00010003010008main.ACT0002020012eActivity Save.EAC0100000017aa differentiation.EAC main.ACT#&,/3" ~[+DW and  its Applicss\4 AuthorUci[] Objective:S In this a ity, we 0study!e consequencesof Mean Valu:WTheorem Rolle's : on1ble Funon6froPalytl,graphl/ geometricpersp. (See notes below for more on the Mean Value <Theorem and Rolle's  )L t Example 1: 1.Let f()=3+2-.VerifyatmvNJholds given functi[-2,1]. Solu':+a)Using a graphical approach,lu% determin slope ofsecant l%D$containing the points (-2,f)),(1 1)). Drag f() to>7"Graph Window" belowTFirst,tap Analysis-GSolveY-YCalT calculate^y-values correspondŽ =-2 and 1.We ob-2)&0To draw secant line,  Sketch-Line.DThen ps 1entervm-2)]~ s1). The equation of t secant line is y=.\Graph Windoẅ26<@[L b) Next, we use aogeometric approach to Cverify existence a number ininterval{(-2,1)at ssfiesconclus MeanValue 3orem for/0obta8da)RWe draw f()GyX andstructtange line to the curve.To animat tangent 1, A Click Edit-A/-  Go(Once).QFromkZion, weN conjecturhat assumes alslope of approxly 1twice in whole cycle. verify this,w}er b>k}}s for,G3gfTap poiLox,usele;apk)the tangent line,slope and select&ble.\Geometry WindowΈ26Ί E `IV`8 <v r  A cL'C uxy q@  umH E@6ۉ$ G !UBgehY  AƒD` `  +@  A޼0" *1*@ k@ x^3+x^2-x9 9c9 5H!C@6WPHd% ٌ ( Du!@5.@!G3a# VxCiV YeRW#pU g|B.@#\r .@  21H2 9 \?p! }[  Based on the table of 'slopes generated, weconclud9hat at Vapproxim3 ly =-1.2V~and.53,yg>nxt linis_1.x valu ` e-2 7.00012500 -1.842105263 5.49595803 %6&% 4.14137557%E15789% 2.936377547J3Ko 809641m1m 0.9751350$ml$2188905$ -0.8947o8 0.38776952'$;44845085OڌQ1523361vO3102427?,1855429ač鉍17J821 0.05263157895 -0.8863976443 0.21*8' 4459425882'3684+: 1440969547Mt` 883720987LK 1.772926m3 2.811722521 4.00009997  c)Analytically,we can culate the existenceGof se numbers by<\solving for inEequation f'(x)=1.We obta0result =-1.2, .54 in the calculation row:R4 solve(diff(33+2P -)=1,)RLxr=]- 15250437y,5= 0 85837704[_Noteat re is a slight ʆncevalues obtained due to?numeri, approxim2s.ad)To visualizcconclusbof}Mean eV| Theorem, we drawf(), the tangent lines  at =-1.2, .54, toge:r withD7secaK in_"Graph < Window".iTap Analysis-Sketch-jkT; q1 to draw. Example 2. Find anpval [g,h] onwhich PN=?4G+3\-)2))-2% satisfies conditionshof Rolle's Theorem.  Solution: a)Graphical and Tabular 2N Approach2 Click "2 9 Window"1fbelowOtap(QButton.Tap Analysis-Gve-}Root to determine thers -21 f()\Thus,3inAvw?desire is [-2,1])Now9y4point satisfyingpYy,oc0?cle dgeneratesthe table of slopes. We conjecture from / !9that f'(x)=0 somewhe9in7terval^(-1.3,25).UuToufirmis information, we dragn) to "2 Graph9Window" and demine itstroot by pping Analysis- GSolve-R)obtax~29Not 4edQ4calcul row a follows:Rdiff(4+3-)2) ) -2,)RH4xT3+3B-2,+1\2 Graph Windoẅ =[h b)Analytical Approach: In the culation row, !we solve f'(x)=0 to \obtain ~-1.29. R{ 8(IIzILxH=]-f 1.288584347, x= 0.2692921734%-34855L06C V,_`+|`^[u Exploratory Exercise: Verify if the function f()=+1 - , [0,2]sO satisfiesVhypo^sesv and conclusno~uMean Value Theorem on^the given interval.   Solution:a) Graphical and Tabular; Approach2FClick p"ExploraT fWindow" belowTap4o button toverifyWat f() is notcontinuous at [0,2].ThtbNumberle or Summary '(1) does`  exist.SThshowsw Mean Value orem$9does not hold for f() on the given interval.\ Explorati'WindowЈ8<BF[Q b)Analytical Approach:xUsing a similar solug@ Example 1a), weaobtaiequof secant line con: (0,-1) andy(2,3) as y= 2x-1./culrow we checkTatY to f'(x)=2 isrealx follows:R solve(diff(+1 - )=2,)RAxI=R1[-c-,6+6[dMoreover, we can verify in thlculation row that!tangent line to@graph of f() at =1_does not exist:WRAH[LineEB,1D Undefined This confirmsfact2that the Mean Value Theorem does not holdfor f() on [0,2]. :\C Notes]a[ieAct020012ellipse-tilted.EAC010000001b47^[\``vrALCuxy1@  H! @6#b (Q Gg16  ! I Yp`g'CrgvU'YH! 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" # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     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 ^ $ `P`P`p`p `(10qy`#YuY ` O[eAct010013shrinkingcircle.ACT000302000fshrink1_mrd.EAC0100000015c8^c[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-->pb4`y`DXBb'd%vr ALCuxy@  H! L@6s PugPpp8Y  @6r%hFTBY`@  " @6XE& eX&( eX&(Y !H ! @6` !A@6 H! E@6'GF @  @    @6r%hFTBY`W2g@@  !H W2g@@  @  H# F@6 fTxD @  H  B@6@    @6 %@  @  @   @65@   SU3Y3   4H'XbYH# 'GF   [ A problem.[We have a fixed circle[C1: (x-1)^2+y^2=1 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) F is the top [intersection of C1 and C2.[3) L is the x-intercept [of the line BF.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point L?[The following matrix is[the coordinates for the[point L as the circle C2[ shrinks to 0.[ 3.4142135620 3.4604411890 3.5037317040 3.5443321250 3.5824495230 3.6182593660 3.6519117310 3.6835359990 3.7132444830 3.7411352510 3.7672943640 3.7917976680 3.8147122360 3.8360975520 3.8560064780 3.8744860570 3.8915781810 3.907320150 3.9217451440 3.9348826190 3.946758640 3.9573961690 3.96681530 3.9750334670 3.9820656070 3.9879243070 3.9926199070 3.9961606680 3.9985528330 3.9997965550[We can see that the [limit (L) approaches [to 4.[Prove anlytically[Let B be the point (0,r)[and (x1,y1) be the [coordinates of F and[(z,0) be the coordinates[of L. [Step 1.[First, we need to[find x1 and y1 in terms[of r.[The point F lies on both[ circles so[x12+y12=r2 (1)[(x1-1)2+y12=1 (2)[It follows from (2) that[x12 -2x1+1+y12=1 and[2x1=x12+y12=r2.[Hence [x1=r22 and[y1=r2-r44=r1-r24.[Step 2.[We need to find z in [ terms of r.[Let F' be on the x-axis[so that FF' is [perpendicular to AL.[By the similar triangles[=====================[ length AL length AB= length F'L length F'F,[=====================[zr=z-x1y1.[This implies that z=rx1r-y1.[Substituting x1 and y1[finally gives z as a [function of r as follows:[CAS capability:R define z(r)=r22-4-r2RdoneR z(r)r0R4R[Remark: [We can change the[size ofthe shrinking [circle butwill not change[the limit solong as the [radius of the shrinking [circle starts out greater [than theradius of the [ fixed circle.[[ Exercise.[What if the radius [of the fixed circle is [greater than the[shrinking circle; try [the fixed circle [(x-1)^2+y^2=4 and see [ what happens.eActzH,r((r^(2))/(2-(4-r^(2))))020010test-April06.EAC010000001cbe^[\a)YcdwpY$`i#UHBwvrALCuxy!H  @6)@4YGT H  C@6 P  `B@6@P@  @   ! @6GTY)@4YAP  H  D@6'Q SRIcw @   @  HW2g@  '@  !!  P 'Q SRIcw @P  &@  '@    .@  ,   A@6.@  .@  @   ! @6)@2YGQ @   RAsolve(0.6039x^2+1.456y^2-1.748xy-0.1382x-0.0864y=0,y)Ry=437x728-17480x+8642-582406039x2-1382x29120+27910,y=437x728+17480x+8642-582406039x2-1382x29120+27910R\Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1\Hy2x\H       ( 4 @ L X d p |    ! " # $ % & ' ( ) * + ,$ -0 .< 0H 1T 2` 3l 4x 5 E F H I J K L M N O P Q R S, T8 ]D ^H _L `P aT bX \ h t   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 ^ $ I 3x437*x/728-((17480*x+864)^2-58240*(6039*x^2-1382*x))^(1/2)/29120+27/910I 3x437*x/728+((17480*x+864)^2-58240*(6039*x^2-1382*x))^(1/2)/29120+27/910`P`P`` `(10qy`#YuY ` Q[eAct02000cversion3.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%$"# #"  )[ A problem.[We have a fixed circle[C1: (x-2)^2+y^2=4 and [we have[another circle[C2: x^2+y^2=2, which [will be shrinking.[[1) A is the center of C2.[2) B is the point of C2[on the y-axis.[3) P is the top [intersection of C1 and C2.[3) Q is the x-intercept [of the line BP.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point Q?[The following matrix is[the coordinates for the[point Q as the circle C2[ shrinks to 0.[ 7.8366652190 7.8476774960 7.8582833660 7.8684861710 7.8782890920 7.8876951530 7.8967072290 7.9053280470 7.9135601940 7.9214061160 7.9288681270 7.935948410 7.9426490230 7.9489718990 7.9549188510 7.9604915680 7.9656916330 7.9705204990 7.9749795340 7.9790699760 7.9827929450 7.9861494860 7.9891405580 7.9917668590 7.9940292420 7.9959283810 7.997464430 7.9986371860 7.9994592530 7.9999221750[We can see that the [limit (L) approaches [to 8.[ Exericse.[Prove anlytically!!eActzH,r((r^(2))/(2-(4-r^(2))))01000cversion2.ACT000502000ba cubic.EAC010000001005^,[Shrinking Circle and [CP 300[\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL: http://www.radford. edu/~wyang[ Objective.[Some problems on finding[limits can be done using[CAS capability. However,[it will be more exciting [to solve them together[with geometry animations.\ Animation! tap-->1p` fT0`0vueFH#"x(vrALCuxy@  H! P@6iPHViY  @6f5 Hb  esD`@    @6r%hFTBY` @  H! O@65AeY  @6r%hFTC  !A@6 @  @   ! @6r%hFTC  @  BJ@  @    @6r%hFTBY` @  H! N@6rIHD 74 @  H  B@6 @    @6  %@  @    1/10x(x-3)(x-5)!H  ! @6` H! E@6'GF @  !@  W2g@"@  #!H W2g@$@  %5@  $&@hq   4H'XbY&H# 'GF!    %[ A problem.[We have a fixed curve[C1: y=110x(x-3)(x-5) and [we have[another circle[C2: x^2+y^2=2, which [will be shrinking.[[1) A is the center of C2.[2) B is the point of C2[on the y-axis.[3) N is the intersection [ of C1 and C2.[3) P is the x-intercept [of the line BN.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point P?[The following matrix is[the coordinates for the[point P as the circle C2[ shrinks to 0.[[We can see that the [limit (P) approaches [to 0.[ 2.5455844120 2.4600001090 2.3744158050 2.2888315020 2.2032471980 2.1176628950 2.0320785910 1.9464942880 1.8609099840 1.7753256810 1.6897413770 1.6041570740 1.518572770 1.4329884670 1.3474041630 1.261819860 1.1762355560 1.0906512520 1.0050669490 0.91948264550 0.83389834190 0.74831403840 0.66272973490 0.57714543140 0.49156112790 0.40597682440 0.32039252090 0.23480821730 0.14922391380 0.06363961030[ Conjecture.[We conjecture the limit[goes to 0 in this case.eActzH,r((r^(2))/(2-(4-r^(2))))02000acubic2.EAC010000001005^,[Shrinking Circle and [CP 300[\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL: http://www.radford. edu/~wyang[ Objective.[Some problems on finding[limits can be done using[CAS capability. However,[it will be more exciting [to solve them together[with geometry animations.\ Animation! tap-->1p` fT0`0vueFH#"x(vrALCuxy@  H! P@6iPHViY  @6f5 Hb  esD`@    @6r%hFTBY` @  H! O@65AeY  @6r%hFTC  !A@6 @  @   ! @6r%hFTC  @  BJ@  @    @6r%hFTBY` @  H! N@6rIHD 74 @  H  B@6 @    @6  %@  @    1/10x(x-3)(x-5)!H  ! @6` H! E@6'GF @  !@  W2g@"@  #!H W2g@$@  %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.EAC010000001568^c[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-->pb4`y`DXBb'd%vr ALCuxy@  H! L@6s PugPpp8Y  @6r%hFTBY`@  " @6XE& eX&( eX&(Y !H ! @6` !A@6 H! E@6'GF @  @    @6r%hFTBY`W2g@@  !H W2g@@  @  H# F@6 fTxD @  H  B@6@    @6 %@  @  5@   SU3Y3   4H'XbYH# 'GF   [ A problem.[We have a fixed circle[C1: (x-1)^2+y^2=1 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) F is the top [intersection of C1 and C2.[3) L is the x-intercept [of the line BF.[[We would like to see[when the radius of C2 [(r)goes to 0, what [happen to the point L?[The following matrix is[the coordinates for the[point L as the circle C2[ shrinks to 0.[ 3.4142135620 3.4604411890 3.5037317040 3.5443321250 3.5824495230 3.6182593660 3.6519117310 3.6835359990 3.7132444830 3.7411352510 3.7672943640 3.7917976680 3.8147122360 3.8360975520 3.8560064780 3.8744860570 3.8915781810 3.907320150 3.9217451440 3.9348826190 3.946758640 3.9573961690 3.96681530 3.9750334670 3.9820656070 3.9879243070 3.9926199070 3.9961606680 3.9985528330 3.9997965550[We can see that the [limit (L) approaches [to 4.[Prove anlytically[Let B be the point (0,r)[and (x1,y1) be the [coordinates of F and[(z,0) be the coordinates[of L. [Step 1.[First, we need to[find x1 and y1 in terms[of r.[The point F lies on both[ circles so[x12+y12=r2 (1)[(x1-1)2+y12=1 (2)[It follows from (2) that[x12 -2x1+1+y12=1 and[2x1=x12+y12=r2.[Hence [x1=r22 and[y1=r2-r44=r1-r24.[Step 2.[We need to find z in [ terms of r.[Let F' be on the x-axis[so that FF' is [perpendicular to AL.[By the similar triangles[=====================[ length AL length AB= length F'L length F'F,[=====================[zr=z-x1y1.[This implies that z=rx1r-y1.[Substituting x1 and y1[finally gives z as a [function of r as follows:[CAS capability:R define z(r)=r22-4-r2RdoneR z(r)r0R4R[Remark: [We can change the[size ofthe shrinking [circle butwill not change[the limit solong as the [radius of the shrinking [circle starts out greater [than theradius of the [ fixed circle.[[ Exercise.[What if the radius [of the fixed circle is [greater than the[shrinking circle; try [the fixed circle [(x-1)^2+y^2=4 and see [ what happens.eActzH,r((r^(2))/(2-(4-r^(2))))020010straightline.EAC0100000007ed^[\ animationu`7&AP`PG&APvr ALCuxy@  H K@6 @Sp @6h426Q #2Q QB"57`@  ! @6 Y@  H! I@6hpvhpv @  H D@6y 2H  H  J@6hpv`hpv`@  ˆ! @6y!A@6 %@  @  @    @6gbYgb @  @   @  H! H@6y`@  @  !`G@6rPP   @6D#4YY IXi @  5H!  ! @6 Y  C@6 %!@  "@    xshBY "! #@  #    [ Conjecture:[If the fixed curve is a line[with positive slope then[the limit tends to 0.[y=xeAct