Let $\alpha(t)$ be a unit-speed curve in $\mathbb R^{3}$ with principal normal $ N(t)$ and binormal $B(t).$ We define a surface by $$ x(t,\theta) = \alpha(t) + a( N(t) \cos({\theta}) + B(t) \sin(\theta)),$$ where $a >0$ is a constant. I have to show that for any fixed $t_0$ the curve $\gamma(\theta) = x(t_0,\theta)$ is geodesic on $x.$

I have no idea from where to begin. Any help. Thank you.


Instead of your notation, I'll use $T_\alpha, N_\alpha$ and $B_\alpha$ for the Frenet trihedron of $\alpha$. Geometrically, the image of $x$ is a tube of radius $a$ around the image of $\alpha$. With this, you can see that $$N(x(t,\theta)) = N_\alpha(t)\cos \theta + B_\alpha(t)\sin \theta$$is a normal field along the surface. Now, if $\gamma(\theta) = x(t_0,\theta)$, then $$\gamma''(\theta) =- N_\alpha(t_0)\cos\theta - B_\alpha(t_0)\sin \theta$$is parallel to $N(\gamma(\theta))$, so $\gamma$ is a geodesic.

  • $\begingroup$ That's a great explanation. Thanks for your time. I have one query, will you please tell me how the surface look like ? If possible a link containing the picture of the surface. Thank you. @IvoTerek $\endgroup$ – hiren_garai Nov 5 '18 at 1:56
  • $\begingroup$ From Google, something like this. (Link for the image magically turned into a youtube link from the video where the figure was taken... go figure) $\endgroup$ – Ivo Terek Nov 5 '18 at 1:59
  • $\begingroup$ One more question, I've read the definition of a geodesic as, a curve $\gamma(t)$ is a geodesic to a surface if the vector field $\dot\gamma(t)$ is parallel along $\gamma(t)$,( dot denotes the derivative) is this the same definition that you're using ? Actually I'm facing some doubts there. @IvoTerek $\endgroup$ – hiren_garai Nov 5 '18 at 2:13
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    $\begingroup$ These definitions are equivalent. You can write $$\alpha''(t) = \frac{D\alpha'}{{\rm d}t}(t) + \alpha''_{\rm nor}(t),$$where $D\alpha'/{\rm d}t$ is the covariant derivative and $\alpha''_{\rm nor}(t)$ is normal to the surface. Then $D\alpha'/{\rm d}t =0$ if and only if $\alpha''(t) = \alpha''_{\rm nor}(t)$ is normal to the surface. $\endgroup$ – Ivo Terek Nov 5 '18 at 2:15
  • $\begingroup$ Now it's fine, but I've one more doubt, pardon my ignorance, how have you found the Normal field along the surface ? according to which formula , I'm unable to understand. @IvoTerek $\endgroup$ – hiren_garai Nov 5 '18 at 2:25

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