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$$k = \frac{y''x' - x''y'}{(x'^{2}+y'^{2})^{\frac{3}{2}}}$$

Are the curvature value of a straight line zero by using this formula?How to prove it?

OGC
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jack fukey
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    Write down the equation of a straight line and plug the expressions in the equation of the curvature ! –  Sep 16 '15 at 06:41

1 Answers1

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Given a straight line $\mathcal{D}$, any point $(x,y)$ of $\mathcal{D}$ must satisfy a relation as the following:

$$ ax + by = c.$$

Thus we have $$ ax' + by' = 0, \qquad ax'' + by'' = 0.$$ with $a, b$ not both zero.

Suppose that $a \neq 0$, then $$ 0 = ax'y'' + by'y'' = ax'y'' - ax''y',$$ which implies $$ 0 = x'y'' - x''y'.$$

So the numerator of the curvature is always $0$.

corindo
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  • You have yet to justify that the denominator is non-zero – Vim Sep 16 '15 at 06:51
  • I don't think the curvature has meaning when the denominator reaches $0$. – corindo Sep 16 '15 at 06:54
  • Thank you very much! But i have used a little differernt formula, which include a scale paramater. The curvature value of a straight line is not zero.I don't why.http://math.stackexchange.com/questions/1437542/invariance-of-curvature-in-curvature-scale-space – jack fukey Sep 16 '15 at 07:15