Figuring out the curvature of a line?

  • Thread starter Thread starter mr_coffee
  • Start date Start date
  • Tags Tags
    Curvature Line
Click For Summary

Homework Help Overview

The discussion revolves around understanding the curvature of a line, particularly in three-dimensional space. The original poster is tasked with hypothesizing about the curvature and expressing a line in parametric vector form to analyze its curvature mathematically.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of curvature and its implications for a straight line, questioning whether the curvature can be considered zero. There are attempts to express the line in parametric form and to apply curvature formulas, with some participants expressing confusion about the next steps.

Discussion Status

Some participants have provided guidance on the mathematical approach to finding curvature, while others have raised questions about the necessity of showing work despite the intuitive understanding that the curvature of a line is zero. The discussion reflects a mix of interpretations and attempts to clarify the problem.

Contextual Notes

There is mention of specific equations and formulas that participants are encouraged to use, as well as a note that the original poster is expected to demonstrate their understanding through calculations, despite the straightforward nature of the problem.

mr_coffee
Messages
1,613
Reaction score
1
Hello everyone, I'm trying to figure out the curvature of a line, First I'm suppose to make a hypothisis on what i think it would be, then I'm suppose to put a line in parametric vector form and find out really what the curvature of the line is. Well a line is pretty straight, so why can't I say the curvature is 0? or very close to 0? Well i got the line in parametric form passing through point (xo,yo,zo):
x = xo + at
y = yo + bt
z = zo + ct
So a line in 3 dimensions pass through (x1,y1) and (x2, y2) has parametric vector equation:
x = x1 + (x2-x1)t

Did i do this part right? Now I'm confused on what I'm suppose to do! Any help would be great!
 
Physics news on Phys.org
mr_coffee said:
Hello everyone, I'm trying to figure out the curvature of a line, First I'm suppose to make a hypothisis on what i think it would be, then I'm suppose to put a line in parametric vector form and find out really what the curvature of the line is. Well a line is pretty straight, so why can't I say the curvature is 0? or very close to 0? Well i got the line in parametric form passing through point (xo,yo,zo):
x = xo + at
y = yo + bt
z = zo + ct
So a line in 3 dimensions pass through (x1,y1) and (x2, y2) has parametric vector equation:
x = x1 + (x2-x1)t

Did i do this part right? Now I'm confused on what I'm suppose to do! Any help would be great!

The curvature of a line? Think about it. What does curvature mean geometrically? What does a line look like? If you understand the concept of curvature you shouldn't need to do any work at all on this problem :smile:
 
well the The curvature is the measure of its deviation from the straightness. So of course its going to be 0 for a line. But he wants us to show what the curvature is, using formula's, so I can't just not show any work.
 
He told us that, if we use the "right" equation, everything will just fall apart?
 
mr_coffee said:
He told us that, if we use the "right" equation, everything will just fall apart?

Use r=<x0+at, y0+bt, z0+ct>

and

[tex]\kappa =\frac{\left|\mathbf{r}'\times\mathbf{r}''\right|}{\left|\mathbf{r}'\right|^{3}}[/tex]
 
Awesome thank u! I ended up with this:


r' = <a,b,c>
r'' = <0,0,0>

|<a,b,c> x <0,0,0>| = 0; so that proves its 0 correct?
 
mr_coffee said:
Awesome thank u! I ended up with this:


r' = <a,b,c>
r'' = <0,0,0>

|<a,b,c> x <0,0,0>| = 0; so that proves its 0 correct?
It certainly does.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
3K
Replies
7
Views
2K
Replies
17
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K