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$\alpha(s)$ natural parameterized regular curve such that $k(s)\neq 0$ (curvature) $\forall s.$ Let curve $\beta(s)=\alpha'(s)$. Prove that $\beta(s)$ is a regular curve. Prove that the curvature of $\beta(s)$ is :

$$(1+\frac{\tau^2}{k^2})^{\frac{1}{2}}$$ where $\tau$ and $k$ are the torsion and curvature of $\alpha(s)$. Find the formula for the torsion of $\beta(s)$ in relation to $k$ and $\tau.$

Well for $\beta(s)$ to be regular then $\beta'(s)\neq 0$ and since $k(s)=\|T'(s)\|=\|\alpha''(s)\|\neq0 \implies \alpha''(s)=\beta'(s) \neq 0$ So there is it's regularity.

I tried figuring out the curvature but didn't come to anything that looked like this expression. I have these to formula's for finding the curvature and torsion of a arbitrary regular curve (that is not necessarily naturally parameterized )

$$k_{\beta}(s)=\frac{\|\beta'(s)\times\beta''(s)\|}{\|\beta'(s)\|^3} \\ t_{\beta}(s)=\frac{[\beta'(s),\beta''(s),\beta'''(s)]}{\|\beta'(s)\times\beta''(s)\|^2}$$

$[\beta'(s),\beta''(s),\beta'''(s)]$- triple product.

I also have the Frenet–Serret formulas:

$$T'=kN \\ N'= -kT+\tau B \\ B'=-\tau N$$

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    $\begingroup$ See another answer here $\endgroup$ – Ng Chung Tak Mar 18 '17 at 0:32
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    $\begingroup$ "I need a complete answer for this." And the answer Ng Chung Tak indicated does not give you that? $\endgroup$ – Did Mar 20 '17 at 13:16
  • $\begingroup$ That completely flew over my head. Apologise. There goes 200 rep for that mistake. Thank you for reminding me. $\endgroup$ – Bozo Vulicevic Mar 20 '17 at 18:20
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Let $\overline{x}(s)$ be a curve in $\textbf{R}^3\equiv E_3$, with $s$ being its canonical parameter and $\overline{x}''(s)\neq 0$. The Frenet vector system of the curve $\overline{x}=\overline{x}(s)$ is given by $\{\overline{t},\overline{h},\overline{b}\}$, where $\overline{t}(s):=\overline{x}'(s)$, $\overline{h}(s):=\frac{\overline{x}''(s)}{|\overline{x}''(s)|}$, $\overline{b}(s):=\overline{t}(s)\times \overline{h}(s)$.

The Frenet vector system of the curve satisfy the equations: $$ \overline{t}'=\frac{\overline{h}}{\rho}\textrm{, }(0<\rho<\infty), $$ $$ \overline{h}'=-\frac{\overline{t}}{\rho}+\frac{\overline{b}}{\tau}\textrm{, }(-\infty\leq\tau\leq+\infty\textrm{, }\tau\neq0), $$ $$ \overline{b}'=-\frac{\overline{h}}{\tau}, $$ where $$ k=\frac{1}{\rho}=|\overline{x}''(s)|\textrm{, }0<\rho\leq+\infty $$ is the curvature$-k$ of the curve and $\rho$ called radious of curvature.

Also $\sigma$ is the torsion of the curve and $\tau$ is the radious of torsion. $$ \sigma=\frac{1}{\tau}=\left\langle \overline{h}'(s),\overline{b}(s)\right\rangle=\frac{(\overline{x}',\overline{x}'',\overline{x}''')}{|\overline{x}''|^2} $$

If $\overline{x}(s)\in\textbf{C}^4$, then $$ \overline{x}'''=-\frac{1}{\rho^2}\overline{t}-\frac{\rho'}{\rho^2}\overline{h}+\frac{1}{\rho\tau}\overline{b}, $$ $$ (\overline{x}',\overline{x}'',\overline{x}''')=\frac{1}{\rho^2\tau}, $$ and $$ (\overline{x}'',\overline{x}''',\overline{x}'''')=\frac{1}{\rho^5}\left(\frac{\rho}{\tau}\right)'. $$ If $\overline{y}=\overline{y}(u)$ with $u$ any parameter and $\dot{\overline{y}}=\frac{d\overline{y}}{du}$, $\ddot{\overline{y}}=\frac{d^2\overline{y}}{du^2}$, $\overline{y}^{(3)}=\frac{d^3\overline{y}}{du^3}$,..., etc, then $$ k=\frac{1}{\rho}=\frac{\left|\dot{\overline{y}}\times \ddot{\overline{y}}\right|}{|\dot{\overline{y}}|^3} $$ and $$ \sigma=\frac{1}{\tau}=\frac{\left(\dot{\overline{y}},\ddot{\overline{y}},\overline{y}^{(3)}\right)}{(\dot{\overline{y}}\times \ddot{\overline{y}})^2} $$ Hence for $\overline{y}=\overline{y}(s)=\overline{x}'(s)$, then $\overline{x}''(s)=\overline{t}'=\frac{\overline{h}}{\rho}$. $$ k_1=\frac{|\overline{x}''(s)\times \overline{x}'''(s)|}{|\overline{x}''(s)|^3}=\rho^2\left|\overline{h}\times\left(-\frac{1}{\rho^2}\overline{t}-\frac{\rho'}{\rho^2}\overline{h}+\frac{1}{\rho\tau}\overline{b}\right)\right|= $$ $$ =\rho^2\left|\frac{1}{\rho^2}\overline{b}+\rho^2\frac{\overline{t}}{\rho\tau}\right|=\sqrt{1+\left(\frac{\rho}{\tau}\right)^2}=\sqrt{1+\frac{\sigma^2}{k^2}} $$ and $$ \sigma_1=\frac{\frac{1}{\rho^5}\left(\frac{\rho}{\tau}\right)'}{\frac{1}{\rho^2}\left(\frac{1}{\rho^2}\overline{b}+\frac{1}{\rho\tau}\overline{t}\right)^2}=\frac{\frac{1}{\rho}\left(\frac{\rho}{\tau}\right)'}{\frac{1}{\rho^2}+\frac{1}{\tau^2}}=\frac{k}{k^2+\sigma^2}\left(\frac{\sigma}{k}\right)' $$

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