# Figuring out the curvature of a line?

1. Oct 3, 2005

### mr_coffee

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!

2. Oct 3, 2005

### amcavoy

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

3. Oct 3, 2005

### mr_coffee

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 forumla's, so I can't just not show any work.

4. Oct 3, 2005

### mr_coffee

He told us that, if we use the "right" equation, everything will just fall apart?

5. Oct 3, 2005

### amcavoy

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

and

$$\kappa =\frac{\left|\mathbf{r}'\times\mathbf{r}''\right|}{\left|\mathbf{r}'\right|^{3}}$$

6. Oct 3, 2005

### mr_coffee

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?

7. Oct 3, 2005

### amcavoy

It certainly does.