# Intuition of position vs time, and complex numbers

I want to differentiate the following equation to obtain an object's velocity, but I can't figure out how the eqation is actually describing the object's position.

It goes as follows: $r = (2.0 m + (3.00 m/s)t)i + (3.0 m - (2.00 m/s^2)t^2)j$, where $r =$ position at time $t$, and all quantities are in SI units.

Is $r$ a vector? And how do I differentiate or even integrate it given the complex $j$ and $i$? How do time fit in this cooridinate system?

• $j$ and $i$ are unit vectors (most likely) corresponding to $\langle0,1\rangle$ and $\langle1,0\rangle$, respectively. This makes $r$ a position vector that depends solely on the variable for time $t$. Feb 28, 2017 at 19:49
• Notice that $i$ and $j$ are unit vectors in the $x$ and $y$ direction respectively. It may be more helpful to rewrite $r$ into its components, i.e. $r_x = 2+3t$ and $r_y = 3-2t^2$. Differentiating both will lead to $v_x = \frac{dr_x}{dt}$ and $v_y = \frac{dr_y}{dt}$, i.e., you'll get the horizontal and vertical velocities. Rewriting in vector notation, we have that $v = v_x +v_y = \frac{dr_x}{dt}i + \frac{dr_y}{dt}j$. Feb 28, 2017 at 19:56
• Thanks! This makes much more sense Feb 28, 2017 at 20:09

I think that's $i$ and $j$ as in the standard unit vectors in the $x$- and $y$-directions, respectively. Not as in complex numbers. The reason I think this is because the scenario involves finding a velocity vector from a position vector, and describing the position in terms of imaginary units $i$ and $j$ is extremely outlandish, if not completely nonsensical. I think this has nothing to do with complex numbers at all.

So we should really be saying $\mathbf i$ and $\mathbf j$ to indicate they're vectors (and therefore avoid this type of confusion).

Anyway, to differentiate a vector, simply differentiate its components individually.

$$\mathbf r(t) = (2 + 3t)\mathbf i + (3-2t^2)\mathbf j$$ Therefore $$\mathbf r'(t) = (2+3t)' \mathbf i + (3-2t^2)' \mathbf j = 3\mathbf i - 4t \mathbf j.$$ Integrating works the same way. Integrate each component independently. And make sure to add an arbitrary constant vector (a vector with two arbitrary constant components) if your integral is indefinite, which would be unusual for vectors given my experience. Unusual but not impossible I guess.

• In quaternions we usually have $i,j,k$ as the three imaginary units, but I'm fairly sure this question is not about quaternions. Feb 28, 2017 at 19:58
• @DavidK, ah, yes, quaternions. Keep forgetting about those. I updated my answer accordingly, slightly.
– user307169
Feb 28, 2017 at 20:03
• Thanks, @tilper! Normally when I see the units $i$ and $j$ I easily jump to the conclusion it's imaginary units. But is bold formatting indicating vectors? Feb 28, 2017 at 20:12
• @user80556: Yes, in this case, they are the basis vectors. Note: $i$ and $j$ can both be imaginary units, but either (a) the subject is quaternions, as David K mentioned; or (b) they can both be $\sqrt{-1}$, but in different contexts: $i$ in most cases, $j$ in electrical engineering contexts where $i$ is likely to be used for current (that's another story for another time). Feb 28, 2017 at 20:19
• @user80556, yes, bold formatting such as $\mathbf x$ indicates something is a vector. But typically when writing by hand, as opposed to typing, vectors will be denoted with an arrow instead of with boldface, like $\vec x$.
– user307169
Feb 28, 2017 at 20:39