Let $\mathcal C$ be the locus of all point $(x,y,z) \in \mathbb R^3$ with $$x^2+y^2+z^2=1 \text{ and } x^2-y^2-z=0.$$ Show that $\mathcal C$ is a smooth curve in the following sense: For each point $p \in \mathcal C$, there exist open sets $U \subset \mathbb R$, $W \subset \mathbb R^3$ and a $\mathcal C^1$ map $f:U \to W$ such that $p \in W$ and $f(U)= \mathcal C \cap W$.
My idea was to define a function $\Phi:\mathbb R^3 \to \mathbb R^2$ by $\Phi(x,y,z)=(x^2+y^2+z^2-1, x^2-y^2-z)$. Then $\Phi^{-1}(0)=\mathcal C$. Now I want to apply the Implicit Function Theorem but I run into a problem:
Let $\phi_1$ and $\phi_2$ denote the coordinate functions of $\Phi$. Then
$$\det \begin{bmatrix} \frac{\partial \phi_1}{\partial y} & \frac{\partial \phi_1}{\partial z} \\ \frac{\partial \phi_2}{\partial y} & \frac{\partial \phi_2}{\partial z} \end{bmatrix} = \det \begin{bmatrix} 2y & 2z \\ -2y & -1 \end{bmatrix} = 4yz-2y = 2y(2z-1). $$
So my determinant is zero when $y=0$ and when $z=\frac{1}{2}$ and I can't apply the Implicit Function Theorem at those points.
I think there might be a way to use the Implicit Function Theorem twice to get two functions $g,h: \mathbb R \to \mathbb R$ such that $\Phi(x,g(x),h(x)) = 0$. Then I would be able to define $f:\mathbb R \to \mathbb R^3$ by $f(x) = (x, g(x), h(x))$. Is this possible? Or do I need to change my function $\Phi$?
The two surfaces are a sphere and a hyperboloid. Their intersection is a perfectly smooth curve, by any reasonable definition of "smooth".
To see what I mean, go to Wolfram Alpha and type in
It seems clear that the curve can be parameterized by polar angle $\theta$ in the $xy$ plane, though finding the parametric equations involves a bit of brute-force algebra. In fact, the parametric equations are:
\begin{align} x &= \sqrt2 w \cos\theta \\ y &= \sqrt2 w \sin\theta \\ z &= 2w^2 \cos 2\theta \end{align}
where $$ w = \frac{1}{\sqrt{1+\sqrt{3+2\cos{4\theta}}}} $$
These functions are clearly $C_\infty$. It is easy to check that $x^2+y^2+z^2=1$ and that $z=x^2-y^2$.