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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[ 66860, 1741]*) (*NotebookOutlinePosition[ 68067, 1787]*) (* CellTagsIndexPosition[ 67927, 1778]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["\<\ Motion Along a Straight Line, Part I: Position \[Rule] Velocity \[Rule] \ Acceleration\ \>", "Title", PageWidth->PaperWidth], Cell[TextData[StyleBox["Chapter 2, Section 2", FontFamily->"Arial", FontSize->16, FontWeight->"Bold"]], "Text"], Cell[BoxData[{ \(\(Off[DiracDelta::"\"];\)\), "\[IndentingNewLine]", \(\(<< Calculus`DiracDelta`;\)\), "\[IndentingNewLine]", \(\(<< Graphics`Arrow`;\)\), "\n", \(\(Off[General::"\"];\)\), "\n", \(\(Off[FindRoot::"\"];\)\[IndentingNewLine]\), "\n", \(\(velocity[fnc_, interval_List, periodic_] := Block[{f, dfdt, d2fdt2, fvalues, dfdtvalues, d2fdt2values, start, dfdtmaxt22, dfdtmin333t22, fmaxt2222, fmint22, fmin, fmax, roots22, maxes, mines, dfdtmax, dfdtmin333, size, p1, p2, p3, ta, to, tf}, \[IndentingNewLine]\[IndentingNewLine]tind = interval[\([1]\)]; \[IndentingNewLine]to = interval[\([2]\)]; \[IndentingNewLine]tf = interval[\([3]\)]; \[IndentingNewLine]\n f[t_] = fnc /. tind -> t; \ndfdt[t_] = D[f[t], t]; \n d2fdt2[t_] = D[dfdt[t], t]; \n\t\t\n fvalues = Table[f[t], {t, to, tf, \((tf - to)\)/500}]; \n dfdtvalues = Table[dfdt[t], {t, to, tf, \((tf - to)\)/500}]; \n d2fdt2values = Table[d2fdt2[t], {t, to, tf, \((tf - to)\)/ 500}]; \n\t\tstart = {}; \[IndentingNewLine]\t\n dfdtmaxt22 = Max[dfdtvalues]; \ndfdtmin333t22 = Min[dfdtvalues]; \n dfdt2maxt = Max[dfdt2values22]; \[IndentingNewLine]dfdt2mint33 = Min[dfdt2values22]; \[IndentingNewLine]\n fmaxt2222 = Max[fvalues]; \nfmint2222 = Min[fvalues]; \n fmin = fmint2222 - Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \n fmax = fmaxt2222 + Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \n\n Do[If[Sign[dfdtvalues[\([k]\)]] != Sign[dfdtvalues[\([k + 1]\)]], start = Append[start, to + \((tf - to)\)/500. *\((k - 1)\)]], {k, 1, 500}]; \n\t roots22 = {}; \n\t Do[roots22 = Append[roots22, \(FindRoot[ dfdt[t] == 0, {t, start[\([kk]\)]}]\)[\([1, 2]\)]], {kk, 1, Length[start]}]; \n\tmaxes = {}; mines = {}; \n\t If[f[tf] > f[tf - 0.00001*\((tf - to)\)], maxes = Append[maxes, {tf, f[tf] + 0.02*\((fmax - fmin)\)}], If[f[tf] < f[tf - 0.00001*\((tf - to)\)], mines = Append[ mines, {tf, f[tf] - 0.03*\((fmax - fmin)\)}]]]; \n\t If[f[to] > f[to + 0.00001*\((tf - to)\)], maxes = Append[maxes, {to, f[to] + 0.02*\((fmax - fmin)\)}], If[f[to] < f[to + 0.00001*\((tf - to)\)], mines = Append[ mines, {to, f[to] - 0.03*\((fmax - fmin)\)}]]]; \n\t If[Length[roots22] != 0, Do[If[dfdt[roots22[\([k]\)] - .0001*\((tf - to)\)] > 0\ && dfdt[roots22[\([k]\)] + .0001*\((tf - to)\)] < 0, maxes = Append[maxes, {roots22[\([k]\)], f[roots22[\([k]\)]] + 0.02*\((fmax - fmin)\)}], If[dfdt[roots22[\([k]\)] - .0001*\((tf - to)\)] < 0\ && dfdt[roots22[\([k]\)] + .0001*\((tf - to)\)] > 0, mines = Append[mines, {roots22[\([k]\)], f[roots22[\([k]\)]] - 0.03*\((fmax - fmin)\)}]]], {k, 1, Length[roots22]}]]; \n\t\t\n dfdtmin333 = fmin - \((\((fmin + fmax)\)/ 2. - \((dfdtmin333t22 + dfdtmaxt22)\)/2. )\); \n dfdtmax = fmax - \((\((fmin + fmax)\)/ 2. - \((dfdtmin333t22 + dfdtmaxt22)\)/2. )\); \n\n Do[Clear[p1, p2]; size = 9; \n\t p1 = Plot[dfdt[t], {t, to - 0.0001*\((tf - to)\), ta}, PlotRange -> {{to, tf}, {dfdtmin333, dfdtmax}}, PlotStyle -> {RGBColor[1, 0, 0]}, AspectRatio -> 1, AxesLabel \[Rule] "\", Epilog -> {RGBColor[1, 0, 0], Text["\", {ta, 0.5*dfdt[ta]}, {\(-1\), 0}], Thickness[0.0125], Arrow[{ta, 0}, {ta, dfdt[ta]}, HeadScaling -> Relative]}, DisplayFunction -> Identity]; \n\t p2 = Plot[{f[t], f[ta] + dfdt[ta]*\((t - ta)\)}, {t, to, tf}, PlotStyle -> {RGBColor[0, 0, 0], {RGBColor[0.316411, \ 0.738293, \ 0.32813], Dashing[{0.02}]}}, PlotRange -> {{to, tf + 1}, {fmin, fmax}}, AspectRatio -> 1, AxesLabel \[Rule] "\", Epilog -> {{RGBColor[0, 0, 1], Text["\<1\>", {ta + .5, f[ta]}, If[dfdt[ta] >= 0, {0, 1}, {0, \(-1\)}]], Line[{{ta + 1, f[ta] + dfdt[ta]}, {ta, f[ta] + dfdt[ta]}}], Line[{{ta, f[ta]}, {ta + 1, f[ta]}, {ta + 1, f[ta] + dfdt[ta]}, {ta, f[ta]}}], RGBColor[1, 0, 0], Text["\", {ta, f[ta] + 0.5*dfdt[ta]}, {1, 0}], Thickness[0.0125], Arrow[{ta, f[ta]}, {ta, f[ta] + dfdt[ta]}, HeadScaling -> Relative]}}, DisplayFunction -> Identity]; \n\t p3 = Plot[0, {t, fmin, fmax}, PlotRange -> {{fmin, fmax}, {fmin, fmax}}, DisplayFunction -> Identity, Axes -> False, Epilog -> {Disk[{\((fmin + fmax)\)/2. , f[ta]}, 0.015*\((fmax - fmin)\)], RGBColor[1, 0, 0], Text["\", {\((fmin + fmax)\)/2. , f[ta] + 0.5*dfdt[ta]}, {1, 0}], Thickness[0.0075], Arrow[{\((fmin + fmax)\)/2. , f[ta]}, {\((fmin + fmax)\)/2. , f[ta] + dfdt[ta]}, HeadScaling -> Relative]}]; \n Show[GraphicsArray[{p3, p2, p1}], ImageSize -> {72*size, 72*size/3}, PlotRegion -> {{0, 1}, {0, 1}}, \ DisplayFunction -> $DisplayFunction], {ta, to, If[periodic == 1, tf - 0.04*\((tf - to)\), tf], 0.04*\((tf - to)\)}];];\)\n\), "\[IndentingNewLine]", \(\(acceleration[fnc_, interval_List, periodic_] := Block[{f, dfdt, d2fdt2, fvalues, dfdtvalues, d2fdt2values, start, dfdtmaxt22, dfdtmin333t22, fmaxt22, fmint22, fmin, fmax, roots22, maxes, mines, dfdtmax, dfdtmin333, size, p1, p2, p3, ta, to, tf}, \[IndentingNewLine]\[IndentingNewLine]tind = interval[\([1]\)]; \[IndentingNewLine]to = interval[\([2]\)]; \[IndentingNewLine]tf = interval[\([3]\)]; \[IndentingNewLine]\n f[t_] = D[fnc /. tind -> t, t]; \ndfdt[t_] = D[f[t], t]; \n ff[t_] = fnc /. tind -> t; \n\t\t\n fvalues = Table[f[t], {t, to, tf, \((tf - to)\)/500}]; \n dfdtvalues = Table[dfdt[t], {t, to, tf, \((tf - to)\)/500}]; \n ffvalues22 = Table[ff[t], {t, to, tf, \((tf - to)\)/500}]; \[IndentingNewLine]\n start = {}; \[IndentingNewLine]\t\ndfdtmaxt22 = Max[dfdtvalues]; \n dfdtmin333t22 = Min[dfdtvalues]; \n dfdt2maxt = Max[dfdt2values22]; \[IndentingNewLine]\[IndentingNewLine]\n fmaxt22 = Max[fvalues]; \nfmint22 = Min[fvalues]; \n fmin = fmint22 - Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \n fmax = fmaxt22 + Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \[IndentingNewLine]ffmax22 = Max[ffvalues22] + Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \[IndentingNewLine]ffmin22 = Min[ffvalues22] - Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \[IndentingNewLine]\n\n Do[If[Sign[dfdtvalues[\([k]\)]] != Sign[dfdtvalues[\([k + 1]\)]], start = Append[start, to + \((tf - to)\)/500. *\((k - 1)\)]], {k, 1, 500}]; \n\t roots22 = {}; \n\t Do[roots22 = Append[roots22, \(FindRoot[ dfdt[t] == 0, {t, start[\([kk]\)]}]\)[\([1, 2]\)]], {kk, 1, Length[start]}]; \n\tmaxes = {}; mines = {}; \n\t If[f[tf] > f[tf - 0.00001*\((tf - to)\)], maxes = Append[maxes, {tf, f[tf] + 0.02*\((fmax - fmin)\)}], If[f[tf] < f[tf - 0.00001*\((tf - to)\)], mines = Append[ mines, {tf, f[tf] - 0.03*\((fmax - fmin)\)}]]]; \n\t If[f[to] > f[to + 0.00001*\((tf - to)\)], maxes = Append[maxes, {to, f[to] + 0.02*\((fmax - fmin)\)}], If[f[to] < f[to + 0.00001*\((tf - to)\)], mines = Append[ mines, {to, f[to] - 0.03*\((fmax - fmin)\)}]]]; \n\t If[Length[roots22] != 0, Do[If[dfdt[roots22[\([k]\)] - .0001*\((tf - to)\)] > 0\ && dfdt[roots22[\([k]\)] + .0001*\((tf - to)\)] < 0, maxes = Append[maxes, {roots22[\([k]\)], f[roots22[\([k]\)]] + 0.02*\((fmax - fmin)\)}], If[dfdt[roots22[\([k]\)] - .0001*\((tf - to)\)] < 0\ && dfdt[roots22[\([k]\)] + .0001*\((tf - to)\)] > 0, mines = Append[mines, {roots22[\([k]\)], f[roots22[\([k]\)]] - 0.03*\((fmax - fmin)\)}]]], {k, 1, Length[roots22]}]]; \n\t\t\n dfdtmin333 = fmin - \((\((fmin + fmax)\)/ 2. - \((dfdtmin333t22 + dfdtmaxt22)\)/2. )\); \n dfdtmax = fmax - \((\((fmin + fmax)\)/ 2. - \((dfdtmin333t22 + dfdtmaxt22)\)/2. )\); \n\n Do[Clear[p1, p2]; size = 9; \n\t p1 = Plot[dfdt[t], {t, to - 0.0001*\((tf - to)\), ta}, PlotRange -> {{to, tf}, {dfdtmin333, dfdtmax}}, PlotStyle -> {RGBColor[1, 0, 0]}, AspectRatio -> 1, AxesLabel \[Rule] "\", Epilog -> {RGBColor[1, 0, 0], Text["\", {ta, 0.5*dfdt[ta]}, {\(-1\), 0}], Thickness[0.0125], Arrow[{ta, 0}, {ta, dfdt[ta]}, HeadScaling -> Relative]}, DisplayFunction -> Identity]; \n\t p2 = Plot[{f[t], f[ta] + dfdt[ta]*\((t - ta)\)}, {t, to, tf}, PlotStyle -> {RGBColor[0, 0, 0], {RGBColor[0.316411, \ 0.738293, \ 0.32813], Dashing[{0.02}]}}, PlotRange -> {{to, tf + 1}, {fmin, fmax}}, AspectRatio -> 1, AxesLabel \[Rule] "\", Epilog -> {{RGBColor[0, 0, 1], Text["\<1\>", {ta + .5, f[ta]}, If[dfdt[ta] >= 0, {0, 1}, {0, \(-1\)}]], Line[{{ta + 1, f[ta] + dfdt[ta]}, {ta, f[ta] + dfdt[ta]}}], Line[{{ta, f[ta]}, {ta + 1, f[ta]}, {ta + 1, f[ta] + dfdt[ta]}, {ta, f[ta]}}], RGBColor[1, 0, 0], Text["\", {ta, f[ta] + 0.5*dfdt[ta]}, {1, 0}], Thickness[0.0125], Arrow[{ta, f[ta]}, {ta, f[ta] + dfdt[ta]}, HeadScaling -> Relative]}}, DisplayFunction -> Identity]; \n\t p3 = Plot[0, {t, ffmin22, ffmax22}, PlotRange -> {{ffmin22, ffmax22}, {ffmin22, ffmax22}}, DisplayFunction -> Identity, Axes -> False, Epilog -> {Disk[{\((ffmin22 + ffmax22)\)/2. , ff[ta]}, 0.015*\((ffmax22 - ffmin22)\)], RGBColor[1, 0, 0], Text["\", {\((ffmin22 + ffmax22)\)/2. , ff[ta] + 0.5*dfdt[ta]}, {1, 0}], Thickness[0.0075], Arrow[{\((ffmin22 + ffmax22)\)/2. , ff[ta]}, {\((ffmin22 + ffmax22)\)/2. , ff[ta] + dfdt[ta]}, HeadScaling -> Relative]}]; \n Show[GraphicsArray[{p3, p2, p1}], ImageSize -> {72*size, 72*size/3}, PlotRegion -> {{0, 1}, {0, 1}}, \ DisplayFunction -> $DisplayFunction], {ta, to, If[periodic == 1, tf - 0.04*\((tf - to)\), tf], 0.04*\((tf - to)\)}];];\)\[IndentingNewLine]\), "\[IndentingNewLine]", \ \(\(posvelacc[fnc_, interval_List, periodic_] := Block[{f, dfdt, d2fdt2, fvalues, dfdtvalues, d2fdt2values, start, dfdtmaxt22, dfdtmin333t22, fmaxt22, fmint22, fmin, fmax, roots22, maxes, mines, dfdtmax, dfdtmin333, size, p1, p2, p3, ta, to, tf}, \[IndentingNewLine]\[IndentingNewLine]tind = interval[\([1]\)]; \[IndentingNewLine]to = interval[\([2]\)]; \[IndentingNewLine]tf = interval[\([3]\)]; \[IndentingNewLine]\n f[t_] = D[fnc /. tind -> t, t]; \ndfdt[t_] = D[f[t], t]; \n ff[t_] = fnc /. tind -> t; \n\t\t\n fvalues = Table[f[t], {t, to, tf, \((tf - to)\)/500}]; \n dfdtvalues = Table[dfdt[t], {t, to, tf, \((tf - to)\)/500}]; \n ffvalues22 = Table[ff[t], {t, to, tf, \((tf - to)\)/500}]; \[IndentingNewLine]\n start = {}; \[IndentingNewLine]\t\ndfdtmaxt22 = Max[dfdtvalues]; \n dfdtmin333t22 = Min[dfdtvalues]; \n dfdt2maxt = Max[dfdt2values22]; \[IndentingNewLine]\[IndentingNewLine]\n fmaxt22 = Max[fvalues]; \nfmint22 = Min[fvalues]; \n fmin = fmint22 - Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \n fmax = fmaxt22 + Max[Abs[{dfdtmin333t22, dfdtmaxt22}]]; \[IndentingNewLine]ffmax22 = Max[ffvalues22] + Max[Abs[{dfdtmin333t22, dfdtmaxt22, fmint22, fmaxt22}]]; \[IndentingNewLine]ffmin22 = Min[ffvalues22] - Max[Abs[{dfdtmin333t22, dfdtmaxt22, fmint22, fmaxt22}]]; \[IndentingNewLine]\n\n Do[If[Sign[dfdtvalues[\([k]\)]] != Sign[dfdtvalues[\([k + 1]\)]], start = Append[start, to + \((tf - to)\)/500. *\((k - 1)\)]], {k, 1, 500}]; \n\t roots22 = {}; \n\t Do[roots22 = Append[roots22, \(FindRoot[ dfdt[t] == 0, {t, start[\([kk]\)]}]\)[\([1, 2]\)]], {kk, 1, Length[start]}]; \n\tmaxes = {}; mines = {}; \n\t If[f[tf] > f[tf - 0.00001*\((tf - to)\)], maxes = Append[maxes, {tf, f[tf] + 0.02*\((fmax - fmin)\)}], If[f[tf] < f[tf - 0.00001*\((tf - to)\)], mines = Append[ mines, {tf, f[tf] - 0.03*\((fmax - fmin)\)}]]]; \n\t If[f[to] > f[to + 0.00001*\((tf - to)\)], maxes = Append[maxes, {to, f[to] + 0.02*\((fmax - fmin)\)}], If[f[to] < f[to + 0.00001*\((tf - to)\)], mines = Append[ mines, {to, f[to] - 0.03*\((fmax - fmin)\)}]]]; \n\t If[Length[roots22] != 0, Do[If[dfdt[roots22[\([k]\)] - .0001*\((tf - to)\)] > 0\ && dfdt[roots22[\([k]\)] + .0001*\((tf - to)\)] < 0, maxes = Append[maxes, {roots22[\([k]\)], f[roots22[\([k]\)]] + 0.02*\((fmax - fmin)\)}], If[dfdt[roots22[\([k]\)] - .0001*\((tf - to)\)] < 0\ && dfdt[roots22[\([k]\)] + .0001*\((tf - to)\)] > 0, mines = Append[mines, {roots22[\([k]\)], f[roots22[\([k]\)]] - 0.03*\((fmax - fmin)\)}]]], {k, 1, Length[roots22]}]]; \n\t\t\n dfdtmin333 = fmin - \((\((fmin + fmax)\)/ 2. - \((dfdtmin333t22 + dfdtmaxt22)\)/2. )\); \n dfdtmax = fmax - \((\((fmin + fmax)\)/ 2. - \((dfdtmin333t22 + dfdtmaxt22)\)/2. )\); \n\n Do[Clear[p1, p2]; size = 4; \n\t p1 = Plot[dfdt[t], {t, to - 0.0001*\((tf - to)\), ta}, PlotRange -> {{to, tf}, {dfdtmin333, dfdtmax}}, PlotStyle -> {RGBColor[1, 0, 0]}, AspectRatio -> 1, AxesLabel \[Rule] "\", Epilog -> {RGBColor[1, 0, 0], Text["\", {ta, 0.5*dfdt[ta]}, {\(-1\), 0}], Thickness[0.0125], Arrow[{ta, 0}, {ta, dfdt[ta]}, HeadScaling -> Relative]}, DisplayFunction -> Identity]; \n\t p2 = Plot[{f[t], f[ta] + dfdt[ta]*\((t - ta)\)}, {t, to, tf}, PlotStyle -> {RGBColor[0, 0, 0], {RGBColor[0.316411, \ 0.738293, \ 0.32813], Dashing[{0.02}]}}, PlotRange -> {{to, tf + 1}, {fmin, fmax}}, AspectRatio -> 1, AxesLabel \[Rule] "\", Epilog -> {{RGBColor[0, 0, 1], Text["\<1\>", {ta + .5, f[ta]}, If[dfdt[ta] >= 0, {0, 1}, {0, \(-1\)}]], Line[{{ta + 1, f[ta] + dfdt[ta]}, {ta, f[ta] + dfdt[ta]}}], Line[{{ta, f[ta]}, {ta + 1, f[ta]}, {ta + 1, f[ta] + dfdt[ta]}, {ta, f[ta]}}], RGBColor[1, 0, 0], Text["\", {ta, f[ta] + 0.5*dfdt[ta]}, {1, 0}], Thickness[0.0125], Arrow[{ta, f[ta]}, {ta, f[ta] + dfdt[ta]}, HeadScaling -> Relative]}}, DisplayFunction -> Identity]; \n\t p3 = Plot[ 0, {t, \(-\((ffmax22 - ffmin22)\)\)/ 2, \((ffmax22 - ffmin22)\)/2}, PlotRange -> {{\(-\((ffmax22 - ffmin22)\)\)/ 2, \((ffmax22 - ffmin22)\)/2}, {ffmin22, ffmax22}}, DisplayFunction -> Identity, Axes \[Rule] {False, True}, AxesOrigin \[Rule] {0, 0}, AxesLabel \[Rule] "\", Epilog -> {Disk[{0, ff[ta]}, 0.01*\((ffmax22 - ffmin22)\)], RGBColor[1, 0, 0], Thickness[0.005], Text["\", {\(- .02\)*\((ffmax22 - ffmin22)\), ff[ta] + 0.5*dfdt[ta]}, {1, 0}], Arrow[{\(- .02\)*\((ffmax22 - ffmin22)\), ff[ta]}, {\(- .02\)*\((ffmax22 - ffmin22)\), ff[ta] + dfdt[ta]}, HeadScaling -> Relative], RGBColor[0, 0, 1], Text["\", {0, ff[ta]*0.5}, {\(-1\), 0}], Arrow[{0, 0}, {0, ff[ta]}, HeadScaling \[Rule] Relative], RGBColor[0, \ 0.679698, \ 0.0156252], Text["\", { .02*\((ffmax22 - ffmin22)\), ff[ta] + 0.5*f[ta]}, {\(-1\), 0}], Arrow[{ .02*\((ffmax22 - ffmin22)\), ff[ta]}, { .02*\((ffmax22 - ffmin22)\), ff[ta] + f[ta]}, HeadScaling \[Rule] Relative]}]; \n Show[p3, \ ImageSize \[Rule] {72*5, 71*5}, DisplayFunction -> $DisplayFunction], {ta, to, If[periodic == 1, tf - 0.04*\((tf - to)\), tf], 0.04*\((tf - to)\)}];];\)\)}], "Input", Editable->False, PageWidth->PaperWidth, CellOpen->False, InitializationCell->True], Cell[CellGroupData[{ Cell["Introduction", "Section", PageWidth->PaperWidth], Cell[TextData[{ "OBJECTIVE: To apply special ", StyleBox["Mathematica", FontSlant->"Italic"], " functions to analyze position, velocity, and acceleration \ simultaneously." }], "Text", PageWidth->PaperWidth], Cell[TextData[StyleBox["Note: There are no executable commands in this \ Introduction.", FontWeight->"Bold"]], "Text", PageWidth->PaperWidth], Cell[TextData[{ StyleBox["Structural engineers must design high-rise buildings to withstand \ earthquakes. As a high-rise building moves back and forth during an \ earthquake, so any single floor also moves back and forth along a straight, \ horizontal line. For earthquake design, the engineer must be able to describe \ completely the position, velocity, and acceleration of each of the floors as \ functions of time, and understand the relationships among these three \ functions. These descriptions of the motion provide the engineer with some of \ the information and understanding that is needed for the seismic design of \ the structure in a high-rise building. (", PageWidth->PaperWidth], "In physics and engineering, the study of motion is called kinematics and \ falls in the realm of mechanics.) \n\n", StyleBox["This module includes three specially designed ", PageWidth->PaperWidth], StyleBox["Mathematica", PageWidth->PaperWidth, FontSlant->"Italic"], StyleBox[" commands that generate animations to help you visualize the \ derivative relations among the position, velocity, and acceleration \ functions. In addition to the seismic vibration of each floor in a building, \ a variety of other motions are investigated, including constant velocity, \ constant acceleration, harmonic oscillation, and decaying oscillation. You \ can also use the specialized ", PageWidth->PaperWidth], StyleBox["Mathematica", PageWidth->PaperWidth, FontSlant->"Italic"], StyleBox[" commands to study other motions that interest you. ", PageWidth->PaperWidth], StyleBox["\n\n", PageWidth->PaperWidth, FontWeight->"Bold", FontSlant->"Italic"], "To measure position along a line, we select a fixed point on the line as a \ reference point or origin, and we scale the line in appropriate units (e.g., \ meters, feet, or miles). ", StyleBox["A distance and a direction specify a position on the line.", PageWidth->PaperWidth], " One direction from the reference point is taken as positive and the other \ direction as negative. Distance is measured using the units of scale along \ the line.\n\nBefore looking at the vibrations of a building shaken by an \ earthquake, let's consider some other motions. \n\nNote: This module uses \ three specially designed functions, \n\n", StyleBox["velocity[s_, timeinterval_, periodic_ ]", FontWeight->"Bold"], ", \n\n", StyleBox["acceleration[s_, timeinterval_, periodic_ ]", FontWeight->"Bold"], ", and \n\n", StyleBox["posvelacc[s_, timeinterval_, periodic_ ]", FontWeight->"Bold"], ". \n\nThese are not built-in ", StyleBox["Mathematica", FontSlant->"Italic"], " functions; and they are only available in this module. The arguments are: \ ", StyleBox["s_", FontWeight->"Bold"], ", a position function; ", StyleBox["timeinterval_", FontWeight->"Bold"], ", the time interval during which the motion occurs; and ", StyleBox["periodic_", FontWeight->"Bold"], ", a flag to indicate whether or not the motion is periodic. In the \ sections that follow, you will see examples of how these functions are used." }], "Text", PageWidth->PaperWidth], Cell[CellGroupData[{ Cell["Technology Guidelines", "Subsection", PageWidth->PaperWidth, CellDingbat->"\[LightBulb]"], Cell[TextData[{ StyleBox["NOTE: If you have just finished a module, restart ", CellFrame->True, Background->None], StyleBox["Mathematica", CellFrame->True, FontSlant->"Italic", Background->None], StyleBox[" before executing a new module.\nTO OPEN CELLS, put your cursor \ on the right cell bracket and double click.", CellFrame->True, Background->None], "\nINITIALIZATION CELLS\n\tWhen asked if you want to \". . . automatically \ evaluate all the initialization cells in the \tnotebook . . . ,\" respond by \ pressing the \"Yes\" button.\nTO STOP AN EXECUTION\n\tSelect the ", StyleBox["Kernel", FontSlant->"Italic"], " pull-down menu and click on ", StyleBox["Abort Evaluation.\n", FontSlant->"Italic"], "ORDER OF EXECUTION\n\tExecute cells in the order given. Do not skip any \ Input cells within a given notebook.\nSAVING NOTEBOOKS\n\tYou can save \ anytime to any directory you choose, and it is wise to save often.\n\t \ However, before you do your final save, delete all your output by selecting \ the \n\t ", StyleBox["Delete All Output", FontSlant->"Italic"], " selection under the ", StyleBox["Kernel", FontSlant->"Italic"], " pull-down menu.\nEXPERIENCING MAJOR PROBLEMS\n\tSave if appropriate, and \ then shut down ", StyleBox["Mathematica", FontSlant->"Italic"], " and start it up again." }], "Text", PageWidth->PaperWidth] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Part I: Constant Velocity", "Section", PageWidth->PaperWidth], Cell["Chapter 2, Section 2", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell[TextData[{ "Suppose that you drive your car along a straight road at a constant speed \ of 60 mph for 4 hours and then turn around and come back at the same speed. \ Let's take the reference point to be your starting position and the direction \ your car travels during the first four hours to be positive. The position of \ your car during the trip is given by the following function of time. (When \ you execute the next command, the function ", StyleBox["s", FontSlant->"Italic"], " is defined, but no output is displayed. That's OK.)" }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(s = Which[t \[LessEqual] 4, 60*t, \ t > 4, 480 - 60*t];\)\)], "Input",\ PageWidth->PaperWidth], 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->"Hyperlink"], ButtonData:>"h1", ButtonStyle->"Hyperlink"]], "Input", PageWidth->PaperWidth, Evaluatable->False, CellTags->"h1b"], Cell[TextData[{ "Note that in mechanics, the symbol ", StyleBox["s", FontSlant->"Italic"], " usually designates position.\n\nNow, we plot the function for the \ duration of the trip." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[s, {t, 0, 8}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The velocity is defined to be the time rate of change of position. That \ is, ", StyleBox["v = ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`ds\/dt\)]], ". Let's find the velocity function and plot it." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(v = D[s, t]\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[v, {t, 0, 8}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->18, 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->"Hyperlink"], ButtonData:>"h2", ButtonStyle->"Hyperlink"]], "Input", PageWidth->PaperWidth, Evaluatable->False, CellTags->"h2b"], Cell[TextData[{ "According to ", StyleBox["Mathematica", FontSlant->"Italic"], ", the velocity, which is the derivative of the position function, is 60 \ mph at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], ". Is that correct? To answer this question, you should look at the two \ one-sided derivatives of ", StyleBox["s", FontSlant->"Italic"], " at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], "." }], "Text", PageWidth->PaperWidth], Cell[TextData[{ "Like position, velocity has a size or magnitude and a direction. The \ magnitude or absolute value of an object's velocity is called its speed. If \ the position function, ", StyleBox["s", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "), is increasing with time, then the velocity is positive, and if it is \ decreasing, then the velocity is negative. Physical quantities that have a \ magnitude and a direction are called vectors, and the position and velocity \ of a moving object are vector quantities.\n\nRecall that the derivative of a \ function at time ", StyleBox["t", FontSlant->"Italic"], " is the slope of the line tangent to the graph of the function at that \ point. Therefore, the velocity at time ", StyleBox["t", FontSlant->"Italic"], " is the slope of the position function at that time. The ", StyleBox["velocity[ ] ", FontWeight->"Bold"], "command in the next executable cell below illustrates this idea by \ producing a sequence of graphs that show: 1) the object as it moves along a \ straight line, 2) the position function, ", StyleBox["s", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "), and 3) the velocity function, ", StyleBox["v", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "). The velocity vector is shown on all three graphs. Note that the value \ for the last argument is 0 because the motion is not periodic. In all of the \ animation graphs, the horizontal axis represents time.\n\n\nTo animate the \ sequence of graphs:\n1. If necessary, widen the notebook window so that in \ each cell all three graphs show across the page.\n2. Put the cursor in the \ cell bracket that contains all the graphics cells, and double click the left \ mouse button. This will collapse all the graphs into one cell, displaying \ only the first graphics cell in the sequence.\n3. Be sure the cell bracket \ that contains the collapsed graphics cells is selected (if not, place the \ cursor in the cell bracket and click once) and then press Ctrl+Y. This will \ play the sequence of graphics slides to generate the animation. \n4. While \ the animation is playing, a control bar appears at the bottom of the notebook \ window. This bar allows you to control the speed and direction of the \ animation.\n\n", StyleBox["Since the following command generates a lot of output, you \ probably won't want to print this notebook until after you have deleted most \ or all of the graphs. We recommend deleting all but a few representative \ cells from the sequence before you print.", CellFrame->True, Background->GrayLevel[0.849989]] }], "Text", PageWidth->PaperWidth, TextJustification->0], Cell[BoxData[ \(\(velocity[s, {t, 0, 8}, 0];\)\)], "Input", PageWidth->PaperWidth, AnimationDisplayTime->0.169], Cell[TextData[{ StyleBox["A note about modeling:", FontWeight->"Bold"], " In the preceding model, the velocity is undefined at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " hours. In reality, however, your car would have to slow down, turn \ around, and accelerate back to cruising speed at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " hours. This maneuver would occur over a very short interval of time (say \ a few minutes) when compared to the four-hour duration of the trip. The sharp \ corner at the turnaround time would in actuality be rounded, and the sudden \ jump from 60 mph to ", Cell[BoxData[ \(TraditionalForm\`\(-\ 60\)\)]], " mph would have a finite negative slope to it. On the scale of eight \ hours, however, the graphs would look much like those depicted above, even if \ the more precise model were used. With these observations in mind, the above \ model is reasonable for describing your trip." }], "Text", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["You Try It: Constant Rate of Change and Extreme Values", "Section", PageWidth->PaperWidth], Cell[TextData[{ "Over each 4-hour subinterval of the motion in Part I, the velocity of your \ car is constant at \[PlusMinus] 60 mph. Each subinterval provides an example \ of a situation where the rate of change or the derivative of a function is \ constant. Write a brief response to each of the following questions.\n\n1. \ On the interval where ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " is positive, what can you say about the value of ", StyleBox["s", FontSlant->"Italic"], "? \n\n2. What can you say about the value of ", StyleBox["s", FontSlant->"Italic"], " on the interval where ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " is negative?\n\n3. When ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " is constant on an interval, what can you say about the relationship \ between ", Cell[BoxData[ \(TraditionalForm\`ds\/dt\)]], " (the slope of the tangents in the interval) and ", Cell[BoxData[ \(TraditionalForm\`\[CapitalDelta]s\/\[CapitalDelta]t\)]], " (the slope of any secant line taken inside the interval)?\n\n4. At ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " hours, the function ", StyleBox["s", FontSlant->"Italic"], " reaches its maximum value of 240 miles. What happens to the derivative, \ ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], ", at this point? What are the values of the two one-sided derivatives of \ ", StyleBox["s", FontSlant->"Italic"], " at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], "? What are the values of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " slightly to the left of ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " and slightly to the right of ", Cell[BoxData[ \(TraditionalForm\`t = \(\(4\)\(?\)\)\)]], "\n\n5. Over the eight-hour interval, what are the largest and smallest \ values of ", StyleBox["s", FontSlant->"Italic"], ",", " and at what times to they occur?\n\n6. What is the value of ", Cell[BoxData[ FormBox[ RowBox[{"a", "=", FractionBox[ RowBox[{ StyleBox[ SuperscriptBox[ StyleBox["d", FontSlant->"Plain"], "2"], FontSlant->"Italic"], StyleBox["s", FontSlant->"Plain"]}], \(dt\^2\)]}], TraditionalForm]]], ", the acceleration, during each of the 4-hour subintervals? In the \ mathematical model, what happens to the acceleration at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " hours? Describe what the acceleration of your car would actually be like \ when you turn around to come back. \n" }], "Text", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["Part II: Constant Acceleration", "Section", PageWidth->PaperWidth], Cell["Chapter 2, Section 2", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell[TextData[{ "Note: ", StyleBox["This exercise generates a lot of graphics and uses a considerable \ amount of computer memory. Before proceeding, pull down the Kernel menu, \ select Delete All Output, and click OK in the resulting dialog box. ", FontWeight->"Plain"] }], "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell["\<\ If you throw an object straight up from the ground with an initial speed of \ 128 ft/s, its height above the ground is given by the following function of \ time.\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(Clear[s];\)\), "\[IndentingNewLine]", \(\(s = 128*t - 16*t^2;\)\), "\[IndentingNewLine]", \(\(Plot[s, {t, 0, 8}, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", PageWidth->PaperWidth], Cell["\<\ Note that the origin is taken at the ground surface and the positive \ direction is upward.\ \>", "Text", PageWidth->PaperWidth], Cell["Now determine the velocity function and graph it.", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(Clear[v];\)\), "\[IndentingNewLine]", \(v = D[s, t]\)}], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[v, {t, 0, 8}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "Let's use ", StyleBox["velocity[ ] ", FontWeight->"Bold"], "to depict the motion. Note that the value for the last argument is 0 since \ the motion is not periodic, but if you change the value to 1, the animation \ will simulate the bouncing motion of a perfectly elastic ball.\n\nYou can \ animate the motion by double clicking the cell bracket that contains all the \ graphics cells in the sequence and then pressing ", "Ctrl+Y", ". \n" }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(velocity[s, {t, 0, 8}, 0]\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "In this example, the rate of change or the slope of the velocity function \ (i.e., the acceleration) is constant at ", Cell[BoxData[ \(TraditionalForm\`a = \(\(-g\) = \(-32\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`ft\/s\^2\)]], ". \n\nAcceleration is also a vector quantity. If the acceleration is in \ the negative direction, the velocity (which can be positive or negative) is \ decreasing with time. If the acceleration is in the positive direction, the \ velocity (which can be positive or negative) is increasing with time. If the \ velocity vector and the acceleration vector are in the same direction, then \ the moving object is speeding up, and if they are in opposite directions, \ then the object is slowing down. " }], "Text", PageWidth->PaperWidth], Cell[TextData[{ "The ", StyleBox["acceleration[ ] ", FontWeight->"Bold"], "command produces a sequence of graphs that show: 1) the object as it moves \ along a straight line, 2) the velocity function, ", Cell[BoxData[ \(TraditionalForm\`v \((t)\)\)]], ", and 3) the acceleration function, ", Cell[BoxData[ \(TraditionalForm\`a \((t)\)\)]], ". The acceleration vector is shown on all three graphs. \n\nYou can \ animate the motion by double clicking the cell bracket that contains all the \ graphics cells in the sequence and then pressing ", "Ctrl+Y", ". " }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(acceleration[s, {t, 0, 8}, 0];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The ", StyleBox["posvelacc[ ]", FontWeight->"Bold"], " command shows the motion of the object, together with its position, \ velocity, and acceleration vectors." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(posvelacc[s, {t, 0, 8}, 0];\)\)], "Input", PageWidth->PaperWidth], Cell["\<\ Note that as the object travels upward, the velocity and acceleration are in \ opposite directions (the velocity is positive and the acceleration is \ negative) and the object is slowing down, whereas, on the way down the \ velocity and acceleration are in the same direction (both negative) and the \ object is speeding up.\ \>", "Text", PageWidth->PaperWidth], Cell[TextData[{ "Before we continue, there is one last thing we need to point out \ pertaining to the physics and mathematics of motion. While the purpose of \ this module is to illustrate the derivative relations between position and \ velocity and between velocity and acceleration, real-world applications \ usually work the other way around. That is, instead of starting with the \ position function and differentiating it to find the velocity and \ acceleration, we usually start with the acceleration function and use it to \ construct the velocity and position functions. The reason for this is that \ Isaac Newton taught us, among other things, that the best way to study \ real-world phenomena is to look at the forces that bring about change. In the \ context of motion problems, this philosophy is embodied in Newton's three \ laws of motion. The second of Newton's three laws states that if the mass of \ an object is constant, then the total of all the forces acting on the object \ is equal to its mass times its acceleration, that is, ", Cell[BoxData[ \(TraditionalForm\`F = m\ a\)]], ". This gives us a way to determine the acceleration function for an object \ by knowing the forces that act on it. Once we know the acceleration, we use \ it to construct the velocity function, and then we use the velocity to \ construct the position function. The problem then is this: if we know the \ derivative of a function, how do we determine what the function is? This \ problem is dealt with in the second big topic of calculus: \ antidifferentiation and integration. You can explore these ideas in a later \ module \"", StyleBox["Motion Along a Straight Line: Acceleration \[Rule] Velocity \ \[Rule] Position", FontWeight->"Bold", FontSlant->"Italic"], ",\" which treats some of the same motions as this module, but starting \ with the acceleration function and then finding the velocity and position \ functions." }], "Text", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["You Try It: Linear Rate of Change and Extreme Values", "Section", PageWidth->PaperWidth], Cell["\<\ Chapter 2, Section 2 Also see Chapter 3, Section 3 \ \>", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell[TextData[{ "The motion in Part II provides an example of a situation where the \ derivative of a function, i.e., the velocity, is a linear function. Write a \ brief response to each of the following questions.\n\n1. On what interval is \ ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], "positive? On what interval is it negative?\n\n2. On the interval where ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " is positive, what can you say about the value of ", StyleBox["s", FontSlant->"Italic"], "? \n\n3. What can you say about the value of ", StyleBox["s", FontSlant->"Italic"], " on the interval where ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " is negative?\n\n4. At ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " seconds, the function ", StyleBox["s", FontSlant->"Italic"], " reaches its maximum value of 256 feet. What happens to the derivative, ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], ", at this point? What is the value of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " just to the left of ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], "? What is its value just to the right of ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], "?\n\n5. What does the value of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " do as ", StyleBox["t", FontSlant->"Italic"], " increases? How is this related to the value of ", Cell[BoxData[ FormBox[ RowBox[{"a", "=", RowBox[{\(dv\/dt\), "=", FractionBox[ RowBox[{ StyleBox[ SuperscriptBox[ StyleBox["d", FontSlant->"Plain"], "2"], FontSlant->"Italic"], StyleBox["s", FontSlant->"Plain"]}], \(dt\^2\)]}]}], TraditionalForm]]], "and the concavity of the graph of ", StyleBox["s", FontSlant->"Italic"], " over the interval of motion?\n\n6. Over the 8-second interval, what are \ the largest and smallest values of ", StyleBox["s", FontSlant->"Italic"], ",", " and at what times to they occur?\n" }], "Text", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["Part III: Oscillations", "Section", PageWidth->PaperWidth], Cell["Chapter 2, Section 4, Simple Harmonic Motion", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell[TextData[{ "Note: ", StyleBox["This exercise generates a lot of graphics and uses a considerable \ amount of computer memory. Before proceeding, pull down the Kernel menu, \ select Delete All Output, and click OK in the resulting dialog box. ", FontWeight->"Plain"] }], "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell["\<\ If you hang a mass from a spring, it will eventually come to rest in an \ equilibrium position, where the weight of the mass is balanced by the tension \ in the spring. Now pull the mass down slightly and let it go. It will \ oscillate (move up and down) about the equilibrium position. If there were no \ air resistance and/or friction in the mass and spring, it would oscillate \ forever with the same amplitude and frequency. Since the motion is periodic, \ it is not surprising that the periodic trigonometric functions can be used to \ describe such a motion. For example, if the amplitude of the vibration is 2 \ centimeters and the period of the oscillations is 4\[Pi]/3 seconds, then the \ motion is described by the following function.\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(Clear[s];\)\), "\[IndentingNewLine]", \(\(s = \(-2\)*Cos[3\ t/2];\)\), "\[IndentingNewLine]", \(\(Plot[s, {t, 0, 8*Pi/3}, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", PageWidth->PaperWidth], Cell["\<\ Now we can calculate the velocity of the mass as it moves up and down and \ graph it.\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(Clear[v];\)\), "\[IndentingNewLine]", \(v = D[s, t]\)}], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[v, {t, 0, 8*Pi/3}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The ", StyleBox["velocity[ ]", FontWeight->"Bold"], " command can be used to depict the motion. Since the motion is periodic, \ we set the last argument of ", StyleBox["velocity[ ]", FontWeight->"Bold"], " to 1 instead of 0. This gives a smooth motion in the animations without a \ hesitation at the wrap-around, end/beginning point of the time interval." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(velocity[s, {t, 0.0, 8.0*Pi/3.0}, 1];\)\)], "Input", PageWidth->PaperWidth], Cell["\<\ The acceleration of the mass is given by the derivative of the velocity.\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(Clear[a];\)\), "\[IndentingNewLine]", \(a = D[v, t]\)}], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[a, {t, 0, 8*Pi/3}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The ", StyleBox["acceleration[ ]", FontWeight->"Bold"], " command illustrates the relationship between the velocity and \ acceleration as the mass oscillates." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(acceleration[s, {t, 0.0, 8.0*Pi/3}, 1];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The ", StyleBox["posvelacc[ ]", FontWeight->"Bold"], " command shows the motion of the object, its position vector, its \ velocity, and its acceleration." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(posvelacc[s, {t, 0.0, 8.0*Pi/3.0}, 1];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "There are some observations to make about this motion. The first is that \ the acceleration is in the direction opposite the position. When the mass is \ below the equilibrium position, the stretch in the spring increases, thus \ exerting a net upward force on the mass, accelerating it in that direction. \ When the mass is above the equilibrium position, the stretch in the spring is \ reduced, resulting in a net downward force on the mass, accelerating it in \ that direction. The second observation is that the acceleration is \ proportional to the position. The first two observations can be demonstrated \ mathematically by noting that if ", Cell[BoxData[ \(TraditionalForm\`s(t) = \(-A\)\ \(cos(\[Omega]\ t)\)\)]], ", then ", Cell[BoxData[ \(TraditionalForm\`\(\(a(t)\)\(=\)\)\)]], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{\(d\^2\), StyleBox["s", FontSlant->"Plain"]}], SuperscriptBox[ StyleBox[ RowBox[{ StyleBox["d", FontSlant->"Italic"], "t"}]], "2"]], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\` = \)]], " ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\^2\)]], Cell[BoxData[ \(TraditionalForm\`\(\(A\)\(\ \)\(cos\)\(\ \)\)\)]], "(\[Omega] t)", Cell[BoxData[ \(TraditionalForm\`\(\(=\)\(\(-\[Omega]\^2\) \(s(t)\)\)\)\)]], ". \n\nAnother thing to note is that the velocity and acceleration are 90 \ degrees or ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " radians out of phase with one another. During the first quarter-cycle of \ the motion, the velocity and acceleration are both positive and the mass is \ speeding up, moving upward. During the second-quarter cycle, the velocity is \ positive and the acceleration is negative, and the mass is still moving \ upward but now it is slowing down. During the third quarter-cycle, the \ velocity and acceleration are both negative, and the mass is speeding up \ moving downward. During the fourth quarter-cycle, the velocity is negative \ and the acceleration is positive, and the mass is slowing down as it moves \ downward. Overlaying the graphs of the velocity and acceleration functions \ shows these relationships." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[{v, a}, {t, 0, 4*Pi/3}];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The velocity is ", Cell[BoxData[ \(TraditionalForm\`v( t)\ = \ \(-\ A\[Omega]\)\ sin\ \((\[Omega]\ t)\)\)]], " and the acceleration is ", Cell[BoxData[ FormBox[ RowBox[{\(a(t)\), " ", "=", " ", RowBox[{\(-\ A\), FormBox[\(\[Omega]\^2\), "TraditionalForm"], "cos", " ", \((\[Omega]\ t)\)}]}], TraditionalForm]]], ". From basic trigonometry, we know that the sine and cosine functions are \ 90 degrees or ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " radians out of phase with one another.\n\nThe final observation is that \ at the instant when the mass passes the equilibrium position, the velocity \ reaches its extreme values (positive on the way up and negative on the way \ down), its speed is at its maximum value, and its acceleration is 0." }], "Text", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["\<\ You Try It: The Second Derivative, Concavity, and Inflection Points\ \>", "Section", PageWidth->PaperWidth], Cell["\<\ Chapter 2, Section 2 Also see Chapter 3, Section 3 \ \>", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell[TextData[{ "After you have viewed the ", StyleBox["velocity[ ] ", FontWeight->"Bold"], "and ", StyleBox["posvelacc[ ] ", FontWeight->"Bold"], "animations, write a brief response to each of the following questions.\n\n\ 1. During intervals when the tangent line is rotating clockwise, the graph \ of ", StyleBox["s", FontSlant->"Italic"], " is said to be \"concave down.\" Over what intervals is the graph of ", StyleBox["s", FontSlant->"Italic"], " concave down? What is the value of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " doing in these intervals? What is the sign of ", Cell[BoxData[ FormBox[ RowBox[{"a", "=", FractionBox[ RowBox[{\(d\^2\), StyleBox["s", FontSlant->"Plain"]}], \(dt\^2\)]}], TraditionalForm]]], " in these intervals?\n\n2. During the intervals when the tangent line is \ rotating counterclockwise, the graph of ", StyleBox["s", FontSlant->"Italic"], " is said to be \"concave up.\" Over what intervals is the graph of ", StyleBox["s", FontSlant->"Italic"], " concave up? What is the value of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " doing in these intervals? What is the sign of ", Cell[BoxData[ FormBox[ RowBox[{"a", "=", FractionBox[ RowBox[{\(d\^2\), StyleBox["s", FontSlant->"Plain"]}], \(dt\^2\)]}], TraditionalForm]]], " in these intervals?\n\n3. An inflection point is a point on the graph of \ ", StyleBox["s", FontSlant->"Italic"], " where the concavity changes from up to down, or vice versa. Describe the \ motion of the green tangent line as it passes over the inflection points on \ the graph of ", StyleBox["s", FontSlant->"Italic"], ". What happens to the value of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], " as the point on the graph passes over the inflection points? What happens \ to the value of ", Cell[BoxData[ FormBox[ RowBox[{"a", "=", RowBox[{\(dv\/dt\), "=", FractionBox[ RowBox[{\(d\^2\), StyleBox["s", FontSlant->"Plain"]}], \(dt\^2\)]}]}], TraditionalForm]]], "as the point on the graph passes over the inflection points?\n\n4. At \ what points on the graph of ", StyleBox["s", FontSlant->"Italic"], " is the slope the steepest? At what points on the graph of ", StyleBox["s", FontSlant->"Italic"], " does the value of ", Cell[BoxData[ \(TraditionalForm\`v = ds\/dt\)]], "reach its extreme (maximum and minimum) values?\n\n5. Describe the motion \ of an object that moves up and down along a straight line for each of the \ following conditions: a) ", Cell[BoxData[ \(TraditionalForm\`v > 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`a > 0\)]], "; b) ", Cell[BoxData[ \(TraditionalForm\`v > 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`a < 0\)]], "; c) ", Cell[BoxData[ \(TraditionalForm\`v < 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`a < 0\)]], "; and d) ", Cell[BoxData[ \(TraditionalForm\`v < 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`a > 0\)]], ". If we take up as positive, specify the direction the object is moving \ and whether it is speeding up or slowing down." }], "Text", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["Part IV: Decaying Oscillations", "Section", PageWidth->PaperWidth], Cell["Chapter 2, Section 4", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell["\<\ If you are getting tired of looking at our animations, you can quit here. \ These next two Parts, IV and V, are just for fun. \ \>", "Text", PageWidth->PaperWidth], Cell["\<\ (Before proceeding with this section, pull down the Kernel menu at the top of \ the screen, select Delete All Output, and click OK to free memory for the \ graphics that will be generated by the commands that follow.) A more realistic model for oscillations of real objects would account for the \ loss of energy that results in a decay in the amplitude of the oscillations. \ In mechanical systems, this energy loss is usually due to friction (objects \ rubbing against one another in some way). For the motion of the mass hanging \ from a spring, a more realistic position function has an amplitude that \ decays exponentially.\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(Clear[s];\)\), "\[IndentingNewLine]", \(\(s = \(-2\)*Exp[\(-t\)/5]*Cos[t];\)\), "\[IndentingNewLine]", \(\(Plot[s, {t, 0, 6*Pi}, PlotRange \[Rule] All, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", PageWidth->PaperWidth], Cell["Now let's find the velocity and acceleration and graph them.", "Text", PageWidth->PaperWidth], Cell[BoxData[ \(v = D[s, t] // Simplify\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[v, {t, 0, 6*Pi}, PlotRange \[Rule] All, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(a = D[v, t] // Simplify\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[a, {t, 0, 6*Pi}, PlotRange \[Rule] All, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth], Cell[TextData[{ "It is apparent that the amplitudes of the velocity and acceleration also \ decay with time.\n\nThe motion can be analyzed using the ", StyleBox["velocity[ ]", FontWeight->"Bold"], ", ", StyleBox["acceleration[ ]", FontWeight->"Bold"], ", and ", StyleBox["posvelacc[ ]", FontWeight->"Bold"], " functions." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(velocity[s, {t, 0, 6*Pi}, 0];\)\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(acceleration[s, {t, 0, 6*Pi}, 0];\)\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(posvelacc[s, {t, 0, 6*Pi}, 0];\)\)], "Input", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["Part V: Earthquake", "Section", PageWidth->PaperWidth], Cell["Chapter 2, Section 2", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell[TextData[{ "Note: ", StyleBox["This exercise generates a lot of graphics and uses a considerable \ amount of memory. Before proceeding, pull down the Kernel menu, select Delete \ All Output, and click OK in the resulting dialog box. ", FontWeight->"Plain"] }], "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell["\<\ During an earthquake, the floors of a building act like masses, and the \ columns between the floors act like elastic springs. Consequently, a single \ floor can be modeled as a mass that is equal to the mass of the floor and a \ spring with elastic stiffness equal to that provided by the columns above and \ below the floor. The motion of the mass in the mass-spring model emulates the \ actual motion of the floor in a building. In part, the motion will consist of oscillations along a straight line. \ Initially, when the earthquake begins, the building is at rest and the ground \ motion puts energy into the building, resulting in back and forth \ oscillations of increasing amplitude. If the earthquake lasts long enough, a \ steady-state condition is reached where the amount of energy that goes into \ the building equals the amount dissipated by friction as parts of the \ building rub against one another. During this phase of the earthquake, the \ building sways with oscillations of constant amplitude. When the earthquake \ stops, the oscillations decay, and the building eventually comes to rest. The following function can be used to describe the position of the floor in a \ building during an earthquake. To see how we obtained this function, see the \ note at the end of this part.\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(s = Which[t < 30, 2*Exp[\(-t\)/5]*Cos[0.9798*t] - 2*Cos[t] + 0.40825*Exp[\(-t\)/5]*Sin[0.9798*t], t \[GreaterEqual] 30, \((\(-698.8\))\)*Exp[\(-t\)/5]*Cos[0.9798*t] + 478.3*Exp[\(-t\)/5]*Sin[0.9798*t]];\)\), "\[IndentingNewLine]", \(\(Plot[s, {t, 0, 70}, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", PageWidth->PaperWidth], Cell["The graphs of the velocity and acceleration look like this.", "Text", PageWidth->PaperWidth], Cell[BoxData[{ \(\(v = D[s, t];\)\), "\[IndentingNewLine]", \(\(Plot[v, {t, 0, 70}, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", PageWidth->PaperWidth], Cell[BoxData[{ \(\(a = D[v, t];\)\), "\[IndentingNewLine]", \(\(Plot[a, {t, 0, 70}, AxesLabel \[Rule] {"\", "\"}];\)\)}], "Input", PageWidth->PaperWidth], Cell[TextData[{ "The ", StyleBox["velocity[ ]", FontWeight->"Bold"], ", ", StyleBox["acceleration[ ]", FontWeight->"Bold"], ", and ", StyleBox["posvelacc[ ]", FontWeight->"Bold"], " functions depict the motion." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(\(velocity[s, {t, 0, 60}, 0];\)\)], "Input", PageWidth->PaperWidth, AnimationDisplayTime->0.2197], Cell[BoxData[ \(\(acceleration[s, {t, 0, 60}, 0];\)\)], "Input", PageWidth->PaperWidth, AnimationDisplayTime->0.2197], Cell[BoxData[ \(\(posvelacc[s, {t, 0, 60}, 0];\)\)], "Input", PageWidth->PaperWidth], Cell["\<\ Note: The position function used for the earthquake motion above was obtained \ by solving the following initial value problem. The ground motion excitation \ occurs between t = 0 and t = 30 seconds, and after t = 30 seconds, the motion \ is free, unforced vibration. (This command takes a while to execute.)\ \>", "Text", PageWidth->PaperWidth], Cell[BoxData[ \(Clear[s, soln]; soln = \(DSolve[{\(s''\)[t] + .4*\(s'\)[t] + s[t] \[Equal] 0.8*Sin[t]*\((1 - UnitStep[t - 30. ])\), s[0] \[Equal] 0, \(s'\)[0] \[Equal] 0}, s[t], t]\)[\([1, 1, 2]\)] // Simplify\)], "Input", PageWidth->PaperWidth], Cell[BoxData[ \(\(Plot[soln, {t, 0, 50}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input", PageWidth->PaperWidth] }, Closed]], Cell[CellGroupData[{ Cell["Part VI: Some Help with Your Homework", "Section", PageWidth->PaperWidth], Cell["Chapter 2, Section 2, Exercises 1 - 4, 10, and 11", "Text", PageWidth->PaperWidth, FontWeight->"Bold"], Cell["\<\ The special commands in this section may be helpful with some of the homework \ exercises from Chapter 2, Section 2. 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", ButtonBox["Go Back.", ButtonData:>"h1b", ButtonStyle->"Hyperlink"] }], "Text", PageWidth->PaperWidth, CellTags->"h1"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " graphs a function by plotting a series of points on the function and then \ connecting them with straight lines, sometimes when they really should not be \ connected. The vertical line at ", Cell[BoxData[ \(TraditionalForm\`t = 4\)]], " hours really shouldn't be there. The upper segment of the graph at ", Cell[BoxData[ \(TraditionalForm\`\(+\ 60\)\ mph\)]], " should not be connected to the lower segment at ", Cell[BoxData[ \(TraditionalForm\`\(-\ 60\)\ mph\)]], ". 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