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{\bf Question}

A curve is given by $x = a \cos \psi, \, y = a \sin \psi, \, z =
\psi,$ where $\psi>0$.  Find the following as functions of $\psi$:
\begin{description}
\item[(a)] the tangent, principal normal and binormal vectors;
\item[(b)] curvature and torsion;
\item[(c)] arc length along the curve.
\end{description}

\vspace{.25in}

{\bf Answer}

$\ds {\bf r} = a \cos \psi {\bf i} + a \sin \psi {\bf j} + \psi
{\bf k}$
\begin{description}
\item[(a)]

 Finding the tangent:
\begin{eqnarray*} \frac{d{\bf r}}{ds} & = & -a \sin \psi \frac{d \psi}{ds}{\bf i}
+ a \cos \psi \frac{d \psi}{ds}{\bf j} + \frac{d \psi}{ds}{\bf k}
\\ {\bf t} = \frac{d {\bf r}}{ds} & = & \frac{d \psi}{ds} \left[-a \sin \psi{\bf i}
+ a \cos \psi {\bf j} + {\bf k}\right] \\ 1 = {\bf t} \cdot {\bf
t} & = & \left( \frac{d \psi}{ds}\right)^2(1+a^2) \Rightarrow
\frac{d \psi}{ds} = \frac{1}{\sqrt{1 + a^2}} \\  {\bf t} & = &
\frac{1}{\sqrt{1 + a^2}}[-a \sin \psi {\bf i} + a \cos \psi {\bf
j} +{\bf k}]
\end{eqnarray*}
where we assume $s$ to increase with $\psi.$

Finding the principal normal:
\begin{eqnarray*}
 \frac{d{\bf t}}{ds} & = & \frac{1}{\sqrt{1 + a^2}}
\frac{d \psi}{ds}[-a \cos\psi {\bf i} - a \sin\psi {\bf j}] \\ & =
& \frac{-a}{1+a^2}[\cos \psi{\bf i} + \sin \psi {\bf j}]
\end{eqnarray*}
Now $\ds\frac{d {\bf t}}{ds} = \kappa{\bf n}$ from the
Serret-Frenet formulae, so

$\ds\kappa = \frac{a}{1 +a^2}$ and ${\bf n} = -[\cos \psi {\bf i}
+ \sin \psi {\bf j}]$


Finding the binormal vector:

\begin{eqnarray*}{\bf m} = {\bf t} \times {\bf n} & = & \left| \begin{array}{ccc}
{\bf i} & {\bf  j} & {\bf k} \\ -a\sin \psi &  a \cos \psi & 1 \\
-\cos \psi & - sin \psi & 0
\end{array} \right| \frac{1}{\sqrt{1 + a^2}} \\ & = & \frac{1}{\sqrt{1 + a^2}}[\sin
\psi {\bf i}- \cos \psi {\bf j} + a(\sin^2 \psi + \cos^2 \psi)
{\bf k}] \\ {\bf m}& = & \frac{1}{\sqrt{1 + a^2}}[\sin \psi {\bf
i} - \cos \psi {\bf j} + a {\bf k}] \end{eqnarray*}


\item[(b)]

Curvature:  $\ds\kappa  =  \frac{a}{1 + a^2}$


Torsion: $(\ds\tau)$

$\ds\frac{d{\bf m}}{ds}  =  \frac{1}{\sqrt{1 + a^1}} \frac{d
\psi}{ds}[\cos \psi {\bf i} + \sin \psi {\bf j}]
=
 \frac{1}{\sqrt{1 + a^2}}[ \cos \psi {\bf i} + \sin \psi {\bf j}]$

Now from the Serret-Frenet formulae $\ds \frac{d{\bf m}}{ds} = -
\tau {\bf n}$ so $\ds\tau  =  \frac{1}{1 + a^2}$

\item[(c)]
Arc length:
\begin{eqnarray*}  \left( \frac{ds}{d \psi}
\right)^2& = & \frac{d {\bf r}}{d \psi}\cdot\frac{d {\bf r}}{d
\psi} = |-a \sin \psi {\bf i} + a \cos \psi {\bf j} + {\bf k}|^2.
\\ {\rm So}\ \ \frac{ds}{d \psi} & = & \sqrt{1 + a^2} \Rightarrow s = \psi
\sqrt{1 + a^2}\end{eqnarray*}

where we have defined $s=0$ at $\psi =0$ and $s$ increasing with
$\psi$

\end{description}


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