Exercícios Resolvidos de Análise Numérica

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Domingos Júnior da Cruz Lopes LMM 2º ano. 
 
 
14. a2-300a+1=0 ã =
300±√(−300)2−4∗1∗1
2∗1
 
 ã =
300±√89996
2
 
 ã=
300±299.993
2
 
 ã1 =
300−299.993
2
 ˅ ã2 =
300+299.993
2
 
 ã1=0.0035 v ã2=299.9965 
Se b2»|4*1*1| há um canselamento subtrativo. 
Se b«0 há um canselamento para o cálculo de a1: 
 a1=>ã1=c/a*a1=1/299.9965=0.00333337222. 
Se b»0 =>ã2 = c/a*a1 
 
15. x=0.57±0.005 ~y=sin(0.57)=0.00994821263 
Y=sin(x ) xϵ|0.57-0.005 , 0.57+0.005| 
|∂y/∂x|=cos(x)≤cos(0.57-0.005)≤0.9999513796 
|E~y|≤μ~xmáxxe~x |∂y/∂x|=0.005*cos(0.57-0.005)≤0.00499975689<0.5*10
-4 
Logo ~y=sen(0.57±0.005) 
 
16. V=hπr2 ṽ=2.00*1.002*3.142=6.284 
 h=2.00±0.001 |Eṽ|≤μ~hmax|∂v/∂h|+μ~rmaxr|∂v/∂r|+μ~πmaxπ|∂v/∂π| 
 r=1.00±0.001 
 π=3.142±0.59*10-3 |∂v/∂h|=πr2=3.14259*1.0012=3.1487881 
 hϵ[2.00-0.001 , 2.00+0.001] |∂v/∂π|=hr2=2.001*1.0012=2.005004 
rϵ[1.00-0.001 , 1.00+0.001] |∂v/∂r|=2hπr=12,58886129 
πϵ[3.142-0.59*10-3 , 3.142+0.59*10-3] 
|Eṽ|≤0.001*3.1487881+0.59*10
-3*2.005004+0.001*12,58886129<o.17*10-1<o.5*10-5 
V=6.284±o.5*10-1