1
$\begingroup$

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 ?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ May 5, 2019 at 4:40
  • $\begingroup$ 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. $\endgroup$ May 5, 2019 at 5:32
  • $\begingroup$ Thanks that was of a great help. But Is it possible to solve it without wolfram alpha ? $\endgroup$ May 5, 2019 at 6:29
  • $\begingroup$ 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. $\endgroup$ May 5, 2019 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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