My question is based on a problem in Do Carmo' Differential Geometry book: ch. 3.2 ex 19. Here is the problem:
Let $S$ a regular surface. Let $C \subseteq S$ be a regular curve in $S$. Let $p \in C$ and $α(s)$ be a parametrization of $C$ in $p$ by arc length so that $α(0) = p$. Choose in $T_p(S)$ an orthonormal positive basis $\{t, h\}$, where $t = α′(0)$. The geodesic torsion $\tau_g$ of $C \subseteq S$ at $p$ is defined by $$\tau_g=\langle \frac{dN}{ds}(0),h \rangle $$ where $N$ denotes the normal vector. If $\tau$ is the torsion of $C$, $n$ is the (principal) normal vector of $C$ and $cos \theta = \langle N, n\rangle $, then prove that $$\frac{d\theta}{ds}=\tau-\tau_g \ \ \ \ \ \ \ \ \ \ (1)$$
Although I can prove it, by observing that $h,N,n$ all belong to the same plane, orthogonal to $a'$, and differentiating some appropriate expressions, I don't have any intuition about why equation (1) holds. Even if $\tau_g$ is zero, say by choosing $a$ to be a line of curvature, why it makes sense for $\tau$ to be the derivative of $\theta$? Can one provide ideally some intuition, or any illuminating example? Both $\tau$ and $\theta$ have clear geometric meaning, but their equality seems to come out of nowhere.