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\bf{Question}

\quad (i) Show that the space curve
$\gamma(t)=(t,t^{-1}(1+t),t^{-1}(1-t^2))$ is in fact planar.

(ii) Find a differential equation that the function $g(t)$ must
satisfy in order for the curve $$\gamma(t)=(\cos t,\sin t,g(t))$$
to be planar.   Hence find the most general such function $g(t)$.

[{\sl Hint: $B(t)$ has constant direction.}]



\bf{Answer}

\begin{description}
\item{(i)}
Observe that $\gamma(t) = \left ( t, \frac{1}{t}+1, \frac{1}{t} -1
\right )$ which lies in $x-y+z=-1$,

OR: Check that $\dot{\gamma} \cap \ddot{\gamma} = \left (
\frac{2}{t^3}, -\frac{2}{t^3}, \frac{2}{t^3} \right ) =
\frac{2}{t^3} (1,-1,1)$

so $B(t) = \dot{\gamma} \cap \ddot{\gamma}/\| \dot{\gamma} \cap
\ddot{\gamma} \| = \frac{1}{\sqrt{3}}(1,-1,1)$, i.e. constant.

\item{(ii)}
Here $\dot{\gamma} \cap \ddot{\gamma} = ( \cos t \ddot{g} + \sin t
\dot{g}, \sin t \ddot{g} - \cos t \dot{g}, 1 )$; then $\gamma$ is
planar precisely when this vector has constant direction,
$=(a,b,1)$ say.

Solving for $(\dot{g}, \ddot{g})$ we find $\dot{g} = a \sin t - b
\cos t$ (and $\ddot{g} = a\cos t + b \sin t$).

So $g(t) = A\cos t + B\sin t + C$ for arbitrary constants $A$, $B$
and $C$.

OR: $I=0$ when $\dot{\gamma} \cap \ddot{\gamma}.\ddot{\gamma} = 0$
which leads to $\ddot{g} + \dot{g}=0$. Hence $\dot{g}=h$ satisfies
$\ddot{h} + h=)$, so $h= \alpha \cos t + \beta \sin t$.

$$\Rightarrow g(t) = A\cos t + B\sin t + C.$$
\end{description}


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