Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
1  
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
1  
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
add comment

1 Answer 1

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 $F=f\textrm{ o }\alpha:\mathbb{R}^3\rightarrow\mathbb{R}$.

The derivate $f(\textrm{ o }\alpha)'$ at the point $p=(x_0,y_0,z_0)$ is given by $f(\textrm{ o }\alpha)'(p)=f'(\alpha(p))\cdot \alpha '(p)$, 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 $$x\cdot\frac{\partial g}{\partial x}+y\cdot\frac{\partial g}{\partial y}=x\cdot\Bigg( \frac{y}{x}\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}+\frac{z}{x}\Bigg)-y\cdot\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}=z=g(x,y)$$

share|improve this answer
    
Thank you for your hints, toypajme. –  Integral Sep 16 '12 at 20:20
1  
no problem, solid work –  toypajme Sep 17 '12 at 16:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.