Question

Find parametric equations for line that is tangent to the curve x=cost, y=sint, z=t at the point
(cos(2π/6),sin(2π/6),2π/6) .

Parametrize the line so that it passes through the given point at t=0. All three answers are required for credit.

Answer 1

$\displaystyle r(t) = (\cos{t}, \sin{t}, t)\\ \textsf{at}\hspace{0.1cm} t = 0, r(0) = (\cos{0}, \sin{0}, 0) = (1, 0, 0)\\ r'(t) = (-\sin{t}, \cos{t}, 1)\\ \textsf{At the point} \hspace{0.1cm}(x(t), y(t), z(t)) = \left(\cos\left(\frac{2\pi}{6}\right), \sin\left(\frac{2\pi}{6}\right), \frac{2\pi}{6}\right),\\ \textsf{ it corresponds to}\hspace{0.1cm} t = \frac{2\pi}{6}.\\ \textsf{Thus, we are looking for the}\\\textsf{tangent vector at}\hspace{0.1cm} t = \frac{2\pi}{6}\\ r'\left(\frac{2\pi}{6}\right)= \left(-\sin\left(\frac{2\pi}{6}\right), \cos\left(\frac{2\pi}{6}\right), 1\right)\\ = \left(-\frac{\sqrt{3}}{2},\frac{1}{2}, 1\right)\\ \therefore \textsf{The parametric equations for the line}\\\textsf{ that is tangent to the curve} \hspace{0.1cm} r(t) \hspace{0.1cm}\textsf{are}\\ x - 1 = -\frac{\sqrt{3}}{2}t, y - 0 = \frac{1}{2}t, z - 0 = t\\ x = -\frac{\sqrt{3}}{2}t + 1, y = \frac{1}{2}t, z = t$