Need to create parametric equation for epitrochoid evolute I am posting this because this equation is too complicated for me to figure out and I looked online finding no solution.
Basically I have a parametric equation for points $(x, y)$ which defines an epitrochoid. I also know the equation to find the evolute, but it's not in parametric form. The dirty form of the equation for the evolute is found in the "Evolute" entry on Xahlee.info.
I have even provided a graph showing the epitrochoid (parametric formula) and its evolute (non-parametric formula) using the equations:
https://www.desmos.com/calculator/itxanasb5v
What I desperately need is the parametric form of the equation. It's going to be huge formula. The formula of graph I need is in red.
I don't remember calculus, but I know I'm going to need first and second derivatives.
Thanks in advance.
Can anybody please help?
 A: From your Desmos project, the epitrochoid is parameterized by $(f(t),g(t))$, where
$$\begin{align}
f(t) &= bc\cos t - h \cos ct \\
g(t) &= bc\sin t - h \sin ct 
\end{align} \tag1$$
and I'm defining $c:=(a+b)/b$ to save myself some typing.
Again from your project, the evolute can be written as
$$\left(f-g' k,\,g + f' k\right) \qquad\text{where}\qquad k = \frac{f'^{\,2}+g'^{\,2}}{f'g''-f''g'} \tag2$$
Calculating the derivatives is straightforward:
$$\begin{align}
f'(t) &= -bc\sin t\,+hc\sin ct \\
f''(t) &= -bc\cos t+hc^2\cos ct \\[6pt]
g'(t) &= \phantom{-}bc\cos t-hc\cos ct \\
g''(t) &= -bc\sin t\,+hc^2\sin ct
\end{align} \tag3$$
Substituting into $(2)$ is tedious, but not terrible (especially with a computer algebra system). We get
$$k = \frac{h^2+b^2 - 2 hb \cos dt}{bp+hcq} \tag4$$
$$d:=c-1=\frac{a}{b} \qquad\qquad p:= b-h\cos dt \qquad\qquad q := h-b\cos dt$$
and, with a bit of simplifying,
$$(f-g'k, g+f'k ) = \left(
\frac{a h (p \cos ct + c q \cos t)}{b p + h c q}, 
\frac{a h (p \sin ct + c q \sin t)}{b p + h c q}\right) \tag5$$
