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

Step 1 ：Given the curve $\overrightarrow{r}(t)=⟨3\cos (4t),3\sin (4t),2t⟩$

Step 2 ：Calculate the derivative of $\overrightarrow{r}(t)$ to get ${\overrightarrow{r}}^{\prime }(t)=⟨-12\sin (4t),12\cos (4t),2⟩$

Step 3 ：Normalize ${\overrightarrow{r}}^{\prime }(t)$ to get the unit tangent vector $\overrightarrow{T}=\frac{{\overrightarrow{r}}^{\prime }(t)}{||{\overrightarrow{r}}^{\prime }(t)||}=⟨-\frac{12\sin (4t)}{\sqrt{144{\sin }^2(4t)+144{\cos }^2(4t)+4}},\frac{12\cos (4t)}{\sqrt{144{\sin }^2(4t)+144{\cos }^2(4t)+4}},\frac{2}{\sqrt{144{\sin }^2(4t)+144{\cos }^2(4t)+4}}⟩$

Step 4 ：Calculate the derivative of $\overrightarrow{T}$ to get ${\overrightarrow{T}}^{\prime }=⟨-\frac{48\cos (4t)}{\sqrt{144{\sin }^2(4t)+144{\cos }^2(4t)+4}},-\frac{48\sin (4t)}{\sqrt{144{\sin }^2(4t)+144{\cos }^2(4t)+4}},0⟩$

Step 5 ：Normalize ${\overrightarrow{T}}^{\prime }$ to get the unit normal vector $\overrightarrow{N}=\frac{{\overrightarrow{T}}^{\prime }}{||{\overrightarrow{T}}^{\prime }||}=⟨-\frac{48\cos (4t)}{\sqrt{2304{\sin }^2(4t)+2304{\cos }^2(4t)}},-\frac{48\sin (4t)}{\sqrt{2304{\sin }^2(4t)+2304{\cos }^2(4t)}},0⟩$

Step 6 ：Calculate the cross product of $\overrightarrow{T}$ and $\overrightarrow{N}$ to get the binormal vector $\overrightarrow{B}=\overrightarrow{T}\times \overrightarrow{N}=⟨0,-\frac{\sqrt{37}}{37},\frac{6\sqrt{37}}{37}⟩$

Step 7 ：Substitute $t=0$ into $\overrightarrow{T}$, $\overrightarrow{N}$, and $\overrightarrow{B}$ to get $\overrightarrow{T}(0)=⟨0,\frac{6\sqrt{37}}{37},\frac{\sqrt{37}}{37}⟩$, $\overrightarrow{N}(0)=⟨-1,0,0⟩$, and $\overrightarrow{B}(0)=⟨0,-\frac{\sqrt{37}}{37},\frac{6\sqrt{37}}{37}⟩$

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