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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[ 28014, 818]*) (*NotebookOutlinePosition[ 29436, 872]*) (* CellTagsIndexPosition[ 29255, 861]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Instructions:", FontWeight->"Bold"], " \n\nIn this lab, you will get some practice with parametric equations, \ and explore polar equations graphically and analytically.\n\nIn Part I, go \ through each section and analyze what each command does.\n\tShow the points \ on the curve where the speed is 0 and where the speed i the maximum.\nIn the \ You Try It for Part I, do the same as above and \n\tindicate why we integrate \ the speed to get the distance traveled around the curve.\nIn Part II, go \ through each section and analyze what each command does.\n\tWhere are the \ places on your petal plot that the speed is zero?\n\tShow the places on the \ curve that we missed when determining where the velocity and acceleration are \ perpendicular.\nIn the You Try It for Part II, replace the terms in red to \ try out a function of your own.\n\tBe careful not to confuse the function in \ parametric form with the function in pure polar form (r as a function of \ \[Theta]).\n\t\n", StyleBox["The heart of the write-up will be a written summary (NOT HAND \ WRITTEN) of your observations, but you may include a few pages of ", FontWeight->"Bold"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" printouts to help clarify your explanations. I don't want just a \ few words written. Reflect on what you have seen in each Part.", FontWeight->"Bold"] }], "Subsubtitle"], Cell[CellGroupData[{ Cell["Parametric and Polar Equations with a Figure Skater", "Title"], Cell[TextData[StyleBox["Chapter 9", FontFamily->"Arial", FontSize->16, FontWeight->"Bold"]], "Text"], Cell[CellGroupData[{ Cell["Part I: Parametric Equations of a Curve in 2-Space", "Section"], Cell[CellGroupData[{ Cell["Defining the Function", "Subsection"], Cell[TextData[{ "First, we 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[{ \(Off[General::spell]\), "\n", \(Off[General::spell1]\), "\n", \(Clear[x, y, t, r]\), "\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. In what direction are you moving \ on the curve as ", StyleBox["t", FontSlant->"Italic"], " increases?" }], "Text"], Cell[BoxData[{ \(\(plotf = ParametricPlot[Evaluate[r[t]], {t, 0, N[2 \[Pi]]}, PlotStyle -> RGBColor[0, 0, 1], AxesLabel -> {"\", "\"}];\)\ \), "\n", \(Print["\", r[t]]\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Taking the Velocity and Acceleration into Account", "Subsection"], 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 do this \ with ", StyleBox["Mathematica", FontSlant->"Italic"], " and plot the velocity in red and the acceleration in green." }], "Text"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h1", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb1"], Cell[BoxData[{ \(\(plotfp = ParametricPlot[Evaluate[\(r'\)[t]], {t, 0, N[2\ \[Pi]]}, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"\", "\"}];\)\), "\n", \(Print["\", \(r'\)[t]]\), "\n", \(\(plotfpp = ParametricPlot[Evaluate[\(\(r'\)'\)[t]], {t, 0, N[2 \[Pi]]}, PlotStyle -> RGBColor[0, 1, 0], AxesLabel -> {"\", "\"}];\)\), "\n", \(Print["\", \(\(r'\)'\)[t] // Simplify]\)}], "Input"], Cell["\<\ Note that the components of the velocity and acceleration functions are more \ complicated than the components of the postion function.\ \>", "Text"], Cell["\<\ Let's look at the speed function and see what it tells us. The following plot \ shows the speed in black, the x-coordinate in orange and the y-coordinate in \ violet. Contrasting that to your parametric plot, identify the places where \ the speed function is 0.\ \>", "Text"], Cell[BoxData[{ \(speed[t_] := \@\(\(r'\)[t] . \(r'\)[t]\)\), "\[IndentingNewLine]", \(Print["\", speed[t]]\), "\n", \(\(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", \(Print["\"]\), \ "\n", \(\(Show[plotf];\)\)}], "Input"], Cell["\<\ Identify the places on your path where the speed is 0 and the speed is a \ maximum.\ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Computing the Distance Traveled on a Curved Path", "Subsection"], Cell["\<\ 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. Why?\ \>", "Text"], Cell[BoxData[{ \(\(Show[plotf];\)\), "\n", \(\(Plot[speed[t], {t, 0, N[2 \[Pi]]}, AxesLabel -> {t, speed}];\)\), "\n", \(Print["\", distance = NIntegrate[speed[t], {t, 0, N[2 \[Pi]]}], "\< units.\>"]\)}], "Input"], Cell[TextData[ "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]] }, Closed]], Cell[CellGroupData[{ Cell["You Try It: Part I", "Section"], Cell[CellGroupData[{ Cell["Defining and Plotting a Function", "Subsection"], Cell[TextData[{ "Select your own functions for ", StyleBox["x[t_]", FontWeight->"Bold"], " and for ", StyleBox["y[t_]", FontWeight->"Bold"], " in the cell below and then execute the command below. You need not select \ a closed path and you may wish to change the bounds for the parameter to \ something other than ", Cell[BoxData[ \(TraditionalForm\`{t, \ 0\ , 10}\)]], ". Change the terms in red." }], "Text"], Cell[BoxData[{\(Clear[x, y, t, r, speed]\), "\n", RowBox[{\(x[t_]\), ":=", StyleBox[\(Sin[t/3.2]\), FontColor->RGBColor[1, 0, 0]]}], "\n", RowBox[{\(y[t_]\), ":=", StyleBox[\(Exp[\(-t\)] t\^2\), FontColor->RGBColor[1, 0, 0]]}], "\n", \(r[t_] = {x[t], y[t]}\), "\n", RowBox[{ RowBox[{"plotf", "=", RowBox[{"ParametricPlot", "[", RowBox[{\(Evaluate[r[t]]\), ",", RowBox[{"{", RowBox[{"t", ",", StyleBox["0", FontColor->RGBColor[1, 0, 0]], StyleBox[",", FontColor->RGBColor[1, 0, 0]], StyleBox["10", FontColor->RGBColor[1, 0, 0]]}], "}"}], ",", \(PlotStyle -> RGBColor[0, 0, 1]\), ",", \(AxesLabel -> {"\", "\"}\)}], "]"}]}], ";"}]}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Computing the Velocity and Acceleration Vectors and Analyzing the Speed\ \>", "Subsection"], Cell[BoxData[{\(Print["\", r[t]]\), "\n", \(Print["\", \(r'\)[ t]]\), "\n", \(Print["\", \ \(\(r'\)'\)[t] // Simplify]\), "\n", \(Print["\", speed[t_] = \@\(\(r'\)[t] . \(r'\)[t]\)]\), "\n", RowBox[{ RowBox[{"Plot", "[", RowBox[{\({speed[t], x[t], y[t]}\), ",", RowBox[{"{", RowBox[{"t", ",", StyleBox["0", FontColor->RGBColor[1, 0, 0]], ",", StyleBox["10", FontColor->RGBColor[1, 0, 0]]}], "}"}], ",", \(PlotStyle -> {RGBColor[0, 0, 0], RGBColor[1, .5, 0], RGBColor[1, 0, 1]}\), ",", \(AxesLabel -> {t, function}\)}], "]"}], ";", "\n", \(Print["\"]\)}], "\n", \(Show[plotf];\)}], "Input"], Cell["\<\ Identify the places on your path where the speed is 0 and the speed is a \ maximum.\ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Finding the Distance Along the Curved Path.", "Subsection"], Cell["\<\ Adjust the values of the parameter in the integration if you did so \ earlier.\ \>", "Text"], Cell[BoxData[ RowBox[{\(Show[plotf]\), ";", "\n", RowBox[{"Plot", "[", RowBox[{\(speed[t]\), ",", RowBox[{"{", RowBox[{"t", ",", StyleBox["0", FontColor->RGBColor[1, 0, 0]], ",", StyleBox["10", FontColor->RGBColor[1, 0, 0]]}], "}"}], ",", \(AxesLabel -> {t, speed}\)}], "]"}], ";", "\n", RowBox[{"Print", "[", RowBox[{"\"\\"", ",", RowBox[{"distance", "=", RowBox[{"NIntegrate", "[", RowBox[{\(speed[t]\), ",", RowBox[{"{", RowBox[{"t", ",", StyleBox["0", FontColor->RGBColor[1, 0, 0]], ",", StyleBox["10", FontColor->RGBColor[1, 0, 0]]}], "}"}]}], "]"}]}], ",", "\"\< units.\>\""}], "]"}]}]], "Input"], Cell["\<\ Think of this answer as either the distance around the curve or as the area \ under the speed function.\ \>", "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Part II: A Figure Skater Tracing a Polar Plot", "Section"], Cell[CellGroupData[{ Cell["Four-Petal Pattern", "Subsection"], Cell[TextData[{ "Think of a figure skater tracing out a four-petal flower on the ice. The \ first set of commands gives the parametric equations of the figure skater in \ terms of a path that would be traced in the ", Cell[BoxData[ \(TraditionalForm\`x - y\)]], " plane. It is easiest to start with the equations in polar form." }], "Text"], Cell[BoxData[{ \(Off[General::spell]\ \), "\n", \(Off[General::spell1]\), "\n", \(Clear[r, \[Theta], t, x, y]\ \), "\n", \(r[t_] := 16 Sin[t]\^2\), "\n", \(\[Theta][t_] := t/2\)}], "Input"], Cell[TextData[{ "You need to load a special graphics package to do polar plots. Execute \ this cell only once BEFORE you try to do a polar plot. \nTo write ", StyleBox["r", FontSlant->"Italic"], " as a function of \[Theta], we had to replace the ", StyleBox["t", FontSlant->"Italic"], " in the ", Cell[BoxData[ \(TraditionalForm\`r[t]\)]], " with ", StyleBox["t", FontSlant->"Italic"], " as a function of \[Theta]. Since", Cell[BoxData[ \(TraditionalForm\`\(\ \[Theta]\ = \ t/2, \ t\ = \ 2 \( \[Theta] . \)\ \)\)]] }], "Text"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h2", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb2"], Cell[BoxData[ \(<< Graphics`Graphics`\)], "Input"], Cell[BoxData[ \(\(pp1 = PolarPlot[r[2 \[Theta]], {\[Theta], 0, N[2 \[Pi]]}, AspectRatio \[Rule] 1, AxesLabel \[Rule] {"\", "\"}, PlotStyle -> RGBColor[0, 0, 1]];\)\)], "Input"], Cell[TextData[{ "For ease in computing equations of motion, we will define our position and \ velocity vectors parametrically in ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], ", using the above parameterizations for ", StyleBox["r ", FontSlant->"Italic"], "and \[Theta]. To show the direction of movement along the flower petals, \ we will place a few arrows on the graph (using the Epilog option). That \ requires a graphics arrow package. You must load the package before you \ execute a command within the package." }], "Text"], Cell[BoxData[ \(<< Graphics`Arrow`\)], "Input"], Cell[BoxData[{ \(Clear[velocity, speed]\), "\n", \(\(parx[t_] = r[t] Cos[\[Theta][t]];\)\), "\n", \(\(pary[t_] = r[t] Sin[\[Theta][t]];\)\), "\n", \(\(position[t_] = {parx[t], pary[t]} // Simplify;\)\), "\n", \(\(velocity[t_] = \(position'\)[t] // Simplify;\)\), "\n", \(\(speed[t_] = \@\(velocity[t] . velocity[t]\) // Simplify;\)\), "\n", \(Print["\", position[t]]\), "\n", \(Print["\", speed[t]]\), "\n", \(\(pp4 = ParametricPlot[{parx[t], pary[t]}, {t, 0, N[4 \[Pi]]}, AspectRatio -> Automatic, PlotStyle -> RGBColor[0, 0, 1], AxesLabel -> {"\", "\"}, Epilog \[Rule] {Arrow[{parx[1], pary[1]}, {parx[1.1], pary[1.1]}, HeadScaling \[Rule] Absolute], Arrow[{parx[2.5], pary[2.5]}, {parx[2.6], pary[2.6]}, HeadScaling \[Rule] Absolute], Arrow[{parx[4], pary[4]}, {parx[4.1], pary[4.1]}, HeadScaling \[Rule] Absolute], Arrow[{parx[5.5], pary[5.5]}, {parx[5.6], pary[5.6]}, HeadScaling \[Rule] Absolute], Arrow[{parx[7], pary[7]}, {parx[7.1], pary[7.1]}, HeadScaling \[Rule] Absolute], Arrow[{parx[8.5], pary[8.5]}, {parx[8.6], pary[8.6]}, HeadScaling \[Rule] Absolute], Arrow[{parx[10], pary[10]}, {parx[10.1], pary[10.1]}, HeadScaling \[Rule] Absolute], Arrow[{parx[11.5], pary[11.5]}, {parx[11.6], pary[11.6]}, HeadScaling \[Rule] Absolute]}];\)\)}], "Input"], Cell["\<\ Can you tell in what direction the skater is moving just before and just \ after passing through at the origin?\ \>", "Text"], Cell["\<\ Let's look at the speed function and see what it tells us. The following plot \ shows the speed in black, the x-coordinate in orange and the y-coordinate in \ violet. Contrasting that to your parametric plot, identify the places where \ the speed function is 0.\ \>", "Text"], Cell[BoxData[{ \(\(tickst = Table[n\ \[Pi]/2, {n, 0, 8}];\)\), "\n", \(\(ticksf = Table[n, {n, \(-10\), 15, 5}];\)\), "\n", \(\(Plot[{speed[t], parx[t], pary[t]}, {t, 0, N[4 \[Pi]]}, PlotStyle -> {RGBColor[0, 0, 0], RGBColor[1, .5, 0], RGBColor[1, 0, 1]}, AxesLabel -> {t, function}, Ticks -> {tickst, ticksf}];\)\), "\n", \(Print["\"]\), \ "\n", \(\(Show[pp4];\)\)}], "Input"], Cell["\<\ Where are the places on your petal plot that the speed is zero?\ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Velocity and Acceleration - When Are They Perpendicular? ", "Subsection"], Cell[TextData[{ "Suppose we wish to determine for which values of ", StyleBox["t", FontSlant->"Italic"], " certain vectors describing the equations of motion are orthogonal. To do \ this, we will use the dot product, 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." }], "Text"], Cell[BoxData[{ \(Clear[acceleration, vdota]\), "\n", \(Print["\", velocity[t]]\), "\n", \(Print["\", acceleration[t_] = \(velocity'\)[t] // Simplify]\), "\n", \(Print["\", vdota = velocity[t] . acceleration[t] // Simplify]\), "\n", \(\(Plot[Evaluate[vdota], {t, 0, N[4 \[Pi]]}, AxesLabel \[Rule] \n\t\t{t, "\"}, Ticks -> {tickst, Automatic}];\)\)}], "Input"], Cell["\<\ As you can see from the graph, there are many times when the velocity and \ acceleration vectors are perpendicular to each other.\ \>", "Text"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h3", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb3"], Cell[TextData[{ "The following commandsl find some of the places where the velocity and \ acceleration vectors are perpendicular. Note that we use seed values in ", StyleBox["FindRoot", FontWeight->"Bold"], " that seem close to some of the places where our function crosses the \ horizontal axis." }], "Text"], Cell[BoxData[{ \(Clear[vperpa]\), "\n", \(vperpa[0] = FindRoot[vdota == 0, {t, 0}]\), "\n", \(vperpa[1] = FindRoot[vdota == 0, {t, 1}]\), "\n", \(vperpa[2] = FindRoot[vdota == 0, {t, 2}]\), "\n", \(vperpa[3] = FindRoot[vdota == 0, {t, 3}]\), "\n", \(vperpa[4] = FindRoot[vdota == 0, {t, 4}]\), "\n", \(vperpa[5] = FindRoot[vdota == 0, {t, 5}]\), "\n", \(vperpa[6] = FindRoot[vdota == 0, {t, 6}]\), "\n", \(vperpa[7] = FindRoot[vdota == 0, {t, 7}]\), "\n", \(vperpa[8] = FindRoot[vdota == 0, {t, 8}]\), "\n", \(vperpa[9] = FindRoot[vdota == 0, {t, 9}]\), "\n", \(vperpa[10] = FindRoot[vdota == 0, {t, 10}]\)}], "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." }], "Text"], Cell[BoxData[{ \(\(special = Table[{parx[t], pary[t]} /. vperpa[i], {i, 0, 10}];\)\), "\n", \(TableForm[special, TableHeadings -> {None, {"\", "\"}}]\)}], "Input"], Cell["\<\ We can now see where those points are relative to our petals.\ \>", "Text"], Cell[BoxData[{ \(\(psp = ListPlot[special, PlotStyle -> PointSize[ .04], DisplayFunction -> Identity];\)\), "\n", \(\(Show[pp1, psp, DisplayFunction -> $DisplayFunction];\)\)}], "Input"], Cell["\<\ Can you see some points that we missed? The dots show where the velocity and acceleration vectors are perpendicular \ to each other. Can you describe what is happening to the figure skater at \ those points?\ \>", "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["You Try It: Part II", "Section"], Cell[TextData[{ "Try your own functions for ", StyleBox["r[t]", FontSlant->"Italic"], " and ", Cell[BoxData[ \(TraditionalForm\`\[Theta][t]\)]], ". Remember to solve for ", StyleBox["t", FontSlant->"Italic"], " as a function of \[Theta] before you attempt to do a polar plot in the \ form of", StyleBox[" r", FontSlant->"Italic"], " as a function of \[Theta]. Replace the terms in red." }], "Text"], Cell[BoxData[{ RowBox[{\(Clear[r, \[Theta], t, x, y]\), " "}], "\n", RowBox[{\(r[t_]\), ":=", StyleBox[\(Cos[t]\^3\), FontColor->RGBColor[1, 0, 0]]}], "\n", RowBox[{\(\[Theta][t_]\), ":=", StyleBox[\(t\^2/8\), FontColor->RGBColor[1, 0, 0]]}], "\n", RowBox[{ RowBox[{"pp1", "=", RowBox[{"PolarPlot", "[", " ", RowBox[{ StyleBox[\(Cos[\@\(8 \[Theta]\)]\^3\), FontColor->RGBColor[1, 0, 0]], ",", \({\[Theta], 0, 3}\), ",", \(AspectRatio \[Rule] 1\), ",", \(AxesLabel \[Rule] {"\", "\"}\), ",", \(PlotStyle -> RGBColor[0, 0, 1]\)}], "]"}]}], ";"}]}], "Input"], Cell[BoxData[{ \(Clear[velocity, speed]\), "\n", \(\(parx[t_] = r[t] Cos[\[Theta][t]];\)\), "\n", \(\(pary[t_] = r[t] Sin[\[Theta][t]];\)\), "\n", \(\(position[t_] = {parx[t], pary[t]} // Simplify;\)\), "\n", \(\(velocity[t_] = \(position'\)[t] // Simplify;\)\), "\n", \(\(speed[t_] = \@\(velocity[t] . velocity[t]\) // Simplify;\)\), "\n", \(Print["\", position[t]]\), "\n", \(Print["\", speed[t]]\), "\n", \(\(Plot[{speed[t], parx[t], pary[t]}, {t, 0, 5}, PlotStyle -> {RGBColor[0, 0, 0], RGBColor[1, .5, 0], RGBColor[1, 0, 1]}, PlotRange -> All, AxesLabel -> {t, function}];\)\), "\n", \(Print["\"]\), \ "\n", \(\(Show[pp1];\)\)}], "Input"], Cell[TextData[{ "What is happening to your speed as ", StyleBox["t", FontSlant->"Italic"], " increases?" }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["\[MathematicaIcon]", FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]] }], "Section", CellDingbat->None], Cell[TextData[{ "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.", "\n", ButtonBox["Go back.", ButtonData:>"hb1", ButtonStyle->"Hyperlink"] }], "Text", CellTags->"h1"], Cell[TextData[{ "If you try to execute the command before loading the package, ", StyleBox["Mathematica", FontSlant->"Italic"], " will be unable to execute the command, even if you try to load the \ package after you have already executed the command and then try re-executing \ the command. The best way to undo your mistake is to EXIT ", StyleBox["Mathematica", FontSlant->"Italic"], " completely and then start it again. Simply closing the notebook will not \ work; do save your notebook before you exit ", StyleBox["Mathematica", FontSlant->"Italic"], ".\n", ButtonBox["Go back.", ButtonData:>"hb2", ButtonStyle->"Hyperlink"] }], "Text", CellTags->"h2"], Cell[TextData[{ "The ", StyleBox["Solve[...]", FontWeight->"Bold"], " command works sometimes, but usually only for algebraic (not \ transcendental) functions. The ", StyleBox["FindRoot", FontWeight->"Bold"], " command works most of the time. Using it is somewhat like using Newton's \ method, so you have to give it a starting point. Note that it yields \ approximate solutions and does not always converge to a solution. 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