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

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?

• Being polynomial, we can write the equations about the point by translating to the origin and back \begin{align}(x-1)^2+(y-2)^2+(z-3)^2+2(x-1)+4(y-2)+6(z-3)&=0\\ (x-1)^3+(y-2)^3+(z-3)^3+3(x-1)^2+6(y-2)^2+9(z-3)^2+3(x-1)+12(y-2)+27(z-3)&=0\end{align} cut the higher order terms, and intersect the tangent cones. \begin{align}&2(x-1)+4(y-2)+6(z-3)&=0\\ \quad\quad\quad\quad&3(x-1)+12(y-2)+27(z-3)&=0\end{align} Now it's linear algebra: \begin{align}(x-1)-3(z-3)&=0\\(y-2)+3(z-3)&=0\\ (z-3)-(z-3)&=0\end{align} And letting $(z-3)=t$ \begin{align}x&=1+3t\\ y&=2-3t\\z&=3+t\end{align} Commented Mar 3 at 7:41

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?

• This sounds better: I get $\phi'(3)$ = 3 and $\psi'(3)$ = -3
– Alex
Commented Mar 2 at 17:48
• Yes, now you are correct. Well done! Commented Mar 2 at 17:49
• thank you so much for your help!
– Alex
Commented Mar 2 at 17:53
• The graph you sent looks awesome! What tool do you use ? :)
– Alex
Commented Mar 2 at 18:04
• Desmos 3D: see desmos.com/3d/993989ce2a Commented Mar 2 at 18:48