Given $\vec r(t)$, what are $\vec v(t), \ v(t), \ \vec a(t), \ a(t)$? I have come to a problem that simply states that we have a parametric curve $$\vec r(t) = (2\sin t, 3\cos t), \ \ t\in \mathbb R$$
and asks that we find $\vec v(t), \ v(t), \ \vec a(t), \ a(t)$.
What do these represent, and how do I find them? (You don't need to use my example; any example will be appreciated.)
 A: Hint: $\vec{v(t)} = \vec{r'(t)}, v(t) = ||\vec{r'(t)}||, \vec{a(t)} = \vec{r''(t)}, a(t)=||\vec{r''(t)}||$. Thus: $\vec{v(t)} = (2\cos t, -3\sin t), v(t) = \sqrt{4\cos^2t+9\sin^2t}$,etc...
A: The notation $\vec{r}(t)$ is very common to denote the position of a particle as a function of time. For example, $\vec{r}(t) = (\cos t, \sin t)$ with $t\in [0,2\pi]$ describes a particle moving counter-clockwise around the unit circle (you can check this with a parametric graph, or noting that the magnitude of $\vec{r}(t)$ (which is the particle's distance from the origin), is
$$\|\vec{r}(t)\|= \sqrt{\cos^2t+\sin^2t}=1.$$
Noting that the particle is always the same distance from the origin but is only ever at the same point at $t=0$ and $t=2\pi$ should make the fact that its path is circular clear. This gives us a decent intuition behind the other symbols:


*

*$\vec{v}(t) = \vec{r}'(t)$ is the velocity; we find it by differentiating each component of $\vec{r}$ with respect to time. For example, if $\vec{r}(t) = (x(t), y(t))$, then $\vec{r}'(t) = (x'(t),y'(t))$.

*The speed of the particle is the magnitude of the velocity; i.e., $v(t) = \|\vec{v}(t)\|$. Note that the velocity is a vector, and the speed is a scalar (thus the arrow drop).

*Likewise, $\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)$ is the acceleration (second time derivative), and $a(t)$ is the magnitude of this.


With our example, $\vec{v}(t) = (-\sin t, \cos t)$ and $v(t) = \|\vec{v}(t)\| = 1$ (in other words, the speed is constant). The acceleration vector is found using a similar computation, and we have $a(t)$ constant as well.
EDIT: Made it more clear what it means to differentiate vector-valued functions.
A: $$\vec v(t) = \dfrac{\text{d} \vec r}{\text{d} t} = (2 \cos t, \, -3 \sin t)$$
$$v(t) = ||\vec v(t) || = \sqrt{2^2 \cos^2 t + (-3)^2 \sin^2 t}$$
$$\vec a(t) = \dfrac{\text{d}^2 \vec r}{\text{d} t^2} = \dfrac{\text{d} \vec v}{\text{d} t}  = (-2 \sin t, \, -3 \cos t)$$
$$a(t) = ||\vec a(t) || = \sqrt{(-2)^2 \sin^2 t + (-3)^2 \cos^2 t}$$
A: $\vec r(t) = (2\sin t, 3\cos t)$
HINT:
$\vec v(t) = \dfrac{d \vec r(t)}{dt}$$= \dfrac {d}{dt}$ $(2\sin t, 3\cos t)$
From this, you can find $v(t)$.
Using Formula : If $\vec r=(a,b)$ then $|\vec r|=r=\sqrt {a^2 + b^2}$
And, $\vec a(t)=\dfrac{dv(t)}{dt}$ .... and you will get $a(t)$.
