Answer
A parametrization of the curve using trigonometric functions:
${\bf{r}}\left( t \right) = \left( {9\cos 2t + 2,3\cos t,3\sin t} \right)$, ${\ \ }$ for $0 \le t \le 2\pi $
Work Step by Step
We have from Exercise 25:
${y^2} - {z^2} = x - 2$, ${\ \ }$ ${y^2} + {z^2} = 9$
The second equation can be parametrized using trigonometric functions:
$y = 3\cos t$, ${\ \ }$ $z = 3\sin t$, ${\ \ }$ for $0 \le t \le 2\pi $,
because these coordinate functions satisfy the equation ${y^2} + {z^2} = 9$.
Now, the first equation can be written $x = {y^2} - {z^2} + 2$. Thus,
$x = 9{\cos ^2}t - 9{\sin ^2}t + 2$
Since $\cos 2t = {\cos ^2}t - {\sin ^2}t$, so $x = 9\cos 2t + 2$.
Hence, a parametrization of the curve is
${\bf{r}}\left( t \right) = \left( {9\cos 2t + 2,3\cos t,3\sin t} \right)$, ${\ \ }$ for $0 \le t \le 2\pi $
