{"id":3968,"date":"2017-11-12T15:00:00","date_gmt":"2017-11-12T20:00:00","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=3968"},"modified":"2026-04-05T20:21:47","modified_gmt":"2026-04-06T01:21:47","slug":"s2eva2012ti_t3_mn-edo-taylor-2-contaminacion-de-estanque","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2012ti_t3_mn-edo-taylor-2-contaminacion-de-estanque\/","title":{"rendered":"s2Eva2012TI_T3_MN EDO Taylor 2 Contaminaci\u00f3n de estanque"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-2eva20\/2eva2012ti_t3_mn-edo-taylor-2-contaminacion-de-estanque\/\" data-type=\"post\" data-id=\"871\">2Eva2012TI_T3_MN EDO Taylor 2 Contaminaci\u00f3n de estanque<\/a><\/p>\n\n\n\n<p>La ecuaci\u00f3n a resolver con Taylor es:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s'- \\frac{26s}{200-t} - \\frac{5}{2} = 0 <\/span>\n\n\n\n<p>Para lo que se plantea usar la primera derivada:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s'= \\frac{26s}{200-t}+\\frac{5}{2} <\/span>\n\n\n\n<p>con valores iniciales de s(0) = 0, h=0.1<\/p>\n\n\n\n<p>La f\u00f3rmula de Taylor para tres t\u00e9rminos es:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s_{i+1}= s_{i}+s'_{i}h + \\frac{s''_{i}}{2}h^2 + error<\/span>\n\n\n\n<p>Para el desarrollo se compara la soluci\u00f3n con dos t\u00e9rminos, tres t\u00e9rminos y Runge Kutta.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">1. Soluci\u00f3n con dos t\u00e9rminos de Taylor<\/h4>\n\n\n\n<p>Iteraciones<\/p>\n\n\n\n<p>i = 0, t<sub>0<\/sub> = 0, s(0)=0<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s'_{0}= \\frac{26s_{0}}{200-t_{0}}+\\frac{5}{2} <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\frac{26(0)}{200-0}+\\frac{5}{2} = \\frac{5}{2} <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s_{1}= s_{0}+s'_{0}h = 0+ \\frac{5}{2}*0.1= 0.25<\/span>\n\n\n\n<p class=\"has-text-align-center\">t<sub>1<\/sub> =&nbsp; t<sub>0<\/sub>+h = 0+0.1 = 0.1<\/p>\n\n\n\n<p>i=1<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s'_{1}= \\frac{26s_{1}}{200-t_{1}}+\\frac{5}{2} <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\frac{26(0.25)}{200-0.1}+\\frac{5}{2} = 2.5325 <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> s_{2}= s_{1}+s'_{1}h = 0.25 + (2.5325)*0.1 = 0.5032<\/span>\n\n\n\n<p class=\"has-text-align-center\">t<sub>2<\/sub> =&nbsp; t<sub>1<\/sub>+h = 0.1+0.1 = 0.2<\/p>\n\n\n\n<p>i=2,<\/p>\n\n\n\n<p>resolver como tarea<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"EDO con Taylor en Python. Ejercicio: Contaminaci\u00f3n de estanque\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4t-oY6vWiq8?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\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p>2. Resolviendo con Python<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>estimado\n&#091;xi,yi Taylor,yi Runge-Kutta, diferencias]\n&#091;&#091; 0.0  0.0000e+00  0.0000e+00  0.0000e+00]\n &#091; 0.1  2.5000e-01  2.5163e-01 -1.6258e-03]\n &#091; 0.2  5.0325e-01  5.0655e-01 -3.2957e-03]\n &#091; 0.3  7.5980e-01  7.6481e-01 -5.0106e-03]\n &#091; 0.4  1.0197e+00  1.0265e+00 -6.7714e-03]\n &#091; 0.5  1.2830e+00  1.2916e+00 -8.5792e-03]\n &#091; 0.6  1.5497e+00  1.5601e+00 -1.0435e-02]\n &#091; 0.7  1.8199e+00  1.8322e+00 -1.2339e-02]\n &#091; 0.8  2.0936e+00  2.1079e+00 -1.4294e-02]\n &#091; 0.9  2.3710e+00  2.3873e+00 -1.6299e-02]\n &#091; 1.0  2.6519e+00  2.6703e+00 -1.8357e-02]\n &#091; 1.1  2.9366e+00  2.9570e+00 -2.0467e-02]\n &#091; 1.2  3.2250e+00  3.2476e+00 -2.2632e-02]\n &#091; 1.3  3.5171e+00  3.5420e+00 -2.4853e-02]\n &#091; 1.4  3.8132e+00  3.8403e+00 -2.7129e-02]\n &#091; 1.5  4.1131e+00  4.1426e+00 -2.9464e-02]\n &#091; 1.6  4.4170e+00  4.4488e+00 -3.1857e-02]\n &#091; 1.7  4.7248e+00  4.7592e+00 -3.4310e-02]\n &#091; 1.8  5.0368e+00  5.0736e+00 -3.6825e-02]\n &#091; 1.9  5.3529e+00  5.3923e+00 -3.9402e-02]\n &#091; 2.0  5.6731e+00  5.7152e+00 -4.2043e-02]]\nerror en rango:  0.04204310894163932<\/code><\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"640\" height=\"480\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2017\/11\/contaminaEstanque02.png\" alt=\"contamina Estanque gr\u00e1fica\" class=\"wp-image-18449\" \/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<h2 class=\"wp-block-heading\">2. Algoritmo en Python<\/h2>\n\n\n<div class=\"wp-block-syntaxhighlighter-code \"><pre class=\"brush: plain; title: ; notranslate\" title=\"\">\n# EDO. M\u00e9todo de Taylor 3 t\u00e9rminos \n# estima la solucion para muestras espaciadas h en eje x\n# valores iniciales x0,y0\n# entrega arreglo &#x5B;&#x5B;x,y]]\nimport numpy as np\n\ndef edo_taylor2t(d1y,x0,y0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,2),dtype=float)\n    # incluye el punto &#x5B;x0,y0]\n    estimado&#x5B;0] = &#x5B;x0,y0]\n    x = x0\n    y = y0\n    for i in range(1,tamano,1):\n        y = y + h*d1y(x,y) # + ((h**2)\/2)*d2y(x,y)\n        x = x+h\n        estimado&#x5B;i] = &#x5B;x,y]\n    return(estimado)\n\ndef rungekutta2(d1y,x0,y0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,2),dtype=float)\n    # incluye el punto &#x5B;x0,y0]\n    estimado&#x5B;0] = &#x5B;x0,y0]\n    xi = x0\n    yi = y0\n    for i in range(1,tamano,1):\n        K1 = h * d1y(xi,yi)\n        K2 = h * d1y(xi+h, yi + K1)\n\n        yi = yi + (K1+K2)\/2\n        xi = xi + h\n        \n        estimado&#x5B;i] = &#x5B;xi,yi]\n    return(estimado)\n\n# PROGRAMA PRUEBA\n# 2Eva_IIT2016_T3_MN EDO Taylor 2, Tanque de agua\n\n# INGRESO.\n# d1y = y' = f, d2y = y'' = f'\nd1y = lambda x,y: 26*y\/(200-x)+5\/2\nx0 = 0\ny0 = 0\nh = 0.1\nmuestras = 20\n\n# PROCEDIMIENTO\npuntos = edo_taylor2t(d1y,x0,y0,h,muestras)\nxi = puntos&#x5B;:,0]\nyi = puntos&#x5B;:,1]\n\n# Con Runge Kutta\npuntosRK2 = rungekutta2(d1y,x0,y0,h,muestras)\nxiRK2 = puntosRK2&#x5B;:,0]\nyiRK2 = puntosRK2&#x5B;:,1]\n\n# diferencias\ndiferencias = yi-yiRK2\nerror = np.max(np.abs(diferencias))\ntabla = np.copy(puntos)\ntabla = np.concatenate((puntos,np.transpose(&#x5B;yiRK2]),\n                        np.transpose(&#x5B;diferencias])),\n                       axis = 1)\n\n# SALIDA\nnp.set_printoptions(precision=4)\nprint('estimado&#x5B;xi,yi Taylor,yi Runge-Kutta, diferencias]')\nprint(tabla)\nprint('error en rango: ', error)\n\n# Gr\u00e1fica\nimport matplotlib.pyplot as plt\nplt.plot(xi&#x5B;0],yi&#x5B;0],'o',\n         color='r', label ='&#x5B;x0,y0]')\nplt.plot(xi&#x5B;0:],yi&#x5B;0:],'-',\n         color='g',\n         label ='y Taylor 2 t\u00e9rminos')\nplt.plot(xiRK2&#x5B;0:],yiRK2&#x5B;0:],'-',\n         color='blue',\n         label ='y Runge-Kutta 2Orden')\nplt.axhline(y0\/2)\nplt.title('EDO: Taylor 2T vs Runge=Kutta 2Orden')\nplt.xlabel('x')\nplt.ylabel('y')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<h3 class=\"wp-block-heading\">Usando Taylor con 3 t\u00e9rminos<\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>estimado \n&#091;xi, yi, d1yi, d2yi ]\n&#091;&#091;0. 0. 2.5 0.325 ]\n &#091;0.1 0.251625 2.53272761 0.32958302]\n &#091;0.2 0.50654568 2.56591685 0.33423301]\n &#091;0.3 0.76480853 2.59957447 0.33895098]\n &#091;0.4 1.02646073 2.63370731 0.34373796]\n &#091;0.5 1.29155015 2.66832233 0.348595 ]\n &#091;0.6 1.56012536 2.70342658 0.35352316]\n &#091;0.7 1.83223563 2.73902723 0.35852351]\n &#091;0.8 2.10793097 2.77513155 0.36359715]\n &#091;0.9 2.38726211 2.81174694 0.36874519]\n &#091;1. 2.67028053 2.84888087 0.37396876]\n &#091;1.1 2.95703846 2.88654098 0.37926901]\n &#091;1.2 3.24758891 2.92473497 0.3846471 ]\n &#091;1.3 3.54198564 2.96347069 0.39010422]\n &#091;1.4 3.84028323 3.00275611 0.39564157]\n &#091;1.5 4.14253705 3.04259931 0.40126036]\n &#091;1.6 4.44880328 3.08300849 0.40696184]\n &#091;1.7 4.75913894 3.12399199 0.41274727]\n &#091;1.8 5.07360187 3.16555827 0.41861793]\n &#091;1.9 5.39225079 3.2077159 0.42457511]\n &#091;2. 5.71514526 0. 0. ]]<\/code><\/pre>\n\n\n\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 2Eva2012TI_T3_MN EDO Taylor 2 Contaminaci\u00f3n de estanque La ecuaci\u00f3n a resolver con Taylor es: Para lo que se plantea usar la primera derivada: con valores iniciales de s(0) = 0, h=0.1 La f\u00f3rmula de Taylor para tres t\u00e9rminos es: Para el desarrollo se compara la soluci\u00f3n con dos t\u00e9rminos, tres t\u00e9rminos y Runge Kutta. [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[48],"tags":[58,54],"class_list":["post-3968","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3968","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=3968"}],"version-history":[{"count":5,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3968\/revisions"}],"predecessor-version":[{"id":23878,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3968\/revisions\/23878"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=3968"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=3968"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=3968"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}