Answer
The arc length parametrization is ${{\bf{r}}_1}\left( s \right) = \left( {1 + 4\cos \frac{s}{4},4 + 4\sin \frac{s}{4},9} \right)$
Work Step by Step
A circle with radius $4$ in the $xy$-plane with center $\left( {1,4} \right)$ is given by
${\left( {x - 1} \right)^2} + {\left( {y - 4} \right)^2} = 16$
This can be parametrized by $x\left( t \right) = 1 + 4\cos t$ and $y\left( t \right) = 4 + 4\sin t$, because it satisfies the equation above.
For the circle in the plane $z=9$ with radius $4$ and center $\left( {1,4,9} \right)$, we have
$x\left( t \right) = 1 + 4\cos t$, ${\ \ }$ $y\left( t \right) = 4 + 4\sin t$, ${\ \ }$ $z\left( t \right) = 9$
Hence, the parametrization is
${\bf{r}}\left( t \right) = \left( {1 + 4\cos t,4 + 4\sin t,9} \right)$, ${\ \ }$ for $0 \le t \le 2\pi $
Next, we find the arc length parametrization.
The derivative is
${\bf{r}}'\left( t \right) = \left( { - 4\sin t,4\cos t,0} \right)$
Evaluate the arc length function
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {16{{\sin }^2}u + 16{{\cos }^2}u} {\rm{d}}u$
$s\left( t \right) = 4\mathop \smallint \limits_0^t {\rm{d}}u = 4t$
The inverse of $s\left( t \right)$ is $t = \frac{s}{4}$. Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {1 + 4\cos t,4 + 4\sin t,9} \right)$ gives
${{\bf{r}}_1}\left( s \right) = \left( {1 + 4\cos \frac{s}{4},4 + 4\sin \frac{s}{4},9} \right)$
Thus, the arc length parametrization is ${{\bf{r}}_1}\left( s \right) = \left( {1 + 4\cos \frac{s}{4},4 + 4\sin \frac{s}{4},9} \right)$
