Calc III help....will give "best answer"?

Question

Find parametric equations for the tangent line at the point
(cos(pi/6), sin(pi/6), (pi/6) on the curve x=cos t, y=sin t z=t

please help...i have tried this a lot and cant get it...i always give best answers for hw help.

Answer 1

The tangent line of a vector valued-function f through a point f(c) has the general form g(t) = f(c) + tf'(c). So in this case f(t) = (cos t, sin t, t), and c=π/6, so f'(t) = (-sin (π/6), cos (π/6), 1) = (-1/2, √3/2, 1). Thus the tangent line g has the form g(t) = (√3/2 - t/2, 1/2 + t√3/2, π/6 + t). Or in parametric form:

x = √3/2 - t/2
y = 1/2 + t√3/2
z = π/6 + t

Answer 2

x' = -sint
y' = cost
z' = 1

substitute the t=pi/6... that is the value of the direction vector [a , b, c]

the parametric equations...

x = at + x0
y = bt + y0
z = ct + z0

(x0 , y0 , z0) is the point... §