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ANIMATION MATHS

I VO D E PAUW AND B IEKE M ASSELIS

ANIMATION

MATHS

Chapter 1, David Ritter; 2, John Evans; 3, Wouter Verweirder; 4, Daryl Beggs, Juan Pablo Arancibia Medina; 5, Stephanie Berghaeuser; 6, Martin Walls; 7, 12, Wouter Tansens; 8, Danie Pratt; 9, Ivo De Pauw; 10, Caetano Lacerda; 11, Ken Munyard; 13, Bieke Masselis; 14, Boke Haide; 15, Waldemar Zielinski; 16, Detje Holger; 17, Cornelia Roessing; p.17, p.??, Wouter Tansens; p.??, Wouter Verweirder; p.??, Leo Storme; p.??, Bieke Masselis; p.??, Yu-Sung Chang; p.??, Angelo Fallein. D/2021/45/69 – ISBN 978 94 014 7495 5 – NUR 918 Cover design: Keppie & Keppie Interior design: Ivo De Pauw, Bieke Masselis © Ivo De Pauw, Bieke Masselis and Lannoo Publishers nv, Tielt, 2021. LannooCampus Publishers is a subsidiary of Lannoo Publishers, the book and multimedia division of Lannoo Publishers nv. All rights reserved. No part of this publication may be reproduced and/or made public, by means of printing, photocopying, microfilm or any other means, without the prior written permission of the publisher. LannooCampus Publishers Vaartkom 41 box 01.02

P.O. Box 23202

B - 3000 Leuven

NL - 1100 DS Amsterdam

Belgium

Netherlands

www.lannoocampus.com

Contents

Ac k n ow l e d g e m e n t s

Chapter 1 · Arithmetic refresher 1.1

1.2

1.3

Algebra Real numbers Real polynomials Equations in one variable Linear equations Quadratic equations Logarithms

Annex D · Real numbers in computers D.1 Scientific notation D.2 The decimal computer D.3 Special values

Annex B · Notations and Conventions B.1 Alphabets Latin alphabet Greek alphabet B.2 Mathematical symbols Sets Mathematical symbols Mathematical keywords Numbers

Annex C · The International System of Units (SI) C.1 C.2 C.3 C.4

SI SI SI SI

Prefixes Base measures Supplementary measure Derived measures

Annex D · Real numbers in computers D.1 Scientific notation

9 10 10 15 17 17 18 23

37 37 37 38

29 29 29 30 30 30 31 32 32

33 33 34 34 35

37 37

A N I M AT I O N M AT H S

D.2 The decimal computer D.3 Special values

37 38

A n n e x E · A n i m a t i o n M a t h s ( 2 0 2 1 ) A n s we r s

Annex F · Animation Maths (2021) Errata

F.1 F.2

Pairs of angles Sum identities

96 96

Acknowledgements

We hereby insist on thanking a lot of people who made this book possible: Prof. Dr. Leo Storme, Wim Serras, Wouter Tansens, Wouter Verweirder, Koen Samyn [??] (credited for the chapters Transformation Analysis, Scene Graphs, View Transformation), Hilde De Maesschalck, Ellen Deketele, Conny Meuris, Hans Ameel, Dr. Rolf Mertig, Dick Verkerk, ir. Gose Fischer, Prof. Dr. Fred Simons, Sofie Eeckeman, Dr. Luc Gheysens, Dr. Bavo Langerock, Wauter Leenknecht, Marijn Verspecht, Sarah Rommens, Prof. Dr. Marcus Greferath, Dr. Cornelia Roessing, Tim De Langhe, Niels Janssens, Peter Flynn, Jurgen Leemans, Stef Lantsoght, Hilde Vanmechelen, Jef De Langhe, Ann Deraedt, Rita Vanmeirhaeghe, Prof. Dr. Jan Van Geel, Dr. Ann Dumoulin, Bart Uyttenhove, Rik Leenknegt, Peter Verswyvelen, Roel Vandommele, ir. Lode De Geyter, Bart Leenknegt, Olivier Rysman, ir. Johan Gielis, Frederik Jacques, Kristel Balcaen, ir. Wouter Gevaert, Bart Gardin, Dieter Roobrouck, Dr. Yu-Sung Chang (Wolfram Demonstrations [??]), Prof. Dr. Sy Blinder (Wolfram Demonstrations), Steven De Keninck, Prof. Dr. Mark McClure (Wolfram Demonstrations), Dr. Felipe Dimer de Oliveira (Wolfram Demonstrations), Steven Verborgh, Ingrid Viaene, Kayla Chauveau, Angelika Kirkorova, Thomas Vanhoutte, Fries Carton, Jef Daels, Andries Geens, Angelo Fallein, Jolan Plaum, Charles Derre, Anna Rich and whomever we might have forgotten!

Hereby our special thanks to Dick Verkerk for supporting our Wolfram Application Server [??]) server in the Netherlands at can.nl

Chapter 1 · Arithmetic refresher

A N I M AT I O N M AT H S

As this chapter offers all the necessary mathematical skills for the full mastery of all further topics explained in this book, we strongly recommend it. To serve its purpose, the successive paragraphs below refresh some required aspects of mathematical language as used on the applied level.

1.1 Algebra Real numbers We typeset the set of: . natural numbers (unsigned integers) as N including zero, . integer numbers as Z including zero, . rational numbers as Q including zero, . real numbers (floats) as R including zero. All the above make a chain of subsets: N ⊂ Z ⊂ Q ⊂ R. To avoid possible confusion, we outline a brief glossary of mathematical terms. We recall that using the correct mathematical terms reflects correct mathematical thinking. Putting down ideas in the correct words is of major importance for profound insight. Sets . We recall writing all subsets in between braces, e.g. the empty set appears as {}. . We define a singleton as any subset containing only one element, e.g. {5} ⊂ N, as a subset of natural numbers. . We define a pair as any subset containing just two elements, e.g. {115, −4} ⊂ Z, as a subset of integers. In programming the boolean values true and false make up a pair {true, f alse} called the boolean set which we typeset as B. . We define Z− = {. . . , −3, −2, −1} whenever we need negative integers only. We express symbolically that −1234 is an element of Z− by typesetting −1234 ∈ Z− . . We typeset the set minus operator to delete elements from a set by using a backslash, e.g. N \ {0} reading all natural numbers except zero, Q \ Z meaning all pure rational numbers after all integer values left out and R \ {0, 1} expressing all real numbers apart from zero and one.

ARITHMETIC REFRESHER

Calculation basics

operation

expression

to add

a+b = c

term

sum

to subtract

a−b = c

term

difference

to multiply

a·b = c

factor

product

to divide

a b

numerator

divisor or denominator

quotient or fraction

to exponentiate

base

exponent

power

to take the root

ab = c √ b a=c

radicand

index

radical

return factorial

n! = c

= c, b 6= 0

n factorial

We define the factorial of a natural argument as the returned product of this argument multiplied with all natural numbers from this number n down to 1. Put in symbols: n! = n · (n − 1) · (n − 2) · . . . · 3 · 2 · 1 restricted to n ∈ N Furthermore we define 1! = 1 and as well 0! = 1. Examples: 2! = 2 · 1 = 2, 3! = 3 · 2 · 1 = 6, 4! = 4 · 3 · 2 · 1 = 24. We write the opposite of a real number r as −r, defined by the sum r + (−r) = 0. We typeset the reciprocal of a nonzero real number r as 1r or r−1 , defined by the product r · r−1 = 1. We define subtraction as equivalent to adding the opposite: a − b = a + (−b). We define division as equivalent to multiplying with the reciprocal: a : b = a · b−1 . When we mix operations we need to apply priority rules for them. There is a fixed priority list ‘PEMDAS’ in performing mixed operations in R that can easily be memorised by ‘Please Excuse My Dear Aunt Sally’. . First process all that is delimited in between Parentheses, . then Exponentiate, . then Multiply and Divide from left to right, . finally Add and Subtract from left to right.

A N I M AT I O N M AT H S

Now we discuss the distributive law ruling within R, which we define as threading a ‘superior’ operation over an ‘inferior’ operation. In conclusion, distributing requires two different operations. Hence we distribute exponentiating over multiplication as in (a · b)3 = a3 · b3 . Likewise rules multiplying over addition as in 3 · (a + b) = 3 · a + 3 · b. However we should never stumble on this ‘Staircase of Distributivity’ by going too fast: (a + b)3 6= a3 + b3 , √ √ √ a + b 6= a + b, p x2 + y2 6= x + y. Fractions A fraction is what we call any rational number written as nt given t, n ∈ Z and n 6= 0, wherein t is called the numerator and n the denominator. We define the reciprocal of a −1 nonzero fraction nt as 1t = nt or as the power nt . We define the opposite fraction as − nt =

−t n

t −n .

We summarise fractional arithmetic: sum

t n

+ ab =

t·b+n·a n·b

difference

t n

− ab =

t·b−n·a n·b

product

t n

· ab =

division

t n a b

exponentiation singular fractions

t·a n·b

= nt · ba m t m = nt m n

1 0

= ±∞ infinity (see page ??)

0 0

=? indeterminate

Powers We define a power as any real number written as gm , wherein g is called its base and m its exponent. The opposite of gm is simply −gm . The reciprocal of gm is g1m = g−m , given g 6= 0.

ARITHMETIC REFRESHER

According to the exponent type we distinguish between: g3 = g · g · g g−3

1 g3

3 ∈ N, 1 g·g·g

= = 1 √ g 3 = 3 g = w ⇔ w3 = g

−3 ∈ Z,

g0 = 1

g 6= 0.

1 3

∈ Q,

Whilst calculating powers we may have to: multiply

g3 · g2 = g3+2 = g5 ,

divide

g3 g2

exponentiate

= g3 · g−2 = g3−2 = g1 , 2 g3 = g3·2 = g6 them.

p We insist on avoiding typesetting radicals like 7 g3 and strongly recommend their contemporary notation using radicand g and exponent 73 , consequently exponentiating g to √ 3 1 g 7 . We recall the fact that all square roots are non-negative numbers, a = a 2 ∈ R+ for a ∈ R+ . As well as knowing the above exponent types, understanding the above rules to calculate them is necessary for using powers successfully. We advise memorising the integer squares running from 12 = 1, 22 = 4, . . ., up to 152 = 225, 162 = 256 and the integer cubes running from 13 = 1, 23 = 8, . . ., up to 73 = 343, 83 = 512 in order to easily recognise them. Recall that the only way out of any power is exponentiating with its reciprocal exponent. For this purpose we need to exponentiate both left hand side and right hand side of any given relation (see also paragraph 1.2). √ 7 Example: Find x when x3 = 5 by exponentiating this power. 3 7 7 3 3 x 7 = 5 ⇐⇒ x 7 = (5) 3 ⇐⇒ x ≈ 42.7494. We emphasise the above strategy as the only successful one to free base x from its exponent, yielding its correct expression numerically approximated if we wish to. Example: Find x when x2 = 5 by exponentiating this power. x2 = 5 ⇐⇒ x2

= (5) 2 or − (5) 2 ⇐⇒ x ≈ 2.23607 or − 2.23607.

We recall the above double solution whenever we free base x from an even exponent, yielding their correct expression as accurately as we wish to.

A N I M AT I O N M AT H S

Mathematical expressions Composed mathematical expressions can often seem intimidating or cause confusion. To gain transparency in them, we firstly recall indexed variables which we define as subscripted to count them: x1 , x2 , x3 , x4 , . . . , x99999 , x100000 , . . ., and α0 , α1 , α2 , α3 , α4 , . . . . It is common practice in industrial research to use thousands of variables, so just picking unindexed characters would be insufficient. Taking our own alphabet as an example, it would only provide us with 26 characters. We define finite expressions as composed of (mathematical) operations on objects (numbers, variables or structures). We can for instance analyse the expression (3a + x)4 by drawing its tree form. This example reveals a Power having exponent 4 and a subexpression in its base. The base itself yields a sum of the variable x Plus another subexpression. This final subexpression shows the product 3 Times a. Let us also evaluate this expression (3a + x)4 . Say a = 1, then we see our expression partly collapse to (3 + x)4 . If, on top of this, we assign x = 2, our expression then finally turns to the numerical value (3 + 2)4 = 54 = 625. When we expand this power to its pure sum expression 81a4 + 108a3 x + 54a2 x2 + 12ax3 + x4 , we did nothing but reshape its pure product expression (3a + x)4 . We warn that trying to solve this expression – which is not a relation – is completely in vain. Recall that inequalities, equations and systems of equations or inequalities are the only objects in the universe we can (try to) solve mathematically. Relational operators We also refresh the use of correct terms for inequalities and equations. We define an inequality as any variable expression comparing a left hand side to a right hand side by applying the ‘is-(strictly)-less-than’ or by applying the ‘is-(strictly)-greaterthan’ operator. For example, we can read (3a + x)4 6 (b + 4)(x + 3) containing variables a, x, b. Consequently we may solve such inequality for any of the unknown quantities a, x or b. We define an equation as any variable expression comparing a left hand side to a right hand side by applying the ‘is-equal-to’ operator. For example (3a + x)4 = (b + 4)(x + 3) is an equation containing variables a, x, b. Consequently we also may solve equations for any of the unknown quantities a, x or b.

ARITHMETIC REFRESHER

We define an equality as a constant relational expression being true, e.g. 7 = 7. We define a contradiction as a constant relational expression being false, e.g. −10 > 5. R e a l p o ly n o m i a l s We elaborate upon the mathematical environment of polynomials over the real numbers in their variable or indeterminate x, a set we denote with R[x] . . Monomials We define a monomial in x as any product axn , given a ∈ R and n ∈ N. We can extend this concept to several indeterminates x, y, z, . . . like the monomials 3(xy)6 and 3(x2 y3 z6 ) are. We define the degree of a monomial axn as its natural exponent n ∈ N to the indeterminate part x. We say constant numbers are monomials of degree 0 and linear terms are monomials of degree 1. We say squares have degree 2 and cubes have degree 3, followed by monomials of higher degree. √ For instance, the real monomial − 12x6 is of degree 6. Extending this concept, the monomial 3(xy)6 is of degree 6 in xy and the monomial 3(x2 y3 z6 )9 is of degree 9 in x2 y3 z6 . We define monomials of the same kind √ as6those having an identical indeterminate part. For instance, both 57 x6 and 12x are of the same kind. Extending the − √ concept, likewise 57 x3 y5 z2 and − 12x3 y5 z2 are of the same kind. All basic operations on monomials emerge simply from applying the calculation rules of fractions and powers. . Polynomials We define a polynomial V (x) as any sum of monomials. We define the degree of V (x) as the maximal exponent m ∈ N to the indeterminate variable x. For instance, the real polynomial √ 1 V (x) = 17x2 + x3 + 6x − 7x2 − 12x6 − 13x − 1, 4 is of degree 6. Whenever monomials of the same kind appear in it, we can simplify the √polynomial. For instance, our polynomial simplifies to V (x) = 10x2 + 14 x3 − 7x − 12x6 − 1. Moreover, we can sort any given polynomial either in an ascending or descending order according to its powers in x. Sorting our √ polynomial V (x) in an ascending 2 + 1 x3 − 12x6 . Sorting V (x) in a descending order yields V (x) = −1 − 7x + 10x 4 √ order yields V (x) = − 12x6 + 14 x3 + 10x2 − 7x − 1.

A N I M AT I O N M AT H S

Eventually we are able to evaluate any polynomial, getting a√numerical value from it. For instance evaluating V√(x) in x = −1, yields V√ (−1) = − 12(−1)6 + 41 (−1)3 + 10(−1)2 − 7(−1) − 1 = − 12 − 14 + 16 = 63 4 − 2 3 ∈ R. . Basic operations Adding two monomials of the same kind: we add their coefficients and keep their indeterminate part 5a2 − 3a2 = (5 − 3)a2 = 2a2 . Multiplying two monomials of any kind: we multiply both their coefficients and their indeterminate parts 7 7 −35 3 4 −5ab · a2 b3 = −5 · · a1+2 b1+3 = a b . 4 4 4 Dividing two monomials: we divide both their coefficients and their indeterminate parts −8 6−4 4−0 −8a6 b4 = a b = 2a2 b4 . 4 −4a −4 Exponentiating a monomial: we exponentiate each and every factor in the monomial 3 − 2a2 b4 = (−2)3 (a2 )3 (b4 )3 = −8a6 b12 . Adding or subtracting polynomials: we add or subtract all monomials of the same kind (x2 − 4x + 8) − (2x2 − 3x − 1) = x2 − 4x + 8 − 2x2 + 3x + 1 = −x2 − x + 9. Multiplying two polynomials: we multiply each monomial of the first polynomial with each monomial of the second polynomial and simplify all those products to the resulting product polynomial (2x2 + 3y) · (4x2 − y) = 2x2 (4x2 − y) + 3y(4x2 − y) = 2x2 · 4x2 + 2x2 · (−y) + 3y · 4x2 + 3y · (−y) = 8x4 − 2x2 y + 12x2 y − 3y2 = 8x4 + 10x2 y − 3y2 .

ARITHMETIC REFRESHER

1.2 Equations in one variable In anticipation of this section, we will refresh the required vocabulary. A solution is any value assigned to the variable that turns the given equation into an equality (being true). The scope of an equation is any number set in which the equation resides, realising it will most likely be R. We define the solution set as the set containing all legal solutions to an equation. This solution set is always a subset of the scope of the equation. L i n e a r e q u at i o n s A linear equation is an algebraic equation of degree one, referring to the maximum natural exponent of the unknown quantity. By simplifying we can always standardise any linear equation to ax + b = 0, (1.1) given a ∈ R\{0} and b ∈ R. We cite 3x+7 = 22, 5x−9d = c and 5(x−4)+x = −2(x+2) as examples of linear equations, and 3x2 + 7 = 22 and 5ab − 9b = c as counter examples. The adjective ‘linear’ originates from the Latin word ‘linea’ meaning (straight) line as referring to the graph of a linear function (see chapter ??). We solve a linear equation for its unknown part by rewriting the entire equation until its shape exposes the solution explicitly. We recall easily the required rules for rewriting a linear equation by the metaphor denoting a linear equation as a ‘pair of scales’. This way we should never forget to keep the equation’s balance: whatever operation we apply, it has to act on both sides of the equals-sign. If we add to (or subtract from) the left hand ‘scale’ than we are obliged to add the same term to (or subtract it from) the right hand ‘scale’. If we multiply (or divide) the left hand side, than we are likewise obliged to multiply (or divide) the right hand side with the same factor. If not, our equation would lose its balance just like a pair of scales would. We realise that our metaphor covers all usual ‘rules’ to handle linear equations. The reason we perform certain rewrite steps depends on which variable we are aiming for. This is called strategy. Solving the equation for a different variable implies a different sequence of rewrite steps.

A N I M AT I O N M AT H S

Example: We solve the equation 5(x − 4) + x = −2(x + 2) for x. Firstly, we apply the distributive law: 5x − 20 + x = −2x − 4. Secondly, we put all terms dependent of x to the left hand side and the constant numbers to the right hand side 5x + x + 2x = −4 + 20. Thirdly, we simplify both sides 8x = 16. Finally, we find x = 2 leading to the solution singleton {2}. Q u a d r at i c e q u at i o n s Handling quadratic expressions and solving quadratic equations are useful basics in order to study topics in multimedia, digital art and technology. . Expanding products We refresh expanding a product as (repeatedly) applying the distributive law until the initial expression ends up as a pure sum of terms. Note that our given polynomial V (x) itself does not change: we just shift its appearance to a pure sum. We illustrate this concept through V (x) = (2x − 3)(4 − x). (2x − 3)(4 − x) = (2x − 3) · 4 + (2x − 3) · (−x) = (8x − 12) + (−2x2 + 3x) = −2x2 + 11x?12. Other examples are 5a(2a2 − 3b) = 5a · 2a2 − 5a · 3b = 10a3 − 15ab and 1 13 13 4 x− x+ = (4x − 2) x + 2 2 2 13 2 = 4x2 − 2x + 26x − 13 = 4x2 + 24x − 13. = (4x − 2) · x + (4x − 2) ·

. Factoring polynomials We define factoring a polynomial as decomposing it into a pure product of (as many as possible) factors. Note that our given polynomial V (x) itself does not change: we just shift its appearance to a pure product. Our trinomial V (x) = −2x2 +11x−12 just shifts its appearance to the pure product V (x) = (2x−3)(4−x) when factored. It merely shows that the product (2x − 3)(4 − x) is a factorisation of the trinomial −2x2 + 11x − 12.

ARITHMETIC REFRESHER

Imagine we had to factor the trinomial −2x2 + 11x − 12 without any hint. This way we realise that factoring generally is a hard job to do. Especially because we do not have any clue about which factors build up the pure product for a polynomial. Many questions arise: how many factors to expect, where to start from, what is the opening step towards factorisation? We observe the need for at least a minimum asset of factoring methods. As an extra motivation, we emphasise the importance of factoring as it reveals all essential building blocks of any polynomial. Knowing the roots of a polynomial gives us a deeper insight. We therefore introduce some factoring basics in the next paragraphs. Common Factor We show how to separate common factors if they appear. For instance 6 + 12x = 6 · 1 + 6 · (2x) = 6 · (1 + 2x) results in a pure product of a number and a linear factor by separating the common factor 6. Another polynomial like 5x + x2 = 5x + xx = (5 + x) · x separates into two linear factors by the use of the common factor x. An example expression like 39x + 3xy = 3 · 13x + 3xy = 3 · x · (13 + y) yields a pure product of a number factor, a linear factor in x and a linear factor in y by separating the common factors 3 and x. Occasionally we may have to factor by grouping. For instance 1 + x + x2 + x3 = (1 + x) + (x2 + x3 ) = (1 + x) + (x2 · 1 + x2 · x) = (1 + x) + x2 (1 + x) = 1 · (1 + x) + x2 (1 + x) = (1 + x2 ) · (1 + x) results stepwise into a pure product of a quadratic and a linear factor in x. Perfect powers Expanding the natural powers of the binomial A + B reveals their corresponding pure sum shapes. (A + B)2 = (A + B) (A + B) = A2 + 2AB + B2 (A + B)3 = (A + B)2 (A + B) = A3 + 3A2 B + 3AB2 + B3 (A + B)4 = (A + B)3 (A + B) = A4 + 4A3 B + 6A2 B2 + 4AB3 + B4 We define a perfect power as any natural exponentiation of a binomial. The important power is (A + B)2 which we define as the perfect square of its binomial A + B.

A N I M AT I O N M AT H S

Those perfect powers of A + B, when ordered to ascending natural exponents, display Pascal’s Triangle for all n ∈ N. 1 1A + 1B 2

1A + 2AB + 1B2 1A3 + 3A2 B + 3AB2 + 1B3 1A4 + 4A3 B + 6A2 B2 + 4AB3 + 1B4 1A5 + 5A4 B + 10A3 B2 + 10A2 B3 + 5AB4 + 1B5 .. . Notice how a coefficient is produced as a sum of its upper two, leading to a symmetric triangle of numbers with the constant ‘1’ on both edges. This ‘triangle’ is named after its explorer Blaise Pascal (1623 –1662). Despite the diminishing need for perfect power formulas in this century of ruling computing power, we do advise you to know at least the perfect square by heart. To put the perfect square into words: ‘The square of a binomial equals the sum of both squares plus two times the product’. (A + B)2 = A2 + 2AB + B2 . We provide a visual aid to help you memorise it. The area of the total square equals (A+B)·(A+B). Alternatively we puzzle this area piece by piece, via adding both white square areas A2 and B2 plus the two grey rectangular areas AB, jointly equalling the perfect square as the trinomial A2 + B2 + 2AB. Consequently we now can explore a new factoring method. For instance, we intend to factor the trinomial 1 − 2x + x2 , whilst we have no guarantee for its pure product shape to even exist.

(1.2) A

Strategically we perform two subsequent checks. 1) Verify whether both squares carry the same sign. 2) Then find 2AB corresponding correctly to the given A and B. Only when both checks hold, are we able to shift the given trinomial to its perfect product (A + B)2 . We give an example of this strategy to the trionomial 1 − 2x + x2 .

ARITHMETIC REFRESHER

We rewrite the trinomial to +(−1)2 − 2x + (x)2 by assigning A = −1 and B = x. By substituting A and B into 2AB , we find +(−1)2 +2(−1)(x)+(x)2 equalling our trinomial. Therefore we confirm A = −1 and B = x, which allows the shift 1 − 2x + x2 = (−1 + x)2 . We realise that alternatively (+1 − x)2 is a correct factorisation as well. Perfect quotient

(A + B)(A − B) = A2 − B2

(1.3)

. Quadratic formula for quadratic equations A quadratic equation is an algebraic equation of degree two in the unknown quantity x that can be reduced to the default shape ax2 + bx + c = 0

(1.4)

given a ∈ R \ {0} and b, c ∈ R. To solve this equation for x we firstly divide both sides of it by a . Dividing by a is valid since a 6= 0. In case of a = 0 we would no longer have a quadratic but a linear equation. c b ax2 + bx + c = 0 ⇐⇒ x2 + x + = 0 a a Secondly, we aim for a perfect square by adding and subtracting the special term 2 b which is again valid since this is equivalent to adding 0. This way we have 2a created a perfect square (A + B)2 , assigning A = x and B =

b 2a .

b b 2 b 2 c x+ − + =0 2a 2a 2a a ! b b 2 b 2 c ⇐⇒ x2 + 2 x+ − + =0 2a 2a 2a a x2 + 2

b 2 b 2 c ⇐⇒ x + − + =0 2a 2a a b 2 b 2 c ⇐⇒ x + = − 2a 2a a The left hand side of this equation is now a square. Before proceeding we make sure that all denominators of the right hand side are equal. b 2 b2 c · 4a b 2 b2 − 4ac x+ = 2− ⇐⇒ x + = 2a 4a a · 4a 2a 4a2

A N I M AT I O N M AT H S

Arithmetically this holds: L2 = R ⇔ L = similar solutions to our equation: r b b2 − 4ac x+ or = 2a 4a2 ⇓ √ b b2 − 4ac x+ = or 2a 2a ⇓ √ b b2 − 4ac x=− + or 2a 2a

√ √ R or L = − R. Hence we reach two r b b2 − 4ac x+ =− 2a 4a2 √ b b2 − 4ac x+ =− 2a 2a √ b b2 − 4ac x=− − 2a 2a

The discriminant of a quadratic equation ax2 + bx + c = 0, given a 6= 0, is the real number to be calculated as D = b2 − 4ac. (1.5) Furthermore, we solve ax2 + bx + c = 0 for x like this: if D < 0 then there are no solutions in R, if D = 0 we find one real root x1 =

−b 2a ,

if D > 0 we have two similar roots √ √ −b + D −b − D x1 = and x2 = . 2a 2a

(1.6)

As a spin-off these roots enable factoring the default left hand side as ax2 + bx + c = a(x − x1 )(x − x2 ) when D > 0

(1.7)

and as ax2 + bx + c = a(x − x1 )2 when D = 0. Examples: Solving the quadratic equation −2x2 + 11x − 12 = 0 for x, we firstly calculate its discriminant D = 112 −4·(−2)·(−12) = 25 to subsequently determine √ √ −11− 25 −11+ 25 3 its roots as x1 = 2·(−2) = 2 and x2 = 2·(−2) = 4. As a bonus, this allows us to factor −2x2 + 11x − 12 as (−2) x − 32 (x − 4). The solution set is the pair { 32 , 4} ⊂ R. We solve 25x2 − 60x + 36 = 0 for x as a next example. In this case the discriminant √ 0 equals zero, yielding a unique root of multiplicity 2 to be found as x = −(−60)± = 2·25 6 6 and thus leading to the solution singleton { } ⊂ R. 5 5 Finally solving also 25x2 + 49x + 36 = 0 for x, we calculate its discriminant as D = √ −1199. It is not possible for us to find any real root due to the fact −1199 ∈ / R, which in this case leads to an empty solution set {} ⊂ R.

ARITHMETIC REFRESHER

Equations of higher degree Solving the polynomial quadratic equation ax2 +bx+c = 0 for x by means of the Quadratic Formula √ −b ± D 2a dates back to Babylonian and Greek times. The next big leap forward for solving equations of higher degree had to wait until the 16th century, till the time of the Renaissance. . Cubic equations Geronimo Cardano (1501–1576) published a similar Cubic Formula for solving polynomial equations of degree three. Despite Cardano publishing it, the Cubic Formula was actually discovered by another Italian. Historians claim that this formula was discovered by the mathematician Niccolo Fontana (1499 –1557) (nicknamed Tartaglia or ‘stutterer’). . Quartic equations Shortly after the former formula, Lodovico Ferrari (1522 – 1565) a pupil of Cardano, and also an Italian mathematician, found the Quartic formula to solve polynomial equations of degree four. . Quintic equations For an apotheosis one needed to wait until the 19th century in France: the very young and brilliant mathematician Evariste Galois (1811–1832) proved the impossibility of finding a similar Quintic Formula for polynomial equations of larger than degree four. Meanwhile, as a workaround (from, among others, Isaac Newton, around 1676) we can solve any polynomial equation numerically, yielding approximations for its solutions. Apart from this modern numerical approach, special subtypes of polynomial equations of larger degree can also still be solved exactly by means of formulas of radical expressions.

1.3 Logarithms . We define the common or Briggsian logarithm as an exponent to base 10, log10 (a) = x ⇔ 10x = a which satisfies the existence condition a ∈ R+ \ {0} with base 10 ∈ R+ \ {0, 1} .

A N I M AT I O N M AT H S

Examples: log10 (100) = 2 log10 (1000) = 3 log10 (100000) = 5

The decimal logarithm is an idea of the Englishman Henry Briggs (1561–1630), contemporary of the Scotsman John Napier (1550 – 1617). Around 1615, both mathematicians agreed that the logarithm base number 10 would offer the better future perspective. Among others, the famous scientist Simon Stevin (1548 – 1620) from Bruges, in Belgium, contributed a lot in establishing decimal numbers worldwide, in line with base number 10. In 1624, Briggs published the very first decimal logarithm table in his book ‘Arithmetica Logaritmica’. Back in 1618, Napier published (unknowingly) the natural base, which was later re-discovered as the transcendental limit value 2.718281828459 . . .

(1.8)

by the Swiss Jakob Bernoulli (1654 –1705) and – similarly to the transcendental number π ≈ 3.14 for circles – by the next Swiss genius Leonhard Euler (1707– 1783) named Euler’s number, simply called by the symbol e ≈ 2.72. . We define the natural logarithm or Napier’s logarithm as an exponent to base e ≈ 2.72, loge (a) = x ⇔ ex = a which satisfies the existence condition a ∈ R+ \ {0} with base e ∈ R+ \ {0, 1} . Examples: loge e1

= 1

= 2

loge e

loge (e ) = 5

John Napier conceived of the natural logarithm around 1594. After decades of calculation, he finally published the first natural logarithm table in his book ’Mirifici Logarithmorum Canonis Descriptio’ in 1614. New mathematical ideas acquire a common status proportional to their ease of use, but Napier’s first design based on 1 e ≈ 0.368 was seemingly less practical. In general, Napier contributed substantially to the popularity and adoption of decimal numbers and the decimal logarithm.

ARITHMETIC REFRESHER

. We define the binary logarithm as an exponent to base 2, log2 (a) = x ⇔ 2x = a which satisfies the existence condition a ∈ R+ \ {0} with base 2 ∈ R+ \ {0, 1} . Examples: log2 (4) = 2 log2 (8) = 3 log2 (32) = 5 log2 (1024) = 10 This binary logarithm is especially applicable in data communication and other binary environments.

Annex A · Real numbers in computers

Real numbers are stored identically into the computer. From the irrational numbers such as π to giant integers such as 1010 , radicals and negative fractions, they all fit into the machine in the same way. Let us also add −1 billion to our example list.

A.1 Scientific notation The storage of real numbers into computers is based on their scientific notation which separates the sign and the precision from the order of magnitude of each exact number x, arranged into the product x = (−1)s × N10 × 10E10 . The first factor (−1)s shows the sign of x, the second factor N10 is the decimal normalised significand lying between 1 and 10 and finally the exponent E10 indicates the decimal order of magnitude of x. exact value x

decimally displayed

1 10

0.1 3.141592653 . . . 0.00001234 −1000000000.

π 0.00001234 −1 billion

decimally scientifically displayed +1. × 10−1 +3.141592653 . . . × 100 +1.234 × 10−5 −1.000000000 × 109

This normalised scientific notation allows us to simulate the storage of our real examples into a decimal machine which allocates a standardised digit sequence for each of them.

A.2 The decimal computer Let us straightforwardly consider a decimal computer which stores one digit denoting the sign, one digit indicating the order of magnitude and stores four significant digits of the original value x.

A N I M AT I O N M AT H S

scientific notation

uniform machine precision

stored machine number x0

+1. × 10−1 +3.141592653 . . . × 100 +1.234 × 10−5 −1.000000000 × 109

(−1)0 × 1. × 10−1 0 (−1) × 3.142 × 100 (−1)0 × 1.234 × 10−5 (−1)1 × 1.000 × 109

(−1)0 × 1.000 × 10−1 (−1)0 × 3.142 × 100 (−1)0 × 1.234 × 10−5 (−1)1 × 1.000 × 109

Our simplified decimal computer stores exact values x ∈ R systematically in a fixed digit sequence x0 containing the sign (1 digit), the exponent (1 digit) and the normalised significand (4 digits). This computer is limited to storing only four significant digits and consequently standardises its stored numbers x0 with fixed machine precision. We call the finite subset of real numbers x0 which are inevitably rounded to fit into the computer, machine numbers. The accompanying figure shows all positive machine numbers, from the smallest to the largest one in R+ in case of 8-bit (which means 2-decimal digit) numbers.

Figure A.1: The subset of (fictitious) 8-bit machine numbers x0 in R+

A.3 Special values Calculations which result in numbers smaller than the smallest machine number, suffer real underflow. Arithmetical outputs which are larger than the largest machine number feature real overflow. For instance, storing the real number zero is an issue, since it would require the exponent E10 = −∞. To be able to store the number zero, and similarly the infinities requiring exponent E10 = +∞ and the indeterminate such as 00 in our decimal machine, we predefine these exceptions respectively as NULL, INFINITY and NAN abbreviating ‘Not A Number’.

Annex B · Notations and Conventions

B.1 Alphabets L at i n a l p h a b e t

meaning constants and coefficients unknown quantities and variables points lines planes vectors unit vectors matrices angles (angles alternatively in Greek) Bezier segment B-spline translations standard rotations standard scalings composite action transformations pivot transformation conventional composite transformations (nonconventional) orbit transformation embedding transformation camera transformation view transformation quaternions normalized quaternions unit quaternions rotation quaternions

symbol a, b, c, . . . x, y, z, . . . P, Q, R, . . . r, s,t, . . . vR , vP , . . . ~v,~w, . . . v̂, ŵ, . . . A, B,C, . . . Â, B̂, Ĉ, . . . α, β , θ , . . . ~b012...n ~s012...n T~c RO SO A PB T RS OB Ei F~c V~c q, p, . . . qn , pn , . . . u u, r, . . .

A N I M AT I O N M AT H S

Greek alphabet Traditionally, we use Greek characters to denote angles (especially in trigonometry). We also choose Greek characters for typesetting mathematical and physical constants. name

Greek character

alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu

α β γ δ, ∆ ε ζ η θ ι κ λ µ

name

Greek character

nu xi omicron pi rho sigma tau upsilon phi chi psi omega

ν ξ o π ρ σ τ υ φ, Φ χ ψ ω

B.2 Mathematical symbols Sets number sets including zero natural numbers (unsigned integers) integer numbers (integers) rational numbers or fractions real numbers (floating points) complex numbers hypercomplex numbers or quaternions We embed these number sets as N ⊂ Z ⊂ Q ⊂ R ⊂ C ⊂ H.

symbol N Z Q R C H

N O TAT I O N S A N D C O N V E N T I O N S

M at h e m at i c a l s y m b o l s

name empty set set minus element of cardinality (number of elements) factorial equal to equivalent with implies distance difference degrees infinity (unbound large value) summation dot product cross product transpose conjugate imaginary unities cartesian coordinates polar coordinates tiny error

symbol {} \ ∈ # ! = ⇔ ⇒ d ∆ ◦

∞ Σ · × T ∗

i, j, k ( )cc ( )pc ε

A N I M AT I O N M AT H S

M at h e m at i c a l k e y w o r d s

name

symbol

logarithm in base b exponential in base e radian sine cosine tangent cotangent arcsine arccosine arctangent extended arctangent determinant absolute value

logb exp rad sin cos tan cot arcsin arccos arctan atan2 det abs

Numbers

name

symbol, (rounded) value

π ≈ 3.1416

radian silver number

1 rad ≈ 57.30◦ √ δ = 1 + 2 ≈ 2.4142

golden number

Φ=

paired golden number

√ 1+ 5 ≈ 1.6180 2 √ Φ0 = 1−2 5 ≈ −0.6180

imaginary unities (quaternions)

i2 = j2 = k2 = −1 and i j = k

natural base

e ≈ 2.7183

acceleration due to gravity (average)

g ≈ 9.8067

Annex C · The International System of Units (SI)

C.1 SI Prefixes

We may use the international default prefixes to specify decimal orders of magnitude.

name

symbol

factor

yotta zetta exa peta tera giga mega kilo hecto deca

Y Z E P T G M k h da

1024 1021 1018 1015 1012 109 106 103 102 101

deci centi milli micro nano pico femto atto zepto yocto

d c m µ n p f a z y

10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24

Examples:

12 km

= 12 × 103 m = 12 000 m

34 mm

= 34 × 10−3 m = 0.034 m

A N I M AT I O N M AT H S

In our modern world, we quantify measures standardised by the SI (the International System of Units). Nature’s base measures length l, mass m and time t are measured in metres m, kilograms kg and seconds s respectively. We may typeset units by putting square brackets around their corresponding measures.

C.2 SI Base measures We mainly use just these three base measures throughout this book; for the few remaining base measures we refer you to the physics literature.

measure

symbol

SI-unit

length

[l] = m

metre

mass

[m] = kg

kilogram

time

[t] = s

second

C.3 SI Supplementary measure

Unlike the real physics units, expressing plane angles in radian is typeset by the supplementary measure or mathematical tag ‘rad’.

measure plane angle

symbol α

SI-unit [α] = rad

radian

T H E I N T E R N AT I O N A L S Y S T E M O F U N I T S ( S I )

C.4 SI Derived measures

Derived measures are composed of the above measures. We mainly use the following derived measures throughout this book; for the remaining ones with special names we refer you to the physics literature.

measure

symbol

SI-unit

width

[b] = m

metre

height

[h] = m

metre

radius

[r] = m

metre

diameter

[d] = m

metre

distance

[d] = m

metre

norm of a location vector

k~sk

[k~sk] = m

metre

area

[area] = m2

square metre

[volume] = m3

cubic metre

volume

speed

volume

[v] =

m s

metre per second

m s2 m s2

metre per second squared

magnitude of acceleration

[a] =

acceleration due to gravity

[g] =

frequency

[ f ] = s−1

Hertz

angular location

[θ ] = rad

radian

angular speed

[ω] =

rad s

radians per second

angular acceleration

[α] =

rad s2

radians per second squared

metre per second squared

Annex D · Real numbers in computers

D.1 Scientific notation The storage of real numbers into computers is based on their scientific notation which separates the sign and the precision from the order of magnitude of each exact number x, arranged into the product x = (−1)s × N10 × 10E10 . The first factor (−1)s shows the sign of x, the second factor N10 is the decimal normalised significand lying between 1 and 10 and finally the exponent E10 indicates the decimal order of magnitude of x. exact value x

decimally displayed

1 10

0.1 3.141592653 . . . 0.00001234 −1000000000.

π 0.00001234 −1 billion

decimally scientifically displayed +1. × 10−1 +3.141592653 . . . × 100 +1.234 × 10−5 −1.000000000 × 109

This normalised scientific notation allows us to simulate the storage of our real examples into a decimal machine which allocates a standardised digit sequence for each of them.

D.2 The decimal computer Let us straightforwardly consider a decimal computer which stores one digit denoting the sign, one digit indicating the order of magnitude and stores four significant digits of the original value x.

A N I M AT I O N M AT H S

scientific notation

uniform machine precision

stored machine number x0

+1. × 10−1 +3.141592653 . . . × 100 +1.234 × 10−5 −1.000000000 × 109

(−1)0 × 1. × 10−1 0 (−1) × 3.142 × 100 (−1)0 × 1.234 × 10−5 (−1)1 × 1.000 × 109

(−1)0 × 1.000 × 10−1 (−1)0 × 3.142 × 100 (−1)0 × 1.234 × 10−5 (−1)1 × 1.000 × 109

Figure D.1: The subset of (fictitious) 8-bit machine numbers x0 in R+

D.3 Special values Calculations which result in numbers smaller than the smallest machine number, suffer real underflow. Arithmetical outputs which are larger than the largest machine number feature real overflow. For instance, storing the real number zero is an issue, since it would require the exponent E10 = −∞. To be able to store the number zero, and similarly the infinities requiring exponent E10 = +∞ and the indeterminate such as 00 in our decimal machine, we predefine these exceptions respectively as NULL, INFINITY and NAN abbreviating ‘Not A Number’.

A n n e x E · A n i m a t i o n M a t h s ( 2 0 2 1 ) A n s we r s

A N I M AT I O N M AT H S

1. Arithmetic Refresher Exercise 1

1) a10

5) a7

2a5

6) a6

3) 0

7) a8

4) a15

8) 14 a6

Exercise 2

6) a2+n+m

1) −a − b 2) 3) 4)

a+b −a−b −c en c 4c+3d 4d 3c 4d

7) −8a6 b9 8) c4 d 2 b2 a10 10) a9

5) −7 Exercise 3

1) −6 26 5

Exercise 4

The odd numbers are 45, 47 and 49. Exercise 5

1) a−12

6) 16a12 b8

1 a

7) b50

3) a−20

8) a23

a6 b9

9) −16a6 b

5) b8

10) 144a10 b20

Exercise 6

1) 0

4) −x12 y6

7) 3x2 − 6x − 3

2) −16x4 y4

5) −x24 y6

8) −8x3 + 38x2 + 6x

3) −1

6) 0

9) 14x2 − 21x + 8

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 7 −5 6 en δ = 0 −5 x = 2 en x = 52 t = −3 2 en t = 2

1) δ =

4) x = −1 en x =

5) t = −2

−1 4

6) t = −3 en t = 2

Exercise 8 V (x, y) = −3(a + 7b)(x − 2y) Exercise 9 K(x) = 9 x − 31

Exercise 10

− 32

4 3

−4

2. Linear Systems Exercise 11

 x    y z    v

=1 =0 =1 =2

Exercise 12

  x1 x  2 x3

= 13 2 = −3 4 = −1

Exercise 13 3 m by 1 m. Exercise 14 123 and 87 Exercise 15 The father is 35 year and his son makes 11 year. Exercise 16 Adison scores 260 points and Valence 210. Exercise 17 In 2 hours 7 minutes 30 seconds the fastest robot produces 16 250

motherboards, while the slowest reaches 12 750. Exercise 18 11 components of type I, 13 components of type II and 21 components

of type III.

A N I M AT I O N M AT H S

Exercise 19 The circumference is 60 cm. Exercise 20 The length of the arm measures 41.41m. Exercise 21 The aeroplane flies at a speed of 583.33 kms per hour and the wind-

speed is 83.33 kms per hour. Exercise 22 The price of one beer is 1.5 EUR. Exercise 23 The parcel’s dimensions are 3 cm by 4.5 cm by 2.5 cm.

3. Trigonometry Exercise 24

β = 42◦ a =p 29 tan 48◦ ≈ 32.21 c = 292 + (29 tan 48◦ )2 ≈ 43.34 √ b = 102 + 122 − 2 · 12 · 10 · cos 65◦ ≈ 11.94 ◦ γ = arcsin 12 sinb 65 ≈ 65.62◦ α = 180◦ − 65◦ − γ ≈ 49.38

8 5 Exercise 25 α = arctan 20 − arctan 20 ≈ 7.77◦

Exercise 26 158.11m 78◦ ≈ 733.42 km Exercise 27 700sinsin 69◦ ◦ ◦ Exercise 28 sin250 23◦ (sin 80 + sin 77 ) ≈ 1253.53 km

√

Exercise 29 45 3 metres Exercise 30

1) α = 60◦ en α = 120◦ 2) α = 45◦ en α = 225◦ Exercise 31

1) x = 8.82 m, and y = 11.46 m 2) h = 13.19 m

3) α = 67, 5◦ en α = 112, 5◦ 4) α = 240◦ en α = 240◦

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

4. Functions Exercise 32

Setting the first slope m = tan α indicates the second slope as n = tan(α − π2 ). Therefore we deduct trigonometrically n = − tan( π2 −α) = − cot α = − tan1 α = − m1 . Exercise 33

1) a > 0, b > 0 and c < 0 2) a < 0, b < 0 and c < 0

3) a > 0, b > 0 and c > 0 4) a < 0, b > 0 and c > 0

Exercise 34 y = 23 x + 20 and y = −3 2 x + 85 Exercise 35 The car collides in the point (1, 1). Exercise 36 (−1, −36) and (4, 24) Exercise 37

. Inverting y = loge (x) by solving x = loge (y) for y yields y = ex = exp(x). . The graphs lie symmetric about the main diagonal y = x in their orthonormal (x, y)−plot frame. exp(x) x

-4

log(x)

-2

-4

A N I M AT I O N M AT H S

Exercise 38

. Composite function f (x) = loge (exp(x)) has domain R, range R and single root x = 0.

. Composite function f (x) = exp(loge (x)) has domain R+ , range R+ \{0} and no roots.

Exercise 39

One sheet of newspaper thickness is 0.08 mm.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 40

1) 2 sin x − π2 + 3 2) 0.5 sin 2 x − π2 − 1 Exercise 41

Approximatively

1 ≈ 3 × 10−6 9!

5. The Golden Section Exercise 42 √ 5 1 and |CB| = 2 √ 2 . Given see Exercise 42 1 = |AS| = |AD| = 5−1 2 Φ.

If |AB| = 1 then is |AC| = conclude that

|CB| = |CD| = |CA| + |AD| we

Exercise 43

If |AB| = 1 then is |AM| =

1 2

and |AT | = 1. Calculating |MT | yields |MT | =

Given |MS| = |MT | we conclude |AS| = |AM| + |MS| =

√ 1+ 5 2

Exercise 44

!2 √ 2 5+1 2 + √ 2 5+1 √ 2 3+ 5 √ + = 2 3+ 5 √ (3 + 5)2 + 22 √ = 2(3 + 5) √ 6(3 + 5) √ = 2(3 + 5) =3

1 Φ + 2= Φ 2

Exercise 45

√ 1 2 1− 5 √ √ = Φ 1+ 5 1− 5 √ 2(1 − 5 = −4 √ 5−1 = 2

= Φ.

√ 5 2 .

A N I M AT I O N M AT H S

√ 5+1 Φ−1 = −1 √2 5−1 = 2 Exercise 46

√ 1 1+ 5 2 √ Φ+ = + Φ 2 1+ 5 √ 2 (1 + 5) + 22 √ = 2(1 + 5 √ 10 + 2 5 √ = 2(1 + 5) √ √ 2 5( 5 + 1) √ = 2(1 + 5) √ = 5

Exercise 47

. for x1 = −1: (−1)3 + 2(−1)2 − 1 = 0

. for x2 = −Φ: −

√ 2 √ 3 √ 1+ 5 5 + 2 − 1+2 5 − 1 = − 16+8 2 8

. for x3 = −Φ0 : −

√ 5

+ 6+22

√ 2 √ 3 √ 1− 5 5 + 2 − 1−2 5 − 1 = − 16−8 2 8

−1 = 0

√ 5

+ 6−22

−1 = 0

Exercise 48

. Φ2 = Φ + 1 . Φ3 = Φ Φ2 = Φ (Φ + 1) = Φ2 + Φ = 2Φ + 1 . Φ4 = Φ Φ3 = Φ (2Φ + 1) = 2Φ2 + Φ = 3Φ + 2 . Φ5 = Φ Φ4 = Φ (3Φ + 2) = 3Φ2 + 2Φ = 5Φ + 3 Exercise 49

month couples of rabbits

0 1

1 1

2 2

3 3

4 5

5 8

6 13

7 21

8 34

9 55

10 89

11 144

12 233

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 50

1) We obtain the silver section by extending one of the sides of an isosceles right triangle by its √ hypotenuse. Hence the line segment |AD| relates to |AB| as √ (1+ 2)|AB| = = 1 + 2. δ = |AD| |AB| |AB|

1 2) Solving the geometric ratio 1x = x−2 mathematically as the quadratic equation 2 x −√2x − 1 = 0 for x, yields a discriminant D = 8 and the two roots x1 = √ 1 + 2 and x2 = 1 − 2.

3) By dividing the definition of the sequence of Pell by pn and subsequently pn substituting its ratio by D = pn−1 , we rewrite the number sequence of Pell in its limit as D = 2 + D1 . Therefore, the standardized equation D2 − 2D − 1 = 0 describes the ratio D for √ incrementing indices without bound. It yields the positive root D1 = 1 + 2 = δ as meaningful answer to the above.

6. Coordinate systems Exercise 51

3) (0, −4) √ cc √ 4) − 2, − 2

1) (0, √1)cc 2)

√ 3 2 −3 2 , 2 2 cc

Exercise 52

√ 1) 2 2, π4 pc 2) 1, 3π 2 pc

3) 1, π3

4) 2, 7π 4

Exercise 53

√ −10 5 5 , 4 2 cc

√ 5)

75 2 , arctan(−

after q rotation becomes √ 75 π 2 , arctan(− 5) + 2

= pc

√ −5 5 −10 , 2 4 cc

A N I M AT I O N M AT H S

Exercise 54

1) tan θ = 3 2) r =

3 2 cos θ +sin θ

3) r = −2 cos θ Exercise 55

√ 1) y = ± 2x + 1 p 2) x2 + y2 = atan2(y, x) Exercise 56

√ √ ! 2 2 The vertex D has coordinates 1 + , . The growing radial coordinates are 2 2 √ √ √ √ the successive square roots of the natural numbers: 2, 3, 4, 5, . . .

The spiral of Theodore of Cyrene, also known as the ‘square root spiral’, is the oldest mathematical spiral. For instance starting from the vertices A(0, 0), B(1, 0) and C(1, 1) it yields in 16 steps the included figure.

2

1 1

2

3

4

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 57

Applying the Pythagorean Identity yields:  √ 2  √   x t  x t   = (cost)2 = cost   t t a a =⇒ =⇒ 2 (x2 + y2 ) = 1 =⇒ 2 r2 = 1, √ √ 2   a a y t     y t = sint  = (sint)2 a a a and solving it for r gives r = √ with its polar angle t ∈ R+ \{0}. t Exercise 58

1) a = 2, b = 0

-4

-2

2) a = 0, b = 3

polar axis

3) a = 2, b = 3

-4

---2

-4

-2

polar axis

4) a = 3, b = 2

polar axis

Exercise 59

1) m = 0 2) a = 2, b = 4, m = 4, n1 = 2, n2 = 2, n3 = 2

-4

-2

polar axis

A N I M AT I O N M AT H S

7. Vectors Exercise 60

k~Fk =

√ √ √ √ (10 − 20 2 + 60 3)2 + (60 − 20 2 + 90 3)2 = 206.22 and √

√

√2+90√3 = 1.14 rad α = arctan 60−20 10−20 2+60 3

Exercise 61

800 cos 240◦ + 90 cos 105◦ 800 sin 240◦ + 90 sin 105◦

~v + ~w =

−423.294 −605.887

ground speed: q (800 cos 240◦ + 90 cos 105◦ )2 + (800 sin 240◦ + 90 sin 105◦ )2 = 739.11 kms per hour direction of ground velocity: arctan

800 sin 240◦ + 90 sin 105◦ = 0.96 + π = 4.10 rad 800 sin 240◦ + 90 sin 105◦ = 235.06◦ (referred to the positive x-axis) = 235.06◦ (referred to the North)

Exercise 62



 99 1)  0  −128   −60 2)  −326  256 3) −428



 112 4)  −22  −92   0 5)  −28  0   104 6)  88  122

Exercise 63

The angle between the vector ~f and the vector from the camera’s point to the object, is less than 90◦ , hence the object is captured.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 64

1) |HB| = 0.3 and A(1.5, 0, 0), B(1.8, 0, −0.4),C(1.8, 1, −0.4), D(1.5, 1, 0), F(1.5, 0.5, 0)   0.4 − → −→  0  2) AB × AD = 0.3 3) G(1.74, 0.5, 0.18) Exercise 65

√ 2390

Exercise 66 70.89◦

8. Parameters Exercise 67

x = r cos θ y = r sin θ

Exercise 68

  x = 1 − 4λ y = −4λ 1)  z = 4 − 8λ       x 1 4 2)  y  =  −2  + λ  3  z 7 1 Exercise 69

      x 4 −1  x = 4−λ 4−x y−2 z−8 y = 2 + 4λ or = = 1)  y  =  2  +λ  4  or  1 4 3 z = 8 + 3λ z 8 3   x = 5 + 2µ x − 5 y − 8 z − 21 y = 8 + 2µ and this is equivalent to 2) = =  2 2 10 z = 21 + 10µ 

3) S (3, 6, 11)

A N I M AT I O N M AT H S

Exercise 70

1) x − 2 = −y + 2 and z = 3   x = 5+λ y = 5+λ 2)  z = 5+µ   8 = 5+λ 8 = 5 + λ has a solution for λ = 1 and µ = −1, hence P ∈ vC  4 = 5+µ Exercise 71



       x 3 −7 2 1)  y  =  1  + λ  0  + µ  8  z 0 1 3   −8 2) ~n =  23  −56 Exercise 72

  x = 3 + λ + 4µ y = 6 + 2λ + 2µ 1)  z = 2 + 3λ + µ 2) z =

11 −4 6 x+ 6 y−7

Exercise 73

  x = λ + 5µ y = 2λ − µ 1)  z = 2λ + µ 2) z =

9 4 11 x + 11 y



 4 3) ~n =  9  −11

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 74

The lines are  not parallel because of their different direction vectors. And since  −8 + 5λ = 4 + 9µ 2 + 2λ = −1 + µ has no solution, the lines are not intersecting as the system  −4 + 3λ = 2 + 6µ well. Hence both lines are skew lines. −79 −61 Exercise 75 S 212 97 , 97 , 97

Exercise 76

The intersection of the three planes is the line by parameter equation x = 23 + λ 31 y = 31 + λ 32 Exercise 77

√ √ √ The intersection consists of the two points S1 1 − 4 2, 2(1 + 2), 3(1 − 2) and √ √ √ S2 1 + 4 2, 2(1 − 2), 3(1 + 2)

9. Kinematics Exercise 78 For the real location function s(t) = 20 + 15t we conclude m 100

-2

-1

its domain s = R, its range s = R, its root lying at t0 = −1.333

A N I M AT I O N M AT H S

Exercise 79

For the real location function s(t) = 20 + 15t + 12 (−9.81)t 2 we conclude its domain s = R, its range s = ]−∞, 31.47], its roots lying at t1 = −1.004,t2 = 4.062 m 30 20 10

-2

-1

-10 -20 -30

Exercise 80 For the velocity function v(t) = 8.2 + (−9.81)t we evaluate

1) v(0.5) = 3.295 2) v(1) = −1.610

m s, m s.

Exercise 81

1 2π y(t) = 3 sin t + arcsin 3 3

Exercise 82

− 1) k→ ω k = ω = 66.67

rad s

− with → ω perpendicular to the wheel(disk),

2) 10.61 (wheel)turns per second, 3) 1000 rad consumed, 4) k~ac k = ac = 1334

m . s2

Exercise 83

1) ~a is in T directed downhill, 2) ~ac is in U directed (centripetally) upwards,

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

3) ~g is in J directed downwards. Exercise 84

1) v = 2.7

m s

and at = 0.27

2) ac = 2.700

m s2

m , s2

and hence k~ak = 2.713

m . s2

Exercise 85 Centering the polar coordinate system (r, θ ) at the Wheel’s axis:

~s = −r~en + 0~et ~v = 0~en + rω~et

~s = r~er + 0~eθ ~v = 0~er + rω~eθ

~a = rω 2~en + 0~et

~a = −rω 2~er + 0~eθ

Exercise 86

1) Both masses land simultaneously (Independence of Motion Principle). 2) This took them 3.375 seconds. 3) As the first mass did not, the second traveled 33.75 meter horizontally. Exercise 87

The rider and his motorcycle reach a maximum height of yT = 20.16 m before they touch down after flying x = 202.1 m horizontally.

10. Collision detection Exercise 88

1) B ((1, 3, 5), 4) 2) B ((3, 6, −3), 2) 3) is not a circle

Exercise 89

√ 1) C (1, 2), 5 2) The center of the circle is the intersection point of the perpendicular bisectors (see page 46) on line segments [AB] and [BC], given A(0, 0), B(2, 0) and C(0, 4).

A N I M AT I O N M AT H S

Exercise 90 (x − 20)2 + (y − 50)2 = 502 Exercise 91 no collision Exercise 92 Both circumscribed circles are tangent in the point T (1, −1).

√ 3 Exercise 93 d = √ ≈ 0.46 14 Exercise 94 d(S, vA ) = 0.07 < 5, hence we have a collision Exercise 95 For any ~v 6= ~o we satisfy (~v ·~p −~v ·~q)(~v ·~r −~v ·~q) 6 0. Hence the point

Q lies in between the points P and R. Exercise 96

1) d(V, vO ) = 11 2) d(V, vO ) = 0 and the point V lies in the polygon with vertices P, Q and O, hence it is a successful landing. Exercise 97 d(S, vA ) = 0 and the point S lies in the polygon with vertices A, B and C, hence the snooker ball lies in the triangle ABC filled with numbered balls. Exercise 98 Collision occurs between the second and the third frame according to:

1) d(P, vA ) = 2) d(P, vA ) =

9 5 4 5

3) d(P, vA ) = 4) d(P, vA ) =

−1 5 −6 5

Exercise 99 In the third frame the squared distance is less than 225, hence we have

a collision. In other words, collision occurs between the second and the third frame.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

11. Matrices Exercise 100



9 1)  20 30

20 280 40

 1 70  1

3 50 4



−7 −10 2)  −10 −440 −40 30  5 10 20  10 200 10 3)   0 10 2 1 20 1

6 50 −2 

 −3 40  −3

  



 10 10 80 30   40 2  50 0

4  10  4)  3 0  −1  0 5)   3 −1  −1  0 6)   3 −1

 0 −10 −120 20   30 0  30 −1  0 −10 −120 20   30 0  30 −1

Exercise 101

8 1

8 4

 3 1 2 2)  −1 0 1  1 2 9 1 3)  0 2

8 1

24 12

−8 2

40 17

8 1

24 12

−8 2

40 17

8 8

1 4

8 8

1 4



−3 3 9

−3 1 −1

 −3 4  14

4) C · B does not exist

Exercise 102

70 30

1 5 −3 10

20 10 −1 5 4 5

5) D−1 does not exist

1 6)

−5 2

−7 5 18 5

A N I M AT I O N M AT H S

−4 3

3 −2



−40 4)  13 5

−19 5 14 5

9 5 −13 10

−19 5 14 5

9 5 −13 10

 9 −3  −1

16 −5 −2

Exercise 103



 













−1 2 1 2

−1 1 −4 x 5 x   2 2 0 · y  =  4  ⇔  y  =   3 3 2 z 1 z 0    5  x 2   y =  −1  2 z −2

7 4 −5 4 −3 4

−1



  5   4 ⇔ · 1   1 1 2

Exercise 104

if k = 1 we have ~f1 = F ~f0 if k = 2 we have ~f2 = F ~f1 = F 2 ~f0 if k = 3 we have ~f3 = F ~f2 = F 3 ~f0 .. . if we assume ~fk = F k ~f0 given k ∈ N ~ = F ~fk = F k+1 ~fk then for k + 1 we get fk+1

Exercise 105

1√

1+ 5 2

√ 1+ 5 2

1− 5 2

√ 1− 5 2

· √ 6+2 5 4

√ 1+ 5 2

0 0

√ 6−2 5 4

0√

Φ0√ −Φ00 5

f1 =

Φ1√ −Φ01 5

f2 = f3 =

Φ2 −Φ02

√ 5 Φ3√ −Φ03 5

= =

1−1 √ 5√

2 5

√ 1+ 5 1− 5 − 2 2 √

√ 5 √ 6+2 5 6−2 5 − 4 4 √

√ 2 5 √2

√ 5 √ (2+ 5)−(2− 5) √ 5

√ 5 , √1 − 1− 2 5 5

0 1 √ · = = F1 1− 5 −1− 5 1 1 1 √ √ − , − 2 2 5 5 √ ! ! 1−√ 5 √1 − 2 5 , 5 1 1 0 1 0 1 √ · = = · = F2 1 2 1 1 1 1 √ 5 , − √1 − −1−

Exercise 106

f0 =

√

5√ 4 5 √4

5√ 2√ 5 5

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 107

1 0 K ·K = 0 0 1 − a (1−a)a b L·L = a  b 1 a 0 M·M =  0 0 0  0 0 1 Exercise 108

0 N ·N = 0 0 P·P = 0  0 R·R =  0 0

0 0 0 0  0 0 0 0  0 0

12. Bezier curves Exercise 109

~b021 (t) = (1 − t)2 ~p0 + 2(1 − t)t ~p2 + t 2 ~p1 ~p0 +~p1 2 + t 2 ~p1 = (1 − t) ~p0 + 2(1 − t)t 2 = (1 − t)2 ~p0 + (1 − t)t (~p0 +~p1 ) + t 2 ~p1 = (1 − 2t + t 2 )~p0 + (t − t 2 ) (~p0 +~p1 ) + t 2 ~p1 ~p0 − 2t ~p0 + t 2 ~p0 + t ~p0 − t 2~p0 + t ~p1 − t 2~p1 + t 2 ~p1 = (1 − t)~p0 + t ~p1 = ~b01 (t) =

Exercise 110

  x(t) = 3t − 12t 2 + 12t 3 ~b0123 (t) = y(t) = 3 − 12t + 27t 2 − 18t 3  z(t) = 3t − 6t 2 + 5t 3

A N I M AT I O N M AT H S

x -5 -5 y

0 0

5 z

-5

Exercise 111

The sum of the coefficients of ~b012 (t) = (1 − t)2 + 2(1 − t)t + t 2 = 1 − 2t + t 2 + 2t − 2t 2 + t 2 = 1, ~ The sum of the coefficients of b0123 (t) = (1 − t)3 + 3(1 − t)2t + 3(1 − t)t 2 + t 3 = 1 − 3t + 3t 2 − t 3 + 3(1 − 2t + t 2 )t + 3(t 2 − t 3 ) + t 3 = 1 − 3t + 3t 2 − t 3 + 3t − 6t 2 + 3t 3 + 3t 2 − 3t 3 + t 3 = 1.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 112 ®

p1 ®

p1 ® p3 ®

p0 ®

p3 ® p0

p3 ®

Exercise 113

1) ~b03 (t) =

2) ~b023 (t) =

x(t) = −2 + 7t y(t) = −1 + t

3) ~b0123 (t) =

x(t) = −2 + 12t − 5t 2 y(t) = −1 + 14t − 13t 2 x(t) = −2 + 6t + 6t 2 − 5t 3 y(t) = −1 + 12t − 3t 2 − 8t 3

-2

A N I M AT I O N M AT H S

Exercise 114

1) ~b0123 (t) = 2) ~b0123 (t) =

x(t) = −2 − 3t + 24t 2 − 19t 3 y(t) = −1 + 12t − 3t 2 − 8t 3

x(t) = −2 + 18t − 39t 2 + 23t 3 y(t) = −1 + 21t − 30t 2 + 10t 3

We notice this second plane cubic Bezier segment has a looped profile. y

-4

-2

x -4

-2

-4

Exercise 115

~b0123 (t) =

~s0123 (t) =

x(t) = −2 − 3t + 24t 2 − 19t 3 y(t) = −1 + 12t − 3t 2 − 8t 3

 3  x(t) = − 35 + 3t + 4t 2 − 19 6t  y(t) =

17 6

+ 72 t − 12 t 2 − 43 t 3

y 6

-4

-2

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

− Exercise 116 Evaluating the initial and final parameter value yields ~s0123 (0) 6= → p0

− and, after simplifying, also ~s0123 (1) 6= → p3 . Exercise 117

~s0123 (t) =

  x(t) = 13 − t − 2t 2 + 53 t 3  y(t) = 7 − 3 t + 1 t 2 6 2 2

Exercise 118

1) ~b01 (t) =

2) ~b012 (t) =

x(t) = −1 + 2t y(t) = 2t

3) ~b0123 (t) =

x(t) = −1 + 4t − 5t 2 y(t) = 4t − 3t 2 x(t) = −1 + 6t − 15t 2 + 10t 3 y(t) = 6t − 9t 2 + 6t 3

4) y = x + 1 5) We obtain the parameter values t of the occasional intersection points by substituting the parametric equation of ~b0123 (t) into the cartesian equation of ~b01 (t) . y = x+1 ⇒ 6t −9t 2 +6t 3 = (−1+6t −15t 2 +10t 3 )+1 ⇒ t = 0∨t = 0∨t = The mathematically found parameter value t = 23 is for this occasion meaningless because of the constraint t ∈ [0, 1] ⊂ R. Substituting the remaining parameter value t = 0 (of multiplicity 2) into the linear Bezier segment ~b01 (t) yields the intersection point ~b01 (0) = (−1, 0) (of multiplicity 2).

13. Transformations Exercise 119

 x0   y0  = 0    z  1



2  0   0 0

0 1 0 0

 0 0 x  y 0 0   0.5 0   z 0 1 1



0 1 0 0

 0 0 0  30 0 0   0, 5 0   −100 0 1 1

2  0   0 0





  −50 −20 0 −100 −40  30 100 0  100 0 = −20 −300   −50 −10 −150 1 1 1 1 1

   

3 2

A N I M AT I O N M AT H S

Exercise 120



cos π4  − sin π 4 0

− sin π4 cos π4 0

 0 0 0  0 1 1

2 1 1

  0 0  1 = 0 1 1

√

√ 2 2 √ 3 2 2

−√ 22

2 2

  

Exercise 121







1 0  y0  =  0 1 1 0 0  1  = 

   

√1 2 √1 2

−1 √ 2 √1 2

3  1   1

√ 2 √1 2

−1 √ 2 √1 2

√ 3− 2 √ 1−2 2 1



√1 2 √1 2

−1 √ 2 √1 2



  1 0  0   0 1 0 0 0 0 1 √    3− 2 x √   y  1−2 2   1 1



  6  1  1

8 2 1

  7  3 = 1

√

  −3 x −1   y  1 1

√

√5 + 3 − 2 2 √ √7 + 1 − 2 2 2

√6 + 3 − 2 2 √ 10 √ +1−2 2 2

√4 + 3 − 2 2 √ 10 √ +1−2 2 2

Exercise 122

1) The plane basic shearing conserves area: shearing transforms rectangles into parallelograms, which have the same formula for area. 2) The inverse basic shearing takes the opposite angle: straightforward matrix calculations prove Sσx · S−σx = I3

and

Sσy · S−σy = I3 .

3) Straightforward matrix calculations prove Sσx ,σy · S−σx ,−σy 6= I3 .

  

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

4) Straightforward matrix calculations also prove Sσx ,σy 6= Sσx · Sσy

and

Sσx ,σy 6= Sσy · Sσx .



 0  1 0   0   0   0 0 1

Exercise 123 

cos π4

  0    − sin π 4  0



√1 2

sin π4

cos π4

√ √3 2 2

1 0  0 cos π6 0     0 sin π6 0   0 0 1

1 √ 2 2

−1 2 √ √3 2 2

√ −3− √ 3 2 2 √ − 3+1 2 √ 1−√ 3 2 2

1 √ 2 2 √ 3 2

   0   −1  √  2 0

0 − sin π6 cos π6 0

 0 1 0 0

0 0 1 0



 

      

0 0 0 1

1 0 0 1

0 1 0 1



 0   0  = 1     1



√1 2

 −1   0 −1  =  −1  −1  √  2 1 0

√ √3 2 2

1 √ 2 2

−1 2 √ √3 2 2

√ −3− √ 3 2 2 √ − 3+1 2 √ 1−√ 3 2 2

1 √ 2 2 √ 3 2

√ −3− 3 2 √ − 3+1 2 √ − 3+1 2

√ −1− √ 3 2 2 √ − 3+1 2 √ − 3−1 2

√ −2− √ 3 2 2

1 2 √ 2−√ 3 2 2

−3 √ 2 2 √ − 3 2 1 √ 2 2

Exercise 124

. The centroid is Z = (1, 5).       1 0 1 0 1 0 1 0 −1 0 1 −4 .  0 1 5   −1 0 0   0 1 −5  =  −1 0 6  0 0 1 0 0 1 0 0 1 0 0 1      0 1 −4 −5 −1 9 6 1 4 5 0  =  11 7 −3  .  −1 0 6   10 0 0 1 1 1 1 1 1 1

Exercise 125



1  0   0 0

0 1 0 0

0 0 1 0

 2 2  2   0 1  0 1 0

0 4 0 0

0 0 3 0

 0 1  0   0 0  0 1 0

0 1 0 0

0 0 1 0

  −2 2  −2  = 0 −1   0 1 0

0 4 0 0

0 0 3 0

 −2 −6   −2  1

       

A N I M AT I O N M AT H S



2  0   0 0

0 4 0 0

0 0 3 0

 2 −2  −6   2 −2   1 1 1

5 1 2 1

5 1 −1 1

  2 5 5 2 2  5 4 1 5  = 2 1 2 4 4   1 1 1 1 1 1

2 2 −1 1

8 −2 4 1

8 −2 −5 1

2 2 8 8 2 14 10 −2 −5 1 4 10 1 1 1 1

 2 14   10  1

Exercise 126



 3 0 −1 3  1 0 1 0 0  −1 6 2 4 0 0  2 2 0 1 1 1

1  0 0  0  1 0

 0 1 0 0  0 1 1 0 0

0 1 0

5 3 1

4 4 1

2 4 1

  −3 −3  =  1   1 4 4 3 = 2 4 1 1 1

0 1 0

 −1 6 0 0  0 1

3 5 1

2 4 1

2 2 1

 3 1  1

Exercise 127

Firstly, we translate over a distance 1 upwards. Secondly, we rotate around the origin O over an angle −θ , given θ = arctan(2). Essentially, we reflect over the x-axis. Nextly, we inversely rotate over the angle θ , given θ = arctan(2). Finally, we inversely translate over the distance 1 downwards. 

1  0 0

   

0 1 0

  √1 0 5  2 −1   √5 1 0

−3 5 4 5

4 5 3 5

4 5 −2 5

−2 √ 5 √1 5

 0  1 0   0 0 1



  4  1  1

5 2 1

6 4 1

  0 0  −1 0    0 1 9 2



√1 5 −2 √ 5

√2 5 √1 5



−0.8 3  =  3.4 1 1





 1  0   0 0 1

−0.6 4.8 1

  0 0  1 1 =  0 1

−3 5 4 5

4 5 3 5

4 5 −2 5

 0.4 0.5 6.8 5  1 1

Exercise 128 Applying row reductions to calculate the inverse matrix operators for

each, yields respectively −1 → . the inverse translation matrix T−→ = T− BA AB

. the inverse basic scale operator 

1 sx

 0  −1 SO =  0 0

0 1 sy

0 0

0 0 1 sz

. the inverse basic rotation operator R−1 O,θ = RO,−θ

0 0 0 1

    

   

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 129

We rotate around the origin O over the angle − π4 . Nextly, we reflect over the x-axis. Finally, we inversely rotate over the angle π4 .

  

√1 2 √1 2

−1 √ 2 √1 2



0  1 0

1 0 0

 0 1  0  0 0 1





0 −1 0 

  √1 0  −12 0  √ 2 1 0

√1 2 √1 2

  0 0   1 0 = 0 1

 1 0 0 0  0 1



0 −3 1 0   1  =  −3  1 1 1

Exercise 130

We rotate around the origin O over the angle −30◦ . Nextly, we reflect over the x-axis. Finally, we inversely rotate over the angle 30◦ .

   

√ 3 2 1 2

−1 √2 3 2

√ 3 2 −1 2

1 √2 3 2

 0 1  0  0 0 1 



0 −1 0 

  √3 0  2 0   −1 2 1 0 

0 3 4.96    4 0.60  =  0 1 1 0 0 1 √ The radius of the circles equals d(P, M) = 2.

 

1 √2 3 2

  √ 3 0 2   −1 0 = 2 1 0

1 √2 3 2

 0  0  1

A N I M AT I O N M AT H S

14. Transformation analysis Exercise 131 We translate the origin O by the displacement~t =



1 the matrix operator  0 0 

−3 4

by applying



0 tx 1 ty  on the left hand side: 0 1      0 −3 1 0 −3  0 1 4 · 0  =  4  1 1 0 0 1

We draw unit disks (to better envision the circles) around the origin O in red, and its translation image O0 in blue.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 132 the square ABCD by applying the standard scaling matrix  We scale 

sx 0 0 operator  0 sy 0  on the left hand side: 0 0 1      5 0 0 1 1 2 2 5  0 2 0 · 1 2 2 1  =  2 0 0 1 1 1 1 1 1

5 4 1

10 4 1

 10 2  1

We draw the square ABCD in red and its standardly scaled image A0 B0C0 D0 in blue.

Exercise 133 We rotate the triangle UVW by applying the standard rotation



cos α matrix operator  sin α 0

− sin α cos α 0

 0 0  on the left hand side: 1

  cos(−23◦ ) − sin(−23◦ ) 0 2 4  sin(−23◦ ) cos(−23◦ ) 0  ·  0 1 0 0 1 1 1 

   4 1.84 4.07 3.29 −1  =  −0.78 −0.64 −2.48  1 1 1 1

We draw the triangle UVW in red and its standardly rotated image U 0V 0W 0 in blue.

A N I M AT I O N M AT H S

Exercise 134

Erratum: the rotation angle is intended to be −23◦ instead of the positive value. Given the translation T −3 , the standard scaling S 5 and the standard rotation 4

RO (−23◦ ) deliver these composite transformation matrices:   4.60 0.78 −3.00 1) the action A = T −3 · RO (−23◦ ) · S 5 =  −1.95 1.84 4.00  4 2 0.00 0.00 1.00   4.60 0.78 −10.68 2) the action B = RO (−23◦ ) · S 5 · T −3 =  −1.95 1.84 13.23  2 4 0.00 0.00 1.00

Exercise 135 We draw the triangle UVW in red, its A-image U 0V 0W 0 in blue and

its B-image U 00V 00W 00 in brown.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 136 Retrieving the ingredients of

1) the TRS-based action transformation A, . read the T −3.00 displacement vector from the last matrix column, 4.00

. calculate column wise the S sx scale factors respectively as sy

sx sy

= k~v1 k = = k~v2 k =

(4.60)2 + (−1.95)2 ≈ 5.00

(0.78)2 + (1.84)2 ≈ 2.00

. and finally retrieve the RO (θ ) rotation angle via the inverse tangent-withquadrant function atan2 which takes two arguments, given sx > 0 θ = atan2 v1y , v1x = atan2 (−1.95, 4.60) ≈ −22.97◦ 2) the loosely composite action transformation B: fail for translation ingredient. Also, in general for various non-TRS composites, beware of a fail. Exercise 137 For the ingredients of the given TRS-based action transformation A,

. read the T 5.00 displacement vector from the last matrix column, 4.00

. calculate column wise the S sx scale factors respectively as sy

sx sy

q (1.73)2 + (1.00)2 ≈ 2.00 q = k~v2 k = (−1.50)2 + (2.60)2 ≈ 3.00

= k~v1 k =

. and finally retrieve the RO (θ ) rotation angle via the inverse tangent-withquadrant function atan2 which takes two arguments, given sx > 0 θ = atan2 v1y , v1x = atan2 (1.00, 1.73) ≈ +30.00◦ Exercise 138

To pivot the square ABCD around Z = 41 (A+B+C +D) its centroid  = (1.5 , 1.5) we ◦ cos(45 ) − sin(45◦ ) 0 ‘sandwich’ the standard rotator  sin(45◦ ) cos(45◦ ) 0  by the appropriate 0 0 1 translators to and fro the standard position in the origin O.

A N I M AT I O N M AT H S

Applying the pivot-operator      1 0 1.5 cos(45◦ ) − sin(45◦ ) 0 1 0 PZ (45◦ ) =  0 1 1.5  ·  sin(45◦ ) cos(45◦ ) 0  ·  0 1 0 0 1 0 0 1 0 0

 −1.5 −1.5  1

onto the vertices of the square ABCD produces the pivoted image square A0 B0C0 D0     1 1 2 2 1.50 0.79 1.50 2.21 PZ (45◦ ) ·  1 2 2 1  ≈  0.79 1.50 2.21 1.50  1 1 1 1 1 1 1 1 We draw the original square ABCD in red and its pivoted image A0 B0C0 D0 in blue.

Exercise 139

To orbit the square ABCD around the origin O we simply apply the standard rotator. Applying this standard orbit-operator   cos(60◦ ) − sin(60◦ ) 0 O(60◦ ) =  sin(60◦ ) cos(60◦ ) 0  0 0 1 onto the vertices of the square ABCD produces the orbited image square A00 B00C00 D00     1 1 2 2 −0.37 −1.23 −0.73 0.13 1.87 2.73 2.23  O(60◦ ) ·  1 2 2 1  ≈  1.37 1 1 1 1 1 1 1 1

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

We draw the original square ABCD in red and its orbited image A00 B00C00 D00 in blue.

Exercise 140 Applying the composite operator

= PZ (45◦ ) · O(60◦ )      1 0 1.5 cos(45◦ ) − sin(45◦ ) 0 1 0 =  0 1 1.5  ·  sin(45◦ ) cos(45◦ ) 0  ·  0 1 0 0 1 0 0 1 0 0

onto the vertices of the square ABCD produces the G-image square     1 1 2 2 0.28 −0.69 −0.95 0.02 G ·  1 2 2 1  ≈  0.09 −0.17 0.79 1.05  1 1 1 1 1 1 1 1

 −1.5 −1.5  · O(60◦ ) 1

A N I M AT I O N M AT H S

We draw the original square ABCD in red and its G-image A0 B0C0 D0 in blue.

Exercise 141

Applying the composite operator H

= O(60◦ ) · PZ (45◦ )  1 0 = O(60◦ ) ·  0 1 0 0

    1.5 cos(45◦ ) − sin(45◦ ) 0 1 0 1.5  ·  sin(45◦ ) cos(45◦ ) 0  ·  0 1 1 0 0 1 0 0

onto the vertices of the square ABCD produces the H-image square     1 1 2 2 0.06 −0.90 −1.16 −0.20 2.40 2.66  H ·  1 2 2 1  ≈  1.70 1.44 1 1 1 1 1 1 1 1

 −1.5 −1.5  1

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

We draw the original square ABCD in red and its H-image A00 B00C00 D00 in blue.

Comparing this H-outcome to the G-outcome of the previous exercise 140, reveals how transformations are not commutative. This is of course due to the fact that the matrix product is not commutative. Exercise 142 → we TRS-wise assemble it as 1) Computing the look-at action matrix A− BT → A− BT

→ · R c · S = T−→ · R c · I3 = T−→ · R c = T− 1 BT BT BT OB OB OB 1

= T

1 7

· RBT c

which will point the isosceles triangle KLM situated in B towards a target positioned in T . We therefore need to determine the shape’s desired direction vector as −→ BT 1 ~ v̂1 = −→ = −→ (~t − b) kBT k kBT k 1 1 −9 1 −10 = p − = √ 6 7 −1 101 (−10)2 + (−1)2

A N I M AT I O N M AT H S

Hence we compose all the above into the look-at operator   −0.995 0.099 1.000 −0.995 → ≈  −0.099 −0.995 7.000  , A− given v̂ ≈ 1 BT −0.099 0 0 1 Applying this look-at action its image pointer K 0 L0 M 0  −2 4 → · 1 0 A− BT 1 1

onto the vertices of the triangle KLM produces   −2 3.09 −2.98 −1  ≈  6.20 6.60 1 1 1

 2.89 8.19  1

2) We draw the original triangle KLM in red and its image pointer K 0 L0 M 0 (from centre B looking at target T ) in blue.

→ of the Exercise 143 To retrieve the ingredients of the look-at action matrix A− BT

previous exercise 143, we

. read the T 1 displacement vector from the last matrix column, 7

. calculate column wise the S sx scale factors respectively as sy

= k~v1 k =

= k~v2 k =

(−0.995)2 + (−0.099)2 ≈ 1.00

(0.099)2 + (−0.995)2 ≈ 1.00

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

. and finally retrieve the RO (θ ) rotation angle via the inverse tangent-withquadrant function atan2 which takes two arguments, given sx > 0 θ = atan2 v1y , v1x = atan2 (−0.099, −0.995) ≈ −174.289◦

Exercise 144 Given is the look-at action matrix computed on page 324 as



→ A− BT

0.93 ≈  0.37 0

 −0.37 5.00 0.93 4.00  . 0 1

→ , we To retrieve the ingredients of this look-at action matrix A− BT

. read the T 5 displacement vector from the last matrix column, 4

. calculate column wise the S sx scale factors respectively as sy

sx sy

= k~v1 k = = k~v2 k =

0.932 + 0.372 ≈ 1.00

(−0.37)2 + 0.932 ≈ 1.00

. and finally retrieve the RO (θ ) rotation angle via the inverse tangent-withquadrant function atan2 which takes two arguments, given sx > 0 θ = atan2 v1y , v1x = atan2 (0.37, 0.93) ≈ +21.80◦

A N I M AT I O N M AT H S

15. Scene Graphs Exercise 145

We model the bones themselves by scaling the blueprint diamond B0 according to their required sizes.   0 0.5 1 0.5 B0 =  0 0.5 0 −0.5  1 1 1 1 1) Firstly, the parent-to-child object tree for this robot arm looks like

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

2) Secondly, the described embedding tranformations that link each successive local space of the scene graph read like     cos 20◦ − sin 20◦ 1 4 0 0 . E1 =  sin 20◦ cos 20◦ 0  hosting B1 =  0 1 0  · B0 , 0 0 1 0 0 1     cos α − sin α 4 3 0 0 cos α 0  hosting limb B2 =  0 1 0  · B0 , . E2 =  sin α 0 0 1 0 0 1     cos β − sin β 3 5 0 0 cos β 0  hosting limb B3 =  0 1 0  · B0 . . E3 =  sin β 0 0 1 0 0 1

3) Finally, implemented in GeoGebra the arm looks like this

A N I M AT I O N M AT H S

Exercise 146

We use the unit circle C0 = C(O; 1) and model by the isosceles triangle the   1.5 −1.5 −1.5 Craft =  0.0 1.0 −1.0  1 1 1 1) Firstly, the parent-to-child object tree for this solar system looks like

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

2) Secondly, the described embedding tranformations read like   2π 2π   cos 350 day − sin 350 day 0 1 0 100   2π 2π  0 1 0  . E1 =  cos 350 day 0   sin 350 day · 0 0 1 0 0 1   15 0 0 hosting the Planet =  0 15 0  ·C0 given the unit circle C0 , 0 0 1     − sin 2π 0 cos 2π 50 day 50 day 1 0 40   2π  0 1 0  . E2 =  cos 2π 0  50 day ·  sin 50 day 0 0 1 0 0 1   5 0 0 hosting its Moon =  0 5 0  ·C0 given the unit circle C0 , 0 0 1     − sin 2π 0 cos 2π 5 day 5 day 1 0 15   2π  0 1 0  . E3 =  cos 2π 0  5 day ·  sin 5 day 0 0 1 0 0 1   1.5 −1.5 −1.5 hosting the satellite Craft =  0.0 1.0 −1.0 . 1 1 1 3) Finally, implemented in GeoGebra this solar system looks like this

A N I M AT I O N M AT H S

Exercise 147

The rectangular blueprint T0 becomes the tank. The blueprint T0 scales non-uniformly by scale factor sy = 0.5 to its square turret. Also T0 scales non-uniformly by scale factor sx = 0.25 to a stretched barrel.   −1 1 1 −1 T0 =  2 2 −2 −2  1 1 1 1 1) The parent-to-child object tree for this layered object tank-turret-barrel is

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

2) Secondly, the described embedding tranformations read like     cos (α) − sin (α) 0 1 0 9     0 1 0  hosting the Tank T0 . E1 =   sin (α) cos (α) 0  · 0 0 1 0 0 1 with its orbit center positioned at a freely chosen distance 9 (in point O),     cos (β ) − sin (β ) 0 1 0 0    . E2 =  0 1 0  ·   sin (β ) cos (β ) 0  hosting, set β = −α 0 0 1 0 0 1   1 0 0 to keep its orientation to the South, the Turret =  0 0.5 0  · T0 , 0 0 1     1 0 0 1 0.25 0 1 0  · T0 . . E3 =  0 1 −2  hosting its fixed Barrel =  0 0 0 1 0 0 1 3) Finally, implemented in GeoGebra this layered tank-turret-barrel system looks like this

A N I M AT I O N M AT H S

16. View Transformation Exercise 148 Given the world space is 100 pixels wide and 50 high and the camera

window is 15 wide and 15 high, and is rotated over a 15◦ angle around its bottom left vertex C in position (87, 10). Performing the required boundary check, takes 1) constructing the camera transformation, F 87 (15◦ ) = T 87 · RO (15◦ ) · S 1 10



cos 15◦

=  sin 15◦ 0

− sin 15◦ cos 15◦ 0

   87 0.97 −0.26 87 10  ≈  0.26 0.97 10  . 1 0 0 1

2) determining the vertices of the camera window,  0 15 15 Cam =  0 0 15 1 1 1

 0 15  1

3) transforming the camera window vertices by the camera transformation, Cam0

= F 87 (15◦ ) ·Cam 10



87.00 ≈  10.00 1

101.50 13.88 1

 97.61 83.12 28.37 24.49  1 1

4) verifying wether the camera window image Cam0 stays within the world boundaries, we discover the second image vertex (101.50, 13.88) to be outside of the given world 100 × 50-rectangle. Exercise 149 A rectangular camera window has a width of 10 by a height of 8 units.

If the left bottom vertex C of the rectangle is located at the point (3, 4) and it is rotated over an angle of 30◦ around C then we determine the 1) camera transformation to put the camera window at this position as, F 3 (30◦ ) = T 3 · RO (30◦ ) without any scaling yet 4

cos 30◦  sin 30◦ = 0 

− sin 30◦ cos 30◦ 0

   3 0.87 −0.50 3 4  ≈  0.50 0.87 4  . 1 0 0 1

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

2) view transformation to return the camera capture horizontally filling our fixed game window measuring 800 by 600 pixels as V 3 (30◦ ) = F 3 (30◦ )−1 4

T 3 · RO (30◦ ) · S sx

−1 ◦ −1 S · R−1 O (30 ) · T 3 sx

−1

= S 1/sx · RO (−30◦ ) · T −3 −4

1/sy



800 10

  0 cos(−30◦ ) − sin(−30◦ ) 0 1 0  ·  sin(−30◦ ) cos(−30◦ ) 0  ·  0 0 0 1 0 1     ◦ ◦ 0 cos 30 sin 30 0 1 0 −3 0  ·  − sin 30◦ cos 30◦ 0  ·  0 1 −4 1 0 0 1 0 0 1    0 0.87 0.50 −4.6 0  ·  −0.50 0.87 1.96  1 0 0 1  40.00 −367.85 64.95 −147.31  0 1  

600 =  0 8 0 0  80 0 =  0 75 0 0  80 0 ≈  0 75 0 0  69.28 =  −37.50 0

Exercise 150

. Verifying that the matrix product of matrices (16.1) and (16.6) returns matrix I3 F 400 (21.8◦ ) ·V 400 (21.8◦ ) 100



100

   0.93 −0.37 400.00 0.93 0.37 −409.00 55.70  ≈  0.37 0.93 100.00  ·  −0.37 0.93 0 0 1 0 0 1   1.00 0.00 0.00 ≈  0.00 1.00 0.00  0 0 1

 0 −3 1 −4  0 1  

A N I M AT I O N M AT H S

. Proving that the matrix product matrices (16.2) and (16.4) yields matrix I3 

sx cos θ

   sx sin θ  

(cos θ )2 + (sin θ )2 0

   =    1 =  0 0

0 1 0

cos θ sy 0

−((cos θ )2 + (sin θ )2 )c1 + c1



(cos θ )2 + (sin θ )2

−((sin θ )2 + (cos θ )2 )c2 + c2

     

0 

−v̂1 ·~c   sx  −v̂2 ·~c    sy 

sin θ sx

sy cos θ

0 

  cos θ sx      sin θ  c2  ·  −   sy  1 0

−sy sin θ



0 0  1

. Proving the matrix product of matrices (16.3) and (16.5) produces matrix I3   v̂1x k~v1 k      v̂  c2  ·  2x    k~v2 k 1 0

v1x

v2x

  F~c (θ ) ·V~c (θ ) =   v1y 

v2y

(v̂1x

)2 + (v̂



   =    

v̂1y k~v1 k v̂2y k~v2 k

v̂1 ·~c  k~v1 k   v̂2 ·~c   −  k~v2 k  −

(v̂1y )2 + (v̂2y )2

1 0



  0    1

(cos θ )2 + (− sin θ )2

(sin θ )2 + (cos θ )2

0  0 0  1

  =    

1  0 = 0

0 1 0



  0    1

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

Exercise 151 Construct an arrow spanning edges between vertices, in their order

{(0, 20), (20, 0), (10, 0), (10, −20), (−10, −20), (-10, 0),(−20, 0), (0, 20)}. 1) Positioning this arrow in the world space implies we need to construct its em1 bedding transformation E. First scale it to 10 of its original size, then rotate ◦ it by an angle of +170 around its pivot and finally position this instanced arrow with its pivot put in point (17, 19). Calculating and visualising the according arrow image positioned in the world space in GeoGebra by setting   0 20 10 10 −10 −10 −20 0 0 0 20  A0 =  20 0 0 −20 −20 1 1 1 1 1 1 1 1 upon which we TRS-place the arrow A0 in world space as pictured underneath.

  1 0 17 cos 170◦ =  0 1 19  ·  sin 170◦ 0 0 1 0   −0.099 −0.017 17 ≈  0.017 −0.099 19  0 0 1 

A00

− sin 170◦ cos 170◦ 0

   0 0.1 0 0 0  ·  0 0.1 0  1 0 0 1

= E · A0   16.65 15.03 16.02 16.36 18.33 17.98 18.97 16.65 ≈  17.03 19.35 19.17 21.14 20.80 18.83 18.65 17.03  1 1 1 1 1 1 1 1

A N I M AT I O N M AT H S

2) Returning the camera captured arrow from world space to the fixed game window, which is requiring a view transformation. Calculating and visualising the according camera capture positioned in the world space in GeoGebra by setting   0 0 15 15 0 Cam =  0 15 15 0 0  1 1 1 1 1 upon which we forwardly TRS-place the camera Cam in world space by the camera transformation F as pictured underneath.



Cam0

 

cos 30◦

1 0 12 =  0 1 14  ·  sin 30◦ 0 0 1 0   0.77 −0.44 12 ≈  0.44 0.77 14  0 0 1

− sin 30◦ cos 30◦ 0

= F ·Cam  12 5.36 16.86 23.50 12 ≈  14 25.50 32.13 20.64 14  1 1 1 1 .1





0  0 ·  1

1 1.13



 0   1

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

3) Stacking the former view transformation V on top of the latter embedding transformation E, to matrix multiply these into one action matrix. Calculating and visualising the subsequent arrow image A000 as brought into the game window in GeoGebra.

A000

= V · E · A0  −1   0.77 −0.44 12 −0.099 −0.017 17 ≈  0.44 0.77 14  ·  0.017 −0.099 19  · A0 0 0 1 0 0 1     0.98 0.57 −19.65 −0.099 −0.017 17 ≈  −0.57 0.98 −6.92  ·  0.017 −0.099 19  · A0 0 0 1 0 0 1   0.98 0.57 −19.65 ≈  −0.57 0.98 −6.92  · A00 0 0 1   6.27 5.99 6.85 8.31 10.04 8.58 9.45 6.27 ≈  0.34 3.52 2.80 4.53 3.07 1.34 0.62 0.34  1 1 1 1 1 1 1 1

A N I M AT I O N M AT H S

Exercise 152 At a certain time, the pivot point P of such a triangular UFO was

located in (299, 99) when its nose point N was in (300, 100). The military camera window measured 15 width by 10 height with the bottom left vertex C of the camera’s rectangle situated in (298, 97) on the UFO’s right wing tip as portrayed.

N P

We model this UFO by the isosceles triangle  300 298 U0 =  100 97 1 1

 297 98  1

1) Given all above, we answer the matrix product to deliver the view transforma−→ tion which brings this UFO horizontally (meaning PN horizontally) into our game window.

A N I M AT I O N M AT H S ( 2 0 2 1 ) A N S W E R S

2) We compute the former matrix product using GeoGebra to answer the UFO view transformation matrix.

  cos(−45◦ ) − sin(−45◦ ) 0 1 0 =  sin(−45◦ ) cos(−45◦ ) 0  ·  0 0 0 0 1 0 0   0.71 0.71 −279.31 ≈  −0.71 0.71 142.13  0 0 1 

U00

 −298 −97  1

= V ·U0   2.12 0.00 0.00 ≈  0.71 0.00 1.41  1 1 1

3) We also append a window-filling scaling for a ground based view port of 60 width by 40 height. This scaling is uniform. We recompute the former matrix product now including the latter scaling using GeoGebra to answer the zoomed UFO view transformation matrix.

60 15

 =   0 0

40 10

 S



 4    0 0 = 0 1

0 4 0

 0 0  1

A N I M AT I O N M AT H S

U000

= S ·U00  8.49 ≈  2.83 1

0.00 0.00 1

 0.00 5.66  1

And as for the zoomed UFO view transformation matrix, we conclude      4 0 0 cos(−45◦ ) − sin(−45◦ ) 0 1 0 Vzoom =  0 4 0  ·  sin(−45◦ ) cos(−45◦ ) 0  ·  0 0 0 0 1 0 0 1 0 0     4 0 0 0.71 0.71 −279.31 ≈  0 4 0  ·  −0.71 0.71 142.13  0 0 1 0 0 1   2.83 2.83 −1117.23 568.51  ≈  −2.83 2.83 0 0 1

 −298 −97  1

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17. Hypercomplex numbers Exercise 153

√ z1 = 2 + 2i = √8(cos 45◦ + i sin 45◦ ) z2 = −1 + i√= 2(cos 135◦ + i sin 135◦ ) z3 = −1 − 3i√= 2(cos 240◦ + i sin 240◦ ) z4 = 3 − 3i = 18(cos 315◦ + i sin 315◦ )

Exercise 154

1) −9 + 3i 2) 16 − 24i 3) −1 + 5i

4) 2 − 10i 5) 31 − 25i 6) −1 + 3i

Exercise 155

1) z = cos 120◦ + i sin 120◦ Via the complex number a referring to the point A, we calculate the image vertices (as a · z, b · z, c · z en d · z): ! √ √ −1 − 3 −1 + 3 0 A = , 2 2 √ B0 = (− 3, 1) C0 = (0, 2) √ √ ! −1 + 3 1 + 3 D0 = , 2 2 2) A complete identical approach applies to r = 3(cos 80◦ + i sin 80◦ ) Exercise 156 √ −1

√ −1 √ √ √ 2−2√5−2−2 2√ 2√ −4 5 √ 5 − 1−2 5 − (1+ 5)(1− 5) 1+ 5 1− 5 1−5 √ √ √ f (−1) = = = = √ =1 5 5 5 5 Analoguously we calculate f (−2) = −1, f (−3) = 2, f (−4) − 3, f (−5) = 5, . . . 1+ 5 2

Exercise 157

1) 3 − 2i − j + 4k 2) 2

4) 4 − 7i − 4 j + k 5) −2 − 10i − 10 j − 6k

3) 2 + 4i − 2k

−1 1 6 i− 3

j + 16 k

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cos θ2 , 0, 0, sin θ2 [0, (x, y, z)] cos θ2 , 0, 0, − sin θ2 = −z sin θ2 , x cos θ2 − y sin θ2 , y cos θ2 + x sin θ2 , z cos θ2 cos θ2 , 0, 0, − sin θ2

Exercise 158

= [0, (x cos a − y sin a, y cos a + x sin a, z)] Exercise 159

 −1 0 0  0 0 0    0 1 0  0 0 1 2) qrq∗ = 0, (0, 1, 0) 

0  1 1)   0 0

  1 0  1 0  = 0   0 1 1

   

Exercise 160



2√  1 0 0 0 2 √  −3 + 3 3  0 −1  3   0  2 2 2      √ 1)    1  =  −1 − 3√3 0 −2 3 −1 0  2  1 0 0 0 1 2 1 " !# √ √ −3 + 3 −1 − 3 3 2) qrq∗ = 0, 2, , 2 2 

Exercise 161





We calculate qrq∗ = 0,

√1 , √2 , √1 6 6 6

       

# " # ∗ 1 2 1 [0, (1, 1, 0)] 0, √6 , √6 , √6 = [0, (0, 1, 1)].

The image point equals (0, 1, 1). Exercise 162

Z + Z∗

1 = 2a 0

0 1

0 −1 Z − Z ∗ = −2b 1 0 a −b a ∗ Z ·Z = · b a −b

b a

a2 + b2 0

0 a2 + b2

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= (a2 +b2 )

1 0

0 1

Annex F · Animation Maths (2021) Errata

Lighthouse photograph by Detje Holger at pixabay.com

A N I M AT I O N M AT H S

3. Trigonometry D ELETION OF ’ COTERMINAL ANGLES ’

F.1 Pairs of angles We briefly explain the properties of two pairs of angles that are useful in this book. Oppositely signed angles; their measurements add up to 0◦ . In other words, if α and β are opposite then α + β = 0◦ or β = −α. The corresponding figure shows how the cosines of opposite angles remain invariant, while their sines receive opposite signs. This leads to the trigonometric formulas cos(−α) = cos α and sin(−α) = − sin α. Complementary angles; their measurements add up to 90◦ . In other words, if α and β are complementary then α + β = 90◦ or β = 90◦ − α. The corresponding figure shows how the sine of α equals the cosine of 90◦ − α and the cosine of α equals the sine of 90◦ − α. This leads to the trigonometric formulas cos(90◦ − α) = sin α and sin(90◦ − α) = cos α.

90°−α α −α

Figure F.1: Opposite and complementary angles

F.2 Sum identities

A N I M AT I O N M AT H S ( 2 0 2 1 ) E R R ATA

In this paragraph we state and prove all trigonometric ratios of a sum of two angles. We firstly emphasise the non-linearity of all trigonometric ratios: e.g. for the sine ◦ and we encounter sin(α + β ) 6= sin α + sin β . Indeed, e.g. for angles α = 60√ ◦ ◦ ◦ ◦ β = 30 the value sin 90 = 1 does not equal the sum sin 60 + sin 30 = 3+1 2 . Given the above inequality, we realise the need for the correct formulas which are stated below. sin(α + β ) = sin α cos β + cos α sin β

sin(α − β ) = sin α cos β − cos α sin β

cos(α + β ) = cos α cos β − sin α sin β

cos(α − β ) = cos α cos β + sin α sin β

tan(α + β ) =

tan α+tan β 1−tan α tan β

tan(α − β ) =

tan α−tan β 1+tan α tan β

I Based upon cos(α − β ) = cos α cos β + sin α sin β , we have less difficulties in proving the five remaining Sum Identities cos(α + β ) = cos (α − (−β )) = cos α cos(−β ) + sin α sin(−β ) = cos α cos β − sin α sin β sin(α − β ) = cos (90◦ − (α − β ))

(opposite angles) (complementary angles)

◦

= cos ((90 − α) + β ) = cos(90◦ − α) cos β − sin(90◦ − α) sin β = sin α cos β − cos α sin β

(complementary angles)

sin(α + β ) = sin (α − (−β )) = sin α cos(−β ) − cos α sin(−β ) = sin α cos β + cos α sin β tan(α ± β ) =

sin(α ± β ) cos(α ± β )

sin α cos β ± cos α sin β cos α cos β ∓ sin α sin β

sin α cos β cos α cos β cos α cos β cos α cos β

tan α ± tan β 1 ∓ tan α tan β

cos α sin β ± cos α cos β sin α sin β ∓ cos α cos β

(opposite angles)

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16. View Transformation M ISSING PICTURES

y' 1000

800

600

400

L(404,242) 200

0 C 0

200

400

Figure 16.4: The game window

600

800

x' 1000

A N I M AT I O N M AT H S ( 2 0 2 1 ) E R R ATA

y' 1000

800

600

400

200

0 C 0

L(202,121)

200

400

600

Figure 16.5: Zooming-out view transformation

800

x' 1000

100

A N I M AT I O N M AT H S

y' 1000

800

600

L(808,484) 400

200

0 C 0

200

400

600

Figure 16.6: Zooming-in view transformation

800

x' 1000

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101

E XERCISES Exercise 151 Construct an arrow spanning edges between vertices, in their order {(0, 20), (20, 0), (10, 0), (10, −20), (−10, −20), (-10, 0),(−20, 0), (0, 20)}. We modelled this arrow around the origin O which we adopt as its pivot point. 1) Positioning this arrow in the world space implies we need to construct its 1 of its original size, then embedding transformation E. First scale it to 10 ◦ rotate it by an angle of +170 around its pivot and finally position this instanced arrow with its pivot put in point (17, 19). Calculate and visualise the according arrow image positioned in the world space in GeoGebra. 2) Return the camera captured arrow from world space to the fixed game window, which is requiring a view transformation. Establish the view transformation corresponding to a camera with its lower left pivot point in (12, 14), rotated by an angle of 30◦ around this pivot and uniformly scaled by factor 1 1.13 with a camera window in pixels of 15 width and 15 height. 3) Stack the former view transformation on top of the latter embedding transformation E, to matrix multiply these into one action matrix. Now calculate and visualise the subsequent arrow image as brought into the game window in GeoGebra. Exercise 152 The Belgian UFO wave was a series of sightings of triangular UFOs over Belgium which lasted from November 1989 until April 1990. At a certain time, the pivot point P of such a triangular UFO was located in (299, 99) when its nose point N was in (300, 100). The military camera window measured 15 width by 10 height with the bottom left vertex C of the camera’s rectangle situated in (298, 97) on the UFO’s right wing tip as portrayed. This isosceles UFO is only principally portrayed, meaning its orientation and proportions may differ from this picture. The picture is solely provided for a better understanding of the above listed points P, N and C.

N P

102

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1) Given these conditions, answer the matrix product as such to deliver the view −→ transformation which brings this UFO horizontally (meaning PN horizontally) into our game window. 2) Compute the former matrix product using GeoGebra to answer the UFO view transformation matrix. 3) Also append a window-filling scaling for a ground based view port of 60 width by 40 height. This scaling is uniform. Recompute the former matrix product now including the latter scaling using GeoGebra to answer the zoomed UFO view transformation matrix.

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