How do you find the unit vectors that are parallel to the tangent line to the curve y = 2sinx at the point (pi/6,1)?

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Answer 1 · 2016-08-24T12:54:24

$+- < cos (pi/6), sin (pi/6) > =+-<1/2, sqrt 3/2>$

y'at $x = pi/6$ is $2 cos (pi/6) = sqrt3$ .

This direction $theta=psi$ is given by $tan psi=sqrt 3$ .

Inversely, $psi = tan^(-1) sqrt 3$ is $pi/6$ . For the opposite direction ,

it is $pi+pi/6$ .

The unit vector in the direction $theta = pi/6$ is

$< cos (pi/6), sin (pi/6) >$ .

For the opposite direction, it is

$< cos(pi+pi/6), sin(pi+pi/6) >$

$=<-cos(pi/6),-sin(pi/6)>$ .