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IX. Opgaver – 1. August 2002 538 2. Angiv den retningsafledede af f i punktet P = (0,2) i retning givet ved enhedsvektoren (3/5,4/5). Løsning 1. Gradienten beregnes fx = 3x 2 /(x 3 + y + 1) fy = 2y + 1/(x 3 + y + 1) ∇f(0,2) = (fx(0,2),fy(0,2)) = (0,13/3) 1.26. Find gradient ☞ Matematik Alfa 1, August 2002 Opgave 5 - løsning
IX. Opgaver – 1. August 2002 539 y ∇f(0,2) (0,2) u 1 x 2. I retning u = (3/5,4/5) er den retningsafledede Duf(0,2) = ∇f(0,2) · u = (0,13/3) · (3/5,4/5) = 52/15 1.27. Find gradient ☞ Matematik Alfa 1, August 2002 Opgave 5 - 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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I. Differentiation - 6. Maksimum/mi
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I. Differentiation - 7. Lagrangemet
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III. Potensrækker - 4. Taylorpolyn
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1. Grafiske/numeriske metoder IV Di
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IV. Differentialligninger - 1. Graf
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1. Vektorer og matricer V Matricer
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V. Matricer - 1. Vektorer og matric
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V. Matricer - 4. Determinanter = (
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1. Egenvektorer VI Egenvektorer og
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VI. Egenvektorer og diagonalisering
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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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