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Can one parametrise any closed curve by the angle its tangent makes to the $x$-axis? I seem to remember that this is only possible for convex curves. Could anyone tell me why, please?

Also is necessarily true that for convex curves, the mean curvature is always positive? I can see why since the mean curvature is kinda like a second derivative and for convex functions the second derivative is positive..

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A parametric representation of a curve $\gamma\subset{\mathbb R}^2$ is an at least continuous function $${\bf f}:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .$$ A priori the independent variable $t$ is not required to have any geometrical interpretation and can be thought of as "time".

It is a fact of life that any curve $\gamma$, which is a static or "drawn" geometrical object, has many equivalent parametric representations, among them some where the independent variable has a geometric meaning connected with the curve; e.g. it could be the $x$-coordinate of the running point, the arc length along the curve, measured from a starting point, etc. One such "geometrical" variable could be the changing tangential angle $\theta={\rm arg}(\dot x,\dot y)$, but it is not suitable for all curves. It is definitely necessary that the above function ${\bf f}$ is well defined, and this means that to any $\theta\in{\mathbb R}$ (or $\in{\mathbb R}/(2\pi)$) should correspond at most one point of the curve having this tangential angle $\theta$. Such is the case for strictly convex curves, but not for curves containing pieces of lines or for a sinusoid $x\mapsto(x,\sin x)$ $\ (-\infty<x<\infty)$.

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"this means that to any $\theta \in \mathbb{R}$ corresponds at most one point of the curve. " - I don't understand. Do you mean that there is only one $\theta$ for each point? The converse is obvious since $f$ is just single-valued? –  TagWoh Jul 7 '12 at 17:24
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You can certainly parametrize a plane curve by its tangential angle. If, for instance, you have the Whewell intrinsic equation $s=f(\varphi)$ ($s$ being the arclength, and $\varphi$ being the tangential angle), you can obtain a corresponding set of parametric equations:

$$\begin{align*}x&=x_0+\int_{\varphi_0}^\varphi f^\prime(u)\cos\,u\;\mathrm du\\y&=y_0+\int_{\varphi_0}^\varphi f^\prime(u)\sin\,u\;\mathrm du\end{align*}$$

where $x_0,y_0,\varphi_0$ are arbitrary.

For instance, the circle involute has the Whewell equation $s=\dfrac{a\varphi^2}{2}$.

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