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$r(t)=t \cos(t) \vec e_1 + t \sin(t) \vec e_2 + (c-dt) \vec e_3$ is a space curve

Fint the torsion $\tau$ and the curvature $\kappa$

Attempt:

I know the formulas but it gets too complicated with standard formulas, can I directly use $\kappa = |r''(t)|$ or should I make it arc-length parameter first?

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  • $\begingroup$ I solved this one. Thanks. $\endgroup$ – MathWizard Dec 5 '20 at 18:31
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$$\kappa =\frac{\sqrt{\left(x' y''-x'' y'\right)^2+\left(x'' z'-x' z''\right)^2+\left(y' z''-y'' z'\right)^2}}{\left(\left(x'\right)^2+\left(y'\right)^2+\left(z'\right)^2\right)^{3/2}}$$ If $\mathbf{r}(t)=(t\cos t;t\sin t;c-dt)$ then the curvature is

$$\kappa=\frac{\sqrt{d^2 \left(t^2+4\right)+\left(t^2+2\right)^2}}{\left(d^2+t^2+1\right)^{3/2}}$$

$\mathbf{r}'(t)=(\cos (t)-t \sin (t),\sin (t)+t \cos (t),-d)$

$\mathbf{r}''(t)=(-2 \sin (t)-t \cos (t),2 \cos (t)-t \sin (t),0)$

$\mathbf{r}'''(t)=(t \sin (t)-3 \cos (t),-3 \sin (t)-t \cos (t))$

The torsion is

$$\tau ={\frac {\det \left({\mathbf {r} ',\mathbf {r} '',\mathbf {r} '''}\right)}{\left\|{\mathbf {r} '\times \mathbf {r} ''}\right\|^{2}}}={\frac {\left({\mathbf {r} '\times \mathbf {r} ''}\right)\cdot \mathbf {r} '''}{\left\|{\mathbf {r} '\times \mathbf {r} ''}\right\|^{2}}}$$

$$\tau=-\frac{d \left(t^2+6\right)}{d^2 \left(t^2+4\right)+\left(t^2+2\right)^2}$$

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    $\begingroup$ @MathWizard I would help you more happily if you upvoted my answer :) $\endgroup$ – Raffaele Dec 5 '20 at 20:00

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