Given a curve $\alpha:\mathbb{R}\to\mathbb{R}^{3}$ defined by $\alpha=\left(\frac{1}{\sqrt{2}}\cos \sqrt{2}s,\frac{1}{2}\sin \sqrt{2}s,\frac{1}{2}\sin \sqrt{2}s\right)$, show that its osculating plane is the same at any two points $s_{1},s_{2}$.
I was only given the definition of the osculating plane at $s$ to be
$P_{s}=\left\{\lambda_{1}n(s)+\lambda_{2}\alpha'(s)+\alpha(s):\lambda_{1},\lambda_{2}\in\mathbb{R}\right\}$,
where $n(s)$ is the normal unit vector. Is this enough to answer the above question though? Any hint on how to start?
Best Answer
For all $t \in \mathbb R$, you have $\alpha_y(t) = \alpha_z(t)$. Therefore $\alpha$ lies in the plane having for equation $y=z$. Moreover, $\alpha^\prime(t), \alpha^{\prime \prime}(t)$ are linearly independent, meaning that the osculating plane is defined for all points of the curve.
We can conclude that the osculating plane of the curve is the plane $y=z$ at any point of the curve.