Abstract Nonsense

Crushing one theorem at a time

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.

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In our last post we discussed a way of measuring the (unsigned) curvature \kappa_\gamma of a curve \gamma. 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 \gamma:I\to\mathbb{R}^2 is unit speed we know that \gamma'(t) is a unit vector, and so is determined uniquely by it’s angle with the positive x-axis. Thus, there exists a function \varphi:I\to\mathbb{R} such that \gamma'(t)=(\cos(\varphi(t)),\sin(\varphi(t)) for all t\in I. That said, it’s obvious that there exists lots of such functions, namely \varphi(t)+2\pi is another such function. The non-obvious, but very important fact is that there exists a unique smooth \varphi:I\to\mathbb{R} which works. This function is what we shall defined as the turning angle. It’s then inuitively obvious that the curvature of \gamma 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.

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New Formulation of Curvature

Let \gamma:I\to\mathbb{R}^2 be a unit speed curve. We define the signed unit normal function of \gamma to be the function n_\gamma(t):I\to\mathbb{R}^2 defined to be equal to

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n_\gamma(t)=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\gamma'(t)

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so that n_\gamma(t) is nothing more than the \frac{\pi}{2} counter-clockwise rotation of \gamma'(t). Since \gamma is unit speed we know that \gamma''(t) is orthogonal to \gamma'(t) and so parallel to n_\gamma(t). Thus, there exists some (obviously unique) scalar c such that \gamma''(t)=c_t \gamma'(t). We call the function \varkappa_\gamma:I\to\mathbb{R} by taking t\mapsto c_t the signed curvature function of \gamma. Note then that since n_\gamma(t) is a unit vector we have

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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 \gamma as the path of a driving car, then \varkappa_\gamma(t)>0 if the car is turning left and k_\gamma(t)<0 if the car is turning right.

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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 \mu of a curve \gamma and say that \gamma''(t) is parallel to n_\mu(t) 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.

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So, as mentioned in the motivation if \gamma:I\to\mathbb{R}^2 is a unit speed curve then \gamma'(t)\in \mathbb{S}^1 for each t\in I and so there exists an angle \varphi(t), unique modulo 2\pi, such that \gamma'(t)=(\cos(\varphi(t)),\sin(\varphi(t)). What we now claim is that there exists a unique smooth functions \varphi:I\to\mathbb{R} with the property that \gamma'(t)=(\cos(\varphi(t)),\sin(\varphi(t)). 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].

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Theorem: Let \gamma:I\to\mathbb{R}^2 be a unit speed curve and t_0\in I and let \varphi_0 be such that \gamma'(t_0)=(\cos(\varphi_0),\sin(\varphi_0)). Then, there exists a unique smooth function \varphi:I\to\mathbb{R} such that \varphi(t_0)=\varphi_0 and \gamma'(t)=(\cos(\varphi(t)),\sin(\varphi(t)) for all t\in I.

Proof: Let f_1(t) and f_2(t) denote the coordinate functions \gamma'(t). Since \gamma is unit speed we have that f_1(t)^2+f_2(t)^2=1 for all t\in I. So, define

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\displaystyle \varphi(t)=\varphi_0+\int_{t_0}^{t}(f_1(t)f_2'(t)-f_1'(t)f_2(t))\; dt

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We know that it satisfies the initial value \varphi(t_0)=\varphi_0. Moreover, since f_1,f_2 being the coordinates of a smooth map are smooth and so \varphi(t) being a constant plus the integral of a smooth function is smooth. What we now claim is that \varphi(t) satisfies the angle condition for \gamma'(t). Indeed, let F(t)=f_1(t)\cos(\varphi(t))+f_2(t)\sin(\varphi(t)) and G=f_1(t)\sin(\varphi(t))-f_2(t)\cos(\varphi(t)). One then easily calculates that F'(t)=(f_1'(t)+f_2(t)\varphi'(t))\cos(\varphi(t))+f_1(t)f_1(t)f_1'(t). That said, using the fact that f_1(t)^2+f_2(t)^2=1 and the obvious consequence that f_1(t)f_1'(t)+f_2(t)f_2'(t)=0 one reduces F'(t) to f_1(t)(f_1(t)f_1'(t)+f_2(t)f_2'(t))=0 and so F'=0 and so F is constant (having zero derivative on connected open subset of \mathbb{R}, this is just the mean value theorem). A similar analysis shows that G'=0 and so G is constant. That said, note that F(t_0)=\cos(t_0)^2+\sin(t_0)^2=1 and similarly G(t_0)=0. Solving then the system of equations

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\begin{cases}f_1(t)\cos(\varphi(t))+f_2(t)\sin(\varphi(t))=1\\ f_1(t)\sin(\varphi(t))-f_2(t)\cos(\varphi(t))=0\end{cases}

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one quickly arrives at f_1(t)=\cos(\varphi(t)) and f_2(t)=\sin(\varphi(t)) which is precisely the condition we are looking for.

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Now, to see that \varphi is unique suppose that \psi was another function satisfying the hypotheses of the theorem. We note then that obviously \varphi(t)-\psi(t)=2\pi n(t) for some integer n(t)\in\mathbb{Z}. We see then that \varphi-\psi is a smooth mapping I\to 2\pi\mathbb{Z} and since 2\pi\mathbb{Z} is a discrete susbspace of \mathbb{R} and I connected we must have that \text{im }(\varphi-\psi) 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, \varphi-\psi is constant. But, since \varphi,\psi agree at t_0 they must coincide for all t. \blacksquare

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Remark: The way F and G 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 f_1(t)+if_2(t)=e^{i\varphi(t)} from where any number of complex ordinary differential equations can be applied.

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We call the unique function \varphi_\gamma determined by \gamma and the initial condition \varphi_\gamma(t_0)=\varphi_0 the turning angle of \gamma with initial condition \varphi(t_0)=\varphi_0.

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So, as promised:

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Theorem: Let \gamma:I\to\mathbb{R}^2 a unit speed curve. Then, \varkappa_\gamma(t)=\varphi'(t).

Proof: Indeed, we have that \gamma'(t)=(\cos(\varphi(t)),\sin(\varphi(t)) and so n_\gamma(t)=(-\sin(\varphi(t)),\cos(\varphi(t)) but since \gamma''(t)=\varphi'(t)(-\sin(\varphi(t)),\cos(\varphi(t)) we may compare these two to arrive at the conclusion. \blacksquare

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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.


September 25, 2011 - Posted by | Differential Geometry | , , , , ,

1 Comment »

  1. […] 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 […]

    Pingback by The Fundamental Theorem of Plane Curve Curvature « Abstract Nonsense | October 5, 2011 | Reply

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