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If we are given a simple closed curve with length $L$ in the plane, and we have a fixed number $r$ such that for each point $x$ on the curve there is a related disc $D(x,r)$ with radius $r$, how can we estimate the area of the region formed by all such discs from above?

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An easy, and in many cases accurate approximation would of course just be L * 2r. –  KSab Apr 8 at 0:55
    
Could you provide more details? Because I know in many cases L*2r is the exact formula, but I want to know a universal estimate and I don't know how to prove it. Thanks. –  Danqing He Apr 8 at 2:38
    
Well I think that the more uniform the curvature the closer this estimate should be. Now that I think about it this could probably be done accurately by integrating over the curve based upon the radii of curvature. Let me think on this and I may post an answer. Out of curiosity how is the curve represented? –  KSab Apr 8 at 2:48
    
Thank you very much for helping me to think about it. Well, there is no specific curve in mind and I am just curious about such a problem because it looks like a classical problem but I cannot find its answer easily. –  Danqing He Apr 8 at 3:03
    
For regular enough closed curve and sufficiently small $r$, the $2rL$ is not an approximation, it is exact! This is a special case of the Weyl's tube formula. I believe the most comprehensive reference on this subject is Alfred Gray's book Tubes. –  achille hui Apr 8 at 4:23

1 Answer 1

Using the following approach it seems that the general answer is indeed just $2rL$.

curve diagram

Assuming the curve has no sharp points, this problem is equivalent to the sum line segments at each point $x_0$ along the curve, each centered at $x_0$, perpendicular to the curve at that point, and of length $2r$. Observing the diagram above, it can be seen that for any radius of the circle not perpendicular to the curve, such as the green one, there must be another point, $x_1$, along the curve closer to its position on the circumference. This is true because obviously any close point along tangential line (the dark red line) in the correct direction will be closer, and since the curve is continuous, in moving closer toward $x_0$ you can get a point $x_1$ arbitrarily close to the tangential line. Another way of thinking about it is to think about $x_1$ as infinitesimally farther down the curve than $x_0$, and since the measure is infinitesimal, the slope is unchanging and it can be thought of as moving down the tangent line.

curve diagram

If we let $R_t$ represent the radius of curvature at a distance $t$ along the curve, we can then do the following.

$$ \begin{align} A & = \frac{\mathrm{d}\theta}{2\pi} \left( \pi(R_t+r)^2 - \pi(R_t-r)^2 \right)\\[2ex] A & = \frac{\mathrm{d}\theta}{2\pi} \left(4\pi rR_t\right)\\[2ex] A & = 2rR_t \:\mathrm{d}\theta \end{align}$$

and since $\mathrm{d}\theta = \mathrm{d}t / R_t$

$$ A = 2r \:\mathrm{d}t\\[2ex] \int_0^L 2r \:\mathrm{d}t = 2rL $$

I realize that this does not hold if the curve has any sharp points. However for any sharp points you can manually add the uncounted circle section ($\frac{\theta}{2}r^2$) as well as subtracting all the overlapping area ($r^2\cot(\frac{\theta}{2})$) for a final answer of

$$2rL + r^2 \sum_{\theta_i} \frac{\theta_i}{2} - \cot\left(\frac{\theta_i}{2}\right) $$

where $\theta_i$ is the angle change at the $i^{th}$ sharp point. The only circumstance I can see in which this would not work is if two points on the curve ever came closer than 2r away from each other and you were counting some area twice.

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Thanks for this answer. This is a very nice situation, but we don't have any restriction about $r$ and $L$. –  Danqing He Apr 8 at 5:01
    
I'm not sure I understand your comment. The diagram given was just a small (technically infinitesimal) section of a potentially more complicated curve with varying curvature. –  KSab Apr 8 at 5:06
    
Yes, you're right. It looks like I misunderstood you. I will rethink about it. Thank you very much. –  Danqing He Apr 8 at 5:10
    
I think this essentially answers the following question: if for any point $x$ we have a related "normal" segment with length $2r$ whose midpoint is $x$, then the resulting region $R$ has the area less than $2rL$. But I am not sure if my question is equivalent to this one. In other words, is every disc $D(x,r)$ contained in $R$? This may not be true if $r$ is large (comparing to your $R_t$). I believe this ares can still be controlled from above by $2rL$, but I don't know how to prove it strictly. –  Danqing He Apr 8 at 5:28
    
You are correct in that that is what my answer shows, which I do believe is equivalent to your question. I'm sorry if I made the jump without explaining myself and I'll edit my post in a minute. –  KSab Apr 8 at 5:41

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