Answer
Result See proof
Work Step by Step
Step 1 Let us denote the arc length parameter as $\boldsymbol{s}$ and some other parameter as $\boldsymbol{t}$ of some fixed curve $\mathcal{L}$. Let $\mathbf{r}(s)$ and $\boldsymbol{\rho}(t)$ be the vector-valued functions that represent the parametrizations of curve $\mathcal{L}$ with respect to parameters $\boldsymbol{s}$ and $\boldsymbol{t}$, respectively. The reason why arc length parametrization is better than the regular one follows from the fact that the vector-valued function that represents the parametrization of the curve $\mathcal{L}$ has a derivative $\mathbf{r'}(s)$ which satisfies that $\left\|\mathbf{r'}(s)\right\| = 1$. However, this need not be the case if we consider the parametrization $\boldsymbol{\rho}(t)$. Step 2 More precisely, by considering the arc length parametrization $\mathbf{r}(s)$, it obtains that $\left\|\mathbf{r'}(s)\right\| = 1$. In this case, the unit tangent vector is given as $\mathbf{T}(s) = \mathbf{r'}(s)$. One more advantage follows from the fact that the vectors $\mathbf{r'}(s)$ and $\mathbf{r''}(s)$ are orthogonal because $\left\|\mathbf{r'}(s)\right\|$ is a constant value for each $\boldsymbol{s}$. Since the vector $\mathbf{r''}(s)$ belongs to the $\mathbf{T}\mathbf{N}$ plane at $\mathbf{r}(s)$, it will be parallel to $\mathbf{N}(s)$. But this need not be the case if we consider the parametrization $\boldsymbol{\rho}(t)$. Also, the curvature $\kappa$ of a curve $\mathcal{L}$ is given as \[ \kappa(s) = \left\|\mathbf{r''}(s)\right\| \] if we consider the arc length parametrization, and it is given as \[ \kappa(t) = \dfrac{\left\|\boldsymbol{\rho'}(t) \times \boldsymbol{\rho''}(t)\right\|} {\left\|\boldsymbol{\rho'}(t)\right\|^3} \] So that, formula (1) is much simpler than formula (2). Therefore, the arc length parametrization made the important formulas much simpler than usual.
