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I played around with computing the curvature of some curves, and found this weird example that is driving me nuts.

Consider the following (Bézier) curve (on a plane, the first point is $[-1,0]$): $$bez(t) = [-1, 0] (1 - t)^3 + 3 [-1, 0] (1 - t)^2 t + 3 [1, 1] (1 - t) t^2 + [1, 0] t^3$$

If I plot this, the curve looks virtually straight near $t=0$, hence I would expect a low curvature. But if I actually plot the curvature (which I've computed in at least four different ways just to be absolutely certain), it's enormous when getting to the tip.

Now, in part, this does not surprise me. I explicitly constructed the curve to have a cusp at $t=0$. So for $t=0$ I would expect the curvature to be infinite. As it's all (mostly) continuous, I guess I should not be surprised that near $t=0$ the curvature is already going up. And of course, as $t$ is nearing zero, the derivative of arclength w.r.t. $t$ is also dropping to zero, so when graphing curvature as a function of $t$, the singularity gets blown up.

Still, the curve looks awfully straight when actually plotting it, and I do not think any sane person who is just shown the curve (for $t\in[0,\varepsilon]$, with $\varepsilon$ very small) would guess that it had huge curvature. I also tried making the endpoint match $[1,1]$ (so you get a straight line segment for $t\in[0,1]$), and this does in fact have zero curvature (even the endpoints have zero curvature in the limit as $t$ approaches them). So why does my (and probably others') intuition not match the objective result here?

For reference, one of the ways I computed the curvature is as follows (where $[bez'(t), bez''(t)]$ is a $2\times2$ matrix): $$curvature(t)=\det([bez'(t), bez''(t)])/\|bez'(t)\|^3$$

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A sane person who is just shown a smooth curve of length $O(\varepsilon)$ will have a very hard time telling it apart from a straight line! The more you zoom in, the harder it is to see the curvature.

Not only does your curve have a cusp at $t=0$, its slope is not constant as it approaches the cusp. Try plotting the slope, $m(t) = y'(t)/x'(t) = (2-3t)/(4-4t)$, against the $x$-coordinate, $x(t)$, over a region close to $t=0$.

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I removed my earlier comments as Rahul has edited the post to incorporate it. This animation should still be helpful to the OP. –  Willie Wong Oct 9 '12 at 10:22
    
Yes, you're right about the zooming part (should've thought of that). The cusp isn't that important though (it's perfectly possible to have a cusp and still have zero curvature everywhere else). Still, you're right that the problem is in having a high curvature over a very small part of the curve, thus leading to very little change in slope. –  Jasper Oct 16 '12 at 11:48

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