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Given the equation $x^2+y^2+z^2=\psi(ax+by+cz)$, with $a,b,c\in\mathbb{R},\ c\neq 0$, and $\psi:\mathbb{R}\to\mathbb{R}$ that satisfies $\psi\in C^2,\ \psi(0)=0,\ \psi'(0)\neq0$, prove that in a neighborhood of $(0,0,0)$ the solutions of the equation can be expressed as $(x,y,f(x,y))$, with $f$ differentiable in that neighborhood, and find $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$.

My idea is to define a function $F:\mathbb{R^3}\to\mathbb{R},\ F(x,y,z)=(x^2+y^2+z^2-\psi(ax+by+cz))$.

In the first part we have to prove that $F$ verifies the Implicit Function Theorem (IFT) in $p$=$(0,0,0)$: $F(0,0,0)=\psi(0)=0,\ F\in C^2(\mathbb R)$ (so it's $C^2$ 'near' $p$), and the third requirement is that $\frac{\partial F}{\partial z}=0$, $\frac{\partial F}{\partial z}=2z-c\psi'\neq 0$ when evaluated in $p$, because $c\neq0$ and $\psi'(0)\neq0$, so applying the IFT we are done with the first part of the exercice.

We can express $F$ as $F(x,y,f(x,y))=(x^2+y^2+(f(x,y))^2-\psi(ax+by+cf(x,y)))$.

Now we want to compute $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$:

$\frac{\partial F}{\partial x}(x,y)=2x+2f(x,y)\frac{\partial f}{\partial x}(x,y)-(a+c\frac{\partial f}{\partial x}(x,y))\frac{\partial\psi}{\partial x}\to\frac{\partial F}{\partial x}(0,0)=-(a+c\frac{\partial f}{\partial x}(0,0))\frac{\partial\psi}{\partial x}$

But I don't know how to continue. Could you help me? Is the first part correctly done? Thanks!

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  • $\begingroup$ I think you basically were almost done except for two things. One is the abuses of notation (like $\dfrac{\partial\psi}{\partial x}$ and treating $F$ as a function defined in both $\mathbb R^2$ and $\mathbb R^3$), the other was what I explained in my answer which says that $\dfrac{\partial F}{\partial x}$ is the null function in a certain neighborhood. $\endgroup$ – Git Gud May 9 '19 at 23:10
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We can express $F$ as $F(x,y,f(x,y))=(x^2+y^2+(f(x,y))^2-\psi(ax+by+cf(x,y)))$.

It's not immediately obvious to me what this means, but even if it meaningful and correct, I don't think it is going in the right direction.

The implicit function theorem tells you that there exist neighborhoods $V$ of $(0,0)$, $W$ of $0$ and a class $C^1$ function $f\colon V\to W$ such that $f(0,0)=(0,0)$ and $$\forall (x,y) \in V\left(F(x,y, f(x,y))=0\right) \tag 1$$

(I'm using the formulation of the IFT used on Wikipedia, which is very similar to the one I wrote here).

Now take a new function $H\colon V\to W$, $(x,y)\to F(x,y,f(x,y))$ and differentiate with respect to the first and second coordinates. Coupling this with $(1)$ should give you the answer.

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    $\begingroup$ The technique suggested by the last paragraph is the same as the one used here, here and here. $\endgroup$ – Git Gud May 9 '19 at 21:31
  • $\begingroup$ If the formulation of the implicit function theorem I'm using here isn't helpful to you, I suggest you provide the one you're using. $\endgroup$ – Git Gud May 9 '19 at 21:42
  • $\begingroup$ So $0=H_x(x,y)=F_x\left(x,y, f(x,y)\right)+F_z\left(x,y,f(x,y)\right)f_x(x,y)=2x-\psi'+(2z-c\psi')f_x(x,y)\to f_x(x,y)=\frac{a\psi'-2x}{2z-c\psi'}$, and if we evaluate it in $(0,0)$, we have that $f_x(0,0)=\frac{a\psi'}{-c\psi'}=\frac{-a}{c}$, and we are done. Now I can calculate $f_y$ in a similar way. Understood. Thanks for your answer! $\endgroup$ – Gibbs May 9 '19 at 22:38
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    $\begingroup$ That's (almost) right (you have a typo along the way where you forget to multiply by $a$, but you included it in the end)! I would just be more precise and write $\psi'(ax+by+cf(x,y))$ instead of $\psi'$. $\endgroup$ – Git Gud May 9 '19 at 23:06

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