Download pdf - Math for smart teachers

Emerges as a response to

academic dishonesty or

academic misconduct in any type

of cheating that occurs in the

schools everywhere.

B”Y

Neftalí Antúnez H. Ph. D. Civil Engineer with a Doctorate Degree in Education

Full time Teacher in the Faculty of Engineering of the UAGro and in CBTis No. 134

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.

ISBN-13:978-1481859462

ISBN-10:1481859463

B”Y

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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

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.

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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 ·

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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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Part I

Introduction

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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: [email protected], 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.


mailto:[email protected]

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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


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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.


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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

mailto:[email protected]

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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.

28

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:

30

31

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:

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 = 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

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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

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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

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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:

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


by


70

71

72

73


by


74

75

76

77

78

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:

81

82

83

84

85

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

term of the binomial


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 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:

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:

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:

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 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

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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:

113

114

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:

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 is obtained by typing:

det([n, 3, 2; -1, -2, -n; 3, 2, 7])

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, 2, 6]^(-1)

The result is:

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 find their roots.

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 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:

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


137

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

138

Their roots are:


139

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:

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" 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:

154

155


156

157


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

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:



169

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:

173

174

175

176

177

178



Solving:

Now, finding:

Results:

minimum

maximum


Tabulating the results using Microsoft Excel®:

179

180

181



Solving:

Now, finding:

Results:

maximum

minimum


182


183

184



Solving:

Now, finding:

Results:

maximum

minimum


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


Example 52. Find the area formed by the intersection of curves:

and

Using Derive® 6.1 to make the operations, we have:

197

198

The resultant are is:

199

200

201

202

203


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

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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 *,’ ’




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:

219

220

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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:

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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:

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(3) Loop cdfec:

Changing by respectively.

The system to solve is:

Solving with Derive®. We have:

224

225