AP Calculus BC

Arc Length vs Distance Travelled

Q: Is speed always the parametric arc length integrand?

Yes. Speed is \sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2}, which is the arc length integrand with time as the tracing variable, so integrating speed is integrating arc length.

Arc length measures how long a curve is, integrating with respect to whatever variable traces it. Distance travelled integrates speed over a time interval, so it counts every retrace. For a particle that never doubles back the two agree, but only distance travelled counts a stretch of path covered twice.

Arc length

Use when: You are handed a curve, as $y = f(x)$ or as parametric equations, and asked how long it is.

Distance travelled

Use when: You are handed motion in time, as $v(t)$ or as a position vector, and asked how far the particle actually went.

Side by side

	Arc length	Distance travelled
What it measures	The length of a curve	The path length a particle covers over a time interval
Integrand	$\sqrt{1 + \left[f'(x)\right]^2}$ for $y = f(x)$	Speed, which is $\lvert v(t) \rvert$ on a line
Variable of integration	Whatever traces the curve: $x$ or $t$	Time, always
When to reach for it	The prompt names a curve and asks how long it is	The prompt names a particle and asks how far it travelled
Common trap	Losing the $1$ under the radical, or using a parametrization that traces the curve twice	Integrating $v(t)$ rather than $\lvert v(t) \rvert$ , which gives displacement instead

The formulas are one object seen from two sides. Arc length for $y = f(x)$ on $[a,b]$ is $\int_a^b \sqrt{1 + \left[f'(x)\right]^2}\,dx$ , and for a parametric curve it is $\int_{t_1}^{t_2} \sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2}\,dt$ . That second integrand is exactly speed, so the distance a particle travels is the arc length of its path, measured in time.

L = \int_a^b \sqrt{1 + \left[f'(x)\right]^2}\,dx \qquad s = \int_{t_1}^{t_2} \sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2}\,dt

The questions still differ. Arc length asks about a set of points, so a curve drawn twice is not twice as long. Distance travelled asks about a journey, so a particle that runs out and comes back covers double. On a line the split is sharpest: the path is a single segment, yet $\int_{t_1}^{t_2} \lvert v(t) \rvert\,dt$ counts each pass over that segment separately.

Trace the curve once

The arc length integral cannot tell whether your parametrization repeats. On $0 \le t \le 4\pi$ the circle $x = \cos t$ , $y = \sin t$ has speed $1$ , so the integral returns $4\pi$ . That is the distance travelled, not the $2\pi$ length of the circle. Restrict the parameter interval to one pass before calling the answer a length.

Frequently asked questions

Is speed always the parametric arc length integrand?

Yes. Speed is $\sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2}$ , which is the arc length integrand with time as the tracing variable, so integrating speed is integrating arc length.

How is this different from displacement?

Displacement integrates velocity itself and can come out zero or negative. Arc length and distance travelled both integrate a quantity that is never negative, so neither can come out negative. Either can still be zero, but only when nothing moves: a particle sitting at rest through $[t_1, t_2]$ has $v(t) = 0$ and travels a distance of $0$ .

Why are arc length integrals so hard to evaluate by hand?

The radical rarely has an elementary antiderivative. On the exam these appear on the calculator-active sections, where setting up the integral correctly is what earns the points.

In the CED: Unit 8: Applications of Integration, Unit 9: Parametric, Polar, and Vector (BC)