This is a question about understanding the concept of curvature. Firstly, what exactly is curvature of a curve (not the formula, what does it actually mean conceptually)? Secondly, I am confused about how one can figure out when a curve would have the maximum curvature. I was told it is when the derivative of the curvature function (k(x)) = 0. But, I don't understand the concept/reason behind it.
In the book I am using there are more than one definitions for curvature. Some of them are:
Number One:
$$k\left(x\right)=\frac{\left|f''\left(x\right)\right|}{\left[1+\left(f'\left(x\right)\right)^2\right]^{\frac{3}{2}}}$$
Number 2:
$$k\left(t\right)=\frac{\left|\vec{r}'\left(t\right)\:\times \:\:\vec{r}''\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|^3}$$
Number 3:
$$k\left(t\right)=\frac{\left|\vec{T}'\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|}$$
Curvature at a point is what it sounds like - a measure of how "curvy" a curve is. How sharply it's bending at that point, if you will. There are many (and well varied) notions of curvature, but by the sounds of your question, this is what you are talking about.
In order to understand your second question, you need to be a bit more precise in telling us what definition of curvature you're using, I think.
EDIT:
From various comments and your edit to your post, my only remark about why the curvature is maximum at the point where the derivative = 0 is that if you imagine the curvature being plotted on a graph (a plot of curvature vs time $t$), there probably be a place where the curvature reaches some local maximum. This will be like a peak - it'll look like the top of a hill. For this reason the derivative with respect to time there will be zero. It depends on the nature of your curve as to whether this is a local maximum or a global maximum.