


\magnification=1095
\font\Head=cmb10 scaled \magstep2
\font\bighead=cmb10 scaled \magstep3
\font\sf=cmss10
\def\bull{\vrule height .9ex width .8ex depth-.1ex }%square bullet
\def\tinycirc{{^{{\,}_{\rm o}}}}
\def\cq{{$''$}}
\centerline{\bf The helix (or spiral) curve}
\bigskip
\bigskip
This is a discussion of an example worked out in class.  The simple
helix or spiral is given by the following parametric form:
\medskip
$${\bf x}(t)~~=~~(\cos t,\sin t, t)$$
It follows that ${\bf x}'(t) = (-\sin t,\cos t,1)$, so that the
length is given by $|{\bf x}'(t)| = \sqrt{2} = s'(t)$ and 
consequently we may take the arc length parametrization to be
give by $s = \sqrt{2}\,t$ or $t = s/\sqrt{2}$.
One then has the following formula for the unit tangent vector:
$${\bf T}'(s)~~=~~{1\over{\sqrt{2}}}\,\left(
-\sin \left[{s\over{\sqrt{2}}}\right],
\cos\left[{s\over{\sqrt{2}}}\right],1
\right)$$
It follows that the curvature is equal to $|{\bf T}'| = {1\over 2}$
and the principal normal is given as follows:
$${\bf N}'(s)~~=~~\left(
-\cos \left[{s\over{\sqrt{2}}}\right],
-\sin\left[{s\over{\sqrt{2}}}\right],0\right)$$
If we compute ${\bf B}(s) = {\bf T}(s)\times {\bf N}(s)$ we 
obtain the following formula:
$${\bf B}(s)~~=~~{1\over{\sqrt{2}}}\,\left(
\sin \left[{s\over{\sqrt{2}}}\right],
-\cos\left[{s\over{\sqrt{2}}}\right],1
\right)$$
It follows that ${\bf B}'(s) = -{1\over 2}\,{\bf N}(s)$, so 
that the torsion is equal to $-{1\over 2}$ at all points.
Therefore the helix is an example of a curve with constant
curvature and torstion.
\medskip
If one changes the radius of the circle and takes the third
coordinate to be an arbitrary nonzero number, then one still
obtains curves with the constant curvatures and (nonzero) 
torsions.  It is an instructive exercise to see  how these
constant values vary as one varies the radius and the vertical
component of the curve.
\medskip
The Fundamental Theorem of Local Curve Theory essentially implies
the following characterization of helix curves:
\medskip
\noindent{\bf PROPOSITION. }
{\it If a regular smooth curve in ${\bf R}^3$ has constant positive
curvature and constant nonzero torsion, then this curve is 
congruent to a helix.}
\medskip
If the Frenet trihedron of the curve at some initial point is
the same as that for the helix, then this follows from the 
results of Unit I; the statement about congruence requires 
material from Section II.4.\bull
\bye

This note provides some additional information about the theorem
on the existence of curves with prescribed curvature and torsion
and an initial Frenet trihedron.  As indicated in the notes, the
first step is to find a solution to the system of 9 linear 
differential equations corresponding to the Frenet-Serret formulas:
$$\matrix{
{\bf T\,}'&=&& &&\kappa\,{\bf N}&&\cr
{\bf N\,}'&=&-&\kappa\,{\bf T} &&&-&\tau\,{\bf B}\cr
{\bf B\,}'&=&& &&\tau\,{\bf N}&&\cr}
$$
Once we have a solution with initial Frenet trihedron
${\bf T}_0$, ${\bf N}_0$, ${\bf B}_0$, we would like to
know that the three functions 
${\bf T}(s)$, ${\bf N}(s)$ and  ${\bf B}(s)$, are
orthonormal everywhere and not just when $s = s_0$.  This is
done on page 310 of do Carmo in a rather terse manner, and
the purpose here is to give another argument which uses matrices
and hopefully sheds more light on why the solutions form an
orthonormal set.
\medskip
Let $A(s)$ be the $3\times 3$ matrix whose columns are the
vector valued functions
${\bf T}(s)$, ${\bf N}(s)$ and  ${\bf B}(s)$.  Then the
Frenet-Serret formulas can be rewritten in the matrix form
$$A'(s)~~=~~C(s)A(s)$$
where differentiation is done coordinatewise
and $C(s)$ is the following matrix:
$$\left(\matrix{
0&\kappa&0\cr
-\kappa&0&-\tau\cr
0&\tau&0\cr}\right)
$$
Note that the matrix $C(s)$ is {\it skew symmetric}; in other
words, it is equal to the negative of its own transpose.  
We shall denote the transpose of a matrix $P$ by $P^*$ here
to avoid typesetting problems.
\medskip
Now define the ``Gram matrix'' $G(s) = A(s)^*\cdot A(s)$.  We
can differentiate $G(s)$ in the usual fashion, and in fact
this differentiation satisfies the following two simple 
conditions:
\medskip
\itemitem{$(i)$}  The derivative of a transpose is the transpose
of the derivative.
\medskip
\itemitem{$(ii)$}  There is a Leibniz rule of the form
$(PQ)' = P'Q+PQ'$.  One must be careful to keep the proper 
ordering of the factors since matrix multiplication is not
commutative.
\medskip
We may now use
both rules to compute this derivative as follows:
$$G'(s)~~=~~A'(s)^*A~+~A(s)^*A'(s)~~=~~
A(s)^*C(s)^*A~+~A(s)^*C(s)A(s)$$
Since $C(s)$ is skew-symmetric, we have $C^* = -C$ and 
therefore the right hand side reduces to zero.
\medskip
It follows that $G'(s)$ is zero and hence $G(s)$ is constant.
Since $G(s_0)$ is the identity, it follows that $G(s) = I$
for all $s$.  By the definition of $G(s)$ in terms of inner
products, this means that the columns of the orignal matrix
$A(s)$ are orthonormal, and therefore it follows that the
solutions to the Frenet-Serret differential equations yield
an orthonormal set of vectors, which is what we wanted to show.\bull
\bye


\bye


From schultz@math.ucr.edu Wed Jan 25 13:42:27 2006
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\magnification=1095
\font\Head=cmb10 scaled \magstep2
\font\bighead=cmb10 scaled \magstep3
\font\sf=cmss10
\def\bull{\vrule height .9ex width .8ex depth-.1ex }%square bullet
\def\tinycirc{{^{{\,}_{\rm o}}}}
\def\cq{{$''$}}
\centerline{\bf The helix (or spiral) curve}
\bigskip
\bigskip
This is a discussion of an example worked out in class.  The simple
helix or spiral is given by the following parametric form:
\medskip
$${\bf x}(t)~~=~~(\cos t,\sin t, t)$$
It follows that ${\bf x}'(t) = (-\sin t,\cos t,1)$, so that the
length is given by $|{\bf x}'(t)| = \sqrt{2} = s'(t)$ and 
consequently we may take the arc length parametrization to be
give by $s = \sqrt{2}\,t$ or $t = s/\sqrt{2}$.
One then has the following formula for the unit tangent vector:
$${\bf T}'(s)~~=~~{1\over{\sqrt{2}}}\,\left(
-\sin \left[{s\over{\sqrt{2}}}\right],
\cos\left[{s\over{\sqrt{2}}}\right],1
\right)$$
It follows that the curvature is equal to $|{\bf T}'| = {1\over 2}$
and the principal normal is given as follows:
$${\bf N}'(s)~~=~~\left(
-\cos \left[{s\over{\sqrt{2}}}\right],
-\sin\left[{s\over{\sqrt{2}}}\right],0\right)$$
If we compute ${\bf B}(s) = {\bf T}(s)\times {\bf N}(s)$ we 
obtain the following formula:
$${\bf B}(s)~~=~~{1\over{\sqrt{2}}}\,\left(
\sin \left[{s\over{\sqrt{2}}}\right],
-\cos\left[{s\over{\sqrt{2}}}\right],1
\right)$$
It follows that ${\bf B}'(s) = -{1\over 2}\,{\bf N}(s)$, so 
that the torsion is equal to $-{1\over 2}$ at all points.
Therefore the helix is an example of a curve with constant
curvature and torstion.
\medskip
If one changes the radius of the circle and takes the third
coordinate to be an arbitrary nonzero number, then one still
obtains curves with the constant curvatures and (nonzero) 
torsions.  It is an instructive exercise to see  how these
constant values vary as one varies the radius and the vertical
component of the curve.
\medskip
The Fundamental Theorem of Local Curve Theory essentially implies
the following characterization of helix curves:
\medskip
\noindent{\bf PROPOSITION. }
{\it If a regular smooth curve in ${\bf R}^3$ has constant positive
curvature and constant nonzero torsion, then this curve is 
congruent to a helix.}
\medskip
If the Frenet trihedron of the curve at some initial point is
the same as that for the helix, then this follows from the 
results of Unit I; the statement about congruence requires 
material from Section II.4.\bull
\bye

This note provides some additional information about the theorem
on the existence of curves with prescribed curvature and torsion
and an initial Frenet trihedron.  As indicated in the notes, the
first step is to find a solution to the system of 9 linear 
differential equations corresponding to the Frenet-Serret formulas:
$$\matrix{
{\bf T\,}'&=&& &&\kappa\,{\bf N}&&\cr
{\bf N\,}'&=&-&\kappa\,{\bf T} &&&-&\tau\,{\bf B}\cr
{\bf B\,}'&=&& &&\tau\,{\bf N}&&\cr}
$$
Once we have a solution with initial Frenet trihedron
${\bf T}_0$, ${\bf N}_0$, ${\bf B}_0$, we would like to
know that the three functions 
${\bf T}(s)$, ${\bf N}(s)$ and  ${\bf B}(s)$, are
orthonormal everywhere and not just when $s = s_0$.  This is
done on page 310 of do Carmo in a rather terse manner, and
the purpose here is to give another argument which uses matrices
and hopefully sheds more light on why the solutions form an
orthonormal set.
\medskip
Let $A(s)$ be the $3\times 3$ matrix whose columns are the
vector valued functions
${\bf T}(s)$, ${\bf N}(s)$ and  ${\bf B}(s)$.  Then the
Frenet-Serret formulas can be rewritten in the matrix form
$$A'(s)~~=~~C(s)A(s)$$
where differentiation is done coordinatewise
and $C(s)$ is the following matrix:
$$\left(\matrix{
0&\kappa&0\cr
-\kappa&0&-\tau\cr
0&\tau&0\cr}\right)
$$
Note that the matrix $C(s)$ is {\it skew symmetric}; in other
words, it is equal to the negative of its own transpose.  
We shall denote the transpose of a matrix $P$ by $P^*$ here
to avoid typesetting problems.
\medskip
Now define the ``Gram matrix'' $G(s) = A(s)^*\cdot A(s)$.  We
can differentiate $G(s)$ in the usual fashion, and in fact
this differentiation satisfies the following two simple 
conditions:
\medskip
\itemitem{$(i)$}  The derivative of a transpose is the transpose
of the derivative.
\medskip
\itemitem{$(ii)$}  There is a Leibniz rule of the form
$(PQ)' = P'Q+PQ'$.  One must be careful to keep the proper 
ordering of the factors since matrix multiplication is not
commutative.
\medskip
We may now use
both rules to compute this derivative as follows:
$$G'(s)~~=~~A'(s)^*A~+~A(s)^*A'(s)~~=~~
A(s)^*C(s)^*A~+~A(s)^*C(s)A(s)$$
Since $C(s)$ is skew-symmetric, we have $C^* = -C$ and 
therefore the right hand side reduces to zero.
\medskip
It follows that $G'(s)$ is zero and hence $G(s)$ is constant.
Since $G(s_0)$ is the identity, it follows that $G(s) = I$
for all $s$.  By the definition of $G(s)$ in terms of inner
products, this means that the columns of the orignal matrix
$A(s)$ are orthonormal, and therefore it follows that the
solutions to the Frenet-Serret differential equations yield
an orthonormal set of vectors, which is what we wanted to show.\bull
\bye


\bye


