This is from Manfredo P. do Carmo's Differential Geometry of Curves and Surfaces.
I have just been introduced to orientations of regular surfaces, the Gauss map and the differential of the Gauss map, $dN_p$.
I have trouble understanding a detail of an example related to the differential of the gauss map.
Example 2. Consider the unit sphere $S^2 = \{(x,y,x)\in \mathbb{R}^3 \colon x^2 + y^2 + z^2 = 1\}$
If $\alpha (t) = (x(t),y(t),z(t))$ is a parameterized curve in $S^2$, then
$$2xx' + 2yy' + 2zz' = 0$$ Which shows that the vector $(x,y,z)$ is normal to the sphere at the point $(x,y,z)$. Thus $\bar{N}=(x,y,z), N > =(-x,-y,-z)$ are fields of unit normal vectors in $S^2\ldots$
I think I'm missing something trivial but I don't understand why the equality $2xx' + 2yy' + 2zz' = 0$ holds.
Since the curve $\alpha$ lies in $S^2$ we have that $x(t)^2 + y(t)^2 + z(t)^2 = 1 \implies \frac{d}{dt} [x(t)^2 + y(t)^2 + z(t)^2] = \frac{d}{dt}(1) \implies 2x(t) + 2y(t) + 2z(t) = 0 \implies x(t) + y(t) + z(t) = 0$
So I think we also have $x'(t) + y'(t) + z'(t) = 0$. Also for any point $(x,y,z) \in S^2$ we have that $x^2+y^2+z^2 = 1$. But how to show that $2(x,y,z)$ and $\alpha'(t)$ are in fact orthogonal to each other? Is it implicitly understood that the point we are interested in lies somewhere on the curve?
Do Carmo seems to think it's a obvious fact so I guess I'm missing something fundamental here. (Disclaimer: I gave up on coffee today.)
You have an error: $\frac{d}{dt}x(t)^2$ is not $2x(t)$, but $2x(t)x'(t)$. So, the whole equation gets to
$$\frac{d}{dt}[x(t)^2+y(t)^2+z(t)^2] = 2 x(t) x'(t) + 2 y(t) y'(t) + 2 z(t) z'(t) = \frac{d}{dt}(1) = 0$$