Theoretical question regarding two elliptic trajectories I am new to Newtonian Mechanics, and I was wondering regarding the following:
Are 
$$\vec{r}(t): a \sin(ωt)\hat x +b\cos(ωt)\hat y$$ 
and 
$$\vec{r}(t): a \sin(ωt^2)\hat x +b\cos(ωt^2)\hat y$$
Identical trajectories?
According to my analysis, they differ in position, velocity and acceleration, but the trajectories themselves are identical. Is that correct?
 A: Mathematically speaking the curve is the application, not the image of this application.
What happens here is that you have two different curves (you should give them different names!) but they have the same image. 
To be more formal, let $I$ and $I'$ be two intervals in which the parameter of the curve varies:
$$ \gamma_1: I \ni t \mapsto (a \sin{\omega t}, b \cos{\omega t}) \in \mathbb{R}^2 $$
is the first curve. In an analogue way: 
$$ \gamma_2: I' \ni t \mapsto (a \sin{\omega t^2}, b \cos{\omega t^2}) \in \mathbb{R}^2 $$
Is the other curve, and they are different. I took two different intervals for the two curves, but this could be unnecessary.
But every curve has an image, that is the set of all points in the target space (in this case $\mathbb{R}^2$) reached by some values of the parameter, and as you found out these curves have the same image, that is they describe the same trajectory.
One way to notice this in a very fast way is to notice that you can get one of the curve by changing the parameterization of the other, that is put $t^2$ in the place of $t$.
Changing the parameterization changes some objects, like the image point for a given value of the parameter (the position) or the rate of change of this image point (the velocity), but leaves unaltered the image of the curve.
Hope this helped.
