# Deriving curvature of plane curve

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.

• 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? Apr 9 '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 $|a\cdot n|$. Apr 9 '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! Apr 9 '19 at 21:58