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I'm looking through my notes and I don't fully understand what this is telling me. What is the purpose of the gradient vector here and why is $S^2$ regular because of the following? I'm looking for the big picture and any underlying propositions or theorems that might be related.


Let $S^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $x^2 + y^2 + z^2 = 1$} be the unit sphere. Then we can write $S^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $f(x,y,z) = 0$} where $f$ : $\Bbb{R}^3 \to \Bbb{R}$ : $f(x,y,z) = (x^2 + y^2 + z^2-1)^2$.

Then...

$\triangledown f(x,y,z) = (\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z})$ = 2($x^2 + y^2 + z^2 - 1$)($2x$, $2y$, $2z$)

$= (0)(2x, 2y, 2z)$ for all $(x, y, z) \in S^2$

$= (0, 0, 0)$ for all $(x, y, z) \in S^2$

$S^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $g(x, y, z) = 0$} where $g(x, y, z)$ = $x^2 + y^2 + z^2 - 1$.

$\triangledown g(x, y, z) = (2x, 2y, 2z) = (0, 0, 0)$ if and only if $(x, y, z) = (0, 0, 0)$ but $(0, 0, 0) \notin S^2$. Therefore, $S^2$ is regular by our proposition.

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The relevant theorem is the level set theorem which in this special case will say if $f: \mathbb{R}^3 \to \mathbb{R}$ is a smooth map and for all $p \in f^{-1}(c)$ we have $Df(p) \not = \textbf{0}$ then $f^{-1}(c)$ is a regular surface.

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  • $\begingroup$ Perhaps you should also explain the significance of the first example $f = (x^2 + y^2 + z^2 -1)^2$. May be this is really where the OP got confused. $\endgroup$ – user99914 Mar 23 '17 at 4:18

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