## More Geometric Way of Viewing the Curvature of a Plane Curve

**Point of Post: **In this post we discuss a nicer, more geometric way of thinking about the curvature of a plane curve via its turning angle.

*Motivation*

In our last post we discussed a way of measuring the (unsigned) curvature of a curve . The definition though, while simple and easy to motivate, lacked the geometric flair that, doing geometry, we’d hope to have. So, we use one of the beautiful aspects of differential geometry, the go-between of simple to define and proof analysis notions and the beautiful and intuitive geometry ones. For those who have taken algebraic topology this could be analogized to the relationship between singular and simplicial homology, where the latter’s existence is almost solely to prove the first makes sense. Regardless, we’d now like t0 formulate a different way of defining curvature which although equivalent to the way defined before (except now we make a sign convention thus arriving at the ‘signed’ curvature) has a much more geometric feel. To see what this definition is we note, using the formulation from last time, that it suffices to give a reinterpretation for unit speed curves. So, what is this super geometric way of viewing the curvature? Well, since is unit speed we know that is a unit vector, and so is determined uniquely by it’s angle with the positive -axis. Thus, there exists a function such that for all . That said, it’s obvious that there exists lots of such functions, namely is another such function. The non-obvious, but very important fact is that there exists a unique smooth which works. This function is what we shall defined as the turning angle. It’s then inuitively obvious that the curvature of should be the rate of change of the turning angle. This, is precisely what we shall prove. This formulation and the associated theorems will have some very interesting consequences that we shall discuss in our next post.

*New Formulation of Curvature*

Let be a unit speed curve. We define the *signed unit normal function *of to be the function defined to be equal to

so that is nothing more than the counter-clockwise rotation of . Since is unit speed we know that is orthogonal to and so parallel to . Thus, there exists some (obviously unique) scalar such that . We call the function by taking the *signed curvature function of *. Note then that since is a unit vector we have

so that, consistent with the naming, the modulus of signed curvature is the unsigned curvature. The sign tells you what direction you are turning in. In other words, if you can think of as the path of a driving car, then if the car is turning left and if the car is turning right.

So, now we wish to extend this new definition of signed curvature to arbitrary regular curves. To do this we’d like to just say select any unit speed reparamaterization of a curve and say that is parallel to and so there exists, etc. This works fine, we just have to show that this is independent of choice of unit speed reparamaterization. But a quick shows that this is the case.

So, as mentioned in the motivation if is a unit speed curve then for each and so there exists an angle , unique modulo , such that . What we now claim is that there exists a unique smooth functions with the property that . Although the proof of this theorem is slightly painful, the idea is pretty simple. Namely, we’ll just find a way to phrase our problem in terms of an ordinary differential equation and then solve in the normal way. Our proof is basically that as on page 36 of [2].

**Theorem: ***Let be a unit speed curve and and let be such that . Then, there exists a unique smooth function such that and for all .*

**Proof: **Let and denote the coordinate functions . Since is unit speed we have that for all . So, define

We know that it satisfies the initial value . Moreover, since being the coordinates of a smooth map are smooth and so being a constant plus the integral of a smooth function is smooth. What we now claim is that satisfies the angle condition for . Indeed, let and . One then easily calculates that . That said, using the fact that and the obvious consequence that one reduces to and so and so is constant (having zero derivative on connected open subset of , this is just the mean value theorem). A similar analysis shows that and so is constant. That said, note that and similarly . Solving then the system of equations

one quickly arrives at and which is precisely the condition we are looking for.

Now, to see that is unique suppose that was another function satisfying the hypotheses of the theorem. We note then that obviously for some integer . We see then that is a smooth mapping and since is a discrete susbspace of and connected we must have that is a point (lest the image is a discrete space with more than two points which is also connected (being the continuous image of a connected set) which is clearly impossible). Thus, is constant. But, since agree at they must coincide for all .

*Remark: *The way and are defined is reminiscent of some kind of Cauchy-Riemann type equation and so one may suspect that there is some kind of complex analysis argument going on in the background. Indeed, one could phrase this as looking for solutions to from where any number of complex ordinary differential equations can be applied.

We call the unique function determined by and the initial condition the *turning angle of with initial condition .*

So, as promised:

**Theorem: ***Let a unit speed curve. Then, .*

**Proof: **Indeed, we have that and so but since we may compare these two to arrive at the conclusion.

**References:**

1. Carmo, Manfredo Perdigão Do. *Differential Geometry of Curves and Surfaces*. Upper Saddle River, NJ: Prentice-Hall, 1976. Print.

2. Montiel, Sebastián, A. Ros, and Donald G. Babbitt. *Curves and Surfaces*. Providence, RI: American Mathematical Society, 2009. Print.

[…] then is smooth (since is) unit speed because which is clearly a unit vector. But, recall that is ‘a’ turning angle for , and by uniqueness ‘the’ turning angle for […]

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