Given $\vec r(t)$, what are $\vec v(t), \ v(t), \ \vec a(t), \ a(t)$?

Question

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.)

Answer 1

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.

Answer 2

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...

Answer 3

$$\vec v(t) = \dfrac{\text{d} \vec r}{\text{d} t} = (2 \cos t, \, -3 \sin t)$$

$$v(t) = ||\vec v(t) || = \sqrt{2^2 \cos^2 t + (-3)^2 \sin^2 t}$$

$$\vec a(t) = \dfrac{\text{d}^2 \vec r}{\text{d} t^2} = \dfrac{\text{d} \vec v}{\text{d} t} = (-2 \sin t, \, -3 \cos t)$$

$$a(t) = ||\vec a(t) || = \sqrt{(-2)^2 \sin^2 t + (-3)^2 \cos^2 t}$$

Answer 4

$\vec r(t) = (2\sin t, 3\cos t)$

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 $a(t)$.