Answer
The angle $\psi$ between the position vector and the tangent vector is constant.
$\psi \simeq 75.964^\circ $
Work Step by Step
Recall that the derivative vector ${\bf{r}}'\left( {{t_0}} \right)$ points in the direction tangent to the path traced by ${\bf{r}}\left( t \right)$ at $t = {t_0}$. For the curve ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t,{{\rm{e}}^t}\sin 4t} \right)$ we get
${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t - 4{{\rm{e}}^t}\sin 4t,{{\rm{e}}^t}\sin 4t + 4{{\rm{e}}^t}\cos 4t} \right)$
Thus, the tangent vector is
${\bf{r}}'\left( t \right) = {{\rm{e}}^t}\left( {\cos 4t - 4\sin 4t,\sin 4t + 4\cos 4t} \right)$
Let $\psi$ denote the angle between the position vector and the tangent vector.
By Theorem 2 of Section 13.3, we obtain
$\cos \psi = \frac{{{\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right)}}{{||{\bf{r}}\left( t \right)||||{\bf{r}}'\left( t \right)||}}$
1. Evaluate ${\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right)$
${\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right) = {{\rm{e}}^{2t}}\left( {\cos 4t,\sin 4t} \right)\cdot$
$\left( {\cos 4t - 4\sin 4t,\sin 4t + 4\cos 4t} \right)$
${\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right) = {{\rm{e}}^{2t}}[\cos 4t\left( {\cos 4t - 4\sin 4t} \right)$
$ + \sin 4t\left( {\sin 4t + 4\cos 4t} \right)]$
${\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right) = {{\rm{e}}^{2t}}[{\cos ^2}4t - 4\cos 4t\sin 4t$
$ + {\sin ^2}4t + 4\sin 4t\cos 4t)]$
${\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right) = {{\rm{e}}^{2t}}$
2. Evaluate $||{\bf{r}}\left( t \right)|{|^2}$
$||{\bf{r}}\left( t \right)|{|^2} = {\bf{r}}\left( t \right)\cdot{\bf{r}}\left( t \right)$
$||{\bf{r}}\left( t \right)|{|^2} = \left( {{{\rm{e}}^t}\cos 4t,{{\rm{e}}^t}\sin 4t} \right)\cdot$
$\left( {{{\rm{e}}^t}\cos 4t,{{\rm{e}}^t}\sin 4t} \right)$
$||{\bf{r}}\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left( {{{\cos }^2}4t + {{\sin }^2}4t} \right) = {{\rm{e}}^{2t}}$
So, $||{\bf{r}}\left( t \right)|| = {{\rm{e}}^t}$.
3. Evaluate $||{\rm{r}}'\left( t \right)|{|^2}$
$||{\rm{r}}'\left( t \right)|{|^2} = {{\rm{e}}^t}\left( {\cos 4t - 4\sin 4t,\sin 4t + 4\cos 4t} \right)\cdot$
${{\rm{e}}^t}\left( {\cos 4t - 4\sin 4t,\sin 4t + 4\cos 4t} \right)$
$||{\rm{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left[ {{{\left( {\cos 4t - 4\sin 4t} \right)}^2} + {{\left( {\sin 4t + 4\cos 4t} \right)}^2}} \right]$
$||{\rm{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}[{\cos ^2}4t - 8\cos 4t\sin 4t + 16{\sin ^2}4t$
$ + {\sin ^2}4t + 8\sin 4t\cos 4t + 16{\cos ^2}4t]$
$||{\rm{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left[ {1 + 16} \right] = 17{{\rm{e}}^{2t}}$
So, $||{\rm{r}}'\left( t \right)|| = \sqrt {17} {{\rm{e}}^t}$.
Substituting the results obtained above in $\cos \psi $ gives
$\cos \psi = \frac{{{\bf{r}}\left( t \right)\cdot{\bf{r}}'\left( t \right)}}{{||{\bf{r}}\left( t \right)||||{\bf{r}}'\left( t \right)||}}$
$\cos \psi = \frac{{{{\rm{e}}^{2t}}}}{{\left( {{{\rm{e}}^t}} \right)\left( {\sqrt {17} {{\rm{e}}^t}} \right)}} = \frac{1}{{\sqrt {17} }}$
$\psi = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {17} }}} \right) \simeq 75.964^\circ $
Hence, the angle between the position vector and the tangent vector is constant.
