# Is arc length displacement or distance traveled?

$$\text{Arc length}= \int_{t_1}^{t_2} \left|R'(t)\right|\,\mathrm dt$$

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?

• In the same way that speed is the magnitude of velocity, distance is the magnitude of a displacement. I am not sure that I understand the question, here... Feb 10 at 11:50
• @XanderHenderson By 'distance' I meant 'distance traveled' and not the magnitude of displacement. I have edited the question accordingly. (Distance traveled is the total length of the path traveled between two positions.) Feb 10 at 12:00
• Yes. Distance is the magnitude of displacement. That integral "adds up" an infinite number of infinitely small distances, each one of which can be thought of as an infinitesimal "magnitude of displacement". Again, I am confused by the question. Feb 10 at 12:05
• @XanderHenderson I'll explain. Suppose a particle starts from point A and goes in a circle and ends at point A itself. Then the magnitude of displacement (distance) should be zero. But the distance traveled and arc length is the circumference of the circle, which is clearly not zero. Feb 10 at 12:09
• Yes, but, again, that integral is adding up an infinite number of infinitesimal distances. Are you familiar with how that integral is derived? Like, do you know where it comes from? Feb 10 at 12:17

$$\text{Arc length}= \int_{t_1}^{t_2}\ |R'(t)|\ dt$$

But arc length, as I understand it, should be 'distance traveled'

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.

This looks similar to the formula for the magnitude of displacement, as the integral gives the area under the velocity-time graph.

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}

This is essentially repeating what the accepted answer already says, only I'm going to skip all of the symbols and boil it down to English.

If you take the velocity vector, integrate that over time to get another vector, and then take the magnitude of that vector, what you get is magnitude of displacement.

But if you take the magnitude of velocity, which is a nonnegative scalar, and integrate that over time to get another scalar, what you get is distance traveled, or curve length. Taking the magnitude before integrating removes the ability of "1 unit left" to cancel with "1 unit right" to make zero; instead you get 1+1 = 2.