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For example, on some platforms, \+ output to a postscript file converts plus-minus signs to plus signs. \+ It is suggested that this manuscript be read from the monitor." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "There ar e a few links in the text to other parts of the text. These function \+ correctly only if all sections are expanded." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 33 "Introduction to Polar Coordinates" }}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 31 "Definition of Polar Coordinates" }}{PARA 0 "" 0 "D efinition of pole" {TEXT -1 71 "A polar coordinate system is defined i n the plane by selecting a point " }{TEXT 260 1 "O" }{TEXT -1 12 " cal led the " }{TEXT 256 5 "pole," }{TEXT -1 50 "and a half-line emanating from the pole, called a " }{TEXT 257 14 "reference line" }{TEXT -1 4 " or " }{TEXT 258 19 "reference direction" }{TEXT -1 116 ". The refere nce line is usually drawn horizontally on the page, pointing towards t he right (like the usual positive " }{TEXT 259 1 "x" }{TEXT -1 19 "-ax is). Each point " }{TEXT 261 1 "P" }{TEXT -1 52 " in the plane is loca ted by specifying the distance " }{TEXT 262 1 "r" }{TEXT -1 6 " from \+ " }{TEXT 269 1 "P" }{TEXT -1 13 " to the pole " }{TEXT 270 1 "O" } {TEXT -1 15 " and the angle " }{XPPEDIT 18 0 "theta" "6#%&thetaG" } {TEXT -1 72 " measured counter-clockwise from the reference line to th e line through " }{TEXT 263 1 "O" }{TEXT -1 5 " and " }{TEXT 264 1 "P " }{TEXT -1 17 ". The quantities " }{TEXT 265 1 "r" }{TEXT -1 5 " and \+ " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 12 " are called " } {TEXT 266 17 "polar coordinates" }{TEXT -1 4 " of " }{TEXT 267 3 "P, \+ " }{TEXT -1 29 "and are listed in the order (" }{TEXT 268 1 "r" } {TEXT -1 1 "," }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 4 " ). 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::::::::::::=Z;Z:R:F::J:@:=::B:;:=::B:=Z;::Z:R:FZ:F:@::Z:Z;::J:<:F::::::::::::::::::::::w<@::::::::::::::::::: ::Z::@::B:@:=::J:@:=::F::Z:>Z;j:::Z:>:@Z;F :B:=Z;::B:;R:F::J:>:F::::::::::::::::::::::^BR::::::::::::::::::::::B: Z;:Z:F:@::Z:>Z;j::J:>Z:Z;j::J:@:=::B:;R:F::Z:F:@::Z:B:R:@j:J:@:=::B:;R:F::Z:F:@:Z;R:=::J:j:::>:F::;B:j :::::::::::::::::::::::v?R:::::::::::::::::::::::>:j:J:F:@::Z:F:@::Z:Z ;:;B:;B::@:=::Z:F:@::j:>Z;j:::B:R:::<:@::J:>Z:B:::::j:< j:@Z;::>:F:::;:=::Z:Z;B:;Z;R::::::::::::::::::::::::W;@::::::::::::::: ::::::::J:B:=J:@Z;F:::;B:=Z;::>:R:::;R:F::Z:>:F::Z:F:@::J:<:@::J:Z;::B :=J::F::Z:>Z;j:::Z;j:::>:F::Z:>Z;j:::<:F::Z::=::B:=Z;::B:R::>Z:>:F::::::::::::: ::::::::::::jLZ;:::::::::::::::::::::::::<:;j:R:B:=Z;::J:j:::<:@::Z:>Z ;j:::>Z;j:::<:@:::;:=:;B:;B:@:=:::::::::::::::::::::::::jTZ;:::::::::: :::::::::::::::Z:J:@:=B:=Z;::J:j:::<:@::Z:>Z;j:::>Z;j:::<:@:::;:=B:;B: ;R:F::::::::::::::::::::::::::JLZ;::::::::::::::::::::::::::>:<:=Z;::B :@:=::B:=Z;::B:;:=::J:@:=::B:;:=::J:>Z:j:R:::::::::::::::::::::::::::v >R:::::::::::::::::::::::::::J:B:=Z;::Z;j:::>Z;j:::>:F: :J:Z;j:::>:;B:;B:F:@::::::::::::::::: ::::::::::::JgZ;:::::::::::::::::::::::::::::Z:>:;:=Z;:Z:FZ;R::::<::@:=:Z:R:F::::F :@::::::::::::::::::::::::::::::::JUZ;:::::::::::::::::::::::::::::::: ::;Z:::::F:@::::::::::::::::::::::::::::::::::J[Z;:::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::>BR:::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::s;@::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::JMZ; :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::><1:" }{TEXT -1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 35 "Non-un iqueness of Polar Coordinates" }}{PARA 0 "" 0 "Extension of polar coor s" {TEXT -1 87 "The above definition of polar coordinates is extended \+ to allow for all real values of " }{TEXT 271 1 "r" }{TEXT -1 5 " and \+ " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 25 ". For example, val ues of " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 14 " greater th an " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 68 " int erpret as one or more complete counterclockwise rotations about " } {TEXT 272 1 "O" }{TEXT -1 20 " before arriving at " }{TEXT 273 1 "P" } {TEXT -1 18 ". Such values of " }{XPPEDIT 18 0 "theta" "6#%&thetaG" } {TEXT -1 161 " do not determine new points; they merely provide additi onal coordinate sets. The point located one unit directly above the p ole would have polar coordinates (" }{XPPEDIT 18 0 "(1,Pi/2)" "6$\"\" \"*&%#PiGF#\"\"#!\"\"" }{TEXT -1 58 " ), but the same point would also have polar coordinates (" }{XPPEDIT 18 0 "1,5*Pi/2)" "6$\"\"\"*(\"\"& F#%#PiGF#\"\"#!\"\"" }{TEXT -1 7 ") and (" }{XPPEDIT 18 0 "1,9*Pi/2" " 6$\"\"\"*(\"\"*F#%#PiGF#\"\"#!\"\"" }{TEXT -1 56 ") as well as infinit ely many others. Negative values of " }{XPPEDIT 18 0 "theta" "6#%&thet aG" }{TEXT -1 46 " interpret as clockwise rotations. The point (" } {XPPEDIT 18 0 "1,Pi/2" "6$\"\"\"*&%#PiGF#\"\"#!\"\"" }{TEXT -1 30 ") a lso has polar coordinates (" }{XPPEDIT 18 0 "1,-3*Pi/2)" "6$\"\"\",$*( \"\"$F#%#PiGF#\"\"#!\"\"F)" }{TEXT -1 50 ") as well as infinitely many others with negative " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Negative values of " }{TEXT 274 1 "r" } {TEXT -1 84 " interpret in terms of backwards (or reverse) motion in t he direction determined by " }{XPPEDIT 18 0 "theta" "6#%&thetaG" } {TEXT -1 42 ". Thus the point with polar coordinates (" }{XPPEDIT 18 0 "1,Pi/2" "6$\"\"\"*&%#PiGF#\"\"#!\"\"" }{TEXT -1 37 ") would also ha ve polar coordinates (" }{XPPEDIT 18 0 "-1,3*Pi/2" "6$,$\"\"\"!\"\"*( \"\"$F$%#PiGF$\"\"#F%" }{TEXT -1 98 ") and infinitely many others obta ined from this one by adding or subtracting integer multiples of " } {XPPEDIT 18 0 "2*pi" "6#*&\"\"#\"\"\"%#piGF%" }{TEXT -1 14 " to the an gle " }{XPPEDIT 18 0 "3*Pi/2" "6#*(\"\"$\"\"\"%#PiGF%\"\"#!\"\"" } {TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 625 "The non-uniqueness of po lar coordinates makes it necessary to be able to determine all polar c oordinates for a given point. For example, suppose one wishes to deter mine whether or not a particular point is on the graph of a certain po lar equation. In Cartesian coordinates, such a determination is simpl e. If the coordinates of the given point satisfy the Cartesian equatio n then the point is on its graph; otherwise it is not. But even when \+ one set of polar coordinates of a given point fails to satisfy a polar equation, the point has infinitely many other sets of polar coordinat es that still might satisfy the equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 276 45 "I f a point has one set of polar coordinates (" }{XPPEDIT 18 0 "r,theta " "6$%\"rG%&thetaG" }{TEXT 277 23 " ), then the others are" }}{PARA 0 "" 0 "" {TEXT 278 1 "(" }{XPPEDIT 18 0 "r,theta+2*k*Pi)" "6$%\"rG,&%&t hetaG\"\"\"*(\"\"#F&%\"kGF&%#PiGF&F&" }{TEXT 279 7 ") and (" } {XPPEDIT 18 0 "-r,theta+(2*k+1)*Pi" "6$,$%\"rG!\"\",&%&thetaG\"\"\"*&, &*&\"\"#F(%\"kGF(F(F(F(F(%#PiGF(F(" }{TEXT 280 3 "), " }{TEXT 275 1 "k " }{TEXT 281 18 " = 0, 1, 2, ... ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "If " } {TEXT 282 1 "r" }{TEXT -1 35 " is required to be nonnegative and " } {XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 47 " is required to lie i n the half open interval [" }{XPPEDIT 18 0 "0,2*Pi)" "6$\"\"!*&\"\"#\" \"\"%#PiGF&" }{TEXT -1 67 ") then there is no ambiguity in the polar c oordinates (except that " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 18 " is arbitrary for " }{XPPEDIT 18 0 "r=0" "6#/%\"rG\"\"!" } {TEXT -1 185 " ). The student may reasonably ask why these restrictio ns are not imposed in order to simplify the treatment. The answer is t hat many applications and contexts require using values of " }{TEXT 283 1 "r" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "theta" "6#%&thetaG" } {TEXT -1 120 " outside of these ranges, so that these restrictions wou ld unduly limit the usefulness of the polar coordinate system. " }}}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 62 " Associated Cartesian Coordinates and Transformation Equations" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 45 " Definition of the Associated Cartesian System" }}{PARA 0 "" 0 "" {TEXT -1 196 "It is very useful (and conventional) to associate, in a \+ natural way, a Cartesian coordinate system with each polar coordinate \+ system. The origin of the associated Cartesian system is taken as the \+ " }{HYPERLNK 17 "pole" 1 "" "Definition of pole" }{TEXT -1 39 " of the polar system, and the positive " }{TEXT 284 1 "x" }{TEXT -1 22 "-axis is taken as the " }{HYPERLNK 17 "reference line" 1 "" "Definition of \+ pole" }{TEXT -1 22 " in the polar system. " }}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 24 "Transforming Coordinates" }}{PARA 0 "" 0 "" {TEXT -1 270 "In this way, every point in the plane has two kinds of coordinate s - Cartesian and polar. It is natural to ask how these are related. G iven coordinates in one of the coordinate systems, how can coordinates be obtained in the other? This determination is referred to as " } {TEXT 285 30 "transformation of coordinates," }{TEXT -1 70 " and the e quations which effect the transformation are referred to as " }{TEXT 286 24 "transformation equations" }{TEXT -1 119 ". These equations can be obtained from the following diagram, in which the geometrical sign ificance of the coordinates " }{XPPEDIT 18 0 "r, theta,x " "6%%\"rG%&t hetaG%\"xG" }{TEXT -1 5 " and " }{TEXT 287 1 "y" }{TEXT -1 168 " has b een shown. The diagram assumes that the point in question is situated in the first quadrant, but the results can be verified for other quad rants, when necessary." }}{PARA 0 "" 0 "" {GLPLOT2D 268 244 244 {PLOTDATA 2 "6,-%%TEXTG6%7$$!+1R`NN!#6$!+2R`NNF)%\"OG-%%FONTG6%%&TIMES G%+BOLDITALICG\"#7-F$6%7$$\"+5gaYN!#5$\"+a\"4Q?%F8%\"rGF--F$6%7$$\"+B0 \"RN(F8$\"+E0\"RN(F8%\"PGF--F$6%7$$\"+j;ASOF8$!+;U4CmF)%\"xGF--F$6%7$$ \"+!\\Q\"ovF8$\"+5MiWOF8%\"yGF--%'CURVESG6$7S7$\"\"!FY7$$\"1nmm;arz@!# Y2aFdoFY7$$\"1mm;zXu9cFdoFY7$$\"1+++]y))GeFdoFY7$$\"1* ***\\i_QQgFdoFY7$$\"1***\\7y%3TiFdoFY7$$\"1****\\P![hY'FdoFY7$$\"1LLL$ Qx$omFdoFY7$$\"1+++v.I%)oFdoFY7$$\"1mm\"zpe*zqFdoFY7$$\"1+++D\\'QH(Fdo FY7$$\"1KLe9S8&\\(FdoFY7$$\"1***\\i?=bq(FdoFY7$$\"1LLL3s?6zFdoFY7$$\"1 ++DJXaE\")FdoFY7$$\"1nmmm*RRL)FdoFY7$$\"1mm;a<.Y&)FdoFY7$$\"1LLe9tOc() FdoFY7$$\"1+++]Qk\\*)FdoFY7$$\"1LL$3dg6<*FdoFY7$$\"1mmmmxGp$*FdoFY7$$ \"1++D\"oK0e*FdoFY7$$\"1++v=5s#y*FdoFY7$$\"\"\"FYFY-%'COLOURG6&%$RGBGF YFYFY-FU6$7SFX7$$\"1h==_:HT:FgnF^x7$$\"1_T\\!4hB)GFgnFax7$$\"1$)3TWS`! 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>>'Fdo7$Fcdl$\"14iK!\\u()>'Fdo7$Fedl$\"1?\"=Sewb?'Fdo7$Fcdl$\"1pc6ffm6 iFdo7$Fcdl$\"1S'4*Q0wUiFdo7$Fcdl$\"1uwC`yVaiFdo7$Fcdl$\"1XZA)p0nE'Fdo7$F cdl$\"1H(=$4Z**yiFdo7$F\\fl$\"1=Ccwr9&G'Fdo7$Fcdl$\"1RDmk\\I\"H'Fdo7$F cdl$\"1_0W$3ouH'Fdo7$F\\fl$\"1s*RFaOOI'Fdo7$Fcdl$\"1NkU_.\")4jFdo7$Fcd l$\"1ixQA&*)fJ'Fdo7$Fcdl$\"1LS`iSmFdo7$Fcdl$\"1.UyF1+E nFdo7$Fcdl$\"1C)4w?eP$oFdo7$Fcdl$\"1ts84LM#*oFdo7$Fcdl$\"1=&*[#p<9&pFd o7$F\\fl$\"1PT81o$)epFdo7$F\\fl$\"1\"=Z7rji'pFdo7$F\\fl$\"1;)RVU)ptpFd o7$Fcdl$\"1Io'>'49\")pFdo7$Fcdl$\"1#yE2M\"f))pFdo7$Fcdl$\"1d@Ex&\\g*pF do7$F\\fl$\"1<&f#)o:N+(Fdo7$Fedl$\"1H'\\/p*)4,(Fdo7$Fcdl$\"1_wg+;Z=qFd o7$Fedl$\"1aKbN9'f-(Fdo7$Fcdl$\"1\"y]@@fM.(Fdo7$Fcdl$\"1/%4t%\\'4/(Fdo 7$Fcdl$\"1vK)zly%[qFdo7$Fcdl$\"1t;R2IN1(Fdo7$Fe dl$\"1wdJ8y1rqFdoFfw-F$6%7$$\"+*H>eQ\"F8$\"+)[^-u&F)%\"qG-F.6$%'SYMBOL GF2-%*AXESSTYLEG6#%%NONEG" 1 2 0 1 10 0 2 9 1 1 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" }}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 23 "From Polar to Cartesian" }}{PARA 0 "" 0 "" {TEXT -1 47 "In the rig ht triangle above, the hypotenuse is " }{TEXT 288 1 "r" }{TEXT -1 20 " , the side opposite " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 4 " is " }{TEXT 289 1 "y" }{TEXT -1 26 " and the side adjacent to " } {XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 4 " is " }{TEXT 290 1 "x " }{TEXT -1 7 ". Thus " }{XPPEDIT 18 0 "cos(theta)=x/r" "6#/-%$cosG6#% &thetaG*&%\"xG\"\"\"%\"rG!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "si n(theta)=y/r" "6#/-%$sinG6#%&thetaG*&%\"yG\"\"\"%\"rG!\"\"" }{TEXT -1 7 " whence" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "x=r*cos(theta)" "6#/%\"xG *&%\"rG\"\"\"-%$cosG6#%&thetaGF'" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "y=r*sin(theta)" "6#/%\"yG*&%\"rG\"\"\"-%$sinG6#%&thetaG F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "These are the transformation equations from polar to Cart esian. They accept values of " }{TEXT 291 1 "r" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 31 " for input and give v alues for " }{TEXT 292 1 "x" }{TEXT -1 5 " and " }{TEXT 293 1 "y" } {TEXT -1 95 " as output. Satisfy yourself that these equations are val id no matter what quadrant the point (" }{XPPEDIT 18 0 "(x,y)" "6$%\"x G%\"yG" }{TEXT -1 32 ") is in and no matter what sign " }{TEXT 294 1 " r" }{TEXT -1 5 " has." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 15 "Example problem" }}{PARA 0 "" 0 "" {TEXT -1 112 "Show that the foregoing equations produce the same Carte sian coordinates for each of the polar coordinate sets (" }{XPPEDIT 18 0 "1,Pi/2" "6$\"\"\"*&%#PiGF#\"\"#!\"\"" }{TEXT -1 4 "), (" } {XPPEDIT 18 0 "-1,-Pi/2" "6$,$\"\"\"!\"\",$*&%#PiGF$\"\"#F%F%" }{TEXT -1 7 ") and (" }{XPPEDIT 18 0 "-1,3*Pi/2" "6$,$\"\"\"!\"\"*(\"\"$F$%#P iGF$\"\"#F%" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT 308 9 "Solution: " }{TEXT -1 19 " For the first set " }{TEXT 309 2 "x " }{TEXT -1 5 "= \+ (1)" }{XPPEDIT 18 0 "cos(Pi/2)" "6#-%$cosG6#*&%#PiG\"\"\"\"\"#!\"\"" } {TEXT -1 9 " = 0 and " }{TEXT 310 1 "y" }{TEXT -1 6 " = (1)" } {XPPEDIT 18 0 "sin(Pi/2)" "6#-%$sinG6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 5 " = 1." }}{PARA 0 "" 0 "" {TEXT -1 19 "For the second set " } {TEXT 311 1 "x" }{TEXT -1 7 " = (-1)" }{XPPEDIT 18 0 "cos(-Pi/2)" "6#- %$cosG6#,$*&%#PiG\"\"\"\"\"#!\"\"F+" }{TEXT -1 6 " = 0, " }{TEXT 312 1 "y" }{TEXT -1 7 " = (-1)" }{XPPEDIT 18 0 "sin(-Pi/2)" "6#-%$sinG6#,$ *&%#PiG\"\"\"\"\"#!\"\"F+" }{TEXT -1 16 " = (-1)(-1) = 1." }}{PARA 0 " " 0 "" {TEXT -1 18 "For the third set " }{TEXT 313 1 "x" }{TEXT -1 7 " = (-1)" }{XPPEDIT 18 0 "cos(3*Pi/2)" "6#-%$cosG6#*(\"\"$\"\"\"%#PiGF( \"\"#!\"\"" }{TEXT -1 6 " = 0, " }{TEXT 314 1 "y" }{TEXT -1 7 " = (-1) " }{XPPEDIT 18 0 "sin(3*Pi/2)" "6#-%$sinG6#*(\"\"$\"\"\"%#PiGF(\"\"#! \"\"" }{TEXT -1 16 " = (-1)(-1) = 1." }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 23 "From Cartesian to Polar" }}{PARA 0 "" 0 "Pairing Problems " {TEXT -1 154 "The problem of transforming Cartesian coordinates to p olar coordinates is a bit more complex than the foregoing. From the Py thagorean theorem we see that " }{XPPEDIT 18 0 "r^2=x^2+y^2" "6#/*$%\" rG\"\"#,&*$%\"xGF&\"\"\"*$%\"yGF&F*" }{TEXT -1 35 " and from the trian gle we see that " }{XPPEDIT 18 0 "tan(theta)=y/x" "6#/-%$tanG6#%&theta G*&%\"yG\"\"\"%\"xG!\"\"" }{TEXT -1 19 ". One solution for " }{TEXT 295 1 "r" }{TEXT -1 4 " is " }{XPPEDIT 18 0 "r=sqrt(x^2+y^2)" "6#/%\"r G-%%sqrtG6#,&*$%\"xG\"\"#\"\"\"*$%\"yGF+F," }{TEXT -1 22 " and one sol ution for " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 4 " is " } {XPPEDIT 18 0 "theta=arctan(y/x" "6#/%&thetaG-%'arctanG6#*&%\"yG\"\"\" %\"xG!\"\"" }{TEXT -1 2 ". " }{TEXT 296 95 "It is not always possible \+ to pair these solutions together to obtain the polar coordinates of ( " }{XPPEDIT 18 0 "(x,y)" "6$%\"xG%\"yG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "To understand why th is is so, the student must review the definition of " }{XPPEDIT 18 0 " arctan(t" "6#-%'arctanG6#%\"tG" }{TEXT -1 23 ". For each real number \+ " }{TEXT 297 1 "t" }{TEXT -1 35 ", there are infinitely many angles " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 12 " satisfying " } {XPPEDIT 18 0 "tan(theta)=t" "6#/-%$tanG6#%&thetaG%\"tG" }{TEXT -1 43 " (the tangent function repeats with period " }{XPPEDIT 18 0 "Pi" "6#% #PiG" }{TEXT -1 19 ") but the function " }{XPPEDIT 18 0 "arctan(t)" "6 #-%'arctanG6#%\"tG" }{TEXT -1 62 " is defined to pick out just one of \+ them - the one satisfying " }{XPPEDIT 18 0 "-Pi/2" "6#,$*&%#PiG\"\"\" \"\"#!\"\"F(" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta" "6#%&thetaG" } {TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi/2" "6#*&%#PiG\"\"\"\"\"#!\"\"" } {TEXT -1 21 ". If these values of " }{XPPEDIT 18 0 "theta" "6#%&thetaG " }{TEXT -1 36 " are paired with positive values of " }{TEXT 298 1 "r " }{TEXT -1 116 ", only points lying in the first and fourth quadrants can be described. To say this another way, polar coordinates (" } {XPPEDIT 18 0 "r,theta" "6$%\"rG%&thetaG" }{TEXT -1 62 ") of points ly ing in the second and third quadrant and having " }{TEXT 299 1 "r" } {TEXT -1 25 " > 0 must have values of " }{XPPEDIT 18 0 "theta" "6#%&th etaG" }{TEXT -1 109 " lying appropriately in the second or third quadr ant, not in the first or fourth quadrant. These same points " }{TEXT 302 3 "can" }{TEXT -1 30 " be described using values of " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 40 " given by the inverse tangent \+ function (" }{TEXT 300 4 "i.e." }{TEXT -1 58 " in the first or fourth \+ quadrant) if they are paired with " }{TEXT 303 8 "negative" }{TEXT -1 11 " values of " }{TEXT 301 1 "r" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "The point of this discuss ion is that it is not possible to write equations for " }{TEXT 304 1 " r" }{TEXT -1 9 " and for " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 13 " in terms of " }{TEXT 305 1 "x" }{TEXT -1 5 " and " }{TEXT 306 1 "y" }{TEXT -1 87 " which can always be used. Instead, the student sh ould begin with the pair of equations" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "r^2=x^2+y^2" "6#/*$%\"rG\"\"#,&*$% \"xGF&\"\"\"*$%\"yGF&F*" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "tan(theta)=y/x" "6#/-%$tanG6#%&thetaG*&%\"yG\"\"\"%\"xG!\"\"" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "and choos e the appropriate combinations of plus or minus signs for " }{TEXT 307 1 "r" }{TEXT -1 19 " and quadrants for " }{XPPEDIT 18 0 "theta" "6 #%&thetaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 16 "Example problems" }}{PARA 0 "" 0 "" {TEXT -1 82 "1. Find a set of polar coordinates for the point with Car tesian coordinates (2,4)." }}{PARA 0 "" 0 "" {TEXT 315 10 "Solution: \+ " }{TEXT -1 23 "From the above, we have" }{TEXT 317 1 " " }{XPPEDIT 18 0 "r^2=20" "6#/*$%\"rG\"\"#\"#?" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "tan(theta)=2" "6#/-%$tanG6#%&thetaG\"\"#" }{TEXT -1 230 ". We can u se Maple to find the inverse tangent of 2, which will be a first quadr ant angle. Because the point with Cartesian coordinates (2,4) is actua lly in the first quadrant, we can pair the inverse tangent of 2 with a positive " }{TEXT 316 1 "r" }{TEXT -1 15 ", in this case " }{XPPEDIT 18 0 "sqrt(20)" "6#-%%sqrtG6#\"#?" }{TEXT -1 6 ". Thus" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "r:=sqrt(20); theta:=arctan(2);" "6#C$ >%\"rG-%%sqrtG6#\"#?>%&thetaG-%'arctanG6#\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG,$*$\"\"&#\"\"\"\"\"#F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG-%'arctanG6#\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 30 "For floating point values, use" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "r:=evalf(r);theta:=evalf(theta);" "6#C$>%\"rG-%&evalfG6 #F%>%&thetaG-F'6#F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"+cf8sW !\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG$\"+=([r5\"!\"*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Note that this value for " } {XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 15 " is in radians." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "2. Find a set of polar coordinates for the point with Cartesian coordinates (-2 ,-4)." }}{PARA 0 "" 0 "" {TEXT 318 9 "Solution:" }{TEXT -1 30 " In thi s case we still obtain " }{XPPEDIT 18 0 "r^2=20" "6#/*$%\"rG\"\"#\"#? " }{TEXT -1 5 " and " }{XPPEDIT 18 0 "tan(theta)=2" "6#/-%$tanG6#%&the taG\"\"#" }{TEXT -1 169 ", but clearly the polar coordinates of (-2,-4 ) must be different than those of (2,4). Because (-2,-4) lies in the \+ third quadrant, we must adjust the value arctan(2) by " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 48 " radians if we wish to use a positive v alue for " }{TEXT 319 1 "r" }{TEXT -1 61 ". Another option would be to keep the value of arctan(2) for " }{XPPEDIT 18 0 "theta" "6#%&thetaG " }{TEXT -1 30 " and use a negative value for " }{TEXT 320 1 "r" } {TEXT -1 50 ". Hence two possible solutions to this problem are" }} {PARA 0 "> " 0 "" {XPPEDIT 19 1 "r:=evalf(sqrt(20)); theta:=evalf(arct an(2)+Pi);" "6#C$>%\"rG-%&evalfG6#-%%sqrtG6#\"#?>%&thetaG-F'6#,&-%'arc tanG6#\"\"#\"\"\"%#PiGF6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"+ cf8sW!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG$\"+s8u[U!\"*" } }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "r:=evalf(-sqrt(20)); theta:= evalf(arctan(2));" "6#C$>%\"rG-%&evalfG6#,$-%%sqrtG6#\"#?!\"\">%&theta G-F'6#-%'arctanG6#\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$!+cf 8sW!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG$\"+=([r5\"!\"*" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "3. Check your solutions to probl em 3." }}{PARA 0 "" 0 "" {TEXT 321 9 "Solution:" }{TEXT -1 145 " We ca n easily check out soultions by transforming the polar coordinates we \+ obtained back to Cartesian. Look at the following Maple calculations: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "r:=4.472135956; theta:=4.248741 372;\nx:=r*cos(theta); y:=r*sin(theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"+cf8sW!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG $\"+s8u[U!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG$!+)*******>!\" *" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG$!+-+++S!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "r:=-4.472135956; theta:=1.107148718 ;\nx:=r*cos(theta); y:=r*sin(theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"rG$!+cf8sW!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG$\"+= ([r5\"!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG$!+++++?!\"*" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG$!+,+++S!\"*" }}}}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 22 "Transforming Equations" }}{PARA 0 "" 0 " " {TEXT -1 86 "Recall what is meant by the graph of an equation. Given and equation in the variables " }{TEXT 322 1 "x" }{TEXT -1 5 " and " }{TEXT 323 1 "y" }{TEXT -1 6 ", the " }{TEXT 324 22 "graph of the equa tion " }{TEXT -1 32 "is the set of all points in the " }{TEXT 325 2 "x y" }{TEXT -1 26 "-plane whose coordinates (" }{XPPEDIT 18 0 "x,y" "6$% \"xG%\"yG" }{TEXT -1 139 ") satisfy the given equation. Going the oth er way, starting with a figure in the plane which is the graph of an e quation in the variables " }{TEXT 326 1 "x" }{TEXT -1 5 " and " } {TEXT 327 1 "y" }{TEXT -1 34 ", we refer to the equation as the " } {TEXT 328 11 "equation of" }{TEXT -1 1 " " }{TEXT 329 10 "the figure" }{TEXT -1 47 ". For example, we would say that the graph of " } {XPPEDIT 18 0 "x^2+y^2=25" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(\"#D" } {TEXT -1 145 " was the circle of radius 5 centered at the origin, and \+ we would also say that the equation of the circle of radius 5 centered at the origin was " }{XPPEDIT 18 0 "x^2+y^2=25" "6#/,&*$%\"xG\"\"#\" \"\"*$%\"yGF'F(\"#D" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 253 "When both polar and Cartesian coordinat e systems are under discussion, we must distinguish between the Cartes ian graph and the polar graph of an equation, and between the Cartesia n equation and the polar equation of a figure. In general, the variab les " }{TEXT 330 1 "x" }{TEXT -1 5 " and " }{TEXT 331 1 "y" }{TEXT -1 65 " will be associated with Cartesian coordinates and the variables \+ " }{TEXT 332 1 "r" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "theta" "6#%&the taG" }{TEXT -1 103 " will be associated with polar coordinates. Any de viation from this convention must be made very clear." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Consider the equation \+ " }{XPPEDIT 18 0 "r=cos(theta)" "6#/%\"rG-%$cosG6#%&thetaG" }{TEXT -1 57 ". One way to graph this equation would be to introduce a " } {XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 13 " axis and an " } {XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 35 " axis and draw the cosine fu nction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot(cos(theta),theta=0..4*Pi);" "6#-%%plotG6$-%$cosG6# %&thetaG/F);\"\"!*&\"\"%\"\"\"%#PiGF/" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6%-%'CURVESG6$7]s7$\"\"!$\"\"\"F(7$$\"1K^\\r* ))QU$!#<$\"1)HaZ1RT***!#;7$$\"1k-*H%zxZoF.$\"1$eH\"HJcw**F17$$\"1S&[9p mr-\"F1$\"1s*4az#HZ**F17$$\"1`!)f)eb&p8F1$\"1u:!RPij!**F17$$\"1zq*GQLV 0#F1$\"1FI_\"zE(*y*F17$$\"11h>x66RFF1$\"1k0h'4.si*F17$$\"1L8'=m^\"*F1$\"1sd(\\ [)e'4'F17$$\"1*QTUBn+0\"!#:$\"1UhCis7v\\F17$$\"1a1=9rK%=\"Fao$\"1xan[R ?pPF17$$\"1>*>T*pe=8Fao$\"1\"[xd.Sa\\#F17$$\"1PXG*3jIW\"Fao$\"1k%R>whQ F\"F17$$\"1b\"\\W=Rvc\"Fao$\"1Rne'eOrD$!#=7$$\"1+`0rpU'p\"Fao$!1dW:BV+ `7F17$$\"1X9mdZJD=Fao$!1MkGEQzFao$!1\"RMD%*f;y$F17$ $\"1j$HUi1>4#Fao$!1(*\\(yoL%y\\F17$$\"1oZwv]xCAFao$!1fF%zC!\\$3'F17$$ \"1u,IFNkdBFao$!1?DD!HK#[#*F17$$\"1]mU=tur GFao$!1@e7)Q@\"Q'*F17$$\"1uR(f?2&RHFao$!1![.YK,lz*F17$$\"1)H@N4ns+$Fao $!1Mb7*y=*4**F17$$\"1g\\HPq9TIFao$!1IN\"4z&f\\**F17$$\"1@'o5)p-vIFao$! 1qIpNK&y(**F17$$\"1$GU[#p!*3JFao$!1Z>C\"peY***F17$$\"1XfhooyUJFao$!0?` #pG******Fao7$$\"1gQpEf!o<$Fao$!1O/gn2!Q***F17$$\"1v4Vn%**F17$$\"1/w#45j)yKFao$!1E\"3C (>$f!**F17$$\"1LM3<7!pM$Fao$!1Ob*F17$$\"1:#H:m3sn$Fao$!1(Q_%**Ra*f) F17$$\"1%>)\\*\\\"H'z$Fao$!18>NHMJKzF17$$\"1srYPVP:RFao$!1AZ:!zHF:(F17 $$\"1s4v\\`(p0%Fao$!1@^E,A([4'F17$$\"1sZ.ijd)>%Fao$!1G)RYK6]\"\\F17$$ \"15XF$*4`=VFao$!11Ed'=n*\\F17$$\"1 fhL=%\\YO&Fao$\"1d%46cV)pgF17$$\"1@s3O**o$\\&Fao$\"1ip:3p.UqF17$$\"1-7 7z,LGcFao$\"1!)[h)ph8$zF17$$\"1%=b@UqHw&Fao$\"1a(Q7x')F17$$\"1NaR&> 7m)eFao$\"1&)obi2!RA*F17$$\"1(oN'oRD5gFao$\"1fFGm([)H'*F17$$\"1\"=nl9O p2'Fao$\"1(HJkKf!)y*F17$$\"1v')\\C$=O9'Fao$\"19UhzKw-**F17$$\"1AWY8%fp <'Fao$\"13$=1LLO%**F17$$\"1p,V-0I5iFao$\"1'=!>#y]M(**F17$$\"1;fR\"fTOC 'Fao$\"19)z9\\#=#***F17$$\"1i;O!o#)pF'Fao$\"17&>xj2)****F17$$\"1GZX,Zh 6jFao$\"1u-))>\"ff***F17$$\"1%zZDsYiM'Fao$\"1H6&z(G7!)**F17$$\"1g3kV(y 3Q'Fao$\"1*GcO!zJ_**F17$$\"1DRtk2^:kFao$\"1:iIUvd7**F17$$\"1c+#p![x%[' Fao$\"1O,^]b\\(z*F17$$\"1(=1\"\\)QSb'Fao$\"1ENf)yGaj*F17$$\"1N0y`ziumF ao$\"1b0\\+ifV#*F17$$\"1$)[Xeq@&z'Fao$\"11SMA&3vr)F17$$\"1b'\\^fca#pFa o$\"1yHCtGP2!)F17$$\"1FW%=8'pbqFao$\"1`m!)\\`ghrF17$$\"1iz>(Fao$ \"10`n!RM1;'F17$$\"1(\\^D]'zCtFao$\"1$yI2*4I[]F17$$\"1)z@!4wUcuFao$\"1 %GCW(GkrQF17$$\"1*4#\\:(e!)e(Fao$\"1Y,&=e)*zi#F17$$\"12vMz&=ar(Fao$\"1 gsuZ3?\"Q\"F17$$\"19H?V%yF%yFao$\"1U\">=k&H?6F.7$$\"1\"*>vd)*=%)zFao$! 1@1rlgS)H\"F17$$\"1n5Is7gD\")Fao$!12_Rk/#Ho#F17$$\"1d,!yrlED)Fao$!1FmL k'f?)QF17$$\"1[#*Hj,tz$)Fao$!1i=XXeg=]F17$$\"1uBB!e)R:&)Fao$!1([lk7cB9 'F17$$\"1*\\lr*p1^')Fao$!1u+[(*RA`rF17$$\"1B%)>ZF+u()Fao$!1Inv1T9czF17 $$\"1Y8B(\\Qp*))Fao$!1+)f7zs*Q')F17$$\"1+TgV(R8.*Fao$!1SAs_U'fB*F17$$ \"1_o(**)4ul\"*Fao$!1%*=\\g5Pm'*F17$$\"12#G$R:(*G#*Fao$!1A)G3$4\"*3)*F 17$$\"1f&z')3-AH*Fao$!1@T^NoC7**F17$$\"1O_Njt\"QK*Fao$!1U=-l!y!\\**F17 $$\"174.QEVb$*Fao$!1,Wlkd'f(**F17$$\"1)e1F\"z/(Q*Fao$!1DjqhI)G***F17$$ \"1lAQ(=j'=%*Fao$!1SYJ[I\")****F17$$\"1rc%*e-r^%*Fao$!1[0zvMP'***F17$$ \"1w!40LdZ[*Fao$!1d%ezw;SmR$*\\F17$$\"1ab\\.**fg5Fd\\m$!1vnH\"Gbzz$F17$$\"1r9`Bg#R2\" Fd\\m$!1$>(=1v;NDF17$$\"1x_c'zTr3\"Fd\\m$!1g&>o$fPQ7F17$$\"1$3*fpvN+6F d\\m$\"10pT[IK,!)Fhp7$$\"1g+[#\\,D6\"Fd\\m$\"1a*pcz%z!H\"F17$$\"1P5O:a kC6Fd\\m$\"1BQ?lPc#[#F17$$\"1IOU@PcQ6Fd\\m$\"1(f%>ehY-QF17$$\"1Ci[F?[_ 6Fd\\m$\"1&)=p+j#)[]F17$$\"1R*\\jsI\\;\"Fd\\m$\"1E#>$Q%R:3'F17$$\"1aO@ D%zt<\"Fd\\m$\"1#\\cSaG,-(F17$$\"1w[lMBl!>\"Fd\\m$\"1G:i\\'e3!zF17$$\" 1)4'4W_#R?\"Fd\\m$\"1Hpd^MgU')F17$$\"1Jk,7\"Hm@\"Fd\\m$\"1&*3&fM+.@*F1 7$$\"1ln$*zHLH7Fd\\m$\"1y=()3ObH'*F17$$\"1uDX+!fhB\"Fd\\m$\"1=;!f@c5z* F17$$\"1#Qo4-&)HC\"Fd\\m$\"1joYUb&p!**F17$$\"1(GE7.)RY7Fd\\m$\"18V\")4 piZ**F17$$\"1\">%[T5\")\\7Fd\\m$\"1#>a/x6n(**F17$$\"1'4U<0CKD\"Fd\\m$ \"154xZi<%***F17$$\"1+++iqjc7Fd\\mF)-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-% +AXESLABELSG6$%&thetaG%!G-%%VIEWG6$;F($\"+iqjc7!\")%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}} {EXCHG {PARA 256 "" 0 "" {TEXT -1 14 "In this plot, " }{XPPEDIT 18 0 " r" "6#%\"rG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "theta" "6#%&thetaG" } {TEXT -1 140 " have been plotted as Cartesian coordinates, not polar c oordinates, so this is the Cartesian graph of the equation and not the polar graph. " }{TEXT 333 17 " If this was what" }{TEXT -1 1 " " } {TEXT 334 138 "was wanted, it would have been referred to as the \"Car tesian graph\" of the equation. Following our convention above, we sh ould interpret " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT 339 4 " and" } {TEXT -1 1 " " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 1 " " } {TEXT 340 89 "as polar coordinates and construct the polar graph. Mapl e will do that for us as follows:" }}{PARA 0 "> " 0 "circle" {XPPEDIT 19 1 "plot([cos(theta),theta, theta=0..2*Pi], coords=polar);" "6#-%%pl otG6$7%-%$cosG6#%&thetaGF*/F*;\"\"!*&\"\"#\"\"\"%#PiGF0/%'coordsG%&pol arG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6#-%'CURVES G6$7]q7$$\"\"\"\"\"!F*7$$\"1X0)p=\"=`**!#;$\"1A$)y74REo!#<7$$\"1/KU[:g 8)*F.$\"1BzwTS\\_8F.7$$\"1)*Q\"*[sn='*F.$\"1(H!)*oa::>F.7$$\"1Z%>ZqW#e $*F.$\"1![Dag`1X#F.7$$\"1XOh%*3t\"**)F.$\"1J/2LR*4,$F.7$$\"1;rp&=NOb)F .$\"13$3GZRt^$F.7$$\"1FCkVUH[!)F.$\"11!**Rf>L'RF.7$$\"1$zsDjjv[(F.$\"1 l`R'Q$GPVF.7$$\"1MO3w>g%)oF.$\"1n(36GG7j%F.7$$\"1X4#*>+sZiF.$\"1BH[)H< =%[F.7$$\"1FxC$)3$pj&F.$\"1K%\\*yfEf\\F.7$$\"1PxU&o&G;]F.$\"1rYdxM(*** \\F.7$$\"1R95Ty\\tVF.$\"1dJ'HA%fg\\F.7$$\"1t#o'*3.6u$F.$\"1g49'GB*Q[F. 7$$\"1Q&=;.q\"4JF.$\"1EVd=!*oGYF.7$$\"1d*\\*eJy5DF.$\"1o&oB![LOVF.7$$ \"1x)Q)y[De>F.$\"1b]'o4[$oRF.7$$\"1t&y/pY$f9F.$\"1,BwyqSINF.7$$\"1lc` \\zN65F.$\"10'y%3X3:IF.7$$\"1Fx7WedxjF1$\"1!zM(f/`VCF.7$$\"1aZRl&Q)ePF 1$\"1di\"fx&)>!>F.7$$\"16^brIR4=F1$\"1U'eX'=\"HL\"F.7$$\"1F7#yggS]%!#= $\"19(4&)*))4'p'F17$$\"1S#[YUn`c$!#A$!1#R*e4(o5(f!#>7$$\"1@U=38S.ZFhr$ !1HRT@v)>%oF17$$\"1zUCz\"ej&=F1$!14XoTox\\8F.7$$\"1X;E(GoY.%F1$!1q9)4/ 5x'>F.7$$\"1A)oG(*zA+(F1$!1XL!)pV&=b#F.7$$\"1c\\k\"GVQ.\"F.$!1DJhD>gWI F.7$$\"1yxc+^jB9F.$!1<3dBeA%\\$F.7$$\"1c%[U*Qc_>F.$!1aDyi%zR'RF.7$$\"1 s1'=L%\\UDF.$!1Dy@uYQaVF.7$$\"1Ed*HnI73$F.$!1uT%))3xrh%F.7$$\"1j:]8LaZ OF.$!1oU5W?h8[F.7$$\"1Ui1(y\"=IVF.$!1s2#oBJ\\&\\F.7$$\"1v)\\kiSe-&F.$! 1)ecdAL***\\F.7$$\"1oC%zYI;k&F.$!1#yBe8g'e\\F.7$$\"18vI7WkZiF.$!1y3o\" yO=%[F.7$$\"1$*[\"H;K.*oF.$!17D+UB*)GYF.7$$\"1wBT@$f$)\\(F.$!1uK!3&Q2J VF.7$$\"1!QpPx@\\.)F.$!1km-.*oN(RF.7$$\"1J1(yW=5_)F.$!1eui7+**\\NF.7$$ \"1l<&Q%3ol*)F.$!1:d#)RcQ$*F.$!11*Ql/L`[#F.7$$\"1R'f wP]>h*F.$!1$HcGX)HJ>F.7$$\"1:Hc!QC\\\")*F.$!13;UOrxZ8F.7$$\"1moWQ;Q^** F.$!1#e(yThrbpF17$$\"1#z(z=Q!*****F.$!1iG2AXH,JFhr7$$\"1!H+Dx)Gc**F.$ \"18$\\\\:\")pf'F17$$\"1Rw3(R9x\")*F.$\"1@_U\"ypxL\"F.7$$\"1)=)H/\")z@ '*F.$\"1a8*)p;h2>F.7$$\"1p2ZmUve$*F.$\"1h_rzru\\CF.7$$\"1iRAVko.!*F.$ \"1zEVS%y]*HF.7$$\"1(ySFo-3e)F.$\"1QG?Vzn*[$F.7$$\"1m0xT=DC[F.7$$\"17a.O/g!p&F.$\"1NHmYu2_\\F. 7$$\"1WBd%z9g0&F.$\"1\"RLWA'o**\\F.7$$\"1hVuzpz]VF.$\"1dA>)Ruw&\\F.7$$ \"1s0E!yR&eOF.$\"1t#yM7)o;[F.7$$\"19h$*z,(*eIF.$\"1Yg\\'>jyg%F.7$$\"1# o4))3(p!\\#F.$\"1*=H(REuCVF.7$$\"1!*>YZ>#)G>F.$\"1/$zQV8c%RF.7$$\"1*Q \")3,)QB9F.$\"12ej2H(R\\$F.7$$\"1P73bz#>-\"F.$\"1,l20o,HIF.7$$\"1A+-:h 80oF1$\"1:HO&>S$=DF.7$$\"1EK\\\"zy,#QF1$\"14lw75$o\">F.7$$\"1#)p!\\tW \"o;F1$\"1$))\\$p)[2G\"F.7$$\"1.my>%ewQ%Fhr$\"1*z)GjCR4mF17$$\"1\"HtTV $fZ$*!#@$\"1]euR\"pt0$Fhr7$$\"1q*)31?PoRFhr$!1Y[Nk(*)pG'F17$$\"1y^%z+x7F.7$$\"1,m%e;orq$F1$!1C\\G>[P*)=F.7$$\"1elDyiBGlF1$!1 #*\\o=VBqCF.7$$\"13X`*3ed-\"F.$!1'=YXTSS.$F.7$$\"1Z[YD\\Lr9F.$!1$3B%=% *QUNF.7$$\"16ig'=d:'>F.$!1jJ5Tr(3(RF.7$$\"1qyZl;I.DF.$!1P[ze:.KVF.7$$ \"1KktVB-,JF.$!155)eH_`i%F.7$$\"1BFPIiTKPF.$!1*)G6z\\lO[F.7$$\"1(*4U9? \"3Q%F.$!18$H/&*)F.$!1#\\pG&4)\\1$F.7$$\"1 !f\\hr,8K*F.$!14!>>%RA:DF.7$$\"1HOVl,:0'*F.$!17bB=yXZ>F.7$$\"11*f\"*zw Z\")*F.$!1#4!zeUI[8F.7$$\"1p5aoxZ`**F.$!1;e4!\\O[!oF17$F($\"1PTpSr8/#) !#D-%'COLOURG6&%$RGBG$\"#5!\"\"F*F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "We shall verify later that this is a circle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 250 "One other point should b e made regarding the definition of the polar graph of an equation. Bec ause points in the plane have many different sets of polar coordinates , we must define the polar graph of an equation to consist of all poin ts in the plane " }{TEXT 336 26 "having at least one set of" }{TEXT -1 1 " " }{TEXT 337 18 "polar coordinates " }{TEXT -1 28 "which satisf y the equation. " }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 23 "From Cartesian to Polar" }}{PARA 0 "" 0 "" {TEXT -1 104 "The rule for transforming a given equation from one coordinate system to another is simple and natural: " }{TEXT 338 37 "always subs titute equals for equals. " }{TEXT -1 78 "Suppose, for example, we wan t to find the polar equation of the vertical line " }{XPPEDIT 18 0 "x= 3" "6#/%\"xG\"\"$" }{TEXT -1 8 ". Since " }{XPPEDIT 18 0 "x=r*cos(thet a)" "6#/%\"xG*&%\"rG\"\"\"-%$cosG6#%&thetaGF'" }{TEXT -1 12 ", we obta in " }{XPPEDIT 18 0 "r*cos(theta)=3" "6#/*&%\"rG\"\"\"-%$cosG6#%&theta GF&\"\"$" }{TEXT -1 28 ", which can also be written " }{XPPEDIT 18 0 " r=3*sec(theta)" "6#/%\"rG*&\"\"$\"\"\"-%$secG6#%&thetaGF'" }{TEXT -1 70 ". As another example, consider the circle of radius 2 centered on \+ the " }{TEXT 335 1 "y" }{TEXT -1 69 "-axis at the point (0, -2). The \+ Cartesian equation of this circle is" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "x^2+(y+2)^2=4" "6#/,&*$%\"xG\"\"#\"\"\"*$,&%\"yGF(F'F(F'F(\"\"%" } {TEXT -1 57 ". To find the polar equation of the circle we substitute \+ " }{XPPEDIT 18 0 "x=r*cos(theta)" "6#/%\"xG*&%\"rG\"\"\"-%$cosG6#%&the taGF'" }{TEXT -1 5 " and " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "y=r*sin(th eta)" "6#/%\"yG*&%\"rG\"\"\"-%$sinG6#%&thetaGF'" }{TEXT -1 11 " to obt ain " }{XPPEDIT 18 0 "r^2*(cos(theta))^2+(r*sin(theta)+2)^2=4" "6#/,&* &%\"rG\"\"#-%$cosG6#%&thetaGF'\"\"\"*$,&*&F&F,-%$sinG6#F+F,F,F'F,F'F, \"\"%" }{TEXT -1 112 ". Technically, this is the polar equation of the circle, but the equation can and should be simplified. In fact," }} {PARA 0 "" 0 "" {XPPEDIT 18 0 "r^2*(cos(theta))^2+r^2*(sin(theta))^2+4 *r*sin(theta)+4=4" "6#/,**&%\"rG\"\"#-%$cosG6#%&thetaGF'\"\"\"*&F&F'-% $sinG6#F+F'F,*(\"\"%F,F&F,-F/6#F+F,F,F2F,F2" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "r^2+4*r*sin(theta)=0" "6#/,&*$%\"rG\"\"#\"\" \"*(\"\"%F(F&F(-%$sinG6#%&thetaGF(F(\"\"!" }{TEXT -1 1 "," }}{PARA 0 " " 0 "" {XPPEDIT 18 0 "r=-4sin(theta)" "6#/%\"rG,$*&\"\"%\"\"\"-%$sinG6 #%&thetaGF(!\"\"" }{TEXT -1 1 "." }}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 23 "From Polar to Cartesian" }}{PARA 0 "" 0 "" {TEXT -1 41 "The simple and natural rule given above (" }{TEXT 341 4 "i.e." }{TEXT -1 199 " s ubstitute equals for equals) is \"easier said than done\" when it come s to transforming polar equations to Cartesian equations. The reason i s that we do not have simple expressions to substitute for " } {XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "theta " "6#%&thetaG" }{TEXT -1 13 " in terms of " }{XPPEDIT 18 0 "x" "6#%\"x G" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "y" "6#%\"yG" }{TEXT -1 2 ", " } {HYPERLNK 17 "as explained above" 1 "" "Pairing Problems" }{TEXT -1 13 ". However we " }{TEXT 342 2 "do" }{TEXT -1 22 " have expressions f or " }{XPPEDIT 18 0 "r^2" "6#*$%\"rG\"\"#" }{TEXT -1 9 " and for " } {XPPEDIT 18 0 "tan(theta)" "6#-%$tanG6#%&thetaG" }{TEXT -1 152 ", and \+ sometimes a polar equation can be manipulated to make use of these, as in the following example. Suppose we want to transform the polar equ ation " }{XPPEDIT 18 0 "r=2sin(theta)-4cos(theta)" "6#/%\"rG,&*&\"\"# \"\"\"-%$sinG6#%&thetaGF(F(*&\"\"%F(-%$cosG6#F,F(!\"\"" }{TEXT -1 56 " to Cartesian form. Multiplying through the equation by " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 7 " gives " }{XPPEDIT 18 0 "r^2=2*r*sin(t heta)-4*r*cos(theta)" "6#/*$%\"rG\"\"#,&*(F&\"\"\"F%F)-%$sinG6#%&theta GF)F)*(\"\"%F)F%F)-%$cosG6#F-F)!\"\"" }{TEXT -1 27 " wherein the subst itutions " }{XPPEDIT 18 0 "r^2=x^2+y^2" "6#/*$%\"rG\"\"#,&*$%\"xGF&\" \"\"*$%\"yGF&F*" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "r*sin(theta)=y" "6#/ *&%\"rG\"\"\"-%$sinG6#%&thetaGF&%\"yG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "r*cos(theta)=x" "6#/*&%\"rG\"\"\"-%$cosG6#%&thetaGF&%\"xG" } {TEXT -1 5 " give" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "x^2+y^2=2y-4x" "6# /,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(,&*&F'F(F*F(F(*&\"\"%F(F&F(!\"\"" } {TEXT -1 29 ". This in turn simplifies to " }{XPPEDIT 18 0 "(x+2)^2+(y -1)^2=5" "6#/,&*$,&%\"xG\"\"\"\"\"#F(F)F(*$,&%\"yGF(F(!\"\"F)F(\"\"&" }{TEXT -1 1 "." }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 26 "Some Standar d Polar Curves" }}{PARA 0 "" 0 "" {TEXT -1 379 "In order to illustrate calculus topics associated with polar coordinates, it is necessary to deal with figures defined by polar equations. Such figures are more a ccessible when the student has a supply of familiar figures and equati ons to draw from. For that reason we introduce and describe a number o f standard examples below. The student should commit this material to \+ memory." }}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "r=c" "6#/%\"rG%\"cG " }}{PARA 0 "" 0 "" {TEXT -1 13 "The graph of " }{XPPEDIT 18 0 "r=c" " 6#/%\"rG%\"cG" }{TEXT -1 8 " (where " }{XPPEDIT 18 0 "c" "6#%\"cG" } {TEXT -1 74 " denotes a constant) is a circle cenetered at the origin \+ (pole) of radius " }{XPPEDIT 18 0 "abs(c" "6#-%$absG6#%\"cG" }{TEXT -1 97 ". As simple as this example is, it merits some elaboration. Not e first that because the variable " }{XPPEDIT 18 0 "theta" "6#%&thetaG " }{TEXT -1 95 " does not appear in the equation, it is unrestricted. \+ That is, when we seek polar coordinates (" }{XPPEDIT 18 0 "r,theta" "6 $%\"rG%&thetaG" }{TEXT -1 27 ") satisfying the equation, " }{TEXT 343 3 "any" }{TEXT -1 10 " value of " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 61 " satisfies the equation so long as the accompanying valu e of " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 4 " is " }{XPPEDIT 18 0 "c" "6#%\"cG" }{TEXT -1 145 ". This is why the graph of the equation c onsists of an entire circle. Secondly, note that the equation has solu tions even for negative values of " }{XPPEDIT 18 0 "c" "6#%\"cG" } {TEXT -1 57 ", since we have assigned a meaning to polar coordinates ( " }{XPPEDIT 18 0 "r,theta" "6$%\"rG%&thetaG" }{TEXT -1 2 ") " } {HYPERLNK 17 "for all values of r." 1 "" "Extension of polar coors" } {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "The following Maple command will plot the graph of " } {XPPEDIT 18 0 "r=-3" "6#/%\"rG,$\"\"$!\"\"" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([-3,theta,theta=0..2*Pi],coords= polar);" "6#-%%plotG6$7%,$\"\"$!\"\"%&thetaG/F*;\"\"!*&\"\"#\"\"\"%#Pi GF0/%'coordsG%&polarG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6#-%'CURVESG6$7S7$$!\"$\"\"!F*7$$!1F$)=7(3>(H!#:$!1$*HnZX $e4%!#;7$$!1MpkH19-HF.$!1vZyyo')*f(F17$$!16$eO%fduFF.$!1+pX(4L49\"F.7$ $!1e!RcF@ff#F.$!1ag$=GEP]\"F.7$$!1l%*****HF.7$$ \"1%\\Cx2Zu3%F1$!1TeG'=C?(HF.7$$\"1o92^kaQzF1$!1xXvk(fI*GF.7$$\"19$>$z E$>8\"F.$!1Moug*f#yFF.7$$\"1\\n3#e%p7:F.$!1$fB>_02f#F.7$$\"1>o!H%o%=\" =F.$!1\\%>:,p5R#F.7$$\"1u#HS!\\zE@F.$!1o-6=;$e6#F.7$$\"1udmh7ErBF.$!18 NH:UpP=F.7$$\"1%3V>-#z(f#F.$!1))z%G7#\\+:F.7$$\"1O4%>)3GpFF.$!1+Y)f(es `6F.7$$\"1s6#G0(3**GF.$!1\"3W)z9`:xF17$$\"1$4De+4@(HF.$!1Hh::]F\"3%F17 $$\"1y\\ysb)***HF.$!1vmZ4$GRI*Fao7$$\"1:ph09`sHF.$\"1xsfobR]SF17$$\"1m gA!HF.$\"12m#[3\\of(F17$$\"1Y*4Rmy*yFF.$\"1`b*31n,8\"F.7$$\"1#[H'R eD-EF.$\"1eg$fEPF\\\"F.7$$\"1xRQJ.#QQ#F.$\"1B*32Gt8#=F.7$$\"1)RF6ypJ8# F.$\"1m/iVVS4@F.7$$\"1``.Hfd9=F.$\"1*GUjd)**)Q#F.7$$\"1miZDl?(\\\"F.$ \"1OS[Bqo*f#F.7$$\"1!=;XWM=8\"F.$\"1\"fS?)**HyFF.7$$\"1Nu(Qu'*f#yF1$\" 1$e\"z3Y7'*GF.7$$\"1NOsD.quQF1$\"1=0#zfs[(HF.7$$\"1PO5)G]@<*Fao$\"1#3x &yf)***HF.7$$!1#p'*3wpJ'QF1$\"1Ca_`D-vHF.7$$!1&z6r.C^m(F1$\"1X0%)))QU+ HF.7$$!1ltm:&R2:\"F.$\"167ldJ_qFF.7$$!1%GHG^!G$=F.$\"1'oav=U]P#F.7$$!1\"GSf6!zH@F.$\"1rl<=m\"G6#F.7$$!1el6 =N0qBF.$\"1#z=bc^#R=F.7$$!1UI'[D,Bg#F.$\"1YRx\"4eE\\\"F.7$$!1]Z'fG+vw# F.$\"1f@,t0*z:\"F.7$$!1gowgmS'*GF.$\"1\"=D!>:b:yF17$$!1u'*ygm3sHF.$\"1 @4W=>!H3%F17$F($!1T#3A9T7Y#!#C-%'COLOURG6&%$RGBG$\"#5!\"\"F*F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}} {SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "theta=c" "6#/%&thetaG%\"cG" }} {PARA 0 "" 0 "" {TEXT -1 13 "The graph of " }{XPPEDIT 18 0 "theta=c" " 6#/%&thetaG%\"cG" }{TEXT -1 61 " is a line through the origin (pole) w hich makes an angle of " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 19 " with the positive " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 37 "-axis (reference direction). Because " }{XPPEDIT 18 0 "r" "6#%\"rG" } {TEXT -1 82 " does not appear in the equation, its value is unrestrict ed (just as the value of " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 127 " is unrestricted in the previous example) and this is what cau ses the graph to be an entire line. Note that negative values of " } {XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 106 " determine points on the li ne lying on the opposite side of the origin from points determined by \+ positive " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "The following Maple \+ command will plot a portion of the line " }{XPPEDIT 18 0 "theta=2*Pi/3 " "6#/%&thetaG*(\"\"#\"\"\"%#PiGF'\"\"$!\"\"" }{TEXT -1 1 "." }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([r,2*Pi/3,r=-10..10],coor ds=polar,scaling=CONSTRAINED);" "6#-%%plotG6%7%%\"rG*(\"\"#\"\"\"%#PiG F*\"\"$!\"\"/F';,$\"#5F-F1/%'coordsG%&polarG/%(scalingG%,CONSTRAINEDG " }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$ 7S7$$\"1)*************\\!#:$!1(QWy.a-m)F*7$$\"1KLLe%G?y%F*$!1/O2`ir#G) F*7$$\"1lmT&esBf%F*$!16\"G_kAU&zF*7$$\"1JL$3s%3zVF*$!12>8Fsz%e(F*7$$\" 1KL$e/$QkTF*$!1nhi\"G%oF*7$$\"1KL3_>f_ PF*$!12C3@*z'*\\'F*7$$\"1***\\(o1YZNF*$!1os7;@QWhF*7$$\"1KL3-OJNLF*$!1 )*Q)zhKpx&F*7$$\"1)**\\P*o%Q7$F*$!1;7/N:m5aF*7$$\"1lmm\"RFj!HF*$!1*z&z 0n!R.&F*7$$\"1KL$e4OZr#F*$!11X7Z31-ZF*7$$\"1*****\\n\\!*\\#F*$!1!\\Qx+ \"[GVF*7$$\"1*****\\ixCG#F*$!1V\"RO@nL&RF*7$$\"1*****\\KqP2#F*$!1B'Ghc v=f$F*7$$\"1KL3-TC%)=F*$!1J.n=lgjKF*7$$\"1mmm\"4z)e;F*$!1v.QqGEtGF*7$$ \"1mmmm`'zY\"F*$!1V\"=))f!fUDF*7$$\"1***\\(=t)eC\"F*$!1h2eO,%z:#F*7$$ \"1mmm;1J\\5F*$!1(4))4Ifu\"=F*7$$\"1!***\\(=[jL)!#;$!1rM.hy*QW\"F*7$$ \"1$***\\iXg#G'Fiq$!1.9hI!z\")3\"F*7$$\"1cmmT&Q(RTFiq$!1Tx@%[P-<(Fiq7$ $\"1km;/'=><#Fiq$!1T2KsL(=w$Fiq7$$\"1BMLLe*e$\\!#=$!1<5?kBA\\&)F^s7$$! 1rm;zRQb@Fiq$\"1amthXBLPFiq7$$!1++](=>Y2%Fiq$\"1@]CauWdqFiq7$$!1smm\"z Xu9'Fiq$\"1;UzW)oZ1\"F*7$$!1)******\\y))G)Fiq$\"1Jl(*penN9F*7$$!1++]i_ QQ5F*$\"1Y2[Kg`)z\"F*7$$!1++D\"y%3T7F*$\"12bi(*=i\\@F*7$$!1****\\P![hY \"F*$\"1+uO#*GWRDF*7$$!1KLL$Qx$o;F*$\"1\\@8%R9(*)GF*7$$!1+++v.I%)=F*$ \"1*3@i)RqjKF*7$$!1lm\"zpe*z?F*$\"1skVUTf-OF*7$$!1+++D\\'QH#F*$\"1v,!e f!4tRF*7$$!1KLe9S8&\\#F*$\"1P;&\\)))p@VF*7$$!1***\\i?=bq#F*$\"1:w-%*\\ 4'o%F*7$$!1KLL3s?6HF*$\"1(3%>'zeB/&F*7$$!1)**\\7`Wl7$F*$\"1>;D!)RL:aF* 7$$!1mmmm*RRL$F*$\"1e6l6MbudF*7$$!1mm;a<.YNF*$\"1g#pM;2>9'F*7$$!1LLe9t OcPF*$\"1+%\\2/>i]'F*7$$!1******\\Qk\\RF*$\"1_>+?Q)4%oF*7$$!1LL$3dg6<% F*$\"1P8@N?mCsF*7$$!1mmmmxGpVF*$\"1>ev/%Gyc(F*7$$!1)**\\7oK0e%F*$\"1*p a'H`rLzF*7$$!1***\\(=5s#y%F*$\"1%e-H!e\"RG)F*7$$!1)*************\\F*$ \"1(QWy.a-m)F*-%'COLOURG6&%$RGBG$\"#5!\"\"\"\"!F`[l-%(SCALINGG6#%,CONS TRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Cur ve 1" }}}}}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "r=2*a*cos(theta)" "6 #/%\"rG*(\"\"#\"\"\"%\"aGF'-%$cosG6#%&thetaGF'" }}{PARA 0 "" 0 "" {TEXT -1 147 "This is a spendid example of a polar equation that can b e effectively analyzed by transforming to Cartesian coordinates. Multi plying both sides by " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 7 " give s " }{XPPEDIT 18 0 "r^2=2*a*r*cos(theta)" "6#/*$%\"rG\"\"#**F&\"\"\"% \"aGF(F%F(-%$cosG6#%&thetaGF(" }{TEXT -1 38 " and substituting for eac h side gives " }{XPPEDIT 18 0 "x^2+y^2=2ax" "6#/,&*$%\"xG\"\"#\"\"\"*$ %\"yGF'F(*&F'F(%#axGF(" }{TEXT -1 16 ". Bringing the " }{XPPEDIT 18 0 "2ax" "6#*&\"\"#\"\"\"%#axGF%" }{TEXT -1 50 " to the left side and c ompleting the square gives " }{XPPEDIT 18 0 "(x-a)^2+y^2=a^2" "6#/,&*$ ,&%\"xG\"\"\"%\"aG!\"\"\"\"#F(*$%\"yGF+F(*$F)F+" }{TEXT -1 19 ". Thus \+ we see that " }{TEXT 344 32 "the polar graph of the equation " } {XPPEDIT 18 0 "r=2*a*cos(theta)" "6#/%\"rG*(\"\"#\"\"\"%\"aGF'-%$cosG6 #%&thetaGF'" }{TEXT 345 23 " is a circle of radius " }{XPPEDIT 18 0 "a bs(a" "6#-%$absG6#%\"aG" }{TEXT 346 14 " centered at (" }{XPPEDIT 18 0 "a,0" "6$%\"aG\"\"!" }{TEXT 347 3 "). " }{TEXT -1 50 "We used Maple \+ to plot an example of this equation " }{HYPERLNK 17 "in an earlier dis cussion" 1 "" "circle" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 399 "Here are some useful remarks about \+ this example. First of all, the student should memorize the informatio n just presented. This makes the sketch of the equation immediately av ailable without having to go through all the steps of transforming th e equation to Cartesian coordinates and then simplifying the result. S econdly, notice that the description of the graph holds even for negat ive values of " }{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 73 ", which is \+ the reason the the radius needs to be described in general as " } {XPPEDIT 18 0 "abs(a)" "6#-%$absG6#%\"aG" }{TEXT -1 17 " instead of ju st " }{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 36 ". Third, the appearanc e of the 2 in " }{XPPEDIT 18 0 "r=2*a*cos(theta)" "6#/%\"rG*(\"\"#\"\" \"%\"aGF'-%$cosG6#%&thetaGF'" }{TEXT -1 66 " is wholly a matter of con venience. The numerical coefficient of " }{XPPEDIT 18 0 "r*cos(theta) " "6#*&%\"rG\"\"\"-%$cosG6#%&thetaGF%" }{TEXT -1 34 " could just as we ll be denoted by " }{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 16 ", instea d of by " }{XPPEDIT 18 0 "2*a" "6#*&\"\"#\"\"\"%\"aGF%" }{TEXT -1 52 " , but in that case the radius would be described by " }{XPPEDIT 18 0 " abs(a/2" "6#-%$absG6#*&%\"aG\"\"\"\"\"#!\"\"" }{TEXT -1 48 " and the c enter would be situated at the point (" }{XPPEDIT 18 0 "a/2,0" "6$*&% \"aG\"\"\"\"\"#!\"\"\"\"!" }{TEXT -1 142 "). So using the 2 in naming \+ the coefficient makes for a \"cleaner\" description of the graph. Fina lly, the point determined by the coordinates (" }{XPPEDIT 18 0 "a,0" " 6$%\"aG\"\"!" }{TEXT -1 135 ") is the same whether the coordinates are interpreted as polar or as Cartesian. In the next example, the coordi nates of the center are " }{TEXT 348 3 "not" }{TEXT -1 108 " the same \+ in either system, so we will have to be careful to specify the system \+ when giving the coordinates." }}{PARA 4 "" 0 "n=1" {XPPEDIT 18 0 "r=2* a*sin(theta)" "6#/%\"rG*(\"\"#\"\"\"%\"aGF'-%$sinG6#%&thetaGF'" }} {PARA 0 "" 0 "" {TEXT -1 36 "Mutiplying through this equation by " } {XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 24 " and substituting gives " } {XPPEDIT 18 0 "x^2+y^2=2a*y" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(*(F'F( %\"aGF(F*F(" }{TEXT -1 21 " which simplifies to " }{XPPEDIT 18 0 "x^2+ (y-a)^2=a^2" "6#/,&*$%\"xG\"\"#\"\"\"*$,&%\"yGF(%\"aG!\"\"F'F(*$F,F'" }{TEXT -1 45 ". Thus the graph is again a circle of radius " } {XPPEDIT 18 0 "abs(a)" "6#-%$absG6#%\"aG" }{TEXT -1 31 ", this time ce ntered at polar (" }{XPPEDIT 18 0 "a,Pi/2" "6$%\"aG*&%#PiG\"\"\"\"\"#! \"\"" }{TEXT -1 3 ") (" }{TEXT 349 4 "i.e." }{TEXT -1 14 " Cartesian ( 0," }{TEXT 350 1 "a" }{TEXT -1 3 "))." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([3*sin(theta),theta,theta=0..Pi],coords=polar);" " 6#-%%plotG6$7%*&\"\"$\"\"\"-%$sinG6#%&thetaGF)F-/F-;\"\"!%#PiG/%'coord sG%&polarG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6#-% 'CURVESG6$7S7$\"\"!F(7$$\"1(\\OQFiVJF,7$$\"1aQ809$*e5!#:$\"1w 4/Z(QhP%F,7$$\"1#=Gu&yp'=\"FG$\"115_!Q;`#eF,7$$\"1z`\"3#R5)H\"FG$\"1e) QbmUQ[(F,7$$\"10!fm#[B'Q\"FG$\"1lP*H`!zp#*F,7$$\"13A@6FG7$$\"1S:c/wO'[\"FG$\"1Kxgw%H#)H\"FG7$$\"1>ufK(****\\\"FG$\"11hYgc *3]\"FG7$$\"1@H9$47g[\"FG$\"1Di)QNsVq\"FG7$$\"1)GxB))HlW\"FG$\"1tNbAt# p*=FG7$$\"1\"FG$\"15MX@M#fS#FG7$$\"1M^04e\"z0\"FG$\"1PY,_uRjD FG7$$\"1lvYw5Z)=*F,$\"1()G$3jIco#FG7$$\"1S*RUhgC](F,$\"1U:(4,'*))z#FG7 $$\"1+I#*z$H'odF,$\"1o/(4WSY)GFG7$$\"1T?#*RdwdQF,$\"1'e5k_V&\\HFG7$$\" 1l!yv]P1/#F,$\"1YD\"H]ag)HFG7$$\"1O$QZ:k>l%!#=$\"1*[#R'y#****HFG7$$!1R ')H%y(>D?F,$\"1d%3Gqli)HFG7$$!1.LTUXU)z$F,$\"1L!)3?)46&HFG7$$!1lxZ/`$3 l&F,$\"1t\\&>L*[*)GFG7$$!1)G!oHjojuF,$\"1TZ\")>z7,GFG7$$!1;Ya.k'o5*F,$ \"1*)>pl,\">p#FG7$$!1L-\"=<-Z0\"FG$\"1*pj0*[emDFG7$$!1X6<)G*\\%>\"FG$ \"1xw^kzG2CFG7$$!1=?u6N%)*H\"FG$\"1L\"QFE.'[AFG7$$!1'H?5**\\\"*Q\"FG$ \"1!4eAAP)*H\"*=FG7$$!1f-'*)HOu[\"FG$\"1 #='G;]t$p\"FG7$$!1T&)G*)H***\\\"FG$\"1_S9vge/:FG7$$!17Fww7^([\"FG$\"1l ^&>^ToI\"FG7$$!1s-UW>@]9FG$\"15W9)zVn6\"FG7$$!11c#)y:E&Q\"FG$\"1uJm@CI Y#*F,7$$!1/MeT@v)H\"FG$\"1\"e`e8\\]\\(F,7$$!1Vtx$4@v=\"FG$\"1Eww+W(f$e F,7$$!1'G)34$3k0\"FG$\"1%f)H?%\\5N%F,7$$!1hRfFyD'>*F,$\"16sT4Ct\\JF,7$ $!1H(p)e/HjuF,$\"1!z%oDP\\))>F,7$$!1&zg]'G&**y&F,$\"1^i " 0 "" {XPPEDIT 19 1 "plot([4*(1+cos(theta)),theta,theta=0..2*Pi],coords=pola r,scaling=constrained);" "6#-%%plotG6%7%*&\"\"%\"\"\",&F)F)-%$cosG6#%& thetaGF)F)F./F.;\"\"!*&\"\"#F)%#PiGF)/%'coordsG%&polarG/%(scalingG%,co nstrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-% 'CURVESG6$7`o7$$\"\")\"\"!F*7$$\"1=+XEx*=(z!#:$\"1dLuLt_na!#;7$$\"1=P& *ob)z)yF.$\"1N-t*))3r3\"F.7$$\"1S=&H6l/x(F.$\"1xZ?q;;Z:F.7$$\"1DqT@(=G h(F.$\"1i\"3g-xN*>F.7$$\"1>V&*o=o*Q(F.$\"16do6x_uCF.7$$\"1gK#em))37(F. $\"1de1>*z\"GHF.7$$\"1Th&R84y!oF.$\"187\\(R_CN$F.7$$\"1iya?QDckF.$\"1_ &R.t\"))RPF.7$$\"1N&Qk&3ysgF.$\"1+OiR87&3%F.7$$\"1*R=6#))ygcF.$\"1]BS( =hpQ%F.7$$\"1W&))oB_zD&F.$\"1]$y(zw%ei%F.7$$\"1NoiTRaR[F.$\"19oQrk!Q#[ F.7$$\"1hHN)>'p%R%F.$\"1L;:Q#QY)\\F.7$$\"1T]RMN-VRF.$\"1r[EMp4+^F.7$$ \"1m#=!)HlSZ$F.$\"1AmQMl\"><&F.7$$\"1aH5')=i3IF.$\"10^XwV9'>&F.7$$\"1h vG+*)Q`DF.$\"1!30,NqV<&F.7$$\"1f(fS`%z6@F.$\"1L*Gf.*y3^F.7$$\"1s\"\\(f thw;F.$\"1,.mtXI/hF 1$\"1'*3K\"p6o\\%F.7$$!1*3ZEw,qQ#!#=$\"1;EWf3h(*RF.7$$!1\"Go_;'Q2ZF1$ \"1%)RPy\")yAMF.7$$!1&4\"G7m\"Qy(F1$\"1J9vr7nOGF.7$$!1.k)>K&*yR*F1$\"1 &y+\\GcmI#F.7$$!1,t/nPG****F1$\"1FVaf,_7Z EE\"F.7$$!1$[d7)G!RD)F1$\"1%)QWQ`N6#)F17$$!1y-)HO#> !#?$\"1T2LXxwlf!#B7$$!1-lTl[%*GOFhu$!1D&[&4*H[%\\Fcr7$$!1M68.[Ch7F1$!1 ))4TURV,LFhu7$$!1k%p67Q)HFF1$!1(oiCP#=56F17$$!1(oym]U,g%F1$!1TcTC!*yQE F17$$!1*H5[(QFGlF1$!1-p.4(pz)\\F17$$!1Ff2qX?=#)F1$!1>$R0NYm7)F17$$!1(3 H%*f'=g&*F1$!1V;zyg7xF1$!16OC+M;aGF.7$$!1`?+2#4!*\\%F1$! 1/u[\\s>aMF.7$$!1#[tVL9#>7Fhu$!1f/@G=v()RF.7$$\"1%*pq&[&=9eF1$!1O7+,v \\xWF.7$$\"1N=T*)f9$G\"F.$!1\"Rh#zbKb[F.7$$\"1$44WK`G7#F.$!1:vBWm)46&F .7$$\"1%)>()*o,ib#F.$!1[zy#RXY<&F.7$$\"1$>Im*3k-IF.$!1Z.un;:'>&F.7$$\" 1+^1*)=)yY$F.$!1Sa&HIWD<&F.7$$\"1*QW>l1n$RF.$!1QZQ3\\Q,^F.7$$\"1`%R:#H $)*R%F.$!1!4]psaI)\\F.7$$\"1/!zOzAd&[F.$!1Vi4-[-<[F.7$$\"1y#Q9L$fj_F.$ !13+e?]%Gi%F.7$$\"1KRf.TeccF.$!1yC;^Ts*Q%F.7$$\"1rDOV'RM3'F.$!1\")f$G/ Lk2%F.7$$\"1%G*Gg-]zkF.$!16&y-!oe;PF.7$$\"1!*fdV97.oF.$!1A$\\X'f;eLF.7 $$\"1b'fZ3ES4(F.$!1gJm#)fJoHF.7$$\"1$*Q]p+WktF.$!1K?)e^v=_#F.7$$\"1$)* [TB'R!f(F.$!1Gwj7J;[?F.7$$\"1+dy'y%HixF.$!1gXkwi\"Qd\"F.7$$\"1%>]S$Hp) )yF.$!1F[(fi3P3\"F.7$$\"1&)))G9e2szF.$!10KQ$*GA]aF17$F($\"15`b7(4Lc'!# C-%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "Once the student knows the general shape of the graph, it is fairly easy to sketch the figure from the four po ints determined by " }{XPPEDIT 18 0 "theta=0,Pi/2,Pi" "6%/%&thetaG\"\" !*&%#PiG\"\"\"\"\"#!\"\"F'" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "3*Pi/2 " "6#*(\"\"$\"\"\"%#PiGF%\"\"#!\"\"" }{TEXT -1 48 ". Notice that the \+ graph is symmetric about the " }{TEXT 352 1 "x" }{TEXT -1 113 "-axis. \+ An interesting phenomenon occurs at the pole. Let's re-plot the curv e around the pole for a closer look." }}{PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([4*(1+cos(theta)),theta,theta=7*Pi/8..9*Pi/8],coords=polar,sca ling=constrained);" "6#-%%plotG6%7%*&\"\"%\"\"\",&F)F)-%$cosG6#%&theta GF)F)F./F.;*(\"\"(F)%#PiGF)\"\")!\"\"*(\"\"*F)F3F)F4F5/%'coordsG%&pola rG/%(scalingG%,constrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$7S7$$!1HpIvc/8G!#;$\"15\"G*p;?l6F*7$$!1Fgg! *H%Qf#F*$\"1mb1rBuA5F*7$$!1FVWMu;3CF*$\"1kp.dSA$3*!#<7$$!19$*\\.DP0AF* $\"1Ty#f#3$*)*yF77$$!1h0wwu23?F*$\"1N\"*3e\")[7oF77$$!1GGCUt%*==F*$\"1 )oH?7`B$eF77$$!1a]$)[>Q];F*$\"1=Wy&*z/5]F77$$!1#Q$Ha_/$[\"F*$\"1)\\m#f ;OUUF77$$!1s4()yd-=8F*$\"1=q(=iOQ`$F77$$!1e%[%4v*>;\"F*$\"1\"pxB(Q`4HF 77$$!1*Qu;nW1,\"F*$\"1SSyF9%yM#F77$$!1)GW@L=E&))F7$\"1A*f+?Um\">F77$$! 1cbR\\fgKvF7$\"1w%oB-Wx\\\"F77$$!1-K([=+tI'F7$\"1gOGU,&H9\"F77$$!12/o5 yxB_F7$\"1Onw*)H>%e)!#=7$$!1xO7]TKCVF7$\"1Y#[m4>mW'F^p7$$!1+Dc0BYhLF7$ \"1bS!*HX`/WF^p7$$!1'='z**[&zj#F7$\"1X$4rqW\\0$F^p7$$!1W'*))GDJ/>F7$\" 1)**\\0bt$p=F^p7$$!1xgs^I*HN\"F7$\"1gqFm)ov6\"F^p7$$!1#*)*z6$R@b)F^p$ \"1'fM!\\>S2c!#>7$$!1_D$f]SE'[F^p$\"1PQADVM,CF]r7$$!1]40:F%H6#F^p$\"1P hVELKso!#?7$$!1qke8!4e,(*F^s7$$!1W]6#fIq/#F^p$!1d0^/99`lFhr7$ $!1!3)R]6&fl%F^p$!1^uZkF^p7$$!1Xec58^a_F7$!1;LF$fF4m)F^p7$$!1&*RV!*e=pjF7$!1\\5'z#)[+ ;\"F77$$!19GlxPb4vF7$!1=RuW)f2\\\"F77$$!1%[DfA/Uz)F7$!1zqGt5L(*=F77$$! 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1&=Pi([5#R*F17$$!1k%*p+qBVcT88![*F17$$\"1V:#4(fybGF.$!10')HBA%fq(F17$$\"1& >q3^gMU$F.$!19vA@k(Hq%F17$$\"1b;G.+++SF.$\"1q*p*e[l\"G$!#C-%'COLOURG6& %$RGBG$\"#5!\"\"F*F*-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 4 "" 0 "" {XPPEDIT 18 0 "r=a*(1-cos(th eta))" "6#/%\"rG*&%\"aG\"\"\",&F'F'-%$cosG6#%&thetaG!\"\"F'" }}{PARA 0 "" 0 "" {TEXT -1 6 "Ditto." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([4*(1-cos(theta)),theta,theta=0..2*Pi],coords=polar,scaling=co nstrained);" "6#-%%plotG6%7%*&\"\"%\"\"\",&F)F)-%$cosG6#%&thetaG!\"\"F )F./F.;\"\"!*&\"\"#F)%#PiGF)/%'coordsG%&polarG/%(scalingG%,constrained G" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6 $7ao7$\"\"!F(7$$\"1?b^@IV5P!#<$\"1&f'*eJcO6&!#=7$$\"1lYTw&HAE\"!#;$\"1 uxxmTT0LF,7$$\"1xcls^!)zFF3$\"1+@>3M3V6F37$$\"1li*[9H?m%F3$\"1+EB7-b+F F37$$\"1Vk\"=0Ghi'F3$\"1=,;'[t]8&F37$$\"1\\k%G$R:l#)F3$\"1l2F$*oFQ#)F3 7$$\"1@UgE1T,&*F3$\"1-@N0$e*G7!#:7$$\"1'))H%*e`*****F3$\"1]-cMl2FHPjWG#FO7$$\"1xZBnZ^]vF3$\"1Q')*)p]$H*GFO7$$\" 1]%Q=1!zcYF3$\"1$)Rn*>#[IMFO7$$!12oecS&)*Q#F/$\"1(G5Wr(Q-SFO7$$!1.Pm3F Z#>'F3$\"1\"f@*Q<&FO7$$!1oK>UQ#R.$FO$\"1)e=*)*y-'>&FO7$$!1h)[njXGX$FO$\"1_LaMk.u^F O7$$!1]I%f68[(QFO$\"1t4+Lo`8^FO7$$!1i\\f'QPUO%FO$\"1OM11.!R*\\FO7$$!1Z cG*yig%[FO$\"1f'\\W6#3@[FO7$$!1-\"pbq$3h_FO$\"1#\\Dn*>=CYFO7$$!1QS9Z%R 2m&FO$\"1NqRmO*pQ%FO7$$!1c4w_QXwgFO$\"1#3O\"oF8#3%FO7$$!1i!*[Ck1jkFO$ \"1F'=Tq&3LPFO7$$!1TQL;xZ*z'FO$\"17JivcfiLFO7$$!1MGS))=y+rFO$\"1]/.trH eHFO7$$!1P(oe]jPP(FO$\"1q9c,?_/DFO7$$!1=<-xT(3g(FO$\"1S9E\\2(G-#FO7$$! 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1dC:A76tPFgn$\"1:_(e=U]P\"F17$$!1[U(*HC`:6F1$\"1KAcXl+QFF17$$!1)o=wRA< v#F1$\"1esv>wr7ZF17$$!1yr+H(4N!\\F1$\"1**))f2M4rkF17$$!1G1.%[KvN)F1$\" 1I(psu]WJ)F17$$!1\"o[&pp%\\A\"F.$\"1$)oZ'=nJ\\*F17$$!1.GgbJHJ`6qBMF.$\"1K)fCa(R,ZF17$$!18]i&evv)RF.$\"1S&3l_wmB\"Fgn7 $$!14$f+7$[)\\%F.$!17eWO)p'HhF17$$!1Dvye]_\\[F.$!12>s/:Tp7F.7$$!12%y#f ;a$*\\F.$!1o8lAG5h;F.7$$!1'R%\\ig<,^F.$!1*3J9MoX2#F.7$$!1@@5ik,r^F.$!1 H(zH^S/_#F.7$$!1d\"z\\!>6'>&F.$!1p*4C:c1)HF.7$$!1VvBgYGv^F.$!1O')pLB() RMF.7$$!1x,AXy73^F.$!1f4e')z*G!RF.7$$!1)**p+st#)*\\F.$!1b0Y50g\\VF.7$$ !1())4Y'35W[F.$!1]$)*p?L,z%F.7$$!1@pP:E\\@YF.$!1*=#3>%GhE&F.7$$!1Y%Q9# )4hM%F.$!1B)=I_:>s&F.7$$!1^ry%eja0%F.$!1?Lu]c#*3hF.7$$!1kL>BC/3'F.7$$\"1b/W^@SyVF.$!1 $\\!fCkvtcF.7$$\"1T,z8J6KYF.$!10E;B\\7Y_F.7$$\"1cyWKhlR[F.$!1mM9&='3,[ F.7$$\"1%pTC)Gq))\\F.$!1D\\NlvT\"Q%F.7$$\"1*\\fz3gu4&F.$!1NleuE#e&RF.7 $$\"1![_#*o;F<&F.$!1af2Z&phY$F.7$$\"1R1t<56'>&F.$!1+nANbW!)HF.7$$\"16< #yy^W<&F.$!14,vc)>Ub#F.7$$\"1:mEL*GV6&F.$!1d&4ssh*R@F.7$$\"1^nXC^'z'[F .$!1)=5%\\Gb88F.7$$\"1pA!z\"R5-XF.$!1v#[z-iZ='F17$$\"1X$=n*******RF.$ \"1S`e`[l\"G$!#C-%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%(SCALINGG6#%,CONSTRA INEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve \+ 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "Any of these four figures i s readily sketched from the four points determined by the values " }} {PARA 0 "" 0 "" {XPPEDIT 18 0 "theta=0,Pi/2,Pi,3*Pi/2" "6&/%&thetaG\" \"!*&%#PiG\"\"\"\"\"#!\"\"F'*(\"\"$F(F'F(F)F*" }{TEXT -1 63 " once the figure is recognized from its equation as a cardioid." }}}}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "r=f(theta-theta[0])" "6#/%\"rG-%\"fG6#, &%&thetaG\"\"\"&F)6#\"\"!!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 216 "In the foregoing examples we have seen similar equations which produce ident ical figures in differing orientations. In this section we explain the reason for this and open up possibilities for even more orientations. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "The s tudent should have learned at some point that for a given function " } {XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 22 " and a given cons tant " }{XPPEDIT 18 0 "x[0]" "6#&%\"xG6#\"\"!" }{TEXT -1 26 ", the Car tesian graphs of " }{XPPEDIT 18 0 "y=f(x)" "6#/%\"yG-%\"fG6#%\"xG" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "y=f(x-x[0])" "6#/%\"yG-%\"fG6#,&%\" xG\"\"\"&F)6#\"\"!!\"\"" }{TEXT -1 46 " are simply related by a horizo ntal shift, or " }{TEXT 356 11 "translation" }{TEXT -1 51 ". The reaso n for this is as follows. If the point (" }{XPPEDIT 18 0 "a,b" "6$%\"a G%\"bG" }{TEXT -1 21 ") is on the graph of " }{XPPEDIT 18 0 "y=f(x)" " 6#/%\"yG-%\"fG6#%\"xG" }{TEXT -1 7 ", then " }{XPPEDIT 18 0 "b=f(a)" " 6#/%\"bG-%\"fG6#%\"aG" }{TEXT -1 18 " which means that " }{XPPEDIT 18 0 "b=f(a+x[0]-x[0])" "6#/%\"bG-%\"fG6#,(%\"aG\"\"\"&%\"xG6#\"\"!F*&F,6 #F.!\"\"" }{TEXT -1 29 ", and this places the point (" }{XPPEDIT 18 0 "a+x[0],b" "6$,&%\"aG\"\"\"&%\"xG6#\"\"!F%%\"bG" }{TEXT -1 18 ") on th e graph of " }{XPPEDIT 18 0 "y=f(x-x[0])" "6#/%\"yG-%\"fG6#,&%\"xG\"\" \"&F)6#\"\"!!\"\"" }{TEXT -1 36 ". The converse is clearly also true. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "rotations" {TEXT -1 62 "A very similar reasoning applies when comparing the graphs of " } {XPPEDIT 18 0 "r=f(theta)" "6#/%\"rG-%\"fG6#%&thetaG" }{TEXT -1 5 " an d " }{XPPEDIT 18 0 "r=f(theta-theta[0])" "6#/%\"rG-%\"fG6#,&%&thetaG\" \"\"&F)6#\"\"!!\"\"" }{TEXT -1 36 ". The point with polar coordinates \+ (" }{XPPEDIT 18 0 "a,b" "6$%\"aG%\"bG" }{TEXT -1 12 ") satisfies " } {XPPEDIT 18 0 "r=f(theta)" "6#/%\"rG-%\"fG6#%&thetaG" }{TEXT -1 48 " e xactly when the point with polar coordinates (" }{XPPEDIT 18 0 "a,b+th eta)" "6$%\"aG,&%\"bG\"\"\"%&thetaGF&" }{TEXT -1 12 ") satisfies " } {XPPEDIT 18 0 "r=f(theta-theta[0])" "6#/%\"rG-%\"fG6#,&%&thetaG\"\"\"& F)6#\"\"!!\"\"" }{TEXT -1 6 ". But " }{XPPEDIT 18 0 "theta" "6#%&theta G" }{TEXT -1 7 " is an " }{TEXT 357 5 "angle" }{TEXT -1 59 ", not a di stance so the result of increasing each value of " }{XPPEDIT 18 0 "the ta" "6#%&thetaG" }{TEXT -1 22 " by a constant amount " }{XPPEDIT 18 0 "theta[0]" "6#&%&thetaG6#\"\"!" }{TEXT -1 12 " produces a " }{TEXT 358 8 "rotation" }{TEXT -1 14 ". In summary, " }{TEXT 359 13 "the grap h of " }{XPPEDIT 18 0 "r=f(theta-theta[0])" "6#/%\"rG-%\"fG6#,&%&theta G\"\"\"&F)6#\"\"!!\"\"" }{TEXT 360 31 " is a rotation of the graph of \+ " }{XPPEDIT 18 0 "r=f(theta)" "6#/%\"rG-%\"fG6#%&thetaG" }{TEXT 361 19 " through the angle " }{XPPEDIT 18 0 "theta[0]" "6#&%&thetaG6#\"\"! " }{TEXT 362 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "The first equation of the cardioid presented above was " }{XPPEDIT 18 0 "r=a*(1+cos(theta))" "6#/%\"rG*&%\"aG\"\"\",&F'F'-%$cos G6#%&thetaGF'F'" }{TEXT -1 61 ". To rotate the figure counter-clockwis e through 90 we write " }{XPPEDIT 18 0 "r=a*(1+cos(theta-Pi/2))" "6#/% \"rG*&%\"aG\"\"\",&F'F'-%$cosG6#,&%&thetaGF'*&%#PiGF'\"\"#!\"\"F1F'F' " }{TEXT -1 53 ". From the trig identity (you should know this one!) \+ " }{XPPEDIT 18 0 "cos(alpha+beta)=cos(alpha)*cos(beta)-sin(alpha)*sin( beta)" "6#/-%$cosG6#,&%&alphaG\"\"\"%%betaGF),&*&-F%6#F(F)-F%6#F*F)F)* &-%$sinG6#F(F)-F36#F*F)!\"\"" }{TEXT -1 9 " we have " }}{PARA 0 "" 0 " " {XPPEDIT 18 0 "cos(theta-Pi/2)=sin(theta)" "6#/-%$cosG6#,&%&thetaG\" \"\"*&%#PiGF)\"\"#!\"\"F--%$sinG6#F(" }{TEXT -1 9 " so that " } {XPPEDIT 18 0 "r=a*(1+sin(theta))" "6#/%\"rG*&%\"aG\"\"\",&F'F'-%$sinG 6#%&thetaGF'F'" }{TEXT -1 74 ". The other two cases of the cardioid ca n be obtained in similar fashion. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 56 "Let's have Maple draw a cardioid at a 60 \+ angle with the " }{TEXT 363 1 "x" }{TEXT -1 6 "-axis." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([1+cos(theta-Pi/3),theta,theta=0..2*Pi ],coords=polar,scaling=constrained);" "6#-%%plotG6%7%,&\"\"\"F(-%$cosG 6#,&%&thetaGF(*&%#PiGF(\"\"$!\"\"F1F(F-/F-;\"\"!*&\"\"#F(F/F(/%'coords G%&polarG/%(scalingG%,constrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$7]o7$$\"1+++++++:!#:\"\"!7$$\"1\"3 \"oMdW)f\"F*$\"1:Kt(*4&H?#!#;7$$\"1HciW(Rvi\"F*$\"1GL]I;cSKF17$$\"1Bd$ HFDvk\"F*$\"13Q.^@R9VF17$$\"1ze_\\7/v:F*$\"1.;,/._f5F*7$$\"1o/Q#)[7A:F*$\"1kC T5kgz6F*7$$\"1h\"\\saE@Y\"F*$\"1'fa.\\^jG\"F*7$$\"1J\"=h5#3#R\"F*$\"1V .8k_b(Q\"F*7$$\"16%G!>#*f48F*$\"1l^zG#*R&[\"F*7$$\"1h:RyPw<7F*$\"1!pg; W9^d\"F*7$$\"1#)Rb]S\"R6\"F*$\"1'oXK13$e;F*7$$\"1;[RBR:-5F*$\"11`(yb/3 t\"F*7$$\"1LUCu].T))F1$\"1F:DR2h\"z\"F*7$$\"1t'=e?Nsg(F1$\"1Q-9y\">.%= F*7$$\"1)[/N?/qH'F1$\"1hQ6azFx=F*7$$\"1\\#z9_C/'\\F1$\"1^NE8tb+>F*7$$ \"1;RglB)Qx$F1$\"1Z'*p#*zf4>F*7$$\"1Yj!>,U**e#F1$\"1[FtHH\"z!>F*7$$\"1 =l6jCat7F1$\"1.%>bp]L*=F*7$$!1$od!G%QS6\"!#=$\"1'p5LOEd'=F*7$$!1>>7v*H [D\"F1$\"1)*ftGxQD=F*7$$!1&*zh%[0'QCF1$\"1uY/dZ8t!>F17$$!1zk\"z1xn;'F1$\"1 f*[*[i;o%)!#<7$$!1$p\"H)*z&o-&F1$\"1EP!>r\"**e:Fhr7$$!1O)\\]?Q5%QF1$!1 ^@?E+$QB&F[x7$$!1ra`V]9tGF1$!1S?8JMu?vF[x7$$!1uG#=762&>F1$!1'))y)\\jBL zF[x7$$!1-!*Q]TEu6F1$!1`-LI^&ft'F[x7$$!1I6k2ZV6hF[x$!1.J4e5[pYF[x7$$!1 3t%\\eKq_#F[x$!1#\\n,;z))\\#F[x7$$!15w&[HEUz%Fhr$!1>,doy)=J'Fhr7$$!1o \"p+89I)G!#A$!1[nk-c%f+&Fgz7$$!1diu)Fhr$!1s#ziNmmB$F[x7$$!1*pmq<[)**)*Fhr$!1n$4J#4z+wF[x7$$!12J%)))G ]\\S!#>$!1vU\"4L$\\C8F17$$\"1=-C+s=ZEF[x$!1CV\\$z&fQ?F17$$\"1?TNj[n@uF [x$!1JV&of/$3GF17$$\"1O(Q=imO]\"F1$!1-dY'GJ-i$F17$$\"1>V#*G?I.DF1$!1%z [x=K?L%F17$$\"1dBULC*oy$F1$!1W6c;![s!\\F17$$\"1:xSy_J*G&F1$!1#[XRuhrC& F17$$\"1a*pBo\\i#oF1$!1D-Zu!HuH&F17$$\"1*p2zPI))p)F1$!1y7&=(ed*)\\F17$ $\"1U\"=48F'R5F*$!1*)epkq0]VF17$$\"1(*)Q(*=4P@\"F*$!15_Jbe-vKF17$$\"1d -a>ymk8F*$!1JUCQ9ru=F17$$\"1*\\52+++]\"F*$\"1Cqor0iI7!#C-%'COLOURG6&%$ RGBG$\"#5!\"\"F+F+-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}}{SECT 1 {PARA 4 "" 0 " " {XPPEDIT 18 0 "r=a+b*cos(theta)" "6#/%\"rG,&%\"aG\"\"\"*&%\"bGF'-%$c osG6#%&thetaGF'F'" }}{PARA 0 "" 0 "" {TEXT -1 39 "The graph of this eq uation is called a " }{TEXT 364 6 "limaon" }{TEXT -1 64 ". Limaons com e in three varieties, depending on the relation of " }{XPPEDIT 18 0 "a bs(a)" "6#-%$absG6#%\"aG" }{TEXT -1 5 " to " }{XPPEDIT 18 0 "abs(b)" "6#-%$absG6#%\"bG" }{TEXT -1 32 ". The simplest case occurs when " } {XPPEDIT 18 0 "abs(a)=abs(b)" "6#/-%$absG6#%\"aG-F%6#%\"bG" }{TEXT -1 23 ". In that case either " }{XPPEDIT 18 0 "a=b" "6#/%\"aG%\"bG" } {TEXT -1 4 " or " }{XPPEDIT 18 0 "a=-b" "6#/%\"aG,$%\"bG!\"\"" }{TEXT -1 38 " so the limaon is a cardioid. In case " }{XPPEDIT 18 0 "abs(a)> abs(b)" "6#2-%$absG6#%\"bG-F%6#%\"aG" }{TEXT -1 14 " the value of " } {XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 207 " can never be 0 and either \+ is always positive or else is always negative. The graphs looks like a cardioid that \"changes it mind\" just before reaching the pole. The \+ following two Maple plots illustrate this." }}{EXCHG {PARA 0 "> " 0 " " {XPPEDIT 19 1 "plot([3+2*cos(theta),theta,theta=0..2*Pi],coords=pola r,scaling=constrained);" "6#-%%plotG6%7%,&\"\"$\"\"\"*&\"\"#F)-%$cosG6 #%&thetaGF)F)F//F/;\"\"!*&F+F)%#PiGF)/%'coordsG%&polarG/%(scalingG%,co nstrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-% 'CURVESG6$7V7$$\"\"&\"\"!F*7$$\"1oH(=-HY$\\!#:$\"1R)37jA3!o!#;7$$\"1B3 fq&*ytZF.$\"1uN'*3u6]7F.7$$\"1Mxz!)HI&[%F.$\"1i&=?*4SW=F.7$$\"13i$F.$\"1qMqg\"Rg!GF.7$$\"1`tn(QH!GJF.$ \"1VEy0v&y6$F.7$$\"1`%H=MdJe#F.$\"1dX3s.=TLF.7$$\"1?Ao)4)Q0?F.$\"1rW5- [ZjMF.7$$\"1ip\\J7\"zV\"F.$\"1q/2p5byMF.7$$\"1Y**[(*Rn^))F1$\"1t8HKlY \"R$F.7$$\"1$Rc@3*G(R%F1$\"13[.#e<$RKF.7$$!1\"yF(eyg!z\"!#=$\"1@HX^_!) )*HF.7$$!1Rc(=Wvhr$F1$\"1R*[z\")o?q#F.7$$!1/x\\c/4QlF1$\"13Rz!*))o#Q#F .7$$!1\"ed?fc?Z)F1$\"1rE.'z9%z?F.7$$!1:Yr:d>/5F.$\"1G+3(eG)>x]1\"F.$ \"1&3\"fM(ysV%F17$$!1&4\">mYPJ5F.$\"1piw'Ql[u#F17$$!15DrHT745F.$\"18HJ U2s&Q\"F17$$!1?a-4[++5F.$\"1^4Ll#R85$Fao7$$!1)Q*>E&))*35F.$!1Nou0g&[P \"F17$$!17>)oyo/.\"F.$!1&3'RDZN(p#F17$$!1)yht7=G1\"F.$!1`)\\Ds9BK%F17$ $!1iPaQdU(4\"F.$!1Z8m)R&=&H'F17$$!12&o3Z?57\"F.$!1&>[N7H_c)F17$$!1HH@A o'>7\"F.$!1)yL()4n%46F.7$$!1RK)H(z'G3\"F.$!1Nmk^4mD9F.7$$!1*HVr2r1***F 1$!1)>Qb\\QZt\"F.7$$!1D!*RB%o:Z)F1$!1IM^+a]z?F.7$$!1IM(3_p\\Y'F1$!1+!> (plX#R#F.7$$!1RAuy82TNF1$!1U00CGs=FF.7$$!1\"HN75bM:*Fao$!1!f(HS7(Q*HF. 7$$\"1Fd['*f\"[>%F1$!1$44]$HUIKF.7$$\"12Jw#Hr2(*)F1$!1Vvd_2Z%R$F.7$$\" 1N.w+l+X9F.$!1GeLT5+zMF.7$$\"1e[#f]];+#F.$!1wd#\\f5RY$F.7$$\"1?5#fO)Gz DF.$!1ksP$=tBM$F.7$$\"1ji[N9zPJF.$!1yp52Yy7JF.7$$\"1+5bg_:$Q$F.$!1WklR mJrHF.7$$\"1^\"p6*)4$=OF.$!1qTvIf%z!GF.7$$\"1k\")GEkXsQF.$!1W2U9A)[f#F .7$$\"1ccw\")Q=2TF.$!1PsR(4YeN#F.7$$\"11yB7TE)H%F.$!1vj!p\"\\r@@F.7$$ \"159qPJ^pWF.$!1+B+\\[:q=F.7$$\"1yn*R+n1w%F.$!1?jGS**f%G\"F.7$$\"1c;i? ?/N\\F.$!116-O/^znF17$F($\"1pqMq&o?5%!#C-%'COLOURG6&%$RGBG$\"#5!\"\"F* F*-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([-6-5*cos(theta),theta,theta=0..2*Pi],coords=polar ,scaling=constrained);" "6#-%%plotG6%7%,&\"\"'!\"\"*&\"\"&\"\"\"-%$cos G6#%&thetaGF,F)F0/F0;\"\"!*&\"\"#F,%#PiGF,/%'coordsG%&polarG/%(scaling G%,constrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$7ao7$$!#6\"\"!F*7$$!1(p(4ZGD'4\"!#9$!1.iL4>l=v!#;7$$!1E )e)>=1&3\"F.$!1g&=/$RT&\\\"!#:7$$!1POX%*HQp5F.$!1O\"4%epAH@F77$$!1f`;h .M[5F.$!1&pp%y,IXFF77$$!16qRr\\`=5F.$!1B:Au@p5MF77$$!1z^;![pf#)*F7$!1b zJJf`SSF77$$!1q*plF%)oS*F7$!1>p$ojYBj%F77$$!18XcnViN*)F7$!1\"zpoD%4w^F 77$$!1R%\\q362U)F7$!1BXK\"=fXm&F77$$!11boH7TmyF7$!1M_D^+E'4'F77$$!1W!3 qENKK(F7$!1#es<#\\%GW'F77$$!1zu*QM(odnF7$!1aFKfBrNnF77$$!1)H>M^%pahF7$ !1M6$\\$z*3)pF77$$!1Ldi#*\\USbF7$!14KysIDmrF77$$!1Xb'Q%4=+\\F7$!1*\\A( **e+&H(F77$$!1N%fK^a=E%F7$!1+eW%3$egtF77$$!1auaiyDMOF7$!1:6Z:2sktF77$$ !1\"yT?8d<-$F7$!1qr,YG95tF77$$!1m#Rzb5T$=F7$!1Eib5hGFqF77$$!1'Ho9Zh'F77$$\"1)482=f3e$!#=$!1#)[.'R8q*fF77$$\"1\\o#e1:nC(F1$! 1F%H<&*f\"p_F77$$\"1`3d\"H&fP7F7$!1\"[2YM#>5XF77$$\"1QZN3y/_:F7$!1g#3( 4qS4QF77$$\"1iJC)*>9a++\"F7$!1U!*e3SQ,JF_r7$$\"1t$fS)ys35F7$\"1uC7SWx $o'F^v7$$\"16+p7c?O5F7$\"1\\%o+BU>T\"F17$$\"1Zn6Z@1D6F7$\"14p2rA'\\%HF 17$$\"1*\\\\k)fbn7F7$\"1oo*=]^\\:&F17$$\"1kYaEkVU9F7$\"1zI(p;xUF)F17$$ \"1!Gz9,T1h\"F7$\"1(eK0k?1B\"F77$$\"1B'o\\;K$Q))*z\"F7$\"1Ua%44`'pBF77$$\"1#pZl]k!\\b\"F7$\"1zKE5Nh4QF77$$\"1'[CIaU\\A\"F7$\"13\"zI`%F77$$\"1zP*R GG`\"pF1$\"1&4m6wq$4`F77$$\"1TO!4Ec(H=F^v$\"1NaX6^o%)fF77$$!1sfw5^Xb&) F1$\"1=+wgga)e'F77$$!1(=N8iO%f=F7$\"1(eBq$\\'f.(F77$$!1ar1%\\Yr.$F7$\" 1kR^C5C7tF77$$!1bt9;o9QOF7$\"11;2v.)[O(F77$$!1-vRy6j`UF7$\"1O5tXV-htF7 7$$!12SA*f?<*[F7$\"1f]%[A\"G'H(F77$$!16$z[N==`&F7$\"1'GV^F77$$!1'z>\\>01S*F7$\"1l$=.RR .k%F77$$!1^6F,F.!z*F7$\"1p'**f$=R'4%F77$$!1QOH<<;:5F.$\"1)4O&y^JwMF77$ $!1_3'zTY`/\"F.$\"1TXwuAs?GF77$$!1>NO;FHo5F.$\"1)GVJjyf;#F77$$!1If6sh: &3\"F.$\"1JK38Dt!\\\"F77$$!1AzX%fwi4\"F.$\"1F_1\"p][\\(F17$F($!1^Nwa3b C!*!#C-%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "In case " }{XPPEDIT 18 0 "abs(a) " 0 "loop" {XPPEDIT 19 1 "plot([1+2*cos(theta),t heta,theta=0..2*Pi],coords=polar,scaling=constrained);" "6#-%%plotG6%7 %,&\"\"\"F(*&\"\"#F(-%$cosG6#%&thetaGF(F(F./F.;\"\"!*&F*F(%#PiGF(/%'co ordsG%&polarG/%(scalingG%,constrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$7[p7$$\"\"$\"\"!F*7$$\"1f0J]DH) )H!#:$\"1rYrC4_\\?!#;7$$\"1]23ZlN`HF.$\"15N4miEqSF17$$\"1]$HJ+%[/HF.$ \"1SA\\W'fIy&F17$$\"1,i#3:H!RGF.$\"1')eWqhfMuF17$$\"1W**pcNfYFF.$\"12Z G7KJ(>*F17$$\"1Eb-&o&eNEF.$\"1&HZq#*yP3\"F.7$$\"1yK&yq\"y1DF.$\"1/Fxe] WM7F.7$$\"1XUR=)>GO#F.$\"1DWAr()po8F.7$$\"1)*zr6al1AF.$\"1xn,jJS%[\"F. 7$$\"1%prs?p*R?F.$\"1!)))pE?#4e\"F.7$$\"14paZ6=y=F.$\"1J&Ry^)Q_;F.7$$ \"1#[\\Rb9:r\"F.$\"1rTgl*[fq\"F.7$$\"1%Q)pLQ-O:F.$\"1CU$=)*=Au\"F.7$$ \"1(elD>m)f8F.$\"19.=il\"*eR!RR'f!#>$\"1]p*)zg0))**F17$$!1Bk#RB2@\"**!#<$ \"1`/D/O>2sF17$$!1#R$ybhsX7F1$\"1M_d4Q#)RXF17$$!1H#)G*H(Qe#*F^s$\"1Z6o ))[TsAF17$$\"1zY))4!RtE%!#=$!1*y+d`F%3tF^t7$$\"1jqk]rfb7F1$!1t.\")\\S* pl\"F17$$\"1.b81\\OiHF1$!1Y3\"f%F1$!1Z+0zh-eNF17 $$\"1qW,.>TPjF1$!1`K6#H20m$F17$$\"1PNZI521rF1$!15A-?v@9NF17$$\"14[gA15 6yF1$!1\\iI0rAaKF17$$\"1cM)owLEY)F1$!1pMd:-OuGF17$$\"1qo*oN+N,*F1$!1=J YmA#))R#F17$$\"1FOJz_%)>%*F1$!1fzu*Q*p#*=F17$$\"1_)3XV*F1$\"1/Q&)z>[q=F17$$\"1&)zolIWV!*F1$\"1')\\:J8@nBF17$$\"1 (e%oyrf=&)F1$\"1uH,\"Q8P$GF17$$\"1A%[C:V$)*yF1$\"1mr3]B87KF17$$\"1BE8Y fdrrF1$\"1,xc)*oa&\\$F17$$\"1K*e(y[6ujF1$\"11dCujRcOF17$$\"1uZ(Q$3$>o% F1$\"1$H6XTfsd$F17$$\"1iL0_OY,IF1$\"1h&)zOY-oHF17$$\"1,LSIJ\\o7F1$\"1: *[\"fw/q;F17$$!1(Gd[jSUH*Fhr$!1\"\\2b@8Qh\"F^t7$$!1C#y&*fa+E*F^s$!1\"p j[7aIF#F17$$!1u%)Ge$QwC\"F1$!18hC0$oqh%F17$$!1]\")f#Gy$z&*F^s$!1F'oVDW ZN(F17$$!1K&*\\U#)oQIF^t$!1*oG\"ze!)Q**F17$$\"1m7A*[p$>;F1$!1V@*fcuqC \"F.7$$\"1Tju8W$om#F1$!1)QMV2k\"f8F.7$$\"1W_N,')ogQF1$!1ZQoE[&3Y\"F.7$ $\"1t5!f,iUD&F1$!1&y.x$G8a:F.7$$\"1zv[O#o%ynF1$!1(o,Hg&)>j\"F.7$$\"1zh G691_$)F1$!1/tIi!\\2p\"F.7$$\"1N`q!R!*4+\"F.$!1rX\"G2TAt\"F.7$$\"1>*Rm qsq<\"F.$!1hpK08nbu\"F.7$$\"1UF>e8$zr\"F.$!1k#*)\\>SUq\"F.7$$\"1ys)32Q/)=F.$ !1kN#4QG:l\"F.7$$\"1!yCC@u#Q?F.$!14$3/Ay)eF17$$\"1R&GM\" 4l`HF.$!1frsBeddSF17$$\"1GjUInO))HF.$!1$QbBPRI/#F17$F($\"1T#3A9T7Y#!#C -%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "Notice that the graph enters and exits th e loop at the pole. If we set " }{XPPEDIT 18 0 "r=0" "6#/%\"rG\"\"!" } {TEXT -1 16 " in the equation" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "r=1+2* cos(theta)" "6#/%\"rG,&\"\"\"F&*&\"\"#F&-%$cosG6#%&thetaGF&F&" }{TEXT -1 8 " we get " }{XPPEDIT 18 0 "cos(theta)=-1/2" "6#/-%$cosG6#%&thetaG ,$*&\"\"\"F*\"\"#!\"\"F," }{TEXT -1 22 " which will occur for " } {XPPEDIT 18 0 "theta=2*Pi/3,4*Pi/3" "6$/%&thetaG*(\"\"#\"\"\"%#PiGF'\" \"$!\"\"*(\"\"%F'F(F'F)F*" }{TEXT -1 6 ". For " }{XPPEDIT 18 0 "theta " "6#%&thetaG" }{TEXT -1 18 " in the interval (" }{XPPEDIT 18 0 "2*Pi/ 3,4*Pi/3" "6$*(\"\"#\"\"\"%#PiGF%\"\"$!\"\"*(\"\"%F%F&F%F'F(" }{TEXT -1 15 ") the value of " }{TEXT 368 1 "r" }{TEXT -1 133 " is negative. \+ This causes the points along the loop to fall in the first and fourth quadrants, even though the associated values of " }{XPPEDIT 18 0 "the ta" "6#%&thetaG" }{TEXT -1 40 " fall in the second and third quadrants ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 164 "Giv en an equation of a limaon, the easiest way to determine which of the \+ three types of limaons its graph will be is to consider the largest an d smallest values of " }{TEXT 369 1 "r" }{TEXT -1 83 ". If these two v alues are either both positive or both negative, then the value of " } {TEXT 370 1 "r" }{TEXT -1 108 " is never zero and the limaon is the fi rst type considered above. If either the maximum or minimum value of \+ " }{TEXT 371 1 "r" }{TEXT -1 74 " is zero, then the limaon is in fact \+ a cardioid. If the maximum value of " }{TEXT 372 1 "r" }{TEXT -1 38 " is positive and the minimum value of " }{TEXT 373 1 "r" }{TEXT -1 41 " is negative, then the limaon has a loop." }}}{PARA 4 "" 0 "" {XPPEDIT 18 0 "r=a+b*sin(theta)" "6#/%\"rG,&%\"aG\"\"\"*&%\"bGF'-%$sin G6#%&thetaGF'F'" }}{PARA 0 "" 0 "" {TEXT -1 41 "Based on the discussio n above concerning " }{HYPERLNK 17 "rotations" 1 "" "rotations" } {TEXT -1 26 " we see that the graph of " }{XPPEDIT 18 0 "r=a+b*sin(the ta)" "6#/%\"rG,&%\"aG\"\"\"*&%\"bGF'-%$sinG6#%&thetaGF'F'" }{TEXT -1 70 " is a rotation of the foregoing. Consider, for example, the graph of " }{XPPEDIT 18 0 "r=4-5*sin(theta)" "6#/%\"rG,&\"\"%\"\"\"*&\"\"&F '-%$sinG6#%&thetaGF'!\"\"" }{TEXT -1 37 ". The largest and smallest va lues of " }{TEXT 374 1 "r" }{TEXT -1 81 " in this equation are 9 and - 1, corresponding to the extreme values -1 and +1 of " }{XPPEDIT 18 0 " theta" "6#%&thetaG" }{TEXT -1 34 ", so this is a limaon with a loop." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([4-5*sin(theta),theta,t heta=0..2*Pi],coords=polar,scaling=constrained);" "6#-%%plotG6%7%,&\" \"%\"\"\"*&\"\"&F)-%$sinG6#%&thetaGF)!\"\"F//F/;\"\"!*&\"\"#F)%#PiGF)/ %'coordsG%&polarG/%(scalingG%,constrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$7jo7$$\"\"%\"\"!F*7$$ \"1v/qGzH'G$!#:$\"1ymMs/7HX!#;7$$\"10l\"o.%>WEF.$\"1Jp(>OzV#pF17$$\"1f -9b[wS>F.$\"154\"zA<1)zF17$$\"1i5Kume#H\"F.$\"1J!y1#>]([(F17$$\"10c3R; #zS(F1$\"1PlpzF%4u&F17$$\"1USo'GH/L$F1$\"1N\"32B*f>LF17$$\"16^sXb17F!# <$\"1$GiUY;z]$FN7$$!1DJh'y(eQ;F1$!1WM<]'z*HGF17$$!14#G^\"oZrBF1$!1/qO. X+PdF17$$!1gR/V2oaBF1$!1xIj\\0$)>qF17$$!1\"3)y$>2h6#F1$!1,1,S&Rx5)F17$ $!17^cG\"R[v\"F1$!16U)z$)[&z))F17$$!18FLK?,%G\"F1$!1^![#4S$)e%*F17$$!1 (o-u0V)H)*FN$!1K.'3Gs_p*F17$$!1E$!0O)*QI*)*****F.7$$\"1u!=.P e0X$FN$!1(Q6y;tS'**F17$$\"1;%QUO)RxnFN$!13CcJ)3!f)*F17$$\"12k\\;)zw(** FN$!1BZXLh_&o*F17$$\"1_lXrZ&*)H\"F1$!1&)***)f<$[W*F17$$\"1_^+7n&R!=F1$ !1c.NXs!zz)F17$$\"1)4)eZKau@F1$!1+z$o\")HZ#zF17$$\"1IG:aijhBF1$!1hF-sn '[&pF17$$\"1olf$R$pyBF1$!1*)*pQNe$QeF17$$\"1O\"*fwAm-;F1$!1p?Q[!)yWFF1 7$$!1#*4H!\\o(*)*)!#=$\"19Y+6sO'=\"FN7$$!1SSFe#*fdLF1$\"1/'>PP\"HSLF17 $$!1Qf!Q\"Hj2uF1$\"1(R%y)f63u&F17$$!1wC_qd=)H\"F.$\"1b].X#e$)\\(F17$$! 1_3%H]zt\">F.$\"1>y9>R5))zF17$$!14@\\!)GyAEF.$\"1>2@s\\=!)pF17$$!1')fb RM#*)G$F.$\"1YF-B>K;XF17$$!1SH5\"HuW)RF.$\"1_k[9ZrN7FN7$$!1J=?)4gAj%F. $!1t\"e$Ra&>J'F17$$!1_!fnx*\\%4&F.$!1I[2yg`L8F.7$$!1v;K.&\\IH&F.$!1s*G $yTtgT,o7sFcF.$!14r'=5T#GKF.7$$!1B*\\C.!HOcF.$!1YK)))=!HYPF.7$$!1HsRHIb !f&F.$!1CKKc!)\\rUF.7$$!1\"HOZY\"[$\\&F.$!19q&oYS0y%F.7$$!1DK0([pSM&F. $!1;6aF<`%G&F.7$$!1Zh>b+E<^F.$!1`x5@([5$eF.7$$!1uiyL'yx#[F.$!1mF*\\ahg N'F.7$$!1cnV/*\\i^%F.$!15(\\FjGI!oF.7$$!1$GmroY'eTF.$!1T0C?3!4A(F.7$$! 1UQ'e$4aHPF.$!1;JLT%y\"HwF.7$$!1Wh]Y!*4cKF.$!1GMhqfq#*zF.7$$!1#3%o=F@$ z#F.$!1]*)ewN9z#)F.7$$!1\\%)>Ijj-BF.$!1wxjsEC@&)F.7$$!1eS88mASkF)F.7$$\"1&pMM19bI$F.$! 1ee89MOezF.7$$\"1#3\"y&Q^F.$!1H_zxEU\"Q\"F.7 $$\"1y7yVV$pj%F.$!1q$)yG_)*pjF17$$\"1Kz*e*******RF.$\"1hA\"H&[l\"G$!#C -%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "r=a*cos(n*theta),n=2,3,4" "6&/%\"rG*& %\"aG\"\"\"-%$cosG6#*&%\"nGF'%&thetaGF'F'/F,\"\"#\"\"$\"\"%" }{TEXT -1 4 " ..." }}{PARA 4 "" 0 "" {XPPEDIT 18 0 "r=a*sin(n*theta), n=2,3,4 " "6&/%\"rG*&%\"aG\"\"\"-%$sinG6#*&%\"nGF'%&thetaGF'F'/F,\"\"#\"\"$\" \"%" }{TEXT -1 4 " ..." }}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 71 "We will discuss these two curves together, since the y are realted by a " }{HYPERLNK 17 "rotation" 1 "" "rotations" }{TEXT -1 197 ". In fact, we will describe first how to sketch the second equ ation, then modify the procedure slightly to sketch the first. Our dis cussion will be easier to follow if we assign specific values to " } {XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "n" "6# %\"nG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The value of " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 35 " should be an integer. The case of " }{XPPEDIT 18 0 "n=1" "6#/%\"n G\"\"\"" }{TEXT -1 20 " has been discussed " }{HYPERLNK 17 "above" 1 " " "n=1" }{TEXT -1 34 " and is a circle, so that case is " }{TEXT 375 3 "not" }{TEXT -1 32 " included here. Suppose we take " }{XPPEDIT 18 0 "a=5" "6#/%\"aG\"\"&" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "n=3" "6#/% \"nG\"\"$" }{TEXT -1 35 ". The resulting figure is called a " }{TEXT 376 13 "3-leaved rose" }{TEXT -1 24 ", and is sketched below." }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "plot([5*sin(3*theta),theta,the ta=0..Pi],coords=polar,scaling=constrained);" "6#-%%plotG6%7%*&\"\"&\" \"\"-%$sinG6#*&\"\"$F)%&thetaGF)F)F//F/;\"\"!%#PiG/%'coordsG%&polarG/% (scalingG%,constrainedG" }}{PARA 13 "" 1 "" {GLPLOT2D 364 364 364 {PLOTDATA 2 "6$-%'CURVESG6$7jp7$\"\"!F(7$$\"1ymy6B!Q7&!#;$\"1a^ez)=]v \"!#<7$$\"1xL_mlc<5!#:$\"1#4ny#>)*ypF/7$$\"1gI76\\wX9F3$\"1dZ4*eF`U\"F ,7$$\"1r9hU!\\'e=F3$\"1]0YB*F,7$$\"1+sum]I_RF3$\"1\"He#)*H?_8F37$$\"1E/,9C([E% F3$\"1pfW*p+Dw\"F37$$\"1gbeauabVF3$\"1@X9VF37$$\"1%y]aS\"H(R%F3$\" 1&pX.$\\ogA%F3$\"1o0b.pS,(\\#F3$\"1yF()\\a])\\#F37$$\"1y&oGW9?:#F3$\"1%Q\">B'H]I#F37$$\" 18E1,%)z,=F3$\"16uA,:cm?F37$$\"1Tzp#R#>l9F3$\"1I0L#\\/hz\"F37$$\"13QR_ f&\\8\"F3$\"1pswj/M)[\"F37$$\"1l'zUQQ%\\%)F,$\"1tzIBDCz6F37$$\"1mGq@s. \"o&F,$\"1i,]p@0\\%)F,7$$\"1&Gs%F,7$$!1d_#R)o5F=F/ $!1XhfL%)f#=$F/7$$!1h`V]/5\"H#F,$!1.(p@\">O$G%F,7$$!1Lf_C^[UTF,$!1pzRe -WO$)F,7$$!19$fKO2,&fF,$!16%)Q8Q>58F37$$!1US'y(ftntF,$!1)*f\"**R\\_y\" F37$$!1WS9RN\"))G)F,$!11j'H$\\e)>#F37$$!1>^iZa1&*))F,$!1->adK()*f#F37$ $!1G$oTwiL?*F,$!1@Vq)pm.-$F37$$!1GJGI#o7:*F,$!1d(ecM1ST$F37$$!1Db0+Qa& y)F,$!1(*4:;SSePF37$$!1AcEjxcN\")F,$!1G%>AHV#oSF37$$!1rO$*Q0!f=(F,$!1+ 48-C-]VF37$$!1(zdhmbq*fF,$!1sh#R%o=&e%F37$$!1)*)pVZ[U)[F37$$!1(pqgy-ia#F,$!1())Q#eKtL\\F37$$!1, %eZO'oMhgp\\F37$$!1hXR:\"*['4*F/$!1%4Yz\"Gq\"*\\F37$$!1XP8> !*>`x!#=$!16(>n))R***\\F37$$\"1X_3)pNN(yF/$!1]A$44)y$*\\F37$$\"18^&=!* Gck\"F,$!16\"z))\\uE(\\F37$$\"1`x'>Bo+\\#F,$!1\"Gf+;&oO\\F37$$\"1v$GmE >NJ$F,$!1#eCKqmf)[F37$$\"13Xj\\\\?wYF,$!1/2sn\"oJw%F37$$\"1ju)yKG!=fF, $!1r([+]$*yf%F37$$\"1;z@v3LI!)F,$!1#oGY*o@1TF37$$\"1`#>kCn)Q()F,$!15'= y@Buy$F37$$\"1fUhN4*49*F,$!1p)Qdy@1V$F37$$\"1'p1-&*\\I F37$$\"1M#yi`[J%*)F,$!1c'3Tf;Nk#F37$$\"1xw%G&H))[$)F,$!1@dRJ>?JAF37$$ \"1,k*>fh+U(F,$!1.!f:))[c!=F37$$\"12nLAp2)*fF,$!1%pM\\x7WK\"F37$$\"1&H sy9>^<%F,$!1([(fw\\;9%)F,7$$\"1%>%\\:_56AF,$!1U,O8.%=7%F,7$$!1K(p(QI`M SF_z$\"1NbgJ5NzpF_z7$$!1pTvum'=t#F,$\"1kQT%R,jP%F,7$$!1G#f@JNEo&F,$\"1 `:HTi5^%)F,7$$!1)GFz=)3V&)F,$\"1N#o[#e%)*=\"F37$$!1G:JwqEa6F3$\"1ze#*) zwv]\"F37$$!1'Rh(ysO$\\\"F3$\"1%31(=\"f.#=F37$$!1d@fjgoQ=F3$\"1[VWp5q$ 4#F37$$!1bzYu(RS;#F3$\"1P'*4-KY7BF37$$!1t@yp9q%[#F3$\"15%fM!)4B\\#F37$ $!1d.)\\T'o!RFF37$$!1rf$e=5zR$F3$\"1E>&oy\"*fz#F37$$!1<' 4n1P@l$F3$\"1bJ^A&[B\"GF37$$!1i%fU4K`)QF3$\"1h2W9u!yy#F37$$!1F37$$!1hJZ([+1E%F3$\"1[>v(z@[v\"F37$$!1 t2-^bWaRF3$\"1\")*)zlOSa8F37$$!1\"=.sv^=S$F3$\"1$**fz6yP1*F,7$$!1lt5>w 7\"4$F3$\"1^f.Rit8sF,7$$!1\\@:MGSXFF3$\"1nIi*3\"=7bF,7$$!1YOkNuTUBF3$ \"1\"yRaeI&**QF,7$$!1m#y3$z!*3>F3$\"11#4Bf8-`#F,7$$!1EF.7$$\"1I%)Rfa`EVF.$\"1dnsp/@1DF.7$ $\"1GvV;g7_RF.$\"1w9,NGziIF.7$$\"1v'>V3(GTNF.$\"1#=Y/Nr(HNF.7$$\"1k'fJ (yAeIF.$\"121wC&fc&RF.7$$\"19P:ydQ0DF.$\"1(f%QpI,FVF.7$$\"17aLA)p+\">F .$\"1;+`bFy?YF.7$$\"1L26QJpi7F.$\"1#pSs1Mz$[F.7$$\"1=c28ToDnF1$\"1*zr= oeX&\\F.7$$!1&=@q)[`&)H!#=$\"1l!e'3\"*****\\F.7$$!1!\\2iz6C\"oF1$\"1M( 4QktL&\\F.7$$!1y&y^2\"4B8F.$\"1h4fuiw@[F.7$$!1!>KbYal)=F.$\"1!R6zEL/j% F.7$$!1\\7\"ojd6_#F.$\"1@$Rl`UyJ%F.7$$!1J!y\"QZu>IF.$\"1\"3*>>]6&)RF.7 $$!1B@Qt\"eYa$F.$\"1zP=IgQENF.7$$!1\"HwF5-@&RF.$\"1ae:#pBG1$F.7$$!1T^! *pLlHVF.$\"1Zmu/-#3]#F.7$$!1F#ol8oah%F.$\"1MV(*fk(G#>F.7$$!1`>qa<\"=$[ F.$\"1ZtI8>#fG\"F.7$$!1*[3(4]^`\\F.$\"1:-ED]7-oF17$$!1I;kaf(***\\F.$\" 17hC=Zl]:!#<7$$!1E:O4!>U&\\F.$!1ia*4Gf1v'F17$$!1wniLF.P[F.$!1ox89:9m7F .7$$!1xl^16jJYF.$!1b#f\"oMl*47e\\F.7$$! 1RRo9&Gn? 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As a n additional aid in sketching the petals, we present the following use ful fact concerning polar graphs." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 39 "Suppose the polar graph of an equation " }{XPPEDIT 18 0 "r=f(theta)" "6#/%\"rG-%\"fG6#%&thetaG" }{TEXT -1 29 " \+ passes through the pole for " }{XPPEDIT 18 0 "theta=theta[0]" "6#/%&th etaG&F$6#\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 "(" } {TEXT 380 4 "i.e." }{TEXT -1 9 " suppose " }{XPPEDIT 18 0 "f(theta[0]) =0" "6#/-%\"fG6#&%&thetaG6#\"\"!F*" }{TEXT -1 17 "). Then the line " } {XPPEDIT 18 0 "theta=theta[0]" "6#/%&thetaG&F$6#\"\"!" }{TEXT -1 74 " \+ is tangent to the graph at the pole. (We are assuming that the functi on " }{XPPEDIT 18 0 "f(theta)" "6#-%\"fG6#%&thetaG" }{TEXT -1 22 " is \+ differentiable at " }{XPPEDIT 18 0 "theta[0]" "6#&%&thetaG6#\"\"!" } {TEXT -1 2 ".)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "As an example of this consider the limaon " }{XPPEDIT 18 0 "r=1+2*cos(theta)" "6#/%\"rG,&\"\"\"F&*&\"\"#F&-%$cosG6#%&thetaGF&F& " }{TEXT -1 9 " graphed " }{HYPERLNK 17 "above" 1 "" "loop" }{TEXT -1 20 ". We saw there that " }{XPPEDIT 18 0 "r=0" "6#/%\"rG\"\"!" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "theta=2*Pi/3,4*Pi/3" "6$/%&thetaG*(\"\"# \"\"\"%#PiGF'\"\"$!\"\"*(\"\"%F'F(F'F)F*" }{TEXT -1 49 ". Suppose we r edraw the limaon and add the lines " }{XPPEDIT 18 0 "theta=2*Pi/3" "6# /%&thetaG*(\"\"#\"\"\"%#PiGF'\"\"$!\"\"" }{TEXT -1 4 " and" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "theta=4*Pi/3" "6#/%&thetaG*(\"\"%\"\"\"%#PiGF' \"\"$!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {GLPLOT2D 315 317 317 {PLOTDATA 2 "6'-%'CURVESG6$7S7$$\"1*************\\#!#;$!1%>A*=q7IVF*7$ $\"1***\\PM@lL#F*$!11TM!otp/%F*7$$\"1*\\i!R%zU>#F*$!1')*fWZ.1!QF*7$$\" 1**\\iSNJM?F*$!1My)3TMN_$F*7$$\"1**\\P%G(Gt=F*$!1G-r`(GYC$F*7$$\"1*\\7 yNEIr\"F*$!18fV'o[q'HF*7$$\"1*\\iS'RWk:F*$!142NJkp4FF*7$$\"1*\\i:]&f59 F*$!1aVjxIAVCF*7$$\"1*\\i:?&[^7F*$!1Go-afjn@F*7$$\"1+DJq^)G4\"F*$!1;)> =kKH*=F*7$$\"1&***\\PaX(H*!#<$!1_d))>lO5;F*7$$\"1'**\\(=2_gyFgn$!1$G#) e7#[h8F*7$$\"1'***\\iD(GC'Fgn$!1rFMYsH\"3\"F*7$$\"1)***\\(=#e=YFgn$!13 Eo2!>'**zFgn7$$\"1(***\\PuF`IFgn$!10PN^;V)G&Fgn7$$\"1+]il2$=j\"Fgn$!1@ lT&z8k#GFgn7$$!1L++]7oSe!#>$\"1'\\\"encj65!#=7$$!1+++](f-\\\"Fgn$\"1%* [Z.c?\"e#Fgn7$$!1)*\\P4^%e:$Fgn$\"1g_DqS3maFgn7$$!1'****\\Pq,j%Fgn$\"1 R-?P.p>!)Fgn7$$!1**\\Pf)QxC'Fgn$\"1$*fo8,9#3\"F*7$$!1**\\7yl/)y(Fgn$\" 1WD]O#H*[8F*7$$!1)**\\P4'>&R*Fgn$\"1lZ9)p&HF;F*7$$!1+vo/h5(3\"F*$\"1Tq `1I#H)=F*7$$!1**\\7y!)HY7F*$\"1Yzo#f^'e@F*7$$!1*\\P%)z`;T\"F*$\"1X8f,h 0XCF*7$$!1*\\i!Rkfb:F*$\"1t%H&o?P%p#F*7$$!1**\\PM%e5r\"F*$\"1en0V,kjHF *7$$!1****\\()emr=F*$\"1%\\V>T?=C$F*7$$!1**\\(o%*)yG?F*$\"1c;#Q`lR^$F* 7$$!1*\\Pfe83=#F*$\"1E-oK*zsx$F*7$$!1**\\7G5h\\BF*$\"1YmtyckpSF*7$$!1) ***\\PIG,DF*$\"14-10$\\BL%F*7$$!1***\\7GDKm#F*$\"18p7*\\TGh%F*7$$!1*\\ PM-p*4GF*$\"1^%)G;\"4q'[F*7$$!1***\\Pp)RqHF*$\"1F7Jc9)[9&F*7$$!1)\\P4^ ]87$F*$\"1C[o&yf\\'F*7$$!1)*\\i:Q_4RF*$\"1SI1#)Q\\rnF*7$$!1*\\Pf[vs1%F* $\"1YJ-!zFZ/(F*7$$!1****\\()GB7UF*$\"1gDYu8!eH(F*7$$!1)*\\7G/PyVF*$\"1 *fpe.gNe(F*7$$!1)****\\#e'p_%F*$\"1gz-8[$4%yF*7$$!1)\\P4^*R&o%F*$\"1?@ q1+N:\")F*7$$!1)\\iSwSq$[F*$\"1L!Q;O+!y$)F*7$$!1)*************\\F*$\"1 (QWy.a-m)F*-%'COLOURG6&%$RGBG\"\"!F^[lF^[l-F$6$7S7$$\"1-++++++DF*$\"1# >A*=q7IVF*7$$\"1-+vV8_OBF*$\"1/TM!otp/%F*7$$\"1-D1R%zU>#F*$\"1&)*fWZ.1 !QF*7$$\"1-]iSNJM?F*$\"1Ky)3TMN_$F*7$$\"1,]P%G(Gt=F*$\"1E-r`(GYC$F*7$$ \"1,D\"yNEIr\"F*$\"17fV'o[q'HF*7$$\"1-D1kRWk:F*$\"132NJkp4FF*7$$\"1,Dc ,bf59F*$\"1`VjxIAVCF*7$$\"1,Dc,_[^7F*$\"1Fo-afjn@F*7$$\"1,DJq^)G4\"F*$ \"1:)>=kKH*=F*7$$\"12+]PaX(H*Fgn$\"1_d))>lO5;F*7$$\"12+v=2_gyFgn$\"1#G #)e7#[h8F*7$$\"1/+]iD(GC'Fgn$\"1rFMYsH\"3\"F*7$$\"10+](=#e=YFgn$\"10Eo 2!>'**zFgn7$$\"1-+]PuF`IFgn$\"1.PN^;V)G&Fgn7$$\"1-]il2$=j\"Fgn$\"1?lT& z8k#GFgn7$$!1S++]7oSeFfp$!1'\\\"encj65Fip7$$!1-++](f-\\\"Fgn$!1$*[Z.c? \"e#Fgn7$$!1-]P4^%e:$Fgn$!1e_DqS3maFgn7$$!1-++v.!)Fgn7 $$!12]Pf)QxC'Fgn$!1#*fo8,9#3\"F*7$$!14]7yl/)y(Fgn$!1WD]O#H*[8F*7$$!16+ v$4'>&R*Fgn$!1lZ9)p&HF;F*7$$!1,vo/h5(3\"F*$!1Tq`1I#H)=F*7$$!1,]7y!)HY7 F*$!1Xzo#f^'e@F*7$$!1,vV)z`;T\"F*$!1W8f,h0XCF*7$$!1,D1Rkfb:F*$!1s%H&o? 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The \+ crucial observation is that " }{TEXT 381 70 "the tangent lines at the \+ pole for the 3-leaved rose will coincide with" }{TEXT -1 1 " " }{TEXT 382 39 "sector lines that we have already drawn" }{TEXT -1 63 ", since our sector lines were determined by setting the factor " }{XPPEDIT 18 0 "sin(3*theta)" "6#-%$sinG6#*&\"\"$\"\"\"%&thetaGF(" }{TEXT -1 16 " equal to zero. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "So here is a system for sketching the 3-leaved rose. Begi n at the pole with " }{XPPEDIT 18 0 "theta=0" "6#/%&thetaG\"\"!" } {TEXT -1 29 ". With increasing values of " }{XPPEDIT 18 0 "theta" "6# %&thetaG" }{TEXT -1 52 ", enter the first sector tangent to the sector line " }{XPPEDIT 18 0 "theta=0" "6#/%&thetaG\"\"!" }{TEXT -1 52 ". Gr adually turn and approach the enveloping circle " }{XPPEDIT 18 0 "r=5 " "6#/%\"rG\"\"&" }{TEXT -1 104 " tangentially at the middle of the se ctor and then turn and reenter the pole tangent to the sector line " } {XPPEDIT 18 0 "theta=Pi/3" "6#/%&thetaG*&%#PiG\"\"\"\"\"$!\"\"" } {TEXT -1 45 ". Now exit the pole entering the next sector " }{TEXT 383 24 "with negative values of " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 360 " and proceed in similar fashion tangential to the enveloping c ircle at the middle of the sector. Then turn around and reenter the p ole tangent to the next sector line, and so on, and so on. By followi ng the sector lines through the pole we are led to describe the three \+ petals of the 3-leaved rose. The following plots illustration the step s described above:" }}{PARA 0 "" 0 "" {GLPLOT2D 268 244 244 {PLOTDATA 2 "6)-%'CURVESG6$7S7$$!1+++++++D!#:$!1$>A*=q7IVF*7$$!1nm;HU,\"R#F*$!1- o`E\"e89%F*7$$!1M$3FH'='H#F*$!1bShA86xRF*7$$!1nmTgBa*=#F*$!1`fc8')R#z$ F*7$$!1nm\"H_\">#3#F*$!1$)3y3:Y1OF*7$$!1M$3_!4Nv>F*$!11!)fI\"39U$F*7$$ !1n;/wfHw=F*$!1.7ag*R)\\KF*7$$!1,]PM.ttsIF*7$$!1n;/,oln;F* $!1\\>**3jY))GF*7$$!1+](oWB>c\"F*$!131_n2L0FF*7$$!1LL$epjJX\"F*$!1**y* GN`p^#F*7$$!1nm\"z/otN\"F*$!1`AcB/.^BF*7$$!1++]P[_\\7F*$!1X#pQ]SU;#F*7 $$!1++]7)Q79\"F*$!1r&>og$ow>F*7$$!1++]i^)o.\"F*$!16V1$yPfz\"F*7$$!1omT 50A@%*!#;$!1l^LfK!=j\"F*7$$!1NLLeaR%H)F`p$!1)=!>N9jO9F*7$$!1LLLLo#)RtF `p$!1r!4%*H&Hr7F*7$$!1,+v$fO%HiF`p$!1!Q!Ho+(*y5F*7$$!1NLL$3`lC&F`p$!1% [S\\]'H(3*F`p7$$!1)**\\P4u\"oTF`p$!1bt;0$*[>sF`p7$$!1***\\7G-89$F`p$!1 :q0`^*3W&F`p7$$!1HL$3Fp)p?F`p$!1q)3@u=^e$F`p7$$!1LL3-$ff3\"F`p$!1q.;'o O4)=F`p7$$!18nm;z%zY#!#=$!1305#=6YF%F^s7$$\"1OLe*)>px5F`p$\"1F$o3G>&F`p$\"1IPSi,o#**)F` 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If " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 29 " is allowed to incre ase from " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "2Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 95 " the rose is traced \+ out a second time. This sort of thing always happens whenever the int eger " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 22 " is odd. Half of th e " }{XPPEDIT 18 0 "2n" "6#*&\"\"#\"\"\"%\"nGF%" }{TEXT -1 23 " sector s (each of size " }{XPPEDIT 18 0 "Pi/n" "6#*&%#PiG\"\"\"%\"nG!\"\"" } {TEXT -1 25 ") will be empty, and the " }{XPPEDIT 18 0 "n" "6#%\"nG" } {TEXT -1 30 " petals will be traced out as " }{XPPEDIT 18 0 "theta" "6 #%&thetaG" }{TEXT -1 21 " increases from 0 to " }{XPPEDIT 18 0 "Pi" "6 #%#PiG" }{TEXT -1 17 ". However, when " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 43 " is even, the petals will be traced out in " }{TEXT 384 3 "all" }{TEXT -1 21 " sectors. Once again " }{XPPEDIT 18 0 "n" "6#%\" nG" }{TEXT -1 30 " petals will be traced out as " }{XPPEDIT 18 0 "thet a" "6#%&thetaG" }{TEXT -1 21 " increases from 0 to " }{XPPEDIT 18 0 "P i" "6#%#PiG" }{TEXT -1 9 ", but as " }{XPPEDIT 18 0 "theta" "6#%&theta G" }{TEXT -1 16 " increases from " }{XPPEDIT 18 0 "Pi" "6#%#PiG" } {TEXT -1 4 " to " }{XPPEDIT 18 0 "2Pi" "6#*&\"\"#\"\"\"%#PiGF%" } {TEXT -1 15 " an additional " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 40 " petals are traced out in the remaining " }{XPPEDIT 18 0 "n" "6#% \"nG" }{TEXT -1 47 " sectors, rather than a retracing of the first " } {XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 57 " petals. The neat part abou t the method outlined above (" }{TEXT 385 4 "i.e." }{TEXT -1 72 " foll owing the sector lines through the pole) is that the generation of " } {XPPEDIT 18 0 "2n" "6#*&\"\"#\"\"\"%\"nGF%" }{TEXT -1 27 " petals for \+ even values of " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 45 " is automa tic. Try it for the example below (" }{XPPEDIT 18 0 "r=sin(2*theta)" " 6#/%\"rG-%$sinG6#*&\"\"#\"\"\"%&thetaGF*" }{TEXT -1 2 ")." }}{PARA 0 " > " 0 "" {XPPEDIT 19 1 "plot(\{[1,theta,theta=0..2*Pi],[sin(2*theta),t heta,theta=0..2*Pi]\},coords=polar,scaling=constrained,axes=none);" "6 #-%%plotG6&<$7%\"\"\"%&thetaG/F);\"\"!*&\"\"#F(%#PiGF(7%-%$sinG6#*&F.F (F)F(F)/F);F,*&F.F(F/F(/%'coordsG%&polarG/%(scalingG%,constrainedG/%%a xesG%%noneG" }{TEXT -1 2 "\n\n" }}{PARA 13 "" 1 "" {GLPLOT2D 268 244 244 {PLOTDATA 2 "6&-%'CURVESG6$7S7$$\"\"\"\"\"!F*7$$\"1#4hRPij!**!#;$ \"1kwb#=y_O\"F.7$$\"18J#))4-Qn*F.$\"1D\\ff*)GLDF.7$$\"1N5')yke[#*F.$\" 1Mj&[K5J!QF.7$$\"1goz=42`')F.$\"1:NXR4U7]F.7$$\"1b](G._U!zF.$\"1`H-qce DhF.7$$\"1]$R'oTd#3(F.$\"1jB*3qU&fqF.7$$\"1G$>jubk6'F.$\"1:7_\\!>8\"zF .7$$\"1GuIc:x5]F.$\"1$>p(Qh-a')F.7$$\"1C3nW'R,#QF.$\"1J+16bcT#*F.7$$\" 1n9AwiQDDF.$\"1&Q\"[M\"oen*F.7$$\"1C^hAo8X8F.$\"1*fVPO<\"4**F.7$$!1qB/ u(p5(f!#>$\"08;t@)******!#:7$$!1)\\T#fB[i8F.$\"1q%>wGZn!**F.7$$!1crN]@ =YEF.$\"1A>=\\D`V'*F.7$$!1zV1J*3Jx$F.$\"1!yAe`m3E*F.7$$!1(\\AOF:B/&F.$ \"1V'yI2&oN')F.7$$!1igNw%*[RgF.$\"1j\")RQ+BqzF.7$$!1XUwYjJ*3(F.$\"1fvO g?x_qF.7$$!1#e_b?/U!zF.$\"14HvXQF.7$$!12RS4Nij'*F.$\"1%p9m#Q%=d#F.7$$!1 xpT>+.2**F.$\"1V?00]Ug8F.7$$!1gKG4>&*****F.$\"1DA\\O%485$!#=7$$!1^Is=! 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180&yw8Ul$Fep7$$!1oknC9?sVF.$!1m=0\\-(\\G)F_p7$$!1^A$pa<)=NF.$!12Gd&>Z NZ\"F.7$$!1z(o/pdWq#F.$!1C]_*38!z\"F.7$$!1g=V.Ox+HF_p$ !1&zR([p\\j%*F_p7$$!1NdX\"H1fQ\"F_p$!1pe&*QjkvmF_p7$$!1pfcth*[k$Fep$!1 -o^HFPpMF_p7$$!1wb'QQ%e8s!#@$!1rp#Q!)R1b\"Fep7$$!1bTwv_9xKFep$\"1E-r.y D\"H$F_p7$$!1n)QF7Y]O\"F_p$\"1]nDL&Qqi'F_p7$$!1F.Xo7SFGF_p$\"1G=F.7$$!18E1$eM!fEF.$\"1ZcD2(H')y\"F.7$$! 1;L(R7$$!16*y*HF+jaF.$!1]=Viq_O6F.7$$!1NU9z\">Qi&F. $!1hIi_T`3BF.7$$!12*HT\"*o*yaF.$!18V=F@PxOF.7$$!1)\\77W$>:]F.$!1YVcRHS p\\F.7$$!1r$*Qb_J!>%F.$!142'*HGPNiF.7$$!1Io7$z#emOF.$!1T>nr.#ez'F.7$$! 1yBK*pb'pIF.$!1M#>M8uUI(F.7$$!1c%\\?**oGP#F.$!1D*=&)=k1x(F.7$$!12iv\") ew5;F.$!1N%3X,G*f\")F.7$$!12\"pN%HpR#)F_p$!1>l`6`u`%)F.7$$\"1])zm_`>& \\Fh[l$!1^G2k`?h')F.7$$\"16f*>LNO&))F_p$!1/[jElNy()F.7$$\"1Vo6I94(y\"F .$!1lt.D'e]z)F.7$$\"1(zM\"fr>!p#F.$!1s9@cYd4()F.7$$\"14P'4zc'*e$F.$!1A j%\\(4?@&)F.7$$\"1(QOWN!>-WF.$!1Ayh/>kd#)F.7$$\"1#*Q7i5P\">&F.$!1X:/-6 *)3zF.7$$\"1O?P&3;\\0'F.$!1&=w?9axS(F.7$$\"1X\"=E*=ehoF.$!1ijS9Nq.oF.7 $$\"1D4pcmGCvF.$!1IZ@Q_D#='F.7$$\"1@7HNZcA\")F.$!1*H/$oc!3\\&F.7$$\"1m rZVZw!o)F.$!1$H(Gr[$[o%F.7$$\"1F5w]hk[\"*F.$!1Jlvhe\"y\"QF.7$$\"1R*)>m NW0&*F.$!1_M#\\D**4%HF.7$$\"1$pUrca#o(*F.$!1%ej=\"zyG?F.7$$\"1R;8_O$=% **F.$!1+x<'o>:-\"F.7$F($\"1@T5r0iI7!#C-%'COLOURG6&%$RGBG$\"#5!\"\"F*F* -%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Wow! This looks like a limaon with a loop, except that the loop i sn't at the origin. I wonder if it really is?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "How about something crazy like " }{XPPEDIT 18 0 "r=cos(theta)*sin(2*theta)" "6#/%\"rG*&-%$cosG6 #%&thetaG\"\"\"-%$sinG6#*&\"\"#F*F)F*F*" }{TEXT -1 1 "?" }}{PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "plot([cos(theta)*sin(2*theta),theta,theta=0..2*P i],coords=polar,scaling=constrained);" "6#-%%plotG6%7%*&-%$cosG6#%&the taG\"\"\"-%$sinG6#*&\"\"#F,F+F,F,F+/F+;\"\"!*&F1F,%#PiGF,/%'coordsG%&p olarG/%(scalingG%,constrainedG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 374 374 374 {PLOTDATA 2 "6$-%'CURVESG6$7[w7$\"\"!F(7$$\"1'>zr;4W$o!#<$\"1-F(4l S4M#!#=7$$\"1$zi@6'))e8!#;$\"1&zR)yD#*>$*F/7$$\"1KFj*3n%=?F3$\"1\\AOsY i!3#F,7$$\"1p`]CvcaEF3$\"1ka.m-[eOF,7$$\"1H[$f*>G$=$F3$\"1Nh\"QSkxN&F, 7$$\"1hF_nCD%o$F3$\"1\"GI2%4kNtF,7$$\"17jK\"RML:%F3$\"18(>@*p0k&*F,7$$ \"1C\\@DKw'e%F3$\"1\")pv=19,7F37$$\"1OXzjbYF]F3$\"1OyQ5j>)\\\"F37$$\"1 GnHM%4[T&F3$\"1)Q(*H*o@8=F37$$\"1M537H`XdF3$\"1oTUXKbS@F37$$\"1)*3$\\t 2s,'F3$\"1'3&R$RNVZ#F37$$\"1>_\"f!>YHiF3$\"1gh2x9y5GF37$$\"1)[>[X#fzjF 3$\"1*GN3W!eTJF37$$\"1%)\\;OIznkF3$\"1@h\"GV@2Y$F37$$\"1@sipq8&\\'F3$ \"1[-yMaSiPF37$$\"1(*)Q#>;$QY'F3$\"1&pnW@)))RSF37$$\"1lJ;vE$oP'F3$\"1L $[$y]l*G%F37$$\"1qZaOFbPiF3$\"1k`&*e0@2XF37$$\"1hiqtP1]gF3$\"1&px.&*Q' )o%F37$$\"1D\\#*Hh@PeF3$\"14@.R[r@[F37$$\"1kf*4\"z+\"f&F3$\"1`I**4Q')= \\F37$$\"1+OJpHvcaF3$\"10A$H]tM&\\F37$$\"1z6%*R+#eJ&F3$\"1w\")p]Dey\\F 37$$\"1v4'QW\"zo^F3$\"1VS\"Hs$4%*\\F37$$\"1L_hw!fi,&F3$\"1J+Hbp%***\\F 37$$\"1#*3\\BEZ&o%F3$\"1a0\\cUI\")\\F37$$\"1F>%\\iH!RVF3$\"1*3z*3!*\\@ \\F37$$\"1 SuwP?DF3$\"1!ysz%f[OSF37$$\"1=6ny#>v<#F3$\"1olk.*f2w$F37$$\"1A'y!46ra= F3$\"1%HdyF_OY$F37$$\"1$RBVS2Ub\"F3$\"1,&yNKd&\\JF37$$\"1\"=Q&[Cvy7F3$ \"1\\0AG>,CGF37$$\"1ynGxvTI5F3$\"1$*\\NG[v#\\#F37$$\"1\")Qs&zAr/)F,$\" 1bE`,x?_@F37$$\"1WLMs$f')4'F,$\"1MNl=p9==F37$$\"1,^H'[F!eWF,$\"1k6?IF! o\\\"F37$$\"1#)*yo;gn6$F,$\"1U.`@#oT>\"F37$$\"1e#pq\\^)H9F,$\"1Jbd$y*4 NsF,7$$\"1>?X&)H_B[F/$\"1)Qit23Lb$F,7$$\"1V0(\\0F>.'!#>$\"1-Ys\"4[v'*) F/7$$!1%G,Nx5yD%!#D$\"1/v#pIK28(!#A7$$!1x'fF[Ed-\"F37$$!1\\.Z\\(fP d$F,$\"1&)o$\\9#R-8F37$$!10>q*\\2)>[F,$\"1'>B76\"er:F37$$!1[o:#\\#G&H' F,$\"1H%*3x,#R&=F37$$!1AZR.J#\\+)F,$\"1c)z(*H5a9#F37$$!1L:\\j]+\\**F,$ \"1J=A@G#>W#F37$$!1e+fn>'fD\"F3$\"1#p3\")eH_z#F37$$!1@VJNY)za\"F3$\"1t BFOmiUJF37$$!1CYjor]o=F3$\"1sx%=**frZ$F37$$!1gzeq'*>9AF3$\"1;E@yJ8#z$F 37$$!1Mj5K9oBDF3$\"1(y%oly&*QSF37$$!1d*fP`<`%GF3$\"19&\\R&[mjUF37$$!1@ =,uU\"e<$F3$\"1cn%QoCIY%F37$$!1AE0&or:^$F3$\"1M\"HA=sTj%F37$$!12'y;aDO !RF3$\"1B[%el@Xz%F37$$!1$Qp&*f]6H%F3$\"1XWK7(o-\"\\F37$$!1.&\\V.qum%F3 $\"15\\X%Qj\"z\\F37$$!11)3r]td-&F3$\"1O\\S_k')**\\F37$$!1lEzXw#o<&F3$ \"1f'o+iaF3$\"1) 3YIUDB&\\F37$$!1Pq[?c)\\f&F3$\"1&e@&o?m<\\F37$$!1'pNL?R*QeF3$\"1g\"Rlj \\3#[F37$$!10Z(G!\\,]gF3$\"1pv3%ow')o%F37$$!1BU6^W/RiF3$\"1x#[R)yQ0XF3 7$$!1qSbc6#*yjF3$\"1+*pF')G`G%F37$$!1SmIZpYlkF3$\"1eY3gX`KSF37$$!1B#zh f*=&\\'F3$\"1zBAS,k^PF37$$!1&>7=\\m\"okF3$\"1s4qE&\\FY$F37$$!1o.qfIY&Q 'F3$\"1+,Ul*\\y:$F37$$!1'>=&*3=hC'F3$\"1TXpO9,UGF37$$!175Zxg!*\\gF3$\" 1S4&z\"e[?DF37$$!1jtu;L8&y&F3$\"10[INrs%=#F37$$!1rsgVs\\gaF3$\"1J9%*)3 vY&=F37$$!1ngcTuMy]F3$\"1boeFQIO:F37$$!1#y@fJ()=k%F3$\"1sOWqgPN7F37$$! 1L#y!Q9n'>%F3$\"1q0_@vo!z*F,7$$!166my*3Fr$F3$\"1_-,Gu#)fuF,7$$!1%zF.`p T>$F3$\"1Dk-$ennR&F,7$$!1Vf;NhmXEF3$\"1i;#*=k+LOF,7$$!1v@L/%)>E?F3$\"1 qY!*\\t%p4#F,7$$!1xE\"zrzVQ\"F3$\"1()\\-4uRw'*F/7$$!1s\\x/GOssF,$\"1.) =(>wR^EF/7$$!1;')*oQHD?'F/$\"1Gy_UbgB>!#?7$$\"1cu$HF*p&H'F,$\"1a6$zdOd )>F/7$$\"1d$3\"))*GOJ\"F3$\"1u7vr?./()F/7$$\"1!f^#4b*>)>F3$\"1@QPG/a/? 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