Question

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

Answer 1

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=(\pi )/(6)`.

We can do that by substituting
`t=(\pi )/(6)` in the given helix equation:

`r((\pi)/(6))=\left \langle cos(7((\pi)/(6))),sin(7((\pi)/(6))),-1((\pi)/(6)) \right \rangle\\r((\pi)/(6))=\left \langle cos((7\pi)/(6)),sin((7\pi)/(6)),-(\pi)/(6) \right \rangle\\`

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

`r((\pi)/(6))=\left \langle -(√(3))/(2),-(1)/(2),-(\pi)/(6) \right \rangle\\`

Answer 2

Consider the helix below:

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

We have to determine the value of helix at t =
`(\pi)/(6)`

So,
`r((\pi)/(6))`

=
`(\cos((7 \pi)/(6)), \sin((7 \pi)/(6)), -(\pi)/(6))`

Consider
`\cos((7 \pi)/(6)) = \cos(\pi + (\pi)/(6)) = - \cos ((\pi)/(6)) = (-\sqrt3)/(2)`

Consider
`\sin((7 \pi)/(6)) = \sin(\pi + (\pi)/(6)) = - \sin ((\pi)/(6)) = (-1)/(2)`

So, the value of helix
`r((\pi)/(6)) = ((-\sqrt3)/(2), (-1)/(2), (- \pi)/(6))`.