Hoja1MTODO DEL TRAPECIOMTODO DE SIMPSONClculo de la
velocidad:SECCIN A - AUn coeficiente (profundidad, velocidad;
materiales del lecho del rio) =0.85iXif(Xi)Cof. Trap.Prod.
F(Xi)*Coef.LONGITUDTIEMPOVELOCIDAD00010.0002043.30.461893764410.500.06040.2402043.060.464468183921.000.15020.3002043.740.457247370831.500.25041.0002031.10.643086816742.000.25020.5002031.620.632511068952.500.34041.3602031.450.635930047763.000.46020.9202044.10.453514739273.500.39041.5602044.380.450653447584.000.36020.7202044.120.453309156894.500.35041.400SUMA
=4.6526145961105.000.40020.800115.500.40041.600VELOCIDAD
=0.5169571773m/seg.126.000.40020.800136.500.40041.600147.000.40020.800Para
hallar el rea nis apoyamos de la siguiente
formula:157.500.40041.600168.000.40010.400SUMA15.600 h =
0.50A=2.600u2SECCIN B - BLos coeficientes del mtodo del trapecio
es:(0.5, 1, 1, 1, , 0.5)iXif(Xi)Cof. Trap.Prod.
F(Xi)*Coef.00010.00010.500.06040.24021.000.13020.26031.500.24040.96042.000.28020.56052.500.31041.24063.000.44020.88073.500.38041.52084.000.34020.68094.500.37041.480105.000.32020.640115.500.33041.320126.000.40020.800136.500.21040.840147.000.18020.360SECCION
A - A157.500.14040.560168.000.01010.010iXif(Xi)Cof. Trap.Prod.
F(Xi)*Coef.SUMA12.35000.0000.50.000 h =
0.50A=2.058u210.50-0.06010.06021.00-0.15010.150SECCION
=2.3291666667u231.50-0.25010.25042.00-0.25010.25052.50-0.34010.34063.00-0.46010.46073.50-0.39010.39084.00-0.36010.36094.50-0.35010.350105.00-0.40010.400115.50-0.42010.420126.00-0.20010.200136.50-0.20010.200147.00-0.18010.180157.50-0.00210.002168.000.0000.50.0004.012AREA
=4.012m2
Q =1.7629273662m3/seg.SECCION B - BiXif(Xi)Cof. Trap.Prod.
F(Xi)*Coef.00.0000.50.00010.50-0.06010.06021.00-0.13010.13031.50-0.24010.24042.00-0.28010.28052.50-0.31010.31063.00-0.44010.44073.50-0.38010.38084.00-0.34010.34094.50-0.37010.370105.00-0.32010.320115.50-0.33010.330126.00-0.40010.400136.50-0.21010.210147.00-0.18010.180157.50-0.14010.140168.00-0.0100.50.005SUMA4.135AREA
=4.135m2
Q =1.8169752391m3/seg.
Hoja2EJEMPLOEJEMPLOResolver la siguiente escuacion diferencial
que representa a la trayectoria de una lanchaResolver la siguiente
escuacion diferencial que representa a la trayectoria de una
lanchaen un intervalo de X = 0 a 2.en un intervalo de X = 0 a
2.
donde Y(0) = 1donde Y(0) = 1MTODO DE EULERPROCESO
INTERATIVOEULERiXiYiHiXiYi0010.2-1.10.780010.780.9036610.20.780.2-0.82680.6146410.20.903660.614640.8334319220.40.614640.2-0.57776160.4990876820.40.833431920.499087680.786077595830.60.499087680.2-0.36932488320.425222703430.60.78607759580.42522270340.757831229540.80.42522270340.2-0.19560244350.386102214740.80.75783122950.38610221470.7461205963510.38610221470.2-0.03861022150.3783801704510.74612059630.37838017040.750622548161.20.37838017040.20.12864925790.404110021961.20.75062254810.40411002190.774431741971.40.40411002190.20.34753461890.473616945771.40.77443174190.47361694570.826382509981.60.47361694570.20.69148074070.611913093981.60.82638250990.61191309390.92643124891.80.61191309390.21.30949402090.87381189891.80.9264312480.8738118981.11860867421020.8738118980.22.53405450431.38062279891021.11860867421.38062279891.5034878628112.2Entonces:
para que la lancha recorra 2 m tendria que pasar 0.8738119
segundos.Entonces: para que la lancha recorra 2 m tendria que pasar
1.11860867 segundos.Comparando los dos metodos el error sera de:
0.2447967762
