ATRACTOR

Let \[\begin{array}{rl} f:\left[-1,1\right] & \rightarrow\mathbb{R}^{2}\\ t & \rightarrow\left(t,\,\left|t\right|\right) \end{array}.\]

Then \[f'(t)=\begin{cases} (1,1) & \mbox{se }t>0\\ (1,-1) & \mbox{se }t<0 \end{cases};v(t)=\left|f'(t)\right|=\sqrt{2}\] and the trace of this curve is:

Note that this function is not a differenciable plane curve, since function \(\left|t\right|\) is not differentiable at \(t=0\).