Arc Length, Curvature and Turning

Measure progress along the path

For a regular curve with $r'(t)\ne0$ , arc length from a starting point is $s(t)=\int_{t_0}^t\|r'(u)\|du$ . The unit tangent is $T=r'/\|r'\|$ . Curvature measures turning per unit distance, rather than per unit time:

\kappa=\left\|\frac{dT}{ds}\right\|=\frac{\|T'(t)\|}{\|r'(t)\|}=\frac{\|r'\times r''\|}{\|r'\|^3}.

The cross-product formula applies to sufficiently smooth regular curves. At zero speed it divides by zero; investigate the geometric curve or choose a regular parametrization instead.

Visual guide

VISUAL GUIDEA tighter circle turns more per unit distance

Radius 1, curvature 1
Radius 2, curvature 1/2

The unit circle has curvature 1; the radius-2 circle has curvature 1/2. Curvature compares a change of tangent direction with distance traveled, so it is independent of how fast a curve is parametrized.

Worked example: a circle

For $r(t)=(R\cos t,R\sin t,0)$ with $R>0$ , speed is $R$ and $\|r'\times r''\|=R^2$ . Thus $\kappa=1/R$ . A smaller circle turns more sharply. Traversing the same circle twice as fast changes velocity and acceleration, but not its curvature.

The normal direction $N=T'/\|T'\|$ exists where the tangent is changing. For a straight line, curvature is zero and this formula does not pick a unique normal direction.

Worked example: the helix

For $r(t)=(a\cos t,a\sin t,bt)$ with $a>0$ , speed is $\sqrt{a^2+b^2}$ . The cross product has magnitude $a\sqrt{a^2+b^2}$ , so $\kappa=a/(a^2+b^2)$ . Increasing the pitch parameter $|b|$ reduces turning per unit distance even though the projection onto the horizontal plane remains a circle.

Writing speed as $v_s=ds/dt$ , acceleration separates into tangential and normal components:

a=\frac{dv_s}{dt}T+\kappa v_s^2N.

The first changes speed; the second changes direction. At constant speed on a circle, acceleration magnitude is $v_s^2/R$ . This is why sharper curves require larger lateral acceleration at the same travel speed.

Practice

Find the curvature of a circle of radius $4$ .
Find the normal acceleration magnitude for speed $6$ on that circle.
What is the curvature of a regular straight line?

Show worked solutions

$1/4$ inverse length units.
$v_s^2/R=36/4=9$ acceleration units.
Zero: its unit tangent is constant, so it does not turn with arc length.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.