Consider
$$\begin{cases}f(x,y,z) = x+y^2 + z^3 \\ g(x,y,z) = ay + z^2 + x^3\end{cases}.$$
- Show that if $a\neq 0$, then there is a unique (up to reparameterization) differentiable curve $\{(x(s), y(s), z(s)): -\epsilon \leq s \leq \epsilon\}$, for some $\epsilon >0$ such that $f(x(s),y(s), z(s)) = 0$, $g(x(s),y(s),z(s))=0$, and $(x(0),y(0),z(0))=0$, and $\frac{dz}{ds}=1$.
- Compute the unit tangent vector to the curve at $(0,0,0)$.
- Show that even if $a=0$ there is still a nontrivial continuous solution to (1) with $(x(0), y(0), z(0)) = 0$, but it is not unique and its image in $\mathbb{R}^3$ does not have a tangent vector at $(0,0,0)$.
My answer: The first two are just simple implicit function theorem, parameterizing the level set $(f,g)=0$ by $z$. For the third one, I am confused which tool to use to show this. Should I be appealing to notions of functional dependence?
I don't need someone to feed me the answer, but where should I be looking?
With $a=0$, the second equation becomes $x = z^{-2/3} $, which is the semicubical parabola with a cusp at the origin. The $y$ coordinate can be written as $y=\pm \sqrt{z^{2/3}-z^3}$ using the first equation. Hence, you have a continuous solution parametrized by $z=s$, and it's not unique due to the choices of sign of $y$. (There are four choices: two for $z<0$ and two for $z>0$.)