00020001010009ropes.ACT0002020012Ladder_April07.EAC010000005c38 Ladder_April07.EAC ropes.ACT#&037 J ~I '[4A lf problemX\ AuthorXx,4 Objective:F[h Example 1.a N A fence qet tall runs par el to a building atdista6of p9from Wthe2. Wh5is length8$stestt* can reachTLground +ov+epwX?'\Example 1=>play the animationΈ%)/ 83SsK `a)Q  fF'% #i#e DvHKr  A kL'C uxy q 9  5@6Wy /  u4Qg` D Length GH: $,H nGnWTHDn ȐCH $5 h %H  7"`@D@G6Np&Dd zH UEC[ @ !  ?, )@`" t9t.4RwyuGSEx  DHE:    @6fF3% $0cuT h%1F.`AB B 6 F-AG0@ @YĘo9"3Ewy "GSEx  FGD: y;}/V `C@6d K@  YoSD9 D1Wy !Y!E6$d Length EH=x:  $5@# 8Ţgɓ_B_SH!_H! d  eQrAI N@ Nl"}Ř88!V"" # #@62p `pJ$`""p%\@*&r @'H! F@6  4(=@  '&%Ў")**} i`zJ+`z",!H-ǘH 2 pQ 1AI#!x  ][[Exploration and Approxim:& L-play the$.collectdata for'point ES7 distance GHTR-dragTx-coordinate of7length5 back to curve, we se mimimum.Ԑ Symbolically ULet AE=p EH=x, AF=qGF=y,n]conside wo simila riangles DEH2GAH.[YWhDEqGA"=HHA? or<qq+y<xx+p. [.,Step 1. We will solve y in terms of x first.RNT Clear_a_z Rdone&N (q/(q+y)=x/(),y)4y=pqN2>define the distance function GH below by using Pythagorean The m.Wf(x =(p+x)^2+^2璟3JsubstituIߍ itosay:= pq xR1f(x,y)+"x+pQ2Z+M+q2[,LStep 4. Find the critical points by settingderivative of f equals to 0. solve(diff(,x)=0 x=-p,x=‹1:3Suppose p is 3.7 aq 6:Bp:=m37]10lq:=6 a x:= pq 213R-666#J5>iyK1113N5nSf(x,y)Y"( 13.5876105`#A5[,Conclusion: The above is t minimum length.4 Discus4GFWhen3GH reaches itsB, willT triangle A+bqnnoceles?,(tan((y+q)/(x+p))/)*180R180 -tan!666-153>5+537#10V111>5y]+6g2Rx+pٮ٘jy+qj[Remark: 3N Notice that x+p is not same as y+q which implies*th3riangle3an U isoceles.[t'ExaF e 2. Replaye wall by a circle\r$Έ.  v3` `R&DVP #f9T 50W DvIKr  A kL'YC uxy qq/9" K2@6XvrX8R QIPi  Length GH: $0@H  IG@P6WR'@0n1CHCbib$2"Pcfi=pt wBP)QsHƇ3V`<6bH2;H1chogA6SplHt%_'BU9Y0Ip7$pH5apTrr 'x6 t+,xw`8XD) DI f @\X0 $SGGp(9T7  `'UH$RUIHXy`ld2`xPX1xf39 16%a&rX8R&`yhI.@*  CGF: bp 4x"jiHQH c&X$1 c Gx"5  @6 ` 6B &CC8 7CDCP 8@Ȏ59H!o5m:D1;"ɫ"I<"X4`=ǘv>xB  ;? @  6@>Ƙ05AXŠ4"B5@!C#hFZ6BWT3Y& 5Y78p DsB83EBr #6T 3EBH AC>FT%H!YGN0D1WG7!PP AVAX H0@ 4DI@! C" J3@ "KEH 4E@U6$h#6Ts {LY5aM898FGE> 761c0;?=132/1(MB>[!OKey: Let HB=x and DE=y, then HF=F=. So we are minimizing x+y but V6,subject to F lying on the circle: (x-r)^2+(y =r^2.R Clear_a_z Rdone[" Note. Consider]triangle CDH:[LCDW2_+HC=H).G So (DE+EC)O +(BC+BH)T and (y+rS+(x=(x+y)Ћ4simplify(solve(ZZ+2 ,y))Ry= r(x+r -r[=RR@y:ArgWdefine f(x,r)=x+yCdone- simplify(/x2+rdiffN,xP-rQ-2Px-usolve(q=0 x=r- 2 r,+[-We let x0=BH and y0=DERIPx0:ER'hJ yJ simplify(:rlx0+r - )i tan(yyWbW$@4M\(What about this?Έ N$U/ H 4RT& `2q1  `EwF  vuX  v r  A 'L' C uxy qqO~9 ! K2@6 A&5ep8H9D(G Length FG: $PH nFns&#!  Ya!1y&nQCG0V4Q6E. R `xCQ00bS !` @63y0 Q 2#(-2TB(BU U08 YpVBYIpBh=WB"XBY"WIfQVWU)RP Hx6pC)$$60)y`<!r`2<vT 0Bb0l<i2#VeD&0p9Et)   6V(9 !PSg$ yh h hE xg@@Q7u(T XrrYBy(e8EL2ueXcw% !6GX yi` Z @%6,c@6YuBA[B9" a1C1&5ep7H9C(C  : _p \g g4G0is sXA vdQU.]B "B YeHacQN7&b&CTZ3^- LA. !P\+ƒ _ .Eqdue i9 q`@  ^]aH"_<][ D*H6b`"6rWIfQVWx U)RP$Hx6p0C)$<6H)yH!r`2<vl 0Bb0i2#VeD&0p9Et) [ n6V(9 !PSg$ yh h hEp  xg@Q7u (T 0rrE0YBy(e8ET2ue`cw%` w!6G`yi`"c@ gQ\deȐ6HYT! Y 8B vi5 FfvPvg6ŠhvvjYU YRV$'hFiv5v_j\ !P WIfQ8 F51y' % :$D^kPH  YB@`6WW{l `CCmC HYT!yY 8 vi5n @ А`o0 pX ˪vYB X1@9h qRĠ`r @  lp s&Ș^jt8'ǘkuR'Rv!H!m~pW2gf@ 4!&18 &P 1WG6 w.x3_\y04zJ4d<"@{12PH |"eh}ŘI~9H D@6RSe3jY F51wYh1@! ~jHH e&@k^vrqponmuts8Vl!o_\PQb][ZXWVUTSRO-qi:[K+We are given the followingree equations:[4)1. E of6 ellipse is asDs d>6.25x^2+2.768y 0.03124 y-11.39$-2847/2.11=0[G2. Line equation DG: y=4.4863c-4.088. We def3 it as f(x,y)RMS(=y>L+LR4doneD@qFD-222.167.8thisg!(F‘Beq:=6.25x^2+2.768y 0.03124 y-11.39$-28472.11=0RM 24+(346y)125+S 781c000o-'11J00-I""H%H1211E( Impdiff(eq,x)y'= -10U+5536y-P|5q+ 11390003124x+553600y-2847R20[8'8Step 2. We define the slope of FG to babove.L~;m(x,y)= -125x/-9yΎ|done[W3. If E=(a,b) is a point on ellipse and.where7*tange,lY(P]@Sellipse) at point E. Now, the line equation FG can be written as[$y-b=m(x-a).6FG:=a,b)*R*0def\h(x,y)*-**done:r 1We use kB0 to. Thk=0.J=6.25x^2+2.768y 0.03124 y-11.39$-0.028472.11getright(ans[1]))[2Q=Step 4. Now we find G, which is the intersection between DG a, FG or f(x,y) h: R'temp1:=solve({5=0,,=0},{G })RKx= 625000000an2,+2768!b"!6+3124@b-569493614544S-2919416`819268~*~7007132++2803724qI 506385821,y /+1N4+14014264ab-5109777000-1277 7b+2328116 *62500*7132> +28037248;I 69506385821R06 G_x:=getright(temp1[1])R\\s2 +2768!b"!6+3124@49361454-29194Њ819268y2 &ݦ+1241E4+14014264ab-5109777000-1277 7b+2328116 *62500*7132> +28037248;I 69506385821R06[?>QStep 5. Now we find F, which is the intersection between FG a, DF or h(x,y)g:t#temp2:=solve({3,(},{A })Rx=  12 +553b -6147928[+b96%p+91233R-4{266a \6033943044lJ ,y= -  2786250000a 2) +12339744b  2+13926792< b-441331L-11038911911242f 5536604'b43165b-50l37RF_x:=getright(temp2[1])R 1ňƠ +ob-6148+b$6248+92Ƈ -4777266y2˥2786250000a2 +12339744b  2+13926792< b-441331L-11038911911242fa 5536604'b43165b-50l37R[Step 5. Set up the function:0%define L(a,b)=(F_x-G_x)^2+ y y R1doneq eq1:=diff(D,a)=01 28037F417248F 4+1401426F5109777G277 7b+23281*H*5625007007132a+2803724800b-569506385821+  278@-a 2) +12339744L  2+1392679wl441331L-1103891 1911242000 5536604'b43160l375#557a{x_- pb2+13926792ab-441331000-11038911911242 5536604)233943167 b-5044037(22+;$56075eg+1401426J b-5109777f*62500713 +28037248b-569506385821f-p< D+1cnj423277 7b+232814Σ>-ďD+1250000a+3124b-569493614544*6270071321280372481 5063858212-;lh63c]7928^ 5536604 -123394316044iU+_ 3 ֆ}29 +0b ba+901 -4777266*(ZB +276800000b26+3124a"-569493614544a-2919416? -5819268G*625007007132a+2803724sK 506385821y 1Ja +553-7928bU6 9289123301 -4777266  \60&943 044Ύ-Uav+3124000000ab-569493614544-2919416 819268*625007007132+ +28037248, 695063858211=0R9?eq2:=diff(L(a,b),b)!cX bu2-+1  4+1401426׊10977-1277 7b+232810ꎌ+78@339742+13926792ab-441331000-11038911911242 5536604'233943165 b-5044037( +L 278625uv2B+`744b 2Ġ+"1401426 +24834496rb-12777'*700713c +28037248- 569506385821e!e488' 5536604a-1233943160b-5044037'-  282480' 500a52>+12415b  4+1401426|109777M77 7b+2328100*62500,7132+ 69506385821+2 1LĠ ++b -6147928H+b6)]+9a01-4266_ه625000000a2 +2768!b"!6+3124@b-569493614544S-2919416`819268g*~7007132++2803724qI 506385821 1233943^ 1ՆѠ +553̠-7928 +bk6+9289t01 -4777266* ^60044ݡa-s'HbE*5625007007132a+2803724800b-569506385821+ 6(51107203 +928912330. 5536604'-94316\044lV` v00a2@+2768!b"!6+3124@4936145429194819268*)y=0R8>[Fsolve({eq1,eq2},{a,b})R6ERROR:OverflowP%[+From Maple: a = 1.784381519,b 2.1230306253[A 1S 3;eActDGXXFy H  x125F_x  EL xzIb \{Ta 2H  ^7 ab yTb X]a  aF_y  =q b ba Ib 8-Na a EL xzIb \{TayG_x  9 rb Ra X fb u4ba Аr@b va *焑aG_y  9 rb Ra  ;Ċ b 8ba 7Jb h0a paH_x%H_yPP  oPx^3LH,a,b(F_x-G_x)^2+(F_y-G_y)^2ans1 X((Fx  9 rb Ra X fb u4ba Аr@b va *焑a$Fy  9 rb Ra  ;Ċ b 8ba 7Jb h0a paeqF d  y  ayx Z}y sdx xeq1 d dF    R 9 rb Ra X fb u4ba Аr@b va *焑a  9 rb Ra v u4b T #a  X] b @x}a EL xzIb \{Ta \{T EL xzIb \{Ta 2H  ^7 ab yTb X]a  a   9 rb Ra X fb u4ba Аr@b va *焑a  EL xzIb \{Ta 2H  ^7 ab yTb X]a  a   R 9 rb Ra  ;Ċ b 8ba 7Jb h0a pa  9 rb Ra h0 8b ;Na \{T =q b ba Ib 8-Na a EL xzIb \{Ta  8-N b  %La EL xzIb \{Ta   =q b ba Ib 8-Na a EL xzIb \{Ta  9 rb Ra  ;Ċ b 8ba 7Jb h0a paeq2 \ \F    r 9 rb Ra  ;Ċ b 8ba 7Jb h0a pa  9 rb Ra  ob 8a xzI =q b ba Ib 8-Na a EL xzIb \{Ta  EL xzIb \{Ta  b a   =q b ba Ib 8-Na a EL xzIb \{Ta  9 rb Ra  ;Ċ b 8ba 7Jb h0a pa    9 rb Ra X fb u4ba Аr@b va *焑a  EL xzIb \{Ta 2H  ^7 ab yTb X]a  a  r 9 rb Ra X fb u4ba Аr@b va *焑a  9 rb Ra f !b u4a xzI EL xzIb \{Ta 2H  ^7 ab yTb X]a  a  EL xzIb \{Ta ^7 b aeq3F  xz  b 4 ba @9b 8aa h a4fH8(x,yy-(4.486*10^(-3))*x+(4.088*10^(-3))f1H$x,y(p+x)^2+(q+y)^2gH(x,yy-(-222.9*x-167.8)hH(x,yy-b-m(a,b)*(x-a)kHL;x,y6.25*x^2+2.768*y^2+0.03124*x*y-11.39*x-0.02847*y-12.11lineFGF  by   8a  rb 4 a 4   rb 4 ab   rb 4 aa  axmHox,y((-1250000*x)/(3124*x+553600*y-2847))-((3124*y)/(3124*x+553600*y-2847))+((1139000)/(3124*x+553600*y-2847))temp1 X((Fx  9 rb Ra X fb u4ba Аr@b va *焑a$Fy  9 rb Ra  ;Ċ b 8ba 7Jb h0a patemp2 HFx  EL xzIb \{Ta 2H  ^7 ab yTb X]a  a(Fy  =q b ba Ib 8-Na a EL xzIb \{Tax0DD  rry0DD  rr02000escenario_1.EAC010000000520 scenario_1.EAC ropes.ACT",/3F ~E '[\ Sh 1 ΈEaP? `F(x .H DvKr mA 'Luxy qq5@  "G@d6"P& ? 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