Answer
(a) $s\left( t \right) = \sqrt {17} {{\rm{e}}^t}$
(b) An arc length parametrization:
${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {17} }}\cos \left( {4\ln \left( {\frac{s}{{\sqrt {17} }}} \right)} \right),\frac{s}{{\sqrt {17} }}\sin \left( {4\ln \left( {\frac{s}{{\sqrt {17} }}} \right)} \right)} \right)$
Work Step by Step
(a) We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t,{{\rm{e}}^t}\sin 4t} \right)$. The derivative is
${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t - 4{{\rm{e}}^t}\sin 4t,{{\rm{e}}^t}\sin 4t + 4{{\rm{e}}^t}\cos 4t} \right)$
${\bf{r}}'\left( t \right) = {{\rm{e}}^t}\left( {\cos 4t - 4\sin 4t,\sin 4t + 4\cos 4t} \right)$
So,
$||{\bf{r}}'\left( t \right)|{|^2} = {\bf{r}}'\left( t \right)\cdot{\bf{r}}'\left( t \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left( {{{\left( {\cos 4t - 4\sin 4t} \right)}^2} + {{\left( {\sin 4t + 4\cos 4t} \right)}^2}} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}({\cos ^2}4t - 8\cos 4t\sin 4t + 16{\sin ^2}4t$
$ + {\sin ^2}4t + 8\sin 4t\cos 4t + 16{\cos ^2}4t)$
$||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left( {1 + 16} \right) = 17{{\rm{e}}^{2t}}$
$||{\bf{r}}'\left( t \right)|| = \sqrt {17} {{\rm{e}}^t}$
Evaluate $s\left( t \right) = \mathop \smallint \limits_{ - \infty }^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \sqrt {17} \mathop \smallint \limits_{ - \infty }^t {{\rm{e}}^u}{\rm{d}}u = \sqrt {17} {{\rm{e}}^u}|_{ - \infty }^t$
$s\left( t \right) = \sqrt {17} {{\rm{e}}^t}$
(b) From part (a) we obtain $s\left( t \right) = \sqrt {17} {{\rm{e}}^t}$. Thus, the inverse of $s\left( t \right)$ is $t = \ln \left( {\frac{s}{{\sqrt {17} }}} \right)$.
Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t,{{\rm{e}}^t}\sin 4t} \right)$ gives the arc length parametrization:
${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {17} }}\cos \left( {4\ln \left( {\frac{s}{{\sqrt {17} }}} \right)} \right),\frac{s}{{\sqrt {17} }}\sin \left( {4\ln \left( {\frac{s}{{\sqrt {17} }}} \right)} \right)} \right)$
