{VERSION 5 0 "SUN SPARC SOLARIS" "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 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Warning" 2 7 1 {CSTYLE "" -1 -1 " " 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 207 "This continues the class \+ discussion of Example 1 of section 1.2 of the text. This is the equat ion which has the function below as its right side. To remi9nd you of its role, we introduce a suggestive name." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "dydt:=(y^2-1)/(t^2+2*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%dydtG*&,&*$)%\"yG\"\"#\"\"\"F+F+!\"\"F+,&*$)%\"tGF*F +F+*&F*F+F0F+F+F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The solution used a factored form of this expression, which we note here." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "factor(dydt);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#**,&%\"yG\"\"\"F&!\"\"F&,&F%F&F&F&F&%\"tGF',&F)F &\"\"#F&F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 542 "If both variables \+ are large, this expression is positive, and the sign changes when you \+ cross the lines y=1, y=-1, t=0, or t=-2. This gives a checkerboard of alternating areas where the soltuions will be increasing or deacreasi ng. Moreover, the derivative is zero on the horizontal lines of this \+ grid, so those lines give constant solutions to the equation. The der ivative is infinite on the vertical lines, so those lines will also ha ve properties similar to graphs of solutions although they are certain ly not graphs of functions of t. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 159 "The equation was separable, so we could \+ find the solutions by integrating. The expression for y in terms of t he independent variable t and the parameter k is:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 29 "sol:=((1+k)*t+2)/((1-k)*t+2);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%$solG*&,&*&,&\"\"\"F)%\"kGF)F)%\"tGF)F)\"\"#F) F),&*&,&F)F)F*!\"\"F)F+F)F)F,F)F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "In order to get a variety of graphs from this expression, we load \+ the plots library." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with( plots):" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Here are the graphs of a few solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot(\{seq(sol,k=[-1,0,1/2,1,2])\},t=-4..2,y=-3..3);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6)-%'CURVESG6$7S7$$!\"% \"\"!\"\"\"F*7$$!3*)*****\\2<#pQ!#iUC#F0F+7$$!3!)***\\7YY08#F0F+7$$!3)******\\ XF*>F0F+7$$!3(*******>#z2)=F0F+7$$!3/++D\"RKvu\"F0F+7$$!3<+++qjeH;F0F +7$$!3()***\\7*3=+:F0F+7$$!3%)***\\PFcpP\"F0F+7$$!3\")****\\7VQ[7F0F+7 $$!3!)***\\i6:.8\"F0F+7$$!31++]P:'H+\"F0F+7$$!3[++]7'pnq)!#=F+7$$!3'3+ +v[G_b(FcpF+7$$!3s)****\\_K:J'FcpF+7$$!35-+++HnE]FcpF+7$$!3y,++D%)opPF cpF+7$$!3F++]78\\DFcpF+7$$!3)3++]x6J?\"FcpF+7$$\"3_y&******Hk-\"!#?F+ 7$$\"30,++]A!eI\"FcpF+7$$\"3E(***\\(=_(zCFcpF+7$$\"3V.++]&*=jPFcpF+7$$ \"3\"f***\\(3/3(\\FcpF+7$$\"3y++]P#4JB'FcpF+7$$\"3W(*****\\KCnuFcpF+7\"3A(***\\(=n#f()FcpF+7$$\"3P+++!)RO+5F0F+7$$\"30++]_!>w7\"F0F+7$$\"3 N++v)Q?QD\"F0F+7$$\"3G+++5jyp8F0F+7$$\"3<++]Ujp-:F0F+7$$\"3++++gEd@;F0 F+7$$\"39++v3'>[!3:++Dr^b^AF07FA!3#****\\7Sw%G@F07FD!3*****\\7;)=,?F0 7FG!3!)***\\i83V(=F07FJ!39+++NkzViUC\"F07FY!3!)*** \\7YY08\"F07Ffn!3s******\\XF**Fcp7Fin!3u*******>#z2))Fcp7F\\o!3 R++]7RKvuFcp7F_o!3s,+++P'eH'Fcp7Fbo!3q)***\\7*3=+&Fcp7Feo!3[)*** \\PFcpPFcp7Fho!3:)****\\7VQ[#Fcp7F[p!32)***\\i6:.8Fcp7F^p!3Wb+++ vhHFiq7Fap\"3]****\\(QIKH\"Fcp7Fep\"38****\\7:xWCFcp7Fhp\"3E,++ vuY)oFcp7F[q\"3*y******4FL(\\Fcp7F^q\"3A)****\\d6.B'Fcp7Faq\"3s ****\\(o3lW(Fcp7Fdq\"35*****\\A))oz)Fcp7Fgq\"3e******Hk-,5F07F[r \"35+++D-eI6F07F^r\"3t***\\(=_(zC\"F07Far\"3M+++b*=jP\"F07Fdr\"3 g***\\(3/3(\\\"F07Fgr\"33++vB4JB;F07Fjr\"3u*****\\KCnu\"F07F]s\" 3r***\\(=n#f(=F07Fs\"3P+++!)RO+?F07Fcs\"30++]_!>w7#F07Ffs\"3N++ v)Q?QD#F07Fis\"3G+++5jypBF07F\\t\"3<++]Ujp-DF07F_t\"3++++gEd@EF0 7Fbt\"39++v3'>[FF07Fet\"37++D6EjpGF07Fht\"\"F*-F[u6&F]uFauF^uF au-F67dp7$$!3)******R.8f*RF0$!3Qq/\"*oAWa>!#97$$!3')*****p1E=*RF0!3 ^wN=Z,@d(*!#:7$$!3%)*****45Rx)RF0$!3]!)4(3.2[\\'Fh_l7$$!3=+++M@l)RF0 !3=#>ft]gj[Fh_l7$$!32+++n^czRF0$!3TFS3jR)[)QFh_l7$$!3/+++,#ya(RF0!3! 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The picture suggests that we examine the behavior of these solutions near t=0 and t=-2. Lets just ask Maple." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "subs(t=0,sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs(t=-2,sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 563 " Thus, all solutions pass through a pair of common points. This is a l ittle misleading since we have been working with an algebraic form of \+ the solution instead of an analytic form. Three solutions are lines, \+ but all other solutions are hyperbolas, and the two components of a hy perbola are connected by algebra, but not by analysis. The analytic a pproach that we will emphasize in this course only show the arc of a s olution moving to the edge of the graphing window. There is no obviou s connection to any arc that crosses another part of the graphing wind ow." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 139 "W e will continue this example using an approach that does not require a formula for the solution, but first notice that we can solve for k." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "solve(y=sol,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,**&%\"yG\"\"\"%\"tGF'F'*&\"\"#F'F&F'F'F(! \"\"F*F+F'F(F+,&F&F'F'F'F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 657 "Th is shows that each point of the plane, except for those on the lines t =0 or y=-1, determines a unique value of k. This means that the curve s with different values of k are disjoint away from these lines. A mo re careful examination of the formula shows that y=-1 is omitted becau se it requires an infinite value of k. The formula also accidentally \+ associates the line t=-2 with the value k=0. In short, the derivation of the formula for the solution made some assumptions that lead to st range results when taken literally. We will take an analytic approach wherever possible that does not require that such results be interpre ted. ...To be continued." }}}}{MARK "15 0 0" 657 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }