Is the following function twice differentiable?

Question

Let $\alpha : I \rightarrow \mathbb{R}^3$ be a twice differentiable curve such that $\alpha ' $ and $\alpha ' ' $ are nonzero everywhere. Let $$s(t):=\int_{t_0}^{t}|\alpha ' (u)|du$$ and let $\beta := \alpha \circ s$. Then $\beta$ is an arc-length parametrization of $\alpha$. Is $\beta$ twice differentiable?

Answer 1

Yes, it is. One differentiation gives $$\beta'(t)=\alpha'(s(t))\cdot s'(t)=\alpha'(s(t))|\alpha'(t)|$$ The function $t\to \alpha'(s(t))$ is differentiable because $\alpha$ is twice differentiable, and the function $|\alpha'(t)|$ can be written as $$\sqrt{\alpha_1'(t)^2+\alpha_2'(t)^2+\alpha_3'(t)^2}$$ where $\alpha(t)=(\alpha_1(t),\alpha_2(t),\alpha_3(t))$. Since $\alpha'(t)$ is nonzero everywhere, its norm is a differentiable function. Hence the product of $\alpha'(s(t))$ and $|\alpha'(t)|$ is also differentiable, so $\beta$ is twice differentiable.