

\section{Arc Length for Curves}

\subsection{Definitions}
In the rest of Chapter 1, we will assume that $\mathbf{r} : I \rightarrow \mathbb{R}^3$ is continuous unless explicitly stated otherwise. The interval $I$ may be open, closed, finite or infinite as needed. We will also need additional terminology, as follows.

\begin{definition}\label{12-def-smooth}
(i) A parametrization $\mathbf{r}:(a,b)\to\rr^n$ of a curve is \textbf{smooth} if $\mathbf{r}(t)$ is $C^1$ and, additionally, $\mathbf{r}' (t)\neq\bf{0}$ for all $t \in (a,b)$. 

\smallskip\noindent
(ii) For a closed interval $[a,b]$, a parametrization $\mathbf{r}:[a,b]\to\rr^n$ of a curve is smooth if $\bfr$
is smooth on $(a,b)$, continuous on $[a,b]$, and the one-sided limits 
$\lim_{t\to a^+}\bfr'(t)$ and $\lim_{t\to b^-}\bfr'(t)$ exist.


\smallskip
\noindent
(iii) A curve in $\rr^n$ is \textbf{smooth} if it admits a smooth parametrization. 

\smallskip
\noindent
(iv) A curve in $\rr^n$ is \textbf{piecewise smooth} if it consists of finitely many curves $\bfr_j:[a_j,b_j]\to\rr^n$, $j=1,2,\dots,k$, such that each $\bfr_j$ is smooth on $[a_j,b_j]$, and 
the curves are joined to each other at endpoints so that $\bfr_j(b_j)=\bfr_{j+1}(a_{j+1})$ for $j=1,\dots,k-1$.



\end{definition}


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/24fig2.pdf}
\caption{An example of a piecewise smooth curve.}
\end{minipage}
\end{figure}




This is purpose-specific terminology that is not universal in mathematics and does not always match terminology used in other contexts or in other textbooks. First, we only require one derivative to exist and be continuous. (In mathematical analysis, for example, a ``smooth" function is usually defined as a function whose all derivatives of all orders are continuous.) Second, we impose the additional requirement that $\bfr'$ should never be zero. This is because we want the underlying geometric object (in this case, the curve) to be smooth in an intuitive sense. Smooth geometric objects are defined here in terms of $C^1$ functions, but with further constraints added to prohibit geometric non-smoothness phenomena. The example below shows what could go wrong if we remove the condition that $\mathbf{r}' (t)\neq\bf{0}$ for a curve. We will see a similar issue later on in connection with parametrized surfaces. 



\begin{example}\label{ex:nonsmooth}
Let $\mathbf{u}$ and $\mathbf{w}$ be nonzero vectors in $\mathbb{R}^2$.
For each $t\in \rr$, define
$$
\mathbf{r}(t) : =
\begin{cases}
t^3 \mathbf{u},  \, t \geq 0 \\
t^3 \mathbf{w} , \, t \leq 0
\end{cases}
$$
Then, the associated curve consists of two half-lines meeting at the origin. Moreover, $\mathbf{r}' (t)$ is defined for all $t$, with $\mathbf{r}' (0) = {\bf 0}$ (see Practice Problems in Section 1.1), even though the curve is not geometrically ``smooth'' if ${\bf u}$ and ${\bf v}$ are not parallel. 
\end{example}


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/12fig1.png}
\caption{The vector-valued function defined in Example \ref{ex:nonsmooth}. In this picture, we have chosen $\mathbf{u} = \langle 1,1 \rangle$, $\mathbf{w} = \langle 1,-1 \rangle$.  The dashed arc indicates the direction of the parametrization.}
\end{minipage}
\end{figure}



If we think of $\mathbf{r}(t)$ as the position of a moving object at time $t$, the above example corresponds to the object moving in the direction of ${\bf w}$ for $t<0$, slowing down as $t\nearrow 0$, coming to a full stop at $t=0$, then starting out again in a new (possibly different) direction. In such cases, the trajectory may have a sharp kink even though $\mathbf{r}(t)$ is still a differentiable function of $t$. 
The condition $\mathbf{r}' (0) \neq {\bf 0} $ in Definition \ref{12-def-smooth} rules out such examples.




We have already noted that the same curve can be parametrized in many different ways. A smooth curve such as a line will also have ``hostile" parametrizations where $\bfr'(t)$ either does not exist at some points (the speed of the moving object changes instantaneously from one value to another) or takes zero values (the object comes to a full stop and then starts moving again). This is fine: for a curve to be smooth, just one smooth parametrization is enough.

Throughout Chapter 1, we will also assume that higher order derivatives of $\mathbf{r}(t)$ exist as needed.
Curves that are not smooth (for example, fractal curves) exist and are important, but the methods used to study them are very different from those covered in this course.



\begin{definition}
The curve $\mathbf{r} : [a,b] \rightarrow \mathbb{R}^3$ is \textbf{closed} if $\mathbf{r} (a) = \mathbf{r}(b)$; it is \textbf{simple and closed} if it has no other self-intersections.
\end{definition}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/12fig2.pdf}
\caption{Two closed curves in the plane; the curve on the left is simple and closed, the one on the right is closed but not simple.}
\end{minipage}
\end{figure}



\subsection{Arc Length of a Curve}
Suppose that some curve $\mathcal{C} \subset \mathbb{R}^n$ is parametrized by a continuous vector-valued function $\mathbf{r}(t)$ for $a \leq t \leq b$. Further suppose that $\mathbf{r}$ is \underline{one-to-one} so  that, as $\mathbf{r}(t)$ travels from $\mathbf{r}(a)$ to $\mathbf{r}(b)$, it never covers the same segment of the curve more than once. Now, subdivide the interval $a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b$, and define $\mathbf{r}_i = \mathbf{r}(t_i)$ and let
$$
S_{t_0,...,t_n} : = \sum\limits_{i=1}^{n} \vert \mathbf{r}_i - \mathbf{r}_{i+1} \vert.
$$
Then, $S_{t_0,...,t_n}$ approximates the length of the curve $\mathcal{C}$ by the sum of the lengths of connected line segments whose endpoints lie on $\mathcal{C}$.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .75\linewidth]{figures/12fig3.pdf}
\caption{A curve $ \mathcal{C}$ approximated by line segments.}
\end{minipage}
\end{figure}

Moreover,  as more points are added to the subdivision $t_0,...,t_n$, the associated length $S_{t_0,...,t_n}$ is \underline{non-decreasing}. You can prove this using the \textit{triangle inequality}: a line segment from $P$ to $Q$ directly is shorter than the triangular path from $P$ to a third point $R$ and then from $R$ to $Q$.


\begin{definition}
Let $\mathcal{C}$ be a curve parametrized by a continuous vector-valued function $\mathbf{r}(t)$ for $t\in I$, where $I$ is an interval.

\begin{itemize}
\item[(i)] Assume that $I$ is compact, so that $I=[a,b]$ for some finite $a,b\in\rr$. 
We say that $\mathcal{C}$ is \textbf{rectifiable} if there is a constant $K > 0$ such that, with the above notation, we have $S_{t_0,...,t_n} \leq K$ for all $n$ and for all possible choices of partitions. The \textbf{smallest such} $K$ is called the \textbf{length} of $\mathcal{C}$.

\item[(ii)] Assume that $I$ is not compact (hence either unbounded or open at one or both ends). We say that $\mathcal{C}$ is rectifiable if, for any compact interval $[a,b]\subset I$, the curve obtained by restricting $\mathcal{C}$ to the values of $t\in[a,b]$ is rectifiable.
\end{itemize}


\end{definition}

Many familiar curves such as a circle, ellipse, or line segment are rectifiable. There are also many curves of infinite length (line, parabola, helix, etc.) that are rectifiable because any bounded part of them is rectifiable. Fractal curves are often unrectifiable. The Koch snowflake curve (see Figure \ref{fig:kochsnowflake}) provides one such example.

\begin{figure}[h!]
        \centering
        \begin{subfigure}[b]{0.45\textwidth}
            \centering
            \includegraphics[width=\textwidth]{figures/koch0}
        \end{subfigure}
        \hspace{.2in}
        \begin{subfigure}[b]{0.45\textwidth}  
            \centering 
            \includegraphics[width=\textwidth]{figures/koch1}
        \end{subfigure}
        \vskip\baselineskip
        \begin{subfigure}[b]{0.45\textwidth}   
            \centering 
            \includegraphics[width=\textwidth]{figures/koch2}
        \end{subfigure}
        \hspace{.2in}
        \begin{subfigure}[b]{0.45\textwidth}   
            \centering 
            \includegraphics[width=\textwidth]{figures/koch3}
        \end{subfigure}
        \caption{\label{fig:kochsnowflake} The Koch snowflake is an example of a \textit{non-rectifiable} curve.  This curve is the boundary of a region (pictured above) with finite area -- and yet, the curve itself has ``infinite'' length.  The Koch snowflake is constructed via an iterative process: a simple triangular pattern is repeated many times to create a more complicated pattern. The first four steps of this pattern-copying process are shown above.  Tools from this class, as well as classes such as MATH 320 and MATH 321, will give you the tools to begin exploring more exotic geometric objects such as this curve!}
    \end{figure}

\subsection{Computing the Length of the Curve} 
We have the following formula for computing the length of curves parametrized by continuously differentiable vector-valued functions.

\begin{theorem}\label{thm:contdifflength}
If $\mathbf{r}(t)$ has a continuous derivative $\mathbf{v}(t)$, then the length of the curve from $t = a$ to $t = b$ is
$$
S = \int_a^b \vert \mathbf{r}' (t) \vert dt = \int_a^b \vert \mathbf{v}(t) \vert dt = \int_a^b v (t) dt.
$$
\end{theorem}

\begin{proof}[Sketch of Proof]
If we take some partition $t_i$ and $\mathbf{r}_i$, let $\Delta \mathbf{r}_i : = \mathbf{r}_i - \mathbf{r}_{i-1}$ so that
\begin{equation}\label{eq:proof1}
S_{t_0,...,t_n} = \sum\limits_{i = 1}^n \big\vert \Delta \mathbf{r}_i \big\vert  
\end{equation}
It is tempting to write $\Delta\bfr_i=\bfr'(t^*_i)\Delta t_i$ for some $t^*_i\in(t_{i-1},t_{i})$ by the Mean Value Theorem, then take the limit of the Riemann sums. This does not quite work because there is no Mean Value Theorem for vector-valued functions (see Practice Problems in Section \ref{sec:lecture11}). One workaround is to 
write the difference $\Delta \mathbf{r}_i$ in terms of its component functions:
$$
\Delta \mathbf{r}_i = \Delta x_i \mathbf{i} + \Delta y_i \mathbf{j} + \Delta z_i \mathbf{k}.
$$
Then equation \eqref{eq:proof1} can be re-written as
\begin{eqnarray*}
S_{t_0,...,t_n} & = & \sum\limits_{i=1}^n  \sqrt{\vert \Delta x_i \vert^2 + \vert \Delta y_i \vert^2 + \vert \Delta z_i \vert^2}
\\
\quad & = &  \sum\limits_{i=1}^n \left[ \sqrt{\left| \frac{\Delta x_i}{\Delta t_i} \right|^2 + \left| \frac{\Delta y_i}{\Delta t_i} \right|^2 + \left| \frac{\Delta z_i}{\Delta t_i} \right|^2}\, \right] \cdot  \Delta t_i \\
\quad & = & \sum\limits_{i = 1}^{n} \bigg[ \sqrt{\vert x' (t_i^{(1)}) \vert^2 + \vert y' (t_i^{(2)}) \vert^2 + \vert z' (t_i^{(3)}) \vert^2 } \,\bigg] \cdot \Delta t_i
\end{eqnarray*}
for some points $t_i^{(1)}, t_i^{(2)}$ and $t_i^{(3)}$ in $I_j : =(t_{i-1}, t_i)$; at the last step, we applied the Mean Value Theorem to the (scalar) component functions.  This is still not quite a Riemann sum since $t_i^{(1)}, t_i^{(2)},t_i^{(3)}$ need not be equal. However, since $x',y'$ and $z'$ are continuous, this does not make much difference and we can still conclude (as in single-variable calculus) that
the above converges to
$$
\int_{a}^{b} \sqrt{\vert x' (t) \vert^2 + \vert y' (t) \vert^2 + \vert z' (t) \vert^2} dt = \int_a^b \vert \mathbf{r}' (t) \vert dt,
$$
where the convergence is over refinements of partitions satisfying $\max_i \Delta t_i \rightarrow 0$.  



\end{proof}

\subsection{Arc-Length Parametrization} 
Let $\mathbf{r} : [a,b] \rightarrow \mathbb{R}^3$ be smooth, and recall that $v(t)=|\mathbf{v}(t)|$.
Define, for $t\in[a,b]$,
$$
s (t) : = \int_a^{t} \vert \mathbf{r}' (\tau) \vert d \tau = \int_a^t \vert v (\tau) \vert d \tau.
$$
This is the length of the curve from $\mathbf{r}(a)$ to $\mathbf{r}(t)$. For this reason, we call $s(t)$ the \textbf{arc-length function}. By the Fundamental Theorem of Calculus, we have
\begin{eqnarray*}
\quad & \quad & \frac{ds}{dt} = \frac{d}{dt} \bigg( \int_a^t v (\tau) d \tau \bigg) = v(t) \\[1ex]
\quad & \Rightarrow & ds = v(t) dt = \vert \mathbf{r}' (t) \vert d t
\end{eqnarray*}
We therefore call the quantity $ds = \vert \mathbf{r}' (t) \vert dt$ the \textbf{arc-length element} associated to $\mathbf{r}$.

Now, suppose we can solve for $t$ as a function of $s$; then, we can use this change of variables to reparametrize the curve in terms of arc length. That is, we want to write $\mathbf{r} = \mathbf{r}(t(s))$ as our new parametrization.  The advantage of this parametrization is that
\begin{equation}\label{eq:arclengthpara}
\Big\vert \frac{d \mathbf{r}}{d s} \Big\vert = \Big\vert \frac{d \mathbf{r}}{dt} \cdot \frac{dt}{ds} \Big\vert = \Big\vert \frac{d \mathbf{r}}{dt} \Big\vert \cdot \Big\vert \underbrace{\frac{1}{ds/dt}}_{(*)} \Big\vert = v(t) \cdot \frac{1}{v(t)} = 1,
\end{equation}
where $(*)$ follows from the Implicit Function Theorem and our assumption that $\frac{ds}{dt} \neq 0$.  In other words, the arc-length parametrization describes an object moving along the given curve with constant speed $ 1$.

\begin{remark}
Even if we cannot solve for $t$ as a function of $s$ \underline{explicitly}, it is still useful to know that such parametrization exists. We will use it in what follows to define various natural quantities associated to smooth curves.
\end{remark}


\begin{example}
A very important special case is when our curve $\mathcal{C}$ is the graph of some function $y = f(x)$ for $a \leq x \leq b$ in the plane.  We may then parametrize our curve by the vector-valued function
$$
\mathbf{r}(t) : = t \mathbf{i} + f(t) \mathbf{j}, \quad a \leq t \leq b,
$$
so that the length is given by the formula
$$
s(t) : = \int_a^b \vert \mathbf{r}' (x) \vert dx = \int_a^b \sqrt{1 + \vert f' (x) \vert^2} dx
$$
\end{example}


\section*{Worked Problems}
\begin{enumerate}[1.]
\item{\it 
Consider the helix
$$
\mathbf{r}(t) : = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j} + b t \mathbf{k}, \quad t \in \mathbb{R}.
$$
where $a,b > 0$ are some real numbers.  


\medskip\noindent
(a) Find the length of part of the helix where $0 \leq t \leq T$ for some $T > 0$.


\medskip\noindent
(b) Reparametrize the helix in terms of arc length.}


\bigskip
\noindent
(a) To find the length, we first need to determine
$
\vert \mathbf{r}' (t) \vert . 
$
Since $\bfr'(t) = - a \sin \mathbf{i} + a \cos (t) \mathbf{j} + b \mathbf{k}$,  we have
$$
\vert \mathbf{r}' (t) \vert = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t + b^2} = \sqrt{a^2 + b^2},
$$
so that
$$
s(T) : = \int_0^T \sqrt{a^2 + b^2} d \tau = (\sqrt{a^2 + b^2}) T.
$$

\medskip\noindent
(b) 
From (a), we have $v(t) : = \vert \bfr'(t) \vert = \sqrt{a^2 + b^2}$. 
Let us use $t_0 = 0$ as our starting point for measuring length, so that
$$
s(t) : = \int_0^t v(\tau) d \tau = \int_0^t \sqrt{a^2 + b^2} \, d \tau = (\sqrt{a^2 + b^2}) t
$$
So, we see that
$
t = t(s) = \frac{s}{\sqrt{a^2 + b^2}}.
$
Performing this change of variables gives us the reparametrization
$$
\mathbf{r}(s) = a \cos \bigg(\frac{s}{\sqrt{a^2 + b^2}} \bigg) \mathbf{i} + a \sin \bigg( \frac{s}{\sqrt{a^2 + b^2}} \bigg)\mathbf{j} + b \bigg( \frac{s}{\sqrt{a^2 + b^2}} \bigg) \mathbf{k}.
$$

\bigskip


\item{\it A curve in $\RR^3$ is given by
\begin{equation}\label{ex:moment}
\mathbf{r}(t) : = 2 t \mathbf{i} + t^2 \mathbf{j} + \frac{t^3}{3} \mathbf{k}.
\end{equation}
Find the length of the curve  from $t = 1$ to $t = 2$.}



\medskip
\noindent
We have $\mathbf{r}' (t) = 2 \mathbf{i} + 2 t \mathbf{j} + t^2 \mathbf{k}$. Therefore
$$
v(t) : = |\mathbf{r}' (t) | = \sqrt{4 + 4 t^2 + t^4} = \sqrt{(2 + t^2)^2} = |t^2 + 2| = t^2 + 2,
$$
since $t^2 + 2  \geq 0$.  Hence,  the length from $t = 1$ to $t = 2$ is given by
$$
 \int_1^2 (t^2 + 2) dt 
= \bigg[\frac{t^3}{3} + 2 t \bigg]_{t = 1}^{2} = \frac{8}{3} + 4 - \frac{1}{3} - 2 = \frac{13}{3}.
$$




\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .8\linewidth]{figures/12fig4.pdf}
\caption{The curve in (\ref{ex:moment}) and its associated arc-length function $s(t)$.}
\end{minipage}
\end{figure}


\end{enumerate}


\begin{remark}
In the last example, we have $v(t) = t^2 + 2$, so that the arc-length function (measuring from $t=0$ as a base point) is
$$
s(t) : = \int_0^t (\tau^2 + 2) d \tau = \frac{t^3}{3} + 2 t.
$$
The function $t \rightarrow s(t)$ is one-to-one (why?), therefore it \underline{indeed has} an inverse function $t = t(s)$. However, to find such a function, we would need to solve the cubic equation appearing previously---a not so easy task. This is typical: in fact, many common arc-length functions cannot be expressed in terms of elementary functions. That will not stop us from working with them!
\end{remark}





\section*{Practice Problems}
\begin{enumerate}[1.]

\item 
{
Let $C$ be the curve $x=e^{2t}$, $y=2e^t $, $z=t$ for $t \in\mathbb{R}$. 
Find the length of the curve from $t=0$ to $t=1$.}

\bigskip

\item 
{ Let $\bfr(t)=e^t\cos(2t) \ii + e^t \sin(2t)\jj$ be a parametrization of a curve $C$ in $\mathbb{R}^2$. 

\medskip
\noindent
(a)
Let $s(b)$ be the length of the curve $\bfr(t)$ from $t=-\infty$ to $t=b$. (This is defined as the limit
as $a\to -\infty$ of the length of $\bfr(t)$ from $t=a$ to $t=b$.) Find $s(b)$ as a function of $b$. }

\medskip
\noindent
{(b) Use part (a) to give an arc-length parametrization of $C$.}

\bigskip

\item 
The paraboloid $z=x^2+y^2+1$ intersects the surface $x^2-2x+y^2=0$ in a curve. Prove that the length of this curve is less than 15.











\end{enumerate}



