Metods Numericos

5.- La distribución de la velocidad de un fluido cerca de una superficie plana está dada por la siguiente tabla:
i y_i (m) u_i (m/s)
0 0.0 0.0
1 0.002 0.006180
2 0.004 0.011756
3 0.006 0.0161804 0.008 0.019021

La Ley de Newton para la tensión superficial está dada por
τ=μ du/dy
Donde µ es la viscosidad que suponemos vale 0.001 Ns/m2. Calcule la tensión superficial en cada puntomediante aproximación por diferencias adelante, hacia atrás y centrada.
format long
>> u=[0.0,0.006180,0.011756,0.016180,0.019021];
>> h=0.002;
>> d1=(u(2)-u(1))/(h)
d1 =
3.090000000000000
>>d2=(u(3)-u(1))/(2*h)
d2 =
2.939000000000000
>> d3=(u(4)-u(2))/(2*h)
d3 =
2.500000000000000
>> d4=(u(5)-u(3))/(2*h)
d4 =
1.816250000000000
>> d5=(u(5)-u(4))/(h)
d5 =
1.420500000000000
>>v=0.001;
>> t=v*d1
t =
0.003090000000000
>> t=v*d2
t =
0.002939000000000
>> t=v*d3
t =
0.002500000000000
>> t=v*d4
t =
0.001816250000000
>> t=v*d5
t =
0.001420500000000

I Yi(m) Ui(m/s)du/dy T
0 0.0 0.0 3.090000000000000 0.003090000000000
1 0.002 0.006180 2.939000000000000 0.002939000000000
2 0.004 0.011756 2.500000000000000 0.002500000000000
3 0.006 0.016180 1.8162500000000000.001816250000000
4 0.008 0.019021 1.420500000000000 0.001420500000000


6.- Dada la función f(x)=xe^x+e^x aproxime f^' (x),f^'' (x) en x=0.6, empleando los valores de h= 0.4, 0.1, 0.0002.Compare con los valores analíticos.
1ra derivada
f' (0.6)= 4.7375
h atrás central adelante
0.4 3.624266927081525 4.963600433907359 6.302933940733193
0.1 4.423081745746216 4.7514884832480905.079895220749964
0.0002 4.736852974118744 4.737508936892843 4.738164899666941

>> f=inline('x*exp(x)+exp(x)')
f =
Inline function:
f(x) = x*exp(x)+exp(x)
>> Xi=0.6
Xi =
0.6000
>>h=[0.4 0.1 0.0002]
h =
0.4000 0.1000 0.0002
>> syms x
>> ve=subs(diff(f(x)),Xi)
ve =
4.7375
>> format long >> for i=1:3
datras(i)=(f(Xi)-f(Xi-h(i)))/h(i);
end, datras
datras =...