
\section{Unit Tangent and Curvature}\label{sec:tangentAndCurvature}
This section seeks to answer the following.

\smallskip
\noindent
\fbox{
\begin{minipage}{\linewidth}
\begin{question}
How quickly does a given curve change direction?
\end{question}
\end{minipage}
}

\smallskip

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .75\linewidth]{figures/13fig1.pdf}
\caption{Two curves; the right-hand curve seems to be changing direction more slowly than the left-hand curve.  The vector overlaid on the points of the curve is the tangent vector at that point.}
\end{minipage}
\end{figure}

For curves, \textit{direction} is represented by a unit tangent vector. (We use the \textit{unit} tangent vector because we are only interested in its direction, not magnitude.) Moreover, we want to make sure that we have a geometric definition for the phrase ``how quickly''.  We say \textit{geometric} because we want this definition to depend solely upon the shape of the curve, and not upon the choice of parametrization (which is a more analytic object).  For this reason, we will use arc-length parametrization, so that motion along the curve has constant speed equal to one.

\begin{remark}
Even when it may be difficult or impossible to \underline{calculate} the arc-length parametrization, we can still use it to \underline{define} new concepts.  Eventually, we will have formulas for the same concepts that do not require us to know an explicit formula for arc length.
\end{remark}

\subsection{Key Ideas and Definitions}
We assume $\mathbf{r} : I \rightarrow \mathbb{R}^3$ is a smooth parametrization of a curve, for some interval $I$. In the sequel, whenever we work both with the initial parametrization $\bfr(t)$ and the arc-length reparametrization of the same curve, we will use $\bfr'$, $\bfr''$, etc.~to denote derivatives with respect to $t$, and $d\bfr/ds$, $d^2\bfr/ds^2$, etc.~to denote differentiation in $s$. In the rest of this chapter, we will always assume that all the needed derivatives exist and are continuous.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .3\linewidth]{figures/13fig2.pdf}
\caption{A tangent vector to some curve.}
\end{minipage}
\end{figure}


We define $\mathbf{T}$ to be the \textbf{unit tangent vector} to the curve. This is a geometric object depending (up to the choice of direction) only on the shape of the curve. In terms of the parametrization $\mathbf{r}$, we have
\begin{equation}\label{eq:tangentvec}
\mathbf{T}(t) : = \frac{\mathbf{v}(t)}{\vert \mathbf{v}(t) \vert} = \frac{\mathbf{r}' (t)}{\vert \mathbf{r}' (t) \vert},  \textrm{ so that } \vert \mathbf{T}(t) \vert = 1,
\end{equation}
with direction matching the direction of increasing $t$. 
If we use arc-length parametrization $s$, then $\vert \frac{d \mathbf{r}}{ds} \vert = 1$, so that
$$
\mathbf{T}(s) = \frac{d \mathbf{r} / d s}{\vert d \mathbf{r} / ds \vert} = \frac{d \mathbf{r}}{d s}.
$$


The \textbf{curvature} $\kappa : = \kappa (s)$ is defined relative to the arc-length parametrization as
$$
\kappa (s) : = \left\vert \frac{d \mathbf{T}(s)}{d s} \right\vert,
$$
assuming that the derivative exists. In words, $\kappa (s)$ measures the rate of change of $\mathbf{T}$ if we move along the curve with unit speed. If $\mathbf{r}$ is any parametrization of the curve, we may use the chain rule to write $\kappa (s) = \kappa(s(t))$ as
\begin{eqnarray*}
\kappa (t) & = & \left\vert \frac{d \mathbf{T}}{ds} \right\vert = \left\vert \frac{d \mathbf{T}}{dt} \cdot \frac{dt}{ds} \right\vert = \frac{\vert d \mathbf{T} / dt \vert }{\vert ds /dt \vert} = \frac{\vert \mathbf{T}' (t) \vert}{\vert s' (t) \vert} \\[1ex]
\quad & = & \frac{\vert \mathbf{T}' (t) \vert}{\vert \mathbf{r}' (t) \vert}.
\end{eqnarray*}

\subsection{Examples, and a Useful Formula for Curvature}

\begin{example}
Consider the straight line $\mathbf{r}(t) : = \mathbf{r}_0 + t \mathbf{v}$. Then,
$$
\mathbf{T} : = \frac{\mathbf{v}}{\vert \mathbf{v} \vert} \textrm{ is constant},
$$
and so $\kappa = \vert \frac{d \mathbf{T}}{d s} \vert = 0$.
\end{example}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .25\linewidth]{figures/13fig3.pdf}
\caption{A line segment in the direction of $\mathbf{v}$.}
\end{minipage}
\end{figure}

\begin{example}
Consider the circle $\mathbf{r}(t) = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j}$ for $a > 0$.  To compute the tangent vector $\mathbf{T}$, we first find
$$
\mathbf{r}' (t) : = -a \sin (t) \mathbf{i} + a \cos (t) \mathbf{j}, \quad \vert \mathbf{r}' (t) \vert = a,
$$
so that
$$
\mathbf{T}(t) = \frac{\mathbf{r}' (t)}{\vert \mathbf{r}' (t) \vert} = -\sin (t) \mathbf{i} + \cos (t) \mathbf{j}.
$$
To compute the curvature, we need
$$
\mathbf{T}' (t) = - \cos (t) \mathbf{i} - \sin (t) \mathbf{j}, \quad \vert \mathbf{T}' (t) \vert = 1,
$$
so that
$$
\kappa = \frac{\vert \mathbf{T}' (t)\vert}{\vert \mathbf{r}' (t) \vert} = \frac{1}{a}.
$$
In particular, we see that as $a \rightarrow 0$, we must have $\kappa \rightarrow \infty$, so that large values of curvature correspond to small values of radius $a > 0$.
\end{example}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .25\linewidth]{figures/13fig4.pdf}
\caption{A circle with associated tangent vector.}
\end{minipage}
\end{figure}

\begin{remark}
For \underline{general} curves, the quantity $\rho : = \frac{1}{\kappa}$ is called the \textbf{radius of curvature}. It is the radius of the circle (in the three-dimensional space) that best approximates the curve at the given point -- more on this later. If the curve is an actual circle, then $\rho = a$ is its radius.
\end{remark}

\begin{example}
Take the helix $\mathbf{r}(t) = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j} + b t \mathbf{k}$ where both $a,b > 0$.  Then, 
$$
\mathbf{r}' (t) = -a \sin (t) \mathbf{i} + a \cos (t) \mathbf{j} + b \mathbf{k}, \quad \vert \mathbf{r}' (t) \vert = \sqrt{a^2 + b^2}.
$$
so that
$$
\mathbf{T}(t) : = \frac{\mathbf{r}' (t)}{\vert \mathbf{r}' (t) \vert} = - \frac{a}{\sqrt{a^2 + b^2}} \sin (t) \mathbf{i} + \frac{a}{\sqrt{a^2 + b^2}} \cos (t) \mathbf{j} + \frac{b}{\sqrt{a^2 + b^2}} \mathbf{k},
$$
$$
\mathbf{T}'(t) : =  - \frac{a}{\sqrt{a^2 + b^2}} \cos (t) \mathbf{i} - \frac{a}{\sqrt{a^2 + b^2}} \sin (t) \mathbf{j}.
$$
Therefore
$$
\kappa (t) : = \frac{\vert \mathbf{T}' (t)\vert}{\vert \mathbf{r}' (t) \vert} = \frac{\frac{a}{\sqrt{a^2 + b^2}}}{\sqrt{a^2 + b^2}} = \frac{a}{a^2 + b^2}
$$
and so our curvature is constant.

\smallskip
Note that if either $a$ is fixed and $b \rightarrow \infty$ (the helix gets stretched out in the $z$-direction), or if 
$a \rightarrow \infty$ as $b$ is fixed (the vertical speed is constant but the radius of the cylinder around which the helix is wrapped increases to infinity), we have $\kappa \rightarrow 0$.
\end{example}

We conclude this section with a formula for computing curvature which is often more practical than the formula given by the definition. As mentioned above, we use primes to denote differentiation in $t$, for example ${\bf T}'=\frac{d{\bf T}}{dt}$. Differentiation in $s$ will always be written as $\frac{d}{ds}$.

\begin{theorem}\label{thm:curvatureformula}
If $\mathbf{r}$ is twice differentiable and $\mathbf{r}' \neq 0 $, then
\begin{equation}\label{eq:curvthm}
\kappa (t) = \frac{\vert \mathbf{r}' (t) \times \mathbf{r}'' (t) \vert}{\vert \mathbf{r}' (t) \vert^3}
\end{equation}
\end{theorem}

\begin{proof}
We begin by computing the cross product $\mathbf{r}' \times \mathbf{r}''$ in terms of $\mathbf{T}$ and the arc-length parametrization.  Recall that $\mathbf{T} : = \frac{\mathbf{r}' (t)}{\vert \mathbf{r}' (t) \vert}$ and $\vert \mathbf{r}' (t) \vert = \frac{ds}{dt}$. 
Therefore
\begin{equation}\label{eq:thm1}
\mathbf{r}' (t) = \vert \mathbf{r}' (t) \vert \mathbf{T}(t) = \frac{ds}{dt} (t) \mathbf{T}(t).
\end{equation}
To find $\mathbf{r}'' (t)$, we differentiate the above with respect to $t$, obtaining
\begin{equation}\label{eq:thm2}
\mathbf{r}'' = \frac{d}{dt} \left(\frac{ds}{dt} \mathbf{T}\right) = \frac{d^2 s}{d t^2} \mathbf{T} + \frac{ds}{dt} \mathbf{T}'.
\end{equation}
Using \eqref{eq:thm1} and \eqref{eq:thm2}, we find that
\begin{eqnarray*}
\mathbf{r}' \times \mathbf{r}'' & = & \bigg( \frac{ds}{dt} \mathbf{T} \bigg) \times  \bigg(\frac{d^2 s}{d t^2} \mathbf{T} + \frac{ds}{dt} \mathbf{T}' \bigg) \\[1ex]
\quad & = & \frac{ds}{dt} \frac{d^2 s}{d t^2} (\mathbf{T} \times \mathbf{T}) + \bigg(\frac{ds}{dt} \bigg)^2\, \mathbf{T} \times \mathbf{T}'
\end{eqnarray*}
From the definition of the cross product, we know that $\mathbf{T} \times \mathbf{T} = 0$. Therefore
\begin{equation}\label{eq:thm3}
 \mathbf{r}' \times \mathbf{r}''  = \bigg( \frac{ds}{dt} \bigg)^2 \, \mathbf{T} \times \mathbf{T}' 
\end{equation}
Now, we know that $\vert \mathbf{T} \vert = 1$ for all values of $t$; this further tells us that $\mathbf{T} \perp \mathbf{T}'$ for all values of $t$ (Worked Problems, Section 1.1).  Consequently, we see that
\begin{equation}\label{eq:thm4}
\vert \mathbf{T} \times \mathbf{T}' \vert = \vert \mathbf{T} \vert \cdot \vert \mathbf{T}' \vert \sin (\frac{\pi}{2}) = \vert \mathbf{T}' \vert.
\end{equation}
Putting together equations \eqref{eq:thm3} and \eqref{eq:thm4} shows that
$$
\vert \mathbf{r}' \times \mathbf{r}'' \vert = \bigg(\frac{ds}{dt} \bigg)^2 \vert \mathbf{T}' \vert = \vert \mathbf{r}' \vert^2 \vert \mathbf{T}' \vert \Rightarrow \vert \mathbf{T}' \vert = \frac{\vert \mathbf{r}' \times \mathbf{r}''\vert}{\vert \mathbf{r}' \vert^2}.
$$
The definition of curvature then gives that
$$
\kappa : = \frac{\vert \mathbf{T}' \vert}{\vert \mathbf{r}' \vert} = \frac{{\vert \mathbf{r}' \times \mathbf{r}'' \vert/\vert \mathbf{r}' \vert^2}}{\vert \mathbf{r}' \vert} = \frac{\vert \mathbf{r}' \times \mathbf{r}'' \vert}{\vert \mathbf{r}' \vert^3}.
$$
\end{proof}

\section*{Worked Problems}
\begin{enumerate}[1.]
\item{\it Compute the curvature $\kappa(t)$ if $\mathbf{r} (t) = \langle 2t, t^2, \frac{t^3}{3}\rangle$ using Theorem \ref{thm:curvatureformula}}.

\medskip
\noindent
We first calculate the following
$$
\mathbf{r}' (t) = \langle 2, 2t, t^2 \rangle, \quad \vert \mathbf{r}' (t) \vert  = \sqrt{4 + 4t^2 + t^4} = t^2 + 2.
$$
$$
\mathbf{r}'' (t) = \langle 0, 2, 2t \rangle
$$
The associated cross product is then given by:
\begin{eqnarray*}
\mathbf{r}' \times \mathbf{r}'' & = &
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 2 t & t^2 \\
0 & 2 & 2t
\end{vmatrix}
= (4t^2 - 2 t^2) \mathbf{i} - 4t \mathbf{j} + 4 \mathbf{k} \\[1ex]
\quad & = & 2 t^2 \mathbf{i} - 4t \mathbf{j} + 4 \mathbf{k},
\end{eqnarray*}
which has length
$$
\vert \mathbf{r}' (t) \times \mathbf{r}'' (t) \vert = 2 \sqrt{t^4 + 4t^2 + 4} = 2 (t^2 + 2).
$$
The curvature is then given by the formula
$$
\kappa (t) = \frac{2 (t^2 + 2)}{(t^2 + 3)^3} = \frac{2}{(t^2 + 2)^2}
$$

\bigskip






\item
{\it Let $\mathcal{C}$ be the curve at the intersection of the cylinder $x^2+y^2=1$ in $\mathbb{R}^3$
with the plane $z=ax$ for some $a>0$. Parametrize the curve, and find its curvature as a function of the parameter.}

\bigskip


The curve can be parametrized as ${\bf r}(t)=\cos t{\bf i}+\sin t{\bf j}+a\cos t {\bf k}$, $0\leq t\leq 2\pi$. Then
$${\bf r}'(t)=-\sin t{\bf i}+\cos t{\bf j}-a\sin  t {\bf k}$$
$${\bf r}''(t)=-\cos t{\bf i}-\sin t{\bf j}-a\cos t {\bf k}$$
$$|{\bf r}'|=\sqrt{\sin^2 t+\cos^2 t + a^2\sin^2  t }=\sqrt{1+a^2\sin^2 t}$$





$${\bf r}'(t)\times {\bf r}''(t)=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
-\sin t& \cos t& -a \sin t\\
-\cos t & -\sin t& -a \cos t\\
            \end{matrix}\right|
 =(-a\cos^2 t -a\sin^2 t){\bf i}$$
 $$
 -(a\cos t\sin t -a\sin t \cos t){\bf j}     + (\sin^2 i +\cos^2 t){\bf k}
 =-a{\bf i}+{\bf k}
 $$
 
 
 $$
 |       {\bf r}'(t)\times {\bf r}''(t)|=\sqrt{a^2+1}
 $$
 $$
\kappa=\frac{|{\bf r}'(t)\times {\bf r}''(t)|}{|{\bf r}'|^3}=\frac{\sqrt{a^2+1}}{(1+a^2\sin^2t)^{3/2}}
$$























\end{enumerate}

\section*{Practice Problems}
\begin{enumerate}[1.]
\item
{Let $\mathcal{C}$ be the curve $x=\frac{t^3}{3}-2t$, $y=\frac{t^3}{3}+2t  $, $z=\sqrt{2}t^2$ for $t \in\mathbb{R}$. 

\medskip\noindent
(a)  Find the arc length of the curve from $t=0$ to $t=1$.}

\medskip\noindent
{(b) Find the curvature at the point where $t=0$.}

\medskip

\item
{
Let $\bfr(t)=t\ii+\frac{t^2}{2}\jj $, $t\in\rr$ (a parabola in $\rr^2$), and let ${\bf T}(t)$ be the unit tangent vector to
$\bfr(t)$. Suppose that $\bfr(u)$, $u\in I$ (for some interval $I$), is a smooth reparametrization of the parabola such that
$$
\left| \frac{d{\bf T}}{du} \right| =1.
$$
Find $\displaystyle \frac{du}{dt}$ as a function of $t$. Prove that the interval $I$ is finite, and find the length of it.}

\medskip


\item {
Let $\bfr:\rr\to \rr^3$ be given by 
$$
\bfr(t)=\cos(t)\ii + \sin(t)\jj + f(t)\kk,
$$
where $f:\rr\to\rr$ satisfies $f(t)=f(t+2\pi)$. Assume that $f$ and all its derivatives are continuous. 
Thus $\bfr(t)$ parametrizes a closed smooth curve that lies on the cylinder $x^2+y^2=1$, with each $t$-interval of length $2\pi$ covering the entire curve. Prove that there is a point $\bfr(t_0)$ on the curve where $\kappa\leq 1$.  }






\medskip

\item 
{ Let ${\bf r}(t)$ be a smooth vector-valued function,
and suppose that there is a point $P$ that lies on every normal plane to ${\bf r}(t)$. Prove
that the curve lies on the surface of a sphere.}


\end{enumerate}


