A little debate arose during discussion today in math, and our GSI sent us an email giving a rather categorical statement that T X N can't change sign for a plane curve.

What this is is essentially differential geometry, though we're discussion it as an extension to multivariable calc, so if this looks like Greek to you (which a letter or two is ), just don't bother thinking about it.


The vectors T, N, and B here are the unit tangent, unit normal, and unit binormal vectors as used by the Frenet-Serret Formulas.

One of our book's exercises was to prove that for any plane curve, the torsion is always zero. One of us noted, however, that as B can seemingly change direction with a plane curve, the torsion, proportional to dB/ds (dB/dt), would be undefined at that point. Our GSI said that T and N can never change orientation, and I sent the following response. I know there are a few math people out there.. thoughts?

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Not to be obstinate, but why can't the orientation of T and N change?

For example, if I had the plane curve r(t) = <t, t³, 0>...

r'(t) = <1, 3t², 0>
T(t) = <1 / √(9t^4 + 1), 3t² / √(9t^4 + 1), 0>
T'(t) = <-18t³(9t^4 + 1)^(-3/2), 6t/√(9t^4 + 1) - 54t^5(9t^4 + 1)^(-3/2), 0>
N(t) = <-18t³(9t^4 + 1)^(-3/2), 6t/√(9t^4 + 1) - 54t^5(9t^4 + 1)^(-3/2), 0> / |T'(t)|

at t = -1, the curve moves to the "right" if you are the particle, looking in the direction of increasing t.

T(-1) = <1 / √(9(-1)^4 + 1), 3(-1)² / √(9(-1)^4 + 1), 0> = <1/√(10), 3/√(10), 0>
N(-1) = <-18(-1)³(9(-1)^4 + 1)^(-3/2), 6(-1)/√(9(-1)^4 + 1) - 54(-1)^5(9(-1)^4 + 1)^(-3/2), 0>
=<18(10)^(-3/2), -6/√(10) + 54(10)^(-3/2), 0> ~= <0.54, -17, 0> / |T'(t)|

T is +x,+y; N is +x, -y; T X N is in negative-z direction

at t = 1, the curve moves to the "left" if you are the particle

T(1) = <1 / √(9(1)^4 + 1), 3(1)² / √(9(1)^4 + 1), 0> = <1/√(10), 3/√(10), 0>
N(1) = <-18(1)³(9(1)^4 + 1)^(-3/2), 6(1)/√(9(1)^4 + 1) - 54(1)^5(9(1)^4 + 1)^(-3/2), 0>
=<-18(10)^(-3/2), 6/√(10) - 54(10)^(-3/2), 0> ~= <-0.54, 17, 0> / |T'(t)|

T is +x,+y; N is -x, +y; T X N is in positive-z direction

The "switch" happens at t=0, where T'(0) = <0, 0, 0>, and therefore N(0) is indeterminate, so neither B(0) nor τ(0) exists.

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Note, a couple of us are of the opinion that since the whole foundation of the theorem assumes that r' is not 0, then its rather irrelevant what undefined points you get when that happens.

You answered your own question right here:


Post has been edited 1 time(s), last time on Feb 23 2008, 9:00 am by cheeze.

Well, technically T and N shift their orientation with respect to each other, since B = <0,0,-1> for t<0, and B = <0,0,1> for t>0, so my point was technically τ(0) is undefined as opposed to zero, but it is indeed sort of irrelevant.

We have issues sometimes communicating mathematically with our GSI

If I understand your question correctly, they don't change orientation because then the line would be heading in the opposite direction. Because the vector points in a certain direction, the tangent, normal, and crossproduct all have to point correspondingly.

I don't really see why you say B can change though, as it is found via the other two.

If N changes direction (T should stay in the same direction), then B does also. If the curve shifts from going "left" to "right", such as the behavior of <t,t^3,0> around t=0, then the direction of B changes from up to down (by the right hand rule).

Oh right, that should have been obvious, (for any inflection point) I'm an idiot.
And in that case I agree with Cheeze.

Post has been edited 1 time(s), last time on Feb 23 2008, 8:29 am by Rantent.

????????????????????????
I won't try to understand till I get older...

What is a GSI?
Other then that, I only understand fractions of what you have posted...
I honestly do not understand it, but then again, im only 12.
I think I may be able to answer your question (or did CheeZe and Rantent already answer it).
Please just list a group of numbers to scale with this equation (eg, Scale 1:3, X=1 --- X=3)

Graduate Student Instructor Fancy title for paid TAs

I essentially made her admit that its irrelevant since the whole orientation of the vectors and the theorem predicates itself on the fact that r'(c) ≠ 0 and r''(c) ≠ 0. She was still wrong, though.

How is she wrong? You're trying to give an example that does not fit the definition of a plane curve (in this context, that is, it's defined for all points) and, obviously, would not fit within the theorem anyway.

Uhh, in the case of <t,t^3,0>, B = <0,0,-1> for t < 0, and <0,0,1> for t > 0. Therefore, B changes.

Note: I agree that τ(t) = 0 for all t (essentially), but she also made the categorical statement that "N and T can't change their relative alignment", but I think it's pretty obvious that T X N changes.

Is DTBK gonna be a prof some day?

This post was strategically placed by DTBK to try to prove he's smart.

Professors are becoming more and more prevalent, pretty soon everyone is going to need that much schooling to get a decent job. Like it is with bachelors degrees now. Back 50 years or so, all you needed was a high school degree, and you could get a good job. (Now that gets you a lowly position in the local mega-mart.)

You really need a Doctorate only for education jobs, usually.

The sad thing is people often go on to get a generic major just so they can have a college degree.

Doctorate is still pretty high. These days, I've heard of the new Consultant or something like that. Those guys lecture you for 4 hours and get paid MILLIONS.

Wow.

Now I'm decent at maths... But, wow.

I'll have to read up on my formulas next time...

Same...

Dudes, this is INSANELY complex.

If you know calculus, then you should be able to understand what he was doing. When you type any mathematical formulas through text, it always looks more complicated. It's just differentiating

Hey, guys, I can do math too.


4rjf83j02093j929j0f3j90f3290j3fj90f3339f r893298328932893289


Or am I doing it wrong..?