{"id":33541,"date":"2024-09-19T17:27:36","date_gmt":"2024-09-19T17:27:36","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=33541"},"modified":"2025-07-26T23:42:57","modified_gmt":"2025-07-26T23:42:57","slug":"brachistochrona-et-aequatio-euler-lagrange-per-calculum-variationum","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/la\/brachistochrona-et-aequatio-euler-lagrange-per-calculum-variationum\/","title":{"rendered":"Brachistochrona et Aequatio Euler-Lagrange per Calculum Variationum"},"content":{"rendered":"<h1 style=\"text-align:center;\">Calculus Variationis in Mechanica Classica et Aequatio Euler\u2011Lagrange<\/h1>\n<p style=\"text-align:center;\"><em><strong>Summarium:<\/strong><br \/>In hac lectione revisuri sumus derivationem aequationis Euler\u2011Lagrange Mechanicae Analyticae per usum technicarum calculi variationis et, ex hoc, ostendetur accurate applicatio eius in solutione problematis Brachistochronae.<\/em><\/p>\n<hr \/>\n<p style=\"text-align:center;\"><strong>Propositi Discendi:<\/strong><br \/>\nAd conclusionem huius lectionis discipulus poterit:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Intellegere<\/strong> principium Hamiltonii minimae actionis<\/li>\n<li><strong>Demonstrationem exhibere<\/strong> aequationis Euler\u2011Lagrange<\/li>\n<li><strong>Solvere<\/strong> problema Brachistochronae utendo aequatione Euler\u2011Lagrange.<\/li>\n<\/ol>\n<hr \/>\n<p style=\"text-align:center;\"><strong><u>INDEX CONTENTORUM<\/u>:<\/strong><br \/>\n<a href=\"#1\">CUR CALCULUS VARIATIONIS IN MECHANICA CLASSICA<\/a><br \/>\n <a href=\"#2\">FORMULATIO PROBLEMATIS VARIATIONALIS<\/a><br \/>\n <a href=\"#3\">AEQUATIO EULER\u2011LAGRANGE<\/a><br \/>\n <a href=\"#4\">PROBLEMA BRACHISTOCHRONAE<\/a><br \/>\n <a href=\"#5\">REPOSITORIUM GITHUB CUM ALGORITHMO WOLFRAM<\/a>\n<\/p>\n<hr \/>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/JyumifjGzM0\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" allowfullscreen><\/iframe><\/center><\/p>\n<hr \/>\n<p><a id=\"1\"><\/a><\/p>\n<h2>Cur calculus variationis in mechanica classica<\/h2>\n<p style=\"text-align:justify;\">Physica Newtoniana multa problemata exhibet quae efficacius tractari possunt uti calculo variationis. Hic accessus fundamentalis est in aequationibus Lagrange et in principio minimae actionis Hamiltonii. Essentialiter, hic modus consistit in inveniendis trajectoriis quae maximant vel minimant aliquam quantitatem. Exempli gratia, quaeri potest trajectoria inter duo puncta quae minimet distantiam percursam aut tempus itineris. Exemplum huius accessus est principium Fermati, quod statuit lucem semper sequi trajectoriam quae tempus itineris minimat, quod rursus ducit ad <a href=\"http:\/\/toposuranos.com\/material\/es\/la-refraccion-de-la-luz-y-la-ley-de-snell\/\" target=\"_blank\" rel=\"noopener\">legem Snellii de refractione lucis<\/a>.<\/p>\n<p style=\"text-align:justify;\">Calculus variationis plures utilitates habet in mechanica classica. Exempli gratia, permittit solutiones analyticas exactas obtinere pro systematibus cum symmetria, et solutiones approximatas per theoriam perturbationum variationalium pro systematibus magis complexis. Praeterea, in adiunctis ubi difficiles sunt vires exprimere per aequationes differentiales, principium minimae actionis efficaciorem methodum praebet ad solvenda problemata in mechanica classica. Summatim, calculus variationis est instrumentum fundamentale quod praebet formulationem alternativam legum Newtonianarum, unificationem legum physicarum, maiorem efficientiam in resolutione problematum et maiorem praecisionem in praedictione eventuum experimentorum.<\/p>\n<p><a id=\"2\"><\/a><\/p>\n<h2>Formulatio problematis variationalis<\/h2>\n<p style=\"text-align:justify;\">Calculus variationis intendit in inveniendi functionem <span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span> quae extremat valorem functional:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">J(x,y(x))=\\displaystyle \\int_{x_1}^{x_2} f\\left(x,y(x),\\frac{dy(x)}{dx}\\right)dx,<\/span>\n<p style=\"text-align:justify;\">ad obtinendum valorem suum maximum vel minimum. In hac aequatione, functional <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> dependet a functione <span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span> et sua derivata <span class=\"katex-eq\" data-katex-display=\"false\">dy(x)\/dx,<\/span> dum limites integrationis manent fixi. Ad extremandum integrale, variationes super functionem <span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span> applicantur, quaerendo functionem quae faciat valorem functional extremum esse. Exempli causa, si contingit integrale valorem suum minimum attingere, quaelibet functio <em>in vicinitate eius<\/em>, quantuncumque propinqua <span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span>, valorem functional auget.<\/p>\n<p style=\"text-align:justify;\">Ad instituendum conceptum <em>functionis vicinae<\/em>, possumus assignare repraesentationem parametricam <span class=\"katex-eq\" data-katex-display=\"false\">y=(\\alpha,x)<\/span> omnibus functionibus possibilibus <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span>, ita ut si <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha=0<\/span>, tum <span class=\"katex-eq\" data-katex-display=\"false\">y(0,x)=y(x)<\/span> est functio quae extremat <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span>. Hoc exprimi potest hoc modo:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha, x) = y(x) + \\alpha \\eta(x),<\/span>\n<p style=\"text-align:justify;\">ubi <span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x)<\/span> est quaedam functio classis <span class=\"katex-eq\" data-katex-display=\"false\">\\mathcal{C}^1<\/span> quae in <span class=\"katex-eq\" data-katex-display=\"false\">x_1<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">x_2<\/span> annihilatur, ita ut functio <span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha,x)<\/span> quae includit hanc variationem idem sit ac <span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span> in punctis initialibus et finalibus trajectoriae integrationis.<\/p>\n<p style=\"text-align:justify;\">Substituendo functionem <span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha,x)<\/span> quae variationem <span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x)<\/span> includit in loco <span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span> in integrale quod functionalem <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> definit, obtinetur novus functional qui a parametro <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span> pendet:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">J(x,y(\\alpha, x)) = \\displaystyle \\int_{x_1}^{x_2} f\\left(x,y(\\alpha,x), \\dfrac{d}{dx}y(\\alpha,x)\\right)dx<\/span>\n<p style=\"text-align:justify;\">Ut extrema localia exsistant, necesse est conditionem impleri:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left.\\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha}\\right|_{\\alpha=0} = 0<\/span>\n<p style=\"text-align:justify;\">pro quavis functione <span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x).<\/span>\n<p><a id=\"3\"><\/a><\/p>\n<h2>Aequatio Euler\u2011Lagrange<\/h2>\n<p style=\"text-align:justify;\">Analysando derivatum <span class=\"katex-eq\" data-katex-display=\"false\">\\partial J(x,y(\\alpha,x))\/\\partial \\alpha<\/span>, obtinetur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n{}\\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha}\n\n&amp;=&amp;\\dfrac{\\partial}{\\partial \\alpha} \\displaystyle \\int_{x_1}^{x_2} f\\left(x,y(\\alpha,x),\\dfrac{dy(\\alpha, x)}{dx}\\right)dx \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\int_{x_1}^{x_2} \\left(\\dfrac{\\partial f}{\\partial x}\\dfrac{\\partial x}{\\partial  \\alpha} + \\dfrac{\\partial f}{\\partial y(\\alpha, x)}\\dfrac{\\partial y(\\alpha, x)}{\\partial  \\alpha}  + \\dfrac{\\partial f }{ \\partial \\frac{dy(\\alpha,x)}{dx}} \\dfrac{\\partial \\frac{dy(\\alpha,x)}{dx}}{\\partial \\alpha}\\right)dx\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Ex hoc puncto notandum est:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n\\dfrac{\\partial x}{\\partial \\alpha} &amp;=&amp; 0 \\\\ \\\\\n\n\\dfrac{\\partial y(\\alpha,x)}{\\partial \\alpha} &amp;=&amp; \\dfrac{\\partial}{\\partial \\alpha} \\left(y(x) + \\alpha \\eta(x) \\right) = \\eta(x) \\\\ \\\\\n\n\\dfrac{\\partial}{\\partial \\alpha}\\left( \\dfrac{dy(\\alpha, x)}{dx} \\right)&amp;=&amp; \\dfrac{\\partial}{\\partial \\alpha} \\left(\\dfrac{dy(x)}{dx} + \\alpha\\dfrac{d\\eta(x)}{dx} \\right) = \\dfrac{d\\eta}{dx}\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Quare expressio reducitur ut infra monstratur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n{} \\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha} &amp;=&amp; \\displaystyle \\int_{x_1}^{x_2} \\left(\\dfrac{\\partial f}{\\partial y(\\alpha,x)}\\eta(x) + \\dfrac{\\partial f}{\\partial  \\frac{dy(\\alpha,x)}{dx}} \\dfrac{d\\eta(x)}{dx} \\right)dx \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\int_{x_1}^{x_2} \\dfrac{\\partial f}{\\partial y(\\alpha,x)}\\eta(x) dx +  \\int_{x_1}^{x_2} \\dfrac{\\partial f}{\\partial  \\frac{dy(\\alpha,x)}{dx}} \\dfrac{d\\eta(x)}{dx} dx\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Deinde, si secundam integralem observamus, videbimus eam posse simpliciter uti integratione per partes:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n\\displaystyle \\int_{x_1}^{x_2} \\dfrac{\\partial f}{\\partial  \\frac{dy(\\alpha,x)}{dx}} \\dfrac{d\\eta}{dx} dx\n\n&amp;=&amp; \\left. \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha,x)}{dx}} \\eta(x)\\right|_{x_1}^{x_2} - \\displaystyle \\int_{x_1}^{x_2}\\eta(x) \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha, x)}{dx}} \\right) dx\\\\ \\\\\n\n&amp;=&amp; - \\displaystyle \\int_{x_1}^{x_2}\\eta(x) \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha, x)}{dx}} \\right)dx\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Et proinde<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n{} \\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha}\n\n&amp;=&amp; \\displaystyle \\int_{x_1}^{x_2} \\left[ \\eta(x) \\dfrac{\\partial f}{\\partial y(\\alpha, x)}  - \\eta(x) \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha,x)}{dx}} \\right) \\right]dx \\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\int_{x_1}^{x_2} \\left[ \\dfrac{\\partial f}{\\partial y(\\alpha, x)}  -  \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha,x)}{dx}} \\right) \\right] \\eta(x) dx\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Itaque, ex conditione quod <span class=\"katex-eq\" data-katex-display=\"false\">\\left.\\dfrac{\\partial J (x,y(\\alpha, x))}{\\partial \\alpha}\\right|_{\\alpha=0} = 0,<\/span> et cum <span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x)<\/span> sit functio quaelibet sub unica condicione se annulandi in <span class=\"katex-eq\" data-katex-display=\"false\">x_1<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">x_2<\/span>, habetur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\dfrac{\\partial f}{\\partial y(0, x)}  -  \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(0,x)}{dx}}\\right) = \\dfrac{\\partial f}{\\partial y(x)}  -  \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(x)}{dx}}\\right) = 0.\n\n<\/span>\n<p style=\"text-align:justify;\">Denique, \u00abnotatione deposita\u00bb in hac ultima expressione pervenitur ad id quod vocatur Aequatio Euler\u2011Lagrange:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\dfrac{\\partial f}{\\partial y}= \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial y^\\prime} \\right)},<\/span>\n<p style=\"text-align:justify;\">et hoc multo simplicius repraesentat conditionem necessariam ut functional <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> valorem extremum attingat.<\/p>\n<p><a id=\"4\"><\/a><\/p>\n<h2>Problema Brachistochronae<\/h2>\n<h3>Formulatio problematis<\/h3>\n<p style=\"text-align:justify;\">Problema brachistochronae est classicum mechanicae physicae quod solvitur per calculum variationis. Situatio posita est haec: Supponamus nos habere objectum materiale quod movetur sub effectu campi virium constantis et quod se transfert ex puncto initiali <span class=\"katex-eq\" data-katex-display=\"false\">(x_1,y_1)<\/span> ad alterum punctum finalem <span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span>, ubi punctum initiale maiori altitudine quam punctum finale sit. Quaestio quae ponitur est: Quae est trajectoria quam particula sequi debet ut ad punctum finalem quam brevissimo tempore perveniat?<\/p>\n<h3>Formulatio solutionis<\/h3>\n<p style=\"text-align:justify;\">Ad resolvendum problema brachistochronae, utile est considerare situationem simpliciter. Itaque punctum originis <span class=\"katex-eq\" data-katex-display=\"false\">(x_1, y_1)<\/span> in origine coordinatarum figi potest, dum punctum destinationis <span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span> ad dextram originis et infra axem <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}<\/span> positum est.<\/p>\n<div style=\"text-align:center;\"><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona.png\" alt=\"c\u00e1lculo variacional - problema de la braquist\u00f3crona\" width=\"711\" height=\"505\" class=\"aligncenter size-full wp-image-28729 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona.png\" alt=\"c\u00e1lculo variacional - problema de la braquist\u00f3crona\" width=\"711\" height=\"505\" class=\"aligncenter size-full wp-image-28729 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona.png 711w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona-300x213.png 300w\" sizes=\"(max-width: 711px) 100vw, 711px\" \/><\/noscript><\/div>\n<p style=\"text-align:justify;\">In hac situatione considerari potest campus vis qui agit deorsum (in directione <span class=\"katex-eq\" data-katex-display=\"false\">-\\hat{y}<\/span>) a gravitate generatus, et supponi motum sine frictione fieri. In hoc contextu, particula variis trajectoriis coercetur quae puncta initii et finis conectunt cum proposito inveniendi quae earum tempus itineris minimat.<\/p>\n<h3>Examinatio energiae<\/h3>\n<p style=\"text-align:justify;\">Ad hoc problema solvendum, uti possumus conservatione energiae systematis gravitationalis. Energia totalis systematis constans manebit, considerata tum energia cinetica <span class=\"katex-eq\" data-katex-display=\"false\">E_{cin}=\\frac{1}{2}mv^2<\/span> tum energia potentiali gravitatoria <span class=\"katex-eq\" data-katex-display=\"false\">E_{pot,g}<\/span>, ubi <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> est massa particulae et <span class=\"katex-eq\" data-katex-display=\"false\">v<\/span> eius velocitas. Pro energia potentiali referentiam sumpsimus originem, ita ut <span class=\"katex-eq\" data-katex-display=\"false\">E_{pot,g}(y=0)=0<\/span>, dum in quacumque alia altitudine <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> habetur <span class=\"katex-eq\" data-katex-display=\"false\">E_{pot,g}(y)=mgy.<\/span>\n<p style=\"text-align:justify;\">Cum particula ab origine cum velocitate nulla incipiat, energia eius totalis aequalis est zero. Tunc habetur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">E_{cin} + E_{pot,g}=0<\/span>\n<p style=\"text-align:justify;\">Cum particula infra punctum referentiae cadat, energia eius potentialis negativa erit et energia cinetica positiva. Hoc modo, velocitatem <span class=\"katex-eq\" data-katex-display=\"false\">v<\/span> ex aequatione conservationis energiae resolvere possumus et obtinere:<\/p>\n<p style=\"text-align:center\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} &amp;\\dfrac{1}{2}mv^2 + (-mgy) = 0 \\\\ \\\\\n\n\\vdash &amp;\\dfrac{1}{2}mv^2 = mgy \\\\ \\\\\n\n\\vdash &amp;v^2 = 2gy \\\\ \\\\\n\n\\vdash &amp;v = \\sqrt{2gy}\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Ita velocitatem particulae in quovis puncto trajectoriae eius in functione altitudinis <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> ubi versatur computare possumus.<\/p>\n<h3>Examinatio temporis trajectoriae<\/h3>\n<p style=\"text-align:justify;\">Cum celeritatem motus obtinuerimus, elementa temporis itineris construere possumus utens elemento displacement <span class=\"katex-eq\" data-katex-display=\"false\">ds=\\sqrt{dx^2 + dy^2}<\/span> hoc modo:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} dt &amp;= \\dfrac{ds}{v} = \\dfrac{\\sqrt{dx^2 + dy^2}}{\\sqrt{2gy}}\\\\ \\\\\n\n&amp;= \\sqrt{\\dfrac{dx^2 + dy^2}{2gy} }\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Ita tempus desplazamenti inter puncta <span class=\"katex-eq\" data-katex-display=\"false\">(x_1,y_1)<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span> obtineri potest integrando<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} t &amp;= \\displaystyle \\int_{(x_1,y_1)}^{(x_2,y_2)} dt \\\\ \\\\\n\n&amp;= \\displaystyle \\int_{(x_1,y_1)}^{(x_2,y_2)} \\sqrt{\\dfrac{dx^2 + dy^2}{2gy}} \\\\ \\\\\n\n&amp;= \\displaystyle \\dfrac{1}{\\sqrt{2g}}\\int_{y_1}^{y_2} \\sqrt{\\dfrac{1+ \\left(\\dfrac{dx}{dy}\\right)^2 }{y}}dy \\\\ \\\\\n\n\\end{array}<\/span>\n<h3>Formulatio problematis variationalis<\/h3>\n<p style=\"text-align:justify;\">Cum hac ultima expressione tempus expressimus ut functional formae<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n{}t = J(y,x(y)) = \\displaystyle \\int_{y_1}^{y_2} f\\left(y,x(y),\\dfrac{dx(y)}{dy} \\right) dy\n\n<\/span>\n<p style=\"text-align:justify;\">ubi<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">f\\left(y,x(y), \\dfrac{dx(y)}{dy}\\right) = \\sqrt{\\dfrac{1+ \\left(\\dfrac{dx(y)}{dy} \\right)^2}{y}} <\/span>\n<p style=\"text-align:justify;\">Hoc loco possumus praeterire factor <span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{2g},<\/span> quia optimizare <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> idem est ac optimizare <span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{2g}J<\/span>.<\/p>\n<p style=\"text-align:justify;\">His praeparatis, nunc aequationem Euler\u2011Lagrange construere possumus sequendo eundem procedendi modum antea usitatum, tandem ad<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{\\partial f}{\\partial x} = \\dfrac{d}{dy} \\dfrac{\\partial f}{\\partial x^\\prime}<\/span>\n<p style=\"text-align:justify;\">Attamen hic videre possumus quod <span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{\\partial f}{\\partial x} = 0,<\/span> unde sequetur<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{d}{dy}\\dfrac{\\partial f}{\\partial x^\\prime} = 0,<\/span>\n<p style=\"text-align:justify;\">sive aliter<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{\\partial f}{\\partial x^\\prime} = \\dfrac{1}{\\sqrt{2a}},<\/span>\n<p style=\"text-align:justify;\">ubi <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> est constans arbitraria ita scripta quia \u00abcommodum\u00bb est ad posteriores evolutiones.<\/p>\n<h3>Resolutio problematis variationalis<\/h3>\n<p style=\"text-align:justify;\">Substituendo functionem <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> in hac ultima expressione habetur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} &amp;\\dfrac{\\partial }{\\partial x^\\prime} \\sqrt{\\dfrac{1+ x^{\\prime 2}}{y}}  = \\dfrac{1}{\\sqrt{2a}} \\\\ \\\\\n\n\\vdash &amp; \\dfrac{1}{2}\\left( \\dfrac{1 + x^{\\prime 2} }{y} \\right)^{-1\/2} \\left(\\dfrac{2x^\\prime}{y} \\right) = \\dfrac{1}{\\sqrt{2a}} \\\\ \\\\\n\n\\vdash &amp; \\dfrac{1}{2}\\sqrt{\\dfrac{y}{1 + x^{\\prime 2}}} \\left(\\dfrac{2x^\\prime}{y} \\right) = \\dfrac{1}{\\sqrt{2a}} \\\\ \\\\\n\n\\vdash &amp; \\sqrt{\\dfrac{4x^{\\prime 2} y}{4y^2 (1 + x^{\\prime 2})} }  = \\sqrt{\\dfrac{1}{2a}} \\\\ \\\\\n\n\\vdash &amp;  \\dfrac{y x^{\\prime 2} }{y^2 (1 + x^{\\prime 2})}   = \\dfrac{1}{ 2a} \\\\ \\\\\n\n\\vdash &amp; 2ayx^{\\prime 2} = y^2 + y^2 x^{\\prime 2} \\\\ \\\\\n\n\\vdash &amp;  x^{\\prime 2} (2ay - y^2) = y^2 \\\\ \\\\\n\n\\vdash &amp; \\left(\\dfrac{dx}{dy}\\right)^2 = \\dfrac{y^2}{2ay - y^2} \\\\ \\\\\n\n\\vdash &amp; \\dfrac{dx}{dy} = \\pm \\sqrt{\\dfrac{y^2}{2ay - y^2}} \\\\ \\\\\n\n\\vdash &amp; dx = \\pm \\dfrac{y\\,dy}{\\sqrt{2ay - y^2}} \\\\ \\\\\n\n\\vdash &amp; x = \\displaystyle  \\pm \\int \\dfrac{y}{\\sqrt{2ay - y^2}}\\,dy\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Ad hanc integralem resolvendam, optio consideranda est substitutio sequens<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} y &amp;= a[1-\\cos(\\theta)] \\\\\n\n dy &amp;= a\\sin(\\theta) d\\theta\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Cum hoc habetur:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} x= &amp; \\pm \\displaystyle \\int \\dfrac{y}{\\sqrt{2ay - y^2}}\\,dy = \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]a\\sin(\\theta)}{\\sqrt{2a^2[1-\\cos(\\theta)] - a^2[1-\\cos(\\theta)]^2 }}\\,d\\theta \\\\ \\\\\n\n&amp; = \\pm \\displaystyle \\int \\dfrac{a^2[1-\\cos(\\theta)]\\sin(\\theta)}{\\sqrt{a^2[1-\\cos(\\theta)]\\left\\{ 2 - [1-\\cos(\\theta)] \\right\\} }}\\,d\\theta \\\\ \\\\\n\n&amp; = \\pm \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]\\sin(\\theta)}{\\sqrt{[1-\\cos(\\theta)]  [1 + \\cos(\\theta)]  }}\\,d\\theta \\\\ \\\\\n\n&amp; = \\pm \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]\\sin(\\theta)}{\\sqrt{ 1-\\cos^2(\\theta)}}\\,d\\theta \\\\ \\\\\n\n&amp; = \\pm \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]\\sin(\\theta)}{\\sin(\\theta)}\\,d\\theta \\\\ \\\\\n\n&amp; = \\pm \\displaystyle \\int a[1-\\cos(\\theta)]\\,d\\theta \\\\ \\\\\n\n&amp; = \\pm a(\\theta - \\sin(\\theta)) + C\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Observare possumus curvam brachistochronam exprimi posse ut curvam parametricam in coordinatis polaribus, quae coincidet cum cycloide quae suum punctum initium in origine habet.<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} x(\\theta) &amp;= \\pm a(\\theta - \\sin(\\theta)) \\\\\n\n   y(\\theta) &amp;= a(1-\\cos(\\theta))\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Constans integrationis <span class=\"katex-eq\" data-katex-display=\"false\">C<\/span> annihilata est ut conditioni initiali satisfiat quod trajectoria in origine incipit. Praeterea, observare possumus par aequationum solutiones possibilis problema dare, ubi constans <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> accommodari potest ut curva transeat per punctum <span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span> in fine itineris. Hae aequationes sunt:<\/p>\n<p style=\"text-align:center;\"><strong>Optio 1:<\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\begin{array}{rl}\n\n{} x(\\theta) &amp;= a(\\theta - \\sin(\\theta)) \\\\\n\n   y(\\theta) &amp;= a(1-\\cos(\\theta))\n\n\\end{array}}<\/span>\n<p style=\"text-align:center;\"><strong>Optio 2:<\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\begin{array}{rl}\n\n{} x(\\theta) &amp;= - a(\\theta - \\sin(\\theta)) \\\\\n\n   y(\\theta) &amp;= a(1-\\cos(\\theta))\n\n\\end{array}}<\/span> <\/p>\n<p style=\"text-align:justify;\">Solutio idonea pro hoc problema datur per secundam optionem, et constantem <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> ut negativum valorem accommodando, accipimus curvam quae condiciones necessarias ad solutionem complendas satisfacit.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis.png\" alt=\"Ejemplo de soluci\u00f3n posible, un arco de cicloide\" width=\"497\" height=\"329\" class=\"aligncenter size-full wp-image-28731 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis.png\" alt=\"Ejemplo de soluci\u00f3n posible, un arco de cicloide\" width=\"497\" height=\"329\" class=\"aligncenter size-full wp-image-28731 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis.png 497w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis-300x199.png 300w\" sizes=\"(max-width: 497px) 100vw, 497px\" \/><\/noscript><\/center><\/p>\n<h3>Aptatio finalis solutionis<\/h3>\n<p style=\"text-align:justify;\">Post ultimos aptationes factas, curva brachistochrona formam parametricam habet:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\nx(\\theta) &amp;= b(\\theta - \\sin(\\theta)) \\\\\n\ny(\\theta) &amp;= -b(1-\\cos(\\theta))\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Substitutum est <span class=\"katex-eq\" data-katex-display=\"false\">a=-b<\/span>, ubi <span class=\"katex-eq\" data-katex-display=\"false\">0\\lt b<\/span>. Curva periodum <span class=\"katex-eq\" data-katex-display=\"false\">2b\\pi<\/span> habet et satisfacere debet conditioni <span class=\"katex-eq\" data-katex-display=\"false\">x_2 \\in ]0,2b\\pi[<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">y_2 \\in ]-2b,0[<\/span>. Hoc ultimum est cruciale, quia postulat ut curva brachistochrona repraesentetur tamquam unus arcus cycloidei, quia solutio invalidabit si particula ad quietem redit redit ad punctum altitudinis zero.<\/p>\n<p style=\"text-align:justify;\">Ad aptandas has aequationes problemati, necesse est invenire valores <span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> qui systemati satisfaciant:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} x_2 &amp;= b(\\theta - \\sin(\\theta))\\\\\n\n y_2 &amp;= - b(1-\\cos(\\theta))\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Hoc systema non lineare solutiones analyticas non habere videtur, itaque methodis numericis in Wolfram Mathematica utimur. Infra, series graduum ad problematis solutionem praebetur:<\/p>\n<h4>Gradus 1: Systema constituere<\/h4>\n<p style=\"text-align:justify;\">Constituere aequationes quae systema solvendum formant<\/p>\n<p><code>eq1 = x2 == b*(theta - Sin[theta])<\/code><br \/>\n<code>eq2 = y2 == -b*(1 - Cos[theta])<\/code><\/p>\n<h4>Gradus 2: punctum destinationis definire<\/h4>\n<p style=\"text-align:justify;\">Definire punctum ad quod particula in fine sui itineris perveniet. Hoc casu id statuimus in <span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)=(1,-2)<\/span>. Hos valores mutare potes ut alias configurationes similes probes.<\/p>\n<p><code>x2val = 1; y2val = -2;<\/code><\/p>\n<h4>Gradus 3: valores quaesitos numerice computare<\/h4>\n<p style=\"text-align:justify;\">Uti functione \u00abFindRoot\u00bb ad solutionem problematis numerice computandam<\/p>\n<p><code>sol = FindRoot[{eq1, eq2} \/. {x2 -> x2val, y2 -> y2val}, {{b,1}, {theta, 1}}]<\/code><\/p>\n<p style=\"text-align:justify;\">Hic usi sumus valoribus <span class=\"katex-eq\" data-katex-display=\"false\">b=1<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">\\theta=1<\/span> ut puncto initii pro approximatione numerica solutionis. Cum hoc, obtinetur solutio <span class=\"katex-eq\" data-katex-display=\"false\">b\\approx 2.4056<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">\\theta \\approx 1.40138<\/span>\n<h4>Gradus 4: Resultatorum corroboratio<\/h4>\n<p style=\"text-align:justify;\">Meminerimus ut hae responsiones sensum physicum habeant, necesse esse <span class=\"katex-eq\" data-katex-display=\"false\">x_2 \\in ]0,2b\\pi[<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">y_2 \\in ]-2b, 0[<\/span>. Cito id confirmare possumus per sequens procedens<\/p>\n<p style=\"text-align:justify;\">Primum extrahimus valores <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span> obtentos ut solutionem<\/p>\n<p><code>bval = sol[[1, 2]]; thetaval = sol[[2, 2]];<\/code><\/p>\n<p style=\"text-align:justify;\">Et deinde iubemus fieri confirmationem<\/p>\n<p><code>If[0 < x2val < 2*Pi*bval &#038;&#038; -2*bval < y2val < 0, \"Valores v\u00e1lidos\", \"Valores inv\u00e1lidos\"]<\/code><\/p>\n<p style=\"text-align:justify;\">Si omnia bene processerunt, debet erui \"valores v\u00e1lidos\" in exitu. Hoc fragmentum codicis adiuvabit te ad explorandum si situatio physica recte est modelata.<\/p>\n<p style=\"text-align:justify;\">His proceduris tandem perfectissime aptata est nostra curva solutio, quae connectit puncta <span class=\"katex-eq\" data-katex-display=\"false\">(x_1,y_1)=(0,0)<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)=(1,-2)<\/span>. Curva resultans est:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} x(\\theta) &amp;\\approx 2.4056(\\theta - \\sin(\\theta)) \\\\\n\n y(\\theta) &amp;\\approx -2.4056(1-\\cos(\\theta))\n\n\\end{array}\\;\\;;\\theta\\in [0, 1.40138]<\/span>\n<p style=\"text-align:justify;\">Quae graphice sic videtur:<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis.png\" alt=\"\" width=\"296\" height=\"345\" class=\"aligncenter size-full wp-image-28733 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis.png\" alt=\"\" width=\"296\" height=\"345\" class=\"aligncenter size-full wp-image-28733 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis.png 296w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis-257x300.png 257w\" sizes=\"(max-width: 296px) 100vw, 296px\" \/><\/noscript><\/center><\/p>\n<p><a id=\"5\"><\/a><\/p>\n<h2>Repositorium Github cum algorithmo Wolfram<\/h2>\n<p style=\"text-align:justify;\">Codex completus solutionis ad problema brachistochronae, incluso algorithmo evoluto in Wolfram Mathematica, praesto est ad download et consultationem in meo reposito GitHub. Hoc repositum includit fasciculum <code>.nb<\/code> cum codice in forma notebook interactivo, itemque versionem in textu plano <code>.m<\/code> pro iis qui codicem directe videre malunt.<\/p>\n<p style=\"text-align:justify;\"><strong>Potes repositum ex GitHub <a href=\"https:\/\/github.com\/girebz\/Braquist-crona\" target=\"_blank\" rel=\"noopener\">hic<\/a> download.<\/strong><\/p>\n<p style=\"text-align:justify;\">Praeter codicem, repositum continet fasciculum \"README\" cum instructionibus accuratis quomodo uti et intelligere algorithmum, itemque explicationem gradatim de solutione problematis brachistochronae. Spero id tibi utile fore!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculus Variationis in Mechanica Classica et Aequatio Euler\u2011Lagrange Summarium:In hac lectione revisuri sumus derivationem aequationis Euler\u2011Lagrange Mechanicae Analyticae per usum technicarum calculi variationis et, ex hoc, ostendetur accurate applicatio eius in solutione problematis Brachistochronae. Propositi Discendi: Ad conclusionem huius lectionis discipulus poterit: Intellegere principium Hamiltonii minimae actionis Demonstrationem exhibere aequationis Euler\u2011Lagrange Solvere problema Brachistochronae utendo [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28745,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":2,"footnotes":""},"categories":[1268,1250],"tags":[],"citadela-post-location":[],"class_list":["post-33541","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanica-classica","category-physica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.0 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Brachistochrona et Aequatio Euler-Lagrange per Calculum Variationum - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Disce aequationem Euler-Lagrange derivare per calculum 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