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Let $S$ be the area (if finite) of the region bounded by a plane curve and a straight line parallel to the tangent line to the curve at a point on the curve, and at a distance $h$ from it. Express $\lim_{h \rightarrow 0} (S^2/h^3)$ in terms of the curvature of the curve.

The first idea occur to me about this problem is that at a given point on the curve we can define an osculating circle, and that the area bounded by the circle and the straight line will tend to $S$ as $h \rightarrow 0$. The latter statement, however, I am not really sure how to prove or even whether it's true.

Based on the assumption that such claim is valid, one can calculate the wanted limit by following expressions (see the following figure) and L'Hospital's rule \begin{gather} h = R(1 - \cos(\theta/2)) \\ S \approx \frac{R}{2}(\theta - \sin(\theta)), \end{gather} where $R = 1/\kappa$, $\kappa$ is the curvature of the curve at the point, and $h(\theta) \rightarrow 0$ as $\theta \rightarrow 0$.

enter image description here

For the justification I owed to everyone:

Reorient the curve to center the given point at the origin of new coordinate system, such that its tangent vector is point at the $+x$-direction and the normal vector the $+y$-dircection. (See the diagram.) Let the straight line be $y = h$ with $0 \leq h \leq R = 1/\kappa$. Locally we can parametrise the curve by its $x$-coordinate, i.e. $\mathbf{r}(x) = (x, f(x))$, for some interval of $x$ containing the origin, such that $f(x)$ is a smooth function. Note that we have (due to our orientation of curve) $f(0) = 0, f'(0) = 0$ and $ f''(0) = \kappa$. Hence we have the Taylor expansion $f(x) = \frac{1}{2}\kappa x^2 + C x^3 + \dots$, for some constant $C$.

For $|x| \leq R$, we have the equation for the circle $g(x) = R - \sqrt{R^2 - x^2}$. Expanded by power series $g(x) = \frac{1}{2R} x^2 + D x^3 + \dots$, where $D$ is some constant. If $y = h$ is close enough to our $x$-axis, by inverse function theorem, we shall find smooth functions $a(h)$ and $b(h)$ such that $f(a(h)) = h = g(b(h))$, and that $a, b \rightarrow 0$ as $h \rightarrow 0$.

Now we can approximate the difference between the area $S'$ bounded by g(x) and $y =h $ and $S$: \begin{multline} |S'-S| = 2\left|\int_{0}^{a(h)} g(x) - f(x) dx + \int_{a(h)}^{b(h)} h - f(x) dx\right| \\ \leq 2\left| \int_{0}^a Ex^3 + \dots dx \right| + 2|h||b(h) - a(h)| \\ \leq 2\left[ Fa^4 + O(a^5) \right] + 2|h||b(h)-a(h)|, \end{multline} where $E ,F$ are constants. This will tend to zero if we let $h$ shrink to zero, which completes my argument.

Kelvin Hsu
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  • What curve is it? sounds like the cycloid. – Narasimham Oct 11 '16 at 04:57
  • @Narasimham The type of the curve is not actually given. I think the result of this problem should be valid for arbitrary smooth curve. – Kelvin Hsu Oct 11 '16 at 16:26
  • Yes, please put $R^2$ on the second line. But you need to justify that the error between the area $S$ and the area of the segment of the osculating disk is $o(h^{3/2})$. It's actually more work to calculate that than to follow the approach I suggested. But I've checked it and that error is $O(h^{5/2})$, so you win. :) ... Also, please add to your discussion what $\theta$ actually is!! – Ted Shifrin Oct 11 '16 at 20:22
  • Nice result for any 2D curve in differential geometry. Hope one of us includes a sketch... suspect it is the situation around sharp cusp of the curve. – Narasimham Oct 11 '16 at 21:33
  • I have no idea what you're talking about @Narasimham. He's talking about a generic point on a smooth curve. $\theta$ is not a parameter for a curve. It's the angle subtended at the center of the osculating circle by the line at distance $h$. – Ted Shifrin Oct 11 '16 at 21:35
  • Dont know how he got the equations. I drew a pic, given below but is not full correct. – Narasimham Oct 11 '16 at 23:09

2 Answers2

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HINT: Reorient the picture so that the tangent line at the given point is the $x$-axis and write the curve in the form $y=\frac12\kappa x^2+\text{higher order terms}$. You will see that the higher order terms disappear in the desired limit.

Ted Shifrin
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  • Actually I have known the technique that first reorient the curve and expand it by power series, which leads one to find out that the graph of the curve will be $f(x) = \kappa x^2/2 + \dots$. However, I've encountered some calculation problems while facing those Big-O's. That's why I am curious to try this argument of approximation by the osculating circle. (For which I have developed a reasoning that explain why the area bounded by the circle and the straight line $S'$ will tend to $S$ as $h \rightarrow 0$. I will post it tomorrow. Would you please to check if there are any problems?) – Kelvin Hsu Oct 11 '16 at 16:35
  • @Kelvin: Unless I'm missing something, the argument that you can ignore the error in using the osculating disk is far more involved than the error analysis in the computation I suggested. But I've done it and you're correct. You of course need to know that the error between $S$ and $S'$ is $o(h^{3/2})$; just saying $S'\to S$ as $h\to 0$ will not do. – Ted Shifrin Oct 11 '16 at 20:24
  • I have added my justification, but maybe with some flaws. Since I currently have other things to do, the diagrams, in which things such as $\theta$ are defined, will be left until I am back home tonight. Thanks!! – Kelvin Hsu Oct 12 '16 at 01:33
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puttng in $R^2$ as coefficient for area $S$,

\begin{gather} h \approx R(1 - \cos(\theta/2)) \approx R \,\theta^2 /8 \\ S \approx \frac{R^2}{2}(\theta - \sin(\theta))\approx R^2 \,\theta^3 /12 \\ R = \frac{9 }{32} \frac{S^2}{h^3} \end{gather}

OP's sketch?

$h,S$ for the above, does not tally with given relations.

$$ h \approx R(1 - \cos \theta) $$ $$ S \approx \frac{R^2}{2}(\theta - \sin\theta\,\cos \theta) $$

Narasimham
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