Topics in
Mathematics for
g Hi h f r Schoo o d l a ls W Volume 2
For Ages 14 to 18
Ron Jarman AWSNA Publications
The Association of Waldorf Schools of North America Publications Office 65-2 Fern Hill Road Ghent, NY 12075
Topics in Mathematics for Waldorf High Schools
Ron Jarman was born in Manchester, England. After obtaining an honors degree (MA) in mathematics and mechanical sciences. He became interested in philosophy and spiritual science from a book about Rudolf Steiner by Rom Landau called God Is My Adventure. At the end of World War II he joined the Steiner Waldorf School in nearby Ilkeston, Michael House School, where he taught mathematics, geography, and gymnastics to adolescents in the upper school, and later became a class teacher. Ron served on the Steiner Schools Fellowship Council and became chairman, a post he held for twenty-five years. He helped bring about the foundation of an International Waldorf Schools Council which meets twice a year in various European capitals and includes schools from all over the six continents of the world. He helped in the development of Emerson College and he became involved with teacher training, suceeding Francis Edmunds as leader of the Education Course. Following visits to Waldorf schools in Slovenia and Croatia he joined the women's movement "Through Heart to Peace," which is very active in former Yugoslavia, especially Bosnia, and became well known for breaking the siege of Sarajevo by Serb extremists. During the last three years (at present he is 88) Ron has given lectures throughout England on the theme of "Destiny, Reincarnation, and Resurrection." He wrote Teaching Mathematics in Rudolf Steiner Schools for Classes I–VIII as well as many articles and essays for magazines.
by Ron Jarman
2
Topics in
M athematic for
Waldorf High S chools Volume 2 For Ages 14 to 18 How to Become Inspired and Wholehearted by
Ron Jarman, M.A.
3
Acknowledgements The author would like to thank Robert Street, B.Sc. (Hons), M.Sc., C. Maths F.I.M.A. of Wynstones School, Gloucester, for checking answers to examples—but for any remaining errors the author is to blame. The author would also like to particularly thank Elisa Wannert for her cartoons, Lynda McGill for her painstaking preparation of the manuscript, and Radek Witek for checking the early chapters, David Mitchell and AWSNA Publications for the layout, redrawing of illustrations, proofreading and finally the publishing and distribution of this book.
Printed with support from the Waldorf Curriculum Fund Published by: AWSNA Publications The Association of Waldorf Schools of North America 65-2 Fern Hill Road Ghent, NY 12075
© 2010 by AWSNA Publications Title: Topics in Mathematics for Waldorf High Schools Author: Ron Jarman, M.A. Editor: David Mitchell Proofreader: Ann Erwin Proofreader/mathematics: Raden Witek Cover: David Mitchell ISBN 978-1-888365-86-3
4
Table of Contents Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
Author’s Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
Chapter 1 Class 9 Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . .
19
Chapter 2 Class 9 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
Chapter 3 Other Mathematical Topics Suitable for Class 9 or Class 10 . . . . . . . . . . . . . . . . . . . . .
63
Chapter 4 Class 10 Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . .
67
Chapter 5 Class 10 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . .
86
Chapter 6 Class 10 Projective Geometry . . . . . . . . . . . . . . . . . . .
99
Chapter 7 The Transition from the First Half to the Second Half of the High School: Logarithms, Spirals and Progressions . . . . . . 112
5
Chapter 8 Class 11 Differential Calculus . . . . . . . . . . . . . . . . . . . 118 Chapter 9 Class 11 Projective Geometry . . . . . . . . . . . . . . . . . . . 135 Chapter 10 Class 11 Imaginary and Complex Numbers . . . . . . . . 149 Chapter 11 Class 12 Integral Calculus . . . . . . . . . . . . . . . . . . . . . . 153 Chapter 12 Class 12 Three Dimensional Projective Geometry . . . 169 Chapter 13
Class 12 Path Curves and Their Applications to Plant Buds and Leaves and to Planetary Alignments . . . . . . . . . . . . . . . . . . 178
Chapter 14 Class 12 Computers, Calculators and Chaos Theory . . . . . . . . . . . . . . . . . . . . . . . . 201 Chapter 15 Concluding Chapter . . . . . . . . . . . . . . . . . . . 220 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
6
Dedication A Dedication to the Origins of Mathematics and Geometry and Some of Their Developers The basic human abilities of counting and measuring by means of our limbs lie at the root of our sciences of mathematics and geometry. That ten is our number base is the result of our having ten fingers. The movements of our arms and legs enable us to compare distances and the lengths of objects. 2500 years ago Pythagoras taught the basic subjects arising from our abilities as follows: arithmetic from number at rest, geometry from measure at rest, music from number in movement and astronomy from measure in movement. These became the quadrivium of the seven liberal arts, to which in medieval universities was added the trivium of “dialectic, rhetoric and grammatic.” Their development came about by following up the work of inspired Greek thinkers like Socrates and Aristotle.
Socrates † 399bc
Aristotle † 321bc
Going back to earlier times, the father of arithmetic was said to be Abraham, but the contributions of the early Chinese played an essential part, too. Geometry was an offshoot of agriculture in early Persian and Egyptian civilizations. Many peoples contributed to the development of music and astronomy. 7
The picture below dates from the early 16th century. It shows the activities attributed to the seven moving stars. It also shows the zodiacal constellations specially related to them, their male or female attributes and other qualities. At the time the Earth was regarded as the center of the Universe with the Sun and all the stars moving around it.
In the course of the 16th and 17th centuries, three scientists brought about revolutionary ways of viewing astronomical phenomena. First Copernicus declared that the Earth rotated around the Sun as center and not vice versa. Later he declared that not only Earth, but all planetary paths circled around the Sun. Later at the 17th century’s beginning, three laws concerning planetary movements were revealed by Kepler: i) The planets move on ellipses for which the Sun is a focus. ii) The areas swept out by a planet’s radius vector from the sun is constant for that planet in any time interval. iii) The cube of a planet’s mean distance from the Sun is equal to the square of the time of the planet’s revolution.
8
Copernicus † 1543
Tycho de Brahe † 1601
Kepler † 1630
Soon afterwards, while not accepting Copernicus’ dictum that the Earth revolved around the Sun, Tycho de Brahe agreed with Kepler’s general laws and proceeded to confirm them by taking thousands of angular measurements of planetary movements. Not only planets but the whole display of fixed stars in the firmament was accurately mapped by Tycho.
9
This casting aside of traditional dogmas in astronomy had its effects on mathematical and geometrical development as a whole. The significant change in interval values in the musical scale at the close of the 16th century due to Zarlino and later by J. S. Bach is not unconnected with these developments. The age of the whole of modern scientific consciousness began with the end of the 15th century. Vasco da Gama’s circumnavigation of the Earth had hammered the last nail in the coffin of the flat earth dogma. The discovery and application of calculus in the 17th century by Isaac Newton and Leibnitz followed. Geometers like Lobachewsky, Gauss, Jakob Steiner and Von Standt finally overcame the problems of the infinitude of space and Projective Geometry was born. So many new sciences came into being. Human physiology, geomorphology, electromagnetism and its applications, radio, sources of power (fossil fuels, tidal and wind power, nuclear power) are among them. How many of them are subject to solely materialistic interpretation has become a vital question in the 21st century. Vast sums of money continue to be spent on the physical exploration of extraterrestrial space. Questions focus upon whether other forms of life exist in our own solar system and possibly that of others. And are there conscious self-aware beings on some other planets with their own psyche or soul? A conception often held before our modern scientific age included the view that more important than the physically observed planets were the invisible spaces between them and the Sun. After a human death the physical body loses its life forces. Does it disappear into such invisible spaces? And what happens to the human psyche and individuality? Just as Tycho de Brahe spent many years of his life recording all the visible stars and constellations in the heavens, so in the 20th century a retired school mathematics teacher, Lawrence Edwards, examined thousands and thousands of buds of many plant species and discovered a special relationship between the form of the plant bud and its earth-moon-planet constellation. Details will be shown in the course of the present book. The relationship is not physical, nor is it due to some gravitational, electromagnetic or
10
other physical effect. There is now clear evidence that when such relationships are used in connection with human ailments and diseases centered i n va r i o u s b o d i l y org a ns, med ici nes ca n be produced which counter such health-destroying tendencies. These healing medicines, mainly h o m e o p at h i c , a re known as anthroposophical (trade names Lawrence Edwards measuring bud growth Weleda and Wala) and are prescribed by qualified doctors and used in anthroposophical hospitals. We may hope that further discoveries (not only in medicine) in the present scientific age will develop in the near future and gain general recognition, combating the materialistic view of human development and that of the universe in which it occurs.
11
12
Author’s Preface In the preface to Volume 1, emphasis was given to the constitution of every human being. Besides having a physical body we are aware that until we die this body is imbued with life. At death the lifeless corpse steadily decays and becomes food for worms, grubs, and so forth, and finally is reduced to dust. Cremation simply hastens the final goal. Another way of describing the situation is to say that during life the physical body perceived by human senses is protected from decay by an invisible life body (or to use alternative terms an etheric or ether body). Plants also have etheric bodies but stones do not. Animals and humans have in addition not only an inner ability to move but also something of a soul nature. The human soul is recognized by its three activities—thinking, feeling and willing—assisted but not caused by the brain. What humans also possess but animals do not are unique individualities which captain the three soul activities. When a person uses the pronoun “I,” he or she can be referring only to himself or herself and to no one else. This fourth part of a human governs the three lower parts and is responsible for everything created or destroyed in the world by him or her. Animals do not have such responsibility. Their actions are instinctive and no moral blame can be laid at their feet as it can for humans, e.g., regarding crime, earth pollution, selfishness and so on. At this point I want to alter what I said in the Preface to Volume 1. Rather than using both masculine and feminine forms on each occasion when referring to human beings (as I have done in the paragraph above), I shall use “he” or possibly “one” unless it is obviously a “she” who is being referred to. For the human individuality is itself neither masculine nor feminine. Writing “he or she” is unnecessarily verbose and “(s)he” looks peculiarly algebraic.
13
In the first seven years of his life, the child gradually transforms the physical body inherited from his parents into his own physical body which will develop thereafter. Such things as formal education, television and videos will harm him in his first seven years, which culminate with the change of teeth, when the inherited milk teeth are pushed out and his own (second) teeth emerge from his gums. Until then the most beneficial things for him are loving guidance, play, and the exercise of imitation and listening to stories told (not read) to him by parents. From the age of 7 to that of 14, the child develops his own life body, leaving aside the one bestowed by his parents (especially his mother). This in turn enhances his physical strength, his memory and his ability to learn the three Rs. Beneficial in this age range is loving but definite authority by parents and teachers as he learns how to behave socially. At 14 he becomes aware that he has a soul with his own thoughts, feelings and actions, often differing from those of his parents. Authority has to be steadily replaced by reason, the moments for such change requiring a fine-tuning judgment by both teachers and parents. Conscious too of growing sexual maturity, for which Germans use the apt equivalent of the adjective “earth-ripe,” girls developing more quickly than boys, teenagers sense the first beginning of adulthood and half conscious questions “Who am I? In which direction is my future life going to be guided?” Especially helpful for them are good biographies and autobiographies for reading matter. My own enquiry as a teenager was particularly helped when our English literature teacher had our class reading and studying John Buchan’s The Path of the King. Succeeding chapters traced the destinies of the descendents of a Norse king through the ages ending up with Abraham Lincoln, each one wearing the same gold ring whose size gradually decreased. Although unintended by the author, I felt the gold ring epitomized the destiny of a single individual. Sixteen-year-olds want to be treated as adults, but they know they are not adults and know that adults know they know this. It is as important for the young person as play is for the young child. Chapter 11 of Volume 1 began with a description of work in mathematics teaching in the higher classes (9, 10, 11 and 12) and
14
the motives underlying it. It seems appropriate here to reproduce these pages with a few modifications and additions before launching into practical detail in the chapters of Volume 2. In high school work it is the activity of thinking which becomes the soul captain’s principal lieutenant in the third phase of growing up. “Thinking,” said Rudolf Steiner, “is the all-one being which permeates everything.” There are four stages in the development and strengthening of the activity of thinking and they correspond to the path running through classes (or grades) 9, 10, 11 and 12. Nevertheless all three parts of soul activity develop strongly during this later phase from age fourteen upwards. As the teenager’s own soul penetrates his bodily activities, he has to experience and overcome an inner soul chaos as he learns to look in a new way at everything in the world and in himself. In mathematics as well as in other subjects, the ninth grader wants to experience everything afresh. He welcomes the change from having a class teacher to being taught each subject by someone who has worked to become an expert in it, assisted by the rigors of a university degree and life experience. Teachers in their twenties and thirties are often the best people to teach young people of high school age. The ninth grader wants to be shown anew how you do addition sums and long divisions for example, but now in a fully conscious way, grasping the reasons for each whim of written layout. No longer is the underlying aim to receive the teacher’s tick at the end of one’s calculations. The pupil wants to be able to know himself whether or not his answers are correct. The authority of the individual teacher is replaced by experience of and guidance by the truth itself. Despite the obvious lies, halftruths, deceptions and hypocrisies in the world and in the young person’s immediate environment, he wants above all to find those realms where he can assert and experience that the world is true. As a young child he needed assurance that the world is fundamentally good. Between roughly 7 and 14 he needed confirmation that the world is beautiful. Already basically strengthened in will and feeling respectively, he now needs this third guarantee to help him find the earthly home for his own personality. In Class 9 the emphasis a teacher can lay is on contrasts. Awareness of contrasts leads to the consciousness in which
15
thinking can grow. In art lessons the best medium is black (charcoal sticks or paint brushes) on white paper. The physics of electromagnetism will contrast the power unit of an electric train with that of the telephone earpiece. The principle is the same in both, but the contrast in current is enormous. The first could be described as allopathic electricity and the second as homeopathic electricity. Contrasts in other subjects are between drama and comedy, plate tectonics and volcanoes, Olympic gymnastics and country dancing. At the same time there are strong common elements. In mathematics there is the contrast between melody and harmony in music. In geometry the conic sections ellipse and hyperbola are contrasted, and comparison can be made between these and the respective form and function relationship between head and limbs. In addition there is the intermediate chest and rhythmic system, represented by the parabola, so that the metamorphosis in both groups can be made apparent. In Class 10, pairs of contrasts and their interactions carries the thinking awareness a stage further. Ancient history in this class culminates in a careful study of Greek civilization and the Socratic dialogues. The Pythagorean-Platonic description of the four elements in terms of the two contrasts moist-dry and warmcold is part of this, and much of the mathematics developed from this can be used. Geometric constructions of the four basic spirals (Archimedean, Equiangular, Reciprocal and Asymptotic Circle) illustrate the 4 progressions of numbers arising therefrom, and how they indicate a progression from material to noetic aspects of the world. It is independent judgment that evolves in Class 10. Perhaps the prime example of this is in trigonometry. To be able to stand far away from a mountain, take three measurements and then be able to calculate the mountain’s height without becoming involved in climbing it, exemplifies the particular ingredient of human intelligence required to make an objective judgment unfettered by personal feelings of involvement—which would be difficult for Class Niners to achieve. By Class 11 the students are ready to plunge into greatest detail in everything they study, whether it be the history of music (one of the main lessons they have), the story of Parzival and its
16
significance, atomic theory, the theory of organic cellular structure, or computer technology. It is the year when analysis is the key word. So in mathematics differential calculus is introduced—the same kind of procedure towards smaller and finally infinitesimal parts is required to do this. Also in geometry the fundamental theorem of Projective Geometry is proved and made use of. At the same time the principle of duality shows how the importance of point-like entities must be complemented by the equally valid and important planar entities which structure and shape the plant and organic world as a whole. The intellect can become finely honed at this age as well as developing an awareness of the noetic dimensions of life. The final school year (Class 12) gives the students an opportunity to gather together the many results obtained through analysis and experience, and discover the higher unities as true progenitors. Intellect has prepared the ground for reason. Analysis gives way to synthesis. The many points of view which scientific and artistic work have revealed help to form the indispensable modesty and wonder which every student requires in beginning to build his own view of the nature both of the world in which he lives and of what being human really means. He becomes aware that he possesses a unique individuality and that it will suffer rust and tarnishing if he allows it to become infused with mere egotism. He will seek a higher synthesis of himself with the best that human society endeavors to achieve. So in physics Class 12 concentrates on understanding the whole world of color. The diverging scientific descriptions of Newton, Goethe, Ostwald and others have to be evaluated and their contributions to the whole subject synthesized, yet not fully determined—only adult life will provide the experiences needed to do that. Renaissance and modern painting, which together with architecture figure in another main lesson, will further succour such endeavors. Class 12 mathematics is again concerned with synthesis. Differential calculus gives rise to integral calculus. Human reason can now grasp chaos theory and fractals. Earthly geometry (what used to be called Euclidean geometry) is seen to be but a limited part of a Projective Euclidean (or parabolic) geometry, which is only one of a whole host of geometries, e.g., hyperbolic, elliptic
17
and polar Euclidean geometries, all of which have their own forms of measurement, be it of distance or angle. Practice in working with money in three different ways back in Class 6 now grows into a deep study of economics; and the realization dawns in Class 12 that the problems of world economy will never be solved until a far more holistic approach to them is made than the world has so far managed to evolve. It is important in older classes just as in younger classes that every student gain a clear impression of what the topics introduced contribute to the whole curriculum of school subjects as well as to the whole mathematics curriculum. The difference in difficulty of problems set for more able and less able students will be become much greater by Class 12. Nonetheless the social value of keeping all students of the same chronological age in the same main lesson block is far reaching. The whole class will experience together the basic qualities of the various topics and their relevance to a wide range of human activities in life—all this in perhaps the initial half hour each day. Then the individual work of students and group work of some of them will follow different directions, some manual, practical and artistic, others of an advanced mathematical, analytical and also synthesizing character.
18
Chapter 1 Class Nine Mathematics (Age 14 to 15) The time allowed for the teaching of mathematics (including geometry) in this school year for the ninth class (or grade) should be 8 or 9 weeks in main lesson blocks comprising the first two hours of each weekday, together with 2 or 3 short practice periods of 45 to 50 minutes each week. It is wise to spend the first seven weeks or so using the practice periods to revise everything the girls and boys have learned (or hope that the class teacher has taught them) in earlier years. For one thing the style of the teacher in approaching and solving problems and the way he writes them out on the blackboard need to be familiar to the pupils, just as the teacher needs to know how the latter have tackled corresponding tasks. He can also help fill in any gaps in their knowledge and methods and show quicker and more elegant ways of obtaining solutions befitting their entry into the upper (or high) school, where clear thinking becomes a primary aim. It is good in these practice periods for the teacher to demonstrate how he works out on the board the answers to a few problems with the help of the class’s suggestions, then give out a duplicated set for everyone to attempt. Textbooks showing such sets are much less valuable than what the teacher has prepared, since the latter will wish to include questions in subsequent sets which enable pupils who found difficulty in some questions in earlier sets to have another go at similar questions. Such a set might be: 1. Add up 32,749, 655, 1982, and 99. 2. Write out the 17 times table from one times to twelve times. 3. Find the difference between 8 x 373 and 2072 ÷ 7 19
4. How many bicycles each of length 3 yd 1 ft could be placed front tire to back tire to stretch for a mile? 5. Find the simple interest on $1,500 invested at 5.6% per annum for nine months. 6. One angle of a triangle is four times bigger than a second angle and the third angle is 9 degrees less than the first angle. How big is the first angle? 7. Two sides of a rectangular field are 65 yd and 72 yd. How far is it to walk directly from one corner of the field to its opposite corner? 8. Solve the equation 9(3x + 2) – 7(2x – 5) = 183. 9. Find the volume of a cylinder 1ft 9in long with a radius of 8in. (Take π = 22/7 and give the answer in cubic inches.) 10. Every edge of both a regular octahedron and a regular icosahedron is 6 cm long. One face of each solid is glued together exactly. How many faces, edges and vertices of the combined solid will be seen from outside? What numerical law do your three answers obey? 11. Transform the formula d = b(x + 4)/3 so that x instead of d becomes its subject. What is the value of x if: (i) d = 2, b = 5? (ii) d = 8, b = 6? (iii) d = 2, b = 0? (iv) d = b?
12. (12 5/6 – 5 1/4) ÷ 7 2/9. Give a decimal answer.
Numerical answers (in wrong order written at the end of the set of questions): -2.8, -1, 0, 1.05, 10, 12, 17, 26, 34, 36, 51, 63, 68, 84, 85, 97, 102, 119, 136, 153, 170, 187, 204, 1320, 4224, 35,485, ∞
20
At the end of the lesson, after helping individual members of the class about how to go about solving some of the problems, the teacher can ask everyone to complete as many questions as possible in half an hour’s homework, collecting their written work (clearly set out) next morning. With other sets of questions the teacher may call out the answers at the end of the lesson and ask the pupils to mark their own work and then go over some of the problems next time. The first main lesson period This may occur in the second half of the first term in Class 9. Remembering the wish of the girls and boys to look forward to the possibilities ahead in their lives, one can begin with the subject of “Permutations and Combinations.” This begins with simple numerical calculations but then leads deep into the heart of algebra. Ask the class to imagine they are going for a holiday starting their journey in London and going by train via Paris. They might go via the Channel Tunnel (alternative A) or cross the sea via Dover or Newhaven (alternatives B and C). From Paris there are 4 train routes through the Alps (P, Q , R and S, say) and going on to Rome. Then to reach their final destination in southern Italy, there are just two alternative train routes (X and Y). Now let each pupil work out how many complete routes there are from London by listing all possibilities. So with care they will make the following lists: BPX CPX APX APY BPY CPY ______________________ AQX BQX CQX AQY BQY CQY ______________________ ARX BRX CRX ARY BRY CRY ______________________ ASX ASY
BSX BSY
CSX CSY
There are 24 different possibilities and no more. 21
With careful observation they will notice that there are 3 columns of results, 4 double rows containing each of the letters P, Q , R and S and alternate single rows contain X and Y. So the total of 24 = 3 x 4 x 2. To obtain the total number of possibilities, it is always necessary to multiply together the number of possibilities for each part. A simple map of Europe could be drawn by the pupils to illustrate these journeys. A different set of routes might be given between modal points a, b, c, d, e and f as follows (or pupils might prefer to design their own sets), e.g. a–
{ } { } { } { } { } 7 routes
–b–
2 routes
–c–
routes
–d –
3 routes
6 – e – routes
–f
with the following questions: How many ways are there from a to c? c to f ? a to f ? (Ans. 1260) d to e and back to d? c to f and back to c? a to c and back to a? a to f and back to a? The last answer is well over a million ways: 1260 x 1260 = 1,587,600 The “multiplication principle” should now be well established and the class will be ready to deal with what used to be called variations. Suppose that a man owns 4 pairs of comfortable and clean shoes and decides he will wear each pair for a whole week in course of a 4-week period. In how many ways can he arrange this period? Call the shoe pairs p, q, r and s. Ask the pupils to write down all possible arrangements, missing out no arrangement and not duplicating one.
22
Thus: pqrs, pqsr, prqs, prsq, psqr, psrq qprs, qpsr, qrps, qrsp, qspr, qsrp rpqs, rpsq, rqps, rqsp, rspq, rsqp spqr, sprq, sqpr, sqrp, srpq, srqp Why are there just 24 arrangements? For the first week there are just 4 possibilities. For the second week there are just 3 possibilities (for one pair has now been used). For the third week there are just 2 possibilities. For the fourth week there is just 1 possibility. Once again multiply the answers 4 x 3 x 2 x 1 = 24.
¬
If you then ask the girls a similar problem—if you have 7 nice dresses and decide to wear just one for every day of a week with no repeats, how many ways of dressing will there be? It will soon dawn that there must be 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 ways. Instead of writing down all these decreasing numbers, the notations of 7 or 7! can be used, the latter being quite dramatic. It is called factorial seven.
13! = 6,227,020,800, already above six billion. 20! has 19 figures in the answer and 50! has 66 figures in it.
We can now pass on to what are called permutations. Suppose a young lady owns a set of jumpers, each one having a different rainbow color. She likes to choose 3 of them to wear on weekends (Friday, Saturday and Sunday, different each day). How many choices has she for each weekend? This is a case of permuting 7 things, 3 at a time and is denoted by 7P3. On Fridays she has to choose from 7 jumpers, on Saturdays from 6 and on Sundays from 5 of them. So 7P3 = 7 x 6 x 5 = 210 choices. Notice that 7P3 is also 7! ÷ 4! Many similar examples can be imagined, e.g., how many numbers are there less than 1000 containing no repeating digits?
23
Not allowed are 999, 434, 001, etc. 36 numbers begin with 0, 28 begin with 1, etc. Answer = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120 Finally we pass on to combinations (not the old-fashioned things which old ladies used to wear). Suppose the young lady above simply chose 3 jumpers for the weekend, not caring about which one she used for Friday or succeeding days. Given the colors red, yellow and blue, that would not constitute a different choice from blue, red and yellow. But the number of arrangements of these colors is 3! So we have to reduce the previous answer by dividing by 3! The number of combinations is thus 7! or 7 x 6 x 5 = 35 and is denoted by C . 7 3 4! x 3! 3x2x1 An alternative symbol is (7C 3). After a few more examples the teacher can set an hour’s written classwork as follows: Calculate: 1. 2. 3. 4. 5. 6. 7. 8.
24
5! 10! P 6 4 C 10 5 C 40 2 How many variations can be obtained by taking five letters in different orders from the set of letters ENOST? How many permutations of four letters, three letters, two letters and one letter can be obtained respectively from ENOST? (4 answers possible) What is the total of the answers to questions 6 and 7?
9. Write down carefully all the possible arrangements of letters used in the last total. Then underline every arrangement which gives an English word (you may use slang words like SNOT). Also underline in a different color any foreign words like NE. Answers not given below. 10. About what proportion of the grand total are actual words? 11. Suppose we had started with all the letters of the alphabet except the vowels A E I O U. How many consonantal variations would we have then? (Do not try to work out the answer; just give the number of digits in the answer.) 12. How many ways are there of choosing a set of half the chocolates when you are offered a box containing a dozen different kinds? 13. In how many ways could you eat those chocolates you had chosen? 14. If you had started from the whole box and chose them one at a time until you had consumed half a dozen, what answer would you get then? 15. If 12 points were arranged in a circle, how many triangles in total could be colored in using only lines joining these points? Draw a circle and all the lines of the triangles. Color 2 overlapping triangles. 16. What is the maximum number of points in which a set of 9 straight lines (as long as you like) could intersect? 17. Suppose you had 13 circles instead of 9 lines. What would the answer be then? 18. A “rowing eight” consists of 8 oarsmen (or oarswomen) and a coxswain. If they form a club and appoint a chairman, a secretary and a treasurer from among themselves, in how many ways could they do so? 19. A grand piano requires 4 people to move it with care. If 14 people volunteer, how many choices from among them can be made?
25
20. Figure (1) shows a network of wires in an early computer in which electricity can flow only downwards. How many circuits from top to bottom are there? Answers (wrong order): 5, 20, 36, 60, 84, 120, 120, 156, 220, 252, 325, 360, 720, 780, 924, 1260, 1001 or 20,349, 665,280, 3,628,800. Fig. 1 Further practice with permutations and combinations occurs in relation to music. In every melody the order of the notes is important, whereas in playing chords the notes sound together; harmony or dissonance arises. In choosing the notes in a musical composition, therefore, melodies have to do with permutations and harmonies have to do with combinations.
Let us consider the notes C D E F G a b c in the scale of C major. Two well known melodies are (i) G F D E C D B C and (ii) c b a G F E D C. How many 8-note sequences with no repeated notes are possible? P = 8! = 40320 8 8 How many 4-note chords can be chosen from the 8 notes of the octave, e.g.: C E G c or D F A b, some harmonious and some dissonant? C =8x7x6x5 8 4 = 70 4x3x2x1 Suppose, however, that repeated notes are allowed. How many 8-note melodies containing C twice, D three times, E twice and G once are possible, e.g., C D E D C D E G? Were the 3 notes D to be replaced by F, a and b, the answer to our question would be 3! times greater.
26
So the actual answer is
8! = 1680. 2! x 3! x 2!
If we include the 5 black notes as well as the 8 white notes of the piano in the whole octave, the number of 3-note chords possible = 13C3 = 286, whereas the number of 6-note melodies with no repeated notes = 13P6 = 1,235,528, but if in each of these melodies one note were allowed to appear 3 times, we would also need to divide by 3! (answer 205,920). A set of examples for the members of the class to work out is as follows. 1. How many 3-note chords on a piano can be chosen using the black notes within an octave? 2. How many 6-note melodies without repeated notes can be composed using the black notes within two octaves? 3. Two of the 8 white notes are chosen to make a simple melody in which each note is to be used 4 times. How many melodies are possible? 4. Four different 4-note chords have to be chosen from the lower set of notes A B C D E F G followed by 4 different 4-note chords from the higher set of notes a b c d e f g. How many arrangements are possible altogether? 5. How many chords can be plucked from the open strings of a guitar if just 3 strings are chosen each time? 6. Using 4/4 common time the same rhythm is repeated for several bars. It consists of a minim, a crotchet and two quavers. How many rhythms could this give? 7. If in the last question the notes do mi so (top do) had to appear in each bar, what total range of possibilities would this now amount to? After checking and making a written comment upon each individual student’s written work and handing back their efforts next school day, the time has come to apply all the numerical work 27
to algebra. Remind the class of what they learned in Class 8 about dissolving and simplifying algebraic expressions, e.g.:
3a(2a – 5b) + b(4a + 7b) – 6(a2 – 2ab + b2) + 8b2 6a2 – 15ab + 4ab + 7b2 – 6a2 + 12ab – 6b2 + 8b2 ab + 9b2
= =
Then take a case where one bracketful has to be multiplied by another one, e.g.: = = =
(4x – 3y)(8x + 5y) 4x(8x + 5y) – 3y(8x + 5y) 32x2 + 20xy – 24xy – 15y2 32x2 – 4xy – 15y2
Now take the case where the two brackets are the same. = = = =
(a + b)2 or write it out as below (a + b)(a + b) a(a + b) + b(a + b) a2 + ab +ab +b2 a2 + 2ab + b2
(a + b)
like
(a + b) ab + b2 a2 + ab a2 + 2ab + b2 23 x 23 69 46 529
After the pupils have copied this, ask them to multiply the answer by (a+b) again.
28
So
a2 +2ab +b2 , for this is (a + b)3 = (a + b)2(a + b) a+b 3 2 a + 2a b + ab2 a2b + 2ab2 + b3 Carry on several more a3 + 3a2b + 3ab2 + b3 steps multiplying by a+b 4 3 2 2 3 (a + b) each time. a + 3a b + 3a b + ab 3 2 2 3 4 a b + 3a b + 3ab + b a4 + 4a3b + 6a2b2 + 4ab3 + b4 = (a + b)4, etc.
A bit later one obtains a7 + 7a6b + 21a5b2 + 35 a4b3 + 35a3b4 + 21a2b5 + 7ab6 + b7 = (a+b)7 Then you can help them to generalize (i.e., make more algebraic still) to obtain (a + b)n = an + na n-1b + n(n – 1) an-2 b2 + 2 … + bn,
n(n – 1)(n – 2) n-3 3 a b + 3!
where the rth term is n(n – 1) … (n – r) n-r+1 r-1 a b (r – 1)! This is called the binomial theorem. For small values of n the whole numbers preceding the letters (including the 1 of the first and last terms) can be arranged in a triangle, called Pascal’s Triangle, which the students can construct [Figure (2)]. These numbers are all the results of working out successive combination formulae which when arranged in a similar triangle beginning: C 1 1 C C 2 1 2 2 C C C 3 1 3 2 3 3 C C etc. 4 1 4 2 and evaluated give exactly the same array of numbers.
29
The pupils should be asked to construct first the large equilateral triangle of side 10 cm, marking centimeter points on these sides and joining up to form the other lines. Then to write in the numbers, noticing that every pair of horizontal numbers adds up to the number in the space below it. More ambitious pupils could be encouraged to take a base of 16 cm and so find they have extended downwards the triangle drawn by the others—provided they can write small enough numbers in the lowest rows in the center, e.g., 16C8 = 12,870. The whole form reminds one of an Egyptian pyramid. Just as the latter can remind one of the hidden secrets of life taught in olden times (the history of Ancient Egypt forms part of the history main lesson in Class 9), so many secrets of numbers can be found in Pascal’s Triangle. Notice the set of number “one” down the left and right edges, then in the next inner layers the ordinary counting numbers. Then taking the next layer the set 1 3 6 10 15 21 28 36…These are called triangular numbers, some of whose secrets were described in Volume 1.
Fig. 2
The number10 was described by Pythagoras, who was very familiar with Egyptian wisdom, as the creative power of 4. The number 28 is the creative power of 7, for if you begin to acquire a new skill (as a dressmaker or a cabinet maker, for example) after one week’s work you will learn something about it, but only after a month will you begin to be creative in it. This is also the reason 30
why in Waldorf schools we teach a main lesson over 4 weeks. 28 days is also the period of ovulation in the female. While if men would observe when they are most imaginative, they would notice that this too has a monthly rhythm, i.e., not a physical bodily rhythm as in women, but a soul rhythm. In contrast many women are imaginative at all times, of course, just as many men need to control regular sexual urges. A further example of triangular numbers can be discerned in Hermes, the scribe of the gods. He wrote down the whole wisdom of the cosmos and needed 78 books to do so. Why? He used the zodiacal power of 12, for 12 x 13 = 78 just as 4 x 5 and 7 x 8 = 28. = 10 2 2 2 A good homework would be to ask the pupils to investigate other secrets of Pascal’s Triangle, such as deeper layers of the numbers, lines of numbers perpendicular to the edges and their sum total, e.g., 3 + 4 + 1, 1 + 6 + 5 + 1. Also to find the number bases giving rise to significant numbers in the Bible, e.g., 153 towards the end of St John’s gospel, 276 in the Acts of the Apostles to do with Paul’s shipwreck, 666 in Revelation. Also if you take symmetrically placed numbers near the edges and look at the numbers contained in a downward pointing triangle of number, e.g.:
5
10
10
15 20
35
5 15
35 70
sometimes they are all divisible by the outside numbers (in this case 5), sometimes not. What is special about numbers like 5 when the divisibility works? They could use different colors in those downward pointing triangles (as far as space allowed) where the property is true.
31
Having considered many examples of the mathematics of possibility, we can also investigate simple examples of the mathematics of probability. Ask sixteen of the class members to each toss a coin 4 times and record the number of times they get “heads” and “tails.” Asking for hands up, make a list on the blackboard of how many students recorded no heads, one head, etc., up to 4 heads. The result may be 1, 3, 7, 5, 0. Repeat the experiment with 16 other pupils. 2, 2, 6, 5, 1 may be the result. Now comes the question, “If we did this hundreds of times, what would be the probable number of times of there being equal numbers of heads and tails in the average result?” The algebraic expression for the tossing process is (h + t)4. The binomial expression is h4 + 4h3t + 6h2t2 + 4ht3 + t4. This shows exactly the average results of tossing, for in tossing the coin 4 times the result means choosing h or t from each of the brackets (h + t)(h + t)(h + t)(h + t) and the total number of ways of doing so is C0 + 4C1 + 4C2 + 4C3 + 4C4
4
= 1 + 4 + 6 + 4 + 1 = 16 = 24.
The probability = possible way = 6 = 3/8 or 0.375. 16 total ways So if a gambler offers you a 50/50, i.e., even, chance of tossing 2 heads and 2 tails when tossing a coin 4 times, you will lose in the long run. Similarly if you ask a larger class to toss a coin 5 times, the probability of tossing 3 heads and 2 tails = C3 = 10 = 0.3125, i.e., less than 1 in 3. 25 32
5
If the teacher of a Class 9 were to ask the 5 tallest members of the class to stand up, assuming the class consisted of 12 boys and 12 girls, what is the probability of all 5 being girls? The number
32
of ways of choosing any 5 young people = 24C5. The number of ways of choosing any 5 girls = 12C5. Is the answer: C5
12
C 24 5
=
12! x 5! x 19! 5! x 7! x 24!
=
12 x 11 x 10 x 9 x 8 ÷ 5! 5! x 24 7 x 23 x 22 x 21 x 20
=
3 = .019 approx? 161
No, of course not, because girls are generally taller than boys at the age of 15. The probability is likely to be nearly 1, i.e., all 5 will be girls or mostly girls. To say that an event has a probability of 1 means that the event is certain. Here is a case where mathematics cannot give a true answer when it is unable to take account of the full living conditions of a problem. We are reminded of the problems of measurement in astronomy. Distances of over 5,000,000,000 light years are estimated to show how far from us are distant galaxies of stars and only when you multiply this enormous number by 186,000 times the number of seconds in a year do you derive the actual distance in miles these galaxies are from us. Also the age of the Earth is calculated to be over 4600 million years. But these huge figures have to be based on very questionable assumptions which take no account of the development of living beings (plants, animals and humans) on our Earth nor upon changes in the nature of lifeless substance in ages past. While mathematics itself exists in a realm of pure truth, the applications of mathematics to physical and other phenomena depend for their veracity upon knowing the full actual conditions under which such phenomena exist and were changed in the past. The next step in algebraic education is the solution of quadratic equations. Hitherto the students have learned how to solve and use linear equations to solve practical problems, e.g., at exactly what time does the minute hand cross the hour hand between 4 o’clock and 5 o’clock? Let this exact time be x minutes past 4. From 4 o’clock the minute hand will have moved through x minutes around the circle
33
of numbers. But the hour hand moves 12 times more slowly, so will have gone only x/12 minutes around since it pointed at 4, which lies 20 minute divisions below 12 at the top. Overtaking means that x x = 20 + 12 So 12x = 240 + x 11x = 240 x = 21 9 . 11 The answer is therefore 21 9 minutes past 4 o’clock. 11 It is best for children up to 14 years old to be set problems for which there is a quite definite single answer like the one above, and not only in mathematics. They expect the authority of adults to supply clear single answers to life’s riddles. With the advent of puberty, however, they become aware that many questions have more than one answer, and they have to begin deciding which answer fits their own situation. Here lies the value of quadratic equations which always have 2 answers. Later they will meet cubic equations (3 answers) and even problems which have an endless set of answers. To a girl the matter of having boy friends poses a question to which her parents, her peers and her teachers may well supply different answers or suggestions. Sink and Toilet
R o a d
Road 34
Fig. 3
It is proposed to erect a shop ABCD within a plot of land PBQR adjoining a crossroads. The plot measures 44 feet by 36 feet. A space is reserved for a toilet and wash bowl as shown. Space for pavements on both roads must be left free. It is wished that two thirds of the plot be used for the shop space. How wide can the 2 pavements of equal width be? NOTE: The widths x are not shown to scale for the solution of the problem, but the given measurements are correct on a scale of 1 cm = 5 ft. (44 – x)(36 – x) = 7 x 12 + 2/3 x 44 x 36, Since 1584 – 80x + x2 = 1056 + 84 x2 – 80x + 444 = 0 We now have to guess two numbers which add up to 80 but multiply to make 444. They are 6 and 74. So
(x – 6)(x – 74) = 0 and x = 6 or 74.
Since 74 is greater than 44, the second answer does not fit the conditions of the problem. Hence the real width x = 6 feet. The above guessing procedure needs practice so it is best to look at simple multiplication of binomial brackets first. Thus (x – 3)(x – 4) = x2 – 7x + 12, combining -3 and -4 to make -7. (x + 3)(x + 5) = x2 + 8x +15, and (x – 3)(x + 5) = x2 +2x – 15, combining -3 and +5 to make +2. (x + 2)(x – 6) = x2 – 4x – 12. Set the class examples like the following: 1. (x – 6)(x – 13) 2. (a + 7)(a + 8) 3. (b – 4)(b – 5) 4. (c + 10)(c – 11)
35
5. (d – 8)(d + 8) 6. (e – 20)( e + 19) 7. (2f + 7)(3f + 5) 8. (4g – 1)(9g + 16) 9. (h + 3)(6h – 17) 10. (8h – 5)2 Then it becomes clear which signs in the answer must be used.
(x + )(x + ) = x2 +… + …
(x – )(x – ) = x2 – … + …
(x + )(x – ) = x2 ± …– …, the first + or – depending on whether the + number in the bracket exceeds the – number or not.
(x – )(x + ) =
We can now set the class many purely algebraic quadratic equations and go on to practical problems using them. Solve: 1. x2 – 9x + 14 = 0 2. x2 – 9x – 22 = 0 Reduce the following to the usual form first, then solve: 3. 5x2 – 35x + 60 = 0 4. x2 = 2x + 35 5. 9 = 6x – x2 6. 7x2 – 13x – 1 = 5x2 + 9x – 21 7. 7x2 + 140x – 6 = 4x – 6 – x2 8. 2(2x2 + 3) – 3(15 – 5x) – 14x = 0 9. 2x2 + 13 = -10 – x + x(x – 7) 6 3 4 10. x(x – 3) – 1/4 = x – 4
36
11. A right-angled triangle has a hypotenuse of 17 cm. Of the other two sides, one is 7 cm longer than the other. Find the length of the shortest side. 12. The diagonal of a rectangular field is a yard longer than the field’s longer side and 18 yards longer than the field’s shorter side. How long is the diagonal? 13. A river flows steadily at 5 mph. One morning I rowed vigorously upstream for 1 mile, then immediately returned, rowing downstream to my starting point. The whole return journey took 9 minutes. What is my rowing speed in still water? 14. Remembering electricity lessons in Class 7, the effect of having 2 resistances R and S in series is equivalent to having one resistance of R+S, but if they are in parallel, the equivalent resistance is RS . R+S If R = S = 5 ohms, what are the two equivalent resistances? If then R is increased by the number of ohms that S is decreased by and the equivalent resistance in parallel is now 1.6 ohms, what were the altered resistances? 15. If a stone is thrown upwards at 44 ft/sec on the edge of a cliff, how long will it take for the stone to rise and then fall to a point 80 feet below the cliff edge? Use the formula d = ut +¹/² at2, where d = distance in feet above the cliff edge, t = time taken in seconds, and a = acceleration of -32 ft/sec/sec due to gravity, and u = initial velocity. The answers are from -22, -17, -5, -3.25, -3, -2, 0, 1, 1.5, 2, 2, 2.5, 3, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 15, 25.
37
Returning to the Binomial Theorem exhibited by Pascal’s Triangle we have: (a + b)n = an + nC1 an-1b + nC2 an-2b2 + … This can save much time when faced with calculations when a is a lot greater than b. For example: Suppose a = 6, b = 0.1 and n = 2 Then 6.12 = 62 + 2 x 6 x 0.1 + 0.12 = 36 + 1.2 + 0.01 The last term is very small and the answer is 37.2 to 3 s.f. Similarly 5.92 = (6 – 0.1)2 = 62 – 2 x 6 x 0.1 + 0.12 = 36 – 1.2 + 0.01
= 34.8 to 3 s.f.
≈ 34.8
Again
12.33 = (12+0.3)3 = 123 + 3 x 122 x 0.3 + 3 x 12 x 0.32 + 0.33
≈ 1728 + 129.6 + 3.24 To 3 s.f. the answer is 1860. We have in this way found a way of obtaining approximations to complicated powers of numbers. The class can then work at getting answers to the following to 3 s.f. 1. 9.12 2. 25.32 3. 25.42 4. 10.23 5. 12.52 (two terms are enough for all the above) 6. 3.053 (to 4 s.f. using three terms) 7. 10.24 (to 5 s.f. using three terms) Answers from 28.37, 82.8, 156, 640, 645, 1060, 10,824
38
Graphical solutions to all equations including quadratic equations are always possible. Example (i)
Solve 7x2 – 17x – 9 = 0
Fig. 4a
x -2 -1.5 -1 -5
7x 28 15.7 7 1.7
-17x 34 25.5 17 8.5
0
5
0 1.7
1
1.5
7 15.7
2 22.5
3
28 43.7 63
0 -8.5 -17 -25.5 -34 -42.5 -51
7x2 – 17x – 9 53 32.2 15 12 -9 -15.8 -19 -18.8 -15 -7.8 3 The parabola meets y = 0 where x = -0.44 or 2.87
Fig. 4b
39
Example (ii) Solve 4x 1 + x3 = 1 + x2 x x3 1 + x3 4x x2 1 + x2 4x/1+x2
-2 -8 -7 -8 4 5 -1.6
-1.5 -3.37 -2.37 -6 2.25 3.25 -1.85
-1 -0.5 0 -1 -0.12 0 0 0.88 1 -4 -2 0 1 0.25 0 2 1.25 1 -2 -1.6 0
5 12 1.12 2 0.25 1.25 1.6
1 1 2 4 1 2 2
1.5 2 3.37 8 4.37 9 6 8 2.25 4 3.25 5 1.85 1.6
The two curves cross where x = -1.41, 0.27, and 1.00. Thus there are 3 solutions. This problem cannot be solved by an algebraic formula, but only by graphs or approximations. y = (cos x)4
(degrees) y=
x
360
Fig. 4c
x Example (iii) Solve cos4x = where x is in degrees. 360 There are 5 solutions: x = 52, 142, 210, 355.5 and 360. So far all the quadratic equations have yielded whole number or fractional solutions, but generally this does not occur. For example take the equation x2 – 8x + 10 = 0. It soon becomes clear that two numbers whose product is 10 cannot be found to add up to 8. A general method of solution, always applicable, can be used as follows—it is called completing the square.
40
Keep the terms containing x on the left, x2 – 8x = -10. To both sides add half the square of the number in front of x (42 = 16) x2 – 8x + 16 = 16 – 10 The left hand side is now a perfect square (x – 4)2 = 6 Take the square root of each side x – 4 = ± √6 So x = 4 + √6 or 4 – √6 We now have the problem of determining the square root of any number. Unless this number is itself a perfect square, we shall soon see that we obtain a decimal whose figures go on forever without any recurring order. The method is found by extending the binomial theorem to a multinomial theorem but only when the 2nd power of a bracket is used. Just as
(a + b)2 = a2 + 2ab + b2,
So (a + b + c + d + e)2 = a2 + b2 + c2 + d2 +e2 + 2ab + 2ac + 2ad + 2ae + 2bc + 2bd + 2be + 2cd + 2ce + 2de = a2 + b(2a + b) + c(2a + 2b + c) + d(2a + 2b + 2c + d) + e(2a + 2b + 2c + 2d +e). We now have the problem of expressing any number we wish in the form of the right hand side and so use the left hand side to write down the square root of the number. For example, find the square root of 331,776. First split it up 2 digits at a time from the decimal point: 5 7 6 33 17 76 52 = 25 817 107 x 7 = 749 6876 1146 x 6 = 6876 Find the nearest whole number square below 33 (25) and write it below 33. Subtract 25 from 33 (8). Bring down the two digits in the next space to follow the 8 (817). Double the 5 and write this (10) on the next line. Now put an underline for the digit
41
which will follow 10. Also write x following the underline and write another underline following the x. For each of these blanks above the underlines, choose the digit which will make the answer (749) just less than 817. This digit (7) can also be written in the emerging answer to the problem. Now subtract 749 from 817 to get 68. Now repeat the process. Bring down the 76 by the side of the 68. Then double 57 to get 114 followed by an underline, then an x and another underline. The two gaps must be filled by the digit 6 since 1146 x 6 = 6876 and as the subtraction this time is 0, the problem is finished by placing 6 at the top next to the 57. One can check that 5762 = 331,776; also by noticing that, referring to the multinomial expression, a = 500, a2 = 25,000, 2a = 1000, b = 70, 2a + b = 1070, b2 = 4900, b(2a + b) = 7490, 2b = 140, c = 6, 2a +2b + c = 1146 The letters d and e, being both zero, are not used. Other examples are 2 3 1 .4 5 35 45 .96 22 = 4 135 43 x 3 = 129 645 461 x 1 = 461 18496 4624 x 4 = 18496 and
.0 0 7 3 .00 00 53 29 72 = 49 429 143 x 3 = 429
42
_
All the numbers so far whose square roots we have found happen to be perfect squares. Let us now find the square root of 2. 1.41421 2.0 2 1 1 100 24 x 4 96 400 281 x 1 281 11900 2824 x 4 11296 60400 28,282 x 2 56564 383600 282,841 x 1 282841 100759 This calculation will grow longer and longer without any end. To 10 significant figures √2 = 1.414213562 and there are no repeating sequences of the figures in the decimal places. This is an example of an irrational number. Another way of finding √2 is to consider the following sequence of numbers: 1, 2, 5, 12, 29, 70, 169, 408 … If you double any number in the series and then add the previous number you always get the following number. Now consider the following mixed numbers (each to not more than 5 d.p.) 1
1 = 2
1
2
1
5
5
12
1.5
=
1.2
=
1.41667
43
1
2 29
1
29
1
70
1
70
169 169 408
=
1.41375
=
1.41429
=
1.41420
=
1.41422
The left hand decimals increase while the right hand ones decrease but get closer to the square root of 2. Why this is so must wait until Class 11 or Class 12. As a class exercise set them to calculate these square roots: 1. 138,384 2. 1,671,849 3. 7140.25 4. 0.004489 5. 610,090,000 6. Write down the next mixed number in the column in the last paragraph and change it into a decimal, showing that it is the same as the square root of 2 correct to 5 decimal places. Answers from 0.067, 1.41421, 84.5, 372, 1293, 24,700 It is good to allow no calculators during the first term of Class 9. Let the pupils see they can do all the usual calculations themselves in mental and written arithmetic. In the second term they can enjoy the delights of our push button age. It is also good to recommend which calculator to buy—or get the school to obtain a good discount for a sufficient quantity of them. Otherwise, the teacher will find himself spending too much time helping individual boys and girls to operate complicated and technical machines.
44
Chapter 2 Class Nine Geometry This new main lesson, in the second or third term, should again last for up to four weeks. In the weekly practice lesson (especially those occurring in the earlier math main lesson on permutations and combinations), a revision of the geometry learned with their class teacher should take place. Sets of questions like the following could be given. 1. Two angles of a triangle are equal and the third one is 38°. What must each of the equal angles be? 2. A regular polygon has equal angles of 172¹/²°. How many sides has it? 3. The side length of a rhombus is 17 cm. One of its diagonals has a length of 30 cm. How long is the other diagonal? 4. The parallel sides of a trapezoid are 13 cm and 19 cm and the distance between them is 17 cm. Calculate the area of the trapezoid. 5. ABCD is a quadrilateral inscribed in a circle. Angles ABC, BCD and BAC are respectively 120, 50 and 24 degrees. Calculate angles ACB and BAD. 6. A triangle ABC has AB = 4", AC = 6" and BC = 5". AB and AC are extended to D and E respectively and BD = 2", CE = 3". How long is DE? 7. A semicircle has an area of 77 sq cm. Taking π = 22/7, calculate the semicircle’s perimeter.
45
8. The diameter DE of a circle is 6" long and is extended to F, where EF = 2". How long will be a tangent drawn to the circle from F? 9. Construct a ∆PQR where PQ = 4", QR = 5" and RP = 6". Then construct the lines bisecting angles PQR and QRP to meet in S. Then construct the line through S perpendicular to QR meeting it in T. Measure ST and show by drawing that this is the radius of the circle which touches all 3 sides of ∆PQR. 10. Construct a ∆ABC where AB = 6 cm, BC = 8 cm, and CA = 7 cm. Also construct three circles each of radius 5 cm, and having A, B and C as their centers. Then draw the three common chords of the circles. What do you notice about these common chords (two things are to be noticed.) 11. Draw an accurate square P1 Q1 R1 S1 of side 8 cm. On P1 Q1 mark P2 where P1 P2 = 1 cm. Similarly on Q1 R1 mark Q2 where Q1Q2 = 1 cm. Carry on similarly to obtain squares P2Q2 R2 S2 , P3Q3R3S3 etc., up to P5Q5R5S5. What areas to the nearest sq cm do these 5 squares have? What kind of forms pass through all the points labelled P? Numerical answers: 1.32, 4, 7.5, 16, 36, 36, 38, 48, 50, 64, 71, 130, 272. Geometrical drawing Geometrical drawing, often requiring legal size paper or larger, needs to be executed with a high standard of accuracy in Classes 9, 10, and 11 and enhanced by the addition of careful coloring to highlight the properties illustrated. A careful watch over the required instruments has to be taken. Sharp #2 black lead pencils and rulers measured with inches and tenths, cm and mm are essential. Pairs of compasses which can stretch to a radius of 6 inches and close up to a radius of 1 cm should be encouraged; also there should be no looseness in the joint between the arms. 46
Full circular protractors are better than semicircular ones. A set square with 30, 60, 90 degree angles is useful. Everyone should have a good clean rubber eraser, too. The achievement of beautiful geometrical drawing, while not justifying the name “works of art,” encourage pupils whose mathematical propensities are on the weak side. While no ruled edge can give rise to a perfectly straight line, nor can any pair of compasses give rise to a perfect circle, the effort to strive as much as possible towards perfection has a very positive learning value. Not only weaker pupils but also very able pupils gain by experiencing beauty in geometry and thereby a deeper kind of knowledge of geometry (and mathematics in general) as such. Only when plane geometry is experienced as a section of 3-dimensional solid geometry does the subject acquire greater reality, for our bodies live in the world which has depth as well as length and breadth. This has already been introduced in Class 8 in which the 5 Platonic solids have been constructed. Also volumes of boxes and cylinders have been calculated there. Perspective drawings have also been made, where space’s 3 dimensions have been represented on a flat picture. A good introduction to the geometry of space in Class 9 is to construct complementary views at right angles of the same simple object. Let us start with a cone. It is worthwhile to take the class into the school’s clay modelling workshop and, after making spheres, cubes and perhaps other Platonic solids, go on to have everyone make models of both long and then short and fat cylinders. Finally have everyone making a right circular cone, ensuring that every imagined line radiating from the cone’s vertex along the cone’s surface goes straight to the base circle. Fingers will need to gently stroke the clay to achieve this. When the clay has begun to set from a fluidic to a solid form, use cheese wire to cut a sloping plane section and remove the conical top. This could all take the first part of the first morning’s geometry main lesson. Then, while the clay models set firm, back in the classroom a revision of volume calculations could be made. For a cylinder V = π r2l, for a sphere V = 4/3 π r3 (but they won’t know why until Class 12) and for an Egyptian pyramid V = 1/3 base area x height (proved in Volume 1). The latter also applies where the base is not a square but any shape including a circle. So for a cone V = 1/3 Ah = 1/3π r2h. 47
Numerical and algebraic examples to set the class: 1. A cone stands 4 ft high and has a base diameter of 2 ft 4 in. Taking π = 22/7, calculate the cone’s volume to 3 s.f. in (i) cubic inches, (ii) cubic feet. 2. The formula for the curved surface area of a cone is C = π rs, where s is a generator, a sloping straight line from vertex to base. Calculate C for the cone in question 1. 3. A frustum of a cone is a cone whose top has been cut off by a plane parallel to its base. Figure (5) shows the elevation of a frustum. Using π = 3.14, calculate the volume of the frustum to 3 s.f. V
P
Q
A
B Fig. 5
4. Also calculate the area of the curved surface of this frustum. 5. Prove the formula C = π rs. 6. If the base radius of a frustum is r1 cm, the cutting radius of it is r2 cm and the distance between the top and bottom circular faces is d cm, prove that the volume of the frustum = 48
π d (r 2 + r r +r 2) cubic cm. 1 2 2 3 1
Numerical answers from 5.70, 188.4, 1055, 2200, 9860. On the following morning we construct large drawings to show 3 views of a cut cone on which are shown a series of straight lines joining the cone’s vertex V to points around the base circle at equal intervals of arc. The cutting plane is shown as a line AC on the front elevation, from which points are projected up to the plane view and across to the side elevation. The two oval curves should be perfect ellipses, but this is not easy to achieve. Color may be added to the drawings by giving alternate spaces between the straight lines on the cone (outside the ellipse on the plane, below the ellipse on the side elevation and below the line AC on the front elevation) a lighter color. Next we construct ellipses in a simpler way, using the basic property which relates two special interior points called foci to every point of the curve. It could be proved later that this applies to the curves already drawn. Imagine a bouquet of flowers enclosed by partially rolling a rectangular sheet of paper to form a conical space. Instead of flowers imagine two balls are dropped inside the inverted cone. Taking a vertical central section cutting the cone in lines m and n, let an interior common tangent of the 2 circular sections of the balls meet the cone in A and B and touch the spheres in F and G. The circles of contact between the spheres and cone are shown perspectively (c,d). Let P be any point on the conic section, its ellipse shown also perspectively through A and B. Draw the line on the cone though P and vertex V and let it meet the circles of contact of spheres and cone in Q and R.
49
Plan
(Front)
(Side)
Elevations
Fig. 6
Since PF and PQ are both tangents to the lower sphere, they are equal in length. Since PG and PR are both tangents to the upper sphere, they are equal in length. Hence PF + PG = PQ + PR = QR = a constant length on the cone’s lines through V touching the spheres. PF + PG is a constant sum on the elliptic sections. F and G are called its foci.
50
Elliptic Section of a Cone
Fig. 7
This last property enables a series of ellipses to be constructed using equal sets of concentric circles whose centers are F and G. Here FG = 4 cm and the circle radii increase by a centimeter. Once the radii have increased to 6 cm, the remaining larger circles only require partial completion. The distribution of points of intersection of the circles giving rise to the ellipses make it clear that as one passes from one point to the next, the radius to F increases by the same amount as the radius to G decreases, i.e., the sum of the focal distances is unchanged on any particular ellipse. A list of these sums for each ellipse is written on the curve. The straight line FG is a special (flat) ellipse, for which of course the sum is the length of FG.
51
Confocal Ellipses
Fig. 8a
The larger the ellipses above become, the more circular they are. Nearer the center the minor (vertical) axis is much less than the major (horizontal) axis. Dividing the length of FG by that of the major axis gives the number which states the ellipse’s eccentricity. Thus the ellipses above starting with the line-ellipse FG have eccentricities 1, 0.8, 0.5, 0.36, 0.29. In a circle the foci coincide in the center, so a circle’s eccentricity = 0, as one might expect. There are various ways of enhancing the above drawing with color. Referring to the top part of Figure (8a) and imagining the circles forming a distorted chessboard, one could color alternate “curved” squares black or any other color. The best practical way of constructing an ellipse is to take a piece of inelastic thread (or string), tie a knot in it, fix 2 drawing pins to paper or board and move a pencil within the triangle, pins and thread as follows [Figure (8b)]:
52
Pin
P + g = AB
Knot in thread
Pencil Point
Pin
Fig. 8b
Another way of constructing ellipses is shown below using just one set of concentric circles where radii increase by 1 cm each time together with a set of vertical straight lines 2 cm apart. The intersection points this time give ellipses all of eccentricity 0.5 and a common focus O. The second focus, varying from ellipse to ellipse is F where, of course, AO = FB in the case of ellipse c. If the points X and Y on the ellipse immediately above O and F are joined and extended 6 cm to the left of X (or 10 cm left of Y) this point Z will lie on one of the straight lines (d3). This line is called a directrix of the conic, related to focus O. Similarly there is a second directrix 6 cm right of Y related to focus F. The ratios OX:XZ, OY:YZ, OW:WU are all equal to the ratio 1:2, i.e., the eccentricity. This provides us with another definition of an ellipse: the locus of a point P, moving so that the ratio of PO to the perpendicular distance of P from the corresponding directrix (d3) is constant (0.5 in this case).
53
Fig. 8c
As a break from accurate geometrical drawing, sets of numerical problems can be placed before Class 9, for example: 1. An ellipse of eccentricity 0.4 has its foci 8 cm apart. A point on the curve is 15 cm from a focus F. How far must it be from the other focus? 2. The area of an = πab, where a = length of semi-major axis and b = length of semi-minor axis. The end of the minor axis is 29 cm from each focus. The foci are 40 cm apart. Calculate the area of the ellipse, taking π = 22/7. 3. A picture of the waning moon shows a semicircle of diameter 10 cm. The thickness of the line picture is 2 cm. Using π = 3.14 calculate the area of the bright part of the picture of the moon.
54
2 cm
10 cm
Fig. 9
4. In an ellipse the foci together with the ends of the minor axis form a square of area 16 sq cm. How far from an end of the major axis must be the nearest focus to 1 d.p.? 5. The column of an elliptic cylinder is 110 cu cm and its major and minor axes are 10 cm and 6 cm. How long must the cylinder be? 6. The flat base of a cone fits exactly onto the plane face of a hemisphere of radius r and the height of the cone’s vertex is also r above it base. Find the formula for the volume of the combined solid. 7. A spheroid has a circular plane of symmetry. All other planes of symmetry are ellipses having the same shape. If the radii of the spheroid range from r to 2r, guess the formula for the spheroid’s volume V. Then calculate V when r = 5¼ inches. Numerical answers from 1.17, 2.33, 5, 15.7, 386, 1973. Another section of a cone is a hyperbola. It has two parts to it, one cutting the cone in generators on one side of the vertex and the other part on the opposite side of the vertex. Again we can begin with two spheres touching the cone but the vertex lying between the spheres. The geometry is almost the same as for the ellipse. Instead of the sum of focal distances we now get the difference of focal distances. PF and PQ , tangents to top sphere are equal. PG and PR, tangents to bottom sphere are equal. PG – PF = PR – PQ = QR, a constant length joining circles d and c through V. For the hyperbola the difference between the focal distances is constant.
55
Hyperbolic Section of a Cone
Fig. 10
Taking two sets of concentric circles again but joining their points of intersection in the opposite way in Figure (11), we can obtain a set of confocal hyperbolae. Moving around each hyperbola the focal radii decrease or increase each time by half a centimeter. So their difference remains constant.
Fig. 11
56
Eccentricity is again measured by dividing FG by the horizontal (shortest) width of the curve. The curves shown have eccentricities of 11/3 and 22/3. Between them lies the double vertical line (e = ∞) and beyond them is the interrupted double line (e = 1). Just as we found another way of constructing ellipses using only one set of concentric circles together with a set of parallel lines, so can we construct hyperbolae. But this time, instead of the distance apart of the lines being larger than that between circles, the distance apart of the lines is smaller than that between the circles. The hyperbolae all have a common focus O. The second focus F varies from hyperbola to hyperbola. For O and F the two directions are labelled d and the hyperbola labelled h.
Fig. 12
Once again, using the construction of ellipses with one set of concentric circles and a set of parallel straight lines, a set of parabolae can be constructed. This time the distance between successive straight lines has to be equal to that between successive circles.
57
Fig. 13
Every parabola has an eccentricity of 1. The directrices and corresponding parabolae are shown with the same suffix.
It is now clear that all conic sections whose
eccentricity is
}
less than 1 are ellipses equal to 1 are parabolae more than 1 are hyperbolae.
The directrices of the parabolae P1…P5 are shown as d1…d5. Using the definition of eccentricity we will evolve the general equation of a conic. Taking rectangular axes x and y, let O be a focus and line x + 1 = O be a directrix. For a point P on the conic OP = e PN.
Fig. 14
58
Hence OP2 = e2 PN2 i.e.,
x2 + y2 = e2 (x + 1) 2 y2 = (e2 – 1) x2 + 2e2 x + e2
Let us take 5 different values of e and sketch the curves. 1. e = 5 y2 = 24x2 + 50x + 25. When y = 0, (6x + 5)(4x + 5) = 0 x = -5/6 or -1∫ When x = 0, y = ±5 When x = -1¹/², y2 = 54 – 75 + 25 = 4 y = ±2 2. e = 2 y² = 3x² + 8x + 4 When y = 0, (3x + 2)(x + 2) = 0 x = -¹/² or -2 When x = 0, y = ±2 When x = -2 2/3, y² = 64/3 – 64/3 + 4 y = ±2 y² = 2x + 1 3. e = 1 When y = 0, x = -¹/² When x = 0, y = ±1 4. e =¹/² y² = -3x²/4 + x/2 + ¼ When y = 0, 3x² – 2x – 1 = 0 (3x + 1)(x – 1) = 0 x = -1/3 or 1 When x = 0, y = ± ¹/² When x = 2/3, y² = -1/3 + 1/3 + ∫ = 0, y = ± ¹/² When x = 1/3 , y² = -1/12 + 1/6 + ∫ = 1/3 ,
y = ± .58
5. e = 1/5 y² = -24/25 x² + 2x/25 + 1/25 When y = 0, 24x² -2x – 1 = 0 (6x + 1)(4x – 1) = 0 x = -1/6 or ∫ When x = 0, y = ± 1/5 59
It would be an excellent exercise to set the class these calculations to determine points on the 5 curves. Other points could also be calculated, but good freehand curves should be well within the pupils’ power at this stage. Once the curves are drawn, a good question to leave with class with is: “Can you see any similarity between this set of curves and the form of the human being?”—to be answered next day. More curves could of course be drawn in the spaces between the given curves (taking more values of e). But the attention needs to be drawn to the metamorphosis which the whole drawing exhibits—how each conic section gradually transforms into each of the others. The five conics are shown in Figure (15) below. Two are hyperbolae, one is a parabola (the only one in the whole array) and two are ellipses.
Common Directory
Fig. 15
Beyond (5) the ellipse shrinks towards a point-circle (the focus O). Before (1) the two hyperbolic branches emerge from a single straight line (the directrix d). The spiky shading shows that the outside of the ellipses corresponds to the inside of the hyperbolae.
60
This is precisely the difference also between the bones of the human skull and the bones of our legs and arms. The spherical head bones enclose the brain’s soft substance, whereas the limb bones are enclosed by the soft muscles. The tubular limb bones are slender in their middle and wider near their ends. The head bones are only parts of the spherical form and are joined together at their edges. The bones of our chest have an intermediate form and the lowest floating ribs have a flattish parabolic curve. Quite apart from the metamorphosis of the physical shapes as one proceeds from head to limbs, the 3 parts of the human soul have activities which conform to the radiant movements indicated by the spikes. Our thinking radiates outwards from our heads, while our will radiates inwards bringing about the practical and purposive movements of our limbs. Our feeling life is more centered in our chest and like the beautiful form of the parabola embraces our inner life with the distant cosmos. Further examples for the class to work at are given below. 1. If any point N on the directrix of a parabola is joined to the focus F and P is the point where the line parallel to the parabola’s axis meets the curve, show that NF bisects angle PFD, DF lying along the axis. 2. Construct the two sets of concentric circles with equal distances between their radii. Coloring all the “squares” of half the chessboard pattern, show how the hyperbolae and ellipses all cross each other at right angles. 3. The line joining the foci F and G of a hyperbola also meets the curve in A and B on the directrices in C and D. FA = BG = 3 cm, AC = BD = 1 cm, and the eccentricity is 3. A line through G perpendicular to BG meets the curve in H and K. The line through H parallel to FG cuts the directrices in U and V. UC is parallel to HG. Calculate the area of rectangle UVDC. 4. Why have all parabolae got the same shape?
61
I have a bad cold and thought he was referring to wiping my nose, but I just realized he was saying, “Dee wiper dee eggs,” No, No! He was saying, “Dee ‘y’ by dee ‘x’.” Just look at the notice on the door!
62
Chapter 3 Other Mathematical Topics Suitable for Class Nine or Class Ten 1. The use of calculators When developing Permutations and Combinations in the first term of Class 9, I strongly recommended that no calculators should be used yet. This is partly to ensure that fractions and decimals, long multiplication and division are continually practiced. After the winter holidays, however, one could give out calculators (carefully chosen oneself) to the whole class and show how to use them. Ask the students to write down the instructions for use, e.g., to calculate
√
i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. xii. xiii.
( 7.45 x 12.09) (27.95 – 13.47)
=
Press “On” Cancel any numbers in the answer space √ ( 7.45 x 12.09 ) ÷ ( 27.95 – 13.47 63
xiv. ) xv. = Such procedure will prove very helpful later when computers are introduced in Class 11. A few of the problems from earlier in the main lesson can now be solved very quickly, e.g., 10C6 and 11P7, and any problem requiring the use of a formula. 2. Development of formulae themselves. For example if u = initial velocity, v = final velocity, t = time taken, a = acceleration which is constant, d = distance travelled, it is clear that, provided the units of measurement conform, d = (u + v) t and a = v – u . t 2 By substitution and changing the subjects of formulae, we soon obtain v = u + at, v2 – u2 = 2da, d = ut + ¹/² at2. 3. Money problems Using compound interest, obtaining loans and mortgages from a bank or building society, interests Class 9 and 10 pupils, especially when they realize with real horror how much money has to be paid in interest before a mortgage can be fully redeemed. A debate on how to change the system is worthwhile in this class. A suggestion might be made that no one lending money should be allowed to reap more interest altogether than half the capital loaned. Again it might be pointed out that money, like everything else in human life, should have to die (e.g., food kept too long will go bad). Unfortunately, no political party will accept such an idea yet. No wonder young people lack enthusiasm for voting in general elections. Returning to modern business arrangements, suppose a young couple needs $120,000 to buy a house and the bank is willing to lend $100,000, provided it is confident that the couple can pay $10,000 back each year to include an annual interest charge of 9% of the remaining loan. (The bank may require a deposit of $20,000, but we will imagine it does not in this case.)
64
After 1 year, the debt is reduced to $100,000 + $9,000 – $10,000 = $99,000. After 2 years, the debt is reduced to $99,000 + $8,910 – $10,000 = $97,910. After 3 years, the debt is reduced to $97,910 + $8,812 – $10,000 = $96,722. The debt hasn’t gone down much so far! But after 26 years it becomes only $6,676 (see later) and 8 months later the loan is completely cleared. It has taken more that 26 x $10,000 in repayments, i.e., $267,000 approximately, or nearly 3 x the loan. However the price of houses has probably gone up by more than 3 times in that period of time, so the couple will find later in life they are actually better off. Perhaps buying a house is a good investment. Let the capital required ($100,000) be P (principal) and the rate of interest per annum (9%) be r/100. Let the annual repayment ($10,000) be R and let 1 + r/100 = x. After one year the debt remaining = After 2 years the debt remaining = After 3 years the debt remaining = After n years the debt remaining =
Px – R dollars Px2 – R (1 + x) Px3 – R (1 + x + x2) Pxn – R xn – 1 x–1
The debt will become zero when
Pxn (x – 1) = R(xn –1) So R = xn(R –P(x –1)), xn = R/R – P(x –1)
10,000 In the numerical case xn = 10,000 –100,000 x 0.09
10,000 = 10,000 – 9000
=
10
65
So n log x = log 10
n = log 10 = 26.7 log 1.09
Check: If n = 26, 26 –10,000 x (1.0926 –1) = 6676 debt = 100,000 x 1.09 0.09
so the debt will be cleared about 8 months after 26 years of mortgage repayments.
66
Chapter 4 Class Ten Mathematics (Age 15 to 16) In Class 9 the great help to develop thinking power in the students is becoming aware of contrasts such as success and failure, life and death, selfish and unselfish deeds, logical and intuitive, reverence and disrespect. In Class 10 this can be taken further with contrasts such as cold and warm and dry and moist to establish the four elements of the Greeks. Earth = cold and dry, Water = cold and moist, Air = warm and moist, Fire = warm and dry. By earth they meant everything of a solid nature, by Water everything liquid, by Air everything gaseous and by Fire everything which can be combustible, including what modern science refers to as fusion and fission of atomic particles. The ancient Greeks had a feeling for each element analogous to human temperaments relating both to individual people and to racial or national qualities. Thus Earth had a melancholic, Water a phlegmatic, Air a sanguine and Fire a choleric flavor. As he gazed into the directions of space, Southwest had an airy quality, Southeast a fiery one, Northeast an earthly one and Northwest a watery one. So the Mediterranean West and North Africa conjured for him mighty winds. Beyond Egypt and towards India were felt the hottest of all lands. Siberia (the word means severe) was felt to be the coldest and driest of lands. And as for those offshore islands to the northwest of continental Europe, everyone knows that it never seems to stop raining there. Associated with the qualitative Earth element, the Greeks pictured sets of points along a line, for example a set of points whose measurements from a fixed point were
3, 10, 17, 24, 31.…
67
Can you suggest what the next number must be? What is it which binds together the numbers at each point? The students of Class 10 will soon declare that each number is 7 more than the one before it. What will be the hundredth number? You can imagine striding out along the line, each stride the same, or imagine it is a line one swims along. The hundredth number will not appear until all the others have come, by which time 7 will have been added 99 times. The hundredth number = 3 + 7 x 99 = 696. A final question: What will the sum total be of all the hundred numbers in the row? Look at the series both backwards and forwards. 696 + 689 + 682…. + 3 3 + 10 + 17 + …. + 696 Instead of casting a horizontal glance at the 2 series, cast a vertical glance. Each column has the same sum, i.e., 699. Since there are 100 columns, the sum total = 69,900, which when halved gives the answer of 34,950. Sequences of numbers which have constant increase between each two terms are called arithmetic progressions. For each sequence below find the next term, the 40th term and the sum total of the 40 terms. 1. 2. 3. 4. 5.
30, 35, 40, … 5, 14, 23, 32, … 1.6, 1.7, … 62, 60.5,… a, a + d, … to n terms
Answers from 1.8, 3.5, 5.5, 41, 45, 59, 142, 225, 356, 1310, 5100, 7220, a + 2d, a + (n – 1)d, n/2(2a + (n – 1)d)
68
Sequences of numbers which have a constant ratio between each two consecutive terms are called geometric progressions. For example consider 7 21 63 The common ratio =3 The next term = 189 x 3 = 567 What would be the hundredth number? Imagine swimming to it.
189….
The answer = 7 x 399 = 1.20 x 1048, a very large number. To find the sum total of the 100 numbers it is best to first consider a general geometrical progression in which the first term is a, the common ratio is r, and the number of terms is n. Thus a ar ar2 ar3… arn-1 Let the sum total be S. So S = a + ar + ar2 + ….+ ar n-1 Hence rS = ar + ar2 + ar 3 + … + arn Subtract S(r – 1) = -a + arn n n So S = a(r – 1) If 0 < r < 1, S = a(1 – r ) r – 1 r – 1 In the numerical case S = 7(3100 – 1) = 7 (5.15 x 10 47– 1) 2 3 – 1 ≈ 1.80 x 1048 Taking a simpler case, 50 + 60 + 72 + …. to 10 terms Since r = 60 = 1.2, 50 next term = 72 x 1.2 = 86.4 tenth term = 72 x 1.29 = 371.5 to 1owest d.p. Sum total = 50(1.210 – 1) = 50 (6.19 – 1) 0.2 1.2 – 1 = 1297.5 to 1 d.p.
Similarly find all 3 answers for each of the geometrical progressions to 7 terms: 1. 3 6 12…. 2. 54 36 24…. 3. 48 60 75…. 69
Answers from 4.7, 16, 24, 93.8, 152.5, 183.1, 192, 381 and 915.5. When r < 1, geometric progressions result in the sum total of their terms tending to a finite limit as n ∞. + 1/8+ 1/6 + ….,
The simplest case is ¹/² + ∫
when clearly the limit is 1, since each term is equal to the difference between the sum so far and unity. Interesting, too, are cases where the common ratio is negative. For example 1 – ¹/² + ∫ – 1/8 … has the sum total of its infinite series = 1 / 1 + ¹/² = 2/3 It is at this point that we can understand Zeno’s famous paradox and its true resolution. The story relates how a tortoise challenged the great Greek runner Achilles to a race provided the tortoise was given a headstart. To make it simple let us suppose that the tortoise had half a mile start and that, while Achilles covered his first half mile, the tortoise managed a yard. Assume that each runner kept up the speeds with which they began. By the time Achilles reached the place a yard ahead of the tortoise’s starting pace, the tortoise had moved further on, in fact
1 800
of a yard. So Achilles had to
1
)2 yards and again the tortoise run further, actually a distance ( 800 was still in front of him. In order for Achilles to win the race, he would have to run altogether 880 + 1 +
1 800
+(
1 800
)2 + (
1 800
)3 + …. yards.
Since the sequence never ends, Zeno argued that that Achilles could never overtake the tortoise. What is illogical about this conclusion? It assumes that Achilles (and also the tortoise) were taking ever smaller steps, each one of the previous one. A quite absurd assumption!
70
1 800
Achilles actually passed his rival after running 880
=
1
1– 800
8802 879
—or just a tad over 881 yards (881.0011377 yd) Here’s a set of examples on progressions: Calculate the limits to the sum of the endless geometrical progressions: 1. 2. 3. 4. 5.
54 + 18 + 6 + 2 + … 3¹/² – 2 5/8 + 1 31/32 – … 3¹/² + 2 5/8 + 1 31/32 + … 10 – 9 + 8.1 – 7.29 + … Figure (16) shows a triangular structure in which AB = 12 m, AC = 13 m, and angle ABC = 90°. Perpendiculars are drawn from B to AC meeting it in D, D to AB meeting it in E, and so on. Calculate the lengths of BD, DE, etc., until the perpendicular length is first less than 2 m. What then is the total length of all the lines in the figure perpendicular to AC? 6. Suppose the construction of perpendicular lengths was continued infinitely. What would the total of all the lengths be then? 7. If in Figure (16) the points C, D, F… had been arranged as an arithmetic progression with CD = DF = etc., all equal to the original length of CD, what would the finite total of all the lengths in the figure perpendicular to AC be then? 8. The first ter m of a geometrical progression is 15 and it continues until a term is reached less than 5. The common ratio is 0.8. How many Fig. 16 terms are there?
71
9. A progression decreases from 7 (first term ) to 3 (last term). It has altogether 10 terms. Find the common difference between the terms if it is an A.P., and find the common ratio of successive terms if it is a G. P. 10. 6 7 9 13 21 … is neither an A.P. nor a G.P. Find the simplest way of continuing it for the next two terms. Answers from: 0.444, 0.91, 2, 5.26, 5, 14, 15, 71, 37, 40.13, 65, 69, 81, 9. The character of 16 year old boys and girls and a lead-in to logarithms There is a marked difference between teaching Class 9 and Class 10. In the latter, acknowledgement by the teacher of the first beginnings of adulthood and objectivity are important. In continental schools it is customary to start addressing 16-yearolds by using the German “Sie” rather than “du” and the French “vous” rather than “tu.” It is also helpful to refer to how, when they were in younger grades, their class teacher used to group the class according to temperaments. “Some of you remember sitting at the back of the classroom and enjoyed so doing. It was because you tended to have something of a phlegmatic temperament.” A description of the 4 temperaments interests 16-year-olds, though you might notice how they speak of their earlier years as though they happened many years ago—or even in a different lifetime. “Others of a choleric temperament tended to sit in the front of the class near the middle where the teacher could keep an eye on them.” It also happens that cholerics can be of great help to the teacher in maintaining order. If the teacher is telling a story about someone with leadership qualities and a few people are talking or interrupting, the helpful cholerics can turn around with, “Shut up! I want to hear the story!” Still other children would be placed on the side of the room away from the windows to help their concentration not be interrupted by what is going on in the playground or beyond—the sanguine children. Finally the melancholics need to sit near the windows, for they need light.
72
It is so healthy and sensible to have younger children and young people within the age range of 6 to 18 in the same school. The older classes enjoy helping with the younger classes, e.g., helping teachers lead them to their seats for an assembly or to watch plays and school festivals. Even if the school does not go above Class 10, this is a task the older boys and girls enjoy with such a responsibility. Class 9 students are not quite ready for that yet; their interest is more in their peers. In arithmetic Class 10 will have noticed that there are three ways of relating numbers to each other. The statement 7 + 3 = 10 can be expressed as 10 = 7 together with 3 or 3 = how much to increase 7 to reach a total of 10. Similarly 24 = 6 x 4 can be expressed as 4 gives rise to 24 by multiplying by 6 or that 24 divided by 4 gives 6. They will also recall that 23 = 8 and that 3 √ 8 = 2. There is a third way of expressing this relationship, i.e., that log 28 = 3 (the logarithm of 8 to the base 2 is 3). Similarly log 381 = 4, log10 1000000 = 6, log 2¹/² = -1, log 17289 = 2, log 2512 = 9, log 4/9 7 = ¹/², log 2929 = 1, log 71 = 0. In the past, logarithm tables to base 10 had to be used to facilitate complicated multiplication and division sums. With the advent of hand calculators. this is no longer necessary. The concept of logarithms will become important in Class 11, but it is helpful to introduce it in Class 10 in connection with a third kind of progression or series. Consider the progression 60 30 20 15 12. It is called a harmonic progression. What number must come next? After a little debate someone will say that 60 is in turn being divided by the set of natural numbers 1, 2, 3, 4, etc. So the next numbers will be
60 and 60 , i.e., 10 and 8 4 . 6 7 7
What will be the 30th number in the progression? This is easy. It will be 60 = 2. 30
73
But what about the sum total of the first 30 numbers? Adding them up we get just over 4. If the base e = 2.71828 … = 1 + ¹/²! + 1/3 ! + ∫ ! + … to infinity is chosen, then log e30 = 3.40 and 1 + log e30 = 4.40. Hence log e30 < 1 + ¹/² + 1/3 + … + 1/3 0 < 1 + log e30 So the infinite sum of the 1 + ¹/² + 1/3 + … is more than the loge of any number, no matter how large. So 1 + ¹/² + 1/3 + ∫ + … is a divergent progression and so must 60 + 60 + 60 + 60 + … be, etc. 2 3 4 Yet a fourth progression was introduced to students in the ancient Pythagorean Mystery School and described as a musical progression, whose prime example was 6
9
8
12
12 9
8 6
Top C
G
F
Bottom C
Fig. 17
Its relevance to music is seen by having 4 equal strings of the same material, cross section and tension as indicated in Figure (17) and then “stopping” them at the distance ratios shown and plucking the lower part of the first 3 strings and the whole fourth string. The pitches of the 4 notes are as shown. The intervals 8 4 Top C down to G = 6 = 3 9 3 Top C down to F = = 6 2
74
12 2 Top C down to Bottom C = = 6 1 G down to F = 9 8 G down to Bottom C = 12 = 3 8 2 F down to Bottom C = 12 = 4 9 3 2 ), Thus we have musical intervals of octave ( 1 perfect fifth ( 3 ) 2 perfect fourth ( 4 ) 3 and major tone ( 9 ). 8 By continuing to use perfect fourths or fifths, the whole monotonic scale was constructed. It is easy to see that a musical progression is neither arithmetical, geometrical or harmonical. How can such a progression be continued? Musically there are many melodies which may be composed, but what about an exact mathematical progression? Pythagoras did not demonstrate one but solely expressed a general one for just four tones, i.e., a, 2ab , a+b , b. a+b 2 a+b It will be recognized that a and b together with 2 form an arithmetic progression While together with 2ab a+b
form a harmonic one, or simply 6 9 12 is arithmetic and 6 8 12 is harmonic (i.e., 24 , 24 , 24 ). 4 3 2 The following may be described as an extended musical progression.
75
a – 1 a2 – 1 a3 – 1 … an – 1 0 a+1 a2+1 a3+1 an+1 For n = 0, the term is 0; For n = -1 it is 1 1 = -a – 1 a –1 a+1 a+1 For n = -2, it is -a2 – 1 , etc. a2+1 Hence preceding the above series we have n 3 2 - an –1 … - a3 –1 , - a2 –1 , - a–1 , a +1 a +1 a +1 a+1 i.e., each term has its negative values below zero at a corresponding ∞, so the fraction equal position to the left of zero. Also as n 1 in the positive direction and -1 in the negative direction and the whole progression becomes -1 ...
n 2 2 n - an –1 , - a –1 , - a –1 , 0, a –1, a2 –1 … an –1 ... 1 a +1 a+1 a+1 a+1 a +1 a +1
The simplest example is where a = 2. Thus -1 … -15 , - 7 , - 3 , - 1 , 0, 1 , 3 , 7 , 15 … 1 17 9 5 3 3 5 9 17 It is also possible to imagine extending this beyond the limits 1 and -1 by including the reciprocals of all the fractions above. To the right of 1 will be… 17 , 9 , 5 , 3 , ∞ , - 3 , - 5 , - 9 15 7 3 1 1 3 7 . .. and so back to -1. This extension of the musical progression also contains unlimited examples of 4-term (Pythagorean) musical progressions, e.g., 1 3
3 5 or 3 5
76
5 3 when a = 2 3 15 17 5 when a = 4, etc. 17 15 3
Going back to the original 4-term Pythagorean musical progression, if a = 1, we have 1, 2b , 1+b , b or 1, c, 1+b 2 1 , c (putting b = c ). 2–c 2 – c 2 – c Exercise: Putting c in terms equal to 16 , 6 , 5 , 4 , 3 , 9 5 4 3 2 8 , 5 , 16 , 9, 16 , 2 5 3 9 5 8 write down each of the eleven progressions. Then taking Middle C = 1, treat each of the other 3 fractions each time as intervals above Middle C and write the appropriate note, using Top C as c, an octave higher as cii, an octave still higher as ciii. The note above G is A. Groves Musical Dictionary gives the following “Just Intonation” intervals: Perfect Fifth 3 Unison 1 2 Minor Tone 10 Minor Sixth 8 9 5 Same key on the piano Major Tone 9 Major Sixth 5 8 3 Minor Third 6 Diminished Seventh 95 or 169 5 5 Major Third Major Seventh 15 4 8 4 Octave 2 Perfect Fourth 3 Play each of the eleven musical progressions on a piano and try to describe your feeling for each 4-note melody. (This is also possible to play on a string or a wind instrument.) Further exercises: By multiplying the appropriate fractions above show the overall interval obtained by
77
1. following a major third by a minor third. 2. following a major third by a perfect fifth. 3. taking 3 successive major thirds. 4. taking 4 successive minor thirds. 5. What do the last two results show? The introduction of spiral forms in Class 10 Each of the 4 progressions gives rise to a quite different and exact spiral. They are known as 1. the Archimedean Spiral 2. the Logarithmic or Equiangular Spiral 3. the Reciprocal Spiral and 4. the Asymptotic Circle Spiral They are respectively related to the lifeless and solid, to the living and liquid, to the gaseous and air currents, e.g., in cyclones and anticyclones, and to the inner and outer relationships of the human being with himself and the cosmos, a fiery relationship. Figure (18) shows four Archimedean Spiral forms. The small cross in the center consists of 4 points each 0.5 cm from the center. Going outwards along the “dashed” radii, every point is 1.5 cm from the previous one. On all other radii successive points are again 1.5 cm apart, but their nearest points from the center are 1 or 1.5 cm from it. If instead of drawing straight lines joining the constructed points as shown, smooth curves either through the points or touching the shown lines are made, beautiful forms result, forms originally made by Archimedes. Notice that not only do every set of points on the radii form arithmetic progressions with a common difference of 1.5 cm, but each set of lines in the 12 sectors of the drawing have lengths which tend more and more to form an arithmetical progression as they get further from the center.
78
Fig. 18 Archimedean Spiral
Fig. 19 Logarithmic Spiral (Equilangular)
79
Figure (19) shows six Logarithmic or Equiangular Spiral forms. Six points A B C D E F are taken on lines radiating from the center at 60°. In the drawing shown they are each 4 cm away, though the chosen measurement is arbitrary. A line is drawn from A towards B to meet the intermediate radius and 5 similar lines from B, C, D, etc., are drawn. From the 6 new points 6 similar but shorter lines are drawn, and so on. The sets of lines can also be continued backwards. Again, instead of drawing lines, smooth curves either through the points constructed can be made or curves touching the shown lines. No matter how many points or lines are used approaching the center, one will never reach it, for all the points on the radii form geometrical progressions. Thus as one moves outwards, 6 sets of spiral forms continue endlessly to expand, but in moving inwards they turn endlessly around the center getting ever close to it. It is interesting to place a drawing pin in the center of such drawings and spin the drawings around it, first in one direction, then in the opposite direction. Try to express your feelings for the movements which the different sets of spirals make.
Reciprocal Spirals (Hyperbolic)
Fig. 20
Figure (20) shows twelve Reciprocal Spiral forms. The points A and B (and then more points beyond the figure) are each 120 cm from the center along the radii, though the chosen 120 cm is
80
arbitrary. Going inward along the radii the following harmonic progressions are measured from the center: 120 , 120 , 120 ,… 2 3 4 and the lines joining up these points are as shown. Paper size does not allow the outward extensions of the spirals to be drawn here, but from A to the left a line runs parallel to the horizontal radius reaching into the infinite, and similar parallels run outwards from B and the other ten points. Whereas in logarithmic spirals the sum total of the endless small lengths running inwards from any point on any of them is finite, the sum total of corresponding lengths in reciprocal spirals is infinite. Again, instead of drawing straight lines, smooth curves can be drawn either through the constructed points or touching the lines joining adjacent points. Each of such curves has a hyperbolic asymptote parallel to a radius.
M. P.s The dots show the position of the asymptotic circle
Fig. 21 Asymtotic Circle Spirals
Figure (21) shows twelve Asymptotic Circle Spirals. Each curve is in two parts. One part approaches the other but can meet the other only on the circle (shown here of radius 3.5 cm) after an endless number of turns. Distant 3 x 3 = 9 cm from the center are points A, B, … along the twelve radii. Going inwards each time are points measured from the center of 3 x 5/3, 3 x 9/7,
81
3 x 17/15…, etc. cm. Also going outwards from the center points are measures of 3 x 1/3, 3 x 3/5, 3 x 7/9 …, etc. cm. The beginning of the line spiral from A to the left parallel to the horizontal is shown and is an asymptote of the spiral touching AK, etc. Using a pin through the center, spinning these spirals first in one direction and then in the other gives a remarkable impression of contrary movements. This has been used to stimulate hypnosis in an observer. Do not try it on with adolescent girls unless accompanied by much joking laughter (in a good spirit) by the class, especially when curves rather than lines have been constructed. Where are these four spirals found wholly or partly in the world? An approximation to an Archimedean Spiral may be seen on top of a backpacker’s knapsack where his sleeping blankets is rolled up tightly. It is the same when observing a roll of toilet paper from the circular end, where the distance between whorls is the small thickness of the paper. An exact example of an Archimedean Spiral, however, can be seen on a sewing machine in the cam which enables cotton to be wrapped perfectly on a spool [see Figure (22)]. Verical Rocker Cotton Source Spool
Cotton Winding
Cam Rotating Round C
Spool Axis
Fig. 22
A is a hook, which slides along radius BC, which rotates around C in the plane in which the cam moves. Its rotation is “worm and 82
pinion” geared (not shown) to the rotation of the spool around its axis. The spool should be pictured above the plane of the cam. A thus moves along alternate parts of two identical Archimedean Spirals enabling the cotton to wrap around the spool alternately upwards and downwards. In modern sewing machines the cam and gearing are enclosed from the viewer. Other examples of Archimedean Spirals are used in engineering and still other examples apart from near their centers can be observed in Catherine wheel fireworks, for example, but never in living forms. Nature provides many examples of the second spiral, the Equiangular or Logarithmic Spiral, in living forms and in the remains of them. Sweet peas and other climbing plants reveal them when viewed downwards along the canes or other vertical supports. A whirlpool in the sea is also a good example. The arrangement of flowers and seeds in inflorescences also follow such spirals, but their forms near the spiral centers are naturally modified because nature cannot manifest in infinitely small areas. Here one requires knowledge of Fibonacci series, but this needs to be left until Class 11. Many varieties of shells, especially sea shells also reveal the second type of spirals, including shells whose spiral wraps around a cone. The reason why snails and similar creatures have this shape of house to live in is that starting very tiny from their eggs they continually enlarge the shell in accordance with the shape of their own growing body. The equiangular nature of doing so conforms with the second spiral. Why does the creature’s home spiral? It is due to the daily rotation of the stars of the heavens around the Earth allied to the outward growth movement. The third spiral is called the Reciprocal Spiral because it is the exact inverse of the Archimedean Spiral. That means the radius to any point multiplied by the corresponding radius to that on the inverse spiral has a constant answer. On a small scale the Reciprocal spiral can be seen frequently in tropical countries, especially in tropical grassland areas. Good examples occur in the Johannesburg region of South Africa where thermals of hot air rise from the sun-warmed ground which is surrounded by vegetation. This can be on quite a small scale too. One can watch leaves, small stones and dust caught up in such vortices. 83
Thermal Spiralling Air
Warm Radiating Earth
Fig. 23
In studying weather maps of Great Britain or Europe (and other continents as well), one can observe cyclones or depressions. Lines of equal atmospheric pressure called isobars surround the point of lowest pressure and winds blow outwards and spiral clockwise in the Northern Hemisphere around it (counterclockwise in the Southern Hemisphere). On such maps the view from space is of many Reciprocal Spiral forms.
High Pressure Low Pressure
Southern Hemisphere Depression
Fig. 24
While an approximation of the fourth spiral—the Asymptotic Circle Spiral—may be seen in the human ear in the cochlea, an impression of the two tendencies of movement can be gained when observing a burning building as the conflagration gets out of hand. Inside the building the flames from burning wooden floors attack everything flammable from clothing and furniture to the walls, while other flames leap into the sky from the burning walls of the building. But the Asymptotic Circle and the spirals born
84
from it do not really appear physically. We are really concerned with the inner and outer activities of the human soul. This is probably why hypnotists have tried to use these inwardly and outwardly rotating spirals. Each of the four spirals can be expressed in polar coordinates, where r is the distance of point P from the center O, and Θ is the angle turned through from the horizontal by the line OP, when the Asymptotic Circle has unit radius. a is a constant. The simplest series of numbers occurs when a = 2.
Spiral Equation Range of - ∞ < Θ < 0,
r as 0<Θ<∞
Archimedean r = aΘ
- ∞ to 0
0 to ∞
Earth
r = aΘ
0 to 1
1 to ∞
Water
a r= Θ
0 to ∞
∞ to 0
Air
r = a Θ –1
0 to 1
1 to ∞
Fire
Equiangular Reciprocal Asymptotic Circle
Element
85
Chapter 5 Class Ten Trigonometry Class 10 is the time to lay down the foundations of trigonometry. How do you find the height of a church tower without climbing it and using a plumb line lowered over the edge of its top and then measuring the line’s length? This corresponds to the growth of objectivity in young people of this age—instead of being bodily involved with a problem (typical of Class 9), they now feel the need to step back and use reason. First measure out a certain length horizontally from the base of the tower (say 40 meters) and then set up a theodolite or simple device for measuring the angle of elevation of the top of the tower’s wall. Measure the height of the point from which the angle is measured. Then make an accurate diagram to a convenient scale and so read off the height of the tower, in this case 36.0 + 2 = 38 m. Tower
Scale: 1 cm to 10 cm
Fig. 25
The relationship between perpendicular distance and base distance has the same ratio no matter what scale the drawing has. The same is true if one is measuring tangents to a circle of unit radius, a geometrical task which can now be set for all members of the class to carry out in their main lesson books as follows. 86
T A B C D E
F
G
H
Radius
2 cm Length Ratio Angle at 0 in cm in degrees TA 0.36 0.18 10 TB 0.72 0.36 20 TC 1.15 0.58 30 TD 1.67 0.85 40 TE 2.38 1.19 50 TF 3.46 1.73 60 TG 5.50 2.75 70 TH 11.34 5.67 80
Fig. 26
As the angle grows from 80 to 90 degrees the measurement and ratio increase without limit. For an angle of θ°, the ratio is called tan θ. Referring back to the tower example, the three sides of the right-angled triangle can be named quite naturally as hypotenuse, base and perpendicular, the second lying between the acute angle being measured and the right angle, and the third lying opposite the acute angle.
n
po
Hy
use
e en
Base
Perpendicular
n
po
Hy
use
e en
adj
Opposite
Fig. 27
There are clearly 6 ways of forming a ratio of 2 of the 3 sides. The six ratios are given the following abbreviations as names. perp = sin θ, base = cos θ, perp = tan θ, hyp hyp hyp base perp hyp = sec θ, base = cot θ. base perp
= cosec θ,
87
These names appearing in front of θ have been derived from the names of lines associated with a circle as follows:
Fig. 28
From tangent comes tan, of course. OT is the unit radius. From secant comes sec. Secant is the name of a line cutting through a circle. The word sine, originally meaning a breast bone, was used in the past to denote half the taut string used by an archer taking aim after stretching his bow (but not necessarily a circular bow). So TC could be a half string, the arrow lying along the circular arc from T to U, with the other hand grasping the bow at U and so firing into the air. The prefix co- in the words cosine, cosecant and cotangent shows the relationship to the other words. Notice that tan and cot ratios are upside down and so are sin and cosec and also cos and sec. The first 3 trigonometric ratios—sin, cos and tan— contain abbreviations having initial letters which are the same as those of the 6 words in “Pat has brought his (or her) pony back”—a way of remembering them. Calculating lengths of sides and angles in right-angled triangles now becomes easy work using hand calculators. For example: 1. If ABC is a triangle with a right angle at B, AB = 7.26 cm and the angle at A is 54.6°, calculate the remaining lengths. CB/7.26 = tan 54.6, so CB = 7.26 x tan 54.6 = 10.2 cm and AC/7.26 = hyp/base = sec 54.6 = 1/cos 54.6, so AC = 7.26/ cos 54.6 = 12.5 cm. 88
2. In a triangle XYZ, if there is a right angle at Y, XY = 24.7 miles and XZ = 45.3 miles, calculate the angle at Z. Sin YΦX = perp/hyp = 24.7/45.3 = 0.545, so YΦΧ = sin-1 0.545 = 33.0° Examples for the class Calculate correct to 3 s.f. the following lengths or angles: 1. EF 2. FD 3. GH (refers to drawings on p. 90) 4. LJ 5. LK 6. MN 7. NP 8. β 9. γ 10. QR 11. RS 12. ST Answers (in wrong order): 2.04, 3.27, 3.86, 4.98, 8.00, 20.1, 24.5, 27.9, 54.7, 64.4, 68.8, 84.0 In Class 10 it is important to provide practical exercises in surveying. This can be done in afternoon blocks during which spinning and weaving, carpentry, cookery and other arts and crafts are taught in smaller groups of girls and boys (not separated into sexes). Surveying is usually confined to a large but flat piece of ground, where accurate measuring of lengths is needed both along a diagonal base line and right-angular offsets from it. Angle checks with a theodolite can be employed, too. In addition to this a spring or summer camp in hilly or mountainous country enables a division of the class into groups each day, one engaged in canoeing, one in cooking meals and campsite tidying up, another in surveying mountain tops. The last can be quite exciting, especially for the volunteer who has to climb them to fix poles on them for a surveyor to line them up with a theodolite. 89
Fig. 29
90
If a peak is in the same vertical plane as two points of the same altitude on level ground, three measurements are needed to calculate the height of the peak above that ground—the angle of elevation from each point to the peak and the distance between the points being known (see drawing and calculation below) Suppose the angles of elevation are 41.4° and 43.6° and the distance AB is 40 meters. Let the height required be h meters.
Mountain Side
Fig. 30
h h AX = and BX = tan 41.4 tan 43.6 h h So 40 = AX – BX = – tan41.4 tan 43.6 40 tan 41.4 x tan 43.6 Hence h = tan 43.6 – tan 41.4 33.58 = = 475 m to 3 s.f. .07067 However, if the mountain top, A and B are not in the same vertical plane, the problem is more difficult.
91
Fig. 31
Find M, the mid-point of AB. Let the 3 elevations from A, M and B be α, γ, and β. AMBX is a horizontal plane. Since AM = MB, Apollonius’ theorem applies (this requires proof); i.e., AX2 + BX2 = 2MX2+ 2AM2 So
h2 h2 2 h2 + 2 = 2 tan α tan β tan2 γ +2d2
Hence So
h2 tan2 γ tan2 γ + h2 tan2 α tan2 γ = 2 h2 tan2 α tan2 β + 2 d2 tan2 α tan2 β tan2γ
2d2 tan2 β tan2 α tan2 γ h2 = 2 tan β tan2 γ + tan2 α tan2 γ – 2 tan2 α tan2 β
If d = 200 ft, α = 23°, β = 23.1°, and γ = 23.07° 475.7 h2 = = 3,531,000 .0001282 h = 1930 ft to 3 s.f.
92
Additional problems for the class to solve in school or at home: 1. The pilot of a plane flying horizontally at 1000 ft observes a landing field at a depression of 25.6° straight ahead. A bit later on the same course, the angle of depression is 31.8°. How far has he flown meanwhile? 2. If his steady speed relative to the ground is 600 mph, how many more seconds will elapse before he is directly above the landing field? 3. A guide-rope 20 ft long attaches a boat to the quayside. The angle to the horizontal of the rope when taut is 42.5°. Some time later due to a rising tide this angle has decreased by 35°. By how much has the sea level risen? 4. How much further from the quay is the boat now? 5. Calculate the area of a regular heptagon, each of its 7 sides being 7 cm. Answers among: 5.33, 10.9, 12.72, 184, 474. The solution of problems for scalene triangles
Fig. 32a
Fig. 32b
The Sine Law
In any triangle ABC let AD be the perpendicular from A to a BC. Then since AD = csinB = bsinC, c = b = sinC sinB sinA similarly. 93
This is the Sine Law. But notice that if the angle C is obtuse, AD = csinB = bsin(180 – C), but sin(180 – c) = sinC, e.g., sin150 = sin30 = 0.5. Trigonometrical values for angles greater than 90°
Fig. 33a
Imagine the radius of a circle rotating counterclockwise starting from the horizontal position along Ox. For OA and angle α the base OP of triangle OAP decreases in the positive x direction and the perpendicular PA increases in the positive y direction. For OB and β the perp is still + but the base is – For OC and γ the perp is now – and the base is – For OD and δ the perp is again – but the base is + Regarding the radius itself, the hypotenuse of each triangle is always + and we see that sin α, cos α and tan α are + but sin β is + but cos β and tan β are –, sin γ and cos γ are – but tan γ is +, sin δ and tan γ are – but cos γ is +. Summing up, the sign depends on the type of angle.
94
Angle acute obtuse simply reflex very reflex ___________________________________________________ sin and cosec + + – – cos and sec + – – + tan and cot + – + – The Cosine Law
Fig. 32a (repeated)
Using the same figures labelled Figure (32a), Pythagoras’ theorem gives in the first figure c2 = AD2 + BD2 = b2 – CD2 + (a – CD) 2 = b2 – CD2 + a2 – 2a. CD +CD2 = b2 + a2 – 2 ab cosC In Figure (32b) c2 = AD2 + BD2 = b2 – CD2 + (a + CD) 2 = b2 – CD2 + a2 + 2a. CD + CD2 = b2 + a2 + 2ab cos (180 – C) When C is obtuse, so the Cosine Law is still and similarly and
cos(180 – C) = -cosC, c2 = b2 + a2 – 2ab cosC a2 = c2 + b2 – 2bc cosA b2 = a2 + c2 – 2ac cosB.
95
The extensions of Pythagoras’ theorem and Apollonius’ theorem
T
A
M
B
N
Fig. 33b
In Figure (33a) TAB is any triangle, AM = MB and TN is perpendicular to AB. Using the geometry above in the Cosine Law,
}
Extensions of AT2 = AM2 + MT2 + 2AM, MN Pythagoras’ and BT2 = BM2 + MT2 – 2BM, MN Theorem Add AT2 + BT2 = 2MT2 + 2AM2 (Apollonius’ Theorem)
Practice of the foregoing laws and theorems Examples: (i) In ∆ ABC a = 11 cm, b = 15 cm, B is 55°. Calculate A a sinB 11 sin55 sin A = = = 0.601, so A = 36.9° b 15 (ii) The sides of a triangle are 7, 8 and 9 cm long. What are its angles? 82 + 92 – 72 96 cos α = = 2x8x9 144
96
cos β =
= .667,
so α = 48.2°.
92 + 72 – 82 66 = = .524, 2x9x7 126
so β = 58.4°.
32 72 + 82 – 92 = = .286, so γ = 73.4°. cos γ = 2x7x8 112 (iii) Calculate the median AX in ∆ABC, X being the middle of BC and AB = 12 cm, BC = 14 cm, and CA = 16 cm. so
2AX2 + 2BX2 = AB2 + AC2, 2AX2 = 122 + 162 – 2 x 72 = 302,
so
AX = √ 151 = 12.3 cm.
(iv) In a ∆ ABC the angle at B is 40°, BA = 77.6" CA = 54.2°. and Calculate the angles at C and A
A 77.6"
B
54.2"
C
40 sin C = 77.6 sin = 0.9203, C = 67° or 113°. 54.2 So A = 180 – 40 – 67 = 73° or 180 – 40 – 113 = 27° This is the ambiguous case when given three data of this kind. When dealing with congruent triangles in geometry, one finds oneself dealing with 2 sides and a non-included angle. Problems to be set to Class 10 (in class or for homework) 1. Calculate FG if EF = 83.9 miles, EG = 66.4 miles and angle FEG = 48.7°. 2. Calculate MN if LM = 14 m, angle LMN = 50.8° and angle MLN = 61.1°. 3. In an isosceles triangle XYZ where XY = YZ = 38.4" and XZ = 19.7", calculate the lengths of the medians passing through X and Y. 4. Find all three angles of a triangle whose sides are 7 cm, 11 cm and 16 cm.
97
5. Do the same when the sides are 55 cm, 48 cm and 73 cm. 6. The perpendicular from P to QR meets it in S between Q and R. Calculate the length of SR if PQ = 9'3", QR = 10'6" and RP = 11'9". (Use an extension of Pythagoras’ theorem.) 7. A, B, C and D are four successive points on a regular decagon of side length 6 cm. Calculate the lengths of AC and AD. Answers (all of them, in wrong order): 7.75, 11.4, 13.2, 15.7, 21.3, 23.7, 34.8, 37.1, 41.1, 48.9, 64.0, 90.0, 123.9. If this is for homework, specify one hour’s work. Only the most fascile will manage to finish all the problems and check that their answers are correct.
98
Chapter 6 Class Ten Projective Geometry Projective Geometry developed from the experience of accessible space Many books have been written in the last hundred years developing the subject of Projective Geometry axiomatically from first principles. Some of them present it in a wholly abstract way in which there is no need to include any drawing at all and instead one is left to contemplate rows and columns of letters and numbers. In most books, however, the texts are illustrated with carefully constructed geometrical drawings although the authors emphasize that points, lines and planes are pure concepts. Nowhere in the physical world are there perfectly flat planes or perfectly straight lines. Tiny dots on a sheet of paper are not perfect points since a geometrical point cannot have area or volume. A geometrical drawing is just an approximation that serves to remind us of the actual geometrical figure we are investigating. While endless lines and unbounded planes are fully justified concepts, difficulties arise when dealing with parallel lines and non-intersecting circles or conics. The axiom that any two distinct lines of a plane are on one and only one point does not hold in Euclidean geometry. Veblen and Young overcome the difficulty in their standard work on projective geometry (first printed in 1910) by attributing to two parallel lines a point of intersection, adding that such a point is not a point in the ordinary sense; it is to be regarded as an ideal point, which we suppose two parallel lines to have in common. They go on to say that this amounts merely to a change in the ordinary terminology, and they call it a “point at infinity” and suppose one such point to exist on every line. Such a procedure is conceptually very unsatisfactory. 99
In his Principles of Geometry (1922), H.F. Baker uses the concept of “postulated points” in an attempt to remove the unsatisfactory supposition. In what follows we need to distinguish between points represented on the drawing and those not necessarily on it by calling the latter “inaccessible” points. Some of the subsequent development is not unlike Baker’s treatment, but goes in a new direction through examining what we really understand by the word “point” and the words “line” and “plane.” A further examination then yields a more satisfactory understanding of “imaginary points.” For the impulse to develop this new direction, I am indebted to the philosopher Rudolf Steiner (1861–1925) who took a similar direction in his description of the nature of the human being, quite different though the subject was. See his books and lectures on Anthroposophy, i.e., the Science of the Spirit. The assumptions of Projective Geometry for accessible space 1. The elements of geometry are points, (straight) lines and (flat) planes. 2. To start with each element has a position in three-dimensional space, i.e., is accessible, but the full extent of each line and plane is unlimited.1 3. Two distinct points determine a unique line. A and B determine line l.
Fig. 34 1. There is a special plane called the infinite plane which has a wholly immaterial existence and has infinite points and lines that lie within it. Every materialist has to deny existence to the infinitive plane, despite its enormous value in projective geometry. Geometrical materialists make up other words to justify its use but fail to ascribe scientific meaning to them.
100
4. A line together with a point not on it determine a unique plane, e.g., l and P determine plane α by joining P to all points on l, all such lines lying in α. 5. When a point lies on each of two distinct planes, there is determined a unique line through the point and lying in both planes, e.g., imagine 2 set squares PQR and PXY touching each other at P and whose edges QR and XY lie in a plane γ. Let the lines QR and XY intersect at N. The line m = PN is the required unique line. 6. Peano’s axiom: If ABC denotes any triangle of points and D lies between A and C, then if AB is extended to any point E, the line DE will intersect BC in an accessible point F lying between B and C. Problem Given a plane containing 2 lines l and m, which only intersect in an inaccessible point J (off the paper) and also two accessible points, one (F) on the plane and one (P) in space above that plane (P), construct the unique lines passing through F and P and the inaccessible J.
Fig. 35
101
Solution i. Take any two points A and B on l and any two points C and D on m. Cross-join lines AD and BC giving point E. (Had D been placed left of C, E might have proved inaccessible, so avoid this possibility.) Extend AF to a point G. FD must intersect GE in accessible H (use Peano’s axiom on triangle AEG). CH must intersect GB in accessible I (use Peano’s axiom on triangle BEG). Then FI is the line required (n). ii. For the dashed lines in space, extend FP to any point Q. AP produced must intersect GQ in accessible R (use Peano’s axiom on triangle FGQ). RE must intersect PD in accessible S (Use Peano’s axiom on triangle ADP). CS must intersect BR in accessible T (use Peanos’ axiom on triangle BER). Then PT is the line required (k). Proof QS must intersect GE (Peano with triangle EGR) and must intersect DF (Peano with triangle DFP). But QS can meet the base plane only once, so it must pass through H. Similarly CS must intersect QI (in T).
PD meets TC in S, so k and m are coplanar lines AP meets BT in R, so k and l are coplanar lines Hence k must contain the inaccessible J. FP meets IT in Q , so k and n are coplanar lines. Hence n, too, must contain J.
Of course, if the lines l, m, n and k were extended onto an adjoining sheet of paper or if l and m were made to intersect in the drawing already, we should immediately recognize the concurrency of all four lines. But drawing verification does not constitute a proof. Instead of denoting points by capital letters, we could denote the five points PQRST by the single numbers 12345 and the tenpoints on the plane by placing together the two numbers of the spatial points which lead to each of them. This shows a beautiful harmonious relationship as follows, whether or not all the points are accessible.
102
Golden-yellow light
Purple light
Black object – 3, 4, 5 Reddish-blue shadow – 13, 14, 15 Yellowish-green shadow – 23, 24, 25
Rest of shadow planes, e.g., – reddish
Fig. 36
The simplest way to draw this is to freely place first the points 1, 12, 23, 34 and 45; then on the lines obtained points 2, 13, 24, 35; then 14, 25 and 15; finally 3, 4 and 5. This drawing also illustrates the phenomenon of colored shadows. Imagine bright light at point 1 (red) and point 2 (orange; also an opaque triangle whose corners are at 3, 4 and 5. The triangle 13, 14 and 15 will then appear green (complementary to red) and the triangle 23, 24 and 25 blue (complementary to orange). The 10 lines of the base plane each of which passes through three of the ten points in it, is called the Desargues Configuration. Each point also has three of the lines passing through it. Desargues’ Theorem states that if two triangles ABC and PQR are such that AP, BQ and CR are concurrent in a point (called a pole), then corresponding sides of the triangles will meet in three collinear points, their common line known as the axis. The converse of this theorem is also true, i.e., if corresponding sides meet in collinear points (the axis), then AP, BQ and CR will be concurrent. These theorems are true in both 2 and 3 dimensions. The easiest case is in 3 dimensions. Imagine a tetrahedral block of cheese having one face on a table, and imagine a hungry knight approaching it with a sharp sword. With a mighty swipe
103
he cuts off a sloping tetrahedral slice. Unfortunately the heavy sword sweeps on and slices through the table too; the whole sweep describes a plane.
Fig. 37
Triangles PQR and ABC have a common pole O. X, Y and Z lie on the common axis. This is a straight line because of the plane cut made by the sword. Thus the 2 triangles ABC and PQR are both coaxial and copolar. The last drawing can now be extended to a set of exercises for Class 10 to undertake on a large scale. Instead of having just two triangles which are both copolar and coaxial (the short words for this are triangles in perspective), we can have a whole (endless) set of such triangles in sequence by placing the point of a new triangle upon the opposite side of the previous triangle and in line with the pole. The triangles could be colored in colors adjacent on the color wheel. Besides the triangles below the axis, another series can be drawn below it and continued above the pole O. To begin, choose any point A and join it to Y and Z, two points on the axis. Their intersections with fixed lines r and q are B and C. BC will give the third point X on the axis. Draw the third fixed line through O, i.e., p = OA. Notice how the triangles close up to both the pole and the axis and decrease in area. There is no limit to the number of triangles
104
in these sequences. The title can be “A HOLOLOGY Homology Sequence of Per spective Triangles.” The shapes within the fixed triang le between q, r and the axis remind one of a leaf metamorphosis, born very small near O, reaching maturity in ABC, and finally flattening and dying as it reaches the axis. When the pole is on the axis, we have a special case. The Fig. 38 homology has become an elation. To begin with the procedure is the same, starting with any triangle ABC if their sides pass through X, Y and Z. Above A the top corners are labelled 1, 2, 3, 4,…. The lines through them joined to X and Y create a mesh whose intersecting points lie on lines through the pole. Transfer the number labels to lines through Y, then to points on p. If we imagine the axis as a horizon and the Sun just appearing over it at the pole, it is as if a sequence of clouds were lit up and somehow causing reflections or shadows below. In this way the term “elation” has arisen. But the lower triangles are simply a continuation of the top ones which have expanded through space and reappeared below; four of them are off the paper. The title is “An Elation Sequence of Perspective Triangles.” The nature of the “Infinite Plane” and its points and lines Having constructed the two previous drawings, Class 10 students are ready to ask about how triangles can disappear in one direction and reappear in the opposite direction. This naturally
105
leads to questions about parallel lines. Euclid defined parallel lines as coplanar lines which do not meet. Some unfortunate teachers proceed to draw a pair of parallel lines, say 10 cm apart, on the blackboard and, appealing to perspective, say: “Can ELATION you see that if these lines were continued Fig. 39 indefinitely, they would reach a vanishing point in the infinite? We call it a ‘point of infinity.’ ” Several or many students will justifiably reply “No!” For they argue that as the lines are 10 cm apart on the board, they will continue to be 10 cm apart no matter how far they continue—for one mile or billions of miles. They feel that Euclid’s definition is fully justified. What helps here are the previously used concepts of accessible and inaccessible points. While many points lying outside the drawing paper cease to remain inaccessible, once a much larger sheet of paper is used so that they have a definite place or position in space and their distance from some accessible point on the paper can be measured, we have proved that two coplanar lines always determine a unique line through every accessible point in space. This is just as true of parallel lines as any other pair of lines. Hence any given accessible point taken together with either a second such point or a pair of coplanar lines will determine a unique line. This ability of either the second accessible point of the pair of coplanar lines is called a point function. The problem of parallel lines is solved by asking a fundamental question: What is a point? It has two distinct attributes in most cases: (i) it has a position in space (and so is subject to distance measurement from other points), and (ii) it has a function (or more precisely a “directing function,” e.g., an accessible point together with a direction in space determine a unique line).
106
Parallel lines determine a point which has a directing function but has no position in space. Only by bearing this statement in mind can one describe the entity that parallel lines determine, is it justified to use the term “infinite point.” “Point at infinity” is unsatisfactory, because the preposition “at” is meaningless here. It is worth comparing the above geometrical conundrum with the problem of death at the end of one’s life. Is it really true that death is the absolute end of human existence? There are countless instances where someone who has “died” has influenced the actions of bereaved friends. Here is a personal example of this. I used to work as a teacher in a school I deeply loved—and my colleagues, the children and students in it. But a time came when the oldest students were unable to continue their education there beyond their 17th birthday, mostly for financial reasons and the need to find employment in the world. I was offered a teaching post in another school which had full classes of 17- and 18-yearold students. I knew that my talents would find better application in this other school, but I could not decide what to do. Could I really leave the school I loved? One dark, foggy night I woke up at 2 a.m. to discover that a revered old lady who had died exactly a year earlier was standing beside my bed. I found myself speaking yet I knew that, while it was my voice, the words were hers. “If you move, others too can move.” I knew then that my heart-rending problem was solved and later when I had changed schools, I soon discovered that other people in consequence were also able to move in their careers. While it is true that after death we no longer have a physical body—just as an infinite point has no spatial position—there is another part of us (especially our individuality) which not only exists but can be effective in helping others—just as a point function can determine a new line through any accessible point. The assumptions of Projective Geometry for the whole of space With the inclusion of infinite points as directing functions, the previously stated 8 axioms for accessible space become much simpler. Two infinite points determine an infinite line and two infinite lines determine the unique infinite plane. 107
1. The elements of geometry are points, lines and planes. 2. Not all elements have position in space, but all elements do have directing functions, and all lines and planes are unlimited in extent. 3. Two distinct points or planes determine a unique line. 4. A line and a point not on it/plane not through it determine a unique plane/point. A special case arises in 4 when both the line and the point have no position in space. The plane determined is the infinite plane. Further axioms will be added in Class 11 geometry when the concept of “ordering function” is developed in connection with conics. We can now complete the list of special homologies and elations arising from the two previous drawings. With only brief indications these tasks can be set to the students, who are to be asked what kind of triangles each sequence produces.
Pole O
AFFINITY OF STRETCH
axis
SIMPLE SHEAR
Pole axis
108
Fig. 40
Pole
ENLARGEMENT or DILATION
Fig. 41 Pole Pole
Pole
Fig. 42
TRANSLATION Pole
Pole
Pole
axis
LINE REFLECTION
Fig. 43
109
Enlargement (dilation) – a homology with the axis an infinite line Similar (equiangular) triangles Translation – an elation with both pole and axis infinite elements Congruent triangles Finally we have a Line Reflection where the pole’s directing function is perpendicular to the axis—an affinity of period 2. There is no endless succession of triangles, but ABC becomes PQR, which becomes ABC again. Students can also be asked to construct a Point Reflection —an enlargement of period 2, where the axis is the infinite line and AB and PQ , AC and PR, BC and QR are 3 pairs of parallel lines, i.e., each pair determines an infinite point on the infinite line. Summing up, the sequences of perspective triangles are related as follows: HOMOLOGY
on
th
s
xi
ea
th
is
Ax
e
on
Pole on the infinite
l Po
e
nit
nfi ei
Affinity Enlargement (Dilation) (stretch) Pole on the axis
Elation
e on s
Ax
xi
is
ea
on
th
th e
in fi
l Po
ni
te
Pole on the axis
finite Axis on the in
Simple Shear
Translation Fig. 44
110
Relationships between the sequences of points and the 4 progressions of numbers: Simple Shear and Translation Enlargement and Affinity Elation Homology
– – – –
Arithmetic Progression Geometric Progression Harmonic Progression Musical Progression
These relationships can be established easily by ordinary Euclidean Geometry. Two different mathematical disciplines introduced in Class 10—that of number sequences in arithmetic and of perspective triangle sequences in geometry—have thus been closely linked. It is through such complementary results that the wholeness of mathematics, not only its quantitative side but its qualitative side too, will appeal to the students of Class 10 age. It is when a wider wholeness can be established among apparently widely different subjects by means of further linking that education will help achieve the social cohesiveness that humanity longs for.
111
Chapter 7 The Transition from the First Half to the Second Half of the High School: Logarithms, Spirals and Progressions As the class enters the third year of the upper school, the students sense the first beginnings of adulthood. Childhood is now well behind and we need to address them as young people as they approach and pass the age of 17 in Class 11. The curriculum in the sciences echoes their developing stage of life and needs. In Class 9 they learned about human physiology and heat engines. A poem composed for “The Ladies’ Diary” in 1725 illustrates the connection between the steam engine and the human body. It is called “The Prize Ænigma” by Henry Beighton. I sprang like Pallas from a fruitful brain About the time of Charles the Second’s reign. My father had numerous progeny And therefore took but little care of me. A hundred children issued from his pate; The number of my birth was sixty eight a My father dead, myself but few did see, Until a warlike man adopted me,b Destroyed what records might disclose my birth, Said he begot me and proclaimed my worth.
112
Begetting me he called a chance— A task easy to him, assisted by a flask.c He then to me strange education gave, Scorched me with heat and cooled me with a wave; More work expected from my single force Than ever was performed by man or horse. To mend my shape he oft deformed it more, Which sometimes made me burst and fret and roar. My heart has ventricles and twice three valves,d Though but one ventricle when made by halves. My vena cava, from my further ends Sucks in what upward my great artery sends. The ventricles receive my pallid blood Alternate and alternate yield the flood. By Vulcan’s art my ample belly’s made; My belly gives the chyle by which I’m fed, From Neptune brought, prepared by Vulcan’s aid. My father (I mean he who claimed by birth) My dwelling fixed in caverns of the earth. And there, he said, I should in strength excel, But there, alas! I was but seldom well. In this sad state, to languish I began, Until a doctor sage, now coming in,e Condemned the methods that were used before And said that I in caves should dwell no more. Then I should dwell in free and open air And gain new vigour from the atmosphere. A house for me he built, did orders give,f
113
I should no weight above my strength receive These gentle means my shape have altered quite. I’m now increased in strength and bulk and height. I can now raise my hand above my head And now at last I by myself am fed. On mighty arms alternately I bear Prodigious weights of water and of air And yet you’ll stop my motion with a hair g He that can find me should rewarded be, By having, from my masters liberty, Whene’er he pleases to make use of me. NOTES: a. a patent number of the steam engine b. Captain Savery c. He claimed that the result of throwing a half-filled long flask of wine into the blazing hearth at a local inn showed him the relationship of steam pressure to atmospheric pressure d. the Kensington engine built by Savery in 1712 e. a dreadful pun on Newcomen, inventor of the vacuum engine (Cornwall) in 1720 f. a beam engine to drain water out of a tin mine g. if a slight gap was present between piston and cylinder Children in the ninth grade usually enjoy learning and reciting this poem. Such steam engines were stationary. The later invention of mobile steam railway engines, beloved of schoolboys, were often compared with mighty beasts—panthers, horses wearing blinkers, oxen (for goods trains)—all full of life, not like modern diesel engines.
114
Class 9 students go on to learn the working of petrol and diesel engines, then jet engines in airplanes and finally rocket propulsion. Calculations of latent heat and entropy belong here and considerations of a final heat death of the Earth in a distant future. Class 10 physics investigates the application of electricity and magnetism to motive force in electric trains and hydro-electric power, a big contrast in the value of electric current used in telephones which were constructed in Class 9. The relationships between metals and their thermal conductivity and specific heat and the planets and time rhythm are also worked at. While Class 9 was concerned with worldwide mountains and species of rocks in the Geography main lesson, Class 10 learns about tides, river systems and ocean currents, fish and again the rhythms of time affecting the planetary system and the living Earth as follows:
Fig. 45
115
Linear Thermal Conductivity Order Silver (highest) Mercury (not solid) Copper Gold Iron Tin Lead (lowest) Specific Heat Order Iron (highest) Copper Silver Tin Gold Mercury Lead (lowest) While official state examinations cannot count as real education in Rudolf Steiner Waldorf schools, by the time Class 11 is reached, it does no harm for capable students who can take it in their stride to pass a few GCSE-level subjects, say, in English language and literature, mathematics and foreign languages. For it is the development of their intellect and reason which is a prime aim at this age. In each main lesson a strong analytical approach is required. This is obviously so in Differential Calculus and the atomic theory in Physics, but also is needed to understand the works of great musical composers and participate in dramatic performances of Shakespeare. Balancing the strong thinking needed in Class 11 is the story of the Quest of the Holy Grail, which arouses deep feeling in the class as the young people read
116
and prepare for conversation works like Eschenbach’s Parzival, Mallory’s Le Morte d’Arthur, and perhaps the Mabinogion and Wagner’s “Ring” and “Parsifal” operas. In Class 12 the analytical approach gives way to its opposite. Reason continues its development and the members of the class who have experienced what their teachers led them to learn in younger classes want to build a synthesis of all this. Main lesson periods in which well known modern writers are studied and views of political and economic problems are expressed, the place of real art in social life can be discussed. With the help of guest presenters, the various religions, not only Christian, can be assessed by our young people. The essential thing is to leave them free to take whatever course they decide upon on leaving school. Of course they also want to know who Rudolf Steiner was and gain some idea of what is meant by his Science of the Spirit. Subject matter in main lessons includes applications of projective geometry to the living world, a thorough grasp of the Phenomenon of Color via practical painting and the theories and discourses of Newton, Goethe, Ostwald, etc. In mathematics as will be seen hereafter, Integral Calculus and Chaos Theory are introduced. Moreover a choice of subjects to try out should be offered to the top class in the school. Some students will want to become expert on computers, others want help with practical skills like cookery, dressmaking, or carpentry. It is also excellent when the formation of a class orchestra is possible. “How can we contribute to a richer social life among our contemporaries, and where can we find the best means of helping others with their social problems?” is frequently just below the surface of Class 12 students in our schools. How different the young people are leaving Class 12 to go into the wide world from the young adolescents who entered Class 9! The transitions in successive years of the individuals experiencing our whole curriculum, especially the changes from the first half to the second half of the 4 years in the upper school, can become a blessing to them.
117
❖
Chapter 8 Class Eleven Differential Calculus (Age 16 to 17) Consider the relationship between x and y if y = x3. When x = 1, y = 1 [position A] When x = 3, y = 27 [position B] As x increases in value, so does y but more quickly. For example, from A to B the increase in x = 2 and the increase in y = 26. These differences are denoted by using the Greek letter δ; thus δx = 2 and δy = 26. δy 26 The ratio of increases = = = 13. δx 2 Taking a position C where x = 2, y = 8, then from A to C 8–1 δy = = 7. δx 2–1 Taking a position D where x = 1.5, y = 3.375, then from A to D
δy 3.375–1 = = δx 1.5–1
2.375 = 4.75 .5
The nearer the position is to A, the smaller the ratio. But y = x3, so and so
118
y + δy = (x + δx) 3 = x3 + 3x2δx + 3xδx2 + δx3 δy = 3x2δx + 3xδx2 + δx3 δy = 3x2 + 3xδx + δx2 δx
δy As δx, δy 0, and δx 0 0 a meaningless fraction, over which Rudolf Steiner suggested that the class should debate at length its numerical meaning,
δy dy (the limit of ) dx δx
= 3x2,
as the remaining terms are each zero. dy = 3 x 12 = 3 At A, x = 1. So dx dy is called the differential coefficient. dx While at A the ratio is 3, at and at and at
B it is 3 x 32 = 27 C it is 3 x 22 = 12 D it is 3 x 1.52 = 9.75.
Exercises: 1. Find the differential coefficient if (i) y = 5x4, (ii) y = 2x2 – 3x dy 2. Calculate when x = 1/3 if y = 6x2 – x3 dx 3. If y = 2x3 – 6¹/² x2 + 6, find for which values of x the differential coefficient is zero. Answers (numerical): 32/3 , 0, 21/6 . dy The differential coefficient dx shows the rate at which y is increasing in comparison to x. d2y If we differentiate a second time—denoted by 2 dx
119
—we find the rate at which the rate of increase is increasing. For example, suppose the distance travelled by a car is s feet in t seconds and the relationship is 3t4 s = 2 The speed achieved after 2 seconds is ds 3 3 = 4t3 x = 4 x 23 x dt 2 2 The acceleration after 2 seconds is
= 48 ft per second.
d2s 3 2 3 = 4 x 3 x 22 x = 72 ft per second per second. 2 = 4 x 3t x dt 2 2 In the exercise 1(ii) above y = 2x2 – 3x. dy d 2y So = 4x –3 and 2 = 4. dx dx It is easy to see, using the binomial theorem, that if y = kxn, dy d2y d3y = kn xn-1, 2 = kn(n – 1) xn-2, 3 = kn(n – 1)(n – 2) xn-3, etc. dx dx dx Fractional or decimal powers The procedure of reducing the power by 1 is unchanged. If y = x3 1/3 , dy/dx = 31/3 x2 1/3 If y = x0.7 , dy/dx = 0.7x -0.3 If y = x-2.4 , dy/dx = -2.4 x-3.4 Differentiations of a product or ratio of variable expressions Let y = uv where u = (2x+3) and v = (3x – 1) y = (2x+3)(3x – 1) = 6x2 + 7x – 3 y + δy = 6 (x+ δx)2 + 7(x+ δx) – 3 δy = 12x δx + 6δx2 + 7δx
120
δy = 12x + 6δx + 7 δx
But
dy = 12x + 7 dx du dv dv du = 2 and = 3 and u + v dx dx dx dx
=
3(2u+3) + 2(3u – 1) = 12x + 7
du dy dv =u +v . Hence dx dx dx u Let y = = u x v-1 v dy = u(-v-2 dv ) + du (v-1) Then dx dx dx du dv So v –u dy dx dx = dx v2 Numerical examples of the above are: (i.) y = (2x2 – 3x + 5)(3x2 + x + 1) for use with product uv.
dy 2 + 1) + (3x2 + x + 1)(4x – 3) dx = (2x 3– 3x + 5)(6x = 12x – 16x2 + 27x + 5 + 12x3 – 5x2 + x – 3 = 24x3 – 21x2 + 28x + 2
d2y/dx2 = 72x2 – 42x + 28
(ii.) y = x+1 / 3x+1 for use with ratio u/v dy/dx = 3x + 1 – (x+1) 3 / (3x+1)2 = -2 / (3x + 1)2 d2y/dx2 = 12 / (3x + 1)3 Examples for the class: 1. Differentiate (i) y = 2x5 – 3x3 + 8x (ii) y = (6x – 1)(4x2 – 3x + 2) (iii) y = 5x – 3 / 5x + 3
121
2. Find the rate of change of y, when x = 1 in the relationship y = (2x – 1)2 3. From a cliff top a stone is thrown down with a speed of 12 ft/sec. The distance s ft fallen after t seconds is given by s = 12t + 16t2. Find the formula for ds / dt and the time taken for the vertical distance dropped to reach 180 feet. 4. Calculate the values of y and dy/dx in each part of question 1 (6 answers) when x = 1. Answers to numerical questions: ¼, 61, 3, 7, 9, 15, 43. Exponential and logarithmic functions The word exponent means power. For example 23 shows the third exponent or power of 2 and means 2 x 2 x 2 = 8. 23 x 24 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 27, which is the same as saying 8 x 16 = 128. We now define the strange symbol exp(x) to be
1 + x + x2 / 2! + x3 / 3! + x4 / 4! … + xn / n! +…
where n → ∞, i.e., a series of additions including whole number powers of x divided by the factorials of the whole numbers, which were met in Class 9, e.g.:
3! = 3 x 2 x 1 = 6, and n! = n(n – 1)(n – 2)… x 2 x 1.
What exp(x) means as a number will emerge below, but if x is put equal to 1, we find that
exp(1) = 1 + 1 + ¹/² + 1/6 + 1/24 + 1/120 + 1/720 + …
Putting these fractions into decimal and adding them one by one, we find the answer increasing as follows:
122
1 2 2 2 2 2 2
• • • • •
5 666,666,667 708,333,334 716,666,667 718,055,556
The last decimal so far was obtained by adding 1/720. After continuing until 1/13! has been added, the nine decimal places answer does not alter at 2.718,281,829. Now exp(x) x exp(y) = (1 + x + x2 / 2 + x3 / 6 + (1 + y + y2 / 2 + y3 / 6 + …) = 1 + x + y + x2 / 2 + xy + y2 / 2 + x3 / 6 + x2y / 2 + xy2 / 2 + y3 / 6 + …
= (1 + (x+y) + (x+y)2 / 2 + (x+y)3 / 6 + …
= exp (x + y)
Hence exp(x) means a power of x and exp(y) a power of y. Since exp1 = the number 2.718…., which we can denote by e, exp(x) = ex and ea x eb = ea+b Definition (Read this as x = log of a to the base e) If ex = a, then x = loge a and if ey = b, then y = loge b. So ab = ex x ey = ex+y. Hence loge (ab) = log ex+y = loge a + loge b ex and loge x are inverse functions. Plus and times (+ and x) become interchanged when one replaces powers of e by logs to the base e. Thus: ea x eb = e(a+b) and loge a + loge b = loge ab.
123
Before the advent of calculating machines and computers, multiplication of complicated numbers was effected by the second of the above relationships using logarithm tables, eg: Number Logarithms to base 10 3.476 0.5411 8.921 0.9504 (log) 1.4915 antilogarithm 31.01 = 101.4915
Logarithms to base e 1.2459 2.1184 (ln) 3.4343 31.01 = e 3.4343
Base 10 logs were used, but the same answer is, of course, found by base e logs, shown on hand calculators or computers as ln. ln means natural log, whereas log means common log, ie: base 10 log. Differentiation of powers and logarithms When the series of terms for ex is differentiated, the result is the same as the original series. So if y = ex , dy/dx = ex = 1 + x + x2/2 + x3/3! + …. , i.e., dy/dx = y = ex . Hence dx / dy = 1 / y But log ex has been defined by noticing that if x and y are interchanged, from x = log ey, we get dx / dy = 1 / y. So swapping x and y we see that dy/dx = 1/x when y = log ex If y = a x
dy/dx If y x so dx/dy
= (e ln a)x = e x ln a , = e x ln a . ln a -----*1 = ln a.ax = log a x , =ay = ln a.ay = ln a.x
Hence dy/dx = 1/ln a.x
124
Function of a function An example of the differentiation of this has already been shown in *1 above. y = f (g (x)) , When or
dy = f 1 (g (x)) x g 1(x) dx d f g(x) dg(x) x dx dx
Example: If y
= √ x2 – 3x + 4,
dy 1 = dx √ x2 – 3x + 4
x (2x – 3)
2x – 3 = 2 √ x – 3x + 4
Further development of trigonometric functions C
B
A
N
Fig. 46
Let point A lie at unit length from a line BC and CÂN = α, BÂN = β. Hence BÂC = α – β, BĈA = 90 – α. Also BN = tan β, BA = 1/cosβ and BC = tan α – tan β.
125
Fig. 47
Using the Sine Law on triangle ABC sin (α – β) sin (90 – α) = tan α – tan β 1 cos β = cos α cos β, since
sin (90 – α) = cos α = x in Figure (47).
Hence sin
(α – β) = cos α cos β (tan α – tan β) sin (α – β) = sin α cos β – sin β cos α
*1, since base cos x tan = = perp = sin. hyp x perp hyp base Put β = -β *2 sin (α + β) = sin α cos β + sin β cos α Also in *1 put α = 90 – α. Sin (90 – α – β) = sin (90 – α) cos β – sin β cos (90 – α) *3
Hence cos(α + β) = cos α cos β – sin β sin α
In *3 put β = -β. *4 cos (α – β) = cos α cos β + sin β sin α *5 If we put β = α in *2, sin 2 α = 2 sin α cos α *6 Doing so in *3 we get cos 2 α = cos 2 α – sin 2 α
126
*7 Using the fact that cos 2α + sin 2α = *8
base2 hyp + perp2 = hyp2 2
base2 + perp2 = hyp2 = 1, hyp2 hyp2 the result *6 can also be written
cos 2α = 2 cos2α – 1
or cos 2α = 1 – 2 sin2α tan (α + β) = sin (α + β) cos (α + β)
= sin α cos β + sin β cos α cos α cos β – sin α sin β
Divide top and bottom by cosα cosβ. *9 So tan (α + β) = tan α + tan β 1 – tan α tan β *10 and tan 2α = 2 tan α 1 – tan2 α
Also since
1 = sin2α + cos2α,
*11 dividing by cos2x, sec2 α = tan2α + 1 *12 and dividing by sin2x, cosec2 α = 1 + cot2 α Differentiation of trigonometric functions If y = sin x, y + δy = sin (x + δx) = sin x, cos δx + sin δx cos x δy = sin x cos δx + sin δx cos x – sin (x + δx) δx δx δx δx When δx → 0, δy = sin x + sin δx δx δx δx
cos x – sin x δx
127
So dy = cos x, dx since, observing the shrinking isosceles triangle OAP, y P
0
N A
x Fig. 48
cos δx = ON/OP → 1, sin δx = PN/OP → 0; sin δx/δx = PN/PA → 1 If y = cos x, y + δy = cos (x + δx) = cos x cos δx – sin x sin δx by *3 δy/δx = cos x cos δx/δx – sin x sin δx/δx – cos (x + δx)/δx When δx → 0, dy/dx = cos x /δx – sin x – cos x/δx = -sin x If y = tan x = sin x/cos x , dy/dx = “v du/ds – u dv/dx / v2” = cos2 x + sin2 x/cos2 x = 1/cos2 x = sec2 x. Exercises: 1. Find formulae for cos 3x and sin 3x in terms of cos x and sin x. Check each formula when x = 20°. 2. Find and check the formula for tan 3x, using x = 40° for the check. 3. Differentiate sin 3x and tan 4x3 and calculate the first answer for x = 1.1 radians (not degrees). 4. Show that sin (α + β) – sin (α – β ) = 2sin βsin (90 – α ) 5. If y = tan x, calculate y, dy and d2y when x = 50°. dx dx2 6. If tan 7x = 7, calculate x in both degrees and radians; there are two answers in degrees between 0 and 360 and two answers in radians. 128
7. Given that sin2θ = 2 – 3cosθ, which angle of θ in degrees between 0 and 90 satisfies this equation? Answers: -2.96, -1.73, .204, .5, 866, 1.19, 1.83, 2.42, 3.35, 5.77, 11.7, 67.5, and 191.7. Application of differential calculus to practical problems The varying character of curves lends itself to accurate examination by the powers of calculus. Questions of curvature and the positions of tangents at every point can be quickly resolved in coordinate geometry. Problems in dynamics concerning acceleration, velocity, distance and time can be resolved quickly by differential calculus and by its companion integral calculus, which will be developed in Class 12. For the present we will just notice that if t = time taken, x = distance covered, s = speed and α = acceleration, then the rate at which these elements change is given by s = dx/ dt and α = d2s/dt2. Moreover a graph plotting the relationship between speed (velocity) and time gives a picture which includes acceleration or deceleration and the distance covered.
Fig. 49
The curve a shows increasing acceleration, the line b shows constant speed and curve c shows decreasing retardation. As will be understood in Class 12, the enclosed area in Figure (49) is equal to the distance covered in the time interval from t0to t1.
129
Example 1 Consider the graph of y = x4 – x2. The table of calculations will include: ±x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.4
1.5
x4
0
0
0
.01
.03
.06
.13
.24
.41
.66
1 1.46 2.07 2.86 3.84
5.06
x
0 0.01 0.04 0.09 0.16 0.25
0.36
0.49
0.64
0.81
1 1.21 1.44 1.69 1.96
2.25
y
0 -0.01 -0.04 -0.08 -0.15 -0.19 -0.23
-0.25 -0.23
-0.15
0 0.25 0.63 1.17 1.88
3.81
2
1.1
1.2
1.3
Also dy/dx = 4x3 – 2x and d2y / dx2 = 12x2 – 2. So, continuing the table, +x 4x3
0 0
.03 0.11 0.26
2x
0 0.2 0.4
dy/dx
0 -0.20 -0.37 -0.49 -0.54
0.6
0.8
0.5
0.86 1.37
2.05
2.92
4 5.32 6.91 8.79 10.98 13.5
1
1.2
1.6
1.8
2 2.2
0.45
1.12
2 3.12 4.51 6.19
-0.5
1.4
-0.49 -0.03
2.4
2.6
2.8
3
8.18
10.5
-x A LL CHANGE SIGNS TO MINUS
±x 12x
0
dy dx2
-2 -1.88 -1.52 -2.92 -.08
2
2
.12
.48 1.08 1.92
Furthermore when
3
4.32
5.88 7.68
3.72
12 14.52 1 7.3 20.3
23.5
27
1
2.32
3.98
9.72
10
21.5
25
d2y = 0, dx2
5.68
12.52 15.3
18.3
12x2 = 2,
n = ± √ 1 = ± 0.41 6 4 2 Then y = x – x = .028 – .038 = -.01 and dy = 4x3 – 2x = -.54 dx At this moment as one views the curve going from left to right, the direction of turning changes from counterclockwise to clockwise when x = -0.41 and at n = +0.41 back to counterclockwise again. At both points the curve is momentarily straight; they are F1 and F2, points of inflection. The tangents there are y + .01= ± 0.54 (x ± 0.41), i.e., y = ± 0.54x + 0.22 – .01, i.e., y = 0.54x + 0.21 and y = -0.54x + 0.21 (lines l1 and l2). 130
The tangent at any other point is easy to have its equation determined, e.g., at P (1.3, 1.17) it is y – 1.17 = 6.19(x – 1.3), i.e., y = 6.19x – 6.78 (line m). y = -0.25 is the only line touching the curve twice; it is called a double tangent.
Fig. 50
Example 2 A light jet plane takes 40 minutes between takeoff and landing on a journey of over 500 miles. Its ground speed at takeoff is 80 feet per second, after which it accelerates at 10 feet per second per second for 2 minutes. It then levels off and flies at a constant cruising speed for half an hour. Finally it slows down for 40 minutes, descending at constant retardation until it has landed. Acceleration and retardation are each calculated as the averages of initial and final speeds. Figure (51) shows the velocity/time graph for the journey. Velocity in feet per second
B
C
A D Time in minutes
Fig. 51
131
Calculate the cruising speed in ft/sec and mph, the total distance flown in miles to 3 s.f., and the retardation in ft/sec/sec during the descent. Solution Cruising speed = 80 + 120 x 10 = 1280 ft/sec = 1280 x 60/88 = 873 mph Distance flown = the sum of the 3 areas shown = (80 + 1280) x 120 / 2 x 5280 = 15 + 436 + 58 = 509 miles Retardation = 1280 – 0 / 2 x 480 = 1.33 ft/sec/sec Example 3 Construct a table of calculations for drawing the ellipse whose equation is y = ± √(1 – x2 ) . 9 Then draw it accurately. Also work out a simplified expression for dy . dx Find the equations of the tangents to the ellipse at the points above the x-axis when x = -2 and x = 1 and calculate the height above the x-axis where each of them crosses the y-axis. Solution dy = (¹/2 1) x 2x x or x2 = 2 2 dx √(1– x ) 9 3 √(9 – x ) 9√(1– x 2) 9 9
132
Tangent at A is y – 0739 = +.301 (x + 2) i.e., y = .301x + .602 + .739 i.e., y = .301x + 1.341 Height = 1.34 Tangent at B is y –.943 + -.118(x – 1) i.e., y = -.118x + 1.06 Height = 1.06 Exercises: 1. A helicopter flying at 1100 ft is called to a place 7 miles from the point O vertically below. At once it drops down towards its goal keeping to a course whose equation is y = 100(11 + x2) 1 + x2 , y being measured in feet and x in miles. Take 1 cm per 100 ft from the ground on the y-axis and 1 cm per mile on the x-axis. Construct the curve, showing a table of values for each mile, including for half a mile and for 1/10 of a mile from the point O. How high above the 7 mile point will the helicopter be hovering? If it continued on the same course what is the smallest height the helicopter would reach? Find expressions for dy and d2y dx dx2 and so find the point of inflexion on the curve. Also find the equations of the tangents to the curve when (i) x = 2, (ii) y = 150. 2. A parabola has the equation y = (x – 2)2. Find the equation of its tangent at a general point U, where x = u. Do the same for a second parabola y = 2x2 at V, where x = v. Hence find 2 common tangents of the 2 parabolae and where they touch the curves. Sketch the 2 curves and their common tangents. Also find the radii of curvature at the 4 points of contact by using the formula r = (1 + (dy )2)3/2 dx2 d 2y dx2 3. A powerful rocket launched from the Earth with great acceleration reaches a height of y thousands of miles after several
133
days denoted by x. y = (x2/1 + x2 – 0.02) x 238 thousandths of miles. Calculate y for values of x: 0.25, 0.5, 1, 2, 4, 8, 16, and 32 days. Find expressions for dy/dx and d2y/dx2 and so find the number of hours in flight in which the acceleration will have become zero. At what height will that have occurred? The rocket will have decelerated from that point and have begun to approach what well known cosmic object? The takeoff of the rocket is below the Earth’s surface. Where approximately? y
B
A O
3
x
-1
Fig. 52
Numerical answers: 1. 1100, 1090, 750, 600, 300, 200, 158, 138, 127, 120, 100, .0205, 1078. 2. 0.25 at (0, 0), 0.05 at (2, 0), 2060 at (4, -32), 0.27 at (0.22, 2.37) 3. 9.24, 13.9, 42.8, 114, 186, 219, 228, 230, 233, 54,700, -4760.
134
Chapter 9 Class Eleven Projective Geometry In Class 10 we worked with points, straight lines and planes and the important theorems of Desargues. In Class 11 we shall be concerned with curved lines and surfaces, especially sections of cones. Moreover we shall meet a third kind of point. So far we have met with familiar points whose two attributes are position in space and function in determining lines and planes. Also we have become acquainted with a second kind of point which has no position in space but has no less power to determine lines and planes by means of the construction of parallel lines and planes. Such points are called infinite points. The third kind of point has neither position in space, nor the functional attribute using parallel lines to determine ordinary points. Instead of direction functions they have ordering functions. Such points are also called imaginary points. Similar remarks can be made about planes, straight lines, curves and curved surfaces. Just as we become aware that there are four kingdoms of Nature—mineral, plant, animal and human—so in geometry there are three kinds of point—ordinary, infinite and imaginary, whose qualities and properties bear a subtle analogy with the first three natural kingdoms, but there seems at first to be no fourth kind of point, but we will leave discussion of this until Class 12. The human kingdom should never be regarded as part of the animal kingdom despite superficial similarities. Plants differ from minerals because they are alive. Animals are not only alive, but can move over the earth instead of being rooted to it. They also have consciousness, which plants lack, but they have no responsibility for the development of the earth. The latter alone among the four kingdoms is the prerogative of human beings. Failure in so
135
many ways has been our lot in the past, but the future beckons us to act differently by taking moral responsibility for what happens to the earth and all its beings. Desargues theorem has shown that whenever 3 fixed lines a, b and c are drawn in a plane through a point in it (called pole) and any 3 points (say P, Q and R) are taken on the 3 fixed lines, then if PQ , QR and RP are drawn to meet any other fixed line in the plane (called axis) in A, B and C, then if line PQ is varied in position through the fixed point C, the point R will continue to lie on c. This construction will now be changed. O, a, b, A and B will remain as before, but C will no longer lie on AB. P and Q will continue lie on a and b and PQ will continue to pass through C. But R, where AP meets BQ , no longer will lie on a line through O, but describe a curve passing through O, A and B. Figure (53) shows part of the homology constructed in Class 10. Varying positions of line CQP give rise to AP and BQ which meet in R. All positions of R lie on line c which passes through the pole. Pole 0 P A
Q
P a
A
Q 0
b
B
C
Fig. 53
Figure (54) shows the changes made by taking O off the line AB. Each different position of the line PCQ gives rise to a point R as before, but the locus of R is no longer a straight line but a curve. In this particular case it is an ellipse. Other choices for the position of C would give rise to circles, hyperbolae, parabolae and longer thinner ellipses.
136
O B
A
Fig. 54
It is easily proved that when PCQ passes through A, it also passes through the point L2 where l2 crosses the ellipse. Similarly when PCQ passes through B it also passes through the point L1 where l1 crosses the ellipse. No matter how the points C, A and B or the pair of lines l1 and l2 are in general chosen, the curve formed by R (and R1) is always a conic section. While this will be proved below in projective geometry, it can be verified by analytical geometry as follows:
B (b, g) A (a, f)
Q C (o, o)
P
R
Fig. 55
137
Referring to Figure (54), let the points and lines be labelled as indicated in Figure (55) by Cartesian coordinates. The varying line through the coordinate origin C is y = Ux.
This crosses l1 where Ux = mx+c. Hence P is
(
Similarly Q is So AP is
c , U – m (
d U–n
Uc ). U–m ,
y–f , = Uc – f U – m
Uα U–n
).
x – a c – a U–m
And (y – f)(c – Ua+ma) = (x – a)(Uc – Uf+mf) Hence U = mfx – cy – may + fc fx – cx – ay + ac
*1
Forming the equation of BQ in a similar way will lead to a second expression for the variable U, thus
U = ngx – dy – nby + gd gx – dx – by + bd
*2
Eliminating U from equations *1 and *2 will lead to an equation in x and y of the second degree, which means that no straight line can cross the curve in more than 2 points. It is a conic section. A better proof arises in projective geometry: Every line AP gives rise to a unique line PCQ , which in turn gives rise to a unique line BQ. This is usually written as A(R) → (P) → C(P) → (Q) →B(R), a one-one correspondence. A similar result occurred in Class 9 when finding the foci of
138
conic sections, each point on the conic arising through a unique generator of the cone. This purely projective geometry method is known as synthetic geometry, in contrast to the analytical (algebraic) geometry above. In Figure (54) we pointed out that AC would meet l2 on the conic and BC would meet l1 on the conic. From this we obtain Pascal’s theorem, which is true of any 6 points of any conic. Relabel the points of this hexagram with numbers 1 to 6, making a continuous figure. After 6 we continue with 1 again (line l1). The three intersection of lines at P, C and Q are collinear. This is the statement of Pascal’s theorem. Another (hexagonal) arrangement of points on a conic (an ellipse is chosen below, but a parabola or hyperbola would be equally effective) is shown in Figure (57). Tangents to conics Figure (56) shows how, given 5 points in a plane (AOBL2L1), there is a unique conic on which they lie. By using the varying Pascal line through C, an endless number of points R can be constructed on this conic.
O B
A
P
C
Q
R
Fig. 56
139
In Figure (58) the 5 points are labelled ABCDE and the tangent at A is constructed.
Let DE and BA meet at P. Let EA and CB meet at Q. Let PQ and DC meet at J. Then JA is the tangent a at point A.
For we are recalling Figure (57) and imagining the points 3 and 4 coinciding to give a tangent instead of a chord. Thus in Figure (58), PQJ, i.e., u, is the Pascal line of hexagon DEAABC.
Fig. 57
Figure (59) shows the dual of Figure (58). The 5 points ABCDE have transformed into the five lines abcde. The 5 chords 12345 above have transformed into the new pentagon corners 12345. Point P has become line p joining 2 to 4. Q , where 3 meets 5, has become the line q joining points 3 and 5. Whereas P joined to Q was line u, now U is the point where p meets q. Finally j is the line joining I to U and it meets a in the point of contact of conic and tangent c. Just as u was the Pascal line of DEAABC, so is U the Brianchon point of hexagon deaabc. Let us now take the cases of 3 pairs of coincident points on a conic: Let A = B, C = D, E = F
140
Fig. 58
The lines joining A to B, C to D, E to F will be tangents to the curve. In this parabola Pascal’s theorem shows that the intersecting points of AB and DE, BC and EF, FA and CD are collinear (x), i.e., the tangents at 3 points of a triangle lying on a conic meet the opposite sides of the triangle in collinear points. Dualizing and so taking 3 pairs of coincident tangents to a conic, let a = b, c = d, e = f. The points where a meets b, c meets d, and e meets f will be the points of contact on the curve.
Fig. 59
141
Fig. 60
In this parabola, Brianchon’s theorem shows that the lines joining these points of contact with the meeting points of the triangle’s other sides are concurrent, or the lines ab/de, cd/fa and ef/bc all pass through a point (O).
Fig. 61
142
Because of the symmetry of statements regarding points and lines in the fundamental propositions of projective geometry, there is no need to prove Brianchon’s theorem. Before the principle of duality was accepted the two Frenchmen—Pascal in the 16th century and Brianchon in the 17th century—proved their theorems independently. Together they may have stated the following: When any six points/tangents are taken on a conic, the three meeting points/joining lines of opposite sides/corners of the hexagon will be collinear/ concurrent (Pascal line)/(Brianchon point). In Figure (62) the two configurations of Pascal and Brianchon are drawn together on the conic, beginning with either the 5 points ABCDE or the 5 tangents abcde.
Fig. 62
143
Notes: (i.) The 5 points JKLMN obtained from TR/PQ = J, etc. are mostly off the paper. (ii.) The 5 lines jklmn here meet in a simple point in the center of the drawing, because a special pentagon (a projectively perfect one) has been chosen. (iii.) Capitals and lower cases of the same letter for points and lines have a special relationship. Each time they are pole and polar with respect to the conic. Special cases of the sets of 5 points or lines arise when 2 or more pairs of them coincide. A very special case arises when b = a, d = c, f = e and E5, C3 and A2 are such that they all three pass through a single point M. This is a special case of Figure (62). It is best to start by drawing any triangle 235 and pick a point M and so draw the dashed lines 2MA, 3MC and 5ME, thus triangle ACE. Make 3 dots at the points where the sides of triangle ACE are crossed by the lines through M. They are roughly middle points. Guess the quarter points and then the eighth points. There are now 7 dots on each of AC, CE, and EA (end of stage I). Now in Figure (61) with ruler passing through 3, join each dot on AC to the line 5C. Again, keeping the ruler passing through 3, join each dot on EC to line C2. Repeat this whole procedure with the ruler passing through 5 and the dots on AE reaching to 3E and the dots on EC reaching E2. Finally repeat the procedure again with the ruler passing through 2 and the dots on AE and AC (end of stage II). The drawing is completed by joining corresponding points on A5 to those on 5C making a set of triangles in space A5C and thus using similar constructions in spaces C2E and E3A (stage III). It is best to use one color for all the lines in stage I, a second color for the lines in stage II, and a third color for the lines in stage III. The third color could also be used to go over the sides of triangle 235 again. Stage III and triangle 235 complete an elliptic envelope.
144
Fig. 63
It would be a good exercise to construct a parabolic envelope and a hyperbolic envelope in a similar manner to the one above. To check the accuracy of the elliptic envelope, place your eye just above the continuation of line M3 off the paper and you should observe a circular envelope. Harmonic ranges and pencils
Fig. 64
When we are given 2 points A and B on a line x and then a further point P on x, we can show that with respect to A and
145
B there is another point Q on x which has a special mating relationship to P. Take any point O not on x and join OP, OA and OB calling these lines p, a and b. Now take any other line through P and let it meet a and b in C and D. Join AD and BC and let them cross at E. Then OE will cross x at Q. P and Q are called harmonic mates with respect to A and B on line x. The symbol #(AB, PQ) denotes this harmonic range of points. The whole procedure is dual with symbol #(ab, pq) a harmonic pencil of lines through O. Notice that instead of starting the construction with line PC, we could have chosen PC1D1, leading to E1. E1 leads to Q just as E does, because of what we learned in Class 10, the two triangles CDE and C1D1E1 being part of a homology. Moreover, were we to take O anywhere else in space, say at / O , but keep P, A and B, we would be working in a new plane UAB having the common line x with OAB. It is easy to prove / / / that the triangles CDE and C D E are in perspective with a new / / / pole O and that O E leads to the same Q on x as before. If we ask what is the harmonic mate of Q with respect to A and B, the dotted lines above through Q , A and P lead to point Y on p, showing that the harmonic mate of Q is in fact P. Also it is easy to show that with respect to P and QA and B are harmonic mates. Hence #(PQ , AB) #(QP, AB), #(AB, PQ) and #(AB, QP) are all true. So are similar examples of harmonic pencils of lines in a point in a plane and of harmonic sheaves of planes in a line. The complete quadrangle and the complete quadrilateral (dual phenomena) In these drawings there are lots of examples of harmonic ranges of points and harmonic pencils of points. A quadrangle consists of 4 points (ABCD) and its 6 diagonals which cross each other in 3 points (XYZ). The sides of this triangle are p, q and r. p crosses the remaining diagonals in I and J, q crosses the remaining diagonals in E and F, r crosses the remaining diagonals in G and H.
146
Fig. 65
A quadrilateral consists of 4 lines (abcd) and their 6 meeting points join up to form 3 lines (xyz). The corners of the triangle are P, Q and R. P joins the remaining meeting points with i and j, Q joins the remaining meeting points with e and f, R joins the remaining meeting points with g and h.
Fig. 66
Exercises: 1. Mark 6 points on a semicircle ABCDEF in that order. Then construct the Pascal line of this hexagon. 2. Draw 6 tangents to a circle, all of whose points of contact lie within a semicircle and touch it in the order a b c d e f. Then construct the Brianchon point of this hexagon. 3. Use graph paper to mark points A(0,0), B(1, 0.5), C(3, 4.5), D
147
(the infinite point on the y axis), E(-2.5, 3.125) and F(-2,2). They lie on a parabola. What is its equation? Construct the Pascal line for the point order A B C D E F. 4. Use graph paper to mark points A (the infinite point on the y axis), B(1,4), C(2,2), D(the infinite point on the x axis), E(-4, -1) and F(-2, -2). They lie on a rectangular hyperbola. What is its equation? Construct the Pascal line for the point order A B C D E F. What is the equation of this Pascal line? 5. Draw 2 perpendicular lines representing x and y axes. Then draw more lines, from (0,5) to (1,0): line a, from (0,4) to (2,0): line b, from (0,3) to (3,0): line c, from (0,2) to(4,0): line d, from (01,1) to (51,0): line e; also call the x-axis line f. These 6 lines are tangents to a parabola. Construct the Brianchon point and show that it lies on line y = x + 1. 6. Construct an equilateral triangle ABC and its center D. Then using the letters of Figure (65), construct the complete quadrangle, indicating by arrows the direction of the 3 infinite points H, E and I. Which other sets of 3 points are collinear (which are not shown and named as such in Figure (65), among the 13 points of the complete quadrangle? 7. Construct the 4 sides of a square abcd. Then using the letters of Figure (66), construct the complete quadrilateral indicating by arrows the 4 sets of 3 parallel lines named by small letters. What name has to be used for the infinite line? 8. Choosing any scalene (but not necessarily obtuse angle) triangle 235 as in Figure (63), and placing M outside the triangle on the opposite side of 2 to 35, construct another conic envelope. If you choose M near to 35, you will find an elliptic envelope being constructed, but if far from 35 a hyperbolic envelope. 9. Draw a circle and a circumscribed pentagon of tangents to it like Figure (59). Find one or more ways of ordering the points 12345 different from the order shown and label the Brianchon point U. 10. Draw Figure (56) again but with a hyperbola instead of an ellipse and OB on one branch and L1 and L2 on the other. What would happen if the hyperbola degenerated into 2 straight lines (a Class 12 theorem)?
148
Chapter 10 Class Eleven Imaginary and Complex Numbers Positive and negative numbers, fractions and either terminating or repeating decimals i.e., rational numbers, and finally irrational (non-recurring decimal) numbers compose the whole domain of real numbers. There are no gaps in this domain where some other kind of number can appear. For example between whole number 3 and 4 there are the rationals 3¾ or 3.75 and 3¹/7 = 3.142857 where these last six digits carry on repeating themselves without stopping. Between any pair of rational numbers there are countless other rationals, but in addition there are the irrational numbers, which when expressed as decimals contain no repeating set of digits, for example √13 = 3.605551275… or 3√40 = 3.419951893… Irrationals also contain special numbers like π = 3.141592654… (quite different from 3¹/7 despite its relative proximity). Another example is e = 2.718281828…, but the set 1828 does not continue to recur thereafter. e and π are transcendental irrationals. All the above can be approximated to any degree of accuracy. So one often uses 3.14 for π, 9.9 or even 10 for π2, 3.6 for √13, 3.42 for 3√40, etc. This cannot be done for what are known as imaginary numbers, for they do not occur on the line of real numbers, for example the square roots of negative numbers such as -9 or -1. 3 x 3 = 9 and -3 x -3 = +9. 1 x 1 = 1 and -1 x -1 = +1. It is usual to denote √-1 by i where i2 = -1. The roots of the equation x 2 + 1 = 0 are + i and -i and of x 2 + 9 = 0 are 3i and -3i. 149
Complex numbers are the sums of real and imaginary numbers. Consider the effect of multiplying two complex numbers, e.g., 1. 2. 3. 4.
(8 + 3i)(3 + 8i) (2 – i)(2 + i) (3i + 4) 2 (a + ib)(c + id) y
= = = =
24 + 64i + 9i – 24 = 73i, 4 + 2i – 2i + 1 = 5, – 9 + 24i + 16 = 7 + 24i, ac – bd + i(bc + ad). P
(imaginary) R
0
(real)
Fig. 67
Any complex number can be represented by a point in Cartesian coordinates on what is called the Argand diagram in which the x-axis exhibits the real part and the y-axis exhibits the imaginary part of the number. So the point (a, b) in the diagram can also be expressed in polar coordinates as [R, θ], where the modulus R is the length of OP and the amplitude θ is the angle at O between the real axis Ox and the radius vector OP. So the point (a, b) in this diagram can also be expressed in polar coordinates as [R, θ], where the modulus R is the length of OP and the amplitude θ is the angle at O between the real axis Ox and the radius vector OP. It follows that R = √(a2 + b2) and tan θ = b/a; also that x = Rcosθ and y = Rsinθ, so the complex number c = a + ib = R(cos θ + i sin θ), also expressed as Rcis θ. This is also called the modulus-amplitude form of the complex number. In the four examples above the resulting forms are, where 0 ≤ θ ≤ 360°, 1. 73 cis 90° 2. 5 cis 0
150
3. 25 cis 73.7°, where tan θ = 24 / 7, since 72 + 242 = 252 4. √((ac – bd) 2 + (bc + ed) 2) cis (tan-1 bc + ad / ac – bd)) This introduction to a new kind of number will seem at first a highly abstract mathematical construct. It will gain pictorial credibility only after the projective geometry in Class 12, when we shall appreciate the relevance of imaginary points, lines and planes—elements whose “ordering function” gives them active geometrical power. Exercises: Express in modulus-amplitude form 1. 3 + 4i 2. (3 + 4i) 2 3. 20 + 21i 4. (1 + i) 3 5. (7 – i) 2 6. If you increase the power of i and simplify: i, i2, i3, i4, i5, i6… and then differentiate successively
cos θ or sin θ, i.e., d , d2 , d3 , dx dx2 dx3 …what rhythm do you notice? What might this suggest? Numerical answers: 2.8, 5, 25, 29, 46.3°, 50, 53.1°,106.3°, 135°, 343.7°.
151
Hey! Are you sure this guy has taken off from the Earth’s surface? (See pp.133–134)
152
Chapter 11 Class Twelve Integral Calculus (Age 18) In Class 11 we were concerned with the analysis of various functions which are differentiated and applied to problems about time, distance covered, speed and acceleration, also to the nature of curves, tangents to them, inflexions and turning points (maxima and minima). All subjects in Class 11 require strong intellectual application and the examination of detailed analysis, whereas in Class 12 we seek to develop the students’ reasoning powers. Reason builds upon what the intellect has achieved for our activity of thinking, but we are now concerned with synthesizing rather than analyzing. Our attention changes from examining details to wholeness and universality. This is reflected in mathematics in the change from differential to integral calculus. In the field of kinematics we begin with perceptions of movement and its two concepts of space and time. Then we forge our ideas of speed or velocity and go further to acceleration and retardation (deceleration) and the increases and decreases in the latter. Three different types of graphs can be constructed:
Fig. 68
If instead of straight line graphs we have curved lines, they will denote changes in the distance covered, the speeds and accelerations. The area in the shape below the form in (ii) shows
153
the total distance covered during the initial and final moments of time. Dividing the total areas in all 3 graphs by the time intervals between O and t will give the average distances, speeds and accelerations in those intervals. In differential calculus we have also learned that if t = time, s = distance, v = velocity and α = acceleration, then ds = v, dv = dt db
d2s = α and dα = rate of change of acceleration dt2 dt
Examples 1. Given that s = 3t3 + 5a2 – 4t + 7, what formulae do v and α have? Answer, formula is: 3t4 5t2 + – 2t2 + 7t 4 3 2. A car journey takes exactly 3 hours. Setting off through a busy town the driver averages 20 mph, but half an hour later has to stop at traffic lights for 10 minutes at a busy junction, then slowly picks up speed at constant acceleration for 10 minutes until the speed becomes 60 mph, which he maintains for 40 minutes. But heavy traffic then causes uniform retardation through 10 minutes down to a speed of 15 mph, where the road is found to be almost free. The driver can now accelerate at a constant rate of 720 mph per hour. = 12 mph per minute until leveling off when the speedometer shows 81 mph on the motorway which the road has become. This speed is maintained for a further 1 hour 2 minutes, when be brakes steadily for 2 minutes and turns off reaching the final destination. Use the v – t curve in Figure (69) to answer the following questions. Speed MPH 80 60 40 20
30
40
50
100
180
Time (Minutes)
Fig. 69
154
(i.) How far from the start is the car after 40 minutes? (ii.) What is the car’s acceleration between A and B in mph per hour, also in mph per minute? (iii.) How many miles are covered between B and C? and C and D, and D and E, E and F? (iv.) Find the decelerations between C and D and F and G, in both mph per hour and mph per minute (v.) Calculate the total length of the journey. (vi.) What is the average speed for the whole journey? 3. Find the area between the curve y = x2 + x, the x-axis and the line x = 2. Numerical answers from: 4, 4.5, 4.66, 6, 10, 45, 45, 50.5, 81, 151, 270, 360 Areas under curved lines There are several ways of determining such areas, of which the most fundamental is that of cutting up the area into 5 thin strips parallel to the y-axis. For example, consider the cubic curve y = x3 and the area between it, the x-axis and the line x = 6 units.
Fig. 70
155
Total area of the 5 rectangular strips = 13 + 23 + 33 + 43 + 53 = 1 + 8 + 27 + 64 + 125 = (1 + 2 + 3 + 4 + 5)2 1∗ = 152 = 225 Area of whole rectangle = 6 x 125 = 750 Ratio of these two results = 225 = 3 = 1 + 1 750 10 4 20 Had we chosen 6 strips, the result would have been 1 + 1 . 4 24 4 strips would have given
1 +1 4 20
3 strips would have given
1 1 + 4 12
2 strips would have given
1 + 1 4 8
1 strip would have given
1 = 1 +1 2 4 4
100 strips would have given
1 + 1 4 400
For ever larger numbers of strips the answer would be just over ¹/⁴. In the limit the answer becomes ¹/⁴ and the area under the curve 3 4 = x Xx = x , i.e., a quarter of the whole rectangle in Figure (70). 4 4 4 a Symbolically this is written A = ∫ 0 x3 dx = a 4
,
NOTE: This only works for the sums of cubes. Other formulae can be found for the sums of other powers. But simple addition of the terms of the series can always be used.
156
where the long ∫ means the limit of the sum of the thin rectangles. Similarly the area A for y = xn = a n+1 , ∫o xn dx = a n +1 as can be confirmed for any numerical value of n. A second (simpler) method for finding areas The opposite of differentiation. We noticed that in v-t curves the distance traveled is equal to the area under the form, i.e., since ds = v, s dt
= ∫ v dt or more particularly the area
= ∫ ab xn dx
n+1 = [ x n+1
]
b = a
bn+1 – an+1 n+1
for any part of the curve as in Figure (71).
y
0
a
b
×
Fig. 71
A few examples will make this clear. 2 1. Since if y = x2, dy = 2x, and if y = x , dy = 2x = x, dx 2 dx 2 2 it follows that ∫ 2x dx = x2 and ∫ x dx = x . 2
157
If y = x2 + 5 (or any constant), dy is still 2x. dx Hence in general integration one should always add a constant. So ∫ 2x dx = x2 + c. 2. If the limits to the integration are given, as in Figure (71), the constant c can be omitted. 3
∫1 2x dx = [x2 ]13 = 9 – 1 = 8 3. If y = 8 + 6x – 12x2 + 3x3 – 35x4 , ∫ y dx = 8x + 3x2 – 4x3 + ¾x4 – 7x5 + c. 9 9 24 x2 dx = [8x3]2 = 8(729 – 8) = 5768 4. ∫ 2 2 2 2 5. ∫ 1 (6x–1) 2 dx = ∫1 (36x2 – 12x + 1)dx = [12x3 – 6x2+ x] 1 = 12 x 8 – 6 x 4 + 2 – 12 – 6 + 1 = 57. 6. Since if y = sin x, dy/dx = cos x, ∫ cosx dx = sin x
and ∫
π 2 π c
cos x dx = [sin x]
π 2 π 6
= 1 – 0.5 = 0.5
Integration “by parts” Another useful method of integrating, when no obvious way is available, arises when a simple integration of one part of a product can be used. This method is derived by transforming the rule by which products are differentiated. Let y = f(x). F (x). Then [f(x). F(x)] = f(x). F1(x) + f (x) .F(x). Rearranging, f(x). F (x) = [f(x). F(x)] -f (x). F(x). /
/
/
/
/
/
Hence putting g(x) for F (x), ∫ g(x) = F(x) and interpreting, we have /
∫f(x)g (x) = f(x)∫ g(x) – f (x)∫∫g(x) /
158
Now let g(x) = cos x, so that ∫ g(n) = sin x; also let f(n) = x, f (x) =1, so x cos x = (x sin x) – 1. sin x /
/
∫x cos x gives, using *, = x sin x + cos x.
Integrating,
Similarly if we let g(x) = sin x, ∫ g(n) = -cos x; also f(n) = x, f (n) = 1, we have on integrating /
i.e.,
∫x sin x = -x cos x + ∫ cos x, ∫x sin x = sin x – x cos x.
Integration by substitution Yet another method of integrating is by substituting another variable from the one given. For example, taking a definite integral, a
∫0 √ a2 – x2 dx, put x = a sin θ, and therefore dx = cosdθ π
∫ 2 √ a2 – a2 sin2 θ (a cos θdθ)
= ∫2
= a2 ∫ 2
2 2 = a ∫0 (1 + cos 2θ) d θ 2
2 = a (θ + sin 2 θ) 2 4 θ
0
π
0
a cos θ. 2 cos θ d θ π
0
cos2 θd θ
π
π
2 = πa 4
This is the area of a quadrant of a circle, radius a, making the area of the whole circle πa2.
159
Circular and hyperbolic functions When the imaginary number i (=√–1) is used in powers of the transcendental number e (= 2.178…), important infinite series arise for cos x and sin x.
2 3 4 5 ex = 1 + x + x + x + x + x + …to ∞ 2 3! 4! 5!
and d ex is clearly the same as ex. dx 2 3 4 5 So eix = 1 + ix – x – ix + x + ix + … 2! 3! 4! 5!
2 3 4 5 and e-ix = 1 – ix x + ix + x – ix – … 3! 4! 5! 2!
Adding, we have 2 4 6 eix + e-ix = 2(1 – x + x – x 4! 6! 2!
+ …).
ix -ix 2 4 6 So e +e = 1 – x + x – x + … = P(x), say 2 2! 4! 6! Subtracting, we have
*1
3 5 eix – e-ix = 2i(x – x + x – … 3! 5! ix -ix 3 5 7 So e – e = x – x + x – x …= Q(x), say 2i 3! 5! 7!
*2
P(x) and Q(x) are called circular functions. |
|
It is easy to see that P (x) = -Q(x) and Q (x) = P(x) and it seems that P(x) = cos x and Q(x) = sin x. But any relationships between P(x) and Q(x) are the same as those between cos x and sin x, by testing out the left or right hand sides of *1 and *2, e.g.,
160
ix -ix 2 ix -ix 2 cos2x + sin2 x = 1 and (e + e ) ≠ (e – e ) 2 2 =e
2ix
+ 2 + e-2ix – e2ix + 2 – e-2ix = 1. 4
ix -ix ix -ix Hence cos x = e + e and sin x = e – e *3 2 2! If we leave i out of results *3 we obtain hyperbolic functions symbolized by cosh and sinh, thus
x -x 2 4 6 cosh x = e +e = 1 + x + x + x + …to ∞ and 2 2! 4! 6!
3 5 x -x sinh x = e – e = x + x + x + …to ∞. 2 3! 5!
*4
It is easy to confirm that cosh2x – sinh2x = 1 or that sin (x + iy) = sin x cosh y + i cosx coshy, for cos ix = cosh x and sin ix = i sinh x. Integration of hyperbolic functions Already referred to are the integrals
∫cosh x dx = sinh x and ∫sinh x dx = cosh x.
2 Just as tan x = sin x and d tan x = cos2x + sin2 x cos x dx cos e
= 12 = sec2x, cos x
and so ∫sec2x dx = tan x,
2 tanh x = sinh x and d tanh x = cosh2x – sinh2x cosh x dx cosh x 1 = = sech2x, 2 cosh x and so, ∫sech2 x dx = tanh x.
161
There are also inverse circular and hyperbolic functions to be considered. dx, dy = 1 If y = sin–1 x, then sin y = x, cos y = cos y dy dx 1 1 = = . 2 √1 –sin y √1 – x2 Similarly if y = sinh-1x, sinh y = x, cosh y = dx , dy dy dx 1 1 = 1 = = 2 cosh y √1+sinh y √1 + x2
Hence
1 ∫1 = sin-1 x and √1– x2 dx √1+x2 dx
= sinh-1 x.
Also if y = cosh-1 x, cosh y = x, sinh y = dx , dy = dy dx
=
1 1 1 . = = 2 2 sinh y √cosh y–1 √x –1
Standard forms worth memorizing: (Always add c to indefinite integrals.) ; ∫xndx = xn+1 / n+1 + c y = xn, dy/dx = nxn-1 x x y = e , dy/dx = e ; ∫exdx = ex + c y = ax, dy/dx = ax/na ; ∫axdx = ax / ln a + c y = ln x, dy/dx = 1/x ; ∫dx = ln x + c y = logex, dy/dx = logae/x ; ∫dx / x ln a = logax + c y = sin x, dy/dx = cos x ; ∫sin x dx = –cos x + c y = cos x, dy/dx = -sin x ; ∫cos x dx = + sin x + c 2 y = tan x, dy/dx = sec x ; ∫sec2xd2 = tan x + c y = sinh x, dy/dx = cosh x ; ∫sinh x dx = cosh x + c y = cosh x, dy/dx – cosh x ; ∫cosh x dx = sinh x + c -1 2 y = sinh x, dy/dx = 1 /√1+x ; ∫dx/√1 + x2 = sinh-1 x + c -1 2 y = cosh x, dy/dx = 1/√x –1 ; ∫dx/√x2 – 1 = cosh-1x + c y = sin-1x, dy/dx = 1/√1– x2 ; ∫dx/√1 – x2 = sin-1x + c 162
Examples for Integration (choose any method) 1. ∫(2x + 1 – 1 ) dx 2x 2. ∫4 cos3 θ d θ [= ∫ (cos 3θ + 3 cosθ) d θ] 3. ∫x2 cos x dx 4. ∫tan2 θ dθ [ = ∫ (sec2 θ – 1) d θ] 5. ∫(cosh x + sinhx) dx in its simplest form 6. ∫x lnx dx 7. ∫10 (2x – 1)2 dx dx 8. ∫.5 .2 √1+x2 dx 9. ∫3 2 2x + 32 + 3x + 4 dx x 10. ∫10 1 log10 x dx
[NB log10 x = ln x/ln 10]
Applications of integral calculus to the calculation of the lengths of curves, the areas of surfaces of revolution and their volumes of revolution The simplest 3-dimensional geometrical example of this is the right circular cone. Take a point (called the vertex) and through it take a straight line (called the axis). Then take any variable line in space through the vertex V making a constant angle (α) with the axis. The rotation of this variable line constructs the cone.
Fig. 72
163
A section of the cone is met by the variable line in P, which describes a circle, recalling Class 9 geometry. P and the circle are drawn in perspective. Let height OV of the cone be h, radius OP be r, and the constant slant height VP be l. Now imagine that the cone surface is cut along VP and opened out flat as in Figure (73).
Fig. 73 |
2 π r, circumference of circle in Figure (72), is equal to arc PP in Figure (73) = l θ. Area of sector PVP = ¹/² l2 θ = ¹/² l2 2π r / l = π r l |
Figure (72) shows a thin slice, a frustum of the cone. Its area, δA = π (y + δy)(m+ δm) – πym = π (y δm – m δy), ignoring δy δm. But δm = δy l/r and m = l/r y. So δA = π (l/r yδy + l/r yδy).= 2π l/r yδy Hence Area A = ∫ ro πl / r 2y dy = π /r [y2] ro= π r l, as expected. The volume of the slice δV = π y2 δx, and since y = rx / h, So the volume of the whole cone 2 = π ∫ho 2 r2 h x dx 2 = 2 π r 2 h ∫ho x dx 2 3 1 = π r2 h = h 3 3 π r2h 164
Fig. 74
The circle and the sphere Let the circle be x2 + y2 = r2 or y = ± √ r2 – x2 . x . So dy = ± 2 dx √ r2 – x 2 Also ds = √ 1 + ( dy ) = dx dx
√ x2 r – x2 + 1 2
= √ n2 + r2 – x2 = r r 2 – x2 √r2 – x2 +r
1. Length of half the circumference = ∫ ds = ∫–r r dx 2 √r – x2 = [r sin-1 x ]–r+r r = r π + 0 +r π + 0 = r π. 2 2 So circumference of circle = 2πr (= π x diameter). 2. Area of semicircle = ∫ y dx = ∫ r – r √ r2 – x2 dx Let x = r sin θ, dx = r cos θ dθ Area = ∫r cos θ. r cos θ dθ = ∫r2 cos2 θ dθ = r2 1 + cos 2 θ 2 dθ 2 θ sin 2θ π π π π = r2 [ + ] – = r2 [ + 0 + + 0 ] = π r 2 4 2 2 4 4 2
So area of whole circle = πr2 165
3. Volume of hemisphere = ∫ ro πy2dx = π∫ro(r2 – x2) dx 2 3 = π [r x – x ] ro 3 3 3 3 = π (r – r ) = 2 π r . 3 3
So volume of whole sphere = 4πr 3
3
4. Surface area of hemisphere ` = ∫2πy ds = ∫2 π √r2 – x2 √1+ (dy/dx)2 dx = 2 π∫ ro √(r2 – x2) (x2/ r2 – x2 +1) dx = 2 π∫ √x2 + r2 – x2 dx = 2 π∫ r dx = 2 πr [x] ro = 2π r2 So the whole curved surface area of sphere = 4πr2 These 4 examples, whose answers are already largely familiar, indicate the general method of tackling the main problems of this whole section of the work. Further examples of volumes and surface areas of solids of revolutions, etc. 1. Sketch the hyperbola y = 1/x and show the lines x = 1 and x = 2. Shade in the solid of revolution about the x-axis bounded by the hyperbola and the two lines. Calculate the volume of this solid. 2. Sketch the parabola y = ± √x between the limits x = 0 and x = 4. Find the volume of the solid formed by rotating the curve about the x axis between the limits of x = 0 and x = 4. 3. In the previous question calculate also the curved surface area of the solid. Answers: 3.62, 6.28, 25.1
166
Exact and good approximations to areas using Simpson’s Law The English mathematician Thomas Simpson (1710–1761) showed that good approximations to
∫b a Ø(x) dx are (b–a) [Ø(a) + Ø (b)]
and (b–a) Ø (a+b) . 2
2
Then, combining them ∫b a Ø(x) dx = (b–a) [Ø (a) + Ø(b) + 4Ø (a+b) ] 6 .2 The last formula is exact up to cubic polynomials such as y = ax3 + bx2 + cx + d and still close for quartics and quintics. For example, suppose Ø(x) = x3 + 2x +1. Then Ø(3) = 34, Ø(4)= 73 and Ø(5) = 136. 4 5 5 By calculus, ∫ 3 (x3 + 2x + 1) dx = [ x + x2 + x] 3 4 = [ 625 + 25 + 5 – 81 – 9 – 3] = 154. 4 4 By Simpson, 5 ∫ 3 (x3 + 2x + 1) dx = 2 (34 + 136 + 4 X 73) = (462) = 154. 6 On the other hand, if Ø(x) = x 4 + 4x 3 + 3x 2 +2x + 1, 3 5 ∫ 1 Ø(x) dx = [ x + x4 + x3 + x2 + x] 3 1 5 = (168.6 – 4.2) = 164.
But
1 x 2(223 + 11 + 4 x 65) = (494) = 164.7. 6 3
167
Further Examples Use both calculus and Simpson’s Law to evaluate: 4
1. ∫ 2 (x3 + x) dx 4
2. ∫ 1 (3x3 – 4x2 + 5) dx 3
3. ∫ 1 (x4 + x2 +1) dx 3
4. ∫ 1 (x3 + x2 + x + 1) dx
Answers: 34.7, 36.8, 59.1, 59.3, 66, 66, 122, 122.3. Final note In 1914 the English mathematician G.H. Hardy wrote his excellent book Pure Mathematics, in which he developed advanced calculus. Commenting on Simpson’s Rule he showed that 1 dx = [tan-1 x]1 = π , ∫ 2 0 0 4 1+x when integrated by the two approximations shown at the beginning of this section, lead to a good approximation to the decimal value of π/4, i.e., 0.7833, and 0.78539 if the interval 0 to 1 is divided into two parts, whereas the accurate value of π/4 to 6 d.p. is 0.785398.
168
Chapter 12 Class Twelve Three Dimensional Projective Geometry Taking up the work in Class 11 in a more fundamental and comprehensive manner, we shall recall and develop the ideas described in Figure (56), relabelling the latter below in Figure (75).
The Pascal Line
Fig. 75
Let 5 points 1, 2, 3, 4 and 5 be chosen randomly in a plane, no three of them being in line. Let lines 12 and 45 intersect in C. Let l1 be a variable line through 1 which intersects line 34 in P. Let PC intersect line 23 in Q and let 5Q (line l2) intersect 1P in point 6. As P varies its position on 34, so do points Q and 6 vary their positions. We have the one-one correspondence (6) → 1(P) → C(P) → (Q) → 5 (Q) → (6), true whether written forwards or backwards. Varying P (or Q) will produce new positions of 6. The totality of the positions of 6 (or its directing functions) defines a conic (shown dashed).
169
Conversely, given any conic (circle, ellipse, etc.) and any 6 points chosen on it (1, 2, 3, 4, 5 and 6), the 3 intersections of opposite sides of such a hexagon (or hexagram), i.e., the points P (where 16 meets 34), C (where 12 meets 45) and Q (where 23 meets 56), lie on a straight line. This is Pascal’s theorem. It is easy to see that had we chosen any 5 of the 6 numbered points we would have ended up with the sixth point; also any seventh point constructed from any 5 points already on the conic would lie on the same conic. Now let us suppose that in the above construction we allowed 3 of the chosen 5 points to lie on a straight line. What difference would this make to the resulting conic? Let 1, 3 and 5 lie on a straight line m.
P C
Q
(line m)
Fig. 76
The lines 12, 23, 34, 45 appear as before. Also 12 meets 45 at C. Points P and Q can be freely chosen on 34 and 23 provided PQ passes through C. Also lines 1P and 5Q give rise to intersection point 6. It will be found that line 42 seems to pass through 6. Why? It is because the conic has degenerated into a pair of straight lines (135 and 264). This also occurs with sectioning a cone in Class 9 work when the cutting plane passes through the vertex of the cone, e.g., in Figure (77).
170
A pair of straight lines (AVB is more vertical cut)
An ellipse (a more horizontal cut)
Cone edges
Fig. 77
The configuration in Figure (76) is the Pappus configuration. As with the Pascal configuration each line contains 3 points and each point contains 3 lines. But whereas there are 10 lines in Pascal’s, there are 9 in Pappus’ theorem. Pappus’ theorem, proved above, may be stated as follows. Given two lines a and b and any three points chosen on each line (X1, X2 and X3 on a) and (Y1, Y2 and Y3 on b), the points of intersection of X1Y2 and X2Y1, X1Y3 and X3Y1 and X2Y3 and X3Y2 are 3 points on a further line (Z1, Z2 and Z3 on c). This is also know as the fundamental theorem of projective geometry. As with Pascal’s theorem, while it is a purely plane theorem, it has important consequences in 3-dimensional projective geometry. Pappus’ theorem is also self-dual in its 9 points and 9 lines, 3 lines per point and 3 points per line, whereas the dual of Pascal’s theorem is Brianchon’s theorem. A second example of Pappus’ theorem is shown in Figure (78), where the dual nature is exemplified in lower case and capital letters.
171
0
C B A P X
U D
M
Z Q
Y
E
G F
Fig. 78a
Pappus’ theorem (the fundamental theorem of Projective Geometry)—the full proof The proof of Pappus’ theorem requires more detail if one has to avoid the assumption that no line can intersect a conic more than twice. No such assumption in the proof sketched below is needed. Given any 2 lines in a plane, each containing any 3 points ABC and DEF, let the intersections of AE / BD be X BC / CE be Z and AF / CD be Y.
Comparison
Fig. 78b
172
We have to prove that X, Y and Z are collinear. (i.) Let OB / UF = G, AG / BD = P and BF / CG = Q. The diagonals of ABGD meet in P. Hence P lies on the harmonic mate of OU with respect to lines UC and UF. So do the diagonals of BCFG and ACFD meet on the same harmonic mate. Thus PYQ is the Pappus line of (ABC, DEF). (ii.) Let AF / BD = M. The range (UEGF) on UF A∫ (BXPM) on BD.} These ranges have B in common. Also (UEGF) C∫ (BZQF) on BF. } Hence XZ passes through PQ / MF, i.e., Y. So XYZ is a straight line, the Pappus line required. The application of Pappus’ theorem to three dimensions Imagine 3 lines in space which are mutually skew, for example the 3 edges AB, FG and DH of a rectangular block. Then imagine a second set of mutually skew lines which intersect all 3 of the first set. For example one such line could be sloping through the middle of AB, up right to meet GF produced and down left to meet HD produced. E
F
A B
H G
D C
Fig. 79
173
But any 3 lines drawn on paper can represent mutually skew lines and any two lines crossing all 3 can be represented on the drawing. Suppose they are a, b, c and t, w.; t must be skew to w, otherwise a, b, and c would be coplanar.
Fig. 80
Let t meet b at O. The Pappus line Z1Z2Z3 arises as before by joining Xs to Ys. These 9 points plus O all lie in a plane, which we can call the ground plane. All other points and lines are not in the ground plane. Imagine them above it; the lines are shown dashed. Let c meet w in L. Since I and Z1 lie on two sides of triangle LX2Y3, by Peano’s axiom, IZ1 must intersect the third side—at point M. Let Y2 M be v. Again by Peano’s axiom, as M and Y2 lie on two sides of triangle X3Z1I, MY2 must meet the third side—at point J. So t, v and w are three mutually skew lines which all meet the original lines a, b and c. Join Z3M. By Peano’s axiom, IZ2 must meet Z3M - at point G, because of triangle MZ3Z1; Y1G because of triangle X3I - at point F, because of triangle IZ2X3; also because of triangleX2M - at point E, because of triangle MZ3X2, X1G, because of triangle Y2M - at point N because of triangle MZ3Y1, 174
also because of triangle Y3L, because of triangle K, because of triangle IY3Z2. This has shown that the one set of 4 skew lines a c d b meets the other set of 4 skew lines t u v w in 4 x 4 = 16 intersection points. This proves the theorem: If two sets of 3 skew lines intersect in 3 x 3 = 9 points, then any other transversal of the first set will meet any other transversal of the second set. This is the fundamental theorem of projective geometry in three dimensions, just as Pappus’ theorem is in two dimensions. The 4 x 4 straight lines in Figure (80) lie in a 3-dimensional surface where each set of lines meets the other set. 2 examples of this are shown below.
Fig. 81
The surface is a hyperboloid and here it is cut off by two planes at top and bottom in circular or elliptic curves. Each set of lines is called a regulus. Notice how the tangents to the hyperbola at left and right have two colors—full and dashed lines. The changeover is at the point of contact of the tangent. Actually the single line is really two lines, e.g., at T the full line runs upwards and towards the observer, whereas the dashed line runs downwards and towards the observer.
175
Fig. 82
The surface is often seen as the cooling towers in a power station. Figure (82) shows the view from above looking inside the hyperboloid. The continuation of each full line and each dashed line goes down below the opposite “color” to a circle below the one drawn and the smaller circle lies half way between the larger circles. Exercises: 1. Repeat Figure (81), but take T, not above C but above D. This will give a narrower hyperboloid. 2. It will be noticed that Figure (82) is a plan view of Figure (81) elevation. By taking 24 equi-spaced points on the circle and drawing twice as many lines, construct a new plan of Figure (81). The resulting figures will not be composed of triangles. 3. So far the hyperboloids have been completely symmetrical. Now take a very general set of lines—2 full lines ABC and PQR and 3 dashed lines AB, BQ and CR crossing them. BR is a chain-dotted line. Choose any point 1 in lR and join it to A. Label its crossing with BR point 2. Let P2 cross BQ in point 3. The line 13 will cross all three dashed lines.
176
By choosing 1 in different places on CR and its continuations up and down, construct many more full lines 13 and so find an ellipse instead of a circle as in Figure (82). (It will, of course, be touched by the 3 dashed lines, too.)
Fig. 83
177
Chapter 13 Class Twelve Path Curves and Their Applications to Plant Buds and Leaves and to Planetary Alignments Once 4 points are chosen anywhere on a plane (but no 3 in line), every other point on that plane can be described by coordinates in relation to the 4 or by linear constructions. For example, if we choose a sequence ABCD, the pairs of coordinates on graph paper and the corresponding linear constructions are easily understood [See Figure (84a)]. In Figure (84b) a projection of Figure (84a) (a shadow of it, for example) shows how the pairs of whole or fractional numbers for each point can be replaced by lines joining the 4 points A, B, C, D and further lines joining already constructed points. XY is the projection of the infinite line. Now suppose we freely choose a fifth point E and find its relation to ABCD both by numerical coordinates and by linear construction, then find a sixth point F whose relationship to BCDE is the same as E’s relationship to Fig. 84a ABCD. Furthermore we could find a seventh point G whose relationship to CDEF was the same. In this way we could go on to find an endless series of points ABCDEFGH… appearing to lie on a curve. Fig. 84b Such a curve is called a path curve. 178
X P B A
0
α C
Y Q D
R
E
F
Fig. 85
Figure (85) shows the 5 points ABCDE and some of the lines joining them to yield points P, Q and X. Just as A → B, B → C, C → D, D → E so AB and CD → BC and DE, i.e., P → Q. Where does X go? Let XQ meet AC at α, Pα meet BC at Y and XY meet AC at O. Range BCQY projects via α to range BAXP, which projects via Y to range CAOα, which projects via X to range CBYQ. Hence by interchanging 2 points B and C of a range and also interchanging the other two points Q and Y, we get an equal range, i.e., BCQY = CBYQ. This is true for any 4 points of any line. Since BCQY also = ABPX, we have ABPX = CBYQ , confirming that AC, PY and XQ all pass through α. Hence X goes to Y. Just as BC meets XD in Q , so CD meets YE in R. Let XR meet BD in θ. Then θD will meet YE in F. Continuing the construction of Figure (85), letting the dashed lines in it now becoming full lines and having new lines as dashed ones, we obtain the following:
179
Fig. 86a
Summarized instructions for Figures (85) and (86a): Given points: A B C D E P Q X Successive points in construction: AC/DQ = Pα/BC = CD/YE = XR/BC = θD/YE = BD/ER = Qβ/CD = DE/ZF = YS/CD = φE/ZF = CE/FS = Rγ/DE = EF/WG = ZT/DE = ψF/WG =
α Y R θ F β Z S φ G γ W T ψ H
Points on the path curve are joined by heavy lines. If we were to continue beyond H, we would find an inward-turning spiral. If we were to continue backwards beyond A, we would find either:
180
(i.) an outward growing spiral, or (ii.) an asymptote with the curve passing through the infinite and then re-emerging from the opposite direction and spiralling inwards, or (iii.) an inflexional point followed by an inward spiralling. The last case would be similar to the eurythmy form called “Schau in dich, schau um dich” or “Look within you, look around you,” as in Figure (87). Other path curves besides A B C D E F G H…of similar character are P Q R S T…, X Y Z W…, α β γ…and θ φ ψ.…
Fig. 86b
Special path curves The 4 successive moves A → B, B → C, C → D, D → E determined the path curve in Figure (86a). We now consider the special case in which the 4 moves are A → A, B → B, N → O, O → P, i.e., A and B become invariant points. AA and BB become tangents to the path curve. Let them intersect at a point C. Also choose any 2 points E and D on AB produced and BA produced. Choose any point 7 on BC. Let D7 meet AC in point 8 and let B8 meet A7 in O. Continue the construction of points 6, 5, …1 and points 9, 10, 11…14 in Figure (87). Now join A to the points with odd numbers and B to the points with even numbers. Let B10 meet A5 in N, B8 meet A7 in O and B6 meet A9 in P. A distorted chessboard has arisen within triangle ABC. N, O and P lie at the corners of this pattern. Moving from the “square” in which N and O are opposite corners to the “square” with O and P as opposite corners would be achieved by a bishop. Join up these opposite corners and those passed by this bishop in its journey both forward and backwards. As the numbers
181
increase in size, the curved form of the path curve will close up to A and B and unite with the tangents AC and BC. The actual curve passing through A…M N O P Q…B is part of a conic. Both the odd-numbered points and the even-numbered ones form musical progressions on BC and AC, each one being a projection of the other.
Fig. 87
Part of another conic path curve polygon is shown with a heavy dashed line. Had D and E been taken further apart, the number of points on the path curve would have been much greater and closer together. Let us now take C as the infinite point perpendicular to AB. With A as origin let the center of this square be (¹/²,¹/²). This selection of path curves all have the parameter λ and have the general equation yλ+1 = xλ (x – 1). The musical progressions on AC and BC have now become geometric progressions. Their common ratios have each become 1.2, but any ratio would result in the same curves. The circle of numbers shows the particular values of λ on the curves where it is written. 182
Fig. 88
There are four cubic curves (λ = 2, ¹/², -¹/², and -2) where chess moves of knights take place. Bishops’ moves take place only where λ = 1 (a perfect semicircle) and λ = -1 (a straight line pair, AB and the central vertical). Quintic (fifth degree) curves occur when λ = 4, ¼, -¼, and -4, and so on for other degrees. Exercises: 1. Show that the path curve for λ = -¹/² could be written as y2 = (x+1)2/x and that for any positive fractional value of λ, the formula y2 = x m / m+n (x+1) n / m+n could be used. 2. Draw 2 straight lines intersecting towards an edge of the paper. On 1 of them mark 3 points A, B and C. On the other line mark 3 points X, Y and Z. Try not to make distances equal. Join all 3 points on one line to all 3 on the other line. Extend all lines to the limits of the paper. Three of these new lines should meet in a point. So should the other 3 in a second point. How are the two new points related to the original 2 lines? 183
A
B
.
P
D
X C
Fig. 89
3. Copy Figure (89). The 4 points ABCD are not in the same plane. Draw the line through P which meets both AB and DC. Also draw the line through P which meets both AD and BC. On BA produced construct the harmonic mate of the intermediate new point on AB with respect to points A and B. Do the same on DA produced to construct the harmonic mate of DA’s intermediate point. Construct the lines through these 2 harmonic points which intersect respectively: (i.) DC and the more horizontal line through P (ii.) BC and the more vertical line through P. Can you prove that the last constructed lines also meet, using harmonic ranges? If so, you have another proof of projective geometry’s Fundamental Theorem. 4. Figure (90) shows a rectangle and a tangent to a conic which also touches the sides of the rectangle. Construct all 5 points of contact of these 5 lines. E
Fig. 90
184
Taking the cubic curve with λ = 2 out of the set of path curves in Figure (88), turning it at right angles and also reflecting in the x-axis, we obtain a typical bud shape, whether of flower or leaf, in Figure (91).
y
x
Fig. 91a
185
From 2 points numbered 5 equidistant from the x-axis on the line through L, draw lines through both A and B. Also join both A and B to the points on the H line where the sloping lines cross it. Continue the process of adding more and more sloping lines as indicated, so that a net of intersecting lines is obtained. Going upwards from the points 5 like knights on a distorted chessboard, label points 4, 3, 2 and 1; similarly going downwards from 5 label points 6, 7, 8 and 9. The whole profile of the bud can now be constructed. The actual bud surface is that obtained by rotating the half profile about AB. Also seen on the surface are spiralling curves (dotted where they move behind the bud). The point 4/ is a quarter of the distance between the 2 points 4 and the point 3// is three quarters of the distance between the 2 points 3. This is because as the actual distance of the moving point on the spiral from the central x line varies between top and bottom, its projection on the drawing depends upon the angle made by the rotating plane to the plane of the drawing. Letting this angle change 60 degrees at a time, the projected distances from AB will be (starting from the top left 1): r1, ¹/² r2, -¹/² r3, -r4, -¹/² r5, ¹/² r6, r7, ¹/² r8, -¹/² r9 and similar distances on other spirals. r stands for radius. The fractions arise because cos 0 = 1, cos 60 = ¹/², cos 120 = -¹/², cos 180 = -1, cos 240 = -¹/², etc.
Spirals on the 3D surface circling around the petals or fruit leaves or pine cone segments
x
186
Fig. 91b
Such spiralling path curves can be observed on flower buds, pine cones, leaf buds and also in animal and human forms. Sea shells are good examples. Bird eggs also reveal such curves if the inside egg surfaces are viewed soon after the liquid yolk and white are removed (e.g., in preparing fried or poached newly laid eggs). The most remarkable example is found in the human heart, where the seven layers of thin muscle surrounding the left ventricle actually exhibit these spiralling three-dimensional path curves. A full description of the latter and many more practical examples of path curves and their occurrences can be found in Lawrence Edwards’ book The Vortex of Life published by Floris Books. Edwards devoted much of the second half of his long life to this scientific research, requiring exact observation and clear thinking, which included a deep understanding of projective geometry. Further examples of his remarkable insights and discoveries are revealed in the pages which follow. Aortic Vestibule Mitra valve
(view from left back)
x
Fig. 91c
187
Fig. 91d
Eighteen-year-olds often show interest in whatever effects the starry worlds may have upon life on Earth. That the Sun’s daily course through the sky affects the human rhythm of sleeping and waking and plant petals’ rhythm of opening and closing is obvious. Effects of the moon, especially full moon, upon dogs and children in need of special care are well known. Nicholas Culpeper’s famous book The Complete Herbal, in which he describes the almost countless remedies available from the plant world for our physical ailments, has been in use since the reign of Queen Elizabeth the First, to whom he was a consultant. The opening paragraph for each plant mentions the particular “moving star” which governs it. The remedies are described thereafter in detail, potions taken internally and externally, plasters and poultices, etc., and how and when they shall be used. Poets, too, have had much to contribute where our soul qualities’ relationship to planets is concerned. The adjectives saturnine and mercurial for human behavior reflect the slowest and quickest planetary orbits. Although the moon is quicker
188
than Mercury, its waxing and waning are compared to amorous relationships. Poets have referred to the oak tree as the “King of the Forest” and the birch tree as the “Lady of the Woods.” These are Mars and Venus qualities and it is noteworthy that their astronomical symbols (E and D respectively) are the same as the symbols for male and female. One pictures the oak with its sturdy limbs often standing alone in the middle of a field and the birch with its slender limbs, rippling sunlight bathing its leaves. The large family of maple trees including plane, sycamore and walnut are sometimes associated with Jupiter qualities—liberality, wisdom and harmony. Jupiter regularly occupies new and successive zodiacal positions each month. The walnut tree’s Latin name is Jovis Glans, which means nuts of Jupiter. By photographing thousands and thousands of flower buds of different species over many years, transforming them to a computer images and measuring the bud widths at various points on the bud axes, Edwards was able to compare bud profiles with the ideal forms like the one in Figure (91) and others with different λ values also constructed on the computer using the appropriate formulae. The Vortex of Life gives the programs used. The correspondences discovered are remarkable, statistical mean differences proving to be extremely small. Buds are formed in the autumn and, contrary to expectation, the λ values of their profiles do not remain constant during the succeeding winter months but increase and decrease in a fortnightly rhythm—just as the visible moon phases do. The volume of the bud does not change during the changes in form. With λ > 1 the top is sharper and the bottom flatter. The full curve of yλ+1 = xλ (x – 1) in the bud’s central vertical plane is shown in Figure (92). The full line shows a cusp at A and a point of inflexion at B. It has zero curvature at A and infinite curvature at B (i.e., is momentarily flat at B). With λ > 1 the reverse is true with a cusp appearing at B and an inflexion at A with the wider part of the bud appearing nearer A instead of B. To give an idea of the accuracy of comparison between ideal mathematical path curves and what is measured in actual plants, Edwards found that out of 150 species 116 of them (about 70%) showed deviations of less than 2%.
189
Fig. 92
Limiting forms for path curves in the range 1 < λ < ∞ are the ellipse or circle when λ = 1 and the triangle as λ tends towards the infinite. When a particular bud or set of buds on the same tree are examined for λ values, they can vary between high and low once in any fortnight. When Edwards first investigated this 20 years ago, he found that the lows took place when the governing planet was in conjunction with or in opposition to the moon (a more relaxed form) and the highs happened when the governing planet was in quadrature with the moon (a tenser form). Since then he became aware that these timings began to lag behind, e.g., the highs and lows occurred a few days late. Only many years later did he realize that this was part of a rhythmic cycle of 18 years before the original alignments with the moon were repeated. So the whole phenomenon is not without its complications. Considerable interest was awakened when articles on this whole subject were published in scientific magazines, e.g., the Scientific American and the New Scientist in England, but no scientific society or wealthy industrial firm offered funding for further
190
research by Lawrence Edwards and his co-workers. Homeopathic enterprises and the special homeopathic medicines produced by anthroposophical firms in medicine factories in many countries lack the funds for supporting the further research needed. Weleda (anthroposophical) pharmacies produce Iscadoron which combats and often cures cancers. It is based upon the properties of mistletoe. It is a sad comment on modern culture that the main financial support for researchers who write articles in scientific magazines is offered by organizations which recognize military applications of the research described. One suspects, too, that the idea that the planets of our solar system can influence plant life can lead to the idea that in that case the planets may influence human life, which can make some people feel uncomfortable and assert that they don’t want to know about such possibilities! The research of Lawrence Edwards has strongly countered the materialistic contention that life on Earth cannot be influenced by the cosmos and its beings. Nearly all plant buds have in common the fact that their forms are good path curves, but there is considerable variety in the λ values of different species. Buds with low λ values include daffodil, hawthorn, blackberry, peach, cabbage, ginger, apple and cherry (between 1 and 1.5). High λ values are found in rose, campion, fuchsia, convolvulus, flax and mallow. Many of the high λ plants provide food, including fruit. There are some discrepancies in comparing the “governing planets” of plants described by Culpeper and the moon-planet correspondences observed by Edwards. Both agree that the ash tree is governed by the Sun. Indeed a simple observation of an ash in winter time, often at considerable distance, shows how its branches curve downwards from the trunk and then turn upwards where the foliage will develop in spring time—a gesture of reverence to the Sun. Agreement is found for the birch tree and Venus, but Culpeper favors Jupiter rather than Mars for the oak tree. Here, however, he may have had in mind the genus to which all deciduous trees belong, just as Saturn governs all conifers and Mars governs the genus of bushes, the Sun all grasses, Venus all mountain plants and the moon all creepers. Edwards was also able
191
to confirm that his tree-planet correspondences were precisely in line with the woods chosen by Rudolf Steiner for the capitals of his stage and great hall in the building he designed and helped build, the Goetheneum in Switzerland. Thus: Planetary Deciduous Plant genus a sphere
Main habitats
G Saturn Hornbeam Conifers
Scandinavia, Canada
F Jupiter Maple Deciduous trees
Western Europe and temperate climates
E Mars
Oak Bushes and thorns
Mediterranean climates
A Sun Ash Herbs and grasses
Meadow lands, African savannah
D Venus Birch Mountain plants (alpines)
Alps, Rockies, Himalayas, Andes
C Mercury Elm Creepers
Equatorial forests b
B Moon
Hot deserts
Cherry
_______________
Cacti
a. This column has been well developed by Frits Julius in his book Metamorphosen. b. The Sun has to be added to the governorship of the 3 planets passing between Earth and Sun.
192
The leaves of deciduous trees The variety of leaf shapes is enormous compared with that of bud shapes, especially in the genus of deciduous trees, but path curves still have their part to play. While we shall not pursue much mathematical analysis of leaf profiles, their geometrical qualities provide an interesting study, particularly the shape metamorphosis as we pass from one species to another. It was Goethe who declared that the whole plant is essentially leaf, all its other parts—cotyledons, calyx, sepals and petals, flowers, roots, stalks and fruits—being transformed leaves. The leaf is the breathing organ of the plant, whereby oxygen and carbon dioxide are interchanged with the air and moisture outside.
Fig. 93
Figure (93) shows the construction of an ideal leaf profile. AB is the leaf ’s center line. Imagine a tetrahedron ABCD, where the base ABC is at the back of your view and its apex D is only a centimeter in front of the base. The tetrahedron therefore has a very small volume and you are looking down upon it. A set of points is constructed on the line CD. They are in musical progression, closer and closer together towards C and towards D [see Figure (94)]. P2 is the middle of CD. Here CP2 =
193
P2D = 3.5 cm, and let this 3.5 be one unit length of the drawing. The whole sequence of points on CD can be represented by the formula for their distances from P2 as 2t – 1/2t + 1 units. As t runs from -∞ through zero to + ∞, these unit distances run from -1 (point C) … -7/9, -3/5, -1/3, 0 at P2, 1/3, 3/5, 7/9…1 (point D). Alternatively this set of points may be constructed as follows: F
G
E
C
J
K
D L
M
N
O
P Fig. 94
Choose the middle point M of CD and the next point N. Draw any triangle CED and produce DE to any point F. Let FN cross CE at 0 (zero) and let M0 meet EF at G. Let dashed line and dashed line and dashed line and dashed line Also let dashed line and dashed line and dashed line and dashed line
194
FM meet G1 meet FL meet G2 meet GN meet F-1 meet GO meet F-2 meet
0C at 1 CD at L 0C at 2 CD at K, etc. 0E at -1 CD at O (capital O) 0E at -2 CD at P, etc.
A closer look at part of Figure (93)
Fig. 95
The points P on CD are joined both to B on plane BCD and to A on plane ACB. The higher curve from C to A is a path curve which can be started from any P. Suppose it is P2. Then draw any straight line to meet the next 2 lines in R2 and R3. Now the position of the path curve is fixed. So join P3R3 and continue to AP5 to get R4 and so on forwards and backwards. Similarly below CD draw a line through P2 to meet the lines to B from P1 and P0 meeting them in Q 0 and Q -1. The path curve Q-1 Q 0 Q 1 Q 2 …is also fixed. Notice that the path curve of Rs comes from C, where it touches CD and ends at A, where it will touch DA. The path curve of Qs comes from B where it touches CB and ends at D where it touches CD. The third curve (the leaf profile) begins along with the Q curve, comes up to cross the Q curve and then approaches the R curve and touches it at A along with line AD.
The 5 points on the leaf profile are the intersection of
R-1B with Q-1A, R0B with Q0A, R1B with Q1A, R2B with Q2A, R3B with Q3A, R4B with Q4A, etc. (S-1) (S0) (S1) (S2) (S3) (S4)
A simple example of this path curve of Rs and the other one of Qs is shown in Figure (96).
195
Fig. 96
Some examples of deciduous tree leaves and their metamorphosis from one species to another On the next 2 pages [Figures (97) and (98)], a series of leaf profiles is sketched showing the changes in shape and arrangement of leaves, where the governing of planets sometimes singly and sometimes combined is indicated by planetary signs. So hornbeam leaves are harder and serrated (Saturn), but beech leaves are softer (combined Saturn and Jupiter) and copper beech leaves are still softer (Jupiter). Ash leaves are pinnate (usually eleven per stalk), but in the oak leaf what were separate seem joined up or webbed rather like what happens in the toes of ducks’ feet. The same sort of “webbing” effect is observable in proceeding from the horse chestnut leaf (A – E ) to the sycamore leaf (F), which are palmate, as that from ash (A) to oak (E), which are pinnate.
196
Fig. 97
Moving on to Figure (98) from Sun to lesser planets, the arrangement of leaves around the stalk becomes less ordered. Birch leaves come alternately, not pinnately, cherry leaves come in bundles. The two parts of the leaf are unsymmetrical when reaching Mercury, a feature of that planet’s orbit around the Sun when viewed from the Earth. Lime and especially elm have this quality. Ginkgo and Alnus are riddles, possibly mixtures of Saturn and moon governance. Ginkgo Biloba is a very old tree. Alder leaves have the bulk of leaf area nearer the top rather than the base as in most leaves.
197
Fig. 98
Geometrical construction of ideal leaf shapes It is possible to develop further the idea used on a previous page by regarding the projection of a 3-dimensional path curve on to one of the faces of an invariant tetrahedron. Let this tetrahedron have A, B, C and D as its invariant points [Figure (99)]. The strong curve APQB represents a path curve within the tetrahedron and its projection from upon plane BCD is the dashed curve C P1 Q1 B.
198
Fig. 99
Once P → Q is chosen, the whole path curve and all other path curves of the set depending on this movement are determined. Generally speaking the view of the strong curve will include one inflexion and the dashed curve will have no inflexion. Let the letters of the four principle points denote their (algebraic) worth in 3 dimensions. Any other point can then be denoted by kA + lB +n C + nD. So the middle point of AB can be denoted by A+B and its infinite point by A-B. One third of the way from A to B will be 2A +B. If the point Z on AB is x units from A and y units from B, it will be denoted by yA + xB. A simple musical progression between A and B containing the two points shown will be A, ….2n A+B,…8A+B, 4A+B, 2A+B, A+B, A+2B, A+4B,…,A + 2nB, …, B. This new kind of algebraic symbolism can become very powerful and the leaf profile which follows can be regarded as the locus of A(23t+1) + B(2-t +1) + C(22t +1). For example in Figure (100), G, the centroid of triangle ABC, is A + B + C. Its distance from C is twice its distance from M, just as the numerical coefficient of M (= A+B) is twice that of C. This advanced kind of geometrical algebra goes beyond Class 12 work, so we leave it here as just an indication for further development.Figure (100) indicates the construction of a sort of convulvulus leaf. The musical progression along line CD is joined
199
to both A and B. The dashed path curve from B through R to D is constructed; also the long and narrow dashed path curve from A through Q to C, although its construction is only partly indicated. P (any point on CD) gives rise to Q and R on the dashed curves. QB meets RA in S and the strong curve is the locus of S, i.e., the half leaf profile. The tip of the leaf is acuminate and its base is cordate. In Figure (101), the two thinner curves are conic path curves and the thicker curve is the leaf profile as usual. For any point P on CD, the lines RA and PB cut the two conics in Q and R. AR cuts BQ in S and the locus of S is the leaf ’s half profile.
200
Fig. 100
Fig. 101
Chapter 14 Class Twelve Computers, Calculators and Chaos Theory In Class 9 hand calculators were introduced to the students in the second term once it was clear that the students were proficient in all the basic operations of arithmetic. This obviated the need to use logarithm and other tables. It also introduced them to the setting up of programs for repeated calculations. In the afternoon arts and crafts blocks the students were introduced to stenography; they learned how to type properly and use word processors. These skills were developed further in Class 10, making tedious calculations in trigonometry unnecessary and the numerical evaluation of formulae for the summation of series, etc., simple matters. Most good schools nowadays have special rooms for learning computer skills, both within school hours and after school. The afternoon craft blocks can now evolve further from word processing to statistical and graphical tasks for which the computer is essential. Such options in the school curriculum are parallel to skills like bookbinding, weaving, cookery and gardening. The arts program—painting, sculpture, music, etc.—still remains essential to all students. It is Class 11 when options should be offered, not earlier. Computer skills do not belong to the main lesson blocks where the subjects (including mathematics and the sciences) have wide cultural content. It is important, however, for all students to have some experience of where only computers and good calculators can solve certain mathematical problems, so Chaos Theory is a helpful part of Class 12 math.
201
Chaos theory It is now over 20 years since James Gleich wrote his book Chaos, since when many new editions by various publishers have followed and many other books on the subject have been written by other authors. The subject has been described as the third breakthrough of the 20th century following on the discoveries of the Theory of Relativity by Einstein and of Quantum Mechanics by others. Gleich graduated from Harvard and for ten years was a reporter and then editor of the New York Times. He was passionately interested in new discoveries in physics and mathematics, especially those which broke new ground and challenged classical foundations of the sciences. He was not one of the discoverers and developers of Chaos Theory but was a brilliant reporter of those who were, describing not only their new ideas but their individual human situations and how they overcame obstacles in their paths. Gleich wrote about many people who made essential contributions to Chaos Theory, starting with Feigenbaum, Lorenz and Smale. Even twenty years ago the output and availability of computers and the speed with which they produced results were very limited compared with today. Often those developing the theory had to wait some time for the answers to come through. This, however, proved to be an advantage as it gave them time to plan the next steps. I have found this too, so that I prefer to work with a good hand calculator rather than a fast computer. This is also a good reason in Class 12 to introduce Chaos Theory through calculators rather than computers. The deepest problems in human life, especially those of a spiritual and moral kind, can be solved only by pure thinking, never by computers; and for this, adequate time has to be given, including time to sleep on the problems. The logistic difference equation While this may sound formidable, it is actually quite a simple idea. Take any number—whole, rational, irrational (e.g √2) or transcendental (e.g., π or e)—and call it x where 0 ≤ x ≤ 1, then multiply it by (1 – x) and also multiply by some constant number k, where 0 ≤ k ≤ 4 and record the result. Thus x2 = k x1 (1 – x1). The original x (now called x1) has given rise to a new number x2.
202
It is easy to see that x1(1 – x1) will always lie between 0 and 1, indeed its greatest value is 0.5 x 0.5 = 0.25, also that since k cannot exceed 4, kx(1 – x) has its greatest value 4 x 0.25 = 1. Let us now take various values for k and keep x1 as 0.5. (i.) Suppose k < 1, say 0.9 = 0.225000000 Then x2 = 0.9 x 0.25 x3 = 0.9 x 0.225 x 0.775 = 0.156937500 x4 = 0.9 x 0.1569 x 0.8431 = 0.119077309 Clearly the further we go, the smaller will the x value become, but we shall simplify the whole process by designing a program on our calculator. (See note about calculators at the end of this chapter.) Let
A = 0.5 B = 0.9 C=0 D = BA(1-A) C+1=C
C ≥ 10?
Put A = D
NO
YES
PRINT D
This loop will make the program go around 10 times and then stop, printing 0.029597779. When entering C = 10 instead of 0 and C ≥100? in the question stage, in less than half a minute later the printed answer will appear, a much smaller number than before. Again having entered C = 100 and C = ≥1000? in the question, an even smaller number will appear for the answer. If this procedure is continued, the final answer will soon show all figures in the decimals as zero. For all values of k less than 2, the sequence of x values decreases to a limit less than 1.
203
(ii.) When k = 2, x2 = 2 x .5 x .5 = .5, so this value does not change. When k = 2.5, x2 = .625 x3 = .586 to 3 d.p. x4 = .607 " " x5 = .597 Alternate values of x regularly decrease and increase towards 0.6 as may be seen at once since 2.5 x 0.6 x 0.4 = 0.6 When k = 2.8, x2 = .700 to 3 d.p. x3 = .588 " " x4 = .678 " " x5 = .611 Decrease and increase occur as before. x11 = .635…x12 = .649… The limit of the sequence is clearly perceived when x27 and x28 are reached at .643… When k = 3, it is at once recognized that 3 x 2/3 x 1/3 = 2/3, so the sequence limit is .6 (decimal 6 recurring). (iii.) A new phenomenon arises once k > 3. When k = 3.1, x2 = .775 x3 = .541 to 3 d.p. x4 = .770 " " x5 = .549 Decrease and increase occur again, but the limits of the decreasing and increasing alternate terms are not the same. Thus x28 = x30 = .7646… While x29 = x31 = .5580… This change from one to two limits is called a bifurcation. When k = 3.2, x2 = 3.2 x 0.5 x 0.5 = 0.8 x3 = .512 x4 = .7995 x5 = .5129 x6 = .79947, writing an increasing number of decimal places as the results become closer. x7 = .51302 x8 = .799458 x9 = .51304 x10 = .799455 Hence to 5 d.p. the two limits are 0.51304…and 0.79945. 204
(iv.) When k= 3.5, another new event has taken place. For if x1 = .5, x2 = 3.5 x .5 x .5 = .875 x5 = 500898…, just a little greater than .5 after 4 iterations from x1. Take k a bit smaller. When k = 3.499 and x1 = .5 then x5 = .50027, which is closer to .5. With k = 3.4985 x5 = .499961, which has overshot .5. So take k = 3.49856 x5 = .499998, then k = 3.498562 x5 = .500000186 k = 3.4985615 x5 = .499999876 k = 3.4985617 x5 = .500000000 These results come from changing the program’s question to ≥4? and so enable the values of x5 to appear on the calculator. x6 = .874640425 x7 = .38359823 x8 = .82723711 completing the four numbers called attractors which continue to remain constant for the value of k. (v.) Once k = 3.4985…has been passed, e.g., k = 3.5 or 3.56, there are eight numbers which are repeated after 4 iterations. So when k = 3.56, x2 = .89, x9 = .492…(too small) Try k = 3.55, x2 = .887 x9 = .504 (too large) k = 3.554, x2 = .888, x9 = .5007 (too large) With k = 3.554640862, x9 = .500000000 and there are 7 other attractors (x10, x11, x12, x13, x14, x15, x16) which remain constant for this value of k. Thereafter with k slightly bigger (say 3.555), another bifurcation takes place with 16 x values in sequence. This whole evolution from k = 3 onwards is called a cascade. At k = 3.566667380 there continue to be 16 attractors but the period thereafter then changes to 32. (vi.) The whole cascade continues with higher values of 2n attractors (e.g., 64, 128, etc.) and the following table and graph [Figures (102) and (103] show the whole phenomenon so far. A table of significant k values at which the number of attractors is doubled is shown below.
205
Number of Nearest result to .5 k value Increase in attractors, t after t iterations k value
1
.5
Ratio of increases to 4 d.p.
2.0
1.236067977
2 .5 3.236067977
0.262493723
4.7089
4 .5 3.498561700
0.056079162
4.6808
8 .5 3.554640862
0.012026518
4.6630
16 .5 3.566667380
0.002576151
4.6684
32 .500000002 3.569243531
0.000551763
4.6692
64 .500000001 3.569795294
0.000118171
(4.669…)
128 .500000006 3.569913465
0.000025309
256
.499999996
?
3.569938774
Fig. 102 The increases contain too few decimal places to give enough accuracy in the last column. Were we using more than the 10 s.f. which most calculators provide, i.e., beyond 20 s.f. on computers, then the last column would show numbers becoming closer and closer to 4.6692016090…, known as the Feigenbaum Constant. Beyond the chaos line (k # 3.57) one at once finds k values which belong to new cascades, but no longer 2n cascades. The graph [Figure (103)] appears at first to be solid chaos alleviated by strips with no chaos. In fact, however, there are no chaos areas. The dark areas could be left uncolored, but with an increasing number (actually infinite) of chaos lines, which have no width, of course. This means that since the x value of any chaos line consists of an endless number of decimal places with no calculable way
206
of detecting rhythmical relationship between them (e.g., with rational numbers), one can choose any rational value less than 4 and determine the cascade in which it lies. For example, take the region 3.57 < k < 3.58. By successive application of the logistic difference equation it is found that there is a small band of 84 attractors, because beginning as usual with 0.5, the 84th iteration of 3.57585190 is .499997999 and the 84th iteration of 3.57585191 is .500000178.
Fig. 103
In Figure (103) the wide strip free of chaos is labelled 3n. This means that it begins on the left with x = 3.83 with 3 attractors and a bit later 6 attractors, then 12, 24, 48…and finally a very large number of attractors which become lost in a chaos line just as the cascade 2n did earlier. Similarly the thinner blank strip beginning with 3.74 begins with 5 attractors and then 10, 20, 40, 80, …
207
before ending in another chaos line. Such chaos-free strips are known as windows. There are an endless number of windows. The cascade in each window has the form shown in Figure (104) for a 3n cascade. The 2n cascade which begins the whole process is similar to what can be perceived in a spray of mistletoe berries. The small area A when enlarged displays a shape very similar to the whole drawing. This is true of any small section, so the whole drawing has the quality of self similarity. Fig. 104 A further development of the ideas visualized in Figure (103) leads to the remarkable drawings in many books on chaos theory. Some remind one of fern leaves, but of course they are only mathematical images and could never be pictures of real, living fern leaves. Others depict what are called Mandelbrot sets (also called gingerbread men), in which a magnification of any small part reveals a copy of a large part of the whole drawing. Once again this reveals the concept of self similarity, a major feature of all chaos theory. Arising from the researches of Mitchell Feigenbaum and others into possible applications of chaos theory, which would have been quite impossible without modern computer technology, thinkers in widely separated scientific disciplines have found a way of overcoming the “crisis of increasing specialization.” In a 1980 lecture by Stephen Hawking called “Is the End in Sight for Theoretical Physics?” he indicated how this crisis could be overcome … and so this has been proved by chaos researches. Applications involve such diverse realms as the astronomy of solar systems, convection currents in heated liquids and gases, cyclical epidemics of diseases like polio and rubella, population biology, the physics of matter at very low temperatures—the list seems endless. Hopes of finding solutions to weather forecasting have not been realized as yet. Nor have efforts to predict the rising and falling of prices on stock exchanges, for which we can be grateful. It is not possible here to go into detail in these realms but an awareness of these modern developments are important requirements at the age of Class 12 students.
208
After viewing a chaos theory graph based upon a trigonometric function instead of the usual logistic difference equation and also an important note about hand calculators, this chapter will conclude with a view of water vortices where λ values are negative and chaos of a deeper kind emerges. Finally a suggestion will be made as to how a non-mathematical experience of chaos—but one requiring pure thinking of the highest order—can lead to spiritual experiences which many young people are beginning to have in our 21st century. Instead of using the logistic difference equation, whose graphic form is a parabola, we could use a sine function, which also produces a form with a single curve. A general case: x → r (1– / sin π(0.5 – x) / n For 0 < x < 0.5, f (x) = r (1 – sinn π (0.5 – x)), f (x) = rn π sin n-1 θ cos θ, where θ = π (0.5 – x). f (x) = rn π2 sin n-2 θ (1 – n cos2 θ). So there is a point of inflexion when /
//
cos2 θ = 1n , i.e., at x = 0.5 – 1π cos-1 √ n1 , provided n ≥1. f(x) then = r [1 – sinn (cos-1 √ 1 )] n n – 1 = r [1 – ( )n/2 ]. 2 As x → 0, f (x) → 0, f (x) → 0, f (x) → rn π2, a minimum point if r and n are +. /
//
As x → 0.5 f(x) → r, f (x) → 0, f (x) → 0 if n >2, → –2 π2r if n = 2 and → ∞ if n < 2. Let r = 1 Examples of points of inflexion for some values of n: /
//
n
1.1
2
coordinates
(.40, .73)
(.25, .5)
A
B
10 (.102, .41) C
209
× Fig. 105
(i.) Above AA , BB and CC the respective curves are wholly convex, but apart from n = 2 the top point is a singular point with zero curvature for n > 2 and infinite curvature for n < 2. (ii.) Cascades in chaos theory can only occur with convex curves. (iii.) The locus ABC… is y = 1 – p 1/1– p2, where p = sin π(¹/² – x). /
/
/
Important note about hand calculators There are now many firms which manufacture hand calculators, not only British but American, Japanese, etc., with books or leaflets of instructions—translated into English where necessary. Each firm can have many models, some specializing in statistics, graph plotting and so on. Often a firm stops making a particular model in favor of something more complicated. For many years I used a Sharp EO-512H accompanied by a clear operation manual. It was excellent for Chaos Theory as it had a single looping function. Unfortunately some operations became faulty after very heavy use and I had to dispose of it. But by then Sharp had declared the model obsolete and had ceased producing it. No other relatively inexpensive model from any of the various firms contained the looping function. The nearest I could find was a Casio fx-83MS, which can be used as follows:
210
Take the case of k = 3.5. I set up the calculation .5 = x [This shows 0.5 and the program reads Ans x] Then continue with (1 – Ans) x 3.5 [The program reads Ans x(1 – Ans) x 3.5, the usual logistic difference equation. 0.5 is still the number shown below the program.] 0.875, which is x2 Press = and you get Repeat and obtain .3828, which is x3 8269, which is x4 . 500898, which is x5. If we continued to press =, we would obtain .500884209 for x17 and .500884210 for x29 and the same answer after every 4 pressings of =. This is not as convenient as pressing some letter which has already been given the number 4 on the program, but the cascade at this point is complete, i.e., .500884210, .874997263, .382819683 and .826940706. Once k is varied the whole program has to be repeated each time. Pressing 4 (or a much larger number, say 100) times is inconvenient, a time waster, but can only be avoided by purchasing a more expensive calculator.
Fig. 106
211
Chaos theory for differing k values (zero being off to the left of the paper and 4 off to the right) This is a further example similar to Figure (103), showing how in the 3-cycle space there are at first 3 attractors, then 6, 12, 24 …, etc. attractors before a chaos line intervenes. k = 3.83 gives a nest (IJKLMNI) of 3 attractors (at .156, .504 and .957). So the logistic difference equation 3.83 x1 (1 – x1) = x2 has the three final attractors above. In this way the geometry of a parabola and the sloping k-line through O illustrate the logistic difference equation’s attractors. In Figure (107) we see how the logistic difference equation x n+1 = kxn (1 – xn) is applied to the parabola y = 4x(1 – x). Starting with N, where x0 = .504, just to the right of H (the top of the parabola) and moving horizontally to the right to meet line y = 4x/k where k = 3.83 in point I, we see that x1 = .957. Then going horizontally from J, where x = .957 meets the parabola again, we reach K on line y = 4x / 3.83. The process continues to reach points L and M and finally to N, the point from which the “nesting” started. After that the 6 arrowed lines repeat the journey infinitely. Hence for k = 3.83 there are just 3 attractors: x = .156, .504 and .957. On the right is shown a series of parabolae indicating how the values of k (3, 3.45, 3.54, …) will lead to k = 3.57, the chaos value. The smaller sketch shows a clearer picture of the fractal development, which looks like a mistletoe plant blown by the wind. The 4 points C bifurcate to give 8 points D, etc., the period doubling with the “twigs” getting closer and closer to the chaos line 3.57 (actually 3.5699456…).
Fig. 107
212
Air/water vortices Positive values of λ, whether greater or less than unity, lead to bud profiles. For example Figure (93) shows profiles for λ = 2 and λ = 3. Were this drawing to be shown upside down, we would need to specify these profiles as λ = ¹/² and λ = ¹/3.. But λ = -2 gave the vortex shown in Figure (93), the vertical axis of the drawing being a tangent to the infinite point below. In Lawrence Edwards’ book The Vortex of Life, reference is made to rotating cylinders of water. At a certain moment in its increased velocity, a vortex of air is formed in its vertical center line. Using a large cylinder up to 2 feet in diameter and a method of allowing water at the tank bottom to be led up and slowly fed into the top of the cylinder near its outside, a remarkably stable vortex in the water/air boundary can be perceived. This experiment was achieved by Alan Hall and is described and illustrated in his book, Water, Electricity and Health, published by Hawthorn Press (Stroud, England). Figure (108) shows the 3 sections of this vortex. At section A, the top of the water surface is smooth and appears stretched like a drum. Section B is twisted as if it were a twisted piece of rope. Section C is narrow and smooth again, but is relatively unstable, its lowest tip subject to upward and downward movement through a fairly small distance. John Blackwood in Sydney, Australia, made several models Fig. 108 of glass cylinders with various dimensions and angular velocities of water flow and concluded that the vortex was remarkably similar to a 3-dimensional path curve having a λ of -1.75. When λ = -2, Section B contains complete chaos with no cascades. This occurs where the angles in the chaos zones are high (>25°) [see Figure (109)]. While the speed of water rotation is low,
213
the small vortex near the surface center is formed. At a critical angular velocity of the water, the vortex deepens quickly towards the base of Section C. In Figure (109) notice how at point P the chaos angle is 31° and lies between PE and tangent PF (which touches the curve at F), also that PD and PE are equi-inclined to the vertical PV. The smaller surface is y2 = 7 -7/2 (x -3/2 + z -3/2) and the larger surface is y2 = x -3/2 + z -3/2, the z-axis being perpendicular to the plane section of the vortices through the origin O. Figure (109) shows a section of a water vortex and includes zones containing complete chaos. To go further into detail and show what Ian Stewart and others have described as “cobweb diagrams” to show graphical iterations of logistic mappings using vortex curves and varying positions of lines through points on the vortex would carry us beyond what was intended in this section of the book. Complete chaos arises when the lines and vortex curves above intersect in imaginary points and so present “cobwebs.”
Fig. 109
214
Chaos theory relevant to the air/water vortex λ = -1.75, n = λ+1 = -0.75 We are no longer concerned with the geometrical relationships between a parabola and lines through the origin, but those between the latter and a higher degree curve y = 1 – (2 x – 1) -0.75 or (y – 1)2 = (2x – 1) -3/2 or (y – 1)4 = (2x – 1) -3.
Fig. 110
The lines through O are y = 4x/k 1. If k = 16, so that 4(1-(2x/ - 1) -0.75) → x2 , choosing any number for x, e.g., 1.5, 1.5 →1.62 →1.82 →2.06 →2.30 →2.47 (5 iterations) → …. 2.669 (13 iterations) → …. 2.669483691 (31 iterations) correct to 9 d.p. 2.
But if k = 12, so that 3(1-(2x/-1)-0.75) →x2, 1.5 → 1.22 → 0.71 → -2.78 →….CHAOS
215
Geometrically this is because the curve and line through O have imaginary points of intersection. The chaos zones are full areas, not a series of momentary veils. Limits of x and corresponding k boundary values are shown in the following examples: Within these restrictions curved chaos areas can be drawn. x
0 to 1.830
0.000121 to 0.741
-0.0000738 to 0.366
0.000444 to 3.663
k
2.53 to 14.08
0.53 to 2.83
0.159 to 0.844
8.96 to 47.2
Non-mathematical experience of chaos On the ceiling of the Sistine Chapel in Rome are seven beautiful paintings, flanked by two others on its slopes, created by Michelangelo. Thus a sevenfold evolutionary time and space sequence is bounded by less form-fulfilling phases. 0. 1. 2. 3. 4. 5. 6. 7. 8.
God creating out of CHAOS God creating Sun and Moon God dividing the waters The creation of Adam The creation of Eve Paradise, temptation and expulsion Noah’s offering to God The ark and the Deluge Noah’s drunkenness (more CHAOS!)
This can be a basis for deep meditation upon the nature of the human being. Qualitatively it reminds one, too, of the seven wandering stars pictured on page (viii). Sun and Moon, the heavenly bodies we first become aware of in life, form anchors for our early earthly development in physical space just as our parents can do so for our soul development. Then the waters which are divided; rivers and oceans give rise to the water vapor above them—mists and clouds. Blood, arterial and veinous, is exchanged through the heart and in the extended capillaries. Blood is the carrier of life. Adam was created by God’s breath and it is breath which stimulates consciousness and feeling. So stages
216
1, 2 and 3 remind us of our human physical body, our life-filled body and our psyche. The fourth principle of human beings is the ego which is responsible for our actions upon Earth. It was Eve who, in the first place, gave way to temptation, but we can also recognize woman as God’s greatest artistic achievement. The first five stages can also be compared to qualities of the five planets Moon, Mercury, Venus, Sun and Mars. Noah’s offering to God has to do with gratitude, an oft-described Jupiter quality of the human soul. Finally the escape from the Deluge by means of the ark is akin to Saturn’s protection from the chaotic incursions of outer space, of which Saturn’s ring is a symbol. The seven stages can also be felt to be allied to the concentric spheres surrounding the surface of our planet—earth, water, air, warmth, light, tone and word—described in old alchemy documents. The words chaos and gas originally denoted the same thing. Between the solid objects of the physical world and ourselves, there is this invisible substance we do not see with our eyes—the air or gas around us. If we had a different set of sense organs, perhaps we could perceive it directly. Carrying this thought further, could it possibly be true that we are surrounded by many kinds of chaos for which we lack perceptive organs? Certainly we sometimes experience in ourselves and others a kind of chaos regarding inner soul life, appearing as insoluble riddles and a longing to find the organs and light to overcome these. We know that no physical perceptions, computer programs or intellectual analysis can help. Yet we feel that somewhere a light can be found that will shine out of the darkness and resolve the chaos. There are quite a number of young people today, and their number is on the increase, who are aware that a different kind of world penetrates our ordinary sense world. Some will speak of an awareness of real but invisible beings. They may call them spiritual or noetic and feel they can offer help. If, however, these young people have been tempted into taking drugs, they feel invisible beings who are hindrances. It is not the task of a teacher in schools such as Steiner/ Waldorf schools to inform 17- or 18-year-olds of the meditative paths of experience and knowledge proposed to adults by
217
Rudolf Steiner and others which lead to healthy results, but if such a young man or woman approaches a teacher in confidence, the teacher ought not withhold advice. Such advice should be based on clear thinking, a rich feeling of companionship for human beings as a whole, and an awareness of what is morally good. Does a mathematical experience of chaos theory help with the deeply human problems indicated above? All mathematics is a great help to achieve clear thinking. Work in chaos theory introduces the students to quite new ways of dealing with numbers and number sequences and of becoming aware of veils. On the path to spiritual experience the student will also find the need to penetrate veils, not of the mathematical kind but of a moral kind, where the aim is always to make oneself a better individual who can be of help to others he meets. So the direct answer to the question posed is “No!” But someone who has had to deal with veils of chaos in mathematics will have experienced the qualitative side of general veiling problems and this will help him in his spiritual efforts. Mathematics is certainly not the key to solving all problems. In trying to climb up a path leading to universal learning about life’s riddles, a moment comes when the being of mathematics declares, “Thus far I have helped you develop imagination and inspired ways of solving problems and intuitive actions to master various processes and fields of enquiry. Here I have to leave you; the further path is yonder. My best wishes to you! A hearty au revoir!” The full answer to the question is “No, but…” For mathematics has not only a quantitative side but also a qualitative side. An example of the latter is furnished by the ancient Egyptian mathematician Ahmes. His compatriots were not very proficient in handling fractions, so he began with the simple truth in whole numbers that 2 = 1 + 1, then proposed the riddle of including denominators: 2/7 = 1/? + 1/?. Of course 2/7 = 1/5 + 1/2 is incorrect and he showed them the correct method: Take the whole number following 7 and halve the result to get the first question mark, then multiply the latter by 7 to get the second question
218
mark. Hence 2/7 = 1/4 + 1/28. The method will work with any odd number as well as 7. Now consider a wholly practical task in life. Suppose someone decides he wants to become proficient in cookery, wood carving or some other skill. After working at it for a week, he feels he knows something about the desired skill, but needs to continue daily for a longer time. Only after working for 4 weeks, i.e., a month or 28 days, will he realize the real nature of the skill he has to acquire. The same kind of example is used in Steiner (Waldorf) schools. The first two hours of each school day is focused on the same particular subject, e.g., writing and reading or arithmetic or geography or whatever. But the main lesson does not stop after one week but continues for a whole month before another subject becomes the focus. This enables children and young people to become fully “at home” in each subject and leads to great economy in learning.
219
Chapter 15 Concluding Chapter As the age of students increases, the question for the teacher is how far to go into detail in a particular subject and how far to briefly summarize. This is as true of mathematic and geometric subjects as historical and other subjects. The choices must depend upon the wishes and interests of the group of students in the class concerned and especially so in Class 12. Early in Class 12 (and this applies to Class 11 to some extent) the teacher needs to survey the possible directions from which to choose, and it would be against the spirit of such enquiries to allow only examination courses to influence choices. In the topics described in this book there are certain omissions. Classes promoting efficiency in the use of computers have been left out largely because they require a different discipline from that needed to develop the pure thinking of mathematics, including geometry. Facing a machine and adapting to its procedures can deny openness and wholeness to the human spirit. As has been suggested, separate periods of study and practice are preferable. Another topic not developed since lower school classes is statistics, but this also benefits from computer techniques. A teacher of upper and high school students may find it more worthwhile to develop the subjects of matrices and vectors, for example from Class 10 geometry where (1 0) (0 1) represents identity in transformations, (3 (0 represents enlargement,
220
0) 1)
(0 (0
2) 1)
translation and (3 1) (1 3) a musical progression, the matrix (a b) (c d) ax + b . representing the general projective transformation y = cx + d Another teacher may find it valuable to extend the work on compound interest and mortgage repayments to familiarize students with the working of the stock exchange and the big issues of economics today, both nationally and internationally. In such a topic the moral and practical qualities of the 3 ways of giving and receiving money and their virtues as introduced in Class 6— initiative in buying and selling, freedom in lending and borrowing and love in giving and receiving—will become integral motives. Another topic in algebra and geometry could be the many kinds of graphic representation: not only Cartesian coordinates, but line coordinates, triangular, aerial and projective coordinates, functions of distances from two foci (e.g., rs = constant for Cassini ovals) as well. Behind all these considerations stands the comprehensive question, “What is the real value of learning mathematics for life itself ?” When one of his students, attempting to answer this question posed by Socrates, referred to the need to become rich, the latter offered him a penny and proceeded to demonstrate that mathematics teaches one how to develop pure thinking, a spiritual activity not dependent on the physical brain. Over a hundred years ago in a modern context Rudolf Steiner wrote his book The Philosophy of Freedom, in which he proved that real thinking is not a material activity, not a product of the brain. His proof is an experience which anyone can attain given sufficient effort. The brain’s real task is to collect the sense impressions conveyed to it from the rest of the body so that our thinking activity can penetrate and cognize them. Knowledge of the world and ourselves is not
221
the result of our cranial nerve activity. The opposite is true—our thinking activity causes changes in the substance and organization of the brain. An offshoot of the latter, but again caused by the former, is the development of our memory of earthly events. In the second half of The Philosophy of Freedom, Steiner showed that not only the thinking activity of the human soul is free from material bodily causes, but the feeling and willing activities of the soul can be equally free, once the interweaving of these 3 is understood. What Steiner went on to achieve in the early quarter of the 20th century in his Science of the Spirit, also called Anthroposophy (the wisdom active in the human being), is not for this book to go into. Its consequences may be found in the many books bearing his name and the many social activities he inspired, such as Rudolf Steiner/Waldorf schools (now over 1000 in the world—on all continents), new impulses in all the arts, biodynamic farming and gardening, medicine and social endeavors of many kinds from economic, political and therapeutic realms, including caregiving of children with special needs. Above all he was a great inspirer of spiritual and moral knowledge and practice.
222
Index A Abraham “Achilles and the tortoise” affinity Ahmes (an ancient Egyptian scribe) amplitude analysis angles > 90° Apollonius’ theorem approximations Archimedes Aristotle arithmetical progressions astronomical measurement attractor
7 70 110, 111 218 150–151 17, 153, 193, 217 90 92, 96 38, 40, 167, 168 78 7 68 8–9 205–206
B Bach, J.S.
10
Baker, H.F.
100
Beighton, H.
112
223
bifurcation
204–205
binomial theorem
29, 37, 41, 120
Brianchon’s theorem
141–143, 147–148, 171
cascade
206–209, 211, 212, 215
C calculators
44, 63, 73, 88, 124, 201–203, 205, 209–211
chaos theory
201–202, 208, 210, 211, 215, 218
chronological age combinations
18 21, 24, 26, 63
complementary perspective views 47, 103–105, 110–111 complete quadrangle and quadrilateral
47
completing the square
40
compound interest cone
64, 221
47–51, 55–56, 83, 135, 139, 163, 164, 170–171
confocal conics Copernicus cosine law
52, 56 8–9 95, 96
D Desargues’ theorem
103
dilation
110
directrix
224
53, 58, 60
E eccentricity Edwards, Lawrence
52, 53, 58 11, 187, 189–191, 213
elation
105, 110, 111
electrical resistance ellipse
37
16, 49, 51–55, 57, 60–61, 136–137, 148, 190
elliptic envelope
144–145, 148
enlargement equations, linear and quadratic
109–111 33, 34, 36, 39, 40, 182
Eschenbach, Wolfram von
117
extensions of Pythagoras’ theorem
96, 98
exponential and logarithmic functions
122
F factorial Feigenbaum and his constant
23, 122 202, 206, 208 220, 221
formulae transformation fractal development
212
function of a cone
47
fundamental theorem of projective geometry
17, 73, 171, 172, 175
225
G Vasco da Gama
10
geometric progressions
69, 70, 111, 182
Gleich, James
202
graphical solutions to equations
39
harmonic progressions
81
H harmonic ranges and pencils
145
Hawking, Stephen
208
Hermes
31
homology sequences of prospective triangles
104–105, 110, 136
hyperbole, constant focal difference hyperbolic functions
160–162
I infinite points, lines, and planes 100, 107, 110, 135, 148, 213 irrational numbers integration
226
43, 149 158–159, 161, 163
J Julius, Frits
192
Kepler, Johannes
8–9
K L leaf types – pinnate, palmate, etc. leaf endings – acuminate, cordate, etc. Leibnitz
196–197 200 10
liberal arts
7
line reflection
110
Lobachewsky, Nicholai
10
logarithms to any base
72, 73, 124
logistic difference equation
202, 207–209, 211–212
M Mallory, Thomas measure at rest and in movement modulus money – its 3 uses mortgages
117 7 150–151 18, 64, 221 64
227
multinomial theorem
41
musical intervals
75
musical progressions
76, 182, 183
Newcomen, Thomas
114
N 10, 17, 117
Newton, Isaac number at rest and in movement
7
numbers – real, rational and transcendental 149, 160, 202, 207
O objective judgment
17
orders of metals in
i) thermal conductivity
115, 116
ii) specific heat
115, 116
P Pappus’ theorem parabola
171, 172, 173, 175 16, 39, 57, 58, 60, 61, 133, 136, 139, 141–142, 209, 211–212, 215
path curves – general linear construction
178–182, 185, 187, 189–191, 193, 195, 198–200
Peano’s axiom 228
101, 102, 174
permutations
21, 23, 26, 63
plan and elevations
50, 92
points, inaccessible and infinite points of inflexion point function probability
100–102, 106, 107, 135, 148 189, 199, 219 106–107 32–33
R radius of curvature
129, 133, 189, 210
S self similarity simple shear Simpson’s rule sine law Smale, Stephen Socrates spirals – 4 main kinds square roots Steiner, Jakob Steiner, Rudolf surveying synthetic geometry
208 110, 111 167 93, 94 202 7, 221 16, 80, 81, 82, 83, 85, 112 41, 43, 44, 149 110 15, 119, 192, 218, 221, 222 89 139 229
T translation triangles, congruent and similar
29, 109, 110, 111 97, 110
triangular numbers
30, 31
tubular and spherical bones
60–61
Tycho de Brahe
9–10
V Variations Veblen and Young volumes of solids
22, 24 99 47, 163–164, 166
W Wagner
117
windows free of chaos
208
world economy
18
Zarlino
10
Zeno
70
Z
230
Topics in
Mathematics for
g Hi h f r Schoo o d l a ls W Volume 2
For Ages 14 to 18
Ron Jarman AWSNA Publications
The Association of Waldorf Schools of North America Publications Office 65-2 Fern Hill Road Ghent, NY 12075
Topics in Mathematics for Waldorf High Schools
Ron Jarman was born in Manchester, England. After obtaining an honors degree (MA) in mathematics and mechanical sciences. He became interested in philosophy and spiritual science from a book about Rudolf Steiner by Rom Landau called God Is My Adventure. At the end of World War II he joined the Steiner Waldorf School in nearby Ilkeston, Michael House School, where he taught mathematics, geography, and gymnastics to adolescents in the upper school, and later became a class teacher. Ron served on the Steiner Schools Fellowship Council and became chairman, a post he held for twenty-five years. He helped bring about the foundation of an International Waldorf Schools Council which meets twice a year in various European capitals and includes schools from all over the six continents of the world. He helped in the development of Emerson College and he became involved with teacher training, suceeding Francis Edmunds as leader of the Education Course. Following visits to Waldorf schools in Slovenia and Croatia he joined the women's movement "Through Heart to Peace," which is very active in former Yugoslavia, especially Bosnia, and became well known for breaking the siege of Sarajevo by Serb extremists. During the last three years (at present he is 88) Ron has given lectures throughout England on the theme of "Destiny, Reincarnation, and Resurrection." He wrote Teaching Mathematics in Rudolf Steiner Schools for Classes I–VIII as well as many articles and essays for magazines.
by Ron Jarman
