Vectors and curves in space · Definition

Arc length

multivariable calculuscurves

Before you start you need to know

It answers how much path a point covers while following a curve, something the distance between its endpoints cannot say.

Definition

The arc length of a parametric curve \(\mathbf r(t)\), with \(a\le t\le b\), is the total distance \(L\) traveled along it, measured along the curve itself.

\[L=\int_a^b\lvert\mathbf r'(t)\rvert\,dt\]

The idea is to approximate the curve by a polygonal path: divide \([a,b]\) into small steps \(\Delta t\), each piece has length about \(|\mathbf r'(t)|\,\Delta t\) (speed times time), and summing and taking the limit gives the integral of the speed \(|\mathbf r'(t)|\). In the plane, with \(\mathbf r(t)=\langle x(t),y(t)\rangle\), the integral is written \(\int_a^b\sqrt{x'(t)^2+y'(t)^2}\,dt\), and in space \(z'(t)^2\) is added under the root.

For example, for the circle \(\langle R\cos t,R\sin t\rangle\) the speed is constant, \(|\mathbf r'|=R\), and the length over \([0,2\pi]\) is \(2\pi R\), the familiar perimeter. The helix \(\langle\cos t,\sin t,t\rangle\) has speed \(\sqrt2\), so one turn has length \(2\sqrt2\,\pi\), more than the base circle because the curve also climbs. When the speed is not constant the integral may have no elementary antiderivative, and it is then computed numerically.

Length depends on the curve and on how many times it is traversed, not on how fast it is traversed: \(\langle\cos2t,\sin2t\rangle\) on \([0,\pi]\) has the same length \(2\pi\) as \(\langle\cos t,\sin t\rangle\) on \([0,2\pi]\). For this to hold one must watch out for repeated pieces: if the curve passes twice over the same arc, that portion counts twice.

The arc length function \(s(t)=\int_a^t|\mathbf r'(u)|\,du\) measures the distance covered from the start up to time \(t\). Reparametrizing by \(s\) gives a curve of unit speed, useful later to define curvature independently of the parametrization.

Think about it

Why can the arc length of a curve never be less than the distance between its endpoints?

Because a straight line is the shortest path between two points. Formally, \(\left|\int_a^b\mathbf r'\,dt\right|\le\int_a^b|\mathbf r'|\,dt\), that is, \(|\mathbf r(b)-\mathbf r(a)|\le L\).

Exercises 10

References

  • James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §13.3
  • Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §3.2
  • Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §13.3