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