(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 104499, 3398]*) (*NotebookOutlinePosition[ 105427, 3429]*) (* CellTagsIndexPosition[ 105383, 3425]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Plotting", "Title", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell["Simplified Version", "Subsubtitle", TextAlignment->Center], Cell[TextData[{ "Sean Mauch\nsean@caltech.edu\n", ButtonBox["http://www.its.caltech.edu/~sean", ButtonData:>{ URL[ "http://www.its.caltech.edu/~sean"], None}, ButtonStyle->"Hyperlink"], "\n", "This work is distributed under the GNU FDL. See ", ButtonBox["license.nb ", ButtonData:>{"license.nb", None}, ButtonStyle->"Hyperlink"], "for details." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Plotting Functions of a Single Variable", "Section"], Cell[CellGroupData[{ Cell["Basic Plotting Options", "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "You call ", StyleBox["Plot[]", "Input"], " with a function and the domain of the independent variable. ", StyleBox["Mathematica", FontSlant->"Italic"], " will automatically choose the range." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Plot[Sin[x], {x, \(-\[Pi]\), \[Pi]}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Note that the output of a ", StyleBox["Plot[]", "Input"], " command is -Graphics- . It is customary to follow ", StyleBox["Plot[]", "Input"], " commands with a semicolon to suppress showing this." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[Sin[x]\/x, {x, \(-2\) \[Pi], 3 \[Pi]}]; \)\)], "Input"], Cell[TextData[{ "When the function you are plotting is singular, ", StyleBox["Mathematica", FontSlant->"Italic"], " will try to choose an appropriate range for the plot." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[1\/x, {x, \(-1\), 1}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can over-ride this default choice of range by using the ", StyleBox["PlotRange", "Input"], " option. As the last argument of ", StyleBox["Plot[]", "Input"], " supply the rule: ", StyleBox["PlotRange->{lower,upper}.", "Input"] }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[1\/x, {x, \(-1\), 1}, PlotRange \[Rule] {\(-5\), 10}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The following function is not singular, but is has a relatively tall hump \ at the origin. ", StyleBox["Mathematica", FontSlant->"Italic"], " will make an educated guess for the y-axis range that you would like to \ see. It tries to show you the interesting part of the function. Below, ", StyleBox["Mathematica", FontSlant->"Italic"], " guesses that you are more interested in the smaller wiggles than in the \ big hump near ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], " and so it crops the picture accordingly. " }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[Sin[x]\/x, {x, \(-6\) \[Pi], 6 \[Pi]}];\)\)], "Input"], Cell[TextData[{ "You can specify that ", StyleBox["Mathematica", FontSlant->"Italic"], " should include the entire range of the sampled function in the plot by \ again using the ", StyleBox["PlotRange", "Input"], " option." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[Sin[x]\/x, {x, \(-6\) \[Pi], 6 \[Pi]}, PlotRange \[Rule] All]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can have a frame drawn around the plot with the ", StyleBox["Frame", "Input"], " option." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[Sin[x]\/x, {x, \(-2\) \[Pi], 3 \[Pi]}, Frame \[Rule] True]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ratio of the height of the plot to the width of the plot has the \ default value of the reciprocal of the golden ratio. 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Then ", StyleBox["AspectRatio", "Input"], " will then be chosen automatically so that the axes have the same scale. \ This is often the desired effect when showing geometric objects." }], "Text", TextJustification->1], Cell[BoxData[ \(\(Plot[x\^2, {x, 0, 2}]; \)\)], "Input"], Cell[BoxData[ \(\(Plot[x\^2, {x, 0, 2}, AspectRatio -> Automatic]; \)\)], "Input"], Cell[TextData[{ "You can plot more than one function by giving ", StyleBox["Plot[]", "Input"], " a list of functions." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[{x, x\^2, x\^3}, {x, \(-1\), 1}, PlotRange \[Rule] All]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can hide the axes by changing the value of the ", StyleBox["Axes", "Input"], " option from ", StyleBox["True", "Input"], " to ", StyleBox["False", "Input"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[{x, x\^2, x\^3}, {x, \(-1\), 1}, PlotRange \[Rule] All, Axes \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 1", "Subsubsection"], Cell["\<\ Draw a unit circle centered about the origin. The circle should \ look like a circle and not an ellipse.\ \>", "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["PlotStyle Plotting Options", "Subsection"], Cell[BoxData[ \(\(?PlotStyle\)\)], "Input"], Cell[TextData[{ "Some ", StyleBox["PlotStyle", "Input"], " options are" }], "Text"], Cell[TextData[{ StyleBox["Dashing[pattern] ", FontWeight->"Bold"], "Draw dashed lines. It takes a list of numbers that specify the length of \ the filled and blank sections of the line pattern.\n", StyleBox["Thickness[line_width]", FontWeight->"Bold"], " Draw lines, where line_width is the fraction of the thickness of the \ line to the width of the graph.\n", StyleBox["GrayLevel[level] ", FontWeight->"Bold"], "Specify the graylevel color of the graphics primitives. The level \ argument must be between 0 and 1. GrayLevel[0] is black; GrayLevel[1] is \ white.\n", StyleBox["RGBColor[r,g,b] ", FontWeight->"Bold"], "Specify a color with its red-green-blue components. 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RGBColor[0,0,0] is black; \ RGBColor[1,1,1] is white; RGBColor[1,0,0] is red; RGBColor[1,0.5,0.5] is \ light red; RGBColor[0.75,0,0] is dark red." }], "Text", TextAlignment->Left, TextJustification->1], Cell["Here is a function plotted with a dashed line.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[Sin[x], {x, \(-2\) \[Pi], 2 \[Pi]}, PlotStyle -> Dashing[{0.02, 0.02}]]; \)\)], "Input"], Cell[TextData[{ "If you are plotting multiple functions you can give ", StyleBox["PlotStyle", "Input"], " a list of two lists of options." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[{Sin[x], Cos[x]}, {x, \(-2\) \[Pi], 2 \[Pi]}, PlotStyle -> {{RGBColor[1, 0.5, 0.5], Thickness[0.01]}, { RGBColor[0, 0.75, 0], Dashing[{0.02, 0.02}]}}]; \)\)], "Input"], Cell[CellGroupData[{ Cell["Exercise 2", "Subsubsection"], Cell["Plot a function in a thick, green, dash-dotted line.", "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Labeling Plots", "Subsection"], Cell[TextData[{ "You can add text to plots with the ", StyleBox["PlotLabel", FontWeight->"Bold"], " and ", StyleBox["AxesLabel", FontWeight->"Bold"], " options." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?PlotLabel\)\)], "Input"], Cell[BoxData[ \(\(?AxesLabel\)\)], "Input"], Cell[BoxData[ \(\(Plot[{Cos[x], Sin[x]}, {x, 0, 2 \[Pi]}, PlotLabel -> "\", AxesLabel -> {"\", "\"}]; \)\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " allows you to customize the text with the ", StyleBox["TextStyle", "Input"], ", ", StyleBox["FontSlant", "Input"], ", and ", StyleBox["FontFamily", "Input"], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?TextStyle\)\)], "Input"], Cell[BoxData[ \(\(Plot[{Cos[x], Sin[x]}, {x, 0, 2 \[Pi]}, PlotLabel -> "\", AxesLabel -> {"\", "\"}, TextStyle -> {FontSize -> 12, FontWeight -> "\", FontSlant -> "\", FontFamily -> "\"}]; \)\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Axes Options", "Subsection"], Cell[TextData[{ "Here is the default way that ", StyleBox["Mathematica", FontSlant->"Italic"], " plots ", Cell[BoxData[ \(TraditionalForm\`2 + sin(x)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[2 + Sin[x], {x, 0, 2 \[Pi]}]; \)\)], "Input"], Cell[TextData[{ "By setting the ", StyleBox["Axes", "Input"], " option to ", StyleBox["False", "Input"], ", you can specify that the axes should not be drawn" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[2 + Sin[x], {x, 0, 2 \[Pi]}, Axes -> False]; \)\)], "Input"], Cell[TextData[{ "By default ", StyleBox["Mathematica", FontSlant->"Italic"], " will put the axes origin at ", Cell[BoxData[ \(TraditionalForm\`\((0, 1)\)\)]], ". 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You can fix this with the ", StyleBox["PlotRange", "Input"], " option." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[2 + Sin[x], {x, 0, 2 \[Pi]}, AxesOrigin -> {0, 0}, PlotRange -> {0, 3}]; \)\)], "Input"], Cell[TextData[{ "You can change the default tick marks with the ", StyleBox["Ticks", "Input"], " option." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[\@x, {x, 0, 5}, Ticks -> {{0, 1, 2, 3, 4, 5}, {0, 1, \@2, \@3, 2, \@5}}]; \)\)], "Input"], Cell["\<\ If you put a frame around the graph, you can also change the \ default frame tick marks.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[\@x, {x, 0, 5}, Frame -> True, FrameTicks -> {{0, 1, 2, 3, 4, 5}, {0, 1, \@2, \@3, 2, \@5}}]; \)\)], "Input"], Cell[CellGroupData[{ Cell["Exercise 3", "Subsubsection"], Cell["Produce the following graph of the cosine and sine.", "Text"], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica 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Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "You can use the ", StyleBox["Show[]", "Input"], " function to redisplay a plot with different options." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[x\ Sin[x], {x, \(-2\)\ \[Pi], 2\ \[Pi]}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Setting ", StyleBox["GridLines", "Input"], " to ", StyleBox["Automatic", "Input"], " will put a grid line at each major tick mark on the axes." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%, Axes \[Rule] False, GridLines \[Rule] Automatic]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can also change the plot range. The ", StyleBox["PlotLabel", "Input"], " function allows you to add a title to your plots." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%%, PlotRange \[Rule] {\(-5\), 5}, PlotLabel \[Rule] "\"]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can specify both the domain and range of the plot with the ", StyleBox["PlotRange", "Input"], " option." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%%%, PlotRange \[Rule] {{\(-\[Pi]\), \[Pi]}, {\(-1\), 3}}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Using ", StyleBox["Show[]", "Input"], " is more efficient than using ", StyleBox["Plot[]", "Input"], " again with different options as ", StyleBox["Mathematica", FontSlant->"Italic"], " does not have to evaluate the function again. It already has all the \ data on the function, using ", StyleBox["Show[]", "Input"], " with different options just changes how the data is displayed." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ "You can also use the ", StyleBox["Show[]", "Input"], " function to display multiple plots together." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(sinPlot = Plot[Sin[x], {x, \(-2\)\ \[Pi], \[Pi]}];\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(cosPlot = Plot[Cos[x], {x, \(-\[Pi]\), 2\ \[Pi]}];\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(Show[sinPlot, cosPlot]; \)\)], "Input", AspectRatioFixed->True], Cell["\<\ When redisplaying multiple plots you can change any of the options \ that you could when redisplaying a single plot.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], 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A contour plot of a function is just like a contour map. The \ lines correspond to constant values of the function whereas in a map they \ correspond to constant altitudes. By default a contour plot is colored in \ shades of grey. Lighter colors correspond to larger function values." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ "You can leave out the greyscale shading by setting the ", StyleBox["ContourShading", "Input"], " option to ", StyleBox["False", "Input"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%, ContourShading \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "By default the ", StyleBox["ContourPlot[]", "Input"], " function samples points on a ", Cell[BoxData[ \(TraditionalForm\`15\[Times]15\)]], " grid." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(ContourPlot[ Exp[\(-\((x\^2 + y\^2)\)\)], {x, \(-2\), 2}, {y, \(-2\), 2}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Sometimes this is not a fine enough grid to give smooth contours. You can \ change the number of sample points with the ", StyleBox["PlotPoints", FontWeight->"Bold"], " option. The sample grid on the plot below is twice as fine." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(ContourPlot[ Exp[\(-\((x\^2 + y\^2)\)\)], {x, \(-2\), 2}, {y, \(-2\), 2}, PlotPoints \[Rule] 30]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "By default the ", StyleBox["ContourPlot[]", "Input"], " function draws 10 contour levels. You can change this by setting the ", StyleBox["Contours", "Input"], " option. You can specify the number of contours with an integer, or you \ can specify that it should draw contours at specific altitudes by giving a \ list of real numbers. " }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(ContourPlot[ Sin[x]\ Sin[y], {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}, ContourShading \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(ContourPlot[ Sin[x]\ Sin[y], {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}, ContourShading \[Rule] False, Contours \[Rule] 20]; \)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(ContourPlot[ Sin[x]\ Sin[y], {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}, ContourShading \[Rule] False, Contours \[Rule] {\(-1\)/2, 0, 1/2}]; \)\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 6", "Subsubsection"], Cell[TextData[{ "Plot the function ", Cell[BoxData[ \(TraditionalForm\`y(x)\)]], " where ", Cell[BoxData[ \(TraditionalForm\`y\)]], " is given by the implicit equation," }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\(\((x + 1)\)\^2\) \((x - 1)\)\^2 + x\ y + y\^2 - 1 = 0. \)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Density Plots"], "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["DensityPlot[]", "Input"], " function divides the domain into a mesh of cells and colors each cell in \ greyscale. Again lighter colors correspond to larger function values." }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(DensityPlot[ Sin[x]\ Sin[y], {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "By default the mesh surrounding the cells in drawn in the plot. You can \ change this with the ", StyleBox["Mesh", "Input"], " option." }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%, Mesh \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "By default the mesh is ", Cell[BoxData[ \(TraditionalForm\`15\[Times]15\)]], ". You change this with the ", StyleBox["PlotPoints", "Input"], " option. The density plot below has a ", Cell[BoxData[ \(TraditionalForm\`60\[Times]60\)]], " grid." }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(DensityPlot[ Sin[x]\ Sin[y], {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}, Mesh \[Rule] False, PlotPoints \[Rule] 60]; \)\)], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Surface Plots"], "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Now for the really spiffy pictures. ", StyleBox["Plot3D[]", "Input"], " draws a surface in 3-dimensional space. ", StyleBox["Mathematica", FontSlant->"Italic"], " colors the surface using a simulated lighting model. " }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Plot3D[ Sin[x]\ Sin[y], {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can remove the black mesh from the mesh by changing the ", StyleBox["Mesh", "Input"], " option to ", StyleBox["False", "Input"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%, Mesh \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell["\<\ You can produce a wireframe of the surface by setting the shading \ to false.\ \>", "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%%, Shading \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can also turn off the lighting and instead use greyscale shading by \ setting the ", StyleBox["Lighting", "Input"], " option to ", StyleBox["False", "Input"], "." }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[%%%, Lighting \[Rule] False]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "There are a variety of options that you can set for ", StyleBox["Plot3D[]", "Input"], ". See a ", StyleBox["Mathematica", FontSlant->"Italic"], " reference for details." }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 7", "Subsubsection"], Cell[TextData[{ "Using the ", StyleBox["Plot3D[]", FontWeight->"Bold"], " and ", StyleBox["Show[]", FontWeight->"Bold"], " functions, display the intersection of the two surfaces:" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm\`z = 2 x\^2 + 3 y\^2, \ \ \ \ z = \(-x\) + y + 5. \)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Find a parametric representation of the intersection curve in terms of ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". Plot the projection of the intersection curve on the ", Cell[BoxData[ \(TraditionalForm\`x\[VeryThinSpace]y\)]], " plane. (That is, plot ", Cell[BoxData[ \(TraditionalForm\`y\)]], " as a function of ", Cell[BoxData[ \(TraditionalForm\`x\)]], ".)" }], "Text"] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Plotting Implicitly Defined Functions", "Section"], Cell[TextData[{ "As you saw in the exercise in the contour plots section, you can plot an \ implicitly defined function by graphing the zero contour of a function of two \ variables. ", StyleBox["Mathematica", FontSlant->"Italic"], " has a built-in function for plotting implicitly defined functions. First \ load the ", StyleBox["Graphics`ImplicitPlot`", FontWeight->"Bold"], " package." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(<< Graphics`ImplicitPlot`\)], "Input"], Cell[TextData[{ "If you call ", StyleBox["ImplicitPlot", FontWeight->"Bold"], " with the equation as the first argument and a range for the \"dependent\" \ variable as the second argument, then ", StyleBox["Mathematica", FontSlant->"Italic"], " will solve your equation using the ", StyleBox["Solve", FontWeight->"Bold"], " function, determine where different branches of the function are \ real-valued and plot the branches on the appropriate domains. Below we plot \ an ellipse and a hyperbola." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(ImplicitPlot[ x\^2\/2 + y\^2\/3 == 1, {x, \(-\@2\), \@2}];\)\)], "Input"], Cell[BoxData[ \(\(ImplicitPlot[x\^2 - y\^2 == 1, {x, \(-3\), 3}]; \)\)], "Input"], Cell[TextData[{ "The ", StyleBox["Solve", FontWeight->"Bold"], " function works really well for algebraic equations, but it can't handle \ transcendental equations very well. Thus if you give ", StyleBox["ImplicitPlot", FontWeight->"Bold"], " an implicit, transcendental equation, you would expect it to choke. \ Below we try to plot ", Cell[BoxData[ \(TraditionalForm\`\(sin\^2\) x + \(sin\^2\) y = 7\/8\)]], ". Since the ", Cell[BoxData[ \(TraditionalForm\`sine\)]], " is ", Cell[BoxData[ \(TraditionalForm\`2 \[Pi]\)]], "-periodic, (and further ", Cell[BoxData[ \(TraditionalForm\`\(sin\^2\) x\)]], " is \[Pi]-periodic), the graph of the solution would \"tile\" the plane. \ That is, it would be the same pattern repeated over and over again." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(ImplicitPlot[Sin[x]\^2 + Sin[y]\^2 == 7/8, {x, \(-\[Pi]\), \[Pi]}]; \)\)], "Input"], Cell[TextData[{ "We see that ", StyleBox["ImplicitPlot", FontWeight->"Bold"], " does in fact choke. It does not show all the branches of the graph and \ it leaves holes in the branches that it does show. This really isn't ", StyleBox["Mathematica", FontSlant->"Italic"], "'s fault. We can't expect it to find an infinite number of solutions in \ closed form. To get around this problem, we suggest that instead of trying \ to solve the implicit equation for ", Cell[BoxData[ \(TraditionalForm\`y\)]], ", ", StyleBox["Mathematica", FontSlant->"Italic"], " should instead plot the zero contour of a function of two variables. We \ suggest this by giving a range to plot in the dependent variable as well." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(ImplicitPlot[ Sin[x]\^2 + Sin[y]\^2 == 7/8, {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}]; \)\)], "Input"], Cell[TextData[{ "Now ", StyleBox["Mathematica", FontSlant->"Italic"], " found all the solutions in the requested range, but the graph isn't very \ smooth. We remedy this by using more points when it does the contour plot." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(ImplicitPlot[ Sin[x]\^2 + Sin[y]\^2 == 7/8, {x, \(-\[Pi]\), \[Pi]}, {y, \(-\[Pi]\), \[Pi]}, PlotPoints -> 100]; \)\)], "Input"], Cell[CellGroupData[{ Cell["Exercise 8", "Subsubsection"], Cell["Plot the solution of", "Text"], Cell[BoxData[ \(TraditionalForm\`\((x\^2 + y\^2)\)\^4 = \(\((x\^2 - y\^2)\)\^2 . \)\)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Parametric Plots", "Section"], Cell[TextData[{ "You can graph functions described parametrically with the ", StyleBox["ParametricPlot[]", "Input"], " function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?ParametricPlot\)\)], "Input"], Cell["Here is an ellipse.", "Text"], Cell[BoxData[ \(\(ParametricPlot[{3 Cos[t], 2 Sin[t]}, {t, 0, 2 \[Pi]}, PlotRange -> {{\(-3\), 3}, {\(-3\), 3}}, AspectRatio -> 1]; \)\)], "Input"], Cell["\<\ You can plot multiple functions. Below is a flower inscribed in a \ circle.\ \>", "Text"], Cell[BoxData[ \(\(ParametricPlot[{{Cos[5 t] Cos[t], Cos[5 t] Sin[t]}, {Cos[2 t], Sin[2 t]}}, {t, 0, \[Pi]}, AspectRatio -> 1]; \)\)], "Input"], Cell[CellGroupData[{ Cell["Exercise 9", "Subsubsection"], Cell["Draw a spiral centered about the origin.", "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["3D Parametric Plots", "Section"], Cell[TextData[{ "You can plot parametrically described curves and surfaces in 3D using the \ ", StyleBox["ParametricPlot3D[]", "Input"], " function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?ParametricPlot3D\)\)], "Input"], Cell["Here is a curve that lies on the unit sphere.", "Text"], Cell[BoxData[ \(\(ParametricPlot3D[{Sin[t/2] Cos[10 t], Sin[t/2] Sin[10 t], Cos[t/2]}, {t, 0, 2 \[Pi]}]; \)\)], "Input"], Cell[TextData[{ "The above curve has sharp corners that are an artifact of sampling the \ function at a finite number of points. If you want a smoother curve, \ increase the number of ", StyleBox["PlotPoints", "Input"], "." }], "Text"], Cell[BoxData[ \(\(ParametricPlot3D[{Sin[t/2] Cos[10 t], Sin[t/2] Sin[10 t], Cos[t/2]}, {t, 0, 2 \[Pi]}, PlotPoints -> 200]; \)\)], "Input"], Cell["\<\ To draw a parametric surface, you specify the range of two \ parameters. Here is a unit sphere.\ \>", "Text"], Cell[BoxData[ \(\(ParametricPlot3D[{Cos[\[Theta]] Sin[\[Phi]], Sin[\[Theta]] Sin[\[Phi]], Cos[\[Phi]]}, {\[Theta], 0, 2 \[Pi]}, { \[Phi], 0, \[Pi]}]; \)\)], "Input"], Cell["Below are two interlocking tori.", "Text"], Cell[BoxData[ \(\(ParametricPlot3D[{{Cos[s] \((3 + Cos[t])\), Sin[s] \((3 + Cos[t])\), Sin[t]}, {3 + Sin[s] \((3 + Cos[t])\), Sin[t], Cos[s] \((3 + Cos[t])\)}}, {s, 0, 2 \[Pi]}, {t, 0, 2 \[Pi]}]; \)\)], "Input"], Cell[TextData[{ "A surface of revolution about the ", Cell[BoxData[ \(TraditionalForm\`z\)]], " axis has the parametrization," }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["x", FontWeight->"Bold"], "(", \(u, v\), ")"}], "=", \((\(f(v)\) cos\ u, \(f(v)\) sin\ u, g(v))\)}], ","}], TraditionalForm]], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`u \[Element] \([0, 2 \[Pi]]\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`v \[Element] \([a, b]\)\)]], ". 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