{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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 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 1 0 0 0 0 0 0 0 1 }{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 "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Outpu t" -1 11 1 {CSTYLE "" -1 -1 "Times" 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 }} {SECT 0 {SECT 0 {PARA 3 "" 0 "" {TEXT -1 34 "First Order Differential \+ Equations" }}{PARA 0 "" 0 "" {TEXT -1 310 "In this section we will inv estigate Maple's ability to symbolically solve differential equations. This entails using the \"dsovle\" command. Next we will plot some in tegral curves for various values of the arbitary constant. Finally we will use this to solve and analyze solutions to initial value proble ms. " }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 20 "The \"dsolve\" command" }}{PARA 0 "" 0 "" {TEXT -1 400 "Recall that initial value problems com e in the general form: dy/dx = F(x,y), y(xo) = yo. Here, the point \+ (xo,yo) describes the \"initial value\". We will first focus on solvi ng the differential equation without an initial condition. The comma nd \"dsolve\" is used to symbolically solve differential equations. F or a differential equation involving the terms y(x) and x, this comma nd has the form:" }}{PARA 0 "" 0 "" {TEXT -1 41 "dsolve( the different ial equation, y(x));" }}{PARA 0 "" 0 "" {TEXT -1 104 "For example, the solution to the differential equation: dy/dx + x y = 2x is solved using the command" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "dsolve ( diff(y(x),x) + x*y(x) = 2*x,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/-%\"yG6#%\"xG,&\"\"#\"\"\"*&-%$expG6#,$*$)F'F)F*#!\"\"F)F*%$_C1GF*F *" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 7 "Notice:" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 59 "An \"= \" sign is used in defining the differential equation. " }}{PARA 0 "" 0 "" {TEXT -1 20 "y is denoted by y(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 "y' (or dy/dx) is denoted by diff(y(x),x). " }}{PARA 0 "" 0 "" {TEXT -1 95 "The \", y(x)\" at the end of the command tells Maple you \+ want it to solve for the function y(x). " }}{PARA 0 "" 0 "" {TEXT -1 58 "The \"_C1\" in the output represents an arbitrary constant. " }}}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 24 "Plotting integral curves" }} {PARA 0 "" 0 "" {TEXT -1 104 "We will now plot a few integral curves f or the differential equation above. First define the solution by" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sol := (x,C1) -> 2 + C1* exp (-1/2*x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solGf*6$%\"xG%#C1G6 \"6$%)operatorG%&arrowGF),&\"\"#\"\"\"*&9%F/-%$expG6#,$*$)9$F.F/#!\"\" F.F/F/F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 199 "This defines a f unction of two variables: x and C1. We can plot these integral curve s by plotting the solution for various values of C1. The example belo w plots the solution for C1 = -1,0, and 1. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot([sol(x,-1),sol(x,0),sol(x,1)],x=0..2,color=[r ed,green,blue]);" }}{PARA 13 "" 1 "" {GLPLOT2D 317 317 317 {PLOTDATA 2 "6'-%'CURVESG6$7S7$$\"\"!F)$\"\"\"F)7$$\"39LLLL3VfV!#>$\"32/A`!y\\4+ \"!#<7$$\"3'pmm;H[D:)F/$\"3;twX'o8/\"F27$$\"3,LL$ezs$HLF;$\"3y<=X8d\"R0\"F27$$\"3_****\\7iI_PF;$ \"3?&)3$*R\"yz1\"F27$$\"3#pmmm@Xt=%F;$\"3u$G24BOR3\"F27$$\"3QLLL3y_qXF ;$\"3SFsg@*y\"*4\"F27$$\"3i******\\1!>+&F;$\"3C))eDvpe<6F27$$\"3()**** **\\Z/NaF;$\"3D;m>s)3t8\"F27$$\"3'*******\\$fC&eF;$\"3'GIY,]%Rd6F27$$ \"3ELL$ez6:B'F;$\"3gh(Gozsk<\"F27$$\"3Smmm;=C#o'F;$\"3&=UVhd%4+7F27$$ \"3-mmmm#pS1(F;$\"3a![ld!R\"3A\"F27$$\"3]****\\i`A3vF;$\"35G+!=9F27$$\"3!****\\PQ#\\\"3\"F2$\"3(HD ts))*yU9F27$$\"3BLL$e\"*[H7\"F2$\"3c$p8eAywY\"F27$$\"3#*******pvxl6F2$ \"3Y)**p'*4RJ\\\"F27$$\"3z****\\_qn27F2$\"3))>LT^bs<:F27$$\"3%)***\\i& p@[7F2$\"3S&3'[3e9T:F27$$\"3#)****\\2'HKH\"F2$\"3_Yx>L^lm:F27$$\"3_mmm wanL8F2$\"3%zSXlBv!*e\"F27$$\"3'******\\2goP\"F2$\"3n^=n]]V7;F27$$\"3C LLeR<*fT\"F2$\"3*[b%Q\"4XIj\"F27$$\"3'******\\)Hxe9F2$\"3auO(yXI\\l\"F 27$$\"3Ymm\"H!o-*\\\"F2$\"3i4)pUTt[n\"F27$$\"3))***\\7k.6a\"F2$\"3_))* GtH<]p\"F27$$\"3emmmT9C#e\"F2$\"3X^B:HW*Rr\"F27$$\"3\"****\\i!*3`i\"F2 $\"3VH!)e'>#3LF2$\"3DU!fJJ.0%=F27$$\"3/++v .Uac>F2$\"3QoA_hb^_=F27$$\"\"#F)$\"3=(QjnrkY'=F2-%'COLOURG6&%$RGBG$\"* ++++\"!\")F(F(-F$6$7S7$F(Fez7$F-Fez7$F4Fez7$F9Fez7$F?Fez7$FDFez7$FIFez 7$FNFez7$FSFez7$FXFez7$FgnFez7$F\\oFez7$FaoFez7$FfoFez7$F[pFez7$F`pFez 7$FepFez7$FjpFez7$F_qFez7$FdqFez7$FiqFez7$F^rFez7$FcrFez7$FhrFez7$F]sF ez7$FbsFez7$FgsFez7$F\\tFez7$FatFez7$FftFez7$F[uFez7$F`uFez7$FeuFez7$F juFez7$F_vFez7$FdvFez7$FivFez7$F^wFez7$FcwFez7$FhwFez7$F]xFez7$FbxFez7 $FgxFez7$F\\yFez7$FayFez7$FfyFez7$F[zFez7$F`zFez7$FezFez-Fjz6&F\\[lF(F ][lF(-F$6$7S7$F($\"\"$F)7$F-$\"3#fzn%>-0**HF27$F4$\"31FBa8Bo'*HF27$F9$ \"3!4?Ic$*=B*HF27$F?$\"3%**Ryf%>8')HF27$FD$\"3%H8b#[,AyHF27$FI$\"3ILRn A(e$pHF27$FN$\"3=@Dg30oeHF27$FS$\"3W#=[lG%3YHF27$FX$\"3![6p+'=-KHF27$F gn$\"3.;F4pP1;HF27$F\\o$\"3\"Gx#Ry5#3!HF27$Fao$\"3w6TuCIT#)GF27$Ffo$\" 3u$Q.y7\"piGF27$F[p$\"3#pp`)*\\0E%GF27$F`p$\"3SQ7<.s_BGF27$Fep$\"39yl& QU0**z#F27$Fjp$\"3Y>XB%4'=zFF27$F_q$\"3!>(**>GQPaFF27$Fdq$\"3[P=8-_'=t #F27$Fiq$\"34mN#[y$o1FF27$F^r$\"3IAVz=)HBo#F27$Fcr$\"3')pW8F7jcEF27$Fh r$\"3!=M$)yhgGj#F27$F]s$\"3U)H\\RTHrg#F27$Fbs$\"3cl8&pa+/e#F27$Fgs$\"3 /Zns7,@dDF27$F\\t$\"3m1j=ua)eCF27$F`u$\"3[`A!o'[MLCF27$Feu$\"31#faMwC4T#F27 $Fju$\"3K[\"G$\\\\c(Q#F27$F_v$\"3)[W:'3\\&pO#F27$Fdv$\"3oDj7U&p]M#F27$ Fiv$\"3g!>IdeE^K#F27$F^w$\"3Y65n-F)\\I#F27$Fcw$\"3c[w%3d0gG#F27$Fhw$\" 3eq>T.y\"pE#F27$F]x$\"3?*z_-z,$\\AF27$Fbx$\"39nB;]`2KAF27$Fgx$\"3:fxdo Ny:AF27$F\\y$\"3I*eOYu3:?#F27$Fay$\"3Q')oR6P'f=#F27$Ffy$\"3#o.,#\\7zs@ F27$F[z$\"3ud4%oo'\\f@F27$F`z$\"3'=tx%QW[Z@F27$Fez$\"3\"GhOKGN`8#F2-Fj z6&F\\[lF(F(F][l-%+AXESLABELSG6$Q\"x6\"Q!Fbhl-%%VIEWG6$;F(Fez%(DEFAULT G" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" " Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 219 "Notice, th e value prescribed to the constant C1 greatly affects the shape of the curve. However, it appears that all solutions tend to y = 2 regardle ss of the value of C1. This can be seen by taking the limit as x -> \+ " }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 16 " of the sol ution" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "limit(sol(x,C1), x =infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 245 "Recall that the constant C1 would be defined o nce an initial value is imposed. The fact that the solution tends to \+ y= 2 regardless of this value is very helpful in determining the end \+ behavior of the solution when the initial value is unknown." }}}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 46 "I nitial value problem, Newton's law of cooling" }}{PARA 0 "" 0 "" {TEXT -1 505 "In this section we will now incorporate an initial value into our differential equation and analyze the solution to an initial value problem for the cooling of a hot cup of coffee left to sit at r oom temperature. Denote the ambient room temperature as Ta and the in itial temperature of the coffee to be To, ie. T(0) = To. Newton's law of cooling states that the rate of temperature change of the coffee i s proportional to the difference between the temperature of the coffee and the ambient temperature:" }}{PARA 0 "" 0 "" {TEXT -1 30 "dT/dt = \+ k(T - Ta), T(0) = To." }}{PARA 0 "" 0 "" {TEXT -1 206 "Here T or T(t) , is the temperature of the coffee at time t, and k is the constant o f proportionality. For now we do not assign a value to Ta,To, or k, \+ but solve the initial value problem symbolically by " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "dsolve(\{diff(T(t),t)=k*(T(t) - Ta),T(0)=To\},T(t)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"TG6#%\"tG,&%#TaG\"\"\"*&-%$ex pG6#*&%\"kGF*F'F*F*,&F)!\"\"%#ToGF*F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 228 "Notice: This is the same command for solving an arbitrar y differential equation but you surround the differential equation and initial condition with curly brackets. Observing the above solution y ields that T(0) = To as desired." }}{PARA 0 "" 0 "" {TEXT -1 341 "We a ssume that we know the values of To and Ta, but this does not give us \+ the value of k. Suppose we take the temperature of the coffee after \+ t1 minutes and denote that temperature as T1, ie T(t1) = T1. It is eas y enough to find the constant k from this data but we might as well ha ve Maple do it. First, let's call the solution sol(t): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol := (t,Ta,To,k) -> Ta + (To - Ta )*exp(k*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solGf*6&%\"tG%#TaG%# ToG%\"kG6\"6$%)operatorG%&arrowGF+,&9%\"\"\"*&,&9&F1F0!\"\"F1-%$expG6# *&9'F19$F1F1F1F+F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "We set th is expression equal to T1 for t=t1 and solve for k. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "solve(sol(t1,Ta,To,k) = T1, k); " } }{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%#lnG6#*&,&%#TaG\"\"\"%#T1G!\"\"F* ,&F)F*%#ToGF,F,F*%#t1GF," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 343 "Assu ming the temperature of the coffee is tending to the ambient temperatu re, you should be able to convince yourself that the above expression for k is negative. Furthermore, since k<0 it is apparent that T(t) \+ -> Ta as t -> infinity. This makes sense right? As time increases t he temperature of the coffee approaches room temperature. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 357 "Variab les such as To,Ta, and k are known as problem parameters, where k may need to calculated from available data. These parameters determine th e behavior of the solution but are often considered constant for any s ingle initial value problem. At this point we are free to vary the pa rameters and see the affect this has on the behavior of the solution. \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 242 "Let 's see what varying the initial temperature does to the solution holdi ng the other parameters constant. We will the assume the ambient temp erature is 70 degrees and the constant of proportionality for coffee \+ is determined to be -0.145. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "k := -.145;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG$!$X\"!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Ta := 70;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#TaG\"# q" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Let's plot some integral cu rves for various initial temperatures = 40,100, and 212 degrees farenh iet:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "plot([sol(t,Ta,40,k ),sol(t,Ta,100,k),sol(t,Ta,212,k),70],t = 0..40,color=[red,green,blue, black]);" }}{PARA 13 "" 1 "" {GLPLOT2D 316 316 316 {PLOTDATA 2 "6(-%'C URVESG6$7U7$$\"\"!F)$\"#SF)7$$\"39LLLL3VfV!#=$\"3M$499(fw$=%!#;7$$\"3H mmmm;')=()F/$\"3]J\\u)HviN%F27$$\"3!*****\\7z>^7!#<$\"3cY^IVVv(\\%F27$ $\"3RLLLe'40j\"F;$\"3NG__I@mJYF27$$\"3ommm;6m$[#F;$\"3_@+l,\\C2\\F27$$ \"3Dnmm;yYULF;$\"3+*Q\\&)GwA:&F27$$\"3ELLLeF>(>%F;$\"3sY(\\G@_wO&F27$$ \"3%omm;>K'*)\\F;$\"3/]NF[.%[a&F27$$\"3g*****\\Kd,\"eF;$\"3)o(**G!4n!3 dF27$$\"3Onmm\"fX(emF;$\"3]fT#>UUw&eF27$$\"3O+++DCh/vF;$\"3YQ/T*y,&*)f F27$$\"3uMLLL/pu$)F;$\"3=n%e1Zt#4hF27$$\"3mnmm;c0T\"*F;$\"3W#R#\\=B&H? 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Fi_m7$FDFi_m7$FIFi_m7$FNFi_m7$FSFi_m7$FXFi_m7$FgnFi_m7$F\\oFi_m7$FaoFi _m7$FfoFi_m7$F[pFi_m7$F`pFi_m7$FepFi_m7$FjpFi_m7$F_qFi_m7$FdqFi_m7$Fiq Fi_m7$F^rFi_m7$FcrFi_m7$FhrFi_m7$F]sFi_m7$FbsFi_m7$FgsFi_m7$F\\tFi_m7$ FatFi_m7$FftFi_m7$F[uFi_m7$F`uFi_m7$FeuFi_m7$FjuFi_m7$F_vFi_m7$FdvFi_m 7$FivFi_m7$F^wFi_m7$FcwFi_m7$FhwFi_m7$F]xFi_m7$FbxFi_m7$FgxFi_m7$F\\yF i_m7$FayFi_m7$FfyFi_m7$F[zFi_m7$F`zFi_m7$FezFi_m7$FjzFi_m7$F*Fi_m-Fb[l 6&Fd[lF)F)F)-%+AXESLABELSG6$Q\"t6\"Q!Facm-%%VIEWG6$;F(F*%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "Notice that in all cases the temperature of the coffee tends toward the ambi ent temperature. " }}}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 10 "Assignmen t" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Problem #1" }}{PARA 0 "" 0 "" {TEXT -1 472 "I was once told by a f riend in Vermont that if you set two pots of hot water out on a cold d ay, it is quite possible that the one which starts out warmer will fre eze sooner. Set up an initial value problem describing this scenerio, plot some solutions with varying initial conditions and convince me t hat his hypothesis is bunk. You may assume the constant of proportion ality is the same as in the coffee example above and the ambient tempe rature is below freezing. " }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Problem #2" }}{PARA 0 "" 0 "" {TEXT -1 556 "On December 24th, Carl os Greene was found in the alley behind Charlie's pub just behind the \+ dumpster. He was dead, strangled by a string of Christmas lights (the \+ kind that blink). The only nice thing about this Christmas was the he at wave that had hit Missoula that week. It was a constant 75 degrees \+ out that evening. When Chief Wigum came to the site, they took Carlos' temperature, it was a cool 85 degrees, suggesting the murder had happ ened fairly recently. Twenty minutes later, they took his temperatur e again and the themometer read 80 degrees. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 294 "Thirty minutes before the init ial temperature was taken, witnesses saw Clint Bronson enter Charlie's pub and stay there, without leaving, for several hours. Clint was \+ the main suspect in the murder. Clint's lawyers claim that he was in \+ the pub when Carlos met that fatal string of lights. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 411 "Your assignment is \+ to determine whether this is a valid argument. Basically you must det ermine the relative time of death by setting up an initial value probl em from Newton's law of cooling. In this case it's Carlos' body that i s cooling. Once you have the solution you must set the body temperatu re equal to 98.6 and solve for time. This will give you enough data to support or deny the lawyers claim. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 226 " You don't really need Maple t o solve this problem but it would be helpful and I want to see the gra ph of Carlos' body temperature justifying the relative time of death a rgument. Good luck, Chief Wigum is counting on you. " }}}}}{MARK " 0 4 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }