Find $\vec{T}, \vec{N}$ and $\vec{B}$ for the curve $\vec{r}(t)=\langle 3 \cos (4 t), 3 \sin (4 t), 2 t\rangle$ at the point $t=0$ Give your answers to two decimal places

Step 1 ：Given the curve \(\vec{r}(t)=\langle 3 \cos (4 t), 3 \sin (4 t), 2 t\rangle\)

Step 2 ：Calculate the derivative of \(\vec{r}(t)\) to get \(\vec{r}'(t) = \langle -12 \sin (4 t), 12 \cos (4 t), 2 \rangle\)

Step 3 ：Normalize \(\vec{r}'(t)\) to get the unit tangent vector \(\vec{T} = \frac{\vec{r}'(t)}{||\vec{r}'(t)||} = \langle -\frac{12 \sin (4 t)}{\sqrt{144 \sin^2 (4 t) + 144 \cos^2 (4 t) + 4}}, \frac{12 \cos (4 t)}{\sqrt{144 \sin^2 (4 t) + 144 \cos^2 (4 t) + 4}}, \frac{2}{\sqrt{144 \sin^2 (4 t) + 144 \cos^2 (4 t) + 4}} \rangle\)

Step 4 ：Calculate the derivative of \(\vec{T}\) to get \(\vec{T}' = \langle -\frac{48 \cos (4 t)}{\sqrt{144 \sin^2 (4 t) + 144 \cos^2 (4 t) + 4}}, -\frac{48 \sin (4 t)}{\sqrt{144 \sin^2 (4 t) + 144 \cos^2 (4 t) + 4}}, 0 \rangle\)

Step 5 ：Normalize \(\vec{T}'\) to get the unit normal vector \(\vec{N} = \frac{\vec{T}'}{||\vec{T}'||} = \langle -\frac{48 \cos (4 t)}{\sqrt{2304 \sin^2 (4 t) + 2304 \cos^2 (4 t)}}, -\frac{48 \sin (4 t)}{\sqrt{2304 \sin^2 (4 t) + 2304 \cos^2 (4 t)}}, 0 \rangle\)

Step 6 ：Calculate the cross product of \(\vec{T}\) and \(\vec{N}\) to get the binormal vector \(\vec{B} = \vec{T} \times \vec{N} = \langle 0, -\frac{\sqrt{37}}{37}, \frac{6 \sqrt{37}}{37} \rangle\)

Step 7 ：Substitute \(t = 0\) into \(\vec{T}\), \(\vec{N}\), and \(\vec{B}\) to get \(\vec{T}(0) = \langle 0, \frac{6 \sqrt{37}}{37}, \frac{\sqrt{37}}{37} \rangle\), \(\vec{N}(0) = \langle -1, 0, 0 \rangle\), and \(\vec{B}(0) = \langle 0, -\frac{\sqrt{37}}{37}, \frac{6 \sqrt{37}}{37} \rangle\)

Step 8 ：Final Answer: The unit tangent vector \(\vec{T}\) at \(t=0\) is \(\boxed{\left(0, \frac{6\sqrt{37}}{37}, \frac{\sqrt{37}}{37}\right)}\), the unit normal vector \(\vec{N}\) at \(t=0\) is \(\boxed{\left(-1, 0, 0\right)}\), and the binormal vector \(\vec{B}\) at \(t=0\) is \(\boxed{\left(0, -\frac{\sqrt{37}}{37}, \frac{6\sqrt{37}}{37}\right)}\)