(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 7.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 60801, 1763] NotebookOptionsPosition[ 39467, 1290] NotebookOutlinePosition[ 56103, 1588] CellTagsIndexPosition[ 56060, 1585] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["Ejercicios propuestos", "Title", CellChangeTimes->{{3.4715936678125*^9, 3.47159367234375*^9}}], Cell[CellGroupData[{ Cell["Ejercicio 1", "Section", CellChangeTimes->{3.469519382875*^9, 3.469519436171875*^9, 3.47159369103125*^9}], Cell[CellGroupData[{ Cell[TextData[{ "Sea f la funci\[OAcute]n definida por : f(x) = ", Cell[BoxData[ FormBox[ StyleBox[ FractionBox["x", RowBox[{"1", "+", SuperscriptBox["x", "5"]}]], FontSize->18], TraditionalForm]]], ", y la curva C de ecuaci\[OAcute]n : y = f(x). Se pide :\n1\.ba.- \ Representar en un mismo gr\[AAcute]fico las funciones f, f' y f''.\n2\.ba.- \ Hallar el \[UAcute]nico extremo relativo de la curva C.\n3\.ba.- Determinar \ el \[UAcute]nico punto de inflexi\[OAcute]n de la curva C.\n4\.ba.- Hallar el \ \[AAcute]rea del recinto limitado por la tangente a la curva C en el origen \ de coordenadas, la curva C y la tangente a la curva C en su punto de inflexi\ \[OAcute]n." }], "Subsection", CellChangeTimes->{3.471593684421875*^9}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593686125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 2", "Section", CellChangeTimes->{3.469519438515625*^9, 3.471593700203125*^9}], Cell[CellGroupData[{ Cell[TextData[{ "Sea f(x) = k ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["x", "2"], "+", "x", "-", "2"}], TraditionalForm]]], " , Siendo k un n\[UAcute]mero natural, \[DownQuestion]Cu\[AAcute]l sera el \ primer valor de k para el que f(2) > 40?. Representar gr\[AAcute]ficamente la \ f(x) correspondiente a dicho valor de k en color y trazo m\[AAcute]s grueso." }], "Subsection", CellChangeTimes->{3.471593695703125*^9}], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593697328125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 3", "Section", CellChangeTimes->{{3.469519464953125*^9, 3.4695194655625*^9}, 3.471593713859375*^9}], Cell[CellGroupData[{ Cell[TextData[{ "Dadas las curvas de ecuaciones :\n\tf(x) = 3 ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["x", "3"], "-", SuperscriptBox["x", "2"], "-", RowBox[{"10", "x"}]}], TraditionalForm]]], " \n\tg(x) = ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"-", SuperscriptBox["x", "2"]}], "+", " ", RowBox[{"2", "x"}]}], TraditionalForm]]], ", \tse pide :\na) Calcular sus puntos de corte.\nb) El \[AAcute]rea \ comprendida entre dichas curvas." }], "Subsection", CellChangeTimes->{3.471593710515625*^9}], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.47159371571875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 4", "Section", CellChangeTimes->{{3.46951951609375*^9, 3.4695195170625*^9}, 3.471593736109375*^9}], Cell[CellGroupData[{ Cell["\<\ Las ecuaciones param\[EAcute]tricas de un cilindro de eje OZ, altura L y \ radio R, utilizando como par\[AAcute]metros el \[AAcute]ngulo polar \[Theta] \ y la cota z, son las siguientes: \t\tx=R cos(\[Theta]) \t\ty=R sen(\[Theta]) \[Theta] \[Epsilon][0, 2\[Pi]], z \[Epsilon][0, L] \t\tz=z Representar, en el mismo gr\[AAcute]fico, dos cilindros de altura L=20 y \ radios R=5 y R=6, respectivamente. Dibujar tambi\[EAcute]n en el mismo gr\ \[AAcute]fico y en plano z=20, una serie de circunferencias de centro (0,0), \ variando sus radios entre 5 y 6. (Con ellas se conseguir\[AAcute] cerrar el \ espacio entre los dos cilindros). P\[OAcute]ngase al gr\[AAcute]fico \ conjunto, el t\[IAcute]tulo \"Dibujo de un tubo\" y elim\[IAcute]nense los \ ejes.\t\ \>", "Subsection", CellChangeTimes->{3.471593732796875*^9}, TextAlignment->Left, TextJustification->1], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593738125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 5", "Section", CellChangeTimes->{{3.46951955253125*^9, 3.469519554015625*^9}, { 3.47159376003125*^9, 3.471593760671875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ " Se considera la serie : \t", Cell[BoxData[ RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"k", "=", "1"}], "\[Infinity]"], FractionBox["1", SuperscriptBox["k", "4"]]}]]], " , y se pide:\n1\.ba.- Obtener su valor exacto, guard\[AAcute]ndolo en la \ variable: var1.\n2\.ba.- Representar gr\[AAcute]ficamente ", Cell[BoxData[ RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"k", "=", "1"}], "n"], FractionBox["1", SuperscriptBox["k", "4"]]}]]], ", para n = 1, 2, ... 20, almacenando previamente estos valores en una lista \ denominada sumpar.\n3\.ba.- Calcular utilizando un ciclo For, cu\[AAcute]ntos \ t\[EAcute]rminos de la serie hay que sumar para que el valor de esta suma \ difiera del valor exacto en una cantidad menor que ", Cell[BoxData[ FormBox[ SuperscriptBox["10", RowBox[{"-", "3"}]], TraditionalForm]]], "." }], "Subsection", CellChangeTimes->{3.471593756125*^9}, TextAlignment->Left, TextJustification->1], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593758234375*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 6", "Section", CellChangeTimes->{{3.469519674078125*^9, 3.4695196749375*^9}, { 3.471593763984375*^9, 3.471593764890625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ " Calcular las matrices B que verifican la siguiente ecuaci\[OAcute]n \ matricial :\n\tA B + ", Cell[BoxData[ FormBox[ SuperscriptBox["Z", "t"], TraditionalForm]]], " = (0)\nsiendo A = ", Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"a", "0"}, {"0", RowBox[{"2", " ", "a"}]} }], "\[NoBreak]", ")"}], Function[BoxForm`e$, MatrixForm[BoxForm`e$]]]]], " , Z = ", Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"b", "0"}, {"c", RowBox[{"3", " ", "b"}]} }], "\[NoBreak]", ")"}], Function[BoxForm`e$, MatrixForm[BoxForm`e$]]]]], ", y (0) la matriz nula." }], "Subsection", TextAlignment->Left, TextJustification->1], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.4715937696875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 7", "Section", CellChangeTimes->{{3.469519789828125*^9, 3.4695197911875*^9}, { 3.471593772609375*^9, 3.4715937735625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Dada la funci\[OAcute]n f(x,y,z) = ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["x", "4"], "+", SuperscriptBox["y", "4"], "+", SuperscriptBox["z", "4"]}], TraditionalForm]]], ", se pide:\na) Obtener el vector gradiente. Dibujar el campo definido por \ dicho vector en [-1,1] x [-1,1] x [-1,1].\nb) Hallar la Laplaciana de f : \ \[CapitalDelta]f =", StyleBox[" ", FontSize->16], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ SuperscriptBox["\[PartialD]", "2"], "f"}], RowBox[{"\[PartialD]", SuperscriptBox["x", "2"]}]], TraditionalForm]], FontSize->16], StyleBox["+ ", FontSize->16], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ SuperscriptBox["\[PartialD]", "2"], "f"}], RowBox[{"\[PartialD]", SuperscriptBox["y", "2"]}]], TraditionalForm]], FontSize->16], StyleBox[" + ", FontSize->16], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ SuperscriptBox["\[PartialD]", "2"], "f"}], RowBox[{"\[PartialD]", SuperscriptBox["z", "2"]}]], TraditionalForm]], FontSize->16], " , mediante la sentencia adecuada de ", StyleBox["Mathematica", FontSlant->"Italic"], " y comprobar que coincide con la divergencia del gradiente." }], "Subsection", CellChangeTimes->{3.4715937770625*^9}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593779046875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 8", "Section", CellChangeTimes->{{3.4695198185*^9, 3.46951981978125*^9}, {3.471593787625*^9, 3.47159378840625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Sea la funci\[OAcute]n :\n \tf(x) =", Cell[BoxData[ FormBox[ TagBox["", #& ], TraditionalForm]]], Cell[BoxData[ FormBox[ TagBox[ FractionBox["x", RowBox[{ SuperscriptBox["x", "2"], "+", "1"}]], #& ], TraditionalForm]]], "\na) Representar gr\[AAcute]ficamente f(x) en el intervalo [-2,2].\nb) \ Calcular los ocho primeros desarrollos ", StyleBox["distintos", FontSlant->"Italic"], " en serie de potencias de f(x) en el origen . \nc) Representar cada uno de \ los polinomios de Taylor correspondientes a los desarrollos anteriores y f(x) \ conjuntamente, con un rango de las ordenadas perteneciente al intervalo \ [-2,2]. A la vista de la gr\[AAcute]fica , \[DownQuestion]En qu\[EAcute] \ intervalo del eje real est\[AAcute] asegurada la convergencia?\nd) \ \[DownQuestion]De qu\[EAcute] grado es el polinomio de Taylor, ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["p", "i"], "(", "x", ")"}], TraditionalForm]]], " que verifica que \[VerticalSeparator]f(0.5)-", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["p", "i"], "("}], TraditionalForm]]], "0.5)\[VerticalSeparator] <", Cell[BoxData[ FormBox[ SuperscriptBox["10", RowBox[{"-", "6"}]], TraditionalForm]]], "? Utilizar el comando For, presentando en cada iteraci\[OAcute]n el grado \ del polinomio y el valor de \[VerticalSeparator]f(0.5)-", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["p", "i"], "("}], TraditionalForm]]], "0.5)\[VerticalSeparator]." }], "Subsection", TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593786171875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 9", "Section", CellChangeTimes->{{3.469519884875*^9, 3.469519885359375*^9}, { 3.471593810125*^9, 3.471593811046875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "a) El plano tangente a una superficie z=f(x,y), en el punto \ (x0,y0,f(x0,y0)) existe si y s\[OAcute]lo ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["\[PartialD]", "x"], " ", RowBox[{"f", "(", RowBox[{"x", ",", "y"}], ")"}]}], " "}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[PartialD]", "y"], " ", RowBox[{"f", "(", RowBox[{"x", ",", "y"}], ")"}]}], TraditionalForm]]], " son continuas en un entorno de (x0,y0). En tal caso, la ecuaci\[OAcute]n \ del plano tangente en el punto (x0,y0,f(x0,y0)) viene dado por: \ z=a(x-x0)+b(y-y0)+c siendo a y b los resultados de evaluar ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["\[PartialD]", "x"], " ", RowBox[{"f", "(", RowBox[{"x", ",", "y"}], ")"}]}], " "}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[PartialD]", "y"], " ", RowBox[{"f", "(", RowBox[{"x", ",", "y"}], ")"}]}], TraditionalForm]]], " en (x0,y0), respectivamente y c=f(x0,y0).\nTeniendo esto en cuenta \ representar la superficie z=xy ", Cell[BoxData[ FormBox[ SuperscriptBox["e", RowBox[{ RowBox[{"-", SuperscriptBox["x", "2"]}], "-", SuperscriptBox["y", "2"]}]], TraditionalForm]]], "en la regi\[OAcute]n [-2,2]x[-1,1], junto con su plano tangente en el \ punto correspondiente a x=1, y=0.5 en la regi\[OAcute]n [0.5,1]x[0.25,0.75].\n\ b) Sea f(x,y,z) = z - xy ", Cell[BoxData[ FormBox[ SuperscriptBox["e", RowBox[{ RowBox[{"-", SuperscriptBox["x", "2"]}], "-", SuperscriptBox["y", "2"]}]], TraditionalForm]]], ". Calcular y representar el gradiente de f en la regi\[OAcute]n \ [-2,2]x[-1,1]. Calcular el vector gradiente en el punto (1,0.5,1) ." }], "Subsection", CellChangeTimes->{3.471593804921875*^9}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.4715938073125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 10", "Section", CellChangeTimes->{{3.469519909734375*^9, 3.469519910265625*^9}, { 3.471593821265625*^9, 3.47159382146875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Dada la funci\[OAcute]n f, definida por :\n\tf(x) = ", Cell[BoxData[ FormBox[ TagBox[ StyleBox[ RowBox[{"{", GridBox[{ { RowBox[{ RowBox[{ RowBox[{"-", "3"}], " ", RowBox[{"sen", "(", "x", ")"}], " ", "x"}], " ", "\[LessEqual]", " ", RowBox[{ RowBox[{"-", " ", "\[Pi]"}], "/", "2"}]}]}, { RowBox[{ RowBox[{ RowBox[{"a", " ", RowBox[{"sen", "(", "x", ")"}]}], "+", "b", " ", "-", " ", RowBox[{"\[Pi]", "/", "2"}]}], "\[LessEqual]", "x", "\[LessEqual]", " ", RowBox[{"\[Pi]", "/", "2"}], " "}]}, { RowBox[{ RowBox[{ RowBox[{"cos", "(", "x", ")"}], " ", "x"}], " ", "\[GreaterEqual]", " ", RowBox[{"\[Pi]", "/", "2"}]}]} }]}], ShowAutoStyles->False], #& ], TraditionalForm]]], ", se pide:\na) Determinar los valores de a y b para que f sea continua en \ todo \[DoubleStruckCapitalR].\nb) Representar conjuntamente la \ funci\[OAcute]n obtenida y la par\[AAcute]bola y = ", Cell[BoxData[ FormBox[ SuperscriptBox["x", "2"], TraditionalForm]]], ".\nc) Obtener los puntos de corte de ambas curvas.\nd) Calcular el \ \[AAcute]rea limitada por ambas entres los puntos del apdo. anterior." }], "Subsection", CellChangeTimes->{{3.46951991365625*^9, 3.46951994125*^9}}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471593824546875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 11", "Section", CellChangeTimes->{{3.469519977734375*^9, 3.46951997846875*^9}, { 3.47159382775*^9, 3.47159382796875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Dada la matriz M=", Cell[BoxData[ FormBox[ RowBox[{"(", GridBox[{ {"1", "2", "2"}, {"2", "1", "2"}, {"2", "2", "1"} }], ")"}], TraditionalForm]]], " \n\na)Obtener el polinomio caracteristico de la matriz M: \ \[VerticalSeparator]M-\[Lambda] I\[VerticalSeparator]\nb)Obtener las ra\ \[IAcute]ces de dicho polinomio.\nc) Obtener los valores propios de M \ utilizando el comando apropiado de ", StyleBox["Mathematica", FontSlant->"Italic"], ".\nd) Diagonalizar la matriz M y definir la matriz de paso P.\nNota: Siendo \ diag la matriz que resulta al diagonalizar M, se cumple:\n\tdiag= ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["P", RowBox[{"-", "1"}]], ".", " ", "M", ".", " ", "P"}], TraditionalForm]], "None", FormatType->"TraditionalForm"], " .\nSe puede utilizar el comando JordanDecomposition." }], "Subsection", CellChangeTimes->{{3.471593844875*^9, 3.4715940558125*^9}}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471594058625*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 12", "Section", CellChangeTimes->{{3.469519995328125*^9, 3.469519997640625*^9}, { 3.4715940618125*^9, 3.471594062625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "a) Obtener los valores de a , b y c de modo que el gr\[AAcute]fico del \ polinomio a", Cell[BoxData[ FormBox[ RowBox[{" ", SuperscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"]}], TraditionalForm]]], "+ b x + c pase por el punto (-1,0) y tenga una tangente horizontal en \ (-2,9).\nb) Representar graficamente el polinomio definido en el apartado \ anterior. \nc) Utilizando el comando adecuado, obtener las ra\[IAcute]ces \ del polinomio. \t" }], "Subsection", CellChangeTimes->{3.47159406775*^9}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471594069125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 13", "Section", CellChangeTimes->{{3.46952001665625*^9, 3.469520017328125*^9}, { 3.471594071265625*^9, 3.47159407290625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ " Sea A el \[AAcute]rea limitada por la recta y=x, la exponencial y= ", Cell[BoxData[ FormBox[ SuperscriptBox["e", RowBox[{ RowBox[{"-", " ", "0.5"}], "x"}]], TraditionalForm]]], " , el eje de abscisas y la recta x=a. Obtener una lista que contenga los \ cinco valores de a que hacen que A tome los valores 1.0, 1.1, 1.2, 1.3, \ 1.4 y 1.5.\t" }], "Subsection", TextAlignment->Left, TextJustification->1], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.47159407715625*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 14", "Section", CellChangeTimes->{{3.46952007946875*^9, 3.4695200803125*^9}, { 3.471594079984375*^9, 3.471594080171875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "a.- Crear una matriz, M, de orden 3 cuyos elementos sean el producto de los \ primeros n\[UAcute]meros enteros pares.\nb.- Obtener el vector, m1, formado \ por los elementos de la diagonal secundaria.\nc.- Obtener el adjunto del \ elemento ", Cell[BoxData[ SubscriptBox["m", RowBox[{"2", ",", "2"}]]]], ".\nd.- Calcular el rango de la matriz de M, obteniendo una matriz \ equivalente a M. \ \ \t \t\t\t\t " }], "Subsection", CellChangeTimes->{ 3.469520071328125*^9, {3.469520108890625*^9, 3.469520120109375*^9}}, TextAlignment->Left, TextJustification->1], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.47159409375*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 15", "Section", CellChangeTimes->{{3.4695201618125*^9, 3.46952016234375*^9}, { 3.471594096140625*^9, 3.471594096296875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Dada la familia de funciones:\n\t\t\tf(x,n) = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ SuperscriptBox["x", "2"]], FontVariations->{"CompatibilityType"->0}], StyleBox[" (1", FontVariations->{"CompatibilityType"->0}], "+ ", Cell[BoxData[ FractionBox["1", RowBox[{"(", RowBox[{"1", "+", SuperscriptBox["x", "2"]}], ")"}]]]], "+", Cell[BoxData[ FractionBox["1", SuperscriptBox[ RowBox[{"(", RowBox[{"1", "+", SuperscriptBox["x", "2"]}], ")"}], "2"]]]], "+...+", Cell[BoxData[ FractionBox["1", SuperscriptBox[ RowBox[{"(", RowBox[{"1", "+", SuperscriptBox["x", "2"]}], ")"}], "n"]]]], StyleBox[")", FontVariations->{"CompatibilityType"->0}], "\na) ", StyleBox["Dibujar en la misma figura en dos lineas distintas las \ gr\[AAcute]ficas de las funciones f(x,n) para los valores de n = 3, ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ SuperscriptBox["3", "2"]]], ", ", Cell[BoxData[ SuperscriptBox["3", "3"]]], " ", StyleBox["y", FontVariations->{"CompatibilityType"->0}], " ", Cell[BoxData[ SuperscriptBox["3", "4"]]], ", ", StyleBox["en el intervalo x\[Epsilon][-1,1].\nb) Hallar la expresi\[OAcute]n \ del \[AAcute]rea encerrada por cada curva de la familia y el eje X, para los \ valores de x comprendidos entre -1 y 1.\nc) Obtener una lista con los valores \ aproximados de tales \[AAcute]reas para n = 3, ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ SuperscriptBox["3", "2"]]], ", ", Cell[BoxData[ SuperscriptBox["3", "3"]]], " ", StyleBox["y", FontVariations->{"CompatibilityType"->0}], " ", Cell[BoxData[ SuperscriptBox["3", "4"]]], ".\nd) ", StyleBox["Hallar mediante un ciclo For el menor valor de n para el que dicha \ \[AAcute]rea sea mayor que 2.3. Obtener tambi\[EAcute]n el valor aproximado \ de tal \[AAcute]rea.", FontVariations->{"CompatibilityType"->0}] }], "Subsection", CellChangeTimes->{{3.46952015484375*^9, 3.469520167328125*^9}, 3.471594100703125*^9}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471594105125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 16", "Section", CellChangeTimes->{{3.46952020771875*^9, 3.469520208921875*^9}, { 3.471594108390625*^9, 3.471594109265625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Resolver el siguiente sistema de ecuaciones no lineales: \n \t", Cell[BoxData[{ FormBox[ RowBox[{ RowBox[{ SuperscriptBox["x", "2"], "+", "x", "-", SuperscriptBox["y", "2"]}], " ", "=", " ", "1"}], TraditionalForm], "\[IndentingNewLine]", FormBox[ RowBox[{ RowBox[{"y", " ", "-", " ", RowBox[{"sen", "(", SuperscriptBox["x", "2"], ")"}]}], " ", "=", " ", "0"}], TraditionalForm]}]], "\t\t\t\t\t\t\t\t\t" }], "Subsection"], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.47159411396875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 17", "Section", CellChangeTimes->{{3.4695202826875*^9, 3.469520283375*^9}, { 3.47159411690625*^9, 3.47159411715625*^9}}], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Dada la matriz A=", FontFamily->"Times New Roman", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ RowBox[{"(", GridBox[{ {"a", "1", "e"}, {"0", "b", "f"}, {"0", "0", "c"} }], ")"}], TraditionalForm]], FontFamily->"Times New Roman"], StyleBox[" \n", FontFamily->"Times New Roman"], StyleBox["a) Hallar los valores de las constantes a, b, c para que la matriz \ A tenga a r=2 como autovalor triple. 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(Como puede observarse el primer dibujo \ obtenido no se corresponde con la realidad). \n b) Calcular el vector \ gradiente de la funci\[OAcute]n f(x,y). 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Representar la curva cuyas ecuaciones param\[EAcute]tricas son: x=tcos(t) e y \ = tsen(t), para t\[Element][0,4\[Pi]]. Representar los puntos de corte de \ la curva anterior con los ejes cartesianos, para \ t\[Element][0,4\[Pi]] y mostrarlos conjuntamente con la curva.\ \>", "Subsection", TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471594153703125*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 20", "Section", CellChangeTimes->{{3.469520596578125*^9, 3.46952059753125*^9}, { 3.471594156953125*^9, 3.471594157359375*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Determinar los coeficientes a,b y c para que la funci\[OAcute]n:\n\n\t\te(", StyleBox["x,y", FontSlant->"Italic"], ")=", Cell[BoxData[ FormBox["a", TraditionalForm]]], "+b ", StyleBox["xy", FontSlant->"Italic"], "+", Cell[BoxData[ FormBox[ FractionBox["1", RowBox[{"c", "+", RowBox[{"x", " ", SuperscriptBox["y", "2"], " "}]}]], TraditionalForm]]], " \nverifique las condiciones siguientes:\na) Valga 3/2 en (2,0).\nb) Su \ vector gradiente en [1,1] sea el vector (3/4,1/2).\n\nRepresentar en el \ dominio [0,3]x[0,1] la funci\[OAcute]n o funciones e(x,y) obtenidas." }], "Subsection"], Cell[BoxData[ RowBox[{"\[IndentingNewLine]", RowBox[{"\[Piecewise]", GridBox[{ {"\[Placeholder]", "\[Placeholder]"}, {"\[Placeholder]", "\[Placeholder]"} }, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxItemSize->{ "Columns" -> {{Automatic}}, "ColumnsIndexed" -> {}, "Rows" -> {{1.}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.84]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}]}]}]], "Input", CellChangeTimes->{{3.471594162921875*^9, 3.47159417125*^9}}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 21", "Section", CellChangeTimes->{{3.469520627859375*^9, 3.469520628953125*^9}, { 3.471594300875*^9, 3.47159430153125*^9}}], Cell[CellGroupData[{ Cell[TextData[{ " ", StyleBox["Consid\[EAcute]rese la matriz tridiagonal definida por:\n ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["a", RowBox[{"i", ",", "j"}]], " ", "="}], TraditionalForm]], "None", FormatType->"TraditionalForm"], StyleBox[" ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ RowBox[{"\[Piecewise]", GridBox[{ {"i", RowBox[{ RowBox[{"si", " ", "i"}], "=", "j"}]}, {"j", RowBox[{ RowBox[{"si", " ", "i"}], "=", RowBox[{"j", "+", "1"}]}]}, {"j", RowBox[{ RowBox[{"si", " ", "i"}], "=", RowBox[{"j", "-", "1"}]}]}, {"0", RowBox[{"en", " ", "otro", " ", "caso"}]} }, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxItemSize->{ "Columns" -> {{Automatic}}, "ColumnsIndexed" -> {}, "Rows" -> {{1.}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.84]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}]}], TraditionalForm]], "None", FormatType->"TraditionalForm"], " ", StyleBox[" siendo 1\[LessEqual] i, j \[LessEqual] n \n\n\ Introducir dicha matriz para n=10. Calcular su determinante y sus autovalores \ y autovectores aproximados . Extraer los elementos de su diagonal principal, \ y los de sus subdiagonales y superdiagonales no nulas.", FontVariations->{"CompatibilityType"->0}] }], "Subsection", CellChangeTimes->{{3.471594176734375*^9, 3.47159429534375*^9}}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471594297890625*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 22", "Section", CellChangeTimes->{{3.46952064325*^9, 3.469520643921875*^9}, { 3.471594309765625*^9, 3.471594311171875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Se quiere encontrar el m\[IAcute]nimo de la funci\[OAcute]n de dos \ variables h(a,b):\n\n\t", StyleBox["h(a,b) ", FontSize->12], StyleBox["=", FontSize->10], Cell[BoxData[ StyleBox[ RowBox[{ SuperscriptBox[ RowBox[{"(", StyleBox[ RowBox[{ RowBox[{"-", "25"}], "-", FractionBox["1", StyleBox[ RowBox[{ RowBox[{"a", " ", SuperscriptBox[ RowBox[{ StyleBox["(", FontSize->14], RowBox[{"Log", "[", "2", "]"}], StyleBox[")", FontSize->14]}], "2"]}], "+", "b"}], FontSize->14]]}], FontSize->12], ")"}], "2"], "+", SuperscriptBox[ RowBox[{ StyleBox["(", FontSize->12], StyleBox[ RowBox[{ FractionBox["1", "3"], "-", FractionBox["1", StyleBox[ RowBox[{ RowBox[{"a", " ", SuperscriptBox[ RowBox[{ StyleBox["(", FontSize->14], RowBox[{"Log", "[", "4", "]"}], StyleBox[")", FontSize->14]}], "2"]}], "+", "b"}], FontSize->14]]}], FontSize->12], ")"}], "2"]}], FontSize->14]], FontSize->10], StyleBox["\n", FontSize->12], "\nPara ello se pide:\na) Representar h(a,b) en el dominio [1,3]x[-2,0], \ mediante el comando que se quiera.\nb) Representar las dos derivadas \ parciales de primer orden de h(a,b) en el dominio [1,3]x[-2,0], mediante el \ comando que se quiera.\nc) Mediante el comando ", StyleBox["FindRoot", "Input"], ", resolver el sistema de ecuaciones al que equivale la condici\[OAcute]n de \ m\[IAcute]nimo relativo, es decir, que las dos derivadas parciales de primer \ orden se anulen. Tomar como punto de partida para la aproximaci\[OAcute]n el \ a=2, b=-1.\nd) Calcular el valor de h en el m\[IAcute]nimo y representar \ h(a,b) en un peque\[NTilde]o entorno cuadrado cercano al punto m\[IAcute]nimo \ obtenido (tomar un radio de 0.001 en cada una de las dos direcciones) ." }], "Subsection", CellChangeTimes->{3.471594315671875*^9}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.471594317421875*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 23", "Section", CellChangeTimes->{{3.46952064325*^9, 3.469520643921875*^9}, { 3.471594309765625*^9, 3.471594311171875*^9}, 3.471594419234375*^9}], Cell[CellGroupData[{ Cell[TextData[{ "Sea f(x) = k ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["x", "2"], "+", "x", "-", "2"}], TraditionalForm]]], " , Siendo k un n\[UAcute]mero natural, \[DownQuestion]Cu\[AAcute]l sera el \ primer valor de k para el que f(2) > 40?. Representar gr\[AAcute]ficamente la \ f(x) correspondiente a dicho valor de k en color y trazo m\[AAcute]s grueso." }], "Subsection", CellChangeTimes->{{3.466494846625*^9, 3.46649485271875*^9}}, TextJustification->1.], Cell[BoxData["\[IndentingNewLine]"], "Input", CellChangeTimes->{3.4715944914375*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ejercicio 24", "Section", CellChangeTimes->{{3.46952064325*^9, 3.469520643921875*^9}, { 3.471594309765625*^9, 3.471594311171875*^9}, 3.471594419234375*^9, { 3.47159453928125*^9, 3.47159454*^9}}], Cell[CellGroupData[{ Cell[TextData[{ " Definir una matriz A de orden 10 donde cada elemento ", Cell[BoxData[ RowBox[{ SubscriptBox["a", "ij"], "=", RowBox[{ StyleBox["{", ShowAutoStyles->False], StyleBox[GridBox[{ { RowBox[{ RowBox[{ FormBox[ RowBox[{ SuperscriptBox["i", "5"], "+", "j"}], TraditionalForm], " ", "i"}], " ", "=", RowBox[{"2", "j", " "}]}]}, { RowBox[{ FormBox[ RowBox[{"i", "+", "j"}], TraditionalForm], " ", "resto"}]} }], ShowAutoStyles->True]}]}]]], " \nCalcular, mediante un ciclo, los valores propios de las siguientes \ potencias de la matriz A: ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["A", "2"], ",", " "}], TraditionalForm]]], Cell[BoxData[ FormBox[ SuperscriptBox["A", "4"], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{",", SuperscriptBox["A", "6"]}], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ SuperscriptBox["A", "8"], TraditionalForm]]], " . 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