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I'm looking for a continuously differentiable parametrization of $x^3+y^2-z^2=1$ but I'm actually totally stuck. If the $x$ term were quadratic instead of cubic, it would be simple: $(x,y,z)=(\sqrt{t^2+1}\cos\theta, \sqrt{t^2+1}\sin\theta, t)$ But with the cubic term there, I'm stuck. I naturally thought about $(x,y,z)=(\sqrt[3]{t^2+1}\cos^{\frac{2}{3}}\theta, \sqrt{t^2+1}\sin\theta, t)$ but this isn't continuously differentiable in $\theta$.

Hints or suggestions?

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    sorry for the delay in replying -- no, I'm not looking for "$c^1$ homeomorphisms" (i.e. I don't care about the inverses). this problem was listed among a set of routine homework problems that asked for parametrizations of very standard surfaces, all of which could be covered by the image of a *single* $C^1$ map defined on an open, connected subset of $\mathbb{R}^2$. I wanted to make sure I wasn't crazy to think this can't be done for this particular surface.2012-10-17

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We are given the function $f(x,y,z):=x^3+y^2-z^2-1$ and have to consider the solution set (a "surface") $S:=\{(x,y,z)\in{\mathbb R}^3\ |\ f(x,y,z)=0\}\ .$ As $\nabla f(x,y,z)=(3x^2,2y,-2z)$ is $\ =(0,0,0)$ only at the origin $O\notin S$, by the implicit function theorem the set $S$ is a smooth surface in the neighborhood of all of its points. Here is a picture of $S$:

enter image description here

For given $y$ and $z$ the equation $f(x,y,z)=0$ has exactly one solution $x=\phi(y,z)\in{\mathbb R}$ which is commonly written as $\phi(y,z)=\root 3\of {1-y^2+z^2}$. Unfortunately along the hyperbola $y^2-z^2=1$ the function $\phi$ is not differentiable as a function of $y$ and $z$.

If we are allowed to use more than one patch to cover all of $S$ we could use three patches as follows: $(x,t)\mapsto\bigl(x,-\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty $(x,t)\mapsto\bigl(x,\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty $(y,z)\mapsto\bigl(\root 3\of{1-y^2+z^2}, y, z\bigr)\qquad\bigl(-\infty

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    @symplectomorphic: As $S$ is diffeomorphic to ${\mathbb R}^2$, in principle you can do it with one patch; but it will be difficult to do it in terms of elementary functions.2012-10-18
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Have you looked for a polynomial parametrization, with x a quadratic in t and y, z being cubics?

Or even simpler,

$x^3 + y^2 - z^2 = 1$

<=> $1 - x^3 = y^2 - z^2$

<=> $(1 - x)(1 + x + x^2) = (y - z)(y + z)$.

If you assume $1 - x = y - z$ and $1 + x + x^2 = y + z$ you can get a simple parametrization by solving the simultaneous equations for $y$ and $z$.

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    wait, wait, that was too hasty on my part -- solving the last two equations for y and z gives y and z as functions of x. the image of that parametrization will only be a curve. but I want a single $C^1$ parametrization that covers the whole surface, if possible.2012-10-17