Show that the unit tangent vector $\mathbf{T}(t)$ to the circle of radius $R$ $$\begin{equation*} \mathbf{r}(t)=R\cos t\mathbf{i}+R\;\sin t\mathbf{j}\quad 0\leq t\leq 2\pi \end{equation*}$$

is everywhere orthogonal to $\mathbf{r}(t).$ Graph $\mathbf{r}=\mathbf{r}( t)$ and $\mathbf{T}=\mathbf{T}( t).$

Solution We begin by finding $\mathbf{r}^{\prime} (t)$ and $\left\Vert \mathbf{r}^{\prime} (t)\right\Vert$: $$\begin{eqnarray*} \mathbf{r}^{\prime} (t)& =&\dfrac{d}{dt}( R\cos t) \mathbf{i}+ \dfrac{d}{dt}( R\;\sin t) \mathbf{j}=-R\;\sin t\mathbf{i}+R\cos t\mathbf{j} \\[6pt] \left\Vert \mathbf{r}^{\prime} (t)\right\Vert & =&\sqrt{(-R\;\sin t)^{2}+(R\cos t)^{2}}=R \end{eqnarray*}$$

Figure 11 $\mathbf{r}( t) =R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}$

Then the unit tangent vector is $$\begin{equation*} \mathbf{T}(t)=\frac{\mathbf{r}^{\prime} (t)}{\left\Vert \mathbf{r}^{\prime} (t)\right\Vert }=\dfrac{-R\;\sin t\mathbf{i}+R\;\cos t\mathbf{j}}{R}=-\!\sin t\mathbf{i}+\cos t\mathbf{j} \end{equation*}$$

To determine whether $\mathbf{r}( t)$ is orthogonal to $\mathbf{T}( t),$ we find their dot product. $$\begin{equation*} \mathbf{T}(t)\,{\cdot}\, \mathbf{r}(t)=( -\!\sin t\mathbf{i}+\cos t\mathbf{j}) \,{\cdot}\, ( R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}) =-R\;\sin t\cos t+R\;\sin t\cos t=0 \end{equation*}$$

for all $t.$ That is, $\mathbf{T}(t)$ is everywhere orthogonal to $\mathbf{r}(t)$, as shown in Figure 11.