# Abstract Nonsense

## Curvature of Plane Curves

Point of Post: In this post we motivate the idea of the curvature of a plane curve in preparation for higher dimensional notions of curvature. We then derive the general formula to calculate the curvature.

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Motivation

We come now to what could be easily stated as the focus of the geometry of curves and surfaces (at least at this level): curvature. I have faith that anyone reading this has at least an intuitive notion of what ‘curvature’ is. It’s how much the curve ‘twists’, or ‘moves around’, or any other equally synonymous term. As always though, we need to find a way to transfer our vague intuitions into hard mathematics. So, let’s see if we can find some way of mathematically describing this ‘twisting’. Well, whatever we may define curvature to be, I doubt that anyone would argue that the curvature of a line should be zero. Thus, the curvature of a curve should mention the deviation that curve takes from a straight line–the question is, which straight line? To make the next leap a little more obvious, let’s take a second and try to relate what we’re doing to a real world scenario. Perhaps where curvature affects us most is when we are driving. No one likes when they accidentally come up to a curve too fast and have difficulty making it. So, what exact property made it hard to make? Well, it’s not hard to see that the difficulty is that we are initially going in a certain direction, and it’s the turning away from this direction (that we are going at some instantaneous point) that’s hard. So, the line we seek to measure deviation from is the ‘trajectory line’, whatever that means. But, a seconds thought gives us that the trajectory line is the line passing through our current position in the direction of our velocity. Aha! That’s it, since velocity is the derivative of position, it’s not hard to take this analysis and see that our ‘trajectory line’ is nothing more than the line spanned by the tangent vector at our current position. Thus, the curvature of a curve $\gamma$ at a point $\gamma(t_0)$ should be the quickeness at which $\gamma$ deviates from $\gamma'(t_0)$. So, how exactly do we do this? There are precisely two unit vectors $n,-n$ perpendicular to $\gamma'(t_0)$, pick one. Since they differ by a sign it’s not important (at this point) which we pick, so let’s say $n$. We then want to measure, for small $\Delta_t$, the change from $d(\ell,\gamma(t_0))$ to $d(\ell,\gamma(t_0+\Delta_t))$ where $\ell$ denotes $\text{span}(n)$ and $d$ denotes the distance between a point and a set. But, since $n$ is normal it’s not hard to see that this difference is equal to $(\gamma(t_0+\Delta_t)-\gamma(t_0))\cdot n$. But, expanding $\gamma(t_0+\Delta_t)$ as a Taylor series we see that it’s equal to

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$\gamma(t_0+\Delta_t)=\gamma(t_0)+\gamma'(t_0)\Delta_t+\frac{1}{2}\gamma''(t_0)\Delta_t^2+O(\Delta_t^3)$

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and so $(\gamma(t_0+\Delta_t)-\gamma(t_0))\cdot n$ is equal to

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$\Delta_t\gamma(t_0)\cdot n+\frac{1}{2}\Delta_t^2\gamma''(t_0)\cdot n+O(\Delta_t^3)$

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Now, since $n$ is perpendicular to $\gamma'(t_0)$ by construction we see that the change in distances is equal to $\frac{1}{2}\Delta_t^2\gamma''(t_0)\cdot n+O(\Delta_t^3)$. Now, if we restrict ourselves to unit speed curves we shall prove that $\gamma'(t_0)$ is perpendicular to $\gamma''(t_0)$ and so consequently $\gamma''(t_0)$ is parallel to $n$. Thus, we may finally conclude that the magnitude of the difference of the distances is $\frac{1}{2}\|\gamma''(t_0)\|\Delta_t^2+O(\Delta_t^3)$. So, forgetting the factor of $\frac{1}{2}$ (we are trying to define a consistent statistic for curves, multiplying by a constant is irrelevant) we see that the only important number as $\Delta_t\to 0$ is $\|\gamma''(t_0)\|$. Ta-da! Thus, the (unsigned) curvature of a unit speed curve $\gamma$ at $\gamma(t_0)$ is just $\|\gamma''(t_0)\|$. Thus, if $\|\gamma''(t_0)\|$ is large then the curve curves ‘a lot’ and conversely. But, you’ll notice that we’ve restricted ourselves to unit speed curves in our definition. But, this is an easy fix. Indeed, as indicated in our last post the only curves we shall really concern ourselves will be regular curves, and we know that every regular curve has a unit speed reparamaterization from where it’s pretty clear that we should just define the curvature of the regular curve to be equal to the curvature of any of its unit speed reparamaterizations. Ostensibly, there may be an issue here, in the sense of ‘which unit speed reparamaterization do we pick’. But, the problem isn’t a problem at all since all unit speed reparamaterizations of a given regular curve will have the same curvature. This make sense intuitively since the curvature of a curve is an intrinsically geometric idea which should be independent of paramaterization. So, with all this in mind, let’s do some math.

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Curvature

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To begin we justify the only unproven fact asserted in the motivation. Namely:

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Theorem: Let $\gamma:I\to\mathbb{R}^n$ be a unit speed curve, then either $\gamma''(t)=0$ or $\gamma'(t)$ and $\gamma''(t)$ are orthongal.

Proof: Note that by definition $\gamma'(t)\cdot\gamma'(t)=1$, differentiating this (recalling the way derivatives work for multilinear forms) we have that $\gamma''(t)\cdot\gamma'(t)+\gamma'(t)\cdot\gamma''(t)=0$, but since $\cdot$ is a bilinear form this may be rewritten as $2(\gamma'(t)\cdot\gamma''(t))=0$. Thus, we either have that $\gamma''(t)=0$ or \gamma'(t)\$ and $\gamma''(t)$ are orthogonal, as desired. $\blacksquare$

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Thus, with this small fact proven we can unabashedly define curvature. In particular, let $\mu:I\to\mathbb{R}^2$ be a unit speed plane curve. We define the curvature function $\kappa_\mu:I\to\mathbb{R}$ to be equal to $\kappa_\mu(t)=\|\mu''(t)\|$. Now, as stated in the motivation we’d like to extend the notion of curvature to arbitrary regular curves. To do this let $\gamma:I\to\mathbb{R}^2$ be any regular curve and $\mu:J\to\mathbb{R}_2$ a unit speed reparamaterization with reparmaterization map $\phi:I\to J$. We then define the curvature function for $\gamma$ to be equal to $\kappa_\mu\circ \phi$. It’s not at all apparent that this definition is independent of our choice of $\phi$, but that’s what the following says:

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Theorem: Let $\gamma:I\to\mathbb{R}^2$ be a regular smooth curve. Then, and let $\mu$ be any reparamaterization of $\gamma$. Then, $\kappa_\gamma(t)$ which is defined to be $\|\mu''(\phi(t)\|$ is equal to.

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$\displaystyle \kappa_\gamma(t)=\frac{\|\gamma'(t)\times\gamma''(t)\|}{\|\gamma'(t)\|^3}$

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Proof: Let $\mu=\gamma\circ\phi$ be a unit speed reparamaterization of  $\gamma$. We then have that

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$\displaystyle \gamma'(\phi(t))=\frac{\gamma'(t)}{\phi'(t)}$

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where one recalls that we have proven that $\phi'(t)$ is non-zero for all $t$. Thus, plugging this into $\kappa_\gamma(t)=\|\mu''(\phi(t)\|$ and apply the quotient rule one finds that

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$\displaystyle \kappa_\gamma(t)=\left\|\frac{\phi'(t)\gamma''(t)-\phi''(t)\gamma'(t)}{\phi'(t)^3}\right\|\quad\textbf{(1)}$

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Recall now that since $\phi$ unit speed reparamaterizes $\gamma$ we have that $\phi(t)=\pm s_\gamma(t)+c$ where $s_\gamma(t)$ is the arc length function of $\gamma$ with respect to any point in $I$. Thus, it’s easy to see from $s_\gamma'(t)=\|\gamma'(t)\|$ that $\phi'(t)^2=\|\gamma'(t)\|^2=\gamma'(t)\cdot\gamma'(t)$. Differentiating this then gives

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$2\phi'(t)\phi''(t)=2(\gamma'(t)\cdot\gamma''(t))$

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and so $\phi'(t)\phi''(t)=\gamma'(t)\cdot \gamma''(t)$. Using these two equations and multiplying top and bottom of $\mathbf{(1)}$ by $\phi'(t)$ gives

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$\displaystyle \kappa_\gamma(t)=\frac{\|(\gamma'(t)\cdot\gamma'(t))\gamma''(t)-(\gamma'(t)\cdot\gamma''(t))\gamma'(t)}{\|\gamma'(t)\|^4}=\frac{\|\gamma'\times(\gamma'(t)\times\gamma''(t))\|}{\|\gamma'(t)\|^4}$

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where the last equality was acheived using the triple product formula. Now, since $\gamma'(t)$ is orthogonal to $\gamma'(t)\times\gamma''(t)$ we know that the magnitude of the numerator is $\|\gamma'(t)\|\|\gamma'(t)\times\gamma''(t)\|$. The conclusion then follows. $\blacksquare$

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Corollary: The definition that $\kappa_\gamma(t)=\kappa_\mu\circ\phi$ for any unit speed reaparamaterization $\gamma=\mu\circ\phi$ is independent of the reparamaterization.

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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.

September 24, 2011 -