Answer
The parametrization:
${\bf{r}}\left( t \right) = \left( {5\cos 4t,5\sin 4t,\frac{{10}}{\pi }t} \right)$
The arc length parametrization:
${{\bf{r}}_1}\left( s \right)=$
$\left( {5\cos \left( {\frac{4}{{\sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}s} \right),5\sin \left( {\frac{4}{{\sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}s} \right),\frac{{10}}{{\pi \sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}s} \right)$
Work Step by Step
A helix of height $20$ cm that makes four full rotations over a circle of radius $5$ cm can be parametrized by
${\bf{r}}\left( t \right) = \left( {5\cos at,5\sin at,bt} \right)$, ${\ \ }$ for $0 \le t \le 2\pi $
Since it makes four full rotations, so
$a\cdot2\pi = 8\pi $, ${\ \ }$ $a=4$
Since the height is $20$, so
$b\cdot2\pi = 20$, ${\ \ }$ $b = \frac{{10}}{\pi }$
Thus, ${\bf{r}}\left( t \right) = \left( {5\cos 4t,5\sin 4t,\frac{{10}}{\pi }t} \right)$.
Finding an arc length parametrization:
Step 1. Evaluate the arc length integral:
$s = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
We have ${\bf{r}}\left( t \right) = \left( {5\cos 4t,5\sin 4t,\frac{{10}}{\pi }t} \right)$. So,
${\bf{r}}'\left( t \right) = \left( { - 20\sin 4t,20\cos 4t,\frac{{10}}{\pi }} \right)$
$||{\bf{r}}'\left( t \right)|| = \sqrt {400{{\sin }^2}4t + 400{{\cos }^2}4t + {{\left( {\frac{{10}}{\pi }} \right)}^2}} = \sqrt {400 + {{\left( {\frac{{10}}{\pi }} \right)}^2}} $
Thus,
$s = \sqrt {400 + {{\left( {\frac{{10}}{\pi }} \right)}^2}} \mathop \smallint \limits_0^t {\rm{d}}u = t\sqrt {400 + {{\left( {\frac{{10}}{\pi }} \right)}^2}} $
Step 2. Write $s = g\left( t \right) = t\sqrt {400 + {{\left( {\frac{{10}}{\pi }} \right)}^2}} $.
The inverse is $t = {g^{ - 1}}\left( s \right)$:
$t = \frac{s}{{\sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}$
Step 3. Substituting the replacement of $t$ in ${\bf{r}}\left( t \right)$ we get the arc length parametrization:
${{\bf{r}}_1}\left( s \right)=$
$\left( {5\cos \left( {\frac{4}{{\sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}s} \right),5\sin \left( {\frac{4}{{\sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}s} \right),\frac{{10}}{{\pi \sqrt {400 + {{\left( {10/\pi } \right)}^2}} }}s} \right)$
for $0 \le s \le 2\pi \sqrt {400 + {{\left( {\frac{{10}}{\pi }} \right)}^2}} $.
