00020001010008main.ACT000202000cImp_Diff.EAC01000000244cÉ ųś [[-3],[-2]]ü÷Ī÷ŹĄ# #`g—u#`g—u ü ų ś5ī’x^2-5ī’y^2-25=0  ÷ ČĄ  $&@hq‚„'GFüųś8ī’x^2+18ī’y^2-144=0 ÷ČČ  @Š6–°2”3xY™Tp–"R ™8gPIc`üųś y=1.5ī’x+2.5÷ČĄ %`@Š6–°Tq“–”) ™2t“S– ™2I—AXüųśy=-0.6667ī’x-4††bVVxXˆ `59RC@†`Œ†Ž0†ˆ:† ˆ<v’K†r†  A† 'LŽ'ˆpC˜ u˜x˜yŽ †qˆq† †Š† †††ŃČČ `Š0ˆ½Š6–°†7ˆ7ŒF†Ļ– ’ Ž ˆ(ˆGˆOˆWˆ_ˆgˆ’E  final.EAC† main.ACT†††&ˆ)†-Š†@ ~E Ž'N[ˆ4Implicit DifferentiationŽ \Œ( Authorķˆ?†CķŽ€†€ˆ§Professor Wei-Chi Yang ŒRadford University Ž9e-†Öl: wy†0@rŒ).edu""URL: http://www.–$/Š6Žf‹Ž¶ Sound TrackŒ» »"Professor Jonathan Lewin Ž Kennesaw Sate University ŽA e-mail: lˆ6s@mindspring.comŽFURL: http://www.m†j-moviesˆ'†Š[Š‘Ž Object†ns:[Ž¬7S1. We will use ClassPad (CP) to explore the graphs of †Ņimplicit equation†d its corresponding derivaˆ}.|T2Š|b one problem ‡4t c‡8be fouˆYn regular Calculus texbooksŠzweŠĮƒ1CPœ½is’\ˆ§extendŠÕresult.‡, [† Note.ˆ 1L We will explore some implicit equations that a†!graphablˆ8thin ClassPad.[ŒT’i Exa†Ie 1Žn= WˆFiˆNŽI ofŠ deriv†kve (dy/dx) for x^2-y†5=0”_Œ»HStep†[ˆĀ drag and dropŠ]¹ intoŠo following Ge†ä try Strip.\ DšE–zĪˆy†}Ī‰J‰Oˆ‹‹R™™™™˜H† `@”Gs#Š Š.†† Ž!Š ŽD‰Žv’K†r† Ž A† †L††ˆ C˜ u˜x˜yŽ ˆ † †JˆH  ŽN†\D€s™€†f ` Œ FT'@Œ7`XpŒ$i!22`Œ03Ž0%YDiPŒT’2YtŽ<upQŽH "ix“$…dŽ` eTcbŠ£`57•b0Š F y– Š ”$”<–T–l–„–œ–“–Ģ–ä–ü——,—D D€s™€ˆ† …f5PŒgDŒ qds0Œ,2x˜pŒ8yu’)`ŒDhQG"@ŒPDI’ŠH`SŠ0`T†„ #gv„Šl`t`IŠH`€pƒ • VŠT`T”RˆÆ` ˜SŠxY™– ™˜$š<˜T˜l˜„–œš“–Ģšä–ü›™,™D—\† ‡z5ÅH ˆ ‰†!‡ŒĘČ ˆA‰LŠ6–°†F y’C‹Q ‡S†CQ††Š†"†ĖČ#… žœ($U‰y&ˆB²("ex•‘$rA7“ˆXY™ˆŒ(`” D‚u† i†$ ™†#†£.Č@ ’Ž$†µ3ÉČ"’ˆ²!†%†,ČČ €Œ@†ŽŠ6–°†åXf2`ŠzbeGCŠ†‡7d‡†’†3ˆĻ.ČH!œ\ˆr'¢ˆˆrˆˆ¢ˆzˆ"‰ˆˆ(‡]1Ä 2ˆPHŽ )†ŹÉȖ"‡6$†…9‹(IsˆÉ‡Ø •——P‹&ˆC@–p€Ž ‰D‰x ƒ1u$v‘PxpŽ0p–se`Ž<e")c0ŽHYgP@PŽTT1i™TI$i0DR–x91•7T4e”T`0•Žœ%CŠ§ˆØ!a‡‰ØVxfegQĢ ˆ•0#UäqePŽT “t5Œš `ARŽ (vUgŒx˜"QQŒähu(…,@17!8ˆ†Qr‹`uphŠ ˆ yR†‹ˆƒbI’ ˆUŠ€ˆ0’r <—veHBT—‹|`Y `!„…P (™€6e`$D†(@0Sf‡’$c Hs#’’<„ ’`•u`$yˆwˆx!eApŽl6slQGTh„…€CŽ“ˆx%6ĢG4i0ŽŲp•cĢ–1#Žš†*†śŹĄ!Œ7ˆ‡ $†…9Šb(IsŠņ†  •——‹"†ˆCŽ–p‹.‰D‰x ƒ1uŽ<v‘PxŠęp–se‹Re")cŠŖYgP@Ž`T1i™TI$i0DR–Ž91•7p4e”T 0•Ž%C@ˆ#ˆ$!a‡‰€Ž0VxfegQŠ%†H ˆ•0#UqePŽT “t5ŒH `ARŽ (vUgŒx˜"QQ`Ž hu(…Š…@17!Ž ††Qr0ˆ`uphŠ ˆ yR†PƒbI’ ˆUŠ€ˆ0’r <—veHBT—‹(ˆY0!„…`(™T6eŠ¼ˆHD†(Sf‡xc ŠĆˆls#’„ •uĄ$y‰«†ä!eAp`6s` QGh’…€CŽ0ˆ$%6 ŽHG4i0ŽTp•c€Ž`–1#Ž$†+†vŹĄ!Œ7ˆ†ˆ$†…9Šb(Is@Ž  •——PŽˆCŽ–pŠb‰D‰x ƒ1uŽ<v‘PxŠęp–seŠęe")cŠŖYgP@Ž`T1i™TI$i0DR–Ž91•7T4e”T`0•ŽĄ%CŠæ†Ģ!a‡‰ØVxfegQŽä ˆ•0#UŽüqeP “t5‹R `AR (vUgŒ ˜"QQ`Œhu(…@Œ$@17!Ž ††:r0ˆ`uphŠ ˆ yR†PƒbI’ ˆU€0’r <—veHBT—pŽ`Y0!„…`(™T6eŠ¼ˆHD†(Sf‡xc ŠĆˆls#’„ •uĄ$y‰'ˆx!eA6slQGTh„…€C‹X`ˆx%6üG4i‘Dp•c‘ –1# †,‡ŖČČ!Œ7†$†…9† (Is@Ž  •——PŽˆCŽ–p€Œ0‰D‰x ƒ1uŽ<v‘PxpŒTp–se`Œ`e")c0ŒlYgP@Ž`T1i™TI$i0DR–Ž91•7T4e”T`0•ŽĄ%CŠæ†Ģ!a‡‰ØVxfegQŽä ˆ•0#UŽüqeP “t5Œš `ARŽ (vUg8˜"QQŒähu(…D@17!P‡opr‹`uphŠ ˆ yR†‹pˆƒbI’ ˆUŠ€ˆ0’r Œ<—ve0`B’ —pY $!„…P0(™€<6e`HD†(@TSf‡’$c ls#’’<„ ’`•u`$yˆ›ˆx!eA6slQGTh„…€CŽŲˆx%6ĢG4iüp•cĢ–1#Žš†"ˆ Ž ,†+†*†)†!†%†ˆ%†)ˆ)Š<‰T[‰\"*Step 2. We construct a tangent to† curve.Ž2-"Step 3. We animate the point alongŠcurve.[RŠ54Š5collectŠ5(x,y) value of˜HndŠVslop’tange†elin†ZŒ[QŠ[5ŠdraŒ}x-ŒTœGback toŠÆ graph we seŒĄsketchˆ†Œ³dy/dx for x^2-y†5=0ŃŽŁProv‰is‡ alytically:ŸTNote‹-can solve y in termsˆõŒdŠB problem. F†vo‡Vr typeŠ'Žsˆ“UP9need to express the original equation by using parametric”[x(t), y†].[Step 1. RˆŒ solve(x^2-y†5=0,y)RŒ7y=-Ž xŒM2ŒU-5(,y=ŗ([R&Š‰2. We demonstrateŠč graph of y=ŗvŒŌ + and its†Iriv‡ve in‹3follow‰"ŒUŽŠeditorŠ~repeat †āČmŠm dy/dx if nec‡žary.\Œ (/Explore y=(x^2-5)^(1/2) and dy/dx here=>Š-†Ź†† ōN&%FinaFormą$N†Graph2D†% 3Š ˆLISTSYSˆ$†@4ˆ< Modify XˆPˆ<STATCALC |ˆdˆŒ< „\ˆx SequenceˆŒ,ˆxSheetˆO |’ŠŒˆ’ olveEqˆ“†€`wrˆ“(Upˆ’tupFLG1 (Š<†Lis†{HD‰ˆPic†ŒÜViewWind°ˆŒ_osve†v‡(Ģxy†^‰Ų †U†y2’ųä(‹h†”ä<‡hŒČäŽP 0‰) †ä† † ‡j<’Š„KH’’T’’`’’l’’x’’„’’’’œ’’ؒ’“’’Ą††Ģ† †Ų†!†ä†"†š†#†ü†,$‘T† %‘h‰^Œ&‘|ˆ‚† '‘,†!(‘¤8ø†ä)†D†  †*†NP†R+†\†ˆ4h†-†t†.†€†0†Œ†1†˜†2†¤†3†°†4†¼†5†ȆE†Ō†F†ą†H†ģ†I†ų†˜J‘@† K†Ē† L‘h† M‘|(† N‘ˆŅ†ĻO†Ę @’‰š†ĒL††Ā†Ö Q††X†[R†_d†S†p†T†|†]‡] ˆ†i ˆ^ŽĢ_††`†”†aŽÜb†œ’ˆ2ŠĖ ’Œ•†¬†Ķ†ø†Ī†ĆŠ’ˆ ‡xׇ»ܒŲ†č†Ł†ō†4Ś‘h‡€ŽŪ‘| ‘ Financial‡ŖFormat †† †Œ†ˆ˜ systemä]Setu’^äa† _LIST’`Œ’†,ˆ†#bŒTˆxšÄ@ŽŽž¾ ž@ž€žĄūÄ@šœ ’’@’x‡øŽÄ®Œ< W£ˆŗ%†† a‡ MatDatab‡*.EAC‹1‹6† ”¤ĶˆĶ †I —p† ††Lī<‰¼¦-^††ś Œ©@ˆ Œö<ö<žxžš’,’h§¤‡‘9’*‡Xc‡p†† †Œ&Œ,’ ’,† †0†!@†@PSheet1|š‘| dŠ2ž3ž 4†5ž ä8ä7Ź|č–|œš"˜|†$ “i™” ™‰G‰K† "“-ˆ‰l~E ‰eŽĄ‰`†$«d«y‰“†`†††°5† Š” †#’$†(œ†AŠ ˆ  3x(x^2-5)^(1/2)†®p† ą:(y1(x),x,1)ˆgŽ#`– PšĘ$“$’į– p”x” –<”  ™€–`’ œ0™L(1…0qy`‰$#Y‡uYƒ‡y ˜–˜œšHŖ„’œ†¶*†† Š – – ²$’ ’¤Č%°lžaž žßžŲ¬š`—t †^’e5‰yˆ” – œ‡Y†‡<‡cX†R[Š[‡C Example 2.ˆ%! We are given two curves, C1 of xŒ*2Œ3-y”Ž -5=0 and C2ˆ64*x 7+9*y”MŒ8 -72=0,ˆ…Ž^Lshall show that these–™Š® orthogonal;ˆ! tangent linˆ"ˆ6o†6Œä.intersections respˆ velyŠūperpendicular. [[ŽšZStep 1.‰*uˆ‹he i‡Eicit differentiaˆ^ to fi‡ˆ•derivaˆi s (dy/dx) for‰Y*„ ˜’.R‹l&Clear_a_z RdoneRˆ Œimpdiff(x^2-y†5=0,x)Š2Œ3y'=Œ>xŒy[ŒO!PWe use the following m1 to denotŒslopˆ!rŠ)tangent line of C1 at (a,b).Š›Žym1:’zaŒĄbŠ›¦’ӐÓ4*†Õ+9*ˆ×72øŲ-4ī’ŽÜ9ī’¾ß2Üß2°ß2’ß-4Žį9’ā -4ī’aˆ 9ī’b[Œ /4Step 2. We find the intersections between C1 a†!C2.RˆRŒL +solve({x^2-y†5=0,4*†+9*ˆ72=0},{x,y})RŒ‡ x=-3,y=-2Œ˜,Žx=Ž¦Š4¢3Œ32Ž×[ŽTMŠą3ŠąshowŠą slopes of mŒŌm2 at (†-2) respˆż vely are negaˆ @†!ciprocalˆEeach o‡9r. — m1|a†ć|b=-2Šü€3ˆ†–.2ž.‘-2Ž3[Œ<:Step 4. We find the tangent lines at (-3,-2) respectively:RˆQŒWy=-2+(3/2)*(x+3)RŒy=Ž 3ī’Œ‹†(œ•-2–TŒT-2/3¾U-2ØVŽ^œV[Ž—6Šś5ŠścreateŠügeometryˆżks for‹poi‡of ’Ū-”ea , C1: x^2-y†5=0†2: wB74*x^2+9*y^2-72=0 and the respective tangent lines of C1Š'C2 at -3 2Ž, y=Ž!3ī’Œ,x+Ž-2Œ5-2 ŽIˆ‡–8-2Ø9Žn-†Š s follows:[Œ-A poi†¹†³intersˆŃon:]†¬Ž&-EquaˆˆŚellipseŠ" Ž"š" hyperbolaŠ$Ž$;šF‰@™5†šr‹U’=‰9aŒžŠHŽ“9!-Equation of the tangent line forŠellipse at a point:][†3Step 6. We dragˆIm intoŠSgeometry strip below:\Œ;?Measur†xho†atw†3–‚ s=>IndeedˆSˆAay perpendicularĪˆ‚ŽŠ[ˆ• ExplorŠÜ:Œ§K G†‘o (above) =>Edit=>Select all=> DŽĻ sŠ†‰items ŒłTaround and notic‡@‡DeTsŠ‰V‹4‡‡dk va‡ accordingly.–Ė–Ķerci‡‚1.Ž{4Repeat the process for showingŠtangent lines †-3Œ-2Ž,š!-Ž"Ž! and¬E”#ˆ…Œa)C1Š1C2 respectively ar†³erpendicular.[Œ’[†Wˆ Exercise 2.ŽNIFollowŠńStep 6 m†Ż ioned aboveŠžexplore‹ExcŽFŒ‰see if X„#‰A”µ‡^”ø intersˆÅons stayœÅ when youžperform horizontal or vertic† shiftings.eActm1(( b€am2@@ € b€asecH t1/ąP(t)020012eActivity Save.EAC01000000244eÉ ųś [[-3],[-2]]ü÷Ī÷ŹĄ# #`g—u#`g—u ü ų ś5ī’x^2-5ī’y^2-25=0  ÷ ČĄ  $&@hq‚„'GFüųś8ī’x^2+18ī’y^2-144=0 ÷ČČ  @Š6–°2”3xY™Tp–"R ™8gPIc`üųś y=1.5ī’x+2.5÷ČĄ %`@Š6–°Tq“–”) ™2t“S– ™2I—AXüųśy=-0.6667ī’x-4††bVVxXˆ `59RC@†`Œ†Ž0†ˆ:† ˆ<v’K†r†  A† 'LŽ'ˆpC˜ u˜x˜yŽ †qˆq† †Š† †††ŃČČ `Š0ˆ½Š6–°†7ˆ7ŒF†Ļ– ’ Ž ˆ(ˆGˆOˆWˆ_ˆgˆ’E  Imp_Diff.EAC† main.ACT†††)ˆ,†0Š†C ~E Ž'N[ˆ4†]licit ˆb erentiationŽ \Œ( Authorķˆ?†CķŽ€†€ˆŖProfessor Wei-Chi Yang ŒRadford University Ž9e-†Öl: wy†0@rŒ).edu""URL: http://www.–$/Š6Žf‹Ž¶ Sound TrackŒ» »"Professor Jonathan Lewin Ž Kennesaw Sate University ŽA e-mail: lˆ6s@mindspring.comŽFURL: http://www.m†j-moviesˆ'†Š[Š‘Ž Object†ns:[Ž¬7S1. We will use ClassPad (CP) to explore the graphs of †Ņimplicit equation†d its corresponding derivaˆ}.|T2Š|b one problem ‡4t c‡8be fouˆYn regular Calculus texbooksŠzweŠĮƒ1CPœ½is’\ˆ§extendŠÕresult.‡, [† Note.ˆ 1L We will explore some implicit equations that a†!graphablˆ8thin ClassPad.[ŒT’i Exa†Ie 1Žn= WˆFiˆNŽI ofŠ deriv†kve (dy/dx) for x^2-y†5=0”_Œ»HStep†[ˆĀ drag and dropŠ]¹ intoŠo following Ge†ä try Strip.\ DšE–zĪˆy†}Ī‰J‰Oˆ‹‹R™™™™˜H† `@”Gs#Š Š.†† Ž!Š ŽD‰Žv’K†r† Ž A† †L††ˆ C˜ u˜x˜yŽ ˆ † †JˆH  ŽN†\D€s™€†f ` Œ FT'@Œ7`XpŒ$i!22`Œ03Ž0%YDiPŒT’2YtŽ<upQŽH "ix“$…dŽ` eTcbŠ£`57•b0Š F y– Š ”$”<–T–l–„–œ–“–Ģ–ä–ü——,—D D€s™€ˆ† …f5PŒgDŒ qds0Œ,2x˜pŒ8yu’)`ŒDhQG"@ŒPDI’ŠH`SŠ0`T†„ #gv„Šl`t`IŠH`€pƒ • VŠT`T”RˆÆ` ˜SŠxY™– ™˜$š<˜T˜l˜„–œš“–Ģšä–ü›™,™D—\† ‡z5ÅH ˆ ‰†!‡ŒĘČ ˆA‰LŠ6–°†F y’C‹Q ‡S†CQ††Š†"†ĖČ#… žœ($U‰y&ˆB²("ex•‘$rA7“ˆXY™ˆŒ(`” D‚u† i†$ ™†#†£.Č@ ’Ž$†µ3ÉČ"’ˆ²!†%†,ČČ €Œ@†ŽŠ6–°†åXf2`ŠzbeGCŠ†‡7d‡†’†3ˆĻ.ČH!œ\ˆr'¢ˆˆrˆˆ¢ˆzˆ"‰ˆˆ(‡]1Ä 2ˆPHŽ )†ŹÉȖ"‡6$†…9‹(IsˆÉ‡Ø •——P‹&ˆC@–p€Ž ‰D‰x ƒ1u$v‘PxpŽ0p–se`Ž<e")c0ŽHYgP@PŽTT1i™TI$i0DR–x91•7T4e”T`0•Žœ%CŠ§ˆØ!a‡‰ØVxfegQĢ ˆ•0#UäqePŽT “t5Œš `ARŽ (vUgŒx˜"QQŒähu(…,@17!8ˆ†Qr‹`uphŠ ˆ yR†‹ˆƒbI’ ˆUŠ€ˆ0’r <—veHBT—‹|`Y `!„…P (™€6e`$D†(@0Sf‡’$c Hs#’’<„ ’`•u`$yˆwˆx!eApŽl6slQGTh„…€CŽ“ˆx%6ĢG4i0ŽŲp•cĢ–1#Žš†*†śŹĄ!Œ7ˆ‡ $†…9Šb(IsŠņ†  •——‹"†ˆCŽ–p‹.‰D‰x ƒ1uŽ<v‘PxŠęp–se‹Re")cŠŖYgP@Ž`T1i™TI$i0DR–Ž91•7p4e”T 0•Ž%C@ˆ#ˆ$!a‡‰€Ž0VxfegQŠ%†H ˆ•0#UqePŽT “t5ŒH `ARŽ (vUgŒx˜"QQ`Ž hu(…Š…@17!Ž ††Qr0ˆ`uphŠ ˆ yR†PƒbI’ ˆUŠ€ˆ0’r <—veHBT—‹(ˆY0!„…`(™T6eŠ¼ˆHD†(Sf‡xc ŠĆˆls#’„ •uĄ$y‰«†ä!eAp`6s` QGh’…€CŽ0ˆ$%6 ŽHG4i0ŽTp•c€Ž`–1#Ž$†+†vŹĄ!Œ7ˆ†ˆ$†…9Šb(Is@Ž  •——PŽˆCŽ–pŠb‰D‰x ƒ1uŽ<v‘PxŠęp–seŠęe")cŠŖYgP@Ž`T1i™TI$i0DR–Ž91•7T4e”T`0•ŽĄ%CŠæ†Ģ!a‡‰ØVxfegQŽä ˆ•0#UŽüqeP “t5‹R `AR (vUgŒ ˜"QQ`Œhu(…@Œ$@17!Ž ††:r0ˆ`uphŠ ˆ yR†PƒbI’ ˆU€0’r <—veHBT—pŽ`Y0!„…`(™T6eŠ¼ˆHD†(Sf‡xc ŠĆˆls#’„ •uĄ$y‰'ˆx!eA6slQGTh„…€C‹X`ˆx%6üG4i‘Dp•c‘ –1# †,‡ŖČČ!Œ7†$†…9† (Is@Ž  •——PŽˆCŽ–p€Œ0‰D‰x ƒ1uŽ<v‘PxpŒTp–se`Œ`e")c0ŒlYgP@Ž`T1i™TI$i0DR–Ž91•7T4e”T`0•ŽĄ%CŠæ†Ģ!a‡‰ØVxfegQŽä ˆ•0#UŽüqeP “t5Œš `ARŽ (vUg8˜"QQŒähu(…D@17!P‡opr‹`uphŠ ˆ yR†‹pˆƒbI’ ˆUŠ€ˆ0’r Œ<—ve0`B’ —pY $!„…P0(™€<6e`HD†(@TSf‡’$c ls#’’<„ ’`•u`$yˆ›ˆx!eA6slQGTh„…€CŽŲˆx%6ĢG4iüp•cĢ–1#Žš†"ˆ Ž ,†+†*†)†!†%†ˆ%†)ˆ)Š<‰T[‰\"*Step 2. We construct a tangent to† curve.Ž2-"Step 3. We animate the point alongŠcurve.[RŠ54Š5collectŠ5(x,y) value of˜HndŠVslop’tange†elin†ZŒ[QŠ[5ŠdraŒ}x-ŒTœGback toŠÆ graph we seŒĄsketchˆ†Œ³dy/dx for x^2-y†5=0ŃŽŁProv‰is‡ alytically:ŸTNote‹-can solve y in termsˆõŒdŠB problem. F†vo‡Vr typeŠ'Žsˆ“UP9need to express the original equation by using parametric”[x(t), y†].[Step 1. RˆŒ solve(x^2-y†5=0,y)RŒ7y=-Ž xŒM2ŒU-5(,y=ŗ([R&Š‰2. We demonstrateŠč graph of y=ŗvŒŌ + and its†Iriv‡ve in‹3follow‰"ŒUŽŠeditorŠ~repeat †āČmŠm dy/dx if nec‡žary.\Œ (/Explore y=(x^2-5)^(1/2) and dy/dx here=>Š-†Ź†† ōN&%FinaFormą$N†Graph2D†% 3Š ˆLISTSYSˆ$†@4ˆ< Modify XˆPˆ<STATCALC |ˆdˆŒ< „\ˆx SequenceˆŒ,ˆxSheetˆO |’ŠŒˆ’ olveEqˆ“†€`wrˆ“(Upˆ’tupFLG1 (Š<†Lis†{HD‰ˆPic†ŒÜViewWind°ˆŒ_osve†v‡(Ģxy†^‰Ų †U†y2’ųä(‹h†”ä<‡hŒČäŽP 0‰) †ä† † ‡j<’Š„KH’’T’’`’’l’’x’’„’’’’œ’’ؒ’“’’Ą††Ģ† †Ų†!†ä†"†š†#†ü†,$‘T† %‘h‰^Œ&‘|ˆ‚† '‘,†!(‘¤8ø†ä)†D†  †*†NP†R+†\†ˆ4h†-†t†.†€†0†Œ†1†˜†2†¤†3†°†4†¼†5†ȆE†Ō†F†ą†H†ģ†I†ų†˜J‘@† K†Ē† L‘h† M‘|(† N‘ˆŅ†ĻO†Ę @’‰š†ĒL††Ā†Ö Q††X†[R†_d†S†p†T†|†]‡] ˆ†i ˆ^ŽĢ_††`†”†aŽÜb†œ’ˆ2ŠĖ ’Œ•†¬†Ķ†ø†Ī†ĆŠ’ˆ ‡xׇ»ܒŲ†č†Ł†ō†4Ś‘h‡€ŽŪ‘| ‘ Financial‡ŖFormat †† †Œ†ˆ˜ systemä]Setu’^äa† _LIST’`Œ’†,ˆ†#bŒTˆxšÄ@ŽŽž¾ ž@ž€žĄūÄ@šœ ’’@’x‡øŽÄ®Œ< W£ˆŗ%†† a‡ MatDatab‡*.EAC‹1‹6† ”¤ĶˆĶ †I —p† ††Lī<‰¼¦-^††ś Œ©@ˆ Œö<ö<žxžš’,’h§¤‡‘9’*‡Xc‡p†† †Œ&Œ,’ ’,† †0†!@†@PSheet1|š‘| dŠ2ž3ž 4†5ž ä8ä7Ź|č–|œš"˜|†$ “i™” ™‰G‰K† "“-ˆ‰l~E ‰eŽĄ‰`†$«d«y‰“†`†††°5† Š” †#’$†(œ†AŠ ˆ  3x(x^2-5)^(1/2)†®p† ą:(y1(x),x,1)ˆgŽ#`– PšĘ$“$’į– p”x” –<”  ™€–`’ œ0™L(1…0qy`‰$#Y‡uYƒ‡y ˜–˜œšHŖ„’œ†¶*†† Š – – ²$’ ’¤Č%°lžaž žßžŲ¬š`—t †^’e5‰yˆ” – œ‡Y†‡<‡cX†R[Š[‡C Example 2.ˆ%! We are given two curves, C1 of xŒ*2Œ3-y”Ž -5=0 and C2ˆ64*x 7+9*y”MŒ8 -72=0,ˆ…Ž^Lshall show that these–™Š® orthogonal;ˆ! tangent linˆ"ˆ6o†6Œä.intersections respˆ velyŠūperpendicular. [[ŽšZStep 1.‰*uˆ‹he i‡Eicit differentiaˆ^ to fi‡ˆ•derivaˆi s (dy/dx) for‰Y*„ ˜’.R‹l&Clear_a_z RdoneRˆ Œimpdiff(x^2-y†5=0,x)Š2Œ3y'=Œ>xŒy[ŒO!PWe use the following m1 to denotŒslopˆ!rŠ)tangent line of C1 at (a,b).Š›Žym1:’zaŒĄbŠ›¦’ӐÓ4*†Õ+9*ˆ×72øŲ-4ī’ŽÜ9ī’¾ß2Üß2°ß2’ß-4Žį9’ā -4ī’aˆ 9ī’b[Œ /4Step 2. We find the intersections between C1 a†!C2.RˆRŒL +solve({x^2-y†5=0,4*†+9*ˆ72=0},{x,y})RŒ‡ x=-3,y=-2Œ˜,Žx=Ž¦Š4¢3Œ32Ž×[ŽTMŠą3ŠąshowŠą slopes of mŒŌm2 at (†-2) respˆż vely are negaˆ @†!ciprocalˆEeach o‡9r. — m1|a†ć|b=-2Šü€3ˆ†–.2ž.‘-2Ž3[Œ<:Step 4. We find the tangent lines at (-3,-2) respectively:RˆQŒWy=-2+(3/2)*(x+3)RŒy=Ž 3ī’Œ‹†(œ•-2–TŒT-2/3¾U-2ØVŽ^œV[Ž—6Šś5ŠścreateŠügeometryˆżks for‹poi‡of ’Ū-”ea , C1: x^2-y†5=0†2: wB74*x^2+9*y^2-72=0 and the respective tangent lines of C1Š'C2 at -3 2Ž, y=Ž!3ī’Œ,x+Ž-2Œ5-2 ŽIˆ‡–8-2Ø9Žn-†Š s follows:[Œ-A poi†¹†³intersˆŃon:]†¬Ž&-EquaˆˆŚellipseŠ" Ž"š" hyperbolaŠ$Ž$;šF‰@™5†šr‹U’=‰9aŒžŠHŽ“9!-Equation of the tangent line forŠellipse at a point:][†3Step 6. We dragˆIm intoŠSgeometry strip below:\Œ;?Measur†xho†atw†3–‚ s=>IndeedˆSˆAay perpendicularĪˆ‚ŽŠ[ˆ• ExplorŠÜ:Œ§K G†‘o (above) =>Edit=>Select all=> DŽĻ sŠ†‰items ŒłTaround and notic‡@‡DeTsŠ‰V‹4‡‡dk va‡ accordingly.–Ė–Ķerci‡‚1.Ž{4Repeat the process for showingŠtangent lines †-3Œ-2Ž,š!-Ž"Ž! and¬E”#ˆ…Œa)C1Š1C2 respectively ar†³erpendicular.[Œ’[†Wˆ Exercise 2.ŽNIFollowŠńStep 6 m†Ż ioned aboveŠžexplore‹ExcŽFŒ‰see if X„#‰A”µ‡^”ø intersˆÅons stayœÅ when youžperform horizontal or vertic† shiftings.eActm1(( b€am2@@ € b€asecH t1/ąP(t)03010b0020000800000000156dxœķŪolē}Ąń;Š”D‰Ši‰iģÄMYGm醒yĒ’­0Č°åŲ‰ģ(–Ż4@Ņˆ±čŠ€žE’c»3 ĖĘU]YP=§0cȋlŠ`0†¼č:£ČŠbK‹ (Ö¦ ¶¼ †¼(æ(Š¼œżž#ļ!u c§MÓ5ż~€G|žßŻs÷4ņ•¤aŌŹ]µ²ķ”ćīr«¹<śąī]£©Äśr2ļ)ē<嬧œń”ÓžrŹSNzŹ¶§lyŹžöŁžöŁžöŁžöŁžöŁžöŁžöŁžöŁžöŁžöŁžöYžöYžöYžöYžöYžöYžöYžöYžöYžöYžöå=ē?ē)g=匧œöŽO9é)Ūž²„Ė.Ź”)S¦ü‡-ĻJ³Ķåį]c£?xč§ģ3Œ½„ŁĀŽ¹™a“æ&‹ūFł¾…Āü”½GŠķ×åG¤Ė%U¬Ć/?B*6ŗüšų#ć*–uSŖī¹ÉŅ±SjO=jå~µŽųį]‡å"°[bSėæwō>™&mI,ØĒFK‹²żĶ)‰ķéŖĒĘJGUģ¤Ū·Æ”Š'.ĶNJģ²Ä¶©Ųćs‹OŌę7æāīcqŖtlé'öŖō×9~Õ°{¶ś^w׫nÖ±7u¬WĒŽÖ±>Ó}«FÜXD÷­z»Ž…tģS:¶EĒīŠ±~Ū¢c ŪŖcC:v§ŽķÓ±»tģ°ŽmÓ± ū“ŽMėŲŻ:vRĒ>£cOėXTĒVtģ³:vYĒ¶ėŲ‹:vŽ]Õ±~»¦cŸÓ±×tģó:ö ū‚ŽUu,¦c×ul‡Ž½«c_tc·wźŲ½:ѱøŽEul@Ēā:6Øc9KčŲ³tlLĒl{TĒ’:6„c)[Ņ±“ŽÕ±[Ö±½:vIĒöéŲ :¶_Ē^Ö±ūuģū:ö€ŽżXĒFuģg:v@ĒŽŅ±ƒ:ök{PĒŽŃ±17ö)æŽ=¤ca;¤cŪtl\Ēb:vXĒR:ö˜Ž Ėæó7āk:¶GĒ×±}:6”c£:VŠ±1{BĒėŲšŽ}U·å;:6©c?ѱyū©ŽÖ±×u¬¢c?×±5ū…Ž]ѱ7tģ%ū„Žéæ§Õ_éŲ«n,Ōi:׳Ł£„‚śc¢®k¦4LŸ|ج\óŌ ō×>jĢę‚ŃīŽs«ó1^\:®Ėr.Ō1Õe9źØĖµ˜.j1]~¢¶Måś«Ć|ņÉ'Ÿ|ž©6®)!uÉ=PXŚSX*źR;8²k·Ń¬«‘½ÕėLÓžL7żŽu ź~‚ėsIjæ{ģ¹¢u;ū Ōc{GļS±—ܘjéŠčžƒü…įō)xØ0;97żJaśx1“P- 'ķ¬­ØŽƒh®ēō;čüŒ:×Įؽ'sr;āÕĻŸ©ŽÆ&āU+^MĘ«©x5³ŽY‰ÄŽx¬ö _2õov¼ßö’īö’{d‹v<ö kąµśõ/ TūÖÕ;>3SX8Œ-Ö2åźIzT=JŸ;¬Žy¦zO¼śz<ę>_Ÿ;Q{pŗa]ՆjųLuk¼śó3µ§BµuöŽ.Ķ-«żqõ ·;\xbŗ¬īŒWćńź@¼ž “¶l÷ÜléčbtäÉ`õQ7ę~#¬ŽƎī{ćÕ}ńź”xu<^=Æī?S}(+Ģ½×–>/Ÿ||Ŗ“(G7ęÖōžŸę±ßęWOøCM_·¼PœÆ[ĪEĢ¤¤WŹ3T‰tŌæi©D6½ŗž‹–ņ—ŪŒŠ;å˜Ł_ 5}Ļ²­|Äģ¬Ä*‘å”öž~õ‹ZŽų~ålyȌWNœ“W>' Łō’ģėōjäܲ/[“×¾O)_5ƕ«>`Eį|“ņbł'¾¹ó™ÆDĢ««‘nżķIųšl5vnØi--–’Ņ~»÷ÆĻöm>¹²ģ«n^˜OÆ\ńU{%³²rÕWķ“Ģå•ūŖ‘ÕČź›¾źķē'{ŒĶĒųŸ*w•{Ō¶®ÉŠwHzMŅI搓URUŅ’®KŗKŅ»’¶É~{;%óiŁiÆlŁ¼[vŚ•Ģgd§½ńÕHåīs’īėĶIä³+ośz÷HfūŹu_ļ˜dī¹ą÷õ>*™ž [|½S’łÜ…øÆwI2Ÿæ0ģė=+™/\8ģė]–LģĀ”Æ÷’dv\8ķė}A2_¼°āė}Y2÷^xĮ×ūżo½Ņé>ī/ūŪŒŽ—»Ś¤or“dŸI”ō–¤„¤_K²$½#É^™ōõł%“”žō…%“’žōm“LZśÓ'§ÖY¹jöUģ刹WŽbß°„öI‡śF%³_:Ō÷UÉÜ/ź›”ĢŅ”¾y鿎oīōõÉ1H‡ś*’9(ź[“ĢƒŅ”¾+’“õ½$™‡¤C}ÆHętØļUɌóuQė{½čØ?/o*÷śÕ7]²ų1Io—|ž.V_«÷źńśIšōI…z/ŸØŸ“µ•e3R~įü²łé\$$”ŸČɊؓżSé[¤_Śż?ēžŪQĒéēŅ·ČŅ·Č>ɼqa‹yv»‰_^ˆ›²Ŗä~%½‹L_8p¦čåHŁ_éXžlyėZOWćŃš³Ó~uX ?3õlB~NśÕÆØd&z7>¾¾mż“ßóē;Ÿ’ zcųĘŁ׎3Žž—“ī… ß;łĢ+õ„ƶX¼qÓ’uÖļtą¶²ß*”Ō%Ź’ęW%ĪżĒ¹’,Ģ®µŪĀŽ»ĶÕ°_Żg–÷ÆD’6lšåŁ†ažļPåäo‡^;ŁŻtćY šęņmķµÖŹ¶ĪĘ:źwŖ+Ń߯8y漍ėļÅŪk÷Øå`ŁośĖf§¤P9l†æuž[k«ŻfūjÜ K,*)!iø<ģ{ļG+aÓ~>l&Ÿš)Iéē£ź6®v‹WśÕķe¹Cv±~S2:ĻÅ;ź÷–O_^·K§Ė'ĶąrGyÕWzn-ŌwƒåŪ*mr/ø+ßqīĖķµ;ĄŹ?Ÿæ& ļ4·”7™ż«÷®¦ž®÷røüŲ·żf臞ńŸŸ’īÅĢĪĖw­¹ü9K÷ÉļĆ”Gœ‹v}yq×Ń„`LŽć‡GŒ5źŻWœ›).©:÷<čY¶§tģ˜\’£µĖqlĻČCī ķīž.Ķ?1W˜”YŃ螦ŗćó ÅĀä¢<2KoT¬/Æ}U$“©½ūź…­ęRĪœ/Xū¹½öėņØūŽĄöxsD½!°>¢®ŽŪeJPŸę9mh¹uŻ <>S\(ÖJ±zŃ)©_į²µw-ŖO'Ō×ājŽT?¢Ŗ'7ÆÆ>vF}ÕØwØøx|z)X=ÆNž©ĒöĻ.-Ģ¦wĻMĒ7ōU}”ź7MŸĻģ5šó2°NŌŽ™0Ōw¤źż SXē/–aČ„Š÷®L{ĢDm`š½aćnCjŖbū@n0•ŹÕ k0aēŻ™ĮDŽj,ČdŻ©ĮT&­X)½ 9˜Ģfō‚¤sŲƒ™¼­dóM ’IwAr0“K7Ųiwē–ģ±± •²…“[]ņvŅ-XƒéT¾Ń®l:ŪXb'u!=˜HŁ%V"Ūč}>Ł“$‘ŅmĪfŅiŻ49bz­¤ģ“©JB÷ĘŹŗ…ŽÄ`>ŸÓĒÆõ’[ “Ų U‹­:hu\­e ÕŃ“8_'8ŠbDZ ”@‹1h1HFµdićGÓĘ?ČõŅ¬ķdėŲĀܱāāāÜBōž¹ŁĀŅTa6:Złäó“÷Y’®ß’T0>ų»¦Wī>ō;wźŠā¦ß„®AŻOtŻś{źż,g}<†×ż;AC{}]µ<*Iī=üƒĘ$ż”ß¹»…}LļÜ=WjŹo½ÕwīźŌ9¾[$õ·8ʤłoF«uźļ:éó¹~™n]øž/oļ{<'$ĶK:+iEŅI/Kŗ&éuIoIŗ®¶mJ}IQI IƒĘ$MHš—tVŅŠłįū‰Wćż8c¢)n¼Og†ÆuŒĘ:ĶłłĘśMł°~².’\°ńžl’ģūÆo4i¼sg„ļÜ] śšÖoĪ7o'Š"žQ­c|DZķė£ŚžK«cÕŸX76ōūt26łp«xóöv·Ķóļ’WĻæõ’ž<’s}>S’oõĻśźYxÆ\®fź’ć­ž±wø[,Ö^›U’|­žqyŸÄbFČgÜęS×7÷æ†e†ęLŲœ‹^”6BmFĄW[čÆÆf˜Ī4ļŖ;å.£q]2¬oÅ–ŌAē9¾Įs|žćßźsšō(—Nē²v*“j„Ņ©\&—Jø[īV]M¤­|&_ßo:?©l.cål+ķVTĮJ[™|¢QŃN'ré¼eg­t2“ĶåķTcŅ„D"•ĻŪi]1)›JXłD½G”Ä`ʶåDęģ¼»uē J’m;„«É”“’vŅŻ_·5(ĖSŁl6—Ņ‘l>›ĪŪv>£·#:‘ĪYɦēóŁd:‘µt­¤Ķå¬|Ī]Ē9 )Ė²³)½œ•#’O7ö.Ķ³2©l²>„ŗ“”°¬DŅŻrz0c„l9[×Jd2Y9 īXėvNa"›Ģ»ƒC"‰¬ŒEĄMµģ\:“I¦Ó†ū7ÓÅ}F=Ļ}Ę×å>Ć`^ļ=Ģė™×’ęõkķļ3Æosē.Ÿ*ō×Ow„ ėŠĄ-\­C¦”[˜„6LDŌœź¦3‘ŠĄ†išÄŻtŽŲ0ĖRÓ³›O³Bęu”[šÖ9ŻY?“ ŻŅD2T»£±Ół¬ž”ŽŹĢ5T»ß‘=&šęvł\&›¶ó‰|£ž-mĻJĶtć<ĖBĀ¶3zĘ%Éf’‰|Ź¶õdü¤2ŅÅLcČȝGJʑngf0HJ%»1ųœ;’D6‘KźzĪL*“”£_ƗLZVŚ¶“‰ęzV>m'ŅM]–[£dVF•{“Ņ%¹{É%ŻóŠ]»cJåSi[ג[­TĘNdóī ӏɁJę2)÷„§~ –L¤¬œ³Žz²©[«čmf÷ĄĶŪŁ½į Øćtó£Ņ½į$Øór󳊽į¤;æŹ7=ėŻ™3^o:Źŗ7 jīĒŒOīżšūś?ĖDPü0411