{"id":8084,"date":"2022-01-25T19:30:53","date_gmt":"2022-01-26T00:30:53","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8084"},"modified":"2026-04-05T20:48:13","modified_gmt":"2026-04-06T01:48:13","slug":"s2eva2021paoii_t2-edo-embudos-conicos-para-llenar-botellas","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2021paoii_t2-edo-embudos-conicos-para-llenar-botellas\/","title":{"rendered":"s2Eva2021PAOII_T2 EDO - Embudos c\u00f3nicos para llenar botellas"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2021PAOII_T2 EDO - Embudos c\u00f3nicos para llenar botellas<\/a><\/p>\n\n\n\n

literal a<\/h3>\n\n\n\n

La expresi\u00f3n dada en el enunciado para EDO, se reordena para definir la funci\u00f3n a usar con Runge-Kutta:<\/p>\n\n\n \\frac{\\delta y(t)}{\\delta t} + \\frac{d^2}{4}\\sqrt{2 g \\text{ }y(t)}\\Bigg[\\frac{tan \\theta}{y(t)} \\Bigg]^2 = 0 <\/span>\n\n\n \\frac{\\delta y(t)}{\\delta t} = - \\frac{d^2}{4}\\sqrt{2 g \\text{ }y(t)}\\Bigg[\\frac{tan \\theta}{y(t)} \\Bigg]^2 <\/span>\n\n\n\n

siendo h = 0.5,  con y(0) = 0.15 m y d= 0.01 m ajustando las unidades de medida.<\/p>\n\n\n \\frac{\\delta y(t)}{\\delta t} = - \\frac{0.01^2}{4}\\sqrt{2 (9.8) \\text{ }y(t)}\\Bigg[\\frac{tan (\\pi\/4)}{y(t)} \\Bigg]^2 <\/span>\n\n\n \\frac{\\delta y(t)}{\\delta t} = - (1.1068e-4) \\sqrt{ y(t)}\\Bigg[\\frac{1}{y(t)} \\Bigg]^2 <\/span>\n\n\n \\frac{\\delta y(t)}{\\delta t}= - (1.1068e-4) \\frac{\\sqrt{ y(t)}}{y(t)^2}<\/span>\n\n\n\n

literal b<\/h3>\n\n\n\n

se inicia el c\u00e1lculo del siguiente punto de la tabla<\/p>\n\n\n\n

i<\/td>t<\/td>y<\/td><\/tr>
0<\/td>0<\/td>0.15<\/td><\/tr>
1<\/td>0.5<\/td>0.1490<\/td><\/tr>
2<\/td>1<\/td>0.1480<\/td><\/tr>
3<\/td>1.5<\/td>0.1471<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

i = 0<\/p>\n\n\n K_1 = h\\Bigg(- (1.1068e-4) \\frac{\\sqrt{ y(t)}}{y(t)^2} \\Bigg)<\/span>\n\n\n K_1 = 0.5\\Bigg(- (1.1068e-4) \\frac{\\sqrt{0.15}}{0.15^2}\\Bigg) = -9.5258e-04 <\/span>\n\n\n K_2 = h\\Bigg(- (1.1068e-4) \\frac{\\sqrt{ y(t)+K_1}}{(y(t)+K_1)^2} \\Bigg)<\/span>\n\n\n K_2 = 0.5\\Bigg(- (1.1068e-4) \\frac{\\sqrt{ 0.15+-9.5258e-04}}{(0.15-9.5258e-04)^2} \\Bigg) <\/span>\n\n\n K_2 = -9.6173e-04<\/span>\n\n\n y_1 = y_0 + \\frac{K_1 + K_2}{2}<\/span>\n\n\n y_1 = 0.15 + \\frac{-9.5258e-04 -9.6173e-04}{2} = 0.149 <\/span>\n\n\n\n

i = 1<\/p>\n\n\n K_1 = 0.5\\Bigg(- (1.1068e-4) \\frac{\\sqrt{0.149}}{0.149^2}\\Bigg) =-9.6177e-04 <\/span>\n\n\n K_2 = 0.5\\Bigg(- (1.1068e-4) \\frac{\\sqrt{ 0.149-9.7120e-04}}{(0.149-9.7120e-04)^2} \\Bigg) <\/span>\n\n\n K_2 = -9.7116e-04<\/span>\n\n\n y_2 = y_1 + \\frac{-9.6177e-04 + -9.7116e-04}{2} = 0.1480<\/span>\n\n\n\n

i = 2<\/p>\n\n\n K_1 = 0.5\\Bigg(- (1.1068e-4) \\frac{\\sqrt{0.1480}}{0.1480^2}\\Bigg) = -9.7120e-04<\/span>\n\n\n K_2 = 0.5\\Bigg(- (1.1068e-4) \\frac{\\sqrt{ 0.1480-9.7120e-04}}{(0.1480-9.7120e-04)^2} \\Bigg) <\/span>\n\n\n K_2= -9.8084e-04 <\/span>\n\n\n y_3 = y_2 + \\frac{-9.7120e-04 + -9.8084e-04}{2} = 0.1471<\/span>\n\n\n\n

\n\n\n\n

literal c<\/h3>\n\n\n\n

Resultados usando Algoritmo, se encuentra que el embudo se vac\u00eda entre 3.15 y 3.20 segundos<\/p>\n\n\n\n

 [ t , y , K1 , K2 ]\n[[ 0.0000e+00  1.5000e-01  0.0000e+00  0.0000e+00]\n [ 5.0000e-01  1.4904e-01 -9.5258e-04 -9.6173e-04]\n [ 1.0000e+00  1.4808e-01 -9.6177e-04 -9.7116e-04]\n [ 1.5000e+00  1.4710e-01 -9.7120e-04 -9.8084e-04]\n [ 2.0000e+00  1.4611e-01 -9.8088e-04 -9.9078e-04]\n [ 2.5000e+00  1.4512e-01 -9.9083e-04 -1.0010e-03]\n...\n[ 3.1000e+01  2.8617e-02 -7.5631e-03 -1.0583e-02]\n [ 3.1500e+01  1.0620e-02 -1.1431e-02 -2.4563e-02]\n [ 3.2000e+01         nan -5.0566e-02         nan]\n [ 3.2500e+01         nan         nan         nan]<\/code><\/pre>\n\n\n\n
\"Embudo<\/figure>\n\n\n\n
Instrucciones Python<\/h2>\n\n\n
\n# 2Eva_2021PAOII_T2 EDO \u2013 Embudos c\u00f3nicos para llenar botellas\nimport numpy as np\n\ndef rungekutta2(d1y,x0,y0,h,muestras):\n    tamano   = muestras + 1\n    estimado = np.zeros(shape=(tamano,4),dtype=float)\n    # incluye el punto [x0,y0,K1,K2]\n    estimado[0] = [x0,y0,0,0]\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[i] = [xi,yi,K1,K2]\n    return(estimado)\n\n# INGRESO\nd = 0.01\ntheta = np.pi\/4\ng = 9.8\nd1y = lambda t,y: -(d**2)\/4*np.sqrt(2*g*y)*(np.tan(theta)\/y)**2\n\nt0 = 0\ny0 = 0.15\nh  = 0.5\nmuestras = 70\n\n# PROCEDIMIENTO\ntabla = rungekutta2(d1y,t0,y0,h,muestras)\n\n# SALIDA\nnp.set_printoptions(precision=4)\nprint('[ t , y , K1 , K2 ]')\nprint(tabla)\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2021PAOII_T2 EDO - Embudos c\u00f3nicos para llenar botellas literal a La expresi\u00f3n dada en el enunciado para EDO, se reordena para definir la funci\u00f3n a usar con Runge-Kutta: siendo h = 0.5,  con y(0) = 0.15 m y d= 0.01 m ajustando las unidades de medida. literal b se inicia el c\u00e1lculo del siguiente […]<\/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-8084","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\/8084","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=8084"}],"version-history":[{"count":6,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8084\/revisions"}],"predecessor-version":[{"id":23902,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8084\/revisions\/23902"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8084"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8084"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8084"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}