0001000601000aicme10.ACT0005020012areaoftriangle.EAC010000001559t^4P[Are the areas the [same?[\AuthorProfessor,Dr. Wei-Chi Yang Department of Maths/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL:www.radford.edu/ ~wyang [[ Objective.[We will explore how to [setup the formula of [finding the area of a [ triangle.[[ Example 1.[Animate the following['Geometry' strip, and [see if the areas of each[triangle CDE should be [ the same.\Area of the triangle``vrALCuxy!@    @6bvWvY Y"AH Ue`c !`A@6` Y!`B@6EP @   @  [  @6bvWvY Y"AH VGWB` H! E@6Witqc`&HD  `D@6  @  @  r@@! H C@6u'7P f86'Y  @6ci rC hpY @  R) @    @6 x"ISYS4YpH)  @  e 5H# vVBA &1WG7    [ Solution:[We record the point C[and the area of CDE[ as follows:[[-2.15-0.9 3.061192603 -1.855263158 -0.8131578947 3.061192603 -1.560526316 -0.7263157895 3.061192603 -1.265789474 -0.6394736842 3.061192603 -0.9710526316 -0.5526315789 3.061192603 -0.6763157895 -0.4657894737 3.061192603 -0.3815789474 -0.3789473684 3.061192603-0.08684210526 -0.2921052632 3.061192603 0.2078947368 -0.2052631579 3.061192603 0.5026315789 -0.1184210526 3.061192603 0.7973684211-0.03157894737 3.061192603 1.092105263 0.0552631579 3.061192603 1.386842105 0.1421052632 3.061192603 1.681578947 0.2289473684 3.061192603 1.976315789 0.3157894737 3.061192603 2.271052632 0.4026315789 3.061192603 2.565789474 0.4894736842 3.061192603 2.860526316 0.5763157895 3.061192603 3.155263158 0.6631578947 3.0611926033.450.75 3.061192603[We note that even if we[moves C along the line[segment AB, the area[of CDE stays the same.[[ Example 2. [Find a formula for the[area of a triangle.[Hint:[1) Open the Geometry[strip above and tap on[line segments DE and AB.[2) Are they parallel? [->Yes, because the angle[between these is 0.[This means the distance [from C to DE stays the[same as C moves along[the line segment AB.[3) To find the distance[tab on C and DE and the[distance measurement.[The distance is 2.193808.[4) Note the followings:[->base of triangle CDE[ =DE=2.790757,[->base of triangle CDE[height= 2.193808.[->Area=3.061192603.[5) Is it right that the[area of a triangle is[(1/2)(base)(height)=R(1/2)*(2.790757)*(2.193808)R 3.061192516[ Exercise.[Draw an arbitray triangle[using Geometry strip. [Verify the formula of the[area of a triangle.\ Any triangle``vrALCuxy@  !`A@6E`PY ! @6D't!Y iRIVYEXxvT`!`C@6H8pwB5H8p`@  {P@   ! @6 %qc2V AW9`!`B@6P 0 @  ` !@  " !! @63aA4UYErb2 TIc$`#@  "!#""[-> Tap a point, say A.[-> Tap the segment [opposite to the point, [say BC.[-> Measure the height [ and the base.[ For example,[ base=5.967354[height=4.934361R(1/2)*(5.967354)*(4.934361)R 14.72253943[By clicking three sides[of the triangle, we see[that the actual area is[the same as what we [ expected.eAct020009deriv.EAC010000003749DkxN[Derivative of Sine [function[\AuthorzProfessor Dr. Wei-Chi Yang Radford University Radford, VA 24142 U.S.A. e-mail: wyang@radford.edu www.radford.edu/~wyang[ Objective[We will use graphical [representation to show [the derivative of [Sine function is Cosine [ function.[[ Example. [ Show that [   ddxsin(x)=cos(x).[ graphically.[Step 1.[-Insert a Geometry strip[below. [-Select the coordinate [system.[-Under 'Draw', select [ function.[-Type in sin(x).[Step 2.[Under Draw, select [construct->tangent to [a curve.[-Selectling the point and[ the curve.[-Edit->Animate->add [animation->go once.[Step 3.[-Select the point on the[ sine curve.[-Select the table [(you should get the[table for y=sin(x)).[-6.47 -0.185729957 -5.799473684 0.4650681779 -5.128947368 0.9144868832 -4.458421053 0.9679231158 -3.787894737 0.602238423 -3.117368421-0.02422186341 -2.446842105 -0.6401938273 -1.776315789 -0.9789551069 -1.105789474 -0.8938184925 -0.4352631579 -0.4216490368 0.2352631579 0.2330989023 0.9057894737 0.7869125563 1.576315789 0.9999847678 2.246842105 0.7800530202 2.917368421 0.2223500815 3.587894737 -0.4316327893 4.258421053 -0.8987141065 4.928947368 -0.9766427299 5.599473684 -0.6316747436.27-0.01318492514[-Select 'slope' of the [tangent line at the point[and the table. This [should show the table [ of slopes.[[Step 4.[Now drag the x-column[and the slope-[column into the graph, [ we should see[the graph of y=cos(x)[(why??)[-6.47 -0.185729957 0.982600828 -5.799473684 0.4650681779 0.8852806827 -5.128947368 0.9144868832 0.4046269738 -4.458421053 0.9679231158 -0.2512344783 -3.787894737 0.602238423 -0.7983087507 -3.117368421-0.02422186341 -0.9997066084 -2.446842105 -0.6401938273 -0.7682054202 -1.776315789 -0.9789551069 -0.2040634768 -1.105789474 -0.8938184925 0.4484401012 -0.4352631579 -0.4216490368 0.9067643853 0.2352631579 0.2330989023 0.9724530345 0.9057894737 0.7869125563 0.6170743608 1.576315789 0.9999847678-5.506936664-3 2.246842105 0.7800530202 -0.6257036712 2.917368421 0.2223500815 -0.9749668915 3.587894737 -0.4316327893 -0.9020440145 4.258421053 -0.8987141065 -0.4385237748 4.928947368 -0.9766427299 0.214881885 5.599473684 -0.631674743 0.7752414216.27-0.01318492514 0.9999130757[-Drag the y-column and[ slope-column.[into the graph, we see [the unit circle. (why?)[Because we are drawing[ x=sin(t) and [ y=cos(t).\ ConstructionsG`rXIf`&rXIfvrALCuxyH  rWYew Hh g1 ## B!4Y@8'0Y xQYPY!d6Y3 0 %V0 Gg 0 "5P 1c'0YqAPY vd')Y1gGCY1Q@Y `( (p bis Q#Dx0Y0PpY pf@Yh T Y4vYHD vC0 rE04P C` Pi6f@Y%p6q Y thPY @PY8R7tY u$! 0up H" G`ysh@`G6`E!0`xxsp`shB`DhBP`wcx`WG@`5&1WY5&1W xsp Wcx$hBPshBXxsp%!0G6Ysh@' `( (p bis Q#Dx0Y0PpY pf@Yh T Y4vYHD vC0 rE04P C` Pi6f@Y%p6q Y thPY @PY8R7tY u$! 0up 5H  H  A@6 2i bu   sin(x) G`&pRcx .@  3     @6'))YI(`4! !s"Y .@   .@   1@  9 [ Exercise 1.[Use the similiar procesure [described in the Example [to showthat [  ddxcos(x)=-sin(x).[\Try your answer here``vrALCuxyH  f!P 6DC! YU)`Y a E60Y ihUYi@Yg'"`YR% @ !0 eW3'` eW3'` !0 R% @ g'"`Yi@Y ihUY a E60YU)`Y6DC! Yf!P  XBuY qdP@Y!5`0YuFPYD6P Av cB 7 gfi  3 `0 `QY 9& Y gffPY cB 7Y9pYD4Ax0Yu0p !7gtp qdP@ XBu H  `G6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&06!0G6 BRc shBG6! XBuY qdP@Y!5`0YuFPYD6P Av cB 7 gfi  3 `0 `QY 9& Y gffPY cB 7Y9pYD4Ax0Yu0p !7gtp qdP@ XBu 5H  H  B@6 Y!`h'   cos(x) Y`&1WG7 .@  3    @6f&08qY"u@4 CD7&`.@  .@  1@  9[ Exercise 2.[#Use the similar procesure described[in the Example to show that[ddx (tan(x))=sec2x.[\Try your solutioin here``vrALCuxyH" 8`````A@YQT uR YIQ` `D0`i@WfYi@Wf D0` YIQ` uR YQTYA@ ``8``$("x# P)#Y@58P#89$`W Qdg0%h%i0dg0W Q 9$`#9@58#Y9pP)x# $("H" `G6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&06!0G6 BRc shBG6!$("x# P)#Y@58P#89$`W Qdg0%h%i0dg0W Q 9$`#9@58#Y9pP)x# $("5H  H  B@6P 2q2(   tan(x)P `&1WG7 !.@  "3! #  @6 tGTtY2h07 $qdE` $.@  "%.H" "!#%$"#&1@   9 #[\A bonus parametric equation' Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind xt1Hyt14 HT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F H( I4 J@ KL LX Md Np 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i Dkx $  4tT(t) 4t(1/P(t))^2`P`P`p`p `(10qy`#YuY #He5RIpryQq ` Q[Note that the parametric[equation of x(t)=tan(t) and[y(t)=sec(t)^2 is a parabola.eAct02000cgeometry.EAC010000005ef0<$a[A Geometry Problem\Author|Professor Dr. Wei-Chi Yang Radford University Radford, VA 24142 U.S.A. e-mail: wyang@radford.edu www.radford.edu/~wyang [[ 1. Objective:[In this e-activity, we [ would like to[explore area of [isoceles ) with [inscribed unit circle.[[ 2. Example:[Find the minimum area of [the triangle with various[angles.[ 3. Exercise:[Find the maximum area [of ) with various [angles. [[Solutions to Example[and Excercise:[(a) With animation:[By 'animate the figure' [below, we may conjecture [when the areacould be [minimum or maximum.\Animate figure -->`"0D f`"yi`7Txw#(gSvr ALCuxy@  H # H@6GuQ  @6 tt44Y#3cq2 YH K@6%P6``H  ?< @    @6 tt44Y#3cq2Y H  L@6%P6`? H ?~` @   ! @6  @!  ?p5@   @6`H  C@6!D@6$@  @  i8wU% VAVA @   @6 tt47 #3cq3 `  @6 `A@6@! B@6%@  H @  ! @6 tt47 #3cq3Y@   H  !@! " @6 #@! $@  %@  "&@  '!`J@6`"(@! ) `I@6``""')*@  +@  G@6fq&xYRa) +,@  -@  F@6$89 f98x -.@  /@  E@6`/0"@!   /-+)'"(&$#   ([-Edit/Animate/Go(once)[[(b) With traditional [ analysis:[It can be shown that the[ Area of ) =[12(1+1sin(x2))2sin(x)(cos(x2))2[(c) With graphical approach:[By Dragging area [function to'Graph [Window' below. We may[use the command ['Analysis/G-Solve/Min' to[anlayize the minimum of[the area function, which [happens at x=60 degree.[\Graph Window -->1 Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y18H@ L X d p |             ! " #$ $0 %< &H 'T (` )l *x + , - . 0 1 2 3 4 5 E F H I J, K8 LD MP N\ Oh Pt Q R S T ] ^ _ ` a b      system]^system^system_`system`asystemabsystemb( LISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLIST abGrapbGraph2D system]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLIST  0@PSheet1 Sheet2 Sheet3 Sheet4 Sheet5 SheetSheet3D 0@PSheet1، 8 Sheet2، 8 Sheet3، 8 Sheet4، 8 Sheet5، 8 Sheetr-value: i <$ $ ( 3x1/2*(1+1/L(x/2))^2*L(x)/(P(x/2))^2`P`P``HHP`P?(10qy`#YuY $gS$gS@UBQ3``&1WG7 P[[Remark: By analyzing [the graph,we see vertical [asymptotesat x=0 and Pi. [We predict maximum [occurs at those two [points.[[(d) With table:[After finished the [aninmationd escribed [in (a).[-Select two sides HK [and HL,and select angle [ and table.[-Next select three sides [of the triangle and [the area of the triangle.[-We get the following list:[(0 2.510073601 130.1816945 9.519763863 112.153776 7.229996322 99.76166621 6.319875664 90.32199451 5.842075726 82.75495461 5.56236532 76.49323976 5.39175351 71.19529043 5.28828893 66.63742715 5.229370728 62.664503 5.201653507 59.16429038 5.196709119 56.05295216 5.208933353 53.2662375 5.23444441 50.75386705 5.270462767 48.47580285 5.31494258 46.39968637 5.366342622 44.49902925 5.423478496 42.75190509 5.485424227 41.13998419 5.551444896 39.64780864 5.62094943 38.26223953 5.693456802 36.97202936 5.768571391 35.76748749 5.845964705 34.64021526 5.925361637 33.58289423 6.006529977 32.5891154 6.089272316 31.65324026 6.173419726 30.77028686 6.258826772 29.9358358 6.345367547 29.14595207 6.432932483 28.39711975 6.521425771 27.68618709 6.610763257 27.01032007 6.700870713 26.36696297 6.791682411 25.75380476 6.883139934 25.1687502 6.975191181 24.60989507 7.067789535 24.07550469 7.160893158 23.56399534 7.254464387 23.07391807 7.348469228[[If we want more detailed list[between the angles and the[areas, we use the stats[application below.[\Lists-->2Graph2Dh Graph3D| LISTCALD LISTSYS4NModify $ STATCALC $NSTATSYS ,\NSequence, Sheet| Sheet3D0| SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict4$ ViewWindX t               (  4 !@ "L #X $d %p &| ' ( ) * + , - . 0 1 2 3 4 5$ E0 F< HH IT J` Kl Lx M N O P Q R S T ] ^ _ ` a b dd e`d      0̒1/2(1+1/sin(x/2))^2sin(x)/(cos(x/2))^2" eActxfeActareafsystem`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemeActmxlisteActmarea=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i <$ $ `P`P`y`@%p)rf`7Q(10qy`#YuY yX6b5p $0<HT`lx     @`uxyyy $0<HT`lxYYcY5Ya`HyTe`aI(X6fGH 3U2F`7sC)E(8xGEaRB'HgC63@TGQC'S8sP06C 62)GU2AX7 T X6b3SX6b5p`%# R[Notice that by carefully[examining the list, we [conjecture that [ a) when x[(angle)is close to 0 or [180 degree,the area is [close to a maximum.[(b) When x is 60 degree, [the area is a minimum.[[Approach with Calculus[We analyze the graph of[the derivative of area[ function.[\Area and its deriv.3`Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y18Hy2T@H            $ 0 < H  T !` "l #x $ % & ' ( ) * + , - . 0 1 2 3, 48 5D EP F\ Hh It J K L M N O P Q R S T ] ^ _  ` a b  ( 4 @ L 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 <$ $ ( 3x1/2*(1+1/L(x/2))^2*L(x)/(P(x/2))^20 3x:(1/2*(1+1/L(x/2))^2*L(x)/(P(x/2))^2,x,1)`P`P`x`xx3vb3vbx`xxx(10qy`x#YuYx `cxsh Q[By examing the graph of[the derivative function,[it crosses from - to +[at x=60, we know that[the minimum area is at[x=60 degrees and[the maximum is when [the angle approaches to[.eActangles  $0<HT`lx @`u8area d $0<HT`lxYYcY5Ya`HyTe`aI(X6fGH 3U2F`7sC)E(8xGEaRB'HgC63@TGQC'S8sP06C 62)GU2AX7 T X6b3SX6b5pm_angle m_area volume D $0<HT`lxY5Ya`HyTe`aI(X6fGH 3U2F`7sC)E(8xGEaRB'HgC63@TGQC'S8sP06C 62)GU2AX7 T X6b3Sx d $0<HT`lx         ( 2 < P d x   xyyyy 020010lineartransf.EAC0100000047f3H]<0[Linear transformation\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail:wyang@radford.edu URL: www.radford.edu/ ~wyang [ Objectives.[1. We will investigate [the relationship[between the linear [transformation and [geometry figures.[2. We extend these ideas[to notions of eigenvalues[and eigenvectors.[[Type I. Transformation[under -1001.[[ Example 1. [ What will the[effect of this 'linear[transformation' on any [ point (x,y)?[ Method 1.[Execute the following[command, we see that[(x,y) is sent to (-x,y),[which means that theR-1001xy+00R-xy[Method 2. Graphical[Representation:[Pick three points and see[what happens under this[transformation.R-1001012023R0-1-2023[[-Drag the matrix[012023 into the [Geometry strip below[to get three points [(0,0), (1,2) and (2,3).[-Drag the matrix[0-1-2023 into the [Geometry strip below to [get (0,0), (-1,2) and [(-2, 3).[-Note that this is a [reflection along the [y-axis.[[Question: Does this[transformation preserve[ the area?\ Along y-axis``vrALCuxy@  H `@6 @  @623x Tp"RYH `@6@   @   @  @6gfYgf gdY H `@6 @   @   @  @6BqYG!5I  @    @  H `@6 @  @623x Tp"R H `@6`@  @   @  @6gfYgfYgd H `@6`@#  @   @! @6BqYG!5IY@   @  H `@6 @  @623x Tp"R H `@6` @   !@  " @  @6gfYgfYgd #H `@6`$@  #"!$"%@  #& @! @6BqYG!5IY#'@  &%oJ'&##&"  [Type II. Transformation[under 100-1.[[ Example 2.[ What will the[effect of this 'linear[transformation' on any [ point (x,y)?[Method 1. Algebraically:R100-1xy+00Rx-y[We see that this [transformation is a [reflection along x-axis.[Method 2. Graphically:[Pick three points (0,0),[(1,2) and (3,2). What[would happen under this[transformation?R100-1013022R0130-2-2R\ Along x-axis(I`y$P`Py$PvrALCuxy)@  *H `@6+ H  @6Tp"RY23xY,H `@6`*-@  ,+)-+.@  ,/ H  @60H `@6`,1@  0/.1/,2@  03 @  @6Bq G!5I *04@  *32430*5@  6H `@67 @  @6Tp"R 23xY8H `@669@  87597:@  8; @  @6`<H `@68=@  <;:=;8>@  <? @  @6BqYG!5I 6<@@  6?>@?<66<?8;7*03,/+[Note that this [transformation preserves[ the area.[[Type III. Transformation[under -100-1.[[ Example 3.[What will happen under[this tranformation?[[We shall see that this is[a relection along the [ origin (0,0).[R-100-1xy+00R-x-yR-100-1013022R0-1-30-2-2R\ Along (0,0)A``vrALCuxyB@  CH `@6D @  @6Tp"RY23x EH `@6``CF@  EDBFDG@  EH @  @6``IH `@6``EJ@  IHGJHEK@  IL @  @6Bq G!5IYCIM@  CLKMLICN@  OH `@6P @" @6Tp"R 23xYQH `@6OR@  QPNRPS@! QT @" @6`UH `@6QVH  UTS#VTQW@  UX H! @6BqYG!5I OUY@  OXWYXUOOUXQTPCILEHD[Note that this [transformation preserves[ the area too.[[ Exercise. [What is the transformation[under -200-2?[ Hint: TryR-200-2013022R0-2-60-4-4R\Preserve area?Z`y$P`y$PvrALCuxy[@  \H `@6] @  @6Tp"RY23x ^H `@6``\_@  ^][:x_]`@  ^a @  @6``bH `@6``^c@  ba`ca^d@  be @  @6Bq G!5PY\bf@  \edfeb\g@  hH `@6i @  @6Tp"R 23xYjH `@6hk@  jigkil@  jm @  @6`nH `@6jo@  nmlomjp@  nq H! @6BqYG!5I hnr@  hqprqnhhnqjmi\be^a][Type IV.[What is the following[transformation??R0-1-10xy+00R-y-x[Step 1. [The matrix 0110[is sending (x,y) to (y,x).[A relection along y=xR0110xyRyx[Step 2. [To go from (y,x) to [(-y,-x), we only need [ to multiply the matrix [0110 by (-1).[R-0110xyR-y-x[So this is simply a [reflection along (0,0).[Try the followings?R0-1-10013012R0-1-20-1-3[\Name the transformation.s``vrALCuxyt@  uH `@6v @  @623xYTp"R wH `@6``ux@  wvt:xvy@  wz @  @6Bq G!5IYG!5I {H `@6``w|@  {zy|zw}@  {~ @  @6gf gfYu{@  u~}~{u@  H `@6 @  @6Tp"R 23xYH `@6@  @  @  @6G!5IYBq G!5IYH `@6@  @  @  @6gfYgf @  u{~wzv[[ Discussion I:[(a) Suppose we start [with an arbitrary matrix[A, what is the [geometric interpretation[of Av?R123101R21R\Where does (0,1) go?Affffr`87 `Affffh87vrALCuxy@   `@6  s@6G!5IYBq  @6@  @ @   `@6 ! r@6`@  [We see that the vector[01 is transformed [to the vector 21. Since R123110R13[Similarly, the vector 10[is transformed to 13.[\Linear transformation%`3"pY%@ C9b'vrALCuxy@   @6  u@6G!5IYBq "@6@  @   @6  t@6`@  0@   `@6  s@6 Hh2Y"wfi @  @   `@6  r@6@  c[[Discussion II.[Now, let's look at some[special cases:ReigVc(1231)R 0.632455532- 0.632455532 0.7745966692 0.7745966692R1231 0.632455532 0.7745966692R 2.18164887 2.671963265\Look at what happens.``vrALCuxy@  !`@6Hgc&P  s@6tYfi'RY2EU1%  @6@  C @   `@62EU2 tYfi ! r@6tYfi$Y2EU2B @  [Notice that these two[vectors are just multiple[ of the other?[What is the 'multiple'?[->Use the measurement[to figure out the number.[The longer vector is[3.44949 times of the [shorter vector, why?ReigVl(1231)R 3.449489743,- 1.449489743[[Remark: [The effect when we apply[the matrix on the eigenvector,[v= 0.632455532 0.7745966692[is the multiplication lamda[ and v, where [lamda=3.449489743.R 3.449489743 0.632455532 0.7745966692R 2.181648871 2.671963265[ Exercise. [Explore the effect ofR1231- 0.632455532 0.7745966692R 0.9167378064- 1.122769927R\Another eigenvector``vrALCuxy@   `@6 sx@ 'ip`  s@6tYfi)h 2EU1W  @6@  ;@   `@62EU2YtYfi  r@6tYfi$Y2EU2BY@  `([eAct02000fprobability.EAC01000000560f^[ A probabilty[problem.[\AuthorProfessor Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, Virginia 24142 USA e-mail: wyang@radford.edu URL: www.radford.edu/ ~wyang[ Objective:[We will use real-life[problems to explore the[maximum or minimum of [some probability[ problems.[[Example.[Suppose we have 50 [white balls and 50 black[balls, which we are [going to put some white[balls and some black [balls into two urns A, [and B.[The rules are that we [have to have at least [one ball in each urn and[the number of black [balls doubles the the [number of white balls in[urn A. A person walks [into the ruoom.[ Our Problems:[(1) Maximize the [probabilitythat this person[draws a white ball.[(2) Minimize the probability[that this person draws[ a white ball.[ Solution:[Let p(x) be the probablity [that a single ball drawn[at random will be white[then we[ R define p(x)=12xx+x2+50-x100-x-x2Rdone[->Open the the graph[editor below and graph[the function p(x).\ Graph of pDGraph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1H4 @ L X d p |            ! " # $$ %0 &< 'H (T )` *l +x , - . /D 0 1 2 3 4 5( E4 F@ HL IX Jd Kp L| M N O P Q R S T ] ^ _ ` a b    $ 0 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xp(x)`P`P`p@`@@PdQ @`@@@(10qy`@#YuY@   $0<HT`lx !X`( TTTU P vv Ps%b Qcccd UUUUV EEEEE`EEEEE`` Q[[Explorations. [a)Find a view window[so you can see the max[and min of the function p.[(Try [-7,10] for x and[[-3.8,3.8] for y).[b)After the plot is done,[we choose the table[mode for the graph;[choosing x from 1 to 10[we can 'roughly' see[where the minimum and[maximum should be. [c) We trace the graph [of pabove to estimate[wherethe max or min of [ p could be.[d) Or we could try the [following values. [Rp(1)R0.5Rp(2)R 0.4219858156Rp(3)R 0.3920454545Rp(4)R0.3875Rp(5)R 0.4047619048Rp(7)R 0.5511363636Rp(8)R 0.8055555556Rp(9)R2.1[ Conjectures: [a) The minimum happens[near x=4. This means[we would put 4 white[balls and 16 black balls[in urn A, and 46 white[balls and 34 black balls[ in urn B.[b) The maximum happens[near x=8, but x=8 is not[suitable because x^2=64,[which is greater than[50. Therefore, when [n=50, we see that the [max of p(x) happens [at x=7. It means that [we have 7 white balls [and 49 black balls in[ urn A and 43 [white and 1 black[balls in urn B.[[Anaytical Solution:[1. First, we observe that[the vertical aysmptotes[for p(x) is at x=0,x=-1,[and at 100-x-x2=0, [which can be solved [below:Rsolve(100-x-x2=0,x)Rx=- 10.5124922,x= 9.512492197[So, the vertical [aysmptote we need is[at x=9.512492197. [2. Next, we observe that[the function p is [increasing from some x to[x=9.512492197, which [can be verified, [alternatively, by drawing[its derivative below:R define r()=p()RdoneRr(x)R-1x+12x2+x-1002x4-174x2-100x-48x3+5025R\-Drag the expressionGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $ `P`P`p`PdQ ``(10qy`#YuY pt` O[-By choose 'root' under['Analysis', we get the[root of r(x) to be[xc=3.709174999115, which[is the minimum of p(x).[-Since p(3)=0.3920454545[and p(4)=0.3875, the[minimum of p is at [x=4.[-Since r(x)>0 in [[xc,9.512492197], we [conclude that the max[of p happens at [somewhere near the end[point, in this case, x=7 is[the point we need.[[ Exercise.[Let's see what[happens when n=100[Rdefine q(x)=12xx+x2+100-x200-x-x2RdoneR\ graph of qGraph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y1H , 8 D P \ h t           ! " # $ % &( '4 (@ )L *X +d ,p -| . 0 1 2 3 4 5 E F H I J K L$ M0 N< OH PT Q` Rl Sx 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xq(x)`P`P``% `(10qy`#YuY `8'x6 ` P[ Explorations:[1) Using 'G-Solve'->Min,[ we found [xc=4.874168397486 to be[the min.[2) By using the tracing,[we estimated that the [max of q is about at[x=11.9. Let's try the[ followings:Rq(3)R 0.3829787234Rq(4)R 0.3666666667Rq(5)R 0.362745098Rq(6)R 0.3688969259Rq(7)R 0.3854166667Rq(8)R 0.4149305556Rq(9)R 0.4636363636Rq(10)R 0.5454545455Rq(11)R 0.6960784314Rq(12)R 1.038461538R[ Conjectures:[1) Since q(5)circle?HGraph2D Graph3D LISTSYS(4NModify \$ STATCALC NSTATSYS \NSequence, Sheet| Sheet3D| SolveEqSolveLwr SolveUpr StupFLG1$(StupListLD StupPict$ ViewWind r1Hr2 Hr3 Hxt4,$Hxt5P,Hxt6|,Hxt7,Hyt4$Hyt5,Hyt6$,Hyt7P,H|              $ 0  < !H "T #` $l %x & ' ( ) * + , - . 0 1 2 3 4 5, E8 FD HP I\ Jh Kt L M N O P Q R S T ] ^ _ ` a b    ( 4 system]listsystem^]=system_^system`_systema`systemb~////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// a9b  system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/system]listsystem^]=system/  0@PSheet1Jl^Sheet2Jl^Sheet3Jl^Sheet4Jl^Sheet5Jl^SheetSheet3D 0@PSheet1Jl^Sheet2Jl^Sheet3Jl^Sheet4Jl^Sheet5Jl^SheetSheet3D i ^ $   5na*P(n) 5na*P(b*n) 5na*L(b*n) 4t(2+P(t))*P(t/a) 4t(1+P(t))*P(t/5+a*P(t)) 4t(1+P(t))*L(t/5+a*P(t)) 4t(1+L(t))*L(t/5+a*P(t)) 4t(2+P(t))*L(t/a) 4t(1+P(t))*L(t/5+a*P(t)) 4t(1+P(t))*P(t/5+a*P(t)) 4t(1+L(t))*P(t/5+a*P(t))`P`P``HH rrrrrs`rrrrrs`#YuY LLL LLL`TTTTTU Z[-Graph the curves[-Open the diamond menu[and select Graph Controller[-Select Manual mode [ -Tap Modify[-Press the right arrow in[the graph window to [advance a; down arrow[ to advance b[\different leaves0Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind xt1$Hyt1@$Hd p |               $ !0 "< #H $T %` &l 'x ( ) * + , - . 0 1 2 3 4 5 E F, H8 ID JP K\ Lh Mt N O P Q R S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ ab  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 ^ $  4t(2+P(t))*P(t/a) 4t(2+P(t))*L(t/a)`P`P`3`33HH 3`333`3#YuY3 ``` ````&1WG7 Q[\ interesting!!@Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind xt1,Hyt1H,Ht               (  4 !@ "L #X $d %p &| ' ( ) * + , - . 0 1 2 3 4 5$ E0 F< HH IT J` Kl Lx M N O P Q R S T ] ^ _ ` a b     , system]listsystem^]=system_^system`_systema`systemb~ ab  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 ^ $  4t(1+L(t))*P(t/5+a*P(t)) 4t(1+L(t))*L(t/5+a*P(t))`P`P`&`&&HH &`&&&`&#YuY& xxx xxx`&1WG7 Q[eActa b x020013Pg 238 Parabola.EAC0100000007a2H]<0[ Parabola [(Pgs. 238-240)[\q2``A"P`6y4tv2vrALCuxy!  )@w`fuvv`ST` r07 (YT#2YXyYV574HYYvS%a)&YvW$YHfUE2 DD8 H`D 15q7B F7#Eq1(f IWU8UtI yW8iDsPr07qRwv10`)VCwit0F)8H)uA'($(x6s W92"u( x`b `3P )(UdY @Yt PYY)HhbY)3@f6a 9Bd3 36 cB 5@! H  E@6Y@`!$uY  @6Hb6 Vvd% YdD  `A@6` !`B@6p0` @   @  r' &1WG7 H"    @6 QB2Y`C&qY!Upx  @6&hp Bq 1EA`H # D@65H3yY'2Tw `C@6 @   @    @6 Vvd& Hb3Y D25H  @  @   H    !H" W2g    [eAct020012Pg 241 Ellipse.EAC0100000009abH]<0[Ellipse (Pgs. 241-244)[\``vrALCuxy! )(4D$X2h3v&gVq!I(C8F82ii& 40H f xEYW1CVYx6@@`3(`Tb`qyA!`Bv$Q`3i`((&!`vI!`dUx'` dB)HI`#I7v`$2Cqp#`$@"S`#1T2`!rpa`ACy3`6&h`di7`Q"5`qqs5`r TP`TTe'`3BX8r`g4`6'hruY2cy&GYIvTs b (AcE9XtfR!EX54D$)( IYRaDW$Y9#s tCb7 VG VU) tt! Sv`% U9t $CE QGghW C2 `D'YGRVDDYvF)QY@Y&YsQYI0fweYiWYd`Ycs`D'`%%v19g`7c"c`P$ea2`cQ$4`v8rP(`1r`vR$`r `1P9`DUx`U9P5`d&`fh2``2!`AAp@s`EG`U6tV%H` PY5H" H! B@6 x B  @6P` PYYy'x `A@6P` PY%@   a  X3 VAVA H     @6Rq0%Y 7d2 A's%EY  @66!$RYqQ8YFB`y$  `C@6``H! D@6@e Y)tG% H! H   ! @68i#IYD`TBY @f47Y@  @  @   @!   @   @68i#IYD`TBX @f47Y@  @     ([eAct020014Pg 245 Hyperbola.EAC0100000007e2H]<0[ Hyperbola [(Pgs. 245-247)[\Qx` %r`rh@cRexw"vrALCuxy! e9SH (DFh dvH W"V)Br8f`X05$`p`ue`r7'7`w$`DPRCc`h`c@0`C Gh`Q`yPG7Tc`X$Q`Ttddu bPffe9SHE`C24sYT&b0WqX4ur`DI%(`U$c` u`%`C)6`f3yg`%w`pVC@pY8&cYF1E#Y9r4IF I!%U6d!gU$&y7`"w`D`5H" H! B@6Q5`Q5f0   @6`E`!`A@6`E`%@   a  4Bbf &1WG7 @!  D@6   @6 Sr '"4Y8igE!Y  @6IxU8 ab7Y2vTQbH! C@6#vR'b5@   @    @6 9HHCX AW@  S` @  @  @   H    !@    @6IxU8 Yab7 2vTQb`@  @        [eAct020018Pg 248 Foc On Ellips.EAC010000001b80H]<0"[Focus on Ellipse[(Pgs. 248-257)[\ Pgs 248-250``vrALCuxy(@    Dfpu iH S'SDw`iH DfpuY1 9W3Dfpu iH S'SDw`iH DfpuY1 9W3&hYI&q$ 0ECP`  @6b8TAYggUY8qVGB6(H  " @6 $"CYwP"cY&`)@   " @6&hYI&q$ 0ECP` @  H  F@6(X2q`` rG9C1  @   `D@6`.@  3    h'8aw H  A@6H  B@6H  C@6h'8aw  @6@  @   @6`r%hFTB @  @  &@  '@  '@  !@  W2g@ .@  .@    1@  9! `E@6   !  [\"Strings & Pins & Foci Pgs. 250-252""XA``"X9vrALCuxy#5J" $H F@6YIAr`x8tQ`% ! h'8aw &H! A@6'H! B@6(H! C@6h'8aw ) @6&'*@ &)+@  ')/3, @6`r%hFTB &(-H  &,.H  (,R/&@  &%0'@  '%1'@  (%2!@  ),W2g@"0' &1WG7 3(@! 4" H @ 5pDYe6tbp` 5pDYH @Y"vS`H @ 5pDYe6tbp` 5pDYH @Y"vS`7VwYp9C!YC7%$`5 @6DQI&& !GGeYS5Dw`6(H  47 @6 wri  &)U`Y5x!A3658)@  49 @67VwYp9C!YC7%$`4:@  ;!`D@6`7<@  $77=.@  >3  %$9?.@  %>@.H# $>=9@?>9A1@  $%9B!`E@6;2.-,+*)10/%('&$&'(%;B$97<:45#[[Albegraic Focus[ Pgs. 252-257[R16x^2+25y^2-400=0R16x2+25y2-400=0Rsolve(16x2+25u-400=0,u)Ru=-16x225+16R (x-3)^2+u+ (x+3)^2+uRx+32+u+x-32+uR expand(ans^2)R2x2-6x+u+9x2+6x+u+9+2x2+2u+18Rx2+6x+u+9|u=-16x225+16R9x225+6x+25Rx2-6x+u+9|u=-16x225+16R9x225-6x+25R simplify((9x225+6x+25)*(9x225-6x+25))R3x-2523x+252625R-3x-253x+25625R-3x-253x+2525R expand(ans)R-9x225+25R2(-9x225+25)+2x2+2u+18R2x2+2u-29x225-25+18R simplify(ans|u=-16x225+16)R100R[eAct020017Pg 258 Eigenvectors.EAC010000001141H]<0[Eigenvectors, [Eigenvalues and[Ellipses (Pgs. 258-263)[R&12.1x^2+17.74y^2-19.35xy-121=0R121x210-387xy20+887y250-121=0Rxyac2c2bxyRax+cy2x+by+cx2yRexpand(ax+cy2x+by+cx2y)Rax2+by2+cxyReigVc(12.1-19.35/2-19.35/217.74)R- 0.6000714432 0.7999464126 0.7999464126 0.6000714432[[- 0.6000714432 0.7999464126, 0.7999464126 0.6000714432[ReigVl(12.1-19.35/2-19.35/217.74R 24.99760016, 4.842399839R 24.99760016x2+ 4.842399839y2-121=0R 24.99760016x2+ 4.842399839y2-121=0R\s`RV3@`sRV3@vrALCuxyB  J `C'@6C`0  B @6'A"rhY Y `A@6B  RB  J `B'@6Fpb  B @6 Y'A$  B    'B   B  Fpb C`0   &B   'B   (J! B  yT d`6yYd`6y yT yT d`6y d`6yYyT  ' YX (J! H  C@62`v(J!  (B!  `B@6 )B  H"  `c`E"@  !`@6d` C ! @6C rYd` @  @  !`@6C Yd` ! @6d`YC rY @  !!@  " @6Y #@  "$@  "% @6YY&@  %'H  %W2g@'&%$#"('@  )  2`v"%*&@  )+'@  )(!+*))   [eAct02000eRieman sum.EAC010000001809k^[ Rieman sum[\Graph.Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y1H , 8 D P \ h t           ! " # $ % &( '4 (@ )L *X +d ,p -| . 0 1 2 3 4 5 E F H I J K L$ M0 N< OH PT Q` Rl Sx T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz aseq_histbNewFolde system]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemzsystem]listsystem^]=systemz  0@PSheet1 Sheet2 Sheet3 Sheet4 Sheet5 SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i k^ $  3xL(x)`P`P`p`p `(10qy`#YuY p 3p ` PRdefine f(x)=sin(x)RdoneR!define right(a,b,j,n)=a+j*(b-a)/nRdoneR$define left(a,b,j,n)=a+(j-1)*(b-a)/nRdoneR8define rsum(a,b,n)=Q(((b-a)/n)*f(right(a,b,j,n)),j,1,n)RdoneRlim(rsum(0,/4,n),n,)R1-22R/4R 0.7853981634R7define lsum(a,b,n)=Q(((b-a)/n)*f(left(a,b,j,n)),j,1,n)RdoneRlim(lsum(0,/4,n),n,)RERROR :Insufficient MemoryRsin(x)04xR1-22ReActfHxL(x)leftH(a,b,j,na+(j-1)*(b-a)/nlsumH<+a,b,n(((b-a)/n)*f(left(a,b,j,n)),j,1,n)rightH$a,b,j,na+j*(b-a)/nrsumH<,a,b,n(((b-a)/n)*f(right(a,b,j,n)),j,1,n)020007abs.EAC0100000017a3t^4Rdefine f(x)=-(x+1)(x-2)(x-3)RdoneR\Compare graphsGraph2D Graph3D LISTSYS4NModify $ STATCALC NSTATSYS \NSequencel, Sheet| Sheet3D| SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict$ ViewWind< y1XHy2pHy3Hy4 H        ( 4 @ L X d p |   ! " # $ % & ' ( ) * + , -$ .0 /< 0P 1\ 2h 3t 4 5 E F H I J K L M N O P Q R( S4 T@ ]L ^P _T `X a\ b` d p |   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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^4 $  3xf(x)  3xf( (x))  3x (f(x)) 3x (f( (x)))`P`P`p`p `(10qy`#YuY 7`77  $0<HT`lx ,8DP\ht`$```&``````````````````cxsh T[[eActfH(x-(x+1)*(x-2)*(x-3)020012areaoftriangle.EAC010000001559t^4P[Are the areas the [same?[\AuthorProfessor,Dr. Wei-Chi Yang Department of Maths/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL:www.radford.edu/ ~wyang [[ Objective.[We will explore how to [setup the formula of [finding the area of a [ triangle.[[ Example 1.[Animate the following['Geometry' strip, and [see if the areas of each[triangle CDE should be [ the same.\Area of the triangle``vrALCuxy!@    @6bvWvY Y"AH Ue`c !`A@6` Y!`B@6EP @   @  [  @6bvWvY Y"AH VGWB` H! E@6Witqc`&HD  `D@6  @  @  r@@! H C@6u'7P f86'Y  @6ci rC hpY @  R) @    @6 x"ISYS4YpH)  @  e 5H# vVBA &1WG7    [ Solution:[We record the point C[and the area of CDE[ as follows:[[-2.15-0.9 3.061192603 -1.855263158 -0.8131578947 3.061192603 -1.560526316 -0.7263157895 3.061192603 -1.265789474 -0.6394736842 3.061192603 -0.9710526316 -0.5526315789 3.061192603 -0.6763157895 -0.4657894737 3.061192603 -0.3815789474 -0.3789473684 3.061192603-0.08684210526 -0.2921052632 3.061192603 0.2078947368 -0.2052631579 3.061192603 0.5026315789 -0.1184210526 3.061192603 0.7973684211-0.03157894737 3.061192603 1.092105263 0.0552631579 3.061192603 1.386842105 0.1421052632 3.061192603 1.681578947 0.2289473684 3.061192603 1.976315789 0.3157894737 3.061192603 2.271052632 0.4026315789 3.061192603 2.565789474 0.4894736842 3.061192603 2.860526316 0.5763157895 3.061192603 3.155263158 0.6631578947 3.0611926033.450.75 3.061192603[We note that even if we[moves C along the line[segment AB, the area[of CDE stays the same.[[ Example 2. [Find a formula for the[area of a triangle.[Hint:[1) Open the Geometry[strip above and tap on[line segments DE and AB.[2) Are they parallel? [->Yes, because the angle[between these is 0.[This means the distance [from C to DE stays the[same as C moves along[the line segment AB.[3) To find the distance[tab on C and DE and the[distance measurement.[The distance is 2.193808.[4) Note the followings:[->base of triangle CDE[ =DE=2.790757,[->base of triangle CDE[height= 2.193808.[->Area=3.061192603.[5) Is it right that the[area of a triangle is[(1/2)(base)(height)=R(1/2)*(2.790757)*(2.193808)R 3.061192516[ Exercise.[Draw an arbitray triangle[using Geometry strip. [Verify the formula of the[area of a triangle.\ Any triangle``vrALCuxy@  !`A@6E`PY ! @6D't!Y iRIVYEXxvT`!`C@6H8pwB5H8p`@  {P@   ! @6 %qc2V AW9`!`B@6P 0 @  ` !@  " !! @63aA4UYErb2 TIc$`#@  "!#""[-> Tap a point, say A.[-> Tap the segment [opposite to the point, [say BC.[-> Measure the height [ and the base.[ For example,[ base=5.967354[height=4.934361R(1/2)*(5.967354)*(4.934361)R 14.72253943[By clicking three sides[of the triangle, we see[that the actual area is[the same as what we [ expected.eAct020012constructderiv.EAC010000002306Dkx[What do we know [about derivative [ functions?\ animation``vrALCuxy5H  H  A@6gB2%Y@   sin(x)`&1WG7 .@  3  !! @6EH%Q"Y)yc VaY .@   .H"   1@  9[-5 0.9589242747 0.2836741735 -4.473684211 0.9716450396 -0.2364321722 -3.947368421 0.7213681192 -0.6925428132 -3.421052632 0.2758366172 -0.9612045355 -2.894736842 -0.2443562981 -0.9696855175 -2.368421053 -0.6984086597 -0.715690469 -1.842105263 -0.9634209364 -0.2679806775 -1.315789474 -0.967661567 0.2522641385 -0.7894736842 -0.7099827291 0.7042279617 -0.2631578947 -0.2601310228 0.965573328 0.2631578947 0.2601310228 0.965573328 0.7894736842 0.7099827291 0.704227964 1.315789474 0.967661567 0.2522641362 1.842105263 0.9634209364 -0.267980683 2.368421053 0.6984086597 -0.7156904715 2.894736842 0.2443562981 -0.969685516 3.421052632 -0.2758366172 -0.961204537 3.947368421 -0.7213681192 -0.6925428132 4.473684211 -0.9716450396 -0.23643217755 -0.9589242747 0.283674169[\ copy to list hGraph2D Graph3D LISTSYS4NModify 4$ STATCALC XNSTATSYS `\NSequence, Sheet| Sheet3Dd| SolveEqSolveLwr SolveUpr StupFLG1(StupList$D StupPicth$ ViewWind y1XHy2H $ 0 < H T ` l x          ! " # $ % & ', (8 )D *P +\ ,h -t . 0 1 2 3 4 5 E F H I J K L M( N4 O@ PL QX Rd Sp T| ]D ^D _D `T aX b\ d`D eD      $ 0 < H T system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem_]=ystemsystem]listsystem_]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i Dkx $ F 3x1.000001469908*L(1.000000514448*x+1.5707963266)+7.375186899999-6 3xP(x)`P`P``y"y" 5g7(f`5&% q(10qy`#YuY gAi  $0<HT`lx G6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&06!0G6 BRc shBG6!  $0<HT`lx XBtp qdP9` !6 uf D5bY@YpY cB 6@Y gfgY ')Y`"Y`"  ') gfg cB 6@ @Yp D5b uf Y!6 Y qdP9`Y XBtpY $0<HT`lxgAsP 6C!r YT( Y a E5PY ihUPYiiYgwPYR&A8P "yap eW3( eW3( "yd R&A6 gYiqPY ihUY a E7YT( Y6C!wPYgAi  $0<HT`lx`G6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&06!0G6 BRc shBG6! $0<HT`lxgAsP 6C!r YT( Y a E5PY ihUPYiiYgwPYR&A8P "yap eW3( eW3( "yd R&A6 gYiqPY ihUY a E7YT( Y6C!wPYgAi FQDHW2f7Q `i3D `VIPPHh X[eAct020014constructriangle.EAC0100000006b1t^4 [Construction of area of[ a triangle\ ->animation``vrALCuxy5H# H E@6YY`  @6xT8 Y  &P !`A@6`E` `B@6PY@   @  eT""""# &1WG7 @   `C@6@`0 ! @6xT8  YpUH H D@6r @ @  oJ @   ! @6 5`w 1YSr vQWgY @  @    @6 D  f W8 @  {P!@    @6xT8 Y  pUH` @  @  [   [1. Draw line segments[AB and CD respectively.[2. Make AB be parallel to[CD.[3. Select a point E on AB.[4. Draw line segments[ CD,DE and EC.eAct02000dcrossprod.EAC0100000005caH]<0 RcrossP(23,4-1)R00-14R1317sin(70.34618)R 14.01676668R142R14R\333334`SEIp`333335SEIsvrALCuxy@   @6 ! u@623xYTp"R )75w @6`@  H" H  @6cg 1 !@   s@6BSV%c p%R  @6 @  @    ! t@6BSV%c p%R 9Tu` @  @  oJ@ @    r@62HWYTp% @      [eAct020009deriv.EAC01000000354bDkxO[Derivative of Sine [function[\AuthorzProfessor Dr. Wei-Chi Yang Radford University Radford, VA 24142 U.S.A. e-mail: wyang@radford.edu www.radford.edu/~wyang[ Objective[We will use graphical [representation to show [the derivative of [Sine function is Cosine [ function.[[ Example. [ Show that [   ddxsin(x)=cos(x).[ graphically.[Step 1.[-Insert a Geometry strip[below. [-Select the coordinate [system.[-Under 'Draw', select [ function.[-Type in sin(x).[Step 2.[Under Draw, select [construct->tangent to [a curve.[-Selectling the point and[ the curve.[-Edit->Animate->add [animation->go once.[Step 3.[-Select the point on the[ sine curve.[-Select the table [(you should get the[table for y=sin(x)).[-6.47 -0.185729957 -5.799473684 0.4650681779 -5.128947368 0.9144868832 -4.458421053 0.9679231158 -3.787894737 0.602238423 -3.117368421-0.02422186341 -2.446842105 -0.6401938273 -1.776315789 -0.9789551069 -1.105789474 -0.8938184925 -0.4352631579 -0.4216490368 0.2352631579 0.2330989023 0.9057894737 0.7869125563 1.576315789 0.9999847678 2.246842105 0.7800530202 2.917368421 0.2223500815 3.587894737 -0.4316327893 4.258421053 -0.8987141065 4.928947368 -0.9766427299 5.599473684 -0.6316747436.27-0.01318492514[-Select 'slope' of the [tangent line at the point[and the table. This [should show the table [ of slopes.[[Step 4.[Now drag the x-column[and the slope-[column into the graph, [ we should see[the graph of y=cos(x)[(why??)[-6.47 -0.185729957 0.982600828 -5.799473684 0.4650681779 0.8852806827 -5.128947368 0.9144868832 0.4046269738 -4.458421053 0.9679231158 -0.2512344783 -3.787894737 0.602238423 -0.7983087507 -3.117368421-0.02422186341 -0.9997066084 -2.446842105 -0.6401938273 -0.7682054202 -1.776315789 -0.9789551069 -0.2040634768 -1.105789474 -0.8938184925 0.4484401012 -0.4352631579 -0.4216490368 0.9067643853 0.2352631579 0.2330989023 0.9724530345 0.9057894737 0.7869125563 0.6170743608 1.576315789 0.9999847678-5.506936664-3 2.246842105 0.7800530202 -0.6257036712 2.917368421 0.2223500815 -0.9749668915 3.587894737 -0.4316327893 -0.9020440145 4.258421053 -0.8987141065 -0.4385237748 4.928947368 -0.9766427299 0.214881885 5.599473684 -0.631674743 0.7752414216.27-0.01318492514 0.9999130757[-Drag the y-column and[ slope-column.[into the graph, we see [the unit circle. (why?)[Because we are drawing[ x=sin(t) and [ y=cos(t).\ ConstructionsG`rXIf`&rXIfvrALCuxyJ  rWYew Hh g1 ## B!4Y@8'0Y xQYPY!d6Y3 0 %V0 Gg 0 "5P 1c'0YqAPY vd')Y1gGCY1Q@Y `( (p bis Q#Dx0Y0PpY pf@Yh T Y4vYHD vC0 rE04P C` Pi6f@Y%p6q Y thPY @PY8R7tY u$! 0up @ G`ysh@`G6`E!0`xxsp`shB`DhBP`wcx`WG@`5&1WY5&1W xsp Wcx$hBPshBXxsp%!0G6Ysh@'rWYew Hh g1 ## B!4Y@8'0Y xQYPY!d6Y3 0 %V0 Gg 0 "5P 1c'0YqAPY vd')Y1gGCY1Q@YH" G`ysh@`G6`E!0`xxsp`shB`DhBP`wcx`WG@`5&1WY5&1W xsp Wcx$hBPshBXxsp%!0G6Ysh@' `( (p bis Q#Dx0Y0PpY pf@Yh T Y4vYHD vC0 rE04P C` Pi6f@Y%p6q Y thPY @PY8R7tY u$! 0up 5H  H  A@6 2i bu   sin(x) G`&pRcx .@  3     @6'))YI(`4! !s"Y .@  .@    1@   9  [ Exercise 1.[Use the similiar procesure [described in the Example [to showthat [  ddxcos(x)=-sin(x).[\Try your answer here``vrALCuxy5H  H  B@6 Y!`h'   cos(x) Y`&1WG7 .@  3    @6f&08qY"u@4 CD7&`.@  .@  1@  9[ Exercise 2.[#Use the similar procesure described[in the Example to show that[ddx (tan(x))=sec2x.[\Try your solutioin here``vrALCuxyH" 8`````A@YQT uR YIQ` `D0`i@WfYi@Wf D0` YIQ` uR YQTYA@ ``8``$("x# P)#Y@58P#89$`W Qdg0%h%i0dg0W Q 9$`#9@58#Y9pP)x# $("H" `G6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&06!0G6 BRc shBG6!$("x# P)#Y@58P#89$`W Qdg0%h%i0dg0W Q 9$`#9@58#Y9pP)x# $("5H  H  B@6P 2q2(  tan(x)P `&1WG7 .@  !3! "  @6 tGTtY2h07 $qdE` #.@  !$.H" ! "$#!"%1@  9"[\A bonus parametric equation& Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind xt1Hyt14 HT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F H( I4 J@ KL LX Md Np 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i Dkx $  4tT(t) 4t(1/P(t))^2`P`P`p`p `(10qy`#YuY #He5RIpryQq ` Q[Note that the parametric[equation of x(t)=tan(t) and[y(t)=sec(t)^2 is a [ parabola.eAct02000adonkey.EAC010000003903 t^;5[A donkey problem\ See the fieldw2%d` 5`w2%fdvr ALCuxy@  H  J@6dH  @65DYH! A@65DYH  B@6s!3&RvU` %@   @   @  H  G@65` @  H  F@6 q7rX @w9`  @65`H# C@65`%@  @   @  H  E@6CH2)5`@  @  H  D@6CH2`)5`@    [The area of ADE is 4.464705.[The area of DFE is 3.467865.[We need to find the proper[circle (the one centered at[(0,0)) so these two[areas are the same.R define A()=r -6sin(n)r3/2 sin(/-6) nRdoneRA(x)-(1/4)**9R 2.356194490x2- 9.000000000sin 0.166666667x+ 4.500000000sin 2.000000000sin 0.166666667x+ 0.500000000sin 0.166666667xx2- 49.48008429Rsolve(A(x)-(1/4)**9=0,x)R 2.356194490x2- 9.000000000sin 0.166666667x+ 4.500000000sin 2.000000000sin 0.166666667x+ 0.500000000sin 0.166666667xx2- 49.48008429= 0.000000000[We plot A(x)-(1/4)**9\Use the plot to find zeroGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i  t^; $ `P`P`p`p `(10qy`#YuY 52` O[We use G-solve to find [&one zero to be at 'xc=4.351115999332'.[[ Exercise. [Redo the problem when the[smaller circle (r=-2u*sin(n) [is of radius 4.R define B()=r -8sin(n)r3/2 sin(/-8) nRdone[Now we should solve[2B=(1/2)*16 or[we plot B(x)-4\use plot to find zeroGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i  t^; $ `P`P`p`p S dI`0DWT(10qy`#YuY `XQO[We find that xc=5.801487999108.[[ Second Method[We use the rectangular[coordinate system to find[ the integral.[First, the larger circle[has the equation:[x^2+y^2=r^2 so[ we have y=-r2-x2 for[the lower half of the circle.[Second, the fixed circle[has the equation of[x^2+(y+3)^2=9; so[the equation is[y=-3-9-x2 . We need to[find the intersection(s).[So, we set these two [equation equal to each[other.Rsolve(-r2-x2-(-3-9-x2)=0,x)R 1.000000000x=- 0.166666667 36.00000000r2-1r4 0.500000000, 1.000000000x= 0.166666667 36.00000000r2-1r4 0.500000000R[ The area is =eActAH@.vZ(Z(r,r,v,-6*L(n)),n,3*/2,M(v/-6) )BH@.vZ(Z(r,r,v,-8*L(n)),n,3*/2,M(v/-8) )020016exponential curves.EAC0100000014ba^[Dynamic graphs\exponenial curvesGraph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y1$H, 8 D P \ h t            ! " # $ %( &4 '@ (L )X *d +p ,| - . 0 1 2 3 4 5 E F H I J K$ L0 M< NH OT P` Ql Rx S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ a(1-^b  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 ^ $  3xa*(1-^((-b)*x))`P`P`f#7f$`3vb3vYtYt `(10qy`#YuY @@@@@@` P[eActa @b 020011famous curves.EAC010000003eb0^[Astroid\parametrically@Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind xt1,Hyt1H,Ht               (  4 !@ "L #X $d %p &| ' ( ) * + , - . 0 1 2 3 4 5$ E0 F< HH IT J` Kl Lx M N O P Q R S T ] ^ _ ` a b     , system]listsystem^]=system_^system`_systema`systemb~ a?b ? 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 ^ $  4ta*(P(t))^3+b*(L(t))^3 4ta*(L(t))^3+b*(P(t))^3`P`P`p`p `(10qy`#YuY L`LLL`LL` Q[Folium\polar equation Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind r18H@ L X d p |             ! " #$ $0 %< &H 'T (` )l *x + , - . 0 1 2 3 4 5 E F H I J, K8 LD MP N\ Oh Pt Q R S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ acos(n)+b 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 ^ $ % 5n(-b)*P(n)+4*a*P(n)*(L(n))^2`P`P``|)) 1`1||(10qy`|#YuY| L`LLL`LL`ccccd P[ Hypocycloids\parametric equationHGraph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind xt10Hyt1L0H|              $ 0  < !H "T #` $l %x & ' ( ) * + , - . 0 1 2 3 4 5, E8 FD HP I\ Jh Kt L M N O P Q R S T ] ^ _ ` a b    ( 4 system]listsystem^]=system_^system`_systema`systemb~ ab  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 ^ $  4t(a-b)*P(t)+b*P((a-b)*t/b) 4t(a-b)*L(t)-b*L((a-b)*t/b)`P`P`p`p$ `$$(10qy`$#YuY$ @@@@@@` Q[** To graph a hypocyloid,[you need two parameters,[a and b and plot the [equation with appropriate[range. In order for the[equations to make sense[in the physical world, we[ need b`"0D f`"yi`7Txw#(gSvr ALCuxy@  H # H@6GuQ  @6 tt44Y#3cq2 YH K@6%P6``H  ?< @    @6 tt44Y#3cq2Y H  L@6%P6`? H ?~` @   ! @6  @!  ?p5@   @6`H  C@6!D@6$@  @  i8wU% VAVA @   @6 tt47 #3cq3 `  @6 `A@6@! B@6%@  H @  ! @6 tt47 #3cq3Y@   H  !@! " @6 #@! $@  %@  "&@  '!`J@6`"(@! ) `I@6``""')*@  +@  G@6fq&xYRa) +,@  -@  F@6$89 f98x -.@  /@  E@6`/0"@!   /-+)'"(&$#   ([-Edit/Animate/Go(once)[[(b) With traditional [ analysis:[It can be shown that the[ Area of ) =[12(1+1sin(x2))2sin(x)(cos(x2))2[(c) With graphical approach:[By Dragging area [function to'Graph [Window' below. We may[use the command ['Analysis/G-Solve/Min' to[anlayize the minimum of[the area function, which [happens at x=60 degree.[\Graph Window -->1 Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y18H@ L X d p |             ! " #$ $0 %< &H 'T (` )l *x + , - . 0 1 2 3 4 5 E F H I J, K8 LD MP N\ Oh Pt Q R S T ] ^ _ ` a b      system]^system^system_`system`asystemabsystemb( LISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLISTLIST abGrapbGraph2D system]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLISTsystem]^system^systemLIST  0@PSheet1 Sheet2 Sheet3 Sheet4 Sheet5 SheetSheet3D 0@PSheet1، 8 Sheet2، 8 Sheet3، 8 Sheet4، 8 Sheet5، 8 Sheetr-value: i <$ $ ( 3x1/2*(1+1/L(x/2))^2*L(x)/(P(x/2))^2`P`P``HHP`P?(10qy`#YuY $gS$gS@UBQ3``&1WG7 P[[Remark: By analyzing [the graph,we see vertical [asymptotesat x=0 and Pi. [We predict maximum [occurs at those two [points.[[(d) With table:[After finished the [aninmationd escribed [in (a).[-Select two sides HK [and HL,and select angle [ and table.[-Next select three sides [of the triangle and [the area of the triangle.[-We get the following list:[(0 2.510073601 130.1816945 9.519763863 112.153776 7.229996322 99.76166621 6.319875664 90.32199451 5.842075726 82.75495461 5.56236532 76.49323976 5.39175351 71.19529043 5.28828893 66.63742715 5.229370728 62.664503 5.201653507 59.16429038 5.196709119 56.05295216 5.208933353 53.2662375 5.23444441 50.75386705 5.270462767 48.47580285 5.31494258 46.39968637 5.366342622 44.49902925 5.423478496 42.75190509 5.485424227 41.13998419 5.551444896 39.64780864 5.62094943 38.26223953 5.693456802 36.97202936 5.768571391 35.76748749 5.845964705 34.64021526 5.925361637 33.58289423 6.006529977 32.5891154 6.089272316 31.65324026 6.173419726 30.77028686 6.258826772 29.9358358 6.345367547 29.14595207 6.432932483 28.39711975 6.521425771 27.68618709 6.610763257 27.01032007 6.700870713 26.36696297 6.791682411 25.75380476 6.883139934 25.1687502 6.975191181 24.60989507 7.067789535 24.07550469 7.160893158 23.56399534 7.254464387 23.07391807 7.348469228[[If we want more detailed list[between the angles and the[areas, we use the stats[application below.[\Lists-->2Graph2Dh Graph3D| LISTCALD LISTSYS4NModify $ STATCALC $NSTATSYS ,\NSequence, Sheet| Sheet3D0| SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict4$ ViewWindX t               (  4 !@ "L #X $d %p &| ' ( ) * + , - . 0 1 2 3 4 5$ E0 F< HH IT J` Kl Lx M N O P Q R S T ] ^ _ ` a b dd e`d      0̒1/2(1+1/sin(x/2))^2sin(x)/(cos(x/2))^2" eActxfeActareafsystem`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemeActmxlisteActmarea=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i <$ $ `P`P`y`@%p)rf`7Q(10qy`#YuY yX6b5p $0<HT`lx     @`uxyyy $0<HT`lxYYcY5Ya`HyTe`aI(X6fGH 3U2F`7sC)E(8xGEaRB'HgC63@TGQC'S8sP06C 62)GU2AX7 T X6b3SX6b5p`%# R[Notice that by carefully[examining the list, we [conjecture that [ a) when x[(angle)is close to 0 or [180 degree,the area is [close to a maximum.[(b) When x is 60 degree, [the area is a minimum.[[Approach with Calculus[We analyze the graph of[the derivative of area[ function.[\Area and its deriv.3`Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y18Hy2T@H            $ 0 < H  T !` "l #x $ % & ' ( ) * + , - . 0 1 2 3, 48 5D EP F\ Hh It J K L M N O P Q R S T ] ^ _  ` a b  ( 4 @ L 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 <$ $ ( 3x1/2*(1+1/L(x/2))^2*L(x)/(P(x/2))^20 3x:(1/2*(1+1/L(x/2))^2*L(x)/(P(x/2))^2,x,1)`P`P`x`xx3vb3vbx`xxx(10qy`x#YuYx `cxsh Q[By examing the graph of[the derivative function,[it crosses from - to +[at x=60, we know that[the minimum area is at[x=60 degrees and[the maximum is when [the angle approaches to[.eActangles  $0<HT`lx @`u8area d $0<HT`lxYYcY5Ya`HyTe`aI(X6fGH 3U2F`7sC)E(8xGEaRB'HgC63@TGQC'S8sP06C 62)GU2AX7 T X6b3SX6b5pm_angle m_area volume D $0<HT`lxY5Ya`HyTe`aI(X6fGH 3U2F`7sC)E(8xGEaRB'HgC63@TGQC'S8sP06C 62)GU2AX7 T X6b3Sx d $0<HT`lx         ( 2 < P d x   xyyyy 020011implicit diff.EAC0100000043f8^[Implicit Differentiations[\``vrALCuxy5@  H  D@6W!rY)W U@  ! %`PYVQ"`xih2W 0P  `A@6%`PY `B@60P H  C@6VQ"`xih2W @6`)5Y USgGG FBd  @   @    ! @6 USgGGY`)5YAYv#b` @   @   &@  '@  'H  !H" W2g@vq%%1 &1WG7 .@  3    @6(gb3i hfY" &vC8PY.@  .@  1@  9 [2.3 -3.227272727 2.004152885 -0.4234666966 1.355667811-0.05598382527 0.4248181994 0.1138611743 -0.687523952 0.2312210449 -1.860819111 0.3363731539 -2.967922541 0.4548812665 -3.888862408 0.6285701917 -4.523840587 1.013463614 -4.804047339 4.430121245 -4.699117915 -0.9275611199 -4.220423047 -0.1934924274 -3.41983675 0.03981282633 -2.384114974 0.1758276078 -1.225494242 0.2837367776-0.06952908927 0.3921565646 0.9585137233 0.5302854651 1.747229805 0.7711173311 2.211149531 1.562645962.3 -3.227272727[\What is the plot?Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind  ( 4 @ L X d p |          ! " # $ % &$ '0 (< )H *T +` ,l -x . 0 1 2 3 4 5 E F H I J K L M, N8 OD PP Q\ Rh St T ]D ^D _ ` a b  d$D ehD      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`QDR)`s9SV!01 uQR`(10qy`#YuY 0"rrrp` $0<HT`lx0ARP5Vg$@ R9RY`y"T`b@`R8@Xp`@G3`iP`"#p`A6u`8A@`"T$ `)pY XQ7#0 tr)P!IS0 $0<HT`lx"rrrp`#Ff`YYRpYt0 1"D 671S TfP (Wp 4ca@C!$P 'VYI$'@Yc0 uv sgw` ed` 0(Te qs1 V&E"rrrp` $0<HT`lx0ARP5Vg$@ R9RY`y"T`b@`R8@Xp`@G3`iP`"#p`A6u`8A@`"T$ `)pY XQ7#0 tr)P!IS0 $0<HT`lx"rrrp`#Ff`YYRpYt0 1"D 671S TfP (Wp 4ca@C!$P 'VYI$'@Yc0 uv sgw` ed` 0(Te qs1 V&E"rrrp``ArW7 QRDsolve(2.268x^2+12.7y^2-7.154xy+0.3049x+10.11y-12.04=0,y)Ry= 0.2816535433x-3.937007874-6- 1.152144+10x2- 1548892000x+71540x-1011002+ 6.11632+100.5- 0.3980314961,y= 0.2816535433x+3.937007874-6- 1.152144+10x2- 1548892000x+71540x-1011002+ 6.11632+100.5- 0.3980314961R\look at this again``vrALCuxy5H# H# A@6 Y " v0.2816535433x+3.937007874-6(-1.152144+10x^2-1548892000x+(71540x-101100)^2+6.11632+10)^0.5-0.3980314961Y`&1WG5 .H  3   "! @64wEY r@ ddwY!.H  ".@   "! #1@  9$" v0.2816535433x-3.937007874-6(-1.152144+10x^2-1548892000x+(71540x-101100)^2+6.11632+10)^0.5-0.3980314961$ [-0.80 -4.473684211 0.94806238 -3.947368421 0.6460102145 -3.421052632 0.5234317015 -2.894736842 0.4454063155 -2.368421053 0.385690755 -1.842105263 0.33467106 -1.315789474 0.28742926 -0.7894736842 0.2405758775 -0.2631578947 0.190862665 0.2631578947 0.133932705 0.7894736842 0.06186808 1.315789474 -0.044999015 1.842105263 -0.2675036733 1.8421052630 1.8421052630 1.8421052630 1.8421052630 1.8421052630 1.8421052630[\%Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind  ( 4 @ L X d p |          ! " # $ % &$ '0 (< )H *T +` ,l -x . 0 1 2 3 4 5 E F H I J K L M, N8 OD PP Q\ Rh St T ]D ^D _ ` a b  d$D ehD      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`Rc`G6!9S% ahYuY`(10qy`#YuY !&0 $0<HT`lxYG6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&0!&0!&0!&0!&0!&0!&0 $0<HT`lx H# FP #CP E@cP iU 4g` B` @WXwP &e 3' h IPYgP6s0Y $0<HT`lxYG6!`shB`BRc `G6 `6!0`!&0`1WG@`G6 YcxpYcxp G6 1WG@!&0!&0!&0!&0!&0!&0!&0 $0<HT`lx H# FP #CP E@cP iU 4g` B` @WXwP &e 3' h IPYgP6s0Y`#bE Q[eAct02000elimitand3d.EAC010000008851^\,}[ The rate of [convergence (II).[\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 e-mail: wyang@radford.edu URL: www.radford.edu/ ~wyang[ Objective:[We encountered a [Calculus problem saying [that[   sin(x)xx0=1.[[In this short note, we[shall use a graphical[method to find an 'a' [such that [sin(x)x approaches[to 1 at the same rate[as xa->0 when x-->0.[In other words, we shall[find a number 'a' such [that the limit[   1-sin(x)xxax0 [is finite and positive.[======================[We Rdefine f(x,a)=1-sin(x)xxaRdoneR[Case 1. When a<2, [we have, for example,Rf(x,1)R-sinxx-1xR f(x,1)x0R0Rf(x,0.5)R-sinxx-1xR f(x,0.5)x0R0[Remark: [We conjecture that the [ limit of [ f(x,a)x0 for a<2 is [ equal to 0.[[Case 2. When a>2,[we have, for example,Rf(x,2.5)R-sinxx-1x52R f(x,2.5)x0R UndefinedR\Graph of f(x,2.5)Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^\, $ `P`P`%xY%x xhh x`xxx(10qy`x#YuYx `1WG6 OR f(x,3)x0RR\Graph of f(x,3)Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^\, $ `P`P`Px`PxxP x`xxx(10qy`x#YuYx `1WG6 O[Remark:[We conjecture that the [ limit of [ f(x,a)x0 for a>2 is [ infinity.[Case 3. When a=2,[Rf(x,2)R-sinxx-1x2R f(x,2)x0R16R\ plot f(x,2)Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^\, $ `P`P``YtYt 5YPP (10qy`#YuY `9G6! O[\ An animationGraph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1(HD P \ h t              ! " #( $4 %@ &L 'X (d )p *| + , - . / 0\ 1h 2t 3 4 5 E F H I J K L M N O P Q( R4 S@ TL ]X ^\ _` `d ah bl p |    system]listsystem^]=system_^system`_systema`systemb~ a1-sin(x)b0 system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^\, $  3x(1-L(x)/x)/( (x))^a`P`P`Y I5I5 `(10qy`#YuY vvv vvv $0<HT`lXR! rgVC)5 e32Cx 0Uh 85ipf 85ipf ` Q[Now, we shall use[the 3d graph to see[what z=f(x,y) look likeRf(x,y)R-sinxx-1xyR\ A 3d graphGraph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind z1H$ 0 < H T ` l x           ! " # $ % &, '8 (D )P *\ +h ,t - . 0 1 2 3 4 5 E F H I J K L( M4 N@ OL PX Qd Rp 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^\, $  x,yf(x,y)`PPP p`p `(10qy`#YuY PYdXd0 ` P[ Exercise 1.[Find a number 'a' such [that the limit[   1-cos(x)xxax0 [is finite and positive.[Hints:[Step 1.Rdefine g(x,a)=1-cos(x)xxaRdoneR g(x,1.5)x0RR g(x,1.6)x0R UndefinedR g(x,1.7)x0R UndefinedR g(x,2.1)x0R UndefinedR g(x,2.5)x0R UndefinedR g(x,2)x0R UndefinedR g(x,1)x0R UndefinedR g(x,3)x0R UndefinedR g(x,0.5)x0R0R g(x,0.7)x0R0R g(x,0.9)x0R0R[Step 2.[ Consider \ A 3d plotGraph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind z1H$ 0 < H T ` l x           ! " # $ % &, '8 (D )P *\ +h ,t - . 0 1 2 3 4 5 E F H I J K L( M4 N@ OL PX Qd Rp 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^\, $  x,yg(x,y)````p`p `(10qy`#YuY rA70YbeQp 7SeY Y` P[What is your conclusion?[ Conjecture.[Case 1. When a1, the[limit is infinity.[Case 2. When a<1, the [ limit is 0.[** There is no a so that[the limit is a positve [number.[ Exercise 2:[The   1-cos(x)xsin(x)ax0 [is a positive number when[a=1.eActa vb vfH4$x,a((1-((L(x))/(x)))/( (x)^(a)))gH4$x,a((((1-P(x))/(x)))/( (x)^(a)))020010lineartransf.EAC0100000047f3H]<0[Linear transformation\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail:wyang@radford.edu URL: www.radford.edu/ ~wyang [ Objectives.[1. We will investigate [the relationship[between the linear [transformation and [geometry figures.[2. We extend these ideas[to notions of eigenvalues[and eigenvectors.[[Type I. Transformation[under -1001.[[ Example 1. [ What will the[effect of this 'linear[transformation' on any [ point (x,y)?[ Method 1.[Execute the following[command, we see that[(x,y) is sent to (-x,y),[which means that theR-1001xy+00R-xy[Method 2. Graphical[Representation:[Pick three points and see[what happens under this[transformation.R-1001012023R0-1-2023[[-Drag the matrix[012023 into the [Geometry strip below[to get three points [(0,0), (1,2) and (2,3).[-Drag the matrix[0-1-2023 into the [Geometry strip below to [get (0,0), (-1,2) and [(-2, 3).[-Note that this is a [reflection along the [y-axis.[[Question: Does this[transformation preserve[ the area?\ Along y-axis``vrALCuxy@  H `@6 @  @623x Tp"RYH `@6@   @   @  @6gfYgf gdY H `@6 @   @   @  @6BqYG!5I  @    @  H `@6 @  @623x Tp"R H `@6`@  @   @  @6gfYgfYgd H `@6`@#  @   @! @6BqYG!5IY@   @  H `@6 @  @623x Tp"R H `@6` @   !@  " @  @6gfYgfYgd #H `@6`$@  #"!$"%@  #& @! @6BqYG!5IY#'@  &%oJ'&##&"  [Type II. Transformation[under 100-1.[[ Example 2.[ What will the[effect of this 'linear[transformation' on any [ point (x,y)?[Method 1. Algebraically:R100-1xy+00Rx-y[We see that this [transformation is a [reflection along x-axis.[Method 2. Graphically:[Pick three points (0,0),[(1,2) and (3,2). What[would happen under this[transformation?R100-1013022R0130-2-2R\ Along x-axis(I`y$P`Py$PvrALCuxy)@  *H `@6+ H  @6Tp"RY23xY,H `@6`*-@  ,+)-+.@  ,/ H  @60H `@6`,1@  0/.1/,2@  03 @  @6Bq G!5I *04@  *32430*5@  6H `@67 @  @6Tp"R 23xY8H `@669@  87597:@  8; @  @6`<H `@68=@  <;:=;8>@  <? @  @6BqYG!5I 6<@@  6?>@?<66<?8;7*03,/+[Note that this [transformation preserves[ the area.[[Type III. Transformation[under -100-1.[[ Example 3.[What will happen under[this tranformation?[[We shall see that this is[a relection along the [ origin (0,0).[R-100-1xy+00R-x-yR-100-1013022R0-1-30-2-2R\ Along (0,0)A``vrALCuxyB@  CH `@6D @  @6Tp"RY23x EH `@6``CF@  EDBFDG@  EH @  @6``IH `@6``EJ@  IHGJHEK@  IL @  @6Bq G!5IYCIM@  CLKMLICN@  OH `@6P @" @6Tp"R 23xYQH `@6OR@  QPNRPS@! QT @" @6`UH `@6QVH  UTS#VTQW@  UX H! @6BqYG!5I OUY@  OXWYXUOOUXQTPCILEHD[Note that this [transformation preserves[ the area too.[[ Exercise. [What is the transformation[under -200-2?[ Hint: TryR-200-2013022R0-2-60-4-4R\Preserve area?Z`y$P`y$PvrALCuxy[@  \H `@6] @  @6Tp"RY23x ^H `@6``\_@  ^][:x_]`@  ^a @  @6``bH `@6``^c@  ba`ca^d@  be @  @6Bq G!5PY\bf@  \edfeb\g@  hH `@6i @  @6Tp"R 23xYjH `@6hk@  jigkil@  jm @  @6`nH `@6jo@  nmlomjp@  nq H! @6BqYG!5I hnr@  hqprqnhhnqjmi\be^a][Type IV.[What is the following[transformation??R0-1-10xy+00R-y-x[Step 1. [The matrix 0110[is sending (x,y) to (y,x).[A relection along y=xR0110xyRyx[Step 2. [To go from (y,x) to [(-y,-x), we only need [ to multiply the matrix [0110 by (-1).[R-0110xyR-y-x[So this is simply a [reflection along (0,0).[Try the followings?R0-1-10013012R0-1-20-1-3[\Name the transformation.s``vrALCuxyt@  uH `@6v @  @623xYTp"R wH `@6``ux@  wvt:xvy@  wz @  @6Bq G!5IYG!5I {H `@6``w|@  {zy|zw}@  {~ @  @6gf gfYu{@  u~}~{u@  H `@6 @  @6Tp"R 23xYH `@6@  @  @  @6G!5IYBq G!5IYH `@6@  @  @  @6gfYgf @  u{~wzv[[ Discussion I:[(a) Suppose we start [with an arbitrary matrix[A, what is the [geometric interpretation[of Av?R123101R21R\Where does (0,1) go?Affffr`87 `Affffh87vrALCuxy@   `@6  s@6G!5IYBq  @6@  @ @   `@6 ! r@6`@  [We see that the vector[01 is transformed [to the vector 21. Since R123110R13[Similarly, the vector 10[is transformed to 13.[\Linear transformation%`3"pY%@ C9b'vrALCuxy@   @6  u@6G!5IYBq "@6@  @   @6  t@6`@  0@   `@6  s@6 Hh2Y"wfi @  @   `@6  r@6@  c[[Discussion II.[Now, let's look at some[special cases:ReigVc(1231)R 0.632455532- 0.632455532 0.7745966692 0.7745966692R1231 0.632455532 0.7745966692R 2.18164887 2.671963265\Look at what happens.``vrALCuxy@  !`@6Hgc&P  s@6tYfi'RY2EU1%  @6@  C @   `@62EU2 tYfi ! r@6tYfi$Y2EU2B @  [Notice that these two[vectors are just multiple[ of the other?[What is the 'multiple'?[->Use the measurement[to figure out the number.[The longer vector is[3.44949 times of the [shorter vector, why?ReigVl(1231)R 3.449489743,- 1.449489743[[Remark: [The effect when we apply[the matrix on the eigenvector,[v= 0.632455532 0.7745966692[is the multiplication lamda[ and v, where [lamda=3.449489743.R 3.449489743 0.632455532 0.7745966692R 2.181648871 2.671963265[ Exercise. [Explore the effect ofR1231- 0.632455532 0.7745966692R 0.9167378064- 1.122769927R\Another eigenvector``vrALCuxy@   `@6 sx@ 'ip`  s@6tYfi)h 2EU1W  @6@  ;@   `@62EU2YtYfi  r@6tYfi$Y2EU2BY@  `([eAct02000flogarithmic.EAC010000009937 y=x^(2)+2``vrALCuxyH ! x^2+2^[About the shiftings of[logarithmic functions.[\AuthorProfessor Dr. Wei-Chi Yang Department of Math/Stats Radford University, Radford, VA 24142 USA e-mail: wyang@radford.edu URL:www.radford.edu/ ~wyang[ Objective:[(1) Traditionally, students[do not have so many [problems to find the [inverses of functions [such as f(x)=ln(x+a)+b; [but we will investigate[ further. [(2) We shall investigate [the effects of the [coefficients a and b on [the graphs of [f(x)=ln(x+a)+b[and its respective [ inverses.[(3) We extend the [observations above to an[arbitruary function and[its inverse (if the [inverse exists).[[Example.[Investigate how the [coefficient 'a' affect[the graphs of y=ln(x+a) [and the its inverse. [Step 1.[First, we find the inverse[ of y=ln(x+a).R clear_a_zRdoneRsolve(y=ln(x+a),x)Rx=-a+yR[Thus, the inverse of[f(x)=ln(x+a) is [g(x)=x-a.[-> We check if [f(g(x))=g(f(x))=x below.[(Be sure the set the[format to be 'Decimal[ Calculations'[under settings.)Rdefine f(x)=ln(x+a)RdoneR define g(x)=x-aRdoneRf(g(x))RlnxRg(f(x))Rx[Step 2.[We will investigate how[y=ln(x+a) is related to[y=ln(x)[\->Open the graph LGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2HHy3dH              ( 4  @ !L "X #d $p %| & ' ( ) * + , - . 0 1 2 3 4$ 50 E< FH HT I` Jl Kx L M N O P Q R S T ] ^ _ ` a b    , 8 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3x_(x)  3x_(x+5)  3x_(x-5)`P`P`p`p `(10qy`#YuY ` R[Remark: We see that [y=ln(x+5) and y=ln(x-5) [are horizontal shiftings[of y=ln(x) to the[left 5 units and right 5[units respectively.[[Step3.[We will investigate how[y=x-a is related to [y=x.[\->Open the graph PGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2LHy3hH              , 8  D !P "\ #h $t % & ' ( ) * + , - . 0 1 2 3 4( 54 E@ FL HX Id Jp K| L M N O P Q R S T ] ^ _ ` a b   $ 0 < 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $   3xexp(x)  3xexp(x)+3  3xexp(x)-3`P`P`p`p `(10qy`#YuY ` R[Remark: [(1) We see that[y=x +3 and y=x-3 are['vertical' shiftings of [y=x up 3 units and [down 3 units respectively.[(2) Consequently, if[y=ln(x) moves to the [ left 5 units [(so the equation becomes [y=ln(x+5)), the [correponding inverse[(y=x-5) will[moves down 5 units.[(3) Similarly, if[y=ln(x) moves to the [right 5 units [(so the equation becomes [y=ln(x-5)), the [correponding inverse[(y= x+5) will[moves up 5 units.[[Now we shall use the[animation to see the[graphs of y=ln(x+a) and[y=x-a.\ Exploration 1 HGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2LHy3hH|              $ 0  < !H "T #` $l %x & ' ( ) * + , - . 0 1 2 3 4 5, E8 FD HP I\ Jh Kt L M N O P Q R S T ] ^ _ ` a b    ( 4 system]listsystem^]=system_^system`_systema`systemb~ a^x-axb system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $   3x_(x+a)  3x^x-a 3xx`P`P`` x)) `xx(10qy`x#YuYx `` ``cxsh R[->Open the the graph[editor above and tap on[ 'graph' icon.[->Ignore the 'error' [message.[->Select the diamond [ shape icon.[->Select 'Modify' and[->Tap on the graph we[observe the followings:[[Remark:[This confirms our previous[observations that if a[function f is being shifted[to the left 'a' unit(s), [then the corresponding[inverse will be shifted[down 'a' units and so on.[Exploration 2.[There is another [interesting way of [finding the inverse[of a function. First open[the graph editor below.\Type a function  Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1Hy28HT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F H( I4 J@ KL LX Md Np 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $   3x_(x-2)  3xexp(x)+2`P`P`p`p `(10qy`#YuY 0% )`` Q[->When a function is[ selected.[->Plot the graph and[go to 'Analysis' icon[->Sketch->Inverse. We [get the inverse of the [ function.[We can guess (by now)[what the inverse should[be.[ Exercise 1.[(1) Investigage how the [variable 'a' will affect[the graph of y=x3+a[and its inverse.[(2) Investigage how the [variable 'a' will affect[the graph of y=(x+a)3[and its inverse.[ Solution:[(1) We first define the[function and find its [inverse below.R clear_a_zRdoneR define f(x)=x3+aRdone[**Basic Format should be[set 'Complex' to get [a symbolic expression.Rsolve(y=x3+a,x)Rx=y-a13,x=-y-a132-3y-a132,x=-y-a132+3y-a132Rdefine g(x)=(x-a)13RdoneRf(g(x))Rx[**Switch the 'Basic[Format' back to 'Decemal'.[Rg(f(x))Rx[Therefore, f and g are[ inverses.\The variable a  Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1Hy24 HT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F H( I4 J@ KL LX Md Np O| P Q R S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ ax-a)^(1/b  system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xx^3+a 3x(x-a)^(1/3)`P`P`p`p `(10qy`#YuY `` Q[Remark:[(1) We see that the [graph of y=x^3+a is [a vertical shifting of [y=x^3.[(2) Specifically, when [y=x^3 moves up 5 units[say, y=x^3+5, the[inverse becomes [y=(x-5)13 , which is a [horizontal shifting of['right 5' from y=x13.[(2) The solution to this[one should be obvious.[The graph of y=(x+a)3 is[the horizontal shiftings[of y=x3 (If a<0, then[it is shifted to the right['a' unit(s), if a>0, then[it is shifted to the left[ 'a' units.[Consequently, the inverse[ of y=(x+a)3 should be[shifted vertically up 'a'[units(if a<0) and[down 'a' units (if a>0).[[ Exercise 2.[(1) Investigage how the [variable 'a' will affect[the graph of y=x2+a, [x>0 and its inverse.[(2) Investigage how the [variable 'a' will affect[the graph of y=(x+a)2,[where x>-a, and its inverse.[[Hints:[(1) We first sketch the[ graph of y=x2+a with [the following geometry[link; replace 'a' with [some positive and negative[numbers and see how 'a'[affects the graph.]\A vertical shifting[(2) Now we will see[if y=x2+a has an inverse[by using the following[ graph editor:\Inverse of y=x^2DGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2HHy3`Hx               ,  8 !D "P #\ $h %t & ' ( ) * + , - . 0 1 2 3 4 5( E4 F@ HL IX Jd Kp L| M N O P Q R S T ] ^ _ ` a b    $ 0 system]listsystem^]=system_^system`_systema`systemb~tttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt aseq_histbNewFolde system]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemt  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xx^2 3xx^2+1 3xx^2-1`P`P`p`p `(10qy`#YuY `a` R[-After graphing y=x^2,[we select Analysis->[Sketch->Inverse. We get[a horizontal parobola,[x=y^2; which is not a [ function.[-Therefore, if we [restrict the domain for[f(x)=x^2 to be x>0, [then the corresponding[inverse will be[ f-1(x)=x.[We leave readers to [verify if f and  f-1 are[inverses graphically and[algebraically.[[(3) Find the inverses for[f(x)=x^2+a with various[ number a.eActa `b fH xx^(3)+agH$x(x-a)^(((1)/(3)))02000cnumber/e.EAC0100000017ed^[Investigate the [number e and related[topicsRdefine f(x)=(1+x)1xRdoneRf(0.1)R 2.59374246Rf(0.01)R 2.704813829Rf(0.003)R 2.714215587R define F(x)=listToMat(f(matToList(x,1))) RdoneR simplify(F(0.10.030.003))R 2.59374246 2.678597691 2.714215587R\f(x)Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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``)) rrrrrs`'''''0 (10qy`#YuY Yyq`TTTTTU OR h-1hh0R1R 2h-1hh0R 0.6931471806R 3h-1hh0R 1.098612289R 3x+h-3xhh0R 1.0986122893xReActFH$x^(f(a(x,1))) fH$x(1+x)^(((1)/(x)))02000dpiecewise.EAC010000002883^Rpiecewise(x<0,x,-x) \Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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 ` ORpiecewise(x<0,x2,-x)R-x[\Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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[[eAct020009polar.EAC010000009485k^5[A Complicated Polar Plot[In this exercise we make [some polar plots of the graph[[For r1, we set the view[windows as follows:[x is between -7 and 7,[y is between -4 and 4,[n is between 0 and 50.[\r1Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind r10H8 D P \ h t             ! " # $( %4 &@ 'L (X )d *p +| , - . 0 1 2 3 4 5 E F H I J$ K0 L< MH NT O` Pl Qx 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 k^ $  5n2-P(7*n)-P(32/31*7*n)`P`P`p3`p33 3`3333#YuY3 ` P[For r2, we set the view[windows as follows:[x is between -30 and 30,[y is between -20 and 20,[n is between 0 and 19.[\r2Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind r2$H, 8 D P \ h t            ! " # $ %( &4 '@ (L )X *d +p ,| - . 0 1 2 3 4 5 E F H I J K$ L0 M< NH OT P` Ql Rx 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 k^ $  5n(n*L(n))^2`P`P`x`xxaa x`xxxx#YuYx `&1WG7 P[\A flowerGraph2D| Graph3D LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceX, Sheet| Sheet3D| SolveEq|SolveLwr SolveUpr StupFLG1(StupListD StupPict$ ViewWind( r1DHr2\(Hr3(H          $ 0 < H T `  l !x " # $ % & ' ( ) * + , - . / D 0d 1p 2| 3 4 5 E F H I J K L M N O P$ Q0 R< SH TT ]` ^d _h `l ap bt x     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 k^ $  5n2 5nL(13*n)+5+`(10,5) 5nL(22*n)+3+`(2)`P`P``aa `#YuY  $0<HT`lx7c)!xhctF(EH11if'e9 'Dx@hUYcw3d()RWh8&%g1uy #t 3 0AbA`%g1uy`Rcx S[\ CochleoidGraph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind r1 H( 4 @ L X d p |           ! " # $ %$ &0 '< (H )T *` +l ,x - . 0 1 2 3 4 5 E F H I J K L, M8 ND OP P\ Qh Rt S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ aټb59740259 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 k^ $  5na*L(n)/n`P`P`d`ddYtYt d`ddd`d#YuYd nnnnnn`cxsh P[(a) Find the area of any[ one loop.[(b) Find the area between[ two loops.[(c) Find an equation of [the tangent line to the[curve at any point.[(d) Find the points on [the curve where the[tangent is vertical,[ horizontal.\ Double FoliumpGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind r10,Hr2\,Hr3H           ( 4 @ L X  d !p "| # $ % & ' ( ) * + , - . 0 1$ 20 3< 4H 5T E` Fl Hx I J K L M N O P Q R S T ] ^ _ `  a$ b( , 8 D P \ system]listsystem^]=system_^system`_systema`systemb~ aPb  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 k^ $  5n4*a*P(n)*(L(n))^2 5n-4*a*P(n)*(L(n))^2  5n1.5*a`P`P`p`p `(10q#YuY LLLLLL` R[(a) Find the area of a[loop.[(b) Find the points on[the curve where the [tangent line is vertical[(c) Find the area [bounded by the loop, a[vertical tangent and the[x-axis.[(d) Find the length of [a loop.[(e) For x=c, 0Open the the graph[editor below and graph[the function p(x).\ Graph of pDGraph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1H4 @ L X d p |            ! " # $$ %0 &< 'H (T )` *l +x , - . /D 0 1 2 3 4 5( E4 F@ HL IX Jd Kp L| M N O P Q R S T ] ^ _ ` a b    $ 0 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xp(x)`P`P`p@`@@PdQ @`@@@(10qy`@#YuY@   $0<HT`lx !X`( TTTU P vv Ps%b Qcccd UUUUV EEEEE`EEEEE`` Q[[Explorations. [a)Find a view window[so you can see the max[and min of the function p.[(Try [-7,10] for x and[[-3.8,3.8] for y).[b)After the plot is done,[we choose the table[mode for the graph;[choosing x from 1 to 10[we can 'roughly' see[where the minimum and[maximum should be. [c) We trace the graph [of pabove to estimate[wherethe max or min of [ p could be.[d) Or we could try the [following values. [Rp(1)R0.5Rp(2)R 0.4219858156Rp(3)R 0.3920454545Rp(4)R0.3875Rp(5)R 0.4047619048Rp(7)R 0.5511363636Rp(8)R 0.8055555556Rp(9)R2.1[ Conjectures: [a) The minimum happens[near x=4. This means[we would put 4 white[balls and 16 black balls[in urn A, and 46 white[balls and 34 black balls[ in urn B.[b) The maximum happens[near x=8, but x=8 is not[suitable because x^2=64,[which is greater than[50. Therefore, when [n=50, we see that the [max of p(x) happens [at x=7. It means that [we have 7 white balls [and 49 black balls in[ urn A and 43 [white and 1 black[balls in urn B.[[Anaytical Solution:[1. First, we observe that[the vertical aysmptotes[for p(x) is at x=0,x=-1,[and at 100-x-x2=0, [which can be solved [below:Rsolve(100-x-x2=0,x)Rx=- 10.5124922,x= 9.512492197[So, the vertical [aysmptote we need is[at x=9.512492197. [2. Next, we observe that[the function p is [increasing from some x to[x=9.512492197, which [can be verified, [alternatively, by drawing[its derivative below:R define r()=p()RdoneRr(x)R-1x+12x2+x-1002x4-174x2-100x-48x3+5025R\-Drag the expressionGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $ `P`P`p`PdQ ``(10qy`#YuY pt` O[-By choose 'root' under['Analysis', we get the[root of r(x) to be[xc=3.709174999115, which[is the minimum of p(x).[-Since p(3)=0.3920454545[and p(4)=0.3875, the[minimum of p is at [x=4.[-Since r(x)>0 in [[xc,9.512492197], we [conclude that the max[of p happens at [somewhere near the end[point, in this case, x=7 is[the point we need.[[ Exercise.[Let's see what[happens when n=100[Rdefine q(x)=12xx+x2+100-x200-x-x2RdoneR\ graph of qGraph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y1H , 8 D P \ h t           ! " # $ % &( '4 (@ )L *X +d ,p -| . 0 1 2 3 4 5 E F H I J K L$ M0 N< OH PT Q` Rl Sx 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xq(x)`P`P``% `(10qy`#YuY `8'x6 ` P[ Explorations:[1) Using 'G-Solve'->Min,[ we found [xc=4.874168397486 to be[the min.[2) By using the tracing,[we estimated that the [max of q is about at[x=11.9. Let's try the[ followings:Rq(3)R 0.3829787234Rq(4)R 0.3666666667Rq(5)R 0.362745098Rq(6)R 0.3688969259Rq(7)R 0.3854166667Rq(8)R 0.4149305556Rq(9)R 0.4636363636Rq(10)R 0.5454545455Rq(11)R 0.6960784314Rq(12)R 1.038461538R[ Conjectures:[1) Since q(5) Clearly, the first [series is an alternating[series and {1/k} is [decreasing to 0. By ['alternating series test',[this series converges['conditionally.[-> We prove the second[series converges by[ 'ratio test':R define a(n)=12nn RdoneR a(n+1)a(n)nR12R[Since the 'ratio' is less[than 1, it follows from[the 'ratio test' that[this series converges.[[(2) Prove graphically[that one series converges[faster than the [other.[->We insert the 'Sequence[Editor' below and select[ 'Explicit'.\ Two seriesGraph2D Graph3D LISTSYS4NModify $ STATCALC DNSTATSYS L\NSequence, Sheet| Sheet3DP| SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPictT$ ViewWindx seq_hist           ( 4 @ L X  d !p "| # $ % & ' ( ) * + , - . 0 1$ 20 3< 4H 5T E` Fl Gx H< IH JT K` Ll Mx N O P Q R S T UVW] ^ _ ` a  b$ ( 4 @ L X d,H(HHsystem]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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^4 $ R`P`P`B.B.B.)) B.B.B.B.(10qy`B.#YuYB. q7"  $0<HT`lx ,8DP\ht(4@LXdp|"" " %" 333333" fffffg" 33333" )ffg" 33333" Tffg" ffffg" X333" YR8 R8" &v" 4R8 R8" u" EcI cI" q" EcI cI" HV" 6T@T@" E5S"  S!x!" @t" 007U8" C" @XpQpR" 3@U" P%7P7" StY" `bP7" c(&P" p!iSyx6" gw)" f$"" i" wT" p6q" hwT" q7" Q((-1)^(k-1)/k,k,1,n),Q(1/(k2^k),k,1,n),1,ln(2)`1WG6 H((-1)^(k-1)/k,k,1,H)H(1/(k*2^k),k,1,H) H_(2)W[-> Seclect the 'table' [icon.[-> Select the 'graph'[mode. (Choose 'G-connect'[under 'Graph' icon).[-> By choosing the [Sequence TableIput to be[from 1 to 20, we [conjecture that the [sequence b(n) converges[faster than a(n) to[the number close to[0.69.[[Remark:[Theorectically, we can[prove that both series[converge to 'ln(2)', [which we will not prove[here.[[(3) Prove algebraically[that one series converges[faster than the other.R define f(n)= (-1)k-1k1nkRdoneR define g(n)= 1k2k1nkRdoneRsolve(f(n)-ln(2)-0.01=0,n)R -1k-1k1nk-1100-ln2=0[->This shows that the['solve' command does not[yield an answer, we try[the following:R f(50)-ln(2)R-9.900019984-3[[Therefore,n=50 is the [smallest integer that [will make the error [between f(n)and ln(2) [less than 0.01.[->Next, we do the same[ for g(n):Rsolve(g(n)-ln(2)-0.01,n)R 12kk1nk-1100-ln2=0Rabs(g(5)-ln(2))R4.605513893-3[[Therefore,n=5 is the [smallest integer making [ g(n)-ln(2) less than[0.01.[[ Exercise 1:[Prove analytically that [the following two series[are convergent:[ (-1)n-12n!4nn!21n and[ (-1)n-122-1nlog(2)1n.[[ Solution:[We R define a(n)=2n!4nn!2RdoneR define b(n)=22-1nlog(2)RdoneR a(n+1)a(n)nR 2n+1!n!242n!n+1!2nR simplify( 2n+1!n!242n!n+1!2n)R2122R[-This shows the first[series is convergent by[ratio test and comparison[test.[-It is easy to see that[the second series is [convergent because it is[an alternating series with[(1/n) decreasing to 0.[(2) Let's see graphically[which one converges[faster.\Open the sequencesGraph2D Graph3D LISTSYS4NModify $ STATCALC NSTATSYS $\NSequence, Sheet| Sheet3D(| SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict,$ ViewWindP seq_histl|              $ 0  < !H "T #` $l %x & ' ( ) * + , - . 0 1 2 3 4 5, E8 FD GP d H I J K L M N O P Q R, S8 TD UP,V|<] ^ _ ` a b       @HLLHsystem]listsystem^]=system_^system`_systema`systemb~jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj aseq_histbNewFolde system]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemj  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^4 $ R`P`P`$gS$gS `(10qy`#YuY t3H@ "2 $0<HT`lx ,8DP\ht(4@LXdp| $0<HT`lx ,8DP\ht(4@LXdp| $0<HT`lx" AwTy"%" gs"7P" Evb"d%" a`3d"bP" CR"WP" @pTT")hu" aCER"f#SQV" f$4u" )4W" Eh!#i"XV" 2&DX"u9wf" 57!5" 79" @fU"0hG6f" (Tr""@GV" G""Pc2(7" "pB"`#2p#" QA6!h"pYp)u" SA1"'vd0" URc"V5PU" !A4$@"0cY" XPRtG"S6g4D6" Q75" 3v%UrC" `FF"0PvfC`" '"UD "@6 @" cHr)p"PHGSTw" 8`a"`85xB" du`)"pFCa'q" vc#U"@("UE" f%gU"D``fI" 6S"Bx5ec" gV6FbD"B"3" v" C`Gh%U" hpipA"0ADbh" Yv3@"@E6"" iqx#RW"P@8'" (%3"`F64`" paydi"p8'G4" '%"GWt" qBF3W"7c2gC" Pa"HpS" rp#C"6TTx" E(f" IuCw%" r r"05Sh" e#V"@Pr`S)" s@ic"P4Yq4R" a76"`QcvTA" svI"p3qE" P`"RI&""(" tF5"2W0" `h1"S)eS6" t3H@ "Q((-1)^(k-1)(2k)!/(4^k(k!)^2),k,1,n)Q((-1)^(k-1)(2)^(1/2)/(((2)^(1/2)-1)klog(2)),k,1,n)`WG6 -H((-1)^(k-1)*(2*k)!/(4^k*(k!)^2),k,1,H);H((-1)^(k-1)*(2)^(1/2)/(((2)^(1/2)-1)*k*`(2)),k,1,H)U[ Exercise 2.[(1) Conjecture if[the following series [ diverges.[[ 4nn!22n!1n.[(2) Prove analytically [that the series diverges.[[ Solution:[For (1), we first andR define g(n)= 4kk!22k!1nkRdoneRsimplify(g(100))R 1194.278335Rsimplify(g(110))R 1376.531746Rsimplify(g(120))R 1567.210587Rsimplify(g(130))R 1765.957792Rsimplify(g(140))R 1972.4582R g(140)-g(130)10R 20.65004079R g(130)-g(120)10R 19.87472056[We compute the following[(by pressing EXE key).R1g(100)2g(110)3g(120)4g(130)5g(140)R1 1194.2783352 1376.5317463 1567.2105874 1765.9577925 1972.4582[->Drag the matrix above[into the 'List Editor' [below.[\ Scatter plot0Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind  ( 4 @ L X d p |          ! " # $ % &$ '0 (< )H *T +` ,l -x . 0 1 2 3 4 5 E F H I J K L M, N8 OD PP Q\ Rh St T ]T ^T _4 `8 a< b@ dDT eT      system]listsystem^]=system_^system`_systema`systemb~jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj aseq_histbNewFolde system]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemj  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^4 $ `P`P` @hh B6 0(10qy`#YuY $X  $0      $0Bx3P7e1t`VrXpvYWy $X  $0 $0Bx3P7e1t`VrXpvYWy $X `C4"PQ[->Draw the scatter plot.[->It follows from the [scatter plots above,[we predict that the series[ diverges.[\ Sequence plotGraph2D| Graph3D LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceX, Sheet| Sheet3D| SolveEq|SolveLwr SolveUpr StupFLG1(StupListD StupPict$ ViewWind( seq_histDT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F G($ HL IX Jd Kp L| M N O P Q R S T U ] ^  _ ` a b  , 8 D P \0Hsystem]listsystem^]=system_^system`_systema`systemb~jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj aseq_histbNewFolde system]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemj  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^4 $ R`P`P`P$gS$gS (10qy`#YuY @$X ""@") $0<HT`lx ,8DP\ht(4@LXdp| $0<HT`lx"Bx3H"!!3H9"#6 "$FGwA"&aCv"(C'eQ"0%q)G"2 ST"3v" 5y dV"7e1tX"9R&th"A@X H"C(g"E&e"GA9"HQHE"PD9"RX!"Twt r" VrXeG"!Xg'sx""`c%$v"#b`d8"$dWYv C"%fU"9&"&hUqfB"'pUg!"(rUxTTg")tW)&w"0vYWy!$"1xbc"2fGx"3q!2T"4vC!"5W"6F2 "7)"8Qxf "9gPB"@$X "@Q(4^k(k!)^2/(2k)!,k,1,n)q`WG6! H(4^k*(k!)^2/(2*k)!,k,1,H)S[->Select the table and[then plot the sequence. [->We conjecture that[the series is divergent.[[(2) Why diverges?[We R define a(k)=4kk!22k! RdoneRsimplify(a(k+1)/a(k))Rk+1k+12Rcombine(simplify(a(k+1)/a(k)))R2k+12k+1[Therefore, a(k+1)/a(k)>1[for each k. Since the[sequence (a(k)} is [increasing and does not[approach to 0 (see [below).R[\a(k) does not goes to 0.|Graph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind 0 < H T ` l x            ! " # $ %, &8 'D (P )\ *h +t , - . 0 1 2 3 4 5 E F GD HH IT J` Kl Lx M N O P Q R S T U] ^ _ ` a b    ( 4 @ L(Hsystem]listsystem^]=system_^system`_systema`systemb~jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj aseq_histbNewFolde system]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemjsystem]listsystem^]=systemj  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^4 $ `P`P`!!!$gS$gS !!!!(10qy`!#YuY! %dQ)"toto2 $0<HT`lx ,8DP\ht(4@LXdp| $0<HT`lx ,8DP\htto"fffffg" "eqBqC"44"C)C)"w8w8" !R)!R" 9f"y"gTcP0"W$"  B3w"0E$"@iC"P!0"`TCW"p6@G"W"#HY"wi7HDP"cFx"VQ" 6 4"0Tg $C"@rTF#"PfYeR"` )T("p %&u8" B " XaH" tbaE" vu0$" euAa"0 a"@sdh"P#I9"`qq"@"pW" b$D"PAbB"Eu%"U" !gb"0VY&BT"@Wdf"P#HS'"` Tt4"p!ii"#Pf"$8wr"%dQ)"4^n(n!)^2/(2n)!`cxsh H4^H*(H!)^2/(2*H)!R[ It follows[from the 'Divergence[Test' that the series is[ divergent.[eActaH0k((4^(k)*(k!)^(2))/((2*k)!)) bH4"n(((2))/(((2)-1)*n*(`(2))))cH4!n(((2))/(((2)-1)*n*(log2)))fH,n(((1)/((k))),k,1,n)gH8(n(((4^(k)*(k!)^(2))/((2*k)!)),k,1,n)n sH$n((1)/(2^(n)*n)) 02000fshortestdis.EAC010000003ae5^O[Finding the shortest[distance from a point[to a given curve.[\AuthorProfessor,Dr. Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 e-mail:wyang@radford.edu URL:www.radford.edu/ ~wyang[ Objective.[We would like to add[an animation to a [common Calculus [ problem. [[Example.[Find the shortest [distancefrom a point to [the curve y=(x-3)^2+3.\ an animation3 `3"d `7 9E(0vrALCuxy5H  H  A@6P %  (x-3)^2+3P cxsi @   ! @6 pY )H !`B@6 @    .@  3   ! @6 vh ' HQS`.@   .@    1@  9 [We collect the x-value, [length AB, slope AB [ and slope of [the tangent below.[012infinity -5.999975267 0.2631578947 10.49360496 39.86315789 -5.4736592 0.5263157895 9.134289281 17.32631579 -4.947343393 0.7894736842 7.925843368 9.989473684 -4.421027605 1.052631579 6.873325879 6.452631579 -3.89471185 1.315789474 5.983042209 4.435789474 -3.368396077 1.578947368 5.261877669 3.178947368 -2.842080233 1.842105263 4.715421916 2.356390977 -2.315764478 2.105263158 4.34469145 1.805263158 -1.789448682 2.368421053 4.142690537 1.435087719 -1.263132927 2.631578947 4.093658013 1.191578947 -0.7368171267 2.894736842 4.176853619 1.040191388 -0.210501285 3.157894737 4.372928675 0.9578947368 0.3158144783 3.421052632 4.668912411 0.9287449393 0.8421302333 3.684210526 5.0597856 0.9413533835 1.368446052 3.947368421 5.547276601 0.9873684211 1.89476184 4.210526316 6.13743403 1.060526316 2.421077638 4.473684211 6.838186773 1.156037152 2.947393403 4.736842105 7.657505818 1.270175439 3.4737091825 8.6023252671.44.000025[[R(1.191578947)*(-0.7368171267)R- 0.877975776[ Method 1.[We use calculus approach[d=sqrt(x^2+y^2), where[y=(x-3)^2+3. To minimize[d is the same as to minimize[d^2. Therefore, we Rdefine f(x)=x^2+((x-3)^2+3)^2RdoneRdefine f1(x)=diff(f(x),x)RdoneRf1(x)R4x3-36x2+122x-144\drawing f and f'Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $ `P`P`T`T Qia```(10qy`#YuY Y yI&`B#yO[By using the G-solve,[we found the min of f is[at x=2.591 which is roughly[what we found from the[ animation.[Method 2 Using the[slope.[Intuitively, the shortest[distance happens when[ the slope of AB is perpendicular[to the slope of the tangent[ at point A.[But slope AB = y/x where[y=(x-3)^2+3; thus[the slope of AB isRdefine s(x)=((x-3)^2+3)/xRdoneR[We need to solve x where[s(x)*g'(x)=-1, or[s(x)*g'(x)+1=0,[where[g(x) is the original [function (x-3)^2+3.Rdefine g(x)=(x-3)^2+3RdoneRdefine g1(x)=diff(g(x),x)RdoneRsolve(s(x)*g1(x)+1=0,x)Rx= 2.590979492R s(x)*g1(x)+1R2x-3x-32+3x+1R[\"dragging the function to the graphGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $ `P`P`p`p `x`(10qy`#YuY Y yI(`shBO[We get the x-intercept is[about x=2.5909794.[ Exercise.[Redo the problem by[replacing the curve to[ y=-(x-3)^2+3eActfH$xx^2+((x-3)^2+3)^2f1H x:(f(x),x)gH x(x-3)^2+3g1H x:(g(x),x)sH x((x-3)^2+3)/x020012slopeoftangent.EAC010000002c0f^[[Slope of the Tangent [Line at a point[\AuthorProfessor, Dr.Wei-Chi Yang Department of Math/Stats Radford University Radford, VA 24142 USA e-mail: wyang@radford.edu URL:www.radford.edu/ ~wyang[ Objective[We shall investigate the[relationship between the[slopes of secant lines [and the slope of the [tangent line at a point.[[Example.[Given the graph of the [function below, find the[slope of the tangent[line at the point [B=(-0.1,0.79) by[using various approaches.\slope of the tangent``vrALCuxy5H  H! C@6  -(x-1)^2+2`Y"iI% @   !% @6 q!Y q! RY H  B@6Y Y H    .@  3     @6 6HAYyG!x ruY.@   .@   1@  9  <[Method 1. Animation.[The lengths BC,[slope of BC, and [the slope of the [tangent line at the[point B is captured [below:[Part 1[derivative from the [left:[ column 1=x[column 2=length BC[column 3=slope BC[<-2 8.0183601814.1 -1.967798305 7.824042579 4.067798305 -1.93559661 7.631765261 4.03559661 -1.903394915 7.441527443 4.003394915 -1.87119322 7.253328315 3.97119322 -1.838991525 7.067167043 3.938991525 -1.806789831 6.883042768 3.906789831 -1.774588136 6.700954602 3.874588136 -1.742386441 6.52090163 3.842386441 -1.710184746 6.342882907 3.810184746 -1.677983051 6.166897456 3.777983051 -1.645781356 5.992944273 3.745781356 -1.613579661 5.821022315 3.713579661 -1.581377966 5.65113051 3.681377966 -1.549176271 5.483267746 3.649176271 -1.516974576 5.317432877 3.616974576 -1.484772881 5.153624718 3.584772881 -1.452571186 4.99184204 3.552571186 -1.420369492 4.832083577 3.520369492 -1.388167797 4.674348016 3.488167797 -1.3559661024.518634 3.455966102 -1.323764407 4.364940122 3.423764407 -1.291562712 4.213264928 3.391562712 -1.259361017 4.063606911 3.359361017 -1.227159322 3.915964509 3.327159322 -1.194957627 3.770336105 3.294957627 -1.162755932 3.626720022 3.262755932 -1.130554237 3.485114521 3.230554237 -1.098352542 3.3455178 3.198352542 -1.066150847 3.207927987 3.166150847 -1.033949153 3.072343141 3.133949153 -1.001747458 2.938761248 3.101747458 -0.9695457627 2.807180214 3.069545763 -0.9373440678 2.677597867 3.037344068 -0.9051423729 2.550011949 3.005142373 -0.872940678 2.424420111 2.972940678 -0.8407389831 2.300819915 2.940738983 -0.8085372881 2.179208821 2.908537288 -0.7763355932 2.05958419 2.876335593 -0.7441338983 1.941943272 2.844133898 -0.7119322034 1.826283209 2.811932203 -0.6797305085 1.712601018 2.779730508 -0.6475288136 1.600893597 2.747528814 -0.6153271187 1.49115771 2.715327119 -0.5831254237 1.383389984 2.683125424 -0.5509237288 1.2775869 2.650923729 -0.5187220339 1.17374479 2.618722034 -0.486520339 1.071859821 2.586520339 -0.4543186441 0.9719279931 2.554318644 -0.4221169492 0.8739451291 2.522116949 -0.3899152542 0.7779068626 2.489915254 -0.3577135593 0.6838086296 2.457713559 -0.3255118644 0.5916456575 2.425511864 -0.2933101695 0.501412953 2.39331017 -0.2611084746 0.41310529 2.361108475 -0.2289067797 0.3267171967 2.32890678 -0.1967050848 0.2422429413 2.296705085 -0.1645033898 0.1596765176 2.26450339 -0.1323016949 0.07901162858 2.232301695-0.10012.416700428-4 2.200100006[Part 2. \derivative from the right``vrALCuxy5H  H  C@6@  -(x-1)^2+2@ YURT#rY@   ! @6sF#D64Y# Q SQdbYH  A@6Y Y@   .@  3  H  B@6Y Y  @6 6HAYyG!x ruY.@  .@  1@  9 1@  9!  y=-(x-1)^2+2!<[ column 1=x[column 2=length BC[column 3=slope BC[<2 2.110473880.1 1.964408475 2.083299182 0.1355915254 1.928816949 2.058328171 0.1711830508 1.893225424 2.035390342 0.2067745763 1.857633898 2.014310467 0.2423661017 1.822042373 1.994910031 0.2779576271 1.786450847 1.977008652 0.3135491525 1.750859322 1.960425452 0.349140678 1.715267797 1.944980342 0.3847322034 1.679676271 1.930495201 0.4203237288 1.644084746 1.916794934 0.4559152542 1.60849322 1.90370839 0.4915067797 1.572901695 1.891069137 0.5270983051 1.537310169 1.878716103 0.5626898305 1.501718644 1.866494069 0.5982813559 1.466127119 1.854254052 0.6338728814 1.430535593 1.841853548 0.6694644068 1.394944068 1.829156692 0.7050559322 1.359352542 1.816034309 0.7406474576 1.323761017 1.802363898 0.7762389831 1.288169492 1.788029543 0.8118305085 1.252577966 1.772921773 0.8474220339 1.216986441 1.756937382 0.8830135593 1.181394915 1.73997921 0.9186050848 1.14580339 1.721955904 0.9541966102 1.110211864 1.702781663 0.9897881356 1.074620339 1.682375973 1.025379661 1.039028814 1.660663331 1.060971186 1.003437288 1.637572978 1.096562712 0.9678457627 1.613038628 1.132154237 0.9322542373 1.586998208 1.167745763 0.8966627119 1.559393598 1.203337288 0.8610711864 1.53017039 1.238928814 0.825479661 1.499277649 1.274520339 0.7898881356 1.466667693 1.310111864 0.7542966102 1.432295876 1.34570339 0.7187050847 1.39612039 1.381294915 0.6831135593 1.358102074 1.416886441 0.6475220339 1.318204238 1.452477966 0.6119305085 1.276392492 1.488069492 0.576338983 1.232634596 1.523661017 0.5407474576 1.186900309 1.559252542 0.5051559322 1.139161254 1.594844068 0.4695644068 1.089390793 1.630435593 0.4339728813 1.037563908 1.666027119 0.3983813559 0.9836570902 1.701618644 0.3627898305 0.9276482383 1.73721017 0.3271983051 0.8695165641 1.772801695 0.2916067797 0.8092425039 1.80839322 0.2560152542 0.7468076365 1.843984746 0.2204237288 0.6821946073 1.879576271 0.1848322034 0.6153870576 1.915167797 0.149240678 0.5463695589 1.950759322 0.1136491525 0.4751275524 1.986350847 0.07805762711 0.4016472918 2.021942373 0.04246610169 0.3259157916 2.0575338986.874576264-3 0.2479207774 2.093125424-0.02871694916 0.1676506412 2.128716949-0.06430847458 0.08509439906 2.164308475-0.099900000012.416517994-42.1999[[By inspection, when the[distance BC gets closer[to 0, the slope of BC[gets closer to the slope[of the tangent line at[ the point B.[[Method 2.Numerically,RDefine f()=-(-1)^2+2Rdone[ We let x=-0.1[R Define Q(h)=f(-0.1+h)-f(-0.1)hRdoneR define F(x)=listToMat(Q(matToList(x,1))) RdoneRF(0.0010.00010.000010.000001-0.001-0.0001-0.00001 -0.000001)R2.1992.19992.199992.1999992.2012.20012.200012.200001R[R[To find the slope of the[tangent line at the point[(-0.1,0.79), we doRdiff(f(x),x,1,-0.1)R2.2[[*the above syntax means[taking derivative of f [with respect to x, [order 1 at x=-0.1 or[we do the followings:Rdefine f1(a)=diff(f(x),x,1,a)RdoneRf1(a)R-2a-1Rf1(-0.1)R2.2R[[[ Exercise 1.[Redo the problem by [changing the point B to[ other points.[[ Exercise 2.[Reconstruct the curve[by using other function,[the secant line and the[ animation.eActFH$x^(Q(a(x,1))) QH,h((f(-0.1+h)-f(-0.1))/(h))fH x-(x-1)^2+2f1H$a:(f(x),x,1,a)020007sum.EAC010000003bbe t^;M[ Objective:[We would like to examine [the rate of convergence[for two infinite series.[Rdefine f(n)=Q(1/(k*2^k),k,1,n)RdoneR"define g(n)=Q((-1)^(k-1)/k,k,1,n)Rdone[First, we try out some[ finite terms.[Rf(100)R 0.6931471806Rg(100)R 0.6881721793Rg(10)R 0.6456349206Rg(12)R 0.6532106782Rg(13)R 0.7301337551Rg(14)R 0.6587051837Rg(15)R 0.7253718504R[Next we draw their [repective partial sums.\plots of f and gXGraph2D0 Graph3DD LISTCALP$ LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1Hy24HL X d p |              ! "$ #0 $< %H &T '` (l )x * + , - . / 0 1 2 3 4 5 E F H I J, K8 LD MP N\ Oh Pt Q R S T ] ^ _ ` a b d e fP g        , 8 D  f(n)g(n)eActnoeActsum1feActsum2fsystem`_systema`systemb~ aseq_histbNewFolde eActmnlisteActmsum1=system1eActmnlisteActmsum1=systemeActmnlisteActmsum2=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i t^; $  3xf(n) 3xg(n)`P`P` @%@% TV)V pDb Dc (10qy`#YuY  $0<HT`lx%  %  %  %  %  hwT   $0<HT`lx@P`p5  $0<HT`lx% )ffg X333 u HV 3@U StY c(&P gw) i p6q q7  $0<HT`lx@P`p5  $0<HT`lx 33333 ffffg 4R8 R8 EcI cI XpQpR %7P7 bP7 !iSyx6 f$" wT hwT I"i B6(U "Gi'4 r6F5 T$eAx `qsx \[We will guess that both[series converge but f[will converge faster than[g.R f(n)nR 121kk1nkn[Now we apply the ratio[test.Rdefine a(n)=1/(n*2^n)RdoneR a(n+1)a(n)nR0.5[This says that the series[of f(n) is convergent.[Next we can apply the [alternating test on the [series of g(n) and [conclude that the series[g(n) is convergent too.[[It can be proved that[both limit(f(n))=limit(g(n))[=ln(2).[[Now we can see how[fast the series f(n) and [g(n) converge to desired[ accuracy.[Rsolve(f(n)-ln(2)=10^(-2),n)R  1.000000000 2.000000000 1.0000000001k 1.000000000k 1.000000000 1.000000000n 1.000000000k- 0.703147181= 0.000000000Rf(50)R 0.693147181R f(50)-ln(2)R 0.000000000Rln(2)R 0.693147181Rf(20)R 0.693147137Rg(20)R 0.668771403Rsolve(g(n)-ln(2)-10^(-3)=0,n)R - 1.000000000 1.000000000k- 1.000000000 1.000000000k 1.000000000 1.000000000n 1.000000000k- 0.694147181= 0.000000000R\Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr@ SolveUprL StupFLG1X(StupListD StupPict$ ViewWind    ( 4 @ L X d p |        ! " # $ % & ' ($ )0 *< +H ,T -` .l 0x 1 2 3 4 5 E F H I J K L M N O, P8 QD RP S\ Th ]t ^x _| ` 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@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3Dg(n)-ln(2)=10^(-3) i  t^; $ `P`P`p`p `(10qy`#YuY ` O[[eActaH n1/(n*2^n)fH(n(1/(k*2^k),k,1,n)gH,n((-1)^(k-1)/k,k,1,n)n s1 l @lP$!x&7, kT  rD0 I oާtJ H׾8S)Kղ~UG pꎷ<w̻&]*nqxڪ8@ފ 4@Q qVu7W1=A7vyLw|'2qFËp)0ipu6ǏP1Fߴj^C0+L$B ڤZZ6{yT]BHC/aаG Vssum1  $0<HT`lx% )ffg X333 u HV 3@U StY c(&P gw) i p6q q7 sum2  $0<HT`lx 33333 ffffg 4R8 R8 EcI cI XpQpR %7P7 bP7 !iSyx6 f$" wT hwT 020010trigformulae.EAC010000003e70<$m[ Trigonometric[Formulae[[ Objective:[We would like to explore[ the formulae[ Formula 1.[c2=a2+b2-2abcos(C).[[We can also write [[ C = cos(a2+b2-c22ab)[[When you know lengths[of sides of a triangle[you can use this formula[to find the angle sizes.[[ Formula 2.[Area =AD*AC*sin(D).[[ Exploration [We shall verify the [formula:[cos(a2+b2-c22ab).[[Tap on the 'Geometry[ Strip' below:\ An animation``vrALCuxyH  H  C@6` Y  @6Thg&i   H  A@6 ` Y@  'r5H!   @6  H  B@60 Y @   @  @8F8F P @    @6 @ fd5#4Y 2"  `D@6 PY@   @    @6 Y@     [-Tap on Edit->Animate[ ->Go once.[-We see that the point C[moves along the line [ segment AB.[-We would like to [examine how the angle [D changes when the [lengths of three sides, [AD, CD and AC change.[-After the animation, we[do the follwoings:[ ->We tap the [measurement box for AD [and capturethe table [for the length of AD.[->Similarly, we capture[the lengths of CD and[AC respectively.[->We last tap on the [lines egments AD and [CD to measure the [angle D, we capture the[table for the angle D.[->We show the table [below; the first column[is AD=a, the second[column CD=b, the third[column AC=c and [the last column is the [angle D.[[-We copy and paste the[matrix from the geometry[strip into the eActivity,[which is shown below.[3.753.75003.75 3.0644126681.625 25.101608713.75 3.162277663.25 55.304846473.75 3.986304684.875 78.055822813.75 5.2021630126.5 91.65230468[-We drag and drop the[above matrix into the [list editor below.\Three sides and an angleGraph2D@ Graph3DT LISTCAL`\ LISTSYS4NModify $ STATCALC NSTATSYS \NSequencex, Sheet| Sheet3D | SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict$$ ViewWindH d p |               $ !0 "< #H $T %` &l 'x ( ) * + , - . 0 1 2 3 4 5 E F, H8 ID JP K\ Lh Mt N O P Q R S T ]T ^(T _|T `T a$T bx |      cos((list1^2+list2^2-list3^2)/(2list1list2))qsystem]listsystem^]=system_^system`_systema`systemb~YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY aseq_histbNewFolde system]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemY  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i <$ $ `P`P`p`p `(10qy`#YuY  $0uuuuu $0uDf"wfch !c  $0bP%PP $0Q`SdpU( R0F $0Q`ASdq8U( R0Fp` P[We shall verify the [formula:[cos(a2+b2-c22ab).[In the list below:[list1=a, list2=b, list3=c,[and list4=list5, which is[what we expected.[[ Example 2.[We will use the animation[above and experiment[the area of the triangle[ACD; [area =AD*AC*sin(D).[ Solution:[-Animate once.[-Capture the lengths of[AD and CD repectively.[-Capture also the angle[of D (by selecting AD [and CD and the angle[measurement box).[-Capture the area of the[ triangle DAC.[3.753.75003.75 3.064412668 25.101608712.43753.75 3.16227766 55.304846474.8753.75 3.98630468 78.055822817.31253.75 5.202163012 91.652304689.75[-Drag and Drop the[matrix to the 'list editor'[below.[\areaGraph2D@ Graph3DT LISTCAL`H LISTSYS4NModify $ STATCALC NSTATSYS \NSequenced, Sheet| Sheet3D | SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict$ ViewWind4 P \ h t               ! "( #4 $@ %L &X 'd (p )| * + , - . 0 1 2 3 4 5 E F H$ I0 J< KH LT M` Nl Ox P Q R S T ]T ^T _hT `T aT bd h t     (1/2)*list1list2sin(list3)system]listsystem^]=system_^system`_systema`systemb~YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY aseq_histbNewFolde system]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemYsystem]listsystem^]=systemY  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i <$ $