Let $\vec{\alpha}:I\longrightarrow \mathbb{R^2}$ be a curve parametrized by arc lenght. If there is a differentiable function $\theta:I\longrightarrow \mathbb{R}$ such that $\theta(s)$ is the angle that the tangent line of $\vec{\alpha}$ at $s$ makes with a fixed direction, show that $\theta'(s)=\pm k(s)$.
My attempt:
Denote by $\vec{d}$ the fixed thirection, we have $\cos \theta(s)=\vec{t}(s)\cdot\vec{d}$, then $$-\sin \theta(s)\theta'(s)=\vec{t'}(s)\cdot\vec{d}=k(s)\vec{n}\cdot\vec{d},$$ but $\vec{n}\cdot\vec{d}=\cos(\theta(s)+\pi/2)=-\sin(\theta(s))$. We obtain $$\theta'(s)=k(s).$$
Where is the mistake?
You could have $\vec n\cdot\vec d = \cos(\theta(s)-\pi/2)$. It depends which direction the curve is bending. In this case, you would have a negative sign.