I am given the function $F:\mathbb{R}^3\to\mathbb{R}$ where $F(x,y,z)=(x^2+y^2+z^2-5)^2+16z^2-16$ and then asked to prove that $M:=F^{-1}(0)$ is a smooth surface. Problem is, I wasn't given a definition of a smooth surface in my lecture and can't seem to find a good one via Googling. I am told that a smooth curve is one for which all higher derivatives exist for each point on the curve, so is this the same for surfaces? Either way could someone let me know how to do this question? Thanks
How to prove a surface is smooth
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differential-geometry
1 Answers
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The easy way would be to show that $0$ is a regular value, i.e, for all $p\in F^{-1}(p)$ the derivative $DF_p$ is surjective.
But I don't know what kind of background we can assume.
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0In your case you have to proof the following: if $F(x,y,z)=0$ then one of the partial derivatives $\partial_xF$,$\partial_yF$ or $\partial_zF$ ist not zero at this point. – 2012-02-06