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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?2012-11-03
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    Yes they are, I will add it in the question. Thanks!2012-11-03
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    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.2012-11-03
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    They are supposed to be continuous but not necessarily differentiable. Would that answer your question?2012-11-03
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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...2012-11-03
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    Sorry, do you mean continuous only in the last sentence? Your question makes me think; a lot of the other theorems etc. assume the curves to be regular, that is they are $C^1$ differentiable and with nowhere vanishing derivative. But the quote is in connection with a definition that applies to continuous functions in general. Perhaps there is a small typo and the clarifier following "i.e." only applies when the function is differentiable. Do you know how one can see that the orthogonal projection is the closest point when the function is differentiable?2012-11-03
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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.2012-11-03
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    I see. Thanks a lot!2012-11-03
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    For instance, I am not sure this would hold for plane curves like $c(t)=(t,pt)$ and $C(t)=(t,t^2 sin(1/t))$ (extended by $(0,0)$ at $t=0$) for $P=(0,0)$.2012-11-03

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