Eingebetteter Torus/Einheitsnormalenfeld/Weingartenabbildung/1/Aufgabe/Lösung

1. Es ist

   ${}h(x,y,z)=x^{2}+y^{2}-2{\sqrt {x^{2}+y^{2}}}R+z^{2}+R^{2}\,$

   und ${}T=h^{-1}(r^{2})$. Der Gradient von ${}h$ ist

   ${}\operatorname {Grad} \,h=2{\begin{pmatrix}x-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\\y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\\z\end{pmatrix}}\,.$

   Das Einheitsnormalenfeld ist daher

   ${}N(x,y,z)={\frac {1}{\sqrt {{\left(x-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}^{2}+{\left(y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}^{2}+z^{2}}}}{\begin{pmatrix}x-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\\y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\\z\end{pmatrix}}\,.$

2. Die zweite partielle Ableitung ist

   ${}{\begin{aligned}&\,\,\,\,\,\,\,{\frac {\partial }{\partial y}}N\\&={\frac {1}{\sqrt {{\left(-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}^{2}+{\left(y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}^{2}+z^{2}}}}{\begin{pmatrix}xyR{\left(x^{2}+y^{2}\right)}^{-{\frac {3}{2}}}\\1-R{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}+y^{2}R{\left(x^{2}+y^{2}\right)}^{-{\frac {3}{2}}}\\0\end{pmatrix}}\\&=+{\frac {2{\left(x-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}{\left(+xyR{\left(x^{2}+y^{2}\right)}^{-{\frac {3}{2}}}\right)}+2{\left(y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}{\left(1-R{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}+y^{2}R{\left(x^{2}+y^{2}\right)}^{\frac {3}{2}}\right)}}{{\left({\left(x-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}^{2}+{\left(y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\right)}^{2}+z^{2}\right)}^{3/2}}}{\begin{pmatrix}x-xR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\\y-yR{\left(x^{2}+y^{2}\right)}^{-{\frac {1}{2}}}\\z\end{pmatrix}}.\,\end{aligned}}$

   Im angegeben Punkt ist dies

Zur gelösten Aufgabe