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If I have a continuous, and smooth curve $\mathcal{C}$, length $\ell$, in $\mathbb{R}^2$ and at each point on the curve I were to draw a line segment, length $d$, normal to the curve centered at the point; would the area covered by all the line segments be $d\cdot\ell$ provided that no two line segments intersect with each other?

Also: if this is true, can this be generalized to more dimensions?

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Sure. Let $d$ or $l$ be $0$. –  Jeremy Dec 18 '13 at 5:00
    
How do you define a normal of a continuous curve? If you curve is smooth, then you can do what you describe, but the area will not be $dl$ in general. –  Andrey Sokolov Dec 18 '13 at 5:10
    
If you draw a line segment at each point, then the only way no two line segments will intersect is if all are parallel, i.e., if $C$ is a straight line. In that one case, the area will indeed be $d\cdot\ell$. –  mjqxxxx Dec 18 '13 at 6:00
    
I am going to add smooth as a criterion. Also, I think that as long as the length d/2 is less than the radius of curvature, no two segments will intersect. –  davik Dec 19 '13 at 0:28
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1 Answer

No. One simple way to see this is to note that there are two choices of normal at any point, one on each side of the curve, and making the two choices won't lead to the same area in general.

To give a concrete example, let $C$ be a circle of radius $r$ (and let $d < r$ if you choose the inward-facing normal). Then the region swept out by the normal lines is an annulus, and you can compute its area for both the inward- and outward-facing normals; you'll see that you don't get the answer $d\ell$ in either case.

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The normal is supposed to be “centered at that point”: I think davik means that the normal of equal length is drawn in both directions to achieve the band. And in that case it's true that the area is $dl$ provided the normals are not too long. –  Michael Hoppe Dec 18 '13 at 10:09
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@MichaelHoppe yes the normal is drawn in both directions (d/2 in both directions so d is the width of the band) –  davik Dec 19 '13 at 0:32
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