Note first that if
${\phi}_{2}\left(\sigma \right)=x+iy$ is a point on the curve
$C,$ then
${v}_{2}\left({\phi}_{2}^{1}(x+iy)\right)=y.$ Then, for any
$\tau \in [a,b],$ we have
$$\begin{array}{ccc}\hfill {v}_{1}^{1}\left({v}_{2}\left({g}^{1}\left(\tau \right)\right)\right)& =& {v}_{1}^{1}\left({v}_{2}\left({\phi}_{2}^{1}\left({\phi}_{1}\left(\tau \right)\right)\right)\right)\hfill \\ & =& {v}_{1}^{1}\left({v}_{2}\left({\phi}_{2}^{1}({u}_{1}\left(\tau \right)+i{v}_{1}\left(\tau \right))\right)\right)\hfill \\ & =& {v}_{1}^{1}\left({v}_{1}\left(\tau \right)\right)\hfill \\ & =& \tau ,\hfill \end{array}$$
showing that
${v}_{1}^{1}\circ {v}_{2}={{g}^{1}}^{1}=g.$ Hence
$g$ is continuously differentiable at every point
$s$ in the subintervals
$({s}_{i1},{s}_{i}).$ Indeed
${g}^{\text{'}}\left(\sigma \right)={{v}_{1}^{1}}^{\text{'}}\left({v}_{2}\left(\sigma \right)\right){v}_{2}^{\text{'}}\left(\sigma \right)$ for all
$\sigma $ near
$s,$ and hence
$g$ is piecewise smooth.
Obviously,
${\phi}_{2}\left(s\right)={\phi}_{1}\left(g\left(s\right)\right)$ for all
$s,$ implying that
${\phi}_{2}^{\text{'}}\left(s\right)={\phi}_{1}^{\text{'}}\left(g\left(s\right)\right){g}^{\text{'}}\left(s\right).$ Since
${\phi}_{2}^{\text{'}}\left(s\right)\ne 0$ for all but a finite number of points
$s,$ it follows that
${g}^{\text{'}}\left(s\right)\ne 0$ for all but a finite number of points,
and the theorem is proved.
Let
${\phi}_{1}$ and
${\phi}_{2}$ be as in the theorem.
Then, for all but a finite number of points
$z={\phi}_{1}\left(t\right)={\phi}_{2}\left(s\right)$ on the curve
$C,$ we have
$$\frac{{\phi}_{1}^{\text{'}}\left(t\right)}{{\phi}_{1}^{\text{'}}\left(t\right)}=\frac{{\phi}_{2}^{\text{'}}\left(s\right)}{{\phi}_{2}^{\text{'}}\left(s\right)}.$$
From the theorem we have that
$${\phi}_{2}^{\text{'}}\left(s\right)={\phi}_{1}^{\text{'}}\left(g\left(s\right)\right){g}^{\text{'}}\left(s\right)={\phi}_{1}^{\text{'}}\left(t\right){g}^{\text{'}}\left(s\right)$$
for all but a finite number of points
$s\in (c,d).$ Also,
$g$ is strictly increasing, so that
${g}^{\text{'}}\left(s\right)\ge 0$ for all points
$s$ where
$g$ is differentiable.
And in fact,
${g}^{\text{'}}\left(s\right)\ne 0$ for all but a finite number of
$s$ 's, because
${g}^{\text{'}}\left(s\right)$ is either
${({v}_{1}^{1}\circ {v}_{2})}^{\text{'}}\left(s\right)$ or
${({u}_{1}^{1}\circ {u}_{2})}^{\text{'}}\left(s\right),$ and these are nonzero except for a finite number of points.
Now the corollary follows by direct substitution.
REMARK If we think of
${\phi}^{\text{'}}\left(t\right)=({x}^{\text{'}}\left(t\right),{y}^{\text{'}}\left(t\right))$ as a vector in the plane
${R}^{2},$ then the corollary asserts that the direction of this vector is
independent of the parameterization, at least at all but a finite number of points.This direction vector will come up again as the unit tangent of the curve.
The adjective “smooth” is meant to suggest that the curve is bending in
some reasonable way, and specifically it should mean that the curve has a tangent, or tangential direction, at each point.We give the definition of tangential direction below, but we note that in the
context of a moving particle, the tangential directionis that direction in which the particle would continue to move if the force
that is keeping it on the curve were totally removed.If the derivative
${\phi}^{\text{'}}\left(t\right)\ne 0,$ then this vector is the velocity vector,
and its direction is exactly what we should mean by the tangential direction.
The adjective “piecewise” will allow us to consider curves that have a finite number
of points where there is no tangential direction, e.g., where there are “corners.”
We are carefully orienting our curves at the moment.
A curve
$C$ from
${z}_{1}$ to
${z}_{2}$ is being distinguished from the same curve
from
${z}_{2}$ to
${z}_{1},$ even though the set
$C$ is the same in both instances.
Which way we traverse a curve will be of great importance at the end of this chapter,when we come to Green's Theorem.

Let
$C,$ the range of
$\phi :[a,b]\to C,$ be a piecewise smooth curve,and let
$z=(x,y)=\phi \left(c\right)$ be a point on the curve.
We say that the curve
$C$ has a tangential direction at
$z,$ relative to the parameterization
$\phi ,$ if the following limit exists:
$$\underset{t\to c}{lim}\frac{\phi \left(t\right)z}{\left\phi \right(t)z}=\underset{t\to c}{lim}\frac{\phi \left(t\right)\phi \left(c\right)}{\left\phi \right(t)\phi (c\left)\right}.$$
If this limit exists, it is a vector of length 1 in
${R}^{2},$ and this unit vector is called the unit tangent (relative to the parameterization
$\phi $ ) to
$C$ at
$z.$
The curve
$C$ has a
unit tangent at the point
$z$ if there exists
a parameterization
$\phi $ for which the unit tangent at
$z$ relative to
$\phi $ exists.
 Restate the definition of tangential direction and unit tangent using the
${R}^{2}$ version of the plane instead of the
$C$ version.
That is, restate the definition in terms of pairs
$(x,y)$ of real numbers instead of a complex number
$z.$
 Suppose
$\phi :[a,b]\to C$ is a parameterization of a piecewise smooth curve
$C,$ and that
$t\in (a,b)$ is a point where
$\phi $ is differentiable
with
${\phi}^{\text{'}}\left(t\right)\ne 0.$ Show that the unit tangent (relative to the parameterization
$\phi $ )
to
$C$ at
$z=\phi \left(t\right)$ exists and equals
${\phi}^{\text{'}}\left(t\right)/\left{\phi}^{\text{'}}\left(t\right)\right.$ Conclude that, except possibly for a finite number of points, the unit tangent to
$C$ at
$z$ is independent of the parameterization.
 Let
$C$ be the graph of the function
$f\left(t\right)=\leftt\right$ for
$t\in [1,1].$ Is
$C$ a smooth curve?
Is it a piecewise smooth curve?Does
$C$ have a unit tangent at every point?
 Let
$C$ be the graph of the function
$f\left(t\right)={t}^{2/3}={\left({t}^{1/3}\right)}^{2}$ for
$t\in [1,1].$ Is
$C$ a smooth curve?
Is it a piecewise smooth curve?Does
$C$ have a unit tangent at every point?
 Consider the set
$C$ that is the right half of the unit circle in the plane.
Let
${\phi}_{1}:[1,1]\to C$ be defined by
$${\phi}_{1}\left(t\right)=(cos\left(t\frac{\pi}{2}\right),sin\left(t\frac{\pi}{2}\right)),$$
and let
${\phi}_{2}:[1,1]\to C$ be defined by
$${\phi}_{2}\left(t\right)=(cos\left({t}^{3}\frac{\pi}{2}\right),sin\left({t}^{3}\frac{\pi}{2}\right)).$$
Prove that
${\phi}_{1}$ and
${\phi}_{2}$ are both parameterizations of
$C.$ Discuss the existence of a unit tangent at the point
$(1,0)={\phi}_{1}\left(0\right)={\phi}_{2}\left(0\right)$ relative to these two parameterizations.
 Suppose
$\phi :[a,b]\to C$ is a parameterization of a curve
$C$ from
${z}_{1}$ to
${z}_{2}.$ Define
$\psi $ on
$[a,b]$ by
$\psi \left(t\right)=\phi (a+bt).$ Show that
$\psi $ is a parameterization of a curve from
${z}_{2}$ to
${z}_{1}.$
 Suppose
$f$ is a
smooth, realvalued function defined on the closed interval
$[a,b],$ and let
$C\subseteq {R}^{2}$ be the graph of
$f.$ Show that
$C$ is a smooth curve, and find a “natural”
parameterization
$\phi :[a,b]\to C$ of
$C.$ What is the unit tangent to
$C$ at the point
$(t,f(t\left)\right)?$
 Let
${z}_{1}$ and
${z}_{2}$ be two distinct points in
$C,$ and define
$\phi :[0,1]\to c$ by
$\phi \left(t\right)=(1t){z}_{1}+t{z}_{2}.$ Show that
$\phi $ is a parameterization of the straight line from the point
${z}_{1}$ to the point
${z}_{2}.$ Consequently, a straight line is a smooth curve.
(Indeed, what is the definition of a straight line?)
 Define a function
$\phi :[r,r]\to {R}^{2}$ by
$\phi \left(t\right)=(t,\sqrt{{r}^{2}{t}^{2}}).$ Show that the range
$C$ of
$\phi $ is a smooth curve, and that
$\phi $ is a parameterization of
$C.$
 Define
$\phi $ on
$[0,\pi /2)$ by
$\phi \left(t\right)={e}^{it}.$ For what curve is
$\phi $ a parametrization?
 Let
${z}_{1},{z}_{2},...,{z}_{n}$ be
$n$ distinct points in the plane, and suppose
that the polygonal line joing these points in order never crosses itself.Construct a parameterization of that polygonal line.
 Let
$S$ be a piecewise smooth geometric set determined by the interval
$[a,b]$ and the
two piecewise smooth bounding functions
$u$ and
$l.$ Suppose
${z}_{1}$ and
${z}_{2}$ are two points in the interior
${S}^{0}$ of
$S.$ Show that there exists a
piecewise smooth curve
$C$ joining
${z}_{1}$ to
${z}_{2},$ i.e., a piecewise smooth function
$\phi :[\widehat{a},\widehat{b}]\to C$ with
$\phi \left(\widehat{a}\right)={z}_{1}$ and
$\phi \left(\widehat{b}\right)={z}_{2},$ that lies entirely in
${S}^{0}.$
 Let
$C$ be a piecewise smooth curve, and suppose
$\phi :[a,b]\to C$ is a parameterization of
$C.$ Let
$[c,d]$ be a subinterval of
$[a,b].$ Show that the range of the restriction of
$\phi $ to
$[c,d]$ is a smooth curve.
Suppose
$C$ is a smooth curve, parameterized by
$\phi =u+iv:[a,b]\to C.$
 Suppose that
${u}^{\text{'}}\left(t\right)\ne 0$ for all
$t\in (a,b).$ Prove that there exists a smooth, realvalued function
$f$ on some closed interval
$[{a}^{\text{'}},{b}^{\text{'}}]$ such that
$C$ coincides with the graph of
$f.$ HINT:
$f$ should be something like
$v\circ {u}^{1}.$
 What if
${v}^{\text{'}}\left(t\right)\ne 0$ for all
$t\in (a,b)?$
Let
$C$ be the curve that is the
range of the function
$\phi :[1,1]\to C,$ where
$\phi \left(t\right)={t}^{3}+{t}^{6}i).$
 Is
$C$ a piecewise smooth curve?
Is it a smooth curve?What points
${z}_{1}$ and
${z}_{2}$ does it join?
 Is
$\phi $ a parameterization of
$C?$
 Find a parameterization for
$C$ by a function
$\psi :[3,4]\to C.$
 Find the unit tangent to
$C$ and the point
$0+0i.$
Let
$C$ be the curve parameterized by
$\phi :[\pi ,\pi \u03f5]\to C$ defined by
$\phi \left(t\right)={e}^{it}=cos\left(t\right)+isin\left(t\right).$
 What curve does
$\phi $ parameterize?
 Find another parameterization of this curve, but base on the interval
$[0,1\u03f5].$