Mid-term MAE 2360 Fall 2020
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School
University of Texas, Arlington *
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Course
2360
Subject
Aerospace Engineering
Date
Dec 6, 2023
Type
docx
Pages
6
Uploaded by MajorWillpowerBeaver8
Mid-term of MAE 2360-001
Note
:
Exam period: 1:00 pm to 1:50pm, Wednesday, 10/28 for all students.
There are five problems, and each is worth 20 points.
For each problem, please
Directly make modifications to the given C program,
highlight your changes with a color, such as
example.
The modified C program is your answer. No need re-writing the C program. No
need running the modified C program. No need giving the output of the modified C
program.
Open-book exam. A student can use and refer to anything during the exam.
During the exam, not allowed to discuss or share information with anybody else.
Please join the new conference to be opened on the Canvas website of MAE 2360-001,
and turn on the camera of your computer, so we can see you on the conference during the
exam period.
Instruction
for submission, which is the same as the way of submitting HWs
Go on Canvas, select course MAE 2360-002 or 003 (depending on which lab session that
you are enrolled in), then click on assignments and select Mid-term to be submitted and
upload the document.
The format in which the mid-term paper is to be submitted is only
docx and doc.
The time period given for the submission is 1-2pm, Wednesday, 10/28. The link for
submission of the mid-term paper will disappear at 2pm and will be considered as “
not
submitted”.
If you get a problem to submit online, please email me immediately at
chengluo@uta.edu
to let me know the issue, with your exam paper attached to your email as the evidence
that you have finished the exam.
Two attempts will be allowed for the submission of the mid-term paper.
Please do not copy others’ works, plagiarism check will be deployed to catch such an
activity.
1
1.
Fix 10 bugs in the following program. Highlight with a color the changes that you have
made.
#include <stdio.h>
#include <math.h>
#define EPS 1.0e-10
#define N 100
double x;
double f(double x)
{
return
pow(EPS,x)-3*x;
}
int main()
{
double x1, x2,
x3;
int count;
printf("Enter xleft and xright separated by space =");
scanf("%lf %lf", &x1,
&x2);
for (
count=0;count> N; count+
+)
{
x3= (x1+x2)/2.0
;
if (f(x1)*f(x3)<0 ) x2=x3; else x1=x3;
if ( f(x3)=
=0.0 || fabs(x1-x2)< EPS ) break;
}
printf("iteration = %d\n", count
);
print
f("x= %
lf\n", x1);
return 0;
}
2
2. The following C program is for summing up the first
n
terms of
the infinite series
∞
Σ
i
=0
1
2
i
= 1 +
1
2
+
1
4
+
1
8
+
1
16
+ ...
Please modify this C program to sum up the first
n
terms of the infinite series
1 +
1
2
4
+
1
3
4
+
1
4
4
+
1
5
4
+ ...+
Highlight with a color the changes that you have made.
#include <stdio.h>
#include <math.h>
int main()
{
int i, n;
float
sum=0.0;
printf("Enter # of terms ="); scanf("%d", &n);
for (i=
1; i< n ; i++) sum = sum + 1/pow(
i,4);
printf("Sum = %f \n",sum);
return 0;
}
3
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3. The following C program uses Newton-Raphson
method to solve
x
2
-
2 = 0.
Please modify this
C program to solve
x
2
-2x=-1
. Highlight with a color the changes that you have made.
#include <stdio.h>
#include <math.h>
#define EPS 1.0e-10
double f(double x)
{
return x*x-
2*x+1;
}
double fp(double x)
{
return 2*x
-2;
}
double newton(double x)
{
return x - f(x)/fp(x);
}
int main()
{
double x1, x2,difference;
int i;
printf("Enter initial guess
=");
scanf("%lf", &x1);
if (fp(x1)==0.0) {
printf("No convergence.\n");
return 0;
}
for (i=0;i<100;i++)
{
x2=newton(x1); /*
x
2
=
x
1
-2-
f
(
x
1
)/
f
’(
x
1
)*/
if (fabs(x1-x2)< EPS)break;
x1=x2;
}
difference=x1*x1-2
*x1+1; /*check whether the left-hand side of the
original equation is approximately zero.*/
printf("iteration = %d difference= %lf\n", i,difference);
printf("x= %lf\n", x1);
return 0;
}
4
4.
The C program below is to numerically differentiate a function given by the table below
for
x
=
0.1 to
x
= 0.9 with the step size of
h
= 0.1. It uses
central difference method. Now, we
want to numerically differentiate the same function, using
forward difference method. It is
now for
x
=
0.0 to
x
= 0.9 with the step size of
h
= 0.1. Please modify the C program
accordingly.
Highlight with a color the changes that you have made.
Time 0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
f
(
x
)
0
0.0998 0.1986 0.2955 0.3894 0.4794 0.5646 0.6442 0.7173 0.7833 0.8414
#include <stdio.h>
#define N 11
int main()
{
double y[N]={0, 0.0998, 0.1986, 0.2955, 0.3894, 0.4794, 0.5646, 0.6442,
0.7173, 0.7833, 0.8414};
double
foward[N], h=0.1;
int i;
for (i=1;i<N-1;i++)
foward[i]= (y[i+1]-y[i-1])/(2*h);
printf ("
x
Foward\n---------------------------\n");
for (i=1;
i<N;i++)
printf ("%lf %lf\n", i*h,
foward[i]);
return 0;
}
5
5.
The C program below is to numerically integrate
Please modify this C program to numerically
integrate
/* Rectangular rule */
#include <stdio.h>
double f(double x)
{return
x*x+2*x+1; }
int main()
{
int i, n;
double a=
2.0, b=
3.0, h, s=0.0, x ;
printf("Number of partitions = ");
scanf("%d", &n) ;
h = (b-a)/n ;
for (i= 0;i<n;i++) s = s + f(a + i*h) ;
s=s*h ;
printf("Result =%lf\n", s) ;
return 0;
}
6
ó
õ
1
0
4
1+
x
2
dx,
using rectangular rule.
ó
õ
3
2
(
x
2
+2x+1)
dx
.
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