Eng Math 4 2 20112012

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UNIVERSITI TUN HUSSEIN ONN MALAYSIA

FINAL EXAMINATION

SEMESTER II

SESSION 2011/2012

COURSE NAME : ENGINEERING MATHEMATICS IV

COURSE CODE : BWM 30603/BSM 3913

PROGRAMME : 2 BDD/BFF

3 BDD/BFF

EXAMINATION DATE : JUNE 2012

DURATION : 3 HOURS

INSTRUCTION : ANSWER ALL QUESTIONS IN PARTAAND TWO (2) QUESTIONS IN PART B.

ALL CALCULATIONS AND ANSWERS

MUST BE IN THREE (3) DECIMALPLACES.

THIS EXAMINATION PAPER CONSISTS OF SEVEN (7) PAGES

CONFIDENTIAL

CONFIDENTIAL

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PART A

Q1 (a) Given the heat equation

,0,20,),(

2),(

2

2

tx

x

txu

t

txu

with the boundary conditions

,0),2(),0( tutu

and the initial condition

).sin()0,( xxu

Find ( ,0.3)u x by using Implicit Crank-Nicolson method with

5.0 hx and .3.0 kt (10 marks)

(b) The longitudinal vibration of a bar with the length of l m is governed by

2 22

2 2c

x t

withE

c

, where ( , )x t is the axial displacement, E is Youngs

modulus and is the mass density of the bar. The boundary conditions and the

initial conditions are given as follows,

(0, ) ( , ) 0t l t for 0 0.04t

0)0,( x and xt

x

)0,(for .200 x

Determine the variation of the axial displacement of the bar by using finite-

difference method with the following data:630 10E , 0.264 , 20l m, 5 hx and .02.0 kt

(15 marks)

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Q2

01 x 12 x 23 x 44 x

Figure Q2

Consider a fin of length 4 unit has four nodes and three elements, as shown in Figure

Q2. The heat flow equation is given by

,0)()(

)()(

xQ

dx

xdTxkxA

dx

dfor ,40 x

with )(xA is the cross-sectional area, )(xk is the thermal conductivity, )(xT is the

temperature at length x and )(xQ is the heat supply per unit time and per unit length.

Find the temperature at each nodal point, 32 , TT and 4T

, if )(xA is 30 unit, )(xk is 10unit and )(xQ is 10 unit. Let the temperature at 0x is 0 unit and the heat flux,

104

xdx

dTk unit.

(25 marks)

PART B

Q3 (a) Table below shows a set of discrete data.

x 1.1 1.3 1.5 1.7

)(xy 0.907 ? 1.355 1.999

Find the missing value of )(xy by using Lagrange interpolation method.

(10 marks)

(b) Use the result in Q4(a) to find )3.1(y and )3.1(y by using any appropriate

difference formulas with the suitable step size ofh.(5 marks)

(c) Find the approximate value of dxe

x

3

0

2

2

3 by using

(i) Simpsons3

8rule with ,6n and

(ii) 3-points Gaussian Quadrature.

(10 marks)

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Q4 (a) Find the largest eigenvalue and its corresponding eigenvector for the

matrix A below by using power method.

210

232

012

A

Use the initial guess for eigenvector Tv 111)0( . Calculate until

.005.01 kk mm (10 marks)

(b) Approximate the solution at )2.1(y for the initial-value problem

12

)(2

2

xxy

yxy with ,1)1( y using the fourth order Runge-Kutta (RK4)

method with the step size .2.0h (7 marks)

(c) Given the boundary-value problem of

3)1(5 xyxyxy

for 20 x with the following boundary conditions

2)0()0(4 yy , 5)2(2)2(3 yy .Derive the system of linear equation by using finite-difference method with grid

size 5.0 xh . DONOT solve the obtained system.(8 marks)

Q5 (a) Locate the positive root of the nonlinear equation 2 4sin( ) 0x x by using

Intermediate value theorem. Hence, solve it by using Bisection method.

(10 marks)

(b) Given the system of linear equations:

2 4 1 9

2 2 5 8 .

3 1 1 9

x

y

z

Solve it by

(i) Doolittle method, and

(ii) Gauss-Seidel iteration method.

(15 marks)

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FINAL EXAMINATION

SEMESTER / SESSION: SEM II/ 2011/2012 PROGRAMME : 2/3 BDD/BFF

COURSE :ENGINEERING MATHEMATICS IV CODE : BWM 30603/ BSM 3913

FORMULAE

Bisection method:2

i i

i

a bc

Doolittle method

11 12 13 1 11 12 13 1

21 22 23 2 21 22 23 2

31 32 33 3 31 32 33 3

1 2 3 1 2 3

1 0 0 0

1 0 0 0

1 0 0

0

1 0 0 0

n n

n n

n n

n n n nn n n n nn

a a a a u u u u

a a a a l u u u

a a a a l l u u

a a a a l l l u

Gauss-Seidel iteration method:

1( 1) ( )

1 1( 1) , 1,2, ,

i nk k

i ij j ij j

j j ik

i

ii

b a x a x

x i na

Lagrange polynomial

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

n

n i iiP x L x f x i n where 0

( )

( ) ( )

nj

ij i jj i

x x

L x x x

Numerical Differentiation:First derivatives:

2-point forward difference:( ) ( )

( )f x h f x

f xh

2-point backward difference:( ) ( )

( )f x f x h

f xh

3-point forward difference:( 2 ) 4 ( ) 3 ( )

( )2

f x h f x h f xf x

h

3-point backward difference:3 ( ) 4 ( ) ( 2 )

( )2

f x f x h f x hf x

h

3-point central difference:( ) ( )

( )2

f x h f x hf x

h

5-point difference:( 2 ) 8 ( ) 8 ( ) ( 2 )

( )12

f x h f x h f x h f x hf x

h

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FINAL EXAMINATION



Second derivatives:3-point central difference:

2

( ) 2 ( ) ( )( )

f x h f x f x hf x

h

5-point difference:2

( 2 ) 16 ( ) 30 ( ) 16 ( ) ( 2 )( )

12

f x h f x h f x f x h f x hf x

h

Numerical Integration:

Simpsons8

3rule

0 1 2 4 5 2 1 3 6 6 33

( ) 3( ) 2( )

8

b

n n n n n

a

f x dx h f f f f f f f f f f f f

Gauss quadrature

For ( )b

af x dx ,

( ) ( )

2

b a t b ax

3-points:

5

5

9

50

9

8

5

3

9

5)(

1

1

gggdttg

Power Method ( 1) ( )

1

1 , 0,1,2, ...k k

k

v Av k m

Initial value problems

Classical 4th order Runge-Kutta method.

1 1 2 3 4

1( 2 2 )

6i i

y y k k k k

where 1 ( , )i ik hf x y 1

2( , )

2 2i i

khk hf x y

2

3( , )

2 2i i

khk hf x y 4 3( , )i ik hf x h y k

Boundary value problems:

Finite difference method

1 1

2

i i

i

y yy

h

, 1 12

2i i i

i

y y yy

h

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FINAL EXAMINATION



Heat Equation: Implicit Crank-Nicolson method

2

,1,,1

2

1,11,1,12

,1,

2

1,

2

22

2

1,

22

2 h

uuu

h

uuuc

k

uu

x

uc

t

u

jijijijijijijiji

jiji

Wave equation: Finite difference method

)(2

)0,(

22

1,1,

2

,1,,12

2

1,,1,

,

2

22

,

2

2

i

jiji

jijijijijiji

jiji

xgk

uu

t

xu

h

uuuc

k

uuu

x

uct

u

Finite element method:

lb FFKT

where dxdx

dN

dx

dNxkxAK

q

p

jiij )()( is stiffness matrix,

iTT

q

p

ibdx

dTxkxANF

)()(

q

p

ildxxQNF )(

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Answer Scheme for Final Exam BWM30603 / BSM3913 Sem II 11/12

Answer Marks Total

2

2( , ) 2 ( , ), 0 2, 0u x t u x t x t

t x

22

22 , where 2

u uc

t x

2

, 1 , 1, 1 , 1 1, 1 1, , 1,

2 2

2 2

2

i j i j i j i j i j i j i j i ju u u u u u u uc

k h h

, 1 , 1, 1 , 1 1, 1 1, , 1,

2 2

2 22

0.3 2 0.5 0.5

i j i j i j i j i j i j i j i ju u u u u u u u

1, 1 , 1 1, 1 1, , 1,1.2 3.4 1.2 1.2 1.4 1.21.2 1.4 1.2

i j i j i j i j i j i ju u u u u uB C

0,1 1,1 2,11.2 3.4 1.2 1.4u u u 1,1 2,11.2(0) 3.4 1.2 1.4u u

1,1 2,13.4 1.2 1.4u u --------- (1)

1,1 2,1 3,11.2 3.4 1.2 0u u u --------- (2)

2,1 3,1 4,11.2 3.4 1.2 1.4u u u 2,1 3,11.2 3.4 1.2(0) 1.4u u

D1

M1

A1

M1

A1

A1

u0,0

u0,1

u2,0 u3,0 u4,0

u1,1 u2,1 u3,1

3.4

1.2 1.4 1.2

Ti,j+1

(A) ui-1,j (B) ui,j (C) ui+1,j

x

t

0 0.5 1 1.

2

0.3

u1,00

u4,1

1.2 1.2

Ti-1,j+1 Ti+1,j

=

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2,1 3,11.2 3.4 1.4u u --------- (3)

In matrix-vector form:

1,1 1,1

2,1 2,1

3,1 3,1

3.4 1.2 0 1.4 0.412

1.2 3.4 1.2 0 00 1.2 3.4 1.4 0.412

u u

u uu u

(using calculator)

A1

Answer

A3

10

Given2 2 2 2

2 2

2 2 2 2, 0 20 , 0c c x t

x t t x

with Ec

62 10636.113 c

, 1 , , 1 1, , 1,6

2 2

2 2113.636 10

i j i j i j i j i j i j

k h

, 1 , , 1 1, , 1,6

2 2

2 2113.636 10

0.02 5

i j i j i j i j i j i j

)2(182.818,12 ,1,,11,,1, jijijijijiji

1,,1,,11, 182.1818364.3634182.1818 jijijijiji ------ (1)

DCBA 182.1818364.3634182.1818

M1

M1

A1

A1

0.3

t

x

0 -0.412 0 0.412 0

0 0.5 1 1.5 2

0 1 0 -1 0

u0,1 u1,1 u2,1 u3,1 u4,1

u0,0 u1,0 u2,0 u3,0 u4,0

1

1,818.182 -3,634.364 1,818.182

i,j+1

=

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(D) i,j-1

xt

x

)0,(

, 1 , 1 , 1 , 1

2 2(0.02)

i j i j i j i jx x

k

, 1 , 1 0.04i j i j x ------ (2)

Substitute eqn. (2) into eqn. (1):

)04.0(182.1818364.3634182.1818 1,,1,,11, xjijijijiji

xjijijiji 02.0091.909182.1817091.909 ,1,,11,

xCBA 02.0091.909182.1817091.909

M1

A1

M1

A1

Initial

+boundary

Value

A1

(A) i-1,j (B) i,j (C) i+1,j

-1

1

909.091 -1,817.182 909.091

=

0.02x

(A) i-1,j (B) i,j (C) i+1,j

i,j+1

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3211

6 points

A1 each

15

Interpolating function for the first and third element

1N 2N

3N 4N

otherwise,0

10,1)(

11

1xxN

xN

otherwise,0

10,111

1 xdx

dN

dx

dN

otherwise,0

10,)(

1

22

xxNxN

otherwise,0

10,1

1

22 x

dx

dN

dx

dN

otherwise,0

42,25.0)(

3

3

3

xxNxN

otherwise,0

42,5.0

3

33 x

dx

dN

dx

dN

D1

M1

M1

M1

3,2 0.04

0.02

0,2

t

x

0 0.1 0. 2 0.3

0 5 10 15 20

0 0 0 0

0,1

0,0

2,24,2

1,1 2,1 3,1 4,1

1,0 2,0 3,0 4,0

1,2

0 0.2 0.4 0.6

0 1

2 4

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otherwise,0

42,15.0)(

3

44

xxNxN

otherwise,0

42,5.03

44 x

dx

dN

dx

dN

Find the components of the stiffness matrix,

q ji

ij p

dNdN

K Ak dxdx dx

For first element, )1(K

1

0

11 300)1)(1(300 dxK , 1

0

1

0

2112 300)1)(1(300)1)(1(300 dxdxKK

1

0

22 300)1)(1(300 dxK

300300

300300)1(K

For second element, )1()2( KK because of the same length.For third element, )3(K

4

2

33 150)5.0)(5.0(300 dxK

4

2

4

2

4334 150)5.0)(5.0(300)5.0)(5.0(300 dxdxKK

4

2

4 150)5.0)(5.0(300 dxK

150150

150150)3(K

Hence assemble K for the whole domain will be

15015000

1501503003000

0300300300300

00300300

K

15015000

1504503000

0300600300

00300300

Find the components of boundary vector, ( )

q

b i

p

dTF N Ak

dx

For first element

1

0)1(

30

30

x

xb

dx

dTk

dx

dTk

F

For second element, )1()2(bb

FF .

A1

M1A1

A1

A1

M1A1

M1A1

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For third element,

4

2)3(

30

30

x

xb

dx

dTk

dx

dTk

F

Hence, assemble

300

0

0

300x

b

dx

dTk

F

Find the components of load vector, q

p

ildxxQNF )(

First Element

5

5

1

101

0

1

0)1(

dxx

dxx

Fl

Second Element,)1()2(

llFF

Third Element,

10

10

15.0

25.0

104

2

4

2)2(

dxx

dxx

Fl

Hence assemblel

F

10

15

10

5

10

105

55

5

lF

Therefore, b lKT F F

10

15

10

5

300

0

0

30

15015000

1504503000

0300600300

003003000

4

3

2

1xdx

dTk

T

T

T

T

1

0T , partition the matrix, the system of linear equations is

M1

A1

M1A1

M1 A1

M1

A1

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290

15

10

1501500

150450300

0300600

4

3

2

T

T

T

733.3

800.1

883.0

4

3

2

T

T

T

M1

A1A1

A1

25

x 1.1 1.3 1.5 1.7

)(xy 0.907 ? 1.355 1.999

Lagrange polynomial:0

( ) ( ) ( ), 0,1,2, ,n

n i i

i

P x L x f x i n

where0

( )( )

( )

nj

i

j i jj i

x xL x

x x

2211003 )3.1( fLfLfLP

)907.0(

)7.11.1)(5.11.1(

)7.13.1)(5.13.1(

)355.1(

)7.11.1)(1.15.1(

))7.13.1)(1.13.1()99.1(

)1.17.1)(5.17.1(

)1.13.1)(5.13.1(

=0.991

M4

A4

Ans A2 10

x 1.1 1.3 1.5 1.7

)(xy 0.907 0.991 1.355 1.999

2-point forward difference:( ) ( )

( )f x h f x

f xh

= 82.1

2.0

991.0355.1)3.1('

y

2-point backward difference:( ) ( )

( )f x f x h

f xh

= 420.0

2.0

907.0991.0)3.1('

y

3-point forward difference:

( 2 ) 4 ( ) 3 ( )( )

2

f x h f x h f xf x

h

= 120.1

)2.0(2

)991.0(3)355.1(4999.1)3.1('

y

3-point central difference:( ) ( )

( )

2

f x h f x hf x

h

= 120.1

)2.0(2

9070.0355.1)3.1('

y

3-point central difference:

A1

A1

A1

A1

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2

( ) 2 ( ) ( )( )

f x h f x f x hf x

h

= 025.61

)2.0(

)907.0991.0(2355.1)3.1(''

2

y

A1

5

i x y=3*exp(-(x^2)/2)

0 0 3.000

1 0.5 2.6472 1 1.820

3 1.5 0.974

4 2 0.406

5 2.5 0.132

6 3 0.033

3.033 5.005 0.974

0 1 2 4 5 2 1 3 6 6 33

( ) 3( ) 2( )8

b

n n n n na

f x dx h f f f f f f f f f f f f

ans= 3.749

Each sum

M3

Ans A2 5

3

0

2

2

3 dxe

x

For ( )b

af x dx , 5.15.12

3)03(

t

tx dtdx 5.1

1

1

2

)5.15.1(3

0

2

22

5.1*33 dtedxe

tx

1

1

5 3 8 5 3( ) 0

9 5 9 9 5f x dx g g g

1

1

2

)5.15.1( 2

5.4 dte

t

= 733.3)029.0(9

5)325.0(

9

8)944.0(

9

55.4

A1M1

M1

A2

5

(i)

2 -1 0

A -2 3 -2

0 -1 2

k vk Avk mk+1

0 1.000 1.000 1.000 1.000 -1.000 1.000 1.000

1 1.000 -1.000 1.000 3.000 -7.000 3.000 -7.000

2 -0.429 1.000 -0.429 -1.857 4.714 -1.857 4.714

3 -0.394 1.000 -0.394 -1.788 4.576 -1.788 4.576

4 -0.391 1.000 -0.391 -1.781 4.563 -1.781 4.5635 -0.390 1.000 -0.390 -1.781 4.562 -1.781 4.562

M1A1

M4

A1

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6 -0.390 1.000 -0.390

Since 001.056 mm , the dominant eigenvalue, 562.46arg mestL and

eigenvector, TestL vv )390.0000.1390.0()4(

arg

A1

A1A1

10

1 1 2 3 4

1

( 2 2 )6i iy y k k k k

where 1 ( , )i ik hf x y 1

2( , )

2 2i i

khk hf x y

2

3( , )

2 2i i

khk hf x y 4 3( , )i ik hf x h y k

i x y k1 k2 k3 k4

0 1 1 0.400 0.371 0.371 0.354

1 1.2 1.373

M4A1

Ans A2 7

3)1(5 xyxyxy 1 12

i i

i

y yy

h

1 12

2i i i

i

y y yy

h

2)0()0(4 yy

5)2(2)2(3 yy

3

11 )4()53()4( iiiiii xyxyxyx

i x 4-x y -3+5x y 4+x y x3

0 0 4.000 y-1 -3.000 y0 4.000 y1 0.000

1 0.5 3.500 y0 -0.500 y1 4.500 y2 0.125

2 1 3.000 y1 2.000 y2 5.000 y3 1.000

3 1.5 2.500 y2 4.500 y3 5.500 y4 3.375

4 2 2.000 y3 7.000 y4 6.000 y5 8.000

7

375.3

1

125.0

8

28000

5.35.45.200

05230

005.45.05.3

000813

4

3

2

1

0

y

y

y

y

y

M3

A5

(A1

each row) 8

Let 2( ) 4sin( )f x x x

(0) 0, (1) 2.3658, (2) 0.3628f f f , since (1) (2) 0,f f so there is a root in the

interval [0, 1] by Intermediate value theorem.

i a c b ( )f a ( )f c ( )f b

0 1 1.5 2 -2.366 -1.740 0.363

1 1.5 1.75 2 -1.740 -0.873 0.363

2 1.75 1.875 2 -0.873 -0.301 0.363

3 1.875 1.938 2 -0.301 0.023 0.363

4 1.875 1.907 1.938 -0.301 -0.139 0.0235 1.907 1.923 1.938 -0.139 -0.057 0.023

6 1.923 1.931 1.938 -0.057 -0.015 0.023

M1M1

M1M1M1

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7 1.931 1.935 1.938 -0.015 0.007 0.023/0.020

8 1.931 1.933 1.935 -0.015 -0.004 0.007

Since 8| ( ) | 0.004 0.005,f c so the root is 8 1.933c .

Or if ( )f c tekan button answer

i a c b ( )f a ( )f c ( )f b

0 1 1.5 2 -2.366 -1.740 0.363

1 1.5 1.75 2 -1.740 -0.873 0.363

2 1.75 1.875 2 -0.873 -0.301 0.363

3 1.875 1.938 2 -0.301 0.020 0.363

4 1.875 1.907 1.938 -0.301 -0.142 0.020

5 1.907 1.923 1.938 -0.142 -0.059 0.020

6 1.923 1.931 1.938 -0.059 -0.017 0.020

7 1.931 1.935 1.938 -0.017 0.004 0.020

Since 7| ( ) | 0.004 0.005,f c so the root is 7 1.935c .

A4

A1

10

2 4 1 9

2 2 5 8 .

3 1 1 9

x

y

z

Step 1: From ,A LU solve forL and U.

2 4 1 1 0 0

2 2 5 1 0 0

3 1 1 1 0 0

d e f

a g h

b c i

=

ichbfcgbebd

hafgaead

fed

Compare element by element .

2d 4e 1f

2ad 1a

2ae g

2g 5af h

4h

3bd 1.5b

1be cg

2.5c

1 ichbf

1 1.5 10 12.5i So LUA ,

2 4 1 1 0 0 2 4 1

2 2 5 1 1 0 0 2 4

3 1 1 1.5 2.5 1 0 0 12.5

Step 2: From ,LY b solve forYby forward substitution.

1

2

3

1 0 0 9

1 1 0 8

1.5 2.5 1 9

y

y

y

9

1

20

Y

Step 3: From YUx , solve forx by backward substitution.

M4

M1

A1

7/28/2019 Eng Math 4 2 20112012

18/18

BWM 30603 /BSM 3913

18

1

2

3

2 4 1 9

0 2 4 1

0 0 12.5 20

x

x

x

3.7

3.7

1.6

x

M1

A1

8

(b)(ii) Rearrange it

3 1 1 9

2 4 1 9 .

2 2 5 8

x

y

z

[ ] [ ] [ 1] [ ] [ 1] [ ]

[ 1] [ 1] [ 1]9 9 2 8 2 2

, ,3 4 5

k k k k k k

k k ky z x z x y

x y z

i x y z

0 0 0 0

1 -3 -3.75 -1.3

2 -3.8167 -3.8333 -1.5933

3 -3.7467 -3.725 -1.6087

4 -3.7054 -3.7006 -1.6020

5 -3.6995 -3.6993 -1.6001

6 -3.6997 -3.7 -1.6

6 -3.7 -3.7 -1.6

Since [7] [6]max{ } 0.0003 0.0005x x , so x=-3.7, y=-3.7 and z=-1.6.

A1

M3

M2

A1

7

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Eng Math 4 2 20112012