Q:

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

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 $\Gamma=0$

A:

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. 

