prove S is a surface

[logistic_guy]

here is the question

Prove that \(\displaystyle S: (x^2 + y^2)^2 + 3z^2 = 1\) is a surface.

Definition

definition of differentiability on a regular surface

all. I am studying the book "Differential Geometry of Curves and Surfaces" written by do Carmo, and there is one thing that confuses me so much: In his book, a regular surface refers to a subset o...

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if i take gradient of the function and set it to zero i get \(\displaystyle x = y = z = 0\). do this proof it is surface? if yes i don't see how. if no, how to proof?

 

[blamocur]

if i take gradient of the function and set it to zero i get x=y=z=0\displaystyle x = y = z = 0x=y=z=0. do this proof it is surface? if yes i don't see how. if no, how to proof?

The point you found is the only point with zero gradient, but that point does not belong to [imath]S[/imath]. I.e., the gradient is non-zero on [imath]S[/imath]. I believe this is sufficient for [imath]S[/imath] to be a smooth surface. Here is a sketch of a proof: for any given point [imath]p[/imath] on [imath]S[/imath] there exists a tangent plane (because the gradient is non-zero). For some small neighborhood [imath]U\subset S[/imath] of [imath]p[/imath] a projection of [imath]U[/imath] on the tangent plane will be a smooth map.
More rigorous proof would probably use the implicit function theorem.

 

[logistic_guy]

The point you found is the only point with zero gradient, but that point does not belong to [imath]S[/imath]. I.e., the gradient is non-zero on [imath]S[/imath]. I believe this is sufficient for [imath]S[/imath] to be a smooth surface. Here is a sketch of a proof: for any given point [imath]p[/imath] on [imath]S[/imath] there exists a tangent plane (because the gradient is non-zero). For some small neighborhood [imath]U\subset S[/imath] of [imath]p[/imath] a projection of [imath]U[/imath] on the tangent plane will be a smooth map.
More rigorous proof would probably use the implicit function theorem.

thank blamocur. i think i'm understand now