UNIT-I (2marks questions)

1. Find the characteristic equation of the matrix 1 2

0 2

� �� �� �

.

Sol.

The characteristic equatin of A is 0A Iλ− =

2

2

1 2 1 00

0 2 0 1

1 20

0 2

(1 )(2 ) 0 0

2 2 0

3 2 0

λ

λλ

λ λλ λ λ

λ λ

� � � �− =� � � �

� � � �−

=−

− − − =− − + =

− + = The required characteristic equation is 2 3 2 0λ λ− + = .

2. Obtain the characteristic equation of 1 2

5 4

−� �� �−� �

.

Sol.

Let A=1 2

5 4

−� �� �−� �

The characteristic equation of A is 21 2 0c cλ λ− + =

1

2

1 4 5

1 2

5 4

4 10

6

c sumof the maindiagonal elements

c A

== + ==

−=

−= −= −

2

2

(5) ( 6) 0

5 6 0

Hencethecharacteristic equationis

λ λλ λ

− + − =− − =

3. Find the sum and product of the eigenvalues of the matrix 1 1 1

1 1 1

1 1 1

−−

−.

Sol.

( 1) ( 1) ( 1)

3

1 1 1

1 1 1

1 1 1

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

1(0) 1( 2) 1(2)

4

sumof theeigenvalues sum ofthe diagonal elements

product of theeigenvalues

== − + − + −= −

−= −

−= − − − − − + += − − − +=

4. Two eigen values of the matrix

11 4 7

7 2 5

10 4 6

− −� � − −� � − −�

are 0 and 1,

find the third eigen value. Sol.

1 2 3

1 2 3

3

3

0, 1, ?

11 ( 2) ( 6)

0 1 3

2

Given

sumof theeigenvalues sum of the main diagonal elements

λ λ λ

λ λ λλλ

= = ==

+ + = + − + −+ + =

=

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5. Verify the statement that the sum of the elements in the diagonal of a matrix is the sum of the eigenvalues of the matrix

2 2 3

2 1 6

1 2 0

.

( 2) (1) (0)

1

2 2 3

2 1 6

1 2 0

2(0 12) 2(0 6) 3( 4 1)

24 12 9

45

sol sum of theeigenvalues sumof the maindiagonal elements

product of theeigenvalues

− −� � −� � − −�

== − + += −

− −= −

− −= − − − − − − += + +=

6. The product of the eigenvalues of the matrix

6 2 2

2 3 1

2 1 3

A

−� � = − −� � −�

is 16, Find the third eigenvalue. Sol.

1, 2 3

1 2

1 2 3

, .

16

6 2 2

2 3 1

2 1 3

let theeigenvalues of the matrix Abe

Given

we knowthat A

λ λ λλ λ

λ λ λ=

=

−= − −

−6(9 1) 2( 6 2) 2(2 6)

6(8) 2( 4) 2( 4)

= − + − + + −= + − + −

3

3

32

16 32

2

λλ

===

7. Two eigenvalues of the matrix

1 2 3

1 2 3

3

3

1 2

8 6 2

6 7 4 3 0. ?

2 4 3

. 3, 0, ?

. .

8 7 3

3 0 18

15

are and what is the product of theeigenvalues of A

sol given

w k tThe sum of theeigenvalues sumof the main diagonal elements

productofeigenvalues

λ λ λ

λ λ λλλ

λ λ

−� � − −� � −�

= = ==

+ + = + ++ + =

== 3 (3)(0)(15) 0λ = =

8. Find the sum and product of the eigen values of the matrix 2 0 1

0 2 0

1 0 2

.

2 2 2

6

2 0 1

0 2 0

1 0 2

2(4 0) 0(0) 1(0 2)

8 2

6

sol sumof theeigenvalues sum of the main diagonal elements

product of theeigenvalues A

� � � � �

== + +==

� � = � � �

= − − + −= −=

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9.Find the characteristic equation of the matrix 1 2

0 2

� �� �� �

and get its

eigenvalues.Sol. Given is a upper triangular matrix. Hence the eigenvalues are 1,2 W.k.t the chacteristic equation of the given matrix is

2

2

2

( ) ( ) 0

(1 2) (1)(2) 0

3 2 0

sumof theeigenvalues product of theeigenvaluesλ λλ λλ λ

− + =− + + =− + =

10.Prove that if λ is an eigenvalues of a matrix A, then 1

λ is the

eigenvalue of 1A−

1

1 1

1

1

1

1

;

,

1

1. ,

proof

If X betheeigenvector corresponding to

then AX X

premultiplying bothsides by A weget

A AX A X

IX A X

X A X

X A X

i e A X X

λλ

λλλ

λ

λ

−

− −

−

−

−

−

=

===

=

=

11.Find the eigenvalues of A given

3 3 3

1 2 3

0 2 7

0 0 3

.

1 2 3

0 2 7

0 0 3

1,2,3

1,2,3

1 ,2 ,

A

sol

A

clearly given Ais aupper triangular matrix

Hencetheeigenvalues are

theeigenvalues of the given matrix Aare

By the property theeigenvalues of the matrix A are

� � = −� � �

� � = −� � �

33 .

12.If α and β are cthe eigen values of 3 1

1 5

−� � −�

form the

matrix whose eigenvalues are 3α and 3β

Sol.

2 2 2

1 7 5

0 2 9 0

0 0 5

(1 )[(2 )(5 ) 0] 7[0 0] 5[0 0] 0

(1 ) (2 )(5 ) 0

1, 2, 5

1 2 5

30

sumof theeigenvalues

λλ

λλ λ λ

λ λ λλ λ λ

−− =

−− − − − − − + − =

− − − == = =

= + +=

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13.Sum of square of the eigenvalues of

1 7 5

0 2 9

0 0 5

� � � � �

is……..

Sol.

The characteristic equatin of A is 0A Iλ− =

2 2 2

1 7 5

0 2 9 0

0 0 5

(1 )[(2 )(5 ) 0] 7[0 0] 5[0 0] 0

(1 ) (2 )(5 ) 0

1, 2, 5

1 2 5

30

sumof theeigenvalues

λλ

λλ λ λ

λ λ λλ λ λ

−− =

−− − − − − − + − =

− − − == = =

= + +=

14 .two eigenvalues of A=

4 6 6

1 3 2

1 5 2

� � � � − − −�

are equal and they are

double the third.Find the eigenvalues of A. Sol.

2 2 2 2

2 ,2

2 2 (4) (3) ( 2)

5 5

1

2,2,1

2 ,2 ,1

Letthethirdeigenvaluebe

Theremainingtwoeigenvaluesare

sumftheeigenvalues sumofthemaindiagonalelements

theeigenvaluesofAare

HencetheeigenvaluesofA are

λλ λ

λ λ λλλ

=+ + = + + −

==

15.show that the matrix 1 2

2 1

−� � �

satisfies its own characteristic

equation.

Sol.

1 2

2 1LetA

−� = �

� The cha.equation of the given matrix is

21 2

1

2

2

2

2

2

0

0

1 1 2

1 21 4 5

2 1

2 5 0

2 5 0

.

1 2 1 2

2 1 2 1

3 4

4 3

3 4 1 2 1 02 5 2 5

4 3 2 1 0 1

0 0

0 0

A I

S S

S sumof main iagonal elements

S A

Thecharacteristic is

Toprove A A I

A A A

A A I

λ

λ λ

λ λ

− =

− + === + =

−= = = + =

− + =− + ==

− −=

− −� �= � �−� �

− − −� � � � � �− + = − +� � � � � �−� � � � � �

� �= � �� �

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16.If A=1 0

4 5

� �� �� �

express 3A in terms of A and I using Cayley

Hamilton theorem.

Sol.The cha.equation of the given matrix is 0A Iλ− =

2

1 0 1 00

4 5 0 1

1 00

4 5

(1 )(5 ) 0 0

(1 )(5 ) 0

6 5 0

λ

λλ

λ λλ λ

λ λ

� � � �− =� � � �

� � � �−

=−

− − − =− − =

− + =

By Cayley Hamilton theorem,

2 2

3 2

3 2

6 5 0, 6 5

6 5 0

6 5

6(6 5 ) 5

36 30 5

31 30

A A I A A I

multiply Aon both sides

A A A

A A A

A I A

A I A

A I

− + = = −

− + == −= − −= − −= −

17.Write the matrix of the quadratic form 2 22 8 4 10 2x z xy xz yz+ + − − .

Sol.

Q=

2

2

2

1 1

2 21 1

2 21 1

2 2

coeff of x coeff of xy coeff of xz

coeff of xy coeff of y coeff of yz

coeff of xz coeff of yz coeff of z

� � � � � � � � �

Q=

2 2 5

2 0 1

5 1 8

� � −� � −�

18.Determine the nature of the following quadratic form

( ) 2 2

1 2 3 1 2, , 2

. .

f x x x x x

sol The matrix of Q F is

= +

Q=

2

2

2

1 1

2 21 1

2 21 1

2 2

coeff of x coeff of xy coeff of xz

coeff of xy coeff of y coeff of yz

coeff of xz coeff of yz coeff of z

� � � � � � � � �

=

1 0 0

0 2 0

0 0 0

� � � � �

There for the eigenvalues are 0,1,2. so find the eigenvalues one eigenvalue is Zero another two eigenvalues are positive .so given Q.F is positive semi definite.

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19. State Cayley Hamilton theorem. Every square matrix satisfies its own characteristic equation.

20. Prove that the Q.F 2 2 22 3 2 2 2x y z xy yz zx+ + + + − .

Sol.The matrix of the Q.F form,

Q=

2

2

2

1 1

2 21 1

2 21 1

2 2

coeff of x coeff of xy coeff of xz

coeff of xy coeff of y coeff of yz

coeff of xz coeff of yz coeff of z

� � � � � � � � �

=

1 1 1

1 2 1

1 1 3

−� �� �� �� �−� �

1 1

1 12

2 2

1 1 1

3 2 2 2

3 3 3

1 1( )

1 1(2 1) 1( )

1 2

1(6 1) 1(3 1) 1(1 2) 2( )

D a ve

a bD ve

a b

a b c

D a b c ve

a b c

= = = +

= = = − = +

= = − − + − + = − −

The Q.F is indefinite.

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UNIT II - SEQUENCES AND SERIESPart A

1. Given an example for (i) convergent series (ii) divergent series (iii) oscillatory series

Solution:

(i) The series

+ is convergent

(ii) 1+2+3+….+n+… is divergent

(iii) 1-1+1-1+…… is oscillatory

2. State Leibnitz’s test for the convergence of an alternating series

Solution:

The series a1-a2+a3-a4+…. In which the terms are alternately +ve and –ve and all ai’s are positive, is convergent if

(i) and

(ii)

3. State the comparison test for convergence of series

Solution:

Let ∑an and ∑bn be any two series and let a

finite quantity ≠ 0, then the two series converges or diverges together

4. State any two properties of an infinite series

Solution:

(i) The converges or diverges of an infinite series is not affected when each of its terms is multiplied by a finite quantity

(ii) If a series in which all the terms are positive is convergent, the series will remain convergent even when some or all of its terms are made negative

5. Define alternating series

Solution:

A series whose terms are alternatively positive and negative is called alternating series

Eg: + is an alternating series

6. Prove that the series is convergent

Solution:

The nth term of the series is an=

Then an+1 =

now = =

= =0(

Hence by D’Alembert’s test, ∑an is convergent

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7. When is an infinite series is said to be (i) convergent (ii) divergent (iii) oscillatory?

Solution:

Let ∑an be an infinite series and let Sn be the sum of the first n terms of an infinite series then

(i) If is finite the series is said to be

convergent

(ii) If If →±∞ the series is said to be

divergent

(iii) If not tend to a definite limit or ±∞, then

the series is oscillatory.

8. State true or false

(i) If ∑an is convergent, ∑an2 is also convergent.

(ii) If the nth term of a series does not tend to zero as n→∞, the series is divergent.

(iii) The convergence or divergence of an infinite seies is not affected by the removal of a finite number of terms from the beginning

(iv) An absolutely convergent series is convergent

Solution:

All are true.

9. Prove that the series is conditionally

convergent

Solution:

The nth term of the series be an=

Then =1/n and =1/n+1

Since , n+1 n ,

an is decreasing and = =0

By Leibnit’z test, the given series is convergent. Also the series formed by the absolute value of its terms is divergent. Hence the series is conditionally convergent.

10. For what values of p, the series + +…+ +… will be

(i) convergent (ii) divergent

Solution:

The p-series is convergent if p 1 and divergent if

UNIT-IIIDIFFERENTIAL CALCULUS

1) Find the curvature of 2 2 4 6 1 0x y x y+ + − − =Solution:

( )3

2 2 2

2 22x y

xx y xy x y yy x

f f

f f f f f f fρ

+=

− +

f = 2 2 4 6 1x y x y+ + − −

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( ) ( )( ) ( )

( ) ( )( ) ( )

3 32 2 2 22 2

2 2 2 2

2 4 2 6

2 2

0

2 4 2 6 2 4 2 6

2 2 6 0 2 2 4 2 2 6 2 4

x y

xx yy

xy

f x f y

f f

f

x y x y

y x y xρ

= + = −

= =

=

� � � + + − + + −� � � = =� − − + + − + +�

( )

( )( )

( )

( )

2 2 312 2 2

32 2 2

1/ 22 2

2 2

2 2 6 (2 4)12 2 6 (2 4)

2 4 (2 6)

2 2 6 (2 4)

1 2

2 6 (2 4)

y xcurvature y x

x y

y x

y x

ρ

ρ

−

−

� − + +� � = = = − + +� � + + −�

� = − + +�

=− + +

2) What is the formula of radius of curvature in Cartesian

form and parametric form? Sol:

Cartesian form:2 3/ 2

1

2

(1 )y

yρ +=

Parametric form:( ) ( )

3/ 22 2' '

' '' ' ''

x y

x y y xρ

� +� =−

3 Find the radius of curvature at x=0 on xy e=Solution: Given xy e=

Radius of curvature ( )3

2 21

2

1 y

yρ

+=

( ) ( )

01 1 0

02 2 0

3/ 2 3/ 221

2

] 1

] 1

1 1 12 2

1

x

xx

xx

y e

y e y e

y e y e

y

yρ

=

=

== � = =

= � = =

+ +∴ = = =

4 Find the radius of curvature of the curve 2 ( , )xy c at c c= Sol:

( ) ( )

2

2

2 2

1 12 2

2 2

2 23 3

3/ 2 3/ 22 3/ 21

2

( , )

1

2 2 2

1 1 1 .2

2 / 2

2.

Given xy c at c c

cy

x

c cy y

x c

c cy y

x c c

y c

y c

c

ρ

ρ

=

=

− −= � = = −

= � = =

+ += = =

=

5 What is the curvature of the curve 2 2 25x y+ = at the point (4,3) on it.

Sol: Since the given curve is a circle & We know that radius of given circle is 5 units radius of curvature of a circle is equal to the radius of the given circle

5

1 1.

5curvature

ρ

ρ

∴ =

= =

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6 Find radius of curvature of the curve cos ,x a θ= siny b θ= at any point ' 'θ

Sol:

( ) ( )

( )

3/ 2 3/ 22 2 2 2 2 2

2 2

3/ 22 2 2 2

2 2

cos sin

' sin ' cos

'' cos '' sin

' ' sin cos

' '' '' ' sin cos

sin cos( sin cos 1)

x a y b

x a y b

x a y b

x y a b

x y x y ab ab

a b

ab

θ θθ θθ θ

θ θρ

θ θ

θ θρ θ θ

= == − == − = −

+ += =

− +

+= + =Q

7 Find the radius of curvature at any point on the curver eθ= .

Sol:

3/ 22 21

2 21 22

r r

r rr rρ

� +� =− +

1 2&

Given r e

r e r e

θ

θ θ

== =

( ) ( )( ) ( )

( )( ) ( ) ( )

( )( )

( )

3/ 2 3/ 22 2 2

2 2 2 2 2

33/ 2 31

22

2

2 2

22 2

2

2.

e e e

e e e e e e e

ee e

e

r

θ θ θ

θ θ θ θ θ θ θ

θθ θ

θ

ρ

ρ

−

� � � +� � � � � � = =− + − +

= = =

=

8 Find the radius of curvature at y=2a on the curve 2 4y ax= Sol:

( )( )

( )

( )

( )

( )

2

3/ 221

2

1

1

1

1

22 1 1 2 1

4 1

1

1 . . ,

2 4

2 2

2

22 1

22 . . ,

0

Given y ax

yFormula

y

diff w r to x

yy a

yy a

ay

y

ay at y a

adiff w r to x

yy y y yy y

ρ

= �

+=

== �

� =

= = =

+ = � = −

( )

21

2

2 2 1/ 2

yy

y

y at y a a

−=

= = −

( )( ) ( ) ( ) ( )

( )

3/ 2 3/ 23/ 2

5/ 2

1 1 22 2

1/ 2 1/ 2

2

aa a

a

ρ+

∴ = = = −− −

= −

i.e. 5/ 22 . 4 2a aρ = =

9 Find the radius of curvature at (a,0) on 3 3

2 a xy

x

−=

Sol: Given 3 3

2 a xy

x

−=

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32 2

3

1 2

3

1 2

2 2

2

ay x

x

ayy x

x

a xy

x y y

= −

−= −

−= −

1

2 3 3

2 2

( ,0)

.2 . 0 3

at a y

dxHence we find dy

xy a x

dx dxx y y x

dy dy

∴ = �

� = −

+ = −

2 2

2 2

2 ( 3 ) 0

2

3

( ,0) 0

dxxy y x

dy

dx xy

dy x y

dxat a

dy

+ + =

−=+

� �� =� �� �

( ) ( )

( )( ) [ ]

( )

[ ]2

2 2

2

22 2 2

22 3

22 42

3/ 22

3/ 2

2

3 2 2 2 6 2

3

3 0 0 2 0 6 2( ,0)

9 33 0

11 0 3

2 23

3

2

dx dxx y y x xy x y

dy dyd x

dy x y

a ad x aat a

dy a aa

dxdy

ad x

ady

a

ρ

ρ

� � � + − − − − +� � � � � � =

+

+ − − − −= = =+

� � �+� � �� � +� −� ∴ = = =

−� �� �� �

=

10 Find the radius of curvature at 2

xπ= on the curve

4sin sin 2y x x= − . Sol:

1

4sin sin 2

4cos 2cos 2

y x x

dyy x x

dx

= −

= = −

2

2 2

1

2

4sin 4sin 2

/ 2, 4(0) 2cos 2

/ 2, 4(1) 4sin 4

d yy x x

dxat x y

at x y

π ππ π

= = − +

= = − == = − + = −

3/ 2 3/ 22 21

2

1 1 (2)

4

y

yρ

� � � + +� � � ∴ = =−

( )

( )

3/ 2 3/ 2 1/ 21 4 5 5.5 5 5

4 4 4 4

5 5

4is veρ ρ

+ −= = = =− − −

∴ = +Q

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11 Define the curvature of a plane curve and what is the curvature of a straight line

Sol:

The curvature of a plane curve at Kd

ds

ψ=

The curvature of a straight line is zero.12 Find the radius of curvature at any point (x,y) on the

curve log secx

y cc

� �� �= � �� �� �� �

Sol:

( )

1

22

3/ 2 3/ 2

2 23/ 221

2 22

3

2

logsec

1 1. tan .sec tansec

1sec

1 tan sec1

1 1sec sec

sec.sec

xy c

c

x x xy c c

x c c c cc

xy

c c

x xy c c

x xyc c c c

xc

cxc

ρ

� �= � �� �� � � � � �= =� � � � � �� � � � � � � �

� �� �� �= � �� �

� � � �� � � �+ � � � �� � � �+ � � � �� � � �∴ = = =� � � �� � � �� � � �

� �� �� �=���

.secx

cc

� �= � �� � ���

13 Find the radius of the curve given by 3 2cos ,x θ= + 4 2siny θ= +

Sol:

( )2

2

23

3/ 2 3/ 22 2 31

3 32

3 2cos 4 2sin

2sin 2cos

2coscot

2sin

1cot

2sin

cos 1cos

2sin 2

1 1 cot cos2

1 1cos cos

2 22

x y

dx dy

d ddy

dx

d y d dy d d

dx d dx dx d

ecec

y ec

y ec ec

θ θ

θ θθ θ

θ θθ

θ θθ θ θ

θ θθ

θ θρθ θ

ρ

= + = +

= − =

= = −−

� �= = −� � −� �−= =

−

� � � + +� � � = = = = −− −

=

14 Write the formula for centre of curvature and equation of circle of curvature.

Sol:

Centre of curvature:( )2

1 1

2

1y yx x

y

+= −

& 2

1

2

(1 )yy y

y

+= +

Circle of curvature: ( ) ( )2 22x x y y ρ− + − =

15 Find the centre of curvature of 2y x= of the origin.Sol:

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The centre of curvature is given by

( )( )( )

( )( )( )

2

1211

2 2

21 2

1

2

2

2

11 ( ) ,

; 2 ; 2.

(0,0), 0

(0,0), 2

01 (0)

2

1 0 1

2 2

(0,0), 0

1(0,0),

21

0,2

yyX x y Y y

y y

Given y x y x y

at y

at y

X x x

Y y y

at X

at Y

Centreof curvature is

+= − + = +

= = ===

∴ = − + =

+= + = +

=

=

� �∴ � �� �16 Write properties of evolutes.

Sol: (i) The normal at any point of a curve touches the evolute at the corresponding Centre of curvature. (ii)The length of an of the evolute is equal to the of curvature at the points on the original curve corresponding to the extremities of the arc

(iii)There is only one evolute, but an infinite number of involutes.

17 Find the envelope of the family of straight lines 2y mx am= + , m being the parameter.

Sol: Given 2y mx am= + Diff. partially w.r.to m, we get,

0 2

2

x am

xm

a

= +−=

2

2

2 2

2

2 2 2

2

2 2

2 4

2 4 4

4

y mx am

x xy x a

a a

x xa

a a

x x xy

a a a

x ay is the required envelope

∴ = +

− −� � � �= +� � � �� � � �−= +

− −= + =

= −18 efine envelope of a family of curves.

Definition: A curve which touches each member of a family of curve is called the envelope of that family curves.

19 efine Evolute and Involute. The locus of the centre of the given curve is called the evolute of the curve. The given curve is called the Involute of its evolute.

20 Find the envelope of the family of lines 2x

yt ct

+ = ,

t being the parameter. Sol: Given family of lines can be written as, 2 2 0yt ct x− + = --------- (1)

The envelope of 2 0At Bt C+ + = is 2 4 0B AC− = From (1) we get A = y, B= -2c, C = x

Putting these values in (2) we get,

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2

2

2

2

( 2 ) 4 0

4 4 0

0

c yx

c yx

c xy

xy c

− − =− =

− ==

This is required envelope.

21 Find the Envelope of the family of Straight linesa

y mxm

= + , where m is a parameter.

Sol:

2

2

(1)

0

aGiven y mx

m

ym m x a

m x ym a

= + − − − − − −

� = +� − + =

This is a quadratic in ‘m2

2

2

4 0

, ,

4 0

4

So the envelope is B AC

Here A x B y c a

y ax

y ax

− == = − =

− ==

22 Find the Envelope of the family of lines cos sin 1,x y

a bθ θ+ = θ

being the parameter Sol:

( )Given, cos sin 1 1

(1) . . ' '

sin cos 0 (2)

x y

a bdiff partially w r to we get

x y

a b

θ θ

θ

θ θ

+ = �

− + = �

2 2

2 22 2

2 22 2

2 2

2 22 2

2 2

(1) (2) ,

cos sin sin cos 1 0

cos sin 2 cos sin0

sin cos 2 cos sin

we get

x y x y

a b a b

x yxy

a b

x yxy

a b

θ θ θ θ

θ θ θ θ

θ θ θ

+

−� � � �+ + + = +� � � �� � � �

+ + �� =��+ + − �

2 22 2 2 2

2 2

2 2

2 2

cos sin sin cos 1

1

x y

y b

x y

a b

θ θ θ θ� � � � + + + =� � �

∴ + =

23 Find the envelope of the straight lines cos sin sec ,x y aα α α+ = whereα is the parameter.

Sol:

Given ( )cos sin sec 1x y aα α α+ = � Dividing equation (1) by cosα we get,

2 2

2 2

sectan sec (1 tan )

cos

tan tan ( ) 0

x y a a a

a y a x

αα α αα

α α

+ = = = +

− + − = Which is a quadratic equation in tanα Here A=a, B=-y, C = (a-x).

2

2

4 0,

4 ( ) 0

B AC

y a a x

− =− − =

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24 Find the envelope of 2 2 2 ,y mx a m b= + + where m is a

parameter. Sol:

2 2 2

2 2 2 2

2 2 2 2 2 2

2 2 2 2 2

( )

2

( ) 2 0

y mx a m b

y mx a m b

y m x mxy a m b

m x a mxy y b

− = +− = ++ − = +

− − + − = Which is a quadratic equation in m. Hence the envelope is 2 4 0B AC− = Here A= ( 2 2x a− ), B=-2xy, C = 2 2y b−

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2

2 2

2 2

4 4( )( ) 0

( )( ) 0

0

. ,

1

x y x a y b

x y x a y b

x y x y b x a y a b

i e b x a y a b

x y

a b

− − − =− − − =

� − − − + =�

+ =

+ =

25 Find the envelope of cos sin ,x y aθ θ+ = where θ is a parameter.

Sol: Given cos sin (1)x y aθ θ+ = � Diff w.r.to θ sin cos 0 (2)x yθ θ− + = � Eliminate θ between (1) and (2)

( ) ( )2 2

2 2 2 2

1 2 ,

( cos sin ) ( sin cos ) 0

we have

x y x y aθ θ θ θ

+

+ + − + = +

2 2 2 2

2

2 2 2 2

cos sin 2 sin cos

sin cos 2 sin cos

x y xya

x y xy

θ θ θ θθ θ θ θ

+ + � =�+ + − �

2 2 2 2 2 2 2

2 2 2

(cos sin ) (sin cos )x y a

x y a

θ θ θ θ+ + + =+ =

26 Find the envelope of the family given by 1

,x mym

= +

m is parameter.Sol:

The given equation can be written as 2 1 0,m y mx− + =

Which is quadratic equation in m ,Here, , , 1A y B x c= = − =

2

2

2

4 0

4 0

4

Hencethe envelopeis B AC

x y

x y

− =− =

=

27 Find the envelope of 21y mx m= + + where m is a

parameter. Sol:

Given 21y mx m= + +

2

2 2

2 2 2 2

2 2 2

1

( ) 1

2 1

( 1) 2 1 0.

y mx m

Squaring onboth sides y mx m

y mxy m x m

m x mxy y

− = +− = +

− + = +− − + − =

2 2

2

2 2 2

1, 2 , 1.

4 0

( 2 ) 4( 1)( 1) 0

Here A x B xy C y

B AC

xy x y

= − = − = −− =

− − − − = 2 2 2 24 4( 1)( 1) 0x y x y− − − =

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UNIT –IV

FUNCTIONS AND SEVARAL VARIABLE

PART-A

1.1 1

cos , . . cot2

x y u uIf u P T x y u

x yx y− � + � �= + =−� � �+� �

.

Proof:

( , ) cos

1hom deg ,

2' .

x yGiven f x y u

x y

As f is ogeneous function of ree n

it is satisfies the Euler s equation

+= =+

=

(cos ) (cos ) 1cos .

2

1( sin ) ( sin ) cos .

2

1 cos.

2 sin

1cot .

2

f fx y nf

x y

u ux y u

x y

u ux u y u u

x y

u u ux y

x y u

u ux y u

x y

� �� + =� �

� �+ =� �� �− + − =� �

� �+ = −� �� �+ = −� �

2.3 3

1tan , . . sin 2x y u u

If u P T x y ux y x y

− � + � �= + =� + � �� .

Solution:

Given 3 3

( , ) tanx y

f x y ux y

+= =+

As f is a homogeneous function of order n=2, it satisfies

the Euler’s theorem.

(tan ) (tan )2 tan .

f fx y nf

x y

u ux y u

x y

� �+ =� �

� �+ =� �

2 2

2

(sec ) (sec ) 2 tan .

sin 12 .

cos sec

u ux u y u u

x y

u u ux y

x y u u

� �+ =� �

� �+ = �� �

2sin2 cos .

cos2sin cos .

sin 2 .

uu

uu u

u ux y u

x y

= �

= �� �+ =� �

3.3 3

log , . . 2x y u u

If u P T x yx y x y

� + � �= + =� + � ��

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Solution: Given 3 3

logx y

ux y

� += � +�

3 3

hom deg 2,

' .

u x yLet f e

x y

As f is ogeneous functionof ree n

it is satisfies the Euler s equation

f fx y nf

x y

+= =+

=

� �� + =� �

( ) ( )

2u u

ue ex y e

x y

� �+ =� �

( ) ( ) 2 .u u uu u

x e y e ex y

� �+ =� �

2.

u ux y

x y

� �+ =� �

Hence the proof.

4. If 2 2( , ) logf x y x y= + , show that 2 2

2 20

f f

x y

� �+ =� �

.

Solution: Given 2 2( , ) logf x y x y= +

( )2 21log

2f x y= +

2 2 2 2

1 2

2

f x x

x x y x y

� = =� + +

( ) ( )

( ) ( )2 22 2 2

2 22 2 2 2 2

.1 2x y x xf y x

x x y x y

+ −� −= =� + +

Similarly, 2

2

f

y

�� ( )

2 2

22 2

x y

x y

−=+

2 2

2 20

f f

x y

� �+ =� �

5. If 1 1sin tan

x yu

y x− −� � � �= +� � � �� �� �

show that 0u u

x yx y

� �+ =� �

Solution: Here u is a homogeneous function of degree n = 0.

Using Euler’s theorem, 0u u

x yx y

� �+ =� �

6. If x y z

uy z x

= + + show that 0u u u

x y zx y z

� � �+ + =� � � .

Solution:x y z

Given uy z x

= + +

2

2

1

.........(1)

1

u z

x y x

u x zx

x y x

u x

y y z

� = −�� = −�� −= +�

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2

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

1

..........(3)

u x yy

y y z

u y

z z xu y z

zz z x

� −= +�

� −= +�� −= +�

.(1), (2) & (3),

0.

Add eqn we get

u u ux y z

x y z

� � �+ + =� � �

7. If ( , ) cos , sinz f x y where x r y rθ θ= = = .Show that22 2 2

2

1z z z z

x y r r θ� �� � � �� � � � � �+ = +� �� � � � � �� � � �� � � � � �� �

Solution: Wkt, z z x z y

r x r y r

� � � � �= +� � � � �

cos sinz z

x yθ θ� �= +

� �

( sin ) ( cos )

z z x z y

x y

z zr r

x y

θ θ θ

θ θ

� � � � �= +� � � � �

� �= − +� �

1

sin cosz z z

r x yθ θ

θ� � �= − +� � �

2 2

2

2 2

222 2

222 2

1. .

cos sin ( sin ) (cos )

cos sin 2 sin cos

sin cos 2 sin cos

z zR H S

r r

z z z z

x y x y

z z z z

x y x y

z z z z

x y x y

z

x

θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

� �� � � �= +� � � �� �� � � �

� � � �� � � �= + + − +� � � �� � � �� � � �

� �� � � �� �= + +� �� �� � � �� � � �

� �� � � �� �+ + −� �� �� � � �� � � �

�� �= � ��

22z

y

� ��+ � �� �� � �

Thus, R.H.S = L.H.S

8. If ( ), , cos , sinu uz f x y x e v y e v= = = show that

2uz z zx y e

v u y

� � �+ =� � � .

Solution: ( ), , cos , sinu uGiven z f x y x e v y e v= = =

z z x z y

u x u y u

� � � � �= � + �� � � � �

cos sinu uz ze v e v

x y

� �= � + �� �

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( )

( )

2 2 2

2 2 2

cos sin

sin cos sin ....(1)

sin cos

sin cos

sin cos cos ....(2)

(1) (2)

u u

u u

u u

u u

u u

z z zy ye v ye v

u x y

z ze v v e v

x y

z z x z y

v x v y v

z ze v e v

x y

z z zx xe v xe v

v x y

z ze v v e v

x y

z zx y

v u

� � �= � + �� � �

� �= +� �

� � � � �= � + �� � � � �

� �= − +� �

� � �= − +� � �

� �= − +� �

+ �� �+� � ( )2 2 2

2

sin cos

.

u

u

ze v v

y

ze

y

Hence proved

�= +�

�=�

9. If log( )u x xy= where 3 3 3 1x y xy+ + = finddu

dx.

Solution:

3 3, log ( ) & 3 1

....(1)

Given u x xy x y xy

du u u dy

dx x y dx

= + + =� �= + �� �

1

( ) log ( )u

x y xyx xy

� = � +�

1 log ( )u

xyx

�∴ = +�

3 3

1

, 3 1

u xx x

y xy y

consider x y xy

� = � � =�

+ + =

Diff. w.r.to ‘x’,

( )

2 2

2 2

3 3 3 3 0

3 3 3 3 0

dy dyx y y x

dx dxdy

x y y xdx

+ + + =

+ + + =

( ) ( )

( )( )

2 2

2 2

2

2

3 3

3 3

(1) 1 log( )

x y x ydy

dx y x y x

x x yduxy

dx y y x

− + − +∴ = =

+ +

+� = + −

+

10. Find 3 3 3dy

when x y axydx

+ =

Solution: Let 3 3( , ) 3f x y x y axy= + −

2 2

2 2

2 2

3 3 ; 3 3

3 3

3 3

f fx ay y ax

x y

fdy x ay x ayx

fdx y ax y axy

� �= − = −� �

�− −�= − = − = −� − −

�

11. Find dy

dx when sin cosy x x y=

Solution:

sin cos

sin cos 0

Given y x x y

y x x y

=� − + =

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( , ) cos sin

cos cos & sin sin

Let f x y x y y x

f fy y x x y x

x y

= −� �∴ = − = − −� �

( )( )

( )

cos cos

sin sin

cos cos

sin sin

fy y xdy x

fdx x y xy

dy y y x

dx x y x

�− −�= − =� − −

�−∴ =

+

12. If 2 2 2u x y z= + + and , sin , cost t tx e y e t z e t= = = find du

dt

with actual substitution.

Solution: Given 2 2 2u x y z= + + , , sin , cost t tx e y e t z e t= = =

[ ]2 2 ( sin cos ) 2 ( cos sin )

2 (sin cos ) ( cos sin )

t t t t t

t

du u dx u dy u dz

dt x dt y dt z dt

x e y e t e t z e t e t

e x y t t z t t

� � �= � + � + �� � �

= + + + −= + + + −

( )

2 2

2 2

2 sin sin cos cos sin cos

2 sin cos

t t t t t t

t t t

e e e t e t t e t e t t

e e e t t

� = + + + −�

� = + +�

2 2t te e� = �

13. Find du

dt if sin ( / )u x y= , where 2,tx e y t= = .

Solution:

. .du u dx u dy

dt t dt y dt

� �= +� �

2

2 2 3

1cos . cos 2

2cos

t

t t t

x x xe t

y y y y

du e e e

dt t t t

� � � � � �−= +� � � � � �� � � � � �� � � �

= −� � � �� � � �

14. If u = f( y –z , z – x , x – y ) find u u u

x y z

� � �+ +� � � .

Solution: ( ), ,Given u f y z z x x y= − − −

,

( 1) (1) .....(1)

Let r y z s z x and t x y

u u r u s u t

x r x s x t xu u

s t

= − = − = −� � � � � � �= � + � + �� � � � � � �

� �= − +� �

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

u u r u s u t

y r y s y t y

u u

r t

� � � � � � �= � + � + �� � � � � � �

� �= + −� �

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

u u r u s u t

z r z s z t zu u

r s

� � � � � � �= � + � + �� � � � � � �

� �= − +� �

(1) (2) (3) 0u u u

x y z

� � �+ + � + + =� � �

15. Find the minimum value of F = x2+y2 subject to the

Constraint x=1.

Solution: Given F = x2+y2

= square of the distance from the origin

The minimum of F is 1.

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16. Define Jacobian.

If u and v are functions of the two independent variables

x and y, then the determinant

u u

x y

v v

y y

� �� �� �� �

is called the Jacobian

of u ,v with respect to x,y.( )( )

,It is denoted by

,

x y

u v

�� .

17. Find the Jacobian( , )

cos , sin( , )

x yif x r y r

rθ θ

θ� = =� .

Solution: Given cosx r θ= siny r θ=

cos

sin

x

ry

rr

θ

θ

� =�� = −�

sin

cos

y

ry

r

θ

θθ

� =�� =�

( )( )

cos sin,

sin cos,

x xrx y r

y y rr

r

θ θθθ θθ

θ

� �−� � �∴ = =

� ��� �

2 2cos sinr rθ θ= +

( )( )( )

2 2cos sin

,

,

r

x yr

r

θ θ

θ

= +

�=

�

18. 2 22 , cos , sin ,If u xy v x y and x r y rθ θ= = − = =

( , )

( , )

u vevaluate

r �

Solution:

( )( )

( )( )

( )( )

2 2

, , ,

, , ,

2

2 2

2 2

u v u v x y

r x y r

u u x xx y rv v y y

x y r

Given u xy v x y

u vy x

x xu v

x yy y

θ θ

θ

θ

� � �= �

� � �

� � � �� � � �= �� � � �� � � �

= = −� �= =� �� �= = −� �

Given cosx r θ= siny r θ=

cos

sin

x

ry

rr

θ

θ

� =�� = −�

sin

cos

y

ry

r

θ

θθ

� =�� =�

( )( )

2 2 cos cos,

2 2 sin cos,

y x ru v

x y rr

θ θθ θθ

−�= �

−�

( ) ( )

( ) ( )2 2 2 2

2 2 2 2

4 4 cos sin

4 cos sin

y x r r

x y r

θ θ

θ θ

= − − � +

= − + � +

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( )( )

2

3

4

,4

,

r r

u vr

r θ

= − ��

= −�

19. If 2 2 ( , )

,( , )

y x u vu v then find

x y x y

�= =�

.

Solution:

2 2

2

2

2

2

2

2

y xGiven u v

x y

u y v x

x x x y

u y v x

y x y y

= =

� �= − =� �� �= = −� �

( )( )

2

2

2

2

2,

, 2

u u y yx yu v x xv vx y x x

x y y y

� �−

� ��= =

� �� −� �

2 2

2 2

2 2y x y x

x y x y

� � � � � �� �= − � − − �� � � � � �� �� � � �� � � �

( )( )

1 4 3

,3

,

u v

x y

= − = −�

= −�

20. If (1 ) ,x u v y uv= − = compute &J J � , and prove . 1J J � = .

Solution: ( )1Given x u v and y uv= − =

( )1x y

v vu u

� �= − =� �

( )( )

,

,

x yu u

v vx x

x y u vJy yu v

u v

� �= − =� �

� �� � �= =

� ��� �

1 v u

v u

− −=

( )( )

( )( )

'

(1 ) ( )

, , 1&

, ,

u v uv

u uv uv

x y u vJ u J

u v x y u

= − − −= − +

� �= = = =

� �

To prove: J .J’ = 1

( )( )

( )( )

'

'

, , 1

, ,

1

x y u vJ J u

u v x y u

J J

� �� = � = �

� �

∴ � =

21. If sin cos , sin sin , cosx r y r z rθ ϕ θ ϕ θ= = = .Find J.

Solution:

sin cos , sin sin , cosGiven x r y r z rθ ϕ θ ϕ θ= = =

( )( )

, ,

, ,

x x x

rx y z y y y

Jr r

y y y

r

θ φ

θ φ θ φ

θ φ

� � �� � �

� � � �= =� � � �

� � �� � �

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( )

2 2 2 2

2 2 2 2

2 2 2 2 2 3 2 2

2 2 2

sin cos cos cos sin sin

sin sin cos sin sin cos

cos sin 0

cos ( cos sin cos cos sin sin )

sin ( sin cos sin sin )

sin cos sin cos sin (sin cos )

sin sin cos

r r

r r

r

r r

r r r

r r

r

θ φ θ φ θ φθ φ θ φ θ φ

θ θ

θ θ θ φ θ θ φθ θ φ θ φ

θ θ φ φ θ φ φ

θ θ

−=

−

� +�= �+ +��

= + + +

= +( )2 sinJ r

θ

θ=

22. Expand ( , ) xyf x y e= in Taylor’s series at (1, 1) up to

second degree.

Solution:

( )( ) ( )( ) ( )( ) ( )( ) ( )2

, int 1, 1

, 1,1

, 1,1

, 1,1

, 1,1

xy

xy

xyx x

xyy y

xyxx xx

Given f x y e and the po a b

f x y e f e

f x y e y f e

f x y e x f e

f x y e y f e

= = =

= � =

= � � =

= � � =

= � � =

( ) ( )

( ) ( )2

, (1) ( ) 1,1 2

, 1,1

xy xyxy xy

xyyy yy

f x y e y e x f e e e

f x y e x f e

= + � = + =

= � =

The Taylor’s series is

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )( ) ( )

2

2

1, , , ,

1

, 2 ,1...

2 ,

x y

xx xy

yy

f x y f a b f a b x a f a b y b

f a b x a f a b y b x a

f a b y b

� = + − + −�

� − + − −� + +� + −�

( ) ( )

( ) ( ) ( ) ( )2 2

1 1 1

11 4 1 1 1

2

xy

x ye e

x x y y

� + − + −� � � �= � �� + − + − − + −� �� �

23. Find the Taylor’s series expansion of ex sin y near the

point 1,4

π� �−� �� � up to the first degree terms.

Solution:

( )

( )

( )

1

1

1

1, sin 1, sin

4 4 2

1, sin 1, sin

4 4 2

1, cos 1, cos

4 4 2

x

xx x

xy y

f x y e y f ee

f x y e y f ee

f x y e y f ee

π π

π π

π π

−

−

−

� �= � − = =� �� �� �= � − = =� �� �� �= � − = =� �� �

The required expansion is

( ) ( ) ( ) ( ) ( ) ( )1, , , ,

1 x yf x y f a b f a b x a f a b y b� = + − + −�

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( ) ( )

( )

, 1, 1 1, 1,4 4 4 4

1sin 1 1

42

x y

x

f x y f x f y f

e y x ye

π π π π

π

� � � � � � � �= − + + − + − −� � � � � � � �� � � � � � � �� � �= + + + −� �� � ��

24. Write condition for finding maxima and minima.

Necessary Conditions:

The necessary conditions for f(x, y) to have a maximum

or minimum at (a, b) are that 0 0 ( , )f f

and at a bx y

� �= =� �

Sufficient Conditions:

( ) ( ) ( ), ; , ,xx xy yyLet r f a b s f a b and t f a b= = =

2( ) 0 0i If rt s and r− > < at (a, b) , then f is maximum and

f (a, b) is maximum value

2( ) 0 0 ( , )ii If rt s and r at a b− > > , then f is minimum and

f(a, b) is minimum value.

2( ) 0iii If rt s− < , then f is neither maximum nor

minimum

at (a, b).

(iv) If rt – s2 = 0 , in this case further investigation are

required.

25. Find the stationary points of

3 3( , ) 3 12 20f x y x y x y= + − − + .

Solution: Given 3 3( , ) 3 12 20f x y x y x y= + − − +

2 23 3 3 12x yf x f y= − = −

2 2

2 2

For stationary points 0, 0

3 3 0 1 1

3 12 0 1 2

x yf f

x x x

y y y

= =

− = � = � = �− = � = � = �

∴The stationary points are (1,2), (1,-2),(-1,2) & (-1,-2).

26. Find the stationary points of 2 2 2z x xy y x y= − + − + .

Solution: Given 2 2 2z x xy y x y= − + − +

2 2 , 2 1x yz x y z x y= − − = − + +

For stationary points 0, 0x yf f= =

2 2 2 1x y and x y− = − = � Solving x =1, y =0

∴ The stationary point is (1,0)

27. Find the maximum and minimum values of

2 2 2x xy y x y− + − +

Solution: 2 2( , ) 2Given f x y x xy y x y= − + − +

2 2 2 1

2 2

1

x y

xx yy

xy

f x y f x y

f f

f

= − − = − + +

= =

= −

At maximum and minimum point: fx = fy = 0

(1,0) may be maximum point or minimum point.

At (1,0): fxx . fyy –( fxy)2 = 4-1 = 3 > 0 & fxx =2 > 0

∴(1,0) is a minimum point

∴ Minimum value = f(1,0) = -1

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28. A flat circular plate is heated so that the temperature

at any point (x,y) is u(x,y) = x2+2y2-x. Find the coldest

point on the plate.

Solution: 2 22Given u x y x= + −

2 1 4

2 4

0

x y

xx yy

xy

u x u y

u u

u

= − =

= =

=

For stationary points 1

0 2 1 02xu x x= � − = � =

0 4 0 0yu y y= � = � =

∴ The point is 1

,02

� �� �� �

At 1

,02

� �� �� �

( ) 28 0& 2 0xx yy xy xxu u u u� − = > = >

The point 1

,02

� �� �� �

is the minimum point.

Hence the point 1

,02

� �� �� �

is the coldest point.

29. Find the shortest distance from the origin to the curve

2 28 7 225x xy y+ + = .

Solution:

Let 2 2 2 2& 8 7 225f x y x xy yφ= + = + + −

( ) ( )2 2 2 28 7 225f x y x xy yλφ λ+ � + + + + −

( )

( )0 1 4 0 (1)

0 4 1 7 0 (2)

x x

y y

f x y

f x y

λφ λ λλφ λ λ

+ = � + + = �

+ = � + + = �

Solving (1) & (2) λ = 1, λ = 19−

21 2 & 5 225

( )

1 2 5, 209

If x y y

no real valueof y

If y x x y

λ

λ

= � = − − =

= − � = � = =

1.1 2

33

x xy

x=

(12. (i) Find the Jacobian ( )( )

, ,

, ,

u v w

x y z

�� , if

, ,x y z u y z u v z u v w+ + = + = =

(ii) If 2 2 , 2 . ( , ) ( , )u x y v xy f x y u vφ= − = = show that

2 2 2 2

2 22 2 2 2

4( )f f

x yx y u v

φ φ� �� � � �+ = + +� �� � � �� �

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UNIT-VMULTIPLE INTEGRALS

PART-A

1. Evaluate 1

0 0

xy xdx e dy� �

Sol:

Let I = 1

0 0

xy xe dydx� �

= 1

0 01

xy x axaxe e

dydx e dxx a

� � �=� ��

� � �� Q

1

00

10

0

1

0

1

0

1

0

( )

( )

( 1)

( 1)

xy x

x x

xe dx

xe xe dx

xe x dx

x e dx

e x dx

� = �

= −

= −

= −

= −

�

�

�

�

�

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12

0

( 1)2

1( 1) 0

2

1

2

xe

e

e

� = − �

�

� = − −� � −=

2. Evaluate 1 1

b a dxdy

xy� � Sol:

[ ] [ ][ ] [ ]

1 1

1 1

1 1

log

log log

log log1 log log1

(log 0)(log 0) ( log1 0)

log log

b a

a b

a b

dxdy dxLet I x

xy x

dx dy

x y

x y

a b

a b

a b

� �= =� �� �� � = � � � �

=

= − −= − − ==

� � �

� �

Q

Q

3. Evaluate 2 2

0 0

a a x

dydx−

� �

Sol:

[ ]

( )

2 2

2 2

0 0

00

2 2

0

2 2

0

2 2 21

0

2 2 2 21 1

1

1

0

sin2 2

sin (1) 0 sin (0)2 2 2

sin 0 0 sin (0) 0,

sin 1 sin (1)2 2

a a x

aa x

a

a

a

Let I dydx

y dx

a x dx

a x dx

x a x a x

a

a a a a a

π π

−

−

−

− −

−

−

=

=

= − −

= −

� −= +� � �

� � � �−= + − +� � � �� � � �� �� �= � =� �� �= � =� �� �

� �

�

�

�

Q

( )

2

2

0 0 02 2

4

a

a

π

π

� �� �= + − +� �� �� �� �

=

4. Evaluate 1

0 0

( )x

xy x y dxdy+� � Sol:

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( )

( ) ( )

1

0 0

12 2

0 0

12 2

0 0

1 2 2 3

0 0

2 321

0

( )

( ) ( )

2 3

2 3

x

x

x

x

Let I xy x y dxdy

x y xy dxdy

x y xy dydx correct form

x y xydx

x x x xdx

= +

= +

= +

� = +�

�

� � = +� � �

� �

� �

� �

�

�

( )1 3 5 2

3 2 5 2

0

1 13 5 2

0 0

1 14 7 2

0 0

2 3

2 3

1 1

2 4 3 7 2

x xdx x x x

x xdx dx

x x

� = + =�

�

= +

� � � = +� � �

� � �

�

� �

Q

( )1 1 1 20 1 0

2 4 3 7

1 2

8 2121 16

16837

168

� � � �= − + −� � � �� � � �

= +

+=

=

5. Evaluate 2

2 20 0

y dxdy

x y+� � Sol:

2

2 20 0

2

2 20 0

21

1 0

21 1

1

21

1

2

1

1tan

1tan tan (0)

1tan (1)

1

4

y

y

y

dxdyLet I

x y

dxdy

x y

xdy

y y

ydy

y y

dyy

dyy

π

−

− −

−

=+

=+

� � �= � � �

� ��

� � �= −� � �

� ��

=

=

� �

� �

�

�

�

�

[ ]

2

1

1

4

log 2 log1 log 2 ( log1 0)4 4

dyy

π

π π

=

= − = =

�

Q

6. Evaluate 2 cos

2

0 0

a

r drdπ θ

θ� � Sol:

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2 cos

2

0 0

cos2 3

0 0

2 3 3

0

2 233

0 0

3

3

3

cos

3

1 3......1,

2cos cos1 33

...... ,2 2

3 1

3 3

2

9

a

a

n

Let I r drd

rd

ad

n nif n is odd

a n ndn n

if n is evenn n

a

a

π θ

θπ

π

π π

θ

θ

θ θ

θ θ θπ

=

� = �

�

=

� − − �� � �� �� �−� �= = � �− −� �� �� �� �−�� �

−� = � �

=

� �

�

�

� �Q

7. Evaluate sin2

0 0

r d dr

πθ

θ� � Sol:

( )

( )

sin2

0 0

sin2

0 0

sin

22

0 0

22

0

222

0 0

2

sin0

2

1 3......1,

1 2sin sin1 32

...... ,2 2

1.

2

n

Let I r d dr

r dr d correct form

rd

d

n nif n is odd

n ndn n

if n isevenn n

πθ

πθ

θπ

π

ππ

θ

θ

θ

θθ

θ θ θπ

=

=

� = �

�

� = −�

� �

� − − �� � �� �� �−� �= = � �− −� �� �� �� �−�� �

=

� �

� �

�

�

� �Q

1.

2 2 8

π π=

8. Evaluate cos

0 0

r dr dθπ

θ� � Sol:

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( )

[ ]

cos

0 0

cos2

0 0

2

0

2

0

2

0

0

0

2

cos0

2

cos

2

1cos

2

1 1 cos 2

2 2

1 sin 2

4 2

1 sin 2 0 10 0

4 2 2 4 4

r

r

Let I r dr d

rd

d

d

d

d

θπ

θπ

π

π

π

π

π

θ

θ

θθ

θ θ

θ θ

θ θ

θθ

π ππ π

=

=

=

� = �

�

� = −�

� �

=

=

+=

� = +� �

� � � � �= + − + = + =� � � �� � � � ��

� �

�

�

�

�

�

9. Why do we change the order of integration in multiple integrals?Justify your answer with an example?

Sol :

Some of the problems connected with double integrals,which seen to be complicated,can be made easy to handle by a change in the order of integration.

Example:

0

y

x

edxdy is difficult to solve

y

� � −

� �

But by changing the order we get,

[ ]0 0

00

0

0

( 0)

y y

yy

y

y

edxdy

y

ex dy

y

ey dy

y

e dy

� −

� −

� −

�−

=

� = �

�

= −

=

� �

�

�

�

( )0

0

0

1

( ) (0 1) 1

y

y

e

e

e e

�−

�−

−�

� = � −�

= −

= − − = − − =

10. Express 2

2 20 ( )

a a

y

xdxdy

x y+� � in polar co-ordinates

Sol:

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The region of integration is bounded by 0, , , .y y a x y x a= = = =

Let us transform this integral in polar co-ordinates by taking cos , sin ,x r y r dxdy rdrdθ θ θ= = = .

Consider the limits , , 0x y x a y= = = .

0 sin 0 0,sin 0

0, 0

If y r r

r

θ θθ

= � = � = =� = =

coscos sin 1

sintan 1

4

coscos

sec

If x y r r

aIf x a r a r

r a

θθ θθθπθ

θθ

θ

= � = � =

� =

� =

= � = � =

� =

( ) ( )

sec2 24

3 22 2 2 20 0 0

4 sec 3 2

3 22 2 20 0

4 sec2

0 0

( cos )

cos sin

cos

(cos sin )

cos

a a a

y

a

a

x r rdrd

x y r r

r drd

r

drd

πθ

π θ

π θ

θ θ

θ θ

θ θθ θ

θ θ

=+ � +�

=� +�

=

� � � �

� �

� �

11 .Find dxdy� � over the region bounded by 0, 0, 1x y x y� � + � Sol: Given 0, 0 & 1x y x y� � + � The region of integration is the triangle.

Here x varies from 0 1x to x y= = − y varies from 0 1y to y= =

[ ]

11

0 0

11

00

1

0

12

0

(1 )

2

1 11

2 2

R

y

y

I dxdy

dxdy

x dy

y dy

yy

−

−

=

=

=

= −

� = −�

�

= − =

� �

� �

�

�

12. Find the area of a circle of radius “a” by double integration in polar Co-ordinates Sol: The equation of circle whose radius is “a” is given by

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2 cosr a θ=

The limits for

: 0 2 cos

: 0 2

r r to r a

to

θθ θ θ π

= == =

2 2 cos

0 0

2 cos2 2

0 0

22 2

0

22 2

0

2

2 2

2

2

22

4 cos

4 cos

2 14

2 2

14

2 2

a

a

Area upper area

rdrd

rd

a d

a d

a

a a

π θ

θπ

π

π

θ

θ

θ θ

θ θ

π

π π

= �

=

� �= � �

� �

=

=

−� �= � �� �� �= =� �� �

� �

�

�

�

13. Define Area in polar Co-ordinates

Sol:

Area=R

rdrdθ� �

14. Express the Volume bounded by 0, 0, 0 1x y z and x y z� � � + + �

in triple integration.

Sol: For the given region

2 22

2 2

2

11 1

0 0 0

var 0 1

var 0 1

var 0 1

x yx

z ies from to x y

y ies from to x

x ies from to

I dzdydx− −−

− −

−

∴ = � � �

15. Evaluate 2 3 2

2

0 1 1

xy z dzdydx� � � Sol:

2 3 22

0 1 1

2 3 22

0 1 1

2 3 22 3 2

0 1 12 3 2

4 27 1 4 10

2 3 3 2 2

26 3(2) 26

3 2

Let I xy zdxdydz

xdx y dy zdz

x y z

=

� � � � = � � � �

� � � �

� � � � � = � � � � �

� � � � �

� � � � � �= − − −� � � � � �� � � � � �� � � �= =� � � �� � � �

� � �

� � �

16. Find the volume of the region bounded by the surface 2 2,y x y x= = and the planes 0, 3z z= = Sol:

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( )

2

2

4

4

3

(1)

(2)

(2) (1)

0

1 0

0,1

y x

x y

Substituting in we get

x x

x x

x x

x

= − − − − −= − − − − −

=� − =

� − =

� =

[ ]

[ ]

2

2

2

2

1 3

0 0

13

00

1

0

1

0

12

0

Re

3

3

3

x

x

x

x

x

x

x

x

quired volume dzdydx

z dydx

dydx

y dx

x x dx

=

=

=

=

� = −�

� � �

� �

� �

�

�

13 2 3

0

13 2 3

0

33 2 3

23

3 3

2(1) 13

3

2 1 1

x x

x x

� = −�

�

� = −�

�

−� = � � = − =

17. Sketch roughly the region of integration for 1

0 0

( , ) .x

f x y dy dx� � Sol:

The region of integration is bounded by 0, 1, 0,x x y y x= = = =

var 0 1

var 0

Here x ies from x to x

y ies from y to y x

= == =

18. Sketch the region of integration

2 2

20

.a a x

ax x

dydx−

−� �

Sol: Given x varies from x = 0 to x = a

y varies from 2 2 2y a x to y ax x= − = − i.e., 2 2y x a+ = which is a circle with centre (0,0) and radius a.

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2 22 2 2

2 22

02 4

. .,2 4

a ax y ax x y

a ai e x y

� �+ = � − − + =� �� �

� �− + =� �� �

This is a circle with centre (a/2,0) and radius a/2.

19. Change the order of integration in 0 0

( , )a x

f x y dydx� � Sol:

Given 0 0

( , )a x

f x y dydx� � The region of integration is bounded by

0, , 0,x x a y y x= = = =

. ., var 0

var 0

i e x ies from x to x a represents Vertical path

y ies from y to y x represents Vertical strip

= == =

Now changing the order of integration we get

var

var 0

x ies from x y to x a represents Horizontal strip

y ies from y to y a represents Horizontal path

= == =

0 0 0

( , ) ( , )a x a a

y

f x y dydx f x y dx dy∴ =� � � �

20. Sketch roughly the region of integration for the following double

integral 2 2

0 0

( , )a a x

f x y dxdy−

� � Sol:

2 2

2 2 2

2 2 2

var 0

var 0

. .,

Given that x ies from x to x a

y ies from y to y a x

i e y x a

x y a

= =

= = −= −+ =

Which is a circle with centre (0,0) and radius a

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21.Change the order of integration in11

0 0

( , )y

f x y dxdy−

� �

Sol:

var 0 1 . ., 1

var 0 1

Given x ies from x to x y i e x y represents Horizontal strip

y ies from y to y represents Horizontal path

= = − + == =

The region of integration is bounded by 0, 1, 0, 1y y x x y= = = + =

var 0 1

var 0 1

x ies from x to x represents Vertical path

y ies from y to y x represents Vertical strip

= == = −

After changing the order of integration limits of x and y becomes0, 1, 0 1x x y and y x= = = = − .

11 1 1

0 0 0 0

. ., ( , ) ( , )y x

i e f x y dxdy f x y dydx− −

=� � � �

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