Question

Consider the helix r(t)=⟨cos(7t),sin(7t),−1t⟩. Compute, at t=π6:

Answer 1

Consider the helix below:

$r(t)= (\cos(7t), \sin(7t), -t)$

We have to determine the value of helix at t = $\frac{\pi}{6}$

So, $r(\frac{\pi}{6})$

=$(\cos(\frac{7 \pi}{6}), \sin(\frac{7 \pi}{6}), -\frac{\pi}{6})$

Consider $\cos(\frac{7 \pi}{6}) = \cos(\pi + \frac{\pi}{6}) = - \cos (\frac{\pi}{6}) = \frac{-\sqrt3}{2}$

Consider $\sin(\frac{7 \pi}{6}) = \sin(\pi + \frac{\pi}{6}) = - \sin (\frac{\pi}{6}) = \frac{-1}{2}$

So, the value of helix $r(\frac{\pi}{6}) = (\frac{-\sqrt3}{2}, \frac{-1}{2}, \frac{- \pi}{6})$.

Answer 2

We have been given the parametric equation of a Helix as shown below:

$r(t)=\left \langle cos(7t),sin(7t),-1t \right \rangle$

We are required to find the value of this helix at $t=\frac{\pi }{6}$.

We can do that by substituting $t=\frac{\pi }{6}$ in the given helix equation:

$r(\frac{\pi}{6})=\left \langle cos(7(\frac{\pi}{6})),sin(7(\frac{\pi}{6})),-1(\frac{\pi}{6}) \right \rangle\\r(\frac{\pi}{6})=\left \langle cos(\frac{7\pi}{6}),sin(\frac{7\pi}{6}),-\frac{\pi}{6} \right \rangle$

Upon simplifying this further by using the values of trigonometric ratios, we get:

$r(\frac{\pi}{6})=\left \langle -\frac{\sqrt{3}}{2},-\frac{1}{2},-\frac{\pi}{6} \right \rangle$