Arc length and curvature [10.3]

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

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).

Arclength

$\myv r(t)$ 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 $\myv b$.

The arclength from 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... $$\begineq L&=&\int_a^b |\myv r'(t)|\,dt\\ &=&\int_a^b \sqrt{\left(\frac{df}{dt}\right)^2 + \left(\frac{dg}{dt}\right)^2 + \left(\frac{dh}{dt}\right)^2 }\,dt \\ &=&\int_a^b \sqrt{\left(f'(t)\right)^2 + \left(g'(t)\right)^2 + \left(h'(t)\right)^2 }\,dt \endeq $$

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 $$ 0 \leq t \leq 2\pi,$$ 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 $$L =\int_0^{2\pi} \sqrt 2 \,dt=2 \sqrt 2 \pi.\nonumber$$

Arclength distance function

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

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

This means that $$\frac{ds}{dt}=|\myv r'(t)|.$$

Parameterizations

[Sorry, I can't bear to write "parametrization" as Stewart does.]

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 $$\myv r_2(u)=\langle e^u,e^{2u},e^{3u} \rangle;\ \ 0\leq u \leq \ln 2$$

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.

Curvature

The question: How to describe "curvature" in a useful way?

The curvature should be...

a scalar number which should be
large when the path is changing direction "quickly",
small when the path is more nearly straight.

The velocity vector $\myv r'(t)$ is tangent to the curve at each position on the curve. So, as the direction of a curve changes,the velocity vector will also change. So our first attempt to quantify the idea of "curvature" might be to look at at the change of the velocity vector, that is, the acceleration: $$\myv a (t)= \frac{d\myv r'(t)}{dt}$$

Can you think of a situation in which the acceleration is not zero, even though the curvature of the path is zero?

Acceleration in a straight line.

The unit tangent $$\uv T(t)=\frac{\myv r'}{|\myv r'|}$$ is a vector quantity that is always the same length, no matter what the velocity $\myv r'$ is. So, what about $$\frac{d\uv T(t)}{dt}?$$ What's the problem with that?

In this picture, the unit tangent vectors are evenly spaced along the curve:
Does this suggest a useful way to think about curvature?

...We should look at how $\uv T$ changes with respect to changes in the arclength, $s$, (instead of w.r.t time $t$):

The curvature, $\kappa$, is $$\kappa = \frac{d\uv T}{ds},$$ the rate of change of unit tanget vector direction with respect to arclength.

By the chain rule, this is $$\kappa = \left|\frac{d\uv T}{dt}\right| \frac{dt}{ds} = \frac{|d\uv T(t)/dt|}{|\myv r'(t)|} .$$

Curvature of a circle

It seems like a circle ought to have a constant curvature. Let's see...

Circle of radius $a$: $$\myv r(t)=a \cos t\uv i + a \sin t\uv j.$$

$$\myv r'(t)=-a\sin t \uv i + a \cos t\uv j.$$

$$|\myv r'(t)| = \sqrt{(-a\sin t)^2+(a\cos t)^2}=a.$$

$$\uv T(t) = \myv r'(t)/|\myv r'(t)|=\myv r'(t)/a=- \sin t \uv i+\cos t\uv j.$$

$$\uv T'(t) =- \cos t \uv i - \sin t\uv j.$$

$$|\uv T'| = \sqrt{(-\cos t)^+(-\sin t)^2}=\sqrt{\cos^2 t+\sin^2 t} = 1.$$

$$\kappa = \frac{|\uv T'(t)|}{|\myv r'(t)|}=\frac{1}{a}.$$

The curvature $1/a$ is constant, independent of $t$.
Small circles have a large curvature, and
large circles have a small curvature.
A straight line has a curvature of 0, because there is no change of the unit tangent vector.

This suggests a geometric definition of curvature...

Osculating circle

The circle with the same curvature (that is, radius=$1/\kappa$) as $\myv r(t)$ at a point $P$ on the curve.

Another way to calculate curvature

We had: $$\kappa = \left|\frac{d\uv T}{dt}\right| \frac{dt}{ds} = \frac{|d\uv T(t)/dt|}{|\myv r'(t)|} = \frac{\left|\frac{d}{dt}\left[ \frac{\myv r'(t)}{|\myv r'(t)|}\right]\right|}{|\myv r'(t)|}.$$

Theorem: The curvature of the curve given by $\myv r(t)$ is given by: $$ \kappa(t)=\frac{|\myv r'(t)\times \myv r''(t)|}{|\myv r'(t)|^3}.$$

Example 4 - group work

...uses the formula above to calculate the curvature of the 'twisted cubic' $\myv r(t)=\langle t,t^2,t^3 \rangle$: $$\kappa(t)=\frac{2\sqrt{1+9t^2+9t^4}}{(1+4t^2+9t^4)^{3/2}}$$ The twisted cubic...[plot the twisted cubic.]

Would you expect the curvature function to be symmetric or not about $t=0$ which corresponds to $(0,0,0)$?
Would you expect there to be a maximum, minimum, or neither at $t=0$?

[Graph $\kappa(t)$ to see...]

Unit tangent, unit normal, binormal

Several quantities, defined in terms of derivatives of $\myv r(t)$:

The unit tangent vector is $$\uv T(t) \equiv \frac {\myv r'(t)}{|\myv r'(t)|}.$$

Under what circumstances is $\uv T'(t)=0$?

(That is to say, $\uv T'\equiv \frac{d}{dt}\uv T$ is not necessarily a unit vector.)

But when $|\uv T'(t)|\neq 0$...

$\uv T(t)$ is a vector whose length (=1) is not changing with time.
Therefore $\uv T'(t)$ must be at right angles to ("normal to") $\uv T$.

The unit normal vector is $$\uv N(t) \equiv \frac {\uv T'(t)}{|\uv T'(t)|}.$$

$\uv N$ points towards the center of the osculating circle for $\myv r(t)$.

What's the magnitude of the cross product of two unit vectors, which are at right angles?

The binormal vector is $$\myv B(t)=\uv T(t)\times\uv N(t).$$

We may as well write it as $\uv B(t)$ because it must have unit length by this definition.

Example

For $\myv r(t)=2\cos t \uv i + 2\sin t \uv j + 3t \uv k$: Find $\uv T$, $\uv N$, and $\uv B$ at $t=3\pi/2$. Sketch these on a plot of $\myv r(t)$.

See 10.3.TBN.nb for calculations

Image credits

Oliver Kurmis