Math for Smart Teachers. Ortho mathetics® for-teachers-stopping-the-cheating-in-science

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.

OrthoMathetics®
For Teachers
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

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OrthoMathetics®
Neftali Antunez H.
“The teachers to teach less and the learner to learn more” Comenius
ISBN-13:978-1481859462
ISBN-10:1481859463

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OrthoMathetics® For Teachers
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.

4

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

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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

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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.

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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

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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.

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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.

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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.

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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

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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

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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.

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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.

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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

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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

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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

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.

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.

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.

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

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

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:

27
Repeat the procedure for the Vb variable. The following window appears.

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
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:

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:

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.

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:

36
Calculating the equation of each side. Using the point-slope form:

38
The equations of the sides AB, BC, and CA respectively are:
The table of results is:

41
Calculating the perimeter and area:

42
The perimeters and areas of the polygon are:

45
Calculating the interior angles:

46
The interior angles A, B, C of the polygon are:

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

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

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:

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 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:

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

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

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:

58
Click again in the Expand button, we have:
The quadratic equation that we will put them so that students solve is:

59
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

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

61
Part II
OrthoMathetics Applications

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
Example 6. Solve the following exponential equation:
Solving with Derive®, the results are:

68
Example 7. Multiply the following polynomials:
The answer is:
Example 8. Divide the following polynomials:

69
by
The answer is: Remainder = 0
by

73
by

79
by
The answer is:
Remainder = 0
Example 12. Multiply the following conjugated binomials:

80
ans.
Example 13. Expanding the binomial using Newton’s theorem:
The results are:

85
Example 14. Find the 12th term of the binomial and find the N-th
term of the binomial

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:

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:

92
The answer 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 -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

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
In the same way, we obtain the answer is:

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

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
By being so obvious answers table is not shown.

97
Example 18. To solve the following system of equations linear with three variables:
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:

98
By being so obvious only some answers are shown.

99

100
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

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

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
To solve the following system of equations linear with three variables:
The results are:

107
Example 21. To solve the following system of equations linear with four variables:

108
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:

109
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

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

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:

112
The next results by rows are:

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
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])

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:

128
Example 26. Find the eigenvalues of the matrix:
With Charpoly find the characteristic equation, which match to zero to find
their roots.

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:

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:

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
following columns:

137
Example 31. For use in numerical methods, find the roots of the following equation:

138
Their roots are:
following columns:

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:

141
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

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.

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

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

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:

147
The results for triangle 3 are: For all N, the angle t is constant:

150
The results for triangle 4 are: For all N, the angle B is constant:

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);" ";

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:

155

157

159
The triangle 4 is isosceles. For all N, the angles are constant: and
. The results for triangle 4 are:

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
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
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

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
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

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:

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

166
Vertex:
focal length: p = 3
Focus:

167
Vertex:
focal length: p = 3
Focus:

168
Vertex:
focal length: p = 5
Focus:

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
answer
answer
Example 44. Derive the product of functions:
answer
Example 45. Derive the quotient of functions:

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:

172
Now, finding:
minimum
maximum
The Graph for N = 30 is:
Tabulating the results:

178
Solving:
Now, finding:
Results:
minimum
maximum
Tabulating the results using Microsoft Excel®:

181
Solving:
Now, finding:
Results:
maximum
minimum

182

184
Solving:
Now, finding:
Results:
maximum
minimum

187
6.3 Definite Integrals
Example 50. Find the definite integral of
from to
The results are:

189
Example 51. Find the area formed by the intersection of curves:
and
Using Derive® 6.1 to make the operations, we have:

196
Example 52. Find the area formed by the intersection of curves:
and
Using Derive® 6.1 to make the operations, we have:

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

205
D N 249.3000
E 80 290.8833
F 40 308.7500
P N 76.2833
Q 60 231.2500
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 *,’ ’
OPEN (7, FILE = ’Addvect.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’)
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

206
END DO
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

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

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

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
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 *,’ ’
OPEN (7, FILE = ’Equivect.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’)
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

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

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

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:

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: .

214
Since at the point P is considered a positive unit charge, each field produced
by the charges are shown in the following figure:
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 *,’ ’
OPEN (7, FILE = ’Efield.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’)
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

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(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

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

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

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
Program Track
REAL h,vb,vc,vd
INTEGER :: N
PRINT *,’ ’
OPEN (7, FILE = ’Track.TXT’, ACCESS = ’APPEND’,STATUS = ’REPLACE’)
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(7)
PRINT *,’ ’
PRINT *," THANK YOU FOR USING Track"
END Program Track
The results table in Microsoft Excel® is:

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:

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:

223
(3) Loop cdfec:
Changing by respectively.
The system to solve is:
Solving with Derive®. We have:

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