Smoothly Equivalent Curves I am studying for an upcoming exam, and I am trying to wrap my head around the following definition:
The two curves $$C_1:z(t),\ a\leq t\leq b$$
and
$$C_2:\omega(t),\ c\leq t\leq d$$
are smoothly equivalent if there exists a 1-1 $C^1$ mapping $\lambda(t):[c,d]\rightarrow[a,b]$ such that $\lambda(c)=a$, $\lambda(d)=b$, $\lambda'(t)>0$ for all $t$, and 
$$\omega(t)=z(\lambda(t)).$$
Could somebody please help me out by providing me with an example of two curves that are smoothly equivalent (and why) and two curves that are not smoothly equivalent?
 A: "$C_{1}$ and $C_{2}$ are smoothly equivalent" means the two curves are $C^{1}$ reparametrizations of each other traced in the same direction (i.e., with the same orientation). Loosely, each curve traces the same set of points in the same direction. (That's not literally correct; a curve might describe the motion of a particle that moves both "forward" and "backward", see $C_{6}$ below.)
(Non-)Examples that trace the unit circle include:
$C_{1}$: $z_{1}(t) = \cos t + i\sin t$, $0 \leq t \leq 2\pi$.
$C_{2}$: $z_{2}(t) = \cos(2\pi t) + i\sin(2\pi t)$, $0 \leq t \leq 1$.
$C_{3}$: $z_{3}(t) = \cos(2\pi t^{2}) + i\sin(2\pi t^{2})$, $0 \leq t \leq 1$.
$C_{4}$: $z_{4}(t) = \cos t + i\sin t$, $0 \leq t \leq 4\pi$.
$C_{5}$: $z_{5}(t) = \cos t - i\sin t$, $0 \leq t \leq 2\pi$.
$C_{6}$: $z_{6}(t) = \cos\bigl(t + (2\sin t)\bigr) + i\sin\bigl(t + (2\sin t)\bigr))$, $0 \leq t \leq 2\pi$.
Of these, $C_{1}$, $C_{2}$, and $C_{3}$ are (mutually) smoothly equivalent. The function "inside" the trig functions of $C_{2}$ (or of $C_{3}$) is the $\lambda$ in the definition.
$C_{4}$ "traces the circle twice", while an arbitrary reparametrization of $C_{1}$ traces the circle once.
$C_{5}$ "traces the circle clockwise", while an arbitrary orientation-preserving reparametrization of $C_{1}$ traces the circle counterclockwise.
(In case you know about winding numbers, $C_{1}$, $C_{2}$, and $C_{3}$ have winding number $+1$ about the origin; $C_{4}$ has winding number $2$, and $C_{5}$ has winding number $-1$.)
$C_{6}$ traces the circle "both clockwise and counterclockwise", making one net turn. Because "there's no way to parametrize away the indecisiveness" (your definition requires $\lambda$ to be injective), $C_{6}$ is not equivalent to $C_{1}$ even though both having winding number $+1$ about the origin.
Two curves with different images are certainly not smoothly equivalent; the preceding examples show that having the same image is not sufficient.
