Exercício de cálculo numérico
Utilizando o método de mínimos quadrados obter os parâmetros “a”, e “b” das seguintes equações: a) y = a+b.x N 1 2 3 4 5 6 7 8 9 10 11 ∑ xi 10 15 20 25 30 35 40 45 50 55 60 385 yi 1,293 1,135 1 0,894 0,8 0,723 0,654 0,608 0,546 0,51 0,46 8,623 xi2 100 225 400 625 900 1225 1600 2025 2500 3025 3600 xi.yi 12,93 17,025 20 22,35 24 25,305 26,16 27,36 27,3 28,05 27,6 yi2 1,671849 1,288225 1 0,799236 0,64 0,522729 0,427716 0,369664 0,298116 0,2601 0,2116 7,489235
16225 258,08
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b) Y= a+b.lnx N 1 2 3 4 5 6 7 8 9 10 11 ∑ xi 10 15 20 25 30 35 40 45 50 55 60 385
∑ ∑
yi 1,293 1,135 1 0,894 0,8 0,723 0,654 0,608 0,546 0,51 0,46 8,623
( )
∑
ln xi 2,302585 2,70805 2,995732 3,218876 3,401197 3,555348 3,688879 3,806662 3,912023 4,007333 4,094345 37,69103
∑ ) ( )
lnxi2 5,301898 7,333536 8,974412 10,36116 11,56814 12,6405 13,60783 14,49068 15,30392 16,05872 16,76366 132,4045
∑
yi.lnxi 2,977243 3,073637 2,995732 2,877675 2,720958 2,570517 2,412527 2,314451 2,135965 2,04374 1,883398 28,00584
( )
∑ (∑
yi2 1,671849 1,288225 1 0,799236 0,64 0,522729 0,427716 0,369664 0,298116 0,2601 0,2116 7,489235
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c) y = a.xb → lnyi = lna + b.lnxi y= ln yi x= ln xi N 1 2 3 4 5 6 7 8 9 10 11 ∑ xi 10 15 20 25 30 35 40 45 50 55 60 385 yi 1,293 1,135 1 0,894 0,8 0,723 0,654 0,608 0,546 0,51 0,46 8,623 A = ln a → a = eA B=b ln(xi) 2,302585 2,70805 2,995732 3,218876 3,401197 3,555348 3,688879 3,806662 3,912023 4,007333 4,094345 37,69103 ln(yi) 0,256965 0,126633 0 -0,11205