Matematik

Funktion af to variable

21. april 2021 af Yaxion - Niveau: A-niveau

Jeg bliver stillet opgaven:

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En funktion f af to variable er bestemt ved

$f(x,y)=x^3 - 3x*y+y^3$

a) Bestem $\frac{\partial f}{\partial x}$ og $\frac{\partial f}{\partial y}$

Det oplyses, at grafen for f har netop to stationære punkter

b) Bestem arten af hvert af de to stationære punkter.

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Jeg forstår ikke helt hvad $\partial$ betyder eller hvordan jeg skal finde det..

Brugbart svar (1)

Svar #1
21. april 2021 af mathon

$\small \small \begin{array}{lllll}&& \frac{\partial f }{\partial x}=f_x{ }'(x,y)&\textup{den delafledede af f mht }x\\\\&& \frac{\partial f }{\partial y}=f_y{ }'(x,y)&\textup{den delafledede af f mht }y \end{array}$

Brugbart svar (1)

Svar #2
21. april 2021 af StoreNord

$\frac{\partial f}{\partial x}$ er den afledte med hensyn til x.

Brugbart svar (1)

Svar #3
21. april 2021 af mathon

$\small \small \begin{array}{llllll}&& \frac{\partial }{\partial x}(f(x,y))=3x^2-3y\\\\&& \frac{\partial }{\partial y}(f(x,y))=-3x+3y^2\\\\& \textup{station\ae re punkter}\\& \textup{kr\ae ver:}\\&& \begin{matrix} 3x^2-3y=0\\ -3x+3y^2=0 \end{matrix}\\& \textup{dvs}\\&& P_1=(0,0)\textup{ og }P_2=(1,1)\\\\\\&& \frac{\partial^2 }{\partial x^2}(f(x,y))=6x\\\\&& \frac{\partial^2 }{\partial x^2}(f(0,0))=0=r\\\\\\&&\frac{\partial^2 }{\partial y^2}(f(x,y))=6y\\\\&& \frac{\partial^2 }{\partial y^2}(f(0,0))=0=t\\\\\\&& \frac{\partial }{\partial y}\left ( \frac{\partial }{\partial x}(f(x,y)) \right )=-3\\\\&& \frac{\partial }{\partial y}\left ( \frac{\partial }{\partial x}(f(0,0)) \right )=-3=s\\\\\\&& r\cdot t-s^2=0\cdot 0-(-3)^2<0\quad P_1(0,0)\textup{ er saddelpunkt} \end{array}$

Brugbart svar (1)

Svar #4
21. april 2021 af mathon

$\small \small \small \small \begin{array}{llllll}&& \frac{\partial^2 }{\partial x^2}(f(x,y))=6x\\\\&& \frac{\partial^2 }{\partial x^2}(f(1,1))=6=r\\\\\\&&\frac{\partial^2 }{\partial y^2}(f(x,y))=6y\\\\&& \frac{\partial^2 }{\partial y^2}(f(1,1))=6=t\\\\\\&& \frac{\partial }{\partial y}\left ( \frac{\partial }{\partial x}(f(x,y)) \right )=-3\\\\&& \frac{\partial }{\partial y}\left ( \frac{\partial }{\partial x}(f(1,1)) \right )=-3=s\\\\\\&& r \cdot t-s^2=6\cdot 6-(-3)^2>0\\\\&& r>0\qquad P_2(1,1) \textup{ er et lokalt minimumspunkt} \end{array}$

Svar #5
21. april 2021 af Yaxion

#4
$\small \small \small \small \begin{array}{llllll}&& \frac{\partial^2 }{\partial x^2}(f(x,y))=6x\\\\&& \frac{\partial^2 }{\partial x^2}(f(1,1))=6=r\\\\\\&&\frac{\partial^2 }{\partial y^2}(f(x,y))=6y\\\\&& \frac{\partial^2 }{\partial y^2}(f(1,1))=6=t\\\\\\&& \frac{\partial }{\partial y}\left ( \frac{\partial }{\partial x}(f(x,y)) \right )=-3\\\\&& \frac{\partial }{\partial y}\left ( \frac{\partial }{\partial x}(f(1,1)) \right )=-3=s\\\\\\&& r \cdot t-s^2=6\cdot 6-(-3)^2>0\\\\&& r>0\qquad P_2(1,1) \textup{ er et lokalt minimumspunkt} \end{array}$

Hvordan får du $f''(1,1)$ til at give 6? For mig giver det bare 0..

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