Bogenparametrisierte Kurve/Ebene/Krümmungskreis/Durchlauf/Fakt/Beweis

Beweis

Wir betrachten den standardorientierten Fall. Es ist

${}\delta (t_{0})=M+r{\begin{pmatrix}\cos \left(\alpha +{\frac {t_{0}}{r}}\right)\\\sin \left(\alpha +{\frac {t_{0}}{r}}\right)\end{pmatrix}}=\gamma (t_{0})+r{\begin{pmatrix}-\gamma _{2}'(t_{0})\\\gamma _{1}'(t_{0})\end{pmatrix}}+r{\begin{pmatrix}\gamma _{2}'(t_{0})\\-\gamma _{1}'(t_{0})\end{pmatrix}}=\gamma (t_{0})\,.$

Ferner ist

${}\delta '(t_{0})={\begin{pmatrix}-\sin \left(\alpha +{\frac {t_{0}}{r}}\right)\\\cos \left(\alpha +{\frac {t_{0}}{r}}\right)\end{pmatrix}}={\begin{pmatrix}\gamma _{1}'(t_{0})\\\gamma _{2}'(t_{0})\end{pmatrix}}=\gamma '(t_{0})\,,$

insbesondere ist ${}\delta$ bogenparametrisiert. Es steht ${}{\begin{pmatrix}\gamma _{1}^{\prime \prime }(t_{0})\\\gamma _{2}^{\prime \prime }(t_{0})\end{pmatrix}}$ senkrecht auf ${}{\begin{pmatrix}\gamma _{1}'(t_{0})\\\gamma _{2}'(t_{0})\end{pmatrix}}$ und ist daher linear abhängig zu dem Orthonormalvektor ${}{\begin{pmatrix}-\gamma _{2}'(t_{0})\\\gamma _{1}'(t_{0})\end{pmatrix}}$. Somit ist

${}{\begin{pmatrix}\gamma _{1}^{\prime \prime }(t_{0})\\\gamma _{2}^{\prime \prime }(t_{0})\end{pmatrix}}=\left\langle {\begin{pmatrix}-\gamma _{2}'(t_{0})\\\gamma _{1}'(t_{0})\end{pmatrix}},{\begin{pmatrix}\gamma _{1}^{\prime \prime }(t_{0})\\\gamma _{2}^{\prime \prime }(t_{0})\end{pmatrix}}\right\rangle {\begin{pmatrix}-\gamma _{2}'(t_{0})\\\gamma _{1}'(t_{0})\end{pmatrix}}={\left(\gamma _{1}'(t_{0})\gamma _{2}^{\prime \prime }(t_{0})-\gamma _{2}'(t_{0})\gamma _{1}^{\prime \prime }(t_{0})\right)}{\begin{pmatrix}-\gamma _{2}'(t_{0})\\\gamma _{1}'(t_{0})\end{pmatrix}}\,$

und daher

${}{\begin{aligned}\delta ^{\prime \prime }(t_{0})&=-{\frac {1}{r}}{\begin{pmatrix}\cos \left(\alpha +{\frac {t_{0}}{r}}\right)\\\sin \left(\alpha +{\frac {t_{0}}{r}}\right)\end{pmatrix}}\\&=-{\left(\gamma _{1}'(t_{0})\gamma _{2}^{\prime \prime }(t_{0})-\gamma _{2}'(t_{0})\gamma _{1}^{\prime \prime }(t_{0})\right)}{\begin{pmatrix}\gamma _{2}'(t_{0})\\-\gamma _{1}'(t_{0})\end{pmatrix}}\\&={\left(\gamma _{1}'(t_{0})\gamma _{2}^{\prime \prime }(t_{0})-\gamma _{2}'(t_{0})\gamma _{1}^{\prime \prime }(t_{0})\right)}{\begin{pmatrix}-\gamma _{2}'(t_{0})\\\gamma _{1}'(t_{0})\end{pmatrix}}\\&={\begin{pmatrix}\gamma _{1}^{\prime \prime }(t_{0})\\\gamma _{2}^{\prime \prime }(t_{0})\end{pmatrix}}\\&=\gamma ^{\prime \prime }(t_{0}).\end{aligned}}$

Zur bewiesenen Aussage