Figuring out the curvature of a line?

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SUMMARY

The curvature of a line in three-dimensional space is definitively zero, as a line is the epitome of straightness. The discussion outlines the parametric vector form of a line, represented as r = , and confirms that the curvature formula, κ = |r' × r''| / |r'|³, yields a result of zero when calculated. The first derivative r' = and the second derivative r'' = <0, 0, 0> lead to the conclusion that the curvature is indeed zero, validating the initial hypothesis.

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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!
 
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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

\kappa =\frac{\left|\mathbf{r}&#039;\times\mathbf{r}&#039;&#039;\right|}{\left|\mathbf{r}&#039;\right|^{3}}
 
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.
 

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