# How to solve $t$ in $x = t - \sin{(t)}$

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.

• You're not going to get a nice analytic solution. – Potato Jun 7 '12 at 3:19
• You have not forgotten how to do it, nor are you blind. As far as I know, there is no way to invert $t-\sin t$ with (a finite number of) elementary functions. Instead (for your vector problem) you should use the $y$ coordinate to find $t$ up to a factor of $2\pi$, and then find this factor of $2\pi$ involving the $x$ coordinate. In your homework do $x$ and $y$ have particular values? – anon Jun 7 '12 at 4:08
• If you are trying to find the area under an arch of the cycloid, or the arclength of part of the cycloid, the parametric equation is nice to work with directly. – André Nicolas Jun 7 '12 at 4:29
• Side remark: This equation appears in study of the two-body problem as "Kepler's equation", $\omega t = \psi - e \sin \psi$. Quoting from Goldstein's "Classical mechanics", at the end of Section 3-8: "The solution of the transcendental Kepler's equation to give the value of $\psi$ corresponding to a given time is a problem that has attracted the attention of many famous mathematicians ever since Kepler posed the question early in the seventeenth century. [...] – Hans Lundmark Jun 7 '12 at 6:38
• (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." – Hans Lundmark Jun 7 '12 at 6:39

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

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

• Hint: $\cos(t) = \cos(2 t/2) = 1 - 2 \sin^2(t/2)$ – Robert Israel Jun 7 '12 at 4:58
• Yes, thank you. Summer has obviously begun and I'm a bit rusty... – math-visitor Jun 7 '12 at 5:05

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.