I was seeing some facts about the arc length function for vector-valued functions of a real variable and I stood upon the fact that "If $\overrightarrow{v}(t) = \overrightarrow{r'}(t)$ is continuous on $t \in [a,b]$ then $\overrightarrow{r}(t)$ is rectifiable and we have $\Lambda(a,b)$ (arc length when $t$ goes from $a$ to $b$) is less than or equal to $\int^b_a||\overrightarrow{v}(t)||dt$" and that "If $||\overrightarrow{v}(t)|| = v(t)$ is continuous, then we have $\Lambda(a,b) = \int^b_av(t)dt$" by the additivity property and the former fact.
Then it came to my mind the question "If $\overrightarrow{v}(t)$ is continuous, of course $v(t)$ is also continuous (since limits can "enter" into square roots and squares since they're both continuous)... but what if we only had $v(t)$ being continuous, would $\overrightarrow{v}(t)$ necessarily be continuous?"
Intuitively, that converse seems to be wrong to me, since, in my head, I couldn't make a way to force each component to be equal to its limit at the point we're analysing, but I also don't know how to make a counter-example to make sure that my intuition is right (or see if it's actually wrong)...
Any help on this approach would be very much appreciated!
Edit: Mark S. made me see the other question, so if I define $\overrightarrow{v}(t)=f(t)(1,1,...,1)$ (in $R^n$) in such way that $f(t) = 1$ to $t \in Q$ and $f(t) = -1$ to $t \in R-Q$, we'd have $v(t) = |f(t)|\sqrt{n} = \sqrt{n}$ which is constant and, by such, is continuous in any interval, thanks so much for the answers!
