Engineering Mathematics

DT022/3 12930MATH 3111 DT023/3 12931MATH 3111 DT024/3 13010CBEH 3111 DT026/3 8819MATH 3111 DT027/3 13121CBEH 3002

DUBLIN INSTITUTE OF TECHNOLOGY

BOLTON STREET, DUBLIN 1

__________

Bachelor of Engineering (Honours)

(Building Services, Civil, Manufacturing,

Mechanical, Structural) _________

THIRD YEAR

SEMESTER 1

JANUARY 2011

ENGINEERING MATHEMATICS

Dr Pat Carroll

TIME : 3 HOURS

Answer ALL of the following 4 questions

All questions carry equal marks.

Attachments: Formula Sheet and Mathematics Handbook

Formula Sheet: Third Year

Q1(iv) : Gregory-Newton forward formula

2 3

0

n(n-1) n(n-1)(n-2)f(a+nh)=f +nΔf(a)+ Δ f(a)+ Δ f(a)+…

2! 3!

Gregory-Newton backward formula

2 3n(n+1) n(n+1)(n+2)f(a+nh)=f(a)+n f(a)+ f(a)+ f(a)

2! 3!

Q3: Central Differences:

i+1 i-1

2

i +1 i i-12 2

y -ydy=

dx 2h

d y 1= y -2y +y

dx h

Q4 FOURIER SERIES Period 2L

0

1

L

0

0

0

Even Function:f(-x)=f(x)

Full Range for even function -L to L

1f(x)= cos

2

2 ( )

L

2 ( )cos

n

n

na a x

L

a f x dx

na f x xdx

L L

1 0

Odd Function:f(-x)=-f(x)

Full Range for odd function -L to L

2f(x)= sin ; b ( )sin

Ln n

n nb x f x xdx

L L

1 (a) (i) Sketch the region of integration covered by the double

integral 2

R

(xy +2)dA where R is the triangle with

vertices (0,0), ( 1,0) and ( 1,3) (2)

(ii) Write the integral with the appropriate limits in both directions

(3)

(iii) Evaluate the integral in either order. (4)

(b) The dominant eigenvalue of the matrix A=

2 7 0

1 3 1

5 0 8

is 9.

(i) Find its corresponding eigenvector (2)

(ii) Use the deflation method to calculate the other 2 eigenvalues.

What is the trace of A? (4)

(c) complete the matrix A= 3 10

a b(find values of a and b) so that A has

eigenvectors 2

1and

5

2. (5)

(d) Form a difference table using the data in the following table and approximate the

value of y at x= 4.7 through a Newton-Gregory procedure. Explain briefly how you

would amend your calculations in order to estimate y at x= 1.3 (5)

x 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

y 43.1 28 19.3 14 10.6 8.2 6.6 5.4 4.5 3.8

2. (a) Find the values of a and b for which the matrix

2 -1 0

A= -1 3 a

0 a b

has

1

0

1

as an eigenvector. (2)

(b) find the eigenvalues and eigenvectors of A. (8)

(c) since A is symmetric comment on one special feature of

the eigenvectors. (2)

(d) Set up a matrix P such that -1P AP is a diagonal matrix D.

Write down this diagonal matrix. (2)

(e) The motion of a system is given by the following equations:

dx= 2x - y

dt

dy= -x + 3y + z

dt

dz= y + 2z

dt

(i) Write this system of equations in matrix form. (3)

(ii) Use the substitution X= PU, where P is the matrix created in (d)

above, to transform these equations to diagonal form. (4)

(iii) Hence solve them. (4)

3.

22

2

d y dy(1-x ) -x +4y=0

dx dx

dy2 x 4; y 7 at x=2; 16 at x=4

dx

(a) State the central difference approximations for both the first and second

derivatives. (2)

(b) Set up a computation scheme to replace the differential equation with a

linear equation at any point xi in the given interval. (8)

(c) Mark out 4 divisions of the given x interval above and recast the

differential equation at each point as a linear equation. (8)

(d) Express these equations in matrix form. (2)

(e) Explain briefly how you would amend the procedure if

(a) y 1 at x=1 and y=7 at x=2 (b) dyy 1 and =2 at x=1

dx (5)

4. (a) Sketch the region of integration covered by the double

integral

2 2 4-x

0 0x dydx . (3)

Calculate the value of the integral by any one of the following methods

(6)

(i) by converting to polar co-ordinates

(ii) as it stands

(iii) by changing the order of integration.

(b) (i) Confirm that 2 2 3F=3x y i+2x yj is a conservative field. (3)

(ii)Find the work done by this force in moving a particle

from A (1,2) to B(2,3). (5)

(c) A rectangle box, open at the top, has a volume of 32 cubic metres.

Find its dimensions in order for the least amount of material to be used in its

construction. (8)

SOLUTIONS

Solution to Q1

1

1 33x 1

2 2 2y

03R 0 0

(xy +2)dA (xy +2)dydx or (xy +2)dxdy

13x

2

00

3x3

3x2 4

00

15

14 2

00

I (xy +2)dydx

xyInner : (xy +2)dy= 2y 9x 6x

3

9xI= (9x 6x)dx= 3x 1.8 3 4.8

5

(b)

1

2 7 0 a a

AX = 1 3 1 b 9 b

5 0 8 c c

2a+7b 9a 7a+7b=0 a=b=1

a+3b+c=9b

a-6b+c=0 1-6+c=0 c=5

1

Eigenvector is 1

5

Deflation:

1 1

1

1

1 1

Set up a new matrix A-X a

X is the eigenvector for =9

and a is first row of matrix A

2 7 0 1

A-X a 1 3 1 1 2 7 0

5 0 8 5

2 7 0 2 7 0

1 3 1 2 7 0

5 0 8 10 35 0

0 0 0

1 4 1

5 35 8

2

0 0 0

1 4 1 0

5 35 8

0

4 10

35 8

( 4 )(8 ) 35 0

4 3 0

( 3)( 1) 0

Eigenvalues are 9, 3, 1

Trace of A is 13

(c) complete the matrix A= 3 10

a b(find values of a and b) below so that A has

eigenvectors 2

1and

5

2.

1

1 1

2

3 10 2A ; one eigenvector is

a b 1

3 10 2 2

a b 1 1

16 28

a+b 1

Also: 2a+b=8

3 10 5A ; other eigenvector is

a b -2

3 10 5 5

a b -2 -2

-5

5a-2b2 2

51

-2

also 5a-2b=-1(-2)=2

we have 2 equations for a and b:

2a+b =8

5a-2b= 2

Hence a=2 and b=4

3 10 A=

2 4

(d)

2 3 4x y y y y y

1.5 43.1

15.1

2 28 6.4

8.7 3

2.5 19.3 3.4 1.5

5.3 1.5

3 14 1.9 0.6

3.4 0.9

3.5 10.6 1 0.7

2.4 0.2

4.0 8.2 0.8 0.2

1.6 0.4

4.5 6.6 0.4

1.2 0.1 0.3

5 5.4 0.3

0.9 0.1

5.5 4.5 0.2

0.7

6 3.8

Backward differences:

2 3n(n+1) n(n+1)(n+2)f(x+nh)=f(x)+n f(x)+ f(x) f(x) ...

2 6

x+nh=4.7

x=4.5

nh=0.2

h=0.5

n=0.4

0.4(1.4)f(4.7)=6.6+0.4(-1.6)+ (0.8)

2

0.4(1.4)(2.4) 0.4(1.4)(2.4)(3.4)( 0.2) 0.7

6 24

6.3

For x=1.8 use forward difference Newton Gregory with n=0.6

Solution to Q2

(a)

2 -1 0 1 1 2

-1 3 a 0 0 1 a

0 a b 1 1 b

=2

and -1+a=0 a=1

and b=2

2 1 0

A= 1 3 1

0 1 2

To get eigenvalues eigenvectors:

2- 1 0

1 3 1 0

0 1 2

2

add R3 to R1

2- 0 2

1 3 1 0

0 1 2

1 0 1

(2 ). 1 3 1 0

0 1 2

=2 is one root

1 0 1 1 0 0

1 3 1 C1 to C3 1 3 2 0

0 1 2 0 1 2

3- 20

1 2-

(3 )(2 ) 2 0

5 4 0

( 4)( 1) 0

4,1

eigenvalues are, 4, 2, 1

1

2 2 2

3

1

2: we know that V 0

1

1

1: AV 1V leads to V 1

-1

1

4: V -2

-1

1 2 3

1 2

2 3

1 3

1 1 1

V 0 ; V 1 ; V -2

1 -1 -1

V .V 0;

V .V 0

V .V 0

1

1 1 1

hence : modal matrix P 0 1 -2

1 -1 1

2 0 0

P AP D 0 1 0

0 0 4

-1 -1

x 2 1 0 x

y A= 1 3 1 y X AX

z 0 1 2 z

u

let X PU where U v

w

X PU

But X AX

PU AX APU

P PU P APU DU

u 2 0 0 u

U DU v 0 1 0 v

w 0 0 4 w

u 6u u Ae2t

t

4t

2t

t

4t

2t t 4t

t 4t

2t t 4t

v v v Be

w 3w w Ce

x 1 1 1 Ae

Since X PU y 0 1 -2 Be

z 1 -1 1 Ce

Solution:

x=Ae +Be +Ce

y= Be -2Ce

z=Ae - Be -Ce

Solution to Q3

2

2

2

d y dy(1-x ) -x +4y=0

dx dx

dy2 x 4; y 7 at x=2; 16 at x=4

dx

This involves 4 points as we have a gradient condition at x=4.

We need to create a virtual point at x=4.5

5

5 5 3

5 3

Call this y

dy y y16

dx 2(0.5)

y y 16

2

+1 12 2

1 1

2

2 1 1 +1 12

2

+1 1 1 1

1= 2

2

1 10.5 1; 4

2

1(1 ) 2 4 0

(0.5) 2(0.5)

4(1 ) 2 4 0

i i i

i i

i ii i i i i i

i i i i i i i i

d yy y y

dx h

y ydy

dx h

hh h

y yx y y y x y

x y y y x y y y

2

1 0

2 1 2 1

2 1 2 1

1 2 3 4

1; x 2.5;(1 2.5 ) 5.25; y 7

4( 5.25) 2 7 2.5 7 4 0

21 2 7 2.5 7 4 0

46y -23.5y +0y +0y =129.5 (1)

i

y y y y

y y y y

1

2

2 3 2 1 2 3 1 2

3 2 1 3 1 2

1 2 3 4

2; x 3;

4(1 ) 2 4 0

32 2 3 4 0

29 68 35 0 0 (2)

i

x y y y x y y y

y y y y y y

y y y y

2

2

3 4 3 2 4 2 3

4 3 2 4 2 3

1 2 3 4

3; x 3.5;

4(1 ) 2 3.5 4 0

45 2 3.5 4 0

0 41.5 94 48.5 0 (3)

i

x y y y y y y

y y y y y y

y y y y

2 5 3

2

4 5 4 3 5 3 4

3 4 3 3 3 4

3 4 4

1 2 3 4

4; x 4; y y 16

4(1 ) 2 4 4 0

60 16 2 4 16 4 0

60 2 16 2 4 16 4 0

0 0 120 124 1024 (4)

i

x y y y y y y

y y y y y y

y y y

y y y y

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

46y -23.5y +0y +0y = 129.5 (1)

29 68 35 0 0 (2)

0 41.5 94 48.5 0 (3)

0 0 120 124 1024 (4)

y y y y

y y y y

y y y y

1

2

3

4

46 23.5 0 0 129.5

29 68 35 0 0

0 41.5 94 48.5 0

0 0 120 124 1024

y

y

y

y

(v) Change of conditions:

(a) Still a Boundary Value Problem

This reduces matrix to a 3 by 3:

as there are just 3 intermediate points.

(b)

This is a Initial Value problem:

Solved by Runge Kutta

Solution to no 4

(a) Sketch the region of integration covered by the double

integral

2 2 4-x

0 0x dydx .

Calculate the value of the integral by any one of the following 3

methods

(i) as it stands

(ii) by changing the order of integration

(iii) by converting to polar co-ordinates

2

2 2

2 4-x

0 0

4-x 4 x 2

00

2

2

0

2

40 41 1 3

2 2 2

4 0 0

I= x dydx

Inner: x dy= xy x 4 x

I= x 4 x

use substitution u=4-x

du2x

dx

du=-2xdx

1xdx=- du

2

new limits:

x=0 u=4

x=2, u=0

1 1 1 2 8I=- .u du= .u du= u

2 2 2 3 3

Polar:

2 2 4-x

0 0

22

0 0

23

22

00

2

20

0

I= x dydx

area is quarter circle of radius 2 centred at origin

limist are : r=0 to 2; 0 to 2

I (rcos ) r drd

r 8Inner: (r cos ) dr cos cos

3 3

8 8 8I= cos d = sin

3 3 3

(b) (i) Confirm that 2 2 3F=3x y i+2x yj is a conservative field.

(ii)Find the work done by this force in moving a particle

from A (1,2) to B(2,3).

2 2 3

B B

2 2 3

A A

B

2 2 3

A

B

A

2 2 3

2 2

F=3x y i+2x yj

Work = F.dr= (3x y i+2x y j).dr

= 3x y dx +2x ydy

= P dx+Q dy

P=3x y ; Q=2x y

6x y ; 6x y

Hence F is conservative,

So work is independent of path.

P Q

y x

P Q

y x

B

2 2 3

A

B B B

A A A

2 2 3

2 2 3 2

3 3 2

3 2

(2,3)3 2

(1,2)

3x y dx +2x y dy

= P dx+Q dy d dx+ dy

3 ; Q= 2

3x y z=x y +c

2x y z=x y +c

z=x y

work= z x y 72 4 68

z zz

x y

z zP x y x y

x x

z

x

z

y

(c) A rectangle box, open at the top, has a volume of 32 cubic metres.

Find its dimensions in order for the least amount of material to be used in its

construction.

S = f(x,y,z) = xy +2xy +2yz

We want to find x,y,z so that S is minimised subject to the

constraints that V =xyz =32.

i.e. find h and r so that

V = f(r,h) = r2 h is maximised subject to the condition that

S = g(r,h)= 2 r2 +2 rh = 100

or g(r,h) = 2 r2 +2 rh -100 =0

We want to find x,y,z so that

S = f(x,y,z) = xy + 2xz + 2yz is minimised subject to

V = g(x,y,z) = xyz - 32 =0

f i j k

g i j k

f

x

f

y

f

z

= (y + 2z)i + (x + 2z)j + (2x + 2y)k

g

x

g

y

g

z

= yzi + xzj + xyk

we want to find x, y, z so that f = g.

i.e. equate co - efficients of i, j,k on both sides

respectively.

y + 2z = yz

x + 2z = xz

2x+2y= xy

Divide each of these across by yz, xz, and xy respectively

so that we get

1 2 1 2 2 2

z y z x y x

Compare (1) with (2) y = x

(2) with (3) y = 2z

x = y = 2z.

We want xyz= 32

x(x)(x / 2) = 32

x

x

x= 4

y= 4

z= 2

3

3

232

64