[quote=MillenniumArmy]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 :P[/quote] [color=#b5b5ff]Yeah, it was just a pretty elementary dispute that I think we settled by her admitting that [b]T[/b] X [b]N[/b] can change sign and that we were really talking about a plane curve with κ>0 for all t.[/color]

Are you honestly trying to create a math discussion or are you just boasting about how you can do high level maths (like you seem to do a lot)?

Are you honestly trying to create a math discussion or are you just boasting about how you can do high level maths (like you seem to do a lot)?

Well it was a math discussion, but it only lasted a few posts. Then just about every post after that was not (thus off topic :P)

Well it was a math discussion, but it only lasted a few posts. Then just about every post after that was not (thus off topic )

Pt1 (3,6) Pt2 (1,4) Work out the graph formula. [Straight Line Graph] A lot of you probably won't remember how to do it because of how long it's been since you've [probably] last done this. Take a crack at it. Cookie for the person who figures it out showing his working here without sketching the graph itself.

Pt1 (3,6) Pt2 (1,4)
Work out the graph formula.
[Straight Line Graph]

A lot of you probably won't remember how to do it because of how long it's been since you've [probably] last done this.

Take a crack at it.

Cookie for the person who figures it out showing his working here without sketching the graph itself.

[color=#b5b5ff]Do you mean y - 4 = x - 1; y = x + 3? But yeah, there actually was a discussion that was over ages ago. This topic should probably be locked :P[/color]

Now I'm no calculus person, but isn't by definition a planar curve a curve with no torsion? Well... Unless it has some like undefined points, right?

Now I'm no calculus person, but isn't by definition a planar curve a curve with no torsion? Well... Unless it has some like undefined points, right?

[color=#b5b5ff]Assumedly not, since we were trying to prove that plane curves have no torsion. I think the formal definition of a plane curve states that for any plane curve [b]r[/b](t) = <x[sub]0[/sub](t),y[sub]0[/sub](t),z[sub]0[/sub](t)> there exists some rotational matrix A such that A([b]r[/b](t)) = <x(t),y(t),0>.[/color]

Curves are one dimensional, Planes are two dimensional. So a curve in a plane may have a torque along whatever plane it is in. (And out of the plane depending of whether or not the plane is Euclidean or not, I believe. Don't quote me on that though.)

Curves are one dimensional, Planes are two dimensional. So a curve in a plane may have a torque along whatever plane it is in. (And out of the plane depending of whether or not the plane is Euclidean or not, I believe. Don't quote me on that though.)

[quote=Rantent]Curves are one dimensional, Planes are two dimensional. So a curve in a plane may have a torque along whatever plane it is in. (And out of the plane depending of whether or not the plane is Euclidean or not, I believe. Don't quote me on that though.)[/quote] But isn't a planar curve just a one dimensional curve extended into 2 dimensions, and therefore lacking any torsion?

Quote from Rantent

Curves are one dimensional, Planes are two dimensional. So a curve in a plane may have a torque along whatever plane it is in. (And out of the plane depending of whether or not the plane is Euclidean or not, I believe. Don't quote me on that though.)

But isn't a planar curve just a one dimensional curve extended into 2 dimensions, and therefore lacking any torsion?