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The question pretty much says it.

I need to solve $t$ in this equation: $ x = t - \sin{(t)} $ Either I've forgotten how to do it, or I am just blind, etc. Anyway, I'm completely stuck at this.

Actually, I need to solve a vector:

$(\; x \; , \; y \;) = (\; t - \sin{(t)} \; , \; 1 - \cos{(t)} \;)$

Inverse of $y$ is trivial: $t = \cos^{-1}{(1 - y)}$. But that doesn't help me much further on.

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    (cont.) Indeed, it can be claimed that the practical need to solve Kepler's equation to accuracies of a second of arc over the whole range of eccentricity fathered many of the developments in numerical mathematics in the eighteenth and nineteenth centuries. A few of the more than 100 methods of solution developed in the pre-computer era are considered in the exercises to this chapter."2012-06-07

2 Answers 2

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The length of the curve is the following:

$ \int_0^{2\pi}\sqrt{(x'(t))^2+(y'(t))^2}dt \\ =\int_0^{2\pi} \sqrt{(1-\cos t)^2 + (\sin t)^2}dt \\ =\int_0^{2\pi}\sqrt{1-2\cos t + \cos^2 t + \sin^2 t}dt \\ =\int_0^{2\pi}\sqrt{2-2\cos t}dt \\ =\int_0^{2\pi}\sqrt{2}\sqrt{1-\cos t}dt. $

Now recall one of the half-angle formulas: $\sin^2 u = \dfrac{1}{2}-\dfrac{1}{2}\cos (2u)$. Plug in $t = 2u$ to obtain $ \sin^2 \left(\frac{t}{2}\right) = \dfrac{1}{2}-\dfrac{1}{2}\cos (t), $ which is the same as $ 2 \sin^2 \left(\frac{t}{2}\right) = 1- \cos (t). $

Returning back to our integral and making appropriate substitutions, we obtain $ \int_0^{2\pi}\sqrt{2}\sqrt{2 \sin^2\left(\frac{t}{2}\right)}dt \\ = \int_0^{2\pi} 2 \sin \left( \dfrac{t}{2}\right) dt\\ = 2\int_0^{2\pi}\sin\left( \frac{t}{2}\right) dt. \\ $

Finally, we finish by making a substitution: let $v = t/2$. Then $dv = dt/2$. This is called a $u$-substitution but in order to avoid confusion, I'm using the letter $v$ instead.

Thus, we conclude

$ 2 \int_0^{\pi}\sin v (2dv) \\ = 4 \int_0^{\pi} \sin v dv \\ = -4 \cos v |_0^{\pi} \\ = -4(-1-1) = -4 (-2) = 8. $

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    Yes, thank you. Summer has obviously begun and I'm a bit rusty...2012-06-07
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To solve for t you would read this http://en.wikipedia.org/wiki/Kepler_equation. It is related to the Kepler equation. It can be done as a series, which may converge fast depending on the values.