Post on 19-Feb-2018
7/23/2019 ejercicio METODDD
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Ejemplo 1:
punto 0 1 2 3
Temperatura 56.5 113 181 214.5
oCPresión 1 5 20 40
atm
Encontrar la temperatura cuando la presión es igual a 2
1) LINEAL
f ( x0 ) = a0 + a1 x0
f ( x1 ) = a0 + a1 x1
x0 = 1
x0 = 5
f ( x0 ) = 56.5
f ( x1 ) = 113
ecuación 1:
ecuación 2 :
a0 + a1 = 56.5
a0 + 5a1 = 113
p( x) = 42.375 +
14.125 x p(2) = 70.625
a0 = 42.375 a1 = 14.125
2)CUADR!ICA 2
f ( x0 ) = a0 + a1 x0 +a 2 x0
f ( x ) = a + a x +a x2
f ( x ) = a + a x +a x2
p( x) = 39.85 + 17.153 x − 0.504 x2
p(2) = 72.14
")#$LIN$%IAL
x0 = 1 | ( x0 ) = Ecuacion 1 : a0 + a1 +a 2= 56.5
x1 = 5 | f ( x1 ) = 113 ecuacion 2 :a0 + 5a1+ 25a 2= 113
x2 = 20
|
f ( x2 ) =
181
Ecuacion3: a0 + 20a1 + 400a 2= 181
a1 = 17.153 a 2= −0.504
1 0 1 1 2 1
2 0 1 2 2 2
7/23/2019 ejercicio METODDD
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2 3
f ( x0 ) = a0+ a1 x
0+a2 x
0 + a3 x
0 x
0 = 1 | f ( x0) = 56.5 ecuacion 1: a
0+ a1+a2+a
3 = 56.
f ( x ) = a + a x +a x2
2+
a x
3
3 x1=
5 | f ( x1 )=
113ecuacion 2 :
a0+
5a1+
25a 2+
125a3=
113 f ( x2 ) = a0+ a1 x2+a 2 x2+ a3 x2 x2 = 20 | f ( x2 ) = 1 ecuacion 3 : a0 + 20a1 + 400a 2+8000a3 = 181
2 3
f ( x3 ) = a0+ a1 x2+a 2 x2 + a3 x3 x3 = 40 | f ( x3 ) = 214.5 cuacion4 : a0+ 40a1 + 1600a 2+64000a3 =
a0 = 38.226605 a1 = 18.65867 a 2= −0.7957017 a3 = 0.01098516
1 0 1 1 2 1 3 1
7/23/2019 ejercicio METODDD
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p( x) = 38.226605 + 18.65867 x − 0.7957017 x2+ 0.01098516 x
3
p(2) = 78.814