Parametric Equations - Finding the smallest interval

Question

Let $\mathcal{G}$ be the graph of the parametric equations \begin{align*} x &= \cos(4t),\\ y &= \sin(6t). \end{align*}What is the length of the smallest interval $I$ such that the graph of these equations for all $t\in I$ produces the entire graph $\mathcal{G}$?

I thought it was a pretty straightforward question, since $x = \cos(4t)$ had a period of $\frac{\pi}{2}$ and $y = \sin(6t)$ had a period of $\frac{\pi}{3}$, the LCM would be $\boxed{\pi}$, hence the interval.

However, the answer keeps on coming as incorrect. Am I overlooking something here?

Answer 1

Evidently by graph they mean the set of points in the coordinate plane which is a curved path from $(-1,1)$ to $(-1,-1)$ in which case the answer would be $\dfrac{\pi}{2}$ since between $t=-\frac{\pi}{4}$ and $t=\frac{\pi}{4}$ the entire set of $(x,y)$ coordinates in the path has been "covered" although this is only half the path.

1. For $t\in \left[-\frac{\pi}{4},-\frac{\pi}{6}\right]$ the path goes from $(-1,1)$ to the point $\left(-\frac{1}{2},0\right)$.
2. For $t\in \left[-\frac{\pi}{6},0\right]$ the path goes from $\left(-\frac{1}{2},0\right)$ to the point $(1,0)$.
3. For $t\in \left[0,\frac{\pi}{6}\right]$ the path goes from $(1,0)$ to the point $\left(-\frac{1}{2},0\right)$.
4. For $t\in \left[\frac{\pi}{6},\frac{\pi}{4}\right]$ the path goes from $\left(-\frac{1}{2},0\right)$ to the point $(-1,-1)$.