{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "restart;\nwith(plots ):\nwith(DEtools):" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 50 "Example 1: Radioactive Decay (Separable Diff. Eq.)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "General form of differential equation" }{MPLTEXT 1 0 1 " \n" }{TEXT -1 21 " k<0: radioactive " }{TEXT 258 5 "decay" }{TEXT -1 22 "\n k>0: exponential " }{TEXT 259 6 "growth" }{MPLTEXT 1 0 29 "\nDE1:=diff(y(x), x) = k*y(x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "k:=-.181; #approximate dec ay constant for radon-222\nDEplot(DE1, y(x), x=-5..20, y=-5..20);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "initialCondition:=[[0,20]]; \nDEplot(DE1, y(x), x=-5..20, initialCondition,y=-5..20,linecolor=blue );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Solve for the " }{TEXT 257 16 "general solution" }{TEXT -1 29 " to the differential equation" } {MPLTEXT 1 0 19 "\ndsolve(DE1, y(x));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Use the initial condition " }{XPPEDIT 18 0 "y(0) = 20;" " 6#/-%\"yG6#\"\"!\"#?" }{TEXT -1 39 " to get the particular solution to the " }{TEXT 256 21 "initial value problem" }{TEXT -1 0 "" }{MPLTEXT 1 0 59 "\nsubs(\{y(x)=20,x=0\},y(x)=C1*exp(-181/1000*x));\nsolve(%,C1) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "y:=x->20*exp(-181/100 0*x);\nplot(y(x),x=-5..20,y=-5..20,title=\"Particular solution for Rad on decay\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 0 "" }{TEXT 261 22 "Half-life of Radon-222" }{MPLTEXT 1 0 1 "\n" }{TEXT -1 56 "Half-life = time at which 1/2 of or iginal amount remains" }{MPLTEXT 1 0 43 "\nsolve(y(x)=(1/2)*y(0),x);\n y(%);\nevalf(%%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "plot( \{y(x),10\},x=-5..20,y=-5..20,title=\"Illustration of half-life of Rad on 222\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 50 "Example 2: Investment growth (Separable D iff. Eq.)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "General form of diffe rential equation" }{MPLTEXT 1 0 1 "\n" }{TEXT -1 21 " k<0: radioact ive " }{TEXT 264 5 "decay" }{TEXT -1 22 "\n k>0: exponential " } {TEXT 265 6 "growth" }{MPLTEXT 1 0 72 "\nk:='k': C1:='C1': y:='y': #cl ear variables\nDE1:=diff(y(x), x) = k*y(x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "k:=.07; #constant interest rate of investment\nDEplot(DE1, y(x), x=-5..20, y=-5..600); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "initialCondition:=[[0, 100]];\nDEplot(DE1, y(x), x=-5..20, initialCondition,y=-5..500,linecol or=blue,title=\"Direction field and Euler's method particular solution \");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Solve for the " }{TEXT 263 16 "general solution" }{TEXT -1 29 " to the differential equation " }{MPLTEXT 1 0 19 "\ndsolve(DE1, y(x));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Use the initial condition " }{XPPEDIT 18 0 "y(0) = 100;" "6#/-%\"yG6#\"\"!\"$+\"" }{TEXT -1 39 " to get the particular solution to the " }{TEXT 262 21 "initial value problem" }{TEXT -1 0 "" } {MPLTEXT 1 0 56 "\nsubs(\{y(x)=100,x=0\},y(x)=C1*exp(7/100*x));\nsolve (%,C1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "y:=x->100*exp(7 /100*x);\nplot(y(x),x=0..20,y=-5..600,title=\"Particular solution to i nvestment initial value problem\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 266 0 "" }{TEXT 267 27 "Doubling time of investment" }{MPLTEXT 1 0 1 " \n" }{TEXT -1 70 "Doubling time = time at which investment is twice o f original amount " }{MPLTEXT 1 0 50 "\nsolve(y(x)=2*y(0),x);\n\"Doubl ing time\" = evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "p lot(\{y(x),200,400\},x=0..20,y=-5..600,title=\"Illustration of doublin g times for 7% interest\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 268 44 "Plot of doubling times versus interest rates" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "plot(ln(2)/r,r=0.03..0.15,y=0..40, labels=[\"Interest rate\",\"Doubling time (yrs)\"],title=\"Doubling ti me decreases as interest rate increases\");" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 37 "Modified to take account of inflation" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 225 "plot([ln(2)/r,ln(2)/((1+r)/(1+0.03)-1)], r=0.03..0.15,y=0..40,labels=[\"Interest rate\",\"Doubling time (yrs)\" ],legend=[\"without inflation\",\"with 3% inflation\"],title=\"Doublin g time vs. interest rate with and without inflation\");" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 64 "Example 3: Newton's Law of \+ Heating/Cooling (Separable Diff. Eq.)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "A differential equation the temperature of an object in ambient space" }{MPLTEXT 1 0 74 "\ny:='y': k:='k': C1:='C1': #Clear variables \nDE3:=diff(y(x),x)=k*(y(x)-TS);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "k:=(1/30)*ln(17/28);\nevalf(k);\nTS:=44;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 288 "k:=(1/30)*ln(17/28);\nDEplot(DE3, \+ y(x), x=0..300,[[0,20]],y=20..80,title=\"Euler's method particular sol ution for Newton's Law of Heating/Cooling\");\nk:=(1/10)*(1/30)*ln(17/ 28):\nDEplot(DE3, y(x), x=0..300,[[0,20]],y=20..80,title=\"Euler's met hod particular solution, smaller (negative) k\");\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "A particular solution plotted on the slop e field" }{MPLTEXT 1 0 209 "\nk:=-.016;\ninitialCondition:=[[0,72],[0, 44],[0,20]];\nDEplot(DE3, y(x), x=0..300,initialCondition,y=20..80,lin ecolor=[blue,red,magenta],title=\"Euler's method for several Heating/C ooling particular solutions\");\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 " " 0 "" {TEXT -1 32 "Example 4: The Logistic Equation" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "A differential equation for logistic population growth " }{MPLTEXT 1 0 80 "\ny:='y': k:='k': C1:='C1': #Clear variabl es\nDE4:=diff(y(x),x)=k*y(x)*(1-y(x)/K);" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 269 19 "Exercise 3, p. 666:" }{TEXT -1 74 "\nThe Pacific halibut fishery has been mode led by the differential equation" }}{PARA 0 "" 0 "" {TEXT -1 20 " \+ " }{XPPEDIT 18 0 "dy/dt = k*y*(1-y/K);" "6#/*&%#dyG\"\" \"%#dtG!\"\"*(%\"kGF&%\"yGF&,&F&F&*&F+F&%\"KGF(F(F&" }{TEXT -1 1 " " } }{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "y(t);" "6#-%\"yG6 #%\"tG" }{TEXT -1 151 " is the biomass (the total mass of the members \+ of the population) in kilograms at time t (measured in years), the car rying capacity is estimated to be " }{XPPEDIT 18 0 "K = 8*10^7;" "6#/% \"KG*&\"\")\"\"\"*$\"#5\"\"(F'" }{TEXT -1 9 " kg, and " }{XPPEDIT 18 0 "k = .71;" "6#/%\"kG-%&FloatG6$\"#r!\"#" }{TEXT -1 11 " per year.\n " }{TEXT 270 3 "(a)" }{TEXT -1 5 " If " }{XPPEDIT 18 0 "y(0) = 2*10^7 ;" "6#/-%\"yG6#\"\"!*&\"\"#\"\"\"*$\"#5\"\"(F*" }{TEXT -1 36 " kg, fin d the biomass a year later.\n" }{TEXT 271 3 "(b)" }{TEXT -1 48 " How l ong will it take for the biomass to reach " }{XPPEDIT 18 0 "4*10^7;" " 6#*&\"\"%\"\"\"*$\"#5\"\"(F%" }{TEXT -1 9 " kg? " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Slope field for Exercise 3: model of a Pacific \+ halibut fishery. " }{MPLTEXT 1 0 268 "\nk:=.71; #relative rate o f increase per year \nK:=8*10^7; #carrying capacity, in kg\ninitial Condition:=[[0,2*10^7]];\nplot1:=DEplot(DE4, y(x), x=0..5,initialCondi tion,y=0..10^8,linecolor=[blue],title=\"Euler's method for halibut log istic growth\"):\ndisplay(plot1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Find the biomass of halibut at the end of the first year if the in itial biomass is 2*10^7." }{MPLTEXT 1 0 33 "\ndsolve(\{DE4,y(0)=2*10^7 \}, y(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "y:=x-> 800000 00/(1+3*exp(-71/100*x)):\nevalf(y(1));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "How log will it take for the biomass to reach 4*10^7 kg? " }{MPLTEXT 1 0 32 "\nsolve(y(x)=4*10^7,x);\nevalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "plot2:=plot(4*10^7,x=0..5,color=ma genta,thickness=3):\ndisplay(\{plot1,plot2\},title=\"Halibut logistic \+ growth from 2*10^7 to 4*10^7\");" }}}}}{MARK "4" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }