Future Engineering Mathematics: Line Integrals, Double Integrals, and Power Method

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This article covers topics such as line integrals, double integrals, and the power method in Future Engineering Mathematics. It includes solved problems and step-by-step solutions.
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2018
INSTITUTIONAL AFFILIATION
FACULTY OR DEPARTMENT
COURSE ID & NAME
TITLE:
FUTURE ENGINEERING MATHEMATICS
STUDENT NAME
STUDENT ID NUMBER
PROFESSOR (TUTOR)
DATE OF SUBMISSION
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QUESTION 1
Part a (2mks)
Sketch the graphs of y=2sin2 ( x ) and y=2cos2 (x) from x=π ¿ x=π
-4 -3 -2 -1 0 1 2 3 4
x
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Y=2sin 2 (x)
-4 -3 -2 -1 0 1 2 3 4
x
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Y=2cos2(x)
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Part b
Using a double integral, calculate the area enclosed by the curves in the interval x=0 ¿ x= π
4

0
π
4
y dy=
0
π
4
2 sin2 ( x ) dx
Computing the indefinite integral
Computing the boundaries:
¿ π
4 1
2 0
Simplifying further,
¿ π
4 1
2

0
π
4
( π
4 1
2 )dx= π22 π
16 0.22415
Part c
Determine area enclosed by the curves in the interval
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π
4
π
2
( 2sin2 x2 cos2 x ) dx
Rewrite or simplifying the function,
2cos 2 x dx
Substitute u=2 x dx =1
2 du
cos (u) du
Now solving:
cos ( u ) du
This is a standard integral,
¿sin ( u )
Redoing the substitution,
¿sin ( 2 x )
The problem is solved as,
2cos 2 x dx=sin 2 x +C
¿ 1
Part d
Infinitesimal elements of volume are expressed in Cartesian coordinates by dV =dxdydz
dv = d xdydz
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¿
0
3 x

2.5
2.5

ex

dv
Part e
Infinitesimal elements of volume are expressed in spherical polar coordinates as
dV =r2 sin (θ) drd θ d φ
QUESTION 2
Part a
Evaluate I=
c

( 2 xy + y ) dxalong the path C described parametrically by x=t2 , y= 1
t2 ¿ t=C ¿ 4
Solution
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¿ r ( t ) ¿= ( dx
dt )2
+ ( dy
dt )2
I =
c

( 2t2 . 1
t2 + 1
t2 )||r ( t )||dt
dx
dt =2t , dy
dt =2
t3
||r ( t )||= ( 2t ) 2+ ( 2
t3 )
2
= 4 t2+ 4
t6
||r ( t )||=2 ( 1+t8
t6 )
I =
c
4
( 2+ 1
t2 )2 ( 1+t8
t6 ) dt
The line integral is given as,
I =
c
4
2 ( 2+ 1
t2 ) ( 1+t8
t6 ) dt
Part b
Calculate the length of the curve, y=ln |sec ( x )|from x=0 ¿ x= π
3
y=ln |sec ( x )|0 x π
3
For the given interval,
|sec ( x )|=sec ( x )
y' = 1
sec ( x ) ( sec x )'=sec x ( tan x
sec x )=tan x
length= ( 1+ ( y' ) 2
) dx
y=ln |sec x|x=0¿ x= π
3
¿ 1+tan2 x dx for thelimits
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¿ sec x dx for thelimita
¿ [ ln ( sec x + tan x ) ] , between x=0 ¿ x= π
3
¿ ln ( sec ( π
3 ) + tan ( π
3 ))
sec ( π
3 )= 1+ tan2 π
3 = 1+3=2
¿ ln ( 2+1 )=ln ( 3 )=1.09
Part c
Find the work done by the force F= ( 3 xy +4 ) i+ y3 j moving anticlockwise around the boundaries
of the triangle defined by the points (0,0), (4,0), and (4,4).
SOLUTION
Denoting the points as C1, C2, and C3.
C1= ( 0,0 ) C2 = ( 4,4 ) C3= ( 4,0 )
Dealing with the line segments
C= ( 0,0 ) ¿ ( 4,4 ) ( 4,4 ) ¿ ( 4,0 ) ( 4,0 ) ¿ ( 0,0 )



fds=
C1

fds+
C2

fds+
C3

fds
F= ( 3 xy +4 ) i+ y3 j
f = ( 3 xy+ 4 , y3 ) =F
f x=3 y
f y=3 x ,3 y2
c1= {x=0. ( 1t ) +2 t
y=0. ( 1t ) +t }
c2= {x=4 . ( 1t )+ 2t
y =4 . ( 1t ) +t }
c3= {x=4 . ( 1t )+ 2t
y=0. ( 1t ) +t }
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fds=
0,0
4,4
+
4,4
4,0
fds+
4,0
0,0
fds
The gradient vector field is,
¿ 3 5 ( 4 t2t2 ) 5 dt + ( 2t2 ) ( t +1 ) 2 2 dt+ ( 1t ) 24 ¿ ¿
¿ 3 5
0
1
t2 dt + 2
0
1
( 36 t ) dt+ 5
0
1
(3+6 t=3 t2 ) dt
¿ 6
Part d
By using Green’s Theorem, verify this result with a double integral over the area of the triangle.
¿
0
4
( 14+10 t ) dt=6
QUESTION 3
Use the power method to find, correct to 3 decimal places, the dominant eigenvalue, and a
corresponding eigenvector of the matrix.
[ 0 11 5
2 17 7
4 26 10 ]T
Start with the initial vector [1,1,1]T and work to 3 decimal places throughout.
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AT = [ 0 11 5
2 17 7
4 26 10 ]T
A=
[ 0 2 4
11 17 26
5 7 10 ]T
initial approximations , x0= [ 1
1
1 ] ,
Taking out the largest element of the resultant matrix,
x1= A ~x1=
( 0 2 4
11 17 26
5 7 10 )(1
1
1 )= ( 6
2
22 )=22 ( 0.2727
0.0909
1.0000 )
Using the resulting matrix as the new initial vector for the next iteration,
x2= A ~x2=
( 0 2 4
11 17 26
5 7 10 )( 0.2727
0.0909
1.0000 )=
( 3.8182
24.5456
10.7272 )=24.5456 (0.1556
1.000
0.4370 )
Repeating the same procedure to obtain an iteration of the sequence,
x3= A ~x3=
( 0 2 4
11 17 26
5 7 10 )(0.1556
1.000
0.4370 )=
( 3.7481
7.3483
12.1481 )=12.1481 ( 0.3085
0.6049
1.000 )
x3= A ~x3=
( 0 2 4
11 17 26
5 7 10 )( 0.3085
0.6049
1.0000 )=
( 2.7902
32.8893
7.3084 )=32.8893 (0.0848
1.000
0.2222 )
x4 =A ~x4 =
( 0 2 4
11 17 26
5 7 10 )(0.0848
1.000
0.2222 )=
(2.8888
12.1556
9.6463 )=12.1556 (0.2377
1.000
0.7936 )
x4 =A ~x4 =
( 0 2 4
11 17 26
5 7 10 )(0.2377
1.000
0.7936 )=
( 1.1743
35.0186
2.1239 )=35.0186 (0.0335
1.000
0.0607 )
x5= A ~x5=
( 0 2 4
11 17 26
5 7 10 )(0.0335
1.000
0.0607 )=
(2.2426
15.7919
7.7742 )=15.7919 (0.1420
1.000
0.4923 )
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x6= A ~x6= ( 0 2 4
11 17 26
5 7 10)(0.1420
1.000
0.4923 )= (0.0308
28.2374
1.3671 )=28.2374 (0.0011
1.0000
0.0484 )
x7= A ~x7= ( 0 2 4
11 17 26
5 7 10 )( 0.0011
1.0000
0.0484 )= (
1.8063
18.2468
6.5104 ) =18.2468 (
0.0990
1.0000
0.3568 )
x8= A ~x8= ( 0 2 4
11 17 26
5 7 10)(0.0990
1.0000
0.3568 )= (0.5728
25.1878
2.9370 )=25.1878 (0.0227
1.0000
0.1166 )
x9= A ~x9= ( 0 2 4
11 17 26
5 7 10)(
0.0227
1.0000
0.1166 )= (
1.5336
19.7816
5.7203 )=19.7816 (
0.0775
1.0000
0.2892 )
x10= A ~x10=
( 0 2 4
11 17 26
5 7 10 )(0.0775
1.0000
0.2892 )= (0.8433
23.6656
3.7207 )=23.6656 (0.0356
1.0000
0.1572 )
To obtain the eigen values and dominant eigen vector, the iterations are carried out
progressively. The dominant eigenvector is obtained as,
23.6656 . dominant eigen value
(0.0356
1.0000
0.1572 )=dominant eigenvector
QUESTION 4
Consider the initial value problem,
dv
dt =100.1 v with vo =50 at to=0
Using the 2nd Runge-Kutta method to calculate the approximate value of v at t=0.4, using step
size h=0.2 and working to six significant figures.
The modified Euler’s method is the second-order Runge Kutta method,
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X n+1=X n+ t
2 [ f ( xn ,tn ) + f ( xn+ tf ( xn ,tn ) , tn + t ) ]
xi+1
¿ =xi t xi
2
To solve the ode using 2nd order RK method,
Initialization,
k =0 , h= bt0
n y ( tk ) = yk
Predict the solution,
yk +1= yk + hf ( tk +1 , yk +1 , p )
2
Increment the value,
tk+1=tk +h , k =k +1
Using the algorithm to find the solution,
dv
dt =100.1 v , , v0 =50 at t0=0 h=0.2
Step 1:
V i+ 1=vi + f ( ti , vi ) h
vi =50+f ( 0,50 ) 0.2
¿ 50+ ( 100.150 ) 0.2
¿ 50+ ( 5 ) 0.2
vi =51
Step 2:
v2=v1 +f ( t1 , v1 ) h
¿ 51+ ( 100.151 ) 0.2
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¿ 51+ ( 0.98 )
v2=51.98 , at t=0.2 , h=0.2
Step 3:
v3 =v2 +f ( t2 , v2 ) h
¿ 51.98+ ( 100.151.98 ) 0.2
v2=52.9404 at t=0.4 , h=0.2
¿ 52.9404 ¿ 6 significant figures
QUESTION 5
The periodic function has a period 2pi. It is defined in the interval 0 x π by
f ( x )=x2
(a) Sketching the function on the interval for two cases,
A function is an even function when f ( x ) =f (x )
-10 -8 -6 -4 -2 0 2 4 6 8
0
10
20
30
40
50
60
70
80
90
A function is said to be odd function when f ( x ) =f ( x)
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-10 -8 -6 -4 -2 0 2 4 6 8
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
(b) Fourier coefficients of the two functions.
The Fourier cosine series describes an even function and the Fourier sine series describes
an odd function. For an odd function
f ( x ) = a0
2 +
n=1

an cos (
L ) x
For an even function,
f ( x ) =
n=1

bn sin (
L ) x
Using the Fourier series, the function in part a is an even function.
a0= 2
1
0
π
x2 dx= 2
3 π3
an=2
0
π
x2 cos ( nπx ) dx
Using integration by parts
an= x2 sin nπx
πn 2 x sin nπx
πn dx
Now solving,
2 x sin nπx
πn dx
Applying linearity,
¿ 2
x sin nπx dx
bn= 2
π
0
π
x sin nx dx
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¿ 2
π ( [ xcos nx
n ]0
π

0
π
cos nx
n dx )
¿ 2
π
( 1
n [ π cos ] + [ sin nx
n2 ] 0
π
)
¿ 2
n cos
¿
{2
n if n is even
2
n if n is odd
(c) Given that
Is a periodic function of period 2pi, by using integration by parts when required, calculate
the Fourier coefficients for the function, g(x), and express it as Fourier series expansion.
g ( x ) =
0
π

π
2 π
( x ) ( x ) dx
The Fourier series is given as
g ( x )=b1 sin x+ b2 sin2 x +b3 sin 3 x +
g ( x )=2 sin x sin 2 x + 2
3 sin 3 x1
2 sin 4 x + 2
5 sin 5 x+
(d) Deduce that π 2
8 = {1+ 1
9 + 1
25 + 1
49 + }=
n=1
1
( 2 n1 ) 2
Let function f(x) be an even function,
f ( x )= a0
2 +
n=1

sin nx
a0= 1
π
π
π
x2 dx = 2
π
0
π
x2 dx
¿ 2
π [ x3
3 ]0
π
=
2
π π3
3 = 2 π2
3
an= 1
π
π
π
x2 cos nx dx= 2
π
0
π
x2 cos nx dx
Using integration by parts,
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¿ 2
π [ x2
n sin nx 2
n x sin nx dx ]0
π
¿ 2
π [ x2
n sin nx 2
n ( x
n cos nx 1
n cos nx dx ) ]0
π
¿ 2
n [ x2
n sin nx+ 2 x
n cos nx 2
n3 sin nx ]0
π
¿ 4
n2 cos = 4
n2 (1 )n
Substitution the values of f(x), a0 and an,
x2= π 2
3 +4
n=1
(1 )n
n2 cos nx
Substituting x= π
2
Substituting x=π,
Adding the above equations
π2
6 π2
12
,
Therefore,
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(e) Write down the values of ¿ ao ,a1 , . , a5 and ¿ b1 , b2 , . ,b5 correct to four significant
figures.
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MATLAB implementation
x1= -pi:pi/10:pi;
gx1=(4/pi)*((sin(x1)/1)+(sin(3*x1)/3)+(sin(5*x1)/5)+(sin(7*x1)/7));
plot(x1,gx1)
grid on
-4 -3 -2 -1 0 1 2 3 4
-1.5
-1
-0.5
0
0.5
1
1.5
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