T2 (B) Eng Math 4 2 20102011

7/30/2019 T2 (B) Eng Math 4 2 20102011

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UNIVERSITI TUN HUSSEIN ONN MALAYSIA

FACULTY OF SCIENCE, ARTS & HERITAGE

SEMESTER II 2010/2011

BSM 3913 / BWM 30603 TEST 2 (20%) – Make Up Test DURATION : 75 MIN

ANSWER ALL QUESTIONS. ALL THE CALCULATION MUST BE IN 3 DECIMAL

PLACES. USE 0.005ε = .

Q1 Refer to the table below, by taking the appropriate h , find )65.2( y′ and )65.2( y ′′ by

using 5 point difference formula.

t 2.55 2.60 2.65 2.70 2.75 2.80

)(t y 3.422 3.301 3.237 3.185 3.104 3.063

(4 marks)

Q2 Evaluate

∫ −

π

0 sin2 t

dt by using the suitable Simpson’s rule with 10 subintervals.

(6 marks)

Q3 Approximate ∫

+

2

1

21

dx x

x using 2-points Gauss quadrature.

(6 marks)

Q4 Find the smallest eigenvalue (in absolute value) smallest λ and its corresponding

eigenvector ( )k

v of matrix

1 2 0

2 1 2

1 3 1

A

= −

by using shifted power method. Given the dominant eigenvalue,Largest

2.932λ = and

trial value of ( ) ( )0

0.543 0.728 1T

= −v , iterate until three iterations.

(7 marks)

Q5 Solve the following first order initial value problem (IVP) 22 xy y y −=′ at

1)2.0(0= x with initial condition 1)0( = y by using

(a) second order Taylor’s series method,

(9 marks)

(b) fourth order Runge-Kutta method (RK4). For this method, solve for )2.0( y ,

)4.0( y and )6.0( y only.

(8 marks)

April 2011 STRIKE FOR SUCCESS [Total marks : 40 ]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

7/30/2019 T2 (B) Eng Math 4 2 20102011


LIST OF FORMULA

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

( )12

f x h f x h f x h f x h f 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 h f xh

− + + + − + − − −′′ ≈

Simpson’s1

3rule:

1 2

0

1 2odd even

( ) 4 23

n nb

n i ia

i ii i

h x dx f f f f

− −

= =

≈ + + +

∑ ∑∫

Simpson’s3

8rule:

[ ]0 1 2 4 5 2 1 3 6 6 3

3( ) 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 − − − −≈ + + + + + + + + + + + + +∫ K K

Gauss quadrature: 2-points:

+

−≈∫

−3

1

3

1)(

1

1

g g dx x f

Shifted Power Method: K,2,1,0,1

)(

)1( ==+

+ k m

A

k

k

Shifted k v

v

Second order Taylor series method:

2

1( ) ( ) ( ) ( )2!i i i i

h

y x y x hy x y x+

′ ′′= + +, 0,1,2,i

=K

Fourth-order Runge-Kutta Method: )22(6

143211 k k k k y y ii ++++=+

where ),(1 ii y xhf k = )2

,2

( 12

k y

h xhf k ii ++=

)2

,2

( 23

k y

h xhf k ii ++= ),( 34 k yh xhf k ii ++=

7/30/2019 T2 (B) Eng Math 4 2 20102011


Answer Scheme of Test 2 (Make-up) Sem II 10/11

Q1

667.403.0

422.3)301.3(16)327.3(30)185.3(16104.3)65.2(

017.16.0

422.3)301.3(8)185.3(8104.3)65.2(

=−+−+−

≈′′

−=+−+−

≈′

y

y

M1A1

M1A1 4

Q2 i t ( )t

t f sin21

−=

0

1

2

3

4

5

6

7

8

9

10

01/π 10/2π 1/π 10/3π

10/4π

10/5π

10/6π

10/7π

10/8π

10/9π π

0.500

0.591

0.708

0.840

0.953

1.000

0.953

0.840

0.708

0.591

0.500

1.000 3.862 3.322

( ) ( )[ ] 418.2322.32862.34000.1103

1

sin20

=++

=

−∫π

π

t

dt

M1

M1

M1

M1

M1

A1

6

Q3

2

2

3

2

12)12(

dt dx

t t x

=

+=

++−=

[ ] 830.4512.5149.42

1

3

2

2

3

2

1

23

2

2

31

1

1

)(

2

1

1

22

1

2

=+=

++

+=

++

+=

+

∫

∫∫

−

−

dt t

t

dt

t

t dx

x x

t g

4 4 34 4 21

M2

A1

M1

M1

A1

6

Q4 Shifted 1 Largestor

1 2 0 1 0 0 1.932 2 0

2 1 2 2.932 0 1 0 2 1.932 2

1 3 1 0 0 1 1 3 1.932

A A I A I λ λ = − −−

= − − = − − −

k ( )

T k

v A shifted ( )k v 1+k m

0 0.543 0.728− 1 2.505− 2.320 3.573− 3.573−

1 0.701 0.650− 1 2.654− 1.854 3.181− 3.181−

2 0.834 0.583− 1 2.777− 1.458 2.847− 2.847−

3 0.975 0.512− 1

A1

M1

A3

7

7/30/2019 T2 (B) Eng Math 4 2 20102011


By three iterations, therefore the largest eigenvalue for matrixShifted A is

Shifted 3 2.847mλ ≈ = − .

Smallest Shifted Largest Shifted Largestor 2.847 2.932 0.085λ λ λ λ λ ∴ = + + = − + =

and its corresponding eigenvector is

( )(3)

Shifted 0.975 0.512 1T

v v≈ = − .

M1

A1

Q5(a) x yy y y y

xy y y

'2'2"

2

2

2

−−=

−=′

)2(2)2(2 222 xy y xy y xy y −−−−=

″+

′+=+ iiii y y y y 02.02.01

ii

x i

y

0 0 1.000

1 0.2 1.460

2 0.4 1.987

3 0.6 2.406

4 0.8 2.535

5 1.0 2.395

M1

M2

A1

M1

A4

9

Q5(b) 22),( xy y y x f −= ( )222.0 xy yk

i−=

x y K1 K2 K3 K4

0 1 0.4 0.451 0.460 0.499

0.2 1.454 0.497 0.497 0.508 0.477

0.4 1.955 0.476 0.396 0.398 0.277

0.6 2.345

M2

M1

A5

each

column

8

T2 (B) Eng Math 4 2 20102011

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