How do you find the interval in which a parametric equation will be traced exactly once I have been all over the internet and I can't find an answer to what seems like a simple question. I want to be able to find the interval for a parametric equation so that it is only traced once. My equations are:
\begin{align}
x &= 11\cos(u) - 4\cos\left(\frac{11u}{2}\right)\\
y &= 11\sin(u) - 4\sin\left(\frac{11u}{2}\right)
\end{align}
After looking at the graph, I realize the answer is $4\pi$, but how would I solve this if I am not able to look at the graph. I have seen solutions where people check one loop at a time until they arrive back at the original starting point, but it always seems as if they know how long one loop is and to me it seems they are just taking arbitrary numbers. In other words, how would I know to check every $\frac{\pi}{4}$ versus every $10\pi$.

You are looking for the smallest value of $p$ such that $x(u)=x(u+p)$ and $y(u)=y(u+p)$ for all $u$. Notice the equations for both contain only trigonometric functions, so this $p$ must be some multiple of $\pi$. Consider $$x(u)=11\cos u-4\cos\left(\frac{11u}{2}\right) $$ The period of the first term is clearly $2\pi$. But for the second term, $$4\cos\left(\frac{11(u+2\pi)}{2} \right)=4\cos\left(\frac{11u}{2} +\pi\right) =\color{red}-4\cos\left(\frac{11u}{2}\right)$$ The smallest rational multiple of $\pi$ that is required is, as you’ll see, $\frac{4\pi}{11}$. Taking the lcm of the two periods gives the overall period: $4\pi$. In general, it’s a good idea to remember that the period of $\cos \left(\frac{ax}{b}\right)$ is $2\pi \left(\frac ba\right)$ (similarly for sin).
The same argument goes for $y(u)$ as well.
