Question

Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.

$\mathbf{r}\left(t\right)=⟨t,5\sin 3t,5\cos 3t⟩,t=\frac{\pi }{6}$

Short answer

Ans: The unit tangent vector to $⟨t,5\sin 3t,5\cos 3t⟩$at $t=\frac{\pi }{6}$ is

Step-by-step solution

Step 1. Given information:

$\mathbf{r}\left(t\right)=⟨t,5\sin 3t,5\cos 3t⟩,t=\frac{\pi }{6}$

Step 2. Simplifying the Unit tangent vectors :

Consider $r\left(t\right)=⟨t,5\sin 3t,5\cos 3t⟩$

First, we compute role="math" localid="1649673535411" $r^{\text{'}}\left(t\right)$

$$\begin{gathered} r^{\text{'}}\left(t\right)=\frac{d}{dt}⟨t,5\sin 3t,5\cos 3t⟩ \\ =⟨1,15\cos 3t,-15\sin 3t⟩ \\ ||r^{\text{'}}\left(t\right)||=\Vert ⟨1,15\cos 3t,-15\sin 3t⟩\Vert \\ =\sqrt{(1{)}^2+(15\cos 3t{)}^2+(-15\sin 3t{)}^2} \\ =\sqrt{1+225{\cos }^{2}3t+225{\sin }^{2}3t} \\ =\sqrt{1+225\left({\sin }^{2}3t+{\cos }^{2}3t\right)} \\ =\sqrt{226} \end{gathered}$$

Step 3. Finding the Unit tangent vectors: 

The unit tangent vector to $r\left(t\right)$is

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨1,15\cos 3t,-15\sin 3t⟩}{\sqrt{226}} \end{gathered}$$

At $t=\frac{\pi }{6}$, the unit tangent vector to $r\left(t\right)$is