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\bf{Question}

\quad The plane curve $\alpha(t)=(2+\cos t,\sin t)$ is a circle.
The space curve $$\gamma(t)=\bigl((2+\cos t)\cos t,(2+\cos t)\sin
t,\sin t\bigr)$$ lies on a torus, thought of as being swept out by
$\alpha$ as the plane of $\alpha$ is spun around the $z$-axis in
$\br^3$.  Show that the curvature of $\gamma$ vanishes at
$(-1,0,0)$.  Find the curvature and the torsion of $\gamma$ at the
point $(3,0,0)$, and find the equation of the osculating plane
there.



\bf{Answer}

If $\underline{u}=(a,b)$ is a unit vector in the $(x,y)$-plane,
then $((2+c)a, (2+c)b, s)$ is a point on a circle in the
$(\underline{u}, x)$-plane (centre $2\underline{u}$, radius $1$).

Thus $\gamma(t)$ is on the torus swept out by these circles as
$\underline{u}$goes around the unit circle in the $(x,y)$-plane.

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\begin{eqnarray*}
\gamma'(t) &amp; = &amp; (-2\sin t - \sin 2t, 2 \cos t + \cos 2t, \cos
t)\\ \gamma''(t) &amp; = &amp; (-2\cos t - 2 \cos 2t, - 2 \sin t-2 \sin
2t, -\sin t)
\end{eqnarray*}

\begin{eqnarray*}
K= 0 &amp; \Rightarrow &amp; x'y''-x''y' =0\\ \textrm{i.e. } ( 2\sin t +
\sin 2t)(-2 \sin t - 2\sin 2t) &amp; = &amp; *2\cos t + 2\cos 2t)(2 \cos t
+ \cos 2t)\\ \Rightarrow -6 &amp; = &amp; 6(\cos t \cos 2t + \sin t \sin
2t)\\ \textrm{i.e. } -1 &amp; = &amp; \cos t\\ t &amp; = &amp; \pi \ \ (+2n\pi, n
\in \textbf{Z})
\end{eqnarray*}

So curvature vanishes at $(-1,0,0)$ (as $C$ passes through the
inner ``equator'').

$$\gamma'''(t) = (2\sin t + 4\sin 2t, -2 \cos t - 4 \cos 2t, -\cos
t)$$ and at $(3,0,0)$ (i.e $t=0$) we have
\begin{eqnarray*}
\gamma' &amp; = &amp; (0,3,1)\\ \gamma'' &amp; = &amp; (-4,0,0)\\ \gamma''' &amp; = &amp;
(0,-6,-1)\\ \gamma' \cap \gamma'' &amp; = &amp;  (0,-4,12)
\end{eqnarray*}

There, $\displaystyle \tau=\frac{\gamma' \cap \gamma'' .
\gamma'''}{\| \gamma' \cap \gamma'' \|^2} = \frac{12}{160} =
\frac{3}{40}$

The binomial $B$ is in the direction of $\gamma' \cap \gamma''$,
so $$B = \frac{1}{\sqrt{160}}(0,-4,12) = \frac{1}{\sqrt{10}}
(0,-1,3).$$

Osculating plane is $\bot B$ so has the equation
$0x-y+3z=\textrm{constant}=0$, since it contains the point
$(3,0,0)$.

So OSC plane is: $y=3z$.

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