Proof of the equation evolute

Question

I have curve $\alpha(t) = (x(t), \quad y(t), \quad 0)$ in arbitrary parametrization.

I trying to prove the formula for evolute $$ \epsilon(t) = \left(x-\dot{y} \frac{(\dot{x})^2+(\dot{y})^2}{\dot{x}\ddot{y}-\dot{y}\ddot{x}},\quad y+\dot{x} \frac{(\dot{x})^2+(\dot{y})^2}{\dot{x}\ddot{y}-\dot{y}\ddot{x}}, \quad 0 \right) $$

I can easily come to the conclusion that $$ \epsilon(s) = \alpha(s) + \frac{1}{\kappa(s)}N(s) $$ in Arclength Parametrization.

I tried to substitute functions that I set before $$ \kappa(t) = \frac{||\dot{\alpha}(t) \times \ddot{\alpha}(t)||}{||\dot{\alpha}(t)||^3} $$ $$ N(t) = \frac{(\dot{\alpha}(t)\times \ddot{\alpha}(t))\times \dot{\alpha}(t)}{||(\dot{\alpha}(t)\times\ddot{\alpha}(t))\times\dot{\alpha}(t)||} $$ but I have not received anything meaningful

$$ \frac{N(s)}{\kappa(s)}=\frac{N(t)}{\kappa(t)}=\frac{\ddot{\alpha}(t)\cdot ||\dot{\alpha}(t)||^4}{||\ddot{\alpha}(t)||} $$

Thanks for help.

Answer 1

Use $$\kappa= \frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}} {((\dot{x})^2+(\dot{y})^2)^{\frac{3}{2}}}$$

and $$N=\frac{1}{((\dot{x})^2+(\dot{y})^2)^{\frac{1}{2}}}(-\dot{y},\dot{x})$$

Answer 2

$$ N(t) = \frac{(\dot{\alpha}(t)\times \ddot{\alpha}(t))\times \dot{\alpha}(t)}{||(\dot{\alpha}(t)\times\ddot{\alpha}(t))\times\dot{\alpha}(t)||} $$ $$ \kappa(t) = \frac{||\dot{\alpha}(t) \times \ddot{\alpha}(t)||}{||\dot{\alpha}(t)||^3} $$ We have $$ \alpha(t) = (x(t), y(t),0) $$ $$ \dot{\alpha}(t)\times \ddot{\alpha}(t) = \begin{vmatrix} \dot{x} & \dot{y} \\ \ddot{x} & \ddot{y}\end{vmatrix}(0,0,1) $$ $$ ||\dot{\alpha}(t)\times \ddot{\alpha}(t) || = |\ddot{y}\dot{x}-\dot{y}\ddot{x}| $$ $$ (\dot{\alpha}(t)\times \ddot{\alpha}(t))\times \dot{\alpha}(t) = (\ddot{y}\dot{x}-\ddot{x}\dot{y})\begin{vmatrix} 0 & 0 & 1 \\ \dot{x} & \dot{y} & 0 \\ i & j & k \end{vmatrix} = (\ddot{y}\dot{x}-\ddot{x}\dot{y})(-\dot{y}, \dot{x},0) $$ $$ ||(\dot{\alpha}(t)\times \ddot{\alpha}(t))\times \dot{\alpha}(t)||= |\ddot{y}\dot{x}-\ddot{x}\dot{y}|\sqrt{(\dot{x})^2+(\dot{y})^2} $$ $$ \frac{1}{\kappa(t)}N(t) = \frac{\left(\sqrt{(\dot{x})^2+(\dot{y})^2}\right)^3}{|\ddot{y}\dot{x}-\dot{y}\ddot{x}|} \cdot \frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{|\ddot{y}\dot{x}-\dot{y}\ddot{x}|\sqrt{(\dot{x})^2+(\dot{y})^2}}(-\dot{y},\dot{x},0)=\frac{(\dot{x})^2+(\dot{y})^2}{\ddot{y}\dot{x}-\dot{y}\ddot{x}}(-\dot{y},\dot{x},0) $$ $$ \epsilon(t)=\alpha(t)+\frac{1}{\kappa(t)}N(t) = \frac{(\dot{x})^2+(\dot{y})^2}{\ddot{y}\dot{x}-\dot{y}\ddot{x}}(-\dot{y},\dot{x},0)+(x,y,0) $$ $$ \epsilon(t) = \left(x-\dot{y} \frac{(\dot{x})^2+(\dot{y})^2}{\dot{x}\ddot{y}-\dot{y}\ddot{x}},\quad y+\dot{x} \frac{(\dot{x})^2+(\dot{y})^2}{\dot{x}\ddot{y}-\dot{y}\ddot{x}}, \quad 0 \right) $$