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How can I parametrize a convex plane curve using the angle $\theta$ between the tangent line and the $x$-axis?

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Only the angle will not be enough for a complete parametrization. Think about circles with different radius. –  Lord Soth Apr 17 '13 at 18:18
    
I'm not used with this parametrization stuff.. How could I use the angle $\theta$ for finding a parametrization? –  PtF Apr 17 '13 at 18:23

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If I understand correctly, you want to find the coordinates of the curve $(x(s),y(s))$, but you know only $\theta(s)$. This angle is defined by

$$\tan\theta=\frac{y'}{x'}\ .$$

(primes are $\partial/\partial s$). Except for points at which $x'=0$ which should be treated separtely (because there $\tan\theta=\pm\infty$), you can parametrize your curve by integrating the differential equation

$$\begin{align} x'(s)&=1\\ y'(s)&=\frac{dy}{dx}\frac{dx}{ds}=\tan\theta(s) \end{align}$$

As pointed out by Lord Soth, for complete parameterization you need some additional data, which is exactly $x(0)$ and $y(0)$ - the initial conditions for this equation. With these initial conditions, the solution is unique.

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