim x 0
0.
110
++
+++
= x x
x x x
x x Lim x 11
11110
( ) ( )[ ] x x x
x x Lim
x ++
+=
11
110
[ ] x x x x
Lim x ++
= 11
20
111
20
=++
= x x
Lim x
Example 2:
Evaluate
(i) x
x Lim
x 0
(ii) [ ] x Lim x 4
Where [x] is the greatest integer less than or equal to x.
Solution:
( ) 11.000
==
== x x x
Lim
x
x Lim
x
x Limhll
( ) 11.000
====+++ x x x
Lim x x
Lim x
x Limhlr
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Since left hand limit of the function at 0 is not equal to the right hand limit of the
function at 0. Hence x
x Lim
x 0 does not exist.
(ii) [ ] 34 = x Lim x since x is coming near to 4 but x is less than 4.
[ ] 44
=+ x Lim
x since x is coming near to 4 but x is more than 4. Hence
[ ] x Lt x 4
doesnt exist since left hand limit is not equal to right hand limit.
Similarly [ ], x Limn x
where Z n , doesnot exist.
Some Important Limits
(a) 1
=
nnn
a xna
a xa x
Lim
(ii) 1sin
0=
x x
Lim x
(iii)( )
11log
0
=+
x
x Lim
x
(iv) 11
0=
xe
Lim x
x
(v) en
Limn
n=
+
11
(vi) a x
a Lim x
xlog1
0=
A real number L is called limit of f(x) as x tends to if f(x) can be made as close
to L as we want by making x sufficiently large. We write ( ) . L x f Lim x
=
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To evaluate these limits we take y
x 1= in the function and ( ) x f Lim
x
becomes
y f Lim
y
10
. In the case of rational function. We divide numerator and
dominator by highest power.
Example
10557432
234
234
++
++++
x x x x x x x
Lim x
20001
00002
10511
57432
42
432=
++
++++=
++
++++=
x x x
x x x x Lim x
Exercise 1.1
Evaluate the following limits
1.292
3 +
x x
Lim x
2.56
2092
2
5 +
+
x x x x
Lim x
3.( ) xaa x
a xa Lim
x +
+
0
4.2255
2
x x
Lim x
5. x x
Lim x 9sin
10sin0
6. x x
Lim x 2tan
cos1 +
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7. x
ee Lim
x x
x
23
0
8. [ ] x x Lim
x 0
9. 222
0
44 x
x x Lim
x
+
Continuous Function
Let f be a real valued function of a real variable x. Let c be any point in thedomain of f. Then f is said to be continuous at x = c if ( ) x f Lim
c x exists and
( ) ( )c f x f Limc x
=
(i.e. the limit of the function at c is equal to the value of the
function at c). If f is not continuous at c then f is called discontinuous at x = c andc is called a point of discontinuity of f.
Definition
A real function is continuous on an interval [a, b] or in (a,b) if it is continuous atevery point of the interval. At the end point a of a closed interval [a,b] f iscontinuous if ( ) ( )a f x f Lim
a x=
+
At the point b,f, is continuous if ( ) ( )b f x f Lt b x
=
A polynomial function f(x) = 011
1 ..... a xa xa xa n
nn
n ++++
is continuous for all
R x .
Theorem 1.2: Let f and g be two real continuous functions then
(i) f + g is continuous(ii) f-g is continuous(iii) f is continuous, where is any real number.
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(iv) fg is continuous
(v) f 1 is continuous for all x such that ( ) 0 x f
(vi)
g
f is continuous for all x such that ( ) 0 xg
(vii) f g
is continuous for all x such that ( ) 0 x f .
Example 1:
x x xe x x x ,1,,cos,sin 2 ++ are continuous for all R x since for any real
number x 0
0
000
,coscossinsin 0,0 x x
x x x x x xee Lim x x Lim x x Lim ===
0020
2
00
,11 x x Lim x x x x Lim x x x x
=++=++
. Limits of each function is equal to
value of that function at x 0 and x 0 is any point. So all these functions arecontinuous for all R x .
Example 2:
f(x) = [x] is not continuous at x = n where n is an integer.
[ ] 1,][ ==+
n x Limn x Limn xn x
so ( ) x f Limn x
doesnt exist. Hence f(x) = [x] is not
continuous at x = n.
Example 3:
( ) 332 622
= xif
x x x x x f
35= if x = 3.
Determine whether the function is continuous at x = 3 ?
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Solution :
f(x) is continuous at x = 3 if limit of the function at x = 3 exists and is equal to
f(3). =
32
62
2
3 x x x x Lt
x( )( )( )( )13
233 +
+ x x
x x Lim x
( )35
345
12
3==
+
+=
f
x x
Lim x
so the function is not continuous at x = 3.
Example 4:
Show that the function defined as
( )
00
0,
==
=
xif
x x x
xF
is not continuous at x = 0.
Solution:
( ) 110000
====++++ x x x x
Lim x x
Lim x x
Lim x f Lim
Since x is approaching zero from right (i.e. x > 0) to |x| = x.
( ) 11)(0000
==
== x x x x
Lim x
x Lim
x x
Lim x f Lim
Since x is approaching zero from left (i.e. x < 0) so |x| = -x. ( ) x f Lim x 0 does
not exist. Hence f(x) is not continuous at x = 0.
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Example 5:
Show that ( ) x
x f 1= is not continuous at x = 0.
Since ( ) == 01
0 f . So f(x) is not continuous at x = 0.
Example 6:
Show that the function f defined as
( )112
112>+=
=
xif x
xif x x f
is not continuous at x = 1
Solution
( ) ( ) 11211
==
x Lim x f Lim x x
( ) ( ) 31211
=+=++
x Lim x f Lim x x
So ( ) x f Lim x 1 does not exist.
f(x) is not continuous at x = 1.
Example 7:
For what value of K the following functions are continuous at x = 0.
(i) ( ) 0sin
2 = xif x x
x f
= K if x = 0
( ) 2sin200
== x
x Lim x f Lim
x x
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f(x) is continuous at x = 0 if ( ) ( )00
f x f Lim x
=
So 2 = K or K =2.
(ii) ( ) ( ) 012 2
= 1
134
14523
2
xat xif x x
xif x x f
3. ( ) ==
= 0
0,0
0,1
sin3 xat xif
xif x
x x f
4. ( ) 1112
+
= xif x
x x f
At x = -1= -2 if x = -1
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5. Find the point of discontinuity of the functions
(i) ( )86
832 +
+=
x x x
x f
(ii) ( ) x x f cot=
Q.6 For what values of K the function f(x) defined as
( )
2,
248
2
3
==
=
xif K
x x x
x f
is continuous at x = 2 ?
Q.7 Let f(x) be a function defined as ( )( )22
sin1
x
x x f
=
when
2 x and
( ) .2 = f Find so that f(x) is continuous at 2 = x .
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Chapter 2
Trigonometric Functions
2.1 Let be any number. Construct the angle whose measure is radianswith vertex at the origin of a rectangular coordinate system and initial sidealong the positive x-axis. Let P (x,y) be any point on the terminal side OAof the angle. Let OP = r.
(i) Since of is denoted by sin and defined asr
y Hypo perp == sin
(ii) Cosine of is denoted by cos defined asr
x Hypo Base == cos
(iii) Tangent of is denoted by tan and defined as
x y
Base perp ===
cossin
tan
(iv) Cotangent of is denoted by cot and defined as
y x
Perp Base ===
tan1cot
(v) Secant of is denoted by sec and defined as
xr
Base Hypo ===
cos1
sec
(vi) Cosecant of is denoted by cosec and defined as
y
r
perp
Hypoec ===
sin
1cos
2.2 Trigonometric Identities : The following identifies can be easily proved.
(i) 1sincos 22 =+
(ii) 22 sectan1 =+
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(iii) 22 coscot1 ec=+
2.3 Sign of Trigonometric Function
In I quadrant x and y are +ve so all trigonometric functions are positive. InII quadrant x is ve and y is positive so sin and cosec are +ve other t-functions are negative. In third quadrant both x and y are negative so tan and cot are positive other t-function are negative. In IV quadrant x is+ve and y is ve so cos and sec are positive other functions arenegative.
2.4 We can prove the following results
(a) ( ) ( ) ( ) tantan,coscos,sinsin ===
( ) ( ) ( ) ecec coscos,secsec,cotcot ===
(b)
cot2
tan,sin2
cos,cos2
sin =
=
=
sec2
cos,cos2
sec,tan2
cot =
=
=
ecec
(c)
sin2
cos,cos2
sin =
+=
+
tan2
cot,cot2
tan =
+=
+
sec2
cos,cos2
sec =
+=
+ ecec
(d) ( ) ( ) ( ) tantan,coscos,sinsin ===
( ) ( ) ( ) == eccos,secsec,cotcot eccos=
(e) ( ) ( ) coscos,sinsin =+=+ ( ) ( ) cotcot,tantan =+=+ ( ) ( ) ecec coscos,secsec =+=+
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(f) ( ) ( ) ( ) tan2tan,cos2cos,sin2sin ===
( ) ( ) ( ) ecec cos2cos,sec2sec,cot2cot ===
(g) ( ) ( ) ( ) tan2tan,cos2cos,sin2sin =+=+=+ ( ) ( ) ecec cos2cos,sec2sec =+=+
2.5 Values of traiogonometric function at different angles. Values of t functionsare given in the following table.
06
4
3
2
sin 0 21 2 / 1 2 / 3 1
cos 1 2 / 3 2 / 1 21
0
tan 0 3 / 1 1 3
Important Formulae
1. sin (A + B) = sin A cos B + cos A sin BTaking B = A, we getsin 2A = 2 sin A cos A
2. sin (A-B) = sin A cos B cos A sin B
3. cos (A+B) = cos A cos B sin A sin BTaking B = A, we getcos 2A = cos 2A sin 2A = 2 cos 2A-1=1-2sin 2A
4. cos (A-B) = cosA cos B + sin A sin B
5. ( ) B A B A
B Atantan1tantan
tan
+=+
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6. ( ) B A B A
B Atantan1
tantantan
=
7. ( ) ( ) B A B A B A += sinsincossin2
8. ( ) ( ) B A B A B A += sinsinsincos2
9. ( ) ( ) B A B A B A ++= coscoscoscos2
10. ( ) ( ) B A B A B A += coscossinsin2
11.2
cos2
sin2sinsin DC DC
DC +
=+
12.2
sin2
cos2sinsin DC DC
DC +
=
13.2
cos2
cos2coscos DC DC
DC +
=+
14.2
sin
2
sin2coscos C D DC
DC +
=
15. 3sin4sin33sin =
16. cos3cos43cos 3 =
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Inverse Trigonometric Functions
x y 1sin = iff y x sin= where [ ]1,1 x and .2
,2
y Domain of the function is
[ ]1,1 and range is
( ) ( )42
1sin,
621
sin,2
1sin,2
1sin,2
,2
1111 =
=
== ,
323
sin 1 =
Other Inverse functions are defiend in the similar way and their domains andranges are given by the following table.
Functions Domain Range
cos -1x 11 x y0
tan -1x
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( ) 11,1cossin 11
= x x
ec x
Similar properties are satisfied by other inverse trigonometric
functions.
(iii) ( ) 0,2
cottan,2
cossin 1111 =+=+ x x x x x
( ) 02
cottan 11 >>
++= xy y x
xy y x
1,0,0,1tan 1
>
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221 11sin x y y x +=
If 1,0,0 22 >+>> y x y x
(viii) ( ) 1,0,)11sinsinsin 22111 =
y x x y y x y x
(ix) ( ) 1,011coscoscos 22111 =+ y xif y x xy y x
( ) 0,111cos2 221 = y x y x xy
(x) 1,111coscoscos 22111 += y x y x xy y
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Chapter 3
3.1 Definition: The derivative of the function y = f(x) at x is denoted by ( ) x f or
dxdyor yor
dxdf and defined as ( ) ( ) ( )
x x f x x f Lt x f
x +=
0, provided the limit
exists. Then f is called differentiable function.
The derivative of f at x = a is defined as ( ) ( ) ( ) x
a f xa f Lt a f x
+=
0
provided the limit exists.
Derivative by Definition: Example 1: Let f(x) = x n where n is any real
number.
( ) ( ) ( ) ( ) x
x x x Lt
x x f x x f
Lt x f nn
x x
+=
+=
00
1= nnx
Example 2:
Let ( ) x x f y sin==
( ) ( ) ( ) x
x f x x f Lt
x y
Lt x f dxdy
x x
+=
==
00
( ) x
x x x Lt x
+=
sinsin0
2
2sin
2cos2
sin2
cos2
000 x
x
Lt x x Lt x
x x x
Lt x x x
+=
+
=
xcos=
Similarly ( ) x xdxd
sincos =
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3.2 Rules of Derivative
1. If C is a constant then 0=dxdC
2. If u is a differentiable function and c is a constant, then ( )dxdu
ccudxd =
3. If u and v are differential functions of x, then ( )dxdv
dxdu
vudxd +=+
If u1, u 2 u n are differentiable function of x, then
( )dx
dudx
dudxduuuu
dxd n
n +++=+++ ........... 2121
4. If u and v differentiable functions of x then ( ) .dxdv
uvdxdu
uvdxd +=
5. If u and v are differentiable functions of x and ( ) 0 xv then
2.
vdxdv
uvdxdu
vu
dxd
=
3.3 Higher Order Derivatives
If y = f(x) then ( ) x f dxdy = is the first order derivative of y with respect
to x.
=
== dxdy
dxd
dx yd
ydx
yd 2
2
is called the second order derivative of
y with respect to x. Similarly third order derivatives
==
2
2
3
3
dx
yd dxd
ydx
yd and ( ) ( )( )1= nn y
dxd
y where ( )n y is the nth order
derivatives of y with respect to x.
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3.4 Formulas of derivative of functions
(i) ( ) 1= nn nx xdxd
(ii) ( ) x x eedxd =
(iii) ( ) x
xdxd 1
ln =
(iv) ( ) aaadxd x x ln=
(v) ( ) x xdxd
cossin =
(vi) ( ) x xdxd
sincos =
(vii) ( ) x xdxd 2sectan =
(viii) ( ) xec xdxd 2coscot =
(ix) ( ) x x xdxd
tansecsec =
(x) ( ) xecxecxdxd
cotcoscos =
(xi) ( )2
1
1
1sin
x x
dxd
=
(xii) ( )2
1
1
1cos
x x
dxd
=
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(xiii) ( ) 21 11
tan x
xdxd
+=
(xiv) ( ) 21 1 1cot x xdxd +=
(xv) ( )1
1sec
2
1
=
x x x
dxd
(xvi) ( )1
1cos
2
1
=
x x xec
dxd
3.5 Chain Rule:
If y is a differential function of u and u is a differential function of x i.e. y= f(u) and u = g(x), then
dxdu
dudy
dxdy
.=
Similarly if y = f(u) , u = g (v) and v = h (x),then
dxdv
dvdu
dudy
dxdy =
Examples
Finddxdy
if 4sintan4 1510 ++++= x x x x y
using the formula ( ) ( ) 9101 10, x xdxd
nx xdud nn ==
( ) ( ) ( ) x xdxd
x xdxd
xdxd 2
455 sectan.5.444 === from (7) of 3.4
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( )2
1
1
1sin
x x
dxd
= from (6) of 3.4
( ) ( ) 040 ==dxd c
dxd
( )4sintan4 1510 ++++= x x x xdxd
dxdy
( ) ( ) ( ) ( )dxd
xdxd
xdxd
xdxd
xdxd ++++= 1510 sintan4 (4)
2
249
11sec2010
x x x x
+++=
2. Finddxdy
if xe y1tan = we write y = e u where u = tan -1x. y is a function of u
and u is a function of x. Using chain ruledxdu
dudy
dxdy = (1)
y = e u
xuand edudy u 1tan ==
211 xdx
du+
= substituting the value ofdudy
anddxdu
in (1)
2
tan
2 111
1
xe
xe
dxdy xu
+=
+=
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3. Finddxdy
if ( )22ln a x x y ++= . We take 22 a x xu ++= then ( )u y ln=
whereudu
dy xa xu
1,22 =++= and
( ) ( ) 11221 2221
22 ++
=++=
a x x xa x
dxdu
22
22
a x
a x x
+
++= . Hence using chain rule, we get
2222
22
2222
22 11.
1
a xa x
a x x
a x xa x
a x xudx
dy
+=
+
++
++=
+
++=
4. If
+
+=
14
tan 22
1
x x
y , then finddxdy
. We take uwand w x x ==
+
+
14
2
2
and
( )14
,tan 22
1
+
+===
x
xwand wuu y
,2
1
2 2
1
,1
1 =+
= wdw
du
udu
dy
( ) ( ) ( ) ( )
( )( ) ( )
( )2222
22
2222
1
4212
1
1414
+
++=
+
++++
= x
x x x x
x
xdxd
x x xdxd
dxdw
( )22 16
+
=
x
x
( )222 16
.2
1.
11
.+
+==
x
x
wudxdw
dwdu
dudy
dxdy
Putting values of u, w in terms of x, we get
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( )222
2
2
21
6
14
2
1
14
1
1+
+
+
+
++
= x
x
x x
x
x
( )2222
2
2
1
3
4
152
1+
+
+
+
+=
x
x
x
x
x
x
( ) ( )( )14523
222 +++
=
x x x
x
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Example 5:
If y = x 4 e x lnx using product rule we get
( ) ( ) ( ) xdxd e xedxd x x xdxd xedxdy x x x lnlnln 444 ++=
( )( ) ( ) x
e xe x x x xe x x x 1ln4ln 443 ++=
x x x e x xe x xe x 343 lnln4 ++=
Example 6:
If x x x x y
cossin
3
2
++= , find
dxdy . Function is in the form ( )
( ) xv xu ,
so by quotient rule
( ) ( ) ( ) ( )
( )23
3223
cos
cossinsincos
x x
x xdxd
x x x xdxd
x x
dxdy
+
++++=
( )( ) ( )( )( )23
223
cos
sin3sincos2cos
x x
x x x x x x x x
+
+++=
( )232234
cos
1sinsin3cos2cos
x x
x x x x x x x x x
+
++++=
( )23
423
cos
1sin2cos2cos
x x
x x x x x x x
+
++=
Theorem: If f(x) is differentiable at a point x = c, then f(x) is continuous at x = c.
Proof: Since f(x) is differentiable at x = c then
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( ) ( ) ( ) ( ) ( )c x
c f x f Lt
xc f xc f
Limc f c x x
=
+=
0is a finite number.
Consider
( ) ( )[ ] ( ) ( )( )
( ) 00 ==
=
c f
c xc x
c f x f Limc f x f Lt
c xc x
( ) ( )c f x f Lt c x
=
. Hence f(x) is continuous at x = C.
Properties of ( ) x f
The derivative of y = f(x) is the slope of the tangent line to the curve at (x,y). If( ) x f is positive, then f(x) is an increasing function of x and if ( ) x f is negative,
then f(x) is decreasing function of x. If ( ) 0> x f on an interval I then the curve y= f(x) is concave upward on I. If ( ) ,0
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( ) x ydxdy
x y 2554 =
x y
x y
dx
dy
54
25
=
Derivative of functions in parametric forms.
If x = f (t) and y = g(t), then( )( )t gt f
dt dx
dt dy
dxdy
== (Provided ( ) 0t g ).
Example
Finddxdy
if x = a (1 cos t) and y = asint. t adt dy
cos= and t adt dx
sin=
.cotsincos
t t at a
dt dxdt dy
dxdy ===
Example of second derivative
If x B x A y 3sin3cos += prove that 0922
=+ ydx
yd
x B x A y 3sin3cos +=
x B x Adxdy
3cos33sin3 += Again differentiating
( ) x B x A x B x Adx
yd 3sin3cos93sin93cos92
2
+==
or
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099 22
2
2
=+= ydx
yd or y
dx yd
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Derivative by Substitution:
Example 1:
If ,11cos 2
21
x x y
+= find
dxdy by simplifying the function.
We take tan= x
( )
2coscossincossincos
costan1tan1
cos 12222
12
21 ==
+
=
+
= z y
( )
2
1
12
tan2
xdxdy
x
+=
=
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Chapter 4
Integration
4.1 If a function f(x) is the derivative of a function F(x) then F(x) is called an
integral of f(x) with respect to x and write as ( ) ( ) ( ) = xF xF dx x f . is also
called antiderivative of f(x) . If ( )[ ] ( ) x f xF dxd = , then for any constant C, we
have ( )[ ] ( )[ ] ( ) x f dxdC
xF dxd
C xF dxd =+=+
( ) ( ) += C xF dx x f . Then function f(x) is called integrand. The functionF(x) is called particular integral of f(x) w.r.t. x . C is an arbitrary constantand called the constant of integration.
Example:
+==
+ C
xdx x xC
xdxd
66
655
6
4.2 Fundamental Integration Formulae.
1. ( ) == C dxC dxd
00
2. ( ) +=== C xdxdx xdxd
11
3. +==
+
++
.1,1
1,1
11
nn x
dx xn xn x
dxd nnn
n
4.( )
( )( )( )
( ) ( )n
nn
baxan
abaxnna
baxdxd +=
+
++=
+
+ +
11
1
1
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( ) ( )( ) +++
=++
C na
baxdxbax
nn
1
1
5. [ ] [ ] x
xdxd x
dxd 1loglog == if x>0,
( )[ ] ( ) x x
xdxd 1
11
log === if x < 0
+= 0log1 xC x x
6. 0,1log
++
= +
baxbaxa
bax
dxd
+=
+
a
bax
baxdx log
7. x x x
aa
aaa
adxd ==
loglog
log += C aa
dxa x
x
log
8. ( ) +== C edxeeedxd x x x x
9. [ ] +== C xdx x x xdxd
sincoscossin
10. ( ) +== C xdx x x xdx
d cossinsincos
11. ( ) +== C x xdx x xdxd
tansecsectan 22
12. ( ) +== C x xec xec xdxd
cotcoscoscot 22
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13. ( ) +== C xdx x x x x xdxd
sectansectansecsec
14. ( ) +== C ecx xdxecx xecxecxdxd coscotcoscotcoscos
15. ( ) ( ) +=
= C x
x
dx
x x
dxd 1
22
1 sin11
1sin .
Similarly +
=
C a x
xa
dx 122
sin
16. ( ) +=++=
C x xdx
x xdxd 1
221
tan111
tan .
Similarly +
=+
C
a x
a xadx 1
22 tan1
17. ( ) ( ) +=
= C x
x x
dx
x x x
dxd 1
22
1 sec11
1sec
Some Properties of Integration
1. If f(x) and g(x) are any integrable functions, then
( ) ( )( ) ( ) ( ) = dx xgdx x f dx xg x f similarly if ( ) ( ) ( ) x f x f x f n....,, 21 are any n integrable functions, then
( ) ( ) ( )[ ] = dx x f x f x f n.....21 ( ) ( ) dx x f xdx f dx x f n.......21 .
2. ( ) ( ) = dx x f adx xaf
3. If ( ) ( ) += C xgdx x f
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Then ( ) ( ) ++=+ C a
baxgdxbax f
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Examples
1. 1. ( ) ( ) ++=+ C xdx x7
47sin47cos
2. 2. ( ) ( ) ( ) +
= C x
dx x x6
63sec63sec63tan
3. Evaluation of xdxand xdx nn cossin
1. +== C x xdx x xdx22sin
21
22cos1
sin 2
since x x 2sin212cos =
2. ( ) +
== dx x
x xdx2
224
216cos
3cos3cos
Since 1cos22cos 2 =
( ) ++= dx x x 16cos26cos41 2
dx x x ++
+= 16cos2
2112cos
41
++= dxdx x x83
6cos21
12cos81
C x x x +++=
83
126sin
9612sin
3.
= dx x xdx x 3sin41
sin43
sin 3
Since 3sin4sin33sin =
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( ) C x xC x x ++=+
= 3cos121
cos43
33cos
41
cos43
4. ( )
+== dx
xdx xdx x
3326
2
12coscoscos
( ) +++= dx x x x 12cos32cos32cos81 23
dx x x
x x
++
+
++= 12cos32
4cos132cos
43
6cos41
81
Since x x x cos3cos43cos 3
=
C x x x x x +++++
=
165
22sin
83
44sin
163
22sin
323
6326sin
C x
x x x ++++=
165
2sin6415
4sin643
1926sin
4.5 Evaluation of nxdxmxdxnxmx coscos,cossin and dxnxmx sinsin
1. ( ) ( )[ ] ++= dx xnm xnmdxnxmx sinsin21
cossin
( ) ( )C
nm xnm
nm xnm +
+
+= coscos
21
2. ( ) ( )[ ] ++= dx xnm xnmdxnxmx coscos2
1coscos
( ) ( )C
nm xnm
nm xnm +
+
+
+= sinsin
21
3. ( ) ( )[ ] += dx xnmdx xnmnxdxmx coscos21
sinsin
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( ) ( )
C nm
xnmnm
xnm ++
+
= sinsin
21
4. = xdx xdx x x 4sin7cos2217cos4sin
( ) ++== C x xdx x x3
3cos1111cos
21
3sin11sin21
5. [ ] = dx x x xdx x x x 4cos3cos2cos221
4cos3cos2cos
[ ] [ ] +=+= dx x x x xdx x x x 4coscos24cos5cos241
4coscos5cos21
( ) C x x x xdx x x x x ++++=+++= sin33sin
55sin
99sin
41
3cos5coscos9cos41
6.( )( ) ++=+
x x x x x x x x
dx x x x x
22
224422
22
66
cossincossincossincossin
cossincossin
( ) ( )C x x x
dx xec xdx x x
+=
+=+= 3cottan
3cossec1cottan 2222
4.6 Integration by substitutions
1.( )( )
dx
xg xg
we take ( ) ( ) dudx xgu xg ==
( ) +=+== C xgC uu
dulnln
2. ( )[ ] ( ) dx xg xg n we take ( ) ( ) dudx xgu xg ==
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( )[ ]C
n xg
C nu
duunn
n ++
=++
== ++
11
11
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Example 1:
u xTakingdx x
x = seclogseclog
tan
dxdu
x x x
=tansecsec
1or tanx dx = du
+== C uu
du x
dx xln
seclogtan
C x += seclogln
Note : ( ) x x logln =
Example 2:
( )
du x
dx
u xTakingdx x
x
=+
=+
2
12
41
1
tan.1
tan
Given integral +
=+==
C x
C u
duu515
4
5tan
5
Example 3:
( )( ) + x x
xedx xe
2cos1 Taking ue x x =
( ) dudx xee x x =+ Given integral
( ) ( ) +=+== C xeC uduu xtantansec 2
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Example 4:
( ) +== C xdx x x
dx x coslogcossin
tan
C x += seclog
Example 5:
+== C x x x
x sinlogsincos
cot
Example 6:
( ) ++
= dx x x
x x xdx x
tansectansecsec
sec
C x x ++= tanseclog since derivative of g(x) = secx + tanx
= secx (secx + tan x)
Similarly += C xecxdxecx cotcoslogcos
Example 7:
(a) xdx x 83 sincos 3 is odd convert all the terms in sinx and except one cos x
( ) dx x x x cossinsin1 82 = let sin x = t cosx dx = dt.
( ) ( ) +=== C t t dt t t dt t t 119
1119
10882
( ) C x xC x x +=+= 29
119 sin91199
sinsin
111
sin91
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Note: Consider xdx x mn cossin where atleast one out of m and n is oddinteger then we use this method.
(b) = dx x x xdx x sincossincossin 4445
( )
dt dx x
t x
dx x x x
=
=
=
sin
cos
sincoscos1 422
Given integral
( ) ( )
c x
x x
ct t t
dt t t t dt t t
++=
++
=
+==
9cos
cos72
5cos
972
5
21
97
5
975
864422
Example 8:
dx x
e x
2
sin
1
1
du x
dx
u x
=
=
2
1
1
sin
+=+==
C eC edue xuu 1sin
Example 9:
Let I=
+=
+dx
ee
eedx
e
e x x
x x
x
x
11
2
2
Let uee x x =
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( ) dudxee x x =+
C eeC uu
du I x x +=+== lnln
Example 10:
( )( ) +
+dx
b xa x
cossin
x + b = udx = du
( ) += duu
baucos
sin
( ) ( ) += duu
baubau
cossincoscossin
( ) ( ) += dubaduuba sintancos ( ) ( ) C ubauba ++= sinseclncos
( ) ( ) ( )( )( ) C b xbab xba ++++= sinseclncos
Example 11:
dx x
x x 2sectan
Let u x =tan
( ) dudx x
x =2
sec 2
Given integral
( ) +=+== C xC uduu 22 tan2
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Example 12:
==+ duadxau x Let a xdx
.22
+=+=+= C u
audu
aauaadu 1
2222 tan1
11
C a x
a+
= 1tan1
4.7 Integration by Partial Fraction
If the integrand is a rational function of the form ( ) ( )( ) xQ xP x f = where P(x) and Q(x)
are polynomials such that degree of P(x) < degree of ( ) xQ . Such a rationalfunction is called proper rational function. If the degree of P(x) > degree of Q(x)then by dividing we can write f(x) = a pro polynomial + a proper rational function.
Method of resolving a proper rational function into partial fractions:
We resolve Q(x) into factors. For each linear factor ax+b corresponding term will
bebax
A+
in the partial fractions. If (ax+b) r is a factor of Q(x) then corresponding
terms in the partial fractions are( )
termsr bax
Bbax
A.....2 ++
++
If C bxax ++2 is a factor of Q(x) then corresponding term iscbxax
B Ax++
+2 If
cbxax ++2 is a factor of ( ) xQ of multiplicity r, then corresponding terms are
( ) r
C bxax Dcx
cbxax B Ax .....222
+++
++++
+ terms.
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We write the function ( ) ( )( ) xQ xP
x f = as the sum of partial fractions and then find the
constants by equating the coefficient of like terms on both sides or by takingparticular values of x.
Some Standard Forms
1. ++
=
C a xa x
aa xdx
log21
22
Let ( ) ( )a x Ba x Aor a x
Ba x
A
a x++=
++
=
1
122
Taking x = a and x = -a, we geta
A2
1=
a B Ba
21
12 ==
+
=
a xa xaa x11
211
22
+=+= a xaa xaa xdx
aa xdx
aa xlog
21
log21
21
211
22
C a xa x
a+
+
= log
21
2. 221
xa
Let( )( )
( ) ( )2222
11 xa
xa B xa A xa
B xa
A xa xa xa
++=
++
=
+=
( ) ( ) ,1=++ xa B xa A taking x = a , x = -a
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++
=
==
xa xaa xaa B
a A
11211
21
,21
22
+
+
=
dx
xadx
xaa xa
dx 11
2
122
( ) C xaa
xaa
+++= log21
log21
C xa xa
a+
+= log
21
Examples
1.( )
( )( ) ++
232
12
x x
dx x
Let( )( ) ( )22 33232
12
+
++
=+
+
x
C x
B x
A
x x
x
( ) ( )( ) ( )
( )( )2
2
32
2323
+
++++=
x x
xC x x B x A
( ) ( )( ) ( )232312 2 ++++=+ xC x x B x A x which is an identity, true for allx. taking x = 3, x = -2 and x = 0 we get 5C = 7, 25A = -3, 9A-6B + 2C=1.
514
2527
1615
146
2527
,5
7,
253 +==+
=
= Bor BC A
2518=
( )( ) ( ) ( ) ( )22 357
3253
2253
3212
253
+
+
+=
++=
x x x x x x B
( )( )( ) ( ) ( )
+
++
=
+
+
22 357
3253
2253
32
12
x
dx x
dx x
dx
x x
dx x
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( )( )
C x
x x +
++=35
73log
253
2log253
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2. ( )( ) ++ dx x x x
112
Let
( )( ) 1111 22
+
+
+
+=
++ x
C
x
B Ax
x x
x
( )( ) ( )11 2 ++++= xC x B Ax x equating coefficients of x 2, x and constantterm on both sides we get A+C = 0 A+B = 1 B+C=0 from first and thirdequation we get A-B = 0 or A = B putting in second equation we get
.21= A
21
21
== C and B
Given integral is equal to
++++=+++
121
121
12
41
121
11
21
222 xdx
x
dx
x
xdx x
dxdx
x
x
( ) C x x x ++++= 1log21
tan21
1log41 12
4.8 Evaluation of the integrals of the form ++ cbxaxdx
2 we write such
integrals in the form + 2222 c xdx
or k x
dx or in the form 22 xa
dx
Example 1:
1. ++=++ 94211882 22 x x dx x x dx
( ) +=++= 521
221
22 udu
s x
dx where u = x +2
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C x
C u +
+
=+
=
5
2tan
52
1
5tan
52
1 11
+=
+
C a
x
aa x
dxSince 1
22 tan
1
Example 2:
( ) =
+=
+ 2222 39254 udu
x
dx
x x
dx where u = x +2
C x x
C uu +
+
=+
+
=
51
log61
33
log61
Note: This integral can be valuated by using partial fractions.
4.9 Evaluation of the integrals of the form( ) ++
+
cbxax
dxedx2 we write
( ) 221 K cbxaxdxd
K edx +++=+ then the integral can be evaluated.
Example 1:
Evaluate the integral
( ) +++
431
2 x xdx x
Let ( ) 221 431 K x xdxd
K x +++=+
Or ( )21
321 121 =++=+ K K xK x and
21
13 221 ==+ K K K (by equating coefficients of x and constant on
both sides).
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( ) ( )
+
+
++
+=
++
+
47
232
143
3221
431
222
x
dx
x x
dx x
x x
dx x
C x
x x +
+
++=
2723
tan7
221
43log21 12 (since in the I integral 2x+3 is
the derivative of the denominator.)
C x
x x +
+
++=
7
32tan
7
143log
21 12
Example 2:
Evaluate ++++
13312
23 x x x
dx x
Let ( )( ) 13113112
13312
2223 ++
+
+=
++
+=
+++
+
xC
x
B Ax
x x
x
x x x
x
or( )( ) ( )11312 2 ++++=+ xC x B Ax x . Equating coefficient of x 2, x and
constant term on both sides, we get
3A+C = 0, A + 3 B = 2 and B + C = 1 BC = 1
107
710
693
13013
=
=
=+
==+
B
B
B A
B Aor B A
103
107
1101
103
107
13 ===
=+= C A A
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Hence given integral
( ) +++
= dx
x x
dx x
13
3
10
1
1
7
10
12
+++++= 133
101
1207
12
201
22 xdx
x
dxdx
x
x
( ) ( ) C x x x +++++= 13log101
tan207
1log201 12
4.10 Integration by Parts
If u and v are functions of x then the formula for integration by parts is
= dxvdxdxdu
vdxuuvdx , u is called I functions and v is called II function.
Rule 1: If both u and v are integrable functions then take that function as the firstfunction which can be finished by repeated differentiation.
Rule 2: If the integral contains logarithmic or inverse trigonometric function thentake this logarithmic or inverse trigometric function as the first function and otherfunction as the II function. If there is no other function then take 1 as the secondfunction.
Rule 3: If both functions are integrable and none can be finished by repeateddifferentiation, then any function can be taken as the I function and other as the IIfunction. Repeat the rule of integration by parts.
Rule 4: That function is taken as the first function which comes first in the wordILATE where I = inverse circular function, L = Logarithmic function, A = Algebraicfunction, T = Trigonometric function, E = Exponential function.
Example 1:
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1. dx x x x xdx x x II I II I
cos2cossin 22 +=
C xdx x x x x ++=
sin.1sin2cos2
[ ] C x x x x x +++= cossin2cos2
2. == dx xee xe xdxe xe xdxe x x x x II
x
I
x
II
x
I 233 23233
+=+= dxe xee xe xdxe xe xe x x x x x x x x 6363 2323
ce xee xe x x x x x ++= 663 23
3. dx x
x x
xdx x x
I II
=
2
21
21
121
sin2
sin (1)
taking d dx x cos,sin ==
( ) === 2
2sin21
2cos121
sin1
2
2
2 d d dx
x
x
[ ] [ ]== 21 1sin21
cossin21
x x x from (1)
++= C x x x x x x x 2112
1 141
sin41
sin2
sin
4. xdx x I II
log3
+== C x x xdx x
x x
x16
log4
1.
4log
4
4444
5. Evaluate bxdxe ax sin
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Let +
== bxdxe
ba
bbx
edxbxe I axaxax coscos
sin
I b
abxe
b
a
b
bxebxdxe
b
a
b
bxe
b
a
b
bxe axax
axaxax
2
2
2 sin
cossin
sincos +
=+
=
or bxeb
abxe
bb
a I axax sincos
11 22
2
+=+
or [ ]bxabxbba
e I
ax
sincos22 ++=
6. ( ) dx x 2log . we take (log x) 2 , I function and 1 as II function
( ) ( ) == II dx
I
x x xdx
x x
x x xlog
2loglog
2.log 22 we take logx, I function and
1, II function.
( ) ( ) [ ] C x x x x xC dx x
x x x x x +=+= log2log1.log2log 22
7.
dx x
x2
1
1
2tan Taking tan= x
( ) === dxdxdx
22tantantan1tan2
tan 121
+== dx
x x
x x xdx 211
1tan2tan2 (Taking 1 = II function).
( ) C x x x ++= 21 1logtan2
4.11 Some standard forms
1. +
=
C a x
xa
dx 122
sin , putting
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+
=+==
= C a x
C d xa
dxa x 1
22sin,sin
2. ( )
++=
C a x xa x
dx 2222
log
We take tansec,sec adxa x ==
( ) ++===
122tanseclogsec
tantansec
C d a
d a
a x
dx
1tan,sec2
2
==
a
x
a
x
( ) C a x xC a
a x xC
a x
a x ++=+
+=+
+= 22
1
22
12
2
loglog1log
3. ( ) +++=+
C a x xa x
dx 2222
log (By taking tana x = )
4. ++= C a xa xa xdx xa 1
22222 sin
22
We denote the given integral by I and integrate by parts taking 22 xa as the I function as 1 as the II function
( ) ( ) = dx x xa x xa x I 22
121
2222
=
+= dx
xa
a xa xa xdx
xa
x xa x
22
22222
22
222
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11222 sin C
a x
a I xa x ++=
or C a
xa xa
x I +
+= 1
222 sin
22
similarly we can derive the following results
5. ( ) ++= C a x xaa x xdxa x 222
2222 log22
6. ( ) +++++=+ C a x xaa x xdxa x 222
2222 log22
4.12 Integrals of the form ++++
cbxaxor cbxax
dx 22
can be evaluated by
writing in any one of the above forms.Example 1:
Evaluate the integral 221 x x
dx
Given integral( ) ( ) +
=++
=22 12122 x
dx
x x
dx
+
+
=+
=
= C x
C u
u
du
2
1sin
2sin
2
11
2
Example 2:
+
=+
1611
254
1
114016 22 x x
dx
x x
dx
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=
+
=2
222
234
1
23
454
1
u
du
x
dx taking u x =+
45
1
22
23
log41
C uu +
+=
1
2
49
45
45
log41
C x x +
+++=
1
2
49
1625
25
45
log41
C x
x x +
++++=
1
2
411401654
log41
C x x x +
+++=
( ) C x x x ++++= 11401654log41 2
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Example 3
( ) ( ) ++=++ dx xdx x x 222 252910
( ) =++= u xduu 52 22
( ) C uuuu +++++= 4log2
22
22
222
( ) C x x x x x x +++++++++= 29105log229102
5 22
Evaluation of the integrals of the form.
4.13( )
++
+
C bxax
dxq px2
and ( ) dxC bxaxq px +++ 2 . In this type of the integrals,
we write ( ) 221 K C bxaxdxd
k q px +++=+
Example 1:
( )++
+
52
22 x x
dx x
( ) 221 522 k x xdxd
k x +++=+
( ) 21 22 k xk ++=
Equating coefficient of x and constant on both sides, we get
1,21
,22,21
21211 ===+= k k k k k . So the given integral
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( )( )
++
+++
+=
222 2152
2221
x
dx
x x
dx x
Putting
I int x x =++ 522 integral ( ) dt dx x =+ 22
Putting x+1 = u in the II integral. Given integral
[ ] ++++=+
+= C uut u
du
t
dt 4log
221 2
22
Putting values of t and u
C x x x x x ++++++++= 521log52 22
Example 2:
( ) + dx x x x 2453
Let ( ) 221 453 K x xdx
d K x +=+
( ) 21 24 K xK += , equating coefficient of x and constant term on bothsides
134,21
12 22111 ==+== K K K K or K
Given integral can be written as
( ) + dx x xdx x x x 22 45452421
In I integral we take 5-4x-x 2= t.
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( ) dx xdt t ++= 22921
In II Integral we take x + 2 = u
++= 1223
9
232
1C duu
t
C u
uu
t +++=
3sin
29
923
1 1223
. Putting values of u and t, we get
( ) ( ) C x
x x
x x +
+
++
+
+=
32
sin29
2922
4531
12
2
3
2
( ) C x x x x x x +
+
++
+=
32
sin29
452
245
31 122
32
4.14 Evaluation of the integrals of the form ++ C xb xadx
cossin. In such
integrals we take t x =2
tan
12
12
tan
2
2sec
22
sec21
222
2
+=
+
===t
dt xdt
xdt
dxor dt dx x
12
12
tan
2tan2
2sec
2tan2
2cos
2sin2sin 2
22 +=
+
===t
t x
x
x
x x x
x
2
2
2
2
22
11
2tan1
2tan1
2sin
2coscos
t
t x
x x x
x+
=
+
==
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Example
Evaluate the Integral
+
+
++
+=++
2
2
2
2
11
12.2
3
12
cossin23t
t
t
t t
dt
x xdx
(Taking t x =2
tan )
++=+++= 4422
14332
222 t t dt
t t t dt
( ) +=+=
++
=++
= C u
u
du
t
dt
t t
dt 1222 tan
11122
( ) C xC t ++=++= 12
tantan1tan 11
4.15 Evaluation of the integral of the form ++
dx xd xa xb xa
cossincossin
Example
Evaluate ++
dx x x x x
cos4sin3cos2sin2
We write ( ) ( ) Dr dxd
K Dr K x x 21cos3sin2 +=+
( ) ( ) x xK x xK sin4cos3cos4sin3 21 ++=
( ) ( ) xK K xK K cos34sin43 2121 ++=
Equating coefficient of sinx and cosx, we get
6129243 2121 == K K or K K
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121216334 2121 =+=+ K K or K K
Adding we get2518
1 =K and putting value of K 1 in I equation, we get
251
254
4242554
222 === K K or K
So given integral =( ) +
+ dx
x x x x
dxcos4sin3sin4cos3
251
2518
C x x x +++= cos4sin2log
251
2518
4.16 Evaluation of the integrals of the form ++++
dx f xe xd C xb xa
cossincossin
we write
( ) ( ) r f xe xd dxd
q f xe xd pc xb xa ++++++=++ cossincossincossin p,q,r
can be evaluated by equating coefficient of sinx, cosx and constant termon both sides.
Given integral
++
++++= f xe xd dx
r f xe xd q pxcossin
cossinlog . The
last integral can be evaluated by the method of 4.14.
4.17 Definite Integral
Let f(x) be continuous function on the interval [a,b]. We divide the interval
into n sub-intervals each of lengthn
abh
=
The definite integral ( )b
a
dx x f
( ) ( ) ( ) ( )( )[ ]hna f ha f ha f a f h Lt h
1.....20
+++++++=
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a is called lower limit and b is called upper limit of the definite integral
( )b
a
dx x f . The definite integral ( )b
a
dx x f represents the area under the
curve y = f(x) from x = a to x = b (or the area bounded above by the curve
y = f (x) below by x-axis on left by x =a on right by x = b)
Fundamental Theorem of Calculus : If f(x) is a continuous function on [a,b]
and ( ) ( ) += ,C xF dx x f then ( ) ( ) ( ) =b
a
aF bF dx x f
Example
]2
0
2
0
coscossin
+= xdx x x x x
] [ ]0sin0cos02
sin2
cos2
sincos 20 ++=+=
x x x
( ) [ ] .10102 =+=
To evaluate ( )b
a
dx x f if we are taking a substitution x = g(t) then find the
values of t corresponding to x=a and x = b. If t = A and t = B when x = aand x = b respectively, then
( ) ( )[ ] ( )
=
b
a
B
Adt t gt g f dx x f
Example
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Evaluate +2
04sin1
cossin2
dx x
x x we take dt xdx xt x == cossin2sin 2
When x = 0, t = sin 2 0 = 0 and when 12sin,22 === t x
The given integral ] =+=1
0
10
12 tan1
t t
dt
( ) ( )4
04
0tan1tan 11 ===
Properties of definite integrals
1. ( ) ( ) =b
a
a
b
dx x f dx x f
2. ( ) ( ) ( )
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7. ( ) ( ) ( ) ( )
( ) ( )
=
==
a
a
a
functionodd anis f ei x f x f if
functionevenanis f ei x f x f if dx x f dx x f
..0
..20
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Examples
1. ( )
672
2 /
67 cossinsin0cossin ==
f ced is an odd function.
2. +=+2
0
2
0 cossinsin
cot1
x x
dx x
xdx
nn
n
n (where n is any real number)
Let
+
=+
=2
0
2
0
2
cos
2
sin
2sin
cossinsin
x x
dx x
x x xdx
I nn
n
nn
n
Using property (4) of
definite integral.
I x x
xdxnn
n
=+
= 2
0 sincoscos
Adding two values of I.
( )
4
2sincoscossin
22
0
2
0
=
==+
+=
I
dx x x
dx x x I nn
nn
Note: If sinx and cosx are replaced by tanx and cotx or by cosecx andsecx respectively in the integral I, then similarly we can evaluate theintegral.
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3. dx x x
x x x
x x
dx x x x I
+
=+
=2
0
2
0 4444
2sin
2cos
2cos
2sin
2sincos
cossin
=+
=2
044 cossin
sincos2
I x x
xdx x x
Adding two integrals, we get
+=
+=
2
0
2
04
2
44 tan1
sectan
2sincos
cossin22
x
xdx x
x x
xdx x I
Taking dt dx x xt x == 22 sectan2,tan
[ ]
==+
=0
01
2 024tan
4121
22
t
t
dt I
16
2
= I
4.
++
+
+
=+
=3
6
33
33
6
33
3
63cos
63sin
63sin
cossinsin
x x
dx x
x x
xdx I
Using property (iii)
=+=
+
=3
6
3
6
33
3
33
3
sincoscos
2cos
2sin
2sin
I x x
xdx
x x
dx x
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Adding given integral and the last integral, we get
12
6632
3
6
=
===
I
dx I
5.
+=
+
1
1
1
06
2
6
2
12
1dx
x
x
x
dx xSince the function is an even function. Taking
323 dt dx xt x ==
( )[ ] ==+=1
0
111
0
12 0tan1tan3
2tan
32
132
t t
dt
60
432 ==
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Chapter 5
Review of two dimensional Geometry
Let OX X and OY Y be two perpendicular straight lines intersecting at O. The lineOX X is taken horizontal and OY Y is taken vertical. XOX is called x-axis and
Y1OY is called y-axis. These lines divide the plane in four parts called thequadrants as shown in the figure. If (x,y) are coordinates of a point P.
In I quadrant both x and y are +ve and in II quadrant x is ve Y is +ve. In IIIquadrant both x and y are ve and in IV quadrant x is positive y is negative.
X O is called ve x-axis and OX is called +ve direction of x-axis.
Similarly OY is +ve direction of Y-axis and Y O is ve direction of y-axis Point(4,3) is in the I quadrant. To plot this point, we take 4 units along +ve x-axis thentake 3 units parallel to +ve y-axis. The point (-4,3) is in II quadrant. We take 4units along ve x-axis and 3 units parallel to positive y-axis. We plot the point (-4,3). Similarly we can plot the point (-4,-3) and (4,-3). On x axis y-coordinate iszero and on y-axis x-coordinate is zero.
5.1 Important Formulae of Coordinate Geometry.
(i) Distance between two points P(x 1, y1) and ( )22 , y xQ is
( ) ( )2122
12 y y x xPQ +=
(ii) The coordinates of a point dividing the line segment joining the points),( 11 y x and ( )22 , y x internally, in the ratio m:n are
++
+
+
nmnymy
nmnxmx 1212 ,
(iii) The coordinates of the mid-point of the line segment joining ),( 11 y x and
( )22 , y x are
++
2,
22121 y y x x
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(iv) Slope of a st line is denoted by m and defined as tan=m where is
the angle between +ve direction of x-axis and the st. line
(v) Slope of a st. line which passes through two points ( )11 , y x and ( )22 , y x is
12
12
x x
y y
.
(vi) The angle between two lines with slopes m 1 and m 2 is given by
21
12
1tan
mm
mm+
=
(vii) Two lines are parallel if m 1=m 2 and lines are perpendicular if m 1m2= -1.
(viii) Equation of a st. line
(a) passing through the point ( 11 , y x ) with slope m is
( )11 x xm y y = .
(b) Passing through two points (x 1, y1) and (x 2,y2) is
( )112
121 x x x x
y y y y
=
(c) With slope m and having intercept c on y-axis is y=mx+c. (meetsy axis at(0,c))
(d) Having intercepts a and b on x-axis and y-axis respectively is
.1=+b y
a x
(e) Whose perpendicular from origin has length p and makes anglewith +ve direction of x-axis is p y x =+ sincos
(f) Any linear equation ax+by + c = represents a st. line if atleast oneof a and b different from zero.
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(ix) Distance of a point P (x 1, y1) from the st. line
22
110ba
cbyaxiscbyax
+
++=++
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Definition:
A circle is the locus of a point which moves in a plane in such a way that itsdistance from a fixed point is constant. The fixed point is called the center and
the constant distance is called radius of the circle.
1. Equation of the circle with radius r centred at (x 0, y0) is
( ) ( ) 2202
0 r y y x x =+ and x2+y2 = r 2 is the equation of the circle
whose centre is at the origin.2. Any equation of second degree of the form x 2 + y 2 + 2 f x + 2 fy + c
=0 represents, a circle if .022 + c f g This equation we can write
as ( ) ( ) c f g f yg x +=+++ 2222 so it represent a circle with
radius c f g + 22 and center at (-g,-f)
Example:
Find center and radius of the circle x 2+y2 2x 6y 6 = 0. We canwrite as (x 2-2x+1) + (y 2 6y + 9) = 6 + 1 + 9 or (x-1) 2 + (y-3) 2 = 4 2.Centre is (1,3) and radius is 4.
Curve Sketching:
An equation of the F(x,y) = C or y = f(x) or x = g(y) represents a curve in the xy-plane. We draw the graph of this curve using some properties of the graph.
1. Symmetry (a) The curve will be symmetrical about x-axis if the equationdoesnt change by taking-y for y. For example y 2 = 4ax is symmetricalabout x-axis. If all the powers of y in the equation are even then the curvewill be symmetrical about x-axis.
2. The curve will be symmetrical about y-axis if the equation does notchange by taking -x for x in the equation. x 2 = 8y is symmetrical about y-axis.
3. The curve will be symmetrical in opposite quadrants if the equation doesnot change by taking x for x and y for y. The curve xy=4 is symmetricalin opposite quadrants.
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5.2. Intercepts on the axes:
We find the points where the curve meets the axes. To find the points on x-axiswe take y = 0 in the equation then solve.
For example y 2 = x (x-1) (x-2) meets the x-axis at the points (0,0) (1,0) and (2,0)
Similarly to find the points where the curve meets to y-axis. We put x = 0 in theequation then we solve the equation for y. If (0,0) satisfies the equation of thecurve then the curve passes through the origin.
Region: We find the value of x and y for which both x and y are real. For exampleconsider the curve y 2 = x-4 if 4 x then y will be real.
Asymptotes: If the distance between the graph of a function and some fixed lineapproaches zero as the distance of the point (x,y) on the curve from origin tends
to , then the line is called an asymptote to the curve. If we write( )( ) x f x f
y2
1= then
we shall get vertical asymptotes by solving f 2 (x) = 0. If we write( )( ) yg yg
x2
1= then
we shall get horizontal asymptotes by solving g 2 (y) = 0 .
Example : Find all asymptotes of the curve 04.4
1 22
=
= x x
y or 2= x are
vertical asymptotes. As .,2 y x Solving for x, we get
( )axis x yso y
y x =
+= 0,
412 is horizontal asymptote.
Regions of rising and region of falling
The curve y = f(x) is rising (f is increasing) if ( ) 0> x f and the curve is falling (f isdecreasing) if ( ) .0
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Extreme Values : These points are very useful for curve sketching. At thesepoints curve changes from rising to falling or falling to rising. f(x) has localmaximum if ( ) ( ) 0,0 x f .
Concavity of the curve : The curve will be concave upward if ( ) 0> x f andconcave downward if ( ) 00 is the parabola whose focus is the point (p,0) and directrix isthe line x = - p
(i) The curve is symmetrical about x-axis(ii) The curve passes through the origin(iii) y is real when 0 x (iv) We consider following points to draw the graph
x 0 p 2p 4p 25p y 0 p2 p22 p4 p10
So the graph is shown as the following figure (0,0) iscalled the vertex of the parabola.
Let p (x,y) be any point on the parabola then usingPF=PD we get (x-p) 2 or (p 2)+y2 = =(x+p) 2 or y 2 = 4px. So we can derive the equations of the parabolaby using definition.
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(b) px y 42 = is the parabola whose directrix is the
line x = p and focus is at (-p,0). This parabola isalso symmetrical about x-axis. Vertex is at (0,0). y isreal when x0 and 0b2). If P(x,y) is any point on the ellipse. Using PF=ePD, where F is (ae,0) and PD is perpendicular to the directrix x = a/e, we get theequation of the ellipse.
Tracing of the ellipse 122
2
2
=+b
y
a
x. It is symmetrical about both axes. If x>a or xb
x
X
y Y
O
O
(-2,3)
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or y < -b, x is not real. So curve lies between the lines y = - b and y = b. Weconsider some points on the ellipse. If
.0,,47
,43
,2
,2
,,0 ======== ya xb ya xb
ya
xb y x so we can
draw the graph of the ellipse.
If the larger number a 2 is below y 2 and smaller b 2 is below x 2 thenmajor axis will be along y axisand foci will be on y-axis and
directrices will be perpendicularto y-axis. So faci are ( )ae,0 and
directrices areea
y = . The
graph of the ellipe ( )2222
2
2
1 baa y
b x >=+ is given in the
adjoining figure.
Graphs of 149
22
=+ y x
and 194
22
=+ y x
are as givenbelow.
If the center of the ellipe is at (h,k) then the equation of the ellipse can be writtenas
( ) ( ) ( ) ( ).11 2
2
2
2
2
2
2
2
=
+
=
+
ak y
bh x
or b
k ya
h x
ea
y =
ea
y =
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Graphs of these ellipse are shown as above.
Example 1: Draw the graph of ellipse x 2 + 2y 2 2x 8y - 7 = 0. We can write theequation as (x 2-2x+1) + 2 (y 2-4y+4) = 7+ 1 + 8 or (x-1) 2 + 2 (y-2) 2 = 16 or
( ) ( )1
82
161 22 =
+
y x. Shifting the origin at
(1,2) .Equation become .1816
22
=+ Y X The
graph is as in the adjoining figure.
5.6 Hyperbola
Hyperbola is a conic for which e>1. A hyperbola has two foci and two directrices.Let F(ae,0) be a focus and corresponding directrix be x = a/e. Let P (x,y) be anypoint on the hyperbola then by definition of the hyprebola.
PF = e PD (perpendicular distance from focus to the directrix)
x
X
y Y
O
O x
X
y Y
O
O
x
X
y Y
O
O
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Or ( )
=+ea
xe yae x 22
Or x 2 + a 2 e 2 - 2xae + y 2 = e 2x2-2aex+a 2
Or (e 2-1) x 2-y2 = a 2 (e 2-1)
Or ( ) 11222
2
2
=
ea y
a x
Or 122
2
2
=b
y
a
x where ( )1222 = eab
Note: We can drive the equation of the hyperbola by taking second focus (-ae,0)and the corresponding directrix x = -a/e and using PF = ePD.
Tracing of the hyperbola 122
2
2
=b
y
a
x
The curve meets x-axis at (a,0) and at (-a,0) and does not meet y-axis the curveis symmetrical about both axes.
We can write the equation as 122
2
2
=a x
b y if a
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122
2
2
=b
x
a
y is the hyperbola whose focis lie on y-axis and
whose graph is given in the adjoining figure.
y-axis is the axis of the hyperbola. Foci are ( )ae,0 and
directrices are ea
y =
, vertices are at (0,a) and (0,-a).
Graphs of 164
22
= y x and 146
22
= x y are as shown in the following figure.
If the center of the hyperbola is at (h,k).
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Equations of the hyperbolas are 122
2
2
=bY
a X
and 122
2
2
=b
X aY
with respect to the
new origin. First equation is the hyperbola whose axis is X-axis and secondequation is the hyperbola whose axis is Y-axis.
These equations w.r. to actual origin becomes
( ) ( ) ( ) ( )11 2
2
2
2
2
2
2
2
=
=
bh x
ak y
and b
k ya
h x
The graphs are as given below
Example 1: Trace the hyperbola 16x2
9y2
= 144 and find its foci and directrices.The equation can be written as
( )116,9.1169
2222222
==== eabba y x
or
35
,925
916
12 ==+= ee
593
53,5
35.3 ====
eaae . So
faci are ( )0,5 and directrices
are .59= x The graph is as
given by the adjoining figure.
X
x
Yy
o
oo
o Xx
y Y
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Note : For the hyperbola 1916
22
= x y graphs is as shown in
the adjoining figure.
Example 2: Sketch the hyperbola
3x2-y2 18x + 4y 25 = 0. We can write the equation as 3(x 2 6x + 9) (y 2 4y
+ 4) = 25 + 27 4 or 3 (x-3) 2 (y-2) 2 = 48 or( ) ( )
148
216
3 22 =
y x
or shifting the origin at (3,2), the equation becomes
14816
22
= Y X
The graphs is shown in the adjoiningfigure a 2 = b 2 (e 2-1) or 16 = 48 (e 2-1) or
3
2342 == ee
3
8
3
2.4==ae
Foci are 0,3
8 == Y X
Or 02,3
83 == y x
coordinates of foci are
2,
3
83 center is 0(3,2)
and vertices are 0,4 == Y X or
( ) ( )2,1,2,702,43 == or y x are the vertices.
Directrices are 32332 === xor ea
X
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Note : Graph of( ) ( )
148
316
2 22 =
x y
is as given in the adjoining figure.
Axis of the hyperbola is Y axis or X = 0 or x 3 = 0 or x = 3. Centre is (3,2).Vertices are (3,6), (3,-2).
5.7 Sketching of the Curves
Example 1 : Sketch the curve x
x y
=
22
(i) The curve passes through the origin
(ii) Extent : when x 2, y is not real. So nocurve when x2.(iii) Symmetry: If we take y = -y in the equation, then the equation doesnot
change. Hence the curve is symmetrical about x-axis.(iv) The st. line x = 2 is an asymptate to the curve since as x 2, then
y .
(v) We consider some points on the curve and draw the graph.
x 0
2
1
1
2
3
1.99 2
y 03
1 1 3 199
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Example 2: Sketch the curve( )
2
22
162
x x x
y
=
(1) Curve is symmetrical about x-axis.(2) y is not real when 4
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x = 1, the curve has point of inflection where curve changes itsconcavity.
5. The curve has no asymptotes. We consider some points to draw
the graph.
x y
dxdy
2
2
dx
yd
Conclusions
X
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has also polar coordinates
+
45
,44
2,447
,4 or or , set of all polar
coordinates is given by }......2,1,0,4
32,4
42,4 =
+
+ nnn
Relations between polar coordinates and Cartesian coordinates : Let (x,y) and( ) ,r be Cartesian and polarcoordinates of a point P respectively.
Thenr y
r x == sin,cos
sin,cos r yr x ==
=+=
x y
y xr 1222 tan,
Using these relations we can changeany equation in polar coordinates intoCartesian coordinates and any equation in Cartesian coordinates in polarcoordinates.
Equation of a line in polar coordinates:
Let ( )00 , pP be the foot of perpendicular fromorigin to a line L and let ( ) ,r P be any point onthe line L. Then from the right triangle
( ) ( ) pr or r
pPOP == 000 coscos,
Which is the equation of the line L. If
pr == cos,00 or x = p. if 2 = we get p yor pr == sin
Equation of a circle Whose center is the point( )00 , r and radius is a.
Let ( ) ,r P be any point on the circle.
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Applying the laws of cosines to triangle OP 0P, we get
( )002022 cos2 += rr r r a (1)
If the circle passes through the origin, then r 0 = a, then equation (1) becomes
( )0cos2 = ar (2)
If the center lies on x-axis then = 00 and equation becomes
cos2ar =
Which we can also obtain from Cartesian equation (x-a)2
+ y2
= a2
or x2
+ y2
2ax =0 by putting .sin,cos r yr x ==
If the center lies on y-axis, then20
= and equation (2) becomes
=2
cos2
ar or
= 2
cos2ar or sin2ar = . Which can be obtained from
the Cartesian equation x 2 + (y-a) 2 = a 2 or x 2 + y 2 2ay= 0 by taking sin,cos r yr x == . If center is at the origin and radius is a then the equation of
the circle is r = a.
Equation of a conic in polar coordinates. Equation of a conic whose one focus isat the origin and corresponding directrix is the line x = K. Let ( ) ,r P be any pointon the conic.
PD = MN = ON- OM = K- cosr .
Using focus-directrix equation( ) cosr K er or PDePF ==
Or cos1 e
eK r
+=
M N
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The equation represents a parabola if e = 1. an ellipse if e1.
Equation of the conic whose one focus is at the origin and whose corresponding
directrix is x = -K is given by
cos1 e
eK r
=
If the directrix is the line y = K and corresponding focus is at the origin, then the
equation of the conic is sin1 e
eK r
+= and if the directrix is the line y=-K and
corresponding focus is at the origin, then equation of the conic is sin1 e
eK r
=
Exercise 5.1 : Sketch the following curves
(a) 44
cos =
r
(b) 53
sin =
+ r
(c) cos8=r
(d) sin6=r
(e) cos2
8+
=r
(f) sin1
4
=r
(g) r = 6
(a) Comparing with ( ) pr = 0cos we recognize that given equationrepresents a st.line. The foot of perpendicular from origin to the line is
4,4
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(b) Given equation we can write as 532
cos =
+
r 56
cos =
r or
56
cos =
r which is a line. Foot of perpendicular from the origin to the
line is
6,5
(c) cos8=r is the equation of the circle with radius 4 whose center lies on x-axis and which passes through the origin.
(d) sin6=r is the equation of the circle whose radius 3, center lies on y-axisand which passes through the origin.
(e) cos
21
1
4cos28
+
=+
=r comparing this equation with cos1 e
eK r
+= , we
get 84,21 === K eK e . So given equation represents the ellipse whose
one focus is at the origin and corresponding directrix is the line x = 8. Sograph is as given below. Distance between focus and corresponding
directrix 818 =
== ee
aor aeea or 8
21
2 =
a
316= a
316
21.
316.22 ==ae , distance beaten two foci
Polar coordinates of second focus are
,3
16 . Center is mid point of the
two foci center is
,38 . Distance between F 1=0 and second directrix
340
21
23
16 =
+=+= aeea
second directrix is3
40= x .
(f) sin1
4
=r comparing with 4,1,sin1
==
= K eeeK
r
. So given equation
represents a parabola whose focus is at the origin and directrix is the line
y = -4. Axis of the parabola is y axis. Vertex is
2,2 or Cartesian
coordinates are (0, -2)
F1=0
X=8
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(g) r = 6 is the equation of the circle whose center is at the origin and whose
radius is 6. If we put sin,cos r yr x == in 3622 =+ y x . We get
63636sincos 22222 ===+ r or r or r r
5.9 Curve Tracing in Polar Coordinates
To draw the graph of ( ) f r = we consider some points on the curve and valuesof for which the curve passes through the pole (origin). We find symmetry andslope of the curve and maximum and minimum values of r.
Symmetry
The curve will be symmetrical
(a) about x-axis if the equation doesnot change by taking = or by taking == ,r r
(b) About y-axis if the equation doesnot change by taking = or bytaking == ,r r
(c) About the origin if the equation does not change by taking r r = or bytaking += .
Example
Sketch the curve ( ) cos1 = ar , where a> 0. The curve is symmetrical about x-axis.
0sin >=
ad dr
if r .0
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Considering these points we draw the portion of the curve which lies above x-axis. Using symmetry about x-axis. We draw portion of the curve which liesbelow x-axis. The curve is called cardioid. Similarly ( ) cos1 += ar is also cardioidwhich passes through origin when = and r = 2a (maximum) when = 0. Sograph of this curve is given in the adjoining figure.
( ) sin1 = ar and ( ) sin1 += ar are also cardioids symmetrical about y-axissince taking = equation doesnot change. We consider some points on thecurve ( ) sin1 = ar
2
4
3
6
7
23
r 0
2
11a
a2
3a
2a
The graph is given below. We draw the graph of the portion which lies in II and IIIquadrant using these points. Then using symmetry, we complete the graph.
The graph of the curve ( ) sin1 += ar can be drawn similarly. For this r = 0 when
22
223
==
== whenar and or
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Example 2: 0,2cos4 22 >= aar is a constant.
Curve is symmetrical about x-axis since equation doesnot change by taking = . It is also symmetrical about the origin since equation does not change by
taking r = -r. when 2,24
< varies from to2
so 2cos is ve r is not
real. So no curve when24
< . We consider some points on the curve. We
shall consider only +ve values of r and then we use symmetry.
012
6
4
r 2a 32a 2a 0
We get half loop in the I quadrant we complete the loop byusing the symmetry about x-axis. Then using symmetry inopposite quadrants we complete the graph. The curve iscalled lemmiscate.
2sin4 22 ar = is also called lemmiscate. This curve is symmetrical about the
origin. When ver =< 2,2
so no curve when
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0
12
6
4
3
125
2
r 0
2
a
2
3a a
2
3a
2
a
0
Considering these points we get a loop in I quadrant using symmetry wecomplete the graph. Graph contains four loops of equal size.
5.11 Three Dimensional coordinate geometry
To plot a point in space we take three mutually perpendicular lines.OY Y OX X , and OZ Z called x-axis, y axis and z-axis respectively. Let (x,y,z) be
the coordinates of a point p in space then z is perpendicular distance PM from Pto XY-plane. Draw perpendicular MN from M to x-axis. So PM is z, MN is y andON is x. The plane containing x-axis and y-axis is called xy-plane. The planecontaining y axis and z-axis is called yz-plane and the plane containing x-axisand z-axis is called xz plane. These three planes divide the three-dimensionalspace into eight octants.
Distance between two points. : Let P 1 (x1, y1, z 1) and P 2 (x2, y2, z 2) be any twopoints. The distance 21 PP between P 1 and P 2 is given by
( ) ( ) ( )2122
122
1221 z z y y x xPP ++=
The distance OP from origin 0 to P(x,y,z) is given by
222 z y xOP ++=
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Sphere
A sphere is the locus of a point P which moves in the space in such a way so thatits distance from a fixed point is constant. The fixed point is called the centre of
the sphere and the constant distance is called the radius of the sphere.
Equation of a sphere
Let P 0 (x0, y 0, z 0) be the centre and a be the radius of the sphere and let P(x,y,z)
be any point on the sphere then P 0P = a or ( ) ( ) ( ) a z z y y x x =++ 202
02
0
squiring on both sides, we get the equation of the sphere whose centre is (x 0, y 0,z0) and whose radius is a. So required equation is (x-x 0)2 + (y-y 0)2 + (z-z 0)2 = a 2.If the centre is at (0,0,0), then its equation becomes x 2 + y 2 + z 2 =a 2. Section ofthe sphere by the plane z = 0 is the circle x 2 + y 2 =a 2. Any equation of the from x 2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0 can be written as (x+u) 2 + (y + v) 2 + (z + w) 2 =u2 + v 2 + w 2 -d. So this equation represents a sphere with centre (-u, -v, -w) and
with radius d wvu ++ 222 , provided u 2 + v 2+ w2 -d > 0
Ellipsoid. The surface 122
2
2
2
2
=++c z
b y
a x
is called ellipsoid. It meets the coordinate
axes at ( ) ( )0,,0,0,0, ba and at ( )c,0,0 respectively. It lies within the rectangularbox defined by c zb ya x ,, . If .a x > say x = 2a then from the equation
of the ellipsoid, we get 322
2
2
=+b
z
b
y which is not possible since r.h. side cannot
be negative.
So there is no surface when a xsoa x > . . Similarly c zb y , . If2c
z = , then
we get the ellipse .1
43
43 2
2
2
2
=+
b
y
a
x Hence sections by the planes ( )cd d z
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coordinate planes are ellipses for example section by z = 0 (xy-plane) is the
ellipse .122
2
2
=+b
y
a
x
Cylinders
An equation in any two of the three variables x,y,z defines a cylinder parallel tothe axis of third variable. Any curve f(x,y) = C in the xy-plane defines a cylinderparallel to the z-axis whose equation is also f(x,y) = C. Similarly f(y,z) = C is acylinder parallel to x-axis which intersects the curve f(y,z) = C in the yz-planes.
Example 1
x2 + y
2 = a
2 is a circle in xy plane. In three dimension x
2 + y
2 = a
2 represents the
cylinder parallel to z-axis which passes through the circle x 2 + y 2 = a 2 . Thecylinder is symmetrical about z-axis so z-axis is the axis of the cylinder. Sectionsby any plane perpendicular to z-axis (z = k) is the circle x 2 + y 2 = a 2 . If we usepolar coordinates then the circle x 2 + y 2 = a 2 becomes r = a. So equation ofcylinder is also r = a parallel to z-axis through the circle r = a shown in theadjoining figure. Similarly y 2 + z 2 = a 2 also represents the cylinder parallel to x-axis (x-axis is the axis of the cylinder) through the circle y 2 + z 2 = a 2 in yz-plane.Sections by the planes perpendicular to x-axis are circles.
Example 2:
1916
22
=+ y x is the equation of the ellipse in xy-plane but in the three dimension
1916
22
=+ y x defines the cylinder parallel to z-axis whose base curve is the ellipse
1916
22
=+ y x .
Example 3:
y = x 2 is a parabola in xy plane but in three dimension y = x 2 is the cylinderparallel to z-axis whose base is the parabola y = x 2. You can visualize thiscylinder by the following process. Take ten pieces of wire of equal size and bend
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each of them into parabolic arc as shown in the figure. Put these 10 parabolicarcs one on the other. Now these parabolic arcs are made complete parabolasand imagine infinite no. of parabolic wires putting second on the first, third on thesecond and so on.
In this way we get the image of the cylinder
Example 4:
(x-a) 2 + y 2 = a 2 or x 2 + y 2 2ax = 0 is the circle in xy-plane whose centre is at(a,0) and radius is a, its equation in polar coordinates is cos2ar = . In three
dimension ( ) cos2222 ar or a ya x ==+ becomes the cylinder parallel to the z-axis which passes through the circle (x-a) 2 + y 2 = a 2.
Paraboloid
2
2
2
2
b y
a x
c z += is the equation of an elliptic paraboloid. If c>0 then 0 z and there
is no surface when z 4.Taking sections by the planes z = 3, z = 2, z = 1, z = 0, z = -1, z = -2, we get
larger and larger circles. The paraboloid opens downward. The section by 3= x plane perpendicular x-axis is the parabola y 2 = 1-z.
So the sections by the planes perpendicular to x-axis are parabolas. Similarlysections by the planes perpendicular to y-axis are parabolas.
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Cones :
The elliptic cone 22
2
2
2
2
c z
b y
a x =+ has vertex at (0,0,0) section by the planes
c zc zc z