Arc length (and curvature) [9.8]


The Nowitna river (Alaska) - How long is it? a

  • Calculating arclength and curvature
  • Arclength is independent of the choice of parameter used to describe a curve.
  • Re-parameterizing a curve - using the arclength as the parameter.
  • Geometric definition of curvature.
  • The TNB frame (Tangent - Normal - Binormal unit vectors).
  • Tangent and Normal components of acceleration

Arclength

You know some special-case formulas for calculating the distance along a straight line between 2 points, or the perimeter of a circle.

But now that you know how to describe *any* continuous curve in space with parametric equations, we'll develop a general way to integrate to find the distance along *any* continuous curve in space.

$\myv r(t)$ is a vector function. $$\myv r(t)=\langle f(t),g(t),h(t)\rangle\nonumber; \ \ t_i\leq t\leq t_f.$$ where $\myv r(t_i)=\myv a$ and $\myv r(t_f)=\myv b$ The tip of the position vector $\myv r(t)$ traces a curve (trajectory) in space which looks schematically like:

  • The arclength is the distance along the curve from $\myv a$ to $\myv b$.
  • Arclength is a scalar.
  • It is not the distance "as the crow flows" of a vector starting at $\myv a$ and terminating at $\myv b$.
  • $\myv a\equiv \myv r(t_i)$ and $\myv b\equiv \myv r(t_f)$.

The arclength from $\bf{a}$ to $\bf{b}$ is approximately equal to the sum of the pink segments: $$\text{arclength}\equiv L \approx \sum \Delta L\nonumber.$$

The derivative $\myv r'(t)$ is a vector tangent to the curve. Its magnitude is the "speed" of a particle moving in time, $t$, according to $\myv r(t)$.

Since distance = speed (times) time, we can approximate the distance $\Delta L$ moved in a time interval $\Delta t$ as $$ \Delta L \approx \left| \myv r'\right| \Delta t\nonumber$$

So, the arclength is approximately $$L\approx \sum |\myv r'(t)|\Delta t\nonumber.$$ In the limit of ever smaller $\Delta t\to dt$, the sum approaches the integral... $$\begineq L&=\int_{t_i}^{t_f} |\myv r'(t)|\,dt\\ &=\int_{t_i}^{t_f} \sqrt{\left(\frac{df}{dt}\right)^2 + \left(\frac{dg}{dt}\right)^2 + \left(\frac{dh}{dt}\right)^2 }\,dt \\ &=\int_{t_i}^{t_f} \sqrt{\left(f'(t)\right)^2 + \left(g'(t)\right)^2 + \left(h'(t)\right)^2 }\,dt \endeq $$

Example

Find the length of the arc of the helix that obeys the equation $$\nonumber \myv r(t)=\langle \cos t, \sin t, t \rangle$$ from the point $(1,0,0)$ to $(1,0,2\pi)$.

It looks like $$ 0 \leq t \leq 2\pi.$$
Estimating first: Before you do the problem, what is the arclength around $\myc{\cos t, \sin t, 0}$ over that same interval $0\leq t\leq 2\pi$? Will the answer to your problem above be greater than or less than this answer?

The derivative is $$ \myv r'(t) = \langle -\sin t, \cos t,1\rangle,\nonumber$$ and so... $$|\myv r'(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2 +1}=\sqrt 2.\nonumber$$

Substituting into the arclength expression $$L =\int_0^{2\pi} \sqrt 2 \,dt=\sqrt 2 2 \pi.\nonumber$$

Arclength distance function

The arclength distance function $s(t)$, where $s(0)=0$,

$$s(t) = \int_0^t |\myv r'(t)|\, dt =\int_0^t\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2 } \,dt$$.

Taking the derivative w.r.t. $t$ of both sides, this means that $$\frac{ds}{dt}=|\myv r'(t)|.$$

Parameterizations

...though some well-regarded textbooks (Stewart!) write "parametrization".]

  • The same curve can be represented in more than one way.
  • E.g. $$\myv r_1(t)=\langle t,t^2,t^3\rangle \nonumber;\ \ 1\leq t\leq 2$$ and $$\myv r_2(u)=\langle e^u,e^{2u},e^{3u} \rangle;\ \ 0\leq u \leq \ln 2$$

The two parametric curves above are plotted here (GeoGebra) (one as a point, and one as the tip of a position vector). Discuss with a partner:

  1. Moving the $t$ and $u$ sliders back and forth... does it seem like the two functions trace out the same trajectory? Change the perspective a few times to make sure...
  2. Next, animate the two functions (press the little play button besided each slider) and watch the two functions. Describe how the motions of the two functions differ.

Re-parameterization

It may be useful (and is certainly beautiful) to re-parameterize a curve in terms of the arclength along a curve

  • is characteristic of the curve itself,
  • does not depend on the choice of parameter used to describe the curve,
  • and does not depend on the coordinate system.

We could use the arclength ("distance along the curve") as the parameter to describe a curve (instead of the time, or angle, or some other parameter).

Example -- $\myv r(t)=\langle \cos t,\sin t,t\rangle$:
Reparametrize with respect to arc length, beginning at (1,0,0) in direction of increasing $t$.

  • $$|\myv r'(t)| = \sqrt{f'^2+g'^2+h'^2}=\sqrt{(-\sin t)^2+(\cos t)^2+1}=\sqrt 2.$$
  • $$s(t)=\int_0^t|\myv r'(u)|\,du=\int_0^t\sqrt 2\,du=\sqrt 2 t.$$
  • Solve for $t$: $t=s/\sqrt 2$,
  • and substitute back into original expression: $$ \myv r(t(s))=\cos(s/\sqrt 2)\uv i+\sin(s/\sqrt 2)\uv j+s/\sqrt 2\uv k. $$

This depends on being able (easily or otherwise) to take $s(t)$ and invert it to $t(s)$.


This is a technique used in General Relativity: We talk about the "proper time" measured along the the space-time trajectory of a particle as a quantity that all observers agree on.

Image credits

Oliver Kurmis