Final Exam Set B

8/12/2019 Final Exam Set B

1/17

UNIVERSITI TUN HUSSEIN ONN MALAYSIA

FINAL EXAMINATIONSEMESTER II

SESSION 2012/2013

COURSE NAME : CIVIL ENGINEERING MATHEMATICS II

COURSE CODE : BFC 14003

PROGRAMME : 1 BFF, 2 BFF

EXAMINATION DATE : JUNE 2013

DURATION : 3 HOURS

INSTRUCTION : ANSWER ALL QUESTIONS IN PART A

AND THREE (3)QUESTIONS IN PART B

THIS QUESTION PAPER CONSISTS OF SIX (6) PAGES

CONFIDENTIAL

CONFIDENTIAL


2/17

BFC 14003

2

PART A

Q1 A periodic function is defined by

2, 2 0

( ) , 0 2

x

f x x x

( ) 4f x f x .

(a) Sketch the graph of the function over 4 4x .(2 marks)

(b) Determine whether the function is even, odd or neither.(1 marks)

(c) Show that the Fourier series of the function )(xf is

21 1

1 13 2 2cos sin

2 2 2

n

n n

n x n x

n n

.

(17 marks)

Q2 (a) Consider the series,

...................125

1

25

1

5

1

(i) Give the sigma notation of the series above.(ii) Determine whether the series converge or diverge.

(6 marks)

(b) Given that

4

1

4

11

lim4

4

)1(lim

4/

4/)1(lim

4

41

4

4

14

n

n

n

n

n

n

n

nnn

n

n.

Determine whether the series

1

4

4nn

nconditionally convergent, absolutely

convergent, or divergent.

(3 marks)

(c) Given

12

)3(

n

n

n

x.


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BFC 14003

3

By using the ratio test for absolute convergence, find

(i) The interval of converges,(ii) The radius of converges. (11 marks)

PART B

Q3 (a) Solve the differential equation by using the method of separation of variables,

2 31 4x dy y xdx .

Hence, find the particular solution when3

(2)5

y .

(5 marks)

(b) By using the substitution of and ,dy dv

y xv x vdx dx

find the solution of

2 2.

dyx y x y

dx

(Hints: 1

2 2sinh ( ) c, c is constant,

dx x

aa x

1 2sinh ( ) ln | 1 |x x x .)

(6 marks)

(c) The temperature of a dead body when it was found at 3 oclock in the morning

is 85 F . The surrounding temperature at that time was 68 F . After two

hours, the temperature of the dead body decreased to 74 F

. Determine the

time of murdered.

(Hints: sdT

k T Tdt

, where

dT

dt is the temperature change in a body , k

is the proportionality constant, T is temperature in the body and sT is the

temperature constant of the surrounding medium. The normal human body

temperature is 98.6 F .)

(9 marks)


4/17

BFC 14003

4

Q4 (a) Use the method of variation of parameters to solve2

2 .( 1)

xey y y

x

(10 marks)

(c) A spring is stretched 0.25 m ( )l when a 4 kg mass ( )M is attached. Theweight is pushed up 1/2 m and released. The damping constant equals 2c . If

the general equation describing the spring-mass system is 0,Mu cu ku

find an equation for the position of the spring at any time t.

(Hints: weight, , 9.8, .W

W Mg g k l

)

(10 marks)

Q5 (a) Find

(i) 2 3 2cos3t te t t e

(ii) 2( 1) ( 2)t H t

(10 marks)

(b) By using Laplace transform, solve

2 5 20, (0) 0, (0) 10.y y y y y

(10 marks)

Q6 (a) Find

(i) 12 2

2 3.

9 4

s

s s

(ii)1

2

1 3.

( 4) 5

s

s

(10 marks)

(b) Find the general solution for the second order differential

equation

3 2 3 10cos ,xy y y e x

by using the undetermined coefficient method.

(10 marks)


5/17

BFC 14003

5

FINAL EXAMINATION

SEMESTER / SESSION : SEM II / 2012/2013 COURSE : 1 BFF / 2 BFF

SUBJECT : CIVIL ENGINEERING MATHEMATIC II SUBJECT CODE : BFC 14003

FORMULA

Second-order Differential EquationThe roots of characteristic equation and the general solution for differential equation

.0 cyybya

Characteristic equation: .02 cbmam Case The roots of characteristic equation General solution

1. Real and different roots: 1m and 2m xmxm BeAey 21

2. Real and equal roots: 21 mmm mxeBxAy )(

3. Complex roots: im 1 , im 2 )sincos( xBxAey x

The method of undetermined coefficients

For non-homogeneous second order differential equation ( ),ay by cy f x the particular

solution is given by ( )py x :

( )f x ( )py x

1

1 1 0( ) n n

n n nP x A x A x A x A

1

1 1 0( )r n n

n nx B x B x B x B

xCe ( )r xx Pe

cos or sinC x C x ( cos sin )rx P x Q x

( ) xn

P x e 11 1 0

( )r n n xn n

x B x B x B x B e

cos( )

sinn

xP x

x

1

1 1 0

1

1 1 0

( )cos

( )sin

r n n

n n

r n n

n n

x B x B x B x B x

x C x C x C x C x

cos

sin

x x

Cex

( cos sin )r xx e P x Q x

cos( )

sin

x

n

xP x e

x

1

1 1 0

1

1 1 0

( ) cos

( ) sin

r n n x

n n

r n n x

n n

x B x B x B x B e x

x C x C x C x C e x

Note : ris the least non-negative integer (r= 0, 1, or 2) which determine such that there

is no terms in particular integral )(xyp corresponds to the complementary function

)(xyc .

The method of variation of parameters

If the solution of the homogeneous equation 0 cyybya is ,21 ByAyyc then the

particular solution for )(xfcyybya is

,21 vyuyy

where ,

)(2AdxaW

xfy

u BdxaWxfy

v )(1

and 122121

21

yyyyyy

yy

W .


6/17

BFC 14003

6

FINAL EXAMINATION

SEMESTER / SESSION : SEM II / 2012/2013 COURSE : 1 BFF / 2 BFF

SUBJECT : CIVIL ENGINEERING MATHEMATIC II SUBJECT CODE : BFC 14003

Laplace Transform

L

0

)()()}({ sFdtetftf st

)(tf )(sF )(tf )(sF

a

s

a )( atH

s

e as

ate as

1 )()( atHatf )(sFe as

atsin 22 as

a

)( at ase

atcos 22 as

s

( ) ( )f t t a ( )ase f a

atsinh 22 as

a

0( ) ( )

t

f u g t u du )()( sGsF

atcosh 22 as

s

( )y t )(sY

nt , ...,3,2,1n

1

!

ns

n ( )y t )0()( yssY

)(tfeat )( asF ( )y t )0()0()(2 ysysYs

)(tftn , ...,3,2,1n )()1( sFds

dn

nn

Fourier Series

Fourier series expansion of periodic

function with period L2

0

1 1

1( ) cos sin2

n n

n n

n x n xf x a a bL L

where

L

Ldxxf

La )(

10

L

Ln dx

L

xnxf

La

cos)(

1

L

Ln dx

L

xnxf

Lb

sin)(

1

Fourier half-range series expansion

11

0 sincos21)(

n

n

n

nL

xnbL

xnaaxf

where

L

dxxfL

a00 )(

2

L

n dxL

xnxf

La

0cos)(

2

L

n dxL

xnxf

Lb

0sin)(

2


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BFC 14003

7

Marking Scheme

(Mmethod, Aanswer) Mark Total

Q1

(a)A2 2

Q1

(b)

NeitherA1 1

Q1 (c)

4x T x 2T L

4 2L 2L

01 ( )LLa f x dx

L

0 2

2 0

12

2dx xdx

2

20

2

0

12

2 2

xx

14 2

2

3

1( )cosLn L

n xa f x dx

L L

0 2

2 0

12cos( ) cos( )

2 2 2

n x n xdx x dx

A1

M1

A1

A1

A1

M1

17


8/17

BFC 14003

8


u dv

x

cos( )2

n x

1 2sin( )

2

n x

n

02 2

4cos( )

2

n x

n

0 2

2 2

2 0

1 4 2 4sin( ) sin( ) cos( )

2 2 2 2

n x x n x n x

n n n

2 2 2 2

1 4 4cos( )

2n

n n

1( )sinLn L

n xb f x dx

L L

0 2

2 0

12sin( ) sin( )

2 2 2

n x n xdx x dx

u dv

x sin( )

2

n x

1 2cos( )

2

n x

n

02 2

4sin( )

2

n x

n

0 2

2 2

2 0

1 4 2 4cos( ) cos( ) sin( )

2 2 2 2

n x x n x n x

n n n

1 4 4 4cos( ) cos( )

2n n

n n n

M1

A1

A1

M1

M1

A1

A1


9/17

BFC 14003

9


2 2 2 2

1

1 4 4cos( ) cos( )

2 23( )

2 1 4 4 4cos( ) cos( ) sin(

2 2

n

n xn

n nf x

n xn n

n n n

2 2 2 21

3 1 4 4 1 4( 1) cos( ) sin( )

2 2 2 2 2

n

n

n x n x

n n n

21 1

1 13 2 2cos sin

2 2 2

n

n n

n x n x

n n

A1

M1

A1

A1

Q2

(a)

(i)

15

1

n n

(ii)By using ratio test,

1

1

1

1 55lim lim1 5 1

5

1

15

nn

nn n

n

so the series converge.

Or the series represent geometric series where r< 1, so converge.

A1A1(1stA1 for

limit, 2ndA1for term)

M1M1

A1

A1

6

Q2

(b)

By ratio test for absolute convergence,

the series converge absolutely as < 1.

M1M1

A1

3

Q2 (c)

(i) Apply the ratio test for absolute convergence.

1 2

1

2

2

2

( 3)lim lim

( 1) ( 3)

lim 31

13 lim 1 3

1

n

n

nn nn

n

n

u x n

u n x

nx

n

x xn

Thus, the series converges absolutely if3 1, or 1 3 1, or 2 4x x x .

M1

A1

M1A1

11


10/17

BFC 14003

10


The series diverges if x< 2 orx> 4.

When 2,x

(usingp-series)

2 2 2 21

( 1) 1 1 11

2 3 4

n

n n

Or the series of absolute values

2 2 2 21

( 1) 1 1 11

2 3 4

n

n n

converge absolutely since 2 1.p

When 4,x

(usingp-series)

2 2 2 21

1 1 1 11

2 3 4

n

n n

also converge since 2 1.p

So, the interval of convergence is [2, 4].

(ii) The radius of convergence isR= 1

M1

A1

M1

A1

A1

A1

A1

Q3

(a)

2 3

1 4x dy xy dx

3 21 4

dy xdx

y x

3 2

1

1 4

xdy dx

y x

assume2

1 4a x

8da

x

dx

3

1 1 1

8dy da

y a

2

1 1(2 )

2 8a k

y

211 4

4x k

Given 3(2)5

y , so

3.45k

A1

M1

A1

A1

5


11/17

BFC 14003

11


2

2

1 1 43.45

2 4

x

y

A1

Q3

(b)

2 2dyx y x y

dx

2 2 2dvx x v xv x x vdx

21xv x v 21x v v

21dv

x v v vdx

21

dvx v

dx

2

1 1

1dv dx

xv

2

1 1

1dv dx

xv

By using the hints,

1sinh ( ) ln | |v x k , kis a constant.

2ln | 1 | ln | |v v x k

2ln | ( ) 1 | ln | |y y

x kx x

A1

M1

A1

A1

M1

A1

6

Q3

(c)

at 3t am, 85T F ; 5t am, 74T F ; ?t ,98.6T F

( )sdT

k T Tdt

( )s

dTkdt

T T

9


12/17

BFC 14003

12


1

s

dT kdt T T

ln | |sT T kt c

, c is constantkt

sT T Ae , cA e

at 3am,t

385 68

kAe

3

17k

Ae

3

17kA e

So3

1768 kt

kT e

e

at 5amt ,

5

3

1774 68 k

ke

e

26 17 ke

2 6

17

ke

1 6ln( )

2 17k

3 6ln( )

2 1717A e

3 6 1 6ln( ) ln( )2 17 2 1768 17

tT e e

3 6 1 6

ln( ) ln( )2 17 2 17

3 6 1 6ln( ) ln( )

2 17 2 17

3 6 1 6ln( ) ln( )

2 17 2 17

98.6 68 17

1.8

ln |1.8 |

1.87

t

t

t

e e

e e

e

t

1:52 am

Therefore, the time of murdered was 1:52am.

M1

A1

A1

A1

A1

M1

A1

M1

A1


13/17

BFC 14003

13


Q4

(a)

Step 1:

21, ( )

( 1)

xea f x

x

Step 2:2

1 2

1 2

2 1 0

1

Thus,

x x

h

x x

x x x

m m

m

y Ae Bxe

y e y xe

y e y xe e

Step 3:

2 2 2 2 .x x

x x x x

x x x

e xeW xe e xe e

e xe e

Step 4:

2

2

2

2

2

( 1)

( 1)1

; 1

1

1 ln | 1|

1

xx

x

exe

xu dx

e

xdx

xu

du u xu

u duu

x Cx

2

2

2

2

2

( 1)

1

( 1)

1 ; 1

1

1

xx

x

ee

xv dxe

dxx

du u xu

u du

Dx

Step 5:

M1

A1

M1A1

M1

M1

A1

M1

A1

10


14/17

BFC 14003

14


1 2

1 1 ( ln | 1| ) ( )

1 1

ln | 1| (1 ).1

x x

x

x x x

y uy vy

x C e D xex x

eCe Dxe e x xx

A1

Q4

(b)

4(9.8) 39.2

39.2156.8

0.25

2

W mg

Wk

l

c

Thus, from 0,mu cu ku

We get 4 2 156.8 0u u u or 2 78.4 0.u u u

The characteristic equation is 22 78.4 0.m m

So, 0.25 6.25 .m i

Therefore, 0.25( ) cos(6.25 ) sin(6.25 ) .tu t e A t B t

0.25

0.25

( ) 6.25 sin(6.25 ) 6.25 cos(6.25 )

0.25 cos(6.25 ) sin(6.25 ) .

t

t

u t e A t B t

e A t B t

The spring is pushed up 1/2 m and released.

So, we have (0) 1/ 2u . (pushed up is negative)

Since the weight is simply released, its initial velocity is zero,

(0) 0u .

These initial conditions give us the equation for the position of the

spring at any time tas

0.25( ) 0.5cos(6.25 ) 0.02sin(6.25 ) .tu t e t t

A1

A1

A1

M1

A1

A1

A1

A1

A1

A1

10

Q5(a)

(i)

2 3 2

2 3 2

2 4

cos3

cos3

2 6

( 2) 9 ( 2)

t t

t t

e t t e

e t t e

s

s s

M1

A1A1

3


15/17

BFC 14003

15


Q5(a)

(ii)

2

2

2 2

2

2

2

2 2 2

2 2 2

3 2

( 1) ( 2)

( 2 1) ( 2)

Let ( 2) 2

2 1 [( 2) 2] 2[( 2) 2] 1

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

( 2) 2( 2) 1

( 2) ( 2) 2 ( 2) ( 2) ( 2)

2 ( 2)

2

s s

s s s

t H t

t t H t

t t

t t t t

t t t

t t

t H t t H t H t

e t e t H t

e e e

s s s

A1

M1

A1

M1

A1A1A1

7

Q5(b)

2

2

2

2 2

2

2 5 20, (0) 0, (0) 102 5 20

20( ) (0) (0) 2 ( ) (0) 5 ( )

20( ) 10 2 ( ) 5 ( )

20( 2 5) ( ) 10

20 10( )

( 2 5) 2 5Byusing partialfraction,

20

( 2 5)

y y y y yy y y

s Y s sy y sY s y Y ss

s Y s sY s Y ss

s s Y ss

Y s

s s s s s

A Bs

s s s s

2

2

1

1

2 2

1

2

1

2

1 1 1

2 2

2 5

20 ( ) (2 ) 5

4, 4, 8

( ) ( )

4 4 8 10

2 5 2 54 4 2

( 1) 4

4 4[( 1) 1] 2

( 1) 4

4 4( 1) 6

( 1) 4 ( 1) 4

C

s s

A B s A C s A

A B C

y t Y s

s

s s s s ss

s s

s

s s

s

s s s

4 4 cos 2 3 sin 2t te t e t

M1

A1

M1

A1

M1

M1

A1

A1A1A1

10


16/17

BFC 14003

16


Q6(a)

(i)

31

2 2

31 1 1

2 2

3

2 3

9 4

2 3

9 4

2sin3 3cosh 2

3

e s

s s s

e s

s s s

e t t

M1

A1A1A1

4

Q6(a)

(ii)

1

2

2

2

2

2 2

1 1

2 2

4 1 4 1

2 2

4

1 3

( 4) 5

1 3( )

( 4) 5

1 3[( 4) 4]( 4) 5

3( 4) 13

( 4) 5

3( 4) 13

( 4) 5 ( 4) 5

3( 4) 13

( 4) 5 ( 4) 5

13

3 5 5

3 co

t t

t

s

s

sF s

s

ss

s

s

s

s s

s

s s

s

e es s

e

413s 5 sin 5

5

tt e t

M1

A1

A1

M1

A1A1

6


17/17

BFC 14003


Q6(b)

2

1 2

2

1

1

1

1

1

1 1 1

3 2 3 10cos , (0) 0, (0) 2

3 2 0

2, 1

( ) 3

0 .......(i)

Compare with , no termalike.Accept (i).

3 2 3

3(

x

x xh

x

r x

p

x

p

h

x

p

x

p

xp p p

x

y y y e x y y

m m

m m

y Ae Be

f x e

y x Pe

r y Pe

y

y Pe

y Pe

y y y e

Pe Pe

1

2

2

2

2

2

2 2 2

) 2 3

6 3

1

2

1

2

( ) 10 cos

cos sin0 cos sin .......(ii)

Compare with , no termalike.Accept (ii).

sin cos

cos sin

3 2 10cos

cos

x x x

x x

x

p

r

p

p

h

p

p

p p p

Pe e

Pe e

P

y e

f x x

y x K x L xr y K x L x

y

y K x L x

y K x L x

y y y x

K x L

2

2

sin 3( sin cos ) 2( cos sin ) 10co

( 7 9 ) cos (9 7 )sin 10cos

7 9,13 13

7 9cos sin

13 13

1 7 9cos sin

2 13 13

p

x x x

x K x L x K x L x

K L x K L x x

K L

y x x

y Ae Be e x x

M1

A1

M1

A1

A1

M1

A1

A1

A1

A1

10

Final Exam Set B

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Transcript of Final Exam Set B