Given $S(t) = (x(t), y(t))$, the curvature at any point on S is given by below formula:

$$K = \dfrac{S'(t) \times S''(t)}{|S'(t)|^{3/2}}$$

Where $S'(t)$ is the first order derivative of $S(t)$ and $S''(t)$ is the second order derivative of $S(t).$

  1. I know that the first order derivative gives the tangent vector function to the curve, but How do i interpret the second order derivative of a parametric curve?

  2. In the above formula for curvature, how does more "perpendicular-ness" between $S'(t)$ and $S''(t)$ increases the curvature of the curve?


Here's an intuitive explanation. The second derivative can be interpreted as acceleration. If you decompose acceleration in a tangential and orthogonal part with respect to the curve then the tangential part does not contribute to a change in direction (it only changes speed along the curve). So the curvature –how much you turn the steering wheel– depends only on the orthogonal part and your speed.

  • $\begingroup$ [+1] Good explanation... and this orthogonal part will be at its maximum when S' and S'' are orthogonal $\endgroup$ – Jean Marie Oct 22 '17 at 6:20

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