Find parametric equations for the tangent line at the point (cos(2π/6),sin(2π/6),2π/6) on the curve x=cost, y=sint, z=t x(t) = y(t)= z(t)= (your line should be parametrized so that it passes through the given point at t=0).

 (i) Note that t = 2π/6 <==> the point (cos 2π/6, sin 2π/6, 2π/6) = (1/2, sqrt(3)/2, π/3). 

(ii) For the direction vector:  
Differentiating the curve yields x' = -sin t, y' = cos t, z' = 1. 

So, the direction vector v for the tangent line is (x', y', z') at t = 2π/6 (where the point is located): 
==> v = (-sin 2π/6, cos 2π/6, 1) = (-sqrt(3)/2, 1/2, 1). 

Hence, the equation of the tangent line at the prescribed point is 
r(t) = (1/2, sqrt(3)/2, π/3) + t(-sqrt(3)/2, 1/2, 1). 

Parametrically, this is given by 
x = 1/2 - t sqrt(3)/2 
y = sqrt(3)/2 + t/2 
z = π/3 + t. 

I hope this helps!