How to prove $x'(s)x''(s)+y'(s)y''(s)=0$ and $\sqrt{x'(s)^2 + y'(s)^2}=1$, s is the arc length in Cartesain coords

Question

A curve embedded in a 2d plane and is referred to Cartesian coordinates x, y. s is the arc length.

How to prove $x'(s)x''(s)+y'(s)y''(s)=0$ and $\sqrt{x'(s)^2 + y'(s)^2}=1$, s is the arc length in Cartesain coords. More specifically,

the contravariant component of normal vector is $N^i=\begin{bmatrix} y'(s) \\ -x'(s)\\ \end{bmatrix}$

the covariant metric tensor is $S_{\alpha\beta}=1$

the contravariant metric tensor is $S^{\alpha\beta}=1$

Area element is $\sqrt{S}=1$

And christoffel symbol $\Gamma=0$

Answer 1

The first statement is simply a differential form of the Pythagorean theorem. Once you have $x'(s)^2+y'(s)^2 = 1$, you can just take the derivative to get the second statement.