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IX. Opgaver – 5. Januar 2005 592 Den retningsafledede i retning u = ( 4 5 , −3 5 ) er ifølge [S] 11.6 Opgave 2. Betragt følgende tre vektorer i R 4 : ∇f(0,0) · u = (1,2) · ( 4 4 6 , −3 ) = − = −2 5 5 5 5 5 . u 1 = (1,2,3,4), u 2 = (−3, −1,5,5), u 3 = (−4, −3,2,1). 1) Undersøg hvilke af følgende udsagn der gælder: 2) Det oplyses, at u 1 ⊥ u 2 , u 2 ⊥ u 3 , u 1 ⊥ u 3 . span(u 1,u 2) = span(u 2,u 3) = span(u 1,u 3). Dette underrum kaldes U. Lad v betegne vektoren v = (30,30,0,0). Angiv den ortogonale projektion projU(v) af vektoren v på U. Løsning. 1) Udregn Så u 1 ⊥ u 3 er det eneste sande udsagn. u 1 · u 2 = −3 − 2 + 15 + 20 = 30 = 0 u 1 · u 3 = −4 − 6 + 6 + 4 = 0 u 2 · u 3 = 12 + 3 + 10 + 5 = 30 = 0
IX. Opgaver – 5. Januar 2005 593 v v u ∈ U ⊥ u = projU(v) Ortogonal projektion på underrum 2) Problemstillingen er illustreret på figuren Da U = span(u 1 ,u 3 ) og u 1 og u 3 er ortogonale fås fra [LA] Sætning 17 projektionen af v = (30,30,0,0) på U projU(v) = proju 1 (v) + proju 3 (v) u1 + u1 · u1 v · u3 u3 · u3 = v · u 1 u 2 30 + 60 −120 − 90 = (1,2,3,4) + (−4, −3,2,1) 1 + 4 + 9 + 16 16 + 9 + 4 + 1 = 3(1,2,3,4) − 7(−4, −3,2,1) = (31,27, −5,5) . Opgave 3. Lad T betegne trekanten i xy-planen afgrænset af x-aksen, y-aksen og linien x + y = 2 U
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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