(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.1' 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[ 197507, 5645]*) (*NotebookOutlinePosition[ 199262, 5698]*) (* CellTagsIndexPosition[ 199158, 5691]*) (*WindowFrame->Normal*) Notebook[{ Cell["Coordenadas curvil\[IAcute]neas", "Title"], Cell["\<\ Francisco Javier P\[EAcute]rez Gonz\[AAcute]lez Departamento de An\[AAcute]lisis Matem\[AAcute]tico Universidad de Granada\ \>", "Author", TextAlignment->Right, FontSize->12], Cell[CellGroupData[{ Cell["Inicializaci\[OAcute]n", "Section"], Cell[BoxData[{ \(Off[ParametricPlot3D::ppcom]\), "\n", \(Off[General::spell1]\), "\[IndentingNewLine]", \(Off[General::spell]\)}], "Input", InitializationCell->True], Cell["<True, CellTags->"S5.32.1"], Cell["<True, CellTags->"S5.32.1"], Cell[BoxData[ \(\(\(vector2D[x_: {0, 0}, y_]\)\(:=\)\(Arrow[x, x + y]\)\(\ \)\( (*\ traslaci\[OAcute]n\ del\ vector\ y\ al\ punto\ x . \ Por\ defecto\ x = {0, 0}*) \)\)\)], "Input", InitializationCell->True], Cell[BoxData[ \(\(\(norma[x_] := N[\@\(x . x\)]\ (*\ norma\ eucl\[IAcute]dea\ del\ vector\ x\ *) \n \(\(normaliza[x_]\ := If[Simplify[x . x] == 0, x, \ x/norma[x]]; \ (*\ vector\ unitario\ en\ la\ direcci\[OAcute]n\ del\ vector\ x\ *) \n normal3D[x_]\ := \ normaliza[ Which[x[\([1]\)] == 0, {1, 0, 0}, \((x[\([2]\)] \[NotEqual] 0)\) \[Or] \((x[\([3]\)] \[NotEqual] 0)\), {0, x[\([3]\)], \(-x[\([2]\)]\)}, \((x[\([2]\)] == 0 \[And] x[\([3]\)] == 0)\), {0, 1, 0}]]\)\(\n\) \)\)\( (*\ un\ vector\ unitario\ ortogonal\ al\ vector\ x\ *) \)\)\)], "Input", InitializationCell->True], Cell[BoxData[ \(\(\(\(arrow3D[x_, y_, color_: RGBColor[0. , 0. , 0. ], grosor_: 0.007] := \n\t\t\t{{color, 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"Input", InitializationCell->True], Cell[BoxData[ \(\(coordesfericas := Show[Graphics[{Line[{{0, 6}, {0, 0}, {2, 4}, {0, 0}, {6, 0}, {0, 0}, {\(-4\), \(-4\)}}], {Dashing[{ .012, .012}], Line[{{0, 5.5}, {2, 4}, {2, \(-1.5\)}, {0, 0}}]}, {Dashing[{ .012, .012}], Line[{{2, \(-1.5\)}, {3.5, 0}}]}, {Dashing[{ .012, .012}], Line[{{2, \(-1.5\)}, {\(-1.5\), \(-1.5\)}}]}, Circle[{0, 0}, 2, {ArcTan[2], \[Pi]/2}], Circle[{0, 0}, .6, {\(-3\) \[Pi]/4, ArcTan[\(- .75\)]}], {Hue[ 0], PointSize[ .02], \ Point[{0.01, 5.5}], Text["\", {1, 5.6}], Text[\*"\"\\"", {2.9, 2.12}], Text["\", {2.2, \(-1.8\)}], Text["\", {\(-1.7\), \(-1.8\)}], Text["\", {3.7, .35}], Text["\<\[Theta]\>", { .485, 2.47}], Text["\<\[Phi]\>", {0, \(-1\)}]}, {Hue[0], PointSize[ .02], \ Point[{3.5, 0}]}, {Hue[0], PointSize[ .02], \ Point[{\(-1.5\), \(-1.5\)}]}, {Hue[0.6], PointSize[ .02], \ Point[{2, 4}]}, {Red, Arrow[{2, 4}, {2, 4} + .4 {2, 4}, HeadLength -> .04], Arrow[{2, 4}, {2, 4} + .6 {2, \(-1.5\)}, HeadLength -> .04], \n\t\t\t\tArrow[{2, 4}, {2, 4} + 1.2 {1.3, .5}, HeadLength -> .04]}, {Blue, Text["\", {2.55, 3.9}]}, Text["\", {\(-3.4\), \(-3.9\)}], Text["\", {0.3, 5.95}], Text["\", {5.95, 0.25}], Text[\*"\"\<\!\(e\_r\)\>\"", {2.9, 5.8}], Text[\*"\"\<\!\(e\_\[Phi]\)\>\"", {3.9, 4.8}], Text[\*"\"\<\!\(e\_\[Theta]\)\>\"", {3.4, 2.9}]}], AspectRatio \[Rule] Automatic, Background \[Rule] RGBColor[1, 1, 0.75], TextStyle \[Rule] {FontSize \[Rule] 12, FontWeight -> "\"}];\)\)], "Input", InitializationCell->True], Cell[BoxData[{ \(Unprotect[Sqrt]; Unprotect[Power]; \@p_\^2 := p /; Positive[p]; \@\(p_*q_\) := \(\@p\) \@q /; Positive[p] && Positive[q]; Protect[Power]; Protect[Sqrt];\), "\n", \(\(Unprotect[Positive];\)\), "\n", \(\(Positive[ Sin[\[Theta]]] = \(Positive[ r] = \(Positive[\[Rho]] = \(Positive[ u] = \(Positive[ v] = \(Positive[ w] = \(Positive[ y] = \(Positive[ x + \@\(x\^2 + y\^2\)] = \(Positive[\@p_] = \ \(Positive[p_\^2] = \(Positive[p_\^2 + q_\^2] = True\)\)\)\)\)\)\)\)\)\);\)\), "\n", \(Positive[\(-\@p_\)] = False; Positive[p_\ q_] = Positive[p] && Positive[q]; Positive[1\/p_] = Positive[p];\), "\n", \(\(Protect[Positive];\)\), "\n", \(Unprotect[ArcTan]; ArcTan[\[Rho]\ Cos[\[Theta]], \[Rho]\ Sin[\[Theta]]] = \[Theta]; ArcTan[r\ Cos[\[Phi]]\ Sin[\[Theta]], r\ Sin[\[Theta]]\ Sin[\[Phi]]] = \[Phi]; Protect[ArcTan];\), "\n", \(Unprotect[ArcCos]; ArcCos[Cos[\[Theta]]] = \[Theta]; Protect[ArcCos];\)}], "Input", InitializationCell->True], Cell[BoxData[ \(todospositivos[x_List] := Positive[x[\([1]\)]] && Positive[x[\([2]\)]] && Positive[x[\([3]\)]]\)], "Input", InitializationCell->True], Cell["<True, CellTags->"S5.29.1"], Cell[BoxData[{ \(\(g[r_, \[Theta]_, \[Phi]_] := {r\ Cos[\[Phi]]\ Sin[\[Theta]], r\ Sin[\[Phi]]\ Sin[\[Theta]], r\ Cos[\[Theta]]};\)\), "\[IndentingNewLine]", \(\(planoscoordenados[ r_, \[Theta]_, \[Phi]_] := {ParametricPlot3D[ Evaluate[g[r, tz, fi]], {tz, 0, \[Pi]}, {fi, 0, 2 \[Pi]}, \[IndentingNewLine]BoxRatios \[Rule] {1, 1, 1}, DisplayFunction \[Rule] Identity], ParametricPlot3D[ Evaluate[g[u, \[Theta], v]], {u, 0, 3.5}, {v, 0, 2 \[Pi]}, \[IndentingNewLine]BoxRatios \[Rule] {1, 1, 1}, DisplayFunction \[Rule] Identity], ParametricPlot3D[ Evaluate[g[\[Rho], th, \(-\[Phi]\)]], {\[Rho], 0, 3.5}, {th, 0, \[Pi]}, \[IndentingNewLine]BoxRatios \[Rule] {1, 1, 1}, DisplayFunction \[Rule] Identity]};\)\), "\[IndentingNewLine]", \(fg[1] = planoscoordenados[1, \[Pi]/3 - .9, \(-\[Pi]\)/2. ]; fg[2] = planoscoordenados[1, \[Pi]/3 - .625, \(-\[Pi]\)/2. ];\), "\n", \(\(fg[3] = planoscoordenados[1, \[Pi]/3 - .4, \(-\[Pi]\)/2. ];\)\), "\n", \(\(fg[4] = planoscoordenados[1, \[Pi]/3 - .125, \(-\[Pi]\)/2. ];\)\), "\n", \(\(fg[5] = planoscoordenados[1, \[Pi]/3. , \(-\[Pi]\)/2. ];\)\), "\n", \(\(fg[6] = planoscoordenados[1, \[Pi]/3. , \(-3\) \[Pi]/8. ];\)\), "\n", \(\(fg[7] = planoscoordenados[1, \[Pi]/3. , \(-\[Pi]\)/4. ];\)\), "\n", \(\(fg[8] = planoscoordenados[1, \[Pi]/3. , \(-\[Pi]\)/8. ];\)\), "\n", \(\(fg[9] = planoscoordenados[1, \[Pi]/3. , 0. ];\)\), "\n", \(\(fg[10] = planoscoordenados[1.25, \[Pi]/3. , 0. ];\)\), "\n", \(fg[11] = planoscoordenados[1.5, \[Pi]/3. , 0. ]; fg[12] = planoscoordenados[1.75, \[Pi]/3. , 0. ];\), "\n", \(\(fg[13] = planoscoordenados[2, \[Pi]/3. , 0. ];\)\), "\n", \(\(esfericasmovie := Table[Show[{Graphics3D[{Thickness[ .01], Line[{{\(-4\), 0, 0}, {4, 0, 0}}], \[IndentingNewLine]Line[{{0, \(-4\), 0}, {0, 4, 0}}], Line[{{0, 0, \(-4\)}, {0, 0, 4}}], Text["\", {4, 0, 0}, {\(-1.25\), 1}], Text["\", {0, 4, 0}, {\(-1\), 1}], \[IndentingNewLine]Text["\", {0, 0, 4}, {0, \(-1.25\)}]}], Graphics3D[{Hue[0], PointSize[ .03], Point[g[2, \[Pi]/3, 0]]}], fg[k]}, Boxed \[Rule] False, Axes -> None, PlotRange \[Rule] {{\(-4\), 4}, {\(-4\), 4}, {\(-4\), 4}}, DisplayFunction \[Rule] $DisplayFunction, ImageSize \[Rule] {527.25, 404.375}], {k, 1, 13}];\)\)}], "Input",\ InitializationCell->True] }, Closed]], Cell[CellGroupData[{ Cell["Coordenadas polares", "Section"], Cell[TextData[{ "La funci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm \`g(\[Rho], \[Theta]) = \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\)]], " es una biyecci\[OAcute]n de", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(\[CapitalOmega] = \), "TraditionalForm"], FormBox[\(\(\[DoubleStruckCapitalR]\^+\)\[Times]\), "TraditionalForm"]}]}], TraditionalForm]]], "]", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\), \[Pi]\)]], "] sobre ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^2\\{\((0, 0)\)}\)]], ". Los n\[UAcute]meros \[Rho] y \[Theta] dados por ", Cell[BoxData[ \(TraditionalForm\`x = \[Rho]\ cos\ \[Theta], \ y = \[Rho]\ sen\ \[Theta]\)]], " donde ", Cell[BoxData[ \(TraditionalForm\`\[Rho] > 0\)]], ", ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(-\[Pi]\), "TraditionalForm"], "<", "\[Theta]", "\[LessEqual]", "\[Pi]"}], TraditionalForm]]], " se llaman las coordenadas polares del punto de coordenadas cartesianas ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], ". " }], "Text"], Cell[BoxData[ \(\(coordpolares;\)\)], "Input"], Cell[TextData[{ "En vez de elegir el intervalo ]", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(-\[Pi]\), \[Pi]\)\)\)]], "] para medir en radianes el ", StyleBox["\[AAcute]ngulo polar", FontWeight->"Bold"], " \[Theta], podemos elegir cualquier otro intervalo semiabierto de longitud \ 2\[Pi], por ejemplo [", Cell[BoxData[ \(TraditionalForm\`0, 2 \[Pi]\)]], "[. La elecci\[OAcute]n m\[AAcute]s conveniente desde un punto de vista \ matem\[AAcute]tico, por razones de c\[AAcute]lculo (definici\[OAcute]n del \ arcotangente) y de simetr\[IAcute]a, es ]", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\), \[Pi]\)]], "]. Observa que lo que hacemos es medir \[AAcute]ngulos en sentido \ antihorario desde la parte negativa del eje de abscisas. Los valores de \ \[Theta] en el intervalo ]", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\), 0\)]], "[ corresponden a puntos situados en el semiplano inferior (", Cell[BoxData[ \(TraditionalForm\`y < 0\)]], ") y valores de \[Theta] en el intervalo ]", Cell[BoxData[ \(TraditionalForm\`0, \[Pi]\)]], "[ corresponden a puntos situados en el semiplano superior (", Cell[BoxData[ \(TraditionalForm\`y > 0\)]], "). Los valores de \[Theta] en el intervalo ]", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\)/2, \[Pi]/2\)]], "[ corresponden a puntos situados en el semiplano de la derecha (", Cell[BoxData[ \(TraditionalForm\`x > 0\)]], ") y valores de \[Theta] en ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(\(\(]\) \(\[Pi]/ 2\)\), \[Pi]\)\(]\)\)\(\[Union]\)\)\(]\)\) - \[Pi], \ \(-\[Pi]\)/\(\(2\)\([\)\)\)]], " corresponden a puntos situados en el semiplano de la izquierda (", Cell[BoxData[ \(TraditionalForm\`x < 0\)]], "). " }], "Text"], Cell["\<\ Los siguientes comandos pasan de coordenadas polares a cartesianas y de \ cartesianas a polares.\ \>", "Text"], Cell[BoxData[{ \(\(polar2cart[{\[Rho]_, \[Theta]_}] = \(g[\[Rho]_, \[Theta]_] := {\[Rho]\ \ Cos[\[Theta]], \ \[Rho]\ Sin[\[Theta]]}\);\)\), "\n", \(cart2polar[{x_, y_}] := {\@\(x\^2 + y\^2\), ArcTan[x, y]}\)}], "Input"], Cell[TextData[{ "El comando ArcTan[x,y] proporciona el valor de \[Theta] teniendo en cuenta \ el cuadrante donde est\[AAcute] situado el punto ", Cell[BoxData[ \(TraditionalForm\`\(\((x, y)\)\ \)\)]], "para lo cual suma o resta \[Pi] al valor de ", Cell[BoxData[ \(TraditionalForm\`ArcTan[y\/x]\)]], ". Para ", Cell[BoxData[ \(TraditionalForm\`x > 0\)]], " se tiene que ", Cell[BoxData[ \(TraditionalForm\`ArcTan[x, y] = ArcTan[y\/x]\)]], " pero esa igualdad no es cierta cuando ", Cell[BoxData[ \(TraditionalForm\`x < 0\)]], ". En muchos textos se afirma que el \[AAcute]ngulo polar, \[Theta], viene \ dado por la igualdad ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = ArcTan[y\/x]\)]], ", debes tener claro que eso es falso cuando ", Cell[BoxData[ \(TraditionalForm\`x < 0\)]], ". Recuerda que la funci\[OAcute]n arcotangente toma valores en el \ intervalo ", Cell[BoxData[ \(TraditionalForm\`\(] \(-\[Pi]\)/2\), \[Pi]/\(2[\)\)]], ". Para ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], " en el segundo cuadrante (", Cell[BoxData[ \(TraditionalForm\`x < 0, y > 0\)]], ") el \[AAcute]ngulo polar est\[AAcute] en el intervalo ", Cell[BoxData[ \(TraditionalForm\`\(] \[Pi]/2\), \(\[Pi][\ \)\)]], "y es igual a ", Cell[BoxData[ \(TraditionalForm\`\[Pi] + ArcTan[y\/x]\)]], ", y para ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], " en el tercer cuadrante (", Cell[BoxData[ \(TraditionalForm\`x < 0, y < 0\)]], ") el \[AAcute]ngulo polar est\[AAcute] en el intervalo ", Cell[BoxData[ \(TraditionalForm\`\(] \(-\[Pi]\)\), \(-\[Pi]\)/\(2[\ \)\)]], "y es igual a ", Cell[BoxData[ \(TraditionalForm\`ArcTan[y\/x] - \[Pi]\)]], "." }], "Text"], Cell[TextData[{ "Cuando se utiliza el sistema de coordenadas polares los vectores se \ refieren a una base ortonormal ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " que se ha representado en la figura anterior. En el lenguaje \ t\[IAcute]pico de los textos de f\[IAcute]sica se dice que el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta \[Rho] manteniendo \[Theta] constante y el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta \[Theta] manteniendo \[Rho] constante. En t\[EAcute]rminos \ matem\[AAcute]ticos, quiz\[AAcute]s m\[AAcute]s precisos, observa que el \ vector de posici\[OAcute]n del punto ", StyleBox["P", FontSlant->"Italic"], " es ", Cell[BoxData[ \(TraditionalForm \`g(\[Rho], \[Theta]) = \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\)]], " su variaci\[OAcute]n con respecto a \[Rho] manteniendo \[Theta] constante \ es la derivada parcial respecto a \[Rho] y su variaci\[OAcute]n con respecto \ a \[Theta] manteniendo \[Rho] constante es la derivada parcial respecto a \ \[Theta]." }], "Text"], Cell[BoxData[{ \(D[g[\[Rho], \[Theta]], \[Rho]]\), "\n", \(D[g[\[Rho], \[Theta]], \[Theta]]\)}], "Input"], Cell["Observa que estos vectores son ortogonales", "Text"], Cell[BoxData[ \(D[g[\[Rho], \[Theta]], \[Rho]] . D[g[\[Rho], \[Theta]], \[Theta]] // Simplify\)], "Input"], Cell[TextData[{ "Para obtener una base ortonormal a partir de ellos todo lo que tenemos que \ hacer es normalizarlos. El primero tiene norma igual a 1 y el segundo tiene \ norma igual a \[Rho]. Por tanto, los vectores ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "=", \((cos\ \[Theta], \ sen\ \[Theta])\)}], ",", RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "=", \((\(-sen\)\ \[Theta], \ cos\ \[Theta])\)}]}], TraditionalForm]]], " forman una base ortonormal. Es a dicha base a la que se refiere un vector \ cuando se usan coordenadas polares. Observa que los vectores de esta base \ dependen de la posici\[OAcute]n del punto, es decir, no se trata de una base \ fija. F\[IAcute]jate en que si ", Cell[BoxData[ \(TraditionalForm\`\(\((\[Rho], \[Theta])\)\ \)\)]], "son las coordenadas polares de un punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], ", se verifica que ", Cell[BoxData[ FormBox[ RowBox[{\((x, y)\), "=", RowBox[{"\[Rho]", " ", StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}]}], TraditionalForm]]], ". En general, la expresi\[OAcute]n en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " de un vector ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["v", FontWeight->"Bold"], "=", \((x, y)\)}], TraditionalForm]]], " se obtiene por el m\[EAcute]todo usual calculando sus proyecciones \ ortogonales sobre los vectores de la base:" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ StyleBox[ \(v = \[LeftAngleBracket]v | e\_\[Rho]\[RightAngleBracket] e\_\[Rho] + \[LeftAngleBracket]v | e\_\[Theta]\[RightAngleBracket] e\_\[Theta]\), FontWeight->"Bold"], TraditionalForm]]], "\t\t(1)" }], "Text", TextAlignment->Center], Cell[TextData[{ "Observa que si escribimos ", StyleBox["v", FontWeight->"Bold", FontSlant->"Italic"], " en la forma ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["v", FontWeight->"Bold"], "=", \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)}], TraditionalForm]]], " entonces ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(\[LeftAngleBracket]v | e\_\[Rho]\[RightAngleBracket]\), FontWeight->"Bold"], "=", "\[Rho]"}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ \(\[LeftAngleBracket]v | e\_\[Theta]\[RightAngleBracket]\), FontWeight->"Bold"], "=", "0"}], TraditionalForm]]], "." }], "Text"], Cell[TextData[{ "Recuerda que la matriz jacobiana de una funci\[OAcute]n de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^2\)]], " en ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^2\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(f(x, y) = \((\(f\_1\)(x, y), \(f\_2\)(x, y))\)\ \)\)]], "es la matriz cuyas filas son los vectores gradiente de las funciones \ componentes de ", StyleBox["f", FontSlant->"Italic"], " . El siguiente comando calcula la matriz jacobiana de un campo vectorial \ de dos variables." }], "Text"], Cell[BoxData[ \(\(matrizjacobiana2[func_]\)[x_, y_] := Module[{u, v}, Outer[D, func[u, v], {u, v}] /. {u -> x, v -> y}]\)], "Input"], Cell[BoxData[ \(\(matrizjacobiana2[g]\)[\[Rho], \[Theta]] // MatrixForm\)], "Input"], Cell[TextData[{ "Las columnas de la matriz jacobiana son las derivadas parciales ", Cell[BoxData[ \(TraditionalForm\`\(\(D[g[\[Rho], \[Theta]], \[Rho]]\)\(\ \)\)\)]], " y", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(D[ g[\[Rho], \[Theta]], \[Theta]]\)\)\)]], ". Las normas eucl\[IAcute]deas de las columnas de la matriz jacobiana se \ llaman ", StyleBox["factores m\[EAcute]tricos ", FontWeight->"Bold"], "o ", StyleBox["factores de escala", FontWeight->"Bold"], " del cambio a cordenadas polares y son ", Cell[BoxData[ \(TraditionalForm\`{1, \[Rho]}\)]], ". " }], "Text"], Cell[CellGroupData[{ Cell["\<\ Expresi\[OAcute]n de la velocidad y la aceleraci\[OAcute]n en cordenadas \ polares\ \>", "Subsection"], Cell[TextData[{ "Consideremos un m\[OAcute]vil cuya trayectoria en el plano viene dada por \ la funci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ RowBox[{\(x(t)\), StyleBox["i", FontWeight->"Bold"]}], "+", RowBox[{\(y(t)\), StyleBox["j", FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". Sean ", Cell[BoxData[ \(TraditionalForm\`\((\[Rho](t), \ \[Theta](t))\)\)]], " las coordenadas polares de ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], TraditionalForm]]], " de forma que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ \((\(\[Rho](t)\) cos\ \(\[Theta](t)\), \(\[Rho](t)\) sen\ \(\[Theta](t)\))\), "=", RowBox[{\(\[Rho](t)\), RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}]}]}], TraditionalForm]]], " donde ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}], "=", \((cos\ \(\[Theta](t)\), sen\ \(\[Theta](t)\))\)}], TraditionalForm]]], ". Observa que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_\[Rho]\%\(\ \[VerticalLine]\)\), FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["\[Theta]", RowBox[{" ", StyleBox["\[VerticalLine]", FontWeight->"Bold"]}]], "(", "t", ")"}], \((\(-sen\)\ \(\[Theta](t)\), cos\ \(\[Theta](t)\))\)}], "=", RowBox[{ RowBox[{ SuperscriptBox["\[Theta]", RowBox[{" ", StyleBox["\[VerticalLine]", FontWeight->"Bold"]}]], "(", "t", ")"}], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}]}]}], TraditionalForm]]], ". Tenemos que " }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{ RowBox[{\(\[Rho]'\), \((t)\), RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{\(\[Rho](t)\), RowBox[{ StyleBox[\(e\_\[Rho]\%\(\ \[VerticalLine]\)\), FontWeight->"Bold"], "(", "t", ")"}]}]}], "=", RowBox[{ RowBox[{\(\[Rho]'\), \((t)\), RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \(\(\[Rho](t)\) \[Theta]\ ' \((t)\)\)}]}]}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "que es la expresi\[OAcute]n de la velocidad en coordenadas polares. \ Derivando la expresi\[OAcute]n anterior y teniendo en cuenta que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_\[Theta]\%\(\ \[VerticalLine]\)\), FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ RowBox[{"-", RowBox[{ SuperscriptBox["\[Theta]", RowBox[{" ", StyleBox["\[VerticalLine]", FontWeight->"Bold"]}]], "(", "t", ")"}]}], RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}]}], TraditionalForm]]], " puedes comprobar que la aceleraci\[OAcute]n viene dada por:" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{ \((\(\[Rho]'\)' \((t)\) - \(\[Rho](t)\) \((\[Theta]\ ' \((t)\))\)\^2)\), RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \((2 \[Rho]' \((t)\) \[Theta]\ '\ \((t)\) + \(\[Rho](t)\) \(\[Theta]\ '\)\ ' \((t)\))\)}]}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ StyleBox["Nota", FontWeight->"Bold"], ". En algunos libros de f\[IAcute]sica se usa la notaci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "=", StyleBox[\(\[Rho]\&^\), FontWeight->"Bold"]}], TraditionalForm]]], " (o ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "=", StyleBox[\(r\&^\), FontWeight->"Bold"]}], TraditionalForm]]], ") y ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "=", StyleBox[\(\[Theta]\&^\), FontWeight->"Bold"]}], TraditionalForm]]], " y se les da a estos vectores el extra\[NTilde]o nombre de ", StyleBox["versores. ", FontSlant->"Italic"], "Todav\[IAcute]a nadie me ha sabido explicar qu\[EAcute] es un versor. " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Expresi\[OAcute]n de la divergencia en cordenadas polares", "Subsection"], Cell[TextData[{ "Sea ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(x, y\), ")"}], "=", \((\(f\_1\)(x, y), \(f\_2\)(x, y))\)}], " "}], TraditionalForm]]], " un campo vectorial de dos variables. La divergencia de este campo es el \ campo escalar dado por ", Cell[BoxData[ FormBox[ RowBox[{"div", FormBox[ RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(x, y\), ")"}], "=", \(\(\[PartialD]f\_1\/\[PartialD]x\) \((x, y)\) + \(\[PartialD]f\_2\/\[PartialD]y\) \((x, y)\)\)}], "TraditionalForm"]}], TraditionalForm]]], ". Consideremos la expresi\[OAcute]n de ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], " en coordenadas polares:" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(\[Rho], \[Theta]\), ")"}], "=", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta]\), ")"}], "=", \((\(f\_1\)(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta]), \(f\_2\)(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta]))\)}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "y calculemos las componentes ", Cell[BoxData[ \(TraditionalForm\`\(F\_\[Rho]\)(\[Rho], \[Theta])\)]], " y ", Cell[BoxData[ \(TraditionalForm\`\(\(F\_\[Theta]\)(\[Rho], \[Theta])\ \)\)]], " de ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(\[Rho], \[Theta]\), ")"}], TraditionalForm]]], " respecto de la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ", esto es, ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(\[Rho], \[Theta]\), ")"}], "=", RowBox[{ RowBox[{\(\(F\_\[Rho]\)(\[Rho], \[Theta])\), StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}], "+", RowBox[{\(\(F\_\[Theta]\)(\[Rho], \[Theta])\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". Sabemos que dichas componentes viene dadas por las correspondientes \ proyecciones ortogonales:" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(F\_\(\(\[Rho]\)\(\ \)\)\)(\[Rho], \[Theta]) \ = \[LeftAngleBracket]\), "TraditionalForm"], RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(\[Rho], \ \ \ \[Theta]\), ")"}]}], "|", StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}], StyleBox["\[RightAngleBracket]", FontWeight->"Bold"]}], "="}], "TraditionalForm"], FormBox[\(\(f\_1\)(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), "TraditionalForm"], "cos", " ", "\[Theta]"}], "+", \(\(\(f\_2\)(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\) sen\ \[Theta]\)}], TraditionalForm]]], "\n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(F\_\(\(\[Theta]\)\(\ \)\)\)(\[Rho], \[Theta]) \ = \[LeftAngleBracket]\), "TraditionalForm"], RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(\[Rho], \ \ \ \[Theta]\), ")"}]}], "|", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "\[RightAngleBracket]"}], "="}], "TraditionalForm"], "-", RowBox[{ FormBox[\(\(f\_1\)(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), "TraditionalForm"], "sen", " ", "\[Theta]"}], "+", \(\(\(f\_2\)(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\) cos\ \[Theta]\)}], TraditionalForm]]] }], "Text", TextAlignment->Center], Cell[TextData[{ "Para obtener la expresi\[OAcute]n de la divergencia de ", StyleBox["f ", FontSlant->"Italic"], " en coordenadas polares debemos expresar la igualdad ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[ RowBox[{"div", StyleBox["f", FontWeight->"Bold", FontSlant-> "Italic"]}]], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\)}], "=", RowBox[{ RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\)}], "+", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\)}]}]}], TextForm]]], " en t\[EAcute]rminos de las funciones ", Cell[BoxData[ \(TraditionalForm\`F\_\[Rho], \ F\_\[Theta]\)]], " y de sus derivadas parciales. Hay dos formas de hacer esto. " }], "Text"], Cell["De forma indirecta", "Text", FontWeight->"Bold"], Cell["\<\ Derivamos en las igualdades anteriores haciendo uso de la regla de la cadena \ (derivaci\[OAcute]n de una funci\[OAcute]n compuesta). Tenemos que:\ \>", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}], \(\[PartialD]\[Rho]\)], \((\[Rho], \[Theta])\)}], "=", RowBox[{ RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), \(cos\^2\), " ", "\[Theta]"}], " ", "+", RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), "cos", " ", "\[Theta]", " ", "sen", " ", "\[Theta]"}], "+", "\[IndentingNewLine]", " ", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), "cos", " ", "\[Theta]", " ", "sen", " ", "\[Theta]"}], "+", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), \(sen\^2\), " ", "\[Theta]"}]}]}], TextForm]], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Theta]"]}], \(\[PartialD]\[Theta]\)], \((\[Rho], \[Theta])\ \)}], "=", RowBox[{ RowBox[{ RowBox[{"-", FormBox[\(\[PartialD]f\_1\/\[PartialD]x\), "TraditionalForm"]}], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), \((\(-\[Rho]\)\ sen\ \[Theta])\), "sen", " ", "\[Theta]"}], " ", "-", RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), "\[Rho]", " ", "cos", " ", "\[Theta]", " ", "sen", " ", "\[Theta]"}], "-", RowBox[{ SubscriptBox[ StyleBox["f", FontSlant->"Italic"], "1"], \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\), "cos", " ", "\[Theta]"}], "+", "\[IndentingNewLine]", " ", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), \((\(-\[Rho]\)\ sen\ \[Theta])\), " ", "cos", " ", "\[Theta]"}], "+", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\), "\[Rho]", " ", \(cos\^2\), " ", "\[Theta]"}], "-", RowBox[{ SubscriptBox[ StyleBox["f", FontSlant->"Italic"], "2"], \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\), "sen", " ", "\[Theta]"}]}]}], TextForm]], "Text"], Cell[TextData[ "Sumando estas igualdades, despu\[EAcute]s de dividir por \[Rho] la segunda \ de ellas, se deduce que"], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}], \(\[PartialD]\[Rho]\)], \((\[Rho], \[Theta])\)}], "+", RowBox[{\(1\/\[Rho]\), FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Theta]"]}], \(\[PartialD]\[Theta]\)], \((\[Rho], \[Theta])\)}]}], "=", RowBox[{ RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)}], "+", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)}], "-", RowBox[{\(1\/\[Rho]\), SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"], \((\[Rho], \[Theta])\)}]}]}], TextForm]], "Text", TextAlignment->Center], Cell[" es decir", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[ RowBox[{"div", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"]}]], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)}], "=", RowBox[{ RowBox[{ RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)}], "+", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)}]}], "=", RowBox[{ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}], \(\[PartialD]\[Rho]\)], \((\[Rho], \[Theta])\)}], "+", RowBox[{\(1\/\[Rho]\), FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Theta]"]}], \(\[PartialD]\[Theta]\)], \((\[Rho], \[Theta])\)}], "+", RowBox[{\(1\/\[Rho]\), SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"], \((\[Rho], \[Theta])\)}]}]}]}], TextForm]], "Text", TextAlignment->Center], Cell["igualdad que suele escribirse en la forma", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"div", " ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"]}], "=", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}], \(\[PartialD]\[Rho]\)], "+", RowBox[{\(1\/\[Rho]\), FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Theta]"]}], \(\[PartialD]\[Theta]\)]}], "+", RowBox[{\(1\/\[Rho]\), SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}]}]}], TextForm]], "Text", TextAlignment->Center], Cell["e incluso m\[AAcute]s condesado, en la forma", "Text"], Cell[BoxData[ FormBox[ FrameBox[ RowBox[{ RowBox[{"div", " ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"]}], "=", RowBox[{ RowBox[{\(1\/\[Rho]\), \(\[PartialD]\/\[PartialD]\[Rho]\), RowBox[{"(", RowBox[{"\[Rho]", " ", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}], ")"}]}], "+", RowBox[{\(1\/\[Rho]\), FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Theta]"]}], \(\[PartialD]\[Theta]\)]}]}]}]], TextForm]], "Text", TextAlignment->Center], Cell[TextData[{ "y se dice que dicha expresi\[OAcute]n proporciona la divergencia de ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], " en polares. " }], "Text"], Cell["De forma directa", "Text", FontWeight->"Bold"], Cell[TextData[{ "El camino que hemos seguido antes para obtener la expresi\[OAcute]n de la \ divergencia en coordenadas polares es un ", StyleBox["camino indirecto", FontSlant->"Italic"], ", pues hemos calculado las derivadas ", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Rho]"]}], \(\[PartialD]\[Rho]\)], TextForm]]], " y ", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["F", FontSlant->"Italic"], "\[Theta]"]}], \(\[PartialD]\[Theta]\)], TextForm]]], " y, al hacerlo, ", StyleBox["nos hemos dado cuenta", FontSlant->"Italic"], " de que pod\[IAcute]amos hacer una operaci\[OAcute]n sencilla con ellas \ para relacionarlas con ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(\[PartialD]f\_1\/\[PartialD]x\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\)}], "+", RowBox[{ FormBox[\(\[PartialD]f\_2\/\[PartialD]y\), "TraditionalForm"], \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\)}]}], TextForm]]], ". Un camino m\[AAcute]s natural consiste en expresar ", Cell[BoxData[ \(TraditionalForm\`\(f\_1\)(x, y)\)]], " y ", Cell[BoxData[ \(TraditionalForm\`\(f\_2\)(x, y)\)]], " en funci\[OAcute]n de ", Cell[BoxData[ \(TraditionalForm\`\(F\_\[Rho]\)(\[Rho], \[Theta])\)]], " y ", Cell[BoxData[ \(TraditionalForm\`\(F\_\[Theta]\)(\[Rho], \[Theta])\)]], "; es decir, se trata de obtener las coordenadas cartesianas ", Cell[BoxData[ \(TraditionalForm\`\((\(f\_1\)(x, y), \(f\_2\)(x, y))\)\)]], " del vector", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(x, y\), ")"}]}], TraditionalForm]]], " en funci\[OAcute]n de las coordenadas de dicho vector respecto de la base \ ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ". Basta para ello tener en cuenta que: " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{\((\(f\_1\)(x, y), \(f\_2\)(x, y))\), "=", RowBox[{ RowBox[{ RowBox[{\(\(F\_\[Rho]\)(\[Rho], \[Theta])\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], StyleBox["\[Rho]", FontWeight->"Bold"]]}], "+", RowBox[{\(\(F\_\[Theta]\)(\[Rho], \[Theta])\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], StyleBox["\[Theta]", FontWeight->"Bold"]]}]}], "=", RowBox[{ RowBox[{"(", GridBox[{ {\(cos\ \[Theta]\), \(\(-sen\)\ \[Theta]\)}, {\(sen\ \[Theta]\), \(cos\ \[Theta]\)} }], ")"}], RowBox[{"(", GridBox[{ {\(\(F\_\[Rho]\)(\[Rho], \[Theta])\)}, {\(\(F\_\[Theta]\)(\[Rho], \[Theta])\)} }], ")"}], " "}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Es decir, la matriz cuyas columnas son los vectores ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\(\(\[Rho]\)\(\ \)\)\), FontWeight->"Bold"], TraditionalForm]]], "y ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], TraditionalForm]]], " es justamente la matiz del cambio de base que necesitamos. Naturalmente, \ en esta igualdad debemos ver a \[Rho] y a \[Theta] como funciones de ", Cell[BoxData[ \(TraditionalForm\`x\)]], " e ", Cell[BoxData[ \(TraditionalForm\`y\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Rho](x, y) = \@\(x\^2 + y\^2\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Theta](x, y) = ArcTan\ [x, y]\)]], ". Vamos a pedirle ayuda a ", StyleBox["Mathematica ", FontSlant->"Italic"], "para hacer los c\[AAcute]lculos. " }], "Text"], Cell[BoxData[{ \(\(\[Rho][x_, y_] := \@\(x\^2 + y\^2\);\)\), "\n", \(\(\[Theta][x_, y_] := ArcTan[x, y];\)\), "\n", \(\(A = Transpose[{{Cos[\[Theta][x, y]], Sin[\[Theta][x, y]]}, {\(-Sin[\[Theta][x, y]]\), Cos[\[Theta][x, y]]}}];\)\), "\n", \(f1[x_, y_] = \((A . {F\[Rho][\[Rho][x, y], \[Theta][x, y]], F\[Theta][\[Rho][x, y], \[Theta][x, y]]})\)[\([1]\)]\), "\n", \(f2[x_, y_] = \((A . {F\[Rho][\[Rho][x, y], \[Theta][x, y]], F\[Theta][\[Rho][x, y], \[Theta][x, y]]})\)[\([2]\)]\)}], "Input"], Cell[TextData[{ "Una vez que disponemos de ", Cell[BoxData[ \(TraditionalForm\`\(f\_1\)(x, y)\)]], " y ", Cell[BoxData[ \(TraditionalForm\`\(f\_2\)(x, y)\)]], " en funci\[OAcute]n de ", Cell[BoxData[ \(TraditionalForm \`\(F\_\[Rho]\)(\@\(x\^2 + y\^2\), \ ArcTan[x, y])\)]], " y de ", Cell[BoxData[ \(TraditionalForm \`\(F\_\[Theta]\)(\@\(x\^2 + y\^2\), \ ArcTan[x, y])\)]], ", todo lo que tenemos que hacer es calcular las derivadas parciales ", Cell[BoxData[ \(TraditionalForm\`\(\[PartialD]f\_1\/\[PartialD]x\) \((x, y)\), \(\[PartialD]f\_2\/\[PartialD]y\) \((x, y)\)\)]], " sumarlas, simplificar y sustituir ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], " por ", Cell[BoxData[ \(TraditionalForm \`\((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)\)]], ". Esto lo hace ", StyleBox["Mathematica", FontSlant->"Italic"], " siempre que le demos una ayudita para que simplifique bien las \ ra\[IAcute]ces cuadradas (por ejemplo para que ", Cell[BoxData[ \(TraditionalForm\`\@\[Rho]\^2\)]], "lo simplifique igual a \[Rho]) y para que sustituya ArcTan[\[Rho] Cos[\ \[Theta]], \[Rho] Sin[\[Theta]]] por \[Theta]. Esta ayudita la hemos incluido \ en las celdas de inicializaci\[OAcute]n." }], "Text"], Cell[BoxData[ \(\((\((D[f1[x, y], x] + D[f2[x, y], y] // Simplify)\) /. {x -> \[Rho]\ Cos[\[Theta]], y -> \[Rho]\ Sin[\[Theta]]})\) // Simplify\)], "Input"], Cell["Que como puedes ver es la misma expresi\[OAcute]n obtenida antes.", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Gradiente y elemento diferencial de longitud en coordenadas polares\ \>", "Subsection"], Cell[TextData[{ "Sea ", StyleBox["f", FontSlant->"Italic"], " un campo escalar de dos variables. Sabemos que el gradiente de ", StyleBox["f", FontSlant->"Italic"], " es el campo vectorial dado por ", Cell[BoxData[ FormBox[ RowBox[{\(\[Del]\(f(x, y)\)\), "=", RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]x\), \((x, y)\), StyleBox["i", FontWeight->"Bold"]}], "+", RowBox[{\(\[PartialD]f\/\[PartialD]y\), \((x, y)\), StyleBox["j", FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". Hagamos en esta igualdad ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\) = \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\)]], " para obtener" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ \), "=", RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]x\), \((\[Rho]\ cos\ \[Theta], \ \ \[Rho]\ sen\ \[Theta])\), StyleBox["i", FontWeight->"Bold"]}], "+", RowBox[{\(\[PartialD]f\/\[PartialD]y\), \((\[Rho]\ cos\ \[Theta], \ \ \[Rho]\ sen\ \[Theta])\), StyleBox["j", FontWeight->"Bold"]}]}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "La expresi\[OAcute]n del gradiente de ", StyleBox["f ", FontSlant->"Italic"], "en polares viene dada por ", Cell[BoxData[ FormBox[ RowBox[{\(\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ \), "=", RowBox[{ RowBox[{\(\(f\_\[Rho]\)(\[Rho], \[Theta])\), StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}], "+", RowBox[{\(\(f\_\[Theta]\)(\[Rho], \[Theta])\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}]}]}], TraditionalForm]]], StyleBox[" ", FontSlant->"Italic"], "donde ", Cell[BoxData[ \(TraditionalForm\`f\_\[Rho]\)]], " y ", Cell[BoxData[ \(TraditionalForm\`f\_\[Theta]\)]], " son las componentes del vector ", Cell[BoxData[ \(TraditionalForm\`\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \ \[Theta])\)\)]], " en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ". Dichas componentes sabemos que vienen dadas por" }], "Text"], Cell[BoxData[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(f\_\[Rho]\)(\[Rho], \[Theta]) = \[LeftAngleBracket]\), "TraditionalForm"], \(\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \[Rho]\ \ sen\ \[Theta])\)\), " ", FormBox[ RowBox[{"|", StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}], "TraditionalForm"]}], "\[RightAngleBracket]"}], "=", \(\(\[PartialD]f\/\[PartialD]x\) \((\[Rho]\ cos\ \[Theta], \ \ \[Rho]\ sen\ \[Theta])\)\ cos\ \[Theta] + \(\[PartialD]f\/\[PartialD]y\) \((\ \[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ sen\ \[Theta]\)}], TraditionalForm], "\n", FormBox[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(f\_\[Theta]\)(\[Rho], \[Theta]) = \ \[LeftAngleBracket]\), "TraditionalForm"], \(\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \ \[Rho]\ sen\ \[Theta])\)\), " ", FormBox[ RowBox[{"|", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "TraditionalForm"]}], "\[RightAngleBracket]"}], "=", \(\(-\(\[PartialD]f\/\[PartialD]x\)\) \((\[Rho]\ cos\ \ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ sen\ \[Theta] + \(\[PartialD]f\/\ \[PartialD]y\) \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ cos\ \ \[Theta]\)}], "TraditionalForm"], TraditionalForm]}], "Text", TextAlignment->Center], Cell["que podemos escribir como sigue", "Text"], Cell[BoxData[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(f\_\[Rho]\)(\[Rho], \[Theta])\)\(=\)\), "TraditionalForm"], \(\[PartialD]f\/\[PartialD]x\), \((\[Rho]\ \ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\), \(\[PartialD]\((\[Rho]\ \ cos\ \ \[Theta])\)\/\[PartialD]\[Rho]\)}], "+", \(\(\[PartialD]f\/\[PartialD]y\) \((\[Rho]\ cos\ \[Theta], \ \ \[Rho]\ sen\ \[Theta])\)\ \[PartialD]\((\[Rho]\ \ sen\ \[Theta])\)\/\ \[PartialD]\[Rho]\)}], "=", \(\(\[PartialD]\/\[PartialD]\[Rho]\) f \((\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)\)}], TraditionalForm], "\n", FormBox[ RowBox[{\(\(f\_\[Theta]\) \((\[Rho], \[Theta])\)\), "=", RowBox[{\(1\/\[Rho]\), FormBox[\(\((\( \(\[PartialD]f\/\[PartialD]x\)\) \((\[Rho]\ cos\ \ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ \[PartialD]\((\[Rho]\ \ cos\ \ \[Theta])\)\/\[PartialD]\[Theta] + \(\[PartialD]f\/\[PartialD]y\) \((\[Rho]\ \ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\ \[PartialD]\((\[Rho]\ \ sen\ \ \[Theta])\)\/\[PartialD]\[Theta])\) = \(1\/\[Rho]\) \ \(\[PartialD]\/\[PartialD]\[Theta]\) \(f(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \ \[Theta])\)\), "TraditionalForm"]}]}], TraditionalForm]}], "Text", TextAlignment->Center], Cell["Hemos obtenido as\[IAcute] que", "Text"], Cell[BoxData[ FormBox[ RowBox[{\(\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\), "=", RowBox[{ RowBox[{\(\[PartialD]\/\[PartialD]\[Rho]\), \(f(\[Rho]\ cos\ \ \[Theta], \[Rho]\ sen\ \[Theta])\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Rho]"]}], "+", RowBox[{\(1\/\[Rho]\), \(\[PartialD]\/\[PartialD]\[Theta]\), \(f(\ \[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Theta]"], " ", \((1)\), " "}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Es conveniente introducir la funci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`h(\[Rho], \[Theta]) = f(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)]], " con lo que la igualdad (1) se escribe mejor en la forma" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{\(\[Del]\(f(\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\), "=", RowBox[{ RowBox[{\(\[PartialD]h\/\[PartialD]\[Rho]\), \((\[Rho]\ , \ \ \[Theta])\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Rho]"]}], "+", RowBox[{\(1\/\[Rho]\), \(\[PartialD]h\/\[PartialD]\[Theta]\), \((\ \[Rho], \[Theta])\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Theta]"], " ", \((2)\), " "}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Observa que en esta igualdad a la izquierda tenemos el gradiente de ", StyleBox["f", FontSlant->"Italic"], " calculado en la expresi\[OAcute]n de ", StyleBox["f", FontSlant->"Italic"], " en coordenadas cartesianas y evaluado en el punto (\[Rho]", " cos \[Theta], \[Rho] sen \[Theta]), y a la derecha lo que tenemos son las \ derivadas parciales de la funci\[OAcute]n compuesta ", Cell[BoxData[ \(TraditionalForm\`h(\[Rho], \[Theta]) = f(\[Rho]\ cos\ \[Theta], \[Rho]\ sen\ \[Theta])\)]], " evaluadas en el punto (\[Rho],\[Theta]). En los textos de f\[IAcute]sica \ es frecuente que no se distinga entre la funci\[OAcute]n ", StyleBox["f", FontSlant->"Italic"], " y la funci\[OAcute]n ", StyleBox["h", FontSlant->"Italic"], " (pues, en definitiva, son la ", StyleBox["misma funci\[OAcute]n ", FontSlant->"Italic"], "expresada en ", StyleBox["distintas", FontSlant->"Italic"], " coordenadas) y que escriban la igualdad (1) en la forma" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ FrameBox[ RowBox[{\(\[Del]f\), "=", RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]\[Rho]\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Rho]"]}], "+", RowBox[{\(1\/\[Rho]\), \(\[PartialD]f\/\[PartialD]\[Theta]\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], \(\(\[Theta]\)\(\ \)\)], " "}]}]}]], " ", \((3)\), " "}], TraditionalForm]], "Text", TextAlignment->Center], Cell["\<\ igualdad que constituye la \"expresi\[OAcute]n del gradiente en polares\". \ \>", "Text"], Cell[TextData[{ "Observa que en la expresi\[OAcute]n anterior del gradiente aparecen los \ inversos de los factores de escala multiplicando a las derivadas parciales a \ las que est\[AAcute] asociado cada uno de ellos. Como sabes, los factores de \ escala son ", Cell[BoxData[ \(TraditionalForm\`{1, \[Rho]}\)]], "; el primero de ellos, 1, est\[AAcute] asociado a la primera columna de la \ matriz jacobiana del cambio de coordenadas que corresponde a la derivaci\ \[OAcute]n parcial respecto a la primera variable, \[Rho]; el segundo de \ ellos, \[Rho], est\[AAcute] asociado a la segunda columna de la matriz \ jacobiana del cambio de coordenadas que corresponde a la derivaci\[OAcute]n \ parcial respecto a la segunda variable, \[Theta]. " }], "Text"], Cell["\<\ Hemos visto antes que la velocidad en coordenadas polares se expresa por\ \>", "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{\(\[Rho]'\), \((t)\), RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \(\(\[Rho](t)\) \[Theta]\ ' \((t)\)\)}]}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], TraditionalForm]]], "\t\t(4)" }], "Text", TextAlignment->Center], Cell["\<\ Esta igualdad suele escribirse con notaci\[OAcute]n m\[AAcute]s \ cl\[AAcute]sica en la forma\ \>", "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"\[DifferentialD]", FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "=", RowBox[{ RowBox[{\(\[DifferentialD]\[Rho]\), " ", StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}], "+", "\[Rho]"}]}], "TraditionalForm"]}], \(\[DifferentialD]\[Theta]\), " ", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], TraditionalForm]]], "\t\t" }], "Text", TextAlignment->Center], Cell[TextData[{ "Imaginemos ahora que ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Plain"], StyleBox["es", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["la", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["funci\[OAcute]n", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["de", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["trayectoria", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["de", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["un", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["m\[OAcute]vil", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox[Cell["y"], FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"]}], TraditionalForm]]], "que ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[" ", FontWeight->"Plain"], Cell[TextData[Cell[BoxData[ FormBox[Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "a", ")"}], TraditionalForm]]]]], TraditionalForm]]]]]}], TraditionalForm]]], "es su punto inicial; entonces la distancia, ", Cell[BoxData[ \(TraditionalForm\`s(t)\)]], ", recorrida por el m\[OAcute]vil en cada momento ", Cell[BoxData[ \(TraditionalForm\`t\)]], " viene dada por " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{\(s \((t)\)\), "=", RowBox[{ RowBox[{\(\[Integral]\_a\%t\), RowBox[{"||", RowBox[{ StyleBox[\(r'\), FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold"], \((t)\)}], "||", \(\[DifferentialD]t\)}]}], "=", RowBox[{ RowBox[{\(\[Integral]\_a\%t\), RowBox[{"||", RowBox[{ FormBox[ RowBox[{ RowBox[{\(\[Rho]'\), \((t)\), RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \(\(\[Rho](t)\) \[Theta]\ ' \((t)\)\)}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "||", \(\[DifferentialD]t\)}]}], "=", \(\[Integral]\_a\%t\(\@\(\((\[Rho]' \((t)\))\)\^2 + \((\(\ \[Rho](t)\) \[Theta]\ ' \((t)\))\)\^2\)\) \(\(\[DifferentialD]\)\(t\)\(\ \ \ \ \ \ \ \ \ \ \)\)\)}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Y, por tanto, ", Cell[BoxData[ \(TraditionalForm\`s' \((t)\) = \@\(\((\[Rho]' \((t)\))\)\^2 + \((\(\ \[Rho](t)\) \[Theta]\ ' \((t)\))\)\^2\)\)]], ". Esta igualdad suele escribirse en la forma" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((\[DifferentialD]s)\)\^2 = \((\[DifferentialD]\ \[Rho])\)\^2 + \(\(\[Rho]\^2\)(\[DifferentialD]\[Theta]\ )\)\^2\)]], "\t" }], "Text", TextAlignment->Center], Cell[TextData[{ "y se llama ", StyleBox["elemento diferencial de longitud", FontSlant->"Italic"], " en coordenadas polares. Observa que aqu\[IAcute] aparecen los factores de \ escala 1 y \[Rho] elevados al cuadrado y multiplicando a las correspondientes \ \"diferenciales\" \[DifferentialD]\[Rho] y \[DifferentialD]\[Theta]. " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Significado de los factores de escala", "Subsection"], Cell["\<\ Consideremos la matriz jacobiana de la funci\[OAcute]n que introduce las \ coordenadas polares.\ \>", "Text"], Cell[BoxData[ \(A = \(matrizjacobiana2[g]\)[\[Rho], \[Theta]]\)], "Input"], Cell[TextData[{ "Esta matriz define una aplicaci\[OAcute]n lineal de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^2\)]], " en ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^2\)]], " que a cada vector ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], " hace corresponder el vector ", Cell[BoxData[ \(TraditionalForm\`A . \((x, y)\)\)]], ". Calculemos la norma eucl\[IAcute]dea de la imagen de un vector en dicha \ transformaci\[OAcute]n." }], "Text"], Cell[BoxData[ \(\@\(\((A . {x, y})\) . \((A . {x, y})\)\) // Simplify\)], "Input"], Cell[TextData[{ "Deducimos que para vectores situados a lo largo del eje de abscisas, es \ decir, de la forma ", Cell[BoxData[ \(TraditionalForm\`\((x, 0)\)\)]], ", se verifica que ", Cell[BoxData[ \(TraditionalForm\`\(\(\(||\)\(A . \((x, 0)\)\)\)\(||\)\) = \(\(||\)\((x, 0)\)\(||\)\)\)]], " y, teniendo en cuenta la linealidad, se sigue que ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(||\)\(A . \((x, 0)\)\)\) - A . \((z, 0)\)\)\(||\)\) = \(\(\(||\)\(\((x, 0)\) - \((z, 0)\)\)\)\(||\)\)\)]], ", esto es, la aplicaci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\) \[Rule] A . \((x, y)\)\)]], " conserva distancias en el eje X. Pues bien, este es el significado de que \ el factor de escala asociado a la primera variable sea igual a 1.\nDeducimos \ tambi\[EAcute]n que para vectores situados a lo largo del eje de ordenadas, \ es decir, de la forma ", Cell[BoxData[ \(TraditionalForm\`\((0, y)\)\)]], ", se verifica que ", Cell[BoxData[ \(TraditionalForm\`\(\(\(||\)\(A . \((0, y)\)\)\)\(||\)\) = \[Rho] || \((0, y)\) || \)]], " y, teniendo en cuenta la linealidad, se sigue que ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(||\)\(A . \((0, y)\)\)\) - A . \((0, w)\)\)\(||\)\) = \[Rho] || \((0, y)\) - \((0, w)\) || \)]], ", esto es, la aplicaci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\) \[Rule] A . \((x, y)\)\)]], " multiplica distancias por \[Rho] en el eje Y. Pues bien, este es el \ significado de que el factor de escala asociado a la segunda variable sea \ igual a \[Rho].\nEn resumen, los factores de escala indican las dilataciones \ a lo largo de los ejes que hace la aplicaci\[OAcute]n lineal asociada a la \ matriz jacobiana de la aplicaci\[OAcute]n", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(g(\[Rho], \[Theta]) = \((\[Rho]\ cos\ \ \[Theta], \ \[Rho]\ sen\ \[Theta])\)\)\)\)]], ". Suele decirse que la aplicaci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " es la que, a ", StyleBox["escala infinitesimal", FontSlant->"Italic"], ", produce esas dilataciones. La expresi\[OAcute]n \"escala infinitesimal\" \ se entiende de la siguiente forma. Supongamos que fijamos valores ", Cell[BoxData[ \(TraditionalForm\`\[Rho] = \[Rho]\_0\)]], " y ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Theta]\_0\)]], " y sea \[Delta] un n\[UAcute]mero \"muy peque\[NTilde]o\" (lo que en los \ siglos XVII y XVIII se llamaba un \"infinit\[EAcute]simo\"; terminolog\ \[IAcute]a precient\[IAcute]fica que todav\[IAcute]a usan algunos textos de f\ \[IAcute]sica). Entonces, por la definici\[OAcute]n de derivada, tenemos que" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(\(\(||\)\(g(\[Rho]\_0 + \[Delta], \[Theta]\_0) - g(\[Rho]\_0, \ \[Theta]\_0)\)\(||\)\(\(\[TildeEqual]\)\(\[Delta]\)\)\(||\)\(\(\[PartialD]g\/\ \[PartialD]\[Rho]\) \((\[Rho]\_0, \[Theta]\_0)\)\)\(||\)\) = \[Delta]\), " ", "\n", RowBox[{ RowBox[{"||", RowBox[{ RowBox[{ StyleBox["g", FontSlant->"Italic"], "(", \(\[Rho]\_0, \ \[Theta]\_0\ + \ \[Delta]\), ")"}], " ", "-", " ", RowBox[{ StyleBox["g", FontSlant->"Italic"], "(", \(\[Rho]\_0, \ \[Theta]\_0\), ")"}]}], "||", " ", \(\(\[TildeEqual]\)\(\ \)\(\[Delta]\)\), "||", \(\(\[PartialD]g\/\[PartialD]\[Theta]\) \((\[Rho]\_0, \ \ \[Theta]\_0)\)\), "||"}], " ", "=", " ", \(\[Delta]\[Rho]\_0\), " "}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "La primera igualdad nos dice que si efectuamos \"incrementos \ infinitesimales\" en la primera variable la aplicaci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " conserva distancias y la segunda igualdad nos dice que si efectuamos \ \"incrementos infinitesimales\" en la segunda variable la aplicaci\[OAcute]n \ ", StyleBox["g ", FontSlant->"Italic"], "multiplica", StyleBox[" ", FontSlant->"Italic"], "las distancias por el correspondiente valor de la primera variable." }], "Text"], Cell[TextData[{ "La expresi\[OAcute]n (5) del \"elemento diferencial de longitud\" en \ coordenadas polares tiene en cuenta dichos factores de escala. \n\ Naturalmente, para calcular la longitud de una curva dada por las ecuaciones \ ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", \((\(\[Rho](t)\) cos\ \(\[Theta](t)\), \ \(\[Rho](t)\) sen\ \(\[Theta](t)\))\)}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`a \[LessEqual] t \[LessEqual] b\)]], ", lo que se hace es integrar la rapidez con que dicha curva se recorre:" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"||", RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "||"}], "=", \(\@\(\((\[Rho]' \((t)\)\ cos\ \(\[Theta](t)\) - \(\[Rho](t)\)\ \[Theta]\ ' \((t)\)\ sen\ \(\[Theta](t)\)) \)\^2 + \((\[Rho]' \((t)\)\ sen\ \(\[Theta](t)\) + \(\[Rho](t)\)\ \[Theta]\ ' \((t)\)\ cos\ \(\[Theta](t)\)) \)\^2\ \ \) = \@\(\[Rho]' \((t)\)\^2 + \(\(\[Rho](t)\)\^2\) \[Theta]\ \ ' \((t)\)\^2\)\)}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "en consecuencia, la longitud de ", StyleBox["r ", FontWeight->"Bold", FontSlant->"Italic"], "viene dada por ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Integral]\_a\%b\), RowBox[{"||", RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "||", \(\[DifferentialD]t\)}]}], "=", \(\[Integral]\_a\%b \(\@\( \[Rho]' \((t)\)\^2 + \(\(\[Rho](t)\)\^2\) \[Theta]\ \ ' \((t)\)\^2\)\) \[DifferentialD]t\)}], TraditionalForm]]], ". ", "Observa que el valor obtenido para ", Cell[BoxData[ FormBox[ RowBox[{"||", RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "||"}], TraditionalForm]]], " pod\[IAcute]amos haberlo calculado directamente usando la igualdad (4)." }], "Text"], Cell[TextData[{ "Al igual que cada factor de escala mide la dilataci\[OAcute]n \ infinitesimal a lo largo de un eje, el producto de los factores de escala, en \ nuestro caso \[Rho], mide el cambio en el \[AAcute]rea de un \ rect\[AAcute]ngulo a escala infinitesimal. El producto de los factores de \ escala es justamente el determinante jacobiano. Recuerda que la \ f\[OAcute]rmula del cambio de variables a coordenadas polares en una integral \ doble afirma que si ", StyleBox["f", FontSlant->"Italic"], " es un campo escalar continuo en un conjunto ", Cell[BoxData[ \(TraditionalForm\`A \[Subset] \[DoubleStruckCapitalR]\^2\)]], " se verifica que\t\t" }], "Text"], Cell[BoxData[ FormBox[ FormBox[\(\[Integral]\(\[Integral]\_A\( f(x, y)\) \(d(x, y)\)\) = \[Integral]\(\[Integral]\_B\ \(\(f(\[Rho]\ cos\ \ \[Theta], \ \[Rho]\ sen\ \[Theta])\) \(\[Rho]\)\(\ \)\(d\) \((\[Rho], \ \[Theta])\)\(\ \)\)\)\), "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "donde", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{"B", "=", RowBox[{"{", RowBox[{\((\[Rho], \[Theta])\), ":", RowBox[{ FormBox[ \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta])\), "TraditionalForm"], "\[Element]", " ", "A"}]}], "}"}]}]}], TraditionalForm]]], ". Observa que ", StyleBox["B", FontSlant->"Italic"], " es la expresi\[OAcute]n del conjunto ", StyleBox["A", FontSlant->"Italic"], " en coordenadas polares, es decir ", Cell[BoxData[ \(TraditionalForm\`A = g(B)\)]], ", y que en la segunda integral se multiplica por \[Rho]. Si \ particularizamos la igualdad anterior para la funci\[OAcute]n constante ", Cell[BoxData[ \(TraditionalForm\`f(x, y) = 1\)]], " obtenemos" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(\[CapitalAAcute]rea \((g(B))\)\), "=", FormBox[\(\[Integral]\(\[Integral]\_A 1 \( d(x, y)\)\) = \[Integral]\(\[Integral]\_B\ \[Rho]\ \ \(d(\[Rho], \[Theta])\)\)\), "TraditionalForm"], " "}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Si ", StyleBox["B", FontSlant->"Italic"], " es un rect\[AAcute]ngulo muy peque\[NTilde]o (un rect\[AAcute]ngulo ", StyleBox["infinitesimal", FontSlant->"Italic"], ") se verifica que ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\(\[Integral]\_B\ \[Rho]\ \(d(\[Rho], \[Theta])\)\)\ \[TildeEqual] \ \[Rho] \(\[Integral]\(\[Integral]\_B\ d(\[Rho], \[Theta])\)\) = \[Rho]\ \(\[CapitalAAcute]rea(B)\)\)]], " y obtenemos ", Cell[BoxData[ \(TraditionalForm \`\[CapitalAAcute]rea(g(B))\ \[TildeEqual] \ \[Rho]\ \(\[CapitalAAcute]rea(B)\)\)]], ". En los libros de f\[IAcute]sica se dice que ", Cell[BoxData[ \(TraditionalForm \`\[Rho] \[DifferentialD]\[Rho] \[DifferentialD]\[Theta]\)]], " es el \"", StyleBox["elemento diferencial de \[AAcute]rea", FontSlant->"Italic"], "\" en coordenadas polares. " }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Coordenadas esf\[EAcute]ricas", "Section"], Cell[TextData[{ "La funci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`g( r, \[Theta], \[Phi]) = \((r\ sen\ \[Theta]\ cos\ \[Phi]\ , \ r\ sen\ \[Theta]\ sen\ \[Phi], r\ cos\ \[Theta])\)\)]], " es una biyecci\[OAcute]n de ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ RowBox[{"\[CapitalOmega]", "=", FormBox[\(\(\[DoubleStruckCapitalR]\^+\)\[Times]\([0, \ \[Pi]]\)\[Times]\), "TraditionalForm"]}], "]"}], "-", "\[Pi]"}], ",", "\[Pi]"}], "]"}], TraditionalForm]]], " sobre ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\\{\((0, 0, 0)\)}\)]], ". Los n\[UAcute]meros ", Cell[BoxData[ \(TraditionalForm\`\((r, \[Theta], \ \[Phi])\)\)]], " dados por ", Cell[BoxData[ \(TraditionalForm\`x = r\ sen\ \[Theta]\ cos\ \[Phi], \ y = \ r\ sen\ \[Theta]\ sen\ \[Phi], \ z = r\ cos\ \[Theta]\)]], " donde ", Cell[BoxData[ \(TraditionalForm\`r > 0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`0 \[LessEqual] \ \[Theta] \[LessEqual] \[Pi]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\) < \ \[Phi] \[LessEqual] \[Pi]\)]], ", se llaman las coordenadas esf\[EAcute]ricas del punto de cordenadas \ cartesianas ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], ". " }], "Text"], Cell[BoxData[ \(\(coordesfericas;\)\)], "Input"], Cell[TextData[{ StyleBox["Nota", FontWeight->"Bold"], ". La notaci\[OAcute]n y el orden en que se escriben las coordenadas esf\ \[EAcute]ricas var\[IAcute]a de unos textos a otros. Las definiciones \ anteriores son las mismas que se usan en ", StyleBox["Mathematica", FontSlant->"Italic"], " y eso nos permitir\[AAcute] aprovechar en lo que sigue los comandos del \ programa. En muchos textos los papeles de \[Phi] y \[Theta] est\[AAcute]n \ intercambiados con respecto a los nuestros. \nPor lo que se refiere al \ intervalo ", Cell[BoxData[ \(TraditionalForm\`\(\(\(] \(-\[Pi]\)\), \[Pi]]\)\ \)\)]], "elegido para medir en radianes el \[AAcute]ngulo \[Phi], podemos hacer las \ mismas observaciones que las hechas para las coordenadas polares. Con \ frecuencia dicho intervalo se sustituye por", Cell[BoxData[ \(TraditionalForm\`\(\ \([0, 2 \( \[Pi][\)\)\)\)]], " lo que, dicho sea de paso, complica las f\[OAcute]rmulas del cambio de \ cartesianas a esf\[EAcute]ricas. Cuando en un libro se usen coordenadas esf\ \[EAcute]ricas debes comprobar c\[OAcute]mo se definen dichas coordenadas." }], "Text"], Cell["\<\ El siguiente comando carga el paquete para an\[AAcute]lisis vectorial.\ \>", "Text"], Cell[BoxData[ \(<< Calculus`VectorAnalysis`\)], "Input"], Cell["Los siguientes comandos no precisan explicaci\[OAcute]n.", "Text"], Cell[BoxData[ \(\(?\ Spherical\)\)], "Input"], Cell[BoxData[ \(Coordinates[Spherical]\)], "Input"], Cell[BoxData[ \(CoordinateRanges[Spherical]\)], "Input"], Cell[BoxData[ \(\(\(CoordinatesToCartesian[{r, \[Theta], \[Phi]}, Spherical]\)\( (*\ paso\ de\ esf\[EAcute]ricas\ a\ cartesianas\ *) \)\(\ \)\)\)], "Input"], Cell[BoxData[ \(\(\(CoordinatesFromCartesian[{x, y, z}, Spherical]\)\( (*\ paso\ de\ cartesianas\ a\ esf\[EAcute]ricas\ *) \)\)\)], "Input"], Cell[TextData[{ "Cuando se utiliza el sistema de coordenadas esf\[EAcute]ricas los vectores \ se refieren a una base ortonormal ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], StyleBox[",", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold"], SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Phi]"]}], "}"}], TraditionalForm]]], " que se ha representado en la figura anterior (trasladada al punto ", StyleBox["P", FontSlant->"Italic"], "). En el lenguaje t\[IAcute]pico de los textos de f\[IAcute]sica se dice \ que el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_r\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta ", StyleBox["r", FontSlant->"Italic"], " manteniendo \[Theta] y \[Phi] constantes, el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta \[Theta] manteniendo ", StyleBox["r", FontSlant->"Italic"], " y \[Phi] constantes y el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta \[Phi] manteniendo ", StyleBox["r", FontSlant->"Italic"], " y \[Theta] constantes. En t\[EAcute]rminos matem\[AAcute]ticos, quiz\ \[AAcute]s m\[AAcute]s precisos, observa que el vector de posici\[OAcute]n \ del punto ", StyleBox["P", FontSlant->"Italic"], " de cordenadas esf\[EAcute]ricas ", Cell[BoxData[ \(TraditionalForm\`\((r, \[Theta], \ \[Phi])\)\)]], " es ", Cell[BoxData[ \(TraditionalForm\`g( r, \[Theta], \[Phi]) = \((r\ cos\ \[Phi]\ sen\ \[Theta], \ r\ sen\ \[Phi]\ sen\ \[Theta], r\ cos\ \[Theta])\)\)]], "; su variaci\[OAcute]n con respecto a ", StyleBox["r", FontSlant->"Italic"], " manteniendo \[Theta] y \[Phi] constantes es la derivada parcial respecto \ a ", StyleBox["r", FontSlant->"Italic"], ", su variaci\[OAcute]n con respecto a \[Theta] manteniendo ", StyleBox["r", FontSlant->"Italic"], " y \[Phi] constantes es la derivada parcial respecto a \[Theta] y su \ variaci\[OAcute]n con respecto a \[Phi] manteniendo ", StyleBox["r", FontSlant->"Italic"], " y \[Theta] constantes es la derivada parcial respecto a \[Phi]. " }], "Text"], Cell[BoxData[{ \(Clear[g]\), "\n", \(g[r_, \[Theta]_, \[Phi]_] := {r\ Cos[\[Phi]]\ Sin[\[Theta]], r\ Sin[\[Phi]]\ Sin[\[Theta]], r\ Cos[\[Theta]]}\), "\n", \(u1 = D[g[r, \[Theta], \[Phi]], r]\), "\n", \(u2 = D[g[r, \[Theta], \[Phi]], \[Theta]]\), "\n", \(u3 = D[g[r, \[Theta], \[Phi]], \[Phi]]\)}], "Input"], Cell["Observa que estos vectores son ortogonales", "Text"], Cell[BoxData[ \({u1 . u2, u1 . u3, u2 . u3} // Simplify\)], "Input"], Cell["\<\ Para obtener una base ortonormal a partir de ellos todo lo que tenemos que \ hacer es normalizarlos. Calculemos sus normas.\ \>", "Text"], Cell[BoxData[ \({norma[u1], norma[u2], norma[u3]} // Simplify\)], "Input"], Cell[TextData[{ "Deducimos que los vectores ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "=", \((cos\ \[Phi]\ sen\ \[Theta], sen\ \[Phi]\ sen\ \[Theta], cos\ \[Theta])\)}], ",", RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "=", \((cos\ \[Phi]\ cos\ \[Theta], sen\ \[Phi]\ cos\ \[Theta], \(-sen\)\ \[Theta])\)}], ",", " ", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "=", \((\(-sen\)\ \[Phi]\ , \ cos\ \[Phi], 0)\)}]}], TraditionalForm]]], " forman una base ortonormal. Es a dicha base a la que se refiere un vector \ cuando se usan coordenadas esf\[EAcute]ricas. Observa que los vectores de \ esta base dependen de la posici\[OAcute]n del punto, es decir, no se trata de \ una base fija. F\[IAcute]jate en que si ", Cell[BoxData[ \(TraditionalForm\`\(\((r, \[Theta], \[Phi])\)\ \)\)]], "son las coordenadas esf\[EAcute]ricas de un punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], ", se verifica que ", Cell[BoxData[ FormBox[ RowBox[{\((x, y, z)\), "=", RowBox[{"r", " ", StyleBox[\(e\_r\), FontWeight->"Bold"]}]}], TraditionalForm]]], ". En general, la expresi\[OAcute]n en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " de un vector ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["v", FontWeight->"Bold"], "=", \((x, y, z)\)}], TraditionalForm]]], " se obtiene por el m\[EAcute]todo usual calculando sus proyecciones \ ortogonales sobre los vectores de la base:" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ StyleBox[ \(v = \[LeftAngleBracket]v | e\_r\[RightAngleBracket] e\_r + \[LeftAngleBracket]v | e\_\[Theta]\[RightAngleBracket] e\_\[Theta] + \[LeftAngleBracket]v | e\_\[Phi]\[RightAngleBracket] e\_\[Phi]\), FontWeight->"Bold"], TraditionalForm]]], "\t\t" }], "Text", TextAlignment->Center], Cell[TextData[{ "Observa que si escribimos ", StyleBox["v", FontWeight->"Bold", FontSlant->"Italic"], " en la forma ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["v", FontWeight->"Bold"], "=", \((r\ sen\ \[Theta]\ cos\ \[Phi]\ , \ r\ sen\ \[Theta]\ sen\ \[Phi], r\ cos\ \[Theta])\)}], TraditionalForm]]], " entonces ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(\[LeftAngleBracket]v | e\_r\[RightAngleBracket]\), FontWeight->"Bold"], "=", "r"}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ \(\[LeftAngleBracket]v | e\_\[Theta]\[RightAngleBracket]\), FontWeight->"Bold"], "=", RowBox[{ StyleBox[ \(\[LeftAngleBracket]v | e\_\[Phi]\[RightAngleBracket]\), FontWeight->"Bold"], "=", "0"}]}], TraditionalForm]]], "." }], "Text"], Cell[TextData[{ "A continuaci\[OAcute]n se definen los vectores ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], TraditionalForm]]], " como funciones que son de \[Theta] y \[Phi]." }], "Text"], Cell[BoxData[{ \(er[\[Theta]_, \[Phi]_] = u1/norma[u1] // Simplify\), "\n", \(e\[Theta][\[Theta]_, \[Phi]_] = u2/norma[u2] // Simplify\), "\n", \(e\[Phi][\[Theta]_, \[Phi]_] = u3/norma[u3] // Simplify\)}], "Input"], Cell[TextData[{ "La siguiente gr\[AAcute]fica muestra la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " para ", Cell[BoxData[ \(TraditionalForm\`{\[Theta], \[Phi]} = {\[Pi]/4, \[Pi]/4}\)]], ". Para visualizarla mejor dicha base se ha desplazado al punto de \ coordenadas esf\[EAcute]ricas ", Cell[BoxData[ \(TraditionalForm\`{1, \[Pi]/4, \[Pi]/4}\)]], "." }], "Text"], Cell[BoxData[{ \(\(punto = CoordinatesToCartesian[{1, \[Pi]/4, \[Pi]/4}, Spherical];\)\), "\n", \(\(graf = Show[Graphics3D[{vector3D[punto], vector3D[punto, .5*er[\[Pi]/4, \[Pi]/4], Blue], vector3D[punto, .5*e\[Theta][\[Pi]/4, \[Pi]/4], Green], vector3D[punto, .7*e\[Phi][\[Pi]/4, \[Pi]/4], Red], Text[\*"\"\<\!\(e\_r\)\>\"", punto + .55*er[\[Pi]/4, \[Pi]/4]], Text[\*"\"\<\!\(e\_\[Theta]\)\>\"", punto + .6* e\[Theta][\[Pi]/4, \[Pi]/ 4]], \n\t\t\t\tText[\*"\"\<\!\(e\_\[Phi]\)\>\"", punto + .78*e\[Phi][\[Pi]/4, \[Pi]/4]]}, Axes -> True, PlotRange -> All, Ticks -> None, TextStyle -> {FontFamily -> "\", FontSize -> 14, FontWeight -> "\"}]];\)\)}], "Input"], Cell[BoxData[{ \(SpinShow[graf]\), "\n", \(\(SelectionMove[EvaluationNotebook[], All, GeneratedCell];\)\), "\n", \(FrontEndTokenExecute["\"]\), "\n", \(FrontEndTokenExecute["\"]\), "\n", \(Clear[graf, punto]\)}], "Input"], Cell[TextData[{ "Recuerda que la matriz jacobiana de una funci\[OAcute]n de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], " en ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], ", ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(x, y, z\), ")"}], "=", \((\(f\_1\)(x, y, z), \(f\_2\)(x, y, z), \(f\_3\)(x, y, z))\)}], " "}], TraditionalForm]]], "es la matriz cuyas filas son los vectores gradiente de las funciones \ componentes de ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], " . El siguiente comando calcula la matriz jacobiana de un campo vectorial \ de tres variables." }], "Text"], Cell[BoxData[ \(\(matrizjacobiana3[func_]\)[x_, y_, z_] := Module[{u, v, w}, Outer[D, func[u, v, w], {u, v, w}] /. {u -> x, v -> y, w -> z}]\)], "Input"], Cell[BoxData[ \(\(matrizjacobiana3[g]\)[r, \[Theta], \[Phi]] // MatrixForm\)], "Input"], Cell[TextData[{ "Las columnas de la matriz jacobiana son las derivadas parciales ", Cell[BoxData[ \(TraditionalForm\`\(u1 = D[g[r, \[Theta], \[Phi]], r]\ \)\)]], ",", Cell[BoxData[ \(TraditionalForm\`\(\ \ u2 = D[g[r, \[Theta], \[Phi]], \[Phi]]\)\)]], " y ", Cell[BoxData[ \(TraditionalForm\`u3 = D[g[r, \[Theta], \[Phi]], \[Theta]]\)]], ". Las normas eucl\[IAcute]deas de las columnas de la matriz jacobiana se \ llaman ", StyleBox["factores m\[EAcute]tricos ", FontWeight->"Bold"], "o ", StyleBox["factores de escala", FontWeight->"Bold"], " del cambio a cordenadas esf\[EAcute]ricas y son ", Cell[BoxData[ \(TraditionalForm\`{1, r, \ r\ sen\ \[Theta]}\)]], ". " }], "Text"], Cell[CellGroupData[{ Cell["\<\ Expresi\[OAcute]n de la velocidad y la aceleraci\[OAcute]n en cordenadas esf\ \[EAcute]ricas\ \>", "Subsection"], Cell[TextData[{ "Consideremos un m\[OAcute]vil cuya trayectoria en el espacio viene dada \ por la funci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ RowBox[{\(x(t)\), StyleBox["i", FontWeight->"Bold"]}], "+", RowBox[{\(y(t)\), StyleBox["j", FontWeight->"Bold"]}], "+", RowBox[{\(z(t)\), StyleBox["k", FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". Sean ", Cell[BoxData[ \(TraditionalForm\`\((r(t), \ \[Theta](t), \[Phi](t))\)\)]], " las coordenadas esf\[EAcute]ricas de ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], TraditionalForm]]], " de forma que" }], "Text"], Cell[TextData[{ " ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ \((\(r(t)\) cos\ \(\[Phi](t)\) sen\ \(\[Theta](t)\), \(r(t)\) sen\ \(\[Phi](t)\) sen\ \(\[Theta](t)\), \(r(t)\)\ cos\ \(\[Theta](t)\))\), "=", RowBox[{\(r(t)\), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}]}]}], TraditionalForm]]] }], "Text", TextAlignment->Center], Cell[TextData[{ "donde ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}], "=", \((cos\ \(\[Phi](t)\) sen\ \(\[Theta](t)\), sen\ \(\[Phi](t)\) sen\ \(\[Theta](t)\), cos\ \(\[Theta](t)\)) \)}], TraditionalForm]]], ". Es f\[AAcute]cil comprobar que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_r\%\(\ \[VerticalLine]\)\), FontWeight->"Bold"], "(", "t", ")"}], "=", RowBox[{ RowBox[{\(\[Theta]\ '\), \((t)\), RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{\(\[Phi]'\), \((t)\), "sen", " ", \(\[Theta](t)\), RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "(", "t", ")"}]}]}]}], TraditionalForm]]], ". Podemos pedirle a ", StyleBox["Mathematica", FontSlant->"Italic"], " que lo haga." }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"(*", " ", RowBox[{ RowBox[{"definimos", " ", "los", " ", "vectores", " ", StyleBox[\(e\_r\), FontWeight->"Bold"], StyleBox[\((t)\), FontWeight->"Plain"]}], ",", RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}], ",", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], StyleBox[\((t)\), FontWeight->"Plain"]}]}], StyleBox[" ", FontWeight->"Plain"], StyleBox["*)", FontWeight->"Plain"]}], "\n", RowBox[{\(er[t_] = er[\[Theta][t], \[Phi][t]]\), ";", "\n", \(e\[Theta][t_] = e\[Theta][\[Theta][t], \[Phi][t]]\), ";", "\n", \(e\[Phi][t_] = e\[Phi][\[Theta][t], \[Phi][t]]\), ";", "\n", RowBox[{\(D[er[t], t]\), "==", RowBox[{ RowBox[{\(\(\[Theta]\ '\)[t]\), RowBox[{ StyleBox["e\[Theta]", FontWeight->"Bold"], "[", "t", "]"}]}], "+", RowBox[{\(\(\[Phi]'\)[t]\), \(Sin[\[Theta][t]]\), RowBox[{ StyleBox["e\[Phi]", FontWeight->"Bold"], "[", "t", "]"}]}]}]}]}], " ", \( (*\ comprobamos\ esta\ igualdad\ *) \)}]], "Input"], Cell["Deducimos que ", "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{ RowBox[{\(r'\), \((t)\), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{\(r(t)\), RowBox[{ StyleBox[\(e\_r\%\(\ \[VerticalLine]\)\), FontWeight->"Bold"], "(", "t", ")"}]}]}], "=", RowBox[{ RowBox[{\(r'\), \((t)\), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \(\(r(t)\) \[Theta]\ ' \((t)\)\)}]}]}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{ \(r(t)\), \(\[Phi]'\), \((t)\), "sen", " ", \(\[Theta](t)\), " ", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "(", "t", ")"}]}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "que es la expresi\[OAcute]n de la velocidad en coordenadas \ esf\[EAcute]ricas. Podemos pedirle a Mathematica que lo compruebe. Para ello \ calculamos las componentes del vector velocidad respecto de la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}], ",", RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}], ",", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}]}], "}"}], TraditionalForm]]], "." }], "Text"], Cell[BoxData[{ \(\(R[t_] := r[t] er[t];\)\), "\n", \({\(R'\)[t] . er[t], \(R'\)[t] . e\[Theta][t], \(R'\)[t] . e\[Phi][t]} // Simplify\)}], "Input"], Cell["\<\ Derivando la expresi\[OAcute]n anterior puedes comprobar que la aceleraci\ \[OAcute]n viene dada por:\ \>", "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{ \((\(r'\)' \((t)\) - \(r(t)\) \[Theta]\ ' \((t)\)\^2 - \(r(t)\) \[Phi]' \(\((t)\)\^2\) sen\^2\ \(\[Theta](t)\)) \), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \((2 r' \((t)\) \[Theta]\ '\ \((t)\) + \(r(t)\) \(\[Theta]\ '\)\ ' \((t)\) - \(r(t)\) \[Phi]' \(\((t)\)\^2\) sen\ \(\[Theta](t)\)\ cos \ \(\[Theta](t)\))\)}]}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{ \((2 r' \((t)\) \[Phi]' \((t)\)\ sen\ \(\[Theta](t)\) + 2 \( r(t)\) \[Phi]' \((t)\) \[Theta]\ ' \((t)\) cos\ \(\[Theta](t)\) + \(r(t)\) \(\[Phi]'\)' \((t)\) sen\ \(\[Theta](t)\))\), RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "(", "t", ")"}]}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "Volvamos a pedirle a Mathematica que lo compruebe calculando las \ componentes del vector aceleraci\[OAcute]n en la base ", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[ RowBox[{"{", RowBox[{ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}], ",", RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}], ",", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], StyleBox["(", FontWeight->"Plain"], StyleBox["t", FontWeight->"Plain"], StyleBox[")", FontWeight->"Plain"]}]}], "}"}], "TraditionalForm"]}], TraditionalForm]]], "." }], "Text"], Cell[BoxData[ \({\(\(R'\)'\)[t] . er[t], \(\(R'\)'\)[t] . e\[Theta][t], \(\(R'\)'\)[t] . e\[Phi][t]} // Simplify\)], "Input"], Cell[TextData[{ StyleBox["Nota", FontWeight->"Bold"], ". En algunos libros de f\[IAcute]sica se usa la notaci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "=", StyleBox[\(r\&^\), FontWeight->"Bold"]}], TraditionalForm]]], " , ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "=", StyleBox[\(\[Theta]\&^\), FontWeight->"Bold"]}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], StyleBox["=", FontWeight->"Bold"], OverscriptBox["\[Phi]", StyleBox["^", FontWeight->"Bold"]]}], TraditionalForm]]], ". " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Expresi\[OAcute]n de la divergencia en cordenadas esf\[EAcute]ricas", "Subsection"], Cell[TextData[{ "Sea ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(x, y, z\), ")"}], "=", \((\(f\_1\)(x, y, z), \(f\_2\)(x, y, z), \(f\_3\)(x, y, z))\)}], " "}], TraditionalForm]]], " un campo vectorial de tres variables. La divergencia de este campo es el \ campo escalar dado por ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"div", FormBox[ RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(x, y, z\), ")"}], "=", \(\(\[PartialD]f\_1\/\[PartialD]x\) \((x, y, z)\) + \(\[PartialD]f\_2\/\[PartialD]y\) \((x, y, z)\)\)}], "TraditionalForm"]}], "+", \(\(\[PartialD]f\_3\/\[PartialD]z\) \((x, y, z)\)\)}], TraditionalForm]]], ". Consideremos la expresi\[OAcute]n de ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], " en coordenadas esf\[EAcute]ricas:" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(r, \[Theta], \[Phi]\), ")"}], "=", RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(r\ sen\ \[Theta]\ cos\ \[Phi]\ , r\ sen\ \[Theta]\ sen\ \[Phi], r\ cos\ \[Theta]\), ")"}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "y sean ", Cell[BoxData[ \(TraditionalForm\`\(F\_r\)(r, \[Theta], \[Phi])\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(\(\(F\_\[Theta]\)( r, \[Theta], \[Phi])\)\(\ \)\)\)]], " y ", Cell[BoxData[ \(TraditionalForm\`\(F\_\[Phi]\)(r, \[Theta], \[Phi])\)]], " las componentes de ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(r, \[Theta], \[Phi]\), ")"}], TraditionalForm]]], " respecto de la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ", esto es, ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(r, \[Theta], \[Phi]\), ")"}], "=", RowBox[{ RowBox[{\(\(F\_r\)(r, \[Theta], \[Phi])\), StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", RowBox[{\(\(F\_\[Theta]\)(\[Rho], \[Theta], \[Phi])\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{\(\(F\_\[Phi]\)(r, \[Theta], \[Phi])\), StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". La expresi\[OAcute]n de la divergencia de ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" ", FontSlant->"Italic"], " en coordenadas esf\[EAcute]ricas consiste en expresar la igualdad" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"div", " ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"], RowBox[{"(", RowBox[{ StyleBox["r", FontSlant->"Italic"], ",", "\[Theta]", ",", "\[Phi]"}], ")"}]}], "=", RowBox[{ RowBox[{"div", " ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], RowBox[{"(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], "=", "\n", "\t", RowBox[{ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["f", FontSlant->"Italic"], "1"]}], RowBox[{"\[PartialD]", StyleBox["x", FontSlant->"Italic"]}]], RowBox[{"(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], "+", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["f", FontSlant->"Italic"], "2"]}], RowBox[{"\[PartialD]", StyleBox["y", FontSlant->"Italic"]}]], RowBox[{"(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], "+", "\[IndentingNewLine]", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", SubscriptBox[ StyleBox["f", FontSlant->"Italic"], "3"]}], RowBox[{"\[PartialD]", StyleBox["z", FontSlant->"Italic"]}]], RowBox[{"(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}]}]}]}], TextForm]], "Text", TextAlignment->Left], Cell[TextData[{ "en t\[EAcute]rminos de las funciones ", Cell[BoxData[ \(TraditionalForm\`\(F\_r\)(r, \[Theta], \[Phi]), \(F\_\[Theta]\)(r, \[Theta], \[Phi]), \(F\_\[Phi]\)(r, \[Theta], \[Phi])\)]], " y de sus derivadas parciales. El camino indirecto consistente en calcular \ derivadas parciales de ", Cell[BoxData[ \(TraditionalForm\`F\_r\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(F\_\[Theta]\ \)\)]], "y ", Cell[BoxData[ \(TraditionalForm\`F\_\[Phi]\)]], " para tratar de relacionarlas con ", Cell[BoxData[ FormBox[ RowBox[{"div", " ", RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(r\ , \ \[Theta], \ \[Phi]\), ")"}]}], TraditionalForm]]], " no se ve nada claro (puedes intentarlo para convencerte). Seguiremos el \ camino directo como hicimos para las coordenadas polares. Para ello \ expresaremos las componentes cartesianas del campo ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], " en funci\[OAcute]n de sus componentes ", Cell[BoxData[ \(TraditionalForm\`F\_r\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(F\_\[Theta]\ \)\)]], "y ", Cell[BoxData[ \(TraditionalForm\`F\_\[Phi]\)]], " en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ". La matriz del cambio de esta base a la base can\[OAcute]nica es la que \ tiene como columnas los vectores ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ","}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ RowBox[{" ", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], TraditionalForm]]], ". En este proceso debemos considerar ", StyleBox["r", FontSlant->"Italic"], ", \[Theta] y \[Phi] como funciones de ", Cell[BoxData[ \(TraditionalForm\`x, \ y, \ z\)]], ". Los c\[AAcute]lculos son largos y es una gran ventaja disponer de ", StyleBox["Mathematica", FontSlant->"Italic"], " para hacerlos. " }], "Text"], Cell[BoxData[{ \(\(r[x_, y_, z_] = \@\(x\^2 + y\^2 + z\^2\);\)\), "\n", \(\(\[Theta][x_, y_, z_] = ArcCos[z\/\@\(x\^2 + y\^2 + z\^2\)];\)\), "\n", \(\(\[Phi][x_, y_, z_] = ArcTan[x, y];\)\), "\n", \(\(er[x_, y_, z_] = er[\[Theta][x, y, z], \[Phi][x, y, z]];\)\), "\n", \(\(e\[Theta][x_, y_, z_] = e\[Theta][\[Theta][x, y, z], \[Phi][x, y, z]];\)\), "\n", \(\(\(e\[Phi][x_, y_, z_] = e\[Phi][\[Theta][x, y, z], \[Phi][x, y, z]];\)\(\n\) \)\t\t\), "\n", \(\(B = Transpose[{er[x, y, z], e\[Theta][x, y, z], e\[Phi][x, y, z]}];\)\), "\n", \(\(f1[x_, y_, z_] = \((B . {F\_r[r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]], F\_\[Theta][ r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]], \n\t\t\tF\_\[Phi][ r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]]})\)[\([1]\)];\)\), "\n", \(\(f2[x_, y_, z_] = \((B . {F\_r[r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]], F\_\[Theta][ r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]], \n\t\t\tF\_\[Phi][ r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]]})\)[\([2]\)];\)\), "\n", \(\(f3[x_, y_, z_] = \((B . {F\_r[r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]], F\_\[Theta][ r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]], \n\t\t\tF\_\[Phi][ r[x, y, z], \[Theta][x, y, z], \[Phi][x, y, z]]})\)[\([3]\)];\)\), "\n", \(\((D[f1[x, y, z], x] + D[f2[x, y, z], y] + D[f3[x, y, z], z] /. {x -> r*Cos[\[Phi]]*\ Sin[\[Theta]], y -> r*\ Sin[\[Phi]]*\ Sin[\[Theta]], z -> r*\ Cos[\[Theta]]\ })\) // Simplify\)}], "Input"], Cell["expresi\[OAcute]n que suele escribirse en la forma", "Text"], Cell[BoxData[ \(\(1\/r\^2\) \[PartialD]\_r\((\ r\^2*\ F\_r[r, \[Theta], \[Phi]])\) + \(1\/\(r* Sin[\[Theta]]\)\) \[PartialD]\_\[Theta]\ \((Sin[\[Theta]]* F\_\[Theta][r, \[Theta], \[Phi]])\) + \(1\/\(r* Sin[\[Theta]]\)\) \[PartialD]\_\[Phi] F\_\[Phi][ r, \[Theta], \[Phi]]\)], "Input"], Cell[BoxData[ \(\(\(Simplify[% == %%]\)\( (*\ comprobamos\ que\ ambas\ expresiones\ son\ iguales\ *) \)\)\)], "Input"], Cell[TextData[{ "Podemos usar comandos de ", StyleBox["Mathematica", FontSlant->"Italic"], " definidos en el paquete \"VectorAnalysis\" para obtener directamente el \ resultado anterior aunque, claro est\[AAcute], eso no te ayudar\[AAcute] a \ comprender el significado de dicho resultado." }], "Text"], Cell[BoxData[ \(Div[{F\_r[r\ , \[Theta], \[Phi]], F\_\[Theta][r\ , \[Theta], \[Phi]], F\_\[Phi][r\ , \[Theta], \[Phi]]}, Spherical[r, \[Theta], \[Phi]]] // Simplify\)], "Input"], Cell["\<\ Tambi\[EAcute]n podemos calcular directamente el rotacional de un campo \ vectorial en coordenadas esf\[EAcute]ricas.\ \>", "Text"], Cell[BoxData[ \(Curl[{F\_r[r\ , \[Theta], \[Phi]], F\_\[Theta][r\ , \[Theta], \[Phi]], F\_\[Phi][r\ , \[Theta], \[Phi]]}, Spherical[r, \[Theta], \[Phi]]] // Simplify\)], "Input"], Cell[CellGroupData[{ Cell["Ejercicio 1", "Exercise"], Cell[TextData[{ "Calcula la expresi\[OAcute]n del rotacional de un campo vectorial en \ coordenadas esf\[EAcute]ricas. Debes hacer t\[UAcute] los c\[AAcute]lculos y \ no usar los comandos espec\[IAcute]ficos de ", StyleBox["Mathematica", FontSlant->"Italic"], " que se definen en el paquete \"VectorAnalysis\"." }], "ExerciseText"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Gradiente y elemento diferencial de longitud en coordenadas esf\[EAcute]ricas\ \ \>", "Subsection"], Cell[TextData[{ "Sea ", StyleBox["f", FontSlant->"Italic"], " un campo escalar de tres variables. Sabemos que el gradiente de ", StyleBox["f", FontSlant->"Italic"], " es el campo vectorial dado por " }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{\(\[Del]\(f(x, y, z)\)\), "=", RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]x\), \((x, y, z)\), StyleBox["i", FontWeight->"Bold"]}], "+", RowBox[{\(\[PartialD]f\/\[PartialD]y\), \((x, y, z)\), StyleBox["j", FontWeight->"Bold"]}], "+", RowBox[{\(\[PartialD]f\/\[PartialD]y\), \((x, y, z)\), StyleBox["k", FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". " }], "Text", TextAlignment->Center], Cell[TextData[{ "La expresi\[OAcute]n del gradiente de ", StyleBox["f ", FontSlant->"Italic"], "en esf\[EAcute]ricas viene dada por" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"\[Del]", RowBox[{"f", "(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], "=", RowBox[{ RowBox[{\(\(f\_r\)(r, \[Theta], \[Phi])\), StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", RowBox[{\(\(f\_\[Theta]\)(\[Rho], \[Theta], \[Phi])\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{\(\(f\_\[Phi]\)(r, \[Theta], \[Phi])\), StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}]}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "donde ", Cell[BoxData[ \(TraditionalForm\`f\_\[Rho]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f\_\[Theta]\)]], " y ", Cell[BoxData[ \(TraditionalForm\`f\_\[Phi]\)]], " son las componentes del vector ", Cell[BoxData[ FormBox[ RowBox[{\(\[Del]f\), RowBox[{"(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], TraditionalForm]]], " en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ". En consecuencia se trata de hacer un simple cambio de coordenadas de la \ base can\[OAcute]nica a la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], ". La matriz de este cambio de base es la matriz inversa de la que tiene \ por columnas los vectores ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], TraditionalForm]]], ", es decir, la matriz ", Cell[BoxData[ FormBox[ RowBox[{"M", "=", RowBox[{"Inverse", "[", RowBox[{"Transpose", "[", FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], "TraditionalForm"], "]"}], "]"}]}], TraditionalForm]]], " y, como los vectores ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], TraditionalForm]]], " son ortonormales dos a dos, se verifca que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"Transpose", "[", FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], "TraditionalForm"], "]"}], "=", RowBox[{"Inverse", "[", FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], "TraditionalForm"], "]"}]}], TraditionalForm]]], ", en consecuencia, M=", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " es la matriz cuyos vectores fila son ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}], TraditionalForm]]], ". Procura entender lo que sigue." }], "Text"], Cell[BoxData[{ \(\(gradf[x_, y_, z_] = {D[f[x, y, z], x], D[f[x, y, z], y], D[f[x, y, z], z]};\)\), "\n", \(\(reglas = {x -> r\ \ Cos[\[Phi]] Sin[\[Theta]], y -> r\ \ Sin[\[Phi]] Sin\ [\[Theta]], z -> r\ Cos[\[Theta]]};\)\), "\n", \(\(M = {er[\[Theta], \[Phi]], e\[Theta][\[Theta], \[Phi]], e\[Phi][\[Theta], \[Phi]]};\)\), "\n", \(fr[r_, \[Theta]_, \[Phi]_] = \((\((M . gradf[x, y, z])\)[\([1]\)] /. reglas)\) // Simplify\), "\n", \(f\[Theta][ r_, \[Theta]_, \[Phi]_] = \((\((M . gradf[x, y, z])\)[\([2]\)] /. reglas)\) // Simplify\), "\n", \(f\[Phi][ r_, \[Theta]_, \[Phi]_] = \((\((M . gradf[x, y, z])\)[\([3]\)] /. reglas)\) // Simplify\)}], "Input"], Cell["\<\ Observando estos resultados nos damos cuenta de que pueden expresarse m\ \[AAcute]s f\[AAcute]cilmente como sigue.\ \>", "Text"], Cell[BoxData[{ \(Simplify[ fr[r, \[Theta], \[Phi]] == D[f[r\ Cos[\[Phi]]\ Sin[\[Theta]], r\ Sin[\[Theta]]\ Sin[\[Phi]], r\ Cos[\[Theta]]], r]]\), "\n", \(Simplify[ f\[Theta][r, \[Theta], \[Phi]] == \(1\/r\) D[f[r\ Cos[\[Phi]]\ Sin[\[Theta]], r\ Sin[\[Theta]]\ Sin[\[Phi]], r\ Cos[\[Theta]]], \[Theta]]]\), "\n", \(Simplify[ f\[Phi][r, \[Theta], \[Phi]] == \(1\/\(r\ Sin[\[Theta]]\)\) D[f[r\ Cos[\[Phi]]\ Sin[\[Theta]], r\ Sin[\[Theta]]\ Sin[\[Phi]], r\ Cos[\[Theta]]], \[Phi]]]\)}], "Input"], Cell["Hemos obtenido as\[IAcute] que", "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{ RowBox[{"\[Del]", RowBox[{"f", "(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], "=", RowBox[{ RowBox[{\(\[PartialD]\/\[PartialD]r\), RowBox[{"f", "(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}], StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/r\), \(\[PartialD]\/\[PartialD]\[Theta]\), RowBox[{"f", "(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}], StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/\(r\ sen\ \[Theta]\)\), \(\[PartialD]\/\[PartialD]\ \[Phi]\), RowBox[{"f", "(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}], StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}]}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Es conveniente introducir la funci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{\(h(r, \[Theta], \[Phi])\), "=", RowBox[{"f", "(", RowBox[{ \(r\ sen\ \[Theta]\ cos\ \[Phi]\), " ", ",", \(r\ sen\ \[Theta]\ sen\ \[Phi]\), ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], TraditionalForm]]], " con lo que la igualdad anterior se escribe mejor en la forma" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{ RowBox[{"\[Del]", RowBox[{"f", "(", RowBox[{ RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "cos", " ", "\[Phi]"}], " ", ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "sen", " ", "\[Theta]", " ", "sen", " ", "\[Phi]"}], ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}], "=", RowBox[{ RowBox[{\(\[PartialD]h\/\[PartialD]r\), \((r, \[Theta], \[Phi])\), StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/r\), \(\[PartialD]h\/\[PartialD]\[Theta]\), \((r, \ \[Theta], \[Phi])\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/\(r\ sen\ \[Theta]\)\), \(\[PartialD]h\/\[PartialD]\ \[Phi]\), \((r, \[Theta], \[Phi])\), StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "\t\t"}]}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "En los textos de f\[IAcute]sica es frecuente que no se distinga entre la \ funci\[OAcute]n ", StyleBox["f", FontSlant->"Italic"], " y la funci\[OAcute]n ", StyleBox["h", FontSlant->"Italic"], " (pues, en definitiva, son la ", StyleBox["misma funci\[OAcute]n ", FontSlant->"Italic"], "expresada en ", StyleBox["distintas", FontSlant->"Italic"], " coordenadas) y que escriban la igualdad anterior en la forma" }], "Text"], Cell[BoxData[ FormBox[ FrameBox[ FormBox[ RowBox[{\(\[Del]f\), "=", RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]r\), StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/r\), \(\[PartialD]f\/\[PartialD]\[Theta]\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/\(r\ sen\ \[Theta]\)\), \ \(\[PartialD]f\/\[PartialD]\[Phi]\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Phi]"], " \t"}]}]}], "TraditionalForm"]], TraditionalForm]], "Text", TextAlignment->Center], Cell["\<\ igualdad que constituye la \"expresi\[OAcute]n del gradiente en cordenadas \ esf\[EAcute]ricas\". \ \>", "Text"], Cell[TextData[{ "Podemos comprobarlo con ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Text"], Cell[BoxData[ \(\(\(Grad[f[r, \[Theta], \[Phi]], Spherical[r, \[Theta], \[Phi]]]\)\(\ \)\)\)], "Input"], Cell[TextData[{ "Observa que en la expresi\[OAcute]n anterior del gradiente aparecen los \ inversos de los factores de escala multiplicando a las derivadas parciales a \ las que est\[AAcute] asociado cada uno de ellos. Como sabes, los factores de \ escala son ", Cell[BoxData[ \(TraditionalForm\`{1, r, r\ sen\ \[Theta]}\)]], "; el primero de ellos, 1, est\[AAcute] asociado a la primera columna de la \ matriz jacobiana del cambio de coordenadas que corresponde a la derivaci\ \[OAcute]n parcial respecto a la primera variable, r; el segundo de ellos, ", StyleBox["r", FontSlant->"Italic"], ", est\[AAcute] asociado a la segunda columna de la matriz jacobiana del \ cambio de coordenadas que corresponde a la derivaci\[OAcute]n parcial \ respecto a la segunda variable, \[Theta], el tercero de ellos, ", Cell[BoxData[ \(TraditionalForm\`\(r\ sen\ \[Theta], \)\)]], " est\[AAcute] asociado a la tercera columna de la matriz jacobiana del \ cambio de coordenadas que corresponde a la derivaci\[OAcute]n parcial \ respecto a la tercera variable, \[Phi]. " }], "Text"], Cell["\<\ Hemos visto antes que la velocidad en coordenadas esf\[EAcute]ricas se \ expresa por\ \>", "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{\(r'\), \((t)\), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \(\(r(t)\) \[Theta]\ ' \((t)\)\)}]}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{ \(r(t)\), \(\[Phi]'\), \((t)\), "sen", " ", \(\[Theta](t)\), " ", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "(", "t", ")"}]}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell["\<\ Esta igualdad suele expresarse con notaci\[OAcute]n m\[AAcute]s \ cl\[AAcute]sica y tambi\[EAcute]n m\[AAcute]s confusa en la forma\ \>", "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"\[DifferentialD]", FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "=", RowBox[{ RowBox[{\(\[DifferentialD]r\), " ", StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", "r"}]}], "TraditionalForm"]}], \(\[DifferentialD]\[Theta]\), " ", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{ "r", " ", "sen", " ", "\[Theta]", " ", \(\[DifferentialD]\[Phi]\), " ", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}]}], TraditionalForm]]], "\t\t(5)" }], "Text", TextAlignment->Center], Cell[TextData[{ "(no te quepa duda, esta igualdad quiere decir exactamente lo mismo que la \ anterior pero en los libros de f\[IAcute]sica acostumbran a escribirla as\ \[IAcute]). Aqu\[IAcute] vuelven a aparecer los factores de escala ", Cell[BoxData[ \(TraditionalForm\`1, \ r, \ r\ sen\ \[Theta]\)]], " multiplicando a las correspondientes \"diferenciales\" ", Cell[BoxData[ \(TraditionalForm\`\[DifferentialD]\[Rho]\ , \[DifferentialD]\[Theta], \ \[DifferentialD]\[Phi]\)]], ". La igualdad anterior es la expresi\[OAcute]n del \"elemento diferencial \ de longitud\" en coordenadas esf\[EAcute]ricas. " }], "Text"], Cell[TextData[{ "Si hacemos el producto escalar ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\[Del]f | \[DifferentialD]r\[RightAngleBracket]\)]], " usando las igualdades anteriores obtenemos lo que sigue:" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(\[LeftAngleBracket]\[Del]f | \[DifferentialD]r\ \[RightAngleBracket]\), "=", RowBox[{ RowBox[{"\[LeftAngleBracket]", RowBox[{ FormBox[ RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]r\), StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/r\), \(\[PartialD]f\/\[PartialD]\[Theta]\), StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{\(1\/\(r\ sen\ \[Theta]\)\), \(\[PartialD]f\/\ \[PartialD]\[Phi]\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Phi]"]}]}], "TraditionalForm"], "|", FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{\(\[DifferentialD]r\), " ", StyleBox[\(e\_r\), FontWeight->"Bold"]}], "+", "r"}], "TraditionalForm"], \(\[DifferentialD]\[Theta]\), " ", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"]}], "+", RowBox[{ "r", " ", "sen", " ", "\[Theta]", " ", \(\[DifferentialD]\[Phi]\), " ", StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"]}]}], "TraditionalForm"]}], "\[RightAngleBracket]"}], "=", \(\(\[PartialD]f\/\[PartialD]r\) \[DifferentialD]r + \(\ \[PartialD]f\/\[PartialD]\[Theta]\) \[DifferentialD]\(\[Theta]++\) \(\ \[PartialD]f\/\[PartialD]\[Phi]\) \[DifferentialD]\[Phi]\)}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "En esta igualdad se entiende que el gradiente est\[AAcute] calculado en \ puntos del camino ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], TraditionalForm]]], ". Con m\[AAcute]s detalle. Si ", Cell[BoxData[ \(TraditionalForm\`\((\[Rho](t), \[Theta](t), \[Phi](t))\)\)]], " son las coordenadas esf\[EAcute]ricas del punto ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], TraditionalForm]]], ", la igualdad anterior quiere decir lo siguiente:" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"\[LeftAngleBracket]", RowBox[{\(\[Del]\(f(\(r(t)\)\ cos\ \(\[Phi](t)\) sen\ \(\[Theta](t)\), \ \(r(t)\)\ sen\ \(\[Phi](t)\) sen\ \(\[Theta](t)\), \(r(t)\) cos\ \(\[Theta](t)\))\)\), "|", RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}]}], "\[RightAngleBracket]"}], "=", RowBox[{ RowBox[{"\[LeftAngleBracket]", RowBox[{ RowBox[{ RowBox[{\(\[PartialD]h\/\[PartialD]r\), \((r(t), \ \[Theta]( t), \[Phi](t))\), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{\(1\/\(r( t)\)\), \(\[PartialD]h\/\[PartialD]\[Theta]\), \((r( t), \ \[Theta](t), \[Phi](t))\), RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{\(1\/\(\(r(t)\)\ sen\ \(\[Theta]( t)\)\)\), \(\[PartialD]h\/\[PartialD]\[Phi]\), \((r( t), \ \[Theta](t), \[Phi](t))\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "\[Phi]"]}]}], "|", RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{\(r'\), \((t)\), RowBox[{ StyleBox[\(e\_r\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", \(\(r(t)\) \[Theta]\ ' \((t)\)\)}], "TraditionalForm"], RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{\(r(t)\), \(\[Phi]'\), \((t)\), "sen", " ", \(\[Theta](t)\), " ", RowBox[{ StyleBox[\(e\_\[Phi]\), FontWeight->"Bold"], "(", "t", ")"}]}]}]}], "\[RightAngleBracket]"}], "=", "\n", "\t\t", \(\(\[PartialD]h\/\[PartialD]r\) \((r(t), \ \[Theta]( t), \[Phi](t))\) r' \((t)\) + \(\[PartialD]h\/\[PartialD]\[Theta]\) \((r( t), \ \[Theta](t), \[Phi]( t))\) \[Theta]\ ' \((t)\) + \(\[PartialD]h\/\[PartialD]\ \[Phi]\) \((r(t), \ \[Theta](t), \[Phi]( t))\) \[Phi]\ ' \((t)\) = \(\[DifferentialD]\/\ \[DifferentialD]t\) \(h(r(t), \ \[Theta](t), \[Phi](t))\)\)}]}], TraditionalForm]], "Text"], Cell["Deducimos que", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Integral]\_a\%b\), RowBox[{ RowBox[{"\[LeftAngleBracket]", RowBox[{ RowBox[{ RowBox[{"\[Del]", StyleBox["f", FontSlant->"Italic"]}], \((\(r \((t)\)\)\ cos\ \(\[Phi] \((t)\)\) sen\ \(\[Theta] \((t)\)\), \ \(r \((t)\)\)\ sen\ \(\[Phi] \((t)\)\) sen\ \(\[Theta] \((t)\)\), \(r \((t)\)\) cos\ \(\[Theta] \((t)\)\))\)}], "|", RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "'"}], \((t)\)}]}], "\[RightAngleBracket]"}], \(\[DifferentialD]t\)}]}], "=", RowBox[{ RowBox[{ RowBox[{ StyleBox["h", FontSlant->"Italic"], \((\[Rho] \((b)\), \[Theta] \((b)\), \[Phi] \((b)\))\)}], "-", \(h \((\[Rho] \((a)\), \[Theta] \((a)\), \[Phi] \((a)\))\)\)}], "=", "\n", "\t\t", RowBox[{"=", RowBox[{ RowBox[{ StyleBox["f", FontSlant->"Italic"], \((r \((b)\)\ cos\ \[Theta] \((b)\)\ sen\ \[Phi] \((b)\), r \((b)\)\ sen\ \[Theta] \((b)\)\ sen\ \[Phi] \((b)\), r \((b)\)\ cos\ \[Phi] \((b)\))\)}], "-", RowBox[{ StyleBox["f", FontSlant->"Italic"], \((r \((a)\)\ cos\ \[Theta] \((a)\)\ sen\ \[Phi] \((a)\), r \((a)\)\ sen\ \[Theta] \((a)\)\ sen\ \[Phi] \((a)\), r \((a)\)\ cos\ \[Phi] \((a)\))\)}]}]}]}]}], TextForm]], "Text"], Cell["\<\ Expresi\[OAcute]n de la integral de l\[IAcute]nea de un gradiente en \ coordenadas esf\[EAcute]ricas que es completamente an\[AAcute]loga a la \ conocida para coordenadas cartesianas. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Significado de los factores de escala", "Subsection"], Cell[TextData[{ "Consideremos la matriz jacobiana en un punto gen\[EAcute]rico ", Cell[BoxData[ \(TraditionalForm\`\((r, \[Theta], \[Phi])\)\)]], " de la funci\[OAcute]n que introduce las coordenadas esf\[EAcute]ricas." }], "Text"], Cell[BoxData[ \(S = \(matrizjacobiana3[g]\)[r, \[Theta], \[Phi]]\)], "Input"], Cell[TextData[{ "Esta matriz define una aplicaci\[OAcute]n lineal de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], " en ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], " que a cada vector ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], " hace corresponder el vector ", Cell[BoxData[ \(TraditionalForm\`S . \((x, y, z)\)\)]], ". Calculemos la norma eucl\[IAcute]dea de la imagen de un vector en dicha \ transformaci\[OAcute]n. A veces ", StyleBox["Mathematica", FontSlant->"Italic"], " se empe\[NTilde]a en simplificar de forma poco apropiada y hay que \ decirle que haga lo que nosotros queremos." }], "Text"], Cell[BoxData[ \(Sqrt[ Collect[Expand[ Simplify[\((S . {x, y, z})\) . \((S . {x, y, z})\)]], \(z\^2\) r\^2] /. \((1\/2 - 1\/2\ Cos[2\ \[Theta]])\) -> Sin[\[Theta]]\^2]\)], "Input"], Cell[TextData[{ "Deducimos que para vectores situados a lo largo del eje X, es decir, de la \ forma ", Cell[BoxData[ \(TraditionalForm\`\((x, 0, 0)\)\)]], ", se verifica que ", Cell[BoxData[ \(TraditionalForm \`\(\( || S . \((x, 0, 0)\)\) || \) = \( || \((x, 0, 0)\) || \)\)]], " y, teniendo en cuenta la linealidad, se sigue que la aplicaci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\) \[Rule] S . \((x, y, z)\)\)]], " conserva distancias en el eje X. Pues bien, este es el significado de que \ el factor de escala asociado a la primera variable sea igual a 1.\nDeducimos \ tambi\[EAcute]n que para vectores situados a lo largo del eje Y, es decir, de \ la forma ", Cell[BoxData[ \(TraditionalForm\`\((0, y, 0)\)\)]], ", se verifica que ", Cell[BoxData[ \(TraditionalForm \`\(\( || S . \((0, y, 0)\)\) || \) = r || \((0, y, 0)\) || \)]], " y, teniendo en cuenta la linealidad, se sigue que la aplicaci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\) \[Rule] S . \((x, y, z)\)\)]], " multiplica distancias por ", StyleBox["r", FontSlant->"Italic"], " en el eje Y. Pues bien, este es el significado de que el factor de escala \ asociado a la segunda variable sea igual a ", StyleBox["r", FontSlant->"Italic"], ".\nDeducimos tambi\[EAcute]n que para vectores situados a lo largo del eje \ Z, es decir, de la forma ", Cell[BoxData[ \(TraditionalForm\`\((0, 0, z)\)\)]], ", se verifica que ", Cell[BoxData[ \(TraditionalForm \`\(\( || S . \((0, 0, z)\)\) || \) = r\ sen\ \[Theta] || \((0, 0, z)\) || \)]], " y, teniendo en cuenta la linealidad, se sigue que la aplicaci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\) \[Rule] S . \((x, y, z)\)\)]], " multiplica distancias por ", Cell[BoxData[ \(TraditionalForm\`r\ sen\ \[Theta]\)]], " en el eje Z. Pues bien, este es el significado de que el factor de escala \ asociado a la tercera variable sea igual a ", Cell[BoxData[ \(TraditionalForm\`r\ sen\ \(\[Theta] . \)\)]], "\nEn resumen, los factores de escala indican las dilataciones a lo largo \ de los ejes que hace la aplicaci\[OAcute]n lineal asociada a la matriz \ jacobiana de la aplicaci\[OAcute]n", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{\(g(r, \[Theta], \[Phi])\), "=", RowBox[{"(", RowBox[{ \(r\ sen\ \[Theta]\ cos\ \[Phi]\), " ", ",", \(r\ sen\ \[Theta]\ sen\ \[Phi]\), ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}]}]}], TraditionalForm]]], ". Suele decirse que la aplicaci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " es la que, a ", StyleBox["escala infinitesimal", FontSlant->"Italic"], ", produce esas dilataciones. Observa que la expresi\[OAcute]n del \ \"elemento diferencial de longitud\" en coordenadas esf\[EAcute]ricas tiene \ en cuenta dichos factores de escala. " }], "Text"], Cell[TextData[{ "Al igual que cada factor de escala mide la dilataci\[OAcute]n \ infinitesimal a lo largo de un eje, el producto de los factores de escala, en \ nuestro caso ", Cell[BoxData[ \(TraditionalForm\`r\^2\ sen\ \[Theta]\)]], ", mide el cambio en el volumen de un ortoedro a escala infinitesimal. El \ producto de los factores de escala es justamente el determinante jacobiano. \ Recuerda que la f\[OAcute]rmula del cambio de variables a coordenadas esf\ \[EAcute]ricas en una integral triple afirma que si ", StyleBox["f", FontSlant->"Italic"], " es un campo escalar continuo en un conjunto ", Cell[BoxData[ \(TraditionalForm\`A \[Subset] \[DoubleStruckCapitalR]\^3\)]], " se verifica que\t\t" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(\[Integral]\(\[Integral]\(\[Integral]\_A\( f(x, y, z)\) \(d( x, y, z)\)\)\)\), "=", RowBox[{"\[Integral]", RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_B\), " ", RowBox[{ RowBox[{"f", "(", RowBox[{\(r\ sen\ \[Theta]\ cos\ \[Phi]\), " ", ",", \(r\ sen\ \[Theta]\ sen\ \[Phi]\), ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}], \(\(\(r\)\(\ \)\)\^2\), "sen", " ", "\[Theta]", " ", "d", \((r, \[Theta], \[Phi])\), " "}]}]}]}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "donde", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{"B", "=", RowBox[{"{", RowBox[{\((r, \[Theta], \[Phi])\), ":", RowBox[{ FormBox[ RowBox[{"(", RowBox[{ \(r\ sen\ \[Theta]\ cos\ \[Phi]\), " ", ",", \(r\ sen\ \[Theta]\ sen\ \[Phi]\), ",", RowBox[{ StyleBox["r", FontSlant->"Italic"], " ", "cos", " ", "\[Theta]"}]}], ")"}], "TraditionalForm"], "\[Element]", " ", "A"}]}], "}"}]}]}], TraditionalForm]]], ". Observa que ", StyleBox["B", FontSlant->"Italic"], " es la expresi\[OAcute]n del conjunto ", StyleBox["A", FontSlant->"Italic"], " en coordenadas esf\[EAcute]ricas, es decir ", Cell[BoxData[ \(TraditionalForm\`A = g(B)\)]], ", y que en la segunda integral se multiplica por ", Cell[BoxData[ \(TraditionalForm\`\(\(r\ \)\^2\) sen\ \[Theta]\)]], ". Si particularizamos la igualdad anterior para la funci\[OAcute]n \ constante ", Cell[BoxData[ \(TraditionalForm\`f(x, y, z) = 1\)]], " obtenemos" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(Volumen \((g(B))\)\), "=", FormBox[\(\[Integral]\(\[Integral]\(\[Integral]\_A 1 \( d(x, y)\)\)\) = \[Integral]\(\[Integral]\(\[Integral]\_B\ \ \(\(\(r\)\(\ \)\)\^2\) sen\ \[Theta]\ \(d(r, \[Theta], \[Phi])\)\)\)\), "TraditionalForm"], " "}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Si ", StyleBox["B", FontSlant->"Italic"], " es un ortoedro muy peque\[NTilde]o (un ortoedro ", StyleBox["infinitesimal", FontSlant->"Italic"], ") se verifica que " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\[Integral]\(\[Integral]\(\[Integral]\_B\ \(\(\(r\)\(\ \ \)\)\^2\) sen\ \[Theta]\ \(d( r, \[Theta], \[Phi])\)\)\)\ \[TildeEqual] \ \ \(\(\(r\)\(\ \ \)\)\^2\) sen\ \[Theta] \(\[Integral]\(\[Integral]\(\[Integral]\_B\ d(\[Rho], \[Theta])\)\)\) = \ \(\(\(r\)\(\ \)\)\^2\) sen\ \[Theta]\ \(Volumen(B)\)\)], "Text", TextAlignment->Center], Cell[TextData[{ " y obtenemos ", Cell[BoxData[ \(TraditionalForm\`Volumen( g(B))\ \[TildeEqual] \ \ \(\(\(r\)\(\ \)\)\^2\) sen\ \[Theta]\ \(Volumen(B)\)\)]], ". ", "Suele decirse que ", Cell[BoxData[ \(TraditionalForm\`\(r\^2\) sen\ \[Theta] \[DifferentialD]r \[DifferentialD]\[Theta] \(\(\ \[DifferentialD]\)\(\[Phi]\)\(\ \)\)\)]], "es el \"elemento diferencial de volumen\" en coordenadas \ esf\[EAcute]ricas. " }], "Text"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " dispone de un comando que proporciona los factores de escala." }], "Text"], Cell[BoxData[ \(ScaleFactors[{r, \[Theta], \[Phi]}, Spherical[r, \[Theta], \[Phi]]]\)], "Input"], Cell["\<\ Para terminar con las coordenadas esf\[EAcute]ricas ejecuta la siguiente \ celda para ver una animaci\[OAcute]n gr\[AAcute]fica en la que aparecen las \ superficies transformadas de los planos coordenados y se muestra c\[OAcute]mo \ un punto en coordenadas esf\[EAcute]ricas se sit\[UAcute]a mediante la \ intersecci\[OAcute]n de dichas superficies.\ \>", "Text"], Cell[BoxData[ \(\(esfericasmovie;\)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Coordenadas cil\[IAcute]ndricas", "Section"], Cell[TextData[{ "La funci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm \`g(\[Rho], \[Theta], z) = \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta], z)\)\)]], " es una biyecci\[OAcute]n de ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ RowBox[{ RowBox[{"\[CapitalOmega]", "=", FormBox[\(\(\[DoubleStruckCapitalR]\^+\)\[Times]\), "TraditionalForm"]}], "]"}], "-", "\[Pi]"}], ",", "\[Pi]"}], "]"}], "\[Times]", "\[DoubleStruckCapitalR]"}], TraditionalForm]]], " sobre ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\\{\((0, 0, 0)\)}\)]], ". Los n\[UAcute]meros ", Cell[BoxData[ \(TraditionalForm\`\((\[Rho], \ \[Theta], \ z)\)\)]], " donde ", Cell[BoxData[ \(TraditionalForm\`x = \[Rho]\ cos\ \[Theta], \ y = \[Rho]\ sen\ \[Theta]\)]], ", con ", Cell[BoxData[ \(TraditionalForm\`\[Rho] > 0\)]], " y ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(-\[Pi]\), "TraditionalForm"], "<", "\[Theta]", "\[LessEqual]", "\[Pi]"}], TraditionalForm]]], ", se llaman las coordenadas cil\[IAcute]ndricas del punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], ". " }], "Text"], Cell[BoxData[ \(\(coordcilindricas;\)\)], "Input"], Cell[BoxData[ \(\(?\ Cylindrical\)\)], "Input"], Cell[BoxData[ \(Coordinates[Cylindrical]\)], "Input"], Cell[BoxData[ \(\(CoordinateRanges[Cylindrical]\ (*\ observa\ que\ Mathematica\ usa\ para\ la\ primera\ variable\ la\ letra\ r\ en\ lugar\ de\ \[Rho]\ *) \)\)], "Input"], Cell[BoxData[ \(\(CoordinatesToCartesian[{\[Rho], \[Theta], \[Phi]}, Cylindrical] (*\ paso\ de\ cil\[IAcute]ndricas\ a\ cartesianas\ *) \ \)\)], "Input"], Cell[BoxData[ \(\(CoordinatesFromCartesian[{x, y, z}, Cylindrical] (*\ paso\ de\ cartesianas\ a\ cil\[IAcute]ndricas\ *) \)\)], "Input"], Cell[TextData[{ "Cuando se utiliza el sistema de coordenadas cil\[IAcute]ndricas los \ vectores se refieren a una base ortonormal ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], StyleBox[",", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold"], SubscriptBox[ StyleBox["e", FontWeight->"Bold"], StyleBox["z", FontWeight->"Bold"]]}], "}"}], TraditionalForm]]], " que se ha representado en la figura anterior (trasladada al punto ", StyleBox["P", FontSlant->"Italic"], "). En el lenguaje t\[IAcute]pico de los textos de f\[IAcute]sica se dice \ que el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta \[Rho] manteniendo \[Theta] y z constantes, el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta \[Theta] manteniendo \[Rho] y z constantes y el vector ", Cell[BoxData[ FormBox[ StyleBox[\(e\_z\), FontWeight->"Bold"], TraditionalForm]]], " es un vector unitario en el sentido en que se mueve el punto ", StyleBox["P", FontSlant->"Italic"], " cuando aumenta z manteniendo \[Rho] y \[Theta] constantes. En \ t\[EAcute]rminos matem\[AAcute]ticos, quiz\[AAcute]s m\[AAcute]s precisos, \ observa que el vector de posici\[OAcute]n del punto ", StyleBox["P", FontSlant->"Italic"], " es ", Cell[BoxData[ \(TraditionalForm \`g(\[Rho], \[Theta], z) = \((\[Rho]\ cos\ \[Theta], \ \[Rho]\ sen\ \[Theta], z)\)\)]], " su variaci\[OAcute]n con respecto a \[Rho] manteniendo \[Theta] y z \ constantes es la derivada parcial respecto a \[Rho], su variaci\[OAcute]n con \ respecto a \[Theta] manteniendo \[Rho] y z constantes es la derivada parcial \ respecto a \[Theta] y su variaci\[OAcute]n con respecto a \[Phi] manteniendo \ \[Rho] y \[Theta] constantes es la derivada parcial respecto a z. " }], "Text"], Cell[BoxData[{ \(Clear[g, u1, u2, u3]\), "\n", \(g[\[Rho]_, \[Theta]_, z_] := {\[Rho]\ Cos[\[Theta]], \[Rho]\ \ Sin[\[Theta]], z}\), "\n", \(u1 = D[g[\[Rho], \[Theta], z], \[Rho]]\), "\n", \(u2 = D[g[\[Rho], \[Theta], z], \[Theta]]\), "\n", \(u3 = D[g[\[Rho], \[Theta], z], z]\)}], "Input"], Cell["Observa que estos vectores son ortogonales", "Text"], Cell[BoxData[ \({u1 . u2, u1 . u3, u2 . u3} // Simplify\)], "Input"], Cell["\<\ Para obtener una base ortonormal a partir de ellos todo lo que tenemos que \ hacer es normalizarlos. Calculemos sus normas.\ \>", "Text"], Cell[BoxData[ \({norma[u1], norma[u2], norma[u3]} // Simplify\)], "Input"], Cell[TextData[{ "Deducimos que los vectores ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], "=", \((cos\ \[Theta], \ sen\ \[Theta], 0)\)}], ",", RowBox[{ StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], "=", \((\(-sen\)\ \[Theta], cos\ \[Theta], 0)\)}], ",", " ", RowBox[{ StyleBox[\(e\_z\), FontWeight->"Bold"], "=", RowBox[{\((0\ , 0, 1)\), "=", StyleBox["k", FontWeight->"Bold"]}]}]}], TraditionalForm]]], " forman una base ortonormal. Es a dicha base a la que se refiere un vector \ cuando se usan coordenadas cil\[IAcute]ndricas. Observa que los vectores de \ esta base dependen de la posici\[OAcute]n del punto, es decir, no se trata de \ una base fija. F\[IAcute]jate en que si ", Cell[BoxData[ \(TraditionalForm\`\(\((\[Rho], \[Theta], z)\)\(\ \)\)\)]], "son las coordenadas cil\[IAcute]ndricas de un punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], ", se verifica que ", Cell[BoxData[ FormBox[ RowBox[{\((x, y, z)\), "=", RowBox[{ RowBox[{"\[Rho]", " ", StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"]}], "+", RowBox[{"z", " ", StyleBox["k", FontWeight->"Bold"]}]}]}], TraditionalForm]]], ". En general, la expresi\[OAcute]n en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox["k", FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " de un vector ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["v", FontWeight->"Bold"], "=", \((x, y, z)\)}], TraditionalForm]]], " se obtiene por el m\[EAcute]todo usual calculando sus proyecciones \ ortogonales sobre los vectores de la base:" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ StyleBox[\(v = \[LeftAngleBracket]v | e\_\[Rho]\[RightAngleBracket] e\_\[Rho] + \[LeftAngleBracket]v | e\_\[Theta]\[RightAngleBracket] e\_\[Theta] + \[LeftAngleBracket]v | k\[RightAngleBracket] k\), FontWeight->"Bold"], TraditionalForm]]], "\t\t" }], "Text", TextAlignment->Center], Cell[TextData[{ "Observa que si escribimos ", StyleBox["v", FontWeight->"Bold", FontSlant->"Italic"], " en la forma ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["v", FontWeight->"Bold"], "=", \((\[Rho]\ cos\ \[Theta]\ , \[Rho]\ sen\ \[Theta]\ sen\ \ \[Phi], z)\)}], TraditionalForm]]], " entonces ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(\[LeftAngleBracket]v | e\_\[Rho]\[RightAngleBracket]\), FontWeight->"Bold"], "=", "\[Rho]"}], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(\[LeftAngleBracket]v | e\_\[Theta]\[RightAngleBracket]\), FontWeight->"Bold"], "=", "0"}], ",", " ", RowBox[{ StyleBox[\(\[LeftAngleBracket]v | k\[RightAngleBracket]\), FontWeight->"Bold"], "=", "z"}]}], TraditionalForm]]], "." }], "Text"], Cell["\<\ A continuaci\[OAcute]n se definen los vectores de la base en coordenadas cil\ \[IAcute]ndricas.\ \>", "Text"], Cell[BoxData[{ \(Clear[er, e\[Theta], e\[Phi]]\), "\n", \(e\[Rho][\[Theta]_] = u1/norma[u1] // Simplify\), "\n", \(e\[Theta][\[Theta]_] = u2/norma[u2] // Simplify\), "\n", \(k = {0, 0, 1}\)}], "Input"], Cell[TextData[{ "La siguiente gr\[AAcute]fica muestra la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_\[Rho]\), FontWeight->"Bold"], ",", StyleBox[\(e\_\[Theta]\), FontWeight->"Bold"], ",", StyleBox["k", FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " para ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Pi]/4\)]], ". Para visualizarla mejor dicha base se ha trasladado al punto de \ coordenadas cil\[IAcute]ndricas ", Cell[BoxData[ \(TraditionalForm\`{1, \[Pi]/4, 1}\)]], "." }], "Text"], Cell[BoxData[{ \(\(punto = CoordinatesToCartesian[{1, \[Pi]/4, 1}, Cylindrical];\)\), "\n", \(\(graf = Show[Graphics3D[{vector3D[punto], vector3D[punto, .5*e\[Rho][\[Pi]/4], Blue], vector3D[punto, .5*e\[Theta][\[Pi]/4], Green], vector3D[punto, .5*k, Red], Text[\*"\"\<\!\(e\_\[Rho]\)\>\"", punto + .6*e\[Rho][\[Pi]/4]], Text[\*"\"\<\!\(e\_\[Theta]\)\>\"", punto + .65*e\[Theta][\[Pi]/4]], \n\t\t\t\tText["\", punto + .6*k]}, Axes -> True, PlotRange -> All, Ticks -> None, TextStyle -> {FontFamily -> Times, FontSize -> 14, FontWeight -> Bold}]];\)\)}], "Input"], Cell[BoxData[{ \(SpinShow[graf]\), "\n", \(\(SelectionMove[EvaluationNotebook[], All, GeneratedCell];\)\), "\n", \(FrontEndTokenExecute["\"]\), "\n", \(FrontEndTokenExecute["\"]\), "\n", \(Clear[graf, punto]\)}], "Input"], Cell[CellGroupData[{ Cell["Ejercicio 2", "Exercise"], Cell["\<\ Completa el estudio de las coordendas cil\[IAcute]ndricas conforme al esquema \ seguido en el estudio de las coordenadas polares y esf\[EAcute]ricas. Es \ decir, debes calcular la velocidad, aceleraci\[OAcute]n, divergencia y \ gradiente en coordenadas cil\[IAcute]ndricas y hacer una \ intrepretaci\[OAcute]n de los factores de escala y de los elementos \ diferenciales de longitud y volumen.\ \>", "ExerciseText"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Coordenadas curvil\[IAcute]neas ortogonales", "Section"], Cell[TextData[{ "Sean ", Cell[BoxData[ \(TraditionalForm\`A\)]], " y ", Cell[BoxData[ \(TraditionalForm\`B\)]], " dos subconjuntos de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], " y sea ", Cell[BoxData[ \(TraditionalForm\`\(\(g : A\)\(\[Rule]\)\(B\)\(\ \)\)\)]], " una biyecci\[OAcute]n de ", Cell[BoxData[ \(TraditionalForm\`A\)]], " sobre ", Cell[BoxData[ \(TraditionalForm\`B\)]], ". Representaremos por ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], " un punto gen\[EAcute]rico de ", Cell[BoxData[ \(TraditionalForm\`B\)]], " y por ", Cell[BoxData[ \(TraditionalForm\`\(\((u, v, w)\)\(\ \)\)\)]], "un punto gen\[EAcute]rico de ", Cell[BoxData[ \(TraditionalForm\`A\)]], ". La biyecci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " permite describir los puntos de ", Cell[BoxData[ \(TraditionalForm\`B\)]], " mediante puntos de ", Cell[BoxData[ \(TraditionalForm\`A\)]], ", pues dado un punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\) \[Element] B\)]], " hay un \[UAcute]nico punto ", Cell[BoxData[ \(TraditionalForm\`\((u, v, w)\) \[Element] A\)]], " tal que ", Cell[BoxData[ \(TraditionalForm\`g(u, v, w) = \((x, y, z)\)\)]], "; diremos que dicho punto ", Cell[BoxData[ \(TraditionalForm\`\((u, v, w)\)\)]], " son las g-coordenadas del punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], " y tambi\[EAcute]n diremos que la aplicaci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " introduce en ", StyleBox["B", FontSlant->"Italic"], " un sistema de coordenadas curvil\[IAcute]neas. Naturalmente, para que \ estas coordenadas definidas por ", StyleBox["g", FontSlant->"Italic"], " sean \[UAcute]tiles hay que suponer que la funci\[OAcute]n ", StyleBox["g ", FontSlant->"Italic"], "tiene buenas propiedades anal\[IAcute]ticas. Suele suponerse como condici\ \[OAcute]n m\[IAcute]nima para ", StyleBox["g", FontSlant->"Italic"], " que sea de clase ", Cell[BoxData[ \(TraditionalForm\`C\^1\)]], ", es decir que sus funciones componentes tengan derivadas parciales \ continuas y tambi\[EAcute]n se supone que el determinante jacobiano de ", StyleBox["g", FontSlant->"Italic"], " no se anula nunca. Estas condiciones aseguran que la biyecci\[OAcute]n \ inversa de ", StyleBox["g", FontSlant->"Italic"], " es una funci\[OAcute]n de clase ", Cell[BoxData[ \(TraditionalForm\`C\^1\)]], " en ", StyleBox["B", FontSlant->"Italic"], ". Las funciones que verifican estas propiedades (biyecciones de clase ", Cell[BoxData[ \(TraditionalForm\`C\^1\)]], "cuya biyecci\[OAcute]n inversa tambi\[EAcute]n es de clase ", Cell[BoxData[ \(TraditionalForm\`C\^1\)]], ") se llaman ", StyleBox["difeormorfismos", FontWeight->"Bold"], ". Suele exigirse tambi\[EAcute]n que la matriz jacobiana de ", StyleBox["g", FontSlant->"Italic"], " calculada en cualquier punto de ", StyleBox["A", FontSlant->"Italic"], " tenga la propiedad de que sus vectores columna sean dos a dos \ ortogonales. Cuando la aplicaci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " verifica estos requisitos (es un difeomorfismo de ", StyleBox["A", FontSlant->"Italic"], " sonbre ", StyleBox["B", FontSlant->"Italic"], " cuya matriz jacobiana tiene columnas ortogonales) se dice que dicha \ aplicaci\[OAcute]n define un sistema de coordenadas curvil\[IAcute]neas \ ortogonales en ", StyleBox["B. ", FontSlant->"Italic"], "Las coordenadas esf\[EAcute]ricas y las cil\[IAcute]ndricas son ejemplos \ ", "de coordenadas curvil\[IAcute]neas ortogonales (y tambi\[EAcute]n lo son \ las coordenadas polares en el plano)." }], "Text"], Cell["Ejemplo", "Example"], Cell["Consideremos la aplicaci\[OAcute]n", "Text"], Cell[BoxData[{ \(Clear[g]\), "\n", \(g[u_, v_, w_] := {u\^2 - v\^2, \ 2 u\ v, \ w\^2}\)}], "Input"], Cell[TextData[{ "Es evidente que dicha aplicaci\[OAcute]n no es biyectiva en todo su \ dominio natural de definici\[OAcute]n que es ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], ". Para obtener un conjunto en el que sea una biyecci\[OAcute]n debemos \ restringir su dominio de definici\[OAcute]n. Para ello vamos a considerar el \ conjunto ", Cell[BoxData[ \(TraditionalForm\`A = {\((u, v, w)\) : \ u > 0, \ v > 0, \ w > 0}\)]], ". En el conjunto ", StyleBox["A", FontSlant->"Italic"], " la funci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " toma valores en el conjunto ", Cell[BoxData[ \(TraditionalForm\`B = {\((x, y, z)\) : y > 0, z > 0}\)]], ". Comprobemos que ", StyleBox["g", FontSlant->"Italic"], " es una biyecci\[OAcute]n de ", StyleBox["A", FontSlant->"Italic"], " sobre ", StyleBox["B", FontSlant->"Italic"], ". Sea ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\)\)]], " un punto cualquiera de ", StyleBox["B", FontSlant->"Italic"], " y veamos si hay un \[UAcute]nico punto de ", StyleBox["A", FontSlant->"Italic"], " cuya imagen por ", StyleBox["g", FontSlant->"Italic"], " es dicho punto. Deberemos seleccionar de las posibles soluciones de la \ ecuaci\[OAcute]n ", Cell[BoxData[ \(TraditionalForm\`g \((u, v, w)\) = \((x, y, z)\)\)]], " la que tenga todas sus coordenadas positivas y debe de haber una sola." }], "Text"], Cell[BoxData[ \(Select[{u, v, w} /. Solve[{u\^2 - v\^2, \ 2 u\ v, \ w\^2} == {x, y, z}, {u, v, w}] // Simplify, todospositivos] // Flatten\)], "Input"], Cell[TextData[{ "Observa que, como ", Cell[BoxData[ \(TraditionalForm\`y > 0, \ z > 0\)]], ", la soluci\[OAcute]n obtenida efectivamente tiene todas sus coordenadas \ positivas. Hemos probado as\[IAcute] que ", Cell[BoxData[ \(TraditionalForm\`g(u, v, w) = {u\^2 - v\^2, 2\ u\ v, w\^2}\)]], " es una biyecci\[OAcute]n de ", StyleBox["A", FontSlant->"Italic"], " sobre ", StyleBox["B", FontSlant->"Italic"], ". Es claro que la funci\[OAcute]n ", StyleBox["g", FontSlant->"Italic"], " es de clase ", Cell[BoxData[ \(TraditionalForm\`C\^\[Infinity]\)]], ". Calculemos su matriz jacobiana y el determinante jacobiano." }], "Text"], Cell[BoxData[{ \(Clear[M]\), "\n", \(M = \(matrizjacobiana3[g]\)[u, v, w]\), "\n", \(Det[M]\)}], "Input"], Cell[TextData[{ "Deducimos que ", StyleBox["g", FontSlant->"Italic"], " es un difeomorfismo de ", StyleBox["A", FontSlant->"Italic"], " sobre ", StyleBox["B", FontSlant->"Italic"], ". Adem\[AAcute]s las columnas de la matriz jacobiana son ortogonales." }], "Text"], Cell[BoxData[ \(\(\({u1, u2, u3} = Transpose[M]\ (*\ las\ columnas\ de\ M\ *) \n {u1 . u2, u1 . u3, u2 . u3}\)\(\ \)\( (*\ las\ columnas\ de\ M\ son\ ortogonales\ *) \)\)\)], "Input"], Cell[TextData[{ "Concluimos que la aplicaci\[OAcute]n ", StyleBox["g ", FontSlant->"Italic"], "introduce un sistema de coordenadas curvil\[IAcute]neas ortogonales en ", StyleBox["B", FontSlant->"Italic"], ". Las g-coordenadas de un punto ", Cell[BoxData[ \(TraditionalForm\`\((x, y, z)\) \[Element] B\)]], " son la \[UAcute]nica terna ", Cell[BoxData[ \(TraditionalForm\`\((u, v, w)\) \[Element] A\)]], " tal que ", Cell[BoxData[ \(TraditionalForm\`g(u, v, w) = \(\((x, y, z)\) . \)\)]], " Definimos a continuaci\[OAcute]n funciones que pasan de unas coordenadas \ a otras." }], "Text"], Cell[BoxData[ \(\(\(gcord2cart[{u_, v_, w_}] := {u\^2 - v\^2, 2\ u\ v, w\^2}\ (*\ pasa\ de\ g - coordenadas\ a\ cartesianas\ *) \n \(\(cart2gcord[{x_, y_, z_}] := {\(\@\(\(-x\) + \@\(x\^2 + y\^2\)\)\ \((x + \@\(x\^2 + y\^2\ \))\)\)\/\(\@2\ y\), \@\(\(-x\) + \@\(x\^2 + y\^2\)\)\/\@2, \@z}\)\(\n\) \)\)\( (*\ pasa\ de\ cartesianas\ a\ g - coordenadas\ *) \)\)\)], "Input"], Cell["Comprobamos que son funciones inversas.", "Text"], Cell[BoxData[{ \(gcord2cart[cart2gcord[{x, y, z}]] // Simplify\), "\n", \(cart2gcord[gcord2cart[{u, v, w}]] // Simplify\)}], "Input"], Cell[TextData[{ "Volvamos a la situaci\[OAcute]n general y supongamos que ", Cell[BoxData[ \(TraditionalForm\`g : A \[Rule] B\)]], " define un sistema de coordenadas curvil\[IAcute]neas ortogonales en ", StyleBox["B", FontSlant->"Italic"], ". En tal caso los vectores columna de la matriz jacobiana de ", StyleBox["g", FontSlant->"Italic"], ", es decir las derivadas parciales de ", StyleBox["g", FontSlant->"Italic"], ", normalizados constituyen una base ortonormal de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], " que notaremos por ", Cell[BoxData[ \(TraditionalForm\`{e\_1, e\_2, e\_3}\)]], ". Se dice que esta base est\[AAcute] asociada al sistema de coordenadas \ dado y es a esa base a la que se refieren los vectores en el sistema de \ coordenadas definido por ", StyleBox["g", FontSlant->"Italic"], ". En general los vectores de la base dependen de las coordenadas, no es \ una base fija. Las normas eucl\[IAcute]deas de los vectores columna de la \ matriz jacobiana de ", StyleBox["g", FontSlant->"Italic"], " se llaman factores de escala o factores m\[EAcute]tricos, los \ representaremos por ", Cell[BoxData[ \(TraditionalForm\`{a, b, c}\)]], " (son funciones de u,v,w). " }], "Text"], Cell[BoxData[ \({a, b, c} = Sqrt[{u1 . u1, u2 . u2, u3 . u3}]\)], "Input"], Cell["Definimos los vectores de la base.", "Text"], Cell[BoxData[{ \(e1[{u_, v_, w_}] = u1/a\), "\n", \(e2[{u_, v_, w_}] = u2/b\), "\n", \(e3[{u_, v_, w_}] = u3/c\)}], "Input"], Cell["Ahora vamos a visualizar la base antes definida.", "Text"], Cell[BoxData[{ \(\(punto = cart2gcord[{0, 1, 1}];\)\), "\n", \(\(graf = Show[Graphics3D[{vector3D[punto], vector3D[punto, .5*e1[punto], Blue], vector3D[punto, .5*e2[punto], Green], vector3D[punto, .5*e3[punto], Red], Text["\", punto + .6*e1[punto]], Text["\", punto + .65*e2[punto]], \n\t\t\t\tText["\", punto + .6*e3[punto]]}, Axes -> True, PlotRange -> All, Ticks -> None, TextStyle -> {FontFamily -> "\", FontSize -> 14, FontWeight -> "\"}]];\)\)}], "Input"], Cell[BoxData[{ \(SpinShow[graf]\), \(SelectionMove[EvaluationNotebook[], All, GeneratedCell]; \n FrontEndTokenExecute["\"]\), \(FrontEndTokenExecute["\"]\), \(Clear[graf, punto]\)}], "Input"], Cell[TextData[{ "Para expresar el vector gradiente de un capo escalar ", StyleBox["f", FontSlant->"Italic"], " dado en coordenadas cartesianas" }], "Text"], Cell[BoxData[ FormBox[ FormBox[ RowBox[{\(\[Del]\(f(x, y, z)\)\), "=", RowBox[{ RowBox[{\(\[PartialD]f\/\[PartialD]x\), \((x\_1, x\_2, x\_3)\), StyleBox["i", FontWeight->"Bold", FontSlant->"Italic"]}], "+", RowBox[{\(\[PartialD]f\/\[PartialD]y\), \((x\_1, x\_2, x\_3)\), StyleBox["j", FontWeight->"Bold"]}], "+", RowBox[{\(\[PartialD]f\/\[PartialD]z\), \((x\_1, x\_2, x\_3)\), StyleBox["k", FontWeight->"Bold"]}]}]}], "TraditionalForm"], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "en g-coordendas, lo que hacemos es expresar las g-componentes del vector \ gradiente en la base ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[\({e\_1\), "TraditionalForm"], ",", \(e\_2\)}], "TraditionalForm"], ",", \(e\_3\)}], "}"}], TraditionalForm]]], " en funci\[OAcute]n de las componentes del vector gradiente en \ cartesianas. Para ello lo que se necesita es la matriz del cambio de base de \ cartesianas a g-coordendas. Dicha matriz es la inversa de la matriz cuyas \ columnas son los vectores ", Cell[BoxData[ \(TraditionalForm\`e\_1, \ e\_2\)]], " y ", Cell[BoxData[ \(TraditionalForm\`e\_3\)]], ". Como se trata de una matriz ortogonal su inversa es su traspuesta. \ Llamemos ", Cell[BoxData[ \(TraditionalForm\`\((f1, f2, f3)\)\)]], " a las componentes del vector ", Cell[BoxData[ \(TraditionalForm\`\[Del]f\)]], " en la base ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[\({e\_1\), "TraditionalForm"], ",", \(e\_2\)}], "TraditionalForm"], ",", \(e\_3\)}], "}"}], TraditionalForm]]], "." }], "Text"], Cell[BoxData[{ \(\(gradf[x_, y_, z_] = {D[f[x, y, z], x], D[f[x, y, z], y], D[f[x, y, z], z]};\)\), "\n", \(\(reglas = {x -> u\^2 - v\^2, y -> 2\ u\ v, z -> w\^2};\)\), "\n", \(\(M = {e1[{u, v, w}], e2[{u, v, w}], e3[{u, v, w}]};\)\), "\n", \(f1[u_, v_, w_] = \((\((M . gradf[x, y, z])\)[\([1]\)] /. reglas)\) // Simplify\), "\n", \(f2[u_, v_, w_] = \((\((M . gradf[x, y, z])\)[\([2]\)] /. reglas)\) // Simplify\), "\n", \(f3[u_, v_, w_] = \((\((M . gradf[x, y, z])\)[\([3]\)] /. reglas)\) // Simplify\)}], "Input"], Cell["Los resultados obtenidos pueden expresarse mejor como sigue", "Text"], Cell[BoxData[{ \(Simplify[\(1\/a\) D[f[u\^2 - v\^2, 2\ u\ v, w\^2], u] == f1[u, v, w]]\), "\n", \(Simplify[\(1\/b\) D[f[u\^2 - v\^2, 2\ u\ v, w\^2], v] == f2[u, v, w]]\), "\n", \(Simplify[\(1\/c\) D[f[u\^2 - v\^2, 2\ u\ v, w\^2], w] == f3[u, v, w]]\)}], "Input"], Cell[BoxData[ \(Simplify[\((gradf[x, y, z] /. reglas)\) \[Equal] f1[u, v, w] e1[{u, v, w}] + f2[u, v, w] e2[{u, v, w}] + f3[u, v, w] e3[{u, v, w}]]\)], "Input"], Cell["Hemos obtenido que", "Text"], Cell[BoxData[ FormBox[ FrameBox[ FormBox[ RowBox[{\(\((\[Del]f)\) \((g(u, v, w))\)\), "=", RowBox[{ RowBox[{\(1\/a\), \(\[PartialD]\/\[PartialD]u\), \(f( g(u, v, w))\), SubscriptBox[ StyleBox["e", FontWeight->"Bold", FontSlant->"Italic"], "1"]}], "+", RowBox[{\(1\/b\), \(\[PartialD]\/\[PartialD]v\), \(f( g(u, v, w))\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "2"]}], "+", RowBox[{\(1\/c\), \(\[PartialD]\/\[PartialD]w\), \(f( g(u, v, w))\), SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "3"]}]}]}], "TraditionalForm"]], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "debes entender bien lo que significa esta igualdad: a la izquierda aparece \ la funci\[OAcute]n gradiente de ", StyleBox["f", FontSlant->"Italic"] }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{"\[Del]", StyleBox["f", FontSlant->"Italic"]}], \((x, y, z)\)}], "=", RowBox[{ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontSlant->"Italic"]}], \(\[PartialD]x\)], \((x, y, z)\), StyleBox["i", FontWeight->"Bold"]}], "+", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontSlant->"Italic"]}], \(\[PartialD]y\)], \((x, y, z)\), StyleBox["j", FontWeight->"Bold"]}], "+", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontSlant->"Italic"]}], \(\[PartialD]z\)], \((x, y, z)\), StyleBox["k", FontWeight->"Bold"]}]}]}], " "}], TextForm]], "Text", TextAlignment->Center], Cell[TextData[{ "evaluada en el punto ", Cell[BoxData[ \(TraditionalForm\`g(u, v, w)\)]], ", a la derecha figuran las derivadas parciales ", StyleBox["de la funci\[OAcute]n compuesta ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`h(u, v, w) = f(g(u, v, w))\)]], " dividida cada una de ellas por el factor de escala que corresponde a su \ variable y multiplicada por el correspondiente vector de la base asociada a \ las nuevas coordenadas (y no olvides que dichos vectores dependen de ", Cell[BoxData[ \(TraditionalForm\`u, v\ , w\)]], "). La expresi\[OAcute]n obtenida para el gradiente es v\[AAcute]lida para \ todo sistema de coordenadas curvil\[IAcute]neas ortogonales." }], "Text"], Cell[TextData[{ "Supongamos que la funci\[OAcute]n de trayectoria de un movil viene dada \ por ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", \(g(u(t), v(t), w(t))\)}], TraditionalForm]]], " donde ", Cell[BoxData[ \(TraditionalForm\`\[Gamma](t) = \((u(t), v(t), w(t))\)\)]], " son las g-coordendas del vector ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], TraditionalForm]]], ". El vector velocidad del m\[OAcute]vil lo obtenemos derivando ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], "(", "t", ")"}], "=", \(g(\[Gamma](t))\)}], TraditionalForm]]], " con lo que resulta que ", Cell[BoxData[ \(TraditionalForm\`r\ ' \((t)\) = \(MJg(\[Gamma]( t))\) . \[Gamma]\ ' \((t)\)\)]], " donde ", Cell[BoxData[ \(TraditionalForm\`MJg(\[Gamma](t))\)]], " es la matriz jacobiana de ", StyleBox["g", FontSlant->"Italic"], " calculada en el punto ", Cell[BoxData[ \(TraditionalForm\`\[Gamma](t)\)]], ". De esta forma obtenemos el vector velocidad en funci\[OAcute]n de las \ derivadas de las funciones ", Cell[BoxData[ \(TraditionalForm\`u(t), \ v(t), \ w(t)\)]], ". Finalmente, expresamos el vector obtenido en la base ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[\({\(e\_1\)(t)\), "TraditionalForm"], ",", \(\(e\_2\)(t)\)}], "TraditionalForm"], ",", \(\(e\_3\)(t)\)}], "}"}], TraditionalForm]]], " (esta es la base ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[\({e\_1\), "TraditionalForm"], ",", \(e\_2\)}], "TraditionalForm"], ",", \(e\_3\)}], "}"}], TraditionalForm]]], " correspondiente al punto ", Cell[BoxData[ \(TraditionalForm\`\((u(t), v(t), w(t))\)\)]], ")." }], "Text"], Cell[BoxData[{ \(Clear[r]\), "\n", \(\(r[t_] = g[u[t], v[t], w[t]];\)\), "\n", \(\(r'\)[t]\)}], "Input"], Cell[BoxData[ \(\(\(Clear[r]\n \({e1[t_], e2[t_], e3[t_]} = {e1[{u[t], v[t], w[t]}], e2[{u[t], v[t], w[t]}], e3[{u[t], v[t], w[t]}]};\)\n \(r[t_] = g[u[t], v[t], w[t]];\)\n {r1[t_], r2[t_], r3[t_]} = \(\({\(r'\)[t] . e1[t], \(r'\)[t] . e2[t], \(r'\)[t] . e3[t]}\)\(//\)\(Simplify\)\( (*\ las\ componentes\ del\ vector\ velocidad\ en\ la\ base\ antes\ \ definida\ *) \)\)\n Simplify[\(r'\)[t] == r1[t] e1[t] + r2[t] e2[t] + r3[t] e3[t]]\)\(\ \)\( (*\ comprobamos\ la\ expresi\[OAcute]n\ de\ \(r'\)[ t]\ en\ la\ base\ {e1[t], e2[t], e3[t]}\ *) \)\)\)], "Input"], Cell[TextData[{ "Hemos obtenido que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold"], "'"}], \((t)\)}], "=", RowBox[{ RowBox[{"a", \((t)\), " ", \(u'\), \((t)\), RowBox[{ StyleBox[\(e\_1\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{"b", \((t)\), " ", \(v'\), \((t)\), RowBox[{ StyleBox[\(e\_2\), FontWeight->"Bold"], "(", "t", ")"}]}], "+", RowBox[{"c", \((t)\), " ", \(w'\), \((t)\), RowBox[{ StyleBox[\(e\_3\), FontWeight->"Bold"], "(", "t", ")"}]}]}]}], TraditionalForm]]], " donde ", Cell[BoxData[ \(TraditionalForm\`a(t), b(t), c(t)\)]], " son los factores de escala calculados en el punto ", Cell[BoxData[ \(TraditionalForm\`\((u(t), v(t), w(t))\)\)]], ". Esta expresi\[OAcute]n de la velocidad es v\[AAcute]lida para todo \ sistema de coordenadas curvil\[IAcute]neas ortogonales. Como la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ RowBox[{ StyleBox[\(e\_1\), FontWeight->"Bold"], "(", "t", ")"}], ",", RowBox[{ StyleBox[\(e\_2\), FontWeight->"Bold"], "(", "t", ")"}], ",", RowBox[{ StyleBox[\(e\_3\), FontWeight->"Bold"], "(", "t", ")"}]}], "}"}], TraditionalForm]]], " es ortonormal, la norma eucl\[IAcute]da del vector ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], " ", "'"}], \((t)\)}], TraditionalForm]]], " viene dada por" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"||", RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], " ", "'"}], \((t)\)}], \( || \^2\)}], "=", RowBox[{ RowBox[{ RowBox[{ StyleBox["r", FontWeight->"Bold"], " ", "'"}], RowBox[{\((t)\), " ", StyleBox[".", FontWeight->"Bold"], " ", RowBox[{ StyleBox["r", FontWeight->"Bold"], " ", "'"}]}], \((t)\)}], "=", \(\((\(a(t)\)\ u' \((t)\))\)\^2 + \((\(b(t)\)\ v' \ \((t)\))\)\^2 + \((\(c(t)\)\ w' \((t)\))\)\^2\)}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell["\<\ igualdad que en los libros de f\[IAcute]sica suelen escribir en la forma\ \>", "Text"], Cell[TextData[Cell[BoxData[ FormBox[ FrameBox[ FormBox[\(ds\^2 = \((a\ du)\)\^2 + \((b\ dv)\)\^2 + \((c\ dw)\)\^2\), "TraditionalForm"]], TraditionalForm]]]], "Text", TextAlignment->Center], Cell["\<\ y dicen que esta es la expresi\[OAcute]n del elemento diferencial de longitud \ de arco en las nuevas coordenadas. Decididamente, los autores de libros de f\ \[IAcute]sica no gustan de las derivadas y prefieren las diferenciales.\ \>", "Text"], Cell[TextData[{ "La divergencia de un campo vectorial ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"], " cuyas componentes en la base ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ StyleBox[\(e\_1\), FontWeight->"Bold"], ",", StyleBox[\(e\_2\), FontWeight->"Bold"], ",", StyleBox[\(e\_3\), FontWeight->"Bold"]}], "}"}], TraditionalForm]]], " asociada a las nuevas coordenadas son los campos escalares ", Cell[BoxData[ \(TraditionalForm\`F\_1, \ F\_2, \ F\_3\)]], "; es decir" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "(", \(g(u, v, w)\), ")"}], "=", RowBox[{ RowBox[{\(\(F\_1\)(u, v, w)\), StyleBox[\(e\_1\), FontWeight->"Bold"]}], "+", RowBox[{\(\(F\_2\)(u, v, w)\), StyleBox[\(e\_2\), FontWeight->"Bold"]}], "+", RowBox[{\(\(F\_3\)(u, v, w)\), StyleBox[\(e\_3\), FontWeight->"Bold"]}]}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell["se prueba que viene dada por", "Text"], Cell[BoxData[ FormBox[ FrameBox[ FormBox[ RowBox[{ RowBox[{"div", " ", StyleBox["F", FontWeight->"Bold"]}], "=", \(\(1\/\(a\ b\ c\)\) \((\(\[PartialD]\/\[PartialD]\ u\) \((b\ c\ F\_1)\) + \(\[PartialD]\/\[PartialD]\ v\) \((a\ c\ F\_2)\) + \(\[PartialD]\/\[PartialD]\ w\) \((a\ b\ \ F\_3)\))\)\)}], "TraditionalForm"]], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "donde se entiende que a la izquierda ", Cell[BoxData[ FormBox[ RowBox[{"div", " ", StyleBox["F", FontWeight->"Bold"]}], TraditionalForm]]], " est\[AAcute] evaluada en el punto ", Cell[BoxData[ \(TraditionalForm\`g \((u, v, w)\)\)]], " y a la derecha las derivadas parciales est\[AAcute]n evaluadas en el \ punto ", Cell[BoxData[ \(TraditionalForm\`\((u, v, w)\)\)]], "." }], "Text"], Cell[TextData[{ "En las mismas condiciones se verifica la siguiente igualdad para el \ rotacional de ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"] }], "Text"], Cell[BoxData[ FormBox[ FrameBox[ FormBox[ RowBox[{"rot", " ", StyleBox["F", FontWeight->"Bold"], " ", FormBox[\(\(=\)\(1\/\(a\ b\ c\)\)\), "TraditionalForm"], "determinante", " ", RowBox[{"(", GridBox[{ { RowBox[{"a", " ", SubscriptBox[ StyleBox["e", FontWeight->"Bold"], "1"]}], RowBox[{"b", " ", SubscriptBox[ StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontWeight->"Bold"]]}], RowBox[{"c", " ", SubscriptBox[ StyleBox["e", FontWeight->"Bold"], StyleBox["3", FontWeight->"Bold"]]}]}, {\(\[PartialD]\/\[PartialD]\ u\), \(\[PartialD]\/\[PartialD]\ v\), \(\[PartialD]\/\[PartialD]\ w\)}, {\(a\ F\_1\), \(b\ F\_2\), \(c\ F\_3\)} }], ")"}]}], "TraditionalForm"]], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "donde se entiende que a la izquierda ", Cell[BoxData[ FormBox[ RowBox[{"rot", " ", StyleBox["F", FontWeight->"Bold"]}], TraditionalForm]]], " est\[AAcute] evaluado en el punto ", Cell[BoxData[ \(TraditionalForm\`g \((u, v, w)\)\)]], " y a la derecha las derivadas parciales est\[AAcute]n evaluadas en el \ punto ", Cell[BoxData[ \(TraditionalForm\`\((u, v, w)\)\)]], "." }], "Text"], Cell[CellGroupData[{ Cell["Ejercicio 3", "Exercise"], Cell["\<\ Completa el estudio general de las coordendas curvil\[IAcute]neas ortogonales \ conforme al esquema seguido en el estudio de las coordenadas polares y esf\ \[EAcute]ricas. Es decir, debes calcular la aceleraci\[OAcute]n y divergencia \ en coordenadas curvil\[IAcute]neas ortogonales y hacer una intrepretaci\ \[OAcute]n de los factores de escala y de los elementos diferenciales de \ longitud y volumen.\ \>", "ExerciseText"] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 4", "Exercise"], Cell[TextData[{ "a) Representa la superficie cuya ecuaci\[OAcute]n en coordenadas esf\ \[EAcute]ricas es ", Cell[BoxData[ \(TraditionalForm\`r = 1\)]], ".\nb) Representa una parte de la superficie cuya ecuaci\[OAcute]n en \ coordenadas esf\[EAcute]ricas es ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Pi]/4\)]], ".\na) Representa una parte de la superficie ecuaci\[OAcute]n en \ coordenadas esf\[EAcute]ricas es ", Cell[BoxData[ \(TraditionalForm\`\[Phi] = \[Pi]/4\)]], ".\na) Representa una parte de la superficie cuya ecuaci\[OAcute]n en \ coordenadas cil\[IAcute]ndricas es ", Cell[BoxData[ \(TraditionalForm\`r = 1\)]], ".\na) Representa una parte de la superficie cuya ecuaci\[OAcute]n en \ coordenadas cil\[IAcute]ndricas es ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Pi]/4\)]], ".\na) Representa una parte la superficie cuya ecuaci\[OAcute]n en \ coordenadas esf\[EAcute]ricas es ", Cell[BoxData[ \(TraditionalForm\`r = 1\)]], "." }], "ExerciseText"] }, Open ]] }, Open ]] }, FrontEndVersion->"5.1 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 685}}, AutoGeneratedPackage->None, WindowSize->{1016, 635}, WindowMargins->{{0, Automatic}, {Automatic, 0}}, PageHeaders->{{Cell[ TextData[ "F. 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