I was working on an assignment and had a good idea to have a water jet model an epitrochoid to create a fountain. While the overall idea was approved by my teachers, I was told that I need to show a proof of the derivation of the parametric equations for the epitrochoid.
the particular Desmos model of what I am modelling and need the equations derived for are below:
https://www.desmos.com/calculator/zb8ccuup1q
Can anyone help with the derivation of the parametric equations?
Thanks
I'll detail the approach in J.G.'s comment. Let's work in the complex plane in order to deal with both coordinates at the same time, as suggested by J.G. From now on, I'll refer to the linked animation in the same comment, with the same initial situation.
Let $R$ and $r$ be the radii of the static big circle (centered at the origin) and the rotating little one respectively. Let $z = x + iy$ be a point of the epitrochoid and $\theta$ the angle between the $x$-axis and the center of the rotating circle, so that $\theta = 0$ initially. Finally, let's denote the distance between the point $z$ and the center of the little by $d$.
In consequence, the position of the center of the little circle is given by $(R+r)e^{i\theta}$. From there, the point $z$ lies at a distance $d$ from this center, hence a translation by the same quantity, but it has to be combined with a rotation of angle $\varphi$ (which is represented by a factor $e^{i\varphi}$ in the complex plane, because the red dot and the center of the little circle don't rotate at the same speed. In consequence, one has $z = (R+r)e^{i\theta} - de^{i\varphi}$. Note that the minus sign comes from the initial position of the red dot, namely on the left of the rotating center (hence an initial phase $e^{i\pi} = -1$).
The big question is : what is $\varphi$ ? Actually, the travelled arc length (on the big circle) corresponding to the angle $\theta$ is given by $R\theta$. Since the little circle possesses a smaller radius, a point inside the latter rotates from an additional angle $\theta' = \frac{R}{r}\theta$ at the same time (because the arc lengths are the same obviously, i.e. $R\theta = r\theta'$). In the end, one has $\varphi = \theta + \theta' = \frac{R+r}{r}\theta$. Finally, separate the real and imaginary parts in order to recover the cartesian parametric equations you are looking for.
Addendum : why $\varphi = \theta + \theta'$ instead of $\varphi = \theta'$ ?
The center of the little circle is subjected to a single rotation, but the motion of an inside point can be decomposed into two rotations : the same one as before, which induces a global phase $e^{i\theta}$, and another rotation with respect to the center of the little circle, hence a relative phase $e^{i\theta'}$.
The first rotation can be seen as a motion of the little circle "as a whole" (actually with respect to the origin / big circle), to which all points of the little circle are subjected, and the second one as a relative motion. This situation is common when considering several frames of reference in physics for instance. The Youtube channel Veritasium made a great video on this specific case.
