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$\newcommand{\Reals}{\mathbf{R}}$So here is an exercise from classical differential geometry;

Prove that if $h_1:(a,b) \to \Reals$ and $h_2:(a,b) \to \Reals$ are $C^2$ functions in a non-zero interval $(a,b) \subset \Reals$, then the parametric equation $r:(a,b) \to \Reals^3$ defined by $r(x) =xi + h_1(x)j + h_2(x)k$ is regular. Explain why every plane $x = c$, with $c \in (a,b)$, intersects with this curve in a single point.

So this is the first part of the exercise. To prove that it's a regular curve, I check the vector of speed which is $$ \frac{dr}{dx} = [1, h_1'(x), h_2'(x)] \neq [0,0,0]. $$ So it's a regular curve.

For the other part is this right? To find their intersection I replace the parameters of the curve in the equation of the plane. So I have that their intersection is $x = c$ and that means that their intersection is the single point $(c,0,0)$ and thus every plane $x = c$ intersects with the curve in a single point. Is that false?

And now the part that I have absolutely no idea what I have to do;

Prove the opposite that if every plane $x = c$ stable intersects with a regular curve in maximum one point transversely, then we can have functions $h_1$, $h_2$ like before, so that the curve can be parameterized through the $x$-axis.

Any help would be appreciated.

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    For some reason the MathJax basic tutorial didn't help with the writting.. Don't know why.. I followed the steps, but the symbols didn't appear.. – 2017-01-15
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    You need dollar signs :) – 2017-01-15
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    @user339970: I took the liberty of adding the formatting and making a couple of changes of wording. If I changed your meaning, please edit to correct. – 2017-01-15

1 Answers 1

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(i) Write your curve $\gamma$ in the form $$\gamma:\quad t\mapsto{\bf h}(t):=\bigl(t,h_2(t),h_3(t)\bigr)\qquad(a

(ii) Intersecting $\gamma$ with the plane $x=c$ enforces $t=c$. Therefore we have exactly one point of intersection, namely the point ${\bf r}_c:=\bigl(c, h_2(c),h_3(c)\bigr)$. (You had this wrong.)

(iii) Now let $$\gamma:\quad t\mapsto{\bf h}(t):=\bigl(h_1(t),h_2(t),h_3(t)\bigr)\qquad(a