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Let $\gamma(s)$ be a unit speed planar curve with $\kappa(s)$ as its curvature. Now what is $\int \frac{1}{\kappa}ds$ and geometrically which things it represents?

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  • $\begingroup$ So $\frac{1}{\kappa}$ by itself is the "radius of curvature". And $\int\;ds$ is arc length. So it seems the quantity $\int \frac{1}{\kappa}\;ds$ has units of area. But the area of what? $\endgroup$ Nov 15, 2021 at 9:42
  • $\begingroup$ Which area it represents? $\endgroup$
    – MAS
    Nov 15, 2021 at 10:18
  • $\begingroup$ Why do you consider this integral? Does it occur from some problem solving? $\endgroup$
    – Zerox
    Nov 15, 2021 at 14:49
  • $\begingroup$ Yes I am trying to solve some problem and stuck here. $\endgroup$
    – MAS
    Nov 15, 2021 at 15:02
  • $\begingroup$ @pinaki - i think the area you are trying to calculate is proportional to the area bounded between a curve and its evolute. Take for example the case of a circle with its evolute being a point - than this area is $2\pi R^2$, twice its area (which is also the area bounded between the circle and its evolute, that is its center). $\endgroup$
    – user2554
    Nov 15, 2021 at 16:06

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A more precise formulation is that it is associated with the area traced out by the segment from a point on the curve to the corresponding centre of curvature as we move along the curve. Even more precisely, it is the area of the image of the mapping $$ \phi(s,t)=\gamma(s)+\frac t{\kappa(s)}\bf{N}(s)$$ where $t$ ranges in the unit interval, $s$ over the required section of the curve.

This can be computed using elementary two dimensional calculus. Note that this allows for changes of sign in the curvature, if we use signed areas.

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  • $\begingroup$ There is a correction factor missing - see @user2554 's comment above. $\endgroup$ Nov 21, 2021 at 20:48
  • $\begingroup$ Yes, $1/2$—comes from $\int_0^1 tdt$. That‘s why I wrote „associated with“ rather than „is“. $\endgroup$
    – hordubal
    Nov 22, 2021 at 11:51

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