Radius of curvature is defined as the radius of a circle that has a section that follows/approximates a function/curve over some interval. Now, it's easy to Google pictures of curves that have osculating circles drawn in and it seems obvious that this is a clever way of defining curvature of a function. But do all curves follow a part of a circle close up? I mean, many curves do look round at some points and it seems reasonable to assume you could fit the curve to a circumference a circle, but is this always true? If so, why are circles so special and can we prove that you can fit a circle into a curve? If not, what are the requirements that a circle fits?
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@GEdgar this should be an answer, not a comment – Eddy Oct 22 '18 at 14:09
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"All curves"? No.
Indeed, non-differentiable curves do not even have a tangent line.
A requirement for osculating circle, will be that the first and second derivatives exist and are non-zero. For the osculating circle, we use the circle with the same values of those two derivatives at that point.
Of course if the second derivative is $0$, then the osculating circle is actually a line.
GEdgar
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Allright, existence of first and second derivatives does rule out many functions. But still, even if both derivatives exist, does that necessarily mean the function/curve resembles a circle close up? – S. Rotos Oct 22 '18 at 14:21
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It resembles a circle close up in the sense that there is a unique circle that matches those two derivatives. That is the unique circle that has "contact of order two" at the point. – GEdgar Oct 22 '18 at 14:30
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And note that if it has an oscillating circle than close up it will resemble a straight line since all circles look like straight lines if you look at a small enough part of them. – Math Man Oct 22 '18 at 20:47