AP Calculus BC glossary

Parametric arc length

Also called: Arc length of a parametric curve, Length of a parametric curve

Parametric arc length is the distance along a curve given by x of t and y of t. Over a parameter interval that traces the curve exactly once it equals the integral of the square root of the sum of the squares of x prime of t and y prime of t, which is the integral of speed.

L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}\,dt

A short piece of the curve is almost a straight segment with horizontal leg $\frac{dx}{dt}\,dt$ and vertical leg $\frac{dy}{dt}\,dt$ , so the Pythagorean theorem supplies the integrand. That integrand is the speed of a particle at $(x(t), y(t))$ , which is why the length of the path and the total distance traveled fall out of one integral. The formula needs $\frac{dx}{dt}$ and $\frac{dy}{dt}$ continuous on the interval, and the interval must trace the curve exactly once. Requiring the two derivatives never to vanish at the same instant is the extra condition that makes the curve smooth, giving it a tangent direction everywhere; the integral itself does not need it, which is why the cycloid $x = t - \sin t$ , $y = 1 - \cos t$ still returns the arch length $8$ despite the cusp where the velocity is zero.

The limits are values of the parameter, never $x$ values, and this is the general form of the arc length formula rather than a separate rule. Setting $x = t$ makes $\frac{dx}{dt} = 1$ and the integrand collapses to $\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}$ , the version used for $y$ as a function of $x$ .

The mistake

Letting the parameter retrace the curve and reporting the result as a length. The integral measures distance traveled, which matches the length of the curve only when the interval traces it exactly once. For $x = \cos t$ and $y = \sin t$ on $0 \le t \le 4\pi$ the integral gives $4\pi$ , two laps of a circle whose length is $2\pi$ .

Appears in: Unit 9: Parametric, Polar, and Vector (BC)