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I am reading 'Differential Geometry of curves and surfaces' by Banchoff and Lovett and I'm confused about the following statement on page 28:

"Let two curves $C_1$ and $C_2$ have a regular point $P$ in common. Given a point $A$ on $C_1$ near $P$, let $D_A$ be the orthogonal projection of $A$ onto $C_2$, i.e. the point on $C_2$ closest to $A$. (...)"

If we were talking about lines I would understand but these are more general, continuous (but not necessarily differentiable) plane curves. I understand orthogonal projection from a point $p$ on $C_1$ as projection along the line orthogonal to the tangent vector at $p$. But this doesn't have to be the shortest distance to $C_2$ it seems? Unless we are assuming something like the curve becoming a line in the limit of being very close to $P$?

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Are the curves planar? –  Olivier Bégassat Nov 3 '12 at 17:29
    
Yes they are, I will add it in the question. Thanks! –  jorgen Nov 3 '12 at 17:36
    
Also how regular are the curves assumed to be? There might be complicated behaviour if they aren't regular enough that may make the statement false. –  Olivier Bégassat Nov 3 '12 at 17:47
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How do you define a regular point for a continuous curve? I as hoping for the curves to be at least continuously differentiable, as I think this might be false for curves that are differentiable only... –  Olivier Bégassat Nov 3 '12 at 17:54
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I'd never heard of orthogonal projection onto a curve, but my guess is that the proof will be relatively straightforward and easy. The definition I know of a regular point is that the curve is differentiable there, and has non-zero derivative. I think one may have to assume continuous differentiablility of the curve to properly define orthogonal projection though. –  Olivier Bégassat Nov 3 '12 at 18:06

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