Let $X$ be a differentiable map from a domain $D$ in the plane into $\mathbb{R}^3$
we suppose $X$ is rotationally symmetric ,Orienting so that the axis of rotation is the $y-$plane , we have the standard parametrisation $X(\theta,s)=(x(s) cos\theta , y(s) , x(s) sin\theta)$
where $\sigma(s)=(x(s),y(s))$ is the unit speed curvature in the xy-plane and $x(s)\geq 0$ for all $s$.
the principales curvature of $X$ are:
$k_{\sigma}=x^{'}y^{''}-x^{''}y^{'} $ and $ k_{\theta}=\dfrac{y^{'}}{x}$
my question is the folowing : Let $\vartheta$ be the angle between the $\sigma^{'}$ and the positive $x-$axis
how can I prouf this $k_{\sigma}=\dfrac{\vartheta(s)-\vartheta(t)}{s-t}$ ?
$\vartheta$ (measured continuously from s=0)
thank you in advance,