The Mathematics That Every Secondary School Math Teacher Needs to Know

472 Chapter 10 Functions and Modeling encryption by multiplying the preceding matrix by EÀ1, where 0 À1=13 À2=13 10=13 1 EÀ1 ¼ @B 3=13 17=65 À59=65 CA: À3=13 À4=65 33=65 The result, as you can verify on a graphing calculator, is 0 À1=13 À2=13 10=13 10 96 10 À1 125 136 À71 1 M ¼ EÀ1F ¼ BB@B 3=13 17=65 À59=65 ACCCB@BB 293 231 413 340 307 306 CACC À3=13 À4=65 33=65 89 68 89 100 101 58 0 16 16 5 15 20 3 1 ¼ BBB@ 18 1 27 27 20 11 CCAC 5 18 20 1 1 27 which is our original message. The decoding is now done by columns. The letters corresponding to 16, 18, 5, 16, 1, 18, 5. . . are PREPARE. . ., and so we recover the original message. This kind of encryption was done by cable companies years ago when people were stealing channels by using illegal cable boxes. The companies then scrambled the signals so that only those with the proper decoder could get cable channels, thus solving a big problem for the cable companies. Of course cable receivers must know the “key” to recover the message, that is, the matrix EÀ1. Notice that this encryption process is really a function from message matrices to encrypted message matrices. More specifically, f(M) = EM = F. To recover the original message matrix, we need to compute the inverse function, which is f À1(F) = EÀ1F. Student Learning Opportunities 1 Given A ¼ ! 04 21 ! 0 1 51 À1 2 3 , B = B@ 0 3 AC; C ¼ 2 1 , and D ¼ @B À3 2 CA: Perform À3 2 4 À5 6 51 4 À1 each of the following operations or tell if an operation is undefined. Do it by hand and then when possible, verify the answer with the calculator. (a)* AB (b)* BA (c) A + B (d) B + D (e)* 4B À 3D (f) 3A (g) det(A) (h) det(C) (i) det(AD) (j)* det(AD)

10.8 Review of Matrices; Functions Defined by Matrices 473 (k) A2 (l) C3 ! ! 4 À3 2* Verify the Associative Law, (AB)C = A(BC) for the matrices A ¼ 21 ;B ¼ ; ! À3 2 21 59 C ¼ : Do it by hand and then check your answer using the calculator. 7 À6 3 A manufacturer makes chairs and tables. Each item must be assembled and finished. The con- struction takes place at two plants, one in New York and another in Chicago. The time in hours for the production of each chair and table is given by the matrix Assembly Finishing 2 ! Chair A¼ 3 12 Table 18 Thus, a chair requires 3 hours of assembly and 2 hours of finishing. The cost per hour for each of these processes, in dollars, is given by the following matrix New York Chicago 14 ! Assembly B¼ 30 32 Finishing 35 That is, it costs 30 dollars per hour to assemble either item in New York and 14 dollars per hour to assemble either item in Chicago. What do the entries of the matrix AB represent? 4 Verify that det(AB) = det A det B for the matrices from question 2. Don’t use the calculator. 0 4 3 À2 1 09 2 11 5 Given A = @B 1 5 8 CA, and B ¼ B@ 4 5 3 AC, use your calculator to compute AB, det(A), À3 3 3 876 det(B), and det(AB), and show that det(AB) = det(A) Á det(B). 6 Solve the following systems of equations by using inverses of the matrices. Check your answer. 2x þ 3y ¼ 5 (a)* 7x À 2y ¼ 7 4x þy Àz ¼ 5 (b) 3x À2y þ4z ¼ 2 x þy þz ¼ 9 7 Determine the inverse of each of the following matrices with the calculator. ! 56 (a) À1 2 0 4 2 À3 1 (b)* B@ 3 1 5 CA 02 0

474 Chapter 10 Functions and Modeling 8 We define a function between matrices, Y = f(X), where f(X) = MX and M i!s some matrix. Write 2 the inverse of each transformation, and then find fÀ1(Y), where Y ¼ : If an inverse func- 1 tion does not exist, show why. !! 15 x (a)* Y ¼ 39 y !! 43 x (b) Y ¼ 32 y !! 48 x (c) Y ¼ 12 y 9 Show that if AB = BA, then both A and B must be square matrices of the same size. 10* Using matrices, find the equation of the parabola passing through the three points (1, 3), (3, 13), and (7, 57). 11 Suppose that we have a 3 × 3 matrix and do the following: first switch rows 1 and 3, then switch rows 2 and 1 in the resulting matrix. How does the determinant of the final matrix compare to that of the original? Explain. 12* Encrypt the message RETREAT using the matrix we used in the final subsection of this chapter. 10.9 Recursive Relations and Modeling LAUNCH A standard roll of paper towels consists of a cardboard tube with outer diameter 4 cm. A typical roll contains 100 sheets of paper each 25 cm long, so the total length of paper is 2500 cm. Imagine the paper being wound onto the cardboard tube. After each complete winding the total diameter of the roll increases by an amount 2t, where t cm is the thickness of the paper. Let Sn cm denote the total length of paper wrapped around the tube when it is wrapped around n times, so that S0 = 0. 1 When the paper is wrapped around the roll the first time (n = 1), what will the outer diameter, D1, of the roll be? 2 What length of paper, C1, is needed to wrap around the roll the first time? 3 When the paper is wrapped around the roll the second time (n = 2), what will the outer dia- meter, D2 of the roll be? 4 What length of paper, C2, is needed to wrap around the roll for the second time?

10.9 Recursive Relations and Modeling 475 5 How much paper, S2, (in centimeters) will you have wrapped altogether, after this second wrapping? 6 After three wrappings, how much paper, S3, will you have wrapped altogether? 7 After n wrappings, how much paper, Sn, will you have wrapped altogether? 8 How does the amount of paper after n wrappings, Sn, compare to the amount of paper after n À 1 wrappings? Write a relationship that expresses it. We hope that when doing the launch problem, you realized that to answer some of the ques- tions, you were using the same operations repeatedly. Did you also notice that the final relationship you expressed involved both Sn and SnÀ1? Without knowing it, you have been examining a recur- sive relation, which is a special type of function. Up until now we have talked about a function as a rule between two sets where each input yields only one output. Sometimes a function is given by a formula, sometimes by a table, and sometimes by a graph. But there is another class of functions that is very useful in mathematical modeling, and this is the class of functions that are defined recursively. In a recursive function (also called a recursive relation or recurrence relation), the value of the function at any natural number n depends on the value of the function at one or more of the function values at numbers less than n. Although they may not realize it, most people encounter recursive relation- ships in the course of their lives. Many practical problems can be solved by using recursive relation- ships, making recursion a very important topic in applied mathematics. Its applications are numerous, ranging from the mundane, like finance, to the obscure, like fractals. Closely related is the topic of mathematical induction, which is a powerful tool both in mathematics and also in the study of recursive relationships. In this section, we take a very close look at recursion in hopes that you will get a deeper perspective on it and appreciate its value. Although we can use function notation in the examples we will give, we will not do so. Instead, we will be using sequen- tial notation since that is how it is seen in secondary school materials and the calculator. We begin by examining a case of recursion that you probably encounter in your daily life. That is, consider what happens when you put money into a bank account and you receive interest. How does your money grow? Example 10.46 You just got a gift of $126, which you put into an account paying annual interest at the rate of 6% per year, hoping that it will grow into $1000 in 10 years. Assuming that you don’t with- draw any of the money, will this be possible? Solution: Let An be the amount in the account at the end of n years, and A0 the initial amount put into the account. So A0 = 126. At the end of year 1, we have in the account our initial amount A0, plus the interest, 0.06A0, for a total amount of 1.06A0. Let us call this amount, A1, to remind us that it is the amount that is in the account after one year, or equivalently, at the beginning of the second year. So, A1 ¼ A0 þ 0:06A0; that is; A1 ¼ 1:06A0:

476 Chapter 10 Functions and Modeling In a similar manner, at the end of year 2, we have our initial amount from the beginning of the year, A1, plus the interest 0.06A1 for a total of 1.06A1. That is, A2 ¼ A1 þ 0:06A1 A2 ¼ 1:06A1: If we let An be the amount in the account at the end of the nth year, we can compute it as follows: It is the amount at the beginning of the year, which is AnÀ1, plus the interest, 0.06AnÀ1 for a total of 1.06AnÀ1. Thus, the amount in the account at the end of n years is given by the equa- tions: An ¼ 1:06AnÀ1 ð10:18Þ A0 ¼ 126: ð10:19Þ The equation (10.18) is an example of a recursive relation. Notice that the value of An is defined in terms of AnÀ1. Equation (10.19) is called an initial condition. Using this recursive relation and the initial condition, we can now generate a sequence of numbers which tells us how much money is in the account after n years. Here is our table: n0 1 2 3 4 5 etc. An 126 133.56 141.57 150.06 159.07 168.61 In each case, we simply multiplied the previous value AnÀ1 by 1.06 to get An. Thus, A3 = 1.06A2 and A4 = 1.06A3, and so on. Finish the table and answer the question asked in Example (10.46). Example 10.47 When you take any medication, say in pill form, written on the bottle are directions for how often you should take the pills. This is because scientists know that with time a certain portion of the medication is excreted from the body. Another pill will have to be taken at a certain time interval in order to maintain the effective level of medication in the bloodstream. Suppose that one begins an initial dosage of 80 milligrams of a medication, and that at the end of each 24-hour period, 40% of what is in the body has been excreted. As a result, a new 80 milligram dose must be taken after each 24-hour period. Write a recursive relation that expresses the amount of medication in the blood- stream after n 24-hour periods right after each new dosage is taken. Solution: Let A0 represent the initial dose. Thus, A0 = 80. After one 24-hour period, 40% of the medication is excreted, thus 60% of the medication stays. So the amount in the system right before we take the new dose is 0.60A0. We then take an 80-milligram dose and the amount of med- ication in our system, represented by A1, is given by A1 ¼ 0:60A0 þ 80: This analysis holds for the other 24-hour periods. In general, the amount of medication in the bloodstream after n days (right after the new dose of medication is taken) is 60% of the amount in the bloodstream after the last dose +80. In symbols, An ¼ 0:60AnÀ1 þ 80: This is our recursive relation, which we will now use to generate some values.

10.9 Recursive Relations and Modeling 477 A1 = 0.60A0 + 80 = 0.60(80) + 80 = 128; A2 = 0.60A1 + 80 = 0.60(128) + 80 = 156.8; and so on. Here is a table with the first 6 values of An: n 01 2 3 4 5 etc. An = 80 128 156.8 174.08 184.45 190.67 In the previous two examples, An was defined in terms of AnÀ1. Now we give you an example of a recursive relation that defines An in terms of both AnÀ1 and AnÀ2. It leads to the famous Fibonacci sequence that we first discussed in Chapter 1. Example 10.48 We start with the two initial conditions F1 = 1 and F2 = 1. We define the nth Fibo- nacci number Fn where n > 2 to be the sum of the previous two Fibonacci numbers. Thus, Fn = FnÀ1 + FnÀ2 when n > 2. Generate a table that gives the first seven Fibonacci numbers. Solution: Our recursive relation, Fn = FnÀ1 + FnÀ2, yields the following table F1 F2 F3 F4 F5 F6 F7 etc. 1 1 2 3 5 8 13 where it is clear that each Fibonacci number is the sum of the previous two. Notice that in the first two examples the An was defined in terms of only one previous value, namely AnÀ1. In this example, An, which is Fn, is defined in terms of two previous values AnÀ1 and AnÀ2, which are FnÀ1 and FnÀ2, respectively. In general, in a recursive relation, it is possible to define An in terms of several of the previous values of A. In each of the previous three examples our recursive relations generated sequences of numbers. However, recursive relations can be defined for other objects as well. Here is an example which we will talk more about when we discuss fractals. Example 10.49 Begin with an equilateral triangle and let’s assume that each side has length 1. Call that triangle, T1. Connect the midpoints of T1 cutting T1 into four congruent equilateral triangles, and then remove the “middle one” which we have shaded leaving us with three equilateral triangles. Call the resulting figure T2. Now do the same thing on each of the remaining three unshaded triangles in T2, namely, join the midpoints of the sides, forming four equilateral triangles, and then remove the middle one in each triangle so divided. Thus, each equilateral triangle will have been divided into three other equilateral triangles. Call this figure T3. On T3 repeat the procedure and call the result T4. Describe the figure T_ 4. Solution: Our first triangle, T1 is shown in Figure 10.38 (a). T4 T1 T2 T3 (a) (b) (c) (d) Figure 10.38

478 Chapter 10 Functions and Modeling We divide this into four equilateral triangles by connecting the midpoints and then removing the middle one. In Figure 10.38(b) we see the resulting figure, T2, where the gray area represents the triangle removed. Now, on each of the remaining (white) triangles in T2 we join the midpoints of the sides and split each such triangle into four triangles and then remove the middle one from each triangle. Figure 10.38(c) shows the resulting figure T3 with the gray representing the trian- gles removed. Now on each white triangle in triangle T3 we perform the same process. We split it into four equilateral triangles and remove the middle one. We get the figure T4 in Figure 10.38(d). If we do this forever, the final figure we get is what is called the Sierpinski triangle. It looks a lot like Figure 10.38(d), only there are many more grey areas. There are some interesting properties of the Sierpinski triangle. First, it is a fractal as described in Chapter 9. That is, it has what is known as infinite complexity. Let’s look at its area and perimeter. In the first step we started pwffiiffiffith an equilateral triangle whose sides were 1, whose perimeter was 3, and whose area we know is 43. In the second step we have three triangles similar to the original, each of whose sides is 1/2 the original. Thus, the perimeter of each of these smaller triangles is 1/2 that of the previous triaÀngÁle. But there are three of them. Thus, the perimeter of this second figure is pl3rai/rmr2igoeertt.hefiTarghtiusuorsefÀ,32.tÁthSh3o(ee3,fi)pa.refAstrteifmrtoetrerhtnee23ristoe(e3cfro)at.hntiAedotnSieitseeartrcphahtienioipstneker,iraittmtrhiioaeenntpe,gerltrehiismeisÀep23tÁtheenre(ri3mli)ismeaÀtni23etÁdr2o(oaf3fs)À.t32nÁhAnegf(t3enet)rseawtlshanrefigggetohu, eritrehsditisosite3eixnr/ap2fitrniteohistnsayit,otwhonefhgtipehcethes- is infinite. What about the sum of the areas of the triangles making up each successivpeffiffififfi gure? Does it also go to infinity? The answer may surprise you. The area of the first triangle is 43. At the second it- eration we have three similar triangles, each of whose sides is 1/2 the side of the original triangle. Hence, the area of each of the smaller similar triangles is 1/4 that of the original triangle. But there are thpreffieffiffi times as mapnffiffiyffi smaller triangles. Thus, the area of the figure at the second iteration is 1 3 3 43. 3 Á 4 Á 4 or just 4 Á After the next iteration, we have three times as many triangles each with area p1/ffi4ffiffi!that of the previous. Thus, the sum of the areas of the triangles in the next figure 3 3 3 is 4 Á 4 Á 4 . Continuing in this manner we can show (by induction) that at the nth step of this iterative process, the sum of the areas of the triangles is 3nÀ1 pffiffiffi Now as n gets large, Á 3. 44 3nÀ1 3nÀ1 pffiffiffi gets small and approaches zero so Á 3, the sum of the areas of the triangles ap- 4 44 proaches zero. Since the Sierpinski triangle is the figure created after infinitely many iterations, we have established that the area of the Sierpinski triangle is zero while its perimeter is infinite! Not only that, the Sierpinski triangle still contains many points even though its area is zero. The vertex of each unshaded triangle at each step is still there! How strange! This kind of property is not unusual for fractals as you will see later in this chapter (see section on fractals).







































498 Chapter 10 Functions and Modeling (c) Figure 10.53 [Hint: Begin with a square and divide it into nine congruent subsquares. Remove the middle square and the one above it. Now repeat the process on each subsquare.]

CHAPTER 11 GEOMETRIC TRANSFORMATIONS 11.1 Introduction In the previous chapter we gave a very general definition of function, along with several examples that were not numerical. Let us briefly review some of those examples. (For more details you should read Sections 10.2 and 10.3 of Chapter 10.) Suppose that, with each point on the earth’s surface, there is associated the ordered pair of numbers (latitude, longitude). This is a function from the points on the earth’s surface to ordered pairs of numbers. As another example, when we assign a set of workers to tasks, the assignment is a function from the set of workers to the set of tasks as- signed. In this function each worker is associated with the task he or she gets. Thus, this function associates people with tasks. In general, if A and B are any sets, then a function from A to B is a rule that associates with each element of the set A, one and only one element of the set B. Thus, in the first example we men- tioned, A is the set of points on the earth’s surface, while B is the set of ordered pairs of numbers consisting of latitude and longitude. With each element of A (that is, with each point on the earth’s surface), we associate one and only one element of B (one and only one latitude and longitude pair). In the second example, A is the set of workers being assigned, and B is the set of tasks to which they are assigned. With each element of A (with each worker), one and only one element of B (one and only one task) is assigned. The only real exposure secondary school students get to this more general concept of function is when they study transformations within the coordinate plane. That is, they study translations, rota- tions, reflections, and dilations. Typically, they are given a point such as (1, 2), and are asked to find its image under a specified translation, rotation, reflection, or dilation. Transformations of this type are exceedingly useful in real-life applications. When you play a game on a computer, and you see an object moving, its motion is really a combination of several different kinds of transformations con- sisting of translations, rotations, reflections, and dilations. Thus, computer animation is one applica- tion of how one uses transformations to give the appearance of motion. Actually, what the computer does to accomplish this is a combination of a large number of matrix multiplications and additions. Another application would be the design of robots that, for example, are used in industry, med- icine, armed forces, and police work to perform tasks that humans would find tedious, sometimes dangerous, and even impossible. Specifically, robots are used in building cars and a variety of man- ufacturing processes, as well as in microscopic surgery. Every robotic movement can be described by some transformation or composition of transformations that is represented by matrices and

500 Chapter 11 Geometric Transformations programmed into the computer. In this chapter we describe how motion can be represented in these ways. We also show how, surprisingly, a calculator uses rotations to compute quantities like sines, cosines, and logarithms. 11.2 Transformations: The Secondary School Level LAUNCH Here you will see a section of the well-known M.C. Escher Symmetry Drawing E45 (Figure 11.1). As you view this drawing, you will probably see examples of many different geometric transformations. On tracing paper, trace the central angel on the picture and keep your tracing paper where it is. Now, do the following: 1 See if you can hold your pencil at some point on the traced angel and turn the paper so the new position of the traced angel exactly overlaps another angel. How many degrees do you think you turned the paper? Where else could you have done this on the drawing? Explain. What type of transformation is this? 2 Holding the tracing paper on top of the original picture so that your traced angel exactly over- laps an angel, fold the tracing paper so that your tracing of the angel overlaps exactly with one of the original angels. Can you find another angel in the picture where you can do a similar process? Explain. What type of transformations are these? 3 Assuming that the picture continues as shown to the right and to the left, is it possible to slide an angel across the drawing so that it overlaps with another angel? Is there more than one way to do this? Explain. What type of transformation is this? 4 In the smaller picture of Escher’s Symmetry Drawing E45 (Figure 11.2), what has changed? What has remained the same? What type of transformation is used to go from one picture to the other? Figure 11.1

11.2 Transformations: The Secondary School Level 501 Figure 11.2 After having explored the many transformations that exist in Escher’s drawing, you probably have a pretty good idea of the types of transformations that will be discussed in this section. As you will see, transformations are not only found in art but in all areas of our lives. 11.2.1 Basic Ideas The most basic transformation is the translation. In this transformation, we essentially slide an object to another position. When we ride in an elevator for example, the elevator translates us from our starting floor to our final floor. Following is a picture of an octagon. If we imagine it made out of metal, so that it is rigid, and then move the whole thing in the direction of the arrow, we have a picture of the octagon after a translation (Figure 11.3). We point out that if the arrow starts at the center of the first octagon and ends at the center of the second octagon, then the center of the first octagon is moved to the center of the second, and the entire figure is moved along. If the arrow starts at some other point on the octagon, then that point is moved along the arrow until it reaches the tip of the arrow, and the entire figure is moved with it. octagon after translation octagon Figure 11.3 Since the arrow shows us the direction in which we moved the object, as well as how far we moved it, we could have drawn the arrow from any vertex of the first octagon to the corresponding vertex of the second octagon to indicate the same translation. The next transformation included in the secondary school curriculum is reflection in a line. Here we are given a line l which acts as a mirror. We simply take the “mirror image” of the

E502 Chapter 11 Geometric Transformations object with respect to our mirror l and that is called the reflection of the object with respect to l or about l. Thus, in the following figure, we have reflected the letter E about the line l (Figure 11.4). E l Figure 11.4 Notice how its mirror image is reversed. Here is a reflection that teachers might find quite meaningful: Figure 11.5 Copyright © 2009 Scott Kim, scottkim.com. All rights reserved. Technically speaking, when we reflect a point P about a line l, we drop a perpendicular from P to l and then extend it the same distance past l to P0. P0 is the reflection of P about l. Figure 11.6 shows this. The length of OP must be the same as the length of OP0 in order for P’ to be the reflec- tion of P. Points on l stay fixed under a reflection. l P O P’ Figure 11.6 Next, there is rotation of a figure about a point O. The easiest way to think of rotating a point P about a point O, say 60 degrees counterclockwise, is to imagine that O is the center of a circle with radius OP. Draw the circle, and using OP as one side of a central angle, draw a central angle of 60°

11.2 Transformations: The Secondary School Level 503 where the other side of the central angle is counterclockwise from P. The point P0 where the other side of the angle intersects the circle is the rotation of P about O at an angle of 60°. (See Figure 11.7.) P’ P 60° O Figure 11.7 If the direction along the circle from P to P0 is counterclockwise, we say that we have rotated P 60° counterclockwise about O or that we have just performed a rotation of 60°. When the rotation is clockwise, we either say the rotation is a clockwise rotation of P about O or that P has been rotated À60°. In general, when we rotate a figure, we rotate every point on that figure the same number of degrees. Next, we see what happens when we rotate the letter F counterclockwise an angle of 60 degrees. Here, every single point on the letter “F” in Figure 11.8 is rotated 60 degrees. FF 60° O Figure 11.8 Finally, there is the dilation. In graphics programs this is known as a “stretch” or a “shrink.” Here we stretch or shrink the figure evenly in all directions, and what we get is a figure similar to the original figure. That is, it has the same shape, but not the same size. When you take a picture of an object, the image you get is a dilation of the object. In the smaller picture in Figure 11.9 you see a picture of a face and the larger picture is a dilation of the face. Figure 11.9

504 Chapter 11 Geometric Transformations Technically speaking, every dilation has a center of dilation (although we didn’t show it in the picture). Thus, when we dilate a point P by a factor of k > 0 where the center of dilation is O, we draw OP, and continue the line to OP0 so that the length of OP0 = k times the length of OP. (See Figure 11.10(a).) We also allow dilations with negative factors. If k < 0, then not only do we stretch or shrink the line OP, but we also reverse its direction. So the the image P0 that we drew in Figure 11.10(a) would be on the opposite side of O, and the length OP = |k| times the length OP. (See Figure 11.10(b).) O P OP’ = k (OP) (k>0) (a) P’ P OP’ = |k|(OP) (k<0) (b) O P Figure 11.10 These are the basic four transformations studied in secondary school. There are more transfor- mations that are used in practice, and we will discuss some of them later when we get to the higher level. For now, we will describe how mathematicians represent these transformations. Student Learning Opportunities 1 (C) Your students make the following comments. In each case state whether what they have said is correct. (a)* Under a translation, when one point of a figure travels a distance of 5 units, all points of the figure travel a distance of 5 units. (b) If we reflect an object touching a line ℓ about ℓ, all points of the object on ℓ are fixed. (c) If we rotate a figure about a point P, we get a congruent figure. (d) If we dilate a figure using a point P as the center of dilation, we get a similar figure. (e) Suppose we have a triangle ABC and we reflect it about a line ℓ to get a new triangle, A0B0C0 where A0, B0, and C0 are the images of A, B, and C respectively. If we have to travel clockwise to go from A to B to C in a triangle, then we have to travel clockwise to go from the image points A0 to B0 to C0. 2 In the following diagrams, the image of a figure is drawn. In each case determine if the image is correct. If it isn’t, draw the correct picture. Check your work with tracing paper if possible.

11.2 Transformations: The Secondary School Level 505 (a) (Translation in the direction of the arrow. The lower triangle is the original and the upper triangle is the image.) Figure 11.11 (b)* (Reflection in a line. The word on the left is the original figure and the word on the right is the image.) HAM HAM Figure 11.12 (c) (Rotation of 90 degrees counterclockwise about P. The figure on the top is the original and the figure on the bottom is the image.) P Figure 11.13 (d) (Dilation of the letter A on the lower left by a factor of 1 where the center of dilation is P.) 2 A P A Figure 11.14

506 Chapter 11 Geometric Transformations 3 If we reflect the point (a, b) about the x-axis, what are the coordinates of the image point? (a) If we reflect the point (a, b) about the y-axis, what are the coordinates of the image point? (b) If we reflect the point (a, b) about the line y = x, what are the coordinates of the image point? 11.3 Bringing in the Main Tool—Functions LAUNCH To create the many transformations that exist in his drawing of Angels and Devils, Escher must have used very specific measurements. He must have known a great deal about the mathematics of trans- formations in order to do that. Let us now examine a very simplified version of his drawing and determine the specific mathematical representations for each of the transformations that exist in this diagram (Figure 11.15). y B CA x D y E x Figure 11.15

11.3 Bringing in the Main Tool—Functions 507 In order to give a frame of reference for units of measurement, we have superimposed a grid on the picture and its smaller image. Use this grid to help you respond to the following questions: 1 How many degrees and about which point do you have to rotate to get from angel A to angel B? 2 Give the equations of two lines of reflection that you notice in the graph. 3 Describe the magnitude and direction of the translation from angel A to angel C. 4 Describe how much larger angel A is than angel E in the second graph. After having completed this launch, you probably have a pretty good idea of how to express transformations in a mathematical way. As you can well imagine, such mathematical representa- tions of transformations are essential for mathematicians. Actually, the simplest way for a mathema- tician to discuss the transformations described earlier is not by what they do to the whole figure, but by what they do to each point in the figure. As you will soon see, this allows them to apply math- ematics to find efficient ways of rotating and translating figures in practice. We will initially only deal with two dimensional concepts, though everything we do can be extended to the real world of three dimensions. Afterwards, we will discuss some of the extensions of this material to three di- mensions. The xy-plane will be denoted by R2, the customary notation. R2 is simply an abbreviation for the set of ordered pairs, (x, y), where x and y are Real, hence the R. Our main concern will be with functions from R2 to R2, that is, rules that associate points in the plane with points in the plane. We will call these rules transformations. We use the word transformation since we will be transforming figures into new figures by transforming all the points in the figure. If under a transformation, a point (x, y) is transformed into another point (z, w), then (z, w) is called the image of (x, y) under the transformation. This is consistent with our terminology for functions in general. The first transformation we discuss is translation. Since it seems clear that moving an entire figure in a certain direction is the same as moving each point in the figure in that direction, we need only define what we mean by the “same direction” and describe what “direction” means. One way of doing that is to imagine an arrow pointing in the direction we want to move. That arrow is called a vector, and every vector has two components, a horizontal one, h, and a vertical one, k. The vector, or equivalently the horizontal and vertical components of the vector (how far the object moves horizontally and vertically) describe the direction representing our translation. In Figure 11.16 we see a vector with horizontal component h and vertical component k. k h Figure 11.16 Suppose we want to move each point (x, y) in the plane in the direction of this arrow. That means that we move the x coordinate of each point the same distance, h, in the horizontal direc- tion, and the y coordinate the same distance, k, in the vertical direction. That is, (x, y) is mapped into, or transformed into the point (x + h, y + k). Or equivalently, (x + h, y + k) is the image of (x, y) under this transformation. Figure 11.17 shows this.

508 Chapter 11 Geometric Transformations y (x + h, y + k) k h (x, y) x Figure 11.17 In symbols this transformation is often denoted by T(h, k), and we denote the image of any point (x, y) under the transformation as T(h,k)(x, y). Thus, T(h,k)(x, y) = (x + h, y + k). The subscript (h, k) tells us the x coordinate is increased by h and the y coordinate is increased by k. (Of course, when h or k is negative, the respective coordinate “increase” really amounts to a decrease. Although we have said that a translation must move points, it is important to extend this idea and include the zero translation that doesn’t move any points. That is the translation when (h, k) = (0, 0). Here there is no “direction” vector. A translation is a function from R2 to R2, as are all the other trans- formations we discuss in the main body of this section. A typical secondary school question is: Example 11.1 (a) Find the image of the point (1, 4) under the translation T(À3, 2). (b) Find the image of the triangle whose vertices are A = (À1, 2), B = (4, 7), and C = (0, 6) under the translation T(À3, 2). Solution: (a) T(À3, 2)(1, 4) = (1 + À3, 4 + 2) = (À2, 6). (b) At first glance, the solution seems simple. We just find the image of each of the three points separately and then connect them. Although that can be done easily, there are some issues that we have to address. One is: How do we know that the image of the triangle is still a triangle? Our intuition says it must be. But does the mathematical definitionof T(h,k) imply this? (See Student Learning Opportunity 10 in this regard.) For now, we just accept that translations map triangles to triangles. We will return to that issue later in Section 11.5.2. Thus, the image of our triangle has vertices A0 = T(À3,2)(À1, 2), B0 = T(À3,2)(4, 7), and C0 = T(À3,2)(0, 6). And this yields A0 = (À4, 4), B0 = (1, 9), and C0 = (À3, 8). In Figure 11.18 we see the original triangle together with its image. C' B' B y A' A C x Figure 11.18

11.3 Bringing in the Main Tool—Functions 509 As we can see, the triangle has been translated parallel to itself, and every point of the triangle has been moved 3 units to the left and 2 units up. The next type of transformation we discuss is the reflection. Initially, we will deal only with the simplest types of reflections, those in the x-axis, the y-axis, and the lines y = x and y = Àx, but will go further in the next section. We use the letter r to represent reflection. Later on when we do rotations, we will use capital R to denote rotation. Thus, reflection will be denoted by a lowercase r and rotation by R. The reflection of a point in the x-axis, which we denote by rx, is easy to describe. (See Figure 11.19.) y-axis P = (x, y) x-axis P' = (x, – y) Figure 11.19 We see immediately that if we reflect a point using the x-axis as a mirror, the image of a point P = (x, y) is the point P0 = (x, Ày). Thus, rxðx; yÞ ¼ ðx; ÀyÞ: ð11:1Þ If we reflect the point (x, y) about the y-axis, and denote that transformation by ry, we can write this as ry(x, y) = (Àx, y). (Draw a picture and see for yourself.) Thus, the reflection about the x-axis of the point (1, 2) is (1, À2), and the reflection of the point (1, 2) about the y-axis is (À1, 2). The reflections about the line y = x and y = Àx require a bit more thinking. We will obtain for- mulas for these later. The next transformation we describe is the dilation function, with the center of dilation at the origin. The dilation function takes a point and multiplies each coordinate by a real number k. We denote the dilation transformation by Dk. Thus, Dk(x, y) = (kx, ky). This will stretch or shrink the figure by a factor of |k|. If k > 1 or k < À1, the figure is enlarged. If À1 < k < 1, the figure is shrunk. Of course, if k = 1, then the figure is left unchanged. So the image of (1, 2) under a dilation with a factor of 5 (that is, D5(1, 2)) will be (5, 10), which is 5 times further from the origin than it was before. If this is applied to every point in a figure, every point will be 5 times further away from the origin than before and the figure will be stretched. If k were À5,) each point would still be 5 times as far from the origin as it was before, only in the opposite direction. So the figure would be stretched and flipped about the origin. If k were 1/5, then each point of the figure would be 1/5 as far from the origin as it was before and the figure would be shrunk. Finally, we discuss rotation about a point O. Initially O will be the origin. (Later we will discuss rotations about points other than the origin.) Here each point, P, is rotated about O a certain number of degrees. To find the image of a point P under such a rotation requires the use of trigo- nometry. Suppose that the starting point P has coordinates (x, y) and that it is a distance r from the

510 Chapter 11 Geometric Transformations origin. Suppose further that the line segment OP makes an angle of α with the positive x-axis. Then using the right triangle in Figure 11.20 y-axis P = (x, y) a r x-axis y x Figure 11.20 we see that cos a ¼ length of adjacent side ¼ x ð11:2Þ length of hypotenuse r and sin a ¼ length of opposite side ¼ y ð11:3Þ length of hypotenuse r and by cross multiplying we get ð11:4Þ x ¼ r cos a and y ¼ r sin a: ð11:5Þ We now rotate OP by an angle of θ to get a new point, (x0, y0). (See Figure 11.21, where θ is the angle between OP and OP0.) P’ = (x’, y’ ) P = (x, y) r y Q q ax O Q’ Figure 11.21 Then using triangle OP0Q0, whose sides are x0 and y0, and the figure, the new point P0 has coor- dinates (x0, y0) where x0 ¼ r cos ða þ yÞ ð11:6Þ and ð11:7Þ y0 ¼ r sin ða þ yÞ:

11.3 Bringing in the Main Tool—Functions 511 But we know from trigonometry that ð11:8Þ cos ða þ yÞ ¼ cos a cos y À sin a sin y and that ð11:9Þ sin ða þ yÞ ¼ sin a cos y þ cos a sin y and when equations (11.8) and (11.9) are substituted into equations (11.6) and (11.7), we get x0 ¼ rð cos a cos y À sin a sin yÞ ¼ r cos a cos y À r sin a sin y ð11:10Þ and ð11:11Þ y0 ¼ rð sin a cos y þ cos a sin yÞ ¼ r sin a cos y þ r cos a sin y: Using the facts that x = r cos α and y = r sin α from equations (11.4) and (11.5), and substituting these in equations (11.10) and (11.11), equations (11.10) and (11.11), respectively, can be written as x0 ¼ r cos a cos y À r sin y sin a ¼ x cos y À y sin y and y0 ¼ r sin a cos y þ r cos a sin y ¼ y cos y þ x sin y ¼ x sin y þ y cos y: In summary, Theorem 11.2 When we rotate a point (x, y) by an angle of θ, the new coordinates of the point are x0 and y0 where x0 ¼ x cos y À y sin y ð11:12Þ and ð11:13Þ y0 ¼ x sin y þ y cos y: Let us illustrate this. Suppose we want to rotate the point P = (x, y) = (1, 2), 90 counterclockwise. Using equations (11.12) and (11.13) we get x0 = 1 cos 90 À2 sin 90 and y0 = 1 sin 90 + 2 cos 90. This gives us (since sin 90 = 1 and cos 90 = 0) that P0 = (x0, y0) = (À2, 1). (See Figure 11.22.) y P (1, 2) P’ (−2, 1) O x Figure 11.22

512 Chapter 11 Geometric Transformations We can check that this is correct. After all, if θ is 90, then the lines OP and OP0 must be perpen- dicular. That means that the slopes of lines OP and OP0 should be negative reciprocals of one another. But the slope of line OP where O is the origin, is 2/1 and the slope of line OP0 is 1/À2 and these are negative reciprocals of one another! As another example, if we took the same point P = (x, y) = (1, 2) and rotated it by an angle of pffiffiffi 30 degrees, then the point P0 = (x0, y0) would be ð1 cos 30 À 2 sin 30; 1 sin 30 þ 2 cos 30Þ ¼ ð 3=2 À pffiffiffi pffiffiffi 1; 1=2 þ 3Þ since sin 30 = 1/2 and cos 30 ¼ 3=2: We are now ready to examine transformations from a somewhat higher level. Student Learning Opportunities 1 Find the image of each of the points under the translation T(À1,2). Then find the reflection of the original points about the x-axis and y-axis. (a)* (2, 4) (b) (À3, 5) (c) (7, 6) 2* Find the image of the point (À2, 3) after a rotation of 30 degrees about the origin. 3 Find the image of the point (5, 9) under a rotation of À60 degrees about the origin. 4 Find the image of the point (2, 5) after a rotation of 90 degrees about the origin. Then rotate the image 30. Show that the result is the same as rotating the original point, (2, 5), 120 about the origin. Use the rotation formulas from this section to do this problem. 5* Find the image of a triangle whose vertices are A(2, 4), B(4, 2), and C(6, 8) under a dilation with a factor of 1/2 where the origin is the center of dilation. Draw the original triangle ABC and its dilated image, triangle A0B0C0 on a coordinate grid. Compare the size and shape of triangle ABC and its image, triangle A0B0C0. 6* If T(h,k)(4, 2) = (7,À5), what point is (h, k)? 7 (C) One of your students asks, “If you are given two transformations, does it matter which one you do first?” How do you reply? Does it matter? Support your answer. 8 (C) A student has a suspicion that if we reflect a point (x, y) about the x-axis and then take the result and reflect it about the y-axis, that we have rotated the point a total of 180 degrees about the origin. How can you use an algebraic argument to help the student understand why this really is the case? 9 The image of a point (x, y) under a rotation about the origin by an angle of 90 degrees yields the point (2, 3). What point is (x, y)? 10* Consider the transformation T such that T(x, y) = (0, 0) for all (x, y). What is the image of any figure under this transformation? 11* (C) You give your students triangle ABC, where to travel from A to B to C we must go counter- clockwise. You ask them to reflect this triangle about a line ℓ, which doesn’t touch the triangle, and ask them to determine if the image triangle, A0B0C0, also has the property that to go from A0 to B0 to C0 we must travel counterclockwise. Answer the question. Then make a statement about the preservation of orientation of the vertices of a triangle after a reflection.

11.4 The Matrix Approach to Transformation Geometry 513 12 (C) How would you help your students understand that translations preserve distance? Speci- fically, give a proof using the distance formula in the Cartesian plane, to show that if A and B are two points, and we translate each of them using T(h,k), then the image points A0 and B0 are the same distance from each other as A and B are. Using this result, explain why under a translation the image of a triangle is congruent to the original image. 13 (C) One of your students, Krysten, claims that if two line segments are parallel to each other, then after a translation, they will also be parallel to each other. Krysten is correct, but wants to know if one can give a proof of this. Make up a proof of this using the slope of the line segment before and after the translation. 14 (C) One of your students, Michael, claims that if two line segments are perpendicular to each other, then after a translation, they will also be perpendicular to each other. He asks your as- sistance in proving this. Give a proof of this using slopes, before you guide him with discovering the proof himself. 15 (C) How would you help your students understand that reflections preserve distance? Specifi- cally, show, using congruent triangles, that if A and B are two points not on a line l, and A0 and B0 are their reflections about l, then the distance from A to B is the same as the distance from A0 to B0. 16 (C) How would you help your students understand that reflections preserve angles between lines? Specifically, show that, if triangle ABC does not touch or cross a line l, and we reflect it about l, the image triangle, A0B0C0, that we get is congruent to ABC, and hence the angles ABC and A0B0C0 are congruent. 17 (C) How would you help your students understand that rotations preserve length? Specifically, using congruent triangles, show that, if AB is rotated about a point P, that the image points A0 and B0 are the same distance from each other as A and B are. 18 Derive the formula which rotates a point P about a point (a, b) where (a, b) is not the origin. (Hint: Translate P and (a, b) simultaneously so that (a, b) is at the origin, then do the rotation, and then translate both points simultaneously so that the rotation point, (a, b), is back where it started.) Using the same idea, tell how to dilate a figure with center at a point other than the origin. 11.4 The Matrix Approach to Transformation Geometry—A Higher Level LAUNCH 1 Perform the following matrix operations: !! 10 x (a) 0 À1 y !! À1 0 x (b) 01 y

514 Chapter 11 Geometric Transformations !! k0 x (c) 0k y !! cos y À sin y x (d) sin y cos y y 2 If we consider ! x as the ordered pair (x, y), what transformations are being performed on y the point (x, y) in examples a, b, c, and d? After having done the launch questions, you might be wondering why we have asked you to represent transformations with matrices. You will learn more about how to use matrices and learn about their value when discussing transformations as you read on. 11.4.1 Reflections, Rotations, and Dilations In this section we expand our study of transformations by examining them from a matrix point of view. Because of the many advantages of using the matrix approach, the latest secondary school texts have been using matrices as well. What are these advantages you might ask? For one, the matrix approach makes computer im- plementation of animation easy (since this requires many transformations and consequently many matrix multiplications, which computers can do effortlessly). Second, and perhaps most important, the matrix approach in this section generalizes to three dimensions and thus gives us the ability to apply this material to many real-life situations that use three-dimensional motion. For example, in practical applications with robotics, we use transformations and the matrices related to them to get the robots to move the way we wish. Third, there is a certain beauty to the approach, as it allows us to connect different areas of mathematics such as, geometry, functions, matrix manipulations, trig- onometry, and algebra. In Chapter 9 we used complex numbers to discuss certain transformations and that is quite a beautiful approach to this subject area. However, complex numbers are a two-dimensional idea, and won’t accomplish our need to work in three dimensions. Matrices, on the other hand, will. Finally, the matrix approach gives us information that we just cannot get otherwise. Thus, matrices not only give us a different perspective on rotations, reflections, and translations, but have genuine advantages over other approaches. And, by the way, we hope you will appreciate how so many branches of mathematics (geometry, functions, matrix operations, trigonometry, and algebra) work together to accomplish the most complex transformations. Normally points in the plane are written as (horizontal) ordered pairs (x, y), but since we will need matrix operations, and to multiply matrices, the matrices have to be the right sizes, we will often x need to write our ordered pairs vertically as y : Initially, we will be dealing only with 2 × 2 matrices, !! ab x and we recall that when we multiply a 2 × 2 matrix by a 2 × 1 matrix , we get a 2 × 1 cd y matrix, which is ax þ dbyy: cx þ

11.4 The Matrix Approach to Transformation Geometry 515 Thus, ! ! ! ! 2 5 1Á5þ2Á6 17 1 3 Á ¼ ¼ : 4 6 3Á5þ4Á6 39 (See Chapter 10, Section 8 for a review of all the matrix manipulations we will do in this chapter.) Let us go back to the reflection transformation, rx which reflects every point about the x-axis. That transformation is given by rx(x, y) = (x,Ày). Using our convention that we will write ordered pairs vertically, this can be written as ! ! x x rx y ¼ : Ày The way to write this using matrix multiplication is ! ! ! ! ð11:14Þ x x 10 x rx y ¼ y ¼ Á Ày 0 À1 as you can verify by multiplying the two matrices on the right. Similarly, ry, the reflection about the y-axis is given by !! !! x Àx À1 0 x ry y ¼ ¼ Á: ð11:15Þ y 01 y The dilation of a point by a factor of k can be written as ! ! ! ! ð11:16Þ x kx k0 x Dk y ¼ ¼ Á : ky 0k y Finally, the rotation transformation Rθ (notice the capital R) that rotates every point about the origin, an angle θ (where θ is always measured from the positive x-axis and θ is positive when we rotate counterclockwise and negative when we rotate clockwise) can be written as ! ! cos y À sin y !! x x cos y À y sin y x Ry y ¼ Á ð11:17Þ ¼ sin y cos y y x sin y þ y cos y which follows from equations (11.12) and (11.13). In each of the transformations we have given, the image of a point after the transformation can be obtained by multiplying the coordinates of the point by a matrix. This matrix is called the matrix of the transformation. Thus, the matrix of the rotation transformation from (11.17) is ! ð11:18Þ cos y À sin y sin y cos y and the matrix of the reflection about the y-axis transformation from equation (11.15) is ! ð11:19Þ À1 0 ; 01

516 Chapter 11 Geometric Transformations while the matrix of the transformation that dilates a point by a factor of k, by equation (11.16), is ! ð11:20Þ k0 : 0k Let us illustrate some of these ideas with one example. Example 11.3 We wish to rotate the three points with coordinates A = (1, 2), B = (À1, 3), and C = (À2,À4), 90 counterclockwise about the origin. (a) Find the image of each point after this counterclock- wise rotation, and then (b) describe a way of rotating all three points at once 90. Solution: The matrix that we need to perform the rotation by equation (11.17) is R90 ¼ ! ! ð11:21Þ cos 90 À sin 90 0 À1 sin 90 cos 90 ¼ : 10 Thus, the images A0, B0, and C0 of the points A = (1, 2), B = (À1, 3), and C = (À2,À4), respectively, are A0 = (À2, 1), B0 = (À3, À1), and C0 = (4, À2), obtained as follows: 1 ! 0 À1 ! 1 ! À2 ! A0 ¼ R90 ¼ Á¼ ; B0 ¼ R90 C0 ¼ R90 2 10 2 1 À1 ! 0 À1 ! À1 ! À3 ! ¼ Á ¼ ; and 3 10 3 À1 À2 ! 0 À1 ! À2 ! 4! ¼ Á¼ À4 10 À4 À2 where we have written our points vertically to do the matrix multiplications. A quick way of finding the images A0, B0, and C0 is to put the three points A, B, and C we want rotated in a matrix whose columns are A, B, and C, and then multiply that matrix by the rotation matrix. Thus, all our work becomes much shorter if we simply multiply !! 0 À1 1 À1 À2 Á 10 2 3 À4 to get ð11:22Þ ! À2 À3 4 : 1 À1 À2 Notice that the columns of this new matrix in (11.22) are A0, B0, and C0 but written vertically. This same method works for all transformations. That is, we can find the image of all points in a figure under a transformation by letting the coordinates of the points of the figure be the columns of a matrix, and then multiplying that matrix by the matrix of the transformation. This is the way computers do it. You might be thinking, “But a figure has infinitely many points on it. How does a computer multiply by an infinite matrix?” The answer is that in a computer’s memory, figures are represented

11.4 The Matrix Approach to Transformation Geometry 517 by a finite number of points, though that finite number may be large. Each point on the computer screen takes up a tiny amount of space called a pixel, and each pixel has a set of coordinates. Each point on a curve takes up one or more of these pixels. Thus, as far as the computer is concerned, the curve itself consists of only a finite number of pixels, and hence a finite number of points. 11.4.2 Compositions of Transformations Computer animations, as well as other real-life applications, require the use of multiple transforma- tions. When several transformations are performed one following another, we say that we have per- formed a composition of transformations. Recall that in secondary school we studied the composition, (fg)(x), of two functions, often abbreviated fg. This was defined as f(g(x)). That is, first we apply g to x to get g(x), (the “inner” function) and then we apply f to the result, to get f(g(x)). So, for example, if f(x) is the function x2 and g(x) is the function 2x À 1, then (fg)(3) = f(g(3)) is obtained by first computing g(3) = 2(3) À 1 = 5, and then computing f(g(3)) = f(5) = 52 = 25. The composition of transformations works in exactly the same way. That is, we can compose two transformations, R and S, to get a new transformation R  S. We interpret the composition the same way, namely, we do the inner one, S, first, and then follow it by the outer one, R. So if we wanted to say, dilate a point (2, 3) by a factor of 5, and then rotate it 90, we would be interested in first performing D5(2, 3) to get (10, 15) and then following it with R90(10, 15), which by equation (11.21) is !! À15 ! 0 À1 10 ¼: 1 0 15 10 Our first theorem discusses the results of composing two translations. Intuitively, if we translate an object twice we have still, ultimately, translated it. To illustrate, if we wanted to compute T(5,6)T(À2,4)) on a point (x, y), the first function would translate the point by the vector (À2, 4) and then the second one would take the result and translate it by the vector (5, 6). So ðTð5;6Þ  TðÀ2;4ÞÞðx; yÞ ¼ ðTð5;6ÞðTðÀ2;4Þðx; yÞÞ ¼ Tð5;6Þðx À 2; y þ 4Þ ¼ ðx À 2 þ 5; y þ 4 þ 6Þ ¼ ðx þ 3; y þ 10Þ and we observe that the result here is T(3,10)(x, y). That is, the net result of composing these two translations, is that we get yet another translation, T(3,10). We can state this more generally as a theorem. Theorem 11.4 The composition of two translations is a translation. More specifically T(a,b)  T(c,d) = T(a+c,b+d). It is also geometrically obvious that when you rotate a point by an angle θ1 about the origin, and then rotate the result by an angle θ2 about the origin, that we are really rotating the initial point through an angle of θ1 + θ2 about the origin. We state this as a theorem also. Theorem 11.5 Rθ2  Rθ1 = Rθ1 + θ2.

518 Chapter 11 Geometric Transformations If our matrix representation is going to be useful at all, then we should be able to derive this theorem using matrices. Let us see how this is accomplished using the matrix approach. Rotating a point (x, y) by θ1 first and then by θ2 next means performing the composition (Rθ2  Rθ1) on (x, y). From the definition of composition, ! !! xx ðRy2  Ry1 Þ y ¼ Ry2 Ry1 y : Using equation (11.17) this can be written as ! !! cos y1 À sin y1 Á x Ry2 sin y1 cos y1 y : Using equation (11.17) again, this is the same as cos y2 ! cos y1 ! !! sin y2 À sin y2 Á sin y1 À sin y1 Á x cos y2 cos y1 y which, by associativity of matrix multiplication gives us cos y2 ! cos y1 À sin y1 !! ! sin y2 À sin y2 Á sin y1 x cos y2 cos y1 Á ð11:23Þ : y Multiplying these matrices we get cos y2 cos y1 À sin y2 sin y1 ! ! sin y2 cos y1 þ cos y2 sin y1 À cos y2 sin y1 À sin y2 cos y1 Á x À sin y2 sin y1 þ cos y2 cos y1 y which from equations (11.8) and (11.9) gives us ! cos ðy2 þ y1Þ À sin ðy2 þ y1Þ sin ðy2 þ y1Þ cos ðy2 þ y1Þ which, of course, is the same as cos ðy1 þ y2Þ ! sin ðy1 þ y2Þ À sin ðy1 þ y2Þ : cos ðy1 þ y2Þ And this is precisely the matrix ðRy1þy2 Þ ! x : y !! ! xx x Thus, we see that ðRy2  Ry1 Þ y ¼ ðRy1þy2 Þ y for all points : Hence, the two transforma- y tions (Rθ2  Rθ1) and (Rθ1+θ2) are the same. Thus, the composition of two rotations is a rotation. Wait a minute! We almost missed something! Look at what (11.23) is saying. It is claiming that when you compose two transformations that can be written in terms of matrices, the matrix of their com- position is obtained by multiplying the matrices associated with each transformation!!! How nice! And, if we compose many transformations, we can get the net result instantly by just multiplying the

11.4 The Matrix Approach to Transformation Geometry 519 matrices of the transformations and applying the product matrix to the point! This is important enough to state as a theorem. Theorem 11.6 If T1, T2, . . .,!Tn are transformations with matrices M!1, M2, . . . , Mn, then when we xx compute ðTn  :::  T2  T1Þ y we get the same as ðMn ::: M2M1Þ y : What this means is that if you perform T1 first, then T2, and so on to Tn, then you multiply the matrices M1, M2, and so on to Mn. Notice that we don’t multiply the matrices M1M2M3 . . . Mn, rather we multiply them in reverse order, Mn . . . M3M2M1 (so M2 multiplies M1 on the left, M3 multiplies the product of M1 and M2 on the left and so on.) Let us give a numerical example. Example 11.7 (a) We wish to take the point (2,3) and do the following: (1) First rotate it 30 counter- clockwise, then (2) reflect the resulting point about the y-axis. Finally, (3) dilate this new point by a factor of 5. Do this using matrices. (b) Then find a single matrix that accomplishes (1)À(3) for any point (x,y). Solution(a): The matrix, M1, associated with our first transformation, a rotation by 30 is, by equa- tion (11.18), M1 ¼ cos 30 À sin 30 ! sin 30 cos 30 Then the image of (2, 3) under this transformation is 2! M1 3 cos 30 À sin 30 ! 2! 3 ¼ sin 30 cos 30 0 pffiffiffi 1 3 À 3 ¼@ 2 A: 1þ 3 pffiffiffi 2 3 (We are using the well-known results that cos 30 ¼ pffiffi and sin 30 ¼ 1 .) The matrix, M2 associated 3 2 2 with our reflection about the y-axis is, by equation (11.19), M2 ¼ ! À1 0 : 01

520 Chapter 11 Geometric Transformations pffiffiffi ! 3 Àp23ffiffiffi , under this transformation is The image of the point we just got, 1 þ 3 3 2 0 pffiffiffi 1 3 À 3 M2@ 2 A 1þ 3 pffiffiffi À1 2 3 !0 pffiffiffi 1 0 3 À 3 @ 2 A ¼ pffiffiffi 01 1þ 3 3 0 pffiffiffi 1 2 À 3 þ 3 A: ¼@ 2 þ 3 pffiffiffi 1 2 3 Finally, the matrix M3, dilating a point by a factor of 5 by (11.20), is ! 50 M3 ¼ 0 5 pffiffiffi ! À 3 þpffiffi23ffi , under this transformation is and the image of the most recent point we got, 1þ 3 3 2 0 pffiffiffi 1 À 3 þ 3 M3@ 2 A pffiffiffi 1 þ 23!03Àpffi3ffiffi 5 0 @ þ 3 1 2 A ¼ pffiffiffi 05 1 þ 3 3 0 pffiffiffi 1 2 À5 3 þ 15 A ¼@ 2 5þ 15 pffiffiffi 2 3 À1:1603 ! %: 17:99 Solution (b): Since in the statement of the example we are doing (1), rotating by 30 counterclock- wise, followed by (2) reflecting the resulting point about the y-axis, followed by (3) dilating the re- sulting point by a factor of 5, we can accomplish this all in one step by using the single matrix M obtained by multiplying the matrices M1, M2, and M3 of the respective transformations, in reverse order. That is, the single matrix that accomplishes all three transformations described in (1)À(3) is M = M3M2M1 or just 5 ! À1 ! cos 30 À sin 30 ! 0 0 05 0 1 sin 30 cos 30 which simplifies to À 5 pffiffiffi ! 2 3 5 M¼ 5 5 p2 ffiffiffi : 2 2 3

11.4 The Matrix Approach to Transformation Geometry 521 This matrix will accomplish (1)À(3) for any point. Let us verify that in the case of the point (2, 3). We have: 2! M 3 0 pffiffiffi 1 À 5 3 5 2! 2 3 ¼ @ 2 A 5 5 pffiffiffi 2 3 0 2pffiffiffi 1 À5 3 þ 15 ¼@ 2 A 5þ 15 pffiffiffi 2 3 ! À1:1603 % 17:99 which is exactly the same result we got earlier. We can if we wish express M in decimal form. In that case, À 5 pffiffiffi ! ! 2 3 À4: 330 1 2: 5 5 2: 5 4: 330 1 M¼ 5 5 p2 ffiffiffi % 2 2 3 2 ! and if you multiply 3 by this, we get À4:3301 2:5 ! 2 ! 2:5 4:3301 3 À1:1602 ! %: 17:99 You might feel it is easier to do this example without the matrix approach and you may be right. But then you will be missing the point, namely that this matrix, M, will transform any point in the same way that we transformed (2, 3). This is important in computer animation when we want to transform a figure that may have thousands of points in it at once. Rather than multiplying each point by three matrices, we multiply each point by one matrix which does the work of three, and this gives us efficiency. 11.4.3 How a Calculator Uses Rotations to Calculate It may surprise you to know that when the TI calculators compute sines, cosines, and logarithms, as well as various other functions, they do it by using rotations! Yes, rotations! It is remarkable that geometry can be used to do these computations and do them so efficiently. We won’t give all the details of how this works in this section since it is well beyond the scope of this book. Rather, we will just give a brief summary of the salient facts to give the reader an idea of what the calculator does, and refer the interested reader to the paper @How Calculators Calculate.@ [See Sultan, 2008] for more details.