Showing a curve to be a geodesic on a surface

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.
• 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 – hiren_garai Nov 5 '18 at 2:13
• 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. – Ivo Terek Nov 5 '18 at 2:15