$$\text{Arc length}= \int_{t_1}^{t_2} \left|R'(t)\right|\,\mathrm dt$$
This looks similar to the formula for the magnitude of displacement, as the integral gives the area under the velocity-time graph.
But arc length, as I understand it, should be 'distance traveled', rather than the magnitude of the displacement vector. Because, for displacement, only the start and end points matter, while for distance traveled, the path taken also matters. Since arc length is the length measured along the curve, I feel it should be equal to distance traveled, rather than displacement.
What is that I'm missing here?
Best Answer
If $R(t)$ gives the position (with respect to a reference point like the origin), then $R'(t)$ is the instantaneous velocity and $|R'(t)|$ is the instantaneous speed, in which case $$\int_{t_1}^{t_2} \left|R'(t)\right|\,\mathrm dt$$ gives $$\text{the average speed over $[t_1,t_2]\:\:\times\:\:$ the time elapsed $(t_2-t_1)$}$$ (the area under the speed-time graph), in other words, the distance travelled over $[t_1,t_2]$ (i.e. arc length), as required.
In contrast, the magnitude of (the displacement over $[t_1,t_2]$) is $$\left|\int_{t_1}^{t_2} R'(t)\,\mathrm dt\right|.$$
Clearly, \begin{align}\Big|\text{displacement over }[t_1,t_2]\Big| &\le \text{distance travelled over }[t_1,t_2]\\ \Big|\text{average velocity}\Big| &\le \text{average speed}.\end{align}
In contrast, by definition, \begin{align}\Big|\text{position}\Big| &= \text{distance from reference point}\\ \Big|\text{instantaneous velocity}\Big| &= \text{instantaneous speed}.\end{align}