Question

Principal unit normal vectors: Find the principal unit normal 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: The principal unit normal vector of $\mathbf{r}\left(t\right)=⟨3\sin 2t,3\cos 2t⟩$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 principal unit normal vector:

The principal unit normal vector of $r\left(t\right)$denoted by $N\left(t\right)$is given by

$N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||}$where $T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||}$

$T\left(t\right)$should be calculated first and then $N\left(t\right)$is to be evaluated. We start by comparing $r^{\text{'}}\left(t\right)$first:

$$\begin{gathered} r^{\text{'}}\left(t\right)=⟨6\cos 2t,-6\sin 2t⟩ \\ ||r^{\text{'}}\left(t\right)||=\sqrt{(6\cos 2t{)}^2+(-6\sin 2t{)}^2} \\ =\sqrt{36\left({\cos }^{2}2t+{\sin }^{2}2t\right)} \\ =6 \end{gathered}$$

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

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

Step 3. Finding the principal unit normal vector:

To find $N\left(t\right)$we divide $T^{\text{'}}\left(t\right)$by its magnitude.

$$\begin{gathered} T^{\text{'}}\left(t\right)=\sqrt{(-2\sin 2t{)}^2+(-2\cos 2t{)}^2} \\ =\sqrt{4{\left({\sin }^{2}2t\right)}^2+(-2\cos 2t{)}^2} \\ =2 \end{gathered}$$

The principal unit normal vector

Thus the principle unit normal vector of $r\left(t\right)$at $t=\frac{\pi }{3}$is