final exam set a

8/12/2019 Final Exam Set A

1/17

UNIVERSITI TUN HUSSEIN ONN MALAYSIA

FINAL EXAMINATION

SEMESTER 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 , 0

( ) 2 , 0

xf x x x

2)( xfxf .

(a) Sketch the graph of the function over x .(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 12 cos 2 sin

2

n

n nnx nx

n n

.

(17 marks)

Q2 (a) Given that !6!4!2

1cos642 xxx

x .

(i) Find the first three nonzero terms of a power series for x

xsin

.

(ii) Hence, evaluate dxx

x

1

0

sinby using the series expansion.

(6 marks)

(b) (i) Determine whether the series

1

2

nne

nconverges or diverges by

using ratio test.

(ii) Determine whether the series

1

)1(

n

n

nconverges

absolutely, converges conditionally, or diverges by using a

suitable convergence test.

(8 marks)

(c) Find the radius of convergence of0

3( 2)n

n

x

.

(6 marks)


3/17

BFC 14003

3

PART B

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

1 1dy

x x y

dx

.

Hence, find the particular solution when (1) 0y .

(7 marks)

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

y xv x vdx dx

find the solution of

2 2

.3

dy x y

dx xy

(6 marks)

(c) In a certain culture of bacteria, the rate of increase is proportional to the

number of present. If it is found that the number doubles in 4 hours, how many

may expected at the end of 12 hours?

(Hints:dN

kNdt

, where N denotes the number of bacteria at time, thours

and kis the proportionality factor.)

(7 marks)

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

sec tany y x x ,

which satisfies the initial conditions (0) 0y and (0) 2y .

(Hints: 1tansec22

xx ,2

sec tan .)x dx x

(10 marks)

(b) A spring is stretched 0.1 m ( )l when a 4 kg mass ( )M is attached. The

weight is then pulled down an additional 0.2 m and released with an upward

velocity of 4 m/s. Neglect damping, c. If the general equation describing the

spring-mass system is


4/17

BFC 14003

4

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) 3 25

4.t

te t

e

(ii) 3cosh 4 ( 3) .t t t t

(iii) 3sinh(3 ) ( 3) .tt e H t

(10 marks)

(b) By using Laplace transform, solve

2 , (0) 2, (0) 3.ty y y e y y (10 marks)

Q6 (a) Find

(i)1

2 2

5 6.

4 ( 2) 4

s

s s

(ii)2

1

2

16.

( 3)( 1)

s

s s

(10 marks)

(b) Find the general solution for the second order differential equation23 2 2 5sin ,y y y x 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 as

e

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 sin

2 n n

n n

n x n xf x a a b

L 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

110

sincos2

1)(

n nn n L

xnb

L

xnaaxf

where

L

dxxfL

a00

)(2

L

n dxL

xnxf

La

0cos)(

2

L

n dxL

xnxf

Lb

0sin)(

2


7/17

BFC 14003

7

Marking Scheme

(Mmethod, Aanswer) Mark Total

Q1

(a)

A2 2

Q1

(b)Neither A1 1

Q1

(c) 2x T x 2T L

2 2L L

0

1( )LLa f x dx

L

0

0

12 2dx xdx

0 2

0

12 x x

2 21 2

3

1( ) cosLn L

n xa f x dx

L L

0

0

12 cos( ) 2 cos( )nx dx x nx dx

A1

M1

A1

A1

A1

M1

17


8/17

BFC 14003

8


u dv

2x cos( )nx

2 1sin( )nx

n

02

1cos( )nx

n

0

2

0

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

xnx nx nx

n n n

2 2

1 2 2cos( )nn n

1( ) sinLn L

n xb f x dx

L L

0

0

12 sin( ) 2 sin( )nx dx x nx dx

u dv

2x sin( )nx

2 1cos( )nx

n

02

1sin( )nx

n

0

2

0

1 2 2 2cos( ) cos( ) sin( )xnx nx nxn n n

2

n

2 213 1 2 2 2

( ) cos( ) cos( ) sin2 n

f x n nx nxn n n

M1

A1

A1

M1

M1

A1

A1

A1M1


9/17

BFC 14003

9


213 2 2

( 1) 1 cos( ) sin( )2

n

nnx nx

n n

21 1

1 13 12 cos 2 sin

2

n

n nnx nx

n n

A1

A1

Q2

(a) (i)

1206

!6

6

!4

4

!2

2

!6!4!21

cossin

53

53

642

xxx

xxx

xxx

dx

d

xdx

dx

So,

1206

1206sin2/92/5

53

xxx

x

xxx

x

x

M1

A1

A1

3

Q2

(a)

(ii)

.621.0

1320

2

42

2

3

2

1320

2

42

2

3

2

1206

sin

1

0

2/112/72/3

1

0

2/92/51

0

xxx

dxxx

xdxx

x

M1

A1

A1

3

Q2

(b) (i)

1

2

nne

n

2

21

2 1 2

2 2

( 1)

( 1)lim lim

1 1 1 1 lim lim 1

1 1 ; the series converges.

nn

nn n

n

n n

n

e ne

n e n

e

n

e n e n

e

M1

A1A1

3


10/17

BFC 14003

10


Q2

(b)

(ii)

5

1

4

1

3

1

2

11

)1(

1n

n

n

Use alternating series test:

(a) 5

1

4

1

3

1

2

11

(b) 01

lim nn

the series converges.

Usep-series:

2

1p 10 p

the series diverges.

Conclusion, the series converges conditionally.

M1

A1

M1

A1

A1

5

Q2

(c)

For 2x , let nn xu )2(3 .Then

|2|

|2|lim)2(3

)2(3limlim

11

x

xx

x

u

u

n

n

n

nn

n

n

Thus, the series converges absolutely if

| 2 | 1x or 1 3x .

The series diverges if 1x or 3x

Therefore, the radius of convergence of the series is 1R .

M1

M1

A1

M1

A1

A1

6

Q3

(a) 1 1

dyx x y

dx

11

dy xdx

xy

1

11

xdy dx

xy

2 1 ln(x 1) k y x , k is constant

Given (1) 0y , so

A1

M1

M2

A1

7


11/17

BFC 14003

11


ln(2) 3k

2 1 ln(x 1) ln(2) 3y x

A1

A1

Q3(b) 2 2

3

dy x y

dx xy

2 2 2

3 ( )

dv x x vx v

dx x xv

21

3

v

v

2

13

dv vx vdx v

21 2

3

v

v

2

3 1

1 2

vdv dx

v x

2

3 1

1 2

vdv dx

v x

assume21 2a v

4da

vdv

1

4da vdv

3 1 1

4da dx

a x

23 ln 1 2 ln4

v x k

23

ln 1 2 ln4

yx k

x

A1

M1

A1

M1

A1

A1

6

Q3

(c) at 0,t 0N N ; 4t , 02N N ; 12t , ?N

dN kNdt

7


12/17

BFC 14003

12


dNkdt

N

1

dN kdt N

ln | |N kt c , c is constant

ktN Ae , cA e

at 0,t 0A N

So 0ktN N e

at 4t , 40 02 kN N e 42 ke

4 ln | 2 |k

1l n | 2 |

4k

So

1ln|2|

4

0

t

N N e

at 12t , 3ln|2|0N N e

08N

At the end of 12 hours, the number of bacteria is 8 times of the

original number.

M1

A1

A1

A1

A1

M1

A1

Q4

(a)

Step 1:

1, ( ) sec tana f x x x

Step 2:2

1 2

1 2

1 0

Thus,

cos sin

cos sin

sin cos

h

m

m i

y A x B x

y x y x

y x y x

M1

A1

10


13/17

BFC 14003

13


Step 3:

2 2cos sin

cos sin 1.sin cos

x xW x x

x x

Step 4:

2

2

sin sec tan

tan

1 sec

( tan )

u x x x dx

x dx

x dx

x x C

cos sec tan

tan

sin

cos

sin

sin

1

ln | cos |

v x x x dx

x dx

xdx

x

x du

u x

duu

x D

Step 5:1 2

(tan )cos ( ln | cos |)sin

cos sin sin cos sin ln | cos |

sin cos cos sin cos sin tan

cos ln | cos |

y uy vy

x x C x D x x

C x D x x x x x x

y C x D x x x x x x x

x x

When (0) 0,y

0.C When (0) 2,y

2.D

Thus,

2sin sin cos sin ln | cos | .y x x x x x x

M1A1

M1

A1

M1

A1

A1

A1


14/17

BFC 14003

14


Q4

(b)

4(9.8) 39.2

39.2392

0.1

0

W mg

Wk

l

c

Thus, from 0,mu cu ku

We get 4 392 0u u or 98 0.u u

The characteristic equation is 2 98 0.m

So, 7 2 .m i

Therefore, ( ) cos(7 2 ) sin(7 2 ).u t A t B t

or ( ) cos(9.9 ) sin(9.9 ).u t A t B t

( ) 7 2 sin(7 2 ) 7 2 cos(7 2 ).u t A t B t

The spring is released after pulling it down 0.2 m.

So, we have (0) 0.2u .

Since its set in motion with an upward velocity of 4 m/s,

we have (0) 4u . (upward is negative)

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

spring at any time tas

( ) 0.2cos(7 2 ) 0.4041sin(7 2 ).u t t t

or

1 2 2( ) cos(7 2 ) sin(7 2 ).

5 7u t t t

A1

A1

A1

M1

A1

A1

A1

A1

A1

A1

10

Q5(a)

(i)

3 2

5

3 2 5

3

4

4

2 4

( 3) 5

t

t

t t

e te

e t e

s s

M1

A1A1

3

Q5(a)

(ii)

3

3

23

2 2

cosh 4 ( 3)

cosh 4 ( 3)

1627

( 16)

s

t t t t

t t t t

se

s

M1

A1A1

3


15/17

BFC 14003

15


Q5(a)

(iii)

3

3

3[( 3) 3]

9 3( 3)

9 3

2

sinh(3 ) ( 3)

sinh(3 ) ( 3)

sinh(3 ) ( 3)

sinh(3 ) ( 3)

3

9 3

t

t

t

t

s

t e H t

t e H t

t e H t

t e e H t

e

s s

M1

A1

A1A1

4

Q5(b)

2

2

2

2

3 2 2

3

2 , (0) 2, (0) 3

2

1

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

1( ) 2 3 2 ( ) 4 ( )

1

12 1 ( ) 2 1

1

1( 1) ( ) 2 1

1

1 2 1( )

( 1) ( 1) ( 1)

1 2( )

( 1)

t

t

y y y e y y

y y y e

s Y s sy y sY s y Y s s

s Y s s sY s Y ss

s s Y s ss

s Y s ss

sY s

s s s

Y ss

2

2 2

1

1

3 2

2

1

( 1)

Byusing partialfraction,

2 1

( 1) 1 ( 1)

2, 1

( ) ( )

1 2 1

( 1) 1 ( 1)

12

2

t t t

s

s

s A B

s s s

A B

y t Y s

s s s

e t e e t

M1

A1

A1

M1

A1

M1

A1

A1A1A1

10

Q6(a)

(i)

1

2 2

1 1

2 2

2

5 6

4 ( 2) 4

65

4 ( 2) 4

5 cosh 2 3 sin 2

t

s

s s

s

s s

t e t

M1

A1A1

3


16/17

BFC 14003

16


Q6(a)

(ii)

21

2

2

2 2

2 2

2 2

21

2

1 1

16

( 3)( 1)

Bypartialfraction,

16

( 3)( 1) 3 1 ( 1)

16 ( 1) ( 1)( 3) ( 3)

16 ( ) (2 2 ) ( 3 3 )

9, 7, 4

16

( 3)( 1)

9 7

3 1

s

s s

s A B C

s s s s s

s A s B s s C s

s A B s A B C s A B C

A B C

s

s s

s s

1

2

3

4

( 1)

9 7 4t t ts

e e e t

M1

M1

A1

A1

A1A1A1

7


17/17

BFC 14003

17


Q6(b)

2

2

1 2

2

2

1

2

1

2

1

1

1

2

1 1 1

3 2 2 5sin , (0) 0, (0) 2

3 2 0

2, 1

( ) 2

0 .......(i)

Compare with , no termalike,accept (i)

2

2

3 2 22 3(2

x xh

r

p

p

h

p

p

p p p

y y y x x y y

m m

m m

y Ae Be

f x x

y x Cx Dx E

r y Cx Dx E

y

y Cx D

y C

y y y xC C

2 2

2 2

2

1

2

2

2

2

) 2( ) 2

2 ( 6 2 ) (2 3 2 3) 2

1 3 11, ,

2 2 4

1 3 11

2 2 4

( ) 5sin

cos sin

0 cos sin ............(ii)

Compare with , no termalike,accept (ii)

sin cos

p

r

p

p

h

p

x D Cx Dx E x

Cx C D x C D E x

C D E

y x x

f x x

y x K x L x

r y K x L x

y

y K x L

2

2 2 2

2

2 2

cos sin

3 2 5sin

cos sin 3( sin cos ) 2( cos sin ) 5sin

( 3 ) cos (3 )sin 5sin

3 1,2 2

3 1cos sin

2 2

1 3 11 3 1cos sin

2 2 4 2 2

p

p p p

p

x x

x

y K x L x

y y y x

K x L x K x L x K x L x x

K L x K L x x

K L

y x x

y Ae Be x x x x

M1

A1

M1

A1

A12

A1

M1

A1

A12

A1

A1

10

final exam set a

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