Why can the length of vector $r(t+dt) - r(t)$ be approximated with the vector $r'(t)dt$? Question: This passus was presented in my textbook. I can't understand their reasoning. Why can the length of vector $r(t+dt) - r(t)$ be approximated with the vector $r'(t)dt$?
From my textbook:
Let $r(t) = \left( x(t), y(t)\right)$. We can then approximate the curve length between the points $r(t)$ and $r(t+dt)$ with the length of the vector $r(t+dt) - r(t)$. 
We can in turn approximate this vector with the length of the vector $r'(t)dt$, i.e:
$$|r'(t)|dt = \sqrt{x'(t)^2 + y'(t)^2}dt$$

 A: Because, by definition,
$$r'(t) = \lim_{dt\to 0}\frac{r(t+dt)-r(t)}{dt}$$
This means that, for very small $dt$, $r'(t)\approx\frac{r(t+dt)-r(t)}{dt}$; that's just how one interprets limits. In other words, $r'(t) dt \approx r(t+dt)-r(t)$.
As a similar use of this reasoning: $e=\sum_{n=0}^\infty\frac{1}{n!}$, and so $e\approx \sum_{n=0}^N \frac{1}{n!}$ for very large $N$. 
A: What is the curve length between $r(t)$ and $r(t+\mathrm dt)$ in the first place? (Not to mention: what is $\mathrm dt$?)
Unless you have a better definition, for any continuous map $r\colon [a,b]\to\Bbb R^n$, the curve length of $r$ is $$\sup\biggl\{\,\sum_{k=1}^m|r(x_{i})-r(x_{i-1})|\biggm|m\in\Bbb N, a=x_0<x_1<\ldots <x_m=b\,\biggr\},$$
provided that supremum exists.
If $r'$ exists, then $|r(x_{i})-r(x_{i-1})|\approx |r'(x_{i-1})|\cdot (x_i-x_{i-1})$. And if additionally $r'$ is continuous then $|r'(x_{i-1})|\approx |r'(a)|$ for all $i$, at least if $b-a\approx 0$ (which we may assume as we are interested in what happens as $\mathrm dt=b-a\to 0$). But then the sum turns into $|r'(a)|\cdot (b-a)=|r'(a)|\,\mathrm dt$. 
This should give you an overview of what happens, but to make this argument more rigorous, you need to invest some work into making the meaning of  $\approx$ as used three times above more precise (i.e., express it in terms of limits).
