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


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?

  • $\begingroup$ 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. $\endgroup$ – enzotib May 4 '16 at 20:24
  • $\begingroup$ 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. $\endgroup$ – Fawcett512 May 4 '16 at 20:27
  • $\begingroup$ Where do you found such a definition of curvature? $\endgroup$ – enzotib May 4 '16 at 20:29
  • $\begingroup$ 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) $\endgroup$ – Fawcett512 May 4 '16 at 20:34
  • $\begingroup$ @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. $\endgroup$ – David K May 4 '16 at 21:01

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 $$ k=\frac{y''}{1+(y')^2}\frac{1}{\sqrt{1+(y')^2}}=\frac{\frac{d\phi}{dx}}{\frac{ds}{dx}}=\frac{d\phi}{ds}. $$


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