Order of points in a curve

1. Jan 4, 2010

Castilla

Hello, happy new year.

I have a question concerning oriented curves described by continuous vectorial functions with domain in an interval [a, b] and range in, say, R^3. I' ve seen this theme in Apostol, Courant and others. To define the length of a rectifiable oriented curve, they inscribe polygons and see if they tend to a limit and they define the length as such limit.

I have a doubt about certain step of their method.

For simplicity let's suppose we have a simple curve (doesn't intersect herself). First they take a partition {a, t1, t2, ..., b} of [a, b]. When they put the correspondent points in the graphic of the curve, f(a) is the first point, then comes f(t1), then f(t2), etc., up to f(b), which is the last point of the curve.

My question is: why f(t1) comes before f(t2) in the oriented curve? Too obvious?Answering "because the curve it is oriented from f(a) to f(b)" seems to be a circular answer.

I think the authors say that the continuity of the function f warrants that f(t1) comes before that f(t2), that f(t2) comes before f(t3), etc., but I fail to see why. I mean, the definition of, say "continuity of f in number t1", is:

"For every little Epsilon there is a little Delta such that if the diference between numbers t1 and any t is less than Delta, then the absolute value of the modulus composed by the components of the points is less than Epsilon".

But this definition does not imply that if I take a "t" such that t1 < t then f(t) will come after f(t1) in the curve. So how we justify this?

2. Jan 4, 2010

LCKurtz

It has to do with continuity implying connectedness. Consider the paramaterized circle:

$$\vec R(t) = \langle \cos t, \sin t,\rangle\, 0\le t \le 2\pi$$

This maps the connected interval $[0,2\pi]$ to the circle, preserving order. Now suppose we take the parameterization:

$$\vec R(t) = \left\{ \begin{array}{rl} \langle \cos t, \sin t\rangle,\ &0\le t \le \pi/2 \\ \langle -\cos t, -\sin t\rangle,\ &\pi/2\le t \le \pi\\ \langle \cos t, \sin t\rangle,\ &\pi\le t \le 3\pi/2\\ \langle -\cos t, -\sin t\rangle,\ &3\pi/2\le t \le 2\pi \end{array} \right.$$

This parameterizes the same circle but the piecewise discontinuities in the components cause the disconnectedness, and hence the order, in the image to be messed up.

3. Jan 4, 2010

Castilla

I know that if a continuous function has a connected domain then it has a connected range.

Applying to my case: the vectorial function f is continuous with [a, c] as interval. Consider a number "b" such that a < b < c.

Imagine a curve in which f(a) comes first, then f(c) and then f(b). But (by the previous theorem) f([a, c]) is connected, and f(b) belongs to it, so there can not be a hole between
f(b) and the curve that goes from f(a) to f(c). Is this explanation okey?

4. Jan 5, 2010

Castilla

Eh, maybe someone can tell me if my proof of the "ordering" of f(a), f(b), f(c) is OK? thanks.

5. Jan 5, 2010

LCKurtz

I think part of the problem here is confusion about the term "before" and "after" when referring to a graph, absent a parameterization of it. Consider the following picture where the same identical graph has been drawn in different orientations:

Does the red point come before or after the green one? My point is, if you are given a parameterization of a curve:

$$\vec R(t) = \langle f(t), g(t), h(t) \rangle,\ a\le t \le b$$

you can define that $\vec R(p)$ precedes $\vec R(q)$ if p < q. But if there is no parameterization there is no "order". In the curve at the upper left you might be tempted to say the green dot precedes the red one, but if you do it is likely because you are implicitly assuming a parameterization in terms of x with the usual positive direction associated. But then what about the lower left one?

I guess what I'm saying is that without parameterization the problem isn't well defined. The graph itself has no defined order.

6. Jan 5, 2010

Castilla

Hi Kurtz. I understand that a "curve" has not implicit order if we consider such curve as a mere assemble of points. But once you have parametrized it, say by a function f, how do we find an order in it? Based on your first hint, I wrote a little proof in my second post. I would like to know if my reasoning was fine, or may be I did not understand what you said.

7. Jan 5, 2010

LCKurtz

See the red above for the definition of how you put an order on the curve given the parameterization of it. Then look at what you wrote that I highlighted blue. The problem is in your notion of "comes first". f(c) can not precede f(b) in the paramaterization because that would require that c < b.

8. Jan 5, 2010

Castilla

So the order in a parametrized curve ("f(p) precedes f(q) if p < q") is just a question of defining it as such? But if it is so, why do we need connectedness to support that order?

9. Jan 5, 2010

LCKurtz

Yes. I don't know how else you would do it. Perhaps that is why such curves are sometimes called trajectories.

You don't. All you need is the parameterization to get the notion of direction or order. Continuity of the functions in the parameterization gives you continuity and connectedness of the curve and avoids the "unnatural" order implied by the circle parameterization I gave earlier.

10. Jan 5, 2010

Castilla

I think I have understood. Only one more question: do things change if we assume that the parametrizacion has derivative in all points and it is different from cero?
I mena, in this case the "order" is a consequence of such conditions?

11. Jan 5, 2010

LCKurtz

No. That condition guarantees a nonzero tangent vector. This prevents the curve from having a sharp corner. Think of an object coming smoothly to a stop then smoothly going in a perpendicular direction. A parametric curve with continuous derivative can do that if R'(t) is allowed to be zero. Consider motion in the plane described by:

$$\vec R(t) = \left\{ \begin{array}{rl} \langle t^2, 0\rangle,\ &t\le 0 \\ \langle 0, t^2\rangle,\ &t > 0 \end{array} \right.$$

This has a continuous derivative which is 0 at (0,0). As t goes from $-\infty$ to $+\infty$ a particle whose motion is described by R(t) comes in along the x axis from the right, executes a sharp corner at (0,0) and goes up the y axis. You don't want to call that a "smooth" curve.

12. Jan 6, 2010

Castilla

Kurtz, here I have try again to prove -without defining it- that if f is a vectorial continuous function with domain [0, 1], f(1/4) precedes f(1/3) and this one precedes f(1/2):

A continuous function with a compact domain makes a compact range. So f([1/4, 1/2]) is compact, so is closed, so every point in it is an acumulation point of it. So f(1/3) is an acumulacion point of f([1/4, 1/2]) and therefore it must be located "inside" the curve that goes from f(1/4) to f(1/2).

Is this okey?

13. Jan 6, 2010

LCKurtz

The notion of one point preceding another on the curve doesn't have anything to do with continuity. It has to do with the parameterization. How do you define it without a parameterization?

Remember that, given a parameterization of a curve:

$$\vec R(t) = \langle f(t), g(t), h(t) \rangle,\ a\le t \le b$$

we define that $\vec R(p)$ precedes $\vec R(q)$ if $$p < q[/itex]. Consider the following two parameterizations of the line x+y=4 in the first quadrant: [tex]\vec R(t) = \langle 4t,4 - 4t \rangle,\ 0\le t \le 1$$

$$\vec S(t) = \langle 4-4t, 4t\rangle,\ 0\le t \le 1$$

By definition R(1/4) = (1,3) precedes R(1/2) =(2,2) for this parameterization.
But S(1/2) = (2,2) precedes S(3/4) = (1,3) for this one.

The parameterization determines the order.

14. Jan 7, 2010

Castilla

Kurtz, thank you for taking time to help me.

15. Jan 11, 2010

Castilla

A question more or less related with these theme. I know that if I have a vectorial continuous function defined on a closed interval, the range will be a continuous curve in some euclidian space. Does the inverse is truth? Does the movement of a fly in this room have necessarily a function the parametrizes that curve?

16. Jan 11, 2010

LCKurtz

In principle, yes. If the curve is rectifiable (fly paths are ) you simply parameterize them in terms of arc length s. So for $0 \le s \le l(C),\ \vec R(s)$ is the point on the curve that is arc length s from R(0). Of course, writing down a formula for it is another question.

17. Jan 11, 2010

Castilla

But we would need some function to be considered as the integrand in the definite integral that defines the arc length. How we obtain such parametrization?
Thanks.

18. Jan 11, 2010

LCKurtz

Just because it may be difficult or impossible to write an "simple" explicit formula for parameterizing a curve in terms of arc length doesn't mean it is not a useful concept. Something like the following makes perfect sense. Let C be a smooth curve with finite arc length l(C). Let $\vec R = \vec R(s)$ be its parameterization in terms of arc length, $0 \le s \le l(C)$.

You will usually see such as that in calculus books where they are discussing velocity, acceleration, and curvature for parametric functions. One example where I have done exactly that is an article about flying an airplane in the wind. If you are interested, look at:

http://math.asu.edu/~kurtz/flywind.html" [Broken]

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19. Jan 11, 2010