I have a curve $C$ given by $\gamma:[0,1] \rightarrow \mathbb R^2$
$$\gamma(t) = \begin{cases} (0,0) & t=0 \\ (\sqrt t \cos(\frac1t),\sqrt t \sin(\frac1t)) & t> 0 \end{cases}$$
I would like to find out whether this curve is rectifiable and continuously differentiable. I already figured, that this function must be continuous, since sine and cosine are limited. However, when it comes down to finding out whether $C$ is rectifiable, i.e :
$$\sup \Big \{ \sum_{i=1}^m \|\gamma(t_i)-\gamma(t_{i-1}) \|: m\in\mathbb N \Big \} < \infty$$
with $\{t_0,....,t_m\}$ denoting a partition of $[0,1]$ seems a bit harder. Intuitively after looking at a plot, it seems that this curve is not rectifiable, but I do not know how to proof it. Any help is appreciated!
With $\tau_i=\frac1{i\pi}$, we have $\gamma(\tau_i)=\bigl((-1)^i\sqrt{1/ i},0\bigr)$ and hence with the partition $\{t_0,t_1,t_2,\ldots,t_{m-1},t_m\}=\{0,\tau_{m-1},\tau_{m-2},\ldots,\tau_1,1\}$ $$\begin{align}\sum_{i=1}^m\|\gamma(t_i)-\gamma(t_{i-1})\|&=\frac1{\sqrt{m-1}}+\sum_{i=1}^{m-1}\left(\frac1{\sqrt {m-i}}+\frac1{\sqrt{m-i+1}}\right) +\|\gamma(\tau_1-\gamma(1)\|\\ &>\sum_{i=1}^{m-1}\frac1{\sqrt i} \end{align}$$ and as $\sum_{i=1}^\infty\frac1{\sqrt i}$ diverges, the supremum we look for cannot be finite.