OrthoMathetics defined as the science of correct
learning, whose primary purpose is to prevent
academic fraud and dishones...
OrthoMathetics®
For Teachers
Stopping the Cheating in Science!
Neftalí Antúnez H.
Civil Engineer with a Master's Degree in...
1
OrthoMathetics®
Neftali Antunez H.
“The teachers to teach less and the learner to learn more” Comenius
ISBN-13:978-14818...
2
This page is intentionally left blank
3
OrthoMathetics® For Teachers
Stopping the Cheating in Science!
NEFTALI ANTUNEZ H.
Civil Engineer with a Master’s Degree ...
4
This page is intentionally left blank
5
INDEX
Chapter 1
Definition
1.1 Roots of OrthoMathetics · 5
1.10 Characteristics of the constructivist student
· 16
1.11 ...
6
This page is intentionally left blank
7
Part I
Introduction
8
Chapter 1
Definition
1.1 Roots of OrthoMathetics
Introducing OrthoMathetics to the scientific community like a new branc...
9
Powerful Ideas, discusses the mathetics approach to learning. By using a mathetic
approach, Papert feels that independen...
10
In the second place, the
influence of Comenius was in
formulating the general theory of
education. In this respect he i...
11
influence in education has now been recognized for half a century. The
educational writings of Comenius comprise more t...
12
Academic dishonesty dates back to the first tests. Scholars note that cheating
was prevalent on the Chinese civil servi...
13
Moreover, there are online services that offer to prepare any kind of
homework of high school and college level and tak...
14
Cheating can take the form of crib notes, looking over someone’s shoulder
during an exam, or any forbidden sharing of i...
15
Everybody does it.
It’s disturbing to discover that young people in middle school and high school
think that it is acce...
16
fine for that purpose. Sadly, many parents and teachers have not learned the
subtleties of electronic cheating
1.8 Solu...
17
the learner, which can only be achieved by the learner that he makes, deals, and
engages in activities of learning. The...
18
In a traditional classroom, can see a teacher who it remains in front of the
group, mainly exposes and tries to fill wi...
19
research, i.e. as they learn how to learn, greater autonomy displayed when you
perform the activities established by th...
20
it proposes strategies that promote active learning, defined as instructional
activities that involve to the students i...
21
and exercises, evaluation of expressions, solving equations, factoring expressions,
operations with functions, write su...
22
The teacher in the classroom or in an examination may not propose problems
to the students, since due to the time, can ...
23
Chapter 2
OrthoMathetics Fundamentals
"We have allowed our schools to remain in the past while our
students were born i...
24
I don’t know that other commercial or free software meets the two
characteristics established. Could comply with these ...
25
acts during impact and Newton’s coefficient of restitution is defined only along
this line.
Specifically, collisions ca...
26
conserved during the collision, determine the velocity of the blocks immediately
after the collision. Consider that the...
27
Repeat the procedure for the Vb variable. The following window appears.
28
29
The final velocities depending on N are:
Is very useful to tabulate the responses using Derive® Version 6.1, the table
...
30
31
32
Elements of a triangle based on its coordinates
Other example, if I want to create a problem of Analytic Geometry of th...
33
To determine the area of a simple polygon whose vertices are described by
ordered pairs in the plane, for a triangle we...
34
where and are two points on the line with . This is
equivalent to the point-slope form above, where the slope is explic...
35
Helping me with Derive® Version 6.1, we use n instead of N.
We first define the variables Mab, Mbc and Mca, so do not c...
36
Calculating the equation of each side. Using the point-slope form:
37
38
The equations of the sides AB, BC, and CA respectively are:
The table of results is:
39
40
41
Calculating the perimeter and area:
42
The perimeters and areas of the polygon are:
43
44
45
Calculating the interior angles:
46
The interior angles A, B, C of the polygon are:
47
48
49
When the angle is negative we add (supplement of the angle) to get the
correct angle:
N Angle A
1 a = 171.5110561
2 a =...
50
4 a = 147.9946167
5 a = 141.2056422
6 a = 135
7 a = 129.3517526
8 a = 124.2157021
9 a = 119.5387822
10 a = 115.2673105
...
51
42 b = 107.0490309
43 b = 108.2775749
44 b = 109.4703020
45 b = 110.6285679
46 b = 111.7536709
47 b = 112.8468537
48 b ...
52
Another window appears to us, we select as the variable x and we give click
on Solve:
53
The answer is:
Select the result and now we give click on Calculus and then on Table:
54
Another window appears to us, we select as the parameter n, we set the
starting value to 1, the ending value equal to t...
55
The full table with all answers is:
N
1 -0.50 1
2 -1.00 2
3 -1.50 3
4 -2.00 4
5 -2.50 5
6 -3.00 6
7 -3.50 7
8 -4.00 8
9...
56
11 -5.50 11
12 -6.00 12
13 -6.50 13
14 -7.00 14
15 -7.50 15
16 -8.00 16
17 -8.50 17
18 -9.00 18
19 -9.50 19
20 -10.00 2...
57
39 -19.50 39
40 -20.00 40
41 -20.50 41
42 -21.00 42
43 -21.50 43
44 -22.00 44
45 -22.50 45
46 -23.00 46
47 -23.50 47
48...
58
Click again in the Expand button, we have:
The quadratic equation that we will put them so that students solve is:
59
The full table with all answers is:
N
1 0.3333 -0.25
2 0.6667 -0.50
3 1.0000 -0.75
4 1.3333 -1.00
5 1.6667 -1.25
6 2.00...
60
26 8.6667 -6.50
27 9.0000 -6.75
28 9.3333 -7.00
29 9.6667 -7.25
30 10.0000 -7.50
31 10.3333 -7.75
32 10.6667 -8.00
33 1...
61
Part II
OrthoMathetics Applications
62
Chapter 3
Applications to Algebra
Mathematics study the number, shape, size and variation. For me everything is based
o...
63
64
65
Example 6. Solve the following exponential equation:
Solving with Derive®, the results are:
66
67
68
Example 7. Multiply the following polynomials:
The answer is:
Example 8. Divide the following polynomials:
69
by
The answer is: Remainder = 0
Example 9. Divide the following polynomials:
by
The table of results is:
70
71
72
73
Example 10. Divide the following polynomials:
by
The table of results is:
74
75
76
77
78
79
Example 11. Divide the following polynomials:
by
The answer is:
Remainder = 0
Example 12. Multiply the following conjug...
80
ans.
Example 13. Expanding the binomial using Newton’s theorem:
The results are:
81
82
83
84
85
Example 14. Find the 12th term of the binomial and find the N-th
term of the binomial
The table of results is:
86
87
88
89
90
3.2 Systems of Equations Linear with Two and Three Variables
Example 15. To solve the following system of equations lin...
91
To use Derive® 6.1 instead of N, we use n. We click on Solve and then click
System, we have:
Another window appears to ...
92
The answer is:
The full table with all answers is:
93
N x y
1 2 -4
2 4 -8
3 6 -12
4 8 -16
5 10 -20
6 12 -24
7 14 -28
8 16 -32
9 18 -36
10 20 -40
11 22 -44
12 24 -48
13 26 -5...
94
29 58 -116
30 60 -120
31 62 -124
32 64 -128
33 66 -132
34 68 -136
35 70 -140
36 72 -144
37 74 -148
38 76 -152
39 78 -15...
95
N x y
1 -3 3
2 -6 6
3 -9 9
4 -12 12
5 -15 15
6 -18 18
7 -21 21
8 -24 24
9 -27 27
10 -30 30
11 -33 33
12 -36 36
13 -39 3...
96
29 -87 87
30 -90 90
31 -93 93
32 -96 96
33 -99 99
34 -102 102
35 -105 105
36 -108 108
37 -111 111
38 -114 114
39 -117 1...
97
Example 18. To solve the following system of equations linear with three variables:
To use Derive® 6.1 instead of N, we...
98
By being so obvious only some answers are shown.
99
Example 19. To solve the following system of equations linear with three variables:
In the same way, we obtain the answ...
100
The full table with all answers is:
N x y z
1 0.13889 0.02778 0.19444
2 0.27778 0.05556 0.38889
3 0.41667 0.08333 0.58...
101
10 1.38889 0.27778 1.94444
11 1.52778 0.30556 2.13889
12 1.66667 0.33333 2.33333
13 1.80556 0.36111 2.52778
14 1.94444...
102
39 5.41667 1.08333 7.58333
40 5.55556 1.11111 7.77778
41 5.69444 1.13889 7.97222
42 5.83333 1.16667 8.16667
43 5.97222...
103
104
105
106
107
Example 21. To solve the following system of equations linear with four variables:
108
To use Derive® 6.1 instead of N, we use n. We click on Solve and then click
System. Another window appears to us, we s...
109
The full table with all answers is:
N x y z t
1 15.0625 -19.75 -39.50 34.8125
2 15.1250 -19.50 -39.00 34.6250
3 15.187...
110
10 15.6250 -17.50 -35.00 33.1250
11 15.6875 -17.25 -34.50 32.9375
12 15.7500 -17.00 -34.00 32.7500
13 15.8125 -16.75 -...
111
39 17.4375 -10.25 -20.50 27.6875
40 17.5000 -10.00 -20.00 27.5000
41 17.5625 -9.75 -19.50 27.3125
42 17.6250 -9.50 -19...
112
The next results by rows are:
113
114
115
Example 23. Raise to square the following matrix A.
The matrices are introduced to Derive® 6.1, by clicking on the opt...
116
117
118
Example 24. Find the value of the determinant of the matrix below:
In Derive® the value of the determinant of a matrix...
119
The next results are:
120
121
122
Example 25. Find the matrix inverse of:
In Derive® the matrix inverse is obtained by typing:
[n, 1, 2; -1, -2, -n; 3, ...
123
The results by rows are:
124
125
126
127
128
Example 26. Find the eigenvalues of the matrix:
With Charpoly find the characteristic equation, which match to zero to...
129
The eigenvalues are:
130
131
3.4 Roots of Equations
3.4.1 Roots of quadratic equations
Example 27. Some additional examples of quadratic equations ...
132
Quadratic Equation Root 1 Root 2
Example 28. Some examples to Completing the Square:
Trinomial Answer
Example 29. For ...
133
134
Example 30. For use in numerical methods, find the roots of the following equation:
135
Their roots are:
The graphic for N = 25 is:
136
It is also very useful to tabulate the responses using Microsoft Excel® or
similar, placing N in first column and the ...
137
Example 31. For use in numerical methods, find the roots of the following equation:
138
Their roots are:
It is also very useful to tabulate the responses using Microsoft Excel® or
similar, placing N in firs...
139
140
Chapter 4
Applications to Geometry and
Trigonometry
4.1 Arc Length
ARC LENGTH PROBLEM S
where θ is the angle in radian...
141
The table of results is:
N S2, S3 x S4
1 0.26667 178 47.46667
2 .53334 176 46.93334
3 0.8 174 46.4
4 1.06667 172 45.86...
142
28 7.46667 124 33.06667
29 7.73334 122 32.53334
30 8 120 32
31 8.26667 118 31.46667
32 8.53334 116 30.93334
33 8.8 114...
143
The results for triangle 1 are: For all N, the angles are constant:
and
N Side b
1 2.23607
2 4.47214
3 6.70820
4 8.944...
144
14 31.30495
15 33.54102
16 35.77709
17 38.01316
18 40.24922
19 42.48529
20 44.72136
21 46.95743
22 49.19350
23 51.4295...
145
43 96.15092
44 98.38699
45 100.62306
46 102.85913
47 105.09519
48 107.33126
49 109.56733
50 111.80340
The results for ...
146
147
The results for triangle 3 are: For all N, the angle t is constant:
148
149
150
The results for triangle 4 are: For all N, the angle B is constant:
151
152
The code in BASIC programming language to solve previous triangles is:
defdbl a-z
cls
pi=3.141593
open "c:exageo.txt" ...
153
a=(58+37/60)
w=90-a
a=(58+37/60)*pi/180
b=i/sin(a)
x=sqr(b^2-i^2)
a=w
ag=int(a)
min=(a-ag)*60
am=int(min)
se=(min-am)*...
154
155
The results for triangle 2 are:
156
157
The results for triangle 3 are:
158
159
The triangle 4 is isosceles. For all N, the angles are constant: and
. The results for triangle 4 are:
160
161
162
Individual responses were obtained using the programming language
FORTRAN 90, whose source code is:
Program Triangles
...
163
mB=int(min)
secB=(min-mB)*60
WRITE(7,*) N,",",dc,"deg",mc,"’",secC,’",’,da,"deg",ma,"’",seca,’",’
WRITE(7,*) N,",",db,...
164
Chapter 5
Applications to Analytic Geometry
5.1 Circle
Example 35. Find the Center and radius of the circle whose gene...
165
Graphing the circle with Algebrator® 5 spanish version, for N = 30
5.2 Parabola
Example 37. Find the vertex, focus, fo...
166
Example 38. Find the vertex, focus, focal length p and the equation of directrix of the
parabola whose general equatio...
167
Example 39. Find the vertex, focus, focal length p and the equation of directrix of the
parabola whose general equatio...
168
Example 40. Find the vertex, focus, focal length p and the equation of directrix of the
parabola whose general equatio...
169
170
Chapter 6
Applications to Calculus
«Mathematics has been, traditionally, the torture of school children in the whole
w...
171
answer
6.2 Maxima and minima
Example 46. Find the maxima and minima of the following function:
Deriving and equating t...
172
Now, finding:
minimum
maximum
The Graph for N = 30 is:
Tabulating the results:
173
174
175
176
177
178
Example 47. Find the maxima and minima of the following function:
Deriving and equating to zero:
Solving:
Now, finding...
179
180
181
Example 48. Find the maxima and minima of the following function:
Deriving and equating to zero:
Solving:
Now, finding...
182
Tabulating the results using Microsoft Excel®:
183
184
Example 49. Find the maxima and minima of the following function:
Deriving and equating to zero:
Solving:
Now, finding...
185
186
187
6.3 Definite Integrals
Example 50. Find the definite integral of
from to
The results are:
188
189
Example 51. Find the area formed by the intersection of curves:
and
Using Derive® 6.1 to make the operations, we have:
190
191
The resultant area is:
192
193
194
195
196
The Graph for N = 30 is:
Example 52. Find the area formed by the intersection of curves:
and
Using Derive® 6.1 to make...
197
198
The resultant are is:
199
200
201
202
203
The Graph for N = 30 is:
204
Chapter 7
Applications to Physics
7.1 Vectors
Example 53. Find the resultant vector and its direction of the following...
205
D N 249.3000
E 80 290.8833
F 40 308.7500
P N 76.2833
Q 60 231.2500
Individual responses were obtained using the progra...
206
END DO
! Close the external file
CLOSE(7)
PRINT *,’ ’
PRINT *," THANK YOU FOR USING VECTRIX 2D"
END Program Vectrix2d
...
207
23 38.216370 52 53 32.402340
24 38.258000 52 43 14.888310
25 38.299980 52 32 58.733830
26 38.342290 52 22 43.938900
27...
208
51 39.507600 48 14 10.822450
52 39.558380 48 4 33.147580
53 39.609460 47 54 56.942140
54 39.660850 47 45 22.233580
55 ...
209
C 45 149.6000
D 100 180+N+18.10
E 70 288.6000
F 50 307.5000
P ? N+30.60
Q ? 180 + N
Individual responses were obtained...
210
CLOSE(7)
PRINT *,’ ’
PRINT *," THANK YOU FOR USING VECTRIX 2D"
END Program Equilibrium
The output file contains the ma...
211
24 29.438990 4.429714
25 31.985730 6.891172
26 34.541360 9.337624
27 37.105010 11.768230
28 39.675990 14.182350
29 42....
212
52 101.515500 65.340510
53 104.012500 67.119040
54 106.496600 68.864300
55 108.966800 70.575560
7.2 Coulomb’s Law and ...
213
To calculate the electric field at a point P due to a group of point charges, we
first calculate the electric field ve...
214
Since at the point P is considered a positive unit charge, each field produced
by the charges are shown in the followi...
215
angZ=atan((15.+n)/(n+4.))
angZ=2*pi-angZ
angT=atan((15.+n)/8.)
angT=pi+angT
! The charges are multiplied by 9x10E09 an...
216
13 1467205.00 -23 50 10.65216
14 1604501.00 -20 3 8.580322
15 1724258.00 -17 16 2.299347
16 1828479.00 -15 7 56.78604
...
217
42 2634005.00 -2 53 33.52936
43 2641808.00 -2 46 54.36951
44 2649119.00 -2 40 38.53729
45 2655976.00 -2 34 44.18633
46...
218
h=(0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/(500*9.8)
vb=sqr((0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/250)
vc=sqr((0.5*500*vb*v...
219
220
221
7.4 Kirchhoff’s Rules
The procedure for analyzing complex electric circuits is greatly simplified if
we use two princi...
222
Applying Kirchhoff’s junction rule to junction c gives:
We now have one equation with three unknowns . There are three...
223
(3) Loop cdfec:
Changing by respectively.
The system to solve is:
Solving with Derive®. We have:
224
225
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Math for Smart Teachers. Ortho mathetics® for-teachers-stopping-the-cheating-in-science

  1. 1. OrthoMathetics defined as the science of correct learning, whose primary purpose is to prevent academic fraud and dishonesty, preventing students from copying answers on science problems and exercises. OrthoMathetics® For Teachers Stopping the Cheating in Science! Neftali Antunez H.
  2. 2. OrthoMathetics® For Teachers Stopping the Cheating in Science! Neftalí Antúnez H. Civil Engineer with a Master's Degree in Education Full time Teacher in the Faculty of Engineering of the UAG Chilpancingo, Gro., México Introducing OrthoMathetics to the scientific community, a new branch of education whose name is derived from Ortho (from the Greek word ὀρθός meaning "straight" or "correct") and Mathetics means the science of learning. Emerges as a response to academic dishonesty or academic misconduct in any type of cheating that occurs in the schools everywhere. ISBN-13:978-1481859462 ISBN-10:1481859463
  3. 3. 1 OrthoMathetics® Neftali Antunez H. “The teachers to teach less and the learner to learn more” Comenius ISBN-13:978-1481859462 ISBN-10:1481859463
  4. 4. 2 This page is intentionally left blank
  5. 5. 3 OrthoMathetics® For Teachers Stopping the Cheating in Science! NEFTALI ANTUNEZ H. Civil Engineer with a Master’s Degree in Education Full time Teacher in the Faculty of Engineering of the Guerrero State Autonomous University (UAG) Chilpancingo, Gro., México Introducing OrthoMathetics to the scientific community like a new branch of education. OrthoMathetics defined as the science of correct learning, whose primary purpose is to prevent academic fraud and dishonesty, preventing students from copying answers on science problems and exercises. Emerges as a response to academic dishonesty or academic misconduct in any type of cheating that occurs in the schools everywhere.
  6. 6. 4 This page is intentionally left blank
  7. 7. 5 INDEX Chapter 1 Definition 1.1 Roots of OrthoMathetics · 5 1.10 Characteristics of the constructivist student · 16 1.11 Constructivist evaluation · 17 1.12 Difference between exercise and problem · 18 1.2 Mathetics in literature · 5 1.3 John Amos Comenius · 6 1.3.1 Educational influence · 6 1.4 Purposes of OrthoMathetics · 8 1.5 History of academic dishonesty · 8 1.6 Academic dishonesty today · 9 1.7 Cheating · 10 1.7.1 The Issue · 11 1.8 Solutions? · 13 1.9 Characteristics of the constructivist teacher · 13 Chapter 2 OrthoMathetics Fundamentals 2.1 How to do problems with features of OrthoMathetics · 21 Chapter 3 Applications to Algebra 3.1 Operations with polynomials · 59 3.2 Systems of Equations Linear with Two and Three Variables · 87 3.3 Matrices and determinants · 108 3.4 Roots of Equations · 128 Chapter 4 Applications to Geometry and Trigonometry 4.1 Arc Length · 137 4.2 Right-angled Triangles · 139 4.3 Non Right-angled Triangles · 150 Chapter 5 Applications to Analytic Geometry 5.1 Circle · 161 5.2 Parabola · 162 Chapter 6 Applications to Calculus 6.1 Derivatives · 167 6.2 Maxima and minima · 168 6.3 Definite Integrals · 184 Chapter 7 Applications to Physics 7.1 Vectors · 201 7.2 Coulomb’s Law and Electric Field · 209 7.3 Conservation of mechanical energy · 215 7.4 Kirchhoff’s Rules · 219 CHAPTERS Chapter 1 Definition · 5 Chapter 2 OrthoMathetics Fundamentals · 20 Chapter 3 Applications to Algebra · 59 Chapter 4 Applications to Geometry and Trigonometry · 137 Chapter 5 Applications to Analytic Geometry · 161 Chapter 6 Applications to Calculus · 167 Chapter 7 Applications to Physics · 201 PARTS Part I Introduction · 4 Part II OrthoMathetics Applications · 58
  8. 8. 6 This page is intentionally left blank
  9. 9. 7 Part I Introduction
  10. 10. 8 Chapter 1 Definition 1.1 Roots of OrthoMathetics Introducing OrthoMathetics to the scientific community like a new branch of education, whose name is Derived from Ortho (from the Greek word ὀρθός meaning "straight" or "correct") and Mathetics means the science of learning. The term Mathetics was coined by John Amos Comenius (1592–1670) in his work Spicilegium didacticum, published in 1680. He understood Mathetics as the opposite of Didactic, the science of teaching. Mathetics considers and uses findings of current interest from pedagogical psychology, neurophysiology and information technology. OrthoMathetics defined as the science of correct learning, whose primary purpose is to prevent academic fraud and dishonesty, preventing students from copying answers on science problems and exercises. Promoting individual real learning, where each of the students have to appropriate the knowledge, because it can not be copied because each student has an unique answer to exercise or problem. The main interest is that this project will contribute to the improvement of scientific education as a new approach to the traditional teaching. 1.2 Mathetics in literature1 Seymour Papert, MIT mathematician, educator, and author, explains the rationale behind the term mathetics in Chapter 5 (A Word for Learning) of his book, The Children’s Machine. The origin of the word, according to Papert, is not from "mathematics," but from the Greek, mathēmatikos, which means "disposed to learn." He feels this word (or one like it) should become as much part of the vocabulary about education as is the word pedagogy or instructional design. In Chapter 6 of The Children’s Machine, Papert mentions six case studies, and all six have their own accompanying learning moral and they all continue his discussion of his views of mathetics. Case study 2 looks at people who use mathematics to change and alter their recipes while cooking. His emphasis here is the use of mathematical knowledge without formal instruction, which he considers to be the central mathetics moral of the study. Papert states "The central epistemological moral is that we all used concrete forms of reasoning. The central mathetics moral is that in doing this we demonstrated we had learned to do something mathematical without instruction – and even despite having been taught to proceed differently" (p. 115).[1] Papert’s 1980 book, Mindstorms: Children, Computers, and 1 From Wikipedia, the free encyclopedia
  11. 11. 9 Powerful Ideas, discusses the mathetics approach to learning. By using a mathetic approach, Papert feels that independent learning and creative thinking are being encouraged. The mathetic approach is a strong advocate of learning by doing. Many proponents of the mathetic approach feel that the best, and maybe the only, way to learn is by self discovery. 1.3 John Amos Comenius (1592–1670)2 John Amos Comenius (Czech: Jan Amos Komenský; Slovak: Ján Amos Komenský; German: Johann Amos Comenius; Polish: Jan Amos Komeński; Hungarian: Comenius Ámos János; Latinized: Iohannes Amos Comenius) (28 March 1592 – 4 November 1670) was a Czech teacher, educator, and writer. He served as the last bishop of Unity of the Brethren, and became a religious refugee and one of the earliest champions of universal education, a concept eventually set forth in his book Didactica Magna. He is considered the father of modern education. He lived and worked in many different countries in Europe, including Sweden, the Polish-Lithuanian Commonwealth,Transylvania, the Holy Roman Empire, England, the Netherlands, and Royal Hungary. 1.3.1 Educational influence The most permanent influence exerted by Comenius was in practical educational work. Few men since his day have had a greater influence, though for the greater part of the eighteenth century and the early part of the nineteenth there was little recognition of his relationship to the current advance in educational thought and practice. The practical educational influence of Comenius was threefold. He was first a teacher and an organizer of schools, not only among his own people, but later in Sweden, and to a slight extent in Holland. In his Didactica Magna (Great Didactic), he outlined a system of schools that is the exact counterpart of the existing American system of kindergarten, elementary school, secondary school, college, and university. 2 From Wikipedia, the free encyclopedia
  12. 12. 10 In the second place, the influence of Comenius was in formulating the general theory of education. In this respect he is the forerunner of Rousseau, Pestalozzi, Froebel, etc., and is the first to formulate that idea of “education according to nature” so influential during the latter part of the eighteenth and early part of the nineteenth century. The influence of Comenius on educational thought is comparable with that of his contemporaries, Bacon and Descartes, on science and philosophy. In fact, he was largely influenced by the thought of these two; and his importance is largely due to the fact that he first applied or attempted to apply in a systematic manner the principles of thought and of investigation, newly formulated by those philosophers, to the organization of education in all its aspects. The summary of this attempt is given in the Didactica Magna, completed about 1631, though not published until several years later. The third aspect of his educational influence was that on the subject matter and method of education, exerted through a series of textbooks of an entirely new nature. The first-published of these was the Janua Linguarum Reserata (The Gate of Languages Unlocked), issued in 1631. This was followed later by a more elementary text, the Vestibulum, and a more advanced one, the Atrium, and other texts. In 1657 was published the Orbis Sensualium Pictus probably the most renowned and most widely circulated of school textbooks. It was also the first successful application of illustrations to the work of teaching, though not, as often stated, the first illustrated book for children. These texts were all based on the same fundamental ideas: (1) learning foreign languages through the vernacular; (2) obtaining ideas through objects rather than words; (3) starting with objects most familiar to the child to introduce him to both the new language and the more remote world of objects: (4) giving the child a comprehensive knowledge of his environment, physical and social, as well as instruction in religious, moral, and classical subjects; (5) making this acquisition of a compendium of knowledge a pleasure rather than a task; and (6) making instruction universal. While the formulation of many of these ideas is open to criticism from more recent points of view, and while the naturalistic conception of education is one based on crude analogies, the importance of the Comenian
  13. 13. 11 influence in education has now been recognized for half a century. The educational writings of Comenius comprise more than forty titles. In 1892 the three-hundredth anniversary of Comenius was very generally celebrated by educators, and at that time the Comenian Society for the study and publication of his works was formed. 1.4 Purposes of OrthoMathetics Emerges as a response to academic dishonesty or academic misconduct in any type of cheating, which occurs in the schools everywhere. Academic dishonesty has been documented in most every type of educational setting, from elementary school to graduate school, and has been met with varying degrees of approbation throughout history. Today, educated society tends to take a very negative view of academic dishonesty. This project is especially directed to the teachers of science in the world. For this reason, dont make an emphasis on theory and full development of the examples, is not because that is not its main objective. We only give examples so that the teacher can develop their own exercises from the examples presented here. This is a dynamic book, which will grow because will be adding more examples by the same author, or, by the contribution of thousands of teachers in the world. Can send their examples to my email: antunezsoftware@gmail.com, or better yet, write their own books according to the rules of OrthoMathetics. Some titles of books could be: OrthoMathetics for Algebra, OrthoMathetics for Calculus, OrthoMathetics for Physics, OrthoMathetics for Chemistry, OrthoMathetics for Numerical Methods, etc. All that is asked is that in the preface, prologue or introduction, you specify that OrthoMathetics was created by the author of this book. 1.5 History of academic dishonesty3 In antiquity, the notion of intellectual property did not exist. Ideas were the common property of the literate elite. Books were published by hand-copying them. Scholars freely made digests or commentaries on other works, which could contain as much or as little original material as the author desired. There was no standard system of citation, because printing—and its resulting fixed pagination—was in the future. Scholars were an elite and a small group, who knew and generally trusted each other. This system continued through the European Middle Ages. Education was in Latin and occasionally Greek. Some scholars were monks, lived in monasteries, and spent much of their time copying manuscripts. Other scholars were in urban universities connected to the Roman Catholic Church. 3 From Wikipedia, the free encyclopedia
  14. 14. 12 Academic dishonesty dates back to the first tests. Scholars note that cheating was prevalent on the Chinese civil service exams thousands of years ago, even when cheating carried the penalty of death for both examinee and examiner. In the late 19th and early 20th centuries, cheating was widespread at college campuses in the United States, and was not considered dishonorable among students. It has been estimated that as many as two-thirds of students cheated at some point of their college careers at the turn of the 20th century. Fraternities often operated so-called essay mills, where term papers were kept on file and could be resubmitted over and over again by different students, often with the only change being the name on the paper.[citation needed] As higher education in the U.S. trended towards meritocracy, however, a greater emphasis was put on anti-cheating policies, and the newly diverse student bodies tended to arrive with a more negative view of academic dishonesty. Unluckily, in some areas academic dishonesty is widely spread and people who do not cheat represent a minority between the class. 1.6 Academic dishonesty today4 Academic dishonesty is endemic in all levels of education. In the United States, studies show that 20% of students started cheating in the first grade. Similarly, other studies reveal that currently in the U.S., 56% of middle school students and 70% of high school students have cheated. Students are not the only ones to cheat in an academic setting. A study among North Carolina school teachers found that some 35 percent of respondents said they had witnessed their colleagues cheating in one form or another. The rise of high-stakes testing and the consequences of the results on the teacher is cited as a reason why a teacher might want to inflate the results of their students. The first scholarly studies in the 1960s of academic dishonesty in higher education found that nationally in the U.S., somewhere between 50%-70% of college students had cheated at least once. While nationally, these rates of cheating in the U.S. remain stable today, there are large disparities between different schools, depending on the size, selectivity, and anti-cheating policies of the school. Generally, the smaller and more selective the college, the less cheating occurs there. For instance, the number of students who have engaged in academic dishonesty at small elite liberal arts college scan be as low as 15%-20%, while cheating at large public universities can be as high as 75%. Moreover, researchers have found that students who attend a school with an honor code are less likely to cheat than students at schools with other ways of enforcing academic integrity. As for graduate education, a recent study found that 56% of MBA students admitted cheating, along with 54% of graduate students in engineering, 48% in education, and 45% in law. 4 From Wikipedia, the free encyclopedia
  15. 15. 13 Moreover, there are online services that offer to prepare any kind of homework of high school and college level and take online tests for students. While administrators are often aware of such websites, they have been unsuccessful in curbing cheating in homework and non-proctored online tests, resorting to a recommendation by the Ohio Mathematics Association to derive at least 80% of the grade of online classes from proctored tests. While research on academic dishonesty in other countries is minimal, anecdotal evidence suggests cheating could be even more common in countries like Japan. The use of crib notes during an examination is typically viewed as cheating. 1.7 Cheating5 Cheating refers to an immoral way of achieving a goal. It is generally used for the breaking of rules to gain advantage in a competitive situation. Cheating is the getting of reward for ability by dishonest means. This broad definition will necessarily include acts of bribery, cronyism, sleaze, nepotism and any situation where individuals are given preference using inappropriate criteria. The rules infringed may be explicit, or they may be from an unwritten code of conduct based on morality, ethics or custom, making the identification of cheating a subjective process. 5 From Wikipedia, the free encyclopedia
  16. 16. 14 Cheating can take the form of crib notes, looking over someone’s shoulder during an exam, or any forbidden sharing of information between students regarding an exam or exercise. Many elaborate methods of cheating have been developed over the years. For instance, students have been documented hiding notes in the bathroom toilet tank, in the brims of their baseball caps, or up their sleeves. Also, the storing of information in graphing calculators, pagers, cell phones, and other electronic devices has cropped up since the information revolution began. While students have long surreptitiously scanned the tests of those seated near them, some students actively try to aid those who are trying to cheat. Methods of secretly signaling the right answer to friends are quite varied, ranging from coded sneezes or pencil tapping to high-pitched noises beyond the hearing range of most teachers. Some students have been known to use more elaborate means, such as using a system of repetitive body signals like hand movements or foot jerking to distribute answers (i.e. where a tap of the foot could correspond to answer "A", two taps for answer "B", and so on). Cheating differs from most other forms of academic dishonesty, in that people can engage in it without benefiting themselves academically at all. For example, a student who illicitly telegraphed answers to a friend during a test would be cheating, even though the student’s own work is in no way affected. Another example of academic dishonesty is a dialogue between students in the same class but in two different time periods, both of which a test is scheduled for that day. If the student in the earlier time period informs the other student in the later period about the test; that is considered academic dishonesty, even though the first student has not benefited himself. This form of cheating—though deprecated— could conceivably be called altruistic. 1.7.1 The Issue6 Cheating in our schools has reached epidemic proportions. Why do students cheat? What can we as parents do to prevent it? Here are some answers to these questions and much more in this article which features an in-depth interview with one of the nation’s top authorities on the subject, Gary Niels. Why do students cheat? 1. Everybody does it. 2. Unrealistic demands for academic achievement by state education boards 3. Expediency or the easy way out 6 http://privateschool.about.com/cs/forteachers/a/cheating.htm
  17. 17. 15 Everybody does it. It’s disturbing to discover that young people in middle school and high school think that it is acceptable to cheat. But it’s our fault, isn’t it? We adults encourage young people to cheat. Take multiple choice tests, for example: they literally invite you to cheat. Cheating, after all, is nothing more than a game of wits as far as teenagers are concerned. Kids delight in outwitting adults, if they can. While cheating is discouraged in private schools by tough Codes of Behavior which are enforced, cheating still exists. Private schools which devise tests requiring written answers rather than multiple guess answers discourage cheating. It’s more work for teachers to grade, but written answers do eliminate an opportunity for cheating. Unrealistic demands for academic achievement by state and federal education authorities. The public education sector is accountable to government, largely as a result of No Child Left Behind. State legislatures, state boards of education, local boards of education, unions, and countless other organizations demand action to correct the real and imagined failings of our nation’s public education system. As a result, students must take standardized tests so that we can compare one school system to another nationally and at the state level. In the classroom these tests mean that a teacher must achieve the expected results or better, or she will be viewed as ineffective, or worse, incompetent. So instead of teaching your child how to think, she teaches your child how to pass the test. No Child Left Behind is driving most of the assessment teaching these days. Educators really have no option but to produce the best possible results. To do that they must teach solely to the test or else. The best antidotes for cheating are teachers who fill children with a love of learning, who impart some idea of life’s possibilities and who understand that assessment is merely a means to an end, not the end itself. A meaningful curriculum will shift the focus from learning boring lists of irrelevant facts to exploring subjects in depth. Expediency or the easy way out Years ago cheaters lifted whole passages from an encyclopedia and called them their own. That was plagiarism. Plagiarism’s newest incarnation is dead easy: you simply point and click your way to the site with the relevant information, swipe and paste it, reformat it somewhat and it’s yours. Need to write a paper in a hurry? You can quickly find a site which provide a paper for a fee. Or go to a chat room and swap papers and projects with students nationwide. Perhaps you’d prefer to cheat using texting or email. Both work just
  18. 18. 16 fine for that purpose. Sadly, many parents and teachers have not learned the subtleties of electronic cheating 1.8 Solutions? Schools need to have zero tolerance policies concerning cheating. Teachers must be vigilant and alert to all the newer forms of cheating, particularly electronic cheating. Cellphones and iPods are powerful tools for cheating with uses limited only by a student’s imagination. How do you fight that kind of brain power? Discuss the issue with both technology-savvy students and adults. Their exploits and perspective will help you fight electronic cheating. Teachers Ultimately the best solution is to make learning exciting and absorbing. Teach the whole child. Make the learning process student-centric. Allow students to buy into the process. Empower them to guide and direct their learning. Encourage creativity and critical thinking as opposed to rote learning. Parents We parents have a huge role to play in combating cheating. That’s because our children mimic almost everything we do. We must set the right sort of example for them to copy. We must also take a genuine interest in our children’s work. Ask to see everything and anything. Discuss everything and anything. An involved parent is a powerful weapon against cheating. Students Students must learn to be true to themselves and their own core values. Don’t let peer pressure and other influences steal your dream. If you are caught, cheating has serious consequences. 1.9 Characteristics of the constructivist teacher7 To be a constructivist teacher, the first requirement is to dominate widely the contents of their subjects, without this, it is impossible to be a good teacher constructivist or traditional. The teacher is a facilitator or Coordinator in learning approach constructivist. The teacher guides the student, stimulating and provoking student’s critical thinking, analysis and synthesis through the learning process. The teacher is also a co-learner. Learning should be an active process. Learning requires a change in 7 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftalí Antúnez H.
  19. 19. 17 the learner, which can only be achieved by the learner that he makes, deals, and engages in activities of learning. The role of the teacher is important to achieve that the students carry out activities that otherwise would not do. The teacher has to involve students in tasks, some of which may include skills acquisition for examples of work. Other tasks include practices of skills to bring them to effective levels, interacting with their peers and the teacher. In a traditional classroom, the teacher’s role is of a transmitter of knowledge and the role of the student is being a passive recipient of such knowledge. In the proposed environment occurs a cooperative egalitarian structure where the ideas and interests of the students are those that drive the learning process. The master serves as a guide, more than source of knowledge. The master involves students helping them organize and assist them in accordance they take the initiative in his self-directed, rather than with authority directing explorations their learning process. Flexibility is the feature most important of the new role that the teacher should play in such environment. Sometimes the teacher will find that his paper tends towards the old model of the teacher as the giver of knowledge, because they sometimes students require guidance and training on a particular task or the contents of a subject. Often, the teacher will walk moving around the classroom, among groups of students, assisting them individually or to the group. The advantage of working in team is not only due to the work cooperative or collaborative, but also that the teacher can give them a more individualized attention to the students. In fact, in constructivism, the interaction between the students is very important to achieve meaningful learning and especially for the social development of the individual. In the Constructivism, the role of the teacher is more complex, since in first of all requires that the teacher be more prepared academically, given that there will be various doubts will be clarified or simply to properly orient the students and able to resolve their questions or problems. The teacher is also a member of the Group and not the focus of the classroom, in fact becomes a learner again, but with the difference of being responsible and driver - facilitator or coordinator- the learning of the group. You have to provide technical assistance and creative consulting, rather than directing students to the creation of tasks closely defined. Students come to the teacher when they need help, but the role of the teacher is more than colleague than of an upper. The master is a friend of students, gives them motivation and confidence, it gets to the level of students and uses the same language. The teacher learn together with their pupils, not so much in knowledge, but if in new ways to do them tasks or do things, so as the way to solve problems. Students need construct their own understandings of each concept, so that the primary role of the teacher is not to teach, explain, or any other attempting to "pass" the knowledge, but the create situations for motivate students to undertake their own constructions mental. The challenge of teaching is to build skills in students, from so that they can continue learning and building your own understandings based on the changing world around them.
  20. 20. 18 In a traditional classroom, can see a teacher who it remains in front of the group, mainly exposes and tries to fill with information your student’s heads, as if they were empty. But this is what they will not if they visit a classroom constructivist; you can see the master moved from one side to another within the group, going from one team to another; mixed with students sometimes it will not be easy to find. There is always a murmur of activity; students working together as a team making a circle or around a roundtable, solving problems, reading ones to others and sharing ideas. Many different activities seem to be carried out at the same time. In constructivist classrooms, teachers describe themselves as contributors, team leaders and guides, not as heads or authoritarian managers. Constructivist teachers asking more than they explain, model rather than teach, they work harder than or equal in a traditional classroom. This means that not always the teacher directs the dialog in the classroom, sometimes the students do, while you listen to them and respond later; is not the only that judged the work of students; students learn to evaluate the work of their peers and be self-evaluated. Education is considered equal to communication. But in one total communication receiver in certain moment turns into transmitter, this is impossible in a traditional education, since it the communication is one-way, where recipients are passive, not they respond messages but that they remain in silence by accepting that is said to them. The constructivist approach accepts the communication complete between the teacher and students; a full dialogue is accepted, since the teacher is one member of the group, although with one of the most important roles, since he is in charge of the learning. To create a good communication, the teacher has that create an atmosphere of trust and harmony in the classroom, a right atmosphere to promote the learning and participation of the students. A constructivist communication system does not mean that the teacher abandons his responsibility. While students have a greater role so that they direct their own learning, they dont are allowed to do whatever they want. The role of the teacher is to guide, orient, sharpen, suggest, organize, select and continuously evaluate the progress of students. Yes, even the role of the teacher is to give direct instruction as in the traditional approach, primarily, when students do not possess the knowledge requirement indispensable to learn new knowledge. The master It also has the responsibility to correct the process when it is not giving the academically relevant results, since has that take the necessary actions to ensure that this happens. The task of the constructivist teacher is to design a series of experiences to the students that will enable them to learn effectively and motivate them to be involved in the relevant activities. The master constructivist sets problems and monitors the search for solution the problems that students perform, guide the search and It promotes new thinking patterns. Class sessions can take unexpected twists as students acquire autonomy for perform their own exploration or
  21. 21. 19 research, i.e. as they learn how to learn, greater autonomy displayed when you perform the activities established by the teacher, which mostly consist of problem solving in mathematics. Constructivist learning is based on the active participation of students in solving problems and in the use of critical thinking regarding the learning activities that are relevant and interesting. They are constructing their own knowledge by testing ideas and approaches based on their knowledge and previous experience, applying them to a new situation, and integrating the new knowledge gained with their pre-existing intellectual constructions. 1.10 Characteristics of the constructivist student8 Constructivist students actively participate and are not limited to passively receiving information. Get involved and are responsible for their learning, investigate, seek, ask, discuss and dialogue with their peers and the teacher. The students read, think and analyze the information and not accept it without thinking, they expose their ideas to the other and work as a team. They perform their tasks and extracurricular work and in the case of math, solve problems and exercises. To learn, a student must be often physically and mentally active. A student learns (this it is, builds structures of knowledge) when it discovers its own answers, solutions, concepts and relationships and creates its own interpretations. Constructivism proposes that when students drive their own learning, discover their own answers and create their own interpretations, their learning is deeper, more comprehensive and lasting, and the learning that takes place actively it leads to think critically. In a constructivist classroom, the students demonstrate their learning and understanding through several forms. They can develop new critical issues, write a script for a video, summarizing the main ideas with your own words, can produce or create something, they can solve problems. It’s about being active, not passive." To change the traditional education, is required to change from an education centered on the teacher to one focused on students, an education where the learners take an active role, i.e. where students participate and responsibly are involved in your learning. Some claim that all learning is inherently active and students are also actively involved while listening the formal presentations in the classroom.. However, the majority agrees that not enough students listen, in addition, they should read, discuss and above all solve problems. Most importantly, must be actively involved, students must perform tasks of higher order thinking such as analysis, synthesis and evaluation. Within this context, be 8 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftalí Antúnez H.
  22. 22. 20 it proposes strategies that promote active learning, defined as instructional activities that involve to the students in doing things and think about what they are doing. 1.11 Constructivist evaluation9 The evaluation and learning are processes, therefore, not should be evaluated only using the test - as some teachers usually do-, but using many more tools and considering all aspects both objective and subjective; In addition, like process students should be continually evaluated and not only in certain isolated moments. Several instruments of assessment must be used and not one instrument or a method most appropriate must select to the activity that is being evaluated. Evaluation has subjective aspects – motivation, dedication, effort, emotions-, therefore not should be evaluated only the learning for students. This should be taken into account when designing our objectives to be achieved and the activities to assess. It is not enough to assess the learning that our students carry out, as teachers we must evaluate our own performance and teaching activities which we plan and develop. Evaluation is an important and fundamental action in the process of learning, but must serve to support this process, so that reinforce learning and help to feed both the teacher as to the students. However, it is not easy to correctly evaluate and more if we use constructivist theory, where the students are gradually building your knowledge. The evaluation must assess different capacities learned: motor, cognitive, affective or emotional balance, relationship interpersonal, and performance and social inclusion. Evaluate the learning acquired by the students is equivalent to clarify until point have developed or learned certain skills as a result of the education received. All teachers are trying to achieve meaningful learning in your students, but how evaluate the meaningful learning? With activities that allow to evaluate the advancement of learning, in the case of mathematics by means of the resolution of problems of difficulty gradual. The significant learning should not be evaluated as all or nothing, but rather as a degree of progress, you must detect the degree of significance of learning done. Constructivist Learning is usually assessed through projects based on performance, instead of traditional paper and pencil tests. The constructivist evaluation focuses on what the student can do with the knowledge. In general, it is recommended that the student demonstrate their learning through the application of knowledge, for example; solving problems 9 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftalí Antúnez H.
  23. 23. 21 and exercises, evaluation of expressions, solving equations, factoring expressions, operations with functions, write summaries and essays, create a product, model or prototype, create a video, writing literature, music or poetry, create or conduct experiments. 1.12 Difference between exercise and problem10 There is a basic difference between the concept ’problem’ and ’exercise’. Not is the same an exercise that to solve a problem. An exercise serves for exercise, practice or reinforce learning of an algorithm or a method, but does not contribute any new knowledge. In change, the problem involves a degree of difficulty and a depth greater, which requires more time to resolve it and forced to perform always an investigation. One thing is to apply an algorithm in form more or less mechanics as it is done in the exercise and another, solve a problem. The answer tends to be unique, but the adjudicative strategy is determined by maturation or other factors. In fact, the exercise does not require the preparation of a plan or a strategy, since it is only necessary to apply the known method to solve it. Problem solving strategy is much richer that the mechanical application of an algorithm, as it involves creating a context where the data store certain coherence. Since this analysis have been establish hierarchies: see which data are priority, reject the distorting elements or distractors, choose transactions to the they relate, to estimate the range of variation, identify the unknown, establish a plan or strategy, execute the plan and check the results obtained. Students who learn math, from a point of view constructivist, must precisely built concepts through the interaction that has with objects and other subjects. It seems so the student can build their knowledge to carry out the active interaction with the mathematical objects, including the reflection that allows you to abstract these objects, it is necessary that these objects are immersed in a problem and not in an exercise. In fact, are these problematic situations which introduce an imbalance in mental structures of the student, in his interest for balance (an accommodation) occurs the construction of the knowledge. The term problem is a complex situation (real or hypothetical) that it involves concepts, objects or mathematical operations. A problem requires time to solve because of its complexity; the student requires your full attention, energy, time and dedication to solve it. Exercise, refers to operations with symbols mathematical only (sums, multiplication, resolution of equations, and so on), implies that the student repeat an algorithm, procedure, or a taught method to directly and quickly get to the solution. 10 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftalí Antúnez H.
  24. 24. 22 The teacher in the classroom or in an examination may not propose problems to the students, since due to the time, can only propose them exercises. The problems are only left like task or homework, because must carry one or more days to resolve them; but such problems should not exceed the level of knowledge of students, but with the knowledge that they possess must be able to solve them, even if they don’t initially know how to start your solution. Is very desirable that the teacher of Mathematics knows the difference between problem and exercise; also, it is important to tell to the students when to them giving a problem, because many of them not knowing and not being able to solve it, they are blocked from this situation and can cause that failing during the course. The exercises are proposed in the classroom or examination. The problems are left of homework for one or more days.
  25. 25. 23 Chapter 2 OrthoMathetics Fundamentals "We have allowed our schools to remain in the past while our students were born in the future. The result is a discrepancy between the educator and learner. But are not students who do not correspond to the schools, but that schools do not correspond to the students." It focuses on preventing academic fraud and academic dishonesty. Its aim is to put an end to copy exams, tests, problems and exercises of basic sciences. It is not intended to be a new theory of teaching or learning, where we know that Constructivism is the current theory that has been tested and accepted throughout the education world. Using OrthoMathetics will stop copying, will promote a real individual learning, because it will force every student to appropriate knowledge, because your problem is personal and has a single answer that only him must find. To use OrthoMathetics you need to use any numerical characteristic unique to each student, as it can be: your list number, your registration, your student number or ID, etc. To this we designate by N. PROPOSITION 1: IN ORTHOMATHETICS, ALL PROBLEMS AND EXERCISES MUST CONTAIN N AND SHOULD GIVE A RESULT UNIQUE FOR EACH STUDENT. THE ANSWERS THE ALL PROBLEMS AND EXERCISES IT SHOULD BE FUNCTION OF N. The teacher must design the problems and exercises to achieve meaningful learning. PROPOSITION 2: N IS INCLUDED AS DATA ON THE PROBLEM, IN A MANNER SUCH THAT SHOULD ALLOW TO EVALUATE ANY DESIRED ASPECT OF LEARNING. ALL PROBLEMS AND EXERCISES MUST STOPPING THE CHEATING. Also, you need to use computer software that meets two main characteristics: 1. ALLOW TO WORK WITH N AS A PARAMETER, IN ADDITION TO BEING ABLE TO WORK WITH THE TRADITIONAL VARIABLES X, Y, Z,... AND WITH THE TRADITIONAL CONSTANTS A, B, C,... 2. MAKE A TABLE OF THE ALL ANSWERS USING THE N PARAMETER FROM 1 TO M, BEING M THE NUMBER OF STUDENTS OR THE NUMERICAL CHARACTERISTIC UNIQUE TO EACH STUDENT. Any software that meets these two conditions will be called: “OrthoMathetics Software” To achieve this, I use Derive® 6.1 of Texas Instruments, which unfortunately already was discontinued, although still being incorporated into their calculators that manufactures.
  26. 26. 24 I don’t know that other commercial or free software meets the two characteristics established. Could comply with these features other software such as: Matlab®, Octave®, Maxima®, Mathematica®, MathCAD®, Maple®, etc. Sometimes, I’ve used a programming language such as Basic, FORTRAN or C, to obtain the custom answers to the problem in question. If you want to run it, copy and paste the code in the compiler, and then run it. All FORTRAN programs can be executed in Microsoft FORTRAN Powerstation or in GNU FORTRAN compiler. The Basic programs can be executed in a free compiler of sourceforge.com. Also, sometimes, the use of some advanced scientific calculator is required to calculate the answers of each student. ONCE THE TEACHER HAS EXPLAINED THE THEORY AND HAS GIVEN GENERAL EXAMPLES OF THE TOPIC, THEN THE TEACHER PROPOSE AN EXERCISE TO THEIR STUDENTS. IN THIS CASE, THE FIRST THING THAT HAS TO REPLACE THE STUDENT IS YOUR NUMBER N IN THE GIVEN EXERCISE AND SHOULD THEN BEGIN TO SOLVE IT AND FIND THEIR PERSONAL ANSWER. 2.1 How to do problems with features of OrthoMathetics Using all the theory of the subject in question, the teacher proposes the problem with some data that include N. As teacher, I have mastered the subject and I must be able to resolve the problem manually or with software. I only propose problems to my students that I myself can resolve, which previously gave them general and complete examples. For example, if I want to create a problem of Physics of the subject of collisions. The theory of this topic is: Collisions in One Dimension We use the law of conservation of linear momentum to describe what happens when two particles collide. We use the term collision to represent an event during which two particles come close to each other and interact by means of forces. The time interval during which the velocities of the particles change from initial to fi- nal values is assumed to be short. The interaction forces are assumed to be much greater than any external forces present, so we can use the impulse approximation. Collisions involve forces (there is a change in velocity). The magnitude of the velocity difference at impact is called the closing speed. All collisions conserve momentum. What distinguishes different types of collisions is whether they also conserve kinetic energy. Line of impact - It is the line which is common normal for surfaces are closest or in contact during impact. This is the line along which internal force of collision
  27. 27. 25 acts during impact and Newton’s coefficient of restitution is defined only along this line. Specifically, collisions can either be elastic, meaning they conserve both momentum and kinetic energy, or inelastic, meaning they conserve momentum but not kinetic energy. An inelastic collision is sometimes also called a plastic collision. A “perfectly-inelastic” collision (also called a "perfectly-plastic" collision) is a limiting case of inelastic collision in which the two bodies stick together after impact. The degree to which a collision is elastic or inelastic is quantified by the coefficient of restitution, a value that generally ranges between zero and one. A perfectly elastic collision has a coefficient of restitution of one; a perfectly-inelastic collision has a coefficient of restitution of zero. Coefficient of restitution The coefficient of restitution (COR) of two colliding objects is a fractional value representing the ratio of speeds after and before an impact, taken along the line of the impact. Pairs of objects with COR = 1 collide elastically, while objects with COR < 1 collide inelastically. The coefficient of restitution is given by for two colliding objects, where is the final velocity of the first object after impact is the final velocity of the second object after impact is the initial velocity of the first object before impact is the initial velocity of the second object before impact If the collision is elastic, both the momentum and kinetic energy of the system are conserved. Therefore, considering velocities along the horizontal direction, we have Total momentum before impact = Total momentum after impact Now we will consider the analysis of a collision in which the two objects do not stick together. In this collision, the two objects will bounce off each other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objects stick together. Example 1. A (N+3)-kg block A moving with a velocity of (N+5) m/s hits a (2N) kg block B moving with a velocity of -N m/s. Assuming that momentum is
  28. 28. 26 conserved during the collision, determine the velocity of the blocks immediately after the collision. Consider that the coefficient of restitution is: Helping me with Derive® Version 6.1, we use n instead of N. We first define the variables Va and Vb, so do not consider them as two variables, but as a single. For this, giving click in Author and then in Variable value, we have: The following window appears. We write Va, leave the value blank and give click OK:
  29. 29. 27 Repeat the procedure for the Vb variable. The following window appears.
  30. 30. 28
  31. 31. 29 The final velocities depending on N are: Is very useful to tabulate the responses using Derive® Version 6.1, the table with three columns N, Va and Vb respectively is:
  32. 32. 30
  33. 33. 31
  34. 34. 32 Elements of a triangle based on its coordinates Other example, if I want to create a problem of Analytic Geometry of the subjects: Perimeter, area, slopes, interior angles, equations of the sides of a triangle based on its coordinates. The theory is: The slope m of the line is: The distance between two points is calculated with: Area of a polygon whose coordinates of vertices are known:
  35. 35. 33 To determine the area of a simple polygon whose vertices are described by ordered pairs in the plane, for a triangle we use the formula: The user cross-multiplies corresponding coordinates to find the area of the polygon. It is also called the surveyor’s formula. Forms for 2D linear equations General (or standard) form: In the general (or standard) form the linear equation is written as: where A and B are not both equal to zero. The equation is usually written so that , by convention. The graph of the equation is a straight line, and every straight line can be represented by an equation in the above form. If A is nonzero, then the x-intercept, that is, the x-coordinate of the point where the graph crosses the x-axis (where, y is zero), is -C/A. If B is nonzero, then the y- intercept, that is the y-coordinate of the point where the graph crosses the y-axis (where x is zero), is -C/B, and the slope of the line is −A/B. Slope–intercept form: where m is the slope of the line and b is the y-intercept, which is the y- coordinate of the location where line crosses the y axis. This can be seen by letting x = 0, which immediately gives y = b. It may be helpful to think about this in terms of y = b + mx; where the line passes through the point (0, b) and extends to the left and right at a slope of m. Vertical lines, having undefined slope, cannot be represented by this form. Point–slope form: where m is the slope of the line and is any point on the line. The point-slope form expresses the fact that the difference in the y coordinate between two points on a line (that is, ) is proportional to the difference in the x coordinate (that is, ). The proportionality constant is (the slope of the line). Two-point form:
  36. 36. 34 where and are two points on the line with . This is equivalent to the point-slope form above, where the slope is explicitly given as: Angle between two lines The angle between two lines in a plane is defined to be: 1. 0, if the lines are parallel; 2. the smaller angle having as sides the half-lines starting from the intersection point of the lines and lying on those two lines, if the lines are not parallel. If denotes the angle between two lines, it always satisfies the inequalities: If the slopes of the two lines are and , the angle is obtained from: This equation clicks in the case that , when the lines are perpendicular. Also, if one of the lines is parallel to y -axis, it has no slope; then the angle must be deduced using the slope of the other line. Example 2. Calculate the slopes and equations of the sides of a triangle based on its coordinates vertices that are: The triangle is shown in the following figure. Too, calculate its perimeter, area and interior angles.
  37. 37. 35 Helping me with Derive® Version 6.1, we use n instead of N. We first define the variables Mab, Mbc and Mca, so do not consider them as three variables, but as a single. For this, giving click in Author and then in Variable value how we did in example 1. By applying the formula of slope, we calculate each one of them. This is:
  38. 38. 36 Calculating the equation of each side. Using the point-slope form:
  39. 39. 37
  40. 40. 38 The equations of the sides AB, BC, and CA respectively are: The table of results is:
  41. 41. 39
  42. 42. 40
  43. 43. 41 Calculating the perimeter and area:
  44. 44. 42 The perimeters and areas of the polygon are:
  45. 45. 43
  46. 46. 44
  47. 47. 45 Calculating the interior angles:
  48. 48. 46 The interior angles A, B, C of the polygon are:
  49. 49. 47
  50. 50. 48
  51. 51. 49 When the angle is negative we add (supplement of the angle) to get the correct angle: N Angle A 1 a = 171.5110561 2 a = 163.2374455 3 a = 155.3583380
  52. 52. 50 4 a = 147.9946167 5 a = 141.2056422 6 a = 135 7 a = 129.3517526 8 a = 124.2157021 9 a = 119.5387822 10 a = 115.2673105 11 a = 111.3509802 12 a = 107.7446716 13 a = 104.4089610 14 a = 101.3099324 15 a = 98.41866287 16 a = 95.71059313 17 a = 93.16489670 18 a = 90.76389846 N Angle B 31 b = 90.71872655 32 b = 92.44949271 33 b = 94.12509877 34 b = 95.74766818 35 b = 97.31924920 36 b = 98.84181456 37 b = 100.3172617 38 b = 101.7474142 39 b = 103.1340223 40 b = 104.4787654 41 b = 105.7832539
  53. 53. 51 42 b = 107.0490309 43 b = 108.2775749 44 b = 109.4703020 45 b = 110.6285679 46 b = 111.7536709 47 b = 112.8468537 48 b = 113.9093060 49 b = 114.9421663 50 b = 115.9465249 51 b = 116.9234251 52 b = 117.8738658 53 b = 118.7988037 54 b = 119.6991544 55 b = 120.5757950 We have two special cases: 1. We give the problem with the N parameter and use the software to find the general answer and tabulate it the results of each student. 2. We propose the answer to the problem and use the software to find the function of the problem and tabulate it the results of each student. Example of case 1 Example 3: Finding the roots of the quadratic equation: To use Derive® 6.1 instead of N, we use n. We introduce the equation; we click on Solve and then click Expression:
  54. 54. 52 Another window appears to us, we select as the variable x and we give click on Solve:
  55. 55. 53 The answer is: Select the result and now we give click on Calculus and then on Table:
  56. 56. 54 Another window appears to us, we select as the parameter n, we set the starting value to 1, the ending value equal to the number of students per group, in this case it is 50, and the step size is taken as 1 and we give click on Simplify: A table with the N parameter and its corresponding response, in this way the teacher knows the answer for each student:
  57. 57. 55 The full table with all answers is: N 1 -0.50 1 2 -1.00 2 3 -1.50 3 4 -2.00 4 5 -2.50 5 6 -3.00 6 7 -3.50 7 8 -4.00 8 9 -4.50 9 10 -5.00 10
  58. 58. 56 11 -5.50 11 12 -6.00 12 13 -6.50 13 14 -7.00 14 15 -7.50 15 16 -8.00 16 17 -8.50 17 18 -9.00 18 19 -9.50 19 20 -10.00 20 21 -10.50 21 22 -11.00 22 23 -11.50 23 24 -12.00 24 25 -12.50 25 26 -13.00 26 27 -13.50 27 28 -14.00 28 29 -14.50 29 30 -15.00 30 31 -15.50 31 32 -16.00 32 33 -16.50 33 34 -17.00 34 35 -17.50 35 36 -18.00 36 37 -18.50 37 38 -19.00 38
  59. 59. 57 39 -19.50 39 40 -20.00 40 41 -20.50 41 42 -21.00 42 43 -21.50 43 44 -22.00 44 45 -22.50 45 46 -23.00 46 47 -23.50 47 48 -24.00 48 49 -24.50 49 50 -25.00 50 Example of case 2 Example 4. We propose the answer to the problem, the value of the roots of the quadratic equation will be: and To use Derive® 6.1 instead of N, we use n. We multiply the roots to find the quadratic equation: (x - n/3)*(x + n/4)=0 We click on Simplify and then click Expand, we have:
  60. 60. 58 Click again in the Expand button, we have: The quadratic equation that we will put them so that students solve is:
  61. 61. 59 The full table with all answers is: N 1 0.3333 -0.25 2 0.6667 -0.50 3 1.0000 -0.75 4 1.3333 -1.00 5 1.6667 -1.25 6 2.0000 -1.50 7 2.3333 -1.75 8 2.6667 -2.00 9 3.0000 -2.25 10 3.3333 -2.50 11 3.6667 -2.75 12 4.0000 -3.00 13 4.3333 -3.25 14 4.6667 -3.50 15 5.0000 -3.75 16 5.3333 -4.00 17 5.6667 -4.25 18 6.0000 -4.50 19 6.3333 -4.75 20 6.6667 -5.00 21 7.0000 -5.25 22 7.3333 -5.50 23 7.6667 -5.75 24 8.0000 -6.00 25 8.3333 -6.25
  62. 62. 60 26 8.6667 -6.50 27 9.0000 -6.75 28 9.3333 -7.00 29 9.6667 -7.25 30 10.0000 -7.50 31 10.3333 -7.75 32 10.6667 -8.00 33 11.0000 -8.25 34 11.3333 -8.50 35 11.6667 -8.75 36 12.0000 -9.00 37 12.3333 -9.25 38 12.6667 -9.50 39 13.0000 -9.75 40 13.3333 -10.00 41 13.6667 -10.25 42 14.0000 -10.50 43 14.3333 -10.75 44 14.6667 -11.00 45 15.0000 -11.25 46 15.3333 -11.50 47 15.6667 -11.75 48 16.0000 -12.00 49 16.3333 -12.25 50 16.6667 -12.50
  63. 63. 61 Part II OrthoMathetics Applications
  64. 64. 62 Chapter 3 Applications to Algebra Mathematics study the number, shape, size and variation. For me everything is based on algebra, nothing more than divided it in subjects like calculus, to facilitate its study. Everything is Algebra. Neftalí Antúnez H. 3.1 Operations with polynomials Example 5. Solve the following exponential equation: Solving with Derive®, the results are:
  65. 65. 63
  66. 66. 64
  67. 67. 65 Example 6. Solve the following exponential equation: Solving with Derive®, the results are:
  68. 68. 66
  69. 69. 67
  70. 70. 68 Example 7. Multiply the following polynomials: The answer is: Example 8. Divide the following polynomials:
  71. 71. 69 by The answer is: Remainder = 0 Example 9. Divide the following polynomials: by The table of results is:
  72. 72. 70
  73. 73. 71
  74. 74. 72
  75. 75. 73 Example 10. Divide the following polynomials: by The table of results is:
  76. 76. 74
  77. 77. 75
  78. 78. 76
  79. 79. 77
  80. 80. 78
  81. 81. 79 Example 11. Divide the following polynomials: by The answer is: Remainder = 0 Example 12. Multiply the following conjugated binomials:
  82. 82. 80 ans. Example 13. Expanding the binomial using Newton’s theorem: The results are:
  83. 83. 81
  84. 84. 82
  85. 85. 83
  86. 86. 84
  87. 87. 85 Example 14. Find the 12th term of the binomial and find the N-th term of the binomial The table of results is:
  88. 88. 86
  89. 89. 87
  90. 90. 88
  91. 91. 89
  92. 92. 90 3.2 Systems of Equations Linear with Two and Three Variables Example 15. To solve the following system of equations linear with two variables:
  93. 93. 91 To use Derive® 6.1 instead of N, we use n. We click on Solve and then click System, we have: Another window appears to us, we select 2 in Number and we give click on OK: We write the equations and selected as variables to x and y and then click in Solve:
  94. 94. 92 The answer is: The full table with all answers is:
  95. 95. 93 N x y 1 2 -4 2 4 -8 3 6 -12 4 8 -16 5 10 -20 6 12 -24 7 14 -28 8 16 -32 9 18 -36 10 20 -40 11 22 -44 12 24 -48 13 26 -52 14 28 -56 15 30 -60 16 32 -64 17 34 -68 18 36 -72 19 38 -76 20 40 -80 21 42 -84 22 44 -88 23 46 -92 24 48 -96 25 50 -100 26 52 -104 27 54 -108 28 56 -112
  96. 96. 94 29 58 -116 30 60 -120 31 62 -124 32 64 -128 33 66 -132 34 68 -136 35 70 -140 36 72 -144 37 74 -148 38 76 -152 39 78 -156 40 80 -160 41 82 -164 42 84 -168 43 86 -172 44 88 -176 45 90 -180 46 92 -184 47 94 -188 48 96 -192 49 98 -196 50 100 -200 Example 16. To solve the following system of equations linear with two variables: In the same way, we obtain the answer is: The full table with all answers is:
  97. 97. 95 N x y 1 -3 3 2 -6 6 3 -9 9 4 -12 12 5 -15 15 6 -18 18 7 -21 21 8 -24 24 9 -27 27 10 -30 30 11 -33 33 12 -36 36 13 -39 39 14 -42 42 15 -45 45 16 -48 48 17 -51 51 18 -54 54 19 -57 57 20 -60 60 21 -63 63 22 -66 66 23 -69 69 24 -72 72 25 -75 75 26 -78 78 27 -81 81 28 -84 84
  98. 98. 96 29 -87 87 30 -90 90 31 -93 93 32 -96 96 33 -99 99 34 -102 102 35 -105 105 36 -108 108 37 -111 111 38 -114 114 39 -117 117 40 -120 120 41 -123 123 42 -126 126 43 -129 129 44 -132 132 45 -135 135 46 -138 138 47 -141 141 48 -144 144 49 -147 147 50 -150 150 Example 17. To solve the following system of equations linear with two variables: In the same way, we obtain the answer is: By being so obvious answers table is not shown.
  99. 99. 97 Example 18. To solve the following system of equations linear with three variables: To use Derive® 6.1 instead of N, we use n. We click on Solve and then click System. Another window appears to us, we select 3 in Number and we give click on OK: We write the equations and selected as variables to x, y and z and then click in solve: The answer is:
  100. 100. 98 By being so obvious only some answers are shown.
  101. 101. 99 Example 19. To solve the following system of equations linear with three variables: In the same way, we obtain the answer is:
  102. 102. 100 The full table with all answers is: N x y z 1 0.13889 0.02778 0.19444 2 0.27778 0.05556 0.38889 3 0.41667 0.08333 0.58333 4 0.55556 0.11111 0.77778 5 0.69444 0.13889 0.97222 6 0.83333 0.16667 1.16667 7 0.97222 0.19444 1.36111 8 1.11111 0.22222 1.55556 9 1.25000 0.25000 1.75000
  103. 103. 101 10 1.38889 0.27778 1.94444 11 1.52778 0.30556 2.13889 12 1.66667 0.33333 2.33333 13 1.80556 0.36111 2.52778 14 1.94444 0.38889 2.72222 15 2.08333 0.41667 2.91667 16 2.22222 0.44444 3.11111 17 2.36111 0.47222 3.30556 18 2.50000 0.50000 3.50000 19 2.63889 0.52778 3.69444 20 2.77778 0.55556 3.88889 21 2.91667 0.58333 4.08333 22 3.05556 0.61111 4.27778 23 3.19444 0.63889 4.47222 24 3.33333 0.66667 4.66667 25 3.47222 0.69444 4.86111 26 3.61111 0.72222 5.05556 27 3.75000 0.75000 5.25000 28 3.88889 0.77778 5.44444 29 4.02778 0.80556 5.63889 30 4.16667 0.83333 5.83333 31 4.30556 0.86111 6.02778 32 4.44444 0.88889 6.22222 33 4.58333 0.91667 6.41667 34 4.72222 0.94444 6.61111 35 4.86111 0.97222 6.80556 36 5.00000 1.00000 7.00000 37 5.13889 1.02778 7.19444 38 5.27778 1.05556 7.38889
  104. 104. 102 39 5.41667 1.08333 7.58333 40 5.55556 1.11111 7.77778 41 5.69444 1.13889 7.97222 42 5.83333 1.16667 8.16667 43 5.97222 1.19444 8.36111 44 6.11111 1.22222 8.55556 45 6.25000 1.25000 8.75000 46 6.38889 1.27778 8.94444 47 6.52778 1.30556 9.13889 48 6.66667 1.33333 9.33333 49 6.80556 1.36111 9.52778 50 6.94444 1.38889 9.72222 Example 20. To solve the following system of equations linear with three variables: To solve the following system of equations linear with three variables: In the same way, we obtain the answer is: The results are:
  105. 105. 103
  106. 106. 104
  107. 107. 105
  108. 108. 106
  109. 109. 107 Example 21. To solve the following system of equations linear with four variables:
  110. 110. 108 To use Derive® 6.1 instead of N, we use n. We click on Solve and then click System. Another window appears to us, we select 4 in Number and we give click on OK: We write the equations and selected as variables to x, y, z and t and then click in solve: In the same way, we obtain the answer is:
  111. 111. 109 The full table with all answers is: N x y z t 1 15.0625 -19.75 -39.50 34.8125 2 15.1250 -19.50 -39.00 34.6250 3 15.1875 -19.25 -38.50 34.4375 4 15.2500 -19.00 -38.00 34.2500 5 15.3125 -18.75 -37.50 34.0625 6 15.3750 -18.50 -37.00 33.8750 7 15.4375 -18.25 -36.50 33.6875 8 15.5000 -18.00 -36.00 33.5000 9 15.5625 -17.75 -35.50 33.3125
  112. 112. 110 10 15.6250 -17.50 -35.00 33.1250 11 15.6875 -17.25 -34.50 32.9375 12 15.7500 -17.00 -34.00 32.7500 13 15.8125 -16.75 -33.50 32.5625 14 15.8750 -16.50 -33.00 32.3750 15 15.9375 -16.25 -32.50 32.1875 16 16.0000 -16.00 -32.00 32.0000 17 16.0625 -15.75 -31.50 31.8125 18 16.1250 -15.50 -31.00 31.6250 19 16.1875 -15.25 -30.50 31.4375 20 16.2500 -15.00 -30.00 31.2500 21 16.3125 -14.75 -29.50 31.0625 22 16.3750 -14.50 -29.00 30.8750 23 16.4375 -14.25 -28.50 30.6875 24 16.5000 -14.00 -28.00 30.5000 25 16.5625 -13.75 -27.50 30.3125 26 16.6250 -13.50 -27.00 30.1250 27 16.6875 -13.25 -26.50 29.9375 28 16.7500 -13.00 -26.00 29.7500 29 16.8125 -12.75 -25.50 29.5625 30 16.8750 -12.50 -25.00 29.3750 31 16.9375 -12.25 -24.50 29.1875 32 17.0000 -12.00 -24.00 29.0000 33 17.0625 -11.75 -23.50 28.8125 34 17.1250 -11.50 -23.00 28.6250 35 17.1875 -11.25 -22.50 28.4375 36 17.2500 -11.00 -22.00 28.2500 37 17.3125 -10.75 -21.50 28.0625 38 17.3750 -10.50 -21.00 27.8750
  113. 113. 111 39 17.4375 -10.25 -20.50 27.6875 40 17.5000 -10.00 -20.00 27.5000 41 17.5625 -9.75 -19.50 27.3125 42 17.6250 -9.50 -19.00 27.1250 43 17.6875 -9.25 -18.50 26.9375 44 17.7500 -9.00 -18.00 26.7500 45 17.8125 -8.75 -17.50 26.5625 46 17.8750 -8.50 -17.00 26.3750 47 17.9375 -8.25 -16.50 26.1875 48 18.0000 -8.00 -16.00 26.0000 49 18.0625 -7.75 -15.50 25.8125 50 18.1250 -7.50 -15.00 25.6250 3.3 Matrices and determinants Example 22. Multiply the following matrices shown in the figure: The matrices are introduced to Derive® 6.1, by clicking on the option Author Matrix located on the toolbar. The first array has a rank of 3 x 4 and the second matrix is 4 x 3. Therefore, the product matrix will have a range 3 x 3. The resulting array with the first tabulated matrices by rows appear in the following figure:
  114. 114. 112 The next results by rows are:
  115. 115. 113
  116. 116. 114
  117. 117. 115 Example 23. Raise to square the following matrix A. The matrices are introduced to Derive® 6.1, by clicking on the option Author Matrix located on the toolbar. The array has a rank of 4 x 4. Multiplying A by A to get A2. The results by rows are:
  118. 118. 116
  119. 119. 117
  120. 120. 118 Example 24. Find the value of the determinant of the matrix below: In Derive® the value of the determinant of a matrix is obtained by typing: det([n, 3, 2; -1, -2, -n; 3, 2, 7])
  121. 121. 119 The next results are:
  122. 122. 120
  123. 123. 121
  124. 124. 122 Example 25. Find the matrix inverse of: In Derive® the matrix inverse is obtained by typing: [n, 1, 2; -1, -2, -n; 3, 2, 6]^(-1) The result is:
  125. 125. 123 The results by rows are:
  126. 126. 124
  127. 127. 125
  128. 128. 126
  129. 129. 127
  130. 130. 128 Example 26. Find the eigenvalues of the matrix: With Charpoly find the characteristic equation, which match to zero to find their roots.
  131. 131. 129 The eigenvalues are:
  132. 132. 130
  133. 133. 131 3.4 Roots of Equations 3.4.1 Roots of quadratic equations Example 27. Some additional examples of quadratic equations and their solutions are:
  134. 134. 132 Quadratic Equation Root 1 Root 2 Example 28. Some examples to Completing the Square: Trinomial Answer Example 29. For use in numerical methods, find the roots of cubic equation: their roots are: It is also very useful to tabulate the responses using Microsoft Excel® or similar, placing N in first column and the responses depending on N in the following columns:
  135. 135. 133
  136. 136. 134 Example 30. For use in numerical methods, find the roots of the following equation:
  137. 137. 135 Their roots are: The graphic for N = 25 is:
  138. 138. 136 It is also very useful to tabulate the responses using Microsoft Excel® or similar, placing N in first column and the responses depending on N in the following columns:
  139. 139. 137 Example 31. For use in numerical methods, find the roots of the following equation:
  140. 140. 138 Their roots are: It is also very useful to tabulate the responses using Microsoft Excel® or similar, placing N in first column and the responses depending on N in the following columns:
  141. 141. 139
  142. 142. 140 Chapter 4 Applications to Geometry and Trigonometry 4.1 Arc Length ARC LENGTH PROBLEM S where θ is the angle in radians. Example 32. INSTRUCTIONS: Using the angle of and its 12 meters of arc, calculate the length of the RADIUS and then CALCULATE THE VALUE OF X, and then the length of each ARC S1, S2 and S4. N is the number of list. Value 2 points. The radius is: The length of ARC S1:
  143. 143. 141 The table of results is: N S2, S3 x S4 1 0.26667 178 47.46667 2 .53334 176 46.93334 3 0.8 174 46.4 4 1.06667 172 45.86667 5 1.3334 170 45.3334 6 1.6 168 44.8 7 1.86667 166 44.26667 8 2.13334 164 43.73334 9 2.4 162 43.2 10 2.6667 160 42.6667 11 2.93334 158 42.13334 12 3.2 156 41.6 13 3.46667 154 41.06667 14 3.73334 152 40.53334 15 4 150 40 16 4.26667 148 39.46667 17 4.53334 146 38.93334 18 4.8 144 38.4 19 5.06667 142 37.86667 20 5.333433 140 37.3334 21 5.6 138 36.8 22 5.86667 136 36.26667 23 6.13334 134 35.73334 24 6.4 132 35.2 25 6.6667 130 34.6667 26 6.93334 128 34.13334 27 7.2 126 33.6
  144. 144. 142 28 7.46667 124 33.06667 29 7.73334 122 32.53334 30 8 120 32 31 8.26667 118 31.46667 32 8.53334 116 30.93334 33 8.8 114 30.4 34 9.06667 112 29.86667 35 9.33334 110 29.33343 36 9.6 108 28.8 37 9.86667 106 28.26667 38 10.13334 104 27.73334 39 10.4 102 27.2 40 10.66667 100 26.66667 41 10.93334 98 26.13334 42 11.2 96 25.6 43 11.46667 94 25.06667 44 11.73334 92 24.53334 45 12 90 24 46 12.26667 88 23.46667 47 12.53334 86 22.93334 48 12.8 84 22.4 49 13.06667 82 21.86667 50 13.33343 80 21.33343 4.2 Right-angled Triangles Example 33. Solve correctly the following right-angled triangles. All N is your number in list. Value 1.25 points each.
  145. 145. 143 The results for triangle 1 are: For all N, the angles are constant: and N Side b 1 2.23607 2 4.47214 3 6.70820 4 8.94427 5 11.18034 6 13.41641 7 15.65248 8 17.88854 9 20.12461 10 22.36068 11 24.59675 12 26.83282 13 29.06888
  146. 146. 144 14 31.30495 15 33.54102 16 35.77709 17 38.01316 18 40.24922 19 42.48529 20 44.72136 21 46.95743 22 49.19350 23 51.42956 24 53.66563 25 55.90170 26 58.13777 27 60.37384 28 62.60990 29 64.84597 30 67.08204 31 69.31811 32 71.55418 33 73.79024 34 76.02631 35 78.26238 36 80.49845 37 82.73452 38 84.97058 39 87.20665 40 89.44272 41 91.67879 42 93.91486
  147. 147. 145 43 96.15092 44 98.38699 45 100.62306 46 102.85913 47 105.09519 48 107.33126 49 109.56733 50 111.80340 The results for triangle 2 are:
  148. 148. 146
  149. 149. 147 The results for triangle 3 are: For all N, the angle t is constant:
  150. 150. 148
  151. 151. 149
  152. 152. 150 The results for triangle 4 are: For all N, the angle B is constant:
  153. 153. 151
  154. 154. 152 The code in BASIC programming language to solve previous triangles is: defdbl a-z cls pi=3.141593 open "c:exageo.txt" for output as #1 for i=1 to 50 r=12./(pi/4) s1=3/4*pi*r s2=i/180*pi*r x=360-(45+2*i+3/4*180) s4=x/180*pi*r b=i*sqr(5) a=atn(b/(2*i)) a=a*180/pi w=90-a ag=int(a) min=(a-ag)*60 am=int(min) se=(min-am)*60 se=int(se*1000+0.5)/1000 print#1,i;" LArc ";r;" ";s1;" ";s2;" ";x;" ";s4; print#1,i;"Triangle 1 ";b;" ";ag;chr$(248);am;"’";se;chr$(34); a=w ag=int(a) min=(a-ag)*60 am=int(min) se=(min-am)*60 se=int(se*1000+0.5)/1000 print#1,ag;chr$(248);am;"’";se;chr$(34) x=sqr(15^2+i^2) a=atn(i/15) a=a*180/pi w=90-a ag=int(a) min=(a-ag)*60 am=int(min) se=(min-am)*60 se=int(se*1000+0.5)/1000 print#1,"Triangle 2 ";x;" ";ag;chr$(248);am;"’";se;chr$(34); a=w ag=int(a) min=(a-ag)*60 am=int(min) se=(min-am)*60 se=int(se*1000+0.5)/1000 print#1,ag;chr$(248);am;"’";se;chr$(34); a=(35+42/60) w=90-a a=(35+42/60)*pi/180 b=2*i*sin(a) x=sqr((2*i)^2-b^2) a=w ag=int(a) min=(a-ag)*60 am=int(min) se=(min-am)*60 se=int(se*1000+0.5)/1000 print#1,"Triangle 3 ";b;" ";x;" ";ag;chr$(248);am;"’";se;chr$(34);" ";
  155. 155. 153 a=(58+37/60) w=90-a a=(58+37/60)*pi/180 b=i/sin(a) x=sqr(b^2-i^2) a=w ag=int(a) min=(a-ag)*60 am=int(min) se=(min-am)*60 se=int(se*1000+0.5)/1000 print#1,"Triangle 4 ";b;" ";x;" ";ag;chr$(248);am;"’";se;chr$(34);" " next i close #1 end 4.3 Non Right-angled Triangles Example 34. Solve correctly the following triangles. All N is your number in list. The results for triangle 1 are: For all N, the angle y is constant:
  156. 156. 154
  157. 157. 155 The results for triangle 2 are:
  158. 158. 156
  159. 159. 157 The results for triangle 3 are:
  160. 160. 158
  161. 161. 159 The triangle 4 is isosceles. For all N, the angles are constant: and . The results for triangle 4 are:
  162. 162. 160
  163. 163. 161
  164. 164. 162 Individual responses were obtained using the programming language FORTRAN 90, whose source code is: Program Triangles REAL :: pi=3.1415923 REAL b,angC,angA,angB, secB,min,secA,secC REAL T INTEGER :: N,dA,dC,mc,ma,mb,dB PRINT *,’ ’ PRINT *, "Programmed by Neftalí Antúnez H. (p)1986" PRINT *, "CEO of Antúnez Software LTD" ! Create an external new data file OPEN (7, FILE = ’c:similar.txt’, ACCESS = ’APPEND’,STATUS = ’REPLACE’) ! Write Writing the header WRITE(7,*) "Triangle 1" PRINT *,’ ’ DO N=1,55,1 x=1.147643*N w=1.562651*N WRITE(7,*) N,",",x,",",w END DO WRITE(7,*) "Triangle 2" DO N=1,55,1 ! x=-N-N**2 b=sqrt((N+1)**2+(N+4)**2+0.4635*(N+1)*(N+4)) angC=acos(((N+1)**2-(N+4)**2-b**2)/(-2*(N+4)*b)) angC=180*angC/pi dC=int(angC) min=(angC-dC)*60 mC=int(min) secC=(min-mC)*60 angA=acos(((N+4)**2-(N+1)**2-b**2)/(-2*(N+1)*b)) angA=180*angA/pi dA=int(angA) min=(angA-dA)*60 mA=int(min) secA=(min-mA)*60 WRITE(7,*)N,",",b,",",dc,"deg",mc,"’",secC,’",’,da,"deg",ma,"’",seca,’"’ END DO WRITE(7,*) "Triangle 3" DO N=1,55,1 ! Changing N integer by T real because cosinus needs a real argument T=N angC=acos(((T+2)**2-(T+1)**2-(T+3)**2)/(-2*(T+1)*(T+3))) angC=180*angC/pi dC=int(angC) min=(angC-dC)*60 mC=int(min) secC=(min-mC)*60 angA=acos(((T+3)**2-(T+1)**2-(T+2)**2)/(-2*(T+1)*(T+2))) angA=180*angA/pi dA=int(angA) min=(angA-dA)*60 mA=int(min) secA=(min-mA)*60 angB=acos(((T+1)**2-(T+2)**2-(T+3)**2)/(-2*(T+2)*(T+3))) angB=180*angB/pi dB=int(angB) min=(angB-dB)*60
  165. 165. 163 mB=int(min) secB=(min-mB)*60 WRITE(7,*) N,",",dc,"deg",mc,"’",secC,’",’,da,"deg",ma,"’",seca,’",’ WRITE(7,*) N,",",db,"deg",mb,"’",secb,’",’,db,"deg",mb,"’",secb,’"’ END DO WRITE(7,*) "Triangle 4" DO N=1,55,1 x=1.08586*N WRITE(7,*) N,",",x end do ! Close the external file CLOSE(7) PRINT *,’ ’ PRINT *," THANK YOU FOR USING Triangles" END Program Triangles
  166. 166. 164 Chapter 5 Applications to Analytic Geometry 5.1 Circle Example 35. Find the Center and radius of the circle whose general equation is: Center: Radius: Graphing the circle with Algebrator® 5 spanish version, for N = 30 Example 36. Find the Center and radius of the circle whose general equation is: Center: Radius:
  167. 167. 165 Graphing the circle with Algebrator® 5 spanish version, for N = 30 5.2 Parabola Example 37. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is: Vertex: focal length: p = 4 Focus: Equation of directrix: Graphing the parabola with Algebrator® 5 spanish version, for N = 24
  168. 168. 166 Example 38. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is: Vertex: focal length: p = 3 Focus: Equation of directrix: Graphing the parabola with Algebrator® 5 spanish version, for N = 24
  169. 169. 167 Example 39. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is: Vertex: focal length: p = 3 Focus: Equation of directrix: Graphing the parabola with Algebrator® 5 spanish version, for N = 24
  170. 170. 168 Example 40. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is: Vertex: focal length: p = 5 Focus: Equation of directrix: Graphing the parabola with Algebrator® 5 spanish version, for N = 30
  171. 171. 169
  172. 172. 170 Chapter 6 Applications to Calculus «Mathematics has been, traditionally, the torture of school children in the whole world, and humanity has tolerated this torture for their children as a suffering inevitable to acquire necessary knowledge; but education should not be a torture, and we would not be good teachers if we don’t interrupt, by all means, transform this suffering into enjoyment, this does not mean absence of effort, but on the contrary, delivery of stimuli and efforts effective and desired». (Puig Adam, Peter 1958) 6.1 Derivatives Example 41. Derive the following function: answer Example 42. Derive the following function: answer Example 43. Derive the following function: answer Example 44. Derive the product of functions: answer Example 45. Derive the quotient of functions:
  173. 173. 171 answer 6.2 Maxima and minima Example 46. Find the maxima and minima of the following function: Deriving and equating to zero: Solving: We give click on Author and then on Function Definition:
  174. 174. 172 Now, finding: minimum maximum The Graph for N = 30 is: Tabulating the results:
  175. 175. 173
  176. 176. 174
  177. 177. 175
  178. 178. 176
  179. 179. 177
  180. 180. 178 Example 47. Find the maxima and minima of the following function: Deriving and equating to zero: Solving: Now, finding: Results: minimum maximum The Graph for N = 25 is: Tabulating the results using Microsoft Excel®:
  181. 181. 179
  182. 182. 180
  183. 183. 181 Example 48. Find the maxima and minima of the following function: Deriving and equating to zero: Solving: Now, finding: Results: maximum minimum The Graph for N = 25 is:
  184. 184. 182 Tabulating the results using Microsoft Excel®:
  185. 185. 183
  186. 186. 184 Example 49. Find the maxima and minima of the following function: Deriving and equating to zero: Solving: Now, finding: Results: maximum minimum Tabulating the results using Microsoft Excel®:
  187. 187. 185
  188. 188. 186
  189. 189. 187 6.3 Definite Integrals Example 50. Find the definite integral of from to The results are:
  190. 190. 188
  191. 191. 189 Example 51. Find the area formed by the intersection of curves: and Using Derive® 6.1 to make the operations, we have:
  192. 192. 190
  193. 193. 191 The resultant area is:
  194. 194. 192
  195. 195. 193
  196. 196. 194
  197. 197. 195
  198. 198. 196 The Graph for N = 30 is: Example 52. Find the area formed by the intersection of curves: and Using Derive® 6.1 to make the operations, we have:
  199. 199. 197
  200. 200. 198 The resultant are is:
  201. 201. 199
  202. 202. 200
  203. 203. 201
  204. 204. 202
  205. 205. 203 The Graph for N = 30 is:
  206. 206. 204 Chapter 7 Applications to Physics 7.1 Vectors Example 53. Find the resultant vector and its direction of the following vectors shown in the figure: Table of forces: Vector Magnitude (Ton) Direction (Degrees) A 50 30.4167 B 60 104.3000 C 75 149.5833
  207. 207. 205 D N 249.3000 E 80 290.8833 F 40 308.7500 P N 76.2833 Q 60 231.2500 Individual responses were obtained using the programming language FORTRAN 90, whose source code is: Program Vectrix2d REAL :: pi=3.1415923 REAL RX,RY,RES,ang,sec REAL MINUT,FX,FY INTEGER MINUTS INTEGER :: N PRINT *, "PROGRAM VECTRIX 2D" PRINT *,"SERVES TO ADD N CONCURRENT VECTOR IN THE XY PLANE" PRINT *,"BY THE METHOD OF RECTANGULAR COMPONENTS" PRINT *,’ ’ PRINT *, "Programmed by Neftalí Antúnez H. (p)1986" PRINT *, "CEO of Antúnez Software LTD" ! Create an external new data file OPEN (7, FILE = ’Addvect.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’) ! Write Writing the header WRITE(7,*) " N MAGNITUDE DIRECTION" PRINT *,’ ’ ! Angles in radians RX=0.0 RY=0.0 a=50;anga=30.4167*pi/180 b=60;angb=104.30*pi/180 c=75;angc=149.5833*pi/180 e=80;ange=290.8833*pi/180 f=40;angf=308.75*pi/180 q=60;angq=231.25*pi/180 RX=a*cos(anga)+b*cos(angb)+c*cos(angc)+e*cos(ange)+f*cos(angf)+q*cos(a ngq) RY=a*sin(anga)+b*sin(angb)+c*sin(angc)+e*sin(ange)+f*sin(angf)+q*sin(a ngq) DO N=1,55,1 d=N;angd=249.30*pi/180 p=N;angp=76.2833*pi/180 FX=RX+d*cos(angd)+p*cos(angp) FY=RY+d*sin(angd)+p*sin(angp) RES=SQRT(FX**2+FY**2) ang=ATAN(FY/FX) ang=180*ang/pi deg=INT(ang) MINUT=abs(ang-deg)*60 MINUTS=INT(MINUT) sec=(MINUT-MINUTS)*60 WRITE(7,*) n,",",res,",",deg,",",minuts,",",sec
  208. 208. 206 END DO ! Close the external file CLOSE(7) PRINT *,’ ’ PRINT *," THANK YOU FOR USING VECTRIX 2D" END Program Vectrix2d The output file contains the magnitude of vector resultant and its direction in degrees, minutes and seconds: N MAGNITUDE Degrees Minutes Seconds 1 37.389270 56 45 23.579410 2 37.423100 56 34 38.338620 3 37.457300 56 23 54.265140 4 37.491860 56 13 11.372680 5 37.526780 56 2 29.674990 6 37.562070 55 51 49.185790 7 37.597710 55 41 9.891357 8 37.633720 55 30 31.819150 9 37.670090 55 19 54.969180 10 37.706810 55 9 19.355160 11 37.743900 54 58 44.977110 12 37.781330 54 48 11.862490 13 37.819130 54 37 39.997560 14 37.857280 54 27 9.409790 15 37.895780 54 16 40.099180 16 37.934630 54 6 12.065730 17 37.973830 53 55 45.323180 18 38.013390 53 45 19.885250 19 38.053290 53 34 55.738220 20 38.093540 53 24 32.923280 21 38.134140 53 14 11.412960 22 38.175080 53 3 51.234740
  209. 209. 207 23 38.216370 52 53 32.402340 24 38.258000 52 43 14.888310 25 38.299980 52 32 58.733830 26 38.342290 52 22 43.938900 27 38.384940 52 12 30.489810 28 38.427940 52 2 18.427730 29 38.471270 51 52 7.738953 30 38.514930 51 41 58.409730 31 38.558930 51 31 50.481260 32 38.603270 51 21 43.939820 33 38.647940 51 11 38.799130 34 38.692940 51 1 35.059200 35 38.738270 50 51 32.720030 36 38.783930 50 41 31.795350 37 38.829920 50 31 32.298890 38 38.876240 50 21 34.230650 39 38.922880 50 11 37.563170 40 38.969850 50 1 42.351380 41 39.017140 49 51 48.567810 42 39.064750 49 41 56.239930 43 39.112690 49 32 5.354004 44 39.160950 49 22 15.910030 45 39.209520 49 12 27.921750 46 39.258420 49 2 41.402890 47 39.307620 48 52 56.339720 48 39.357150 48 43 12.759700 49 39.406990 48 33 30.635380 50 39.457140 48 23 49.994200
  210. 210. 208 51 39.507600 48 14 10.822450 52 39.558380 48 4 33.147580 53 39.609460 47 54 56.942140 54 39.660850 47 45 22.233580 55 39.712550 47 35 49.008180 Example 54. Find the 2 unknown forces P and Q acting on a body that is in equilibrium:, as shown in the figure: Table of forces: Vector Magnitude (Ton) Direction (Degrees) A 80 30.6000 B 90 103.9000
  211. 211. 209 C 45 149.6000 D 100 180+N+18.10 E 70 288.6000 F 50 307.5000 P ? N+30.60 Q ? 180 + N Individual responses were obtained using the programming language FORTRAN 90, whose source code is: Program Equilibrium REAL :: pi=3.1415923 REAL RX,RY REAL FX,FY INTEGER :: N PRINT *, "PROGRAM Equilibrium" PRINT *,"SERVES to Find 2 unknown forces P and Q acting on a body" PRINT *,"that is in equilibrium, BY THE METHOD OF RECTANGULAR COMPONENTS" PRINT *,’ ’ PRINT *, "Programmed by Neftalí Antúnez H. (p)1986" PRINT *, "CEO of Antúnez Software LTD" ! Create an external new data file OPEN (7, FILE = ’Equivect.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’) ! Write Writing the header WRITE(7,*) "N, P, Q," PRINT *,’ ’ ! Angles in radians RX=0.0 RY=0.0 a=80;anga=30.60*pi/180 b=90;angb=103.90*pi/180 c=45;angc=149.60*pi/180 e=70;ange=288.60*pi/180 f=50;angf=307.50*pi/180 RX=a*cos(anga)+b*cos(angb)+c*cos(angc)+e*cos(ange)+f*cos(angf) RY=a*sin(anga)+b*sin(angb)+c*sin(angc)+e*sin(ange)+f*sin(angf) DO N=1,55,1 d=100;angd=(198.10+n)*pi/180 angp=(N+30.60)*pi/180 angq=(180+N)*pi/180 a11=cos(angp);a12=cos(angq) a21=sin(angp);a22=sin(angq) delta=a11*a22-a21*a12 FX=RX+d*cos(angd) FY=RY+d*sin(angd) p=(a12*FY-a22*FX)/delta q=(a21*FX-a11*FY)/delta WRITE(7,*) n,",",p,",",q END DO ! Close the external file
  212. 212. 210 CLOSE(7) PRINT *,’ ’ PRINT *," THANK YOU FOR USING VECTRIX 2D" END Program Equilibrium The output file contains the magnitude of the vectors P and Q: N P Q 1 -24.959500 -54.570990 2 -22.822000 -51.976650 3 -20.658940 -49.379410 4 -18.470970 -46.780050 5 -16.258800 -44.179420 6 -14.023050 -41.578240 7 -11.764500 -38.977410 8 -9.483756 -36.377650 9 -7.181504 -33.779720 10 -4.858502 -31.184490 11 -2.515388 -28.592660 12 -0.1529727 -26.005130 13 2.228091 -23.422630 14 4.627101 -20.845910 15 7.043258 -18.275820 16 9.475905 -15.713070 17 11.924200 -13.158540 18 14.387470 -10.612950 19 16.864980 -8.077031 20 19.355900 -5.551652 21 21.859570 -3.037486 22 24.375120 -0.5353943 23 26.901880 1.953956
  213. 213. 211 24 29.438990 4.429714 25 31.985730 6.891172 26 34.541360 9.337624 27 37.105010 11.768230 28 39.675990 14.182350 29 42.253440 16.579160 30 44.836600 18.957950 31 47.424760 21.318070 32 50.016990 23.658690 33 52.612590 25.979160 34 55.210760 28.278780 35 57.810780 30.556900 36 60.411700 32.812680 37 63.012800 35.045510 38 65.613310 37.254710 39 68.212520 39.439720 40 70.809430 41.599650 41 73.403380 43.733970 42 75.993530 45.842000 43 78.579140 47.923110 44 81.159530 49.976810 45 83.733660 52.002200 46 86.300900 53.998820 47 88.860400 55.966020 48 91.411440 57.903220 49 93.953360 59.809960 50 96.485140 61.685420 51 99.006090 63.529110
  214. 214. 212 52 101.515500 65.340510 53 104.012500 67.119040 54 106.496600 68.864300 55 108.966800 70.575560 7.2 Coulomb’s Law and Electric Field Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects using the torsion balance, which he invented. Coulomb confirmed that the electric force between two small charged spheres is proportional to the inverse square of their separation distance r—that is: From Coulomb’s experiments, we can generalize the following properties of the electric force between two stationary charged particles. The electric force • is inversely proportional to the square of the separation rbetween the particles and directed along the line joining them; • is proportional to the product of the charges on the two particles; • is attractive if the charges are of opposite sign and repulsive if the charges have the same sign; • is a conservative force. We will use the term point charge to mean a particle of zero size that carries an electric charge. The electrical behavior of electrons and protons is very well described by modeling them as point charges. From experimental observations on the electric force, we can express Coulomb’s law as an equation giving the magnitude of the electric force (sometimes called the Coulomb force) between two point charges: where k is a constant called the Coulomb constant. The electric field vector E at a point in space is defined as the electric force acting on a positive test charge placed at that point divided by the test charge:
  215. 215. 213 To calculate the electric field at a point P due to a group of point charges, we first calculate the electric field vectors at P individually using Equation: And then add them vectorially. In other words, at any point P, the total electric field due to a group of source charges equals the vector sum of the electric fields of all the charges.This superposition principle applied to fields follows directly from the superposition property of electric forces, which, in turn, follows from the fact that we know that forces add as vectors. Thus, the electric field at point P due to a group of source charges can be expressed as the vector sum: Where: is the distance from the ith source charge to the point P and is a unit vector directed from toward P. Example 55. Find the total electric field at point P, produced by the charges shown in Figure. The distances are in centimeters and must be divided by 100 to convert them in meters. The charges are on microcoulombs and is used: .
  216. 216. 214 Since at the point P is considered a positive unit charge, each field produced by the charges are shown in the following figure: Individual responses were obtained using the programming language FORTRAN 90, whose source code is: Program Efield implicit none REAL :: pi=3.1415923 REAL DPB,DPD REAL angZ,angT,ang REAL ea,eb,ec,ed REAL r,sfx,sfy REAL MINUT,sec INTEGER deg,MINUTS INTEGER :: n PRINT *, "PROGRAM EFIELD" PRINT *,"USED TO CALCULATE THE TOTAL ELECTRIC FIELD" PRINT *,’ ’ PRINT *, "Programmed in Microsoft Powerstation by Neftalí Antúnez H. (p)1986" PRINT *, "CEO of Antúnez Software LTD" PRINT *,’ ’ ! Create an external new data file OPEN (7, FILE = ’Efield.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’) ! Write Writing the header WRITE(7,*) " N TOTAL ELECTRIC FIELD DIRECTION" DO n=1,55,1 DPB=sqrt((n+4.)**2+(n+15.)**2) DPD=sqrt(64.+(n+15.)**2) ! angles are in radians
  217. 217. 215 angZ=atan((15.+n)/(n+4.)) angZ=2*pi-angZ angT=atan((15.+n)/8.) angT=pi+angT ! The charges are multiplied by 9x10E09 and by 10000 due to conversion ! from centimeters to meters. ea=-450000000/(n+4)**2 ec=2812500. eb=360000000/DPB**2 ed=270000000/DPD**2 sfx=eb*cos(angZ)+ed*cos(angT)+ea+ec sfy=eb*sin(angZ)+ed*sin(angT) r=sqrt(sfx**2+sfy**2) ang=atan(sfy/sfx) ang=180*ang/pi deg=INT(ang) MINUT=abs(ang-deg)*60 MINUTS=INT(MINUT) sec=(MINUT-MINUTS)*60 WRITE(7,*) n,",",r,",",deg,",",minuts,",",sec END DO PRINT *,’ ’ ! Close the external file CLOSE(7) PRINT *," Finished" END Program Efield The output file contains the total electric field and its direction in degrees, minutes and seconds: N Total Electric Field Degrees Minutes Seconds 1 15310950000000.00 7 25 14.88762 2 9799622.00 10 12 26.75446 3 6488275.00 13 41 19.13361 4 4356343.00 18 16 36.75522 5 2920362.00 24 45 12.36649 6 1935090.00 34 39 26.41388 7 1282441.00 50 54 6.17157 8 929416.60 76 38 50.06287 9 861157.90 -73 21 55.75012 10 972141.60 -51 1 47.18536 11 1140858.00 -37 31 10.66956 12 1311528.00 -29 15 5.740356
  218. 218. 216 13 1467205.00 -23 50 10.65216 14 1604501.00 -20 3 8.580322 15 1724258.00 -17 16 2.299347 16 1828479.00 -15 7 56.78604 17 1919313.00 -13 26 34.77539 18 1998725.00 -12 4 18.13408 19 2068420.00 -10 56 8.198776 20 2129837.00 -9 58 42.36786 21 2184184.00 -9 9 38.30315 22 2232470.00 -8 27 13.17604 23 2275539.00 -7 50 10.76385 24 2314096.00 -7 17 33.33481 25 2348739.00 -6 48 36.30707 26 2379968.00 -6 22 44.70966 27 2408212.00 -5 59 30.68207 28 2433831.00 -5 38 31.74133 29 2457136.00 -5 19 29.53583 30 2478395.00 -5 2 8.921127 31 2497836.00 -4 46 17.28539 32 2515656.00 -4 31 44.04396 33 2532032.00 -4 18 20.20695 34 2547111.00 -4 5 58.13885 35 2561027.00 -3 54 31.26846 36 2573892.00 -3 43 53.93784 37 2585812.00 -3 34 1.212616 38 2596872.00 -3 24 48.81821 39 2607156.00 -3 16 12.97234 40 2616731.00 -3 8 10.37905 41 2625663.00 -3 0 38.09509
  219. 219. 217 42 2634005.00 -2 53 33.52936 43 2641808.00 -2 46 54.36951 44 2649119.00 -2 40 38.53729 45 2655976.00 -2 34 44.18633 46 2662414.00 -2 29 9.654236 47 2668470.00 -2 23 53.42995 48 2674171.00 -2 18 54.16328 49 2679545.00 -2 14 10.6216 50 2684615.00 -2 9 41.68659 51 2689404.00 -2 5 26.34974 52 2693933.00 -2 1 23.6772 53 2698218.00 -1 57 32.82784 54 2702279.00 -1 53 53.02139 55 2706129.00 -1 50 23.55034 7.3 Conservation of mechanical energy Example 56. A wagon of mass m = 500 kg is released from point A and slides on the frictionless track shown in Figure. Determine (a) the height h and (b) the a wagon’s speed at points B, C, D. Assume that your speed is (N+3) m/s and that in point E stops . Uses the equations of conservation of mechanical energy. The code in BASIC programming language to solve previous problem is: cls print"PROGRAM TO CALCULATE SPEEDS AND HEIGHTS ON THE TRACK" pi=3.1415923 open "c:track.txt" for output as #1 print#1, "conservation of mechanical energy: " for N=1 to 55
  220. 220. 218 h=(0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/(500*9.8) vb=sqr((0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/250) vc=sqr((0.5*500*vb*vb-500*9.8*(10+N))/250) vd=sqr((0.5*500*vc*vc+500*9.8*(10+N))/250) print#1, N;"h = ";h;" vb = ";vb;" vc = ";vc;" vd = ";vd next N close #1 end Too individual responses were obtained using the programming language FORTRAN 90, whose source code is: Program Track REAL h,vb,vc,vd INTEGER :: N PRINT *,’ ’ PRINT *, "Programmed by Neftalí Antúnez H. (p)1986" PRINT *, "CEO of Antúnez Software LTD" ! Create an external new data file OPEN (7, FILE = ’Track.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’) ! Write Writing the header WRITE(7,*) "N, h, Vb, Vc, Vd" PRINT *,’ ’ DO N=1,55,1 h=(0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/(500*9.8) vb=sqrt((0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/250) vc=sqrt((0.5*500*vb*vb-500*9.8*(10+N))/250) vd=sqrt((0.5*500*vc*vc+500*9.8*(10+N))/250) WRITE(7,*) N,",",h,",",vb,",",vc,",",vd END DO ! Close the external file CLOSE(7) PRINT *,’ ’ PRINT *," THANK YOU FOR USING Track" END Program Track The results table in Microsoft Excel® is:
  221. 221. 219
  222. 222. 220
  223. 223. 221 7.4 Kirchhoff’s Rules The procedure for analyzing complex electric circuits is greatly simplified if we use two principles called Kirchhoff’s rules: 1. Junction rule.The sum of the currents entering any junction in a circuit must equal the sum of the currents leaving that junction: 2. Loop rule.The sum of the potential differences across all elements around any closed circuit loop must be zero: PROBLEM-SOLVING HINTS Kirchhoff’s Rules • Draw a circuit diagram, and label all the known and unknown quantities. You must assign a direction to the current in each branch of the circuit. Although the assignment of current directions is arbitrary, you must adhere rigorously to the assigned directions when applying Kirchhoff’s rules. • Apply the junction rule to any junctions in the circuit that provide new relationships among the various currents. • Apply the loop rule to as many loops in the circuit as are needed to solve for the unknowns. Toapply this rule, you must correctly identify the potential difference as you imagine crossing each element while traversing the closed loop (either clockwise or counterclockwise). Watch out for errors in sign! • Solve the equations simultaneously for the unknown quantities.Do not be alarmed if a current turns out to be negative; its magnitude will be correct and the direction is opposite to that which you assigned. Example 57. Find the currents in the circuit shown in Figure:
  224. 224. 222 Applying Kirchhoff’s junction rule to junction c gives: We now have one equation with three unknowns . There are three loops in the circuit—abdca, cdfec, and abfea.We therefore need only two loop equations to determine the unknown currents. (The third loop equation would give no new information.) Applying Kirchhoff’s loop rule to loops abdca and cdfec and traversing these loops clockwise, we obtain the expressions (2) Loop abdca:
  225. 225. 223 (3) Loop cdfec: Changing by respectively. The system to solve is: Solving with Derive®. We have:
  226. 226. 224
  227. 227. 225

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