## Thomas' Calculus 13th Edition

$v=\lt \dfrac{\sqrt 3}{2}, \dfrac{1}{2}\gt$
The components of vector $v$ are: $v=\lt v_x,v_y\gt$ Thus, $v_x=(1) \cos (\dfrac{\pi}{6})=\dfrac{\sqrt 3}{2}$ and $v_y=(1) \sin (\dfrac{\pi}{6})=\dfrac{1}{2}$ Hence, $v=\lt \dfrac{\sqrt 3}{2}, \dfrac{1}{2}\gt$