FIRST SEMESTER MATHEMATICS-1 TWO MARKS WITH ANSWERS REGULATION 2013

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UNIT-I (2marks questions)

12

1. Find the characteristic equation of the matrix2.

Sol.0

A I 0

The characteristic equatin of A is

1210

021 0

0

1 2 0

02

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

The required characteristic equation is 2 3 2 0 .

12

2. Obtain the characteristic equation of5.

Sol.4

12

Let A=5

4

The characteristic equation of A is 2 c1 c2 0

c1 sumof the maindiagonal elements

1 4 5

c2A2

1

54

4 10

6

Hencethecharacteristic equationis 2 (5) (6) 0 2 5 6 0

3. Find the sum and product of the eigenvalues of the matrix 1 1 1111 .

111

Sol.

sumof theeigenvalues sum ofthe diagonal elements

(1) (1) (1)

3

1 1 1 product of theeigenvalues 1 1 1 1 1 1

1(1 1) 1( 1 1) 1(11)

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

4

1147

4. Two eigen values of the matrix72

5 are 0 and 1,

1046

find the third eigen value.

Sol.

Given 1 0, 2 1, 3 ?

sumof theeigenvalues sum of the main diagonal elements

123 11 ( 2) (6)

0 1 3 3

3 2



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 223

21

6

120

sol .sum of theeigenvalues sumof the maindiagonal elements

(2) (1) (0)

1

223

product of theeigenvalues 216

120

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

24 12 9

45

622

23

6. The product of the eigenvalues of the matrix A 1

213

is16, Find the third eigenvalue. Sol.

let theeigenvalues of the matrix Abe 1, 2 , 3.

Given12 16

we knowthat 12 3 A2

62

231

213

6(9 1) 2( 6 2) 2(2 6)

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

32

163 32

3 2

7. Two eigenvalues of the matrix

862

674 are 3and 0.what is the product of theeigenvalues of A?

24

3

sol .given1 3, 2 0, 3 ?

w.k .tThe sum of theeigenvalues sumof the main diagonal elements 1 2 3 8 7 3

3 0 3 18

3 15

productofeigenvalues 123 (3)(0)(15) 0

8. Find the sum and product of the eigen values of the matrix

201

020

102

sol .sumof theeigenvalues sum of the main diagonal elements

2 2 2

6

product of theeigenvalues A1

20

02

0

102

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

8 2

6



9.Find the characteristic equation of the matrix

eigenvalues.Sol.Given is a upper triangular matrix. Hence the eigenvalues are 1,2

1 2 and get its 0 2

11.Find the eigenvalues of A given

123

A 027

003

sol.

1 2 3

W.k.t the chacteristic equation of the given matrix is

2 (sumof theeigenvalues ) ( product of theeigenvalues) 0 2 (1 2) (1)(2) 0 2 3 2 0

10.Prove that if is an eigenvalues of a matrix A, then1is the

eigenvalue of A1

proof ;

If X betheeigenvector corresponding to

then AX X

premultiplying bothsides by A 1, weget

A 1 AX A 1 X

IX A 1 X

X A 1 X

1X A 1 X

i.e, A 1 X 1X

027

A

003

clearly given Ais aupper triangular matrix Hencetheeigenvalues are1,2,3 theeigenvalues of the given matrix Aare1,2,3By the property theeigenvalues of the matrix A 3 are13 ,23 , 33.

31

12.If and are cthe eigen values of15form the

matrix whose eigenvalues are 3and 3

1 75

02 9 0

005

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

(1 )(2 )(5 ) 0

1, 2, 5

sumof theeigenvalues 12 22 52

30



175

13.Sum of square of the eigenvalues of02

9 is..

005

Sol.

1 The characteristic equatin of A isA I 0

75

02 9 0

005

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

1, 2, 5

sumof theeigenvalues 12 22 52

30

466

132

14 .two eigenvalues of A=are equal and they are

1 5 2

double thethird.Find the eigenvalues of A. Sol.Letthethirdeigenvaluebe

Theremainingtwoeigenvaluesare2 ,2 sumftheeigenvalues sumofthemaindiagonalelements

2 2 (4) (3) (2) 5 5

1 theeigenvaluesofAare2,2,1

HencetheeigenvaluesofA2 are 22 ,22 ,12

15.show that the matrix12

2satisfies its own characteristic

equation.1

Sol.12

LetA

21

The cha.equation of the given matrix is

A I 0

2 S S2 0

1

S1 sumof main iagonal elements

1 1 2

S2 A1 2 1 4 5

21

Thecharacteristicis 2 2 5 0

Toprove A2 2A 5I 0

A2 A.A

1212

2121

34

43

A2 2A 5I 3 4 21 21 0

5

43

2 10 1

00

00



10

16.If A=45express A3 in terms of A and I using Cayley

Hamilton theorem.A I 0

Sol.The cha.equation of the given matrix is

1010

451 0

0

1 0 0

5

(1 )(5 ) 0 0 (1 )(5 ) 0 2 6 5 0

By Cayley Hamilton theorem,

A 2 6A 5I 0, A 2 6A 5I multiply Aon both sides

A3 6A 2 5A 0

A3 6A 2 5A

6(6A 5I ) 5A

36A 30I 5A

31A 30I

17.Write the matrix of the quadratic form

2x 2 8z 2 4xy 10xz 2 yz .

Sol.11

coeff of x2

2 coeff of xy2 coeff of xz

11

coeff ofy2

Q=2 coeff of xy2 coeff of yz

1 coeff of xz1 coeff of

yzcoeff of z2

22

225

Q=20

1

518

18.Determine the nature of the following quadratic form

f x1 , x2 , x3 x12 2x22

sol .The matrix of Q.F is

1Q=1

coeff of x2


1 coeff of xy1 coeff of

coeff ofy 2yz

22


yzcoeff of z2

22

100

02

=0

000

There for the eigenvalues are0,1,2. so find the eigenvalues one

eigenvalue is Zero another two eigenvalues are positive .so given Q.F is positive semi definite.



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

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

Sol.The matrix of the Q.F form,

coeff of x211


11

coeff ofy2

Q=2 coeff of xy2 coeff of yz


yzcoeff of z2

22

111

121

=

113

D1 a1

D2 a1 a2

a1 D3 a2 a3

1 1(ve)

b1 1 1 (2 1) 1(ve) b2 1 2

b1 c1

b2 c2 1(6 1) 1(3 1) 1(1 2) 2(ve) b3 c3

The Q.F is indefinite.



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 Leibnitzs 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 ais are positive, is convergent if

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

(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:

The nth term of the series is an=

Then an+1 =

now = =

= =0(

Hence by DAlemberts test, an is convergent



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.

The nth term of the series be an=

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

Since , n+1n ,

an is decreasing and = =0

By Leibnitz 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 p1 and divergent if

(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:

UNIT-IIIDIFFERENTIAL CALCULUS

1) Find the curvature of x 2 y 2 4 x 6 y 1 0

Solution:

f x3

2 f y2 2

f xx f y 2 2 f xy f x f y f yy fx2

f = x2 y2 4x 6 y 1



f x 2x 4f y 2 y 6

f xx 2f yy 2

fxy 0

223223

22

2x 4 2 y 6 2x 4 2 y 6

2 2 y 6 2 0 2 2x 422 2 y 6 2 2x 4 2

12 2 y 62 (2x4)2

curvature 3 2 2 y 6 2 (2x 4

2 (2 y 6)22

2x 4

21/ 2

2 2 y 6 (2x4)2

12 2 y 6 2 (2x 4)2

2) What is the formula of radius of curvature in Cartesian form and parametric form? Sol:

Cartesian form: (1 y12 )3/ 2 y2

Parametric form: x ' 2 y ' 2 3/ 2 x ' y '' y ' x ''

3 Find the radius of curvature at x=0 on y ex

Solution:

Given y ex

3Radius of curvature 1 y12 2 y2

y ex

y e xy ]x0e01

11

y 2 e xy 2 ]x0 e01

1 y12 3/ 2113/ 2

2 2

y21

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

Sol:

Given xy c 2 at ( c , c)

y c2

x

y c 2yc21

x 2c2

11

y 22c 2y22c22

x 3c 3c

1 y12 3/ 2113/ 2c.23/ 2

y 22 / c2

c2.

What is the curvature of the curve x 2 y2 25 at the

point (4,3) on it. Sol:

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

curvature 1 15 .



6 Find radius of curvature of the curve x a cos , y b sin at any point ' '

Sol:

x a cosy bsin

x ' a siny ' b cos

x '' a cosy '' b sin

x '2 y '2 3/ 2 a 2 sin2 b2 cos2 3/ 2

x ' y '' x '' y 'ab sin2 ab cos2

a 2 sin2 b2 cos2 3/ 2(Qsin2 cos2 1)

ab

Find the radius of curvature at any point on the curve r e .

Sol:3/ 2 r 2 r12 r 2 rr1 2r22

Given r er e&r e

12

e2 e 23/ 2 e 23/ 2

2

e 2 e e 2 e 2 e2 e 2 2 e 2

3/ 233

2 e 22 1 ee

2

e 2

2

2.r

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

Sol:

Given y 2 4ax1

Formula 1 y12 3/ 2 y2

diff 1 w.r .to x, 2 yy1 4ayy1 2a 2

y2a

1y

y1 at y 2a 2a 1

diff 2 w.r .to x,2a

yy2 y y 0 yy2y 2

111

y2 y 2

1

y

y 2 at y 2a 1/ 2a

113/ 223/ 2 2a 23/ 2

1/ 2a

1/ 2a

25/ 2 a

i.e. 25/ 2.a 4a

2

9 Find the radius of curvature at (a,0) on y2 a 3 x3 xSol: Given y2 a 3 x3 x


y 2 a3 x2

x

2 yya3 2x

x2

1

ya 3x

2x 2 yy

1

at (a ,0) y1

Hence we finddxdy

xy 2 a 3 x3

x. 2y y 2 . dydx 0 3x2 dydx 2xy ( y 2 3x2 ) dydx 0

dx2xy

dy3x 2 y2

at (a,0)dx0

dy


3x2 y2 2 ydxdx 2 y

d2xdy2x 2xy 6xdy

dy2 3x 2 y2 2

at (a,0)d 2 x 3a 2 0 0 2a 0 6a32

dy 2 3a2 029a 43a

1dx2 3/ 2

dy1 03/ 23

a

d2x22

dy23a

32 a

10 Find the radius of curvature at x 2 on the curve

y 4sin x sin 2x .

Sol:

y 4sin x sin 2x

y1 dydx 4cos x 2cos 2x

y 2 d 2 y 4sin x 4sin 2x dx2at x / 2, y1 4(0) 2cos 2 at x / 2, y2 4(1) 4sin 43/ 21 (2)23/ 2

1 y12

y24

1 43/ 253/ 25.51/ 2 5

5

4444

5

5Q is ve

4



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

The curvature of a plane curve at K dds

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

curve y c logx

sec

c

Sol:

x

y c logsec

c

y1 c.1xx 1 tanx

tan. sec c

xc

secc cc

c

y2 1sec2x

cc

tan2x 3/ 22 x3/ 2

23/ 21sec

cc

1 y1

y21sec2x1sec2x

ccc

c

sec3x

x

c

c. c.sec

xc

sec2

c

13 Find the radius of the curve given by x 3 2cos ,

4 2sin

Sol:

x 3 2cosy 4 2sin

dx2sindy 2cos

d

2cosd

dy cot

dx2sin

d 2 yddyd d cot 1

2

dxdxd2sin

d dx

cos ec21 cos ec3

2sin2

1 y 23/ 23/ 2

1 cot2 3

1cos ec 2

y2 1313

2cos ec 2cos ec

2

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

Centre of curvature: x x y1 1 y12 y2(1 y 2 )

y y

&1

y2

Circle of curvature: x x 2 y y 2 2

15 Find the centre of curvature of y x2 of the origin.

Sol:



The centre of curvature is given by

y11 y1 2

X x 1 ( y1 )2 , Y y

y2

y 2

Given y x 2 ; y 2x ;y2 2.

1

at (0,0),y1 0

at (0,0),y2 2

x 02 1 (0)2 x

X

y 1 0 2 y 1

Y

22

at (0,0),X 0

at (0,0), 1

Y

2

0,1

Centreof curvatureis2

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 y mx am2 , m being the parameter.

Sol:

Given y mx am2

Diff. partially w.r.to m, we get,

0 x 2am

2ax

y mx am2

xx2

y 2ax a

2a

x2 ax2

2a4a2

y x 2x 2x2

2a4a4a

x 2 4ay 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 xt yt 2c ,

t being the parameter. Sol:

Given family of lines can be written as, yt 2 2ct x 0 --------- (1)The envelope of At 2 Bt C 0 is B 2 4AC 0 From (1) we get A = y, B= -2c, C = x

Putting these values in (2) we get,



(2c ) 2 4 yx 0 4c 2 4 yx 0 c 2 xy 0

(1)2 (2)2 we get,

xy2xy222

cos sinsin cos 1 0

abb

a

xy c2

This is required envelope.

21 Find the Envelope of the family of Straight lines a

x 2cos2 y2sin2

a 2b2

x 2 sin2 y2 cos2 a 2 b2

2xy cos sin 0 2xy cos sin

y mx m , where m is a parameter.Sol:

Given y mx a(1)

m

x 2cos2 sin2

y 2

x 2 y2 1 a 2 b2

y2sin2 cos2 1

b2

ym m 2 x a

m 2 x ym a 0 This is a quadratic in m

So the envelope is B 2 4AC 0

Here A x , B y , c a

y 2 4ax 0

y 2 4ax

22 Find the Envelope of the family of lines ax cos by sin 1, being the parameter

Sol:

Given,xcos ysin 11

ab

' ' we get

diffpartially (1) w.r .to

x sin ycos 0(2)

ab

23 Find the envelope of the straight lines

x cos y sin a sec , where is the parameter.

Sol:

Given x cos y sin a sec 1 Dividing equation (1) by cos we get,

x y tan a cossec a sec2 a(1 tan2 ) a tan2 y tan2 (a x) 0

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

B 2 4AC 0,

y 2 4a ( a x) 0



24 Find the envelope of y mx a 2 m 2 b2 , where m is a

parameter.Sol:

mx a 2 m 2 b2 ( y mx ) 2 a 2 m 2 b2 2 m 2 x 2 2mxy a 2 m 2 b2

m 2 (x 2 a 2 ) 2mxy y 2 b2 0

Which is a quadratic equation in m. Hence the envelope is B 2 4AC 0

Here A= ( x 2 a2 ), B=-2xy, C = y 2 b2

4x 2 y 2 4(x 2 a 2 )( y 2 b2 ) 0

x 2 (cos2 sin2 ) y 2 (sin2 cos2 ) a2 x 2 y 2 a2

26 Find the envelope of the family given by x my m1 ,

m is parameter. Sol:

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

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

Hencethe envelopeis B 2 4AC 0

x 2 y 2 (x 2 a 2 )( y 2 b2 ) 0

x 2 y 2 x 2 y 2 b 2 x 2 a 2 y 2 a 2 b2 0

i.e, b 2 x 2 a 2 y 2 a 2 b2

x 2y2 1

22

x 2

27 Find the envelope of

parameter.Sol:

4 y 0 x 2 4 yy mx 1 m2 where m is a

ab

25 Find the envelope of x cos y sin a, where is

a parameter.

Sol:Given x cos y sin a(1)

Diff w.r.to

x sin y cos 0(2)

Eliminate between (1) and (2)

1 2 22, we have

(x cos y sin )2 ( x sin y cos )2 a2 02

x 2 cos2 y 2 sin2 2xy sin cos

2222 a2

xsin ycos

2xy sin cos

Given y mx 1 m2 y mx 1 m2

Squaring onboth sides ( y mx ) 2 1 m2 y 2 2mxy m 2 x 2 1 m2m 2 (x 2 1) 2mxy y2 1 0.

Here A x 2 1, B 2xy , C y2 1. B 2 4AC 0(2xy ) 2 4(x 2 1)( y2 1) 0 4x 2 y 2 4(x 2 1)( y2 1) 0



UNIT IV

FUNCTIONS AND SEVARAL VARIABLE

PART-A

1x yuu1

1. If u cos, P.T . x ycot u .

xy2

x y

Proof:

Given f ( x , y ) cosu x y

xy

As f is hom ogeneous function of deg ree n 1,

2

it is satisfiesthe Euler ' s equation.

xf yfnf

xy

x(cos u ) y(cos u)1 cos u .

x

y2

x ( sin u )u y ( sin u )u1 cos u.

xy

2

xu yu1 cos u .

xy

2 sin u

xu yu1 cot u.

xy

2

As f is a homogeneous function of order n=2, it satisfies the Eulers theorem.

xf yf nf

y

x

x(tan u ) y(tan u) 2 tan u.

xy

x (sec2 u )u y (sec2 u )u 2 tan u.

xy

xu yu 2 sin u1.

xysec2 u

cos u

2 sin ucos2 u.

cos u

2sin u cos u.

xu yu sin 2u.

xy

3. If u logx 3 y 3u yu 2

, P.T. x

x yxy

2. If u tan1x 3 y 3u yu sin 2u .

, P.T . x

xy

x y

Solution:

Givenf ( x , y ) tan u x 3 y3

x y



3322222

x yx y.1 x2x

fyx

Solution:Given u log

x y

2 x 2 y 2 2 x 2 y2 2

x

Similarly,2 fx 2 y2

x 2 y2 2

y2

Letf eux 3 y3

x y

Asf is hom ogeneous functionof deg ree n 2,

it is satisfies the Euler ' s equation.

xf yfnf

xy

2 f2 f 0

x 2y2

5. If u sin1x tan1yxu yu 0

show that

yxxy

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

x(e u ) y(eu ) 2eu

xy

x ( e u )u y ( e u )u 2eu .

y

x

xu yu 2.

y

x

Hence the proof.

4. If f ( x , y ) log x 2 y2 , show that

Solution: Given f ( x , y ) log x 2 y2 f 12 log x 2 y2

f1 2xx

x2x 2 y 2x 2 y2

2 f2 f 0 .

x 2y2

Using Eulers theorem,xu yu 0

xy

6. If u xyzshow that xu yu zu 0 .

yzxz

xy

Solution: Given u xyz

yz

x

u1z

xyx2

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

xyx

ux1

yy2z



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

yyz

z21z2

R.H .S

rr2

uy 1

zz 2

x

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

zzx

Add eqn. (1),(2) & (3), we get

zz

cos sin

xy

z22z

cos

xy

2zz

( sin ) (cos )

xy

2zz

sin2 2sin cos

x

y

xu yu zu 0.

xyz

z

x

22z

sin

y

2zz

cos2 2sin cos

x

y

7. If z f ( x , y )

z2z

wherex r cos , y r sin .Show that

2z21z 2

2

z2

x

2zy

xy

Solution:Wkt,

z

1 z r

rr

zzxz y

rxryr

zcos zsin

xy

zxzy

xy

z( r sin ) z(r cos )

x

y

zsin zcos

y

x

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

8.If z f x , y , x e u cos v ,y e u sin v show that

xz yz e2uz

.

vuy

Solution: Givenz f x , y , x e u cos v , y e u sin v

zzxzy

uxuyu

ze u cos v ze u sin v

xy



yzzye u cos v zye u sin v

uxy

e 2 u sin v cos vz e 2 u sin 2 vz....(1)

xy

zzxzy

vxvyv

zu sin v zucos v

ee

xy

xzzxe u sin v zxe ucos v

v

xy

e 2 u sin v cos vz e 2 ucos 2 vz....(2)

xy

(1) (2)

zz2 u sin22z

x yev cosv

vuy

e2u z y Hence proved. 9. If u x log( xy) wherex 3 y 3 3 xy 1 find du .

dx

Solution:

Given , u x log ( xy ) & x 3 y 3 3 xy 1

duuudy....(1)

dxxy

dx

u x1( y ) log ( xy)

xxy

u 1 log (xy)

x

u x1xx

y

xyy

consider , x 3 y 3 3xy 1

Diff. w.r.to x,

3x 2 3y 2 dy 3y 3x dy 0

dxdx

3x 2 3y 3y 2 3x dydx 0

dy 3 x 2 3y x 2 y

dx3 y 2 3xy 2 x

dux x 2 y

(1)dx1 log( xy) y y 2 x

10. Finddywhenx 3 y 3 3axy

dx

Solution:

Letf ( x , y ) x 3 y 3 3axy

f 3x 2 3ay ;f 3y 2 3ax

xy

dyf x3x 2 3ay x 2 ay

dxf3y 2 3axy 2 ax

y

11. Finddywheny sin x x cos y

dx

Solution:

Giveny sin x x cos y

y sin x x cos y0



Let f ( x , y) x cos y y sin x

f cos y y cos x &f x sin y sin x

x

y

dyfx cos y y cos x

dxfy x sin y sin x

dy cos y y cos x dx x sin y sin x

12.If u x2 y2 z2andx et , y et sin t , z et cos t finddu

dt

withactual substitution.

Solution: Given u x2 y2 z2 , x et , y et sin t, z etcos t

duudxudyudz

dtxdtydtz

dt

2x et 2 y (et sin t et cos t) 2z (et cos t et sin t)

2 et x y (sin t cos t) z (cos t sin t)

2 et et etsin2 t et sin t cos t et cos2 t et sin t cos t

2etet etsin2 t cos2 t

2et2et

13. Find duif u sin (x / y) , where x et , y t2 .

dt

Solution:

duu. dx u. dy

dttdtydt

x1

cos.et cosxx2t

2

yy

yy

duetet2et

dt cost22t3

t

uuu

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

xyz

Solution: Given u f y z, z x, x y

Let r y z, s z x and t x y

uurus ut

xrxs

xtx

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

st

uurusut

yrysyty

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

r

t

uurusut

zrzsztz

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

rs

(1) (2) (3)uuu0

xyz

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.



16. Define Jacobian.

If u and v are functions of the two independent variables

uu

x and y, then thedeterminantxyis called the Jacobian

vv

yy

18. If u 2xy , v x2 y2and x r cos , y r sin ,

evaluate(u, v)

(r, )

Solution:

u, vu, vx, y

r,x, yr,

of u ,v with respect to x,y.

(x, y)17. Find the Jacobian (r, )

It is denoted by x, y

.

u, v

if x r cos , y r sin .

uuxx

xy

r

vvyy

xyr

Given u 2xyv x2 y2

Solution: Given x r cos

x cos

r

y r sin

r

x, yxx

r

r, y

y

r

x, y r, r sin

y sin

r

y r cos

cosr sin

sinr cos

r cos2 r sin2

r cos2 sin2

r

u 2 yv2x

xx

u 2xv 2 y

yy

Given x r cosy r sin

x cosy sin

rr

y r siny r cos

r

u, v2 y2xcosr cos

r, 2x2 ysinr cos

4 y2 4x2 r cos2 r sin2

4x2 y2 r cos2 sin2



4 r 2 r

u , v 4r3

r,

19. If u y 2, v x 2thenfind(u , v).

xy(x , y)

Solution:

Givenu y 2v x2

xy

uy 2v2x

xx 2xy

u2 yvx2

yxyy2

u , vuuy 22 y

xyx

x 2

x , yvv2xx2

xyyy2

y 2x 22 y2x

x2y2xy

1 4 3

u , v 3

x , y

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

Solution: Given x u 1 v and y uv

x 1 v y v

uu

x uyu

v

v

x , yxx

J uv

u , v

yy

uv

1 v

v u

u (1 v ) (uv)

u uv uvu x , y ' u , v1

J u& J u

u , v x , y

To prove: J .J = 1

' x , y u , v1

J Ju

u , v x ,y

u

J J '1

21. If x r sin cos , y r sin sin ,z r cos .Find J.

Solution:

Given x r sin cos , y r sin sin , z r cos

xxx

x , y , z r

J yyy

r , , r

yyy

r



sin cosr cos cosr sin sin

sin sinr cos sinr sin cos

cosr sin0

cos (r 2 cos sin cos2 r2 cos sin sin2 ) r sin (r sin2 cos2 r sin2 sin2 )

r 2 sin cos2 sin2 cos2 r2 sin3 (sin2 cos2 )

r2 sin sin2 cos2

J r2 sin

22. Expand f ( x , y ) exy in Taylors series at (1, 1) up to

second degree.

f x , y f a , b 1f x a , b x a f y a , b y b

1

1f xx a , b x a 2 2 f xy a , b y b x a

...

22

f yy a , b y b

1 x 1 y 1

exy1

ex 12 4x 1y 1y 12

2

23. Find the Taylors series expansion of ex sin y near the

up to the first degree terms.

point 1,4

Solution:

Solution:

Givenf x , y e xy and the po int a 1,b 1

fx , y e xyf1,1e

fxx , y e xy yfx1,1e

f y x , y e xy xf y 1,1e

f x , y exsin y1,

f4

f x x , y exsin y1,

f x4

f y x , y excos y1,

f y4

e1 sin

4 e1 sin

4 e1 cos

4

e12e12e12

f xx x , y e xy y 2f xx 1,1 e

f xy x , y e xy (1) y e xy (x )f xy 1,1 e e 2e

f yy x , y e xy x 2f yy 1,1e

The Taylors series is

The required expansion is

f x , y f a , b 1f x a , b x a f y a , b y b

1



fx , y f 1,x 1ff

x 1,yy1,

4444

exsin y11x 1 y

4

e2

24. Write condition for finding maxima and minima. Necessary Conditions:

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

ff

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

xy

Sufficient Conditions:

Let r fxx a,b ; s fxy a,b and t f yy a,b

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

f (a, b) is maximum value

(ii) If rt s2 0 and r 0 at (a,b) , then f is minimum and f(a, b) is minimum value.

(iii) If rt s2 0 , then f is neither maximum nor minimumat (a, b).

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

required.

25. Find the stationary points of

f (x, y) x3 y3 3x 12 y 20 .

Solution: Given f (x, y) x3 y3 3x 12 y 20

fx 3x2 3 fy 3y2 12 For stationary points fx 0, f y 03x2 3 0 x2 1 x 1 3y2 12 0 y2 1 y 2

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

26. Find the stationary points of z x2 xy y2 2x y .

Solution: Given z x2 xy y2 2x y

zx 2x y 2 , zy x 2 y 1 For stationary points fx 0, f y 0

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

The stationary point is (1,0)

27. Find the maximum and minimum values of x2 xy y2 2x y

Solution: Given f (x, y) x2 xy y2 2x y

fx 2x y 2f y x 2 y 1

fxx 2f yy 2

fxy 1

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



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: Givenu x 2 2 y 2 x

u x 2 x 1u y 4 y

u xx 2uyy 4

uxy 0

For stationary points u x 02x 1 0x1

2

u y 0 4 y 0 y 0

The point is1,0

2

At1,0uu uxy 280&u2 0

xxyyxx

2

The point1,0

2is the minimum point.

1,0

Hence the point2is the coldest point.

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

x 2 8xy 7 y2 225 .

Solution:

Let f x 2 y 2 & x 2 8xy 7 y2 225

f x 2 y 2 x 2 8xy 7 y2 225

f x x 0 1 x 4 y 0(1)

fyy 04 x 1 7y 0(2)

Solving (1) & (2) = 1, = 19

If 1 x 2 y & 5y2 225 (no real valueof y)If 1y 2x x

95, y 20

1.y3x1 x2

x3

(12. (i) Find the Jacobian u , v , w, if

x , y , z

x y z u , y z u v , z u v w

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

2 f2 f2222

4(x y)

x2y2u22

v



UNIT-VMULTIPLE INTEGRALSPART-A

1. Evaluate 1dx x e yx dy

00

Sol:= 1x e y x dydx

Let I

00

=1e y x xaxdx eax

0dydxQ e

1 x 0a

1

xe y x xdx

0

0

1 (xe x x xe 0 )dx

0

1 (xe x ) dx

0

1 x ( e 1) dx

0

(e 1)1 x dx

0


(e 1) x2 1 2 0

(e 1) 1 0 2

e 1

2

2. Evaluate b a dxdy

1 1xy

Sol:

b adxdydx

Let I Q log x

1 1xyx

adx bdy

y

1x 1

log x1a log y1b

log a log1 log b log1

(log a 0)(log b 0) (Qlog1 0)

log a log b

3. Evaluate a a 2 x2 dydx

0

Sol:


Let I a a 2 x2 dydx

0

a y0 a 2 x2 dx

0

a 0dx

a2 x2

0

a

a2 x2 dx

0

x a

x a2 x2a21

sin

22

a

0

2 a222

a a a sin10 a sin1 (0)

(1)

222

sin1(0)0,

Qsin 0 0

sin1sin1(1)

22

a2 0 0

0 22

a2

4

4. Evaluate 1

x xy(x y) dxdy

00

Sol:


Let I 1 x xy(x y) dxdy

0 0

1 x x2 y xy2 dxdy

0 0

1

x (x2 y xy2 ) dydx (correct form)

00

xy3

1x2 y2x

dx

23

00

x2 2x 3

1xx

dx

23

0

1

x3x5 2Q x3 25 2

dxx x

23

0

1x3dx 1 x5 2dx

0203

1x11x7 21

4

23

4 07 20

11 012 0

2431

7

18 212

2116

168

16837


2ydxdy

5. Evaluate 0 0 x2 y2

Sol:y

2dxdy

Let I 0 0 x2 y2

2y

0 0 x2 dxy2dy

21tan1x

1yy

211y

1tan

yy

21tan1 (1) dy

1y

21 dy

1y 4

21dy

41y

log 2 log1

4

2 a cos

6. Evaluater2 drd

00

Sol:

ydy0 tan1 (0) dy

4 log 2 (Qlog1 0)


2 a cos

Let I r2drd

00

2r3 a cos

d

03 0

2a3 cos3 d

03

a3 2

cos3 d

30

a33 1

33

2a3

9

sin

7. Evaluate2

r d dr

00

Sol:


Let I 2 sin r d dr

0

2 sin r dr d correct form

0

sin

2r2

n 1 n 3d

2......1,if nis odd020

n n 2

Qcosn

n 1 n 32 sin 2

0......,if nis even

0d

n n 222

0

n 1 n 3

12 2......1,if nis odd

0 sin20 sinnn n 2

2 dQn 1 n 3

......,if niseven

n n 22

1.1 .

2228

cos

8. Evaluater dr d

00

Sol:



cos

Let I r dr d

00r2 r cos

d

2

0r 0 0 d

cos 2

2

0

cos2 d

02

12

2cos d

0

1 1 cos 2 d

2 02

1sin 2

42

0

1sin 20

4202

1 0

44

e ydxdy is difficult to solve

0 x y

But by changing the order we get, y e y dxdy 0 0 y

eyy x0y dy

0

e y( y 0) dy

0 y

e y dy

0

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:

e y

10

e y 0

(e e0 ) (0 1) 1

10. Express a a2 x22)dxdyin polar co-ordinates

0 y (x y

Sol:



The region of integration is bounded by y 0, y a , x y , x a.Let us transform this integral in polar co-ordinates by taking

x r cos , y r sin , dxdy rdrd .

Consider the limitsx y , x a , y 0 .

Ify 0r sin0r0,sin 0

r0, 0

If x yr cosr sincos1

sin

tan1

a4

If x ar cosar

cos

ra sec

a aasec

x 24(r cos )2 rdrd

2 y22 r sin 2 3 2

0 y x00 r cos

4 asecr 3 cos2 drd

r2(cos2 sin23 2

00 )

4 asec

cos2 drd

00

11 .Find dxdy over the region bounded by x 0, y 0, x y 1

Sol:

Given x 0, y 0 & x y 1

The region of integration is the triangle.

Here x varies from x 0 to x 1 y

y varies from y 0 to y 1

I dxdy

R

1 1 y

dxdy

0 0

1 x 10 y dy

0

1 (1 y ) dy

0

y y2 1 2 0

1 12 12

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


r 2a cos

The limits for

r : r 0 to r 2a cos

: 0to 2

Area 2 upper area

22 a cos

2rdrd

0 2r22 a cos

202d

0

2

4a 2 cos2 d

0

2

4a 2 cos2 d

0

4a22 1

22

4a 2 1 a2 2 2

13. Define Area in polar Co-ordinates Sol: Area= rdrd

R

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

in triple integration.

Sol:For the given region


z var ies from 0 to 1 x 2 y2

y var iesfrom 0 to1 x2

x var iesfrom 0 to 1

1 x 2 y2

I 112

xdzdydx

000

15. Evaluate2 3 2xy 2 z dzdydx

011

Sol:

2 32

Let I xy2 zdxdydz

01 1

232dy2

xdxyzdz

011

x2 2y33 z 22

23

021

1

4 027141

2332

2

(2) 26 3 26 3 2

16. Find the volume of the region bounded by the surface y x2 , y x2 and the planes z 0, z 3

Sol:


y 2 x (1) x 2 y (2)

Substituting (2) in (1) we get x 4 xx 4 x 0

xx3 10

x0,1

13

x

Re quired volume dzdydx

0x20

1

x z 30dydx

0x2

1

x 3dydx

0 x2

1

3 y x2xdx

0

31 x 2 dx

x

0

x 3 2x31

3

3 23

0

2x 3 21

x3

33

3 0

3 2(1) 1 3

2 1 1


17. Sketch roughly the region of integration for 1 x f ( x , y ) dy dx.

0 0

Sol:

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

Here x var ies from x 0 to x 1 y var ies from y 0 to y x

a a 2 x2

18. Sketch the region of integrationdydx.

0ax x2

Sol:

Givenx varies from x = 0 to x = a

y varies from y

a 2 x 2to y ax x2

i.e., y 2 x 2 a which is a circle with centre (0,0)

andradius a.



x2 y2 axa 2a2 y20

x24

a 22a2

i.e., x 2 y4

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

19. Change the order of integration in a x f ( x , y ) dydx

0 0

Sol:

Given a x f ( x , y ) dydx

0 0

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

i.e., x var iesfrom x 0 to x a represents Vertical path

y var iesfrom y 0 to y x represents Vertical strip

Now changing the order of integration we get

x var ies from x y to x a represents Horizontal strip y var ies from y 0 to y a represents Horizontal path

a x f ( x , y ) dydx a a f ( x , y ) dx dy

0 00 y

20. Sketch roughly the region of integration for the following double integral a a 2 x2 f ( x , y ) dxdy 0

Sol:Given that x var ies from x 0 to x a

y var ies from y 0 to y a 2 x2 i.e., y 2 x 2 a2

x 2 y 2 a2

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



11 y

21.Change the order of integration inf ( x , y ) dxdy

00

Sol:

Given x var ies from x 0 to x 1 y i .e., x y 1 represents Horizontal y var ies from y 0 to y 1 represents Horizontal path

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

x var ies from x 0 to x 1 represents Vertical path

y var ies from y 0 to y 1 x represents Vertical strip

After changing the order of integration limits of x and y becomes x 0, x 1, y 0 and y 1 x .11 y1 1 x

i.e.,f ( x , y ) dxdy f ( x , y ) dydx

0000


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FIRST SEMESTER MATHEMATICS-1 TWO MARKS WITH ANSWERS REGULATION 2013