LoesningEksamenBasisMat1F19.pdf - Løsning til 3-timers prøve i 01901 BasisMat1 F19 Opgave 1(a Brøken forlænges med nævnerens konjugerede z= 9 2i(9

# LoesningEksamenBasisMat1F19.pdf - Løsning til 3-timers...

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Løsning til 3-timers prøve i 01901 BasisMat1 F19 Opgave 1(a) Brøken forlænges med nævnerens konjugerede z = 9 + 2 i 4 - i = ( 9 + 2 i )( 4 + i ) ( 4 - i )( 4 + i ) = 36 + 9 i + 8 i + 2 i 2 4 2 - i 2 = 36 + 17 i - 2 16 + 1 = 34 + 17 i 17 = 2 + i . Opgave 1(b) Vi finder w = 2e - i π 12 e - i π 6 = 2e - i π 12 e i π 6 = 2e i π 12 s˚a w 3 = ( 2 ) 3 e i π 12 · 3 = 2 2e i π 4 = 2 2 ( cos ( π 4 )+ i sin ( π 4 )) = 2 2 ( 2 2 + i 2 2 ) = 2 + 2 i . Opgave 2(a) For f ( t , u ) = t sin ( u ) - 4cos ( 2 u ) finder man f t ( t , u ) = sin ( u ) og f u ( t , u ) = t cos ( u ) - 4 ( - sin ( 2 u )) · 2 = t cos ( u )+ 8sin ( 2 u ) . Vi finder f t ( 0 , π ) = sin ( π ) = 0 og f u ( 0 , π ) = 0cos ( π )+ 8sin ( 2 π ) = 0 · ( - 1 )+ 8 · 0 = 0 . Hermed er det vist, at ( t , u ) = ( 0 , π ) er et stationært punkt for funktionen f . 1

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Opgave 2(b) Ud fra at f ( 2 , π 6 ) = 2sin ( π 6 ) - 4cos ( 2 · π 6 ) = 2 · 1 2 - 4 · 1 2 = - 1 , at f t ( 2 , π 6 ) = sin ( π 6 ) = 1 2 , og at f u ( 2 , π 6 ) = 2cos ( π 6 )+ 8sin ( 2 · π 6 ) = 2 · 3 2 + 8 · 3 2 = 5 3 f˚as, at lineariseringen af f i udviklingspunktet ( 2 , π 6 ) er ˜ f 1 ( t , u ) = f ( 2 , π 6 )+ f t ( 2 , π 6 ) · ( t - 2 )+ f u ( 2 , π 6 ) · ( u - π 6 ) = - 1 + 1 2 ( t - 2 )+ 5 3 ( u
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