This post goes back to high school: the area of the symmetric difference of two -dimensional disks when one center is shifted a little bit. Let’s start with . So we have two intervals and . It is easy to see the symmetric difference is of length .

Then we move to : two disks and of radius and center distance . So the angle satisfies .

The symmetric difference is the union of and , which have the same area: . Note that the limit

.

So .

I didn’t try for . Looks like it will start with a linear term .

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Now let be an ellipse with , and be the tangent vector at . Let be the angle from to .

Let be the arc-length parameter and be the curvature at . Alternatively we have .

The following explains the geometric meaning of curvature:

, or equivalently, . .

Proof. Viewed as functions of , it is easy to see that is equivalent to .

Note that . Taking derivatives with respect to , we get

. Then is equivalent to

, or

. Note that . Therefore , which completes the verification.