Answer
(a) The helix can be parametrized by
${\bf{r}}\left( t \right) = \left( {10\cos t,10\sin t,\frac{{34t}}{{2\pi }}} \right)$.
(b) The arc length of one full turn of the helix:
$s \simeq 71.44$ Å
Work Step by Step
(a) We have the radius of the helix $10$ Å. Thus, $x = 10\cos t$ and $y = \sin t$. Since parametrization is not unique, let $z = \alpha t$, where $\alpha$ is a scalar. Let $t$ start from zero. So, we may parametrize the curve by ${\bf{r}}\left( t \right) = \left( {10\cos t,10\sin t,\alpha t} \right)$. A full turn occurs when $t = 2\pi $.
Since one full turn of the helix has a height of $34$ Å, so $\alpha \cdot2\pi = 34$. It follows that $\alpha = \frac{{34}}{{2\pi }}$. Thus, the parametrization is ${\bf{r}}\left( t \right) = \left( {10\cos t,10\sin t,\frac{{34t}}{{2\pi }}} \right)$.
(b) From part (a) we get the parametrization of the helix ${\bf{r}}\left( t \right) = \left( {10\cos t,10\sin t,\frac{{34t}}{{2\pi }}} \right)$. So,
$x\left( t \right) = 10\cos t$, ${\ \ }$ $y\left( t \right) = 10\sin t$, ${\ \ }$ $z\left( t \right) = \frac{{34t}}{{2\pi }}$
$x'\left( t \right) = - 10\sin t$, ${\ \ }$ $y'\left( t \right) = 10\cos t$, ${\ \ }$ $z'\left( t \right) = \frac{{34}}{{2\pi }}$
By Theorem 1, the arc length of one full turn of the helix is
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( { - 10\sin t} \right)}^2} + {{\left( {10\cos t} \right)}^2} + {{\left( {\frac{{34}}{{2\pi }}} \right)}^2}} {\rm{d}}t$
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {100{{\sin }^2}t + 100{{\cos }^2}t + {{\left( {\frac{{34}}{{2\pi }}} \right)}^2}} {\rm{d}}t$
$s = \sqrt {100 + {{\left( {\frac{{34}}{{2\pi }}} \right)}^2}} \mathop \smallint \limits_0^{2\pi } {\rm{d}}t = 2\pi \sqrt {100 + {{\left( {\frac{{34}}{{2\pi }}} \right)}^2}} $
$s \simeq 71.44$ Å
