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Given the equation $(x - h)^2 + (y - k)^2 = r^2$ representing the family of all circles of radius r at the point $(h,k)$ if we try to form the differential equation representing this family we find an equation of the form

$$\kappa = \frac{1}{r} = \frac{y''}{\sqrt{(1 + y'^2)^3}}$$

which is surprisingly the equation for the curvature of a plane curve (ignoring absolute values which arise in the derivation).

But this expression was derived interpreting $y$ as part of a circle, how in the world can one justify plugging in other functions & saying this represents the curvature of that curve at the point? I know you're trying to say that, locally, the curve moves as though it were moving along the arc of a circle of radius $r$, but there seems to be a jump in my mind as to how you get that interpretation out of what I've written.

bolbteppa
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2 Answers2

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One definition of the curvature of a plane curve at a given point is $\tfrac{1}{\rho}$ where $\rho$ is the radius of the osculating circle to the curve at that point.

Consider a smooth curve in the plane, say $C$, and a fixed point on that curve, say $p$. There are lots of circles tangent to $C$ at $p$. In fact, you find that all of these circle have one important thing in common: all of their centres lie on the normal line to $C$ at $p$. (If you draw the tangent line to $C$ at $p$ then the normal line is the line through $p$ that is perpendicular to this tangent line.)

One of these tangent circles is different to all of the rest: it has higher order contact with $C$ at $p$. All but one of the circles are simply tangent to $C$. They meet $C$ like the the parabola $y=x^2$ meets the $x$-axis at the origin. As far as the first derivative is concerned, the circle and the curve are the same at $p$.

A single circle, called the osculating circle, has higher order contact with $C$ at $p$. Ordinarily, it meets $C$ like $y=x^3$ meets the $x$-axis at the origin. As far as the first two derivatives are concerned, the circle and the curve are the same at $p$.

Although, if $p$ is a vertex ($\kappa \neq 0$ and $\kappa' = 0$) then it meets $C$ like $y=x^4$ meets the $x$-axis at the origin. As far as the first three derivatives are concerned, the circle and the curve are the same at $p$.

(When I say "meets like", I mean up to diffeomorphism, i.e. up to smooth changes of coordinates.)

The radius of this osculating circle, $\rho$, is called the radius of curvature of $C$ at $p$ and $\tfrac{1}{\rho}$ turns out to be $\kappa.$ Interestingly, if $p$ is an ordinary inflection then $\kappa = 0$ (and $\kappa' \neq 0$) and so $\rho = \infty$. The osculating circle is centred at infinity and the circle becomes a line, i.e. the curve meets it tangent line line $y=x^3$ meets the $x$-axis at the origin.

The beauty of working with the osculating circles and not the horrible formula for $\kappa$ is that they are totally independent of coordinates. There is something natural and uncontrived about the contact between circles and curves. Moreover, circles and lines are the orbits of Euclidean transformations.

Fly by Night
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  • Although I do really appreciate your intuition on the subject, it unfortunately doesn't address what I asked. My question is basically about the justification for your very first sentence. My derivation derives the circle of osculation, but I'm asking about why one can feel justified in calling this the curvature of a plane curve in the first place using the thinking in my op as much as possible, let alone be justified in thinking this very derivation results in the osculating circle & not one of the other circles you've mentioned, see what I mean? – bolbteppa Aug 19 '13 at 01:08
  • @bolbteppa: perhaps you should go to the intuitive definition of the osculating circle (circle passing through 3 "infinitesimally" close points on the curve) and formalise it in terms of 3 real points on the curve. You should be able to show that as you bring the two outer points together, the resulting circles converge; would this satisfy you? – Anthony Carapetis Aug 19 '13 at 01:11
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    @bolbteppa My reply does address your question. As I said: in the case of an osculating circle, as far as the first two derivatives are concerned, the circle and the curve are the same. You will get the same value for your expression on the right-hand side. Other than that, I don't see what you're asking. Perhaps you want to know why we define curvature in terms of circles. Again, I eluded to this in my reply. Circle and lines are homogeneous spaces of the action of the Euclidean transformations on the plane. It gets quite heavy, but one consequence is constant differential invariants. – Fly by Night Aug 19 '13 at 01:23
  • Apologies I get what you're saying now, I wasn't paying full attention to your point about order of contact. I was first introduced to curvature by deriving the formula then arbitrarily defining it's reciprocal to be the radius of curvature, then a comment about it relating to a circle. Now following your idea mixed with my derivation I've got a beautiful interpretation for the osculating circle, basically the circle of curvature, however going one step further to call it's reciprocal the curvature is where I get lost, just seems arbitrary in the opposite direction this time. – bolbteppa Aug 19 '13 at 02:03
  • Is there a better way to relate all this to the basic concept of curvature apart from the fact that it's a differential invariant? I mean, if I were to explain to a person on the street that there is a special circle that hugs a curve more closely than any other curve I would be at a loss as to how one could then say that the reciprocal of the radius of this circle is, in fact, the very essence of the idea of curvature they've known all their life. Thanks! – bolbteppa Aug 19 '13 at 02:05
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    @bolbteppa: The "idea of curvature they've known all their life" is not a quantitative thing - we just have a notion that flat planes "are not curved" (have zero curvature) and that locally sphere-like surfaces have higher curvature when they are more tightly curved; i.e. have lower radii of curvature. So any function $\kappa(r)$ decreasing to zero as $r \to \infty$ would fit our intuition; the particular choice $\kappa = 1/r$ is just a convenient quantity to work with. – Anthony Carapetis Aug 19 '13 at 03:25
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    @bolbteppa There are lots of ways. Another way to define curvature is as the angular velocity of the frame ${{\bf T},{\bf N}}$. For a line, this frame doesn't rotate and so $\kappa = 0$. For a circle $\kappa$ is constant. So what does $\kappa = 5$ look like? Well, it's a circle with radius $\tfrac{1}{5}$. That's why we look at contact with circles. If a circle has constant curvature and a curve agrees with a circle up to a certain order then the curvature of the curve is $\kappa = 5$ at that point. Another way is to look at prolonged group actions on jet spaces. Again $\kappa$ drops out. – Fly by Night Aug 19 '13 at 17:56
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    @bolbteppa P.S. The contact idea is very general. In the world of affine differential geometry, where we allow linear transformations and not just rigid motions, affine curvature is defined in terms of contact with conics. This time the osculating conic almost always has order five contact. (More degrees of freedom in the group action.) Conics are also the homogeneous spaces of the group action and so have constant differential invariants with respect to the action of the affine group. – Fly by Night Aug 19 '13 at 17:59
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Try

$$ \kappa = \frac{1}{r} = \frac{ \dot{y} \ddot{x} - \ddot{y} \dot{x} }{ \left( \dot{x}^2 + \dot{y}^2 \right)^\frac{3}{2} } $$

John Alexiou
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    This is merely the definition of the curvature of a plane curve defined parametrically, how does that address my question? – bolbteppa Aug 19 '13 at 01:09