# What Are Vector Functions?

Theory

### VectorFunctions

A vector function is a function that is written as a vector. It looks like this:

 $\phantom{\rule{-0.17em}{0ex}}\left(x\phantom{\rule{-0.17em}{0ex}}\left(t\right),y\phantom{\rule{-0.17em}{0ex}}\left(t\right)\right)$

$x\phantom{\rule{-0.17em}{0ex}}\left(t\right)$ is the $x$-coordinate of the vector written as a function of $t$, and $y\phantom{\rule{-0.17em}{0ex}}\left(t\right)$ is the $y$-coordinate of the vector also written as a function of $t$.

An important area of use for vector functions is position, velocity, acceleration and speed. The connection between these hasn’t changed from what you have learned earlier, but now you’re going to look at them as vectors. The only newcomer is speed, but don’t worry, it’s just the length of the velocity vector. If you want to know how fast the acceleration is, you can find that out by calculating the length of the acceleration vector.

Theory

### TheVectorFunctions

The position vector:

$\stackrel{\to }{r}\phantom{\rule{-0.17em}{0ex}}\left(t\right)=\phantom{\rule{-0.17em}{0ex}}\left(x\phantom{\rule{-0.17em}{0ex}}\left(t\right),y\phantom{\rule{-0.17em}{0ex}}\left(t\right)\right)$

The velocity vector:

$\stackrel{\to }{v}\phantom{\rule{-0.17em}{0ex}}\left(t\right)={\stackrel{\to }{r}}^{\prime }\phantom{\rule{-0.17em}{0ex}}\left(t\right)=\phantom{\rule{-0.17em}{0ex}}\left({x}^{\prime }\phantom{\rule{-0.17em}{0ex}}\left(t\right),{y}^{\prime }\phantom{\rule{-0.17em}{0ex}}\left(t\right)\right)$

The speed:

$\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}\phantom{\rule{-0.17em}{0ex}}\left(t\right)|=\sqrt{{x}^{\prime }\phantom{\rule{-0.17em}{0ex}}{\left(t\right)}^{2}+{y}^{\prime }\phantom{\rule{-0.17em}{0ex}}{\left(t\right)}^{2}}$

The acceleration vector:

$\stackrel{\to }{a}\phantom{\rule{-0.17em}{0ex}}\left(t\right)={\stackrel{\to }{v}}^{\prime }\phantom{\rule{-0.17em}{0ex}}\left(t\right)={\stackrel{\to }{r}}^{″}\phantom{\rule{-0.17em}{0ex}}\left(t\right)$

The acceleration:

$a=\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}\phantom{\rule{-0.17em}{0ex}}\left(t\right)|=\sqrt{{x}^{″}\phantom{\rule{-0.17em}{0ex}}{\left(t\right)}^{2}+{y}^{″}\phantom{\rule{-0.17em}{0ex}}{\left(t\right)}^{2}}$

Example 1

An insect flying away from a flower is in the position $\stackrel{\to }{r}\phantom{\rule{-0.17em}{0ex}}\left(t\right)=\phantom{\rule{-0.17em}{0ex}}\left({t}^{2},2{t}^{2}-t+1\right)$. Find the velocity vector, the speed, the acceleration vector and the size of the acceleration, all when $t=2$.

You can find the velocity vector by differentiating the position vector $\stackrel{\to }{r}\phantom{\rule{-0.17em}{0ex}}\left(t\right)$. That gives you

 $\stackrel{\to }{v}\phantom{\rule{-0.17em}{0ex}}\left(t\right)=\phantom{\rule{-0.17em}{0ex}}\left(2t,4t-1\right).$

Next, you can insert $t=2$ into $\stackrel{\to }{v}\phantom{\rule{-0.17em}{0ex}}\left(t\right)$ to find the velocity vector at $t=2$.

 $\stackrel{\to }{v}\left(2\right)=\phantom{\rule{-0.17em}{0ex}}\left(2\cdot 2,4\cdot 2-1\right)=\phantom{\rule{-0.17em}{0ex}}\left(4,7\right)$

From here it’s easy to find the speed as well, as it’s just the length of the velocity vector $\stackrel{\to }{v}\phantom{\rule{-0.17em}{0ex}}\left(t\right)$. That means you have

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}\left(2\right)|& =\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(4,7\right)|=\sqrt{{4}^{2}+{7}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\sqrt{16+49}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\sqrt{65}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}\left(2\right)|=\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(4,7\right)|=\sqrt{{4}^{2}+{7}^{2}}=\sqrt{16+49}=\sqrt{65}.$

Moving on, the acceleration is the derivative of the velocity vector $\stackrel{\to }{v}\phantom{\rule{-0.17em}{0ex}}\left(t\right)$. You will see that there are no variables left after you perform this differentiation, which means that the acceleration is constant. This is the result:
 $\stackrel{\to }{a}\phantom{\rule{-0.17em}{0ex}}\left(t\right)=\phantom{\rule{-0.17em}{0ex}}\left(2,4\right)$

The length of that vector is

 $a=\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}\phantom{\rule{-0.17em}{0ex}}\left(2\right)|=\sqrt{{2}^{2}+{4}^{2}}=\sqrt{4+16}=\sqrt{20},$

giving you an acceleration of $\sqrt{20}$.