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here is a problem im trying to solve for a few days, and Im not getting sucess.

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $f\in C^1$ and $F(x,y,z)=f(y/x,z/x)$. Consider a level surface $S$ defined by $F(x,y,z)=0$ and $(x_0,y_0,z_0)\in S$. What are the conditions, in a neighborhood of $(x_0,y_0,z_0)$, that we can write $S$ in the form $z=g(x,y)$ ? Also, show that at this conditions, its true that $$x\frac{\partial g}{\partial x}(x,y)+y\frac{\partial g}{\partial y}(x,y)=g(x,y)$$

Well, here is what I've done: To use the Implicit Function Theorem I choose $F(x_0,y_0,z_0)=0$ and I have to show that $\frac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$. $$\frac{\partial F}{\partial z}(x_0,y_0,z_0)=\textrm{lim}_{t\rightarrow 0}\frac{F(x_0,y_0,z_0+t)-F(x_0,y_0,z_0)}{t}=\textrm{lim}_{t\rightarrow 0}\frac{F(x_0,y_0,z_0+t)}{t}=\textrm{lim}_{t\rightarrow 0}\frac{f(y_0/x_0,(z_0+t)/x_0)}{t}\neq 0$$

Here I tried something that I dont know if is right, and even if is right, i couldnt conclude the right answer from this. $$\textrm{lim}_{t\rightarrow 0}\frac{f(y_0/x_0,(z_0+t)/x_0)}{t}=\textrm{lim}_{t\rightarrow 0}\frac{f(y_0/x_0,(z_0/x_0)+(t/x_0))}{t}=\frac{1}{x_0}\frac{\partial f}{\partial y}(y_0/x_0,z_0/x_0)$$ I did this variable change $u=t/x_0$ with $u\rightarrow 0$ and calculate the limit. But Im really not sure about it and I didnt found the last equality they are asking.

Thank you everyone one more time.

share|cite|improve this question
You know $f$ is $C^1$ so there's no reason to use the definition of derivative. Just use the chain rule. For the second part I'd use the extra bit of the Implicit Function theorem about a formula for the derivatives of $g$ and I'm sure the answer will just fall out. – toypajme Sep 16 '12 at 15:32
thanks. I ll try this way. Later I ll write here my results. – Integral Sep 16 '12 at 15:40
How am i suposed to use the chain rule? I mean, i can only think about $Fof$ and $foF$ and both dont make sense. – Integral Sep 16 '12 at 18:09
Just came out other idea. I can define $F$ doing the following map: $(x,y,z)\mapsto (y/x,z/x)\mapsto f(y/x,z/x)$ Is that the right way? – Integral Sep 16 '12 at 18:19
up vote 5 down vote accepted

Its possible to define $F:\mathbb{R}^3\rightarrow\mathbb{R}$ through the map $(x,y,z)\mapsto(y/x, z/x)\mapsto f(y/x, z/x)$. We can define $\alpha:\mathbb{R}^3\rightarrow\mathbb{R}^2$ such that $\alpha(x,y,z)=(y/x, z/x)$, this way we have $F=f\circ\alpha:\mathbb{R}^3\rightarrow\mathbb{R}$.

The derivate $(f\circ\alpha)'$ at the point $p=(x_0,y_0,z_0)$ is given by $(f\circ\alpha)'(p)=f'(\alpha(p))\cdot \alpha '(p)$, by the chain rule. We compute this: $$\left( \begin{array}{cc} \frac{\partial f}{\partial x}(\alpha(p)), & \frac{\partial f}{\partial y}(\alpha(p)) \end{array} \right)\cdot \left( \begin{array}{ccc} -\frac{y_0}{x_0^2}, & \frac{1}{x_0}, & 0\\ -\frac{z_0}{x_0^2}, & 0, & \frac{1}{x_0} \end{array} \right)=\\ = \left( \begin{array}{ccc} -\frac{y_0}{x_0^2}\cdot\frac{\partial f}{\partial x}(\alpha(p)) -\frac{z_0}{x_0^2}\cdot\frac{\partial f}{\partial y}(\alpha(p)), \frac{1}{x_0}\cdot\frac{\partial f}{\partial x}(\alpha(p)), \frac{1}{x_0}\cdot\frac{\partial f}{\partial y}(\alpha(p)) \end{array} \right) $$

To apply the Implicit Function Theorem is necessary that $\frac{\partial F}{\partial z}(p)=\frac{1}{x_0}\cdot\frac{\partial f}{\partial y}(\alpha(p))$ be $\neq 0$, this way we need that $x_0\neq 0$ and $\frac{\partial f}{\partial y}(\alpha(p))\neq 0$.

And to finalize...

By the Implicit Function Theorem, we can write $z=g(x,y)$ in a neighborhood of $p=(x_0,y_0,z_0)$ such that $F(x,y,g(x,y))=0$ for all points in this neighborhood. Also, we have that $$\frac{\partial g}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}=\frac{y}{x}\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}+\frac{z}{x}$$ $$\frac{\partial g}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$

Therefore $$x\cdot\frac{\partial g}{\partial x}+y\cdot\frac{\partial g}{\partial y}=x\cdot\Bigg( \frac{y}{x}\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}+\frac{z}{x}\Bigg)-y\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}=z=g(x,y)$$

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Thank you for your hints, toypajme. – Integral Sep 16 '12 at 20:20
no problem, solid work – toypajme Sep 17 '12 at 16:13

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