I'm currently working on a problem related to particle motion in the plane $( \mathbf{R}^2 )$ and could use some guidance. The particle moves along a curve, and at any given time $( t )$ seconds after it starts, its position is described by the vector $( \mathbf{r}(t) = (2 \cos \pi t, \sin \pi t) )$, with the units in meters.
I have two specific questions:
a) What type of curve does the particle move along? I’m trying to visualize the path based on the parametric equations given but could use some confirmation or guidance on how to accurately describe the curve.
b) How do I determine the particle's velocity, speed, and acceleration at the specific time $( t = 2 )$ seconds? I understand that velocity is the derivative of the position vector, and speed is the magnitude of the velocity, but I'm having trouble applying these concepts to find the acceleration as well and to calculate these quantities at $( t = 2 )$.
The curve parametrized by $$ \mathbf{r}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \begin{pmatrix} 2\cos(\pi t) \\ \sin(\pi t) \end{pmatrix} $$ describes an ellipse with half-axes $2$ and $1$ respectively, because $\frac{x^2}{4} + y^2 = 1$. The velocity and the acceleration are computed straightforwardly by differentiating component-wise; one has thus : $$ \mathbf{v}(t) = \mathbf{\dot{r}}(t) = \frac{\mathrm{d}}{\mathrm{d}t}\mathbf{r}(t) = \begin{pmatrix} \dot{x}(t) \\ \dot{y}(t) \end{pmatrix} = \begin{pmatrix} -2\pi\sin(\pi t) \\ \pi\cos(\pi t) \end{pmatrix} $$ and $$ \mathbf{a}(t) = \mathbf{\dot{v}}(t) = \begin{pmatrix} -2\pi^2\cos(\pi t) \\ -\pi^2\sin(\pi t) \end{pmatrix} = -\pi^2\,\mathbf{r}(t), $$ which is actually the equation of motion of a harmonic oscillator of frequency $\pi$.