Ebene differenzierbare Kurve/Krümmung/Beschreibung/Fakt/Beweis

Beweis

Wegen

{}{\begin{aligned}{\frac {\left\langle \alpha ^{\prime \prime }(t),N(t)\right\rangle }{\Vert {\alpha '(t)}\Vert ^{2}}}&={\frac {\left\langle \alpha ^{\prime \prime }(t),{\frac {1}{\Vert {\alpha '(t)}\Vert }}{\begin{pmatrix}-\alpha _{2}'(t)\\\alpha _{1}'(t)\end{pmatrix}}\right\rangle }{\Vert {\alpha '(t)}\Vert ^{2}}}\\&={\frac {\left\langle {\begin{pmatrix}\alpha _{1}^{\prime \prime }(t)\\\alpha _{2}^{\prime \prime }(t)\end{pmatrix}},{\begin{pmatrix}-\alpha _{2}'(t)\\\alpha _{1}'(t)\end{pmatrix}}\right\rangle }{\Vert {\alpha '(t)}\Vert \cdot \Vert {\alpha '(t)}\Vert ^{2}}}\\&={\frac {\alpha _{1}'(t)\alpha _{2}^{\prime \prime }(t)-\alpha _{2}'(t)\alpha _{1}^{\prime \prime }(t)}{{\left(\alpha _{1}'(t)^{2}+\alpha _{2}'(t)^{2}\right)}^{3/2}}}\end{aligned}}

stimmen die beiden Terme überein. Es sei

{}\alpha (t)=\gamma (\beta (t))\,

mit einer Umparametrisierung ${}\beta$ und der bogenparametrisierten Kurve ${}\gamma$ . Dann ist

{}{\begin{aligned}&\,\,\,\,\,\,\,{\frac {\alpha _{1}'(t)\alpha _{2}^{\prime \prime }(t)-\alpha _{2}'(t)\alpha _{1}^{\prime \prime }(t)}{{\left(\alpha _{1}'(t)^{2}+\alpha _{2}'(t)^{2}\right)}^{3/2}}}\\&={\frac {\gamma _{1}'(\beta (t))\beta '(t){\left(\gamma _{2}^{\prime \prime }(\beta (t))\beta '(t)^{2}+\gamma _{2}'(\beta (t))\beta ^{\prime \prime }(t)\right)}-\gamma _{2}'(\beta (t))\beta '(t){\left(\gamma _{1}^{\prime \prime }(\beta (t))\beta '(t)^{2}+\gamma _{1}'(\beta (t))\beta ^{\prime \prime }(t)\right)}}{{\left(\gamma _{1}'(\beta (t))^{2}\beta '(t)^{2}+\gamma _{2}'(\beta (t))^{2}\beta '(t)^{2}\right)}^{3/2}}}\\&={\frac {\gamma _{1}'(\beta (t)){\left(\gamma _{2}^{\prime \prime }(\beta (t))\beta '(t)^{2}+\gamma _{2}'(\beta (t))\beta ^{\prime \prime }(t)\right)}-\gamma _{2}'(\beta (t)){\left(\gamma _{1}^{\prime \prime }(\beta (t))\beta '(t)^{2}+\gamma _{1}'(\beta (t))\beta ^{\prime \prime }(t)\right)}}{\beta '(t)^{2}{\left(\gamma _{1}'(\beta (t))^{2}+\gamma _{2}'(\beta (t))^{2}\right)}^{3/2}}}\\&={\frac {\gamma _{1}'(\beta (t))\gamma _{2}^{\prime \prime }(\beta (t))\beta '(t)^{2}-\gamma _{2}'(\beta (t))\gamma _{1}^{\prime \prime }(\beta (t))\beta '(t)^{2}}{\beta '(t)^{2}}}\\&=\gamma _{1}'(\beta (t))\gamma _{2}^{\prime \prime }(\beta (t))-\gamma _{2}'(\beta (t))\gamma _{1}^{\prime \prime }(\beta (t))\\&=\kappa (\beta (t)).\,\end{aligned}}