continuously differential parametrization of a tricky surface 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?
 A: 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 $S$:

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:
$$(x,t)\mapsto\bigl(x,-\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty<x<1, \ -\infty<t<\infty)\ ,$$
$$(x,t)\mapsto\bigl(x,\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty<x<1, \ -\infty<t<\infty)\ ,$$
$$(y,z)\mapsto\bigl(\root 3\of{1-y^2+z^2}, y, z\bigr)\qquad\bigl(-\infty<z<\infty,\ |y|<\sqrt{1+z^2}\bigr)\ .$$
A: 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 $z$.
