math zc161 lecture notes

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


31/106

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


33/106

( ) 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


34/106

099 22

2

2

=+= ydx

yd or y

dx yd


35/106

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

+=

=


36/106

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


37/106

( ) ( )( ) +++

=++

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


38/106

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


39/106

Then ( ) ( ) ++=+ C a

baxgdxbax f


40/106

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 =


41/106

( ) 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


42/106

( ) ( )

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 ==


43/106

( )[ ]C

n xg

C nu

duunn

n ++

=++

== ++

11

11


44/106

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


45/106


46/106

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


47/106

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 =


48/106

( ) 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


49/106


50/106

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.


51/106

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


52/106

++

=

==

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


53/106

( )( )

C x

x x +

++=35

73log

253

2log253


54/106

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


55/106

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).


56/106

( ) ( )

+

+

++

+=

++

+

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


57/106

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:


58/106

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


59/106

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


60/106

+

=+==

= 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


61/106

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


62/106

=

+

=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


63/106

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


64/106

( )( )

++

+++

+=

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.


65/106

( ) 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+

=

+

==


66/106


67/106

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 ==


75/106

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

+

++=++


78/106

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


80/106

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.


81/106

(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)


85/106

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 =


86/106

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


87/106

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).


89/106

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


90/106

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


92/106

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


94/106

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.


95/106

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


96/106

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


97/106

(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

math zc161 lecture notes

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