# Differential Geometry Proof Regarding Arclength, Tangents, Curvature, and Parameters

Consider a regular curve q(t) with arclength parameter s. Show that if $T(t_{n}) \neq T(t_{0})$ and $t_{n} \rightarrow t_{0}$, then $$1 = lim_{t_{n} \rightarrow t_{0}} \frac{|\theta(t_{n}) - \theta(t_{0})|}{arccos(T(t_{n})) . T(t_{0}))}$$ and $$\kappa(t_{0}) = 1 = lim_{t_{n} \rightarrow t_{0}} \frac{arccos(T(t_{n}) . T(t_{0}))}{|s(t_{n}) - s(t_{0})|}$$

Notation-wise, $\theta$ denotes the parameter for the unit tangent T such that $\frac{d \theta}{dt} = |\frac{dT}{dt}|$ and $\kappa$ is the curvature defined as $\frac{d\theta}{ds}. S$ denotes the arclength parameter with $\frac{ds}{dt} = |\frac{dv}{dt}|$ for $v$ the velocity of the curve.

Actually I have been able to show the first part of the question using the squeeze lemma but am not sure if it will help solve the question to have that information. I have proven the fact that the second limit is equal to the curvature $\kappa$ at $t_{0}$ but am not able to show that it equals 1.

Thanks for any help in advance.

The teacher just told me that it was a mistake on the text, and $\kappa\neq$ 1.