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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 "Normal " -1 256 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 "Normal" -1 257 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 }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 49 "Applications of 1st order differe ntial equations " }}{PARA 0 "" 0 "" {TEXT -1 37 "by Peter Stone, Nanai mo, B.C., Canada" }}{PARA 0 "" 0 "" {TEXT -1 19 "Version: 27.3.2007" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "load " }{TEXT 0 7 "desolve" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 17 "The Maple m-file " } {TEXT 403 7 "DEsol.m" }{TEXT -1 32 " is required by this worksheet. 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esgI)Fh]l7$$!+()z)4/\"Fdo$!+F<$z%yFh]l7$$!+3$4z3\"Fdo$!+YqevuFh]l7$$!+ uEHT6Fdo$!+UF)Q?(Fh]l7$$!+d)4!*>\"Fdo$!+53lVqFh]l7$$!+d)f(e7Fdo$!+\">z 7+(Fh]l7$$!+G1;=8Fdo$!+2rXyqFh]l7$$!+^S%[P\"Fdo$!+?x5ssFh]l7$$!+N.bE9F do$!+*y5Xd(Fh]l7$$!+:\"=7Z\"Fdo$!+e/htzFh]l7$$!+Em12:Fdo$!+&)e\\`%)Fh] l7$$!+-nmK:Fdo$!+\\b.&**)Fh]l7$$!+\\x*pa\"Fdo$!+w*Rmd*Fh]l7$$!+L%)[\\: Fdo$!+VA^<5Fdo7$$!+e%R+a\"Fdo$!+JBmw5Fdo7$$!+Dv->:Fdo$!+LhvK6Fdo7$$!+9 .H([\"Fdo$!+ntb$=\"Fdo7$$!+'3$4Y9Fdo$!+\\2/F7Fdo7$$!+Y#yqR\"Fdo$!+KFZh 7Fdo7$$!+h)*>U8Fdo$!+A1[&G\"Fdo7$$!+edk$G\"Fdo$!+,t5)H\"Fdo7$$!+0.vB7F do$!+$Q\\))H\"Fdo7$$!+W8!\\;\"Fdo$!+#GxwG\"Fdo7$$!+**\\W46Fdo$!+(RO]E \"Fdo7$$!+P@ff5Fdo$!+Y$H=B\"Fdo7$$!+O-L<5Fdo$!+\"**z$*=\"Fdo7$$!+\\9WV )*Fh]l$!+a1QR6Fdo7$$!+S\"*[>'*Fh]l$!+\\Y#Q3\"Fdo7$$!+4PP5&*Fh]l$!+@o# \\-\"Fdo7$$!+W_W?&*Fh]l$!+&Q_.l*Fh]l-%*AXESSTYLEG6#%%NONEG" 1 2 0 1 10 0 2 9 1 1 2 1.000000 46.000000 44.000000 0 0 "Curve 1" "Curve 2" "C urve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" }}}{PARA 0 "" 0 "" {TEXT -1 89 "Consider a series circuit containing a resistor, a coil ( inductor), and a voltage source." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 30 "The voltage drop (measured in " }{TEXT 258 5 "volts" }{TEXT -1 11 ") across a " }{TEXT 259 8 "resistor" } {TEXT -1 45 " is proportional to the current (measured in " }{TEXT 258 7 "amperes" }{TEXT -1 4 " or " }{TEXT 258 4 "amps" }{TEXT -1 23 ") through the resistor:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "V[res]=I[res]*R" "6#/&%\"VG6#%$resG*&&%\"IG6#F'\"\"\"%\"RGF," } {TEXT -1 13 " ------- (i)." }}{PARA 0 "" 0 "" {TEXT -1 13 "The constan t " }{TEXT 293 1 "R" }{TEXT -1 15 " is called the " }{TEXT 259 10 "res istance" }{TEXT -1 40 " of the resistor, and it is measured in " } {TEXT 258 4 "ohms" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 96 "The voltage drop across a coil is proport ional to the rate of change of the current in the coil:" }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "V[coil]=L" "6#/&%\"VG6#%%coilG% \"LG" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI[coil]/dt;" "6#*&&%#dIG6#%%coi lG\"\"\"%#dtG!\"\"" }{TEXT -1 14 " ------- (ii)." }}{PARA 0 "" 0 "" {TEXT -1 14 "The constants " }{TEXT 294 1 "L" }{TEXT -1 15 " is called the " }{TEXT 259 10 "inductance" }{TEXT -1 36 " of the coil, and it i s measured in " }{TEXT 258 6 "henrys" }{TEXT -1 68 ", and the equation (ii) holds when the time is measured in seconds. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "According to " }{TEXT 259 15 "Kirchoff's laws" }{TEXT -1 103 ", the current through the resi stor in this circuit is equal to the current in the coil, so we can wr ite" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{TEXT 290 1 "I" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "I[coil] = I[res]" "6#/&%\"IG6#%%coilG&F%6#%$resG " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 119 "Also the sum of the \+ voltage drops across the resistor and coil is equal to the voltage gen erated by the voltage source:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "V[coil]+V[res] = V(t);" "6#/,&&%\"VG6#%%coilG\"\"\"&F&6 #%$resGF)-F&6#%\"tG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 37 "Th is gives the differential equation:" }}{PARA 256 "" 0 "" {TEXT -1 2 " \+ " }{TEXT 295 1 "L" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt + R*I=V(t)" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&%\"RGF'%\"IGF'F'-%\"VG6#%\"tG" } {TEXT -1 16 " ------- (iii)." }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {TEXT 291 9 "_________" }{TEXT -1 16 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "This equation can be written in the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+``(R/L)*I=V(t)/L" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&-%!G6#* &%\"RGF'%\"LGF)F'%\"IGF'F'*&-%\"VG6#%\"tGF'F0F)" }{TEXT -1 16 " ----- -- (iv), " }}{PARA 0 "" 0 "" {TEXT -1 22 "and it is therefore a " } {TEXT 259 40 "first order linear differential equation" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 85 "We can attempt to solve this differe ntial equation by the integrating factor method. " }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 259 18 "integrating factor" }{TEXT -1 4 " is \+ " }{XPPEDIT 18 0 "exp(Int(``(R/L),t))=exp(R*t/L)" "6#/-%$expG6#-%$IntG 6$-%!G6#*&%\"RG\"\"\"%\"LG!\"\"%\"tG-F%6#*(F.F/F2F/F0F1" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "Mul tiplying equation (iv) by this integrating factor gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(R*t/L)" "6#-%$expG6#*(%\"RG \"\"\"%\"tGF(%\"LG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+ ``(R/L )*exp(R*t/L)*I=1/L" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*(-%!G6#*&%\"RGF'% \"LGF)F'-%$expG6#*(F/F'%\"tGF'F0F)F'%\"IGF'F'*&F'F'F0F)" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(R*t/L)*V(t)" "6#*&-%$expG6#*(%\"RG\"\"\"%\"tGF) %\"LG!\"\"F)-%\"VG6#F*F)" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "d/dt " "6#*&%\"dG\"\"\"%#dtG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "[exp(R*t /L)*I]=1/L" "6#/7#*&-%$expG6#*(%\"RG\"\"\"%\"tGF+%\"LG!\"\"F+%\"IGF+*& F+F+F-F." }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(R*t/L)*V(t)" "6#*&-%$exp G6#*(%\"RG\"\"\"%\"tGF)%\"LG!\"\"F)-%\"VG6#F*F)" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "exp(R*t/L)*I = 1/L" "6#/*&-%$expG6#*(%\"RG\"\"\"%\"t GF*%\"LG!\"\"F*%\"IGF**&F*F*F,F-" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int( exp(R*t/L)*V(t),t)" "6#-%$IntG6$*&-%$expG6#*(%\"RG\"\"\"%\"tGF,%\"LG! \"\"F,-%\"VG6#F-F,F-" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 69 " To proceed any further we need to know more about the voltage source \+ " }{XPPEDIT 18 0 "V(t)" "6#-%\"VG6#%\"tG" }{TEXT -1 2 ". " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 13 "Suppose that \+ " }{XPPEDIT 18 0 "L = 2" "6#/%\"LG\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "R = 4" "6#/%\"RG\"\"%" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "V(t) = PIECEWISE([0, t < 0],[4, 0 <= t]);" "6#/-%\"VG6#%\"tG-%*PIECEWISEG6 $7$\"\"!2F'F,7$\"\"%1F,F'" }{TEXT -1 21 " in equation (iii)." }} {PARA 0 "" 0 "" {TEXT -1 132 "Such a source voltage could be obtained \+ by closing a switch at time to connect the circuit to a constant volta ge source of 4 volts. " }}{PARA 0 "" 0 "" {TEXT -1 26 "Suppose also th at at time " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 53 ", ther e is no current flowing in the circuit so that " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\" \"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "V := t -> piecewise(t<0,0,4):\n'V(t)'=V( t);\nplot(V(t),t=-2..4,thickness=2,labels=[`t`,`voltage V(t)`]);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"VG6#%\"tG-%*PIECEWISEG6$7$\"\"!2F 'F,7$\"\"%%*otherwiseG" }}{PARA 13 "" 1 "" {GLPLOT2D 542 194 194 {PLOTDATA 2 "6&-%'CURVESG6$7hn7$$!\"#\"\"!$F*F*7$$!3!******\\2<#p=!#iUCFH F+7$$!3k***\\7GVS(=FHF+7$$!3B++]7YY08FHF+7$$!3/,+vo4F+7$$!3w*** *\\7ep$H'F[oF+7$$!38**\\PM#eMg%F[oF+7$$!3\\)**\\il?K\"HF[oF+7$$!3=)\\( =n=5o?F[oF+7$$!3&y*\\7yI)HA\"F[oF+7$$!3(oZPf$oB/!)!#?F+7$$!3Gv\\i!*Gky PF^pF+7$$\"3Gj_(oa5&pW!#@$\"\"%F*7$$\"3%z-+++XDn%F^pFfp7$$\"3s-++]im%> 'F[oFfp7$$\"3C++++y?#>\"FHFfp7$$\"3h****\\(3wY_#FHFfp7$$\"3F)******HOT q$FHFfp7$$\"3I,+](3\">)*\\FHFfp7$$\"3_,+]isVIiFHFfp7$$\"3&=++](o:;vFHF fp7$$\"3#>++v$)[op)FHFfp7$$\"3W*****\\i%Qq**FHFfp7$$\"3&****\\(QIKH6F/ Ffp7$$\"3#****\\7:xWC\"F/Ffp7$$\"37++]Zn%)o8F/Ffp7$$\"3y******4FL(\\\" F/Ffp7$$\"3#)****\\d6.B;F/Ffp7$$\"3(****\\(o3lWw7$F/Ffp7$$\"3O++v)Q?QD$F/Ffp7$$\"3G+++5jypLF/Ffp7$$\"3<+ +]Ujp-NF/Ffp7$$\"3++++gEd@OF/Ffp7$$\"39++v3'>$[PF/Ffp7$$\"37++D6EjpQF/ Ffp7$FfpFfp-%'COLOURG6&%$RGBG$\"#5!\"\"F+F+-%*THICKNESSG6#\"\"#-%+AXES LABELSG6$%\"tG%-voltage~V(t)G-%%VIEWG6$;F(Ffp%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "t>=0" "6#1\"\"!%\"t G" }{TEXT -1 13 " the current " }{TEXT 305 1 "I" }{TEXT -1 38 " satis fies the differential equation:" }}{PARA 256 "" 0 "" {TEXT -1 3 " 2 " }{XPPEDIT 18 0 "dI/dt+4*I = 4;" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"% F'%\"IGF'F'F+" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+2*I = 2;" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"#F'%\"IGF'F'F+" }{TEXT -1 1 "." } }{PARA 0 "" 0 "" {TEXT -1 26 "The integrating factor is " }{XPPEDIT 18 0 "exp(Int(2,t)) = exp(2*t);" "6#/-%$expG6#-%$IntG6$\"\"#%\"tG-F%6# *&F*\"\"\"F+F/" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 85 "Multiplying both sides of the differentia l equation by this integrating factor gives:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t);" "6#-%$expG6#*&\"\"#\"\"\"%\" tGF(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+2*exp(2*t)*I = 2*exp(2*t); " "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*(\"\"#F'-%$expG6#*&F+F'%\"tGF'F'%\"I GF'F'*&F+F'-F-6#*&F+F'F0F'F'" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 48 "The left side is the der ivative with respect to " }{TEXT 306 1 "t" }{TEXT -1 4 " of " } {XPPEDIT 18 0 "exp(2*t)*I" "6#*&-%$expG6#*&\"\"#\"\"\"%\"tGF)F)%\"IGF) " }{TEXT -1 13 ", so we have " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "exp(2*t)*I =Int(2*exp(2*t),t)" "6#/*&-%$expG6#*&\"\"#\" \"\"%\"tGF*F*%\"IGF*-%$IntG6$*&F)F*-F&6#*&F)F*F+F*F*F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I=exp(2*t)+c" "6#/*&-%$expG6#*&\"\"# \"\"\"%\"tGF*F*%\"IGF*,&-F&6#*&F)F*F+F*F*%\"cGF*" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\" !" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 18 ", it follows that " }{XPPEDIT 18 0 "0=1+c" "6#/\"\"!,&\"\"\"F&% \"cGF&" }{TEXT -1 10 ", so that " }{XPPEDIT 18 0 "c=-1" "6#/%\"cG,$\" \"\"!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 5 "Thus " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I = exp(2*t) -1" "6#/*&-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%\"IGF*,&-F&6#*&F)F*F+F*F*F*! \"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I=1-exp(-2*t)" "6#/% \"IG,&\"\"\"F&-%$expG6#,$*&\"\"#F&%\"tGF&!\"\"F." }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "This mean s that the current increases towards1 amp." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 119 "plot([1-exp(-2*t ),1],t=0..2.5,color=[COLOR(RGB,.5,0,1),black],\n linestyle=[ 1,3],labels=[`t`,`current I(t)`]);" }}{PARA 13 "" 1 "" {GLPLOT2D 501 270 270 {PLOTDATA 2 "6&-%'CURVESG6%7U7$$\"\"!F)F(7$$\"3ALL$3FWYs#!#>$ \"3&)G%)H!RvMI&F-7$$\"3WmmmT&)G\\aF-$\"3q#=WoAoD.\"!#=7$$\"3O**\\7`p)* >yF-$\"3gaH65F5$\"3kj&y'**p&Q%=F57$$\"3omm\"z> )G_:F5$\"3q&3DM@'))oEF57$$\"3-nmT&QU!*3#F5$\"3\"Q*[X!o;^T$F57$$\"3HL$e RZXKi#F5$\"33R7]$>qB3%F57$$\"3xm;z>,_=JF5$\"3Ku;*zNW/k%F57$$\"3v**\\7G $[8j$F5$\"3yGW)3r**G;&F57$$\"35n;z%*frhTF5$\"3#zmVwn7(\\cF57$$\"3A+]il FQ!p%F5$\"3Ol.'p*3?'3'F57$$\"3@ML$3_\"=M_F5$\"3Q$HL=,M&*['F57$$\"3HnmT g(fJr&F5$\"3&4*)\\x>;-\"oF57$$\"3k++]7eP_iF5$\"3!QHq828j8(F57$$\"3Q++] Pf!Qz'F5$\"3r2HV92@IuF57$$\"3@++](=ubJ(F5$\"3?je$4yf[o(F57$$\"37n;zW(* Q*y(F5$\"3'pnO'e`<%*yF57$$\"3#QLL3F-GN)F5$\"3@!RSl3%e=\")F57$$\"3=MLL$ e'3I))F5$\"3U/lKhF()*G)F57$$\"3?+]7.f+'p%)F57$$\"3)HLL $eMsw)*F5$\"3G\"*p#Q4lGh)F57$$\"3;+DJ&H\"fT5!#<$\"3Y*fC\\dnYv)F57$$\"3 5+v$f)[$H4\"Fbr$\"3%*3Y#y`+i())F57$$\"3cL$ek`1l9\"Fbr$\"3SVvh@?Q!**)F5 7$$\"3OLe*[.-d>\"Fbr$\"3ai(fJ!)))\\3*F57$$\"3km;/Egw[7Fbr$\"3Iwo)op@r< *F57$$\"3zm\"z%*f%)QI\"Fbr$\"3Ozj#>K7IE*F57$$\"3/+voza'=N\"Fbr$\"3:\\# p7hZ/L*F57$$\"3(om\"zWho.9Fbr$\"3g#>iG)fO'R*F57$$\"3-++]i>Ad9Fbr$\"3Yr Wy!>ewX*F57$$\"32+]i:jf4:Fbr$\"3z-k\\^Of6&*F57$$\"39+DJ&>r-c\"Fbr$\"39 KT?bxme&*F57$$\"3++]P4q`;;Fbr$\"37w%[*\\\"Rcg*F57$$\"3;LL$eM%4n;Fbr$\" 3\\\"QPl2lNk*F57$$\"37++v$4v5s\"Fbr$\"3'>*Q)o&=/!o*F57$$\"3cm\"zWn*)*p Fbr$\"3C!ydi&*)y(y*F57$$\"3CLL3 -=!y(>Fbr$\"3jv58A*G&3)*F57$$\"3))*\\7G8O;.#Fbr$\"3BSw3?L2G)*F57$$\"3! pmm;*\\[$3#Fbr$\"3$*=2l#[3]%)*F57$$\"3*pmT&Qz]O@Fbr$\"3;eKUBFbr$\"3J6ero(Rw! **F57$$\"3/+DJqJ8&R#Fbr$\"3TcOoiv*o\"**F57$$\"3G+voa-oXCFbr$\"3?,$H`.) )[#**F57$$\"3++++++++DFbr$\"3DX\"4+`?E$**F5-%&COLORG6&%$RGBG$\"\"&!\" \"F($\"\"\"F)-%*LINESTYLEG6#Fi[l-F$6%7S7$F(Fh[l7$F1Fh[l7$F \+ " 0 "" {MPLTEXT 1 0 103 "i := 'i':\nde := 2*diff(i(t),t)+4*i(t)=piecew ise(t<0,0,4);\nic := i(0)=0;\nsimplify(dsolve(\{de,ic\},i(t)));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&*&\"\"#\"\"\"-%%diffG6$-%\"iG 6#%\"tGF0F)F)*&\"\"%F)F-F)F)-%*PIECEWISEG6$7$\"\"!2F0F77$F2%*otherwise G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG-%*PIECEWISEG6$7$\"\"!2F'F,7$, &\"\"\"F0-%$expG6#,$*&\"\"#F0F'F0!\"\"F71F,F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 13 "Suppose that " } {XPPEDIT 18 0 "L = 2" "6#/%\"LG\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "R = 4" "6#/%\"RG\"\"%" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "V(t) = PI ECEWISE([0, t < 0],[8*exp(-t), 0 <= t]);" "6#/-%\"VG6#%\"tG-%*PIECEWIS EG6$7$\"\"!2F'F,7$*&\"\")\"\"\"-%$expG6#,$F'!\"\"F11F,F'" }{TEXT -1 22 " in equation (iii). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "Suppose also that at time " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 53 ", there is no current flowing in the cir cuit so that " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "V := t -> piecewise(t<0,0,8*exp(-t)):\n'V(t)'=V(t);\nplot(V(t),t=-2..4,thick ness=2,labels=[`t`,`voltage V(t)`]);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/-%\"VG6#%\"tG-%*PIECEWISEG6$7$\"\"!2F'F,7$,$*&\"\")\"\"\"-%$expG6#, $F'!\"\"F2F2%*otherwiseG" }}{PARA 13 "" 1 "" {GLPLOT2D 542 194 194 {PLOTDATA 2 "6&-%'CURVESG6$7jn7$$!\"#\"\"!$F*F*7$$!3!******\\2<#p=!#iUCFH F+7$$!3k***\\7GVS(=FHF+7$$!3B++]7YY08FHF+7$$!3/,+vo4F+7$$!3w*** *\\7ep$H'F[oF+7$$!38**\\PM#eMg%F[oF+7$$!3\\)**\\il?K\"HF[oF+7$$!3=)\\( =n=5o?F[oF+7$$!3&y*\\7yI)HA\"F[oF+7$$!3(oZPf$oB/!)!#?F+7$$!3Gv\\i!*Gky PF^pF+7$$\"3Gj_(oa5&pW!#@$\"3MYW]!>Dk*zF/7$$\"3%z-+++XDn%F^p$\"3'H'p%[ $oqizF/7$$\"3s-++]im%>'F[o$\"3LX@0`TY>vF/7$$\"3C++++y?#>\"FH$\"3awL`( \\%*35(F/7$$\"3%****\\P%>We=FH$\"3W;;&3tBKk'F/7$$\"3h****\\(3wY_#FH$\" 31W6$FH$\"3y&*HW#zG\"feF/7$$\"3F)******HOTq$ FH$\"3%oa9-X*eBbF/7$$\"3I,+](3\">)*\\FH$\"3qmh!e3BJ&[F/7$$\"3_,+]isVIi FH$\"3wM0NqlZ!H%F/7$$\"3&=++](o:;vFH$\"3I`]ux<$Gx$F/7$$\"3#>++v$)[op)F H$\"3!3EKzqoEN$F/7$$\"3W*****\\i%Qq**FH$\"3q,JXeVw^HF/7$$\"3&****\\(QI KH6F/$\"3W'QU;4;ge#F/7$$\"3#****\\7:xWC\"F/$\"38G2Jw>t/BF/7$$\"37++]Zn %)o8F/$\"3\"yStYU,_.#F/7$$\"3y******4FL(\\\"F/$\"3s*fw$f)3)*y\"F/7$$\" 3#)****\\d6.B;F/$\"3oFWUhzRy:F/7$$\"3(****\\(o3lWw7$F/$\"3#pX!z$)3w0NFH7$$\"3O++v)Q?QD$F/$ \"3#4T.4j3,4$FH7$$\"3G+++5jypLF/$\"3-QX-[*e b(\\&[cG4CFH7$$\"3++++gEd@OF/$\"37krCR@FH7$$\"39++v3'>$[PF/$\"3YcRT ZQe%)=FH7$$\"37++D6EjpQF/$\"3+(\\mqr#Gp;FH7$$\"\"%F*$\"3cM()466Dl9FH-% 'COLOURG6&%$RGBG$\"#5!\"\"F+F+-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$%\"t G%-voltage~V(t)G-%%VIEWG6$;F(Fh[l%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 5 "When " }{XPPEDIT 18 0 "t>=0" "6#1\"\"!%\"tG" }{TEXT -1 13 " the current " }{TEXT 303 1 "I" }{TEXT -1 37 " satisfies the diffe rential equation:" }}{PARA 256 "" 0 "" {TEXT -1 3 " 2 " }{XPPEDIT 18 0 "dI/dt+4*I = 8*exp(-t);" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"%F'%\" IGF'F'*&\"\")F'-%$expG6#,$%\"tGF)F'" }{TEXT -1 2 ", " }}{PARA 0 "" 0 " " {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+2*I = 4*exp(-t);" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"#F' %\"IGF'F'*&\"\"%F'-%$expG6#,$%\"tGF)F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 37 "Asc before the integrating factor is " }{XPPEDIT 18 0 "exp(Int(2,t)) = exp(2*t);" "6#/-%$expG6#-%$IntG6$\"\"#%\"tG-F%6#*&F *\"\"\"F+F/" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Multiplying both sides of the differential equa tion by this integrating factor gives:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t);" "6#-%$expG6#*&\"\"#\"\"\"%\"tGF(" } {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+2*exp(2*t)*I = 4*exp(t);" "6#/,&* &%#dIG\"\"\"%#dtG!\"\"F'*(\"\"#F'-%$expG6#*&F+F'%\"tGF'F'%\"IGF'F'*&\" \"%F'-F-6#F0F'" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 48 "The left side is the derivative with resp ect to " }{TEXT 304 1 "t" }{TEXT -1 4 " of " }{XPPEDIT 18 0 "exp(2*t)* I" "6#*&-%$expG6#*&\"\"#\"\"\"%\"tGF)F)%\"IGF)" }{TEXT -1 13 ", so we \+ have " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I = 4*Int(exp(t),t);" "6#/*&-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%\"IGF**&\"\"% F*-%$IntG6$-F&6#F+F+F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 16 "It follows that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I = 4*exp(t)+c;" "6#/*&-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%\"I GF*,&*&\"\"%F*-F&6#F+F*F*%\"cGF*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 18 ", it follow s that " }{XPPEDIT 18 0 "0 = 4+c;" "6#/\"\"!,&\"\"%\"\"\"%\"cGF'" } {TEXT -1 10 ", so that " }{XPPEDIT 18 0 "c = -4;" "6#/%\"cG,$\"\"%!\" \"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 5 "Thus " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I = 4*exp(t)-4;" "6#/* &-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%\"IGF*,&*&\"\"%F*-F&6#F+F*F*F/!\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = 4*exp(-t)-4*exp(-2*t); " "6#/%\"IG,&*&\"\"%\"\"\"-%$expG6#,$%\"tG!\"\"F(F(*&F'F(-F*6#,$*&\"\" #F(F-F(F.F(F." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "plot(4*exp(-t)-4*exp(-2*t), t=0..4.5,color=COLOR(RGB,.5,0,1),\n linestyle=[1,3],labels=[ `t`,`current I(t)`]);" }}{PARA 13 "" 1 "" {GLPLOT2D 542 223 223 {PLOTDATA 2 "6&-%'CURVESG6$7\\o7$$\"\"!F)F(7$$\"3!****\\P%)z@X#!#>$\"3 ;B=W@*>ZX*F-7$$\"3z****\\(ofV!\\F-$\"33I)ofh!zA=!#=7$$\"3Q++DJ&RlN(F-$ \"3K#*ea.DxNEF57$$\"3e*****\\P>(3)*F-$\"3ct&zD=Q!)Q$F57$$\"3%*\\7`%zMU >\"F5$\"3E%eyEage*RF57$$\"3%**\\i:l(f29F5$\"3Yi'ph.sCc%F57$$\"32]Pf30' 4i\"F5$\"3%RG0)*=G**3&F57$$\"3?+]ilLKM=F5$\"3#)Q#)=6;>!e&F57$$\"3C+v$4 1@UJ#F5$\"3sDk&\\5_lb'F57$$\"3E++Dc(=Tz#F5$\"3!z!R9d4)QP(F57$$\"3;+++D v>xKF5$\"3;+Km-X/a!)F57$$\"31++v$Hw-w$F5$\"3z*Q#RZfY2')F57$$\"3x*\\PM2 f5C%F5$\"3-26')yl)p/*F57$$\"3\\**\\7`=%=s%F5$\"3+mR/&\\R&)Q*F57$$\"3s* *\\PM!*en^F5$\"3wj&3F[\"*yi*F57$$\"3'***\\i:iL8cF5$\"3DJe(zK%[,)*F57$$ \"3q**\\7.;)[2'F5$\"3=7c!Q=9+#**F57$$\"3V**\\i!*pUOlF5$\"3#>TO++jP)**F 57$$\"3/**\\iSC4vnF5$\"3#*zGUcf^(***F57$$\"3x**\\i!*yv8qF5$\"3?^?'\\WG $****F57$$\"3]+]iSLU_sF5$\"3G:$pu^B+***F57$$\"35+]i!z)3\"\\(F5$\"3u&oi qdz.(**F57$$\"3k**\\P%))))o'zF5$\"3qR&y*))*\\K!**F57$$\"3>**\\7y*)oU%) F5$\"3GN3fq)zK!)*F57$$\"39,+]Pn_@%*F5$\"3W\"3(3S789&*F57$$\"3'***\\(ov o$G5!#<$\"3SlkX3Mx)=*F57$$\"39++DYwUD6Ffs$\"3%HQ&y(y\"=o()F57$$\"3#*** *\\(o])GA\"Ffs$\"3ucfk@1\")3$)F57$$\"38++v`L!oJ\"Ffs$\"3en]wJc'o%yF57$ $\"3$**\\iS:!4-9Ffs$\"3)4hVm_F5U(F57$$\"3-++v3W].:Ffs$\"3_NyKz*4k\"pF5 7$$\"3-+++&e:%*e\"Ffs$\"3%f+xgM@k\\'F57$$\"3'**\\ilq]$*o\"Ffs$\"3%>v+s #p!>-'F57$$\"3))****\\A-\"yx\"Ffs$\"3jr&*Q'Qnxh&F57$$\"3?+DcJV'[(=Ffs$ \"3y4CEHe1%>&F57$$\"3C+vo%z#Gn>Ffs$\"3=W5?&\\%G6[F57$$\"3O+]il)*fyo_4*=F57$$\"3I++vo^$z4$Ffs$\"3(3w TmazTs\"F57$$\"3y*\\iST\")f=$Ffs$\"3]u2G(=a^e\"F57$$\"3E++D;#RAG$Ffs$ \"3U/I;t8QX9F57$$\"3q*\\ilI5GP$Ffs$\"3l%[q(fVoC8F57$$\"3%)*\\7G>$[nMFf s$\"3uCYIj:*)37F57$$\"3/++vVK/gNFfs$\"3%=cO`#f:06F57$$\"3!)*\\i!R]%pl$ Ffs$\"3Bn1qv4!e+\"F57$$\"3]+++&)HF]PFfs$\"3Q(H1@qEDwF-7$$\"3x**** \\K(Rt-%Ffs$\"3$QPwn2E;+(F-7$$\"3p**\\(oDAq7%Ffs$\"37OmNSZD[jF-7$$\"3W +++&\\zh@%Ffs$\"3H'3:dZH&F-7$$ \"3))*\\P%eWA-WFfs$\"3xO!*>V$)**R[F-7$$\"3++++++++XFfs$\"3)3DiOpMUR%F- -%*LINESTYLEG6#\"\"\"-%&COLORG6&%$RGBG$\"\"&!\"\"F($Fe_lF)-%+AXESLABEL SG6$%\"tG%-current~I(t)G-%%VIEWG6$;F($\"#XF\\`l%(DEFAULTG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Note that " }{XPPEDIT 18 0 "dI/dt=-4*ex p(-t)+8*exp(-2*t)" "6#/*&%#dIG\"\"\"%#dtG!\"\",&*&\"\"%F&-%$expG6#,$% \"tGF(F&F(*&\"\")F&-F-6#,$*&\"\"#F&F0F&F(F&F&" }{TEXT -1 9 " so that \+ " }{XPPEDIT 18 0 "dI/dt+2*I=-4*exp(-t)+8*exp(-2*t)+8*exp(-t)-8*exp(-2* t)" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"#F'%\"IGF'F',**&\"\"%F'-%$exp G6#,$%\"tGF)F'F)*&\"\")F'-F16#,$*&F+F'F4F'F)F'F'*&F6F'-F16#,$F4F)F'F'* &F6F'-F16#,$*&F+F'F4F'F)F'F)" }{XPPEDIT 18 0 "``=4*exp(-t)" "6#/%!G*& \"\"%\"\"\"-%$expG6#,$%\"tG!\"\"F'" }{TEXT -1 76 ", which checks that \+ the given solution satisfies the differential equation. " }}{PARA 0 " " 0 "" {TEXT -1 60 "The maximum point on the curve for the current occ urs where " }{XPPEDIT 18 0 "dI/dt=0" "6#/*&%#dIG\"\"\"%#dtG!\"\"\"\"! " }{TEXT -1 18 ". This holds when " }{XPPEDIT 18 0 "-4*exp(-t)+8*exp(- 2*t)=0" "6#/,&*&\"\"%\"\"\"-%$expG6#,$%\"tG!\"\"F'F-*&\"\")F'-F)6#,$*& \"\"#F'F,F'F-F'F'\"\"!" }{TEXT -1 10 ", that is " }{XPPEDIT 18 0 "8*ex p(-2*t)=4*exp(-t)" "6#/*&\"\")\"\"\"-%$expG6#,$*&\"\"#F&%\"tGF&!\"\"F& *&\"\"%F&-F(6#,$F-F.F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 51 "On multiplying both sides of this last equation by " }{XPPEDIT 18 0 "exp(2*t)" "6#-%$expG6#*&\"\"#\"\"\"%\"tGF(" }{TEXT -1 11 " we obtai n " }{XPPEDIT 18 0 "8=4*exp(t)" "6#/\"\")*&\"\"%\"\"\"-%$expG6#%\"tGF' " }{TEXT -1 9 " so that " }{XPPEDIT 18 0 "exp(t)=2" "6#/-%$expG6#%\"tG \"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "t=ln*2" "6#/%\"tG*&%#lnG\" \"\"\"\"#F'" }{TEXT -1 1 " " }{TEXT 311 1 "~" }{TEXT -1 10 " 0.69315. \+ " }}{PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "t=ln*2" "6#/%\" tG*&%#lnG\"\"\"\"\"#F'" }{TEXT -1 6 " (and " }{XPPEDIT 18 0 "exp(t)=2 " "6#/-%$expG6#%\"tG\"\"#" }{TEXT -1 3 "), " }{XPPEDIT 18 0 "exp(-t)=1 /2" "6#/-%$expG6#,$%\"tG!\"\"*&\"\"\"F+\"\"#F)" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "exp(-2*t)=1/4" "6#/-%$expG6#,$*&\"\"#\"\"\"%\"tGF*!\"\" *&F*F*\"\"%F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 4 "The " } {TEXT 259 15 "maximum current" }{TEXT -1 4 " is " }{XPPEDIT 18 0 "4*`` (1/2)-8*``(1/4)=1" "6#/,&*&\"\"%\"\"\"-%!G6#*&F'F'\"\"#!\"\"F'F'*&\"\" )F'-F)6#*&F'F'F&F-F'F-F'" }{TEXT -1 6 " amp. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Using " }{TEXT 309 1 "i" } {TEXT -1 12 " instead of " }{TEXT 310 1 "I" }{TEXT -1 35 " for the cur rent as before Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 13 " gives . . \+ . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "i := 'i':\nde := 2*diff(i(t),t)+4*i(t)=piecewise(t<0 ,0,8*exp(-t));\nic := i(0)=0;\nsimplify(dsolve(\{de,ic\},i(t)));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&*&\"\"#\"\"\"-%%diffG6$-%\"iG 6#%\"tGF0F)F)*&\"\"%F)F-F)F)-%*PIECEWISEG6$7$\"\"!2F0F77$,$*&\"\")F)-% $expG6#,$F0!\"\"F)F)%*otherwiseG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG -%*PIECEWISEG6$7$\"\"!2F'F,7$,&*&\"\"%\"\"\"-%$expG6#,$F'!\"\"F2F2*&F1 F2-F46#,$*&\"\"#F2F'F2F7F2F71F,F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 9 "Example 3" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "; " }}}{PARA 0 "" 0 "" {TEXT -1 13 "Suppose that " }{XPPEDIT 18 0 "L = 2 " "6#/%\"LG\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "R = 4" "6#/%\"RG\" \"%" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "V(t) = PIECEWISE([0, t < 0], [26*cos*3*t, 0 <= t]);" "6#/-%\"VG6#%\"tG-%*PIECEWISEG6$7$\"\"!2F'F,7$ **\"#E\"\"\"%$cosGF1\"\"$F1F'F11F,F'" }{TEXT -1 20 " in equation (iii )." }}{PARA 0 "" 0 "" {TEXT -1 26 "Suppose also that at time " } {XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 53 ", there is no curren t flowing in the circuit so that " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"! " }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 119 "V := t -> piecewise(t<0,0,t>=0,26*cos(3*t)):\n'V(t)' =V(t);\nplot(V(t),t=-1..3.3,thickness=2,labels=[`t`,`voltage V(t)`]); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"VG6#%\"tG-%*PIECEWISEG6$7$\" \"!2F'F,7$,$*&\"#E\"\"\"-%$cosG6#,$*&\"\"$F2F'F2F2F2F21F,F'" }}{PARA 13 "" 1 "" {GLPLOT2D 372 247 247 {PLOTDATA 2 "6&-%'CURVESG6$7\\r7$$!\" \"\"\"!$F*F*7$$!3wLL$3PAF1*!#=F+7$$!3_m;H<@?Z#)F/F+7$$!3F/F+7$$!3+gmmTy2s**!#>F+7 $$!3W%**\\7)\\'G&eFKF+7$$!3()GL$37_Ot\"FKF+7$$!3yXP%['[*R:\"FKF+7$$!3& oiT&)3wLu&!#?F+7$$!3S6Pf3)*3XGFXF+7$$\"3z2/UNrk>`!#A$\"3q%e!*o'*****f# !#;7$$\"3c>@I^F[^HFX$\"3g^e.z!)*)*f#F\\o7$$\"3+N+DJ!p(\\eFX$\"3of?kQ'* f*f#F\\o7$$\"3ppT5:9HW7F/$\"3[9$=5oryU#F\\o7$$\"3s++]7iM&o\"F/$\"3w#4 l)=OpuAF\\o7$$\"3s++]-wy#e#F/$\"3+Kn#=^#zd=F\\o7$$\"3)Q$3x\")*o-*HF/$ \"3&)pyfDY7A;F\\o7$$\"3/n;/h.v(R$F/$\"31G$eGPYAO\"F\\o7$$\"3#3]PM8&G#) QF/$\"3-KB.aD5F5F\\o7$$\"3cML$e!*>oO%F/$\"3?(HE7*f%Hq'!#<7$$\"3uLLe9WG xZF/$\"3G0$pcdtX`)pF^s7$$\"3]mT&)QWK llF/$!3FW)=!fch45F\\o7$$\"37LLL[V'z)pF/$!3j^xD*o([/8F\\o7$$\"3qmT&QJn; X(F/$!3-95A#zZPg\"F\\o7$$\"3E+]Pz-P:zF/$!3i=iHUC-s=F\\o7$$\"3U++ve^#pN )F/$!3SdX9)*R$R4#F\\o7$$\"3c+]7Q+[)z)F/$!311V/Hi:zAF\\o7$$\"3onTgKi>f# *F/$!3FB1IbC!)HCF\\o7$$\"3\"[L$3FC\"*>(*F/$!3?;?#)o^5MDF\\o7$$\"3O&ek. 2`9$**F/$!3-I:@!f\"*ec#F\\o7$$\"3f$ek8P*H95F^s$!3iuT^zuM(e#F\\o7$$\"3& fkGNSw[-\"F^s$!3uQLWWG<%f#F\\o7$$\"3`3FpNMXN5F^s$!39[y?`kQ)f#F\\o7$$\" 37rn&yYIg/\"F^s$!3]r!RO1%)**f#F\\o7$$\"3qL3-+vgc5F^s$!3!H'zHnS'*)f#F\\ o7$$\"3q(=<\"ei,o5F^s$!3o&3d5lI\\f#F\\o7$$\"3#>a8i,D%z5F^s$!3+T@cJx&ye #F\\o7$$\"39'*)4VxL34\"F^s$!3S]**[$f`xd#F\\o7$$\"39]iSKDC-6F^s$!3E$*[0 s+jkDF\\o7$$\"3Oe*)f[+1D6F^s$!3.#o@`))*QHDF\\o7$$\"3em;zkv(y9\"F^s$!3; i&[OB-B[#F\\o7$$\"3i;z%z`z_>\"F^s$!3?@%)*o(4jZBF\\o7$$\"3pmT56:oU7F^s$ !3_['f*[Ycl@F\\o7$$\"3ILL3o]%RG\"F^s$!3wm)HDvH8(>F\\o7$$\"3#**\\i]i3_K \"F^s$!31lRp*\\Cpu\"F\\o7$$\"3E$3_qRu(p8F^s$!37\")>Q9Bku9F\\o7$$\"3gm; /p,M99F^s$!3'yq9$z,/w6F\\o7$$\"3=L3Fs4Qg9F^s$!3a#Hb5uhdX)F^s7$$\"3y*** *\\vPY^:F^s$!3eS\\UX\")42:F^s7$$ \"3%)**\\(Gm0lf\"F^s$\"3.:f&HhSL+#F^s7$$\"3&)\\iSfg3S;F^s$\"3I6Fm#QhdO &F^s7$$\"3')*\\PfXmOo\"F^s$\"3Zk(pyP%fO')F^s7$$\"3**\\7.c]0Ko-!z\"F\\o7$$\"3?LL$[F-u'=F^s$\"3M71=<%)3??F\\o7$$\"39++Dh\" \\-'>F^s$\"3]8\"QY?$G#R#F\\o7$$\"3J$3x1q:B+#F^s$\"3%[yHdRH9]#F\\o7$$\" 3ZmT5SAQW?F^s$\"33%\\6#3)*yqDF\\o7$$\"3#e*)4)*pze0#F^s$\"35j=c(4jEe#F \\o7$$\"3=Dc^frPn?F^s$\"3IFFqMSY\"f#F\\o7$$\"33a8A>Y()y?F^s$\"3U(4%)* \\@=(f#F\\o7$$\"3U$3F*y?P!4#F^s$\"37$>+6k5)*f#F\\o7$$\"3y7GjQ&p=5#F^s$ \"3!e]\"=\"QY$*f#F\\o7$$\"3pT&Q$)*pO8@F^s$\"3OXX^A**y&f#F\\o7$$\"3/rU/ eW'[7#F^s$\"3'\\$[)e\\X\"*e#F\\o7$$\"3S++v<>OO@F^s$\"35TRa05UzDF\\o7$$ \"3a;/,sZjz@F^s$\"34V`7lBX:DF\\o7$$\"3;L3FEw!HA#F^s$\"3qOx&Hm]\"4CF\\o 7$$\"3;+voGGP8BF^s$\"3)e(yBm*e)e?F\\o7$$\"3s;a8kffdBF^s$\"3ZyK')o;tI=F \\o7$$\"3GLLe*4>=S#F^s$\"3Wc8\\&eG/d\"F\\o7$$\"3S;/,Cm6[CF^s$\"3=)*4tt ^Uo7F\\o7$$\"3)**\\P%[TT%\\#F^s$\"3%Ra7<(p\"*>%*F^s7$$\"3a$3_q;/!RDF^s $\"31+Sr6hp.hF^s7$$\"35nmm&=%f$e#F^s$\"3yY\")GxeTyEF^s7$$\"3'p;z*>R>HE F^s$!3%p.C*e!*QM()F/7$$\"3#om\"HaOzuEF^s$!3V,_E4O(*3WF^s7$$\"3>]7y*z:+ s#F^s$!3Q?6c>\"\\S$yF^s7$$\"3cL3FXzBlFF^s$!3t/];)p::6\"F\\o7$$\"3+rl\"F\\o7$$\"3 !p;z/&H(f*GF^s$!3\\W'e4A:b#>F\\o7$$\"3RL$ea/*fVHF^s$!3!edciCqY:#F\\o7$ $\"3O+Dc#R'>')HF^s$!3*=7huU[DK#F\\o7$$\"3(omm'RPzGIF^s$!3a-d!46\\DX#F \\o7$$\"3CvV)zc-:0$F^s$!3!*>@Z:oh0DF\\o7$$\"3i$3-jR6U2$F^s$!3'*R(oL!*f qa#F\\o7$$\"3)>z>YA?p4$F^s$!3c$)Rd-howDF\\o7$$\"3P+v$H0H'>JF^s$!3;#G8Y 'zN%f#F\\o7$$\"3VD\"ygn'\\IJF^s$!35%eC1if&)f#F\\o7$$\"3\\](=#*Hk89$F^s $!3G9^*)*Q****f#F\\o7$$\"35v$fB#>B_JF^s$!3im=;Udn)f#F\\o7$$\"3;++]X&*4 jJF^s$!3KHxE%3!f%f#F\\o7$$\"3%)\\7y\"zM[=$F^s$!3a(41)QI:yDF\\o7$$\"3'* *\\i!Q+d1KF^s$!3#*)e!f:\"e2b#F\\o7$$\"3!*\\7.>]G`KF^s$!3#ph@e*3SbCF\\o 7$$\"3#)*************H$F^s$!3oSf#o*p*=J#F\\o-%'COLOURG6&%$RGBG$\"#5F)F +F+-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$%\"tG%-voltage~V(t)G-%%VIEWG6$; F($\"#LF)%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "; " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "For \+ " }{XPPEDIT 18 0 "t>=0" "6#1\"\"!%\"tG" }{TEXT -1 13 " the current " } {TEXT 301 1 "I" }{TEXT -1 37 " satisfies the differential equation:" } }{PARA 256 "" 0 "" {TEXT -1 3 " 2 " }{XPPEDIT 18 0 "dI/dt+4*I = 26*cos *3*t;" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"%F'%\"IGF'F'**\"#EF'%$cosG F'\"\"$F'%\"tGF'" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 33 "subj ect to the initial condition " }{XPPEDIT 18 0 "I(0)=0" "6#/-%\"IG6#\" \"!F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 97 "The differentia l equation can be written in the form (standard form for a first order linear DE):" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt +2*I = 13*cos*3*t;" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"#F'%\"IGF'F'* *\"#8F'%$cosGF'\"\"$F'%\"tGF'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 26 "The integrating factor is " }{XPPEDIT 18 0 "exp(Int(2,t)) = exp(2*t);" "6#/-%$expG6#-%$IntG6$\"\"#%\"tG-F%6#*&F*\"\"\"F+F/" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Multiplying both sides of the differential equation by th is integrating factor gives:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "exp(2*t);" "6#-%$expG6#*&\"\"#\"\"\"%\"tGF(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+2*exp(2*t)*I = 13*exp(2*t)*cos*3*t;" "6#/, &*&%#dIG\"\"\"%#dtG!\"\"F'*(\"\"#F'-%$expG6#*&F+F'%\"tGF'F'%\"IGF'F'*, \"#8F'-F-6#*&F+F'F0F'F'%$cosGF'\"\"$F'F0F'" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 48 "The left sid e is the derivative with respect to " }{TEXT 302 1 "t" }{TEXT -1 4 " o f " }{XPPEDIT 18 0 "exp(2*t)*I" "6#*&-%$expG6#*&\"\"#\"\"\"%\"tGF)F)% \"IGF)" }{TEXT -1 13 ", so we have " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "exp(2*t)*I = 13*Int(exp(2*t)*cos*3*t,t);" "6#/*&-%$e xpG6#*&\"\"#\"\"\"%\"tGF*F*%\"IGF**&\"#8F*-%$IntG6$**-F&6#*&F)F*F+F*F* %$cosGF*\"\"$F*F+F*F+F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The integral " }{XPPEDIT 18 0 "Int (exp(2*t)*cos*3*t,t)" "6#-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$c osGF,\"\"$F,F-F,F-" }{TEXT -1 53 " can be found using the integration \+ by parts formula " }{XPPEDIT 18 0 "Int(u*``(dv/dt),t) = u*v-Int(v*``(d u/dt),t);" "6#/-%$IntG6$*&%\"uG\"\"\"-%!G6#*&%#dvGF)%#dtG!\"\"F)%\"tG, &*&F(F)%\"vGF)F)-F%6$*&F4F)-F+6#*&%#duGF)F/F0F)F1F0" }{TEXT -1 3 ". \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*cos* 3*t,t)" "6#-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$cosGF,\"\"$F,F- F,F-" }{TEXT -1 9 " ... " }{XPPEDIT 18 0 "PIECEWISE([u = exp(2*t), v = sin*3*t/3],[du/dt = 2*exp(2*t), dv/dt = cos*3*t]);" "6#-%*PIECEWI SEG6$7$/%\"uG-%$expG6#*&\"\"#\"\"\"%\"tGF./%\"vG**%$sinGF.\"\"$F.F/F.F 4!\"\"7$/*&%#duGF.%#dtGF5*&F-F.-F*6#*&F-F.F/F.F./*&%#dvGF.F:F5*(%$cosG F.F4F.F/F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "Int(u*``(dv/dt),t) = u*v-Int(v*``(du/dt),t)" "6#/-%$Int G6$*&%\"uG\"\"\"-%!G6#*&%#dvGF)%#dtG!\"\"F)%\"tG,&*&F(F)%\"vGF)F)-F%6$ *&F4F)-F+6#*&%#duGF)F/F0F)F1F0" }{TEXT -1 10 " becomes: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*cos*3*t,t)=1/3" "6 #/-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF-F-%$cosGF-\"\"$F-F.F-F.*&F-F -F0!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*sin*3*t - 2/3" "6#, &**-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%$sinGF*\"\"$F*F+F*F**&F)F*F-!\"\"F/ " }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*sin*3*t,t)" "6#-%$IntG6 $**-%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$sinGF,\"\"$F,F-F,F-" }{TEXT -1 15 " ------- (I). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "The residual integral " }{XPPEDIT 18 0 "Int(exp(2*t)*sin* 3*t,t)" "6#-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$sinGF,\"\"$F,F- F,F-" }{TEXT -1 34 " can be treated in a similar way. " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*sin*3*t,t);" "6#-%$ IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$sinGF,\"\"$F,F-F,F-" }{TEXT -1 9 " ... " }{XPPEDIT 18 0 "PIECEWISE([u = exp(2*t), v = -cos*3*t /3],[du/dt = 2*exp(2*t), dv/dt = sin*3*t]);" "6#-%*PIECEWISEG6$7$/%\"u G-%$expG6#*&\"\"#\"\"\"%\"tGF./%\"vG,$**%$cosGF.\"\"$F.F/F.F5!\"\"F67$ /*&%#duGF.%#dtGF6*&F-F.-F*6#*&F-F.F/F.F./*&%#dvGF.F;F6*(%$sinGF.F5F.F/ F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I nt(u*``(dv/dt),t) = u*v-Int(v*``(du/dt),t)" "6#/-%$IntG6$*&%\"uG\"\"\" -%!G6#*&%#dvGF)%#dtG!\"\"F)%\"tG,&*&F(F)%\"vGF)F)-F%6$*&F4F)-F+6#*&%#d uGF)F/F0F)F1F0" }{TEXT -1 10 " becomes: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*sin*3*t,t)=-1/3" "6#/-%$IntG6$** -%$expG6#*&\"\"#\"\"\"%\"tGF-F-%$sinGF-\"\"$F-F.F-F.,$*&F-F-F0!\"\"F3 " }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*cos*3*t + 2/3" "6#,&**-%$ex pG6#*&\"\"#\"\"\"%\"tGF*F*%$cosGF*\"\"$F*F+F*F**&F)F*F-!\"\"F*" } {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*cos*3*t,t)" "6#-%$IntG6$** -%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$cosGF,\"\"$F,F-F,F-" }{TEXT -1 16 " \+ ------- (II). " }}{PARA 0 "" 0 "" {TEXT -1 17 "Substituting for " } {XPPEDIT 18 0 "Int(exp(2*t)*sin*3*t,t)" "6#-%$IntG6$**-%$expG6#*&\"\"# \"\"\"%\"tGF,F,%$sinGF,\"\"$F,F-F,F-" }{TEXT -1 26 " in (I) using (II) gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2 *t)*cos*3*t,t) = 1/3" "6#/-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF-F-%$ cosGF-\"\"$F-F.F-F.*&F-F-F0!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp( 2*t)*sin*3*t+2/9" "6#,&**-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%$sinGF*\"\"$F *F+F*F**&F)F*\"\"*!\"\"F*" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*co s*3*t-4/9" "6#,&**-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%$cosGF*\"\"$F*F+F*F* *&\"\"%F*\"\"*!\"\"F1" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*co s*3*t,t)" "6#-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF,F,%$cosGF,\"\"$F, F-F,F-" }{TEXT -1 1 "." }}{PARA 257 "" 0 "" {TEXT -1 5 "Hence" }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "13/9;" "6#*&\"#8\"\" \"\"\"*!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*cos*3*t,t) \+ = 1/3" "6#/-%$IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF-F-%$cosGF-\"\"$F-F. F-F.*&F-F-F0!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*sin*3*t+2/ 9" "6#,&**-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%$sinGF*\"\"$F*F+F*F**&F)F*\" \"*!\"\"F*" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*cos*3*t" "6#**-%$ expG6#*&\"\"#\"\"\"%\"tGF)F)%$cosGF)\"\"$F)F*F)" }{TEXT -1 4 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "After add ing a constant of integration, we see that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(exp(2*t)*cos*3*t,t) = 3/13;" "6#/-% $IntG6$**-%$expG6#*&\"\"#\"\"\"%\"tGF-F-%$cosGF-\"\"$F-F.F-F.*&F0F-\"# 8!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*sin*3*t+2/13;" "6#,&* *-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%$sinGF*\"\"$F*F+F*F**&F)F*\"#8!\"\"F* " }{TEXT -1 1 " " }{XPPEDIT 18 0 "cos*3*t+c;" "6#,&*(%$cosG\"\"\"\"\"$ F&%\"tGF&F&%\"cGF&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Int(exp(2*t)*cos(3*t),t);\n value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&-%$expG6#,$%\" tG\"\"#\"\"\"-%$cosG6#,$F+\"\"$F-F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #,&*&-%$expG6#,$%\"tG\"\"#\"\"\"-%$cosG6#,$F)\"\"$F+#F*\"#8*(#F0F2F+F% F+-%$sinGF.F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "Returning to the solution of the differential equation we now have the equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I = 3*e xp(2*t)*sin*3*t+2*exp(2*t)*cos*3*t+c[1];" "6#/*&-%$expG6#*&\"\"#\"\"\" %\"tGF*F*%\"IGF*,(*,\"\"$F*-F&6#*&F)F*F+F*F*%$sinGF*F/F*F+F*F**,F)F*-F &6#*&F)F*F+F*F*%$cosGF*F/F*F+F*F*&%\"cG6#F*F*" }{TEXT -1 2 ", " }} {PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "c[1] = 13*c;" "6#/ &%\"cG6#\"\"\"*&\"#8F'F%F'" }{TEXT -1 10 ". Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = 3*sin*3*t+2*cos*3*t+c[1]*exp(- 2*t);" "6#/%\"IG,(**\"\"$\"\"\"%$sinGF(F'F(%\"tGF(F(**\"\"#F(%$cosGF(F 'F(F*F(F(*&&%\"cG6#F(F(-%$expG6#,$*&F,F(F*F(!\"\"F(F(" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Since \+ " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 14 ", we see that " }{XPPEDIT 18 0 "0=2+c[1]" "6#/\"\"!,&\"\"#\"\"\"&%\"cG6#F'F'" }{TEXT -1 10 ", so th at " }{XPPEDIT 18 0 "c[1]=-2" "6#/&%\"cG6#\"\"\",$\"\"#!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 37 "We now have the particular so lution: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = 3*sin *3*t+2*cos*3*t-2*exp(-2*t);" "6#/%\"IG,(**\"\"$\"\"\"%$sinGF(F'F(%\"tG F(F(**\"\"#F(%$cosGF(F'F(F*F(F(*&F,F(-%$expG6#,$*&F,F(F*F(!\"\"F(F4" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "The term " }{XPPEDIT 18 0 "-2*exp(-2*t);" "6#,$*&\"\"#\"\" \"-%$expG6#,$*&F%F&%\"tGF&!\"\"F&F-" }{TEXT -1 47 ", involving the ex ponential, tends to zero as " }{XPPEDIT 18 0 "t -> infinity" "6#f*6#% \"tG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 20 ", and \+ is called the " }{TEXT 259 9 "transient" }{TEXT -1 22 " part of the so lution." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I[trans] = -2*exp(-2*t);" "6#/&%\"IG6#%&transG,$*&\"\"#\"\"\"-%$expG6#,$*&F*F+% \"tGF+!\"\"F+F2" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 91 "The effect of this term is eventually neg lible, so if we subtract it, we are left with the " }{TEXT 259 12 "ste ady-state" }{TEXT -1 22 " part of the solution." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I[steady_state] = 3*sin*3*t+2*cos*3*t; " "6#/&%\"IG6#%-steady_stateG,&**\"\"$\"\"\"%$sinGF+F*F+%\"tGF+F+**\" \"#F+%$cosGF+F*F+F-F+F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 332 "sstate := t -> 3*sin( 3*t)+2*cos(3*t):\n`steady state current`=sstate(t);\ntrans := t -> -2* exp(-2*t):\n`transient current`=trans(t);\nplot([trans(t)+sstate(t),tr ans(t),sstate(t)],t=0..3.5,\n color=[red,blue,magenta],labels=[`t`,`c urrent I(t)`],\nlegend=[`complete solution`,`transient part of solutio n`,`steady state part of solution`]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%5steady~state~currentG,&*&\"\"#\"\"\"-%$cosG6#,$*&\"\"$F(%\"tGF(F (F(F(*&F.F(-%$sinGF+F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2transie nt~currentG,$*&\"\"#\"\"\"-%$expG6#,$*&F'F(%\"tGF(!\"\"F(F/" }}{PARA 13 "" 1 "" {GLPLOT2D 513 301 301 {PLOTDATA 2 "6'-%'CURVESG6%7[q7$$\"\" !F)F(7$$\"3]mm;z>]9Q!#>$\"3+VdC0U!Qw%!#=7$$\"3+LLLeR+HwF-$\"3n)R@U'=) \\6*F07$$\"31+vVt\")z%4\"F0$\"3m49tJV3a7!#<7$$\"3'omT5&fpE9F0$\"39!36J q\\7c\"F;7$$\"3MLL3xM?t@F0$\"3-ue%HeN]6#F;7$$\"3kLLL39$*[DF0$\"3Db:\\< z()=BF;7$$\"3oLLeR$fY#HF0$\"3axOY$Rc7Z#F;7$$\"3G'GU0 yFDF;7$$\"3K+Dc,:g)H$F0$\"3/aL![Sw9d#F;7$$\"3O$3_Desb[$F0$\"3w\"p!>dbQ -EF;7$$\"3'pmTNmVDn$F0$\"3][_[CKe?EF;7$$\"3GLLe*y*)e%QF0$\"3&pAEE F;7$$\"3:+]i:fB>SF0$\"3=@u+yH3@EF;7$$\"3/nmmT?e#>%F0$\"3jZJGZ\"*Q0EF;7 $$\"3OL$3x;GfO%F0$\"3%4`v)yMFzDF;7$$\"3XZ9)ekBF;7$$\"3;M$3FR-k#eF0$\"3M_J,#e$*p(>F; 7$$\"3)***\\(=(e`mlF0$\"3_)QnSf-!\\9F;7$$\"3nnm;HT&yK(F0$\"34%>KAcI:#z F07$$\"3Q,](o*)QJm(F0$\"3?c3Q95AjZF07$$\"3'RL$ekOU)*zF0$\"3C1U&)eom$\\ \"F07$$\"3wn;/,\\(eP)F0$!3qCB?Zx&QF#F07$$\"3a,+]PhK`()F0$!3#))G2I-(e\" 3'F07$$\"3%4++]AFB8*F0$!3ytia>d=\"*)*F07$$\"3L++]7$G8^*F0$!3!QOc;o8LO \"F;7$$\"3a++](3ml()*F0$!3@%*za(*3f7?\\:g$F; 7$$\"30+vVQ%RRJ\"F;$!3Yi+>DD4!p$F;7$$\"3k\"zp)**)R6L\"F;$!3]iItB;D9PF; 7$$\"3C$3-8OS$[8F;$!3B/I6O(f!HPF;7$$\"3/vVtA3al8F;$!3RYG;!HfWt$F;7$$\" 3kmm;%GTFQ\"F;$!3'))*f:u$=/t$F;7$$\"39vV[;Hh,9F;$!3,wD)=Q<^r$F;7$$\"3T $3-)[X[?9F;$!3gEk*G)GZ)o$F;7$$\"3p\"z>6=c$R9F;$!3Kd!R0Z\\0l$F;7$$\"3=+ vV8yAe9F;$!32\\q5]'[9g$F;7$$\"37+](oKoT\\\"F;$!3Qs<8e*\\vZ$F;7$$\"3F+D JS)3,`\"F;$!3j\"*Qx0I&\\J$F;7$$\"3&omT5:4^g\"F;$!3M#[wqf,$fGF;7$$\"3#o ;a)[G)Rn\"F;$!3sj9\"4xO%=BF;7$$\"31]ilUw76F;7$$\"3IL$ ekVs#[S7F;7$$\"3QL3FR%Q a#=F;$!37\"p!pk,*yM)F07$$\"3ummTb]-f=F;$!3BiXNh'fTt%F07$$\"33+Dcr;h#*= F;$!3S^ZuxU_z5F07$$\"3o;a8Ph))G>F;$\"3qUG\"Q(fIrGF07$$\"3[L$3Fgg^'>F;$ \"3Y\"4f\")QY4y'F07$$\"3'p;/^nNE+#F;$\"3ecHEUi\"G2\"F;7$$\"3******\\Z2 6S?F;$\"3SB'*RmuK`9F;7$$\"3')*\\(o%zsn2#F;$\"30zXB'fhs!=F;7$$\"3>+](=% [V8@F;$\"3^cJjUn!)Q@F;7$$\"3G+vVt'zV=#F;$\"3)4$ea)H#F;$\"3,Vc(*o4$4M$F;7$$\"3Umm;%3K RL#F;$\"3)Q?SNWw;Y$F;7$$\"39+v$fMDGN#F;$\"3()ogjW$=,^$F;7$$\"3UL$3xg=< P#F;$\"3]YjETF?ZNF;7$$\"3qm\"z%p=h!R#F;$\"3a![z+k8Gd$F;7$$\"3V++DJ^]4C F;$\"3/Ib>#yroe$F;7$$\"3i3_]M_iECF;$\"3i4%yv=(f*e$F;7$$\"3!oTgxLXPW#F; $\"3c:?*oG'z#e$F;7$$\"3)\\i:5Wl3Y#F;$\"3/2$yu,*[mNF;7$$\"3=L3FWb)zZ#F; $\"3&y8v/K?2a$F;7$$\"3h;/,M\">a^#F;$\"3]2NY!\\!z^MF;7$$\"3]++vBF&Gb#F; $\"3V%zoNPO\">LF;7$$\"3emT50pHBEF;$\"3%4zjLpss&HF;7$$\"35+v=s8$pp#F;$ \"3O8#Q\\&G(zV#F;7$$\"3V$3_v%p#Ht#F;$\"3^z-$F;$!3cCUF!Qh<'))F07$$\"3;nT5 g&GZ1$F;$!33\"RN(>avl7F;7$$\"3-%3-Q&>b)4$F;$!3f1'f#o/8,;F;7$$\"3Y++]Z` PKJF;$!3D()e\"=h%4?>F;7$$\"3\"pm\"z*>1*4KF;$!3O\"y71=J>d#F;7$$\"3[LLL= 2DzKF;$!3@1*G2C/(QIF;7$$\"33+vVQk=`LF;$!33y\\6bL&=R$F;7$$\"3S++](Rp&)Q $F;$!3v%)G]$F;7$$\"3I+DccB&RU$F;$!3_&=6'>7_uNF;7$$\"3Av=UnU'HW$F; $!3i3/#GKTkf$F;7$$\"39]7Gyh(>Y$F;$!3V![&R3Zn1OF;7$$\"33D19*3))4[$F;$!3 #\\Exqt(=0OF;7$$\"3++++++++NF;$!33/\"Q)**[)>f$F;-%'COLOURG6&%$RGBG$\"* ++++\"!\")F(F(-%'LEGENDG6#%2complete~solutionG-F$6%7W7$F($!\"#F)7$F+$! 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39&*p!f8h,f$F;-F^il6&F`ilFailF(Fail-Feil6#%>steady~state~part~of~solut ionG-%+AXESLABELSG6$%\"tG%-current~I(t)G-%%VIEWG6$;F($\"#N!\"\"%(DEFAU LTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 1 "complete solution" "transient part of solution" "steady state part of solution " }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Usin g " }{TEXT 312 1 "i" }{TEXT -1 12 " instead of " }{TEXT 313 1 "I" } {TEXT -1 35 " for the current as before Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 13 " gives . . . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "i := 'i':\nde := 2*diff(i(t),t)+4*i (t)=26*cos(3*t);\nic := i(0)=0;\ndsolve(\{de,ic\},i(t));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#deG/,&*&\"\"#\"\"\"-%%diffG6$-%\"iG6#%\"tGF0F )F)*&\"\"%F)F-F)F),$*&\"#EF)-%$cosG6#,$*&\"\"$F)F0F)F)F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG,(*&\"\"#\"\"\"-%$cosG6#,$*&\"\"$F+F'F+F+ F+F+*&F1F+-%$sinGF.F+F+*&F*F+-%$expG6#,$*&F*F+F'F+!\"\"F+F;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 4" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 13 "Suppose tha t " }{XPPEDIT 18 0 "L = 2" "6#/%\"LG\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "R = 4" "6#/%\"RG\"\"%" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "V(t) = PIECEWISE([0, t < 0],[4, 0 <= t and t <= 2],[0, 2 < t]);" "6#/-%\"V G6#%\"tG-%*PIECEWISEG6%7$\"\"!2F'F,7$\"\"%31F,F'1F'\"\"#7$F,2F3F'" } {TEXT -1 22 " in equation (iii). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 193 "Such a source voltage could be obtained by closing a switch to connect the circuit to a constant source volta ge for 2 seconds, at which time the switch is opened to cut off the vo ltage source. " }}{PARA 0 "" 0 "" {TEXT -1 26 "Suppose also that at ti me " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 53 ", there is no \+ current flowing in the circuit so that " }{XPPEDIT 18 0 "I=0" "6#/%\"I G\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "V := t -> piecewise(t<0,0,t<2,4,0):\n'V(t)'=V(t );\nplot(V(t),t=-1..5,thickness=2,labels=[`t`,`voltage V(t)`]);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"VG6#%\"tG-%*PIECEWISEG6%7$\"\"!2F 'F,7$\"\"%2F'\"\"#7$F,%*otherwiseG" }}{PARA 13 "" 1 "" {GLPLOT2D 542 194 194 {PLOTDATA 2 "6&-%'CURVESG6$7co7$$!\"\"\"\"!$F*F*7$$!3/+++]2<#p )!#=F+7$$!3[++]7bBavF/F+7$$!3++++D$3XF'F/F+7$$!3c*****\\F)H')\\F/F+7$$ !3J++]i3@/PF/F+7$$!3V++]7F+7$$!33')****\\7;)=\"!#?F+7$$\"3W7+DJ&Rox#FI$\"\"%F*7$$\"3$4,+DJS=u 'FIFM7$$\"3%4+v$4Toq5FEFM7$$\"3!3++v=%=n9FEFM7$$\"3]++vVV=gAFEFM7$$\"3 ?++++X=`IFEFM7$$\"3f****\\7[=RYFEFM7$$\"3**)****\\7&=DiFEFM7$$\"3y(*** **\\d=(R*FEFM7$$\"3m****\\P'=pD\"F/FM7$$\"3A++vVrZ4>F/FM7$$\"3y+++]c.i DF/FM7$$\"3;+++DMe6PF/FM7$$\"32,++]>q0]F/FM7$$\"3h******\\U80jF/FM7$$ \"3'4+++0ytb(F/FM7$$\"3w****\\(QNXp)F/FM7$$\"3.+++XDn/5!#6F\\qFM7$$\"3'****\\(3wY_7F\\qFM7$$\"3#)******HOTq8F\\qFM7$$\"37+ +v3\">)*\\\"F\\qFM7$$\"3:++DEP/B;F\\qFM7$$\"3=++](o:;v\"F\\qFM7$$\"3=+ +v$)[op=F\\qFM7$$\"32+]7t;OL>F\\qFM7$$\"3%*****\\i%Qq*>F\\qFM7$$\"3Z7y ]bB<,?F\\qF+7$$\"3+Dc^[iI0?F\\qF+7$$\"3`PM_T,W4?F\\qF+7$$\"31]7`MSd8?F \\qF+7$$\"36voa?=%=-#F\\qF+7$$\"3<+Dc1'4,.#F\\qF+7$$\"3G]Pfy^kY?F\\qF+ 7$$\"3%***\\i]2=j?F\\qF+7$$\"3s*\\(o%*=D'4#F\\qF+7$$\"3&****\\(QIKH@F \\qF+7$$\"3#******\\4+p=#F\\qF+7$$\"3#****\\7:xWC#F\\qF+7$$\"37++]Zn%) oBF\\qF+7$$\"3y******4FL(\\#F\\qF+7$$\"3#)****\\d6.BEF\\qF+7$$\"3(**** \\(o3lWFF\\qF+7$$\"3!*****\\A))ozGF\\qF+7$$\"3e******Hk-,IF\\qF+7$$\"3 6+++D-eIJF\\qF+7$$\"3u***\\(=_(zC$F\\qF+7$$\"3M+++b*=jP$F\\qF+7$$\"3g* **\\(3/3(\\$F\\qF+7$$\"33++vB4JBOF\\qF+7$$\"3u*****\\KCnu$F\\qF+7$$\"3 s***\\(=n#f(QF\\qF+7$$\"3P+++!)RO+SF\\qF+7$$\"30++]_!>w7%F\\qF+7$$\"3O ++v)Q?QD%F\\qF+7$$\"3G+++5jypVF\\qF+7$$\"3<++]Ujp-XF\\qF+7$$\"3++++gEd @YF\\qF+7$$\"39++v3'>$[ZF\\qF+7$$\"37++D6Ejp[F\\qF+7$$\"\"&F*F+-%'COLO URG6&%$RGBG$\"#5F)F+F+-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$%\"tG%-volta ge~V(t)G-%%VIEWG6$;F(F\\y%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "0<=t" "6#1\"\"!%\"tG" }{XPPEDIT 18 0 "``<2" "6#2%!G\"\"#" }{TEXT -1 13 " the current " }{TEXT 292 1 "I" } {TEXT -1 37 " satisfies the differential equation:" }}{PARA 256 "" 0 " " {TEXT -1 3 " 2 " }{XPPEDIT 18 0 "dI/dt+4*I = 4;" "6#/,&*&%#dIG\"\"\" %#dtG!\"\"F'*&\"\"%F'%\"IGF'F'F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 48 "From the solution obtained in example 1 we have " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = 1-exp(-2*t)" "6#/ %\"IG,&\"\"\"F&-%$expG6#,$*&\"\"#F&%\"tGF&!\"\"F." }{TEXT -1 13 " ---- --- (I) " }}{PARA 0 "" 0 "" {TEXT -1 5 "for " }{XPPEDIT 18 0 "0<=t" " 6#1\"\"!%\"tG" }{XPPEDIT 18 0 "`` <= 2;" "6#1%!G\"\"#" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 13 "Substituting " }{XPPEDIT 18 0 "t=2" " 6#/%\"tG\"\"#" }{TEXT -1 20 " in (I) we see that " }{XPPEDIT 18 0 "I=1 -exp(-4)" "6#/%\"IG,&\"\"\"F&-%$expG6#,$\"\"%!\"\"F," }{TEXT -1 6 " wh en " }{XPPEDIT 18 0 "t=2" "6#/%\"tG\"\"#" }{TEXT -1 2 ". " }}{PARA 0 " " 0 "" {TEXT -1 94 "After this time the voltage source becomes 0, and \+ the corresponding differential equation is: " }}{PARA 256 "" 0 "" {TEXT -1 3 " 2 " }{XPPEDIT 18 0 "dI/dt+4*I = 0;" "6#/,&*&%#dIG\"\"\"%# dtG!\"\"F'*&\"\"%F'%\"IGF'F'\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI /dt+2*I=0" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"#F'%\"IGF'F'\"\"!" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "Multiplying both sides of this differential equation by t he integrating factor " }{XPPEDIT 18 0 "exp(2*t)" "6#-%$expG6#*&\"\"# \"\"\"%\"tGF(" }{TEXT -1 8 " gives: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t);" "6#-%$ex pG6#*&\"\"#\"\"\"%\"tGF(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+2*exp( 2*t)*I = 0;" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*(\"\"#F'-%$expG6#*&F+F'% \"tGF'F'%\"IGF'F'\"\"!" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 48 "The left side is the derivative w ith respect to " }{TEXT 314 1 "t" }{TEXT -1 4 " of " }{XPPEDIT 18 0 "e xp(2*t)*I" "6#*&-%$expG6#*&\"\"#\"\"\"%\"tGF)F)%\"IGF)" }{TEXT -1 12 " , so we have" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "exp( 2*t)*I = c;" "6#/*&-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%\"IGF*%\"cG" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 13 "Substituting " } {XPPEDIT 18 0 "t=2" "6#/%\"tG\"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "I=1-exp(-4)" "6#/%\"IG,&\"\"\"F&-%$expG6#,$\"\"%!\"\"F," }{TEXT -1 24 " in this equation gives " }{XPPEDIT 18 0 "exp(4)*(1-exp(-4))=c" "6 #/*&-%$expG6#\"\"%\"\"\",&F)F)-F&6#,$F(!\"\"F.F)%\"cG" }{TEXT -1 9 " s o that " }{XPPEDIT 18 0 "c=exp(4)-1" "6#/%\"cG,&-%$expG6#\"\"%\"\"\"F* !\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 12 "We now have " } }{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(2*t)*I=exp(4)-1 " "6#/*&-%$expG6#*&\"\"#\"\"\"%\"tGF*F*%\"IGF*,&-F&6#\"\"%F*F*!\"\"" } {TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I=exp(4-2*t)-exp(-2*t)" "6#/%\"I G,&-%$expG6#,&\"\"%\"\"\"*&\"\"#F+%\"tGF+!\"\"F+-F'6#,$*&F-F+F.F+F/F/ " }{TEXT -1 16 " ------- (II), " }}{PARA 0 "" 0 "" {TEXT -1 4 "for " }{XPPEDIT 18 0 "2<=t" "6#1\"\"#%\"tG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 42 "Combining the results (I) and (II) gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = PIECEWISE([1-exp(-2*t), 0 <= t and t <= 2],[exp(4-2*t)-exp(-2*t), 2 < t]);" "6#/%\"IG-%*PIECE WISEG6$7$,&\"\"\"F*-%$expG6#,$*&\"\"#F*%\"tGF*!\"\"F231\"\"!F11F1F07$, &-F,6#,&\"\"%F**&F0F*F1F*F2F*-F,6#,$*&F0F*F1F*F2F22F0F1" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 190 "g := t -> piecewise(t<=2,1-exp(-2*t),t>2,exp(4-2*t)- exp(-2*t)):\n`I(t)`=g(t);\nplot(g(t),t=0..5,color=COLOR(RGB,.5,0,1),nu mpoints=65,\n labels=[`t`,`current I(t)`]) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%%I(t)G-%*PIECEWISEG6$7$,&\"\"\" F*-%$expG6#,$*&\"\"#F*%\"tGF*!\"\"F21F1F07$,&F+F2-F,6#,&*&F0F*F1F*F2\" \"%F*F*2F0F1" }}{PARA 13 "" 1 "" {GLPLOT2D 503 213 213 {PLOTDATA 2 "6& -%'CURVESG6#7io7$$\"\"!F)F(7$$\"3%)***\\iSmp3%!#>$\"3!H$Q>Jgy[yF-7$$\" 3m****\\7G$R<)F-$\"3*RIvSw`\"3:!#=7$$\"3/](oH/)*H<\"F5$\"3A13>#\\D64#F 57$$\"3)**\\(o/GgG:F5$\"3H1$QgwvSj#F57$$\"3#*\\7y]v^G>F5$\"3.)\\buu`acYF57$$\"3&**\\P4@o[$RF5$\"3CxGJv izZaF57$$\"3g*\\(oz,yxYF5$\"3e=YScOKwgF57$$\"3!**\\(=#\\AqW&F5$\"3>#*= >#HKej'F57$$\"34+v=#*RdUiF5$\"3NC#=ip$pIrF57$$\"32+vV[TdNqF5$\"3'R\\jw /8:b(F57$$\"3A++D\"Gs7&yF5$\"3!*4B0$\\x+#zF57$$\"3Q+]iS'R(p&)F5$\"3ya( f)[%o%)>)F57$$\"3Q++v=Pcy$*F5$\"3uF.1qHan%)F57$$\"3*)**\\i!*32>5!#<$\" 3Saz+FWH(p)F57$$\"3-+]7GhL(4\"Fip$\"3;![5WN\\g)))F57$$\"3+](=z'*F57 $$\"3=]PM_Ib$z\"Fip$\"3NcCBso@B(*F57$$\"33+D1R!\\J(=Fip$\"3-W(3%>.&Rw* F57$$\"3\"\\iS\"pz[9>Fip$\"3wi]'*49o#y*F57$$\"3'*\\(=#**o#e&>Fip$\"3U7 )o\\9E**z*F57$$\"35voHH(>Q(>Fip$\"3wMZ;Iz*p!)*F57$$\"3-+]PfD\"=*>Fip$ \"3C1[$pv>Q\")*F57$$\"3#\\7`%*Q0)4?Fip$\"3i+b!f2.ii*F57$$\"3%)\\7`>#)z F?Fip$\"3!\\V![=l&fG*F57$$\"3![Pf$=Pmm?Fip$\"3=c/Am-\\\"f)F57$$\"3>+v= <#Hb5#Fip$\"3Crkxo3'*[zF57$$\"31](o/3\"oX@Fip$\"3!)>QjjteNtF57$$\"3#** **\\P%H$e=#Fip$\"3a%*30.UapnF57$$\"3C]Pf3P6DAFip$\"3[KqU\"\\s!eiF57$$ \"37+vVtWRkAFip$\"3a`4EM\\C&y&F57$$\"30DJ?L1S-BFip$\"3Y.*3)3Fzh`F57$$ \"3)*\\(oHz1/M#Fip$\"3?Y\"z%e^Lp\\F57$$\"3y*\\iS^0[U#Fip$\"3-Z%R^$z\\( >%F57$$\"3u***\\(=:k+DFip$\"3DfL=KVy1OF57$$\"3'***\\iSEh\"e#Fip$\"37b6 >w^`nIF57$$\"3%)\\(=<^%)\\l#Fip$\"3M&G!3[L')[EF57$$\"3W+](oM*>NFFip$\" 3!RrP5!GCcAF57$$\"3(*\\(oaDv1\"GFip$\"3ry,@e'=,%>F57$$\"3;]PMF$p&*)GFi p$\"3=*zp^)3#pl\"F57$$\"3u**\\7.FqmHFip$\"3uNX)R1`+U\"F57$$\"3/](=#*>a u/$Fip$\"38%[n'4?F37F57$$\"38++]([F_7$Fip$\"3TCKHaq@M5F57$$\"3D+D\"y!> w/KFip$\"3%yR3t`E7#))F-7$$\"3A](oHuPOG$Fip$\"3uwhi2I(Q`(F-7$$\"3S++vVk 6cLFip$\"3'*RMjBdC#\\'>bF-7$$\"3y****\\ 7H[8NFip$\"3e!3fez$[dZF-7$$\"3I](oav*p#f$Fip$\"3SHt%\\<:/1%F-7$$\"3=]7 .#Q?&oOFip$\"3Qd\"R'H05*[$F-7$$\"3;++vV%4Fv$Fip$\"3%[99)Q%4%[HF-7$$\"3 l\\i:&)4hDQFip$\"3mP@d,DR[DF-7$$\"34](=n)[N3RFip$\"3a<(yi=7(f@F-7$$\"3 #)*\\7.(pw$)RFip$\"3mq3_2,Nd=F-7$$\"3)*\\7GQfDmSFip$\"3%eO0,@q[d\"F-7$ $\"3*)*\\(=n*Hu8%Fip$\"3=S?Bq`\"fO\"F-7$$\"31+]7y))[=UFip$\"39ySJbQ\\h 6F-7$$\"3k+]P%y(y'H%Fip$\"3*GN(Qx&G8$**!#?7$$\"3g]P%)Rb.vVFip$\"3iD(\\ I?IE\\)Ff`l7$$\"3#**\\PfV&*HX%Fip$\"3SPZzH1`msFf`l7$$\"3w\\P%)*[!*y_%F ip$\"3e>LE`PnbiFf`l7$$\"3w*\\il-a)3YFip$\"3J.$G9rq/K&Ff`l7$$\"3c]ilZ\" =go%Fip$\"3eJ@#Gy$ffXFf`l7$$\"3u*\\i:vHsw%Fip$\"3!Q!fNUgM$Ff`l7$$\"3=+]i:[*>#\\Fip$\"3$QeE\"e`>WGFf` l7$$\"\"&F)$\"3#Q(Q!pC_LV#Ff`l-%&COLORG6&%$RGBG$Facl!\"\"F($\"\"\"F)-% +AXESLABELSG6$%\"tG%-current~I(t)G-%%VIEWG6$;F(F`cl%(DEFAULTG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Using " }{TEXT 315 1 "i" }{TEXT -1 12 " instead of " }{TEXT 316 1 "I" }{TEXT -1 35 " \+ for the current as before Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 13 " gives . . . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "i := 'i':\nde := 2*diff(i(t),t)+4*i(t)=piecewise (t<=2,4,0);\nic := i(0)=0;\ndsolve(\{de,ic\},i(t));" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%#deG/,&*&\"\"#\"\"\"-%%diffG6$-%\"iG6#%\"tGF0F)F)*& \"\"%F)F-F)F)-%*PIECEWISEG6$7$F21F0F(7$\"\"!%*otherwiseG" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG-%*PIECEWISEG6$7$,&\"\"\"F--%$expG6#,$*& \"\"#F-F'F-!\"\"F42F'F37$,&F.F4-F/6#,&*&F3F-F'F-F4\"\"%F-F-1F3F'" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 5 " }}{PARA 0 " " 0 "" {TEXT 354 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 67 "An LR circuit has a variable inductor with the inductance given by " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "PIECEWI SE([1-t/10, 0 <= t and t < 10],[0, 10 <= t]);" "6#-%*PIECEWISEG6$7$,& \"\"\"F(*&%\"tGF(\"#5!\"\"F,31\"\"!F*2F*F+7$F/1F+F*" }{TEXT -1 1 " " } }{PARA 0 "" 0 "" {TEXT -1 22 "henrys where the time " }{TEXT 358 1 "t " }{TEXT -1 24 " is measured in seconds." }}{PARA 0 "" 0 "" {TEXT -1 43 "Find a (piecewise) formula for the current " }{TEXT 355 1 "I" } {TEXT -1 6 " when " }{XPPEDIT 18 0 "0<=t" "6#1\"\"!%\"tG" }{TEXT -1 38 " if the resistance of the resistor is " }{XPPEDIT 18 0 "1/5" "6#*& \"\"\"F$\"\"&!\"\"" }{TEXT -1 62 " ohms and the voltage source is cons tant at 4 volts, that is, " }{XPPEDIT 18 0 "V(t)=4" "6#/-%\"VG6#%\"tG \"\"%" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t>=0" "6#1\"\"!%\"tG" } {TEXT -1 20 ". Suppose also that " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"! " }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 30 ". Plot a graph of the current " }{TEXT 356 1 "I" }{TEXT -1 13 " ve rsus time " }{TEXT 357 1 "t" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 359 8 "Solution" }{TEXT -1 2 ": " }} {PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "0<=t" "6#1\"\"!%\"t G" }{XPPEDIT 18 0 "`` < 10;" "6#2%!G\"#5" }{TEXT -1 27 " the different ial equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 365 1 "L" } {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+R*I=V(t)" "6#/,&*&%#dIG\"\"\"%#dt G!\"\"F'*&%\"RGF'%\"IGF'F'-%\"VG6#%\"tG" }{TEXT -1 13 " ------- (i)" }}{PARA 0 "" 0 "" {TEXT -1 8 "becomes " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``(1-t/10);" "6#-%!G6#,&\"\"\"F'*&%\"tGF'\"#5!\" \"F+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+``(1/5)*I = 4;" "6#/,&*&%# dIG\"\"\"%#dtG!\"\"F'*&-%!G6#*&F'F'\"\"&F)F'%\"IGF'F'\"\"%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 " Dividing the left and right sides of this differential equation by" } {XPPEDIT 18 0 "``(1-t/10)" "6#-%!G6#,&\"\"\"F'*&%\"tGF'\"#5!\"\"F+" } {TEXT -1 7 " gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+``(1/5)/(1-t/10)" "6#,&*&%#dIG\"\"\"%#dtG!\"\"F&*&-%!G6#*&F&F &\"\"&F(F&,&F&F&*&%\"tGF&\"#5F(F(F(F&" }{TEXT -1 1 " " }{XPPEDIT 18 0 "I = 4/(1-t/10);" "6#/%\"IG*&\"\"%\"\"\",&F'F'*&%\"tGF'\"#5!\"\"F,F," }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt+``(2/(10-t))*I = 40/(10-t );" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&-%!G6#*&\"\"#F',&\"#5F'%\"tGF)F)F '%\"IGF'F'*&\"#SF',&F1F'F2F)F)" }{TEXT -1 16 " ------- (ii). " }} {PARA 0 "" 0 "" {TEXT -1 67 "Since this equation is linear we construc t the integrating factor: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "exp( Int(2/(10-t),t))=exp(-2*ln(10-t))" "6#/-%$expG6#-% $IntG6$*&\"\"#\"\"\",&\"#5F,%\"tG!\"\"F0F/-F%6#,$*&F+F,-%#lnG6#,&F.F,F /F0F,F0" }{XPPEDIT 18 0 "``=exp(ln((10-t)^(-2)))" "6#/%!G-%$expG6#-%#l nG6#),&\"#5\"\"\"%\"tG!\"\",$\"\"#F0" }{XPPEDIT 18 0 "``=1/(10-t)^2" " 6#/%!G*&\"\"\"F&*$,&\"#5F&%\"tG!\"\"\"\"#F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 64 "Multiplying both sides of (ii) by this integrat ing factor gives " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "1/((10-t)^2)" "6#*&\"\"\"F$*$,&\"#5F$%\"tG!\"\"\"\"#F)" }{TEXT -1 1 " " }{XPPEDIT 18 0 " dI/dt+2/((10-t)^3)" "6#,&*&%#dIG\"\"\"%#dtG!\"\"F& *&\"\"#F&*$,&\"#5F&%\"tGF(\"\"$F(F&" }{TEXT -1 1 " " }{XPPEDIT 18 0 "I = 40/((10-t)^3);" "6#/%\"IG*&\"#S\"\"\"*$,&\"#5F'%\"tG!\"\"\"\"$F," } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "Now " }{XPPEDIT 18 0 "Diff(1/((10-t)^2),t)=(-2)/(10-t)^3" "6#/-%%DiffG6$*&\"\"\"F(*$,&\"#5F(%\"tG!\"\"\"\"#F-F,*&,$F.F-F(*$,&F+F (F,F-\"\"$F-" }{TEXT -1 1 " " }{TEXT 360 1 "." }{XPPEDIT 18 0 " ``(-1) =2/((10-t)^3)" "6#/-%!G6#,$\"\"\"!\"\"*&\"\"#F(*$,&\"#5F(%\"tGF)\"\"$F )" }{TEXT -1 53 ", so the left side is the derivative with respect to \+ " }{TEXT 361 1 "t" }{TEXT -1 16 " of the product " }{XPPEDIT 18 0 "1/( (10-t)^2)" "6#*&\"\"\"F$*$,&\"#5F$%\"tG!\"\"\"\"#F)" }{TEXT -1 1 " " } {TEXT 362 1 "." }{TEXT -1 1 " " }{TEXT 363 1 "I" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 28 "Thus we obtain the equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "d/dt" "6#*&%\"dG\"\"\"%#dtG !\"\"" }{XPPEDIT 18 0 "``(I/((10-t)^2)) = 40/((10-t)^3);" "6#/-%!G6#*& %\"IG\"\"\"*$,&\"#5F)%\"tG!\"\"\"\"#F.*&\"#SF)*$,&F,F)F-F.\"\"$F." } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I/((10-t)^2) = 20*Int(2/((10-t)^3) ,t);" "6#/*&%\"IG\"\"\"*$,&\"#5F&%\"tG!\"\"\"\"#F+*&\"#?F&-%$IntG6$*&F ,F&*$,&F)F&F*F+\"\"$F+F*F&" }{TEXT -1 1 " " }{TEXT 364 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "We have just observed that " } {XPPEDIT 18 0 "Diff(1/((10-t)^2),t) = 2/((10-t)^3);" "6#/-%%DiffG6$*& \"\"\"F(*$,&\"#5F(%\"tG!\"\"\"\"#F-F,*&F.F(*$,&F+F(F,F-\"\"$F-" } {TEXT -1 64 ", which allows the integral to be obtained immediately to give " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I/((10-t)^ 2) = 20/((10-t)^2)+c;" "6#/*&%\"IG\"\"\"*$,&\"#5F&%\"tG!\"\"\"\"#F+,&* &\"#?F&*$,&F)F&F*F+F,F+F&%\"cGF&" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 21 "which implies that " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "I=20+c*(10-t)^2" "6#/%\"IG,&\"#?\"\"\"*&%\"cGF'*$,& \"#5F'%\"tG!\"\"\"\"#F'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "I=0" "6#/%\"IG\"\"!" }{TEXT -1 6 " when \+ " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 18 ", it follows that " }{XPPEDIT 18 0 "0=20+100*c" "6#/\"\"!,&\"#?\"\"\"*&\"$+\"F'%\"cGF'F '" }{TEXT -1 9 " so that " }{XPPEDIT 18 0 "c=-1/5" "6#/%\"cG,$*&\"\"\" F'\"\"&!\"\"F)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = 20-(10-t)^2/5 ;" "6#/%\"IG,&\"#?\"\"\"*&,&\"#5F'%\"tG!\"\"\"\"#\"\"&F,F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 4 "for " }{XPPEDIT 18 0 "0<=t" "6#1\" \"!%\"tG" }{XPPEDIT 18 0 "``<10" "6#2%!G\"#5" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 50 "Since the inductance of the inductor is z ero when " }{XPPEDIT 18 0 "t>=10" "6#1\"#5%\"tG" }{TEXT -1 83 " the di fferential equation (i) reverts to the ordinary (non-differential) equ ation " }{XPPEDIT 18 0 "I*R=V(t)" "6#/*&%\"IG\"\"\"%\"RGF&-%\"VG6#%\"t G" }{TEXT -1 11 ", that is, " }{XPPEDIT 18 0 "``(1/5)*I = 4;" "6#/*&-% !G6#*&\"\"\"F)\"\"&!\"\"F)%\"IGF)\"\"%" }{TEXT -1 14 ", which gives " }{XPPEDIT 18 0 "I=20" "6#/%\"IG\"#?" }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 39 "The formula for the current in amps is " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I = PIECEWISE([20-(10-t)^2/5, 0 \+ <= t and t < 10],[20, 10 <= t]);" "6#/%\"IG-%*PIECEWISEG6$7$,&\"#?\"\" \"*&,&\"#5F+%\"tG!\"\"\"\"#\"\"&F0F031\"\"!F/2F/F.7$F*1F.F/" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" }{TEXT -1 81 ": The differential equation (ii) can also be so lved by separating the variables. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "g := t -> piecewise(t<10,20 -(10-t)^2/5,20):\n'i(t)'=g(t);\nplot(g(t),t=0..15,color=COLOR(RGB,.5,0 ,1),labels=[`t`,`current I(t)`]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/ -%\"iG6#%\"tG-%*PIECEWISEG6$7$,&\"#?\"\"\"*&\"\"&!\"\",&\"#5F.F'F1\"\" #F12F'F37$F-%*otherwiseG" }}{PARA 13 "" 1 "" {GLPLOT2D 390 230 230 {PLOTDATA 2 "6&-%'CURVESG6#7S7$$\"\"!F)F(7$$\"3')*****\\7t&pK!#=$\"3[a 0;L!\\kG\"!#<7$$\"3$****\\(=7T9hF-$\"3oZgRQC*4P#F07$$\"3X****\\(=HPJ*F -$\"3%*>QCs0+_NF07$$\"3;++DJaU`7F0$\"3[a\"fEm'[*p%F07$$\"3)***\\P%GZRd \"F0$\"3N87Fhn7$$ \"3c****\\iNGwSF0$\"3/Y=Or;>)H\"Fhn7$$\"37++]7XM*Q%F0$\"3!Gs6+!4Tq8Fhn 7$$\"3/+](o%QjtYF0$\"3l()o0skfK9Fhn7$$\"32++]i8o6]F0$\"3x-aTVNL-:Fhn7$ $\"3i******\\>0)H&F0$\"3)Q--2pLyb\"Fhn7$$\"3Y**\\(=-p6j&F0$\"342:p0=Fhn7$$\"3q+]P4A@urF0$\"3f$4ZsY)HS=Fhn7$$\"3I++Dchf#\\(F0$\"3WapG >&eU(=Fhn7$$\"3))**\\(of2L#yF0$\"31eNk.-C0>Fhn7$$\"3M**\\7yG>6\")F0$\" 3^&3FJ:['G>Fhn7$$\"3w++voo6A%)F0$\"3#RL#['p0-&>Fhn7$$\"3q*****\\xJLu)F 0$\"3#p]X%*p:%o>Fhn7$$\"3W++v$*ydd!*F0$\"3OEIZ6oB#)>Fhn7$$\"3#***\\(=< F;O*F0$\"3#RBYEg\\=*>Fhn7$$\"35***\\i0A#*p*F0$\"3cq5d_1>)*>Fhn7$$\"3*) ****\\2mD+5Fhn$\"#?F)7$$\"3*****\\i0XE.\"FhnFeu7$$\"3%**\\(o/Q*>1\"Fhn Feu7$$\"3=++vQ(zS4\"FhnFeu7$$\"3***\\(=-,FC6FhnFeu7$$\"33+v$4tFe:\"Fhn Feu7$$\"3!****\\73\"o'=\"FhnFeu7$$\"3-+voz;)*=7FhnFeu7$$\"31+++&*44]7F hnFeu7$$\"35+]7jZ!>G\"FhnFeu7$$\"34+v=(4bMJ\"FhnFeu7$$\"3;++]xlWU8FhnF eu7$$\"39+]i&3ucP\"FhnFeu7$$\"3\"******\\;$R09FhnFeu7$$\"38+v=-*zqV\"F hnFeu7$$\"33+D\"G:3uY\"FhnFeu7$$\"#:F)Feu-%+AXESLABELSG6$%\"tG%-curren t~I(t)G-%&COLORG6&%$RGBG$\"\"&!\"\"F($\"\"\"F)-%%VIEWG6$;F(Fex%(DEFAUL TG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "(1-t/10)*diff(i(t),t)+i(t)/5=4,i(0)=0;\ndsolve(\{%\},i(t));" } }{PARA 11 "" 1 "" {XPPMATH 20 "6$/,&*&,&\"\"\"F'*&\"#5!\"\"%\"tGF'F*F' -%%diffG6$-%\"iG6#F+F+F'F'*&#F'\"\"&F'F/F'F'\"\"%/-F06#\"\"!F9" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG,&\"#?\"\"\"*&\"\"&!\"\" ,&F'F*\"#5F-\"\"#F-" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 40 "An object falling i n a resisting medium " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" } }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 12 "First model " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "A mass " }{TEXT 367 1 "m " }{TEXT -1 131 " falling in a resisting medium will experience a forc e opposing its motion which usually increases as the speed of fall inc reases. " }}{PARA 0 "" 0 "" {TEXT -1 74 "We could suppose that the res isting force is proportional to the velocity " }{TEXT 368 1 "v" } {TEXT -1 14 " of the body. " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {GLPLOT2D 140 149 149 {PLOTDATA 2 "6)-%)POLYGONSG6$7U7$$\"\"&!\"\"$\" \"!F,7$$\"+1Ndg\\!#5$\"+!ohmE'!#67$$\"+1e\"H%[F0$\"+O%\\MC\"F07$$\"+IC ))[YF0$\"+kFiS=F07$$\"++M`\"Q%F0$\"+q$o(3CF07$$\"+s\\3XSF0$\"+ii#*QHF0 7$$\"+OJ%[k$F0$\"+IbtAMF07$$\"+[*>r=$F0$\"+9ic_QF07$$\"+wR8zEF0$\"+F'R ;A%F07$$\"+fk*)G@F0$\"+i_8CXF07$$\"+p\\3X:F0$\"+#e#GbZF07$$\"+5d1p$*F3 $\"+aiV6\\F07$$\"+if_RJF3$\"+UO8!*\\F07$$!+$)f_RJF3Fco7$$!+Id1p$*F3F^o 7$$!+r\\3X:F0Fin7$$!+ck*)G@F0$\"+j_8CXF07$$!+tR8zEF0$\"+G'R;A%F07$$!+Y *>r=$F0$\"+;ic_QF07$$!+UJ%[k$F0$\"+DbtAMF07$$!+v\\3XSF0$\"+di#*QHF07$$ !+-M`\"Q%F0$\"+m$o(3CF07$$!+JC))[YF0$\"+gFiS=F07$$!+1e\"H%[F0$\"+K%\\M C\"F07$$!+2Ndg\\F0$\"+];mmiF37$$!\"&F*$!+3Q.^?!#>7$$!+1Ndg\\F0$!+!phmE 'F37$Fbr$!+O%\\MC\"F07$$!+IC))[YF0$!+kFiS=F07$$!++M`\"Q%F0$!+q$o(3CF07 $$!+t\\3XSF0$!+gi#*QHF07$$!+QJ%[k$F0$!+GbtAMF07$$!+W*>r=$F0$!+=ic_QF07 $$!+qR8zEF0$!+J'R;A%F07$$!+`k*)G@F0$!+k_8CXF07$$!+n\\3X:F0$!+$e#GbZF07 $$!+!pl!p$*F3$!+aiV6\\F07$$!+Uf_RJF3$!+UO8!*\\F07$$\"+/g_RJF3Fiv7$$\"+ ]d1p$*F3Fdv7$$\"+t\\3X:F0$!+\"e#GbZF07$$\"+ek*)G@F0$!+i_8CXF07$$\"+vR8 zEF0$!+G'R;A%F07$FN$!+9ic_QF07$$\"+VJ%[k$F0$!+CbtAMF07$$\"+w\\3XSF0$!+ bi#*QHF07$$\"+/M`\"Q%F0$!+k$o(3CF07$$\"+KC))[YF0$!+eFiS=F07$$\"+2e\"H% [F0$!+I%\\MC\"F07$$\"+2Ndg\\F0$!+I;mmiF37$F($\"+:w1-TF`s-%&COLORG6&%$R GBG$\"\"(F*FhzFhz-%'CURVESG6'7$7$F+F(7$F+$\"#IF*7%7$$!+++++IF0$\"#DF*F _[l7$$\"+++++IF0Ff[l-%&STYLEG6#%,PATCHNOGRIDG-%'COLOURG6&FgzF,F,F,-%*T HICKNESSG6#\"\"#-F[[l6'7$7$F+F\\s7$F+$!#IF*7%7$Fi[l$!#DF*Fj\\l7$Fd[lF_ ]lF[\\lF_\\lFb\\l-%%TEXTG6$7$$\"#9F*$!#AF*Q#mg6\"-Fc]l6$7$Ff]l$\"#AF*Q #kvF[^l-%*AXESSTYLEG6#%%NONEG-%(SCALINGG6#%,CONSTRAINEDG" 1 2 0 1 10 0 2 9 1 1 1 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curv e 3" "Curve 4" "Curve 5" }}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 150 "In this case, applying Newton's second law of motion (which says \+ that the force on a moving object is equal to its mass times the accel eration) gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 369 1 "m" } {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = m*g-k*v" "6#/*&%#dvG\"\"\"%#dtG !\"\",&*&%\"mGF&%\"gGF&F&*&%\"kGF&%\"vGF&F(" }{TEXT -1 15 " ------- ( i), " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 370 1 "g" }{TEXT -1 40 " is the acceleration due to gravity and " }{TEXT 371 1 "k" } {TEXT -1 16 " is a constant. " }}{PARA 0 "" 0 "" {TEXT -1 44 "The equa tion (i) can be written in the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt+k*v/m = g;" "6#/,&*&%#dvG\"\"\"%#dtG!\"\"F'*( %\"kGF'%\"vGF'%\"mGF)F'%\"gG" }{TEXT -1 16 " ------- (ii). " }}{PARA 0 "" 0 "" {TEXT -1 118 "Equation (ii) is a first order linear differen tial equation. Hence it can be solved by the integrating factor method . " }}{PARA 0 "" 0 "" {TEXT -1 26 "The integrating factor is " } {XPPEDIT 18 0 "exp(Int(k/m,t)) = exp(k*t/m);" "6#/-%$expG6#-%$IntG6$*& %\"kG\"\"\"%\"mG!\"\"%\"tG-F%6#*(F+F,F/F,F-F." }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 65 "Multiplying both sides of (ii) by this in tegrating factor gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "exp(k*t/m);" "6#-%$expG6#*(%\"kG\"\"\"%\"tGF(%\"mG!\"\" " }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt+k/m" "6#,&*&%#dvG\"\"\"%#dtG! \"\"F&*&%\"kGF&%\"mGF(F&" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(k*t/m)*v = g*exp(k*t/m);" "6#/*&-%$expG6#*(%\"kG\"\"\"%\"tGF*%\"mG!\"\"F*%\"vG F**&%\"gGF*-F&6#*(F)F*F+F*F,F-F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "d/dt" "6#*&%\"dG\"\"\"%#dtG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "[ exp(k*t/m)*v] = g*exp(k*t/m);" "6#/7#*&-%$expG6#*(%\"kG\"\"\"%\"tGF+% \"mG!\"\"F+%\"vGF+*&%\"gGF+-F'6#*(F*F+F,F+F-F.F+" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "exp(k*t/m)*v = g*Int(exp(k*t/m),t);" "6#/*&-%$expG6# *(%\"kG\"\"\"%\"tGF*%\"mG!\"\"F*%\"vGF**&%\"gGF*-%$IntG6$-F&6#*(F)F*F+ F*F,F-F+F*" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 " so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(k*t/m)*v = m*g/ k;" "6#/*&-%$expG6#*(%\"kG\"\"\"%\"tGF*%\"mG!\"\"F*%\"vGF**(F,F*%\"gGF *F)F-" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(k*t/m)+c;" "6#,&-%$expG6#*( %\"kG\"\"\"%\"tGF)%\"mG!\"\"F)%\"cGF)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 19 "If we suppose that \+ " }{XPPEDIT 18 0 "v=v[0]" "6#/%\"vG&F$6#\"\"!" }{TEXT -1 6 " when " } {XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 12 " this gives " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v[0]=m*g/k+c" "6#/&% \"vG6#\"\"!,&*(%\"mG\"\"\"%\"gGF+%\"kG!\"\"F+%\"cGF+" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "c=v[0]-m*g/k" "6#/%\"cG,&&%\"vG6#\"\"!\"\"\"*(% \"mGF*%\"gGF*%\"kG!\"\"F/" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(k *t/m)*v = m*g/k;" "6#/*&-%$expG6#*(%\"kG\"\"\"%\"tGF*%\"mG!\"\"F*%\"vG F**(F,F*%\"gGF*F)F-" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(k*t/m)+v[0]-m *g/k;" "6#,(-%$expG6#*(%\"kG\"\"\"%\"tGF)%\"mG!\"\"F)&%\"vG6#\"\"!F)*( F+F)%\"gGF)F(F,F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 11 "Thi s gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v=m*g/k+` `(v[0]-m*g/k)*exp(-k*t/m)" "6#/%\"vG,&*(%\"mG\"\"\"%\"gGF(%\"kG!\"\"F( *&-%!G6#,&&F$6#\"\"!F(*(F'F(F)F(F*F+F+F(-%$expG6#,$*(F*F(%\"tGF(F'F+F+ F(F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 3 "As " }{XPPEDIT 18 0 "t->infinity" "6#f*6#%\"tG7\"6$%)operatorG%&arrowG6\"%)infinityGF *F*F*" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "exp(-k*t/m)" "6#-%$expG6#,$*(% \"kG\"\"\"%\"tGF)%\"mG!\"\"F," }{TEXT -1 22 " approaches 0 so that " } {XPPEDIT 18 0 "v->m*g/k" "6#f*6#%\"vG7\"6$%)operatorG%&arrowG6\"*(%\"m G\"\"\"%\"gGF-%\"kG!\"\"F*F*F*" }{TEXT -1 62 ". Thus the velocity appr oaches a limiting velocity called the " }{TEXT 259 17 "terminal veloci ty" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 56 "Typically the veloc ity increases form the initial value " }{XPPEDIT 18 0 "v[0]" "6#&%\"vG 6#\"\"!" }{TEXT -1 91 " towards the terminal velocity, so it is probab ly better to write the solution in the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = m*g/k-``(m*g/k-v[0])*exp(-k*t/m);" "6#/%\"vG,&*(%\"mG\"\"\"%\"gGF(%\"kG!\"\"F(*&-%!G6#,&*(F'F(F)F(F*F+F(& F$6#\"\"!F+F(-%$expG6#,$*(F*F(%\"tGF(F'F+F+F(F+" }{TEXT -1 2 ". " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 372 17 "_________________" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "The differential equation (ii) can also be solved by separating the variables. " }}{PARA 0 "" 0 "" {TEXT -1 53 "The differential equation can be written in the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt=g-k*v/m" " 6#/*&%#dvG\"\"\"%#dtG!\"\",&%\"gGF&*(%\"kGF&%\"vGF&%\"mGF(F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = ``(-k/m)*(v-m*g/k);" "6#/*&%#dv G\"\"\"%#dtG!\"\"*&-%!G6#,$*&%\"kGF&%\"mGF(F(F&,&%\"vGF&*(F0F&%\"gGF&F /F(F(F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " } {XPPEDIT 18 0 "v[1]=m*g/k" "6#/&%\"vG6#\"\"\"*(%\"mGF'%\"gGF'%\"kG!\" \"" }{TEXT -1 44 ", so that the differential equation becomes " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = ``(-k/m)*(v-v [1]);" "6#/*&%#dvG\"\"\"%#dtG!\"\"*&-%!G6#,$*&%\"kGF&%\"mGF(F(F&,&%\"v GF&&F26#F&F(F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 38 "Then s eparate the variables to obtain " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "1/(v-v[1])" "6#*&\"\"\"F$,&%\"vGF$&F&6#F$!\"\"F)" } {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt=-k/m" "6#/*&%#dvG\"\"\"%#dtG!\"\" ,$*&%\"kGF&%\"mGF(F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 39 " Integrating both sides with respect to " }{TEXT 382 1 "t" }{TEXT -1 8 " gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(1/(v -v[1]),t)=-Int(k/m,t)" "6#/-%$IntG6$*&\"\"\"F(,&%\"vGF(&F*6#F(!\"\"F-% \"tG,$-F%6$*&%\"kGF(%\"mGF-F.F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(abs(v-v[1])) = -k*t/m+c" "6#/-%#lnG6#-%$absG6#,&%\"vG\"\"\"&F+6# F,!\"\",&*(%\"kGF,%\"tGF,%\"mGF/F/%\"cGF," }{TEXT -1 2 ". " }{TEXT 379 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "abs(v-v[1])=exp(-k*t/m+c)" "6#/-%$absG6#,&%\"vG\"\"\"&F(6#F)!\"\"-%$expG6#,&*(%\"kGF)%\"tGF)%\"mG F,F,%\"cGF)" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " } }{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v-v[1]=A*exp(-k*t/m) " "6#/,&%\"vG\"\"\"&F%6#F&!\"\"*&%\"AGF&-%$expG6#,$*(%\"kGF&%\"tGF&%\" mGF)F)F&" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " } {XPPEDIT 18 0 "A=``" "6#/%\"AG%!G" }{TEXT 380 1 "+" }{TEXT -1 1 " " } {XPPEDIT 18 0 "exp(c)" "6#-%$expG6#%\"cG" }{TEXT -1 2 ". " }}{PARA 0 " " 0 "" {TEXT -1 5 "Thus " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "v=v[1]+A*exp(-k*t/m)" "6#/%\"vG,&&F$6#\"\"\"F(*&%\"AGF( -%$expG6#,$*(%\"kGF(%\"tGF(%\"mG!\"\"F3F(F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "v=m*g/k+A*exp(-k*t/m)" "6#/%\"vG,&*(%\"mG\"\"\"%\"gGF(% \"kG!\"\"F(*&%\"AGF(-%$expG6#,$*(F*F(%\"tGF(F'F+F+F(F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Given " }{XPPEDIT 18 0 "v=v[0]" "6#/% \"vG&F$6#\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\" \"!" }{TEXT -1 18 ", it follows that " }{XPPEDIT 18 0 "v[0]=m*g/k+A" " 6#/&%\"vG6#\"\"!,&*(%\"mG\"\"\"%\"gGF+%\"kG!\"\"F+%\"AGF+" }{TEXT -1 9 " so that " }{XPPEDIT 18 0 "A=v[0]-m*g/k" "6#/%\"AG,&&%\"vG6#\"\"!\" \"\"*(%\"mGF*%\"gGF*%\"kG!\"\"F/" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 46 "This gives the same result as before, namely " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = m*g/k-``(m*g/k-v[0])*ex p(-k*t/m);" "6#/%\"vG,&*(%\"mG\"\"\"%\"gGF(%\"kG!\"\"F(*&-%!G6#,&*(F'F (F)F(F*F+F(&F$6#\"\"!F+F(-%$expG6#,$*(F*F(%\"tGF(F'F+F+F(F+" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 381 17 "__________ _______" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "m := 'm': g := 'g': v := 'v':\nm*di ff(v(t),t)=m*g-k*v(t),v(0)=v0;\ndsolve(\{%\},v(t));" }}{PARA 11 "" 1 " " {XPPMATH 20 "6$/*&%\"mG\"\"\"-%%diffG6$-%\"vG6#%\"tGF-F&,&*&F%F&%\"g GF&F&*&%\"kGF&F*F&!\"\"/-F+6#\"\"!%#v0G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"tG,&*(%\"gG\"\"\"%\"kG!\"\"%\"mGF+F+*(-%$expG6#,$*(F ,F+F.F-F'F+F-F+,&*&F.F+F*F+F+*&F,F+%#v0GF+F-F+F,F-F-" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "m*diff(v( t),t)=m*g-k*v(t),v(0)=v0;\ndesolve(\{%\},v(t),method=linear,info=true) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/*&%\"mG\"\"\"-%%diffG6$-%\"vG6#% \"tGF-F&,&*&F%F&%\"gGF&F&*&%\"kGF&F*F&!\"\"/-F+6#\"\"!%#v0G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%0Lin ear~DE~.~.~~G/,&-%%DiffG6$-%\"vG6#%\"tGF,\"\"\"*(%\"kGF-%\"mG!\"\"F)F- F-%\"gG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%9Integrating~factor~.~.~~G-%$expG6#-%$IntG6$*&%\"kG\" \"\"%\"mG!\"\"%\"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%5~~~~~~~~~~~~~ ~~~=~~~G-%$expG6#*(%\"kG\"\"\"%\"mG!\"\"%\"tGF)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&-%\"vG6#%\"tG \"\"\"-%$expG6#*(%\"kGF)%\"mG!\"\"F(F)F)-%$IntG6$*&%\"gGF)F*F)F(" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/*&-%\"vG6#%\"tG\"\"\"-%$expG6#*(%\"kGF)%\"mG!\"\"F(F)F),&**%\"gGF)F. F0F/F)F*F)F)&%\"CG6#F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%EApplying~the~initial~condition~.~.~~ G/&%\"CG6#\"\"\"*&,&*&%\"mGF(%\"gGF(!\"\"*&%\"kGF(%#v0GF(F(F(F0F." }} {PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/-%\"vG6#%\"tG,&*(%\"gG\"\"\"%\"kG!\"\"%\"mGF+F+*(-%$expG6#,$*(F,F+F. F-F'F+F-F+,&*&F.F+F*F+F-*&F,F+%#v0GF+F+F+F,F-F+" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 1" }}{PARA 0 "" 0 "" {TEXT 375 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 154 "A skydiv er with a mass of 60 kg jumps from a high-flying plane. Due to air res istance she experiences a resisting force which opposes the motion giv en by " }{XPPEDIT 18 0 "12*v;" "6#*&\"#7\"\"\"%\"vGF%" }{TEXT -1 16 " \+ Newtons, where " }{TEXT 373 1 "v" }{TEXT -1 55 " is her vertical veloc ity downwards in metres per sec. " }}{PARA 0 "" 0 "" {TEXT -1 79 "Assu me that at the instant she leaves the plane her vertical velocity is z ero. " }}{PARA 0 "" 0 "" {TEXT -1 36 "(a) Find a formula for the veloc ity " }{TEXT 376 1 "v" }{TEXT -1 33 " downwards as a function of time \+ " }{TEXT 374 1 "t" }{TEXT -1 13 " in seconds. " }}{PARA 0 "" 0 "" {TEXT -1 43 "(b) Sketch a graph of velocity versus time " }{TEXT 377 1 "t" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 32 "(c) Find her term inal velocity. " }}{PARA 0 "" 0 "" {TEXT -1 59 "Take the gravitational acceleration to be 9.807 metres per " }{XPPEDIT 18 0 "sec^2" "6#*$%$s ecG\"\"#" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 378 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 185 "This problem could be solved by substituting appropriate numbers in the general expression given above. However to illuminate \+ the previous discussion we go through all the steps again. " }}{PARA 0 "" 0 "" {TEXT -1 74 "(a) The differential equation which governs the motion of the skydiver is " }}{PARA 256 "" 0 "" {TEXT -1 4 " 60 " } {XPPEDIT 18 0 "dv/dt = ``(60)*`.`*``(9.807)-12*v;" "6#/*&%#dvG\"\"\"%# dtG!\"\",&*(-%!G6#\"#gF&%\".GF&-F,6#-%&FloatG6$\"%2)*!\"$F&F&*&\"#7F&% \"vGF&F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt+v/5 = 9.807;" " 6#/,&*&%#dvG\"\"\"%#dtG!\"\"F'*&%\"vGF'\"\"&F)F'-%&FloatG6$\"%2)*!\"$ " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 73 "Multiplying both sid es of this linear equation by the integrating factor " }{XPPEDIT 18 0 "exp(Int(1/5,t)) = exp(t/5);" "6#/-%$expG6#-%$IntG6$*&\"\"\"F+\"\"&!\" \"%\"tG-F%6#*&F.F+F,F-" }{TEXT -1 8 " gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(t/5);" "6#-%$expG6#*&%\"tG\"\"\"\" \"&!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt+1/5;" "6#,&*&%#dvG\"\" \"%#dtG!\"\"F&*&F&F&\"\"&F(F&" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(t/5 )*v = 9.807*exp(t/5);" "6#/*&-%$expG6#*&%\"tG\"\"\"\"\"&!\"\"F*%\"vGF* *&-%&FloatG6$\"%2)*!\"$F*-F&6#*&F)F*F+F,F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "d/dt" "6#*&%\"dG\"\"\"%#dtG!\"\"" }{TEXT -1 1 " " } {XPPEDIT 18 0 "[exp(t/5)*v] = 9.807*exp(t/5);" "6#/7#*&-%$expG6#*&%\"t G\"\"\"\"\"&!\"\"F+%\"vGF+*&-%&FloatG6$\"%2)*!\"$F+-F'6#*&F*F+F,F-F+" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "exp(t/5)*v = 9.807*Int(exp(t/5), t);" "6#/*&-%$expG6#*&%\"tG\"\"\"\"\"&!\"\"F*%\"vGF**&-%&FloatG6$\"%2) *!\"$F*-%$IntG6$-F&6#*&F)F*F+F,F)F*" }{TEXT -1 1 " " }}{PARA 0 "" 0 " " {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "exp(t/5)*v = ``(9.807)*`.`*``(5)*exp(t/5)+c;" "6#/*&-%$ expG6#*&%\"tG\"\"\"\"\"&!\"\"F*%\"vGF*,&**-%!G6#-%&FloatG6$\"%2)*!\"$F *%\".GF*-F16#F+F*-F&6#*&F)F*F+F,F*F*%\"cGF*" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "exp(t/5) *v = 49.035*exp(t/5)+c;" "6#/*&-%$expG6#*&%\"tG\"\"\"\"\"&!\"\"F*%\"vG F*,&*&-%&FloatG6$\"&N!\\!\"$F*-F&6#*&F)F*F+F,F*F*%\"cGF*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "v=0" "6#/% \"vG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" } {TEXT -1 2 ", " }{XPPEDIT 18 0 "0 = 49.035+c;" "6#/\"\"!,&-%&FloatG6$ \"&N!\\!\"$\"\"\"%\"cGF+" }{TEXT -1 9 " so that " }{XPPEDIT 18 0 "c = \+ -49.035;" "6#/%\"cG,$-%&FloatG6$\"&N!\\!\"$!\"\"" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "exp(t/5)*v = 49.035*exp(t/5)-49.035;" "6#/*&-%$expG6 #*&%\"tG\"\"\"\"\"&!\"\"F*%\"vGF*,&*&-%&FloatG6$\"&N!\\!\"$F*-F&6#*&F) F*F+F,F*F*-F16$F3F4F," }{TEXT -1 3 ", " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = 49.035-49.035*exp(-t/5);" "6#/%\"vG,&-%&FloatG6$\"&N!\\!\"$\"\"\"*& -F'6$F)F*F+-%$expG6#,$*&%\"tGF+\"\"&!\"\"F6F+F6" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "v = 49.035*(1-exp(-t/5));" "6#/%\"vG*&-%&FloatG6$\"&N! \\!\"$\"\"\",&F+F+-%$expG6#,$*&%\"tGF+\"\"&!\"\"F4F4F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "(b) \+ " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "60*diff(v(t),t)=60*9.80 7-12*v(t),v(0)=0;\ndsolve(\{%\},v(t));\ng := unapply(rhs(%),t):\nplot( g(t),t=0..50,labels=[`t`,`velocity v(t)`]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/,$*&\"#g\"\"\"-%%diffG6$-%\"vG6#%\"tGF.F'F',&$\"'?%)e! \"$F'*&\"#7F'F+F'!\"\"/-F,6#\"\"!F9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/-%\"vG6#%\"tG,&#\"%2)*\"$+#\"\"\"*&#F*F+F,-%$expG6#,$*&\"\"&!\"\"F'F ,F5F,F5" }}{PARA 13 "" 1 "" {GLPLOT2D 364 296 296 {PLOTDATA 2 "6%-%'CU RVESG6$7Z7$$\"\"!F)F(7$$\"3\\LL$3FWYs#!#=$\"32PGjd\"f0g#!#<7$$\"3)pmm; a)G\\aF-$\"3[b/L+$)>j]F07$$\"3#*****\\7G$R<)F-$\"3/^K9#)>B&R(F07$$\"3S LLL3x&)*3\"F0$\"3\"yjZ:$oe.'*F07$$\"3)***\\i!R(*Rc\"F0$\"3]zrbtp4<8!#; 7$$\"3zmm\"H2P\"Q?F0$\"3;:Y:V4cT;FE7$$\"37+]PMnNrDF0$\"35iW@;-_r>FE7$$ \"3XLL$eRwX5$F0$\"3)3'*Q(yE5oAFE7$$\"3fLLL$eI8k$F0$\"3EY1.w%ej`#FE7$$ \"3=ML$3x%3yTF0$\"3+M#Rwk3tx#FE7$$\"3R+]PfyG7ZF0$\"3e$3Q+(zv#*HFE7$$\" 3gmm\"z%4\\Y_F0$\"3S(>+Udvj=$FE7$$\"34LLeR-/PiF0$\"3#o#oSLw(\\\\$FE7$$ \"3;++DcmpisF0$\"3)3i')\\m,iv$FE7$$\"3vLLe*)>VB$)F0$\"3M&zTaj7b(RFE7$$ \"3o++DJbw!Q*F0$\"3k(f^'G0R_TFE7$$\"3%ommTIOo/\"FE$\"3%*>%\\2[B#*H%FE7 $$\"3^LL3_>jU6FE$\"3bgbAqDe/WFE7$$\"3E++]i^Z]7FE$\"3&34mp`y8]%FE7$$\"3 /++](=h(e8FE$\"3;;uwM=ozXFE7$$\"3A++]P[6j9FE$\"3qB$Rpcy1k%FE7$$\"3[L$e *[z(yb\"FE$\"3;==+>W0'o%FE7$$\"3+nm;a/cq;FE$\"35'3kMaH*HZFE7$$\"3mmmm; t,mFE$\"3'f1Qej\\\"4[FE7$$\"3M+]i!f#=$3#FE$\"3:A?bIRXF[FE7$$\"37+](=xp e=#FE$\"394))GhCdT[FE7$$\"3-nm\"H28IH#FE$\"3IRC&)Gs^`[FE7$$\"3%om\"zpS S\"R#FE$\"3wI5QncWi[FE7$$\"3cLL3_?`(\\#FE$\"3Gc/6%*pHq[FE7$$\"3fL$e*)> pxg#FE$\"3i(pw')fmo([FE7$$\"3D+]Pf4t.FFE$\"3GeW#)*e<:)[FE7$$\"3ZLLe*Gs t!GFE$\"3u;Mg>Hj&)[FE7$$\"39+++DRW9HFE$\"3mLX-3r2*)[FE7$$\"3:++DJE>>IF E$\"3c$orP:.=*[FE7$$\"35+]i!RU07$FE$\"3!y?S&\\#\\R*[FE7$$\"3$)***\\(=S 2LKFE$\"3-$egK0ue*[FE7$$\"3nmmm\"p)=MLFE$\"3Ogl^1.F(*[FE7$$\"3U++](=]@ W$FE$\"3UjN[H,[)*[FE7$$\"36L$e*[$z*RNFE$\"3G&\\$)[>s$**[FE7$$\"3e++]iC $pk$FE$\"3g-ao5r;+\\FE7$$\"3Sm;H2qcZPFE$\"3))Q#R=su2!\\FE7$$\"3Y+]7.\" fF&QFE$\"3sH$**4y\"H,\\FE7$$\"3amm;/OgbRFE$\"3Yn&\\$>Bq,\\FE7$$\"3I+]i lAFjSFE$\"3xmD1&e]?!\\FE7$$\"3)RLLL)*pp;%FE$\"3Zs\\1j?K-\\FE7$$\"3WLL3 xe,tUFE$\"3&*fZ1yra-\\FE7$$\"3Wn;HdO=yVFE$\"3vbTn;zs-\\FE7$$\"3a+++D># [Z%FE$\"3?%yCsgjG!\\FE7$$\"3)om;aG!e&e%FE$\"3-DK'o0!*H!\\FE7$$\"3wLLL$ )Qk%o%FE$\"3(yuJ7r\"3.\\FE7$$\"3m+]iSjE!z%FE$\"3aj,bj8;.\\FE7$$\"3u+]P 40O\"*[FE$\"3L[@)RNBK!\\FE7$$\"#]F)$\"394WW\"QxK!\\FE-%'COLOURG6&%$RGB G$\"#5!\"\"F(F(-%+AXESLABELSG6$%\"tG%.velocity~v(t)G-%%VIEWG6$;F(Ff\\l %(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "C urve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "(c) Since " }{XPPEDIT 18 0 "exp(-t/5);" "6#-%$expG6#,$*&%\"tG\"\" \"\"\"&!\"\"F+" }{TEXT -1 15 " tends to 0 as " }{XPPEDIT 18 0 "t->infi nity" "6#f*6#%\"tG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" } {TEXT -1 53 ", the terminal velocity is 49.035 metres per second. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 13 "Second model " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "In many situations it may be m ore realistic to suppose that the force resisting the motion is propor tional to the " }{TEXT 259 22 "square of the velocity" }{TEXT -1 1 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 134 167 167 {PLOTDATA 2 "6*-%)POLYGONSG6$7U7$$\"\"&!\"\"$\"\"!F,7$$\"+1Ndg\\!#5$\"+!ohmE'!#6 7$$\"+1e\"H%[F0$\"+O%\\MC\"F07$$\"+IC))[YF0$\"+kFiS=F07$$\"++M`\"Q%F0$ \"+q$o(3CF07$$\"+s\\3XSF0$\"+ii#*QHF07$$\"+OJ%[k$F0$\"+IbtAMF07$$\"+[* >r=$F0$\"+9ic_QF07$$\"+wR8zEF0$\"+F'R;A%F07$$\"+fk*)G@F0$\"+i_8CXF07$$ \"+p\\3X:F0$\"+#e#GbZF07$$\"+5d1p$*F3$\"+aiV6\\F07$$\"+if_RJF3$\"+UO8! *\\F07$$!+$)f_RJF3Fco7$$!+Id1p$*F3F^o7$$!+r\\3X:F0Fin7$$!+ck*)G@F0$\"+ j_8CXF07$$!+tR8zEF0$\"+G'R;A%F07$$!+Y*>r=$F0$\"+;ic_QF07$$!+UJ%[k$F0$ \"+DbtAMF07$$!+v\\3XSF0$\"+di#*QHF07$$!+-M`\"Q%F0$\"+m$o(3CF07$$!+JC)) [YF0$\"+gFiS=F07$$!+1e\"H%[F0$\"+K%\\MC\"F07$$!+2Ndg\\F0$\"+];mmiF37$$ !\"&F*$!+3Q.^?!#>7$$!+1Ndg\\F0$!+!phmE'F37$Fbr$!+O%\\MC\"F07$$!+IC))[Y F0$!+kFiS=F07$$!++M`\"Q%F0$!+q$o(3CF07$$!+t\\3XSF0$!+gi#*QHF07$$!+QJ%[ k$F0$!+GbtAMF07$$!+W*>r=$F0$!+=ic_QF07$$!+qR8zEF0$!+J'R;A%F07$$!+`k*)G @F0$!+k_8CXF07$$!+n\\3X:F0$!+$e#GbZF07$$!+!pl!p$*F3$!+aiV6\\F07$$!+Uf_ RJF3$!+UO8!*\\F07$$\"+/g_RJF3Fiv7$$\"+]d1p$*F3Fdv7$$\"+t\\3X:F0$!+\"e# GbZF07$$\"+ek*)G@F0$!+i_8CXF07$$\"+vR8zEF0$!+G'R;A%F07$FN$!+9ic_QF07$$ \"+VJ%[k$F0$!+CbtAMF07$$\"+w\\3XSF0$!+bi#*QHF07$$\"+/M`\"Q%F0$!+k$o(3C F07$$\"+KC))[YF0$!+eFiS=F07$$\"+2e\"H%[F0$!+I%\\MC\"F07$$\"+2Ndg\\F0$! +I;mmiF37$F($\"+:w1-TF`s-%&COLORG6&%$RGBG$\"\"(F*FhzFhz-%'CURVESG6'7$7 $F+F(7$F+$\"#IF*7%7$$!+++++IF0$\"#DF*F_[l7$$\"+++++IF0Ff[l-%&STYLEG6#% ,PATCHNOGRIDG-%'COLOURG6&FgzF,F,F,-%*THICKNESSG6#\"\"#-F[[l6'7$7$F+F\\ s7$F+$!#IF*7%7$Fi[l$!#DF*Fj\\l7$Fd[lF_]lF[\\lF_\\lFb\\l-%%TEXTG6$7$$\" #9F*$!#AF*Q#mg6\"-Fc]l6$7$Ff]l$\"#AF*Q#kvF[^l-Fc]l6%7$$Fe\\lF,Ff[lQ\"2 F[^l-%%FONTG6$%*HELVETICAG\"\")-%(SCALINGG6#%,CONSTRAINEDG-%*AXESSTYLE G6#%%NONEG" 1 2 0 1 10 0 2 9 1 1 1 1.000000 44.000000 45.000000 0 0 "C urve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" }}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 65 "The differential equation governi ng the motion then has the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {TEXT 383 1 "m" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = m*g-k*v^2;" "6 #/*&%#dvG\"\"\"%#dtG!\"\",&*&%\"mGF&%\"gGF&F&*&%\"kGF&*$%\"vG\"\"#F&F( " }{TEXT -1 17 " ------- (iii), " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt=g-k*v^2/m" "6 #/*&%#dvG\"\"\"%#dtG!\"\",&%\"gGF&*(%\"kGF&*$%\"vG\"\"#F&%\"mGF(F(" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "This differential equation is not linear but it can be so lved by separating the variables. " }}{PARA 0 "" 0 "" {TEXT -1 21 "It \+ can be written as " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt=``(-k/m)*(v^2-a^2)" "6#/*&%#dvG\"\"\"%#dtG!\"\"*&-%!G6#,$*&%\"k GF&%\"mGF(F(F&,&*$%\"vG\"\"#F&*$%\"aGF4F(F&" }{TEXT -1 16 " ------- ( iv), " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "a=sqrt(m* g/k)" "6#/%\"aG-%%sqrtG6#*(%\"mG\"\"\"%\"gGF*%\"kG!\"\"" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/(v^2-a^2)" "6#*&\"\"\"F$,&*$%\"vG\"\"#F$*$%\"a GF(!\"\"F+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt =-k/m" "6#/*&%#dvG\" \"\"%#dtG!\"\",$*&%\"kGF&%\"mGF(F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 " " {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(1/(v^2-a^2),v)=-Int(k/m,t)" "6#/-%$IntG6$*&\"\"\"F(,&*$%\"vG \"\"#F(*$%\"aGF,!\"\"F/F+,$-F%6$*&%\"kGF(%\"mGF/%\"tGF/" }{TEXT -1 13 " ----- (v). " }}{PARA 0 "" 0 "" {TEXT -1 4 "Now " }{XPPEDIT 18 0 "1/ (v-a) -1/(v+a)=(``(v+a)-(v-a))/((v-a)*(v+a))" "6#/,&*&\"\"\"F&,&%\"vGF &%\"aG!\"\"F*F&*&F&F&,&F(F&F)F&F*F**&,&-%!G6#,&F(F&F)F&F&,&F(F&F)F*F*F &*&,&F(F&F)F*F&,&F(F&F)F&F&F*" }{XPPEDIT 18 0 "``=2*a/(v^2-a^2)" "6#/% !G*(\"\"#\"\"\"%\"aGF',&*$%\"vGF&F'*$F(F&!\"\"F-" }{TEXT -1 44 ", so w e have the partial fraction expansion " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/(v^2-a^2)=1/(2*a)" "6#/*&\"\"\"F%,&*$%\"vG\"\" #F%*$%\"aGF)!\"\"F,*&F%F%*&F)F%F+F%F," }{XPPEDIT 18 0 "``(1/(v-a) -1/( v+a))" "6#-%!G6#,&*&\"\"\"F(,&%\"vGF(%\"aG!\"\"F,F(*&F(F(,&F*F(F+F(F,F ," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 18 "Hence (v) becomes \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/(2*a)" "6#*&\" \"\"F$*&\"\"#F$%\"aGF$!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(``(1/ (v-a)-1/(v+a)),v)=-Int(k/m,t)" "6#/-%$IntG6$-%!G6#,&*&\"\"\"F,,&%\"vGF ,%\"aG!\"\"F0F,*&F,F,,&F.F,F/F,F0F0F.,$-F%6$*&%\"kGF,%\"mGF0%\"tGF0" } {TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(``(1/(v-a)-1/(v+a)),v) = -Int(b,t )" "6#/-%$IntG6$-%!G6#,&*&\"\"\"F,,&%\"vGF,%\"aG!\"\"F0F,*&F,F,,&F.F,F /F,F0F0F.,$-F%6$%\"bG%\"tGF0" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "b = 2*a*k/m;" "6#/%\"bG**\"\"#\"\" \"%\"aGF'%\"kGF'%\"mG!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 16 "It follows that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(abs(v-a))-ln(abs(v+a)) = -b* t+c;" "6#/,&-%#lnG6#-%$absG6#,&%\"vG\"\"\"%\"aG!\"\"F--F&6#-F)6#,&F,F- F.F-F/,&*&%\"bGF-%\"tGF-F/%\"cGF-" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(abs((v-a)/(v+a))) = -b*t+c;" "6#/-%#lnG6#-%$absG6#*&,&%\"vG\" \"\"%\"aG!\"\"F-,&F,F-F.F-F/,&*&%\"bGF-%\"tGF-F/%\"cGF-" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 59 "Converting to the corresponding exp onential equation gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "(v-a)/(v+a) = A*exp(-b*t);" "6#/*&,&%\"vG\"\"\"%\"aG!\" \"F',&F&F'F(F'F)*&%\"AGF'-%$expG6#,$*&%\"bGF'%\"tGF'F)F'" }{TEXT -1 15 " ------- (vi), " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "A=``" "6#/%\"AG%!G" }{TEXT 384 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(c);" "6#-%$expG6#%\"cG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 27 "Equation (vi) has the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(v-a)/(v+a)=E" "6#/*&,&%\"vG\"\"\"%\"aG!\"\"F ',&F&F'F(F'F)%\"EG" }{TEXT -1 17 " ------- (vii), " }}{PARA 0 "" 0 " " {TEXT -1 6 "where " }{XPPEDIT 18 0 "E = A*exp(-b*t);" "6#/%\"EG*&%\" AG\"\"\"-%$expG6#,$*&%\"bGF'%\"tGF'!\"\"F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 32 "We can solve equation (vii) for " }{TEXT 385 1 "v " }{TEXT -1 13 " in terms of " }{TEXT 395 1 "E" }{TEXT -1 5 " and " } {TEXT 396 1 "a" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 7 "Indeed \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v-a=E*v+E*a" "6#/ ,&%\"vG\"\"\"%\"aG!\"\",&*&%\"EGF&F%F&F&*&F+F&F'F&F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v-E*v = a+E*a;" "6#/,&%\"vG\"\"\"*&%\"EGF&F%F&! \"\",&%\"aGF&*&F(F&F+F&F&" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v(1-E) \+ = a*(1+E);" "6#/-%\"vG6#,&\"\"\"F(%\"EG!\"\"*&%\"aGF(,&F(F(F)F(F(" } {TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = a*``((1+E)/(1-E));" "6# /%\"vG*&%\"aG\"\"\"-%!G6#*&,&F'F'%\"EGF'F',&F'F'F-!\"\"F/F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 26 "Hence equation (vi) gives " } }{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = a*``((1+A*exp(-b *t))/(1-A*exp(-b*t)));" "6#/%\"vG*&%\"aG\"\"\"-%!G6#*&,&F'F'*&%\"AGF'- %$expG6#,$*&%\"bGF'%\"tGF'!\"\"F'F'F',&F'F'*&F.F'-F06#,$*&F4F'F5F'F6F' F6F6F'" }{TEXT -1 19 " ------- (viii). " }}{PARA 0 "" 0 "" {TEXT -1 3 "If " }{XPPEDIT 18 0 "v=v[0]" "6#/%\"vG&F$6#\"\"!" }{TEXT -1 6 " whe n " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 22 ", equation (vi) gives " }{XPPEDIT 18 0 "A=(v[0]-a)/(v[0]+a)" "6#/%\"AG*&,&&%\"vG6#\" \"!\"\"\"%\"aG!\"\"F+,&&F(6#F*F+F,F+F-" }{TEXT -1 3 ". " }}{PARA 0 " " 0 "" {TEXT -1 22 "Hence the solution is " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = a*``((1+``((v[0]-a)/(v[0]+a))*exp(-b*t))/ (1-``((v[0]-a)/(v[0]+a))*exp(-b*t)));" "6#/%\"vG*&%\"aG\"\"\"-%!G6#*&, &F'F'*&-F)6#*&,&&F$6#\"\"!F'F&!\"\"F',&&F$6#F4F'F&F'F5F'-%$expG6#,$*&% \"bGF'%\"tGF'F5F'F'F',&F'F'*&-F)6#*&,&&F$6#F4F'F&F5F',&&F$6#F4F'F&F'F5 F'-F:6#,$*&F>F'F?F'F5F'F5F5F'" }{TEXT -1 2 ", " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 398 18 "__________________" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "a = sqrt(m*g/k)" " 6#/%\"aG-%%sqrtG6#*(%\"mG\"\"\"%\"gGF*%\"kG!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "b=2*a*k/m" "6#/%\"bG**\"\"#\"\"\"%\"aGF'%\"kGF'%\"mG! \"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 3 "As " }{XPPEDIT 18 0 "t->infinity" "6#f*6#%\"tG7\"6$%)operatorG%&arrowG6\"%)infinityGF *F*F*" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "exp(-b*t);" "6#-%$expG6#,$*&% \"bG\"\"\"%\"tGF)!\"\"" }{TEXT -1 20 " tends to 0 so that " }{XPPEDIT 18 0 "v->a" "6#f*6#%\"vG7\"6$%)operatorG%&arrowG6\"%\"aGF*F*F*" } {TEXT -1 33 ". Hence the terminal velocity is " }{XPPEDIT 18 0 "a = sq rt(m*g/k)" "6#/%\"aG-%%sqrtG6#*(%\"mG\"\"\"%\"gGF*%\"kG!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 33 "This can also be seen by putt ing " }{XPPEDIT 18 0 "dv/dt=0" "6#/*&%#dvG\"\"\"%#dtG!\"\"\"\"!" } {TEXT -1 61 " in the original differential equation (iii) and solving \+ for " }{TEXT 394 1 "v" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 34 "Note that if the initial velocity " } {XPPEDIT 18 0 "v[0]" "6#&%\"vG6#\"\"!" }{TEXT -1 12 " is 0, then " } {XPPEDIT 18 0 "A=-1" "6#/%\"AG,$\"\"\"!\"\"" }{TEXT -1 51 " in (viii) so the solution has the simpler form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = a*``((1-exp(-b*t))/(1+exp(-b*t)));" "6#/% \"vG*&%\"aG\"\"\"-%!G6#*&,&F'F'-%$expG6#,$*&%\"bGF'%\"tGF'!\"\"F4F',&F 'F'-F.6#,$*&F2F'F3F'F4F'F4F'" }{TEXT -1 3 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 386 11 "___________" }{TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT -1 15 "Note that with " }{XPPEDIT 18 0 "a = sqrt(m*g/k)" "6#/%\"aG-%%sqrtG6#*(%\"mG\"\"\"%\"gGF*%\"kG!\"\"" }{TEXT -1 5 " and \+ " }{XPPEDIT 18 0 "b=2*a*k/m" "6#/%\"bG**\"\"#\"\"\"%\"aGF'%\"kGF'%\"mG !\"\"" }{TEXT -1 55 " the original differential equation (iii) has the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "2*a" "6#*& \"\"#\"\"\"%\"aGF%" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = -b*(v^2-a^ 2)" "6#/*&%#dvG\"\"\"%#dtG!\"\",$*&%\"bGF&,&*$%\"vG\"\"#F&*$%\"aGF/F(F &F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "m := 'm': a := 'a': v := 'v':\n2*a*diff( v(t),t)=-b*(v(t)^2-a^2),v(0)=0;\ndsolve(\{%\},v(t)):\nsimplify(convert (expand(%),exp));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/,$*(\"\"#\"\"\"- %%diffG6$-%\"vG6#%\"tGF.F'%\"aGF'F',$*&%\"bGF',&*$)F+F&F'F'*$)F/F&F'! \"\"F'F8/-F,6#\"\"!F<" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"t G*(%\"aG\"\"\",&-%$expG6#*&F'F*%\"bGF*F*F*!\"\"F*,&F*F*F,F*F1" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 2" }}{PARA 0 "" 0 "" {TEXT 389 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 27 "A skydiver with a mass of " }{XPPEDIT 18 0 "3600/49" "6#*&\"%+ O\"\"\"\"#\\!\"\"" }{TEXT -1 2 " " }{TEXT 397 1 "~" }{TEXT -1 134 " \+ 73.4694 kg jumps from a high-flying plane. Due to air resistance he ex periences a resisting force which opposes the motion given by " } {XPPEDIT 18 0 "v^2/5;" "6#*&%\"vG\"\"#\"\"&!\"\"" }{TEXT -1 16 " Newto ns, where " }{TEXT 387 1 "v" }{TEXT -1 55 " is his vertical velocity d ownwards in metres per sec. " }}{PARA 0 "" 0 "" {TEXT -1 78 "Assume th at at the instant he leaves the plane his vertical velocity is zero. \+ " }}{PARA 0 "" 0 "" {TEXT -1 36 "(a) Find a formula for the velocity \+ " }{TEXT 390 1 "v" }{TEXT -1 33 " downwards as a function of time " } {TEXT 388 1 "t" }{TEXT -1 13 " in seconds. " }}{PARA 0 "" 0 "" {TEXT -1 43 "(b) Sketch a graph of velocity versus time " }{TEXT 391 1 "t" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 32 "(c) Find his terminal ve locity. " }}{PARA 0 "" 0 "" {TEXT -1 57 "Take the gravitational accele ration to be 9.8 metres per " }{XPPEDIT 18 0 "sec^2" "6#*$%$secG\"\"# " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 392 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 161 "As before the problem could be solved by substituting appropriate numbers in the general expression given above. Nevertheless we go thr ough all the steps again. " }}{PARA 0 "" 0 "" {TEXT -1 74 "(a) The dif ferential equation which governs the motion of the skydiver is " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``(3600/49);" "6#-%!G 6#*&\"%+O\"\"\"\"#\\!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = ``( 3600/49)*`.`*``(9.8)-v^2/5;" "6#/*&%#dvG\"\"\"%#dtG!\"\",&*(-%!G6#*&\" %+OF&\"#\\F(F&%\".GF&-F,6#-%&FloatG6$\"#)*F(F&F&*&%\"vG\"\"#\"\"&F(F( " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = 9.8-49*v^2/18000;" "6#/*&%#dvG\"\"\"%#dtG!\"\",&-%&FloatG6$\"#)*F(F&*(\"#\\F&*$%\"vG\"\"# F&\"&+!=F(F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = -49/18000;" " 6#/*&%#dvG\"\"\"%#dtG!\"\",$*&\"#\\F&\"&+!=F(F(" }{XPPEDIT 18 0 "``(v^ 2-3600);" "6#-%!G6#,&*$%\"vG\"\"#\"\"\"\"%+O!\"\"" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 32 "Separating the variables gives: " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/(v^2-3600);" "6#*& \"\"\"F$,&*$%\"vG\"\"#F$\"%+O!\"\"F*" }{TEXT -1 1 " " }{XPPEDIT 18 0 " dv/dt = -49/18000;" "6#/*&%#dvG\"\"\"%#dtG!\"\",$*&\"#\\F&\"&+!=F(F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(1/(v^2-3600),v) = -Int(49/18 000,t);" "6#/-%$IntG6$*&\"\"\"F(,&*$%\"vG\"\"#F(\"%+O!\"\"F.F+,$-F%6$* &\"#\\F(\"&+!=F.%\"tGF." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "Now " }{XPPEDIT 18 0 "1/(v-60)-1/(v +60) = 120/(v^2-3600" "6#/,&*&\"\"\"F&,&%\"vGF&\"#g!\"\"F*F&*&F&F&,&F( F&F)F&F*F**&\"$?\"F&,&*$F(\"\"#F&\"%+OF*F*" }{TEXT -1 4 " so " } {XPPEDIT 18 0 "1/(v^2-3600)=1/120" "6#/*&\"\"\"F%,&*$%\"vG\"\"#F%\"%+O !\"\"F+*&F%F%\"$?\"F+" }{XPPEDIT 18 0 "``(1/(v-60)-1/(v+60))" "6#-%!G6 #,&*&\"\"\"F(,&%\"vGF(\"#g!\"\"F,F(*&F(F(,&F*F(F+F(F,F," }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 15 "Thus we obtain " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "1/120" "6#*&\"\"\"F$\"$?\"!\"\"" } {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(``(1/(v-60)-1/(v+60)),v) = -Int(49/ 18000,t);" "6#/-%$IntG6$-%!G6#,&*&\"\"\"F,,&%\"vGF,\"#g!\"\"F0F,*&F,F, ,&F.F,F/F,F0F0F.,$-F%6$*&\"#\\F,\"&+!=F0%\"tGF0" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "Int(``(1/(v-60)-1/(v+60)),v)=-Int(49/150,t)" "6#/-%$Int G6$-%!G6#,&*&\"\"\"F,,&%\"vGF,\"#g!\"\"F0F,*&F,F,,&F.F,F/F,F0F0F.,$-F% 6$*&\"#\\F,\"$]\"F0%\"tGF0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(ab s(v-60))-ln(abs(v+60))=-49*t/150+c" "6#/,&-%#lnG6#-%$absG6#,&%\"vG\"\" \"\"#g!\"\"F--F&6#-F)6#,&F,F-F.F-F/,&*(\"#\\F-%\"tGF-\"$]\"F/F/%\"cGF- " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(abs((v-60)/(v+60)))= -49*t/150+ c" "6#/-%#lnG6#-%$absG6#*&,&%\"vG\"\"\"\"#g!\"\"F-,&F,F-F.F-F/,&*(\"# \\F-%\"tGF-\"$]\"F/F/%\"cGF-" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(v-60)/(v+60)=A*exp(-49*t/150)" "6#/*&,&%\"vG\"\"\"\"#g!\"\"F',&F&F 'F(F'F)*&%\"AGF'-%$expG6#,$*(\"#\\F'%\"tGF'\"$]\"F)F)F'" }{TEXT -1 2 " , " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "A=``" "6#/% \"AG%!G" }{TEXT 393 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(c)" "6#- %$expG6#%\"cG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "v=0" "6#/%\"vG\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 13 " we see that " }{XPPEDIT 18 0 "A=-1" "6#/%\"AG,$\"\"\"!\"\"" }{TEXT -1 13 " which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(v-60)/(v+60) = -exp(-49*t/ 150)" "6#/*&,&%\"vG\"\"\"\"#g!\"\"F',&F&F'F(F'F),$-%$expG6#,$*(\"#\\F' %\"tGF'\"$]\"F)F)F)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 5 "Th en " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v-60=-v*exp(-4 9*t/150)-60*exp(-49*t/150)" "6#/,&%\"vG\"\"\"\"#g!\"\",&*&F%F&-%$expG6 #,$*(\"#\\F&%\"tGF&\"$]\"F(F(F&F(*&F'F&-F,6#,$*(F0F&F1F&F2F(F(F&F(" } {TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 7 "giving " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v+ v*exp(-49*t/150)=60-60*exp(-49* t/150)" "6#/,&%\"vG\"\"\"*&F%F&-%$expG6#,$*(\"#\\F&%\"tGF&\"$]\"!\"\"F 0F&F&,&\"#gF&*&F2F&-F)6#,$*(F-F&F.F&F/F0F0F&F0" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 16 "It follows that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v = 60*``((1-exp(-49*t/150))/(1+exp(-49 *t/150)));" "6#/%\"vG*&\"#g\"\"\"-%!G6#*&,&F'F'-%$expG6#,$*(\"#\\F'%\" tGF'\"$]\"!\"\"F5F5F',&F'F'-F.6#,$*(F2F'F3F'F4F5F5F'F5F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "(b) " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 166 "3600/49*diff(v(t),t)=3 600/49*9.8-v(t)^2/5,v(0)=0;\ndsolve(\{%\},v(t)):\nsimplify(convert(%,e xp));\ng := unapply(rhs(%),t):\nplot(g(t),t=0..50,labels=[`t`,`velocit y v(t)`]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/,$*&#\"%+O\"#\\\"\"\"-% %diffG6$-%\"vG6#%\"tGF0F)F),&$\"+++++s!\"(F)*&#F)\"\"&F)*$)F-\"\"#F)F) !\"\"/-F.6#\"\"!F?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"tG,$ *(\"#g\"\"\",&-%$expG6#,$*(\"#\\F+\"$]\"!\"\"F'F+F+F+F+F4F+,&F+F+F-F+F 4F+" }}{PARA 13 "" 1 "" {GLPLOT2D 364 296 296 {PLOTDATA 2 "6%-%'CURVES G6$7Y7$$\"\"!F)F(7$$\"3)pmm;a)G\\a!#=$\"3b=twMbCE`!#<7$$\"3SLLL3x&)*3 \"F0$\"3UTA:8.#p0\"!#;7$$\"3)***\\i!R(*Rc\"F0$\"34VS`LbA+:F67$$\"3zmm \"H2P\"Q?F0$\"3FF67$$\"37+]PMnNrDF0$\"3yRiF$GB:Q#F67$$\"3XLL$e RwX5$F0$\"3c]XYVb+1GF67$$\"3fLLL$eI8k$F0$\"3)H9tWPA)*>$F67$$\"3=ML$3x% 3yTF0$\"3QA!fkPW&eNF67$$\"3R+]PfyG7ZF0$\"3g[jSzzV!)QF67$$\"3gmm\"z%4\\ Y_F0$\"3'RDRI_D!oTF67$$\"3S***\\Pfl`S%F67$$\"34LLeR- /PiF0$\"33=MOK'fgh%F67$$\"3;++DcmpisF0$\"3+pC7XaXw\\F67$$\"3vLLe*)>VB$ )F0$\"3a7nJ'y`wD&F67$$\"3o++DJbw!Q*F0$\"3+>?,PMzkaF67$$\"3%ommTIOo/\"F 6$\"3rp#*4J?w>cF67$$\"3^LL3_>jU6F6$\"3H*zv\\QW&>dF67$$\"3E++]i^Z]7F6$ \"3!p#R+.+W,eF67$$\"3/++](=h(e8F6$\"3YF)*RV!3*feF67$$\"3A++]P[6j9F6$\" 3'=$pcp\"Q+!fF67$$\"3[L$e*[z(yb\"F6$\"3@X(oh'z[EfF67$$\"3+nm;a/cq;F6$ \"3)=F&e'zH!\\fF67$$\"3mmmm;t,mF6$\"3'HqR\"*Q;6)fF67$$\"3M+]i!f#=$3#F6$ \"3#[91r\"pr')fF67$$\"37+](=xpe=#F6$\"3)e+MLL*\\!*fF67$$\"3-nm\"H28IH# F6$\"3A[!ptV.L*fF67$$\"3%om\"zpSS\"R#F6$\"3g5+(pSV^*fF67$$\"3cLL3_?`( \\#F6$\"3S$fMY%ec'*fF67$$\"3fL$e*)>pxg#F6$\"3y9!RD7/w*fF67$$\"3D+]Pf4t .FF6$\"3M1$pGv[#)*fF67$$\"3ZLLe*Gst!GF6$\"3kf]!ym^()*fF67$$\"39+++DRW9 HF6$\"3ESWY$3?\"**fF67$$\"3:++DJE>>IF6$\"3kJ)H:0v$**fF67$$\"35+]i!RU07 $F6$\"3y9vC'=^&**fF67$$\"3$)***\\(=S2LKF6$\"35)**o9C*o**fF67$$\"3nmmm \"p)=MLF6$\"3!Giualw(**fF67$$\"3U++](=]@W$F6$\"3WKKBJI%)**fF67$$\"36L$ e*[$z*RNF6$\"39D*\\%of))**fF67$$\"3e++]iC$pk$F6$\"3CemF$f>***fF67$$\"3 Sm;H2qcZPF6$\"3Os&z17U***fF67$$\"3Y+]7.\"fF&QF6$\"3!e)4L_*e***fF67$$\" 3amm;/OgbRF6$\"3!eCZ^mq***fF67$$\"3I+]ilAFjSF6$\"3cu)RNOz***fF67$$\"3) RLLL)*pp;%F6$\"3A?-9$H&)***fF67$$\"3WLL3xe,tUF6$\"39CJ/*f*)***fF67$$\" 3Wn;HdO=yVF6$\"3SWf7BE****fF67$$\"3a+++D>#[Z%F6$\"3U5]7?Y****fF67$$\"3 )om;aG!e&e%F6$\"3i:!yLD'****fF67$$\"3wLLL$)Qk%o%F6$\"3[59>*G(****fF67$ $\"3m+]iSjE!z%F6$\"3u,T@!3)****fF67$$\"3u+]P40O\"*[F6$\"3#)3G:?')****f F67$$\"#]F)$\"3oj.QK!*****fF6-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLA BELSG6$%\"tG%.velocity~v(t)G-%%VIEWG6$;F(Fa\\l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "(c) Since " } {XPPEDIT 18 0 "exp(-49*t/150);" "6#-%$expG6#,$*(\"#\\\"\"\"%\"tGF)\"$] \"!\"\"F," }{TEXT -1 15 " tends to 0 as " }{XPPEDIT 18 0 "t->infinity " "6#f*6#%\"tG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 49 ", the terminal velocity is 60 metres per second. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 42 "Populat ion growth - the logistic equation " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 40 "Simplest mo del giving exponential growth" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 261 "A simple model for the growth of a population of organisms, from the lowliest bacteria, to complex \+ creatures such as ourselves, is given by assuming that the rate of cha nge of the population is proportional to the size of the population at a particular instant." }}{PARA 0 "" 0 "" {TEXT -1 37 "This gives the \+ differential equation:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt=k*x" "6#/*&%#dxG\"\"\"%#dtG!\"\"*&%\"kGF&%\"xGF&" }{TEXT -1 13 " ------- (i)," }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 285 1 "x" }{TEXT -1 39 " is the size of the population at time " } {TEXT 270 1 "t" }{TEXT -1 6 ", and " }{TEXT 269 1 "k" }{TEXT -1 15 " i s a constant." }}{PARA 0 "" 0 "" {TEXT -1 63 "This equation can be sol ved by separating the variables to give" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/x" "6#*&\"\"\"F$%\"xG!\"\"" }{TEXT -1 1 " " } {XPPEDIT 18 0 "dx/dt= k" "6#/*&%#dxG\"\"\"%#dtG!\"\"%\"kG" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 5 " so " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(1/x,x)=Int(k,t)" "6#/-%$IntG6$*&\"\"\"F(% \"xG!\"\"F)-F%6$%\"kG%\"tG" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(x)=k *t +C[1]" "6#/-%#lnG6#%\"xG,&*&%\"kG\"\"\"%\"tGF+F+&%\"CG6#F+F+" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "x = exp(k*t+C[1])" "6#/%\"xG-%$expG6# ,&*&%\"kG\"\"\"%\"tGF+F+&%\"CG6#F+F+" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x = C[2]*exp(k*t)" "6#/%\"xG*&&%\"CG6#\"\"#\"\"\"-%$expG6#*&%\"kGF*% \"tGF*F*" }{TEXT -1 16 " ------- (ii), " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "C[2]= exp(C[1])" "6#/&%\"CG6#\"\"#-%$expG6# &F%6#\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 41 "If the ini tial size of the population is " }{XPPEDIT 18 0 "x[0]" "6#&%\"xG6#\"\" !" }{TEXT -1 9 " at time " }{XPPEDIT 18 0 "t = 0" "6#/%\"tG\"\"!" } {TEXT -1 14 ", we see that " }{XPPEDIT 18 0 "C[2]=x[0]" "6#/&%\"CG6#\" \"#&%\"xG6#\"\"!" }{TEXT -1 21 ", and so (ii) becomes" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x = x[0]*exp(k*t)" "6#/%\"xG*&&F$6 #\"\"!\"\"\"-%$expG6#*&%\"kGF)%\"tGF)F)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 85 "This is the classic case of exponential growth, as f ormulated by T.R.Malthus in 1798." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 40 "The logistic model for population growth" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 202 " The model for population growth given in the previous subsection is no t very realistic, because in most situations there will be factors whi ch tend to inhibit growth if the population becomes too large." }} {PARA 0 "" 0 "" {TEXT -1 68 "A more realistic model in some situations is given by the so called " }{TEXT 259 17 "logistic equation" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt = k* x*(a-x);" "6#/*&%#dxG\"\"\"%#dtG!\"\"*(%\"kGF&%\"xGF&,&%\"aGF&F+F(F&" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 289 1 "x " }{TEXT -1 27 " is the population at time " }{TEXT 286 2 "t," }{TEXT -1 5 " and " }{TEXT 287 1 "k" }{TEXT -1 5 " and " }{TEXT 288 1 "a" } {TEXT -1 15 " are constants." }}{PARA 0 "" 0 "" {TEXT -1 12 "The facto r (" }{XPPEDIT 18 0 "a-x;" "6#,&%\"aG\"\"\"%\"xG!\"\"" }{TEXT -1 17 ") tends to 0, as " }{TEXT 271 1 "x" }{TEXT -1 10 " tends to " }{TEXT 272 1 "a" }{TEXT -1 118 ". This causes the rate of growth of the popul ation also to tend to 0, as the population approaches the critical val ue " }{XPPEDIT 18 0 "a=x[crit]" "6#/%\"aG&%\"xG6#%%critG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 72 "This differential equation is not \+ linear because the dependent variable " }{TEXT 273 1 "x" }{TEXT -1 25 " occurs with the power 2." }}{PARA 0 "" 0 "" {TEXT -1 16 "However, we can " }{TEXT 259 22 "separate the variables" }{TEXT -1 22 " to give t he equation:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/(x* (x-a));" "6#*&\"\"\"F$*&%\"xGF$,&F&F$%\"aG!\"\"F$F)" }{TEXT -1 1 " " } {XPPEDIT 18 0 "dx/dt = -k;" "6#/*&%#dxG\"\"\"%#dtG!\"\",$%\"kGF(" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "Hence" }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(1/(x*(x-a)),x) = -Int(k,t);" "6#/ -%$IntG6$*&\"\"\"F(*&%\"xGF(,&F*F(%\"aG!\"\"F(F-F*,$-F%6$%\"kG%\"tGF- " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "1/(x-a)-1/x=(x-(x-a))/(x*(x-a))" "6#/,&*&\"\"\"F&,&%\"xGF&%\"aG!\" \"F*F&*&F&F&F(F*F**&,&F(F&,&F(F&F)F*F*F&*&F(F&,&F(F&F)F*F&F*" } {XPPEDIT 18 0 "``=a/(x*(x-a))" "6#/%!G*&%\"aG\"\"\"*&%\"xGF',&F)F'F&! \"\"F'F+" }{TEXT -1 48 ", we obtain the partial fraction decomposition : " }{XPPEDIT 18 0 "1/(x*(x-a)) = 1/a;" "6#/*&\"\"\"F%*&%\"xGF%,&F'F%% \"aG!\"\"F%F**&F%F%F)F*" }{TEXT -1 1 " " }{XPPEDIT 18 0 "``(1/(x-a)-1/ x);" "6#-%!G6#,&*&\"\"\"F(,&%\"xGF(%\"aG!\"\"F,F(*&F(F(F*F,F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1/a " "6#*&\"\"\"F$%\"aG!\"\"" } {XPPEDIT 18 0 "Int(``(1/(x-a)-1/x),x) = -Int(k,t);" "6#/-%$IntG6$-%!G6 #,&*&\"\"\"F,,&%\"xGF,%\"aG!\"\"F0F,*&F,F,F.F0F0F.,$-F%6$%\"kG%\"tGF0 " }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(``(1/(x-a)-1/x),x)=-Int(k*a,t) " "6#/-%$IntG6$-%!G6#,&*&\"\"\"F,,&%\"xGF,%\"aG!\"\"F0F,*&F,F,F.F0F0F. ,$-F%6$*&%\"kGF,F/F,%\"tGF0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 11 "This gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(abs(x-a))-ln(abs(x))=-k*a*t+c" "6#/,&-%#lnG6#-%$absG6#,&%\"xG\"\" \"%\"aG!\"\"F--F&6#-F)6#F,F/,&*(%\"kGF-F.F-%\"tGF-F/%\"cGF-" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "ln(abs((x-a)/x))=-k*a*t+c" "6#/-%#lnG 6#-%$absG6#*&,&%\"xG\"\"\"%\"aG!\"\"F-F,F/,&*(%\"kGF-F.F-%\"tGF-F/%\"c GF-" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 42 "The corresponding exponential equation is " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "abs((x-a)/x) =exp(-k*a*t+c)" "6#/-%$absG6#*&,&%\"xG\"\" \"%\"aG!\"\"F*F)F,-%$expG6#,&*(%\"kGF*F+F*%\"tGF*F,%\"cGF*" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(x-a)/x=A*exp(-k*a*t)" "6#/*&,&%\"xG\"\"\"%\" aG!\"\"F'F&F)*&%\"AGF'-%$expG6#,$*(%\"kGF'F(F'%\"tGF'F)F'" }{TEXT -1 15 " ------- (i), " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "A=``" "6#/%\"AG%!G" }{TEXT 326 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(c)" "6#-%$expG6#%\"cG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 31 "The last equation has the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(x-a)/x = E;" "6#/*&,&%\"xG\"\"\"%\"aG! \"\"F'F&F)%\"EG" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 6 "where \+ " }{XPPEDIT 18 0 "E = A*exp(-k*a*t);" "6#/%\"EG*&%\"AG\"\"\"-%$expG6#, $*(%\"kGF'%\"aGF'%\"tGF'!\"\"F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x-a=E*x" "6#/,&%\"xG\"\"\"%\"aG!\"\"*&%\"EGF&F%F&" }{TEXT -1 3 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x-E*x=a" "6#/,&%\"xG\"\"\"*&%\"EGF&F%F&!\"\"%\"a G" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x*(1-E)=a" "6#/*&%\"xG\"\"\",&F&F& %\"EG!\"\"F&%\"aG" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 4 "and \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x=a/(1-E)" "6#/% \"xG*&%\"aG\"\"\",&F'F'%\"EG!\"\"F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 17 "Hence (i) gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "x=a/(1-A*exp(-k*a*t))" "6#/%\"xG*&%\"aG\"\"\",&F'F'*&% \"AGF'-%$expG6#,$*(%\"kGF'F&F'%\"tGF'!\"\"F'F2F2" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 31 "Suppose that the population is " } {XPPEDIT 18 0 "x=x[0]" "6#/%\"xG&F$6#\"\"!" }{TEXT -1 6 " when " } {XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 31 ". Then equation (i) \+ shows that " }{XPPEDIT 18 0 "A=(x[0]-a)/x[0])" "6#/%\"AG*&,&&%\"xG6#\" \"!\"\"\"%\"aG!\"\"F+&F(6#F*F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Hence we obtain the solution " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x=a/(1+``((a-x[0])/x[0])*exp(-k*a*t))" "6#/% \"xG*&%\"aG\"\"\",&F'F'*&-%!G6#*&,&F&F'&F$6#\"\"!!\"\"F'&F$6#F1F2F'-%$ expG6#,$*(%\"kGF'F&F'%\"tGF'F2F'F'F2" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 327 14 "______________" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "diff(x(t),t)=k*x(t)*(a-x(t)),x(0)=x0;\ndsolve(\{%\},x(t));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$/-%%diffG6$-%\"xG6#%\"tGF**(%\"kG\"\" \"F'F-,&%\"aGF-F'!\"\"F-/-F(6#\"\"!%#x0G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG*&%\"aG\"\"\",&F*F**(-%$expG6#,$*(F'F*%\"kGF*F)F*! \"\"F*,&F)F*%#x0GF3F*F5F3F*F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 1 " }}{PARA 0 "" 0 "" {TEXT 330 8 "Question" } {TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 29 "Suppose that a populati on of " }{TEXT 328 1 "x" }{TEXT -1 29 " organisms has a growth rate " }{XPPEDIT 18 0 "dx/dt" "6#*&%#dxG\"\"\"%#dtG!\"\"" }{TEXT -1 9 " given by" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt = k*x*(a -x);" "6#/*&%#dxG\"\"\"%#dtG!\"\"*(%\"kGF&%\"xGF&,&%\"aGF&F+F(F&" } {TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 34 "where the critical popul ation is " }{XPPEDIT 18 0 "x[crit]" "6#&%\"xG6#%%critG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "a = 500;" "6#/%\"aG\"$+&" }{TEXT -1 18 " and the \+ constant " }{TEXT 329 1 "k" }{TEXT -1 14 " is given by " }{XPPEDIT 18 0 "k = 1/2000;" "6#/%\"kG*&\"\"\"F&\"%+?!\"\"" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 54 "Suppose also that we have an initial popu lation of 40." }}{PARA 0 "" 0 "" {TEXT -1 49 "(a) Find a formula for t he population at time t. " }}{PARA 0 "" 0 "" {TEXT -1 51 "(b) Plot a g raph of the population versus time for " }{XPPEDIT 18 0 "0<=t" "6#1\" \"!%\"tG" }{XPPEDIT 18 0 "``<=30" "6#1%!G\"#I" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 331 8 "Solution " }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 68 "(a) We could repeat \+ all the steps used to obtain the formula above. " }}{PARA 0 "" 0 "" {TEXT -1 71 "However let's start from the equation (i) which in the cu rrent case is " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(x- 500)/x = A*exp(-t/4);" "6#/*&,&%\"xG\"\"\"\"$+&!\"\"F'F&F)*&%\"AGF'-%$ expG6#,$*&%\"tGF'\"\"%F)F)F'" }{TEXT -1 15 " ------- (i). " }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 18 0 "x=40" "6#/%\"xG\"#S" } {TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 18 " we see that that " }{XPPEDIT 18 0 "A=-3/2" "6#/%\"AG,$*&\"\"$\"\"\" \"\"#!\"\"F*" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "(x-500)/x = -23/2;" "6#/*&,&%\"xG\"\"\"\"$+&!\"\"F'F&F) ,$*&\"#BF'\"\"#F)F)" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(-t/4)" "6#-%$ expG6#,$*&%\"tG\"\"\"\"\"%!\"\"F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 12 "Solving for " }{TEXT 332 1 "x" }{TEXT -1 73 " using simil ar steps to those used previously for the general case gives " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x=500/(1+``(23/2)*exp (-t/4))" "6#/%\"xG*&\"$+&\"\"\",&F'F'*&-%!G6#*&\"#BF'\"\"#!\"\"F'-%$ex pG6#,$*&%\"tGF'\"\"%F0F0F'F'F0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "de := diff(x(t),t)=x(t)*(500-x(t))/2000;\nic := x( 0)=40;\ndsolve(\{de,ic\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%# deG/-%%diffG6$-%\"xG6#%\"tGF,,$*&F)\"\"\",&\"$+&F/F)!\"\"F/#F/\"%+?" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"xG6#\"\"!\"#S" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,$*&\"\"\"F*,&F*F**&#\"#B\"\"#F*- %$expG6#,$F'#!\"\"\"\"%F*F*F5\"$+&" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "de := diff(x(t),t)=x(t)*(500-x(t))/2000;\nic := x(0)=40;\ndesol ve(\{de,ic\},x(t),method=sepvar,info=true);\ng := unapply(rhs(%),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/-%%diffG6$-%\"xG6#%\"tGF,,$*& F)\"\"\",&\"$+&F/F)!\"\"F/#F/\"%+?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#icG/-%\"xG6#\"\"!\"#S" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#%DThe~DE~has~separable~variables~.~.~G " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$-%$IntG6$*&\"\"\"F)*&%\"xGF),&! $+&F)F+F)F)!\"\"F+!%+?%\"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%#lnG6#%\"xG\"\"\"-F&6#,&!$+&F)F(F )!\"\",&%\"tG#F)\"\"%&%\"CG6#F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#% !G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%\"xG\"\"\",&!$+&F&F%F&!\"\"* &&%\"CG6#\"\"#F&-%$expG6#,$%\"tG#F&\"\"%F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%EApplying~the~initial~condition~.~.~~G/&%\"CG6#\"\"##! \"#\"#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,$*&-%$expG6#,$F'#\"\"\"\"\"%F/,&\"#BF/*& \"\"#F/F*F/F/!\"\"\"%+5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#% \"tG6\"6$%)operatorG%&arrowGF(,$*&-%$expG6#,$9$#\"\"\"\"\"%F4,&\"#BF4* &\"\"#F4F.F4F4!\"\"\"%+5F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 60 "(b) We can plot this solution along with \+ the gradient field." }}{PARA 0 "" 0 "" {TEXT -1 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AF2F[g^lFja_l7'7$$\"3K;$pr*\\:PBF2F[f^l7$$\"3?TQ43ro*R#F2F`f^l7$$\"37_ o,ca-oBF2Fff^l7$$\"3#zPk=crUS#F2F[g^lFgb_l7'7$$\"3Y@9,M(\\]\\#F2F[f^l7 $$\"3MYf$\\%=edDF2F`f^l7$$\"3Cd*eG>?f_#F2Fff^l7$$\"31$[1()Hm@c#F2F[g^l Fdc_l7'7$$\"3eEN&3ZWHl#F2F[f^l7$$\"3Z^!yF6Fgg_lF]h_l7' 7$$\"3IG^NP7C3HF6Fhf_l7$$\"3L#3HjB[vS$F6F]g_l7$$\"3kXTOv\\?7JF6Fbg_l7$ $\"3%=h%fC_o'[$F6Fgg_lFjh_l7'7$$\"3'4=wdg)=([%F6Fhf_l7$$\"35M,v/c\\') \\F6F]g_l7$$\"3S(>&yVB:\"p%F6Fbg_l7$$\"3;jc,$fKc1&F6Fgg_lFgi_l7'7$$\"3 /Ms>uf8mgF6Fhf_l7$$\"3?(=rJ(HWllF6F]g_l7$$\"3_]i?7(*4qiF6Fbg_l7$$\"3D; nVh*zXk'F6Fgg_lFdj_l7'7$$\"3;(G=EM$3XwF6Fhf_l7$$\"3HSAfT.RW\")F6F]g_l7 $$\"3g.ti!3Z!\\yF6Fbg_l7$$\"3Cqx&)Ht_B#)F6Fgg_lFa[`l7'7$$\"3DS$R5rISA* F6Fhf_l7$$\"3R$H8+rPLs*F6F]g_l7$$\"3#eN[!\\W*zU*F6Fbg_l7$$\"3MB)y#)puC !)*F6Fgg_lF^\\`l7'7$$\"3MRg%z!yH!3\"F2Fhf_l7$$\"3lMM%y]G-8\"F2F]g_l7$$ \"3*3%pu\"=%p+6F2Fbg_l7$$\"3l()*pm?U\"Q6F2Fgg_lF[]`l7'7$$\"3lW\")yWD>Q 7F2Fhf_l7$$\"3'*RboWK7)G\"F2F]g_l7$$\"3?Y!*e=*)ee7F2Fbg_l7$$\"3'H47N%p .'H\"F2Fgg_lFh]`l7'7$$\"3'*\\-j\"G(3'R\"F2Fhf_l7$$\"3GXw_\")z,Y9F2F]g_ l7$$\"3^^6VbO[;9F2Fbg_l7$$\"3E)>a.oJRX\"F2Fgg_lFe^`l7'7$$\"3GbBZ=?)Rb \"F2Fhf_l7$$\"3e](p$=F\"Rg\"F2F]g_l7$$\"3#oDtARyVd\"F2Fbg_l7$$\"3S.j>< k#=h\"F2Fgg_lFb_`l7'7$$\"3SgWJbn(=r\"F2Fhf_l7$$\"33c=@bu!=w\"F2F]g_l7$ $\"3Ji`6HJFK-(>>F2F]g_l7$$\"3Wnu&f'y;!*=F2Fbg_l7$$\"3-90)3*ehF>F2 Fgg_lF\\a`l7'7$$\"3mq')**GimF?F2Fhf_l7$$\"3Mmg*)Gpfx?F2F]g_l7$$\"3es&* z-E1[?F2Fbg_l7$$\"3;>EsF1^&3#F2Fgg_lFia`l7'7$$\"3!exSe'4c&=#F2Fhf_l7$$ \"3Zr\"Qdm\"\\NAF2F]g_l7$$\"3rx;kRt&f?#F2Fbg_l7$$\"3GCZck`SVAF2Fgg_lFf b`l7'7$$\"3$4)Go-dXVBF2Fhf_l7$$\"3gw-e-kQ$R#F2F]g_l7$$\"3%Gy$[w?&QO#F2 Fbg_l7$$\"3THoS,,I,CF2Fgg_lFcc`l7'7$$\"31')\\_R/N,DF2Fhf_l7$$\"3u\"QA% R6G^DF2F]g_l7$$\"3(z)eK8ou@DF2Fbg_l7$$\"3cM*[#Q[>fDF2Fgg_lF`d`l7'7$$\" 3?\"4nj4K\"*Rr\"GF2Fhf_l7$$\"3+#f1Jhqq'GF2F ]g_l7$$\"3C)45qGOv$GF2Fbg_l7$$\"3#[9L>J%)\\(GF2Fgg_lFje`l7'7$$\"3Y,80] Y.vHF2Fhf_l7$$\"38(p[*\\`'\\-$F2F]g_l7$$\"3O.A&Q-Ja*HF2Fbg_l7$$\"3%*\\ _x[!zG.$F2Fgg_lFgf`l-%&COLORG6&F][l$\"\"&!\"\"F)\"\"\"-%+AXESLABELSG6% %\"tG%%x(t)G-%%FONTG6#%(DEFAULTG-%%VIEWG6$;$!+++++:!\"*$\"++++]JF`[l;$ !+++++NF`[l$\"++++]t!\"(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 122 "The population increases rapidly at firs t, but then the growth rate slows as the population approaches the cri tical level." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" }{TEXT -1 142 ": If we attempt to solve the differential \+ equation with an initial population equal to the critical population, \+ we obtain a constant solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "de := diff(x(t),t)=x(t)*(500 -x(t))/2000;\nic := x(0)=500;\ndsolve(\{de,ic\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/-%%diffG6$-%\"xG6#%\"tGF,,$*&F)\"\"\",&\"$+ &F/F)!\"\"F/#F/\"%+?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"xG6 #\"\"!\"$+&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG\"$+&" }} }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} }{SECT 1 {PARA 4 "" 0 "" {TEXT -1 14 "An RC circuit " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 256 "" 0 "" {TEXT -1 1 " " } {GLPLOT2D 329 193 193 {PLOTDATA 2 "6)-%'CURVESG6#717$$!\"&!\"\"\"\"!7$ F+$\"#:F*7$\"\"\"$!#:F*7$\"\"#F-7$\"\"$F17$\"\"%F-7$\"\"&F17$\"\"'F-7$ \"\"(F17$\"\")F-7$\"\"*F17$\"#5F-7$\"#6F17$\"#7F-7$$\"$D\"F*F+-F$6+7$7 $$\"%a7!\"#F+7$\"#CF+7$7$$!$D\"F*F+F'7$FS7$FT$!#%*F*7$7$FT$!$1\"F*7$FT !#?7$7$\"#?Fen7$\"#GFen7$7$F_oFin7$FaoFin7$FV7$FW!\"(7$7$FW!#87$FWF\\o 7$F[pF[o-%%TEXTG6$7$$F>F+$F8F+Q\"R6\"-F^p6$7$$\"$v#F*$!#lF*Q\"CFdp-F^p 6$7$FW$!#5F+Q%V(t)Fdp-F$6#7C7$$!#&*F*F`q7$$!+n-!)f&*!\"*$!+3?*RS*F\\r7 $$!+=q\"ot*F\\r$!+t\\uJ))F\\r7$$!+;$*R-5!\")$!+!esgI)F\\r7$$!+()z)4/\" Fgr$!+F<$z%yF\\r7$$!+3$4z3\"Fgr$!+YqevuF\\r7$$!+uEHT6Fgr$!+UF)Q?(F\\r7 $$!+d)4!*>\"Fgr$!+53lVqF\\r7$$!+d)f(e7Fgr$!+\">z7+(F\\r7$$!+G1;=8Fgr$! +2rXyqF\\r7$$!+^S%[P\"Fgr$!+?x5ssF\\r7$$!+N.bE9Fgr$!+*y5Xd(F\\r7$$!+: \"=7Z\"Fgr$!+e/htzF\\r7$$!+Em12:Fgr$!+&)e\\`%)F\\r7$$!+-nmK:Fgr$!+\\b. &**)F\\r7$$!+\\x*pa\"Fgr$!+w*Rmd*F\\r7$$!+L%)[\\:Fgr$!+VA^<5Fgr7$$!+e% R+a\"Fgr$!+JBmw5Fgr7$$!+Dv->:Fgr$!+LhvK6Fgr7$$!+9.H([\"Fgr$!+ntb$=\"Fg r7$$!+'3$4Y9Fgr$!+\\2/F7Fgr7$$!+Y#yqR\"Fgr$!+KFZh7Fgr7$$!+h)*>U8Fgr$!+ A1[&G\"Fgr7$$!+edk$G\"Fgr$!+,t5)H\"Fgr7$$!+0.vB7Fgr$!+$Q\\))H\"Fgr7$$! +W8!\\;\"Fgr$!+#GxwG\"Fgr7$$!+**\\W46Fgr$!+(RO]E\"Fgr7$$!+P@ff5Fgr$!+Y $H=B\"Fgr7$$!+O-L<5Fgr$!+\"**z$*=\"Fgr7$$!+\\9WV)*F\\r$!+a1QR6Fgr7$$!+ S\"*[>'*F\\r$!+\\Y#Q3\"Fgr7$$!+4PP5&*F\\r$!+@o#\\-\"Fgr7$$!+W_W?&*F\\r $!+&Q_.l*F\\r-%*AXESSTYLEG6#%%NONEG" 1 2 0 1 10 0 2 9 1 1 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "The voltage drop (measured in " }{TEXT 258 5 "volts" } {TEXT -1 11 ") across a " }{TEXT 259 8 "resistor" }{TEXT -1 45 " is pr oportional to the current (measured in " }{TEXT 258 4 "amps" }{TEXT -1 23 ") through the resistor:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "V[res]=I[res]*R" "6#/&%\"VG6#%$resG*&&%\"IG6#F'\"\"\"% \"RGF," }{TEXT -1 13 " ------- (i)." }}{PARA 0 "" 0 "" {TEXT -1 13 "Th e constant " }{TEXT 296 1 "R" }{TEXT -1 15 " is called the " }{TEXT 259 10 "resistance" }{TEXT -1 40 " of the resistor, and it is measured in " }{TEXT 258 4 "ohms" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "The voltage drop " }{XPPEDIT 18 0 "V[cap]" "6#&%\"VG6#%$capG" }{TEXT -1 10 " across a " }{TEXT 259 9 "ca pacitor" }{TEXT -1 45 " is proportional to the instantaneous charge " }{TEXT 297 1 "Q" }{TEXT -1 14 " (measured in " }{TEXT 258 8 "coulombs " }{TEXT -1 19 ") on the capacitor." }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "V[cap]=Q/C" "6#/&%\"VG6#%$capG*&%\"QG\"\"\"%\"CG!\" \"" }{TEXT -1 14 " ------- (ii)," }}{PARA 0 "" 0 "" {TEXT -1 6 "where \+ " }{TEXT 300 1 "C" }{TEXT -1 15 " is called the " }{TEXT 259 11 "capac itance" }{TEXT -1 31 " of the capacitor, measured in " }{TEXT 258 6 "f arads" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "The current " }{XPPEDIT 18 0 "I[cap]" "6#&%\"IG6#%$cap G" }{TEXT -1 92 " passing through the capacitor is, by definition, the rate of change of the charge, that is," }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "I[cap]=dQ/dt" "6#/&%\"IG6#%$capG*&%#dQG\"\"\"%# dtG!\"\"" }{TEXT -1 15 " ------- (iii)." }}{PARA 0 "" 0 "" {TEXT -1 117 "When the time is measured in seconds and the charge is measured i n coulombs, equation (ii) gives the current in amps." }}{PARA 0 "" 0 " " {TEXT -1 34 "Thus 1 amp = 1 coulomb per second." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "According to " }{TEXT 259 15 "Kirchoff's laws" }{TEXT -1 108 ", the current through the resi stor in this circuit is equal to the current in the capacitor, so we c an write" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "I(t);" " 6#-%\"IG6#%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "I[cap] = I[res];" " 6#/&%\"IG6#%$capG&F%6#%$resG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 119 "Also the sum of the voltage drops across the resistor and coil is equal to the voltage generated by the voltage source:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "V[res]+V[cap] = V(t);" "6#/ ,&&%\"VG6#%$resG\"\"\"&F&6#%$capGF)-F&6#%\"tG" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 50 "Substituting from (i) and (ii) gives the \+ equation:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "I*R+Q/C \+ = V(t);" "6#/,&*&%\"IG\"\"\"%\"RGF'F'*&%\"QGF'%\"CG!\"\"F'-%\"VG6#%\"t G" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 299 1 "R" }{TEXT -1 1 " " }{XPPEDIT 18 0 "d Q/dt + Q/C=V(t)" "6#/,&*&%#dQG\"\"\"%#dtG!\"\"F'*&%\"QGF'%\"CGF)F'-%\" VG6#%\"tG" }{TEXT -1 14 " ------- (vi)." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 298 9 "_________" }{TEXT -1 18 " " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 8 "Example " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 28 "As an example, suppose that " } {XPPEDIT 18 0 "R = 10" "6#/%\"RG\"#5" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "C=1/10" "6#/%\"CG*&\"\"\"F&\"#5!\"\"" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "V(t) = 30*exp(-3*t);" "6#/-%\"VG6#%\"tG*&\"#I\"\"\"-%$expG6#,$*& \"\"$F*F'F*!\"\"F*" }{TEXT -1 18 " in equation (vi):" }}{PARA 256 "" 0 "" {TEXT -1 4 " 10 " }{XPPEDIT 18 0 "dQ/dt + 10*Q=exp(-3*t)" "6#/,&* &%#dQG\"\"\"%#dtG!\"\"F'*&\"#5F'%\"QGF'F'-%$expG6#,$*&\"\"$F'%\"tGF'F) " }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 59 "Suppose also that th e initial charge on the capacitor is 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "de := 10*diff(Q(t),t)+ 10*Q(t)=exp(-3*t);\nic := Q(0)=0;\ndesolve(\{de,ic\},Q(t),method=linea r,info=true);\nq := unapply(rhs(%),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"QG6#%\"tGF-\"#5*&F.\"\"\"F*F0F0-%$expG6#, $F-!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"QG6#\"\"!F)" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 $%0Linear~DE~.~.~~G/,&-%%DiffG6$-%\"QG6#%\"tGF,\"\"\"F)F-,$-%$expG6#,$ F,!\"$#F-\"#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%9Integrating~factor~.~.~~G-%$expG6#-%$IntG6$\"\"\"% \"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%5~~~~~~~~~~~~~~~~=~~~G-%$expG 6#%\"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&-%\"QG6#%\"tG\"\"\"-%$expGF'F)-%$IntG6$,$-F+6#,$F(! \"##F)\"#5F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#/*&-%\"QG6#%\"tG\"\"\"-%$expGF'F),&-F+6#,$F(!\"##!\" \"\"#?&%\"CG6#F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%EApplying~the~initial~condition~.~.~~G/&%\"CG6 #\"\"\"#F(\"#?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"QG6#%\"tG,&-%$expG6#,$F'!\"$#!\"\"\"#?*&#\"\" \"F0F3-F*6#,$F'F/F3F3" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qGf*6#%\" tG6\"6$%)operatorG%&arrowGF(,&-%$expG6#,$9$!\"$#!\"\"\"#?*&#\"\"\"F5F8 -F.6#,$F1F4F8F8F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "We can plot a graph to show how the charge on the ca pacitor varies with time." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(q(t),t=0..6,labels=[`t`,`q(t)` ]);" }}{PARA 13 "" 1 "" {GLPLOT2D 462 256 256 {PLOTDATA 2 "6%-%'CURVES G6$7_o7$$\"\"!F)F(7$$\"3$*****\\ilyM;!#>$\"3GBPJ_3F#e\"!#?7$$\"3')**** *\\7t&pKF-$\"3]K]J_r:jIF07$$\"3z****\\(ofV!\\F-$\"3Avn.As#zW%F07$$\"3s ******\\i9RlF-$\"3))p6)Q\\!fTdF07$$\"3m****\\7G$R<)F-$\"3#>=e0,5*[pF07 $$\"3e*****\\P>(3)*F-$\"3wC#)*\\K9W2)F07$$\"3&****\\Pf]V9\"!#=$\"3vKd) *G)3C7*F07$$\"3%*******\\#HyI\"FM$\"3Z_$)y`')p45F-7$$\"37+]PfIJ#f\"FM$ \"3iHBUu`$H;\"F-7$$\"3-++voozw=FM$\"3S2$y%H?-(H\"F-7$$\"3#***\\7y1Gh@F M$\"3#GOEnq5PT\"F-7$$\"33++]([kdW#FM$\"3:F<]h\\g9:F-7$$\"3K++D\"3Gc3$F M$\"3%z*zsMk@\"p\"F-7$$\"3++++v;\\DPFM$\"3qd)*fCvj4=F-7$$\"3A++++nfpVF M$\"3%=5!\\*[s?)=F-7$$\"3W+++DF-7$$\"3M+Dc^E'R<& FM$\"3pM`%=7L9#>F-7$$\"3D+]7yNAM`FM$\"3\\hb\")fpvB>F-7$$\"39+vo/X[%\\& FM$\"3zP,&)Q3]C>F-7$$\"31++DJauacFM$\"3,$pI!3BwB>F-7$$\"3))**\\P%Gn_(f FM$\"3&=(eT(p.#=>F-7$$\"3o****\\P\"*y&H'FM$\"3h+7^:nw2>F-7$$\"3i****\\ 7(=,*oFM$\"3yA5Lywbx=F-7$$\"3e****\\(G[W[(FM$\"32&3[25[g$=F-7$$\"3i*** *\\()fB:()FM$\"3#*f&G%4EdDq0]\"F\\t$\"3Of]1%zk& f5F-7$$\"3'******\\U80j\"F\\t$\"3\\K7dBz&fT*F07$$\"35+++0ytb#p(3\"Q)F07$$\"3)****\\(QNXp=F\\t$\"3%)*\\LJcTq_(F07$$\"3.+++XDn/?F \\t$\"3/2Mn)[2Ih'F07$$\"3.+++!y?#>@F\\t$\"31E*\\Z4*e>fF07$$\"3'****\\( 3wY_AF\\t$\"31e'zn'3'))>&F07$$\"3#)******HOTqBF\\t$\"3^^(\\[H48j%F07$$ 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R7Fb[l-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$%\"tG%%q(t)G-%%V IEWG6$;F(F_`l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 31 "The rate o f a chemical reaction" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" } }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 259 20 "law of mass reactio n" }{TEXT -1 106 " states that, if the temperature is kept constant, t he rate of a chemical reaction is proportional to the " }{TEXT 259 29 "product of the concentrations" }{TEXT -1 37 " of the substances that \+ are reacting." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Suppose that a chemical reaction of two substances A and \+ B" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 263 1 "A" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "B -> C" "6#f*6#%\"BG7\"6$%)operatorG%&arrowG6\"%\"C GF*F*F*" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 9 "combines " } {TEXT 264 1 "a" }{TEXT -1 36 " moles per litre of substance A and " } {TEXT 265 1 "b" }{TEXT -1 57 " moles per litre of substance B to produ ce a substance C." }}{PARA 0 "" 0 "" {TEXT -1 3 "If " }{TEXT 266 1 "x " }{TEXT -1 101 " is the number of moles per litre which have reacted \+ at time t, the rate of the reaction is given by " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt = k*(a-x)*(b-x)" "6#/*&%#dxG\"\" \"%#dtG!\"\"*(%\"kGF&,&%\"aGF&%\"xGF(F&,&%\"bGF&F-F(F&" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 69 "We consider the solution of this diff erential equation assuming that " }{XPPEDIT 18 0 "a <> b" "6#0%\"aG%\" bG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 67 "de := diff(x(t),t)=k*(a-x(t))*(b-x(t));\ndes olve(de,x(t),info=true);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/-%% diffG6$-%\"xG6#%\"tGF,*(%\"kG\"\"\",&%\"aGF/F)!\"\"F/,&%\"bGF/F)F2F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%DThe~DE~has~separable~variables~.~.~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%\"kG!\"\"-%$IntG6$*&\"\"\"F+*&,&%\"aGF+%\"xGF&F+,&% \"bGF+F/F&F+F&F/F+%\"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/*(,&*(%\"kG!\"\",&%\"aG\"\"\"%\"bGF(F (-%#lnG6#,&F*F(%\"xGF+F+F+*(F'F(F)F(-F.6#,&F,F(F1F+F+F(F+F'F+F)F+,&*(% \"tGF+F'F+F)F+F+&%\"CG6#F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&,&%\"aG\"\"\"%\"xG!\"\"F',&%\"bGF' F(F)F)*&&%\"CG6#\"\"#F'-%$expG6#*(%\"tGF'%\"kGF',&F&F'F+F)F'F'" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/-%\"xG6#%\"tG*&,&%\"aG!\"\"*(&%\"CG6#\"\"#\"\"\"-%$expG6#*(F'F1%\"kG F1,&F*F1%\"bGF+F1F1F8F1F1F1,&F+F1*&F-F1F2F1F1F+" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "For an example, suppose t hat " }{XPPEDIT 18 0 "a = 300" "6#/%\"aG\"$+$" }{TEXT -1 2 ", " } {XPPEDIT 18 0 "b = 700" "6#/%\"bG\"$+(" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "k = 1/250" "6#/%\"kG*&\"\"\"F&\"$]#!\"\"" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 17 "Measure the time " }{TEXT 274 1 "t" } {TEXT -1 61 " from the instant when the reaction starts, so that x(0) \+ = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "de := diff(x(t),t)=(300-x(t))*(700-x(t))/250;\nic := x(0)=0;\ndesolve(\{de,ic\},x(t),info=true);\nc := unapply(rhs(%),t); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/-%%diffG6$-%\"xG6#%\"tGF,,$ *&,&\"$+$\"\"\"F)!\"\"F1,&\"$+(F1F)F2F1#F1\"$]#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"xG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%DThe~DE~has~separable~varia bles~.~.~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$-%$IntG6$*&\"\"\"F)*& ,&!$+$F)%\"xGF)F),&!$+(F)F-F)F)!\"\"F-\"$]#%\"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%#lnG6#,&!$+ $\"\"\"%\"xGF*!\"\"-F&6#,&!$+(F*F+F*F*,&%\"tG#\"\")\"\"&&%\"CG6#F*F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&,&!$+$\"\"\"%\"xGF'!\"\",&!$+(F'F(F'F'*&&%\"CG6#\"\"#F'-%$expG6# ,$%\"tG#\"\")\"\"&F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%EApplying~the ~initial~condition~.~.~~G/&%\"CG6#\"\"##\"\"(\"\"$" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG ,$*&,&!\"\"\"\"\"-%$expG6#,$F'#\"\")\"\"&F,F,,&!\"$F,*&\"\"(F,F-F,F,F+ \"%+@" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cGf*6#%\"tG6\"6$%)operato rG%&arrowGF(,$*&,&!\"\"\"\"\"-%$expG6#,$9$#\"\")\"\"&F0F0,&!\"$F0*&\" \"(F0F1F0F0F/\"%+@F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 83 "We can plot a graph to show how the concentration \+ of substance C changes with time." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(c(t),t=0..3,labels=[`t` ,`c(t)`]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-% 'CURVESG6$7W7$$\"\"!F)F(7$$\"3')*****\\7t&pK!#>$\"39N)H'>!>td#!#;7$$\" 3s******\\i9RlF-$\"3KXmLQyn`[F07$$\"33++vVV)RQ*F-$\"3aI&3%eDXEmF07$$\" 3/++vVA)GA\"!#=$\"3\"R&=E=/\\K#)F07$$\"3;+]iSS\"Ga\"F>$\"3)3PgCAua')*F 07$$\"3+++]Peui=F>$\"3Dy?xlT*Q8\"!#:7$$\"37+++]$)z%=#F>$\"3k2!)y@L>o7F K7$$\"3A++]i3&o]#F>$\"3mcN`*)[$\"3D8F^hI& >g\"FK7$$\"3z***\\P9CAu$F>$\"3+A*4$)=Pyw\"FK7$$\"3!)***\\P*zhdVF>$\"3X QyUPks9>FK7$$\"31++v$>fS*\\F>$\"3'*fg-c!*)[/#FK7$$\"3$)***\\(=$f%GcF>$ \"3%[8\")pw=l:#FK7$$\"3Q+++Dy,\"G'F>$\"3gKGUX]qbAFK7$$\"33++]7$ \"3w`74D3-KBFK7$$\"3`+++v4&G](F>$\"3,^z!p-\\uS#FK7$$\"3!)*****\\7nD:)F >$\"3M9sH&G_OZ#FK7$$\"3[+++D!*oy()F>$\"3*Qx\"=x_sHDFK7$$\"3))***\\Ppns M*F>$\"37Ax))3W'\\d#FK7$$\"3,++]siL-5!#<$\"3^#)egFUlAEFK7$$\"3-+++!R5' f5Fjq$\"3E%fd1T<&eEFK7$$\"3)***\\P/QBE6Fjq$\"3[_?9fDn&p#FK7$$\"3!***** *\\\"o?&=\"Fjq$\"3)3mY6**e\\s#FK7$$\"31+]Pa&4*\\7Fjq$\"3Y(*)eK*Qn`FFK7 $$\"33+]7j=_68Fjq$\"396$z#e53yFFK7$$\"33++vVy!eP\"Fjq$\"3T]I,mB&3!GFK7 $$\"34+](=WU[V\"Fjq$\"3()4>,U8h>GFK7$$\"3)****\\7B>&)\\\"Fjq$\"3k)\\J- ?$zPGFK7$$\"3)***\\P>:mk:Fjq$\"31LALgQnaGFK7$$\"3'***\\iv&QAi\"Fjq$\"3 QpL_Zg)y'GFK7$$\"31++vtLU%o\"Fjq$\"3-/i#z*4x!)GFK7$$\"3!******\\Nm'[Fjq$\"3sw+&y/`:#HFK7$$ \"3z*****\\@80+#Fjq$\"31**f^3*Q*GHFK7$$\"31++]7,Hl?Fjq$\"3!\\i!eQp/OHF K7$$\"3()**\\P4w)R7#Fjq$\"3Tp.*=1j=%HFK7$$\"3;++]x%f\")=#Fjq$\"3g!QlGN 5w%HFK7$$\"3!)**\\P/-a[AFjq$\"39'\\P^<#\\_HFK7$$\"3/+](=Yb;J#Fjq$\"3mT %zVP/r&HFK7$$\"3')****\\i@OtBFjq$\"3]w60yi+++!*>=+DFjq$\"3k&)*phji$oHFK7$$\"3-++DE&4Qc#Fjq $\"3iyrd'yY9(HFK7$$\"3=+]P%>5pi#Fjq$\"3[?H\"R[1U(HFK7$$\"39+++bJ*[o#Fj q$\"3YRz&p>0l(HFK7$$\"33++Dr\"[8v#Fjq$\"3Q2uQMw))yHFK7$$\"3++++Ijy5GFj q$\"35J'oP673)HFK7$$\"31+]P/)fT(GFjq$\"3MEG#eMqE)HFK7$$\"31+]i0j\"[$HF jq$\"3_a6Bg%zU)HFK7$$\"\"$F)$\"3E/w=E>%e)HFK-%'COLOURG6&%$RGBG$\"#5!\" \"F(F(-%+AXESLABELSG6$%\"tG%%c(t)G-%%VIEWG6$;F(Fi[l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Tasks" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q1 " }}{PARA 0 "" 0 "" {TEXT -1 11 "The vo lume " }{TEXT 267 1 "x" }{TEXT -1 51 " cubic meters of carbon dioxide \+ present in a room, " }{TEXT 276 1 "t" }{TEXT -1 76 " minutes after ope ning a window, is determined by the differential equation " }{XPPEDIT 18 0 "dx/dt=1-5*x" "6#/*&%#dxG\"\"\"%#dtG!\"\",&F&F&*&\"\"&F&%\"xGF&F( " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 10 " (a) Find " }{TEXT 275 1 "x" }{TEXT -1 18 " as a function of " }{TEXT 325 1 "t" }{TEXT -1 4 " if " }{TEXT 268 2 "x " }{TEXT -1 13 "= 0.35, when " }{XPPEDIT 18 0 "t = 0" "6#/%\"tG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 86 " (b) Find the volume of carbon dioxide present after 10 seconds and after 10 minutes. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "de := diff(x(t),t)=1-5*x(t) ;\nic := x(0)=7/20;\ndsolve(\{de,ic\},x(t));\ng := unapply(rhs(%),t): \ng(1/6);\n``=evalf(%);\ng(10);\n``=evalf(%);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/-%%diffG6$-%\"xG6#%\"tGF,,&\"\"\"F.*&\"\"&F.F)F. !\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"xG6#\"\"!#\"\"(\"# ?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&#\"\"\"\"\"&F**&# \"\"$\"#?F*-%$expG6#,$*&F+F*F'F*!\"\"F*F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&#\"\"\"\"\"&F%*&#\"\"$\"#?F%-%$expG6##!\"&\"\"'F%F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G$\"+8t*=l#!#5" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,&#\"\"\"\"\"&F%*&#\"\"$\"#?F%-%$expG6#!#]F%F%" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G$\"+++++?!#5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q2 " }} {PARA 0 "" 0 "" {TEXT -1 47 "The differential equation relating the cu rrent " }{TEXT 319 1 "I" }{TEXT -1 46 " in amps, in a particular RL ci rcuit, to time " }{TEXT 278 1 "t" }{TEXT -1 17 " in seconds is 2 " } {XPPEDIT 18 0 "dI/dt+I=12" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'%\"IGF'\"#7 " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 22 " (a) If the current " }{TEXT 317 1 "I" }{TEXT -1 80 " is zero when a switch in the circui t is closed, solve this equation to find the" }}{PARA 0 "" 0 "" {TEXT -1 19 " current " }{TEXT 318 1 "I" }{TEXT -1 23 " as a funct ion of time " }{TEXT 277 1 "t" }{TEXT -1 55 ", measured from the inst ant when the switch is closed." }}{PARA 0 "" 0 "" {TEXT -1 64 " (b) \+ Find the current 0.3 seconds after the switch is closed. " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "de := 2*diff(y(t),t)+y(t)=12;\nic := y(0)=0;\ndsolve(\{de,ic\},y(t ));\ng := unapply(rhs(%),t):\ng(0.3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&*&\"\"#\"\"\"-%%diffG6$-%\"yG6#%\"tGF0F)F)F-F)\"#7" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&\"#7\"\"\"*&F)F*-%$expG6#,$*&\"\"# !\"\"F'F*F2F*F2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"*G/:n\"!\")" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_______________________________ ______" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 3 "Q3 " }}{PARA 0 "" 0 "" {TEXT -1 98 "Let the proportion o f a forest destroyed in a forest fire, at a particular instant, be den oted by " }{TEXT 342 1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 22 "The rate of change of " }{TEXT 341 1 "x" }{TEXT -1 22 " with respe ct to time " }{TEXT 340 1 "t" }{TEXT -1 16 ", is called the " }{TEXT 259 16 "destruction rate" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 66 "Investigations show that the destruction rate is proportional to \+ " }{XPPEDIT 18 0 "x*(1-x)" "6#*&%\"xG\"\"\",&F%F%F$!\"\"F%" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 236 "A particular fire is discovered \+ when half of the forest has been destroyed, and it is found that the d estruction rate is such that, if it remained constant after that time, the forest would be completely destroyed in a further 24 hours." }} {PARA 0 "" 0 "" {TEXT -1 13 "Show that 12 " }{XPPEDIT 18 0 "dx/dt=x*(1 -x)" "6#/*&%#dxG\"\"\"%#dtG!\"\"*&%\"xGF&,&F&F&F*F(F&" }{TEXT -1 101 " , and deduce that approximately 73% of the forest is destroyed 12 hour s after it is first discovered." }}{PARA 0 "" 0 "" {TEXT 343 4 "Hint" }{TEXT -1 88 ": The differential equation which determines the proport ion of the forest burnt at time " }{TEXT 346 1 "t" }{TEXT -1 14 " has \+ the form " }{XPPEDIT 18 0 "dx/dt=k*x*(1-x)" "6#/*&%#dxG\"\"\"%#dtG!\" \"*(%\"kGF&%\"xGF&,&F&F&F+F(F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 12 "If the time " }{TEXT 344 1 "t" }{TEXT -1 72 " in hours is measured from the instant when the fire is discovered then " } {XPPEDIT 18 0 "x=1/2" "6#/%\"xG*&\"\"\"F&\"\"#!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "dx/dt=1/48" "6#/*&%#dxG\"\"\"%#dtG!\"\"*&F&F&\"#[F( " }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 52 ". This information can be used to find the constant " }{TEXT 345 1 "k" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 37 "________________ _____________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_______________________________ ______" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q4 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 100 "Suppose that a student carrying a flu vi rus returns to an isolated college campus of 1000 students. " }}{PARA 0 "" 0 "" {TEXT -1 131 "Assuming that the rate at which the virus spre ads depends not only on the number of students already infected at a p articular time " }{TEXT 335 1 "t" }{TEXT -1 110 ", but also depends on the number of students not infected, so that if the number of student s infected at time " }{TEXT 333 1 "t" }{TEXT -1 4 " is " }{TEXT 334 1 "x" }{TEXT -1 6 " then " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt=k*x*(1000-x)" "6#/*&%#dxG\"\"\"%#dtG!\"\"*(%\"kGF&%\"xGF&, &\"%+5F&F+F(F&" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 6 "where \+ " }{TEXT 336 1 "k" }{TEXT -1 76 " is a constant. It is observed that \+ 50 students are infected after 4 days. " }}{PARA 0 "" 0 "" {TEXT -1 46 "(a) Find a formula for the number of students " }{TEXT 337 1 "x" } {TEXT -1 16 " infected after " }{TEXT 339 1 "t" }{TEXT -1 7 " days, " }}{PARA 0 "" 0 "" {TEXT -1 69 "(b) Sketch a graph which shows the numb er of students infected after " }{TEXT 338 1 "t" }{TEXT -1 10 " days f or " }{XPPEDIT 18 0 "0<=t" "6#1\"\"!%\"tG" }{XPPEDIT 18 0 "`` <= 12;" "6#1%!G\"#7" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 73 "(c) Find the number of students infected after 6 days and after 12 days. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "An s " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 268 "unassign('k'):\ndiff(x(t),t)=k*x(t)*(1000-x(t)),x(0) =1;\ndsolve(\{%\},x(t));\nsol := %: k=solve(eval(subs(\{x(t)=50,t=4\}, sol)));\nassign(%):\nsol;\ng := unapply(rhs(%),t):\nplot(g(t),t=0..12, labels=[`t`,`no. of infected students x(t)`]);\nfloor(evalf(g(6)));\nf loor(evalf(g(12)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/-%%diffG6$-%\" xG6#%\"tGF**(%\"kG\"\"\"F'F-,&\"%+5F-F'!\"\"F-/-F(6#\"\"!F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,$*&\"%+5\"\"\",&F+F+*&\"$***F +-%$expG6#,$*(F*F+%\"kGF+F'F+!\"\"F+F+F5F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"kG,$*&#\"\"\"\"%+SF(-%#lnG6##\"#>\"$***F(!\"\"" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,$*&\"%+5\"\"\",&F+F+*& \"$***F+-%$expG6#,$*&#F+\"\"%F+*&-%#lnG6##\"#>F.F+F'F+F+F+F+F+!\"\"F+ " }}{PARA 13 "" 1 "" {GLPLOT2D 515 363 363 {PLOTDATA 2 "6%-%'CURVESG6$ 7S7$$\"\"!F)$\"\"\"F)7$$\"3*)*******\\ech#!#=$\"3S3.%4N#Q&H\"!#<7$$\"3 ;+++v*G:*[F/$\"3>I=2#*RUA;F27$$\"3++++]L)4X(F/$\"37sET?3j*3#F27$$\"34+ ++X.u-5F2$\"3GJ?I,*Hbp#F27$$\"3%*****\\Fy:f7F2$\"3ykRV%o+BZ$F27$$\"3#* ****\\d'*)o\\\"F2$\"3[rY#Q&yD!R%F27$$\"3$*****\\(>ZIu\"F2$\"3uN&Hv#=z& f&F27$$\"3-++]xOi(*>F2$\"3Eh'f33n#*=(F27$$\"3%*****\\FPQ^AF2$\"35G-Ojy *[A*F27$$\"3;+++IrS7DF2$\"3mWh:-_W\">\"!#;7$$\"3.+++&o;Bu#F2$\"3KAS)*3 xi\"\\\"Fjn7$$\"3@+++!RS6+$F2$\"3X,OQD/>>>Fjn7$$\"3#*******\\o-hKF2$\" 3q*4$GwcxoCFjn7$$\"3>+++5cZ6NF2$\"30#y*p!ex?9$Fjn7$$\"3'*****\\xq!*QPF 2$\"3W7%RilF]!RFjn7$$\"31+++!4X$4SF2$\"3$yGFJgaT/&Fjn7$$\"30+++g:WQUF2 $\"3K$RZ;zl)[iFjn7$$\"3#*****\\<_$\\]%F2$\"3KMvwVh#f)zFjn7$$\"3m****** fs#3u%F2$\"370nPFHM!))*Fjn7$$\"3E++]<#Q'**\\F2$\"3'zj)oA6%4C\"!#:7$$\" 3I++]_u3Y_F2$\"3i$*\\Sd\">:`\"F]r7$$\"3P+++v8B.bF2$\"3'pM(47bw\"*=F]r7 $$\"3R++]n(p$RdF2$\"3?_&HXS.oF#F]r7$$\"3))*****\\#p2%*fF2$\"37=R&4'G]] FF]r7$$\"3!*****\\xgkeiF2$\"32ulzc?Y-LF]r7$$\"3%)****\\-V&*)['F2$\"3x4 J5n0/DQF]r7$$\"3E+++&\\$pPnF2$\"3E72xY>E@WF]r7$$\"3e******>am%*pF2$\"3 E'[]rkJ]0&F]r7$$\"3k*****\\JigC(F2$\"3cT#)f9q[tcF]r7$$\"3%*****\\Pl'=yF]r7$$\" 3Y****\\P/&f\\)F2$\"3)H06\"p#f$*=)F]r7$$\"3q+++5zj_()F2$\"3N'QS%\\)zj` )F]r7$$\"3=****\\<3;%**)F2$\"3RPn'[$Hx5))F]r7$$\"3;++]Z=iY#*F2$\"3%y& \\h/%y)[!*F]r7$$\"3[******\\'[M\\*F2$\"3'4:dN/#\\R#*F]r7$$\"3W****\\PM &=v*F2$\"3i)zu^\"G&4S*F]r7$$\"32+++'zs++\"Fjn$\"3Pzy_W4jD&*F]r7$$\"3,+ +]5Q_D5Fjn$\"3)*F] r7$$\"3++++KXJC6Fjn$\"3c;2(HjHm&)*F]r7$$\"3/++v@Rm\\6Fjn$\"3;>*>v@5\") ))*F]r7$$\"3/++DAl#R<\"Fjn$\"3hpf*G\\.=\"**F]r7$$\"#7F)$\"3Z*3n^tT<$** F]r-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$%\"tG%>no.~of~infec ted~students~x(t)G-%%VIEWG6$;F(Ffz%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$w#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$$**" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_______________________________ ______" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 3 "Q5 " }}{PARA 0 "" 0 "" {TEXT -1 47 "The differential equ ation relating the current " }{TEXT 348 1 "I" }{TEXT -1 43 " in amps, \+ in a certain RL circuit, to time " }{TEXT 347 1 "t" }{TEXT -1 17 " in \+ seconds is 2 " }{XPPEDIT 18 0 "dI/dt+4*I = V(t);" "6#/,&*&%#dIG\"\"\"% #dtG!\"\"F'*&\"\"%F'%\"IGF'F'-%\"VG6#%\"tG" }{TEXT -1 3 ", " }}{PARA 0 "" 0 "" {TEXT -1 28 "where the voltage source is " }{XPPEDIT 18 0 "V (t)=PIECEWISE([0,t<0],[4+8*exp(-t),t>=0])" "6#/-%\"VG6#%\"tG-%*PIECEWI SEG6$7$\"\"!2F'F,7$,&\"\"%\"\"\"*&\"\")F1-%$expG6#,$F'!\"\"F1F11F,F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 25 "Suppose that the curre nt " }{TEXT 349 1 "I" }{TEXT -1 11 " is 0 when " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 57 "(a) Fi nd a formula for the current as a function of time " }{TEXT 350 1 "t" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "t>=0" "6#1\"\"!%\"tG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 34 "(b) Sketch a graph of the current \+ " }{TEXT 351 1 "I" }{TEXT -1 14 " against time " }{TEXT 352 1 "t" } {TEXT -1 5 " for " }{XPPEDIT 18 0 "0<=t" "6#1\"\"!%\"tG" }{XPPEDIT 18 0 "``<=4" "6#1%!G\"\"%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 70 "(c) Find the maximum value of the current and the time when it occ urs." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 129 "de := 2*diff(i(t),t)+4*i(t)=4+8*exp(-t);\nic := i( 0)=0;\ndsolve(\{de,ic\},i(t));\ng := simplify(unapply(rhs(%),t));\nplo t(g(t),t=0..4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&*&\"\"#\" \"\"-%%diffG6$-%\"iG6#%\"tGF0F)F)*&\"\"%F)F-F)F),&F2F)*&\"\")F)-%$expG 6#,$F0!\"\"F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"iG6#\"\" !F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG,(\"\"\"F)*&\"\"% F)-%$expG6#,$F'!\"\"F)F)*&\"\"&F)-F-6#,$*&\"\"#F)F'F)F0F)F0" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,( \"\"\"F-*&\"\"%F--%$expG6#,$9$!\"\"F-F-*&\"\"&F--F16#,$*&\"\"#F-F4F-F5 F-F5F(F(F(" }}{PARA 13 "" 1 "" {GLPLOT2D 542 226 226 {PLOTDATA 2 "6%-% 'CURVESG6$7fn7$$\"\"!F)F(7$$\"3emmm;arz@!#>$\"3C4kMhUVq7!#=7$$\"39LLLL 3VfVF-$\"3Qv\"47,![oCF07$$\"3s******\\i9RlF-$\"3]AAQ![^7F0$\"3/B3C@o$\\O'F07$$\"3RLLL e'40j\"F0$\"3=z;-dZL&*yF07$$\"3/++](Q&3d?F0$\"3(Q79,&))GF%*F07$$\"3mmm m;6m$[#F0$\"3QfR(\\k\"FT7$$\"3fm mm;yYULF0$\"3i`o+/$36I\"FT7$$\"3/++](GI)pPF0$\"3k!44%4nA\"R\"FT7$$\"3% HLL$eF>(>%F0$\"3+T)>`Ig\"p9FT7$$\"3Qmmm\">K'*)\\F0$\"3YAEsG0AN< FT7$$\"3.*****\\U7Y](F0$\"3s+lj0G(Rx\"FT7$$\"3Wmm;H9lRzF0$\"3-31p7YX'y \"FT7$$\"3'QLLLV!pu$)F0$\"3#RLt\"3$>Yz\"FT7$$\"3K+++DI(yv)F0$\"35d=5UJ j)z\"FT7$$\"3xmmm;c0T\"*F0$\"3A&)pmrh***z\"FT7$$\"3#*******H,Q+5FT$\"3 9n9ETjz%z\"FT7$$\"3)*******\\*3q3\"FT$\"3YT)=Gk&H!y\"FT7$$\"3)*******p =\\q6FT$\"3'\\U#4dEpfFT$\"3_>!=H==:[\"FT7$$\"3im mmTc-)*>FT$\"3!=NHWYq/X\"FT7$$\"3Mmm;f`@'3#FT$\"3J*>I$33b>9FT7$$\"3y** **\\nZ)H;#FT$\"3I&\\?E.AQR\"FT7$$\"3YmmmJy*eC#FT$\"3sT!p[MFtO\"FT7$$\" 3')******R^bJBFT$\"3p^Hs+HRT8FT7$$\"3f*****\\5a`T#FT$\"3/k\"R2YQuJ\"FT 7$$\"3o****\\7RV'\\#FT$\"3)\\(4*))4#e&H\"FT7$$\"3k*****\\@fke#FT$\"3)[ oI![X!GF\"FT7$$\"3/LLL`4NnEFT$\"3q8H+[pj`7FT7$$\"3#*******\\,s`FFT$\"3 8o0\")R&zWB\"FT7$$\"3[mm;zM)>$GFT$\"3PW9KH+C=7FT7$$\"3$*******pfaX-g6FT7$$\"3#)**** \\7yh]KFT$\"3))[N*o7$\\Z6FT7$$\"3xmmm')fdLLFT$\"3C;&o-F,j8\"FT7$$\"3bm mm,FT=MFT$\"3bjX`s+pD6FT7$$\"3FLL$e#pa-NFT$\"3@M9ARg%f6\"FT7$$\"3!**** ***Rv&)zNFT$\"3^!oC*)HKw5\"FT7$$\"3ILLLGUYoOFT$\"33ibKos!))4\"FT7$$\"3 _mmm1^rZPFT$\"3Cai<$33:4\"FT7$$\"34++]sI@KQFT$\"3gdZZl.I%3\"FT7$$\"34+ +]2%)38RFT$\"3%*yDHq!>z2\"FT7$$\"\"%F)$\"3=CaTU_er5FT-%'COLOURG6&%$RGB G$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"t6\"Q!F_^l-%%VIEWG6$;F(F`]l%(DEFAULT G" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" } }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "solve(D(g)(t));\ng(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-% #lnG6##\"\"#\"\"&!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"*\"\"& " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(ln(5/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$ \"+>t!H;*!#5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________ ________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_______________________________ ______" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q6 " }}{PARA 0 "" 0 "" {TEXT -1 127 "A sky diver with a mass of 73 kg (including pack) opens his/her parachute wh en he/she has reached a speed of 10 metres per sec. " }}{PARA 0 "" 0 " " {TEXT -1 154 "Find his/her subsequent velocity as a function of time with each of the following assumptions regarding the air resistance a fter the parachute has opened." }}{PARA 0 "" 0 "" {TEXT -1 39 "(a) Sup pose that the air resistance is " }{XPPEDIT 18 0 "150*v;" "6#*&\"$]\" \"\"\"%\"vGF%" }{TEXT -1 15 " Newtons where " }{TEXT 366 1 "v" }{TEXT -1 35 " is the velocity in metres per sec." }}{PARA 0 "" 0 "" {TEXT -1 39 "(b) Suppose that the air resistance is " }{XPPEDIT 18 0 "30*v^2 " "6#*&\"#I\"\"\"*$%\"vG\"\"#F%" }{TEXT -1 15 " Newtons where " } {TEXT 353 1 "v" }{TEXT -1 35 " is the velocity in metres per sec." }} {PARA 0 "" 0 "" {TEXT -1 60 "Take the acceleration due to gravity to b e 9.807 metres per " }{XPPEDIT 18 0 "sec^2" "6#*$%$secG\"\"#" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 4 "(a) " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "73*diff(v(t),t)=73*9.807-150*v(t),v(0)=10 ;\ndsolve(\{%\},v(t));\nevalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/ ,$*&\"#t\"\"\"-%%diffG6$-%\"vG6#%\"tGF.F'F',&$\"'6fr!\"$F'*&\"$]\"F'F+ F'!\"\"/-F,6#\"\"!\"#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\" tG,&#\"'P'Q#\"&++&\"\"\"*&#\"'j8EF+F,-%$expG6#,$*(\"$]\"F,\"#t!\"\"F'F ,F7F,F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"tG,&$\"+++usZ! \"*\"\"\"*&$\"+++EF_F+F,-%$expG6#,$*&$\"+@Xza?F+F,F'F,!\"\"F,F," }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "m := 73;\ng := 9.807;\nk := 150:\nevalf(m*g/k):\na := evalf(%,5); \nevalf(k/m):\nb := evalf(%,5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"mG\"#t" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG$\"%2)*!\"$" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\"&Fx%!\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"&[0#!\"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "eval(a-(a-10)*exp(-b*t));\ng := unapply(%,t):\nplot(g(t),t=0..2,0..10,labels=[`t`,`velocity v(t)`] );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&$\"&Fx%!\"%\"\"\"*&$\"&tA&F&F' -%$expG6#,$*&$\"&[0#F&F'%\"tGF'!\"\"F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "(b) " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "m := 73;\ng := 9.807;\nk := 30:\nevalf(sqrt(m*g/k)): \na := evalf(%,5);\nevalf((10-a)/(10+a)):\nA := evalf(%,5);\nevalf(2*a *k/m):\nb := evalf(%,5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\"#t " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG$\"%2)*!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\"&])[!\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG$\"&jV$!\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"&^,%! \"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "eval(a*((1+A*exp(-b*t))/(1-A*exp(-b*t))));\ng := una pply(%,t):\nplot(g(t),t=0..2,0..10,labels=[`t`,`velocity v(t)`]);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*($\"&])[!\"%\"\"\",&F(F(*&$\"&jV$! \"&F(-%$expG6#,$*&$\"&^,%F'F(%\"tGF(!\"\"F(F(F(,&F(F(*&$F,F-F(F.F(F6F6 F(" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "__ ___________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q7 " }} {PARA 0 "" 0 "" {TEXT -1 47 "The differential equation relating the cu rrent " }{TEXT 321 1 "I" }{TEXT -1 46 " in amps, in a particular RL ci rcuit, to time " }{TEXT 320 1 "t" }{TEXT -1 17 " in seconds is 2 " } {XPPEDIT 18 0 "dI/dt+4*I = V(t);" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\" \"%F'%\"IGF'F'-%\"VG6#%\"tG" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 28 "where the voltage source is " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "V(t) = PIECEWISE([0, t < 0],[26*cos*3*t, 0 <= t and \+ t <= Pi/2],[0, Pi/2 < t]);" "6#/-%\"VG6#%\"tG-%*PIECEWISEG6%7$\"\"!2F' F,7$**\"#E\"\"\"%$cosGF1\"\"$F1F'F131F,F'1F'*&%#PiGF1\"\"#!\"\"7$F,2*& F8F1F9F:F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 25 "Suppose th at the current " }{TEXT 322 1 "I" }{TEXT -1 11 " is 0 when " } {XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 69 "(a) Find a (piecewise) formula for the current as a fun ction of time " }{TEXT 399 1 "t" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "t >=0" "6#1\"\"!%\"tG" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" {TEXT -1 34 " (b) Sketch a graph of the current " }{TEXT 400 1 "I" }{TEXT -1 14 " ag ainst time " }{TEXT 401 1 "t" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "0<=t " "6#1\"\"!%\"tG" }{XPPEDIT 18 0 "``<=4" "6#1%!G\"\"%" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT 323 4 "Hint" }{TEXT -1 21 ": From example 3 \+ for " }{XPPEDIT 18 0 "0<=t" "6#1\"\"!%\"tG" }{XPPEDIT 18 0 "``<=Pi/2" "6#1%!G*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 25 " the current is given by " }{XPPEDIT 18 0 "I = 3*sin*3*t+2*cos*3*t-2*exp(-2*t)" "6#/%\"IG,(** \"\"$\"\"\"%$sinGF(F'F(%\"tGF(F(**\"\"#F(%$cosGF(F'F(F*F(F(*&F,F(-%$ex pG6#,$*&F,F(F*F(!\"\"F(F4" }{TEXT -1 18 ". This means that " } {XPPEDIT 18 0 "I=-3-2*exp(-Pi)" "6#/%\"IG,&\"\"$!\"\"*&\"\"#\"\"\"-%$e xpG6#,$%#PiGF'F*F'" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=Pi/2" "6#/% \"tG*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 4 "For " }{XPPEDIT 18 0 "t>Pi/2" "6#2*&%#PiG\"\"\"\"\"#!\"\"%\"tG" }{TEXT -1 37 " the differential equation becomes 2 " }{XPPEDIT 18 0 "d I/dt+4*I=0" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&\"\"%F'%\"IGF'F'\"\"!" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 36 "To find the formula for the current " }{TEXT 324 1 "I" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t >Pi/2" "6#2*&%#PiG\"\"\"\"\"#!\"\"%\"tG" }{TEXT -1 35 " solve the diff erential equation 2 " }{XPPEDIT 18 0 "dI/dt+4*I=0" "6#/,&*&%#dIG\"\"\" %#dtG!\"\"F'*&\"\"%F'%\"IGF'F'\"\"!" }{TEXT -1 34 " subject to the ini tial condition " }{XPPEDIT 18 0 "I=-3-2*exp(-Pi)" "6#/%\"IG,&\"\"$!\" \"*&\"\"#\"\"\"-%$expG6#,$%#PiGF'F*F'" }{TEXT -1 6 " when " }{XPPEDIT 18 0 "t=Pi/2" "6#/%\"tG*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "An s " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 191 "de := 2*diff(i(t),t)+4*i(t)=piecewise(t%#deG/,&*&\"\"#\"\"\"-%%diff G6$-%\"iG6#%\"tGF0F)F)*&\"\"%F)F-F)F)-%*PIECEWISEG6$7$,$*&\"#EF)-%$cos G6#,$*&\"\"$F)F0F)F)F)F)2F0,$*&F(!\"\"%#PiGF)F)7$\"\"!%*otherwiseG" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG-%*PIECEWISEG6$7$,(*&\"\"#\"\"\"-%$c osG6#,$*&\"\"$F/F'F/F/F/F/*&F5F/-%$sinGF2F/F/*&F.F/-%$expG6#,$*&F.F/F' F/!\"\"F/F?2F',$*&F.F?%#PiGF/F/7$,&*&F.F/F:F/F?*&F5F/-F;6#,&*&F.F/F'F/ F?FCF/F/F?1FAF'" }}{PARA 13 "" 1 "" {GLPLOT2D 491 252 252 {PLOTDATA 2 "6&-%'CURVESG6#7go7$$\"\"!F)F(7$$\"39LLLL3VfV!#>$\"3:_3CN.V6a!#=7$$\"3 Hmmmm;')=()F-$\"3$Q#*RE#HzF5!#<7$$\"3-++]7z>^7F0$\"3IbsuXTM.9F67$$\"3R LLLe'40j\"F0$\"3$*G41nSjJ&*pQ(f#F67$$\"3KLL3_ !\\hb$F0$\"3g4r6eIu5EF67$$\"3Sm;zp'*)Hm$F0$\"3\")Rx_X*f*>EF67$$\"3/++] (GI)pPF0$\"3?N+H)**e]i#F67$$\"37L$3_!4nwQF0$\"3wdO)f8lgi#F67$$\"3Amm\" H_6N)RF0$\"3()f*y]%o+BEF67$$\"3')**\\iS@N!4%F0$\"3$=Q`J;;fh#F67$$\"3%H LL$eF>(>%F0$\"3%e*\\K'*)\\F0$\"3w)Q;'RJ?-CF67$$\"3P*****\\Kd,\"eF0$\"35m'\\;s.r)>F 67$$\"39KLLe9XMiF0$\"3d-&>vwj9q\"F67$$\"3-mmm\"fX(emF0$\"3M\\6J'Q^[P\" F67$$\"3_KLL3!z;3(F0$\"3E#o;LV^[,\"F67$$\"3.*****\\U7Y](F0$\"3!3*3hs_h siF07$$\"3Wmm;H9lRzF0$\"3wAEv))[?t?F07$$\"3'QLLLV!pu$)F0$!3yu$4;$R%>E# 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***\\7RV'\\#F6$!3#f(zq#>Fo%[F07$$\"3k*****\\@fke#F6$!3c]PI4b?[SF07$$\" 3/LLL`4NnEF6$!3]sXI)G1NW$F07$$\"3#*******\\,s`FF6$!3)o#H6*\\?s*GF07$$ \"3[mm;zM)>$GF6$!3K*Gks^QuZ#F07$$\"3$*******pfa]\"F07$$ \"3;LLL$)G[kJF6$!3S61BkD3u7F07$$\"3#)****\\7yh]KF6$!3'>\"RCK:Ys5F07$$ \"3xmmm')fdLLF6$!3'GLBg'\\+&3*F-7$$\"3bmmm,FT=MF6$!3)[Liz^-sm(F-7$$\"3 FLL$e#pa-NF6$!3+:!odLf(zkF-7$$\"3!*******Rv&)zNF6$!3c')Qm\"em9b&F-7$$ \"3ILLLGUYoOF6$!3G^9_(=9*\\YF-7$$\"3_mmm1^rZPF6$!33itBwlLoRF-7$$\"34++ ]sI@KQF6$!3s:,[/#48N$F-7$$\"34++]2%)38RF6$!3[3(Q&Q@!3&GF-7$$\"\"%F)$!3 [IM?%zVfR#F--%+AXESLABELSG6$%\"tG%-current~I(t)G-%&COLORG6&%$RGBG$\"\" &!\"\"F($\"\"\"F)-%%VIEWG6$;F(Febl%(DEFAULTG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q8 " }} {PARA 0 "" 0 "" {TEXT -1 32 "Solve the differential equation " }} {PARA 256 "" 0 "" {TEXT -1 4 " L " }{XPPEDIT 18 0 "dI/dt + R*I=V(t)" "6#/,&*&%#dIG\"\"\"%#dtG!\"\"F'*&%\"RGF'%\"IGF'F'-%\"VG6#%\"tG" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 16 "for the current " } {TEXT 402 1 "I" }{TEXT -1 30 " in a series RL circuit given " } {XPPEDIT 18 0 "L = 1" "6#/%\"LG\"\"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "R = 3" "6#/%\"RG\"\"$" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "V(t) = \+ sin*2*t;" "6#/-%\"VG6#%\"tG*(%$sinG\"\"\"\"\"#F*F'F*" }{TEXT -1 29 ", \+ with the initial condition " }{XPPEDIT 18 0 "I(0) = 2;" "6#/-%\"IG6#\" \"!\"\"#" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Plot a graph \+ of the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 " " 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "de := diff(i(t),t)+3*i(t)=sin(2*t) ;\nic := i(0)=2;\ndsolve(\{de,ic\},i(t));\ng := unapply(rhs(%),t);\npl ot(g(t),t=0..6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6 $-%\"iG6#%\"tGF-\"\"\"*&\"\"$F.F*F.F.-%$sinG6#,$*&\"\"#F.F-F.F." }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"iG6#\"\"!\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#%\"tG,(*&#\"\"#\"#8\"\"\"-%$cosG6#,$*& F+F-F'F-F-F-!\"\"*&#\"\"$F,F--%$sinGF0F-F-*&#\"#GF,F--%$expG6#,$*&F6F- F'F-F3F-F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#%\"tG6\"6$%)op eratorG%&arrowGF(,(*&#\"\"#\"#8\"\"\"-%$cosG6#,$*&F/F19$F1F1F1!\"\"*&# \"\"$F0F1-%$sinGF4F1F1*&#\"#GF0F1-%$expG6#,$*&F;F1F7F1F8F1F1F(F(F(" }} {PARA 13 "" 1 "" {GLPLOT2D 624 346 346 {PLOTDATA 2 "6%-%'CURVESG6$7\\o 7$$\"\"!F)$\"\"#F)7$$\"3')*****\\7t&pK!#>$\"3#*3*Q>BuT\"=!#<7$$\"3s*** ***\\i9RlF/$\"3a:p?c:uZ;F27$$\"3e*****\\P>(3)*F/$\"3a?Q8op())\\\"F27$$ \"3%*******\\#HyI\"!#=$\"3%ph=kM*R.yAn+\"F27$$\"3K++D\"3Gc3$F@$\"3c3+x=/c: ')F@7$$\"3++++v;\\DPF@$\"3!esc(R=.yuF@7$$\"3A++++nfpVF@$\"3g@&3A#Hj)e' F@7$$\"3W+++DA(Gy #)\\F@7$$\"3e****\\(G[W[(F@$\"3e$QLH!oaoWF@7$$\"3i****\\()fB:()F@$\"3^ 1;]6G$Q6%F@7$$\"39++](Q=\"))**F@$\"3sUz/X1s8QF@7$$\"3(****\\P'=pD6F2$ \"3iz>v5)*4(\\$F@7$$\"33+++lN?c7F2$\"3wVf)y%*)4*4$F@7$$\"3-++]U$e6P\"F 2$\"3?*)4;F6umEF@7$$\"36+++&>q0]\"F2$\"3/9/@F2$!34S;,4:$QJ\"F@7$$\"3'****\\(3wY_AF2$!3y+0RzPI;>F@7$$\"3#)***** *HOTqBF2$!3.mC`u;$HL#F@7$$\"37++v3\">)*\\#F2$!3.)*Q!oS*3PEF@7$$\"39++] <9VhDF2$!3l5`=%f-+s#F@7$$\"3:++DEP/BEF2$!3Sp4Dfpf@AL5F@7$$\"37++]Zn%)oLF2$!3E1Rm(p\"p#o$F/7$$\"3y**** **4FL(\\$F2$\"3v\\!GqqK>U$F/7$$\"3#)****\\d6.BOF2$\"3hVV$*=J%f,\"F@7$$ \"3(****\\(o3lWPF2$\"384?&zwDvg\"F@7$$\"3!*****\\A))ozQF2$\"3!y>H'[aB_ @F@7$$\"3e******Hk-,SF2$\"3%fFKkhU&4DF@7$$\"3S****\\FL!e1%F2$\"3=^DCmq 3TEF@7$$\"36+++D-eITF2$\"3'\\fsF#4PGFF@7$$\"3Y**\\(=sx#*=%F2$\"3MEYci' )*zw#F@7$$\"3u***\\(=_(zC%F2$\"3)3!*y(\\h_pFF@7$$\"3/+](o3Z@J%F2$\"3_% *4Su!*eFFF@7$$\"3M+++b*=jP%F2$\"32[._gsySEF@7$$\"3g***\\(3/3(\\%F2$\"3 eiRrdkMhBF@7$$\"33++vB4JBYF2$\"3A=Wug%*3B>F@7$$\"3u*****\\KCnu%F2$\"3I pWn9L]w8F@7$$\"3s***\\(=n#f([F2$\"3Y$p.qyt]:(F/7$$\"3P+++!)RO+]F2$\"3e Ga)f?oLM$Fbr7$$\"30++]_!>w7&F2$!35S)yh3E)emF/7$$\"3O++v)Q?QD&F2$!3vY$= $)\\crJ\"F@7$$\"3G+++5jyp`F2$!3R:!3d\">#H%=F@7$$\"3<++]Ujp-bF2$!3?9?f- YoABF@7$$\"3++++gEd@cF2$!3lq,z`0L9EF@7$$\"31+]PMh%\\o&F2$!3R4tuL,U5FF@ 7$$\"39++v3'>$[dF2$!35%H#G-c-jFF@7$$\"3p******4h(*3eF2$!3m()3e))='=x#F @7$$\"37++D6EjpeF2$!3nHTj!4b*RFF@7$$\"3^+]i0j\"[$fF2$!3;Z-s7uzgEF@7$$ \"\"'F)$!3kNP%ps\"[ODF@-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6 $Q\"t6\"Q!F``l-%%VIEWG6$;F(Fa_l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q9 " }} {PARA 0 "" 0 "" {TEXT -1 336 "Suppose that two substances A and B reac t to form a substance C and that the initial concentrations of A and B are 100 and 90 moles per litre respectively. Suppose that the reactio n rate in moles per litre per hour is proportional to the product of t he concentrations of the the two substances, where the constant of pro portionality is " }{XPPEDIT 18 0 "k = 1/100" "6#/%\"kG*&\"\"\"F&\"$+\" !\"\"" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Find a formula for the concentration of substance C \+ after " }{TEXT 279 1 "t" }{TEXT -1 88 " hours, and plot a graph to sho w how the concentration of substance C changes with time." }}{PARA 0 " " 0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 4 "Q10 " }} {PARA 0 "" 0 "" {TEXT -1 272 "Suppose that two substances A and B reac t to form a substance C. If the initial concentrations of A and B are \+ 100 and 90 moles per litre respectively, and after 1 hour the concentr ation of C is 45 moles per litre, find a formula for the concentration of substance C after " }{TEXT 280 1 "t" }{TEXT -1 7 " hours." }} {PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 4 "Q11 " }}{PARA 0 "" 0 "" {TEXT -1 15 "A body of mass " }{TEXT 283 1 "m " }{TEXT -1 168 " falling through a viscous medium encounters a resist ing force proportional to its instantaneous velocity. In this situatio n the differential equation for the velocity " }{XPPEDIT 18 0 "v = v(t )" "6#/%\"vG-F$6#%\"tG" }{TEXT -1 9 " at time " }{TEXT 281 1 "t" } {TEXT -1 4 " is " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 282 1 "m" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dv/dt = m*g-k*v;" "6#/*&%#dvG\"\"\"%#d tG!\"\",&*&%\"mGF&%\"gGF&F&*&%\"kGF&%\"vGF&F(" }{TEXT -1 2 ", " }} {PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 284 1 "k" }{TEXT -1 44 " is a positive constant of proportionality. " }}{PARA 0 "" 0 "" {TEXT -1 57 "Solve the differential equation subject to the condition " } {XPPEDIT 18 0 "v(0) = v[0];" "6#/-%\"vG6#\"\"!&F%6#F'" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 51 "What is the limiting velocity of the \+ falling body? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "g := 'g':\nde := m*diff(v(t),t)=m*g-k*v( t);\nic := v(0)=v0;\n#dsolve(\{de,ic\},v(t));\ndesolve(\{de,ic\},v(t), info=true);\nsimplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/*& %\"mG\"\"\"-%%diffG6$-%\"vG6#%\"tGF/F(,&*&F'F(%\"gGF(F(*&%\"kGF(F,F(! \"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"vG6#\"\"!%#v0G" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #%DThe~DE~has~separable~variables~.~.~G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%\"mG\"\"\"-%$IntG6$*&F&F&,&*&F%F&%\"gGF&F&*&%\"kGF&%\"vGF&! \"\"F1F0F&%\"tG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*&%\"mG\"\"\"-%#lnG6#,&*&F&F'%\"gGF'F'*&%\"kGF' %\"vGF'!\"\"F'F1,&*&%\"tGF'F/F'F'&%\"CG6#F'F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/),&*&%\"mG\"\" \"%\"gGF(F(*&%\"kGF(%\"vGF(!\"\",$F'F-*&&%\"CG6#\"\"#F(-%$expG6#*&%\"t GF(F+F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%EApplying~the~initial~co ndition~.~.~~G/&%\"CG6#\"\"#),&*&%\"mG\"\"\"%\"gGF-F-*&%\"kGF-%#v0GF-! \"\",$F,F2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"tG*&,&-%$expG6#,$*&,&*&F'\"\"\"%\"kGF1F1*&% \"mGF1-%#lnG6#,&*&F4F1%\"gGF1F1*&F2F1%#v0GF1!\"\"F1F=F1F4F=F=F=F9F1F1F 2F=" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"tG*&,&-%$expG6#,$*& ,&*&F'\"\"\"%\"kGF1F1*&%\"mGF1-%#lnG6#,&*&F4F1%\"gGF1F1*&F2F1%#v0GF1! \"\"F1F=F1F4F=F=F=F9F1F1F2F=" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "____________ _________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 " ;" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {SECT 1 {PARA 0 "" 0 "" {TEXT -1 25 "Code for drawing pictures" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 0 "" 0 " " {TEXT -1 10 "LR circuit" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 623 "coil := PLOT(CURVES([evalf(seq([24 -sin(r/5),cos(r/5)+r/15-15],r=16..158))])):\nsource :=\nPLOT(CURVES([e valf(seq([3*cos(r/5)-12.5,3*sin(r/5)-10],r=0..32))])):\nwire := PLOT(C URVES([[12.54,0],[24,0]],[[-12.5,0],[-0.5,0]],\n[[24,0],[24,-3.6]],[[2 4,-15.0],[24,-20]],[[-12.5,0],[-12.5,-7]],\n[[-12.5,-13],[-12.5,-20]], [[-12.5,-20],[24,-20]])):\nresistor := PLOT(CURVES([[-0.5,0],[0,1.5],[ 1,-1.5],[2,1.5],\n[3,-1.5],[4,1.5],[5,-1.5],[6,1.5],[7,-1.5],[8,1.5], \n[9,-1.5],[10,1.5],[11,-1.5],[12,1.5],[12.5,0]])):\nt1:=plots[textplo t]([[7,4,`R`],[27.5,-6.5,`L`],[-12.5,-10,`V(t)`]]):\nplots[display](\{ coil,wire,source,resistor,t1\},axes=NONE);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 12 "Falling body" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 330 "p1 := plottools[disk ]([0,0],.5,color=COLOR(RGB,.7,.7,.7)):\np2 := plottools[arrow]([0,.5], [0,3],0,.6,.2,arrow,color=black,thickness=2):\np3 := plottools[arrow]( [0,-.5],[0,-3],0,.6,.2,arrow,color=black,thickness=2):\nt1 := plots[te xtplot]([[1.4,-2.2,`mg`],[1.4,2.2,`kv`]]):\nplots[display]([p1,p2,p3,t 1],axes=none,scaling=constrained);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 388 "p1 := plottools[disk]([0,0 ],.5,color=COLOR(RGB,.7,.7,.7)):\np2 := plottools[arrow]([0,.5],[0,3], 0,.6,.2,arrow,color=black,thickness=2):\np3 := plottools[arrow]([0,-.5 ],[0,-3],0,.6,.2,arrow,color=black,thickness=2):\nt1 := plots[textplot ]([[1.4,-2.2,`mg`],[1.4,2.2,`kv`]]):\nt2 := plots[textplot]([2,2.5,`2` ],font=[HELVETICA,8]):\nplots[display]([p1,p2,p3,t1,t2],axes=none,scal ing=constrained);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 10 "R C circuit" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 587 "source :=\nPLOT(CURVES([evalf(seq([3*cos(r/5)-12.5,3 *sin(r/5)-10],r=0..32))])):\nwire := PLOT(CURVES([[12.54,0],[24,0]],[[ -12.5,0],[-0.5,0]],\n[[24,0],[24,-9.4]],[[24,-10.6],[24,-20]],[[20,-9. 4],[28,-9.4]],\n[[20,-10.6],[28,-10.6]],[[-12.5,0],[-12.5,-7]],\n[[-12 .5,-13],[-12.5,-20]],[[-12.5,-20],[24,-20]])):\nresistor := PLOT(CURVE S([[-0.5,0],[0,1.5],[1,-1.5],[2,1.5],\n[3,-1.5],[4,1.5],[5,-1.5],[6,1. 5],[7,-1.5],[8,1.5],\n[9,-1.5],[10,1.5],[11,-1.5],[12,1.5],[12.5,0]])) :\nt1:=plots[textplot]([[7,4,`R`],[27.5,-6.5,`C`],[-12.5,-10,`V(t)`]]) :\nplots[display](\{wire,source,resistor,t1\},axes=NONE);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }} }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "4 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }