Parametrization of Lines and Curves

Question

Let $D \subseteq \mathbb{R}$ and let $C_1 : D \rightarrow \mathbb{R}^n$ and $C_2 : D \rightarrow \mathbb{R}^n$ be parametrizations of curves. In rigorous terms (unlike what is found in most calculus books), what does it mean for two parametrizations $C_1(t)$ and $C_2(t)$ to represent the same curve? My guess is that the images $C_1(D)$ and $C_2(D)$ are equal. Does this sound right?

Also, I am trying to prove that $L_1(t)$ and $L_2(s)$ are parametrizations of a line, then $L_1$ and $L_2$ are the same line if and only if they intersect at one point and their direction/slope vectors are parallel.

EDIT:

I have a related issue concerning equality of functions. To illustrate this issue, consider the simple example of showing that $r_1(t) = (3,-1,4) + t(8,12,-6)$ and $r_2(s) = (11,11,-2) + s(4,6,-3)$. Clearly if you let $t = \frac{s+2}{2}$, then $r_1(\frac{s+2}{2}) = r_2(s)$ every $s \in \mathbb{R}$. Does this imply they are the same curve? From my recollection, to functions $h,k : A \rightarrow B$ are equal if and only if $h(a) = k(a)$ for every $a \in A$, but perhaps I am confusing several concepts.

But this problem isn't merely about vector-valued functions. For instance, suppose that we have $f,g : \mathbb{R} \rightarrow \mathbb{R}$ are functions, and suppose that through some deductions we showed that $f(x) = g(2x)$ for all $x \in \mathbb{R}$. Would this imply that $f = g$? I am thoroughly confused about these subtle points.

What is the root of my problem? How can I clear up this problem?

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$Strictly speaking, two functions $f_{1}:X_{1} \to Y_{1}$ and $f_{2}:X_{2} \to Y_{2}$ are equal precisely when:

• The domains are the same set: $X_{1} = X_{2}$.

• The codomains are the same set: $Y_{1} = Y_{2}$.

• The values are identical at each point: $f_{1}(x) = f_{2}(x)$ for all $x$ in $X_{1} = X_{2}$.

Consequently, your parametrized lines $r_{1}$ and $r_{2}$ are not equal as functions.

Similarly, if $f(x) = g(2x)$ for all real $x$, it does not follow that $f$ and $g$ are equal as functions. (It could happen in this circumstance that $f(x) = g(x)$ for all $x$, e.g., if $f$ and $g$ are constant.)

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There is no universal definition of what it means to say two curves $C_{1}:[a_{1}, b_{1}] \to \Reals^{n}$ and $C_{2}:[a_{2}, b_{2}] \to \Reals^{n}$ are "the same". Definitions in use (in roughly decreasing order of likelihood) include:

1. One curve is an orientation-preserving reparametrization of the other: There exists an increasing bijection $\tau:[a_{1}, b_{1}] \to [a_{2}, b_{2}]$ such that $C_{1} = C_{2} \circ \tau$ as functions. (This is common when working with vector line integrals.)

2. One curve is a reparametrization of the other: There exists a continuous bijection $\tau:[a_{1}, b_{1}] \to [a_{2}, b_{2}]$ such that $C_{1} = C_{2} \circ \tau$ as functions. (This is common when working with scalar line integrals.)

3. The images are the same: $C_{1}([a_{1}, b_{1}]) = C_{2}([a_{2}, b_{2}])$. (I suspect this is uncommon, but it could be useful in, say, computer graphics.)

Conceivably, "the same" could mean "equal as functions", in which case the requirement would be much stronger: $a_{1} = a_{2}$, $b_{1} = b_{2}$, and $C_{1}(x) = C_{2}(x)$ for all $x$ in $[a_{1}, b_{1}] = [a_{2}, b_{2}]$. (Again, I suspect this is unusual.)

Answer 2

This is mainly to supplement Andrew D. Hwang's good answer.

Although $r_1 \neq r_2$ because $r_1(s) = r_2(s)$ doesn't hold for every $s$, what you wrote can be framed as an equality of functions if instead of comparing $r_2$ with $r_1$ we compare it with the composition of $r_1$ with the function $f(s) = \frac{s+2}{2}$ (that is, the function $r_1 \circ f$). If we do that, then we can say $(r_1 \circ f) = r_2$ as functions precisely because $(r_1 \circ f)(s) = r_1(\frac{s+2}{2}) = r_2(s)$ for every $s$.

Put another way: when we see $r_1 \left(\frac{s+2}{s} \right) = r_2(s)$, we're tempted to say, "Oh! Well then $r_1$ must equal $r_2$". But we're forgetting that when we say "$r_1$", we are really referring to the function $s \mapsto r_1(s)$. The function $s \mapsto r_1 \left(\frac{s+2}{2} \right)$ is related to $r_1$ (indeed, it is the composition of $r_1$ with the function $s \mapsto \frac{s+2}{2}$), but they are not the same function, and thus $(r_1 \circ f) = r_2$, but $r_1 \neq r_2$.