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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
SEMESTER / SESSION: SEM II/ 2011/2012 PROGRAMME : 2/3 BDD/BFF
COURSE :ENGINEERING MATHEMATICS IV CODE : BWM 30603/ BSM 3913
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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BWM 30603 /BSM 3913
7
FINAL EXAMINATION
SEMESTER / SESSION: SEM II/ 2011/2012 PROGRAMME : 2/3 BDD/BFF
COURSE :ENGINEERING MATHEMATICS IV CODE : BWM 30603/ BSM 3913
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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12
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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BWM 30603 /BSM 3913
15
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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16
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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