{"id":8754,"date":"2023-01-24T12:15:12","date_gmt":"2023-01-24T17:15:12","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8754"},"modified":"2026-04-05T20:46:08","modified_gmt":"2026-04-06T01:46:08","slug":"s2eva2022paoii_t1-altura-de-cohete-en-30-segundos","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2022paoii_t1-altura-de-cohete-en-30-segundos\/","title":{"rendered":"s2Eva2022PAOII_T1 Altura de cohete en 30 segundos"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2022PAOII_T1 Altura de cohete en 30 segundos<\/a><\/p>\n\n\n\n

literal a<\/h2>\n\n\n v = u \\ln\\Big(\\frac{m_0}{m_0-qt}\\Big) - gt <\/span>\n\n\n v = 1800 \\ln\\Big(\\frac{160000}{160000-2500t}\\Big) - 9.8t <\/span>\n\n\n\n

Seleccionando el m\u00e9todo de Simpson de 3\/8, se requieren al menos 3 tramos o segmentos para usarlo, que generan 4 muestras. El vector de tiempo se obtiene como:<\/p>\n\n\n\n

v = lambda<\/span> t: 1800*np.log(160000\/(160000-2500*t))-9.8*t\na = 0\nb = 30\ntramos = 3\nh = (b-a)\/tramos\nti = np.linspace(a,b,tramos+1)\nvi = v(ti)<\/code><\/pre>\n\n\n\n
siendo los vectores:<\/p>\n\n\n\n
ti = [ 0. 10. 20. 30.]\nvi = [ 0. 207.81826623 478.44820899 844.54060574]<\/code><\/pre>\n\n\n\n
la aplicaci\u00f3n del m\u00e9todo de Simpson de 3\/8 es:<\/p>\n\n\nI = \\frac{3}{8}(10) \\Bigg(1800 \\ln\\Big(\\frac{160000}{160000-2500(0)}\\Big) - 9.8(0) <\/span>\n\n\n+3(1800 \\ln\\Big(\\frac{160000}{160000-2500(10)}\\Big) - 9.8(10)) <\/span>\n\n\n+3(1800 \\ln\\Big(\\frac{160000}{160000-2500(20)}\\Big) - 9.8(20)) <\/span>\n\n\n+1800 \\ln\\Big(\\frac{160000}{160000-2500(30)}\\Big) - 9.8(30) \\Bigg) = <\/span>\n\n\nI = \\frac{3}{8}(10) \\Big(v(0)+3(v(10))+3(v(20))+v(30) \\Big)<\/span>\n\n\nI = \\frac{3}{8}(10) \\Big(0+3(207.81)+3(478.44)+844.54 \\Big)<\/span>\n\n\nI = 10887.52<\/span>\n\n\n\n
literal b<\/h2>\n\n\n\n
para el primer segmento se usa t entre [0,10]<\/p>\n\n\nx_a = \\frac{0+10}{2} + \\frac{1}{\\sqrt{3}}\\frac{10-0}{2} = 7.88 <\/span>\n\n\nx_b = \\frac{0+10}{2} - \\frac{1}{\\sqrt{3}}\\frac{10-0}{2} = 2.11<\/span>\n\n\nI = \\frac{10-0}{2}\\Big(v(7.88)+v(2.11)\\Big)=995.79<\/span>\n\n\n\n
para el 2do segmento se usa t entre [10,20]<\/p>\n\n\nx_a = \\frac{10+20}{2} + \\frac{1}{\\sqrt{3}}\\frac{20-10}{2} = 17.88 <\/span>\n\n\nx_b = \\frac{10+20}{2} - \\frac{1}{\\sqrt{3}}\\frac{20-10}{2} = 12.11<\/span>\n\n\nI = \\frac{20-10}{2}\\Big(v(17.88)+v(12.11)\\Big) =3368.42<\/span>\n\n\n\n
para el 3er segmento se usa t entre [20,30]<\/p>\n\n\nx_a = \\frac{20+30}{2} + \\frac{1}{\\sqrt{3}}\\frac{30-20}{2} = 27.88 <\/span>\n\n\nx_b = \\frac{20+30}{2} - \\frac{1}{\\sqrt{3}}\\frac{30-20}{2} = 22.11<\/span>\n\n\nI = \\frac{30-20}{2}\\Big(v(27.88)+v(22.11)\\Big) = 6515.23<\/span>\n\n\nAltura = 995.79+ 3368.42 + 6515.23 = 10879.44<\/span>\n\n\n\n
literal c<\/h2>\n\n\n\n
el error es la diferencia entre los m\u00e9todos
error_entre = |10887.52-10879.44| = 8.079<\/p>\n\n\n\n
Resultados con algoritmo<\/h2>\n\n\n\n
M\u00e9todo de Simpon 3\/8\nti\n[ 0. 10. 20. 30.]\nvi\n[ 0. 207.81826623 478.44820899 844.54060574]\nAltura con Simpson 3\/8 : 10887.52511781406\nsegmento Cuad_Gauss :    [995.792, 3368.421, 6515.231]\nAltura Cuadratura Gauss: 10879.445437288954\ndiferencia s3\/8 y Cuad_Gauss: 8.079680525106596\n>>><\/code><\/pre>\n\n\n\n
Instrucciones en Python<\/h3>\n\n\n
\n# 2Eva_2022PAOII_T1 Altura de cohete en 30 segundos\nimport numpy as np\n\n# INGRESO\nv = lambda t: 1800*np.log(160000\/(160000-2500*t))-9.8*t\na = 0\nb = 30\ntramos = 3\n\n# PROCEDIMIENTO literal a\ndef integrasimpson38_fi(xi,fi,tolera = 1e-10):\n    ''' sobre muestras de fi para cada xi\n        integral con m\u00e9todo de Simpson 3\/8\n        respuesta es np.nan para tramos desiguales,\n        no hay suficientes puntos.\n    '''\n    n = len(xi)\n    i = 0\n    suma = 0\n    while not(i>=(n-3)):\n        h  = xi[i+1]-xi[i]\n        h1 = (xi[i+2]-xi[i+1])\n        h2 = (xi[i+3]-xi[i+2])\n        dh = abs(h-h1)+abs(h-h2)\n        if dh<tolera:# tramos iguales\n            unS38 = fi[i]+3*fi[i+1]+3*fi[i+2]+fi[i+3]\n            unS38 = (3\/8)*h*unS38\n            suma = suma + unS38\n        else:  # tramos desiguales\n            suma = 'tramos desiguales'\n        i = i + 3\n    if (i+1)<n: # incompleto, tramos por calcular\n        suma = 'tramos incompletos, faltan '\n        suma = suma +str(n-(i+1))+' tramos'\n    return(suma)\n\nh = (b-a)\/tramos\nti = np.linspace(a,b,tramos+1)\nvi = v(ti)\naltura = integrasimpson38_fi(ti,vi)\n\n# SALIDA\nprint('M\u00e9todo de Simpon 3\/8')\nprint('ti')\nprint(ti)\nprint('vi')\nprint(vi)\nprint('Altura con Simpson 3\/8 :',altura)\n\n# PROCEDIMIENTO literal b\n# cuadratura de Gauss de dos puntos\ndef integraCuadGauss2p(funcionx,a,b):\n    x0 = -1\/np.sqrt(3)\n    x1 = -x0\n    xa = (b+a)\/2 + (b-a)\/2*(x0)\n    xb = (b+a)\/2 + (b-a)\/2*(x1)\n    area = ((b-a)\/2)*(funcionx(xa) + funcionx(xb))\n    return(area)\n\narea = 0\narea_i =[]\nfor i in range(0,tramos,1):\n    deltaA = integraCuadGauss2p(v,ti[i],ti[i+1])\n    area = area + deltaA\n    area_i.append(deltaA)\n# SALIDA\nprint('segmento Cuad_Gauss :   ', area_i)\nprint('Altura Cuadratura Gauss:', area)\n\nprint('diferencia s3\/8 y Cuad_Gauss:',altura-area)\n\nimport matplotlib.pyplot as plt\nplt.plot(ti,vi)\nplt.plot(ti,vi,'o')\nplt.title('v(t)')\nplt.xlabel('t (s)')\nplt.ylabel('v (m\/s)')\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2022PAOII_T1 Altura de cohete en 30 segundos literal a Seleccionando el m\u00e9todo de Simpson de 3\/8, se requieren al menos 3 tramos o segmentos para usarlo, que generan 4 muestras. El vector de tiempo se obtiene como: siendo los vectores: la aplicaci\u00f3n del m\u00e9todo de Simpson de 3\/8 es: literal b para el primer […]<\/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":[49],"tags":[58,54],"class_list":["post-8754","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8754","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=8754"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8754\/revisions"}],"predecessor-version":[{"id":23896,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8754\/revisions\/23896"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8754"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8754"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8754"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}