Tarea métodos numericos
Donde V= volumen (m3), h=profundidad del agua en el tanque (m), y R=radio del tanque (m). Si R=3m, ¿a qué profundidad debe llenarse el tanque de modo que contenga 30 m3? Maneje una tolerancia de 0.0001.Primero graficamos la función:
VENTANA DE COMANDOS
>> f=inline('((9*pi*h.^2-pi*h.^3)/3)-30')
f =
Inline function:
f(h) = ((9*pi*h.^2-pi*h.^3)/3)-30
>> h=0:0.1:15;
>> plot(h,f(h)),grid
Encontramos dos raices una entre 2.02 y 2.04
Y la otra entre 8.6 y 8.65
Para la raíz 1 tenemos a=2.02, b=2.04 y n=8.
n
a
r
b
F(a)
F(r)
F(b)
error
1
2.02
2.03
2.04
-+
+
0.01
2
2.02
2.025
2.03
-
-
+
0.005
3
2.025
2.0275
2.03
-
+
+
0.0025
4
2.025
2.0263
2.0275
-
-
+
0.0012
5
2.0263
2.0269
2.0275
-
-
+
0.000625
6
2.0269
2.0272
2.0275
-
+
+
0.0003125
7
2.0269
2.0270
2.0272
-
+
+
0.00015625
8
2.0269
2.02695
2.0270
-
+
+
0.000078125
Para este caso profundidad del agua en el tanque h=2.026953125 mVENTANA DE COMANDOS
>> a=2.02;b=2.04;
>> n=ceil((log((b-a)/0.0001))/log(2))
n =
8
>> r=(a+b)/2
r =
2.0300
>> f(a)
ans =
-0.1746
>> f(r)
ans =
0.0783
>> f(b)
ans =
0.3318
>> e=(b-a)/2
e =
0.0100
>> b=r
b =
2.0300
>> r=(a+b)/2
r =
2.0250
>> f(a)
ans =
-0.1746
>> f(r)
ans =
-0.0482
>> f(b)
ans =
0.0783
>> e=(b-a)/2
e =0.0050
>> a=r
a =
2.0250
>> r=(a+b)/2
r =
2.0275
>> f(a)
ans =
-0.0482
>> f(b)
ans =
0.0783
>> f(r)
ans =
0.0150
>> b=r
b =
2.0275
>> r=(a+b)/2
r =
2.0263
>> f(a)
ans =
-0.0482
>> f(r)
ans =
-0.0166
>> f(b)
ans =
0.0150
>> e=(b-a)/2
e =
0.0012
>> a=r
a =
2.0263
>> r=(a+b)/2
r =
2.0269
>> b
b =
2.0275
>>f(a)
ans =
-0.0166
>> f(r)
ans =
-7.7741e-004
>> f(b)
ans =
0.0150
>> e=(b-a)/2
e =
6.2500e-004
>> a=r
a =
2.0269
>> r=(a+b)/2
r =
2.0272
>> f(a)
ans =
-7.7741e-004
>> f(r)
ans =
0.0071
>> f(b)
ans =
0.0150
>> e=(b-a)/2
e =
3.1250e-004
>> b=r
b =
2.0272
>> r=(a+b)/2
r =
2.0270
>> f(a)
ans =
-7.7741e-004
>> f(r)
ans =0.0032
>> f(b)
ans =
0.0071
>> e=(b-a)/2
e =
1.5625e-004
>> b=r
b =
2.0270
>> r=(a+b)/2
r =
2.0270
>> a
a =
2.0269
>> b
b =
2.0270
>> f(a)
ans =
-7.7741e-004
>> f(r)
ans =
0.0012
>> f(b)
ans =
0.0032
>> e=(b-a)/2
e =
7.8125e-005
>> format long
>> r=(a+b)/2
r =
2.02695312500000
>> a
a =
2.02687500000000
>> b
b=
2.02703125000000
>> r=(a+b)/2
r =
2.02695312500000
Para la raíz 2 tenemos a=8.6, b=8.65, n=9
n
a
r
B
F(a)
F(r)
F(b)
error
1
8.6
8.625
8.65
+
-
-
0.025
2
8.6
8.6125
8.625
+
+
-
0.0125
3
8.6125
8.61875
8.625
+
-
-
0.00625
4
8.6125
8.615625
8.61875
+
-
-
0.003125
58.6125
8.614062
8.615625
+
-
-
0.0015625
6
8.6125
8.613281
8.614062
+
+
-
0.00078125
7
8.613281
8.613671
8.614062
+
+
-
0.000390625
8
8.613671
8.613867
8.614062
+
+
-
0.00019531
9
8.613867
8.613964
8.614062
+
-
-
0.00009765624
Para este caso profundidad del agua en el tanque h=8.613964 m
VENTANA DE COMANDOS
>> a=8.6;b=8.65;
>>n=ceil((log((b-a)/0.0001))/log(2))
n =
9
>> r=(a+b)/2
r =
8.62500000000000
>> f(a)
ans =
0.98029235460012
>> f(r)
ans =
-0.78686987550585
>> f(b)
ans =
-2.57612142895747
>> e=(b-a)/2
e =
0.02500000000000
>> b=r
b =
8.62500000000000
>> r=(a+b)/2
r =
8.61250000000000
>> b
b =
8.62500000000000
>> f(a)
ans =
0.98029235460012
>> f(r)
ans =...
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