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$$\dfrac{dT}{ds} = \kappa N$$

$\kappa$ = curvature

$T$ = unit tangent vector

$N$ =unit normal vector

How can i show this equation? I don't know why direction of $\dfrac{dT}{ds}$ is direction of $N$.

Steven Alexis Gregory
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Dobu
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  • Try looking up the definitions of $T, N$ and $B$. – Steven Alexis Gregory Jul 03 '17 at 05:14
  • Since, by definition, $T$ is a unit vector $T \circ T = 1$, Taking the derivative of both sides wrt $s$, we get $2T \circ \dfrac{dT}{ds} = 0$. That is, $\dfrac{dT}{ds} \perp T$. We define $N$ to be the unit vector parallel to $\dfrac{dT}{ds} $ and we define the curvature, $\kappa$ to be $\left| \dfrac{dT}{ds} \right|$. – Steven Alexis Gregory Jul 03 '17 at 05:21

1 Answers1

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That $dT/ds$ is parallel to $N$ is true by definition. $N$ is the unit vector with the same direction of $dT/ds$, and $\kappa$ is its scaling factor. Intuitively, within infinitesimal time, the particle is undergoing a circular motion, and it is easy to see the derivative of the tangent line must be a vector pointing to the center of circle.

No offense, but you may want to learn to Google. Even Wikipedia is providing more information than you need to know in this case: Curvature, first section.

Violapterin
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