Let $f:S^1\rightarrow \mathbb{R}^3$ be a smooth map with $df_x \neq 0$, for all $x \in S^1$.

Question

Let $f:S^1\rightarrow \mathbb{R}^3$ be a smooth map with $df_x \neq 0$, for all $x \in S^1$. Show that there is a plane $C$ through the origin of $\mathbb{R}^3$ such that $p\circ f:S^1\rightarrow C$ has $d(p\circ f)_x \neq 0$, for all $x \in S^1$, where $p$ denotes orthogonal projection of $\mathbb{R}^3$ onto $C$.

Answer 1

Consider the function $g$ that sends any $x\in S^1$ to the unit tangent vector to the curve $f$ at $x$ (up to notational issues, $g(x)=df_x/\Vert df_x\Vert$). This is a smooth map from $S^1$ to $S^2$. So there is a vector $v\in S^2$ such that neither $v$ nor $-v$ is in the image of $g$. Choose the plane $C$ to be perpendicular to such a $v$, so that your projection $p$ is in the direction of $v$.