Printable View
Let http://data.artofproblemsolving.com/...1c4df4304f.gif be the graph of the parametric equations
http://data.artofproblemsolving.com/...d0ea6ee9f9.gif
What is the length of the smallest interval http://data.artofproblemsolving.com/...63c10b675d.gif such that the graph of these equations for all http://data.artofproblemsolving.com/...f9566ea488.gif produces the entire graph http://data.artofproblemsolving.com/...1c4df4304f.gif?
Cos[4t] needs (2pi)/4 = pi/2Quote:
Originally Posted by orange
Sin[6t] needs (2pi)/6 = pi/3
what's the least common multiple of 1/2 and 1/3 ? So what's the minimum interval needed for a full cycle?
1 is their LCM. So I guess the minimum interval is just pi?
give that man a kewpie doll!
Hello, orange!
I "eyeballed" the problem.
I think I have the solution.
Quote:
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.
Attachment 29944
