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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[ 26197, 834]*) (*NotebookOutlinePosition[ 62708, 2081]*) (* CellTagsIndexPosition[ 62600, 2074]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Lab 4B - Line Integrals of Vector Fields", FontSize->24, FontWeight->"Bold", FontVariations->{"Underline"->True}], "\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n\ Questions to: swenson@math.umn.edu" }], "Text", TextAlignment->Center, FontColor->GrayLevel[1], Background->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell["Introduction", "Section", Background->None], Cell[TextData[{ "In last week's lab, we looked at arc length. You should remember from \ class that the arc length of a curve is calculated with a line integral: the \ arc length of the curve C is ", Cell[BoxData[ \(\(\[Integral]\_C\%\ \)1 \[DifferentialD]L\)], "Text"], ", which is the same as ", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b\), RowBox[{"\[DoubleVerticalBar]", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}], \((t)\)}], "\[DoubleVerticalBar]", \(\[DifferentialD]t\)}]}]], "Text"], " when C is parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t), a\[LessEqual]t\[LessEqual]b", ". This time, we're looking at something which is related, but distinct: \ line integrals of vector fields. You should clearly separate these two \ concepts in your mind, and ask yourself, each time you calculate a line \ integral, \"What kind of function am I integrating?\" In your textbook, Barr \ often helps by writing line integrals of scalar functions with the \ differential \"\[DifferentialD]L,\" and line integrals of vector fields with \ the differential \" \[FilledVerySmallSquare] \[DifferentialD]", StyleBox["x ", FontWeight->"Bold"], ".\"" }], "Text"], Cell[TextData[{ "On the other hand, you should not let yourself be confused by the symbols \ ", Cell[BoxData[ FormBox[ SubscriptBox["\[Integral]", RowBox[{" ", StyleBox["C", FontSlant->"Italic"]}]], TraditionalForm]]], " and ", Cell[BoxData[ \(TraditionalForm\`\[ContourIntegral]\_\(\(\ \)\(C\)\)\)]], "- they mean the same thing! The ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(\(\ \)\(C\)\)\)]], " symbol can always be used in place of ", Cell[BoxData[ \(TraditionalForm\`\[ContourIntegral]\_\(\(\ \)\(C\)\)\)]], ". We write ", Cell[BoxData[ \(TraditionalForm\`\[ContourIntegral]\_\(\(\ \)\(C\)\)\)]], " when we want an extra reminder that C is a closed path (it ends at the \ same point where it began). Of course, we don't write ", Cell[BoxData[ \(TraditionalForm\`\[ContourIntegral]\_\(\(\ \)\(C\)\)\)]], " if C is not closed!" }], "Text"], Cell["\<\ Finally, remember that \"line integral\" and \"path integral\" are \ synonymous -- for this course, there is absolutely no difference between the \ two terms.\ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Example - Integration Using Mathematica", "Section", Background->None], Cell[TextData[{ "Here's a sample problem to illustrate some of the ", StyleBox["Mathematica", FontSlant->"Italic"], " commands we'll need this week. You should evaluate each \"evaluatable\" \ cell as you come to it." }], "Text"], Cell[TextData[{ "Calculate the integral ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]], "Text"], ", where ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(", Cell[BoxData[ \(TraditionalForm\`z\^3, y\^2, x\)]], ") and C is parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t)=(t,sin(t),1), 0\[LessEqual]t\[LessEqual]\[Pi]." }], "Text"], Cell["First, let's define and graph our functions.", "Text"], Cell[BoxData[{ \(\(F[x_, y_, z_] = {z^3, y^2, x};\)\), "\[IndentingNewLine]", \(\(\(f[t_] = {t, Sin[t], 1};\)\(\[IndentingNewLine]\) \)\), "\[IndentingNewLine]", \(\(\(vf1 = PlotVectorField3D[F[x, y, z], {x, 0, Pi}, {y, \(-1\), 1}, {z, 0, 2}, VectorHeads \[Rule] True]\)\(\[IndentingNewLine]\) \)\), "\[IndentingNewLine]", \(c1 = ParametricPlot3D[f[t], {t, 0, Pi}]\), "\[IndentingNewLine]", \(Show[vf1, c1, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\", "\"}]\)}], "Input"], Cell["\<\ If you've grown to prefer an interactive view of things, evaluate \ the next cell.\ \>", "Text"], Cell[BoxData[ \(ShowLive[vf1, c1, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\", "\"}]\)], "Input"], Cell[TextData[{ "Which direction does C go? Look at the definition of ", StyleBox["f,", FontWeight->"Bold"], " and ask your TA if you're not sure. Then evaluate the cell below to \ create an animation and check your answer. (This is the same sort of \ animation that you created at the end of Lab 2A; ask your TA if you don't \ remember how to create and view them.) Should we expect the value of ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]]], " to be positive or negative?" }], "Text"], Cell[BoxData[ \(PathAnimate3D[f[t], {t, 0, \[Pi]}, 40]\)], "Input", GeneratedCell->False, CellAutoOverwrite->False], Cell[TextData[{ "Now that we know what we're working with, let's calculate. By definition, \ ", Cell[BoxData[ RowBox[{ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}], StyleBox["=", FontWeight->"Bold"], RowBox[{\(\[Integral]\_a\%b\), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], \((t)\)}], ")"}], "\[CenterDot]", RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}]}], \((t)\), \(\[DifferentialD]t\)}]}]}]], "Text"], ", so we must first plug ", StyleBox["f", FontWeight->"Bold"], " into ", StyleBox["F,", FontWeight->"Bold"], " and find the derivative of ", StyleBox["f", FontWeight->"Bold"], ". Here are the appropriate commands." }], "Text"], Cell[BoxData[{ \(Apply[F, f[t]]\), "\[IndentingNewLine]", \(D[f[t], t]\)}], "Input"], Cell[TextData[{ "Remember that we use the command ", StyleBox["Apply[F,f[t]]", FontFamily->"Courier", FontWeight->"Bold"], ", where you'd think that we could just use ", StyleBox["F[f[t]]", FontFamily->"Courier", FontWeight->"Bold"], ". We mentioned before that this is because ", StyleBox["Mathematica", FontSlant->"Italic"], " reads ", StyleBox["F[f[t]]", FontFamily->"Courier", FontWeight->"Bold"], " as ", StyleBox["F[{t,Sin[t],1}]", FontFamily->"Courier", FontWeight->"Bold"], ", but what we really want is ", StyleBox["F[t,Sin[t],1]", FontFamily->"Courier", FontWeight->"Bold"], ". Look closely until you see the difference!" }], "Text", CellFrame->True, Background->GrayLevel[0.849989]], Cell["\<\ Next, we need to take a dot product. To do this, we just put a dot \ (a period) between the two vectors:\ \>", "Text"], Cell[BoxData[ \({1, Sin[t]^2, t} . {1, Cos[t], 0}\)], "Input"], Cell["Finally, we integrate.", "Text"], Cell[BoxData[ \(Integrate[1 + Cos[t]\ Sin[t]\^2, {t, 0, Pi}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["The Gravitational Force, Near Earth", "Section", Background->None, CellTags->"flatearth"], Cell[TextData[{ "In this example, we consider the gravitational force field for small \ objects near Earth. We approximate Earth by the ", StyleBox["xy-", FontSlant->"Italic"], "plane, and assume that the acceleration due to gravity is the constant ", StyleBox["g", FontSlant->"Italic"], "=9.8 ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["m", FontSlant->"Plain"], SuperscriptBox[ StyleBox["s", FontSlant->"Plain"], "2"]], TraditionalForm]]], ". Newton's first law of motion, ", StyleBox["F=ma", FontSlant->"Italic"], ", gives an expression for the gravitational force on an object: ", StyleBox["mg", FontSlant->"Italic"], ", where ", StyleBox["m", FontSlant->"Italic"], " is the mass of the object. This force pulls straight down, so the \ gravitational force field is ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(0, 0, -", StyleBox["mg", FontSlant->"Italic"], "), which should not be a surprise. Here's a graph, for a specific value \ of ", StyleBox["m.", FontSlant->"Italic"] }], "Text", CellTags->"flatearth"], Cell[BoxData[{\(m = 1;\), "\[IndentingNewLine]", \(g = 9.8;\), "\[IndentingNewLine]", RowBox[{" ", StyleBox[\(F[x_, y_, z_] = {0, 0, \(-m\)*g};\), FontFamily->"Courier", FontWeight->"Bold"], "\[IndentingNewLine]"}], "\[IndentingNewLine]", \(vf2 = PlotVectorField3D[ F[x, y, z], {x, \(-50\), 50}, {y, \(-50\), 50}, {z, \(-50\), 50}, \ VectorHeads \[Rule] True, ScaleFactor \[Rule] Max, ViewPoint -> {0, \(-3\), 0}]\)}], "Input", CellTags->"flatearth"], Cell[TextData[{ StyleBox["\"Points for Style\"", FontWeight->"Bold"], ": Since we're picturing Earth as flat, it would make more sense for the \ gravitational force to stop at Earth's surface. The ", StyleBox["If", FontFamily->"Courier", FontWeight->"Bold"], " command will help (you can look it up in the Help Browser). Try it by \ replacing the above definition of ", StyleBox["F", FontWeight->"Bold"], " with ", StyleBox["F[x_,y_,z_]:={0,0,-m*g}*If[(z\[LessEqual]0),0,1]", FontFamily->"Courier", FontWeight->"Bold"], "." }], "Text", CellFrame->True, Background->GrayLevel[0.849989], CellTags->"flatearth"], Cell[TextData[{ "Let ", StyleBox["f", FontWeight->"Bold"], "(t), a\[LessEqual]t\[LessEqual]b, parametrize a curve C. Then ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]], "Text"], " represents the work done by gravity on a particle moving from the \ beginning of C to the end. In other words, this is the gravitational \ potential energy that the particle loses (or \"uses up\") by following the \ curve C. If you've studied physics, you should expect that this equals ", StyleBox["mg\[CapitalDelta]h = mg", FontSlant->"Italic"], "(", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["f", FontSlant->"Plain"], "3"], TraditionalForm]]], "(a)-", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["f", FontSlant->"Plain"], "3"], TraditionalForm]]], "(b)), where ", StyleBox["f", FontWeight->"Bold"], "=(", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["f", FontSlant->"Plain"], "1"], TraditionalForm]]], ",", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["f", FontSlant->"Plain"], "2"], TraditionalForm]]], ",", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["f", FontSlant->"Plain"], "3"], TraditionalForm]]], "). Let's prove this." }], "Text", CellTags->"flatearth"], Cell[TextData[{ "By definition, ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]]], "=", Cell[BoxData[ RowBox[{\(\[Integral]\_a\%b\), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], \((t)\)}], ")"}], "\[CenterDot]", RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}]}], \((t)\), \(\[DifferentialD]t\)}]}]], "Text"], ", and since ", StyleBox["F", FontWeight->"Bold"], " is constant, ", StyleBox["F", FontWeight->"Bold"], "(", StyleBox["f", FontWeight->"Bold"], "(t)) is (0,0,-", StyleBox["mg", FontSlant->"Italic"], ")=-", StyleBox["mg", FontSlant->"Italic"], StyleBox["k,", FontWeight->"Bold"], " where ", StyleBox["k", FontWeight->"Bold"], "=(0,0,1)", ". Then we have ", Cell[BoxData[ RowBox[{ RowBox[{\(\[Integral]\_a\%b\), RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"-", StyleBox[ RowBox[{ StyleBox["mg", FontSlant->"Italic"], StyleBox["k", FontWeight->"Bold"]}]]}], ")"}], "\[CenterDot]", RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}]}], \((t)\), \(\[DifferentialD]t\)}]}], "=", RowBox[{ RowBox[{ RowBox[{"-", StyleBox["mg", FontSlant-> "Italic"]}], \(\[Integral]\_a\%b f\_3' \((t)\) \ \[DifferentialD]t\)}], "=", RowBox[{ RowBox[{"-", StyleBox["mg", FontSlant-> "Italic"]}], \((\(f\_3\) \((b)\) - \(f\_3\) \ \((a)\))\)}]}]}]], "Text"], ", using the Fundamental Theorem of Calculus." }], "Text", CellTags->"flatearth"], Cell[TextData[{ StyleBox["Exercise 1", FontWeight->"Bold"], ": Use ", StyleBox["Mathematica", FontSlant->"Italic"], " to check this calculation, for the path parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t)=(", Cell[BoxData[ \(TraditionalForm\`50 \( \[ExponentialE]\^\(-t\)\) \(cos(t)\), 50 \( \[ExponentialE]\^\(-t\)\) \(sin(t)\), \ t\^2 + 5 t\)]], "), 0\[LessEqual]t\[LessEqual]5. [Specifically: Use ", StyleBox["ParametricPlot3D", FontFamily->"Courier", FontWeight->"Bold"], " and ", StyleBox["Show", FontFamily->"Courier", FontWeight->"Bold"], " to plot the path and the vector field together, then evaluate the \ integral. Finally, check that the answer is what you expected! Use the \ values ", StyleBox["m", FontSlant->"Italic"], "=1, ", StyleBox["g", FontSlant->"Italic"], "=9.8, as given above.]" }], "Text", CellFrame->True, Background->RGBColor[1, 0.501961, 0.501961], CellTags->"flatearth"] }, Closed]], Cell[CellGroupData[{ Cell["The Gravitational Force, Farther From Earth", "Section", Background->None, CellTags->"spaceflight"], Cell[TextData[{ "Newton's law of gravitation shows how the force of Earth's gravity on an \ object diminishes as the object moves farther from Earth. If the center of \ the earth is at the origin (0,0,0), then the exact formula is ", StyleBox["F", FontWeight->"Bold"], "(", StyleBox["v", FontWeight->"Bold"], ")=", Cell[BoxData[ FormBox[ StyleBox[\(\((GmM\/\(\(\[DoubleVerticalBar]\)\(v\)\( \ \[DoubleVerticalBar] \^2\)\))\) \((\(-\(v\/\(\(\[DoubleVerticalBar]\)\(v\)\(\ \[DoubleVerticalBar]\)\)\)\))\)\), FontSize->16], TraditionalForm]]], ".\n\nHere's what the letters mean:\n", StyleBox["F", FontWeight->"Bold"], " is the gravitational force; a vector\n", StyleBox["v", FontWeight->"Bold"], " is the position of the object -- the vector (x,y,z)\nG is the universal \ gravitational constant, 6.67\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`10\^\(-11\)\)]], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ StyleBox["N", FontSlant->"Plain"], StyleBox["\[CenterDot]", FontSlant->"Plain"], SuperscriptBox[ StyleBox["m", FontSlant->"Plain"], "2"]}], \(kg\^2\)], TraditionalForm]]], "\nm is the mass of the object\nM is the mass of Earth, 5.98\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`10\^24\)]], "kg\n\nIn this formula, the first factor, ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["(", FontSize->16], StyleBox[\(GmM\/\(\(\[DoubleVerticalBar]\)\(v\)\( \ \[DoubleVerticalBar] \^2\)\)\), FontSize->16], StyleBox[")", FontSize->18]}], TraditionalForm]]], ", is a scalar, which measures how strong the gravitational force will be. \ The second factor, ", Cell[BoxData[ \(TraditionalForm\`\((\(-\(v\/\(\(\[DoubleVerticalBar]\)\(v\)\(\ \[DoubleVerticalBar]\)\)\)\))\)\)]], ", is the unit vector pointing from the object toward Earth; naturally, \ this is the direction of the gravitational force!" }], "Text", CellTags->"spaceflight"], Cell[TextData[{ "Suppose that the position of our object at time t is given by ", StyleBox["f", FontWeight->"Bold"], "(t), a\[LessEqual]t\[LessEqual]b. Then the work done by Earth's gravity \ between time a and time b is given by the integral ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]], "Text"], ", where C is the curve parametrized by ", StyleBox["f", FontWeight->"Bold"], "." }], "Text", CellTags->"spaceflight"], Cell[TextData[{ StyleBox["Exercise 2", FontWeight->"Bold"], ": Let C be parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t)=(0,0,", Cell[BoxData[ \(TraditionalForm\`R\/\(1 - t\)\)]], "), 0"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]]], " , the work done by Earth's gravity, as a TA with mass 90 kg travels along \ C. [Hint: Your answer should be negative: this represents work against \ gravity.]\nc) Calculate the work done by gravity as a Boeing 747-400 with \ mass 396890 kg climbs to a cruising altitude of 10000 m. [Hint: you may use \ any path with initial height R and final height R+10000.]\nd) Compare your \ answers to parts b) and c). What could you do to the TA with the energy from \ 57,285 gallons of jet fuel?\n[For airliner-related questions, see ", ButtonBox["http://www.boeing.com/commercial/747family/technical.html", ButtonData:>{ URL[ "http://www.boeing.com/commercial/747family/technical.html"], None}, ButtonStyle->"Hyperlink"], ".]" }], "Text", CellFrame->True, Background->RGBColor[1, 0.501961, 0.501961], CellTags->"spaceflight"] }, Closed]], Cell[CellGroupData[{ Cell["A Radical New Vector Field", "Section", Background->None], Cell[TextData[{ "In the ", ButtonBox["first gravitational-force example", ButtonData:>"flatearth", ButtonStyle->"Hyperlink"], ", we found that the value of the line integral of the gravitational force \ field depended only on the heights of the endpoints. The same is true in the \ ", ButtonBox["second gravitational-force example", ButtonData:>"spaceflight", ButtonStyle->"Hyperlink"], ", as is shown in most textbooks. When a vector field's line integrals \ depend only on the endpoints (and not on the route connecting them), the \ vector field is called ", StyleBox["conservative", FontSlant->"Italic"], ". From the first two examples, a person might get the idea that this \ always happens, but it's actually pretty rare. In the following exercise, \ for example, ", StyleBox["F", FontWeight->"Bold"], " is not conservative (hence the title of this section)." }], "Text"], Cell[TextData[{ StyleBox["Exercise 3", FontWeight->"Bold"], ": Let ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(y,-x,z). For each curve C: first, animate the curve using the ", StyleBox["PathAnimate3D", FontFamily->"Courier", FontWeight->"Bold"], " and/or ", StyleBox["PathTangentAnimate3D", FontFamily->"Courier", FontWeight->"Bold"], " commands. Then find the endpoints of the curve (specify which is the \ beginning and which is the end), and evaluate ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]]], ".\n\na) C is the upper half of the unit circle in the ", StyleBox["xy", FontSlant->"Italic"], "-plane -- i.e. the half where ", StyleBox["y\[GreaterEqual]", FontSlant->"Italic"], "0. [", StyleBox["Hint", FontWeight->"Bold"], ": All points in the ", StyleBox["xy", FontSlant->"Italic"], "-plane have the same ", StyleBox["z", FontSlant->"Italic"], "-value -- what is it?]\nb) C is parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t)=(cos t, -sin t, sin 4t), 0\[LessEqual]t\[LessEqual]\[Pi].\nc) C is \ contained in both the plane ", StyleBox["y", FontSlant->"Italic"], "=0 and the parabolic sheet ", StyleBox["z=", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2 - 1\)]], "; it begins at ", StyleBox["x", FontSlant->"Italic"], "=1 and ends at ", StyleBox["x", FontSlant->"Italic"], "=-1.\nd) C is parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t)=(-t, ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[ExponentialE]\^\((\(-1\)\/\(1 - t\^2\))\)\) cos\ \((20 t)\)\)\(,\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(\[ExponentialE]\^\((\(-1\)\/\(1 - t\^2\))\)\) sin\ \((20 t)\)\)]], "), -1\[LessEqual]t\[LessEqual]1. [", StyleBox["Hint", FontWeight->"Bold"], ": This is a hard integral to compute, even for ", StyleBox["Mathematica", FontSlant->"Italic"], ". Instead of using the command ", StyleBox["Integrate", FontFamily->"Courier", FontWeight->"Bold"], " in this part, use ", StyleBox["NIntegrate", FontFamily->"Courier", FontWeight->"Bold"], ", which approximates the value of the integral. Notice that ", StyleBox["f", FontWeight->"Bold"], "(t)", " is not actually defined at the two endpoints, so you'll have to integrate \ from \"close to -1\" to \"close to 1.\" Try integrating from -.99 to .99, \ then -.999 to .999, and so on, to see if the values approach something. Ask \ your TA about using limits to evaluate improper integrals if this doesn't \ seem familiar.]" }], "Text", CellFrame->True, Background->RGBColor[1, 0.501961, 0.501961]], Cell[CellGroupData[{ Cell["Credits", "Subsection"], Cell["\<\ This lab was written from scratch in 2002 by James Swenson. In \ Spring 2004 Jonathan Rogness went through and updated a few minor things to \ reflect the use of our math2374.nb file. This lab is copyright 2002 by James Swenson (swenson@math.umn.edu) and is \ protected by the Creative Commons Attribution-NonCommercial-ShareAlike \ License. You can find more information on this license at \ http://creativecommons.org/licenses/by-nc-sa/1.0/. Although it's not specifically required by the license, I'd appreciate it if \ you let me know at rogness@math.umn.edu if you use parts of our labs, just so \ I can keep track of it. Please send me any questions or comments!\ \>", \ "Text"] }, Closed]] }, Closed]] }, FrontEndVersion->"5.0 for X", ScreenRectangle->{{0, 1024}, {0, 768}}, ScreenStyleEnvironment->"Working", WindowSize->{652, 651}, WindowMargins->{{146, Automatic}, {Automatic, 46}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "PostScriptOutputFile":>FrontEnd`FileName[{"user002", "rogness"}, \ "Newlab.nb.ps", CharacterEncoding -> "iso8859-1"], "Magnification"->1}, CellLabelAutoDelete->True, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions", "Subtitle"], Cell["\<\ Modify the definitions below to change the default appearance of \ all cells in a given style. Make modifications to any definition using \ commands in the Format menu.\ \>", "Text"], Cell[CellGroupData[{ Cell["Style Environment Names", "Section"], Cell[StyleData[All, "Working"], PageWidth->WindowWidth, CellLabelMargins->{{12, Inherited}, {Inherited, Inherited}}, ScriptMinSize->9], Cell[StyleData[All, "Presentation"], PageWidth->WindowWidth, CellLabelMargins->{{24, Inherited}, {Inherited, Inherited}}, ScriptMinSize->12], Cell[StyleData[All, "Condensed"], PageWidth->WindowWidth, CellLabelMargins->{{8, Inherited}, {Inherited, Inherited}}, ScriptMinSize->8], Cell[StyleData[All, "Printout"], PageWidth->PaperWidth, CellLabelMargins->{{2, Inherited}, {Inherited, Inherited}}, ScriptMinSize->5, PrivateFontOptions->{"FontType"->"Outline"}] }, Closed]], Cell[CellGroupData[{ Cell["Notebook Options", "Section"], Cell["\<\ The options defined for the style below will be used at the \ Notebook level.\ \>", "Text"], Cell[StyleData["Notebook"], PageHeaders->{{Cell[ TextData[ { CounterBox[ 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FontSize->12], Cell[StyleData["Section", "Printout"], CellMargins->{{13, 0}, {7, 22}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 20}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, FontFamily->"Times", FontSize->14, FontWeight->"Bold"], Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->22], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", 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Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, TextJustification->0.5, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Input/Output", "Section"], Cell["\<\ The cells in this section define styles used for input and output \ to the kernel. 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Closed]], Cell[StyleData["InputOnly"], Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, LinebreakAdjustments->{0.85, 2, 10, 0, 1}, CounterIncrements->"Input", StyleMenuListing->None, FontWeight->"Bold"], Cell[CellGroupData[{ Cell[StyleData["Output"], CellMargins->{{47, 10}, {7, 5}}, CellEditDuplicate->True, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, CounterIncrements->"Output"], Cell[StyleData["Output", "Presentation"], 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FontWeight->"Bold", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[StyleData["SO10"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "Printout"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->7, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "EnhancedPrintout"], StyleMenuListing->None, FontFamily->"Futura", FontSize->7, FontWeight->"Plain", FontSlant->"Italic"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Formulas and Programming", "Section"], Cell[CellGroupData[{ Cell[StyleData["InlineFormula"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, LanguageCategory->"Formula", ScriptLevel->1, SingleLetterItalics->True], Cell[StyleData["InlineFormula", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["InlineFormula", "Condensed"], 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The \"Hyperlink\" style is for links within the same Notebook, \ or between Notebooks.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Hyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonNote->ButtonData}], Cell[StyleData["Hyperlink", "Presentation"], FontSize->16], Cell[StyleData["Hyperlink", "Condensed"], FontSize->11], Cell[StyleData["Hyperlink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell["\<\ The following styles are for linking automatically to the on-line \ help system.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["MainBookLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { 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StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "RefGuide", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["RefGuideLink", "Presentation"], FontSize->16], Cell[StyleData["RefGuideLink", "Condensed"], FontSize->11], Cell[StyleData["RefGuideLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GettingStartedLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "GettingStarted", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["GettingStartedLink", "Presentation"], FontSize->16], Cell[StyleData["GettingStartedLink", "Condensed"], 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