METODO DE LAS DERIVADAS PARCIALES

METODO DE LAS DERIVADAS PARCIALES1. Consideremos el sistema.

F(x, y) = 9X2 -Y = 0 .. (1)

G(x, y) = X - 9Y2 = 0 .. (2)

2. Graficamos las funciones

9X2 -Y = 0 9X2 = eY Y =LN (9X2)

X - 9Y2 = 0 Y = (eX/9)1/2Tabulamos

Xy1=LN (9X2)y2 =(eX/9)1/2

0.1-2.410.35

0.2-1.020.37

0.50.810.43

0.81.750.5

12.20.55

1.53.010.71

23.580.91

Graficamos

3. Localizamos una Aproximacin

; , Po (Xo,Yo)

Xo = 0.5 Yo =0.43 Po = (0.5, 0.43)

4. Hallamos las derivadas parcialesF(x, y) = 9X2 -Y = 0

G(x, y) = X - 9Y2 = 0

5. Usando el siguiente proceso Iterativo se Obtiene la solucin.

Operando Po = (0.5, 0.43)

1 Iteracin

X1 = 0.50 ( 9(0.50)2- e0.43 ) = 0.420806

18(0.50)

X1 = 0.43 ( e0.50-9(0.43)2 ) = 0.428013 P1 = (0.420806, 0.428013)

-18(0.43)2 Iteracin

X2 = 0.420806 ( 9(0.420806)2- e0.428013 ) = 0.412952

18(0.420806)

Y2 = 0.428013 ( e0.420806-9(0.428013)2 ) = 0.411715 ; P2 = (0.412952, 0.411715)

-18(0.428013)3 Iteracin

X3 = 0.412952 ( 9(0.412952)2- e0.411715 ) = 0.409540

18(0.412952)

Y3 = 0.411715 ( e0.412952-9(0.411715)2 ) = 0.409082 ; P2 = (0.409540, 0.409082)

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