"}]\)}], "Input"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["(b)", "Section"], Cell["What happens to the quantity as it gets close to q = 0?", "Text"], Cell[CellGroupData[{ Cell["Solution to (b)", "Subsection"], Cell["\<\ The quantity is still decreasing, but the slope is getting very small because \ q is very small. So the graph is \"flattening.\"\ \>", "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["(c)", "Section"], Cell[TextData[{ "Find an analytical solution to the conditions q[0] = 4 and\n\n", Cell[BoxData[ \(dq\/dt\)]], " = ", Cell[BoxData[ \(\(-\(q\/2\)\)\)]], ". (See Chapter 8. This is not needed for the sketch.)" }], "Text"], Cell[CellGroupData[{ Cell["Solution to (c)", "Subsection"], Cell[TextData[{ "From Section 8.2 (Exercise Set 8.2 number 1(d)) we know that our solution \ is q = ", Cell[BoxData[ \(4\ e\^\(k\ t\)\)]], " where k = ", Cell[BoxData[ \(\(-\(1\/2\)\)\)]] }], "Text"], Cell[TextData[{ "We can check this solution.\n\n\tq[0] = ", Cell[BoxData[ \(4\ e\^0\)]], "= 4\n\t\n\t", Cell[BoxData[ \(dq\/dt\)]], " = 4 k ", Cell[BoxData[ \(\(\ e\^\(k\ t\)\)\)]], "\t = ", Cell[BoxData[ \(\(-\(q\/2\)\)\)]] }], "Text"] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ "Exercise 2) Sketch Solutions", "Subtitle"]], "Subtitle"], Cell[TextData[{ "Sketch the solutions of\n\n\t\t", Cell[BoxData[ \(dy\/dt\)]], " = y(4 - y)\n\t\t\nand\t\t", Cell[BoxData[ \(dy\/dt\)]], " = 10y(4 - y)\nthat begin with" }], "Text"], Cell["(a)\ty[0] = 1", "Text"], Cell["(b)\ty[0] = 2", "Text"], Cell["(c)\ty[0] = 6", "Text"], Cell["(d)\ty[0] = 4", "Text"], Cell["(e)\ty[0] = 0", "Text"], Cell[CellGroupData[{ Cell[TextData[{ " Solution for ", Cell[BoxData[ \(dy\/dt\)]], " = y(4 - y)" }], "Subsection"], Cell[TextData[{ "First let's compute the second derivative.\n\n\t\t", Cell[BoxData[ \(\(d\^2\ y\)\/dt\^2\)]], "= ", Cell[BoxData[ \(\(dy\ \((4 - y)\)\)\/dt + \(y\ \((\(-\ dy\))\)\)\/dt\)]], "\n\t\t\n\t\t\n\t\t= ", Cell[BoxData[ \(\(dy\ \((4 - \ 2 y)\)\)\/dt\)]], " = ", "y (4-y) (4-2 y)" }], "Text"], Cell["\<\ (a)\tFor y[0] = 1 we know the graph starts at (0, 1) and the inital slope is \ 1(4 - 1) = 3 which means that y is increasing. As y approaches 4, the slope \ is still positive but becomes very small, so our graph flattens out as t gets \ as large as we want. The second derivative is initially 1(4 - 1)(4 - 2) = 6. \ It remains positive (a smile) until y = 2. For all y between 2 and 4 the \ second derivative is negative (frown). \ \>", "Text"], Cell[BoxData[{ \(Needs["\"]; \n eulerApprox[f_, {t0_, x0_}, dt_, final_]\ := \n Module[{steps, G}, \nsteps\ = \ Floor[final/dt]; \n G[{tt_, xx_}]\ := \ {tt + dt, xx\ + \ f[{tt, xx}]\ dt}; \n NestList[\ G\ , \ {t0, x0}\ , \ steps\ ]]\), \(Clear[t0, x0, G, f]; \nG[{t_, x_}] := {t + dt, x + f[t, x]\ dt}; \n Clear[f]\), \(f[t_, y_] := y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 1.0; \ndt = 0.1; \n finalTime = 2; \nsolnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \nplot1 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["\<\ (b)\tFor y[0] = 2 we have the same situation as (a) except that our graph \ starts at (0, 2) and the second derivative is initially zero, but since y is \ increasing, the second derivative quickly becomes negative (frown).\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 2.0; \ndt = 0.1; \n finalTime = 2; \nsolnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \nplot2 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["\<\ (c)\tFor y[0] = 6 we know the graph starts at (0, 6) and the inital slope is \ 6(4 - 6) = -12 which means that y is decreasing. As y approaches 4, the \ slope is still negative but becomes very small, so our graph flattens out as \ t gets as large as we want. The second derivative is initally 6(4 - 6)(4 - \ 12) = -96 and remains negative (frown) for all y between 6 and 4.\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 6.0; \ndt = 0.1; \n finalTime = 2; \nsolnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \nplot3 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["\<\ (d)\tFor y[0] = 4 we know the graph starts at (0, 4) and the inital slope is \ 4(4 - 4) = 0 which means that y starts at 4. The second derivative is \ initially also 0 which means that the first derivative does not change so our \ graph is a horizontal line, y = 4.\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 4.0; \ndt = 0.1; \n finalTime = 2; \nsolnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \nplot4 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["\<\ (e)\tFor y[0] = 0 we know the graph starts at (0, 0) and the inital slope is \ 0(4 - 0) = 0 which means that y is initially flat. The second derivative is \ initially also 0 which means that the first derivative does not change so our \ graph is a horizontal line, y = 0.\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 0.0; \ndt = 0.1; \n finalTime = 2; \nsolnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \nplot5 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["Now all 5 graphs together.", "Text"], Cell[BoxData[ \(\(\(Show[plot1, plot2, plot3, plot4, plot5]; \)\n\)\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ " Solution for ", Cell[BoxData[ \(dy\/dt\)]], " = 10 y(4 - y)" }], "Subsection"], Cell[TextData[{ "First let's compute the second derivative.\n\n\t\t", Cell[BoxData[ \(\(d\^2\ y\)\/dt\^2\)]], "= 10[", Cell[BoxData[ \(\(dy\ \((4 - y)\)\)\/dt + \(y\ \((\(-\ dy\))\)\)\/dt\)]], "]\n\t\t\n\t\t\n\t\t= 10[", Cell[BoxData[ \(\(dy\ \((4 - \ 2 y)\)\)\/dt\)]], "] = 10 y (4-y) (4-2 y)" }], "Text"], Cell["\<\ (a)\tFor y[0] = 1 we know the graph starts at (0, 1) and the inital slope is \ 10(4 - 1) = 30 which means that y is increasing even faster than in the first \ part. As y approaches 4, the slope is still positive but becomes very small, \ so our graph flattens out as t gets as large as we want. The second \ derivative is initially 10(4 - 1)(4 - 2) = 60. It remains positive (a smile) \ until y = 2. For all y between 2 and 4 the second derivative is negative \ (frown). \ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := 10\ y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 1.0; \ndt = 0.01; \nfinalTime = 2; \n solnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \n plot6 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Blue}]; \)}], "Input"], Cell["\<\ (b)\tFor y[0] = 2 we have the same situation as (a) except that our graph \ starts at (0, 2) and the second derivative is initially zero, but since y is \ increasing, the second derivative quickly becomes negative (frown).\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := 10\ y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 2.0; \ndt = 0.01; \nfinalTime = 2; \n solnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \n plot7 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Blue}]; \)}], "Input"], Cell["\<\ (c)\tFor y[0] = 6 we know the graph starts at (0, 6) and the inital slope is \ 60(4 - 6) = -120 which means that y is decreasing. As y approaches 4, the \ slope is still negative but becomes very small, so our graph flattens out as \ t gets as large as we want. The second derivative is initally 60(4 - 6)(4 - \ 12) = -960 and remains negative (frown) for all y between 6 and 4.\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := 10\ y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 6.0; \ndt = 0.01; \nfinalTime = 2; \n solnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \n plot8 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Blue}]; \n\)}], "Input"], Cell["\<\ (d)\tFor y[0] = 4 we know the graph starts at (0, 4) and the inital slope is \ 40(4 - 4) = 0 which means that y starts at 4. The second derivative is \ initially also 0 which means that the first derivative does not change so our \ graph is a horizontal line, y = 4.\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := 10\ y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 4.0; \ndt = 0.01; \nfinalTime = 2; \n solnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \n plot9 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Blue}]; \n\)}], "Input"], Cell["\<\ (e)\tFor y[0] = 0 we know the graph starts at (0, 0) and the inital slope is \ 0(4 - 0) = 0 which means that y is initially flat. The second derivative is \ initially also 0 which means that the first derivative does not change so our \ graph is a horizontal line, y = 0.\ \>", "Text"], Cell[BoxData[{ \(Clear[f]\), \(f[t_, y_] := 10\ y\ \((4\ - \ y)\); \nt0 = 0; \ny0 = 0.0; \ndt = 0.01; \nfinalTime = 2; \n solnList = NestList[G, {t0, y0}, Floor[finalTime\/dt]]; \n plot10 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Blue}]; \)}], "Input"], Cell["\<\ So the graphs of the second part are similar to those of the first part \ except that they flatten out faster. Now all 10 graphs.\ \>", "Text"], Cell[BoxData[ \(\(\(Show[plot1, plot2, plot3, plot4, plot5, plot6, plot7, plot8, plot9, plot10]; \)\n\n\)\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ "Exercise 3) Euler's Approximation", "Subtitle"]], "Subtitle"], Cell[TextData[{ "Suppose y[0] = ", Cell[BoxData[ \(1\/4\)]], " and ", Cell[BoxData[ \(dy\/dt = \@y\)]], ". Why is the tangent line to the exact solution y = y[t] below the curve \ at t = 0? If Euler's approximate solution ", Cell[BoxData[ \(TraditionalForm\`y\_approx\)]], " [\[Delta]t] is low at the first step, why is the slope f[ ", Cell[BoxData[ \(TraditionalForm\`y\_approx\)]], " [\[Delta]t]] below the slope of y[t] at t = \[Delta]t ?" }], "Text"], Cell[CellGroupData[{ Cell[" Solution", "Subsection"], Cell[TextData[{ "First let's compute the second derivative.\n\n\t", Cell[BoxData[ \(\(d\^2\ y\)\/dt\^2\)]], " = ", Cell[BoxData[ \(1\/2\)]], " ", Cell[BoxData[ \(1\/\@y\)]], " ", Cell[BoxData[ \(dy\/dt\)]], " = ", Cell[BoxData[ \(1\/2\)]], " ", Cell[BoxData[ \(1\/\@y\)]], " ", Cell[BoxData[ \(\@y\)]], " = ", Cell[BoxData[ \(1\/2\)]], "\n\t\nSince the second derivative is always positive, the function is \ smile shaped. We know y[0] = ", Cell[BoxData[ \(1\/4\)]], " and ", Cell[BoxData[ \(\(dy\/dt\)[0] = \(\@0\ = \ \ 0\)\)]], ". Our function is smile shaped and the tangent at 0 is flat, so it is \ beneath the function. In fact, since our function is always smile shaped, \ the tangent is always below the function." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ "Exercise 4) Solutions to the S-I-S Equations", "Subtitle"]], "Subtitle"], Cell[TextData[{ "An S-I-S disease is one that does not confer immunity; you are either \ susceptible or infectious. If s is the fraction of the population that is \ susceptible, the spread of the disease in time t is given by\n\t\t", Cell[BoxData[ \(ds\/dt = b\ \((1 - s)\)\ \((1 - c\ s)\)\)]], "\n\t\t\nwhere b is the reciprocal of the infectious period and c is the \ contact number. Assume that c \[Succeeds] 1 (so that each infectious person \ contacts more than one other person) and sketch the graphs of the solutions \ for various initial conditions between 0 and 1." }], "Text"], Cell[CellGroupData[{ Cell[" Solution", "Subsection"], Cell[TextData[{ Cell[BoxData[ \(ds\/dt\)]], " = 0 when s = 1 and when s = ", Cell[BoxData[ \(1\/c\)]], ". For 0 \[Precedes] s \[Precedes] ", Cell[BoxData[ \(1\/c\)]], ", the first derivative is positive and for ", Cell[BoxData[ \(1\/c\)]], " \[Precedes] s \[Precedes] 1, it is negative. The \n\nsecond derivative \ is ", Cell[BoxData[ \(b\^2\)]], "(s - 1)(1 - c s)((1 + c) - 2 c s ) so it is negative (frown) for 0 \ \[Precedes] s \[Precedes] ", Cell[BoxData[ \(1\/c\)]], ", positive (smile) for ", Cell[BoxData[ \(1\/c\)]], "\[Precedes] s \[Precedes] ", Cell[BoxData[ \(\(c + 1\)\/\(2\ c\)\)]], " and negative for ", Cell[BoxData[ \(\(c + 1\)\/\(2\ c\)\)]], " \[Precedes] s \[Precedes] 1." }], "Text"], Cell["\<\ Evaluate the cells below for graphs with b = 1/4 and c = 3. \ \>", "Text"], Cell[BoxData[{ \(Clear[f, b, c]\), \(b\ = \ 1/4; \nc\ = \ 3; \n f[t_, s_] := b\ \((1 - s)\)\ \((1 - c\ s)\); \nt0 = 0; \ns0 = 0.1; \n dt = 0.1; \nfinalTime = 20; \n solnList = NestList[G, {t0, s0}, Floor[finalTime\/dt]]; \n plots1 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["", "Text"], Cell[BoxData[{ \(Clear[f, b, c]\), \(b\ = \ 1/4; \nc\ = \ 3; \n f[t_, s_] := b\ \((1 - s)\)\ \((1 - c\ s)\); \nt0 = 0; \ns0 = 0.3; \n dt = 0.1; \nfinalTime = 20; \n solnList = NestList[G, {t0, s0}, Floor[finalTime\/dt]]; \n plots2 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["", "Text"], Cell[BoxData[{ \(Clear[f, b, c]\), \(b\ = \ 1/4; \nc\ = \ 3; \n f[t_, s_] := b\ \((1 - s)\)\ \((1 - c\ s)\); \nt0 = 0; \ns0 = 0.7; \n dt = 0.1; \nfinalTime = 20; \n solnList = NestList[G, {t0, s0}, Floor[finalTime\/dt]]; \n plots3 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["", "Text"], Cell[BoxData[{ \(Clear[f, b, c]\), \(b\ = \ 1/4; \nc\ = \ 3; \n f[t_, s_] := b\ \((1 - s)\)\ \((1 - c\ s)\); \nt0 = 0; \ns0 = 0.9; \n dt = 0.1; \nfinalTime = 20; \n solnList = NestList[G, {t0, s0}, Floor[finalTime\/dt]]; \n plots4 = ListPlot[solnList, PlotJoined \[Rule] True, PlotStyle -> {Red}]; \)}], "Input"], Cell["", "Text"], Cell[BoxData[ \(\(Show[plots1, plots2, plots3, plots4]\n\)\)], "Input"], Cell[TextData[{ "Notice that all graphs approach a final value of ", Cell[BoxData[ \(1\/c\)]], ".\n" }], "Text"] }, Closed]] }, Closed]] }, Open ]] }, FrontEndVersion->"Microsoft Windows 3.0", ScreenRectangle->{{0, 640}, {0, 424}}, ScreenStyleEnvironment->"Working", WindowSize->{612, 358}, WindowMargins->{{2, Automatic}, {Automatic, 2}}, PrivateFontOptions->{"FontType"->"Outline"}, Magnification->1.5, StyleDefinitions -> "Report.nb" ] (*********************************************************************** Cached data follows. 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