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

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


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?

  • $\begingroup$ I solved this one. Thanks. $\endgroup$ – MathWizard Dec 5 '20 at 18:31

$$\kappa =\frac{\sqrt{\left(x' y''-x'' y'\right)^2+\left(x'' z'-x' z''\right)^2+\left(y' z''-y'' z'\right)^2}}{\left(\left(x'\right)^2+\left(y'\right)^2+\left(z'\right)^2\right)^{3/2}}$$ 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

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

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

  • 1
    $\begingroup$ @MathWizard I would help you more happily if you upvoted my answer :) $\endgroup$ – Raffaele Dec 5 '20 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.