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.