91Ó°ÊÓ

To find \(\mathbf{r}^{\prime}(t)\), we simply differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\): \(\mathbf{r}(t) = \langle 2 \cos t, 8 \sin t, 0 \rangle\) \(\mathbf{r}^{\prime}(t) = \langle -2 \sin t, 8 \cos t, 0 \rangle\)

Two vectors are orthogonal if their dot product is 0. Therefore, we want to find the values of \(t\) where: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = 0\) So, \((2\cos t)(-2\sin t) + (8\sin t)(8\cos t) + (0)(0) = 0\) Simplifying, we get: \(-4\sin t\cos t + 64\sin t\cos t = 0\) Combining the terms, we get: \(60\sin t\cos t = 0\) Since \(\sin t\) and \(\cos t\) will always be non-zero for \(0 \leq t \leq 2\pi\), we can divide both sides by 60: \(\sin t\cos t = 0\)

We now solve the equation \(\sin t\cos t = 0\) for \(t\): There are two cases to consider: - Case 1: \(\sin t = 0\) At \(t = 0, \pi\), we get \(\sin t = 0\) - Case 2: \(\cos t = 0\) At \(t = \frac{\pi}{2}, \frac{3\pi}{2}\), we get \(\cos t = 0\) We then have four solutions for \(t\): 0, \(\pi\), \(\frac{\pi}{2}\), and \(\frac{3\pi}{2}\).

Finally, we plug these values of \(t\) back into the original function \(\mathbf{r}(t)\) to find the corresponding points on the ellipse: - For \(t = 0\), \(\mathbf{r}(0) = \langle 2\cos 0, 8\sin 0, 0\rangle = \langle 2, 0, 0 \rangle\) - For \(t = \pi\), \(\mathbf{r}(\pi) = \langle 2\cos \pi, 8\sin \pi, 0\rangle = \langle -2, 0, 0 \rangle\) - For \(t = \frac{\pi}{2}\), \(\mathbf{r}(\frac{\pi}{2}) = \langle 2\cos \frac{\pi}{2}, 8\sin \frac{\pi}{2}, 0\rangle = \langle 0, 8, 0 \rangle\) - For \(t = \frac{3\pi}{2}\), \(\mathbf{r}(\frac{3\pi}{2}) = \langle 2\cos \frac{3\pi}{2}, 8\sin \frac{3\pi}{2}, 0\rangle = \langle 0, -8, 0 \rangle\) Thus, the points on the ellipse where \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal are: \(\langle 2, 0, 0 \rangle\), \(\langle -2, 0, 0 \rangle\), \(\langle 0, 8, 0 \rangle\), and \(\langle 0, -8, 0 \rangle\).