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We know that $\langle \dot{c},\dot{c} \rangle=1(\because$ parameter used here is the arc length). differenting with respect to $t$. We get $\langle \ddot{c},\dot{c} \rangle=0\implies \dot c \perp \ddot c \implies \forall t\in I, $ we get a real value for that. since $\ddot c || n$.

Q1. Why does $\ddot c$ always pointing inwards in the given figure?

Q2. How does the curvature measure of how much a curve deviates from a straight line?

I know that if $\kappa=0 \implies \ddot c=0 \implies c(t)=at+b$. Which represents a straight line. I really don't understand for other curves.

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It points inward because that is the direction that the tangent to $c$ is bending towards as we move along the curve. Since $\ddot{c}\approx\frac1{\Delta t}\left(\dot{c}(t+\Delta t)-\dot{c}(t)\right)$ it is a rescaled difference between two nearby tangents, and indicates the direction of their bending.

The curvature measures deviation from the straight line for the same reason. It is the magnitude of $\ddot{c}$, i.e. the measure of how fast the tangents bend, regardless of which direction the bending takes. Since the parameter is the arclength the speed with which we move along the curve is always constant, so the only thing that contributes is how fast the direction of the tangent is changing due to the curve's shape.

It is zero over an interval when no bending takes place at all, i.e. when we have a straight line segment. It can also be zero at inflection points, like $c(t_2)$ in the middle of the picture, where the curve "straightens out". There is no bending to the first order of approximation at that point, only higher order terms contribute to it. If the curvature stays small over an interval the curve can not bend much, it waves only very mildly, like a circle of a very large radius. By the way, the curvature is the reciprocal of the radius of a circle that can be most closely fit to the curve at the point, it is called the osculating ("kissing") circle. And the reciprocal of $R=\frac1\kappa$ is called the radius of curvature.

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  • $\begingroup$ division of vectors are not defined. How to define $\kappa(t)$? $\endgroup$ – Unknown x Jul 7 at 11:20
  • $\begingroup$ Ussually, $\kappa(t)=||\ddot c||$ in most of the text book. $\endgroup$ – Unknown x Jul 7 at 11:21
  • $\begingroup$ @Unknownx Division is defined when the vectors are collinear. With $\kappa(t)=||\ddot c||$ we assume that the curvature is always positive, which is the more standard convention. Then $n$ will always point inward. Some authors, like the one you are quoting, make $n$ always face the same direction relative to $\dot{c}$, so that they form a right pair. In that case, curvature will change sign at inflection points, and negative sign corresponds to $\dot{c}$ and $n$ facing in opposite directions, as at $c(t_1)$ in your picture. $\endgroup$ – Conifold Jul 8 at 17:59

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