{"id":3223,"date":"2019-02-14T12:50:17","date_gmt":"2019-02-14T17:50:17","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=3223"},"modified":"2026-04-05T21:01:14","modified_gmt":"2026-04-06T02:01:14","slug":"s3eva2018tii_t2-drenar-tanque-cilindrico","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva20\/s3eva2018tii_t2-drenar-tanque-cilindrico\/","title":{"rendered":"s3Eva2018TII_T2 Drenar tanque cil\u00edndrico"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2018TII_T2 Drenar tanque cil\u00edndrico<\/a><\/p>\n\n\n\n

La ecuaci\u00f3n a desarrollar es:<\/p>\n\n\n \\frac{\\delta y}{\\delta t} = -k\\sqrt{y} <\/span>\n\n\n\n

con valores de k =0.5, y(0)=9<\/p>\n\n\n\n

\n\n\n\n

Formula de Taylor con t\u00e9rmino de error:<\/p>\n\n\n P_{n}(x) = \\sum_{k=0}^{n} \\frac{f^{(k)}(x_0)}{k!} (x-x_0)^k <\/span>\n\n\n P_{n}(x) = f(x_0)+\\frac{f'(x_0)}{1!} (x-x_0) +<\/span>\n\n\n + \\frac{f''(x_0)}{2!}(x-x_0)^2 + <\/span>\n\n\n + \\frac{f'''(x_0)}{3!}(x-x_0)^3 + \\text{...} <\/span>\n\n\n\n

\n\n\n\n

Se requiere la 2da y 3ra derivadas:<\/p>\n\n\n \\frac{\\delta^2 y}{\\delta t^2} = -k\\frac{1}{2} y^{(\\frac{1}{2}-1)} = -\\frac{k}{2} y^{-\\frac{1}{2}}<\/span>\n\n\n \\frac{\\delta^3 y}{\\delta t^3} = -\\frac{k}{2}\\Big(-\\frac{1}{2}\\Big) y^{(-\\frac{1}{2}-1)} = \\frac{k}{4} y^{-\\frac{3}{2}}<\/span>\n\n\n\n

con lo que inicia las iteraciones y c\u00e1lculo del error, con avance de 0.5 para t.<\/p>\n\n\n\n

\n\n\n\n

t=0 , y(0) = 9<\/p>\n\n\n\n

\n\n\n\n

t = 0.5<\/p>\n\n\n \\frac{\\delta y(0)}{\\delta t} = -(0.5)\\sqrt{9} = -1.5<\/span>\n\n\n \\frac{\\delta^2 y(0)}{\\delta t^2} = -\\frac{0.5}{2} 9^{-\\frac{1}{2}} = - 0.08333 <\/span>\n\n\n \\frac{\\delta^3 y(0)}{\\delta t^3} = \\frac{0.5}{4} 9^{-\\frac{3}{2}} = 0.004628<\/span>\n\n\n P_{2}(0.5) = 9 - 1.5 (0.5-0) + \\frac{-0.08333}{2}(0.5-0)^2 <\/span>\n\n\n P_{2}(0.5) = 8.2395<\/span>\n\n\n\n

Error orden de:<\/p>\n\n\n Error = \\frac{0.004628}{3!}(0.5-0)^3 = 9.641 . 10^{-5}<\/span>\n\n\n\n

\n\n\n\n

t = 1<\/p>\n\n\n \\frac{\\delta y(0.5)}{\\delta t} = -(0.5)\\sqrt{8.2395} = -1.4352<\/span>\n\n\n \\frac{\\delta^2 y(0.5)}{\\delta t^2} = -\\frac{0.5}{2} (8.2395)^{-\\frac{1}{2}} = - 0.08709 <\/span>\n\n\n \\frac{\\delta^3 y(0.5)}{\\delta t^3} = \\frac{0.5}{4} (8.2395)^{-\\frac{3}{2}} = 0.005285<\/span>\n\n\n P_{2}(1) = 8.2395 - 1.4352(1-0.5) + \\frac{-0.08709}{2}(1-0.5)^2 <\/span>\n\n\n P_{2}(1) = 7.5110<\/span>\n\n\n\n

Error orden de:<\/p>\n\n\n Error = \\frac{0.005285}{3!}(1-0.5)^3 = 4.404 . 10^{-4}<\/span>\n\n\n\n

\n\n\n\n

t = 1.5<\/p>\n\n\n \\frac{\\delta y(1)}{\\delta t} = -(0.5)\\sqrt{7.5110} = -1.3703 <\/span>\n\n\n \\frac{\\delta^2 y(1)}{\\delta t^2} = -\\frac{0.5}{2} (7.5110)^{-\\frac{1}{2}} = - 0.09122 <\/span>\n\n\n \\frac{\\delta^3 y(1)}{\\delta t^3} = \\frac{0.5}{4} (7.5110)^{-\\frac{3}{2}} = 0.006072<\/span>\n\n\n P_{2}(1.5) = 7.5110 - 1.3703(1.5-1) + \\frac{-0.09122}{2}(1.5-1)^2 <\/span>\n\n\n P_{2}(1.5) = 6.8144 <\/span>\n\n\n\n

Error orden de:<\/p>\n\n\n Error = \\frac{0.006072}{3!}(1.5-1)^3 = 1.4 . 10^{-4}<\/span>\n\n\n\n

\n\n\n\n

t = 2<\/p>\n\n\n \\frac{\\delta y(1.5)}{\\delta t} = -(0.5)\\sqrt{6.8144} = -1.3052 <\/span>\n\n\n \\frac{\\delta^2 y(1.5)}{\\delta t^2} = -\\frac{0.5}{2} (6.8144)^{-\\frac{1}{2}} = - 0.09576 <\/span>\n\n\n \\frac{\\delta^3 y(1.5)}{\\delta t^3} = \\frac{0.5}{4} (6.8144)^{-\\frac{3}{2}} = 0.007026 <\/span>\n\n\n P_{2}(2) = 6.8144 - 1.3052 (2-1.5) - \\frac{0.09576}{2}(2-1.5)^2 <\/span>\n\n\n P_{2}(2) = 6.1498 <\/span>\n\n\n\n

Error orden de:<\/p>\n\n\n Error = \\frac{0.007026}{3!}(2-1.5)^3 = 1.4637 . 10^{-4}<\/span>\n\n\n\n

\n\n\n\n

Se estima que el pr\u00f3ximo t\u00e9rmino pasa debajo de 6 pies.
Por lo que estima esperar entre 2 y 2.5 minutos.<\/p>\n\n\n\n

resultados usando el algoritmo:
<\/strong><\/em><\/p>\n\n\n\n

ti, p_i,  error\n[[0.00000000e+00 9.00000000e+00 0.00000000e+00]\n [5.00000000e-01 8.23958333e+00 9.64506173e-05]\n [1.00000000e+00 7.51107974e+00 1.10105978e-04]\n [1.50000000e+00 6.81451855e+00 1.26507192e-04]\n [2.00000000e+00 6.14993167e+00 1.46391550e-04]\n [2.50000000e+00 5.51735399e+00 1.70751033e-04]]<\/code><\/pre>\n\n\n\n
\"3e2018tii_t2<\/figure>\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# 3Eva_IIT2018_T2 Drenar tanque cil\u00edndrico\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\ny0 = 9\nt0 = 0\nbuscar = 6\nk = 0.5\nh = 0.5\n\ndy  = lambda t,y: -k*np.sqrt(y)\nd2y = lambda t,y: -(k\/2)*(y**(-1\/2))\nd3y = lambda t,y: (k\/4)*(y**(-3\/2))\n\n# PROCEDIMIENTO\nresultado = [[t0,y0,0]]\nyi = y0\nti = t0\nwhile not(yi<buscar):\n    ti = ti+h\n    dyi = dy(ti,yi)\n    d2yi = d2y(ti,yi)\n    d3yi = d3y(ti,yi)\n    p_i = yi +dyi*(h) + (d2yi\/2)*(h**2)\n    errado = (d3yi\/6)*(h**3)\n    yi = p_i\n    resultado.append([ti,p_i,errado])\nresultado = np.array(resultado)\n\n# SALIDA\nprint('ti, p_i,  error')\nprint(resultado)\n\n# Grafica\nplt.plot(resultado[:,0],resultado[:,1])\nplt.ylabel('nivel de agua')\nplt.xlabel('tiempo')\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2018TII_T2 Drenar tanque cil\u00edndrico La ecuaci\u00f3n a desarrollar es: con valores de k =0.5, y(0)=9 Formula de Taylor con t\u00e9rmino de error: Se requiere la 2da y 3ra derivadas: con lo que inicia las iteraciones y c\u00e1lculo del error, con avance de 0.5 para t. t=0 , y(0) = 9 t = 0.5 Error […]<\/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":[51],"tags":[58,54],"class_list":["post-3223","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3223","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=3223"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3223\/revisions"}],"predecessor-version":[{"id":23928,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3223\/revisions\/23928"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=3223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=3223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=3223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}