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I'm struggling with this problem:

"Find a parametrization of the first quadrant part of the circular arc $x^2 + y^2 = a^2$ in terms of the angle between the tangent line and the positive x-axis, oriented counterclockwise."

I drew a sketch of the first quadrant in the $xy$-plane, and specified an arbitrary point $(x, y)$ on the arc. Then I sketched the tangent at that point, making an obtuse angle $\theta$ with the positive $x$-axis.

I'm not sure how to do the parametrization though, any ideas/help?

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  • $\begingroup$ There is a triangle with angles $90^\circ, 180^\circ-\theta$ and $\alpha$ where $\theta$ is the given angle. Using $\alpha$, you should easily find a parameterisation. $\endgroup$
    – AlexR
    Mar 4 '15 at 13:47
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enter image description here

Look at the figure. The angles $\theta$ are the same because formed by orthogonal straight lines. Now looking at the blue circle you see that: $ OA= \sin \theta $ and $ XA= \cos \theta $

Using this result you can parametrize the arc as $(\sin t, -\cos t)$ with $ 0 \le t \le \theta $.

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  • $\begingroup$ Thanks a lot. Can I ask what program you used to draw that? I'm looking for a geometry-like program which is compatible with LaTeX aswell. $\endgroup$
    – Kamil
    Mar 4 '15 at 15:31
  • $\begingroup$ Welcome. I use LaTex with the pakage Pgfplots: pgfplots.sourceforge.net $\endgroup$ Mar 4 '15 at 15:37
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Call the tangent point $P$, the origin $O$ and the point of intersection on the $x$-Axis $Q$. The triangle $\Delta OQP$ is a right triangle (Because of the tangent property, $\angle OPQ = 90^\circ$). We know that the arc is parameterised by $$\phi(\xi) = (\cos \xi, \sin \xi) \qquad \xi\in [0, \angle QOP]$$ But $\angle QOP = 90^\circ - \underbrace{\angle PQO}_{=180^\circ - \theta} = \theta - 90^\circ$, where $\theta$ is the given obtuse angle.

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