St. JOSEPHS COLLEGE OF ENGINEERING, CHENNAI- 119

St. JOSEPHS INSTITUTE OF TECHNOLOGY, CHENNAI- 119

I-YEAR B.E./ B.TECH. ( COMMON TO ALL BRANCHES )

MATHEMATICS I (MA6151)

CYCLE TEST III (QUESTIONS WITH ANSWERS)

UNIT -1 MATRICES

1. If

1

2

1

is an eigen vector of

021

612

322

, find the corresponding eigen value.

Ans: (A - I)X = 0

21

612

322

0

0

0

x

x

x

3

2

1

21

612

322

0

0

0

1

2

1

(2)(1)+2(2)+(-3)(-1) = 0 = 5.

2. Determine so that (x2 + y2 +z2) + 2xy 2xz + 2zy is positive definite.

Ans: The matrix of the given quadratic form is

11

11

11

A

D1 = , D2 = )1)(1(11

1 2

& D3 = |A| = (+1)2( -2)

The Quadratic form is positive definite if D1, D2 & D3> 0 > 2

3. State Cayley Hamilton theorem.

Ans: Every square matrix satisfies its own characteristics equation.

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

312

301

221

Ans: Sum of the eigen values = Sum of the main diagonal elements = 1+0+3 = 4

Product of the eigen values = |A|= 13

5. One of the eigen values of

814

184

4-47

is -9, Find the other two eigen values.

Ans: If 1 , 2be the other two eigen values, then

1 + 2 9 = 7 8 8 = 9 (since sum of the eigen values = sum of the leading diagonal elements) 1 + 2 = 0 =>1 = 2 (1) 912 = |A| = 441 ( since product of the eigen values = | A | )

12 = 49 =>2

1

49

(2)

substitute in (1) we get, 72 (1) => 71 . Hence the other two eigen values are 7 and -7.

UNIT -2 SEQUENCES & SERIES

6. Test the convergence of 1

1

n log n

Ans: The series 1

1

n log n

is a positive term series decreases as n increases after n 2.

So we can apply integral test.

22 2

1 (1/ x)dx dx log (log x)

x log x log x

By integral test, the series diverges.

7. Define Monotonically increasing and Monotonically decreasing sequence with examples

Ans: A sequence { an } is said to be monotonically increasing if an an+1 , for every n

A sequence { an } is said to be monotonically decreasing if an+1 an , for every n

Eg. an = { n } is monotonically increasing

an = { n } is monotonically decreasing

8. Define Absolute convergence.

Ans: If the series of arbitrary terms 1 2 3 .. .nu u u u be such that the series

1 2 3 . nu u u u is convergent, then the series nu is said to be absolute convergent

9. Test the convergence of

n

n1

n! 2

n

Ans: Let

n (n 1)

n n 1n (n 1)

(n 1) nn 1

(n 1) nn nn

n

n

n

n

n! 2 (n 1)! 2u , u

n (n 1)

u (n 1)! 2 nlim lim

u (n 1) n! 2

nlim 2

n 1

1 2lim 2 1

1 e1n

By Ratio Test,

n

n

n! 2

n

converges.

10. Define Conditional convergence with example

Ans: If nu is convergent and | |nu is divergent, then nu is said to be conditionally convergent.

ex: 1 1 1

1 ...2 3 4

UNIT III APPLICATIONS OF DIFFERENTIAL CALCULUS

11. Find the centre of curvature of the curve y = x2 at the point (1,1)

Ans: 2121

1

2 2

1 31 4;

2

yyx x y y y

y y

Centre of curvature is 3

4,2

12. Find the envelope of the family of lines cos sin 1x y

a b , where being the parameter.

Ans: (x/a) cos + (y/b) sin =1 --------(1)

Diff. w.r.to partially, we get

(x/a) sin + (y/b) cos =0 --------(2)

(1)2 + (2)2 2

2

a

x (cos2 + sin2) + 2

2

b

y(sin2 + cos2) =1 2

2

2

2

b

y

a

x = 1

13. Find the curvature of the curve 2 22 2 5 2 1 0x y x y

Ans: Radius of the circle 2 2 25 1 1 21

16 4 2 4f g c radius of curvature

Curvature1 4

21

14.

Find the evolute of the curve whose centre of curvature of the curve is

2 32 3 , 2x a at y at

Ans: 3 2

3262 27 4 23 2

x a yt a y x a

a a

Equation of the evolute is 3227 4 2ay x a

15. For the catenary y = c cosh (x/c), find the curvature.

Ans: 1 sinhx

yc

, 2

1cosh

xy

c c

3 / 2

2

2

2

1 sinh1

1cosh

x

c y c

x c y

c c

UNIT IV DIFFERENTIAL CALCULUS OF SEVERAL VARIABLES

16.

If , ,x y z u y z uv z uvw find ( , , )

, ,

x y z

u v w

Ans: 1 ; 1 ;y uv z uv w x u uv u v z uvw

J =

uvuwvw

uvwuwv

uv

w

z

v

z

u

zw

y

v

y

u

yw

x

v

x

u

x

wvu

zyx

)1()1(

01

),,(

),,( = u2v

17. If cos , sinx r y r find ( , )

( , )

r

x y

Ans:

cos sin( , )

sin cos( , )

rx yr

rr

( , ) 1 1

( , )( , )( , )

r

x yx y rr

18. If y xx y c , find

dx

dy

Ans: Let , y xf x y x y c

/

/

dy f x

dx f y

1

1

log

log

y x

y x

yx y y

x x xy

19. Find Taylors series expansion of e

xsin y near the point 1,4

up to first degree terms.

Ans: f (x , y) = ex sin y , f x(x , y) = ex sin y, fy(x ,y) = e

x cosy

f(1,4

) =2

1

e, fx(1,

4

) =2

1

e, fy(1,

4

)=2

1

e .

f(x ,y) = f(a ,b) + [(xa) f x(a ,b) + (yb) f y(a ,b)] =

4)1(1

2

1 yx

e

20. Find the maxima and minima of f(x, y) = 3x2 + y2 + 12x +36

Ans: fx= 6x +12 = 0 x = 2; fy= 2y = 0 y = 0 .

The stationary point is (2,0).

A = f xx = 6 ,B = f xy = 0 , C = f yy = 2, AC B2 = 12 > 0 and A > 0 .

f is minimum at (2,0) and the minimum value is f (2,0) = 24.

UNIT V MULTIPLE INTEGRALS

21. Evaluate .

22 x

1 0

x dy dx

Ans:

2

22 2

01 0 1

xx

x dy dx x y dx = 2

3

1

15

4x dx

22. Find the area bounded by the lines x = 0, y = 1, y = x using double integration.

Ans: Area = 2

1

2

1

0

21

0

0

1

0 0

ydyxdxdydydx

yy

R

23. Evaluate

a b

2 2

dxdy.

xy

Ans: I = 1 log22

bax

y dy =

y

dylog2][logb

a

2

= ydy

2

blog

a

2

=

2log

b

a2logy

=

2log

b]2log[log a =

2log

b

2log

a

24. Shade the region of integration in

2 2

2

a a x

0 ax x

dxdy.

Ans: y = 2xax x2 + y2 ax 0 which is a circle

with centre at ( a 2 , 0) and radius a 2

y = 22 xa x2 + y2 = a2 which is a circle

with centre at (0,0) and radius a

25. Change the order of integration

1

0

x2

x2

dxdyy),(xf .

Ans:

Given, I = 1

0

x2

x 2

dxdy)y,x(f

After changing order of integration

y 2 y1 2

0 0 1 0

I f(x,y) dx dy f(x,y) dx dy

26. Evaluate R

22 dx)dyy(x over the region R for which x, y 0, x+y 1.

Ans: The region of integration is the triangle bounded by the

lines x 0, y 0 and x y 1

Limits of y : 0 to 1x ; Limits of x : 0 to 1

R

22 dxdy)yx( = 1 1

2 2

0 0

x

x y dydx

= dx3y

yx1

0

x1

0

32

1 32

0

(1 )(1 )

3

xx x dx

13 4 4

0

(1 )

3 4 12

x x x

=

6

1

12

1

4

1

3

1

27. Change the order of integration in

2 2a a x2 2

a 0

(x y )dxdy.

Ans: I = xddyyxax

ax

xay

y

22

0

22 )( ( Correct Form )

order)thechanging(afterdydx)y(x

ay

0y

yax

yax

22

22

22

28. Change the order of integration in

2 y1

0 y

xy dx dy

.

Ans: Given, I = 1

0

2 y

y

dydxxy

After changing order of integration

2

1

2

0

1

0 0

xxdxdyxydxdyxyI

29.

Compute the area enclosed by y2 = 4x, x + y = 3 and y = 0.

Ans: Area A =

R

dydx =

2

0

3

42y

y

yx

dydx = 22

3

/ 40

y

yy

x dy

2

0

322

0

2

1223

43

yyydy

yy

y

= 6 2 12

8 = 4

3

2 =

3

10

30. Evaluate

a sin

0 0

rdr d.

Ans: a

0

sin

0

drdr =

dra

sin

00

2

2

=

ad

0

2

2

sin

=

2

sin2aa

4

1

31. Evaluate

52

0 0

r sin dr d.

Ans: I =

ddrr

0

5

0

2sin =

dr

0

5

0

22

2sin =

d

0

2sin2

25

=

d 0

2cos14

25 =

02

2sin

4

25

=

0

2

2sin

4

25 =

4

25

32. Evaluate

2

2 2 20 0

r dr d.

(r a )

Ans: I =

2

0 0222 )(

ar

ddrr =

22 12 2 22 ( )0 0

d rd

r a =

2

0 022

1

2

1

dar

=

2

02

10

2

1

da

= 2

2 0

1 1

2 a

= 24a

33. Evaluate / 2 sin

0 0

.

r dr d

Ans: / 2 sin

0 0

I r dr d

sin/ 2 2

0 02

rd

/ 2 2

0

sin0

2d

/ 2

2

0

1sin

2d

1 1

.2 2 2

8

34.

Evaluate

cos

0 0

r drd .

Ans: I = cos

2

20 0

r

d = 1

2

0

2cos d = 2

0

221 cos2

d = 1

2 2 4

35. Transform the integration

0

y

0

dydx into polar coordinates.

Ans: Let x = r cos and y = r sin , dxdy = r drd

2

4

00 0

r

yddrrdydx

36. Compute the entire area bounded by r2 = a2 cos2. Ans:

Area A = R

ddrr

4/

0

2cos

0

4

a

r

drdr

cos2/ 4 2

0 0

42

a

rd

= 4

4

0

2

2

2cos

d

a4

2 2

0

sin 22

2a a

37. Transform the integration from Cartesian to polar co-ordinates

22ax x2a 2 2(x y )dxdyx 0 y 0

.

Ans: Let x = r cos and y = r sin,dxdy = r drd

a

x

xax

y

dydxyx2

0

2

0

222

)( =2 cos/ 2

3

0 0

a

r dr d

38. Express the region bounded by 2 2 20, 0, 0, 1x y z x y z as a triple integral.

Ans: Here z varies from 0 to 2 21 x y , y varies from 0 to 21 x , x varies from 0 to 1

2 22 111

0 0 0

x yxI dz dy dx

39. Evaluate

1 1 1x y ze dxdydz.

0 0 0

Ans: I =

1

0

1

0 0

zyx dxdydze1

= dydzee zyzy 1

0

1

0

1

= 1

2 1

0

2z z ze e e dz 323 1133 eeee

40.

Evaluate

x y4 x

0 0 0

z dxdy dz.

Ans: I =

4

0

x

0

yx

0

dxdydzz = dydxzx

yx

4

00 0

2

2 = dydxyx

x 4

002

1

= 24

0 0

1

2 2

x

yxy dx

=

242

0

1

2 2

xx dx

= 1634

3

4

34

0

34

0

2

xdxx