Find the principal unit normal vector $\mathbf{N}(t)$ to the circle $\mathbf{r}( t) =R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}$, $0\leq t\leq 2\pi .$ (Refer to Example 2.) Graph $\mathbf{r}=\mathbf{r}( t),$ and show a unit tangent vector and a unit normal vector.

Solution From Example 2, the unit tangent vector $\mathbf{T}(t)$ is $$\begin{equation*} \mathbf{T}(t)=-\!\sin t\mathbf{i}+\cos t\mathbf{j} \end{equation*}$$

Figure 13 $\mathbf{r}( t) =R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j,} \\ 0\leq t\leq 2\pi$

Since $\mathbf{T}^{\prime} ( t) =-\cos t\mathbf{i}-\sin t\mathbf{j},$ the principal unit normal vector $\mathbf{N}(t)$ is $$\begin{equation*} \mathbf{N}(t)=\frac{\mathbf{T}^{\prime} (t)}{\left\Vert \mathbf{T}^{\prime} (t)\right\Vert }=\frac{-\!\cos t\mathbf{i}-\sin t\mathbf{j}}{\sqrt{(-\!\cos t)^{2}+(-\!\sin t)^{2}}}=-\cos t\mathbf{i}-\sin t\mathbf{j} \end{equation*}$$

Since $\mathbf{r}( t) = R ( \cos t\mathbf{i}+\sin t\mathbf{j}),$ the vector $\mathbf{N}( t)$ is parallel to $-\mathbf{r}(t)$. That is, $\mathbf{N}(t)$ is a unit vector opposite in direction to the vector $\mathbf{r}(t)$, so $\mathbf{N}$ is directed toward the center of the circle, as shown in Figure 13.