Hello, orange!
I "eyeballed" the problem.
I think I have the solution.
Let $\displaystyle G$ be the graph of the parametric equations: .$\displaystyle \begin{Bmatrix}x &=& \cos(4t) \\ y &=& \sin(6t)\end{Bmatrix}$
What is the length of the smallest interval $\displaystyle I$ such that the graph
of the equations for all $\displaystyle t \in I$ produces the entire graph of $\displaystyle G$ ?
When $\displaystyle t = 0\!:\;\begin{Bmatrix}x &=& \cos(0) &=& 1 \\ y &=& \sin(0) &=& 0\end{Bmatrix}$
At the "start", the graph is at $\displaystyle (1,0).$
Here is what I found:
. . $\displaystyle \begin{array}{c|c|c|} t & \cos(4t) & \sin(6t) \\ \hline 0 & 1 & 0 \\ \frac{\pi}{4} & \text{-}1 & \text{-}1 \\ \frac{\pi}{2} & 1 & 0 \\ \frac{3\pi}{4} & \text{-}1 & 1 \\ \hline \pi & 1 & 0 \\ \vdots & \vdots & \vdots\end{array}$
. . and the cycle repeats.
Therefore: .$\displaystyle I \,=\,[0,\,\pi]$
Hi,
If a given parametric curve is periodic, finding its period is a mystery to me. Here's a problem that I tackled some time ago with its solution; you might try your hand at it (my proof was somewhat long and involved). I wish I knew some general procedure to find the period of such periodic functions.