Suppose we have a smooth curve $\vec{r}:[0,+\infty)\to\mathbb{R}^2$ that satisfies: 1) being bounded, i.e., $\exists M>0$ such that $\|\vec{r}(t)\|<M$ for all $t\in[0,+\infty)$; 2) its velocity tends to zero, i.e., $\lim_{t\to+\infty} r'(t)=0$. My question is: Can we prove $\vec{r}(t)$ has finite length, i.e., $\int_0^{+\infty}\|r'(t)\|dt<\infty$?
I found that some famous examples of bounded curves with infinity length do not satisfy the above conditions. For example, the Koch curve has infinity length but it is not smooth nor its velocity tends to zero; the graph of the function defined by $f(x) = x \sin(1/x)$ around zero has infinity length and is smooth but its velocity does not tend to zero.
I wonder is there a counterexample for my case, or it is sufficient for finite length?
Any ideas and suggestions are welcomed. Thanks in advance!
Consider your second example $f(x) = x\sin\left(\frac{1}{x}\right)$ parametrized as
$$x(t) = 1/t\qquad (t\geq 1)$$
We compute
$$\|r'(t)\| = \sqrt{\left(\frac{df}{dt}\right)^2 + \left(\frac{dx}{dt}\right)^2} = \sqrt{\frac{(t\cos(t)-\sin(t))^2 +1}{t^4}}$$ which can be approximated by $\frac{1}{t}$ as $t\to \infty$, hence velocity tends to zero. As you've pointed out, the other condition is satisfied and the curve has infinite length.
More generally, it should be possible to generate other counterexamples by taking functions whose smooth graphs are bounded with infinite length, and parametrizing them to slow down as they approach the appropriate $x$ coordinate.