1. **Problem statement:** Given the vector-valued function $$\mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + 2t \mathbf{k},$$ find (i) $$\mathbf{r}'(t) \cdot \mathbf{r}''(t)$$ and (ii) $$\mathbf{r}'(t) \times \mathbf{r}''(t)$$. 2. **Step 1: Find the first derivative $$\mathbf{r}'(t)$$.** $$\mathbf{r}'(t) = \frac{d}{dt}(\cos t) \mathbf{i} + \frac{d}{dt}(\sin t) \mathbf{j} + \frac{d}{dt}(2t) \mathbf{k} = -\sin t \mathbf{i} + \cos t \mathbf{j} + 2 \mathbf{k}$$ 3. **Step 2: Find the second derivative $$\mathbf{r}''(t)$$.** $$\mathbf{r}''(t) = \frac{d}{dt}(-\sin t) \mathbf{i} + \frac{d}{dt}(\cos t) \mathbf{j} + \frac{d}{dt}(2) \mathbf{k} = -\cos t \mathbf{i} - \sin t \mathbf{j} + 0 \mathbf{k} = -\cos t \mathbf{i} - \sin t \mathbf{j}$$ 4. **Step 3: Compute the dot product $$\mathbf{r}'(t) \cdot \mathbf{r}''(t)$$.** $$\mathbf{r}'(t) \cdot \mathbf{r}''(t) = (-\sin t)(-\cos t) + (\cos t)(-\sin t) + (2)(0) = \sin t \cos t - \sin t \cos t + 0 = 0$$ 5. **Step 4: Compute the cross product $$\mathbf{r}'(t) \times \mathbf{r}''(t)$$.** Using the determinant formula: $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin t & \cos t & 2 \\ -\cos t & -\sin t & 0 \end{vmatrix}$$ Calculate each component: - $$\mathbf{i}: (\cos t)(0) - (2)(-\sin t) = 0 + 2 \sin t = 2 \sin t$$ - $$\mathbf{j}: -\left((-\sin t)(0) - (2)(-\cos t)\right) = -\left(0 + 2 \cos t\right) = -2 \cos t$$ - $$\mathbf{k}: (-\sin t)(-\sin t) - (\cos t)(-\cos t) = \sin^2 t + \cos^2 t = 1$$ So, $$\mathbf{r}'(t) \times \mathbf{r}''(t) = 2 \sin t \mathbf{i} - 2 \cos t \mathbf{j} + 1 \mathbf{k}$$ **Final answers:** (i) $$\mathbf{r}'(t) \cdot \mathbf{r}''(t) = 0$$ (ii) $$\mathbf{r}'(t) \times \mathbf{r}''(t) = 2 \sin t \mathbf{i} - 2 \cos t \mathbf{j} + \mathbf{k}$$