(1 point) Consider the helix $\mathbf{r}(t)=(\cos (2 t), \sin (2 t),-3 t)$. Compute, at $t=\frac{\pi}{6}$ :
A. The unit tangent vector $\mathbf{T}\left(\frac{\pi}{6}\right)=($
B. The unit normal vector $\mathbf{N}\left(\frac{\pi}{6}\right)=($
C. The unit binormal vector $\mathbf{B}\left(\frac{\pi}{6}\right)=$
D. The curvature $\kappa\left(\frac{\pi}{6}\right)=$

Step 1 ：Given the helix \(\mathbf{r}(t)=(\cos (2 t), \sin (2 t),-3 t)\), we are asked to compute the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), the unit binormal vector \(\mathbf{B}\), and the curvature \(\kappa\) at \(t=\frac{\pi}{6}\).

Step 2 ：The unit tangent vector \(\mathbf{T}(t)\) is given by the derivative of the position vector \(\mathbf{r}(t)\) divided by its magnitude.

Step 3 ：The unit normal vector \(\mathbf{N}(t)\) is given by the derivative of the unit tangent vector \(\mathbf{T}(t)\) divided by its magnitude.

Step 4 ：The unit binormal vector \(\mathbf{B}(t)\) is the cross product of the unit tangent vector \(\mathbf{T}(t)\) and the unit normal vector \(\mathbf{N}(t)\).

Step 5 ：The curvature \(\kappa(t)\) is the magnitude of the derivative of the unit tangent vector \(\mathbf{T}(t)\) divided by the magnitude of the derivative of the position vector \(\mathbf{r}(t)\).

Step 6 ：After performing the necessary calculations, we find that the unit tangent vector at \(t=\frac{\pi}{6}\) is \(\mathbf{T}\left(\frac{\pi}{6}\right)=\boxed{(-0.480384461415261, 0.277350098112615, -0.832050294337844)}\).

Step 7 ：The unit normal vector at \(t=\frac{\pi}{6}\) is \(\mathbf{N}\left(\frac{\pi}{6}\right)=\boxed{(-0.5, -0.866025403784439)}\).

Step 8 ：The unit binormal vector at \(t=\frac{\pi}{6}\) is \(\mathbf{B}\left(\frac{\pi}{6}\right)=\boxed{(-0.720576692122892, 0.416025147168922, 0.554700196225229)}\).

Step 9 ：The curvature at \(t=\frac{\pi}{6}\) is \(\kappa\left(\frac{\pi}{6}\right)=\boxed{0.307692307692308}\).