Calculate the period of the motion of P

Question

A point P runs through a curve K given by the parametric representation: $x(t)=\sin(3x)$ and $y(t)=\cos(2x)$

Calculate the period of the motion of P.

So I can't really figure out how to solve this. I know that calculating the period of a trigonometric function would go by: $\frac{2\pi}{c}$

That means we get the period $\frac{2\pi}{3}$ for $x(t)$ and $\frac{2\pi}{2}$ for $y(t)$.

Somehow the answer is $2\pi$ and I don't get how. Any idea guys?

Answer 1

As you stated, the periods of motion in the $x$ and $y$ direction are different. So after $\frac{2\pi}3$, while the $x$ co-ordinate is back to the same value, the $y$ co-ordinate is not. What they are asking you for is the period of the total motion around the curve, which is when the $x$ and $y$ co-ordinates have rotated back to the same co-ordinate from which they started. This is found by finding the lowest common multiple of the two periods you have computed. This gives $2\pi$.

Answer 2

If $t$ goes from $0$ to $\frac{2\pi}{3},$ then $x(t) = \sin(3t)$ goes through one full cycle and is ready to repeat. But at $t = \frac{2\pi}{3},$ $y(t) = \cos(2t)$ is only partway through a cycle, so $P$ has not returned to its starting point and it cannot have completed its full period.

If you instead go to $t = \frac{2\pi}{2}$, then $y(t) = \cos(2t)$ is at the start of a new cycle but $x(t) = \sin(3t)$ is just halfway through a cycle. $P$ is at its starting point (coincidentally) but going in a different direction than at the start, so it is not at the end of a period.

But if you go to $t = 2\pi,$ then $x(t) = \sin(3t)$ will have completed $3$ periods exactly and $y(t) = \cos(2t)$ will have completed $2$ periods exactly, and everything is ready to repeat exactly as before. So you have completed one full period. You don't get to this condition anytime between $t= 0$ and $t = 2\pi,$ so $2\pi$ is the smallest possible period.