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IX. Opgaver – 2. Januar 2003 550 3) Angiv en enhedsvektor u så at den retningsafledede Duf(5,2) er 0. 2.4. Find gradient ☞ Matematik Alfa 1, Januar 2003 Opgave 1 - løsning 1) De partielle afledede beregnes 2) Gradienten angives fx = fy = −y − 1 (x + y) 2 x − 1 (x + y) 2 fx(5,2) = − 3 49 ∇f(5,2) = (fx(5,2),fy(5,2)) = (− 3 4 , 49 49 ) 2.5. Find gradient ☞ Matematik Alfa 1, Januar 2003 Opgave 1 - løsning
IX. Opgaver – 2. Januar 2003 551 7 ∇f (5,2) u 5 6 x 3. I retning u = (u1,u2) er den retningsafledede Duf(5,2) = ∇f(5,2) · u = (− 3 4 , ) · (u1,u2) 49 49 = 1 49 (−3u1 + 4u2) 2.6. Find gradient ☞ Matematik Alfa 1, Januar 2003 Opgave 1 - løsning Den retningsafledede for En løsning er Duf(5,2) = 1 49 (−3u1 + 4u2) = 0 −3u1 + 4u2 = 0 u 2 1 + u 2 2 = 1 u = ( 4 3 , 5 5 ) 2.7. Find gradient ☞ Matematik Alfa 1, Januar 2003 Opgave 1 - ekstra
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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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I. Differentiation - 1. Kontinuitet
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1. Grafiske/numeriske metoder IV Di
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V. Matricer - 4. Determinanter = (
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IX. Opgaver - 5. Januar 2005 600 Gr
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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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Identitetsmatricen, 383 Identitetsm
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etningsafledede, 89, 102 retningsfe
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