Answer
The answer is
(c) ${{\bf{r}}_3}\left( t \right) = \left( {4\sin \frac{t}{4},4\cos \frac{t}{4}} \right)$
Work Step by Step
By definition, an arc length parametrization has unit speed. Now, consider:
(a) ${{\bf{r}}_1}\left( t \right) = \left( {4\sin t,4\cos t} \right)$
The derivative is ${{\bf{r}}_1}'\left( t \right) = \left( {4\cos t, - 4\sin t} \right)$. The speed is
$||{{\bf{r}}_1}'\left( t \right)|| = \sqrt {{{\left( {4\cos t} \right)}^2} + {{\left( { - 4\sin t} \right)}^2}} $
$||{{\bf{r}}_1}'\left( t \right)|| = \sqrt {16{{\cos }^2}t + 16{{\sin }^2}t} = 4$
Since the speed is not $1$, it is not an arc length parametrization.
(b) ${{\bf{r}}_2}\left( t \right) = \left( {4\sin 4t,4\cos 4t} \right)$
The derivative is ${{\bf{r}}_2}'\left( t \right) = \left( {16\cos 4t, - 16\sin 4t} \right)$. The speed is
$||{{\bf{r}}_2}'\left( t \right)|| = \sqrt {{{\left( {16\cos 4t} \right)}^2} + {{\left( { - 16\sin 4t} \right)}^2}} $
$||{{\bf{r}}_2}'\left( t \right)|| = \sqrt {{{16}^2}{{\cos }^2}4t + {{16}^2}{{\sin }^2}4t} = 16$
Since the speed is not $1$, it is not an arc length parametrization.
(c) ${{\bf{r}}_3}\left( t \right) = \left( {4\sin \frac{t}{4},4\cos \frac{t}{4}} \right)$
The derivative is ${{\bf{r}}_3}'\left( t \right) = \left( {\cos \frac{t}{4}, - \sin \frac{t}{4}} \right)$. The speed is
$||{{\bf{r}}_3}'\left( t \right)|| = \sqrt {{{\left( {\cos \frac{t}{4}} \right)}^2} + {{\left( { - \sin \frac{t}{4}} \right)}^2}} $
$||{{\bf{r}}_3}'\left( t \right)|| = \sqrt {{{\cos }^2}\frac{t}{4} + {{\sin }^2}\frac{t}{4}} = 1$
Since the speed is $1$, it is an arc length parametrization.
