Find the radius of the osculating circle of the ellipse traced out by the vector function $$\begin{equation*} \mathbf{r}(t)=3\sin t\mathbf{i}+4\cos t\mathbf{j} \qquad 0\leq t\leq 2\pi \end{equation*}$$

(a) At $t=0$

(b) At $t=\dfrac{\pi }{2}$

Solution To find the radius of the osculating circle, we need to find the curvature $\kappa$. We treat the ellipse as a curve in space with a $\mathbf{k}$ component of $0$ and find $\mathbf{r}^{\prime} (t),$ $\left\Vert \mathbf{r}^{\prime} (t) \right\Vert, \mathbf{r}^{\prime \prime} (t),$ $\mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t),$ and $\left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert$. $$\begin{eqnarray*} \mathbf{r^{\prime} }(t) &=& 3\cos t\mathbf{i}-4\sin t\mathbf{j}+0\mathbf{k} \qquad \left\Vert \mathbf{r^{\prime} }(t) \right\Vert\;=\; \sqrt{9\cos ^{2}t+16\sin ^{2}t}\\[9pt] \mathbf{r^{\prime \prime} }(t) &=& -3\sin t\mathbf{i}-4\cos t\mathbf{j}+0\mathbf{k} \end{eqnarray*}$$

$$\begin{eqnarray*} \mathbf{r^{\prime} }(t)\times \mathbf{r^{\prime \prime} }(t)= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 3\cos t & -4\sin t & 0 \\[3pt] -3\sin t & -4\cos t & 0 \end{array}\right| &=&( -12\cos ^{2}t-12\sin ^{2}t) \mathbf{k}=-12\mathbf{k}\\[9pt] \left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r} ^{\prime \prime} (t) \right\Vert &=&12 \end{eqnarray*}$$

Using formula (4), the curvature $\kappa$ of $C$ is $$\begin{equation*} \kappa\;=\;\dfrac{\left\Vert \mathbf{r^{\prime} }(t) \times \mathbf{ r^{\prime \prime} }(t) \right\Vert }{\left\Vert \mathbf{r^{\prime} } (t) \right\Vert ^{3}}=\dfrac{12}{( 9\cos ^{2}t+16\sin ^{2}t) ^{3/2}} \end{equation*}$$

The radius $\rho$ of the osculating circle is $$\begin{equation*} \rho\;=\;\frac{1}{\kappa }=\frac{(9\cos ^{2}t+16\sin ^{2}t)^{3/2}}{12} \end{equation*}$$

(a) At $t=0$, the curvature of $C$ is $\kappa\;=\;\dfrac{4}{9}$, so the radius of the osculating circle is $\rho\;=\;\dfrac{9}{4}$.

(b) At $t=\dfrac{\pi }{2},$ the curvature of $C$ is $\kappa\;=\;\dfrac{3}{16}$, so the radius of the osculating circle is $\rho\;=\;\dfrac{16}{3}$.