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I am considering curvature of a plane curve as covered in chapter 2 of Elementary Differential Geometry by Pressley. For a curve $c(t)$, we are considering the calculation of curvature and it is said that as we go from $c(t)$ to $c(t+dt)$, the curve moves away from the tangent at $c(t)$ by a distance of $(c(t+dt)-c(t)).n$ where $n$ is the normal to the tangent vector at $c(t)$. I just don't see how that expression gives that distance though. Is there some fact about vectors and dot products that I'm missing here? Any hints are much appreciated.

C.F.G
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Salazar_3854708
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  • It helps me to think of it in extreme cases. In other words, if the curve moves away very, very gradually then the quantity $c(t+dt) - c(t)$ will be largely in the direction of the tangent vector, implying its dot product, or projection, with $n$ would be very minimal. Also, if it moves away very sharply, then $c(t+dt) - c(t)$ will be largely in the direction of the normal. Does that help some? – JacobCheverie Apr 09 '19 at 19:33
  • The distance from the point (vector) $a$ to the line $n\cdot x=0$ is length of the projection of $a$ on the normal vector $n$ of the line. When $n$ is a unit vector, this is just $|a\cdot n|$. – Ted Shifrin Apr 09 '19 at 21:34
  • Ah, yes the distance between a point and line is what I've been missing. This clears it up. Thanks Prof. Shifrin! – Salazar_3854708 Apr 09 '19 at 21:58

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I made a comment to your question. Does the following picture help? I apologize for the poor quality.

enter image description here

JacobCheverie
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  • Thanks for the answer Jacob. What I want to understand is why the expression itself gives the distance of deviation of the curve from the tangent vector. Is it because projecting c(t+dt)-c(t) onto n gives a vector of magnitude (c(t+dt)-c(t) . n)/|n| and then |n|=1 since it's a unit vector? – Salazar_3854708 Apr 09 '19 at 21:26