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I. Differentiation – 5. Gradient 108 y 1 0 1 Skalerede gradienter 0.1∇z for z = xe y . 5.34. Opgave ☞ Matematik Alfa 1, August 2002 Opgave 5 Betragt funktionen f(x,y) = y 2 + ln(x 3 + y + 1). 1. Angiv gradientvektoren ∇f(0,2). 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) x
I. Differentiation – 5. Gradient 109 fy = 2y + 1/(x 3 + y + 1) ∇f(0,2) = (fx(0,2),fy(0,2)) = (0,13/3) 5.35. Opgave ☞ Matematik Alfa 1, August 2002 Opgave 5 - fortsat 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
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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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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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