Question

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

$\mathbf{r}\left(t\right)=⟨3\sin 2t,3\cos 2t⟩,t=\frac{\pi }{3}$

Short answer

Ans: Thus the unit tangent vector to $⟨3\sin 2t,3\cos 2t⟩$at $t=\frac{\pi }{3}$is

Step-by-step solution

Step 1. Given information: 

$\mathbf{r}\left(t\right)=⟨3\sin 2t,3\cos 2t⟩,t=\frac{\pi }{3}$

Step 2. Simplifying the Unit tangent vectors :

Consider $r\left(t\right)=⟨3\sin 2t,3\cos 2t⟩$

First, we compute

$$\begin{gathered} =6 \\ since{\sin }^{2}2t+{\cos }^{2}2t=1 \end{gathered}$$

Step 3. Finding the Unit tangent vectors: 

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

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}\text{'}}\left(t\right)||} \\ =\frac{6⟨\cos 2t,-\sin 2t⟩}{6} \\ =⟨\cos 2t,-\sin 2t⟩ \end{gathered}$$

At $t=\frac{\pi }{3}$, the unit tangent vector is

Thus the unit tangent vector to $⟨3\sin 2t,3\cos 2t⟩$at $t=\frac{\pi }{3}$ is