Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
You're not going to get a nice analytic solution. –  Potato Jun 7 '12 at 3:19
1  
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
1  
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
2  
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
1  
(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
show 3 more comments

2 Answers 2

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

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

share|improve this answer
1  
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
1  
Many thanks! I'll check up with another friend and accept your answer. –  polemon Jun 7 '12 at 5:08
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.