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Web skærmformat. - Aarhus Universitet

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IX. Opgaver – 4. August 2004 585<br />

Løsning. 1) De partielle afledede er<br />

hvorfra gradienten angives<br />

fx = ln(1 + x + y) +<br />

∇f(x,y) = (fx,fy) = (ln(1 + x + y) +<br />

2) Den retningsafledede i retning u = 1<br />

√ 2 (1,1) er<br />

x<br />

1 + x + y , fy<br />

x<br />

=<br />

1 + x + y ,<br />

x<br />

1 + x + y ,<br />

x<br />

) .<br />

1 + x + y<br />

∇f(0,1) · u = (ln 2,0) · 1 √ 2 (1,1) = 1 √ 2 ln 2 .<br />

Opgave 3. Det oplyses, at vektoren v = (1,2) er en egenvektor for en matrix A af formen<br />

hvor a er et vist reelt tal.<br />

1) Angiv den tilhørende egenværdi<br />

2) Beregn tallet a.<br />

Løsning. 1) Udregningen<br />

A v =<br />

2 1<br />

a 3<br />

A =<br />

1<br />

2<br />

2 1<br />

a 3<br />

<br />

=<br />

<br />

,<br />

4<br />

a + 6<br />

<br />

<br />

1<br />

= λ<br />

2

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