
\section{Velocity and Acceleration}
We have now constructed and examined the Frenet frame $\mathbf{T}$, $\mathbf{N}$, $\mathbf{B}$ for the past few sections. For this section, we wish to write these vectors in terms of well-known kinematic functions---specifically \textbf{velocity} and \textbf{acceleration}.

\subsection{Deriving Equations for Velocity and Acceleration}
Assume that $\mathbf{r} (t)$ is twice differentiable, with $\mathbf{r}' \neq 0$.  Recall the following formulas
\begin{eqnarray}\label{16-formula1}
\mathbf{v} (t) = \mathbf{r}' (t) & \quad & \textrm{(Velocity)} \\
v(t) = \vert \mathbf{v}' (t) \vert & \quad & \textrm{(Speed)} \\
\mathbf{a}(t) = \mathbf{v}' (t) = \mathbf{r}'' (t) & \quad & \textrm{(Acceleration)}
\end{eqnarray}
Our goal is to derive formulas for $\mathbf{v}$ and $\mathbf{a}$ in terms of the vectors $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$.  For velocity, we have the straightforward identity
\begin{equation}\label{1.6eq-velocity}
\mathbf{v} (t) = \mathbf{r}' (t) = \vert \mathbf{r}' (t) \vert \mathbf{T}(t) = v(t) \mathbf{T} (t).
\end{equation}
For acceleration, we proceed in a few steps.  First, note that
\begin{equation}\label{1.6eq-acceleration1}
\mathbf{a} (t) = \mathbf{v}' (t) = \frac{d}{dt} (v \mathbf{T}) = \frac{dv}{dt} \mathbf{T} + v \frac{d \mathbf{T}}{dt},
\end{equation}
where we used (\ref{1.6eq-velocity}) in the second equality.  Now, recalling that
$$
\mathbf{N} = \frac{\mathbf{T}'}{\vert \mathbf{T}' \vert} \Rightarrow \mathbf{T}' = \vert \mathbf{T}' \vert \mathbf{N},
$$
equation \eqref{1.6eq-acceleration1} becomes
\begin{equation}\label{1.6eq-acceleration2}
\mathbf{a} = \frac{dv}{dt} \mathbf{T} + v \vert \mathbf{T}' \vert \mathbf{N}.
\end{equation}
Lastly, we use that
$$
\kappa = \frac{\vert \mathbf{T}' \vert}{\vert \mathbf{r}' \vert} = \frac{\vert \mathbf{T}' \vert}{v} \Rightarrow \vert \mathbf{T}' (t) \vert = \kappa v (t),
$$
to again rewrite \eqref{1.6eq-acceleration2} as
\begin{equation}\label{1.6eq-accelerationfinal}
\boxed{
\mathbf{a} = \frac{dv}{dt} \mathbf{T} + v^2 \kappa \mathbf{N}.
}
\end{equation}
We call the scalar-valued functions $a_T = v' (t)$ and $a_N = \kappa v^2 (t)$ the \textbf{tangential component} and \textbf{normal component} of acceleration, respectively.  With this notation, we have
$$
\mathbf{a} (t) = a_T (t) \mathbf{T} (t) + a_N (t) \mathbf{N} (t).
$$
Two observations now follow immediately:
\begin{itemize}
\item Acceleration $\mathbf{a}$ has no $\mathbf{B}$ component in the Frenet frame.  Geometrically, this means that acceleration is always parallel to the osculating plane (which is spanned by $\mathbf{T}$ and $\mathbf{N}$).
\item Normal (or \textit{centripetal}) acceleration is always directed towards the centre of curvature (the centre of the osculating circle); this component of acceleration represents the change in direction.
\end{itemize}

\subsection{Alternative Formulas for Normal and Binormal Vectors} Let us derive new formulas for $\mathbf{B}$ and $\mathbf{N}$ in terms of velocity and acceleration.

To start,  notice that \eqref{1.6eq-velocity} and \eqref{1.6eq-accelerationfinal} imply that both $\mathbf{v}$ and $\mathbf{a}$ are parallel to the osculating plane.  So, their cross-product should be a scalar multiple of $\mathbf{B}$.  In fact:
\begin{equation}\label{1.6eq-add1}
\begin{split}
\mathbf{v} \times \mathbf{a} & =  v \mathbf{T} \times (v' \mathbf{T} + v^2 \kappa \mathbf{N}) = v^3 \kappa (\mathbf{T} \times \mathbf{N} ) \\
 & =  v^3 \kappa \, \mathbf{B}.
\end{split}
\end{equation}
SInce $\mathbf{B}$ is a unit vector, this means that 
\begin{equation}\label{1.6eq-add2}
|\mathbf{v} \times \mathbf{a}|= v^3\kappa
\end{equation}
 (we already knew this from an earlier formula for $\kappa$), and that
$$
\mathbf{B} = \frac{\mathbf{v} \times \mathbf{a}}{v^3 \kappa} = \frac{\mathbf{v} \times \mathbf{a}}{\vert \mathbf{v} \times \mathbf{a} \vert}.
$$
Now that we have this, we can compute $\mathbf{N}$ from the identity $\mathbf{N} = \mathbf{B} \times \mathbf{T}$. This calculation is often simpler than differentiating ${\bf T}$. Additionally, we can find the scalar functions $a_T$ and $a_N$ from the formulas
$$
a_T = \mathbf{a} \cdot \mathbf{T} = \frac{\mathbf{a} \cdot \mathbf{v}}{\vert \mathbf{v} \vert}
$$
$$
a_N = \mathbf{a} \cdot \mathbf{N} = \sqrt{\vert \mathbf{a} \vert^2 - \vert a_T \vert^2}.
$$




\subsection{Optional: A Formula for Torsion}
We can also use our functions $\mathbf{v}$ and $\mathbf{a}$ to find an easily computable formula for torsion: 

\begin{equation}\label{1.6eq-torsionformulaaccel}
\tau = \frac{\mathbf{a}' \cdot (\mathbf{v} \times \mathbf{a})}{\vert \mathbf{v} \times \mathbf{a} \vert^2}
\end{equation}

\begin{proof}
We write $\mathbf{a}'$ in terms of the Frenet frame: $\mathbf{a}'=\lambda{\bf T}+\mu{\bf N}+\nu{\bf B}$, 
and then pay particular attention to the last component. 
To begin, we differentiate 
\begin{eqnarray*}
\mathbf{a}' (t) & = & \frac{d}{dt} (v' \mathbf{T} + v^2 \kappa \mathbf{N}) \\
\quad & = & v'' \mathbf{T} + v' \underbrace{\mathbf{T}'(t)}_{|{\bf T}'|{\bf N}}
+ \frac{d}{dt} (v^2 \kappa) \mathbf{N} + \underbrace{v^2 \kappa \mathbf{N}'(t)}_{(*)}
\end{eqnarray*}
The first 3 parts are linear combinations of just ${\bf T}$ and ${\bf N}$. 
For the last part $(*)$, we convert to arc length parametrization with $\frac{ds}{dt} = v$, and then use the the Frenet-Serret equations:
$$
(*) = v^2 \kappa \frac{d \mathbf{N / ds}}{dt/ds} =v^2 \kappa v (\tau \mathbf{B} - \kappa \mathbf{T}) = \dots + v^3 \kappa \tau \mathbf{B},
$$
where $\dots$ denotes some multiple of ${\bf T}$ that we are not interested in.
So, in particular, we have
\begin{equation*}
\mathbf{a}' (t) = \lambda \mathbf{T} + \mu \mathbf{N} + v^3 \kappa \tau \mathbf{B},
\end{equation*}
for some scalar-valued  functions $\lambda,\mu$. The exact form of $\lambda,\mu$ does not matter because we will dispose of them in the next step. Recall from (\ref{1.6eq-add1}) that
$$\mathbf{v} \times \mathbf{a} = v^3 \kappa \mathbf{B}.$$
Since the Frenet frame vectors are orthonormal, we get
\begin{align*}
\mathbf{a}' \cdot (\mathbf{v} \times \mathbf{a})
&= 
\Big(\lambda \mathbf{T} + \mu \mathbf{N} + v^3 \kappa \tau \mathbf{B}\Big) \cdot (v^3 \kappa \mathbf{B})\\
& =(v^3 \kappa \tau)(v^3 \kappa)
\\
&= (v^3\kappa)^2 \tau
\\
&= \vert \mathbf{v} \times \mathbf{a}\vert^2 \tau,
\end{align*}
where we used (\ref{1.6eq-add2}) to replace $v^3\kappa$ by $\vert \mathbf{v} \times \mathbf{a}\vert $. Solving for $\tau$, we get
(\ref{1.6eq-torsionformulaaccel}).

\end{proof}

\begin{remark} Let $\bfr_0=\bfr(t_0)$ be a point on the curve. For simplicity, we will assume that $t_0=0$. Consider the Taylor expansion at this point:
\begin{align*}
\bfr(t)&\approx \bfr(0)+ t \bfr'(0)+\frac{t^2}{2}\bfr''(0)+\frac{t^3}{6}\bfr'''(0)
\\
&= \bfr_0+ t{\bf v}(0)+\frac{t^2}{2}{\bf a}(0)+\frac{t^3}{6}{\bf a}'(0).
\end{align*}
Using the formulas we derived for ${\bf v}$, ${\bf a}$, and ${\bf a}'$ (can you find them?), we get
$$
\bfr{t}\approx \bfr_0+\dots + \frac{t^3}{6} v^3\kappa\tau{\bf B}(0),
$$
where $\dots$ stands for some linear combination of ${\bf T}(0)$ and ${\bf N}(0)$. Notice that the $\bfr_0+\dots$ part lies in the osculating plane $\Pi_0$ at $t=0$, and ${\bf B}(0)$ is perpendicular to that plane. Thus the distance from the curve near $\bfr_0$ to $\Pi_0$ is about $v^3\kappa\tau t^3/6$, and this simplifies additionally to $\kappa\tau s^3/6$ if we use the arclength parametrization with $v=1$.  (For more precision, we could use the Taylor expansion with error estimates.) This justifies our earlier (and more vague) statements that $\tau$ measures how close the curve stays to a plane.

\end{remark}




\section*{Worked Problems}
\begin{enumerate}[1.]
\item{\it Find $\mathbf{T}, \mathbf{N}, \mathbf{B}$ and $\kappa$ if the curve is given by $\mathbf{r} (t) = \langle 2t , t^2 , \frac{t^3}{3} \rangle$.}

\medskip

From earlier examples, we know that
$$
\mathbf{v} = \mathbf{r}' = \langle 2, 2t, t^2 \rangle \Rightarrow v (t) = t^2 + 2
$$
$$
\mathbf{a} = \mathbf{v}' = \langle 0,2,2t \rangle.
$$
We next compute
$$
\mathbf{v} \times \mathbf{a} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 2t & t^2 \\
0 & 2 & 2t
\end{vmatrix}
=
2t^2 \mathbf{i} - 4 t \mathbf{j} + 4 \mathbf{k}
$$
and also
$$
\vert \mathbf{v} \times \mathbf{a} \vert = \sqrt{4(t^4 + 4 t^2 + 4)} = 2(t^2 + 2).
$$
We can now use the formulas from this section, to determine that
\begin{eqnarray*}
\kappa & = & \frac{\vert \mathbf{v} \times \mathbf{a} \vert}{v^3} = \frac{2(t^2 + 2)}{(t^2 + 2)^3} = \frac{2}{(t^2 + 2)^2} \\[1ex]
\mathbf{T} & = & \frac{\mathbf{v}}{v} = \frac{1}{t^2 + 2} \langle 2, 2t, t^2 \rangle \\[1ex]
\mathbf{B} & = & \frac{\mathbf{v} \times \mathbf{a}}{\vert \mathbf{v} \times \mathbf{a} \vert} = \frac{1}{t^2 + 2} \langle t^2, -2t, 2 \rangle \\[1ex]
\mathbf{N} & = & \mathbf{B} \times \mathbf{T} = \frac{1}{(t^2 + 2)^2}
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
t^2 & -2t & 2 \\
2 & 2t & t^2
\end{vmatrix}
\\
\quad & = & \frac{1}{(t^2 + 2)^2} \big[(-2 t^3 - 4t) \mathbf{i} - (t^4 - 4)\mathbf{j} + (2t^3 + 4t)\mathbf{k} \big]
\end{eqnarray*}

\medskip












\item{\it Let $\mathbf{r}(t) = \langle \cos t, t, t^2 \rangle$. Find $\kappa$, $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$ at $t = 0$.}

\medskip

The advantage of using our new formulas is that as soon as we have computed $\mathbf{v}$ and $\mathbf{a}$, we can plug in $t = 0$ and then do the rest of the calculation with $t = 0$ plugged in. No further differentiations will be needed!

\smallskip
We have ${\bf v}(t)=\langle -\sin t,1,2t\rangle$ and ${\bf a}(t)=\langle -\cos t,0,2\rangle$, so that 
${\bf v}(0)=\langle 0,1,0\rangle$ and ${\bf a}(0)=\langle -1,0,2\rangle$.
Then
$$
\mathbf{v}(0) \times \mathbf{a}(0) =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 0 \\
-1 & 0 & 2
\end{vmatrix}
=
2 \mathbf{i} + \mathbf{k}, \,
\vert \mathbf{v} (0) \times \mathbf{a}(0) \vert = \sqrt{4+1} = \sqrt{5}
$$
and $v(0) = 1$. At time $t = 0$, we have
$$
\kappa (0) = \frac{\vert \mathbf{v}(0) \times \mathbf{a}(0) \vert}{v^3} = \frac{\sqrt{5}}{1} = \sqrt{5}
$$
For the Frenet frame, we have
\begin{eqnarray*}
\mathbf{T}(0) & = & \frac{\mathbf{v}(0)}{v(0)} = \langle 0,1,0 \rangle = \mathbf{j} \\[1ex]
\mathbf{B}(0) & = & \frac{\mathbf{v}(0) \times \mathbf{a}(0)}{\vert \mathbf{v}(0) \times \mathbf{a}(0) \vert} = \frac{1}{\sqrt{5}} (2 \mathbf{i} + \mathbf{k}) \\[1ex]
\mathbf{N}(0) & = & \mathbf{B}(0) \times \mathbf{T}(0) = \frac{1}{\sqrt{5}}
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\[1ex]
2 & 0 & 1 \\
0 & 1 & 0
\end{vmatrix}
=
\frac{1}{\sqrt{5}}(-\mathbf{i} + 2 \mathbf{k})
\end{eqnarray*}








\item 
{\it (\underline{Banking the road}) 
A level, curved road has radius of curvature $R$ (not necessarily constant). Assuming that cars will likely travel with constant speed
$v$ (in meters per second), at what angle $\theta$ should the road be banked so that the resultant of the
centrifugal and gravitational forces acting on the vehicle is perpendicular to the surface of the road?
Compute $\theta$ at a point where $R=100$m, assuming that $v=80$km/h. What if $R=200$m?}

\begin{figure}[htbp]
\centering\includegraphics[width=3in]{figures/banking-the-road.png}
\end{figure}

The gravitational force acting on the car is ${\bf F}_g = -mg\kk$, where $m$ is the mass of the car and $g=9.8$ m/s$^2$ is the gravitational acceleration, and the centrifugal force
is ${\bf F}_c = -ma_N {\bf N}$, where $a_N=v^2\kappa = v^2/R$. We want to choose $\theta$ so that 
$$
\tan\theta = \frac{ma_N}{mg}
$$
(see picture). Thus $\tan\theta = a_N/g = v^2/(9.8R)$, with meters and seconds as units.

If $v = 80$km/h $\approx 22.2$m/s and $R=100m$, then, again using meters and seconds as units, 
$$
\tan\theta\approx \frac{(22.2)^2}{100 \cdot 9.8} \approx 0.5039
$$
so that $\theta \approx \tan^{-1}(0.5039) \approx 26.7$ degrees.

If $v$ is the same and $R=200 m$, then using the same calculation, $\theta\approx \tan^{-1}(0.25195) \approx 14.1$ degrees. 

\medskip

\item {\it (\underline{Acceleration and skidding})
Consider a car of mass $m$ travelling along a curved but level road. To avoid skidding, the road
must supply a frictional force ${\bf F}=m{\bf a}$, where ${\bf a}$ is the car's acceleration
vector. The estimated magnitude of the frictional force is $\mu mg$ where $\mu$ is the coefficient of friction and $g=9.8$m/s$^2$. Let $v$ be the car's speed in meters per second.}

\begin{enumerate}

\smallskip

\item 
{\it Show that the car will not skid if the curvature of the road is such that (with $R=1/\kappa$)
$$(v')^2+\Big( \frac{v^2}{R} \Big)^2 \leq (\mu g)^2.$$
Note that braking ($v'<0$) and speeding up ($v'>0$) contribute equally to skidding.}



\medskip

We need to have $|m{\bf a} |\leq \mu mg$, or equivalently, $|{\bf a}|\leq \mu g$, 
$|{\bf a}|^2\leq (\mu g)^2$. Writing out the components of acceleration
$${\bf a}= v'{\bf T}+v^2 \kappa {\bf N}$$
we get the condition
$$
|{\bf a}|^2= (v')^2 + (v^2\kappa)^2 = (v')^2+\Big( \frac{v^2}{R} \Big)^2 \leq (\mu g)^2$$
as required.

\medskip

\item 
{\it Suppose that the approximate radius of curvature along a stretch of a curved highway is $R=180$m. How fast can an automobile travel (at constant speed) along that part of the highway
if the coefficient of friction is $\mu=0.5$? What if $R=100m$?}

\medskip

If $R=180$m, $\mu=0.5$, and $v$ is constant so that $v'=0$, we get the condition
$$
\frac{v^2}{R}  \leq \mu g = 0.5 \cdot 9.8 \hbox{m/s}^2 = 4.9 \hbox{m/s}^2.
$$
Plugging in $R=180$m, we get
$$
v^2\leq 180 \hbox{m} \cdot 
4.9 \hbox{m/s}^2 = 882 \hbox{(m/s)}^2
$$
so that $v\leq 29.7 \hbox{m/s}\approx 106.9 \hbox{km/h}$. 

If $R=100$m, $\mu=0.5$, and $v$ is constant, we get instead 
$$
v^2\leq 100 \hbox{m} \cdot 
4.9 \hbox{m/s}^2 = 490 \hbox{(m/s)}^2
$$
so that $v\leq 22.136 \hbox{m/s}\approx 79.7 \hbox{km/h}$. 

\end{enumerate}
\end{enumerate}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\section*{Practice Problems}

\begin{enumerate}[1.]




\item 
{The position of an object is given by a smooth vector-valued function ${\bf r}(t)$. Let ${\bf v}(t)=\bfr'(t)$, and ${\bf a}(t)=\bfr''(t)$. Assume that $
{\bf a}(t) =\kk \times {\bf v}(t)$ for all $t$.

\medskip

(a) Prove that the speed $v$ is constant.

\medskip

(b) Assuming that $v$ is nonzero, prove that the angle $\theta(t)$ between ${\bf v}(t)$ and $\kk$ is constant.

}


\bigskip


\item 
{Assume that ${\bf a}(t)=c\,\bfr(t)$ for some constant $c$. (This describes the motion of an object under a central force.)

\medskip

(a) Prove that ${\bf w}:=\bfr\times{\bf v}$ is constant. 

\medskip

(b) Assuming that ${\bf w}\neq {\bf 0}$, prove that the curve $\bfr(t)$ lies in the plane passing through the origin and perpendicular to ${\bf w}$.}

\bigskip

\item 
{ Suppose that a curve $\mathcal{C}$ in $\rr^3$ has a smooth parametrization $\bfr(t)$,
the curvature $\kappa$ is a positive constant, and the torsion $\tau$ is zero. Prove, without using the Fundamental Theorem of Calculus for curves, that $\mathcal{C}$ is a circle. (Hint: how would you find the center of the circle? You may also want to use the Frenet-Serret formulas.)}



\bigskip




\item 
{Let $C$ be a curve on the sphere $\mathcal{S}$ of radius 1 centered at ${\bf 0}$. Suppose that $C$ has a smooth parametrization ${\bf r}(t)$ such that the velocity ${\bf r}'(t)$ is never zero and the acceleration $\bfr''(t)$ is always perpendicular to $\mathcal{S}$ at $\bfr(t)$. Prove that the curve is contained in a great circle of $\mathcal{S}$ (that is, a circle where $\mathcal{S}$ intersects a plane through ${\bf 0}$).}


\bigskip



\item {
Let $\bfr(s):\R\to\R^3$ be a smooth curve, parametrized by arc length.
Assume that $\frac{d^2\bfr}{ds^2}=\bfr\times {\bf w}$ for all $s$, where ${\bf w}$ is a fixed nonzero vector. Prove that this is only possible when $\frac{d^2\bfr}{ds^2}=\bfr\times {\bf w}={\bf 0}$ for all $s$.


\medskip\noindent
(Hint: suppose that there is a value of $s$ where $\bfr\times {\bf w}\neq {\bf 0}$. Prove first that there are scalar functions $\lambda(s),\mu(s)$ such that $\frac{d\bfr}{ds}=\lambda\bfr + \mu {\bf w}$ near that point. What does this imply about $\frac{d^2\bfr}{ds^2}$?)}












\end{enumerate}



