Intuition about derivative of x^2 at 0

[Deter Pinklage]

Homework Statement

So my problem is mainly intuitive one, in that this *feels* wrong, and am mostly looking for insight.

If we have uniform 1D motion of a particle along $x$ with constant velocity $v$, what is the rate of change (first derivative with respect to time) of the variable $x^2$, particularly when evaluated at $t=0$?

Homework Equations

Well, pretty simply:
$\frac {d(x^2)}{dt} = 2v^2t$

The Attempt at a Solution

Now, I get that, but that would mean that at $t=0$, $x$ is increasing at a rate of $v$, whereas $x^2$ is not increasing at all. This confuses me because when you increase some positive number you must increase its square as well, right? Am I missing something obvious or something about the nature of infinitesimals?

 

[vela]

It looks like you've assumed $x=vt$, so when $t=0$, you have $x=0$, which isn't a positive number. Does that clear up your confusion?

 

[Deter Pinklage]

Right, of course, since when $x$ goes negative, $x^2$ will still have to increase so you got to have the local minimum. Idk why when you actually think about it physically it was weird. Thanks!