00020004010008main.ACT0001020012eActivity Save.EAC01000000046d straightline.EAC version2.ACT$'148K^E '[\ animat[ΈC_Hduk `qyHq PxUwǘDvr m A 'L{xy qq@  H!I@g6 a(67 UBT"d&  40T" U&Y`T `@  + @:60*4hm!)AC]/? ȆCHP ފ oဏ"@. ?5H!2 D <Dy' !i 5Y   )C@06; % R@  Yo  K xshB YhH!zER5GE&bX! B2 wXH {F{`{`wUoˆ!@6 y %@  + Ę0 !@HF87RYcaEU rrrz Ȉ. ΈBF/7?gocgE [ Conjecture: 0If the fixed curve is a line with positi slope6n Blimit tends to 0.R;A factor(eAct010013shrinkingcircle.ACT0003020009case1.EAC01000000141e  case1.EAC shrinkingcircle.ACT$'148  ^E 'd[4SX CY and CP 300k\3 AuthorGGProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University, VA 24142 USA e-mail: wyang@r7.edu URL: http://www. /~0 Objectp.Some problems on finding limits can be done us 9CAS capability. However,[it will be more exciting !to solve them toge r?with geometry animations.\`EA-! tap--> (x+1.3)^2+y^2=2.1^2=> corresponds ucase3 in 3D.Έ BY5W `v `uRE5P A901@ DvQr ^A 'L'8C uxy qq @  uH AA64g``t9  !  @067uw#`B R80E Yu"3.^c@#ByȨXr%hFTBYhF @F H Zs1si%(F2 Y ȐYX 7P Y] 2]YĶ` @# * @96 DSP =l!uACC %@ YĘ0ˈ"0)`"NӈF/\vvȈ*5 !s` +H! 4E@;6<'GF Sn@ CYo h$T \2e1Y RH RH W2ca`a&! HYG72o]USaSp SeCY]!CVC "@  !#.Ř $P `YT@`6 II@p)& `uVw~Y%2e&H OU 7Tw#TYU'C#P& `(Vs'P C(C@#U r%hFTA  *N7X* v fY+@ ),*@ * '-LŘ"&"@^.L/f0Ȑ@6r%hFTBY`׊F1\!H 0W2gZ23EE4.E5H#f'GF K0*)&'$ !    K2 \a51(x-2.3)^2+y^2=2.2^2=> corresponds to case2 in 3D.Έ96EΌ\XVTq 1U`2t6 `"A 'AasVwDvQrˈ^ A 'L'C uxy qq7b@ K8tH #]K@6xcH(2RΓH 9U@6 c)Td dqAS p A%f2.`3:8@  8;NX'4hF<\=H nIMr6q Y9>?ȐYCQ( Q'Y]E2@ A#aJĺ_aB@#UCÈ  @6  $Q( Q' D9 BACC?EG%@ YCFĘ0GŘAHJȈ 0)AFIv=vJHvK5L^ g`rMk*HG N(@  DLOM )1)1c K\VAVAYNlZHPxQH G@6MQ( Q(;GRS `F]¦]T.eEFC? 8 -Se=A9><8(K [. A problem.?We have a fixed circleC1: (x-1)^2+y^2=1 and <wpoint L?WThe following matrix isv3 coordinat`forN ascircle shrink.[ 3.414213562060441189 5037317044 544332125N 582449523h 618259366651911k683535993.7132444830 4113525167294364491797668N 814712236h 83609755285600647N7448605789157818 3.9073201592174514 934882619M46758957396163 3.915h975033P98206 9879243070 3.992619907616066848552833N9796555h[qWe can see thate limit (L) approachesto 4. Prove anlytically_Let B b] e point (0,r) and (x1,y1)& coordinatuof F and(z,01)0L.JVStep 1dFirst, we need to fix1i y1 in termUUrE%The point F lies on both[ circlso[x12'+y1=r( (1)N(x1-1)QeRQ=1@(2)It follows from thatY -2x1+1__and)=.DHence [WTx1=LrJ*y1=4 - r 44=r&1129.[TStep 2lWe need to find z in  terms of r0Let F' be on the x-axisOso tha#$isIperpendicularlALByHsimiriangleS==== [$ lengthRB3=* *F'+F&,uze z-x1 y .[This implies that z=0r:r-  ^I 'c[4SX CY and CP 300k\3 AuthorGGProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University, VA 24142 USA e-mail: wyang@r7.edu URL: http://www. /~0 Objectp.Some problems on finding limits can be done us CAS capability. However,[ it will be more exciting !to solve them toge r?with geometry animations.\`A ! tap-->Έ  pbT `u'pA YDXB vf DvQr ^ A 'L'&C uxy qq @  uH!L@62s PugPpp8Y;  @6 r%hFTBY"` 9@ BO"XXE&p eX&( YF \!H \  ! !Aˠߠ ,H!5EC'GF 6 6Po Ȑ r%hFTB Y` $W2g@9@  $ !H  iEEcH#uF6 fTxD  BY>Y]h `Ql%ȓy  @  < @K6RB2F\5v SU3Y3  Z 4H'Xb YJH#J 'GFt    )19:=ڍJ[ A problem.aWe have a fixed circleC1: (x-1)^2+y^2=1 and <we have[another circlC2: x^2+y^2=2, which 3will be shrinking.MU 1) A is Z cent`of C2)2) B!point on;y-axis^3) F5topiasecti8h1 ando:Lx-0cept line BF We would!k{ se@wheradius+(r)goe o 0Gathapp;oL?The following matrixRthe coordinates for [ point L as circle C2  shrinks to 0.[; 3.414213562M060441189 5037317044 544332125N 582449523h 618259366651911k68353599713i48h74125N76729436 7917976688147122383609755R0 3.8560064787448605791578181 3.90732015M 921745144g 934882619M467586357396163 3.9668153975033P9820698792439490N9985528399796555[We can see thate limit (L) approaches [to 4. Prove anlytically&Let B be the poinP0,r)F and (x1,y1)&a coordinatuof F and}(z,01)0L.JStep 1First, we need tofix1i y1 in termUUrETF lion both circl/so[3x1>2F+y1=r( (1)N7-1)QeRf2 =1 (2)[It follows from that[2x1<2 -2x1+1+y1__and=)=x';<=rO.Hence [Wx1=ry1=Ֆ4-P44 11e9wStep 2We need to find z in terms of r!Let F' be on the x-axis[soa#$is perpendicular to AL.3ByHsimiriangleS==== [s lengthRB@=* *F'+F,uzrez-x1 y .[ This implies*zr:r-<[Substituting x1 and y1yfinally gives z as a1function of r as follows:[CAS capability:R define z(r)=r2'2-24-RbJdone[ [TQn014 Remark: We n change the)size shrinking6circle butwill notBgI limit solong7]wradiusRaastarts out greatertJ fixed .[ Exercise.What if the radius 4ofixed circle igreater:an?k shrinking4; tryUR(x-1)^2+y^2=4 and se w happens.eActzH,r((r^(2))/(2-(4-r^(2))))02000cversion3.EAC010000001025^*[Shrinking Circle and [CP 300[\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL: http://www.radford. edu/~wyang[ Objective.[Some problems on finding[limits can be done using[CAS capability. However,[it will be more exciting [to solve them together[with geometry animations.\Animation! tap-->U#cRD`4u``  2df#FvrALCuxy@  H! Q@6WVV#%A5Y" @6Quv(4 Aw%f a' `@    @6r%hFTBY` @  H# N@64Ei0Y # @6r%hFT@  @  !A@6 @  H  P@6` G'@  H  B@6 2@! H" O@6` G'`H! " @64Ei0Y %@  @  @    @6  %@  @  @  @  @  @    @6r%hFTBY` !!H " ! @6` #H! E@6'GF$@  "%@  #" W2g@&@  '!H "W2g@(@  )5@  "$&@hq   4H'XbY*H# #'GF%$"# #"  )[ A problem.[We have a fixed circle[C1: (x-2)^2+y^2=4 and [we have[another circle[C2: x^2+y^2=2, which [will be shrinking.[[1) A is the center of C2.[2) B is the point of C2[on the y-axis.[3) P is the top [intersection of C1 and C2.[3) Q is the x-intercept [of the line BP.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point Q?[The following matrix is[the coordinates for the[point Q as the circle C2[ shrinks to 0.[ 7.8366652190 7.8476774960 7.8582833660 7.8684861710 7.8782890920 7.8876951530 7.8967072290 7.9053280470 7.9135601940 7.9214061160 7.9288681270 7.935948410 7.9426490230 7.9489718990 7.9549188510 7.9604915680 7.9656916330 7.9705204990 7.9749795340 7.9790699760 7.9827929450 7.9861494860 7.9891405580 7.9917668590 7.9940292420 7.9959283810 7.997464430 7.9986371860 7.9994592530 7.9999221750[We can see that the [limit (L) approaches [to 8.[ Exericse.[Prove anlytically!!eActzH,r((r^(2))/(2-(4-r^(2))))01000ftwo circles.ACT000102000ftwo circles.EAC010000000560 two circles.EACACT&)36:M~E '[f Try this\ΈB^Gc `@Gs# .Q! Dvr mA 'Luxy qq5H#  Q@h6#!= HY5pW 1p` fT0`0vueFH#"x(vrALCuxy@  H! P@6iPHViY  @6f5 Hb  esD`@    @6r%hFTBY` @  H! O@65AeY  @6r%hFTC  !A@6 @  @   ! @6r%hFTC  @  BJ@  @    @6r%hFTBY` @  H! N@6rIHD 74 @  H  B@6 @    @6  %@  @    1/10x(x-3)(x-5)!H  ! @6` H! E@6'GF @  !@  W2g@"@  #!H W2g@$@  %5@  $&@hq   4H'XbY&H# 'GF!    %[ A problem.[We have a fixed curve[C1: y=110x(x-3)(x-5) and [we have[another circle[C2: x^2+y^2=2, which [will be shrinking.[[1) A is the center of C2.[2) B is the point of C2[on the y-axis.[3) N is the intersection [ of C1 and C2.[3) P is the x-intercept [of the line BN.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point P?[The following matrix is[the coordinates for the[point P as the circle C2[ shrinks to 0.[[We can see that the [limit (P) approaches [to 0.[ 2.5455844120 2.4600001090 2.3744158050 2.2888315020 2.2032471980 2.1176628950 2.0320785910 1.9464942880 1.8609099840 1.7753256810 1.6897413770 1.6041570740 1.518572770 1.4329884670 1.3474041630 1.261819860 1.1762355560 1.0906512520 1.0050669490 0.91948264550 0.83389834190 0.74831403840 0.66272973490 0.57714543140 0.49156112790 0.40597682440 0.32039252090 0.23480821730 0.14922391380 0.06363961030[ Conjecture.[We conjecture the limit[goes to 0 in this case.eActzH,r((r^(2))/(2-(4-r^(2))))02000bellipse.EAC010000000ec4^,[Shrinking Circle and [CP 300[\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL: http://www.radford. edu/~wyang[ Objective.[Some problems on finding[limits can be done using[CAS capability. However,[it will be more exciting [to solve them together[with geometry animations.\ Animation! tap-->1p` fT0`0vueFH#"x(vrALCuxy@  @  H! O@65AeY  @6r%hFTC !A@6 @    @   ! @6r%hFTC  @   BJ @    @6r%hFTBY`@    @6r%hFTBY`!H  ! @6`H! E@6'GF@  @  W2g@@  !H W2g@@  5@  H  B@6 $&@hq   4H'XbYH# 'GF%@    @6 @   [ A problem.[We have a fixed curve[C1: (x-1)^2+y^2/4=1and [we have[another circle[C2: x^2+y^2=1, which [will be shrinking.[[1) A is the center of C2.[2) B is the point of C2[on the y-axis.[3) N is the intersection [ of C1 and C2.[3) P is the x-intercept [of the line BN.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point P?[The following matrix is[the coordinates for the[point P as the circle C2[ shrinks to 0.[[We can see that the [limit (P) approaches [to 0.[ 2.5455844120 2.4600001090 2.3744158050 2.2888315020 2.2032471980 2.1176628950 2.0320785910 1.9464942880 1.8609099840 1.7753256810 1.6897413770 1.6041570740 1.518572770 1.4329884670 1.3474041630 1.261819860 1.1762355560 1.0906512520 1.0050669490 0.91948264550 0.83389834190 0.74831403840 0.66272973490 0.57714543140 0.49156112790 0.40597682440 0.32039252090 0.23480821730 0.14922391380 0.06363961030[ Conjecture.[We conjecture the limit[goes to 0 in this case.eActzH,r((r^(2))/(2-(4-r^(2))))02000cellipse2.EAC010000001b71^$[We have a fixed curve[ C1: (x-1)2+y24 =1, which[ implies y2 =4-4(x-1)2[and a shrinking circle[C2: x2+y2=r2Rsimplify(solve(x2+4-4(x-1)2=r2,x))Rx=--12r2+64-86,x=-12r2+64+86R define a(r)=4-16-3r23RdoneR define b(r)= 4-4(a(r)-1)2RdoneRa(0.5)R 0.03162505402Rb(0.5)R 0.4989988537R define l(r)=-r*a(r)b(r)-rRdoneRsimplify(l(r))R--3r2+16-4r3r--4-3r2+163-132+4Rl(1)R 15.14507448Rl(0.04)R 15.99870023Rl(0.03)R 15.99714324Rl(0.02)R 15.98748438R l(r)r0+R16Rl(0.01)R 16.02572035Rl(0.005)R 15.62492459R\4Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y1`Hh t                ( !4 "@ #L $X %d &p '| ( ) * + , - . 0 1 2 3 4 5 E$ F0 H< IH JT K` Ll Mx N O P Q R S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1 ^Sheet2 ^Sheet3 ^Sheet4 ^Sheet5 ^SheetSheet3D 0@PSheet1 ^Sheet2 ^Sheet3 ^Sheet4 ^Sheet5 ^SheetSheet3D i ^ $ N 3x(-((-3*x^2+16)^(1/2)-4))*x/(3*(x-(-4*((-3*x^2+16)^(1/2)/3-1/3)^2+4)^(1/2)))`P`P`=d=d=dI5I5 =d``=d=d=d(10qy`=d#YuY=d QR `(W$0`#hBRc P[eActaH,r((4-(16-3*r^(2)))/(3))bH(r(4-4*(a(r)-1)^(2))lH(r-((r*a(r))/(b(r)-r))02000cellipse3.EAC010000000f91  ellipse3.EAC version2.ACT #-04G^E ' R^d?    T̰ج` P: e5y  8Z[\Έ #R@)eY P e)4hIp Dvdr AkLxC u x y   !@  4B !@Q6C!evrXc Y W!iQ  :v `CC(P&W CH BW߆C뤆 Ęڈ H o~  Y )F%XՎh ddXD#G%YdG @    < DW2gf@+I7 C'Cs(PW #G%YdWM j&ʘ&'Ƙ<"@=.Ș.O,$ jA 6[.H!o.@  "6@ .*D D@[6b!evrXm Y W!iI Q J``z"2 !![\Έ L# RVVQYRXG`%ee 5%t Dvxrr AvL'C uxy  !@  ( #@76>P!Xx7I Y ghX "&aY3jH!TCC5PC `CBpC8C@"ڈވHoO!{ 'Y 9F2Vh"dd#XD%)d $7Fp#Œ!$ H" "!34W2gf@+#7.%O'@C&e#Y5Pl %)d $7Fpp00 j'&@ &('H <").@.*1,+$ A 6-,o!oa-)-,*.H /2 @A6HP!Xx8S Y gh` Qk +J0`@J`z"2"&![FeAct020010straightline.EAC01000000045d straightline.EAC version2.ACT$'148K^E '[\ animat[ΈC_Hduk `)`t5 PxtDvr m A 'L{xy qq@  H!I@g6 a(67 UBT"d&  40T" U&Y`T `@  + @:60*4hm!)AC]/? ȆCHP ފ oဏ"@. ?5H!2 D <Dy' !i 5Y   )C@06; % R@  Yo  K xshB YhH!zER5GE&bX! B2 wXH {F{`{`wUoˆ!@6 y %@  + Ę0 !@HF87RYcaEU rrrz Ȉ. 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