Math for smart teachers

  • Published on
    22-Jul-2016

  • View
    225

  • Download
    1

DESCRIPTION

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.

Transcript

  • Emerges as a response to

    academic dishonesty or

    academic misconduct in any type

    of cheating that occurs in the

    schools everywhere.

    BY

  • Neftal Antnez 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., Mxico

    Introducing OrthoMathetics to the scientific community, a new branch of education whose name is derived from Ortho (from the Greek word

    meaning "straight" or "correct") and Mathetics means the science of learning.

    ISBN-13:978-1481859462

    ISBN-10:1481859463

    BY

  • 1

    OrthoMathetics Neftali Antunez H.

    The teachers to teach less and the learner to learn more Comenius

    ISBN-13:978-1481859462

    ISBN-10:1481859463

  • 2

    This page is intentionally left blank

  • 3

    OrthoMathetics For Teachers

    Stopping the Cheating in Science!

    NEFTALI ANTUNEZ H.

    Civil Engineer with a Masters Degree in Education

    Full time Teacher in the Faculty of Engineering of the

    Guerrero State Autonomous University (UAG)

    Chilpancingo, Gro., Mxico

    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

    This page is intentionally left blank

  • 5

    INDEX

    Chapter 1

    Definition

    1.1 Roots of OrthoMathetics 5 1.10 Characteristics of the constructivist student

    16 1.11 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 Coulombs Law and Electric Field 209 7.3 Conservation of mechanical energy

    215 7.4 Kirchhoffs 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

  • 6

    This page is intentionally left blank

  • 7

    Part I

    Introduction

  • 8

    Chapter 1

    Definition

    1.1 Roots of OrthoMathetics

    Introducing OrthoMathetics to the scientific community like a new branch of education, whose name is Derived from Ortho (from the Greek word meaning "straight" or "correct") and Mathetics means the science of learning. The term Mathetics was coined by John Amos Comenius (15921670) 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 Childrens Machine. The origin of the word, according to Papert, is not from "mathematics," but from the Greek, mathmatikos, 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 Childrens 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] Paperts 1980 book, Mindstorms: Children, Computers, and

    1 From Wikipedia, the free encyclopedia

  • 9

    Powerful Ideas, discusses the mathetics approach to learning. By using a mathetic approach, Papert feels that independent learning and creative thinking are being encouraged. The mathetic approach is a strong advocate of learning by doing. Many proponents of the mathetic approach feel that the best, and maybe the only, way to learn is by self discovery.

    1.3 John Amos Comenius (15921670)2

    John Amos Comenius (Czech: Jan Amos Komensk; Slovak: Jn Amos Komensk; German: Johann Amos Comenius; Polish: Jan Amos Komeski; Hungarian: Comenius mos Jnos; Latinized: Iohannes Amos Comenius) (28 March 1592 4 November 1670) was a Czech teacher, educator, and writer. He served as the last bishop of Unity of the Brethren, and became a religious refugee and one of the earliest champions of universal education, a concept eventually set forth in his book Didactica Magna. He is considered the father of modern education. He lived and worked in many different countries in Europe, including Sweden, the Polish-Lithuanian Commonwealth,Transylvania, the Holy Roman Empire, England, the Netherlands, and Royal Hungary.

    1.3.1 Educational influence

    The most permanent influence exerted by Comenius was in practical educational work. Few men since his day have had a greater influence, though for the greater part of the eighteenth century and the early part of the nineteenth there was little recognition of his relationship to the current advance in educational thought and practice. The practical educational influence of Comenius was threefold. He was first a teacher and an organizer of schools, not only among his own people, but later in Sweden, and to a slight extent in Holland. In his Didactica Magna (Great Didactic), he outlined a system of schools that is the exact counterpart of the existing American system of kindergarten, elementary school, secondary school, college, and university.

    2 From Wikipedia, the free encyclopedia

  • 10

    In the second place, the influence of Comenius was in formulating the general theory of education. In this respect he is the forerunner of Rousseau, Pestalozzi, Froebel, etc., and is the first to formulate that idea of education according to nature so influential during the latter part of the eighteenth and early part of the nineteenth century. The influence of Comenius on educational thought is comparable with that of his contemporaries, Bacon and Descartes, on science and philosophy. In fact, he was largely influenced by the thought

    of these two; and his importance is largely due to the fact that he first applied or attempted to apply in a systematic manner the principles of thought and of investigation, newly formulated by those philosophers, to the organization of education in all its aspects. The summary of this attempt is given in the Didactica Magna, completed about 1631, though not published until several years later.

    The third aspect of his educational influence was that on the subject matter and method of education, exerted through a series of textbooks of an entirely new nature. The first-published of these was the Janua Linguarum Reserata (The Gate of Languages Unlocked), issued in 1631. This was followed later by a more elementary text, the Vestibulum, and a more advanced one, the Atrium, and other texts. In 1657 was published the Orbis Sensualium Pictus probably the most renowned and most widely circulated of school textbooks. It was also the first successful application of illustrations to the work of teaching, though not, as often stated, the first illustrated book for children.

    These texts were all based on the same fundamental ideas: (1) learning foreign languages through the vernacular; (2) obtaining ideas through objects rather than words; (3) starting with objects most familiar to the child to introduce him to both the new language and the more remote world of objects: (4) giving the child a comprehensive knowledge of his environment, physical and social, as well as instruction in religious, moral, and classical subjects; (5) making this acquisition of a compendium of knowledge a pleasure rather than a task; and (6) making instruction universal. While the formulation of many of these ideas is open to criticism from more recent points of view, and while the naturalistic conception of education is one based on crude analogies, the importance of the Comenian

  • 11

    influence in education has now been recognized for half a century. The educational writings of Comenius comprise more than forty titles. In 1892 the three-hundredth anniversary of Comenius was very generally celebrated by educators, and at that time the Comenian Society for the study and publication of his works was formed.

    1.4 Purposes of OrthoMathetics

    Emerges as a response to academic dishonesty or academic misconduct in any type of cheating, which occurs in the schools everywhere. Academic dishonesty has been documented in most every type of educational setting, from elementary school to graduate school, and has been met with varying degrees of approbation throughout history. Today, educated society tends to take a very negative view of academic dishonesty. This project is especially directed to the teachers of science in the world. For this reason, dont make an emphasis on theory and full development of the examples, is not because that is not its main objective. We only give examples so that the teacher can develop their own exercises from the examples presented here.

    This is a dynamic book, which will grow because will be adding more examples by the same author, or, by the contribution of thousands of teachers in the world. Can send their examples to my email: antunezsoftware@gmail.com, or better yet, write their own books according to the rules of OrthoMathetics. Some titles of books could be: OrthoMathetics for Algebra, OrthoMathetics for Calculus, OrthoMathetics for Physics, OrthoMathetics for Chemistry, OrthoMathetics for Numerical Methods, etc. All that is asked is that in the preface, prologue or introduction, you specify that OrthoMathetics was created by the author of this book.

    1.5 History of academic dishonesty3

    In antiquity, the notion of intellectual property did not exist. Ideas were the common property of the literate elite. Books were published by hand-copying them. Scholars freely made digests or commentaries on other works, which could contain as much or as little original material as the author desired. There was no standard system of citation, because printingand its resulting fixed paginationwas in the future. Scholars were an elite and a small group, who knew and generally trusted each other. This system continued through the European Middle Ages. Education was in Latin and occasionally Greek. Some scholars were monks, lived in monasteries, and spent much of their time copying manuscripts. Other scholars were in urban universities connected to the Roman Catholic Church.

    3 From Wikipedia, the free encyclopedia

  • 12

    Academic dishonesty dates back to the first tests. Scholars note that cheating was prevalent on the Chinese civil service exams thousands of years ago, even when cheating carried the penalty of death for both examinee and examiner. In the late 19th and early 20th centuries, cheating was widespread at college campuses in the United States, and was not considered dishonorable among students. It has been estimated that as many as two-thirds of students cheated at some point of their college careers at the turn of the 20th century. Fraternities often operated so-called essay mills, where term papers were kept on file and could be resubmitted over and over again by different students, often with the only change being the name on the paper.[citation needed] As higher education in the U.S. trended towards meritocracy, however, a greater emphasis was put on anti-cheating policies, and the newly diverse student bodies tended to arrive with a more negative view of academic dishonesty. Unluckily, in some areas academic dishonesty is widely spread and people who do not cheat represent a minority between the class.

    1.6 Academic dishonesty today4

    Academic dishonesty is endemic in all levels of education. In the United States, studies show that 20% of students started cheating in the first grade. Similarly, other studies reveal that currently in the U.S., 56% of middle school students and 70% of high school students have cheated.

    Students are not the only ones to cheat in an academic setting. A study among North Carolina school teachers found that some 35 percent of respondents said they had witnessed their colleagues cheating in one form or another. The rise of high-stakes testing and the consequences of the results on the teacher is cited as a reason why a teacher might want to inflate the results of their students.

    The first scholarly studies in the 1960s of academic dishonesty in higher education found that nationally in the U.S., somewhere between 50%-70% of college students had cheated at least once. While nationally, these rates of cheating in the U.S. remain stable today, there are large disparities between different schools, depending on the size, selectivity, and anti-cheating policies of the school. Generally, the smaller and more selective the college, the less cheating occurs there. For instance, the number of students who have engaged in academic dishonesty at small elite liberal arts college scan be as low as 15%-20%, while cheating at large public universities can be as high as 75%. Moreover, researchers have found that students who attend a school with an honor code are less likely to cheat than students at schools with other ways of enforcing academic integrity. As for graduate education, a recent study found that 56% of MBA students admitted cheating, along with 54% of graduate students in engineering, 48% in education, and 45% in law.

    4 From Wikipedia, the free encyclopedia

  • 13

    Moreover, there are online services that offer to prepare any kind of homework of high school and college level and take online tests for students. While administrators are often aware of such websites, they have been unsuccessful in curbing cheating in homework and non-proctored online tests, resorting to a recommendation by the Ohio Mathematics Association to derive at least 80% of the grade of online classes from proctored tests. While research on academic dishonesty in other countries is minimal, anecdotal evidence suggests cheating could be even more common in countries like Japan.

    The use of crib notes during an examination is typically viewed as cheating.

    1.7 Cheating5

    Cheating refers to an immoral way of achieving a goal. It is generally used for the breaking of rules to gain advantage in a competitive situation. Cheating is the getting of reward for ability by dishonest means. This broad definition will necessarily include acts of bribery, cronyism, sleaze, nepotism and any situation where individuals are given preference using inappropriate criteria. The rules infringed may be explicit, or they may be from an unwritten code of conduct based on morality, ethics or custom, making the identification of cheating a subjective process.

    5 From Wikipedia, the free encyclopedia

  • 14

    Cheating can take the form of crib notes, looking over someones 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 students 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 cheatingthough deprecatedcould 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 nations 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

  • 15

    Everybody does it.

    Its disturbing to discover that young people in middle school and high school think that it is acceptable to cheat. But its our fault, isnt 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. Its 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 nations 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 lifes 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. Plagiarisms 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 its 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 youd prefer to cheat using texting or email. Both work just

  • 16

    fine for that purpose. Sadly, many parents and teachers have not learned the subtleties of electronic cheating

    1.8 Solutions?

    Schools need to have zero tolerance policies concerning cheating.

    Teachers must be vigilant and alert to all the newer forms of cheating, particularly electronic cheating. Cellphones and iPods are powerful tools for cheating with uses limited only by a students 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. Thats 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 childrens 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. Dont

    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 students 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. Neftal Antnez H.

  • 17

    the learner, which can only be achieved by the learner that he makes, deals, and engages in activities of learning. The role of the teacher is important to achieve that the students carry out activities that otherwise would not do. The teacher has to involve students in tasks, some of which may include skills acquisition for examples of work. Other tasks include practices of skills to bring them to effective levels, interacting with their peers and the teacher.

    In a traditional classroom, the teachers 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.

  • 18

    In a traditional classroom, can see a teacher who it remains in front of the group, mainly exposes and tries to fill with information your students 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

  • 19

    research, i.e. as they learn how to learn, greater autonomy displayed when you perform the activities established by the teacher, which mostly consist of problem solving in mathematics.

    Constructivist learning is based on the active participation of students in solving problems and in the use of critical thinking regarding the learning activities that are relevant and interesting. They are constructing their own knowledge by testing ideas and approaches based on their knowledge and previous experience, applying them to a new situation, and integrating the new knowledge gained with their pre-existing intellectual constructions.

    1.10 Characteristics of the constructivist student8

    Constructivist students actively participate and are not limited to passively receiving information. Get involved and are responsible for their learning, investigate, seek, ask, discuss and dialogue with their peers and the teacher.

    The students read, think and analyze the information and not accept it without thinking, they expose their ideas to the other and work as a team. They perform their tasks and extracurricular work and in the case of math, solve problems and exercises. To learn, a student must be often physically and mentally active. A student learns (this it is, builds structures of knowledge) when it discovers its own answers, solutions, concepts and relationships and creates its own interpretations.

    Constructivism proposes that when students drive their own learning, discover their own answers and create their own interpretations, their learning is deeper, more comprehensive and lasting, and the learning that takes place actively it leads to think critically. In a constructivist classroom, the students demonstrate their learning and understanding through several forms. They can develop new critical issues, write a script for a video, summarizing the main ideas with your own words, can produce or create something, they can solve problems. Its about being active, not passive."

    To change the traditional education, is required to change from an education centered on the teacher to one focused on students, an education where the learners take an active role, i.e. where students participate and responsibly are involved in your learning. Some claim that all learning is inherently active and students are also actively involved while listening the formal presentations in the classroom.. However, the majority agrees that not enough students listen, in addition, they should read, discuss and above all solve problems. Most importantly, must be actively involved, students must perform tasks of higher order thinking such as analysis, synthesis and evaluation. Within this context, be

    8 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftal Antnez H.

  • 20

    it proposes strategies that promote active learning, defined as instructional activities that involve to the students in doing things and think about what they are doing.

    1.11 Constructivist evaluation9

    The evaluation and learning are processes, therefore, not should be evaluated only using the test - as some teachers usually do-, but using many more tools and considering all aspects both objective and subjective; In addition, like process students should be continually evaluated and not only in certain isolated moments. Several instruments of assessment must be used and not one instrument or a method most appropriate must select to the activity that is being evaluated.

    Evaluation has subjective aspects motivation, dedication, effort, emotions-, therefore not should be evaluated only the learning for students. This should be taken into account when designing our objectives to be achieved and the activities to assess. It is not enough to assess the learning that our students carry out, as teachers we must evaluate our own performance and teaching activities which we plan and develop.

    Evaluation is an important and fundamental action in the process of learning, but must serve to support this process, so that reinforce learning and help to feed both the teacher as to the students. However, it is not easy to correctly evaluate and more if we use constructivist theory, where the students are gradually building your knowledge. The evaluation must assess different capacities learned: motor, cognitive, affective or emotional balance, relationship interpersonal, and performance and social inclusion. Evaluate the learning acquired by the students is equivalent to clarify until point have developed or learned certain skills as a result of the education received.

    All teachers are trying to achieve meaningful learning in your students, but how evaluate the meaningful learning? With activities that allow to evaluate the advancement of learning, in the case of mathematics by means of the resolution of problems of difficulty gradual. The significant learning should not be evaluated as all or nothing, but rather as a degree of progress, you must detect the degree of significance of learning done. Constructivist Learning is usually assessed through projects based on performance, instead of traditional paper and pencil tests.

    The constructivist evaluation focuses on what the student can do with the knowledge. In general, it is recommended that the student demonstrate their learning through the application of knowledge, for example; solving problems

    9 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftal Antnez H.

  • 21

    and exercises, evaluation of expressions, solving equations, factoring expressions, operations with functions, write summaries and essays, create a product, model or prototype, create a video, writing literature, music or poetry, create or conduct experiments.

    1.12 Difference between exercise and problem10

    There is a basic difference between the concept problem and exercise. Not is

    the same an exercise that to solve a problem. An exercise serves for exercise, practice or reinforce learning of an algorithm or a method, but does not contribute any new knowledge. In change, the problem involves a degree of difficulty and a depth greater, which requires more time to resolve it and forced to perform always an investigation. One thing is to apply an algorithm in form more or less mechanics as it is done in the exercise and another, solve a problem. The answer tends to be unique, but the adjudicative strategy is determined by maturation or other factors. In fact, the exercise does not require the preparation of a plan or a strategy, since it is only necessary to apply the known method to solve it.

    Problem solving strategy is much richer that the mechanical application of an algorithm, as it involves creating a context where the data store certain coherence. Since this analysis have been establish hierarchies: see which data are priority, reject the distorting elements or distractors, choose transactions to the they relate, to estimate the range of variation, identify the unknown, establish a plan or strategy, execute the plan and check the results obtained. Students who learn math, from a point of view constructivist, must precisely built concepts through the interaction that has with objects and other subjects. It seems so the student can build their knowledge to carry out the active interaction with the mathematical objects, including the reflection that allows you to abstract these objects, it is necessary that these objects are immersed in a problem and not in an exercise. In fact, are these problematic situations which introduce an imbalance in mental structures of the student, in his interest for balance (an accommodation) occurs the construction of the knowledge. The term problem is a complex situation (real or hypothetical) that it involves concepts, objects or mathematical operations.

    A problem requires time to solve because of its complexity; the student requires your full attention, energy, time and dedication to solve it. Exercise, refers to operations with symbols mathematical only (sums, multiplication, resolution of equations, and so on), implies that the student repeat an algorithm, procedure, or a taught method to directly and quickly get to the solution.

    10 The effectiveness of the constructivist teaching of the arithmetic and Algebra in high school. Neftal Antnez H.

  • 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 dont 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 dont 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, Ive 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 Newtons 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 surveyors 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.

    Slopeintercept 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.

    Pointslope 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

  • 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

    The full table with all answers is:

    N

    1 0.3333 -0.25

    2 0.6667 -0.50

    3 1.0000 -0.75

    4 1.3333 -1.00

    5 1.6667 -1.25

    6 2.0000 -1.50

    7 2.3333 -1.75

    8 2.6667 -2.00

    9 3.0000 -2.25

    10 3.3333 -2.50

    11 3.6667 -2.75

    12 4.0000 -3.00

    13 4.3333 -3.25

    14 4.6667 -3.50

    15 5.0000 -3.75

    16 5.3333 -4.00

    17 5.6667 -4.25

    18 6.0000 -4.50

    19 6.3333 -4.75

    20 6.6667 -5.00

    21 7.0000 -5.25

    22 7.3333 -5.50

    23 7.6667 -5.75

    24 8.0000 -6.00

    25 8.3333 -6.25

  • 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. Neftal Antnez 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

    Example 9. Divide the following polynomials:

    by

    The table of results is:

  • 70

  • 71

  • 72

  • 73

    Example 10. Divide the following polynomials:

    by

    The table of results is:

  • 74

  • 75

  • 76

  • 77

  • 78

  • 79

    Example 11. Divide the following polynomials:

    by

    The answer is:

    Remainder = 0

    Example 12. Multiply the following conjugated binomials:

  • 80

    ans.

    Example 13. Expanding the binomial using Newtons theorem:

    The results are:

  • 81

  • 82

  • 83

  • 84

  • 85

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

    term of the binomial

    The table of results is:

  • 86

  • 87

  • 88

  • 89

  • 90

    3.2 Systems of Equations Linear with Two and Three Variables

    Example 15. To solve the following system of equations 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:

    The full table with all answers is:

  • 93

    N x y

    1 2 -4

    2 4 -8

    3 6 -12

    4 8 -16

    5 10 -20

    6 12 -24

    7 14 -28

    8 16 -32

    9 18 -36

    10 20 -40

    11 22 -44

    12 24 -48

    13 26 -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

    Example 16. To solve the following system of equations linear with two variables:

    In the same way, we obtain the answer is:

    The full table with all answers is:

  • 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

    Example 17. To solve the following system of equations linear with two variables:

    In the same way, we obtain the answer is:

    By being so obvious answers table is not shown.

  • 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

    Example 19. To solve the following system of equations linear with three variables:

    In the same way, we obtain the answer is:

  • 100

    The full table with all answers is:

    N x y z

    1 0.13889 0.02778 0.19444

    2 0.27778 0.05556 0.38889

    3 0.41667 0.08333 0.58333

    4 0.55556 0.11111 0.77778

    5 0.69444 0.13889 0.97222

    6 0.83333 0.16667 1.16667

    7 0.97222 0.19444 1.36111

    8 1.11111 0.22222 1.55556

    9 1.25000 0.25000 1.75000

  • 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

    Example 20. To solve the following system of equations linear with three variables:

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

    In the same way, we obtain the answer is:

    The results are:

  • 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:

    In the same way, we obtain the answer is:

  • 109

    The full table with all answers is:

    N x y z t

    1 15.0625 -19.75 -39.50 34.8125

    2 15.1250 -19.50 -39.00 34.6250

    3 15.1875 -19.25 -38.50 34.4375

    4 15.2500 -19.00 -38.00 34.2500

    5 15.3125 -18.75 -37.50 34.0625

    6 15.3750 -18.50 -37.00 33.8750

    7 15.4375 -18.25 -36.50 33.6875

    8 15.5000 -18.00 -36.00 33.5000

    9 15.5625 -17.75 -35.50 33.3125

  • 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:

  • 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

    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:

  • 137

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

  • 138

    Their roots are:

    It is also very useful to tabulate the responses using Microsoft Excel or similar, placing N in first column and the responses depending on N in the following columns:

  • 139

  • 140

    Chapter 4

    Applications to Geometry and Trigonometry

    4.1 Arc Length

    ARC LENGTH PROBLEM S

    where 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

    The table of results is:

    N S2, S3 x S4

    1 0.26667 178 47.46667

    2 .53334 176 46.93334

    3 0.8 174 46.4

    4 1.06667 172 45.86667

    5 1.3334 170 45.3334

    6 1.6 168 44.8

    7 1.86667 166 44.26667

    8 2.13334 164 43.73334

    9 2.4 162 43.2

    10 2.6667 160 42.6667

    11 2.93334 158 42.13334

    12 3.2 156 41.6

    13 3.46667 154 41.06667

    14 3.73334 152 40.53334

    15 4 150 40

    16 4.26667 148 39.46667

    17 4.53334 146 38.93334

    18 4.8 144 38.4

    19 5.06667 142 37.86667

    20 5.333433 140 37.3334

    21 5.6 138 36.8

    22 5.86667 136 36.26667

    23 6.13334 134 35.73334

    24 6.4 132 35.2

    25 6.6667 130 34.6667

    26 6.93334 128 34.13334

    27 7.2 126 33.6

  • 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

    The results for triangle 2 are:

  • 156

  • 157

    The results for triangle 3 are:

  • 158

  • 159

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

  • 160

  • 161

  • 162

    Individual responses were obtained using the programming language FORTRAN 90, whose source code is:

    Program Triangles

    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 Neftal Antnez H. (p)1986"

    PRINT *, "CEO of Antnez Software LTD"

    ! Create an external new data file

    OPEN (7, FILE = c:similar.txt, ACCESS = APPEND,STATUS = REPLACE)

    ! Write Writing the header

    WRITE(7,*) "Triangle 1"

    PRINT *,

    DO N=1,55,1

    x=1.147643*N

    w=1.562651*N

    WRITE(7,*) N,",",x,",",w

    END DO

    WRITE(7,*) "Triangle 2"

    DO N=1,55,1

    ! x=-N-N**2

    b=sqrt((N+1)**2+(N+4)**2+0.4635*(N+1)*(N+4))

    angC=acos(((N+1)**2-(N+4)**2-b**2)/(-2*(N+4)*b))

    angC=180*angC/pi

    dC=int(angC)

    min=(angC-dC)*60

    mC=int(min)

    secC=(min-mC)*60

    angA=acos(((N+4)**2-(N+1)**2-b**2)/(-2*(N+1)*b))

    angA=180*angA/pi

    dA=int(angA)

    min=(angA-dA)*60

    mA=int(min)

    secA=(min-mA)*60

    WRITE(7,*)N,",",b,",",dc,"deg",mc,"",secC,",,da,"deg",ma,"",seca,"

    END DO

    WRITE(7,*) "Triangle 3"

    DO N=1,55,1

    ! Changing N integer by T real because cosinus needs a real argument

    T=N

    angC=acos(((T+2)**2-(T+1)**2-(T+3)**2)/(-2*(T+1)*(T+3)))

    angC=180*angC/pi

    dC=int(angC)

    min=(angC-dC)*60

    mC=int(min)

    secC=(min-mC)*60

    angA=acos(((T+3)**2-(T+1)**2-(T+2)**2)/(-2*(T+1)*(T+2)))

    angA=180*angA/pi

    dA=int(angA)

    min=(angA-dA)*60

    mA=int(min)

    secA=(min-mA)*60

    angB=acos(((T+1)**2-(T+2)**2-(T+3)**2)/(-2*(T+2)*(T+3)))

    angB=180*angB/pi

    dB=int(angB)

    min=(angB-dB)*60

  • 163

    mB=int(min)

    secB=(min-mB)*60

    WRITE(7,*) N,",",dc,"deg",mc,"",secC,",,da,"deg",ma,"",seca,",

    WRITE(7,*) N,",",db,"deg",mb,"",secb,",,db,"deg",mb,"",secb,"

    END DO

    WRITE(7,*) "Triangle 4"

    DO N=1,55,1

    x=1.08586*N

    WRITE(7,*) N,",",x

    end do

    ! Close the external file

    CLOSE(7)

    PRINT *,

    PRINT *," THANK YOU FOR USING Triangles"

    END Program Triangles

  • 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

    Example 38. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is:

    Vertex:

    focal length: p = 3

    Focus:

    Equation of directrix:

    Graphing the parabola with Algebrator 5 spanish version, for N = 24

  • 167

    Example 39. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is:

    Vertex:

    focal length: p = 3

    Focus:

    Equation of directrix:

    Graphing the parabola with Algebrator 5 spanish version, for N = 24

  • 168

    Example 40. Find the vertex, focus, focal length p and the equation of directrix of the parabola whose general equation is:

    Vertex:

    focal length: p = 5

    Focus:

    Equation of directrix:

    Graphing the parabola with Algebrator 5 spanish version, for N = 30

  • 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 dont interrupt, by all means, transform this suffering into enjoyment, this does not mean absence of effort, but on the contrary, delivery of stimuli and efforts effective and desired. (Puig Adam, Peter 1958)

    6.1 Derivatives

    Example 41. Derive the following function:

    answer

    Example 42. Derive the following function:

    answer

    Example 43. Derive the following function:

    answer

    Example 44. Derive the product of functions:

    answer

    Example 45. Derive the quotient of functions:

  • 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

    Example 47. Find the maxima and minima of the following function:

    Deriving and equating to zero:

    Solving:

    Now, finding:

    Results:

    minimum

    maximum

    The Graph for N = 25 is:

    Tabulating the results using Microsoft Excel:

  • 179

  • 180

  • 181

    Example 48. Find the maxima and minima of the following function:

    Deriving and equating to zero:

    Solving:

    Now, finding:

    Results:

    maximum

    minimum

    The Graph for N = 25 is:

  • 182

    Tabulating the results using Microsoft Excel:

  • 183

  • 184

    Example 49. Find the maxima and minima of the following function:

    Deriving and equating to zero:

    Solving:

    Now, finding:

    Results:

    maximum

    minimum

    Tabulating the results using Microsoft Excel:

  • 185

  • 186

  • 187

    6.3 Definite Integrals

    Example 50. Find the definite integral of

    from to

    The results are:

  • 188

  • 189

    Example 51. Find the area formed by the intersection of curves:

    and

    Using Derive 6.1 to make the operations, we have:

  • 190

  • 191

    The resultant area is:

  • 192

  • 193

  • 194

  • 195

  • 196

    The Graph for N = 30 is:

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

    and

    Using Derive 6.1 to make the operations, we have:

  • 197

  • 198

    The resultant are is:

  • 199

  • 200

  • 201

  • 202

  • 203

    The Graph for N = 30 is:

  • 204

    Chapter 7

    Applications to Physics

    7.1 Vectors

    Example 53. Find the resultant vector and its direction of the following 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

    Individual responses were obtained using the programming language FORTRAN 90, whose source code is:

    Program Vectrix2d

    REAL :: pi=3.1415923

    REAL RX,RY,RES,ang,sec

    REAL MINUT,FX,FY

    INTEGER MINUTS

    INTEGER :: N

    PRINT *, "PROGRAM VECTRIX 2D"

    PRINT *,"SERVES TO ADD N CONCURRENT VECTOR IN THE XY PLANE"

    PRINT *,"BY THE METHOD OF RECTANGULAR COMPONENTS"

    PRINT *,

    PRINT *, "Programmed by Neftal Antnez H. (p)1986"

    PRINT *, "CEO of Antnez Software LTD"

    ! Create an external new data file

    OPEN (7, FILE = Addvect.TXT, ACCESS = APPEND,STATUS = REPLACE)

    ! Write Writing the header

    WRITE(7,*) " N MAGNITUDE DIRECTION"

    PRINT *,

    ! Angles in radians

    RX=0.0

    RY=0.0

    a=50;anga=30.4167*pi/180

    b=60;angb=104.30*pi/180

    c=75;angc=149.5833*pi/180

    e=80;ange=290.8833*pi/180

    f=40;angf=308.75*pi/180

    q=60;angq=231.25*pi/180

    RX=a*cos(anga)+b*cos(angb)+c*cos(angc)+e*cos(ange)+f*cos(angf)+q*cos(a

    ngq)

    RY=a*sin(anga)+b*sin(angb)+c*sin(angc)+e*sin(ange)+f*sin(angf)+q*sin(a

    ngq)

    DO N=1,55,1

    d=N;angd=249.30*pi/180

    p=N;angp=76.2833*pi/180

    FX=RX+d*cos(angd)+p*cos(angp)

    FY=RY+d*sin(angd)+p*sin(angp)

    RES=SQRT(FX**2+FY**2)

    ang=ATAN(FY/FX)

    ang=180*ang/pi

    deg=INT(ang)

    MINUT=abs(ang-deg)*60

    MINUTS=INT(MINUT)

    sec=(MINUT-MINUTS)*60

    WRITE(7,*) n,",",res,",",deg,",",minuts,",",sec

  • 206

    END DO

    ! Close the external file

    CLOSE(7)

    PRINT *,

    PRINT *," THANK YOU FOR USING VECTRIX 2D"

    END Program Vectrix2d

    The output file contains the magnitude of vector resultant and its direction in degrees, minutes and seconds:

    N MAGNITUDE Degrees Minutes Seconds

    1 37.389270 56 45 23.579410

    2 37.423100 56 34 38.338620

    3 37.457300 56 23 54.265140

    4 37.491860 56 13 11.372680

    5 37.526780 56 2 29.674990

    6 37.562070 55 51 49.185790

    7 37.597710 55 41 9.891357

    8 37.633720 55 30 31.819150

    9 37.670090 55 19 54.969180

    10 37.706810 55 9 19.355160

    11 37.743900 54 58 44.977110

    12 37.781330 54 48 11.862490

    13 37.819130 54 37 39.997560

    14 37.857280 54 27 9.409790

    15 37.895780 54 16 40.099180

    16 37.934630 54 6 12.065730

    17 37.973830 53 55 45.323180

    18 38.013390 53 45 19.885250

    19 38.053290 53 34 55.738220

    20 38.093540 53 24 32.923280

    21 38.134140 53 14 11.412960

    22 38.175080 53 3 51.234740

  • 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

    Individual responses were obtained using the programming language FORTRAN 90, whose source code is:

    Program Equilibrium

    REAL :: pi=3.1415923

    REAL RX,RY

    REAL FX,FY

    INTEGER :: N

    PRINT *, "PROGRAM Equilibrium"

    PRINT *,"SERVES to Find 2 unknown forces P and Q acting on a body"

    PRINT *,"that is in equilibrium, BY THE METHOD OF RECTANGULAR

    COMPONENTS"

    PRINT *,

    PRINT *, "Programmed by Neftal Antnez H. (p)1986"

    PRINT *, "CEO of Antnez Software LTD"

    ! Create an external new data file

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

    ! Write Writing the header

    WRITE(7,*) "N, P, Q,"

    PRINT *,

    ! Angles in radians

    RX=0.0

    RY=0.0

    a=80;anga=30.60*pi/180

    b=90;angb=103.90*pi/180

    c=45;angc=149.60*pi/180

    e=70;ange=288.60*pi/180

    f=50;angf=307.50*pi/180

    RX=a*cos(anga)+b*cos(angb)+c*cos(angc)+e*cos(ange)+f*cos(angf)

    RY=a*sin(anga)+b*sin(angb)+c*sin(angc)+e*sin(ange)+f*sin(angf)

    DO N=1,55,1

    d=100;angd=(198.10+n)*pi/180

    angp=(N+30.60)*pi/180

    angq=(180+N)*pi/180

    a11=cos(angp);a12=cos(angq)

    a21=sin(angp);a22=sin(angq)

    delta=a11*a22-a21*a12

    FX=RX+d*cos(angd)

    FY=RY+d*sin(angd)

    p=(a12*FY-a22*FX)/delta

    q=(a21*FX-a11*FY)/delta

    WRITE(7,*) n,",",p,",",q

    END DO

    ! Close the external file

  • 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 Coulombs Law and Electric Field

    Charles Coulomb (17361806) 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 rthat is:

    From Coulombs 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 Coulombs 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:

    Individual responses were obtained using the programming language FORTRAN 90, whose source code is:

    Program Efield

    implicit none

    REAL :: pi=3.1415923

    REAL DPB,DPD REAL angZ,angT,ang

    REAL ea,eb,ec,ed

    REAL r,sfx,sfy

    REAL MINUT,sec

    INTEGER deg,MINUTS

    INTEGER :: n

    PRINT *, "PROGRAM EFIELD"

    PRINT *,"USED TO CALCULATE THE TOTAL ELECTRIC FIELD"

    PRINT *,

    PRINT *, "Programmed in Microsoft Powerstation by Neftal Antnez H.

    (p)1986"

    PRINT *, "CEO of Antnez Software LTD"

    PRINT *,

    ! Create an external new data file

    OPEN (7, FILE = Efield.TXT, ACCESS = APPEND,STATUS = REPLACE)

    ! Write Writing the header

    WRITE(7,*) " N TOTAL ELECTRIC FIELD DIRECTION"

    DO n=1,55,1

    DPB=sqrt((n+4.)**2+(n+15.)**2)

    DPD=sqrt(64.+(n+15.)**2)

    ! angles are in radians

  • 215

    angZ=atan((15.+n)/(n+4.))

    angZ=2*pi-angZ

    angT=atan((15.+n)/8.)

    angT=pi+angT

    ! The charges are multiplied by 9x10E09 and by 10000 due to conversion

    ! from centimeters to meters.

    ea=-450000000/(n+4)**2

    ec=2812500.

    eb=360000000/DPB**2

    ed=270000000/DPD**2

    sfx=eb*cos(angZ)+ed*cos(angT)+ea+ec

    sfy=eb*sin(angZ)+ed*sin(angT)

    r=sqrt(sfx**2+sfy**2)

    ang=atan(sfy/sfx)

    ang=180*ang/pi

    deg=INT(ang)

    MINUT=abs(ang-deg)*60

    MINUTS=INT(MINUT)

    sec=(MINUT-MINUTS)*60

    WRITE(7,*) n,",",r,",",deg,",",minuts,",",sec

    END DO

    PRINT *,

    ! Close the external file

    CLOSE(7)

    PRINT *," Finished"

    END Program Efield

    The output file contains the total electric field and its direction in degrees, minutes and seconds:

    N Total Electric Field Degrees Minutes Seconds

    1 15310950000000.00 7 25 14.88762

    2 9799622.00 10 12 26.75446

    3 6488275.00 13 41 19.13361

    4 4356343.00 18 16 36.75522

    5 2920362.00 24 45 12.36649

    6 1935090.00 34 39 26.41388

    7 1282441.00 50 54 6.17157

    8 929416.60 76 38 50.06287

    9 861157.90 -73 21 55.75012

    10 972141.60 -51 1 47.18536

    11 1140858.00 -37 31 10.66956

    12 1311528.00 -29 15 5.740356

  • 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 wagons 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 FORTRAN 90, whose source code is:

    Program Track

    REAL h,vb,vc,vd

    INTEGER :: N

    PRINT *,

    PRINT *, "Programmed by Neftal Antnez H. (p)1986"

    PRINT *, "CEO of Antnez Software LTD"

    ! Create an external new data file

    OPEN (7, FILE = Track.TXT, ACCESS = APPEND,STATUS = REPLACE)

    ! Write Writing the header

    WRITE(7,*) "N, h, Vb, Vc, Vd"

    PRINT *,

    DO N=1,55,1

    h=(0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/(500*9.8)

    vb=sqrt((0.5*500*(3+N)*(3+N)+500*9.8*(15+N))/250)

    vc=sqrt((0.5*500*vb*vb-500*9.8*(10+N))/250)

    vd=sqrt((0.5*500*vc*vc+500*9.8*(10+N))/250)

    WRITE(7,*) N,",",h,",",vb,",",vc,",",vd

    END DO

    ! Close the external file

    CLOSE(7)

    PRINT *,

    PRINT *," THANK YOU FOR USING Track"

    END Program Track

    The results table in Microsoft Excel is:

  • 219

  • 220

  • 221

    7.4 Kirchhoffs Rules

    The procedure for analyzing complex electric circuits is greatly simplified if we use two principles called Kirchhoffs 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

    Kirchhoffs 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 Kirchhoffs 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 Kirchhoffs junction rule to junction c gives:

    We now have one equation with three unknowns . There are three loops in the circuitabdca, 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 Kirchhoffs 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:

  • 224

  • 225