{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Ti mes" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }{PSTYLE "Author" -1 257 1 {CSTYLE "" -1 -1 "Ti mes" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 8 8 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 2" -1 258 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 259 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 260 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "The Cycloid" }}{PARA 257 "" 0 "" {TEXT -1 8 "PROBLEM:" }}{PARA 0 "" 0 "" {TEXT -1 217 "Consider a circle of radius a that rolls without sliding along the horizontal \+ x-axis at a constant speed v. We want to find the trajectory traced by a point on the boundary of this circle, as it rolls over the x-axis. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "with(plots): read \"h: \\\\cygwin\\\\home\\\\carlos\\\\2004\\\\cycloid-figs.mpl\";" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "display(fig1);" }}{PARA 13 " " 1 "" {GLPLOT2D 559 559 559 {PLOTDATA 2 "6--%'CURVESG6&7$7$$\"\"!F)F( 7$$\"+aEfTJ!\"*$\"\"\"F)7%7$$\"+0K(f&GF-$\"+L'z?3\"F-F*7$$\"+0K(f&HF-$ \"+uO?zw!#5-%&STYLEG6#%,PATCHNOGRIDG-%'COLOURG6&%$RGBGF($\"*++++\"!\") F(-F$6&7$F*7$F+$\"\"#F)7%7$$\"+aEf\"4$F-$\"++++D>F-FJ7$$\"+aEf\">$F-FQ F<-FA6&FCF(F(FD-F$6&7$7$F(F.F'7%7$$\"+++++]!#6$\")+++vF-F'7$$!+++++]Fj nF[oFW\"!#?$\"3w)zNM3KF5#!#>7$$\"3W7!oak[ 4Q*Fdp$\"35vjUwSQ*G(Fgp7$$\"3/yO)espLG$Fgp$\"37V;SJu(Rm\"!#=7$$\"3;\"e /\"3/M#*yFgp$\"3l-^dWt;WHFbq7$$\"3TM.+F@'\\`\"Fbq$\"3cxi9_YC/XFbq7$$\" 3d,%>7_Cr_#Fbq$\"330(R,131:'Fbq7$$\"39-DYY]MlJFbq$\"3sUDLU(f$fqFbq7$$ \"39Xvq&Q2G*QFbq$\"3a(p#p8#ob*zFbq7$$\"3W^$>Ge!RTZFbq$\"3%3[c!ziP$)*)F bq7$$\"3_tH?$4<$*o&Fbq$\"35JBh>kL\")**Fbq7$$\"3:5*=.\\%eLnFbq$\"3#y*e \"*=#Hw4\"!#<7$$\"3S'[(*Qj*fwyFbq$\"3[nZu,h:'>\"F[t7$$\"3!p;<\\n?Q:*Fb q$\"39;m$f&GY&H\"F[t7$$\"3++p&4HLJ0\"F[t$\"3E;$='**ym\"R\"F[t7$$\"3Iqu YG*zC=\"F[t$\"3!>STB#>.t9F[t7$$\"30Z1)zY.!>8F[t$\"3C:D\"pnU0b\"F[t7$$ \"3Gl0p!)ph];F[t$\"3&HZ0^#)Rxq\"F[t7$$\"3CSH*o#f&G,#F[t$\"3[kV'*o)\\i$ =F[t7$$\"3S7;s[`c$Q#F[t$\"3([yny0\"HF>F[t7$$\"3[4_2gx2dDF[t$\"3S&>9$\\ R)p&>F[t7$$\"3Fj;Q()H!Ht#F[t$\"3WIma\"Q[!z>F[t7$$\"3+=XSF8.WHF[t$\"3Cj b0;s6&*>F[t7$$\"3Cq.c@=FcJF[t$\"3oYD8lI(***>F[t7$$\"3l>))e_.0OLF[t$\"3 'p>!HX&p_*>F[t7$$\"3/'*QpwI/:NF[t$\"3mx#R(oe^#)>F[t7$$\"3h$QnCz58s$F[t $\"3'pS$34@pd>F[t7$$\"3H$>t\\XHW#RF[t$\"3()\\VRs#*QA>F[t7$$\"3K6q!Qx-I F%F[t$\"3n!yX*o/XN=F[t7$$\"3kcP`2DhLYF[t$\"3*4[Q>n2tq\"F[t7$$\"3'3?fV# f-]\\F[t$\"3')pO![jA\"e:F[t7$$\"3yeb'**evqC&F[t$\"3E!*oGtuL!Q\"F[t7$$ \"37xop8?0r`F[t$\"3;)GC6'Q1$H\"F[t7$$\"3%3vt%*y7o[&F[t$\"33%G,!p>F.7F[ t7$$\"3Koja*GSAg&F[t$\"3bB')Q!>%[/6F[t7$$\"3$=Is$)=Exq&F[t$\"3n=K$[&>l /5F[t7$$\"3*R6LzT7n!eF[t$\"3y&)RwPu=4!*Fbq7$$\"3iIzlHq'\\*eF[t$\"35S:6 H4a#)zFbq7$$\"3O\\g\"zT%=jfF[t$\"3*>'*))3.sh5(Fbq7$$\"3rfbi*RxN-'F[t$ \"3Yl59n@\\_%Fbq7$$\"30R0;b0m.iF[ t$\"3Golq-.ieHFbq7$$\"3?&G470$=\\iF[t$\"3@m:Faa_-!ejk^siF [t$\"3g@^.')*\\o$zFgp7$$\"3[D3B#Qi?G'F[t$\"3ErY!*ydW!y\"Fgp7$$\"3Qxi<> $[IG'F[t$\"3%zh'*>JE`Q%Fdp7$$\"3!f8yyI&=$G'F[t$\"3-V%pzzw(*H\"!#B7$$\" 3'G>>dEpLG'F[t$\"3,/Z'R3lsL&Fdp7$$\"39X9Kb0i%G'F[t$\"3Ie&*yn>C'4#Fgp7$ $\"3%y/*e%ygHH'F[t$\"3g)Rg.3,3\\(Fgp7$$\"3)zkUL09qJ'F[t$\"3Y[T_w?$op\" Fbq7$$\"3e&*\\f\")39gjF[t$\"3eyUqi\"yl*GFbq7$$\"3*zJ?c;(GKkF[t$\"3+Qy! 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G'F[tFaam7$$\"3P4GaKw&>T'F[tFfam7$$\"3Yz*\\x^l9.K2(F[tFdcm7$$\"3o+1!)e!>1:(F[tFicm7$$\"3=p6]-ao0s F[tF^dm7$$\"3]s=5%>a'[sF[tFcdm7$$\"3#Q9w2'3)QF(F[tFhdm7$Fg^nF[emF]em-% +AXESLABELSG6%Q!6\"Fjgn-%%FONTG6#%(DEFAULTG-%(SCALINGG6#%,CONSTRAINEDG -%%VIEWG6$F_hnF_hn" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 317 "The f igure above displays a rolling unit circle at t = 0, t = Pi, and t = 2 Pi. As the circle moves to the right, the blue arrow initially at 6 o 'clock, turns around tracing the black archs that form the cycloid. To find out the equation of the cycloid we notice that the position vect or at time t, R(t), is given by," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "R := t -> C(t) + d(t):" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "where C(t) represents the position vector of the center of the circle at time t," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "C := t -> v *t*i + a*j:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 257 "which encodes the given fact that the center of the circle moves with constant speed v \+ along the x-axis starting from (0,a). Recall that a is the radius of t he circle. The blue vector d(t) just turns around the center of the ci rcle. Hence, d(t) is given by:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "d := t -> a*cos(b-w*t)*i + a *sin(b-w*t)*j:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Clearly d(t) = d(t+T), i.e., it is a periodic function of t with period T obtained f rom the periodicity of sines and cosines. We find T by," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "T := s olve(b - w*(t+T) = b - w*t - 2*Pi, T);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"TG,$**\"\"#\"\"\"%#PiGF(%\"aGF(%\"vG!\"\"F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "and since we know that the circle of rad ius a is rolling at constant speed v along the x-axis, we deduce that \+ it takes," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 14 "T := 2*Pi*a/v;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %\"TG,$**\"\"#\"\"\"%#PiGF(%\"aGF(%\"vG!\"\"F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "units of time to make one complete turn. Thus," }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "w := v/a:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The value of b \+ is chosen so that d(0) = -a j. This requires cos(b) = 0 and sin(b) = - 1. Therefore," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "b := -Pi/2:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 79 "and we obtain the vector position funtion for the \+ trajectory of the cycloid as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "`R(t)` := collect(R(t),[i,j]);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%%R(t)G,&*&,&*&%\"vG\"\"\"%\"tGF*F**&%\"aGF*-%$sinG6#*(F)F*F-!\"\"F +F*F*F2F*%\"iGF*F**&,&F-F**&F-F*-%$cosGF0F*F2F*%\"jGF*F*" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT 256 37 "THE LE NGTH OF ONE ARCH OF THE CYCLOID" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 " Clearly the speed v only affects the parametrization but not the resul ting curve and we can always assume v = 1. We have," }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "x := t - \+ a*sin(t/a); y := a*(1-cos(t/a));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"xG,&%\"tG\"\"\"*&%\"aGF'-%$sinG6#*&F&F'F)!\"\"F'F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG*&%\"aG\"\"\",&F'F'-%$cosG6#*&%\"tGF'F&!\"\"F. F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "The components of the veloc ity vector are then," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "dx/dt = diff(x,t); dy/dt = diff(y,t);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%#dxG\"\"\"%#dtG!\"\",&F&F&-%$cosG6 #*&%\"tGF&%\"aGF(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%#dyG\"\"\"% #dtG!\"\"-%$sinG6#*&%\"tGF&%\"aGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "the speed at time t is then given by the length of the velocity vector. i.e.," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 40 "ds/dt = sqrt(diff(x,t)^2 + diff(y,t)^2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%#dsG\"\"\"%#dtG!\"\"*$,&*$),&F&F&- %$cosG6#*&%\"tGF&%\"aGF(F(\"\"#F&F&*$)-%$sinGF0F4F&F&#F&F4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "the length of one arch is then," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "L := Int(sqrt(diff(x,t)^2 + diff(y,t)^2),t=0..2*Pi*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG-%$IntG6$*$,&*$),&\"\"\"F--%$cosG6#*&%\"tGF- %\"aG!\"\"F4\"\"#F-F-*$)-%$sinGF0F5F-F-#F-F5/F2;\"\"!,$*(F5F-%#PiGF-F3 F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "expanding the square insi de the square root we obtain," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,*\" \"\"F$*&\"\"#F$-%$cosG6#*&%\"tGF$%\"aG!\"\"F$F-*$)F'F&F$F$*$)-%$sinGF) F&F$F$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "and L becomes," }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "L := Int(sqrt(2)*sqrt(1-cos(t/a)),t=0..2*Pi*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG-%$IntG6$*&\"\"##\"\"\"F),&F+F+-%$cosG6#*&% \"tGF+%\"aG!\"\"F3F*/F1;\"\"!,$*(F)F+%#PiGF+F2F+F+" }}}{PARA 11 "" 1 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "assume(a>0 ): L := int(sqrt(2)*sqrt(1-cos(t/a)),t=0..2*Pi*a);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"LG,$*&\"\")\"\"\"%#a|irGF(F(" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 0 "" }{TEXT 257 38 "THE AREA UNDER ONE ARCH OF THE CYCLOID" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Solving \+ for t in terms of y we get," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "y := 'y': t := solve(y = a*( 1-cos(t/a)),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"tG,$*&%#a|irG\" \"\",&%#PiG!\"\"-%'arccosG6#*&,&%\"yGF(F'F+F(F'F+F(F(F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "so x as a function of y is given by," }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%#a| irG\"\"\",&F%F%-%$cosG6#*&%\"tGF%F$!\"\"F,F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "x := t - a*sin(t/a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG,&*&%#a|irG\"\"\",&%#PiG!\"\"-%'arccosG6#*&,&%\"y GF(F'F+F(F'F+F(F(F+*&F'F(,&F(F(*&F0\"\"#F'!\"#F+#F(F5F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "We compute the area under one arch by ad ding horizontal stripes for each value of y from 0 to 2a, like in the \+ following figure:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 14 "display(fig2);" }}{PARA 13 "" 1 "" {GLPLOT2D 559 559 559 {PLOTDATA 2 "6)-%'CURVESG6%7$7$$\"+aEfTJ!\"*$\" \"!F,7$F($\"\"#F,-%'COLOURG6&%$RGBGF,F,F,-%*LINESTYLEG6#\"\"$-F$6$7$7$ $\"+eB;-b!#5$\"+A<0#z*F>7$F(F?-F16&F3$\"*++++\"!\")F+F+-F$6$7$7$$\"+M, 0D]F>$\"+$)zi#H*F>7$F(FMFB-F$6$7S7$F+F+7$$\"3e;cW/cSxU!#@$\"3%R#3*QgiP O*!#?7$$\"3!e+5`Gi4z#FZ$\"3Sn)o<,z>E$!#>7$$\"3=%\\Q;ZV:#)*FZ$\"3!)\\'* Q6_89vFjn7$$\"3s^>#3I_\"zBFjn$\"3#)RJ?\"3HpM\"!#=7$$\"3]e/y%)H\\tYFjn$ \"3eW\\7nzu&4#Feo7$$\"3=rfe)=K:y(Fjn$\"3[]1OJeUQA>'\\WFeo$\"3?x[QxJ'[l)Feo7$$\"3x'ph27ORr&Feo$\"3B/u(p5(f+5!#<7$$ \"3['zTJP\\z;(Feo$\"3q\\T#fB[i8\"F^r7$$\"3>&p6Ev5Du)Feo$\"3[:d.:#=YE\" F^r7$$\"3/d\"GX^+;.\"F^r$\"3%zV1J*3Jx8F^r7$$\"3:(HTIn>dB\"F^r$\"3u\\AO F:B/:F^r7$$\"3^+4r2)>AU\"F^r$\"3O1cjZ*[Rg\"F^r7$$\"37$*QIe8]`;F^r$\"3] CknM;$*3b5msp=F^r$\"3Ie_b?/U!z\"F^r7$$\"3]Y*4&3/k<@F^r$ \"3'z-\")RnIf'=F^r7$$\"3!fRu^npAO#F^r$\"3>XOJFO4B>F^r7$$\"3p<*)\\F3ICE F^r$\"3q!RS4Nij'>F^r7$$\"3Ixg&G[%3pGF^r$\"3!ypT>+.2*>F^r7$$\"3KE`uH+RN JF^r$\"3+E$G4>&****>F^r7$$\"3#\\*)H\"4E.7MF^r$\"3O0B(=!Q%3*>F^r7$$\"3' G]t4$o$3l$F^r$\"3?b`sYlSn>F^r7$$\"3j:6V=/d/RF^r$\"3@:LI@iKE>F^r7$$\"3% HKb(Qt(*fTF^r$\"3gF)4Kh=u'=F^r7$$\"3LY8W$y`6S%F^r$\"3K#*z7xng%z\"F^r7$ $\"3\\h+c,S_CYF^r$\"3mK\"4Pfc5r\"F^r7$$\"3Mm!QO\\L\"f[F^r$\"3Q%y6Ike[g \"F^r7$$\"3I$Hx9UFk0&F^r$\"35Aa#=%)o!*\\\"F^r7$$\"3!)>%>7\\L;D&F^r$\"3 '*f?<[\"ysP\"F^r7$$\"3]uoz;T%QT&F^r$\"3wZ#H\"el'3E\"F^r7$$\"3!=\\OGp% \\ubF^r$\"33@T_xm:H6F^r7$$\"3:mWQ=pK4dF^r$\"3!zOHMQdI+\"F^r7$$\"3$o77H 1%>LeF^r$\"3spxOYnF7()Feo7$$\"3(z@TH0mv$fF^r$\"3ungHa'e\\W(Feo7$$\"35X GdRicHgF^r$\"3C:@W9;?khFeo7$$\"3Z/qowO?-hF^r$\"3k(Q-RU*p'*\\Feo7$$\"31 UI8X3JhhF^r$\"3!3v60g\\1*QFeo7$$\"3H)*HI_+11iF^r$\"3#fRKNh*p+HFeo7$$\" 3]'=)=I4JOiF^r$\"3]S\"yig@)*4#Feo7$$\"35r[&G$G'*fiF^r$\"32g)*y$[icK\"F eo7$$\"3/L^'=)R*GF'F^r$\"3-s;%yY!**\\xFjn7$$\"3+\"*=`-Y9!G'F^r$\"3OR8x 286`MFjn7$$\"3]w>MG;w#G'F^r$\"3Hz?W.kW/$*FZ7$$\"3C'ezrI&=$G'F^rF+F0-%+ AXESLABELSG6$Q!6\"Fd]l-%(SCALINGG6#%,CONSTRAINEDG-%%VIEWG6$%(DEFAULTGF ]^l" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1 " "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Area := 2*Int(Pi*a-x,y=0..2*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%AreaG,$*&\"\"#\"\"\"-%$IntG6$,(*&%#PiGF(%#a|irGF(F(*&F/F(,&F. !\"\"-%'arccosG6#*&,&%\"yGF(F/F2F(F/F2F(F(F(*&F/F(,&F(F(*&F7F'F/!\"#F2 #F(F'F(/F8;\"\"!,$*&F'F(F/F(F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Area := 2*int(Pi*a-x,y=0..2*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%AreaG,$*(\"\"$\"\"\"%#PiGF()%#a|irG\"\"#F(F(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 281 "This integration is nowadays done by mighty Maple in a fraction of a second but it was historicaly a so urce of pride for the few mathematicians that were able to obtain it. \+ The problem becomes simpler with the help of Green's theorem that allo ws to compute areas by line integrals." }}}}{MARK "41 0 0" 97 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }