Question

Could somebody help me with a Differential Geometry question? I'm trying to make sense of "Natural Parametrization" or at least how some definitions/theorems result from it, and I don't understand the math they are doing. Posted below shortly.

Answer 1

(Having trouble with LaTeX, one moment)

Answer 2

"Consider a plane regular curve C, defined by equation\[C = \left\{(x_{1},x_{2}) \in U |F(x_{1},x_{2})=0 \right\}.\]

Answer 3

(Alright, good so far. Posting the rest, just tedious LaTeX)

Answer 4

"It is well known that Grad F, computed at the points of C, is orthogonal to the curve. If one considers any parametric representation, x = x(t), then the vector dx/dt is tangent to the curve. Using the natural parametrization, it follows from

Answer 5

\[\bigg|\frac{dx}{ds}(s)\bigg|=1\]for all s, that

Answer 6

that the vector dr/ds is of unit norm. In addition, (right here is where I get lost) \[\frac{d^2x}{ds^2} \cdot \frac{dx}{ds}=0\]

Answer 7

Okay, so, I get how you get the natural parametrization, it's that integral that's kind of like arc length in 3D and you describe your natural parametrization as s(t); how the hell fo you go from s(t) to phrasing your function x in terms of s?

Answer 8

\[s(t)=\pm \int\limits_{0}^{t}|x'(\tau)|d \tau\]

Answer 9

That's your natural parametrization, how the heck do you go from s(t) (tau in the above is just a dummy variable) to describing x in terms of like, literal terms of s?

Answer 10

Like, sample natural parametrization problem, to get a handle on that, just so we're on the same page

Answer 11

Forget trying to think one up, just going to take a book example: "Compute the natural parametrization of the following plane curve:
\[x_{1}(t)=t, \ \ \ x_{2}=\log (t)\](EU text, log = ln)

Answer 12

\[x(t)=(t, \ \log(t))\]

Answer 13

So \(x(t)\) is a vector representing displacement and \(s(t)\) is going to be distance.

Answer 14

Well, its not really distance, it's not arc length, but it's a parametrization based on arc length, like,
\[l=\int\limits_{a}^{b}\sqrt{x'(t)\cdot x'(t)}dt=\int\limits_{a}^{b}|x'(t)|dt\]for arc length,

Answer 15

I think we're on the same page but I'm just confused about what you wrote, oh well.

Answer 16

\[s(t)= \pm \int\limits_{0}^{t}|x'(\tau)|d \tau\]For the natural parametrization of at least any plane curve, I think curves in R3 and otherwise but am not sure

Answer 17

When you find \(s(t)\), you can remember chain rule:\[
\frac{dx}{dt}= \frac{dx}{ds}\frac{ds}{dt}\implies \frac{dx}{ds}= \frac{\frac{dx}{dt}}{\frac{ds}{dt}}
\]

Answer 18

Okay, that makes infinitely more sense, hah.

Answer 19

And one of the book definitions states that, for the natural parametrization, at least, \[\bigg| \frac{dx}{ds}(s)\bigg|=1 \ \text{for all }s.\]So should be able to test that to see if I'm both getting the natural parametrizsation correctly, *and* that I'm computing dx/dt

Answer 20

I wish I had a PDF so I could show you exactly what I'm looking at, anyways. I think I'm going to bike home now, but I might hop on when I get home, possibly. Thanks for pointing that out, for some reason I didn't think of phrasing x as a function of s, but thought of phrasing both independently as functions of t, I'm still hazy as to relating x and s in formula.

Answer 21

If you go from point A to B and back, your displacement will be 0 (a zero vector), while your distance will be twice the length of AB. So \(x\) is your displacement, or basically were you are on the map, while \(s\) is your odometer.

So rather than change of position with respect to time, we're looking at change in position with respect to total distanced traveled.

Answer 22

I think that: \[
\left|\frac{dx}{ds}\right|=1\implies \left|\frac{dx}{dt}\right|=\frac{ds}{dt}
\]Which means the magnitude of our "velocity" is equal to our "speed".

Answer 23

If \(s(t)\) is just some parametrization, then I guess we're checking if it has similar properties as arclength.

Answer 24

Another advice I can give is that:\[
\frac{d}{ds}\bigg(\ldots\bigg)=\frac{\frac{d}{dt}\bigg(\ldots\bigg)}{\frac{ds}{dt}}
\]

Answer 25

So in terms of \(t\):\[
\frac{d^2x}{ds^2}=\frac{\frac{d}{dt}\bigg(\frac{dx}{ds}\bigg)}{\frac{ds}{dt}}=\frac{\frac{d^2x}{dt^2}\frac{ds}{dt}-\frac{dx}{dt}\frac{d^2s}{dt^2}}{\left(\frac{ds}{dt}\right)^3}
\]