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A cycloid is given by the parametric equations: $ x = 2 - \pi \cos(t)$ and $ y = 2t - \pi \sin(t)$.

The problem asks for the slope of the tangents on the cycloid at a point where the cycloid intersects itself. That point is not given, but it lies on the x - axis.

I wanted to find that point by cancelling the parameter, $t$, but I couldn't come up with important elimination. Is there such way of solving the problem ?

Rafa Budría
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Amanuel Getachew
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1 Answers1

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At $t$ and $t'$ the coordinates repeat: $x(t)=x(t')\land y(t)=y(t')\implies2t-\pi\sin t=2t'-\pi\sin t'\land\cos t=\cos t'$

$\cos t=\cos t'\implies t'=t+k2\pi;\in\mathbb Z-\{0\}\lor (t'=-t+k'2\pi,t\neq n\pi);k,n\in\mathbb Z$

a) $2t-\pi\sin t=2(t+k2\pi)-\pi\sin (t+k2\pi)$

$\sin(t+k2\pi)-\sin t=4k$, no solutions.

b) $2t-\pi\sin t=2(-t+k'2\pi)-\pi\sin (-t+k'2\pi)$

$4t=\pi(4k'+\sin t-\sin(-t+k'2\pi))$

$4t=\pi(4k'+\sin t+\sin t)$

$t/\pi=k'+(1/2)\sin t$

Having as a valid solution (Wolfram Alpha) $t=\pi/2,t'=-\pi/2$ with $k'=0$ Only this value is needed as the function is periodic and th tangents have the same slope at the other intersection points.

Now the slope at $t$ is $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2-\pi\cos t}{\pi\sin t}$

And at the intersection $\left.\dfrac{dy}{dx}\right|_{t=\pi/2}=2/\pi;\left.\dfrac{dy}{dx}\right|_{t=-\pi/2}=-2/\pi$

Rafa Budría
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  • Thanks,but If we end up with an equation with structure $t = a + bsin(t)$, just like you did before the link to 'wolfram alpha' what about just $2t - \pi sin(t) = 0$ and hence $ 2t = \pi sin (t)$, at the intersection point... I mean why all those steps, if we used Wolfram to solve the equation. Appreciate your answer by the way. – Amanuel Getachew May 05 '19 at 04:40
  • With those steps we know why $y=0$. If you don't need this, you can use WA directly for that transcendental equation, of course. – Rafa Budría May 05 '19 at 05:32
  • Thanks that was of a great help. But Is it possible to solve it without wolfram alpha ? – Amanuel Getachew May 05 '19 at 06:29
  • Yes, by approximate methods as the equation is a trascendental one (e.g. here). Nevertheless, considering the nice the solution is, maybe it is meant you have to get the solution by guessing. – Rafa Budría May 05 '19 at 07:12