(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 4.2' 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[ 55884, 1711]*) (*NotebookOutlinePosition[ 56514, 1733]*) (* CellTagsIndexPosition[ 56470, 1729]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["LAB 5: Polar Coordinates", "Subtitle", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Introduction ", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["The graph of an equation in polar coordinates consists of points \ P with polar coordinates (r, \[Theta]), where r is the distance from the \ origin O to the point P along the ray OP, and \[Theta] is the angle \ from the positive x-axis to the ray OP. While every equation relating ", Evaluatable->False, AspectRatioFixed->True], StyleBox["r", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[" and \[Theta] has a graph, in practice (as with equations written \ in rectangular/Cartesian coordines ", Evaluatable->False, AspectRatioFixed->True], StyleBox["x", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[" and ", Evaluatable->False, AspectRatioFixed->True], StyleBox["y", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[") it is generally much easier to graph an equation in which one \ variable has been solved for in terms of the other (i.e, the one is \ considered dependent and the other independent). General practice is to \ write (when possible) r as a function (dependent on) \[Theta]:\n\t", Evaluatable->False, AspectRatioFixed->True], StyleBox["r = f ", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox["(\[Theta]).\nOur objectives in this lab are:\n\n1. To get a \ little practice graphing polar functions in ", Evaluatable->False, AspectRatioFixed->True], StyleBox["Mathematica", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[".\n\n2. To refamiliarize ourselves with how to define functions \ and solve equations in ", Evaluatable->False, AspectRatioFixed->True], StyleBox["Mathematica", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[".\n\n3. To get more intuition as to why it can be difficult to \ find the points of intersection between two polar curves.\n\n4. To further \ develop our intuition about what we expect of polar functions. We have seen \ that their graphs need not pass the vertical line test. We will look at the \ relationship between derivatives and slopes of tangents.", Evaluatable->False, AspectRatioFixed->True] }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell["Basic Plotting ", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["The PolarPlot Command", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[" does not have a built-in command for plotting polar curves, but \ there is one available in the package ", Evaluatable->False, AspectRatioFixed->True], StyleBox["Graphics`Graphics`", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox[". It has the general form ", Evaluatable->False, AspectRatioFixed->True], StyleBox["PolarPlot[", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox["r", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[",", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["{", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox["t, tmin, tmax", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox["}]", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox[".", Evaluatable->False, AspectRatioFixed->True, FontWeight->"Bold"], StyleBox[" ", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox["This command produces a polar plot as the angle t varies from \ t = ", Evaluatable->False, AspectRatioFixed->True], StyleBox["tmin", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[" to t = ", Evaluatable->False, AspectRatioFixed->True], StyleBox["tmax", Evaluatable->False, AspectRatioFixed->True, FontSlant->"Italic"], StyleBox[". The \"radius\" r is written as a function of t. The \ command always plots the x and y axes at the same scale; that is, the \n\ ", Evaluatable->False, AspectRatioFixed->True], StyleBox["AspectRatio", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", Evaluatable->False, AspectRatioFixed->True, FontSize->9], StyleBox["\[Rule]", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", Evaluatable->False, AspectRatioFixed->True, FontSize->9], StyleBox["Automatic", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" option is built in to this command. Notice that t is used \ instead of \[Theta], since the standard keyboard has no \[Theta] key.", Evaluatable->False, AspectRatioFixed->True] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["The next few cells show some easy examples. Enter each of them \ in turn, and try to decide how they were traced out as the angle t was \ increased. The first cell plots the graph of the ", Evaluatable->False, AspectRatioFixed->True], StyleBox["cardiod", Evaluatable->False, AspectRatioFixed->True, FontWeight->"Bold"], StyleBox["\t\t", Evaluatable->False, AspectRatioFixed->True] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["r = 1 + sin t ", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox[" ", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" for 0 ", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox["\[LessEqual] ", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox[" t ", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox["\[LessEqual]", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox[" 2", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", CellMargins->{{144, Inherited}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox[". 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Enter the next cell to watch the graph of r = 1 + a sin t ", Evaluatable->False, AspectRatioFixed->True], StyleBox[" ", Evaluatable->False, AspectRatioFixed->True, FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" for 0 \[LessEqual] t \[LessEqual] 2\[Pi] change as the \ constant a assumes the values a = .5, .75, 1, 1.25, 1.5, 1.75, and 2. \ The plots can be animated.\nJust surround the ", Evaluatable->False, AspectRatioFixed->True], StyleBox["PolarPlot", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" command with ", Evaluatable->False, AspectRatioFixed->True], StyleBox["Do", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" and give the range and increment for the parameter a. 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o`030000003oool0oooo00@0oooo00<000000?ooo`3oool0"], ImageRangeCache->{{{0, 229.312}, {171.75, 0}} -> {-1.0566, -0.537507, \ 0.00919289, 0.00919289}}], Cell[CellGroupData[{ Cell["Points of Intersection", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ Enter the next cell to plot the graphs of r = 1 + cos t (a \ cardiod) and r = 3 cos t (a circle) on the same coordinate system.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(PolarPlot[{1 + Cos[t], 3\ Cos[t]}, {t, 0, 2\ \[Pi]}, PlotStyle \[Rule] {Hue[0], Hue[ .7]}]\)], "Input", AspectRatioFixed->True], Cell["\<\ Examination of the graph leads us to think that there are three \ points of intersection for these curves. Try to find them by solving 1 + \ cos t = 3 cos t . \ \>", "Text", CellMargins->{{Inherited, 22}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Solve[1 + Cos[t] == 3\ Cos[t], t]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["We don't get all the solutions. A bit of algebra applied to the \ equation 1 + cos t = 3 cos t gives cos t = 1", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox["/", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["2, so that if 0 \[LessEqual] t \[LessEqual] 2\[Pi], then t = \ \[Pi]", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox["/", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["3 or t = 5\[Pi]", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], StyleBox["/", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["3. ", CellMargins->{{Inherited, 94}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True] }], "Text", CellMargins->{{Inherited, 22}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], Cell["\<\ But the circle and the cardiod also intersect at (0, 0). Why \ didn't we get this point of intersection when we solved 1 + cos t = 3 cos t \ for t? Entering the next cell shows why.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Do[\[IndentingNewLine]DisplayTogether[\[IndentingNewLine]Plot[ 0, {x, \(-0.5\), 3}, AspectRatio \[Rule] Automatic, PlotRange \[Rule] {\(-1.5\), 1.5}], PolarPlot[{1 + Cos[t], 3\ Cos[t]}, {t, 0, c}, PlotStyle \[Rule] {Hue[0], Hue[0.7]}]], \n\ \t{c, \[Pi]\/6, \[Pi], \[Pi]\/6}]\)], "Input", CellGroupingRules->"InputGrouping", AutoSpacing->False, DelimiterMatching->None], Cell[TextData[{ StyleBox["We see that as the curves are traced out, we have a \"collision\" \ point when t = ", Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", Evaluatable->False, AspectRatioFixed->True], StyleBox["/", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["3", Evaluatable->False, AspectRatioFixed->True], StyleBox[", but when t = 2", Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", Evaluatable->False, AspectRatioFixed->True], StyleBox["/", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["3", Evaluatable->False, AspectRatioFixed->True], StyleBox[", tracing the circle reaches the origin, but the tracing of the \ cardiod has only reached (0, 1). When t = ", Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", Evaluatable->False, AspectRatioFixed->True], StyleBox[", the tracing of the cardiod has reached the origin, but the \ circle has been completely traced out. If we vary t from ", Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", Evaluatable->False, AspectRatioFixed->True], StyleBox[" to 2", Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", Evaluatable->False, AspectRatioFixed->True], StyleBox[", we would complete the cardiod and trace over the circle a \ second time. ", Evaluatable->False, AspectRatioFixed->True] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Do[\[IndentingNewLine]DisplayTogether[\[IndentingNewLine]Plot[ 0, {x, \(- .5\), 3}, \[IndentingNewLine]AspectRatio \[Rule] Automatic, \[IndentingNewLine]PlotRange \[Rule] {\(-1.5\), 1.5}], PolarPlot[{1 + Cos[t], 3\ Cos[t]}, {t, \[Pi], c}, PlotStyle \[Rule] {Hue[0], Hue[ .7]}]], \[IndentingNewLine]{c, \(7\ \[Pi]\)\/6, 2\ \[Pi], \[Pi]\/6}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["At t = 5", Evaluatable->False, AspectRatioFixed->True], StyleBox["\[Pi]", Evaluatable->False, AspectRatioFixed->True], StyleBox["/", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Courier"], StyleBox["3", Evaluatable->False, AspectRatioFixed->True], StyleBox[" we have another \"collision\", but no collision ever occurs at \ (0, 0). ", Evaluatable->False, AspectRatioFixed->True] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ In general we may have points of intersection that are not \ \"collision\" points; that is, the curves arrive at these other \"crossing \ points\" at different times. The graphs must be plotted to see where these \ crossing points are.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Exercise 2", "Subsection"], Cell["\<\ (a) Find all points of intersection of the two curves r = 1 and \ r = 2 sin 2t, 0 \[LessEqual] t \[LessEqual] 2\[Pi].\ \>", "Text"], Cell["\<\ (b) Which of the intersection points you found in (a) are \ \"collision\" points? Entering the next cell may help you decide.\ \>", "Text"], Cell[BoxData[ \(Do[PolarPlot[{1, 2 Sin[2 t]}, {t, 0, c}, \n\t\t\t\t\t\ \ \ \ \ \ \ \ \ PlotStyle \[Rule] {Hue[0], Hue[ .7]}, \n\t\t\t\t\t\ \ \ \ \ \ \ \ \ PlotRange -> {{\(-2\), 2}, {\(-2\), 2}}], {c, \[Pi]\/12, 2\ \[Pi], \[Pi]\/12}]\)], "Input", AspectRatioFixed->True, FontSize->10] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Tangents to a Curve", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Suppose we have a polar equation written so that ", StyleBox["r", FontSlant->"Italic"], " is a function of \[Theta], that is\n\t", StyleBox["r = f", FontSlant->"Italic"], "(\[Theta]).\nWhat do you expect to observe on the graph when the \ derivative of ", StyleBox["f", FontSlant->"Italic"], " is zero? Test out your answer with the following exercise" }], "Text"], Cell[CellGroupData[{ Cell["Exercise 3", "Subsection"], Cell[TextData[{ "(a) Consider once again the cardioid\n\t", StyleBox["r", FontSlant->"Italic"], " = 1 + sin \[Theta].\nDefine a function ", StyleBox["f", FontSlant->"Italic"], " of \[Theta] (or ", StyleBox["t", FontSlant->"Italic"], ", since that is on the keyboard) that represents this expression (just the \ right-hand side). Then use the ", StyleBox["Solve", FontWeight->"Bold"], " command to find where ", StyleBox["f ' (t)", FontSlant->"Italic"], " is zero.\n(Once you've defined ", StyleBox["f", FontSlant->"Italic"], ", ", StyleBox["Mathematica", FontSlant->"Italic"], " recognizes ", StyleBox["f ' [t]", FontSlant->"Italic"], " as the derivative at ", StyleBox["t", FontSlant->"Italic"], ".) ", StyleBox["Mathematica", FontSlant->"Italic"], " should be able to find two zeros (ones you should be able to find \ relatively easily yourself algebraically), though, in fact, there are \ infinitely many. (Do you know what they are?)" }], "Text"], Cell[TextData[{ "(b) Now plot (use PolarPlot) ", StyleBox["f", FontSlant->"Italic"], " in a region of \[Theta]-values that includes one of the zeros of the \ derivative. (You may wish for this to be a \"smallish\" neighborhood of the \ zero.) Is what you expected confirmed? Is it confirmed at both of the zeros \ you found?" }], "Text"], Cell["\<\ (c) Using the notion that a zero derivative indicates a momentary \ halt in the change of the dependent variable, as well as the notion that r \ and \[Theta] represent magnitude and direction, write a brief explanation of \ how one should interpret a zero derivative along a polar curve.\ \>", "Text"], Cell[TextData[{ "(d) Recall that the relationship between rectangular coordinates (", StyleBox["x, y", FontSlant->"Italic"], ") and polar coordinates (", StyleBox["r", FontSlant->"Italic"], ", \[Theta]) may be expressed by the equations\n\t", StyleBox["x = r", FontSlant->"Italic"], " cos \[Theta]\n\t", StyleBox["y = r", FontSlant->"Italic"], " sin \[Theta].\nIf we are dealing with a polar equation where ", StyleBox["r", FontSlant->"Italic"], " is written as a function of \[Theta] (i.e., ", StyleBox["r = f", FontSlant->"Italic"], "(\[Theta])), then we may substitute to eliminate r from the above \ expressions:\n\t", StyleBox["x = f (\[Theta])", FontSlant->"Italic"], " cos \[Theta]\n\t", StyleBox["y = f (\[Theta])", FontSlant->"Italic"], " sin \[Theta],\nturning them into parametric equations with parameter \ \[Theta]. You may remember from Math 161 we learned that the slope of the \ tangent to a parametric curve was given by\n\t", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ x\)\ = \ \(\(\(d\ y\ /\ d\ \[Theta]\)\/\(d\ x\ /\ d\ \[Theta]\)\)\(.\)\)\)]], "\nUse the product rule as necessary to write out an expression for ", StyleBox["dy/dx", FontSlant->"Italic"], " in terms of \[Theta]. (Do this for the \"general\" parametric equations\ \n\t", StyleBox["x = f (\[Theta])", FontSlant->"Italic"], " cos \[Theta]\n\t", StyleBox["y = f (\[Theta])", FontSlant->"Italic"], " sin \[Theta],\nnot for any specific polar equation just yet.) Check with \ me that you have the correct expression." }], "Text"], Cell[TextData[{ "(e) Using your expression from part (d) now for the specific polar curve\n\ \t", StyleBox["f (\[Theta])", FontSlant->"Italic"], " = 1 + sin \[Theta],\nfind all \[Theta]-values which result in horizontal \ tangents. Use plots of the graph to confirm your answers. Can you also \ locate (algebraically) the vertical tangents?" }], "Text"] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"4.2 for X", ScreenRectangle->{{0, 1024}, {0, 768}}, WindowSize->{520, 600}, WindowMargins->{{Automatic, 192}, {Automatic, 47}} ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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