Resolucao Passo a Passo Questoes Prova Calculo L11s2s3 e L1

Faça a resolução das questões com passo a passo PROVA Questão para estudar para a prova realizar passo a passo dos cálculos a L11s2s3 1s2s3 As2 Bs3 1 As3 Bs2 1 ABs 3A 2B A B 0 3A 2B 1 A 25 B 25 L11s2s3 25 e2t 25 e3t b L11s2s1 1s2s1 A5 Bs2 Cs1 1 Ass1 Bs2 Cs2 ABCs2 As 1 A 1 ABC 0 A 1 B 1 C 0 1s2s1 15 1s2 L11s2s1 L115 1s2 1 t c ss1s21 As1 Bs Cs2 1 s As21 Bs Cs1 s As2 A Bs21 Bs Cs1 s A Bs2 B Cs A C A B 0 B C 1 A C 0 A C 0 B 1 L1ss1s21 L11s2 1 sent 2 a yt 2yt yt et y0 1 y0 1 s2 Ys sy0 y0 2sYs y0 Ys 1s1 s2 Ys 2s Ys Ys s 2 1s1 Ys s2 s 1s12 s2 s 1s12 As1 Bs12 s2 s 1 As1 B A 1 B 2 entao Ys 1s1 2s12 Yt et 2t et b Yt 2Yt 1 Y0 3 dYtdt 2 Y1 1 s Ys Y0 2 Ys 15 s Ys 3 2 Ys 15 s2 Ys 15 3 Ys 1 ss2 3 s2 Yt L1 1 5s2 3 s2 15 1 s2 3 1 s2 Yt 12 L1 15 L1 1 s2 3 L1 1 s2 Yt 12 1 e2t 3 e2t c Yt 3 Yt et Y0 2 s Ys Y0 3 Ys 1 s1 Y0 2 s Ys 2 3 Ys 1 s1 Ys s3 3 s1 Ys 3 s1s3 3 s1s3 A s1 B s3 A1 B1 Ys 1 s1 1 s3 Yt L1 1 s1 1 s3 et e3t Yt et e3t d Yt 5 Yt 6 Yt 0 Y01 Y00 s2 Ys s Y0 Y0 5s Ys Y0 6 Ys 0 s2 Ys 5 5 s Ys 6 Ys 0 Ys s2 5 s 6 5 Ys s s2 5 s 6 s s2 5 s 6 A s3 B s2 s A s2 B s3 A B 1 2 A 3 B 0 A 3 B 4 s s2 5 s 6 3 s3 4 s2 L1 3 s3 4 s2 3 e3t 4 e2t e yt yt 2 cost y0 3 y0 4 s2 Ys s Y0 y0 Ys 2 s s1 s2 Ys 3s 4 Ys 2 5 51 Yss2 1 2 5 s2 1 3s 4 Ys 5s 4 s2 1 L1 5s 4 s2 1 L1 5 s s2 1 4 s2 1 Yt 5 cost 4 sint 3 a yt 0 to t yμdμ 1 y0 2 5 Ys Y0 1s Ys 1s Y0 2 5 Ys 2 1s Ys 1s Ys 1s 5 s2 1 2 5 s2 1 L1 1s 5 s2 1 2 5 s2 1 1 sin t 2 cos t Yt sin t 2 cos t b yt yt 6 0 to yμ du 12 e 3t y0 3 5 Ys Y0 Ys 6 1s Ys 12 s 3 5 Ys 3 Ys 6s Ys 12 s 3 Ys5 1 6s 12 s 3 3 Ys 35s2 21s 5 s 3s 2 35s2 21s 5 s 3s 2 As Bs3 Cs2 35ss7 As3s2 B s s2 C s s3 A0 B2125 C5425 Yt 2125 153 5425 152 Yt 2125 e3t 5425 e2t c Yt Yu du sen 2 t 0 t Yt 15 Yt 252 4 Yt 21 15 52 4 Yt 2 25 52 4 252 4 2 5 52 4 2 552 4 As Bs C52 4 A 25 B 45 C 0 Yt 25 cos 2t 45 sen 2 t 25 et PROVA L ditdt Rit 1C iτdτ Et 0 L 01H R 2Ω C 01F i0 0 Et 120t 120t u1t Aplicando Laplace em ambos os lados LEt L dditdt R L it 1C L iτdτ L 0 t iτdτ 1s L it L s L it 2i0 R L it 1C L its 12052 1205 L it 2 s R 1C s L i0 12052 1205 L it 2i02s R 1Cs 120s2 2s R 1Cs 12052s R 1Cs Aplicando frações parciais L i0 2 s R 1C s As Bs C 2 s R 1C s L i0 A2 s R 1C s Bs C 120s2 1205 A 2 s R 1C s B s C 120s2 1205 A 0 B 120L C 120L A R C 120L L i0 Substituindo A B e C 0 R 120L 120L 120L L i0 R 1 2i0120 Então it 120L 2i0120 120L etRC Substituindo valores it 1200 1200 e5t