Tensor calculus - curve definition doesn't quite make sense This is probably a rather stupid question, and I'm probably just looking at this wrong, but I can't get this definition to make sense.
I'm reading Tensor Calculus by Synge and Schild, and it notes

A curve is defined as the totality of points given by the equations
$$ x^r = f^r(u)    (r = 1,2,..., N)$$
Here $u$ is a parameter and $f^r$ are $N$ functions.

Earlier it introduced superscripts (as opposed to subscripts) as a way to notate numerical labels, so this a series of $N$ equations. I just don't understand how this produces a "curve". 
 A: You can think of $u$ as "time" and the $x^r$ coordinates as "space." So $ f^1$ gives the value of the $x^1$-coordinate at time $u$, and so on.  
Working with $N = 2$, say, the set of points $\{(u, (x^1, x^2)) : x^1 = f^1(u), x^2 = f^2(u) \}$ is a 1-dimensional curve parametrized by $u$ in the $x^1x^2$-plane. 
P.S. There are no such thing as stupid questions! 
A: The curve is on the $N$-dimensional space $\mathbb R^N$. That is, a map
$\mathbb R^1\cdot\!\!\dashrightarrow\mathbb R^N$
given by an assignment 
$$u\longmapsto
\left(\begin{array}{c}f^1(u)\\\vdots\\f^N(u)\end{array}\right),
$$
from which, choosing different instances for $N$ you get:


*

*A parametrization for plane curve $\mathbb R^1\cdot\!\!\dashrightarrow\mathbb R^2$

*For spatial curve:  $\mathbb R^1\cdot\!\!\dashrightarrow\mathbb R^3$

*For spacetime curve:  $\mathbb R^1\cdot\!\!\dashrightarrow\mathbb R^4$
etc. 
Usually the maps ought to be injective and has at least one derivative each and
different from zero, to associate the notion of tangent direction at point in the curve.
