(* Content-type: application/vnd.wolfram.mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 10.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 158, 7] NotebookDataLength[ 41846, 1194] NotebookOptionsPosition[ 39988, 1128] NotebookOutlinePosition[ 40330, 1143] CellTagsIndexPosition[ 40287, 1140] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[BoxData[ RowBox[{ RowBox[{"Quit", "[", "]"}], ";"}]], "Input", CellChangeTimes->{{3.7311072152584*^9, 3.7311072163036003`*^9}}], Cell[CellGroupData[{ Cell["Mec\[AAcute]nica Cl\[AAcute]sica 2018 1c - Turno A", "Chapter", CellChangeTimes->{{3.7311487609214*^9, 3.7311487993634*^9}, { 3.7311557874128*^9, 3.7311557877404003`*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "El problema es encontrar la superficie de revoluci\[OAcute]n de \ \[AAcute]rea m\[IAcute]nima definida por la curva ", 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es\n\n", Cell[BoxData[ RowBox[{ RowBox[{"A", "[", "y", "]"}], "=", " ", RowBox[{"2", "\[Pi]", RowBox[{ SubsuperscriptBox["\[Integral]", RowBox[{"-", "L"}], "L"], RowBox[{"y", RowBox[{"(", "x", ")"}], " ", SqrtBox[ RowBox[{"1", " ", "+", " ", RowBox[{ RowBox[{"y", "'"}], SuperscriptBox[ RowBox[{"(", "x", ")"}], "2"]}]}]], RowBox[{"\[DifferentialD]", "x"}]}]}]}]}]]], ".\n\nVimos en clase que las soluciones catenarias del problema de \ extremizaci\[OAcute]n son de la forma\n\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"y", "(", "x", ")"}], "=", " ", RowBox[{ FractionBox["1", "\[Alpha]"], "Cosh", " ", RowBox[{"(", RowBox[{"\[Alpha]", " ", "x"}], ")"}]}]}], TraditionalForm]], "Subtitle"], ".\n\nDebido a las condiciones en los extremos del intervalo, \[Alpha] debe \ ser soluci\[OAcute]n de la ecuaci\[OAcute]n\n\n", Cell[BoxData[ FormBox[ RowBox[{"\[Alpha]", " ", "=", " ", RowBox[{"Cosh", " ", RowBox[{"(", RowBox[{"\[Alpha]", " ", "L"}], ")"}]}]}], TraditionalForm]]], ".\n\nDependiendo del valor de ", Cell[BoxData[ FormBox["L", TraditionalForm]]], ", esta ecuaci\[OAcute]n puede tener dos, una o ninguna soluci\[OAcute]n, \ como es f\[AAcute]cil de ver gr\[AAcute]ficamente:" }], "Subchapter", CellChangeTimes->{{3.7323130920355997`*^9, 3.7323131750106*^9}, { 3.7323156030053997`*^9, 3.7323156052704*^9}}], Cell[BoxData[{ RowBox[{ RowBox[{ RowBox[{"{", RowBox[{"L1", ",", " ", "L2", ",", " ", "L3"}], "}"}], " ", "=", " ", RowBox[{"{", RowBox[{"0.5", ",", " ", "0.6627", ",", "0.8"}], "}"}]}], ";"}], "\[IndentingNewLine]", RowBox[{"Plot", "[", RowBox[{ RowBox[{"{", RowBox[{"\[Alpha]", ",", " ", RowBox[{"Cosh", "[", RowBox[{"\[Alpha]", " ", "L1"}], "]"}], ",", " ", RowBox[{"Cosh", "[", RowBox[{"\[Alpha]", " ", "L2"}], "]"}], ",", " ", RowBox[{"Cosh", "[", RowBox[{"\[Alpha]", " ", "L3"}], "]"}]}], "}"}], ",", " ", RowBox[{"{", RowBox[{"\[Alpha]", ",", " ", "0", ",", " ", "5"}], "}"}], ",", " ", "\[IndentingNewLine]", RowBox[{"PlotStyle", "\[Rule]", RowBox[{"{", RowBox[{ 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Eso da dos ecuaciones. En particular, el producto ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["x", "c"], " ", "=", " ", RowBox[{ SubscriptBox["L", "c"], SubscriptBox["\[Alpha]", "c"]}]}], TraditionalForm]], FormatType->"TraditionalForm"], " satisface la ecuaci\[OAcute]n\n\n", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["x", "c"], " ", "=", " ", RowBox[{"Coth", " ", RowBox[{ SubscriptBox["x", "c"], "."}]}]}], TraditionalForm]], FormatType->"TraditionalForm"], "\n\nLa soluci\[OAcute]n para ", Cell[BoxData[ FormBox[ SubscriptBox["x", "c"], TraditionalForm]], FormatType->"TraditionalForm"], " puede hallarse mediante FindRoot o usando el comando FixedPoint, ya que el \ problema est\[AAcute] expresado como ", Cell[BoxData[ FormBox[ RowBox[{"x", " ", "=", " ", RowBox[{"f", "(", "x", ")"}]}], TraditionalForm]], FormatType->"TraditionalForm"], "." }], "Subchapter", CellChangeTimes->{{3.7323096647436*^9, 3.7323097097416*^9}, { 3.7323099236876*^9, 3.7323100445916*^9}, {3.7323156487576*^9, 3.7323156757692003`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"FindRoot", "[", RowBox[{ RowBox[{"x", " ", "\[Equal]", " ", RowBox[{"Coth", "[", "x", "]"}]}], ",", " ", RowBox[{"{", RowBox[{"x", ",", " ", "1"}], "}"}]}], "]"}], "[", RowBox[{"[", "1", "]"}], "]"}], "[", RowBox[{"[", "2", "]"}], "]"}]], "Input", CellChangeTimes->{{3.7323100466266003`*^9, 3.7323100555866003`*^9}, { 3.7323100885906*^9, 3.7323101081745996`*^9}, {3.7323101599236*^9, 3.7323101839826*^9}, {3.7323103054556*^9, 3.7323103169066*^9}}], Cell[BoxData[ RowBox[{"xc", " ", "=", " ", RowBox[{"FixedPoint", "[", RowBox[{"Coth", ",", " ", "1."}], "]"}]}]], "Input", CellChangeTimes->{{3.7323100616366*^9, 3.7323100779625998`*^9}, { 3.7323101138386*^9, 3.7323101199316*^9}, {3.7323101725116*^9, 3.7323102040626*^9}, {3.7323103098546*^9, 3.7323103102855997`*^9}, { 3.7323114404406*^9, 3.7323114414986*^9}}] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Este valor del producto ", Cell[BoxData[ FormBox[ RowBox[{"L", " ", "\[Alpha]"}], TraditionalForm]], FormatType->"TraditionalForm"], " corresponde a \n\n", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["L", "c"], "\[TildeEqual]"}], TraditionalForm]], FormatType->"TraditionalForm"], " 0.662743\n\n", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[Alpha]", "c"], "\[TildeEqual]", "1.8101705806989774`", " "}], TraditionalForm]], FormatType->"TraditionalForm"] }], "Subchapter", CellChangeTimes->{{3.7323113443816*^9, 3.7323114236726*^9}, 3.7323132863356*^9, 3.7323156891102*^9}], Cell["En efecto:", "Text", CellChangeTimes->{{3.7323132913668003`*^9, 3.7323132943464003`*^9}}], Cell[BoxData[{ RowBox[{ RowBox[{"xc", " ", "=", " ", RowBox[{"FixedPoint", "[", RowBox[{"Coth", ",", " ", "1."}], "]"}]}], ";"}], "\[IndentingNewLine]", RowBox[{"Lc", " ", "=", " ", SqrtBox[ RowBox[{ SuperscriptBox["xc", "2"], "-", "1"}]]}], "\[IndentingNewLine]", RowBox[{"\[Alpha]c", " ", "=", " ", RowBox[{"xc", "/", "Lc"}]}]}], "Input", CellChangeTimes->{{3.7323114335386*^9, 3.7323114701736*^9}, { 3.732317117318*^9, 3.7323171191588*^9}}] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Podemos definir funciones que encuentren num\[EAcute]ricamente las ra\ \[IAcute]ces de la ecuaci\[OAcute]n \n\n\[Alpha] - Cosh (\[Alpha] L) = 0, \n\n\ en el caso en que existan. Como hay dos ra\[IAcute]ces ", Cell[BoxData[ FormBox[ SubscriptBox["\[Alpha]", "1"], TraditionalForm]], FormatType->"TraditionalForm"], " y ", Cell[BoxData[ FormBox[ SubscriptBox["\[Alpha]", "2"], TraditionalForm]], FormatType->"TraditionalForm"], ", tales que ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[Alpha]", "1"], "<", " ", SubscriptBox["\[Alpha]", "c"], "<", " ", SubscriptBox["\[Alpha]", "2"]}], TraditionalForm]], FormatType->"TraditionalForm"], ", el problema se complica un poco, porque el ", StyleBox["Mathematica", FontSlant->"Italic"], " busca una sola ra\[IAcute]z, y a veces puede ir a parar a una, y a veces a \ la otra. \n\nSi el valor tentativo para iniciar la b\[UAcute]squeda se fija \ en ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[Alpha]", "pr"], "=", " ", "1"}], TraditionalForm]], FormatType->"TraditionalForm"], ", FindRoot encuentra siempre la ra\[IAcute]z m\[AAcute]s peque\[NTilde]a, ", Cell[BoxData[ FormBox[ SubscriptBox["\[Alpha]", "1"], TraditionalForm]], FormatType->"TraditionalForm"], ". Para encontrar la segunda ra\[IAcute]z, podemos definir una nueva funci\ \[OAcute]n que no tenga una ra\[IAcute]z en ", Cell[BoxData[ FormBox[ SubscriptBox["\[Alpha]", "1"], TraditionalForm]], FormatType->"TraditionalForm"], " pero que s\[IAcute] la siga teniendo en ", Cell[BoxData[ FormBox[ SubscriptBox["\[Alpha]", "2"], TraditionalForm]], FormatType->"TraditionalForm"], ". Eso puede conseguirse buscando las ra\[IAcute]ces de \n", Cell[BoxData[ FormBox[ RowBox[{"\[IndentingNewLine]", RowBox[{ FractionBox[ RowBox[{"\[Alpha]", " ", "-", " ", RowBox[{"Cosh", " ", RowBox[{"(", RowBox[{"\[Alpha]", " ", "L"}], ")"}]}]}], RowBox[{"\[Alpha]", "-", SubscriptBox["\[Alpha]", "1"]}]], " ", "=", " ", "0"}]}], TraditionalForm]], FormatType->"TraditionalForm"], "\n\n(Est\[AAcute]mos usando impl\[IAcute]citamente que el cero de la funci\ \[OAcute]n \[Alpha] - ", Cell[BoxData[ FormBox[ RowBox[{"Cosh", "(", RowBox[{"\[Alpha]", " ", "L"}]}], TraditionalForm]], FormatType->"TraditionalForm"], ") es simple, de modo que en la vecindad de ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[Alpha]", "1"], " "}], TraditionalForm]], FormatType->"TraditionalForm"], "la funci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{"\[Alpha]", " ", "-", " ", RowBox[{"Cosh", "(", RowBox[{"\[Alpha]", " ", "L"}], ")"}]}], TraditionalForm]], FormatType->"TraditionalForm"], " se comporta linealmente).\n\nAs\[IAcute], para encontrar la segunda ra\ \[IAcute]z tenemos que usar el resultado obtenido para la primera." }], "Subchapter", CellChangeTimes->{{3.7323118127216*^9, 3.7323123003546*^9}, { 3.732315705383*^9, 3.732315709049*^9}, {3.7323157539849997`*^9, 3.7323157734236*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"(*", "Auxiliare", "*)"}], "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"xc", " ", "=", " ", RowBox[{"FixedPoint", "[", RowBox[{"Coth", ",", " ", "1."}], "]"}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"Lc", " ", "=", " ", SqrtBox[ RowBox[{ SuperscriptBox["xc", "2"], "-", "1"}]]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"\[Alpha]c", " ", "=", " ", RowBox[{"xc", "/", "Lc"}]}], ";"}], "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"(*", RowBox[{"La", " ", "primera", " ", "ra\[IAcute]z"}], " ", "*)"}], "\[IndentingNewLine]", RowBox[{"(*", " ", RowBox[{"?", RowBox[{ "NumericQ", " ", "incluido", " ", "en", " ", "la", " ", 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Independientemente del valor de ", Cell[BoxData[ FormBox["L", TraditionalForm]]], ", el \[AAcute]rea de esta superficie es ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["A", RowBox[{"G", " "}]], "=", " ", RowBox[{"2", "\[Pi]"}]}], TraditionalForm]]], ", el \[AAcute]rea de los dos discos de radio 1.\n\nLa funci\[OAcute]n ", Cell[BoxData[ FormBox[ SubscriptBox["y", "G"], TraditionalForm]]], " queda afuera de la familia de funciones continuas y derivables \ contempladas en el c\[AAcute]lculo variacional, pero genera de todos modos \ una superficie admisible. \n\nSi graficamos ahora las tres \[AAcute]reas, ", Cell[BoxData[ FormBox[ SubscriptBox["A", "1"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ SubscriptBox["A", "2"], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ SubscriptBox["A", "G"], TraditionalForm]]], ", nos encontramos con una situaci\[OAcute]n no prevista: existe un ", Cell[BoxData[ FormBox["L", TraditionalForm]]], " m\[AAcute]ximo, menor que ", Cell[BoxData[ FormBox[ SubscriptBox["L", "c"], TraditionalForm]]], ", tal que la soluci\[OAcute]n de Goldschmidt da un \[AAcute]rea menor que \ la de las catenarias. 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