{"id":824,"date":"2017-12-11T15:40:36","date_gmt":"2017-12-11T20:40:36","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=824"},"modified":"2026-04-05T06:10:50","modified_gmt":"2026-04-05T11:10:50","slug":"3eva2011ti_t3_mn-perimetro-y-area-de-arco-semieliptico","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-3eva20\/3eva2011ti_t3_mn-perimetro-y-area-de-arco-semieliptico\/","title":{"rendered":"3Eva2011TI_T3_MN Per\u00edmetro y \u00e1rea de arco semiel\u00edptico"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">3ra Evaluaci\u00f3n I T\u00e9rmino 2011-2012. 13\/Septiembre\/2011. ICM02188 M\u00e9todos Num\u00e9ricos<\/h2>\n\n\n\n<p><strong>Tema 3<\/strong>. (35 puntos) Una empresa debe construir un arco con forma semiel\u00edptica como se indica en la figura.<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"512\" height=\"369\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2017\/12\/SydneyHarbourBridge01.png\" alt=\"Sydney Harbour Bridge\" class=\"wp-image-17732\" style=\"width:350px\" \/><\/figure>\n\n\n\n<p>Modelo El\u00edptico:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 <\/span>\n\n\n\n<p>Longitud:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> 2 \\int_0^2 \\sqrt{1+(y')^2} \\delta x <\/span>\n\n\n\n<p>Para asignar recursos se debe calcular su longitud con las dimensiones mostradas en el diagrama:<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"393\" height=\"82\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2017\/12\/modeloeliptico01.png\" alt=\"modelo el\u00edptico\" class=\"wp-image-17734\" \/><\/figure>\n\n\n\n<p>a. Encuentre la longitud del arco mediante una aproximaci\u00f3n de la integral con el m\u00e9todo de la<strong> Cuadratura de Gauss <\/strong>con <strong>n=2 <\/strong>subintervalos<\/p>\n\n\n\n<p>b. Encuentre el \u00e1rea bajo la curva y la cota del error, utilizando el m\u00e9todo de <strong>Simpson 1\/3<\/strong>, <strong>n=4<\/strong><\/p>\n\n\n\n<p><em><strong>R\u00fabrica<\/strong><\/em>: Planteamiento (5 puntos), literal a (15 puntos), literal b (15 puntos).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p><em><strong>Referencia<\/strong><\/em>: The Impossible Bridge | National Geographic.<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-4-3 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"The Impossible Bridge | National Geographic\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/-yLZYETYlmM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>3ra Evaluaci\u00f3n I T\u00e9rmino 2011-2012. 13\/Septiembre\/2011. ICM02188 M\u00e9todos Num\u00e9ricos Tema 3. (35 puntos) Una empresa debe construir un arco con forma semiel\u00edptica como se indica en la figura. Modelo El\u00edptico: Longitud: Para asignar recursos se debe calcular su longitud con las dimensiones mostradas en el diagrama: a. Encuentre la longitud del arco mediante una aproximaci\u00f3n [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[28],"tags":[59],"class_list":["post-824","post","type-post","status-publish","format-standard","hentry","category-mn-3eva20","tag-integracion-numerica"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/824","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=824"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/824\/revisions"}],"predecessor-version":[{"id":17735,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/824\/revisions\/17735"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=824"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=824"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=824"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}