0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.