(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' 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[ 44012, 1255]*) (*NotebookOutlinePosition[ 44782, 1281]*) (* CellTagsIndexPosition[ 44738, 1277]*) (*WindowFrame->Normal*) Notebook[{ Cell["Lab 7: Dick and Jane have fun with Parameterizations", "Title", Background->RGBColor[0, 0, 1]], Cell[TextData[{ "Math 233\tFall 2004\t", StyleBox["Chapter 17, Sections 1 - 3", FontFamily->"Arial", FontSize->16, FontWeight->"Bold"], " " }], "Subtitle"], Cell[CellGroupData[{ Cell["Initialization Cells", "SmallText"], Cell[BoxData[{ \(\(BeginPackage["\"];\)\n\n (*\ Author : \ \ Bradford\ Kline\ *) \[IndentingNewLine] (*\ Last\ \(Revision : \ \ October\ 1998\)\ *) \n (*\ Rewritten\ by\ Mark\ Parker\ *) \[IndentingNewLine] (*\ September\ 2001\ *) \[IndentingNewLine]\), "\n", \(\(Off[General::spell];\)\), "\n", \(\(\(Off[General::spell1];\)\(\n\) \)\), "\n", \(Needs["\"]\), "\n", \(\(\(Needs["\"]\)\(\n\) \)\), "\n", \(\(\(DirectedCurvePlot::usage = \n"\";\)\(\n\) \)\), "\n", \(\(\(PositionPlot::usage = \n"\";\)\(\n\) \)\), "\n", \(\(\(VelocityPlot::usage = \n"\";\)\(\n\) \)\), "\n", \(\(\(AccelerationPlot::usage = \n"\";\)\(\n\) \)\), "\n", \(\(\(FieldCurvePlot::usage = \n"\";\)\(\n\) \)\), "\n", \(\(\(Begin["\<`Private`\>"];\)\(\n\) \)\), "\n", \(\(\(DirectedCurvePlot[position_, \ {t_, \ t0_, \ t1_}, \ num_Integer, \ opts___]\ := \ \n\ \ Module[{curveplot, \ arrow, \ arrows, \ tail, \ tip, \ j, \ a, \ dt}, \n\ \ \ \ curveplot = \n\ \ \ \ \ \ ParametricPlot[ position, \ {t, \ t0, \ t1}, \ DisplayFunction -> Identity, \ opts]; \n\n\ \ \ \ Do[ a\ = \ N[ t0\ + \ \((t1\ - \ t0)\)* j/num]; \n\ \ \ \ \ \ \ dt\ = \ \((t1\ - \ t0)\)/ 100; \n\ \ \ \ \ \ \ tail\ = \ position /. t -> a; \n\ \ \ \ \ \ \ tip\ = \ position /. t -> \((a\ + \ dt)\); \n\ \ \ \ \ \ \ arrow[ j]\ = \ \n\ \ \ \ \ \ \ \ \ Graphics[{Hue[0], Arrow[tail, tip, \n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ HeadScaling -> Absolute, HeadWidth -> .75]}], \n\ \ \ \ {j, 1, num}]; \n\t\t\n\ \ \ \ arrows = Table[arrow[j], {j, 1, num}]; \n\n\ \ \ \ Show[curveplot, \ arrows, \ FilterOptions[Graphics, \ opts], \ DisplayFunction -> $DisplayFunction]\n\ \ ]\)\(\n\) \)\), "\n", \(\(\(DirectedCurvePlot[position_, \ tRange_List, \ opts___Rule]\ := \n\ \ \ DirectedCurvePlot[position, \ tRange, \ 5, \ opts]\)\(\n\) \)\), "\n", \(\(\(PositionPlot[position_, \ {t_, \ t0_, \ t1_}, \ num_Integer, \ opts___]\ := \ \n\ \ Module[{curveplot, \ arrow, \ arrows, \ tail, \ tip, \ j, \ a, \ velocity}, \n\ \ \ \ curveplot = \n\ \ \ \ \ \ ParametricPlot[ position, \ {t, \ t0, \ t1}, \ DisplayFunction -> Identity, \ opts]; \n\n\ \ \ \ Do[ a\ = \ N[ t0\ + \ \((t1\ - \ t0)\)* j/num]; \n\ \ \ \ \ \ \ tail\ = \ {0, 0}; \n\ \ \ \ \ \ \ tip\ = \ position /. t -> a; \n\ \ \ \ \ \ \ arrow[ j]\ = \ \n\ \ \ \ \ \ \ \ \ Graphics[{Hue[0.4], Arrow[tail, tip, \n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ HeadScaling -> Absolute, HeadWidth -> .75]}], \n\ \ \ \ {j, 1, num}]; \n\t\t\n\ \ \ \ arrows = Table[arrow[j], {j, 1, num}]; \n\n\ \ \ \ Show[curveplot, \ arrows, \ FilterOptions[Graphics, \ opts], \ DisplayFunction -> $DisplayFunction]\n\ \ ]\)\(\n\) \)\), "\n", \(\(\(PositionPlot[position_, \ tRange_List, \ opts___Rule]\ := \n\ \ \ PositionPlot[position, \ tRange, \ 5, \ opts]\)\(\n\) \)\), "\n", \(\(\(VelocityPlot[position_, \ {t_, \ t0_, \ t1_}, \ num_Integer, \ opts___]\ := \ \n\ \ Module[{curveplot, \ arrow, \ arrows, \ tail, \ tip, \ j, \ a, \ velocity}, \n\ \ \ \ curveplot = \n\ \ \ \ \ \ ParametricPlot[ position, \ {t, \ t0, \ t1}, \ DisplayFunction -> Identity, \ opts]; \n\n\t\tvelocity\ = \ D[position, t]; \n\t\t\n\ \ \ \ Do[ a\ = \ N[ t0\ + \ \((t1\ - \ t0)\)*j/num]; \n\ \ \ \ \ \ \ tail\ = \ position /. t -> a; \n\ \ \ \ \ \ \ tip\ = \ position + velocity /. t -> a; \n\ \ \ \ \ \ \ arrow[ j]\ = \ \n\ \ \ \ \ \ \ \ \ Graphics[{Hue[0], Arrow[tail, tip, \n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ HeadScaling -> Absolute, HeadWidth -> .75]}], \n\ \ \ \ {j, 1, num}]; \n\t\t\n\ \ \ \ arrows = Table[arrow[j], {j, 1, num}]; \n\n\ \ \ \ Show[curveplot, \ arrows, \ FilterOptions[Graphics, \ opts], \ DisplayFunction -> $DisplayFunction]\n\ \ ]\)\(\n\) \)\), "\n", \(\(\(VelocityPlot[position_, \ tRange_List, \ opts___Rule]\ := \n\ \ \ VelocityPlot[position, \ tRange, \ 5, \ opts]\)\(\n\) \)\), "\n", \(\(\(AccelerationPlot[position_, \ {t_, \ t0_, \ t1_}, \ num_Integer, \ opts___]\ := \ \n\ \ Module[{curveplot, \ arrow, \ arrows, \ tail, \ tip, \ j, \ a, \ acceleration}, \n\ \ \ \ curveplot = \n\ \ \ \ \ \ ParametricPlot[ position, \ {t, \ t0, \ t1}, \ DisplayFunction -> Identity, \ opts]; \n\n\t\tacceleration\ = \ D[position, {t, 2}]; \n\t\t\n\ \ \ \ Do[ a\ = \ N[ t0\ + \ \((t1\ - \ t0)\)*j/num]; \n\ \ \ \ \ \ \ tail\ = \ position /. t -> a; \n\ \ \ \ \ \ \ tip\ = \ position + acceleration /. t -> a; \n\ \ \ \ \ \ \ arrow[ j]\ = \ \n\ \ \ \ \ \ \ \ \ Graphics[{Hue[0.7], Arrow[tail, tip, \n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ HeadScaling -> Absolute, HeadWidth -> .75]}], \n\ \ \ \ {j, 1, num}]; \n\t\t\n\ \ \ \ arrows = Table[arrow[j], {j, 1, num}]; \n\n\ \ \ \ Show[curveplot, \ arrows, \ FilterOptions[Graphics, \ opts], \ DisplayFunction -> $DisplayFunction]\n\ \ ]\)\(\n\) \)\), "\n", \(\(\(AccelerationPlot[position_, \ tRange_List, \ opts___Rule]\ := \n\ \ \ AccelerationPlot[position, \ tRange, \ 5, \ opts]\)\(\n\) \)\), "\n", \(\(\(FieldCurvePlot[position_, \ {t_, \ t0_, \ t1_}, vectorfield_, {x_, y_}, \ num_Integer, \ opts___]\ := \ \n\ \ Module[{curveplot, \ arrow, \ arrows, \ tail, \ tip, \ j, \ a, fieldvector}, \n\ \ \ \ curveplot = \n\ \ \ \ \ \ DirectedCurvePlot[ position, \ {t, \ t0, \ t1}, 3, \ DisplayFunction -> Identity, \ opts]; \n\n\t\ fieldvector = vectorfield /. {x -> position[\([1]\)], y -> position[\([2]\)]}; \n\t\t\n\ \ \ \ Do[ a\ = \ N[ t0\ + \ \((t1\ - \ t0)\)*j/num]; \n\ \ \ \ \ \ \ tail\ = \ position /. t -> a; \n\ \ \ \ \ \ \ tip\ = \ position + fieldvector /. t -> a; \n\ \ \ \ \ \ \ arrow[ j]\ = \ \n\ \ \ \ \ \ \ \ \ Graphics[{Hue[0.9], Arrow[tail, tip, \n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ HeadScaling -> Absolute, HeadWidth -> .75]}], \n\ \ \ \ {j, 1, num}]; \n\t\t\n\ \ \ \ arrows = Table[arrow[j], {j, 1, num}]; \n\n\ \ \ \ Show[curveplot, \ arrows, \ FilterOptions[Graphics, \ opts], \ DisplayFunction -> $DisplayFunction]\n\ \ ]\)\(\n\) \)\), "\n", \(\(\(FieldCurvePlot[position_, \ tRange_List, \ vectorfield_, vars_List, \ opts___Rule]\ := \n\ \ \ FieldCurvePlot[position, \ tRange, \ vectorfield, \ vars, \ \ 5, \ opts]\)\(\n\) \)\), "\n", \(\(\(End[\ ];\)\(\n\) \)\), "\n", \(\(\(Protect[DirectedCurvePlot, \ PositionPlot, \ VelocityPlot, \ AccelerationPlot, \ FieldCurvePlot];\)\(\n\) \)\), "\n", \(EndPackage[\ ]\)}], "Input", InitializationCell->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Instructions!", FontColor->RGBColor[1, 0, 0]]], "Section"], Cell[TextData[{ "Work in ", StyleBox["groups of 2 or 3", FontWeight->"Bold"], " today, going through the entire notebook. If you have questions that you \ can't work through, ask and I'll (hopefully) get over to you. You'll hand in \ a group lab report that will be ", StyleBox["due on Tuesday (16 November)", FontWeight->"Bold"], ". \n\n", StyleBox["Group Report:", FontWeight->"Bold"], "\n", StyleBox["The heart of the report should be a summary of your observations \ - you may cut and paste from ", FontWeight->"Bold"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" to help clarify your explanations. For this lab, you should \ answer the questions I've given you and then use your experiences in the ", FontWeight->"Bold"], StyleBox["You Try It", FontWeight->"Bold", FontVariations->{"Underline"->True}], StyleBox[" sections (along with the questions you answered from the \ previous sections) to write a write a summary of your observations. Bottom \ Line: I don't want just a few words written. Reflect on what you have seen \ in each section.\n\n", FontWeight->"Bold"], "The easiest way to create a group report is to open up a Word document \ while doing the lab. When you come to a question, you can highlight the text \ and copy and paste it into the Word document. Your group can then type your \ answers into the Word document. If you want to use a graph in your report to \ help clarify an idea or to answer a question, you can click on the graph (so \ that it is surrounded by a box) and then copy and paste it into the Word \ document. Once you've finished the You Try It section, go back to your \ report and write a summary of your experiences - this summary can either go \ in front of or behind the questions you've already answered. \n" }], "Subsubtitle"] }, Closed]], Cell[CellGroupData[{ Cell["Part I: Interpreting Lines Using Their Vector Definition", "Section"], Cell[TextData[{ "Let's start in the plane. A line, you should recall, has infinite \ length... whereas a vector has a finite length (or magnitude). We will use \ ", StyleBox["position vectors", FontWeight->"Bold"], " to help us describe the direction or orientation of the line. A position \ vector is simply a vector from the ", StyleBox["origin", FontWeight->"Bold"], " to a point on the line. Suppose you want to write the equation of the \ line segment between the points (1,5) and (4,2). Let's look at what we are \ working with. \nIn ", StyleBox["Mathematica", FontSlant->"Italic"], ", you can graph points using the ", StyleBox["ListPlot", FontWeight->"Bold"], " command - simply enter them as lists (ordered pairs). I set the ", StyleBox["PlotJoined ", FontWeight->"Bold"], "argument to ", StyleBox["True", FontWeight->"Bold"], " to connect the points in a line (or vector...). The only problem is that \ ", StyleBox["ListPlot", FontWeight->"Bold"], " doesn't give us anyway to have the arrows that we see on the head of a \ vector. There is an extra package called ", StyleBox["Arrow", FontWeight->"Bold"], " that we can load into ", StyleBox["Mathematica", FontSlant->"Italic"], " to allow us to draw vectors:" }], "Text"], Cell[TextData[{ "Note: When you execute this cell, you'll be asked if you want to execute \ the initialization cells - ", StyleBox["SAY YES!", FontWeight->"Bold"] }], "Text", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(<< Graphics`Arrow`\)], "Input"], Cell[TextData[{ "You'll only need to execute that cell once this session (unless you have \ to ", StyleBox["Stop", FontWeight->"Bold"], " the ", StyleBox["Kernel", FontWeight->"Bold"], " at some point)... so I can draw an ", StyleBox["Arrow", FontWeight->"Bold"], " connecting {1,5} (the vector tail) and {4,2} (the vector head). We use \ ", StyleBox["{ }", FontWeight->"Bold"], " to denote vectors in ", StyleBox["Mathematica", FontSlant->"Italic"], " (instead of ", StyleBox["< >", FontWeight->"Bold"], " like we do on paper)." }], "Text"], Cell[BoxData[{ \(Off[General::spell]\), "\[IndentingNewLine]", \(Off[General::spell1]\), "\[IndentingNewLine]", \(Clear[x, y, t]\), "\[IndentingNewLine]", \(\(line = ListPlot[{{1, 5}, {4, 2}}, PlotJoined -> True, AxesOrigin \[Rule] {0, 0}, DisplayFunction \[Rule] Identity];\)\), "\[IndentingNewLine]", \(\(vectors = Graphics[{Hue[0], Arrow[{0, 0}, {4, 2}], Hue[0.25], Arrow[{0, 0}, {1, 5}]}];\)\), "\[IndentingNewLine]", \(\(Show[{line, vectors}, DisplayFunction \[Rule] $DisplayFunction];\)\)}], "Input"], Cell[TextData[{ StyleBox["Notice that I named 2 different plots and used the ", FontFamily->"Arial"], StyleBox["Show ", FontFamily->"Arial", FontWeight->"Bold"], StyleBox["command to print them all on the same graph. I got rid of the \ output from the individual graphs by using the ", FontFamily->"Arial"], StyleBox["DisplayFunction->Identity", FontFamily->"Arial", FontWeight->"Bold"], StyleBox[" command. This is handy for getting rid of clutter sometimes. \ The ", FontFamily->"Arial"], StyleBox["Hue[x]", FontFamily->"Arial", FontWeight->"Bold"], StyleBox[" lets me choose the color of the graph - values for ", FontFamily->"Arial"], StyleBox["x", FontFamily->"Arial", FontWeight->"Bold"], StyleBox[" range from 0 to 1 and they cause our color to change from ", FontFamily->"Arial"], StyleBox["Red to Green to Blue and back to Red again as ", FontFamily->"Arial"], StyleBox["x", FontFamily->"Arial", FontWeight->"Bold"], StyleBox[" increases.", FontFamily->"Arial"] }], "Text", FontColor->RGBColor[0, 0, 1]], Cell["\<\ In the graph above, we are interested in the line segment in black - it is \ the line from (1,5) to (4,2). The position vectors for (1,5) and (4,2) are \ in green and red respectively. We could create a vector in the same \ direction as the line by finding the displacement vector from (1,5) to (4,2):\ \ \>", "Text"], Cell[BoxData[{ \(\(p1 = {1, 5};\)\), "\[IndentingNewLine]", \(\(p2 = {4, 2};\)\), "\[IndentingNewLine]", \(\(dir = p2 - p1;\)\), "\[IndentingNewLine]", \(Print["\", \ dir]\)}], "Input"], Cell[TextData[StyleBox["Note: The resulting vector dir IS NOT A UNIT VECTOR! \ it won't matter for our application here, but this is an exception to the \ rule that you need a unit vector when discussing direction.", FontFamily->"Arial"]], "Text", FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ "Now we can write any other point on the line as the position vector to \ that point plus a multiple (", StyleBox["t", FontSlant->"Italic"], ") of the vector in the direction of the line (which we called ", StyleBox["dir", FontWeight->"Bold"], " above)", ". We plot points along the line corresponding to different values of the \ parameter ", StyleBox["t", FontSlant->"Italic"], ". " }], "Text"], Cell[BoxData[{ \(\(origin = {0, 0};\)\), "\n", \(Print["\<{x,y} = \>", eq = \((p1 - origin)\) + t\ dir]\), "\n", \(\(ParametricPlot[Evaluate[eq], {t, \(-10\), 10}, AxesLabel -> {x, y}];\)\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["You Try It - Part I", "Section"], Cell[TextData[{ "Write and plot the line connecting the points ", Cell[BoxData[ \(TraditionalForm\`\((\(-23\), \ \(-5\))\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((10, \ 12)\)\)]], " in parametric form. I've gotten you started below, all you need to do is \ identify and enter the new points for ", StyleBox["p1", FontSlant->"Italic"], " and ", StyleBox["p2", FontSlant->"Italic"], " on the line and re-execute the input cell. ", StyleBox["REMEMBER: ", FontWeight->"Bold"], "in ", StyleBox["Mathematica", FontSlant->"Italic"], ",", " you need to enter points as lists - which means use curly brackets! " }], "Text"], Cell[BoxData[{\(origin = {0, 0};\), "\n", StyleBox[\(\(\(p1\)\(=\)\);\), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[\(\(\(p2\)\(=\)\);\), FontColor->RGBColor[1, 0, 0]], "\n", \(Print["\", dir = p2 - p1]\), "\n", \(Print["\<{x,y} = \>", eq = \((p1 - origin)\) + t\ dir]\), "\n", \(plot1 = ParametricPlot[Evaluate[eq], {t, \(-10\), 10}, AxesLabel -> {x, y}];\)}], "Input"], Cell[TextData[{ " For the preceding function, match the parametric plot with the standard \ Cartesian form and plot. You need to solve for ", StyleBox["y", FontSlant->"Italic"], " as a function of ", StyleBox["x", FontSlant->"Italic"], " and put that into the expression for ", Cell[BoxData[ \(TraditionalForm\`f[x_]\)]], ". Replace the expression in red with the appropriate expression." }], "Text"], Cell[BoxData[{ RowBox[{\(f[x_]\), "=", " ", StyleBox["x", FontColor->RGBColor[1, 0, 0]]}], "\n", \(pf = Plot[f[x], {x, \(-300\), 300}, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"\", "\"}];\), "\n", \(Show[plot1, pf];\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Part II: Different Parametric Forms for the Same Line", "Section"], Cell[TextData[{ "Write the equation of the line ", Cell[BoxData[ \(TraditionalForm\`y\ = \ 3 x\ - \ 5\)]], " in parametric form. \nYou may be accustomed to seeing equations of lines \ in the standard slope intercept form. That form is a unique representation. \ In contrast, the parametric form can appear in different forms; however, the \ set of points represented is the same." }], "Text"], Cell[TextData[{ "If we take the direction vector from the slope to be ", Cell[BoxData[ \(TraditionalForm\`\((1, 3)\)\)]], " and use the ", StyleBox["y", FontSlant->"Italic"], "-intercept ", Cell[BoxData[ \(TraditionalForm\`\((0, 5)\)\)]], " as our given point on the line, we get the following." }], "Text"], Cell[BoxData[{ \(Clear[x, f]\), "\n", \(\(origin = {0, 0};\)\), "\n", \(\(dir = {1, 3};\)\), "\n", \(\(p1 = {0, \(-5\)};\)\), "\n", \(Print["\<{x,y} = \>", eq = \((p1 - origin)\) + t\ dir]\), "\n", \(\(plot1 = ParametricPlot[Evaluate[eq], {t, \(-3\), 5}, AxesLabel -> {x, y}, PlotStyle -> RGBColor[0, 1, 0]];\)\)}], "Input"], Cell[TextData[{ "Had we used a different point on the line, for example,", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\((2, 1)\)\)\)]], ", and written the slope as ", Cell[BoxData[ \(TraditionalForm\`\(\((2, \ 6)\)\(,\)\)\)]], " the equation would look a bit different, although it would produce the \ same line." }], "Text"], Cell[BoxData[{ \(Clear[x, f]\), "\n", \(\(dir = {2, 6};\)\), "\n", \(\(p2 = {2, 1};\)\), "\n", \(Print["\<{x,y} = \>", eq = \((p2 - origin)\) + t\ dir]\), "\n", \(\(plot2 = ParametricPlot[Evaluate[eq], {t, \(-1\), 3}, AxesLabel -> {x, y}, PlotStyle -> RGBColor[0, 0, 1]];\)\)}], "Input"], Cell["We can look at these graphs together.", "Text"], Cell[BoxData[ \(\(Show[plot1, plot2];\)\)], "Input"], Cell[TextData[{ "Note that the parameter ", StyleBox["t", FontSlant->"Italic"], " is not the same in the two equations. However, all the points lie on the \ same line, even though we are plotting only segments of that line.\n\n For \ the preceding function, match the parametric plot with the standard Cartesian \ form and plot." }], "Text"], Cell[BoxData[{ \(f[x_] := 3 x - 5\), "\n", \(\(pf = Plot[f[x], {x, \(-1\), 5}, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"\", "\"}];\)\), "\n", \(\(Show[plot1, plot2, pf];\)\)}], "Input"], Cell[TextData[StyleBox["Question 1:\tDo they match?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]] }, Closed]], Cell[CellGroupData[{ Cell["You Try It - Part II", "Section"], Cell[TextData[{ "Write the line ", Cell[BoxData[ \(TraditionalForm\`4 x\ + \ 5 y\ = \ 12\)]], " in parametric form. \nYou need to identify a point on the line and a \ vector in the direction of the line. Substitute these for ", StyleBox["dir", FontWeight->"Bold"], " and ", StyleBox["p1", FontWeight->"Bold"], " (items in red), and re-execute the input cell. " }], "Text"], Cell[BoxData[{ StyleBox[\(\(\(dir\)\(=\)\)\ \ ;\), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[\(\(\(p1\)\(=\)\)\ ;\), FontColor->RGBColor[1, 0, 0]], "\n", \(Print["\<{x,y} = \>", eq = \((p1 - origin)\) + t\ dir]\), "\n", \(plot1 = ParametricPlot[Evaluate[eq], {t, \(-3\), 5}, AxesLabel -> {x, y}, PlotStyle -> RGBColor[0, 1, 0]];\)}], "Input"], Cell[TextData[{ "For the preceding function, match the parametric plot with the standard \ Cartesian form and plot. \nYou need to solve for ", StyleBox["y", FontSlant->"Italic"], " as a function of x and put that into the expression for ", Cell[BoxData[ \(TraditionalForm\`f[x_]\)]], ". Replace the red with the appropriate expression." }], "Text"], Cell[BoxData[{ RowBox[{\(f[x_]\), ":=", StyleBox["x", FontColor->RGBColor[1, 0, 0]]}], "\n", \(pf = Plot[f[x], {x, \(-1\), 5}, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"\", "\"}];\), "\n", \(Show[plot1, pf];\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Part III: Parametric Equations of a Curve in 2-Space", "Section"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Let's define a function!", FontWeight->"Bold"]], "Subsubsection"], Cell[TextData[{ "Let's define the ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " coordinates parametrically. Suppose that time,", StyleBox[" t", FontSlant->"Italic"], ", is the independent variable. Once ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " are defined, we can write the position vector ", StyleBox["r(t)", FontSlant->"Italic"], "." }], "Text"], Cell[BoxData[ \(\(\(Clear[x, y, t, r]\n x[t_] = Cos[t]\^3\n y[t_] = 1 - Exp[Sin[t]]\[IndentingNewLine] r[t_] = {x[t], y[t]}\n Print["\", r[t]]\)\(\ \)\)\)], "Input"], Cell[TextData[{ "Now we plot the resulting curve in blue. To do this, we use the ", StyleBox["Mathematica", FontSlant->"Italic"], " ", "command ", StyleBox["ParametricPlot", FontWeight->"Bold"], " - here is some information about it:" }], "Text"], Cell["\<\ Note: A single question mark before a command gives us a summary of \ information about that command. You can enter 2 question marks to get more \ information - e.g. ??ParametricPlot \ \>", "Text", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(\(?ParametricPlot\)\)], "Input"], Cell[TextData[{ "Note: The ", StyleBox["PlotStyle->RGBColor[0,0,1]", FontWeight->"Bold"], " option specifies the Red, Green, and Blue relative magnitudes in the \ color of the plot - so here, we will have a blue one!" }], "Text", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[{ \(\(plotf = ParametricPlot[Evaluate[r[t]], {t, 0, N[2 \[Pi]]}, PlotStyle -> RGBColor[0, 0, 1], AxesLabel -> {"\", "\"}];\)\ \), "\n", \(Print["\", r[t]]\)}], "Input"], Cell[TextData[{ StyleBox["Question 2:\tLook closely at the functions describing motion in \ the x- and y-directions. Estimate both the starting point and the direction \ of motion you will follow along the path as ", FontWeight->"Bold"], StyleBox["t", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" increases? Why do you think so?", FontWeight->"Bold"] }], "Text", Background->RGBColor[0, 1, 1]], Cell[TextData[{ "You can check your answer by using the ", StyleBox["DirectedCurvePlot", FontWeight->"Bold"], " command. Note that this ", StyleBox["is not ", FontWeight->"Bold"], "a standard ", StyleBox["Mathematica", FontSlant->"Italic"], " command. It is one that I've defined and made available in this \ notebook. If you are interested, you can see the implementation details in \ the section titled \"Initialization\"\nHere is some information about ", StyleBox["DirectedCurvePlot", FontWeight->"Bold"] }], "Text"], Cell[BoxData[ \(\(?DirectedCurvePlot\)\)], "Input"], Cell[BoxData[{ \(\(plotf1 = DirectedCurvePlot[r[t], {t, 0, N[2 \[Pi]]}, 8, AxesLabel -> {"\", "\"}];\)\ \), "\n", \(Print["\", r[t]]\)}], "Input"], Cell[TextData[StyleBox["Question 3:\tWhere you correct in your estimate from \ Question 2? If not, try to find the flaw in your analysis above.", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Taking the Velocity and Acceleration into Account", FontWeight->"Bold"]], "Subsubsection"], Cell[TextData[{ "If you consider the parametric equation as a vector equation for the \ motion of a particle, the velocity vector is found by differentiating each \ component of the position vector. Similarly, the acceleration vector is found \ by differentiating the components of the position vector twice. We can do \ this automatically with the supplemental commands ", StyleBox["VectorPlot", FontWeight->"Bold"], " and ", StyleBox["AccelerationPlot", FontWeight->"Bold"], "." }], "Text"], Cell[TextData[{ "Note: In ", StyleBox["Mathematica", FontSlant->"Italic"], ", the derivative can be found with either the derivative command ", StyleBox["D[function,variable to differentiate with respect to] ", FontWeight->"Bold"], "or the prime (single quote), ", StyleBox["f'[x]", FontWeight->"Bold"], ". For the prime to work, the function must be defined with an argument (", StyleBox["f[x]", FontWeight->"Bold"], "), whereas the differentiation command can be applied to a function in any \ form, with or without an argument. You could then manually graph the \ derivative function via the ", StyleBox["ParametricPlot", FontWeight->"Bold"], " command as shown below:\n", "ParametricPlot[Evaluate[r'[t]],{t,0,N[2 \ \[Pi]]},AxesLabel->{\"x\",\"y\"}];" }], "Text", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(\(?VelocityPlot\)\)], "Input"], Cell[BoxData[{ \(\(plotv = VelocityPlot[r[t], {t, 0, N[2\ \[Pi]]}, 8, AxesLabel -> {"\", "\"}];\)\), "\n", \(Print["\", \(r'\)[t]]\)}], "Input"], Cell[BoxData[ \(\(?AccelerationPlot\)\)], "Input"], Cell[BoxData[{ \(\(plota = AccelerationPlot[r[t], {t, 0, N[2 \[Pi]]}, 8, AxesLabel -> {"\", "\"}];\)\), "\n", \(Print["\", \(\(r'\)'\)[t] // Simplify]\)}], "Input"], Cell[TextData[{ "Note that the components of the velocity and acceleration functions are \ more complicated than the components of the postion function. Let's take a \ look at both of them on the same set of axes. The ", StyleBox["Show", FontWeight->"Bold"], " command allows us to view previously named graphs, or to view multiple \ graphs on the same set of axes:" }], "Text"], Cell[BoxData[ \(\(Show[plotv, plota];\)\)], "Input"], Cell[TextData[{ "Let's look at the ", StyleBox["speed", FontWeight->"Bold"], " function and see what it tells us. " }], "Text"], Cell["\<\ Note: Another way to find the magnitude of a vector v(t) is to take the \ SquareRoot of (v dot v).\ \>", "Text", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[{ \(speed[t_] := \@\(\(r'\)[t] . \(r'\)[t]\)\), "\[IndentingNewLine]", \(Print["\", speed[t]]\)}], "Input"], Cell[TextData[{ "Now, we can plot the speed as a function of ", StyleBox["t", FontSlant->"Italic"], ". Here is some basic information about the ", StyleBox["Plot ", FontWeight->"Bold"], "command:" }], "Text"], Cell[BoxData[ \(\(?Plot\)\)], "Input"], Cell["\<\ The following plot shows the speed in black, the x-coordinate of the path in \ orange and the y-coordinate of the path in violet.\ \>", "Text"], Cell["\<\ Note: We want to plot 3 functions, not 1... we can accomplish this by \ enclosing all 3 of the functions in a single set of curly brackets { }. The \ three different RGBColor options specify a different color for each of the 3 \ graphs we wish to represent.\ \>", "Text", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[{ \(\(Plot[{speed[t], x[t], y[t]}, {t, 0, N[2 \[Pi]]}, PlotStyle -> {RGBColor[0, 0, 0], RGBColor[1, .5, 0], RGBColor[1, 0, 1]}, AxesLabel -> {t, function}];\)\), "\n", \(Print["\"]\)}], "Input"], Cell[TextData[StyleBox["Question 4:\t Execute the cell below. Contrast the \ plot above with your parametric plot below to identify the places on your \ path where the speed function is 0 and where the speed is a maximum. Discuss \ how this compares to, or contrasts with, your intuition.", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]], Cell[BoxData[ \(\(Show[plotf];\)\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Computing the Distance Traveled on a Curved Path", FontWeight->"Bold"]], "Subsubsection"], Cell[TextData[StyleBox["Question 5:\t Suppose that you are walking along the \ path given above. The distance traveled can be found by integrating the speed \ function over a particular interval. Think about this and explain why this \ integral gives you the distance traveled.", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]], Cell["\<\ Here is the parameterized path and a graph of speed versus time while walking \ along the path:\ \>", "Text"], Cell[BoxData[{ \(\(Show[plotf1];\)\), "\n", \(\(Plot[speed[t], {t, 0, N[2 \[Pi]]}, AxesLabel -> {t, speed}];\)\)}], "Input"], Cell[TextData[{ "Note: I'm using the ", StyleBox["NIntegrate", FontWeight->"Bold"], " command to numerically integrate the speed function as the variable ", StyleBox["t", FontSlant->"Italic"], " ranges from 0 to 2 Pi. All that the ", StyleBox["Print", FontWeight->"Bold"], " command does is spit out to the screen any text between \" \" and the \ numerical values of any ", StyleBox["Mathematica ", FontSlant->"Italic"], "variables (like distance)." }], "Commentary", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(Print["\", distance = NIntegrate[speed[t], {t, 0, N[2 \[Pi]]}], "\< units.\>"]\)], "Input"], Cell["\<\ Think of this answer as either the distance around the curve or as the area \ under the speed function over the interval t from 0 to 2\[Pi].\ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Velocity and Acceleration - When Are They Perpendicular?", FontWeight->"Bold"], " " }], "Subsubsection"], Cell["\<\ What is the relationship between the acceleration and velocity vectors? As \ we saw in class, sometimes they are perpendicular to each other (to see for \ yourself, parameterize a circle and then graph the velocity and acceleration \ vectors. They are perpendicular (or orthogonal, or normal) everywhere on the \ curve! But circles are special functions, so is there a way that we can look \ at a curve and determine when the acceleration and velocity vectors are \ perpendicular?\ \>", "Text"], Cell[TextData[StyleBox["Question 6:\tExecute the cell below to see, yet \ again, the parameterized path. Think about it and then explain where on the \ graph you think the acceleration and velocity vectors will be perpendicular.", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]], Cell[BoxData[ \(\(Show[plotf1];\)\)], "Input"], Cell[TextData[{ "Back to the task at hand... We can use the dot product to help us figure \ this out, since perpendicular vectors yield a dot product of 0. Here we will \ examine when the velocity and acceleration vectors are perpendicular to one \ another. We begin by computing the dot product of the two vectors, and we \ plot the resulting function of ", StyleBox["t", FontSlant->"Italic"], " to get an idea of when the dot product might be 0.\nJust to remind us, \ here are a few of the values:" }], "Text"], Cell[BoxData[{ \(Print["\", \(r'\)[t]]\), "\n", \(Print["\", \(r''\)[t] // Simplify]\)}], "Input"], Cell[TextData[{ "Now, let's see what the dot product is. We can either use the ", StyleBox["Mathematica", FontSlant->"Italic"], " ", "command ", StyleBox["Dot", FontWeight->"Bold"], ", or we can use a period", ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Print["\", vdota = \(r'\)[t] . \(r''\)[t] // Simplify]\)], "Input"], Cell[TextData[{ "We should note that velocity dot acceleration ", StyleBox["isn't", FontWeight->"Bold"], " a constant, but rather a function of ", StyleBox["t", FontSlant->"Italic"], ". So, we can plot it and see where it is 0: " }], "Text"], Cell[BoxData[ \(\(Plot[Evaluate[vdota], {t, 0, N[2 \[Pi]]}, AxesLabel \[Rule] {t, "\"}];\)\)], "Input"] }, Open ]], Cell[TextData[{ "As you can see from the graph, there are many times when the velocity and \ acceleration vectors are perpendicular to each other. It would be nice to \ know what those points look like on our path, so that we can perhaps \ characterize when acceleration and velocity will be perpendicular to each \ other. \nFirst, we'll need to know ", StyleBox["when", FontSlant->"Italic"], " these ", StyleBox["roots ", FontWeight->"Bold"], "are occuring. We can use the ", StyleBox["Mathematica", FontSlant->"Italic"], " ", "command ", StyleBox["FindRoot", FontWeight->"Bold"], " to search for these roots:" }], "Text"], Cell[TextData[{ "Note: ", StyleBox["Roots", FontWeight->"Bold"], " of a function are those values of the domain where the range is 0." }], "Commentary", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(\(?FindRoot\)\)], "Input"], Cell[TextData[{ "Note that we are able to give approximate root values to help ", StyleBox["FindRoot", FontWeight->"Bold"], " locate 0's of a function. Use the graph above and the template below to \ find roots of the dot product results (which we called vdota). It looks \ like there are 8 of them, we'll store these values in an array called vperpa \ (which has indices from 0 to 7):" }], "Text"], Cell[BoxData[{ \(Clear[vperpa]\), "\n", \(vperpa[0] = FindRoot[vdota == 0, {t, 0.75}]\), "\n", \(vperpa[1] = FindRoot[vdota == 0, {t, 1.5}]\), "\n", \(\)}], "Input"], Cell["OK here is your part!", "Text"], Cell[BoxData[{ \(vperpa[2] = FindRoot\), "\n", \(vperpa[3] = FindRoot\), "\[IndentingNewLine]", \(vperpa[4] = FindRoot\), "\n", \(vperpa[5] = FindRoot\), "\n", \(vperpa[6] = FindRoot\), "\n", \(vperpa[7] = FindRoot\)}], "Input"], Cell[TextData[{ "Now we evaluate", StyleBox[" x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " at the values of ", StyleBox["t", FontSlant->"Italic"], " we have found. The ", StyleBox["Table", FontWeight->"Bold"], " command allows us to create a table of values... this is a cool command \ for organizing lots o' output:" }], "Text"], Cell[BoxData[ \(\(?Table\)\)], "Input"], Cell[BoxData[ \(\(?TableForm\)\)], "Input"], Cell[BoxData[{ \(\(special = Table[{x[t], y[t]} /. vperpa[i], {i, 0, 7}];\)\), "\n", \(TableForm[special, TableHeadings -> {None, {"\", "\"}}]\)}], "Input"], Cell[TextData[{ "We can now see where those points are relative to our original graph. \ The", StyleBox[" Listplot ", FontWeight->"Bold"], "command allows us to plot a list of data points, which are contained in \ the list we called \"special\":" }], "Text"], Cell[BoxData[ \(\(?ListPlot\)\)], "Input"], Cell[BoxData[{ \(\(psp = ListPlot[special, PlotStyle -> PointSize[ .04], DisplayFunction -> Identity];\)\), "\n", \(\(Show[plotf1, psp, DisplayFunction -> $DisplayFunction];\)\)}], "Input"], Cell["\<\ The dots on the graph above show where the velocity and acceleration vectors \ are perpendicular to each other. \ \>", "Text"], Cell[TextData[StyleBox["Question 7:\tWhat is happening along the path at the \ points where the acceleration and velocity vectors are perpendicular to each \ other? Does this match your intuition from Question 6? If not, where did \ intuition conflict with reality?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Part IV: Parametric Equations of a Curve in 3-Space", "Section"], Cell[TextData[{ "In the last section, we looked at parametric equations of a curve in 2 \ dimensions. Here, we'll make the jump to 3-space. First, we define the ", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], ", and ", StyleBox["z", FontSlant->"Italic"], " coordinates for motion parametrically. " }], "Text"], Cell[BoxData[{ \(Clear[x, y, z, t]\), "\n", \(x[t_] = Cos[t]\), "\n", \(y[t_] = Sin[t]\), "\n", \(z[t_] = 4 - t\^2/25\)}], "Input"], Cell[TextData[{ "Next, we plot the resulting curve in blue. Here, instead of ", StyleBox["ParametricPlot", FontWeight->"Bold"], ", we can use ", StyleBox["ParametricPlot3D", FontWeight->"Bold"], " for 3-dimensional plots." }], "Text"], Cell[BoxData[ \(\(plotf = ParametricPlot3D[{x[t], y[t], z[t], RGBColor[0, 0, 1]}, {t, 0, 10}, AxesLabel -> {"\", "\", "\"}];\)\)], "Input"], Cell[TextData[{ StyleBox["Question 8:\tAs ", FontWeight->"Bold"], StyleBox["t", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" increases, what direction are you moving?", FontWeight->"Bold"] }], "Text", Background->RGBColor[0, 1, 1]], Cell["\<\ As before, the derivative of the position vector gives us the velocity \ vector, and the its second derivative is the acceleration vector. Recall \ that we differentiate component by component. Here are the velocity and \ acceleration vectors:\ \>", "Text"], Cell[BoxData[ \(\(\(\ \ \)\(Print["\", {\(x'\)[t], \(y'\)[ t], \(z'\)[t]}]\n \(plotf' = ParametricPlot3D[{\(x'\)[t], \(y'\)[t], \(z'\)[t], RGBColor[1, 0, 0]}, {t, 0, 10}, AxesLabel -> {x, y, z}];\)\n Print["\", {\(\(x'\)'\)[t], \(\(y'\)'\)[ t], \(\(z'\)'\)[t]}]\n \(\(plotf'\)' = ParametricPlot3D[{\(\(x'\)'\)[t], \(\(y'\)'\)[t], \(\(z'\)'\)[t], RGBColor[0, 1, 1]}, {t, 0, 10}, AxesLabel -> {x, y, z}];\)\)\)\)], "Input"], Cell[TextData[{ "The following command plots all three functions together, with the \ position in blue, the velocity in red, and the acceleration in aqua. You \ should be able to see that the acceleration in the vertical (z) direction is \ a negative constant, so the velocity in that direction is decreasing at a \ linear rate, causing the motion to spiral down at a rate proportion to ", Cell[BoxData[ \(TraditionalForm\`t\^2\)]], ". The motion, velocity, and acceleration in the ", StyleBox["x ", FontSlant->"Italic"], "and ", StyleBox["y", FontSlant->"Italic"], " directions are circular." }], "Text"], Cell[BoxData[ \(\(Show[plotf, plotf', \(plotf'\)'];\)\)], "Input"] }, Closed]] }, FrontEndVersion->"5.0 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 692}}, AutoGeneratedPackage->None, WindowToolbars->{"RulerBar", "EditBar"}, CellGrouping->Manual, WindowSize->{1016, 644}, WindowMargins->{{0, Automatic}, {Automatic, 0}}, StyleDefinitions -> "DemoText.nb" ] (******************************************************************* 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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