Find the equation of the tangent to a system using implicit function theorem

Question

Here is my system of equations:

$C: \begin{cases}x^2 + y^2 +z^2 = 14\\ x^3+y^3+z^3=36 \end{cases}$

Firstly I managed to show that for all $a \in C$, the implicit function theorem applies to express $x$ and $y$ as a function of $z$. Which implies the existence of $\phi$ such that, for all $(x,y,z)$ close enough to $a$, we get:

$f(x,y,z)=0 \Leftrightarrow \phi(z) = (x,y)$

(with f being the function: $f(x,y,z) = (x^2+y^2+z^2-14, x^3+y^3+z^3-36)$

What I do not undesrstand is how to find the equation of the tangent at the point $(1,2,3) \in C$.

What I had done was to start from the realtion given by the theorem: $f(\phi(z),z)=0$

By deriving both sides, I get: $\frac{\partial f}{\partial x \partial y}(\frac{\partial\phi(z)}{\partial z},z) \circ \frac{\partial\phi(z)}{\partial z} + \frac{\partial f(\phi(z),z)}{\partial z} = 0$

Which I rearrange:

$\frac{\partial\phi(z)}{\partial z} = \frac{\partial f^{-1}}{\partial x \partial y}(\frac{\partial\phi(z)}{\partial z},z) \circ \frac{\partial f(\phi(z),z)}{\partial z}$

If this is anywhere near to the anwser, where do I plug the values $(1,2,3)$ in?

Answer 1

Your system in a neighbourhood of the point $P=(1,2,3)$ defines a differentiable curve $\gamma$. By using the Implicit Function Theorem, we may find the tangent line to $\gamma$ at $P$: there exist two functions $x=\phi(z)$ and $y=\psi(z)$ such that $\phi(3)=1$ and $\psi(3)=2$ and the desired tangent line has the following parametric equation: $$\begin{cases} x=1+\phi'(3)t\\ y=2+\psi'(3)t\\ z=3+t\end{cases}$$ You may find $\phi'(3)$ and $\psi'(3)$, by deriving the equations in the system with respect to $z$ and solving it: $$\begin{cases}\phi(z)^2 + \psi(z)^2 +z^2 = 14\\ \phi(z)^3+\psi(z)^3+z^3=36 \end{cases} \implies \begin{cases} 2\cdot 1\cdot \phi'(3) + 2\cdot 2\cdot \psi'(3) +2\cdot 3 = 0\\ 3\cdot 1^2\cdot \phi'(3) + 3\cdot 2^2\cdot \psi'(3) +3\cdot 3^2 = 0 \end{cases}$$ Can you take it from here?