# Studying regular space curves when restricted to two differentiable functions

a) Let $\alpha:I \mapsto \mathbb{R^3}$ be a regular curve. Prove that $\forall t_0 \in I$ there is an open interval $J$ that contains $t_0$ on which $\alpha$ is injective.

b) Prove that if $\alpha: I \mapsto \mathbb{R^3}$ is a regular curve, then $\forall t_0 \in I$, there is an open interval $J$ that contains $t_0$ and there exist differentiable functions $F$ and $G$ such that the trace of $\alpha$ when restricted to $J$ is contained in the set $\{(x, y, z) \in \mathbb{R^3} ; F(x, y, z) = G(x, y, z) = 0 \}$.

I can't see any starting points. I'd appreciate any help.

EDIT: Now that I think about it, isn't a) already proven by the inverse function theorem? Also, my intuition tells me that, for b), somehow applying the implicit function theorem will solve it, but I currently can't see how.

Second update: Now that I've had some help on another website, I think I've made some progress: if we let $\alpha(t) = (x(t), y(t),z(t))$, then by the inverse function theorem, there are open intervals containing $t_0,$ $A$, $B$ and $C$ such that $x(t), y(t), z(t)$ are all invertible on $A$, $B$ and $C$, respectively. If we take $J = A \cap B \cap C$, the exercise is done. I'm not sure if this is alright (and also, I can't quite see how $J$ is necessarily non empty).

For $b)$, the proper manner to apply the implict function theorem still eludes me.

## 1 Answer

I've got it (thanks a lot, /u/fattymattk):

a) Let $\alpha(t) = (x(t), y(t), z(t))$. Since the curve is regular, $||\alpha'(t_0)|| \neq 0$, so that at least one of the components has non vanishing derivative at $t = t_0$, so we can apply the inverse function theorem to say that there exists $J$, which contains $t_0$, where, WLOG, $x(t)$ is invertible and therefore injective. Further, on $J$, $\alpha(t) = \alpha(s)$ implies $x(t) = x(s)$, which implies $t = s$. Therefore $\alpha(t)$ is injective on $J$.

b) Let $\alpha(t) = (w(t), g(t), h(t))$. By the inverse function theorem, one of the components is invertible on some $J$ containing $t_0$. WLOG, let that component be $w(t)$. By the implicit function theorem, there is $J$ containing $t_0$ such that $w(t) = f$ on J. Then $t = w^{-1}(f)$. The curve is then reparameterized as $\alpha(f) = (f, g(w^{-1}(f)), h(w^{-1}(f)))$. Now consider the functions $F(x, y, z) = y -g(w^{-1}(f))$ and $G(x, y, z) = z -h(w^{-1}(f))$. Then $F(x, y, z) = G(x, y, z) = 0$ implies $y =g(w^{-1}(f))$ and $z = h(w^{-1}(f))$, so that the point $(x,y,z)$ is on $\alpha$, as we wanted to prove.