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 $\mathcal{G}$?
That answer keeps on coming as wrong, I had gotten that answer earlier. I am not sure what is wrong :(
its pi/2
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Well, it looks like $\mathcal{G}$ is doing some fancy dance moves with $\cos(4t)$ and $\sin(6t)$. I bet it's practicing for the next "Graph's Got Talent" competition.
To find the length of the smallest interval $I$ that produces the entire graph $\mathcal{G}$, we need to determine the period of both $\cos(4t)$ and $\sin(6t)$.
The period of $\cos(4t)$ is $\frac{2\pi}{4} = \frac{\pi}{2}$. This means that $\cos(4t)$ will complete one full cycle in $\frac{\pi}{2}$ units of $t$.
Similarly, the period of $\sin(6t)$ is $\frac{2\pi}{6} = \frac{\pi}{3}$. This means that $\sin(6t)$ will complete one full cycle in $\frac{\pi}{3}$ units of $t$.
To make sure that both functions complete their respective cycles, we need to find the least common multiple of $\frac{\pi}{2}$ and $\frac{\pi}{3}$. The least common multiple of $\frac{\pi}{2}$ and $\frac{\pi}{3}$ is $\pi$, which means that both functions will complete their cycles after $\pi$ units of $t$.
Hence, the length of the smallest interval $I$ is $\pi$. So, the graph $\mathcal{G}$ will be fully produced for all $t \in [0, \pi]$. Just sit back and enjoy the show, folks!
To find the length of the smallest interval $I$ such that the graph of the parametric equations produces the entire graph $\mathcal{G}$, we can start by finding the period of both $\cos(4t)$ and $\sin(6t)$.
The period of $\cos(4t)$ is given by $T_1 = \frac{2\pi}{4} = \frac{\pi}{2}$. This means that the graph of $\cos(4t)$ repeats every $\frac{\pi}{2}$.
Similarly, the period of $\sin(6t)$ is given by $T_2 = \frac{2\pi}{6} = \frac{\pi}{3}$. This means that the graph of $\sin(6t)$ repeats every $\frac{\pi}{3}$.
To ensure that the entire graph $\mathcal{G}$ is covered, we need to find the smallest interval $I$ that includes an integer number of periods for both $\cos(4t)$ and $\sin(6t)$.
The smallest common multiple of $T_1$ and $T_2$ will give us the length of $I$. In this case, the smallest common multiple of $\frac{\pi}{2}$ and $\frac{\pi}{3}$ is $\pi$.
Therefore, the length of the smallest interval $I$ is $\pi$. In other words, the graph of the parametric equations covers the entire graph $\mathcal{G}$ for $t\in [0, \pi]$ (or any other interval of length $\pi$).
since cos(4t) has period pi/2
and sin(6t) has period pi/3
the LCM is pi