Suppose I give you a position function $x(t)$ where t is time and x is position. Then i ask you to find the total distance travelled from $t = 0$ to $t = t_f$.

Which formula would give you that distance?

$$\int_{0}^{t_f} \sqrt{1 + [x'(t)]^2} dt$$

$$\int_{0}^{t_f} |x(t)| dt$$

I know the first one gives you the arc length of the path (which is the total length of the path) and the second one gives me the signed distance, but why are they fundamentally different again?

  • 1
    $\begingroup$ Comment to new v3: Why did you remove the prime in the second formula? $\endgroup$
    – Qmechanic
    Jan 1, 2012 at 11:17

1 Answer 1


The first formula gives you the arc length of the curve that describes the particle's motion in the $x$-$t$ plane. It counts not only the accumulation of changes in position, but also the accumulation of changes in time. These two kinds of accumulations correspond to the two terms under the square root: the $1$ represents "motion" in time and the $[x'(t)]^2$ represents motion in space. Even while the particle is stopped or slowly moving through space, it is still advancing in time, and the first integral takes that into account.

If you remove the $1$ which corresponds to the time, then you get the second integral, which only takes into account spatial motion. That's the one that gives you the distance you're looking for.

  • $\begingroup$ Just a wording confusion. When you say "the latter corresponds to the '1' under the...". Do you mean "one" or the "1" in the arclength formula? $\endgroup$
    – Lemon
    Dec 31, 2011 at 23:08
  • $\begingroup$ I mean the number $1$. Of the two terms under the square root, $1$ and $x'(t)^2$, the first corresponds to "motion" in time and the second corresponds to motion in space. If I meant "the one under..." in the sense of "the thing under..." it's grammatically incorrect to write it as a number. (Perhaps I can edit that to make it a bit more clear, though.) $\endgroup$
    – David Z
    Dec 31, 2011 at 23:11

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