Tangent planes passing through coordinates origin.

The problem from Differential Geometry:

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

Can anyone give me a hint?
 

AKG

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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 [itex]\sigma : V \to \mathbb{R}^3[/itex] such that [itex]\sigma (V) = f^{-1}(0)[/itex]. 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.
 
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?
 
Last edited:

AKG

Science Advisor
Homework Helper
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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?
 
Thanks for support. I think now I can catch what is going on. I was little bit confused :blushing:
 

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