metodos

1. CAPITULO 1. CALCULO DE ERRORES
1. Redondear los siguientes números a tres cifras significativas
c. 0,169124 x 102
= 16,9
2. Determinar la cantidad de cifras significativas para los siguientes números
c. 263,3±0,01
Cifras significativas = 4
3. Calcule el error absoluto y relativo en las aproximaciones deA por a:
c. A=e10
, a=22 000
∆ 𝑎= |𝐴 − 𝑎|
∆ 𝑎= |e10
− 22 000|
∆ 𝑎= 26.46579481
𝛿𝐴 =
∆ 𝑎
𝐴
𝛿𝐴 =
26.46579481
e10
𝛿𝐴 = 0,0012
𝛿𝐴 ≅ 0,12%
4. Encuentre el intervalo más grande en que debe encontrarse a para que aproxime A con un
error relativo màximo de 10-4
para cada valor de A
c. √17
𝐴 = 𝑎(1 ± 𝛿𝑎)
𝑎 =
𝐴
(1 ± 𝛿𝑎)
𝑎1 =
𝐴
(1 + 𝛿𝑎)
𝑎1 =
√17
(1 + 10−4 )
𝑎1 = 4.1226
𝑎2 =
𝐴
(1 − 𝛿𝑎)
𝑎2 =
√17
(1 − 10−4 )
𝑎2 = 4.1235
4.1226 ≤ 𝑎 ≤ 4.1235
5. Calcular los errores de las siguientes expresiones:
c. 𝑥 =
𝐴2
√ 𝐵
𝐶3 √ 𝐷
4
+𝐸 4 √ 𝐹
3 +
𝐺2
√ 𝐻+𝐼5
√ 𝐽3
𝐾3 √ 𝐿
4
A = 85,63 ± 0,001 𝐴2
= (85,63)2
= 7332,4969

2. B = 326,8 ± 0,02 √𝐵 = √326,8 = 18.07761
C = 3,451 ± 0,001 𝐶3
= (3,451)3
= 41.09934
D = 1875,2 ± 0,03 √𝐷
4
= √1875,2
4
= 6.58055
E = 2,481 ± 0,002 𝐸 4
= (2,481)4
= 37.88847
F = 825,7 ± 0,02 √𝐹
3
= √825,7
3
= 9.38154
G = 10,36 ± 0,001 𝐺2
= (10,36)2
= 107.3296
H = 37,42 ± 0,001 √𝐻 = √37,42 = 6.11719
I = 1,534 ± 0,002 𝐼5
= (1,534)5
= 8.49428
J = 475,21 ± 0,003 √𝐽
3
= √475,21
3
= 7.803603
K = 2,932 ± 0,001 𝐾3
= (2.932)3
= 25.205302
L = 1796,1 ± 0,02 √𝐿
4
= √1796.1
4
= 6.51002
X=
(7332,4969)(18.07761)
(41.09934)(6.58055)+(37.88847)(9.38154)
+
(107.3296)(6.11719)+(8.49428)(7.803603)
(25.205302)(6.51002)
𝑋 =
132554.0193
270.45626 + 355.4521968
+
656.55556 + 66.28599
164.087021
𝑋 =
132554.0193
625.9084568
+
422.84155
164.087021
X = 211.7786041 + 2.576934772
X = 214.3555389
𝛿𝑎𝑏 = 2 (
0,001
85,63
) +
1
2
(
0,02
326,8
) = 5,33956 𝑥 10−5
𝛿𝑐𝑑 = 3 (
0,001
3,451
) +
1
4
(
0,03
1875.2
) = 8.73312815 𝑥 10−4
∆𝑐𝑑 = (8.73312815 𝑥 10−4)(270.45626) = 0,236192918
𝛿𝑒𝑓 = 4 (
0,002
2.481
) +
1
3
(
0,02
825,7
) = 3.2325802 𝑥 10−3
∆𝑒𝑓 = (3.2325802 𝑥 10−3)(355.4521968) = 1.149027735

3. ∆𝑐𝑑𝑒𝑓 = ∆𝑐𝑑 + ∆𝑒𝑓 = 0,236192918 + 1.149027735 = 1.385220653
𝛿𝑐𝑑𝑒𝑓 = (
∆𝑐𝑑𝑒𝑓
𝐶𝐷𝐸𝐹
) = (
1.385220653
625.9084568
) = 2.21313618 𝑥 10−3
𝛿𝑎𝑏𝑐𝑑𝑒𝑓 = 𝛿𝑎𝑏 + 𝛿𝑐𝑑𝑒𝑓 = 5,33956 𝑥 10−5
+ 2.21313618 𝑥 10−3
= 2.26653178 𝑥 10−3
∆ 𝑎𝑏𝑐𝑑𝑒𝑓 = (2.26653178 𝑥 10−3
)(211.7786041) = 0,4800029372
𝛿𝑔ℎ = 2 (
0,001
10.36
) +
1
2
(
0,001
37.42
) = 2.06412031 𝑥 10−4
∆𝑔ℎ = (2.06412031 𝑥 10−4)(656.55556) = 0.135520967
𝛿𝑖𝑗 = 5 (
0,002
1.534
) +
1
3
(
0,003
475.21
) = 6.52100915 𝑥 10−3
∆𝑖𝑗 = (6.52100915 𝑥 10−3)(66.28599) = 0,4322515478
∆𝑔ℎ𝑖𝑗 = ∆𝑔ℎ + ∆𝑖𝑗 = 0.135520967 + 0,4322515478 = 0,5677725148
𝛿𝑔ℎ𝑖𝑗 = (
∆𝑔ℎ𝑖𝑗
𝐺𝐻𝐼𝐽
) = (
0,5677725148
422.84155
) = 1.34275478 𝑥 10−3
𝛿𝑘𝑙 = 3 (
0,001
2.932
) +
1
4
(
0,02
1796,1
) = 1.02597617 𝑥 10−3
𝛿𝑔ℎ𝑖𝑗𝑘𝑙 = 𝛿𝑔ℎ𝑖𝑗 + 𝛿𝑘𝑙 = 1.34275478 𝑥 10−3
+ 1.02597617 𝑥 10−3
= 2.36873095 𝑥 10−3
∆ 𝑔ℎ𝑖𝑗𝑘𝑙 = (2.36873095 𝑥 10−3
)(2.576934772) = 6.10406516 𝑥 10−3
∆𝑎𝑏𝑐𝑑𝑒𝑓 𝑔ℎ𝑖𝑗𝑘𝑙 = ∆ 𝑎𝑏𝑐𝑑𝑒𝑓 + ∆ 𝑔ℎ𝑖𝑗𝑘𝑙 = 0,4800029372 + 6.10406516 𝑥 10−3
= 0.4861070024
↓
X = (214.355 ± 0.486)