From String Art to Caustic Curves: Envelopes in Symbolic Geometry

Transcription

1 From Sring r o ausic urves: nvelopes in Symbolic eomery Philip Todd phil@salire.com Salire Sofware, PO 565, eaveron, OR 97223, U.S.. bsrac: In his paper, we use he symbolic geomery program eomery xpressions o analyze hree problems involving envelope curves. Firs we examine he envelopes of families of lines passing hrough poins which are equally spaced on a pair of line segmens. We use a combinaion of symbolic geomery and algebra o develop an expression for he area of he void in a popular sring ar figure consising of 3 parabolas inscribed in a riangle. We use an envelope approach o reduce a popular calculus problem - ha of finding he longes ladder which fis around an asymmeric corner o an algebra problem which is readily solved using S. Finally we sudy he causic curves generaed by reflecing a poin ligh source in a shiny cylinder. We analyze hese boh experimenally and heoreically, and focus on deermining he parameric and aresian locaions of he cusps. These examples illusrae how symbolic geomery echnology can be used o make mahemaics fun, accessible and challenging.. Inroducion The envelope is ha phanom curve your eye picks ou when you look a a family of sraigh lines. Perhaps he mos familiar example of he envelope is found in sring ar, or mahemaical embroidery [], where a family of lines is creaed by aaching srings o arfully locaed ses of pins (Fig.). nvelopes also occur naurally as ligh causics, here he family of lines is a family of refleced rays, and he causic appears as a fringe of ligh. eomery xpressions [4] ( ) is a symbolic geomery sysem. s such i is able o generae symbolic expressions from geomeric figures. mongs oher hings, eomery xpressions is able o generae he envelope curve of a family of lines or circles, and o derive boh parameric and implici equaions for he curve. In his paper, we will use envelopes in he symbolic geomery program eomery xpressions o explore a variey of mahemaical problems ranging from sring ar o ligh causics. We will noe how he use of a symbolic geomery sysem in concer wih a S allows a mode of hinking which is par geomeric, par algebraic, and which faciliaes problem solving, exploraion and discovery. 2. Sring r Mahemaically, he envelope of a one parameer family of lines is he curve which is muually angen o ha family [3]. In eomery xpressions a locus curve is specified by selecing a poin and specifying he parameer which is o vary o creae he locus. n envelope curve is creaed in a similar fashion excep ha insead of a poin, a line, line segmen or circle is seleced. 97

2 Figure : We ask he quesion, wha is he area of he void in he cener of his sring figure? In Figure 2, we model a ypical sring ar configuraion where pins are evenly spaced along a pair of lines. is specified o lie proporion along while F is proporion along. The envelope of F is consruced wih respec o parameer. (c,d) - (a,b) F b 2 c 2-2 a b c d+a 2 d 2 +Y 2 a 2 +2 a c+c 2 +X Y (-2 a b-2 b c-2 a d-2 c d)+x 2 b 2 +2 b d+d 2 +Y 2 a b c-2 b c 2-2 a 2 d+2 a Figure 2: eomery xpressions model of he envelope curve creaed by a sring ar model The equaion of he curve is second order and hence clearly a conic. Inspecion of he coefficiens shows ha he second order erm in Y and he erm in XY disappear if a= -c. Hence, if poins and are posiioned equal disances on eiher side of he y axis, he parabola will be verical (fig. 3) xamining he parameric equaion for he envelope (fig 3b), we see ha X=0 occurs when he b d parameer = ½. his locaion he Y coordinae is, or he midpoin of he median. 4 98

3 In he siuaion of figure 2, he riangle median is clearly verical, as is he axis of symmery of he envelope parabola. Hence, if we consider he general riangle (figure ), we observe ha he axis of symmery of he parabola is parallel o he median of he riangle. (-a,d) (-a,d) 4 Y a 2 -a 2 b-a 2 d+x 2 (-b-d)+x (-2 a b+2 a d)=0 0, b+d 2 - (a,b) - X=a (-+2 ) Y=d-2 d + 2 (b+d) (a,b) Figure 3: (a) Implici equaion of he envelope, where poins and are equal disances along he x axis from. (b) parameric equaion of he envelope, along wih he median of riangle. 2. reas We address he quesion: wha is he area of he region bounded by he hree envelope curves in he sring ar of figure? We firs ask he quesion: wha is he area bounded by a single curve? We could do his using inegraion, bu here we develop a geomerical argumen. (This is essenially rchimedes mehod of parabola quadraure oulined in [2]). I is convenien iniially o ransform our piece of sring ar so ha he parabola is of he form y=ax 2 (fig 4). x a h3 4 h+x x a h3 8 h+x H Figure 4: (a) rea of he riangle formed by a chord of he parabola and he angens a is end depends only on he widh of he chord, bu no is locaion. (b) rea of H is half he area of 99

4 xamining he riangle formed by he poin on he parabola wih x-coordinae x, he poin wih x- coordinae x+h, and he poin a he inersecion of he angens o he parabola a hose poins, we see ha is area is cubic in h, bu independen of x. If we now examine he riangle H, where H is he cener of he median, we observe ha is area is half ha of. 3 h ah Now if we add a poin I on he parabola a x coordinae x, he area HI will be and as he h ah formula is independen of x, adding a poin J a x locaion x, he area HJ will also be a h3 32 x J h+x I H Figure 5: rea of IHJ is 5/4 ha of H If he area of H is, we see ha he area of IHJ is. 4 oninuing he process, we see ha he area beween he chord and he parabola is: Now we look a hree parabolas drawn on he sides of he same riangle. To calculae he poins of inersecion beween he parabolas, we copy heir equaions ino an algebra sysem and solve We find 3 roos: 4 c 4 a 4 b { Y 0, X 0 }, { Y, X }, { Y 4 3 RooOf ( _Z 2 _Z, label _L4 ) c, X 4 3 ( a RooOf ( _Z 2 _Z, label _L4 ) 5 b RooOf ( _Z 2 _Z, label _L4 ) 5 a 4 b ) ( 4 RooOf ( _Z 2 _Z, label _L4 ) ) } The second is he ineresing one. From is form, we see ha i is 8/9 of he way along he median. We can herefore draw he sraigh edged riangle covering his area by placing poins 8/9 of he way along he medians (fig 6): 200

5 (b,c) (b,c) 4 (a+b) 9, 4 c 9 H a+b 2, c 2 a c 8 I K H (a,0) L J (a,0) Figure 6: (a) The inersecion of he parabolas is 8/9 of he way along he medians of he riangle. (b) rea KLM is /9 he area of s he area of he riangle is ac/2, we see ha he area of HKL is /9 he area of he original riangle. Now we can verify ha K and L lie a equal disance o he median (fig 7). Now le R be midpoin of he median. R is he poin a which he parabola crosses he median. From he above, he area of he parabola segmen KRL is 4/3 of he area of he riangle KLR, and ac hence is. 8 (b,c) -a c 3 a 2 +2 a b+b 2 +c 2 a c 8 I K H a c 3 a 2 +2 a b+b 2 +c 2 L J (a,0) Figure 7: K and L lie a equal disances from he median. 20

6 I K H R a c 08 L J (a,0) Figure 8: Segmen KRL has area /54 ha of he original riangle dding hree such areas ono he area of he riangle HKL gives a oal area for he void of: V= 5 ac 54 or, if T is he area of he original riangle, 5T V= Ladder Problem classic problem, usually posed as an opimizaion problem in calculus, is ha of finding he longes ladder which fis round a wall. Use of an envelope curve, however, can reduce his o a problem in algebra, readily solved wih he aid of a S. I can be used o illusrae he relaionship beween geomeric incidence beween a poin and a curve and he soluion of an algebraic equaion. X L Y Figure 9: The envelope of a ladder of lengh L and he corner beween corridors of widh X and Y. The problem is o deermine he lengh of he longes ladder which will fi round a corner beween wo corridors of widh X and Y. The siuaion is modeled in figure 9, and he envelope of he family of locaions of he ladder is shown. To creae his model in eomery xpressions, a line 202

7 is drawn such ha is consrained o lie on he y-axis and on he x-axis. The lengh of he line is consrained o be L, and he locaion of is consrained o be disance from he y-axis. The envelope curve is creaed simply by selecing he line and specifying which variable () should vary. key observaion is ha he ladder iself says below is envelope curve, and herefore as long as he envelope curve does no inersec he corner, hen he ladder will clear. In order o find he longes ladder which clears he corner, one wans o find he value for L such ha he poin lies on he envelope curve. X L Y L 6-3 L 4 X 2 +3 L 2 X 4 -X 6-3 L 4 Y 2-2 L 2 X 2 Y 2-3 X 4 Y 2 +3 L 2 Y 4-3 X 2 Y 4 -Y 6 =0 Figure 0: quaion of he envelope curve is compued by eomery xpressions Using eomery xpressions, we can generae he equaion of he envelope curve simply by selecing he curve and requesing he implici equaion. We noice ha he curve equaion is 6 h order in L, bu ha i only includes even powers. Hence i can be considered as a cubic in L 2. We would herefore expec 6 soluions, bu ha hey would be presen as posiive / negaive pairs. opying he equaion from eomery xpressions ino an algebra sysem allows us o solve for L. > solve(l^6-3*x^2*l^4+3*x^4*l^2-x^6-3*y^2*l^4-2*y^2*x^2*l^2-3*y^2*x^4+3*y^4*l^2-3*y^4*x^2-y^6 = 0,L); ( Y X 2 ) ( / 3 ) ( Y 2 ( Y X 2 ) ( / 3 ) 3 Y ( Y X 2 ) ( 2/ 3 ) 3 Y X 2 X 2 ( Y X 2 ) ( / 3 ) ) ( Y X 2 ) ( / ) 3, Four of he resuling soluions are complex, and of he remaining wo real soluions, only one is posiive. This is he one o copy back ino eomery xpressions. We observe ha wih his lengh ladder, he poin lies on he envelope curve and he ladder only jus makes i around he corner. s a final observaion, we noe ha he soluion for L is no symmeric in X and Y. However, he geomery problem clearly is symmeric: he longes ladder which can be dragged from a corridor of widh X ino a corridor of widh Y is obviously he same lengh as he longes ladder which can be dragged from a corridor of widh Y ino one of widh X. I would herefore seem reasonable ha some simplificaion could be done on he soluion o expose is essenial symmery. This could be se as an exercise o be compleed by hand, or one could coerce simplificaion in he algebra sysem by assering ha boh X and Y are posiive. > simplify(%) assuming X>0, Y>0; 203

8 X ( 2/ 3) Y ( 2/ 3 ) ( 3/ 2) ( ) X Y 3 X 2 Y+X Y +3 X Y +X Y X 3 3 Y Y Figure : Soluion is pased back from Maple ino eomery xpressions, and we observe ha he corner poin lies on he envelope. 4. Ligh ausics When ligh reflecs off a convex curved surface, he refleced rays form a curve of brigh ligh called a causic. This curve is exacly he envelope curve of he refleced rays. Figure 2 shows he causic curve formed by ligh reflecing in a wedding ring. Figure 2: ausic curves formed by ligh reflecing inside a gold ring 204

9 We will sudy he form of he causic curve, when ligh from a poin source reflecs in a cylinder. To acquire experimenal daa agains which o compare our heoreical findings, we use he apparaus shown in figure 3. Perspex cylinder is used as our reflecing surface. small L lamp is used o posiion our ligh source inside he cylinder. circular paper disk is marked in concenric rings, each ring being /0 of he radius of he cylinder. This allows us o observe he locaion of feaures of he curve as a proporion of he cylinder radius. Figure 3: pparaus for sudying causic curves wih a ligh source inside he reflecing cylinder. In figure 4 we see he causic curve when he ligh source is a he circumference of he cylinder, and when he ligh source moves closer ino he cener. video of his experimen may be seen a: hp:// Figure 4: ausic curves observed wih (a) he ligh a he circumference of he cylinder and (b) he ligh source inerior o he cylinder. We can model he behavior of a single beam of ligh in eomery xpressions. Firs we creae a circle, and consrain is cener o be and is radius o be. We hen consrain he parameric locaion of on he curve o be. This has he effec of specifying he line o be angle (in radians) from he x axis. eomery xpressions allows you o reflec in a line, bu no in a curve. However reflecion of ligh in a cylinder is equivalen o reflecion in he angen o he 205

10 cylinder. Hence we draw a line hrough and consrain i o be angen o he circle. We now creae poin, consrain is locaion o be (0,-a), and draw an infinie line hrough and. We complee he model by reflecing in he angen line. (0,-a) Figure 5: eomery xpressions model of a beam of ligh refleced in a cylinder. The causic curve is he envelope of he family of refleced lines. We are ineresed in he locaion of he cenral cusp. Firs in he siuaion where a = (fig. 6) (0,-) X= 2 cos()3 3 (+sin()) Y= 2 sin() + cos(2 ) 3 3 Figure 6: Parameric equaion of he causic wih ligh source on he circumference 206

11 The parameer specifies he locaion of poin on he circle as an angle in radians. When is a he locaion (,0) will have he value 0, when is a he locaion (0,) will have he value π/2, ec. learly when is verically above, he ligh from will reflec sraigh back, and will ouch he curve a he cusp. To find he locaion of he cusp, we need simply subsiue = π/2 ino he Y componen of he curve equaion in figure 6: 2sin 2 cos Looking a he picure in figure 4a, we see ha he ligh cusp is somewhere around he hird circle or approximaely 0.33 radii from he cener 4. usp on he ircumference s he ligh source moves owards he cener of he cylinder, he causic curve is no longer fully conained in he cylinder. However, a some poin, i re-eners o give he shape in Figure 7. Using he circular graph paper, we can esimae he locaion of he cenral cusp a his poin o be on he second circle. Figure 7: The causic when he far cusp ouches he circumference. In eomery xpressions, we can change he coordinaes of back o (0,-a) and drag owards he cener of he circle unil he second cenral cusp appears: 207

12 F -a 0, -+2 a a 0, +2 a 3 π 2 π 2 (0,-a) 2 a 2 cos() 3 X= +2 a 2 +3 a sin() Y= a (2+3 a sin()+a sin(3 )) 2 +2 a 2 +3 a sin() Figure 8: urve equaions for general a. Poins and F are pu on he curve and consrained o be a parameric locaions π/2 and 3π/2 In figure 8, poins and F are posiioned a parameric locaions π/2 and 3π/2 on he curve. oordinaes for hese poins have been compued by he sofware. For F o lie on he circumference, we need: a 2a Solving for a yields: a 3 Subsiuing ino he expression for he y coordinae of gives xaminaion of figure 7 shows he cusp lies on he second ring of he circular graph paper, or /5 of he way ou from he cener o he circumference, which corresponds exacly wih our heoreical predicions. 4.2 Furher Quesions Le s assume we move our ligh source a long way from he cylinder, so ha he disance a is effecively infinie. Our cusp will now have locaion: a lim a 2a 2 208

13 This is he siuaion observed in he wedding ring (fig. 2). Having analyzed he cenral cusps, we are lef wih he quesion wha are he coordinaes of he cusps which do no lie on he y axis? For a curve (X(),Y()), cusps occur where he derivaives of X and Y wih respec o simulaneously vanish [3]. The causic s curve equaion can be copied ino Maple, differeniaed and solved as follows: 2 cos( ) 3 a 2 X := + 2 a sin ( ) a Y := ( sin ( ) a + sin ( 3 ) a) a 2 ( + 2 a sin ( ) a) > solve({diff(x,)=0,diff(y,)=0},); ì í = - î 2 p ü ý þ, ì í = î 2 p ü ý þ, { = arcan ( -a, RooOf ( a _Z 2 ))} > allvalues(arcan(-a, RooOf(a^2-+_Z^2))); arcan -a, - a 2 ( ), arcan -a, - - a 2 ( ) rawing poins on he curve wih hese parameric values (copying he soluions back from Maple ino eomery xpressions), we see hey lie on he cusps (fig. 9a). π+arcan a -a 2 F arcan -a -a 2 F -2 a 2 -a 2,a-2 a 3 2 a 2 -a 2,a-2 a 3 (0,-a) arcan -a -a 2 (0,-a) -a 2,-a Figure 9: (a) is he poin on he circumference of he circle wih he same parameric locaion as he righ hand cusp. (b) has he same y coordinae as. oordinaes of he cusps are also given. Where does poin lie on he circle when he refleced ray goes hrough he cusp? One way of finding ou is o pu a poin on he circle and give i he arcan() as is value. Now, hiding he 209

14 parameric locaions of,f and, we can display heir coordinae locaions (fig 9b). We see ha has he same y coordinae as. We have discovered ha cusps occur when he original ray emanaes from poin eiher in a verical or a horizonal direcion. The verical seems geomerically inuiive, his is he direcion a which he ray reflecs back on iself, and one migh expec singular behavior. Wha, if anyhing, is special abou he siuaion where he original ray is horizonal? To answer his quesion, we consider he angle beween he inciden ray and he refleced ray. learly when he inciden ray is verical, his angle is 0 and hence a a minimum. We ask he quesion, wha direcion for yields a maximum angle? Firs, we noe ha his is equivalen o finding a maximum value for. We could, of course generae an equaion for his angle in eomery xpressions, hen differeniae and solve o yield he appropriae value for : (0,-a) (0,-a) Figure 20: (a) is inside he circumcircle of, hence angle < angle. (b) The circumcircle of is angen o he original circle. This corresponds o a maximum of angle. However, a geomeric approach o he problem can be aken by observing ha all he poins such ha = lie on he circumcircle of. In figure 20a we observe ha if his circumcircle inersecs he original circle in wo places, and if lies on he porion of he circumference of cu off by i hen >. Hence, can only be maximal if he circumcircle o is angenial o circle (fig 20b). In his case i is easy o see ha is a diameer of he circumcircle and hence he angle is righ. We have hence shown ha he cusps of he causic occur a exrema of he angle beween inciden and refleced rays. 5. onclusion We have examined hree siuaions where he auomaic derivaion of envelope curves wihin a symbolic geomery program (eomery xpressions) allow problems which would normally be addressed using calculus o be analyzed using a combinaion of geomery and algebra. In addiion, 20

15 resuls obained symbolically, using a compuer algebra sysem, moivaed furher geomerical sudy. Wihou he use of echnology, envelope curves are no accessible a a high school level. Wih he use of echnology hey can become a rich source of example problems and illusraive scenarios. In shor, we have made some ineresing and fun mahemaics accessible hrough echnology. Supplemenary ocumens eomery xpression file for experimenaion in Secion 2 eomery xpression file for experimenaion in Secion 3 eomery xpression file for experimenaion in Secion 4 References []. Pedoe (976) eomery and he Visual rs, over, New York [2] H. orrie (965) 00 rea Problems of lemenary Mahemaics, over, New York [3] J Sruik (950) Lecures on lassical ifferenial eomery, over, NY [4] eomery xpressions websie 2