
\section{The Principal Normal and The Osculating Plane}
In this section, we introduce many new vector-valued functions and definitions associated to a curve $\mathbf{r} (t)$ in three-dimensional space. Each of these will have a natural geometric interpretation, motivated by the following general question.

\smallskip
\noindent
\fbox{
\begin{minipage}{\textwidth}
\begin{question}\label{1.4q-mainquestion}
Given a curve in $\mathbb{R}^3$, how can we measure the ``three-dimensionality'' of the that curve?
\end{question}
\end{minipage}
}

\smallskip

\noindent
We will see that the \textbf{osculating plane} will be the plane of ``best fit'' for the curve at any given point, whereas a quantity called \textbf{torsion} will measure how far the curve is from fitting into said plane.  These two objects give an answer to Question \ref{1.4q-mainquestion}. Here and in the rest of this chapter, we will assume that $\bfr$ has as many derivatives as we need.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .35\linewidth]{figures/14fig1.png}
\caption{A curve in $\mathbb{R}^3$, together with its osculating plane (given by the equation $z = 0$) and its projection (i.e. \textit{shadow}) onto that plane.}
\end{minipage}
\end{figure}

\subsection{Unit Principal Normal} Recall that the unit tangent vector is given by $\mathbf{T} = \frac{d \mathbf{r}}{d s}$ in the arclength parametrization. It has the properties
$$
\mathbf{T} = \frac{\mathbf{r}'}{\vert \mathbf{r}' \vert}, \, \vert \mathbf{T} \vert = 1.
$$
Assuming that the curvature $\kappa=|\frac{d \mathbf{T}}{d s}|$ is nonzero, 
we define 
\begin{eqnarray}\label{1.4eq-normalvector}
\mathbf{N} (s) : = \frac{d\mathbf{T}/ds}{\vert d\mathbf{T}/ds\vert}  & \quad & \textrm{(Normal Vector)}.
\end{eqnarray}
This is a new vector-valued function, normalized so that $\mathbf{N} (s)$ is always a unit vector. Geometrically, $\mathbf{N}$ indicates the direction in which $\mathbf{T} (s)$ is bending as we travel along the curve. We call $\mathbf{N}$ the \textbf{unit principal normal vector} or just the \textbf{principal normal} to $\mathbf{T}$.

\begin{proposition}
We always have $\mathbf{N} \perp \mathbf{T}$.
\end{proposition}

\begin{proof}
We know that $\vert \mathbf{T} (t) \vert = 1$ for all  $t$, and we proved earlier (Section 1.1) that this implies $\mathbf{T} (t) \perp \mathbf{T}' (t)$.  Hence, by \eqref{1.4eq-normalvector}, we have $\mathbf{N} \perp \mathbf{T}$.  
\end{proof}


In order to actually compute $\mathbf{N}$, it is typically easier to use the original $\mathbf{r}(t)$ parametrization rather than arc length. Our formula for this is
$$
\boxed{
\mathbf{N}(t) = \frac{d \mathbf{T} / d s}{\vert d\mathbf{T} / ds \vert} = \frac{\frac{d \mathbf{T}}{d t} \cdot \frac{dt}{ds}}{\vert \frac{d \mathbf{T}}{d t} \cdot \frac{dt}{ds} \vert} = \frac{\mathbf{T}' (t)}{\vert \mathbf{T}' (t) \vert}.
}
$$
Notice that, since $\frac{ds}{dt} : = \vert \mathbf{r}' (t) \vert > 0$, we always have $\frac{dt}{ds} = \vert \frac{dt}{ds} \vert$.


\begin{figure}[h]
    \centering
    \begin{minipage}{0.45\textwidth}
        \centering
        \includegraphics[width=0.5\textwidth]{figures/14fig2.pdf}
        \caption{{\bf T} and {\bf N} at a point}
    \end{minipage}\hfill
    \begin{minipage}{0.5\textwidth}
        \centering
        \includegraphics[width=0.5\textwidth]{figures/14fig3.pdf}
        \caption{{\bf T} and {\bf N} along the curve}
    \end{minipage}
\end{figure}

\subsection{The Osculating Plane and Circle}

Equipped with our definitions of the tangent vector $\mathbf{T}(t)$ and the normal vector $\mathbf{N} (t)$ to a curve $\mathbf{r} (t)$, we define natural geometric objects to answer Question \ref{1.4q-mainquestion}.

The \textbf{osculating plane} at a point $\bfr(t)$ is the plane passing through $\bfr(t)$ and parallel to the vectors $\mathbf{T}$ and $\mathbf{N}$. 
This is the plane that offers the ``best fit'' (among planes) for the curve $\mathbf{r}$ at the given point, in the following sense. Assume that $\kappa(t)\neq 0$, and let $t_1,t_2,t_3$ be distinct and close enough to $t$. It can then be proved that 
$\bfr(t_1),\bfr(t_2),\bfr(t_3)$ are not collinear, so that there is a unique plane $\Pi$ passing through all three points. Further, if we take the limit as $t_1,t_2,t_3\to t$, then $\Pi$ converges to the osculating plane at $\bfr(t)$. See the Practice Problems for a specific example of this.

Given a parametrization $\bfr(t)$, we can find the equation of the osculating plane using the formula (\ref{eq:thm3}). We recall that formula here:
$$
 \mathbf{r}' \times \mathbf{r}''  = \big( \frac{ds}{dt} \big)^2 \, \mathbf{T} \times \mathbf{T}' .
$$
The osculating plane, by the definition above, should be perpendicular to both ${\bf T}$ and ${\bf N}= \frac{\mathbf{T}'}{\vert \mathbf{T}' \vert}$. Hence the normal to it should be parallel to the cross product ${\bf T}\times {\bf T}'$, therefore also to $ \mathbf{r}' \times \mathbf{r}'' $. This offers a way to find the equation of the osculating plane directly by computing $\mathbf{r}' \times \mathbf{r}''$ and using it as a normal vector. We also note the physical interpretation of this: at any given point on the curve, the osculating plane is parallel both to the velocity $\bfr'$ and the acceleration $\bfr''$. We will have more to say about this in Section 1.6.



Within the osculating plane spanned by $\mathbf{T} (t)$ and $\mathbf{N}(t)$,  we also construct the \textbf{osculating circle}.  The osculating circle is tangent to the curve at the given point $\mathbf{r} (t)$, with radius $\rho = \frac{1}{\kappa}$ and center at $\mathbf{r} (t) + \rho \mathbf{N} (t)$. 

The osculating circle has the same tangent vector and osculating plane as the curve $\mathbf{r} (t)$.  It also has the same \textit{curvature} as $\mathbf{r} (t)$ for the given value of $t$ (see Worked Problem 1 below). Geometrically, it is the circle that gives the best approximation (among all circles) to the curve at the given point. 


\begin{figure}[H]
\centering\includegraphics[width=\textwidth]{figures/14fig5.png}
\caption{The osculating plane and osculating circle for the helix.  The red vector is the tangent vector and the blue is the normal vector. See Worked Problem 2 for the calculation.}
\end{figure}




\section*{Worked Problems}
\begin{enumerate}[1.]
\item{\it Calculate $\mathbf{T} (t)$, $\mathbf{N} (t)$, and the osculating plane of the circle $\mathbf{r} (t) = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j}$. Then, show that $\mathbf{r} (t)$ is its own osculating circle.}

\medskip

First, we calculate $\mathbf{T} (t)$ and $\mathbf{N} (t)$ for arbitrary $t$.  From an earlier example, the unit tangent vector is 
$$
\mathbf{T} (t) = - \sin (t) \mathbf{i} + \cos (t) \mathbf{j}.
$$
Thus, 
$$
\mathbf{T}' (t) = - \cos (t) \mathbf{i} - \sin (t) \mathbf{j} \Rightarrow \vert \mathbf{T}' (t) \vert = 1.
$$
Therefore, the principal unit normal vector at time $t$ is given by the formula
\begin{equation}\label{1.4eq-circle}
\mathbf{N} (t) = \frac{\mathbf{T}' (t)}{\vert \mathbf{T}' (t) \vert} = - \cos (t) \mathbf{i} - \sin (t) \mathbf{j} = - \frac{1}{a} \mathbf{r} (t).
\end{equation}
In particular,  we see that $\mathbf{N}$ is a unit vector directed towards the center of the circle $\mathbf{r} (t)$.

The osculating plane spanned by $\mathbf{T}$ and $\mathbf{N}$ is given by the $xy$-plane.  This follows since the vectors $\langle - \sin (t), \cos (t) \rangle$ and $- \langle \cos (t), \sin (t) \rangle$ form a basis for $\mathbb{R}^2$.  Moreover,  since the curvature of the circle is given by $\kappa = \frac{1}{a}$, we know that $\rho = a$ (so that the osculating circle has radius equal to the circle $\mathbf{r} (t)$).  We also see from (\ref{1.4eq-circle}) that the center of the osculating circle is at $\mathbf{r} (t) + \rho \mathbf{N} (t)={\bf 0}$.
Since the osculating circle has the same center and radius as $\mathbf{r} (t)$, and lies in the same plane, this proves that $\mathbf{r}$ is its own osculating circle.

\medskip

\item{\it Find $\mathbf{T} (t)$, $\mathbf{N} (t)$, and the center and radius of the osculating circle at a general point $\mathbf{r}(t)$ of the helix
$$
\mathbf{r}(t) = a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j} + b t \mathbf{k}, \quad a > 0, \, b \neq 0,
$$ 
Does the helix have any points where the osculating plane is parallel to the $xy$-plane? Does the center of the osculating circle lie on the $z$-axis?
}

\medskip

From Section \ref{sec:tangentAndCurvature}, we already know that 
\begin{align*}
\mathbf{T} (t) & = - \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} 
\\
 \vert \mathbf{T}' (t) \vert  &= \frac{a}{\sqrt{a^2 + b^2}}.
 \\
\kappa (t) & = \frac{a}{a^2 + b^2}
\end{align*}
Therefore, the unit principal normal vector is
$$
\mathbf{N} (t) = \frac{\mathbf{T}'(t) }{|\mathbf{T}'(t) |}=
- \cos (t) \mathbf{i} - \sin (t) \mathbf{j},
$$
the same as for the circle (Worked Problem 1 above). The radius of curvature is given by
$$
\rho = \frac{1}{\kappa} = \frac{a^2 + b^2}{a},
$$
and the center of the osculating circle is located at 
\begin{align*}
\mathbf{r} (t) + \rho \mathbf{N} (t)&= a \cos (t) \mathbf{i} + a \sin (t) \mathbf{j} + b t \mathbf{k}
+ \frac{a^2 + b^2}{a}( - \cos (t) \mathbf{i} - \sin (t) \mathbf{j})
\\
&=-\frac{b^2}{a} \cos(t) \ii  -\frac{b^2}{a} \sin(t) \jj +bt\kk.
\end{align*}
In particular, the center of the osculating circle {\it does not} lie on the $z$-axis unless $b=0$. Further,
 because of the factor of $\frac{b}{\sqrt{a^2 + b^2}}$ appearing on the $\mathbf{k}$ term in $\mathbf{T}$, the osculating plane is never parallel to the $xy$-plane. 


\bigskip



\item 
{\it 
If $\bfr(t)=t \ii + at^2 \jj$ for some $a>0$ (a parametrization of the parabola $y=ax^2$ in $\mathbb{R}^2$), find ${\bf T}(t)$, ${\bf N}(t)$ and $\kappa(t)$ at a general point of the curve.}

\medskip
We have
$\bfr'(t)=\ii+2at\jj$, $|\bfr'(t)|=\sqrt{1+4a^2t^2}$, and 
\begin{align*}
{\bf T}(t)& =(1+4a^2 t^2 )^{-1/2} \  \ii + 2at (1+4a^2 t^2)^{-1/2} \ \jj,
\\
{\bf T}'(t)&= - \frac{1}{2}(1+4a^2 t^2)^{-3/2} (8a^2t)   \ii 
+ \left[ 2a (1+4a^2 t^2)^{-1/2}
 -2at \frac{8a^2t}{2 (1+4 a^2 t^2)^{3/2}} \right]  \jj
\\
&= -\frac{4a^2t}{ (1+4a^2 t^2)^{3/2} }\ \ii
+ \frac{ 2a(1+4a^2 t^2) - 8a^3 t^2}{ (1+4a^2 t^2)^{3/2} }\ \jj
\\
&= -\frac{4a^2t}{ (1+4a^2 t^2)^{3/2} }\ \ii
+ \frac{ 2a}{ (1+4a^2 t^2)^{3/2} }\ \jj,
\\
|{\bf T}'(t)|&= \frac{ 2a}{ (1+4a^2 t^2)^{3/2} } \sqrt{4a^2t^2+1} =  \frac{ 2a}{ 1+4a^2 t^2 },
\\
{\bf N}(t) & = \frac{ {\bf T}'(t) }{ |{\bf T}'(t)| }
= -\frac{2at}{ \sqrt{1+4a^2 t^2 }}\ \ii + \frac{ 1}{ \sqrt{1+4a^2 t^2} }\ \jj,
\\
\kappa(t) &= \frac{ |{\bf T}'(t)| }{ |\bfr'(t)| } = \frac{ 2a/ (1+4a^2 t^2 )}{ \sqrt{1+4a^2t^2 }}
= \frac{2a}{(1+4a^2 t^2)^{3/2}}.
\end{align*}


\bigskip







\end{enumerate}

\section*{Practice Problems}
\begin{enumerate}[1.]
\item{Let $${\bf r}(t)=\langle\sqrt{2}t^{3/2},(t+2)^{3/2},(t-2)^{3/2}\rangle, \textrm{ for } t\in(2,\infty).$$
\noindent Find the unit tangent vector ${\bf T}(t)$, the principal normal vector ${\bf N}(t)$, and 
the curvature $\kappa(t)$. Find the limit of ${\bf N}(t)$ as $t\to 2^+$. (No need for an $\epsilon-\delta$ 
proof here, it suffices to use the general rules for limits.)}

(The calculations in this question, as well as in Worked Problem 3 above, are a little bit messy. Later on, we will see another way of finding ${\bf N}$ that leads to simpler calculations in many cases.)


\bigskip

\item {\it
Let $\bfr(t)=t\ii + t^2 \jj +t^3\kk$ for $t\in\rr$.


\medskip

(a) 
Find the equation of the osculating plane at the point $t=0$. }

\medskip


{(b) For $s,t\in\rr$ such that $s,t,0$ are all distinct, find the unit vector ${\bf n}(t,s)$ normal to the plane that passes through the points $\bfr(t)$, $\bfr(s)$, $\bfr(0)$. (There are two such vectors, pointing in opposite directions; choose the one that has a positive $\kk$-component.) Prove that 
$$
\lim_{s,t\to 0,\ s\neq t} {\bf n}(t,s) 
$$
is perpendicular to the osculating plane found in (a).}




\bigskip


\item
{ Let $\bfr(t)=t\ii + at^2 \jj $ for some $a>0$.

\medskip
(a) 
Find the center $Q$ and radius $\rho$ of the osculating circle at the point $(0,0)$. }


\medskip

{(b) For $t>0$, find the center $Q(t)$ and radius $R(t)$ of the circle in the $xy$-plane that passes through the points $\bfr(t)$, $\bfr(-t)$, $\bfr(0)$. Prove that $Q(t)\to Q$ and $R(t)\to \rho$ as $t\to 0$, where $Q$ and $\rho$ are the answers in (a).)}







\item 
{ Let $\bfr$ be a parametrized curve in $\rr^3$ such that $\bfr$ and its derivatives of all orders
are continuous, $\bfr'(t)\neq 0$ for $t$ in some interval $I$, and $|\bfr (t)|=r$ for
all $t\in I$ (i.e. the curve lies on the sphere $S$ of radius $r$ centered
at the origin $O$). Let $P=\bfr(t)$ be a point on
the curve.  
\medskip


(a) Prove that $\rho=r \cos\theta$, where $\theta$ is the angle between 
$\vec{PO}$ and ${\bf N}$. Use this to prove that $\kappa\geq 1/r$.

\smallskip

(Hint: Use arclength parametrization. We have $\bfr(s)\cdot \frac{d\bfr}{ds}=0$ (why?).
If we differentiate this in $s$ and use that $|\frac{d\bfr}{ds}|=1$, what does it say about 
the dot product $\vec{PO}\cdot {\bf N}$?)}


\medskip

 
{(b)
Prove that for any point $P=\bfr(t)$ on the curve, the osculating circle for the curve at that point
lies on the sphere $S$. }

\bigskip




\item {\it 
Let $\bfr$ be a closed, simple parametrized curve in $\rr^3$.
We use a ``wrap-around" parametrization: if ${\bf r}(t)$ for $t\in[0,a]$ parametrizes one full loop around $C$ returning to the starting point, we extend ${\bf r}$ to a periodic function $\bfr:\rr\to\rr^3$ so that $\bfr(t+a)=\bfr(t)$ for all $t$. (This is similar to parametrizing the unit circle by $\bfr(t)=\cos(t)\ii + \sin(t)\jj$, with period $2\pi$.)

Assume that $\bfr$ and its derivatives of all orders
are continuous, $\bfr'(t)\neq 0$, and $|\bfr (t)|\leq 1$ for
all $t$ (i.e. the curve lies in the closed ball of radius $1$ centered
at the origin $O$). 
Prove that there is a point $\bfr(t_0)$ on the curve where $\kappa(t_0)\geq 1$.  

(Hint: Let $f(t)=|\bfr(t)|^2$, then $f(t)$
attains its maximum at some point $t=t_0$, What does this say about $f'(t_0)$ and $f''(t_0)$?
It may be helpful to use arclength parametrization, starting from $t=t_0$ so that $s(t_0)=0$.)}







\end{enumerate}

