Question

Find a tangent vector of unit length at the point with the given value of the parameter t.
r(t) = 2 sin(t)i + 9 cos(t)j
t = π/6

Answer 1

find r'(pi/6), then divide the tangent vector by its length

Answer 2

r'(pi/6)= 2cos(pi/6)i -9sin(pi/6) j
=2(sqrt3/2) - 9(1/2)
=sqrt3i - 9/2 j
?? and I am lost after this,

Answer 3

sin(pi/6) = 1/2
cos(pi/6) = sqrt(3)/2

<(2)(1/2) , 9 sqrt(3)/2>= <1, (9/2)sqrt(3)>

L^2 = 1^2 + [(9/2)sqrt(3)]^2
L = sqrt(61.75)

(1/sqrt(61.75)) <1, (9/2)sqrt(3)>