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I'm trying to prove that if $\vec{x}:I\rightarrow\mathbb{R}^2$ is a curve parametrized by arc length and $\theta(t)$ is the angle between the tangent line to $\vec{x}$ at point $t$ and the $x$ axis, then $\kappa=\theta'$, where $\kappa$ denotes the curvature.

I know that for a curve in $\mathbb{R}^2$, the curvature is given by

$\kappa=\frac{1}{|\vec{x}'|^3}det(\vec{x}'\vec{x}'')$.

I also know that the tangent line to $\vec{x}$ at point $\alpha\in\mathbb{R}$ is given by $y(t)=\vec{x}(\alpha)+t\vec{x}'$.

I can picture the problem, but can't write the solutions. Any ideas?

Fawcett512
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  • The curvature at a point is an intrinsic property, so it does not change when the curve is rigidly moved/rotated. I deduce that it cannot depend on the position of the $x$-axis. – Vincenzo Tibullo May 04 '16 at 20:24
  • No, it is the angle between the tangent line and the x-axis. So when you move along the curve, the angle is changing. And if you picture it, it makes sense that it is equal to the curvature. – Fawcett512 May 04 '16 at 20:27
  • Where do you found such a definition of curvature? – Vincenzo Tibullo May 04 '16 at 20:29
  • From the Frenet equations you get that $\kappa=\frac{1}{|\vec{x}'|}<e'_1, e'_2>$, where $e'_1, e'_2$ are the vectors from the Frenet Frame (all in $\mathbb{R}^2$ of course) – Fawcett512 May 04 '16 at 20:34
  • @enzotib If you take the angle with the $x$-axis, the angle with the $y$-axis, and the angle with the line $y=mx + b$ ($m\neq 0$) you will get three different angles, but if you take the rate of change of each of those angles with respect to path length (which is $t$ in this question) you will get the same value in all three cases. So I think the definition of curvature in the question is independent of the position of the $x$-axis. – David K May 04 '16 at 21:01
  • http://mathworld.wolfram.com/Curvature.html shows the two definitions are equivalent. It starts with the $\theta'$ definition (but without assuming $t$ is path length) and transforms it to the $\vec x', \vec x''$ definition, but you should be able to follow the steps backwards, especially if you add the assumption that the curve is parameterized by path length. – David K May 04 '16 at 21:19
  • @DavidK Thanks. I'm going to try the steps backwards. – Fawcett512 May 04 '16 at 21:25

2 Answers2

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Assume a planar curve (for simplicity function) $y=f(x)$, i.e. $\overline{x} (t)=(t,f(t))$, then $$ k=\frac{\left(\dot{\overline{x}},\ddot{\overline{x}}\right)}{|\dot{\overline{x}}|^3} $$ or equivalently $$ k=\frac{y''}{\left(1+(y')^2\right)^{3/2}} $$ Suppose that $\phi$ is the angle between the tangent line $(\epsilon)$ and the $x$ axis. We have $\frac{dy}{dx}=\tan \phi$ or equivalently $\phi=\arctan\left(\frac{dy}{dx}\right)$. Taking derivatives in both parts of the last equation we have $$ \frac{d\phi}{dx}=\frac{y''}{1+(y')^2} $$ But $$ \frac{ds}{dx}=\sqrt{1+(y')^2} $$ Hence $$ k=\frac{y''}{1+(y')^2}\frac{1}{\sqrt{1+(y')^2}}=\frac{\frac{d\phi}{dx}}{\frac{ds}{dx}}=\frac{d\phi}{ds}. $$

Nikos Bagis
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A simple alternative is to use the tangent vector to the curve $$\vec{t} = (\cos(\theta), \sin(\theta))$$ and plug it in into $$ \kappa = \left|\frac{d\vec{t}}{ds}\right| $$ to get $$ \kappa = \frac{d\theta}{ds} $$ via chain rule.

Oscillon
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