Vtyjkyuk

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: $frac2pic$

That means we get the period $frac2pi3$ for $x(t)$ and $frac2pi2$ for $y(t)$.

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

As you stated, the periods of motion in the $x$ and $y$ direction are different. So after $frac2pi3$, 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 $2pi$.

If $t$ goes from $0$ to $frac2pi3,$
then $x(t) = sin(3t)$ goes through one full cycle and is ready to repeat.
But at $t = frac2pi3,$ $y(t) = cos(2t)$ 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 $t = frac2pi2$,
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 $t = 2pi,$ 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 $t = 2pi,$
so $2pi$ is the smallest possible period.

We need that

• $x(t+T_1)=sin(3t+3T_1)=sin (3t)implies 3T_1=2hpi implies
  T_1=frac23 hpi$


• $y(t+T_2)=cos(2t+2T_2)=cos(2t)implies 2T_2=2jpiimplies T_2=jpi$

then

$$T_1=T_2implies frac23 h=j implies h=3k quadland quad h=2k $$

that is

$$T_1=T_2=2pi cdot k implies T=2pi$$