Suppose $S$ is a minimal surface, show that the gaussian curvative is negative on all interior points

Question

I am trying to improve my proving skills, started learning by myself, can anybody help me with this?

thanks you

Answer 1

One definition of a minimal surface is that its mean curvature is zero at each of its points. If $\kappa_1$ and $\kappa_2$ are the principal curvatures then $\tfrac{1}{2}(\kappa_1+\kappa_2)=0$.

Recall that the Gaussian curvature is given by the product $\kappa_1\kappa_2$.

If $\tfrac{1}{2}(\kappa_1+\kappa_2)=0$ then $\kappa_2 = -\kappa_1$ and so the Gaussian curvature is $\kappa_1\kappa_2 = -\kappa_1^2 \le 0$.

The case $\kappa_1=\kappa_2=0$ gives a plane and both the mean and the Gaussian curvatures are zero.

Answer 2

Disclaimer added at the advice of Javier Badia; see comments below: this answer was posted to a version of the question which was such a radical revision of the current (rolled back) question that it is really about a completely different topic. The question I answered was, basically: what is the derivative of $x^2 + x$? Apparently the OP put it through some rather severe revisions. So forgive me if this answer seems, well, rather curious for the current version of the question. Thanks for your patience. Robert Lewis

Oh well, while we're at it, might as well answer the new question, since I'm on this page and it's not too hard:

If

$y(x) = x^2 + x, \tag{1}$

then

$y'(x) = 2x + 1; \tag{2}$

and just for the sake of completeness:

given (1), we have

$y(x + h) = (x + h)^2 + (x + h); \tag{3}$

$y(x + h) = x^2 + 2xh + h^2 + x + h; \tag{4}$

$y(x + h) - y(x) = x^2 + 2xh + h^2 + x + h - x^2 - x; \tag{5}$

$y(x + h) - y(x) = 2xh + h^2 + h; \tag{6}$

$(y(x + h) - y(x)) / h = 2x + h + 1 = 2x + 1 + h; \tag{7}$

$y'(x) = \lim_{h \to 0}(y(x + h) - y(x)) / h = \lim_{h \to 0}(2x + h + 1) = 2x + 1; \tag{8}$

QED!!!

Hope this helps answer the new question! Cheerio,

and as always,

Fiat Lux!!!