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\bf{Question}

\quad Let $\gamma$ be the space curve given by $$\gamma(t)=(\cosh
t,\sinh t,2t).$$ Show that $\gamma$ has a vertex ($\dot\kappa=0$)
at $P=(1,0,0)$ and at points $A,B$ given by solutions $t$ to
$\cosh2t=13/8$.  Calculate the torsion at $P,A$ and $B$.



\bf{Answer}

\begin{eqnarray*}
\gamma(t) & = & (\cosh t, \sinh t, 2t)\\ \dot{\gamma}(t) & = &
(\sinh t, \cosh t, 2)\\ \dot{\gamma} \cap \ddot{\gamma} & = &
(-2\sinh t, 2 \cosh t, -1)\\ \dot{\gamma} \cap
\ddot{\gamma}.\stackrel{...}{\gamma} & = & -2\sinh^2 t + 2 \cosh^2
t =2
\end{eqnarray*}
Hence $$K^2= \|\dot{\gamma} \cap \ddot{\gamma} \|^2 \|
\dot{\gamma} \|^{-6} = (1+ 4 \cosh 2t)( 4 + \cosh 2t)^{-3}$$ Note
that $K\ne 0$.

Differentiate: $$2K\dot{K} = \cdots = 2(t + \cosh 2t)^{-4}
[4(4+\cosh 2t) - 3(1 + 4\cosh 2t)] \sinh 2t$$ This is $=0$ when
$\sinh 2t =0$ or $13-8\cosh 2t =0$.

Now, $\tau = (\dot{\gamma} \cap
\ddot{\gamma}.\stackrel{...}{\gamma}) \| \dot{\gamma} \cap
\ddot{\gamma} \| ^{-2} = 2(1+4\cosh 2t)^{-1}$.

Hence at $P$ ($t=0$) we have $\tau = \frac{2}{5}$

while at $A$,$B$ ($\cosh 2t = \frac{13}{8}$) we have $\tau = 2 (1
+ \frac{13}{2})^{-1} = \frac{4}{15}$.



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