Answer:
[tex]\vec u = (\frac{1}{2},\frac{\sqrt{3} }{2} )[/tex]
Step-by-step explanation:
The slope of the tangent line at any point is found by the deriving the curve. Hence:
[tex]y' = 2 \cdot \cos x[/tex]
The slope at [tex]x = \frac{\pi}{6}[/tex] is:
[tex]y' = \sqrt{3}[/tex]
Vectorially speaking, the slope has the following expression:
[tex]\vec m = (1, \sqrt{3} )[/tex]
The magnitude of this vector is:
[tex]||\vec m|| = 2[/tex]
The unit vector is finally obtained by dividing each component by magnitude:
[tex]\vec u = (\frac{1}{2},\frac{\sqrt{3} }{2} )[/tex]
