Tangent planes passing through coordinates origin.

[Alteran]

The problem from Differential Geometry:

Let $$\gamma : R -> R$$ is smooth function and $$U = {(x,y,z) \in R^3 : x \ne 0}$$ - open subset.
Function $$f : U -> R$$ is defined as $$f(x,y,z) = z - x\gamma(y/x)$$ and this is smooth function.
Proof that for surface$$S = f^{-1}(0)$$ all tangent planes passing through coordinates origin.

Can anyone give me a hint?

 

[AKG]

You mean to say:

"Prove that for the surface f^-1(0), all tangent planes pass through the origin"

What have you tried so far? What definitions are you working with? I defined a set V = {(x,y) in R² : x not equal to 0}. Then I found a smooth regular surface patch $\sigma : V \to \mathbb{R}^3$ such that $\sigma (V) = f^{-1}(0)$. Writing it this way, I found it easy to find the tangent planes to the surface, and then easily showed that each of those tangent planes pass through the origin.

 

[Alteran]

Yes, exactly - "Prove that for the surface f-1(0), all tangent planes pass through the origin".

I have found tangent planes for this surface, but how to show that they are passing through the origin? I need to show, that these planes are defined in (0,0,0)? or something else?

 

[AKG]

A plane is just a set of points. Show that for each tangent plane, (0,0,0) is in that plane, i.e. (0,0,0) is in that set of points. For example, the point (1,0,y(0)) [I'm writing y instead of gamma] is in the surface. The tangent plane at this point is the plane:

{a(1,0,y(0)) + b(0,1,y'(0)) + (1,0,y(0)) : a, b in R}

I need to show that the origin (0,0,0) is in this plane, i.e. that there exist real a and b such that:

a(1,0,y(0)) + b(0,1,y'(0)) + (1,0,y(0)) = (0,0,0)

But that's easy:

a=-1, b=0.

Is there any part of this you didn't understand?

 

[Alteran]

Thanks for support. I think now I can catch what is going on. I was little bit confused