yüzeyin teğet düzlemi

[Lokman Gökçe]

Soru (L. Gökçe):

$z=x^2-xy+y^3$ yüzeyinin $(1,1,1)$ noktasındaki teğet düzleminin denklemi nedir?

$
\textbf{a)}\ x+y+z=3
\qquad\textbf{b)}\ 2x-y+z=2
\qquad\textbf{c)}\ x+2y-z=2
\qquad\textbf{d)}\ 2x+3y-z=4
\qquad\textbf{e)}\ -x+2y+2z=3
$

[cunomat]

Yanıt: $\boxed {C}$

$f\left( x,y,z\right) =z-x^{2}+xy-y^{3}$

$f_{x}\left( x,y,z\right) =-2x+y \Rightarrow f_{x}\left( 1,1,1\right) =-1$

$f_{y}\left( x,y,z\right) =x-3y^2 \Rightarrow f_{y}\left( 1,1,1\right) =-2$

$f_{z}\left( x,y,z\right) =1\Rightarrow f_{z}\left( 1,1,1\right) =1$ olur. Teğet düzleminin denklemi

$f_{x}\left( 1,1,1\right)  \cdot \left( x-1\right) +f_{y}\left( 1,1,1\right) \cdot \left( y-1\right) +f_{z}\left( 1,1,1\right) \cdot \left( z-1\right) =0$

$\Rightarrow -x+1-2y+2+z-1=0$

$\Rightarrow x+2y-z=2$ dir.