We are tasked with finding the curvature $\kappa$ of the vector-valued function $\mathbf{r}(t) = \sqrt{5}t \, \hat{i} + e^t \, \hat{j} + e^{-t} \, \hat{k}$ at the point $P(0, 1, 1)$.

Step 1: Find $t$ corresponding to point $P(0, 1, 1)$

Given that: $\mathbf{r}(t) = \sqrt{5}t \, \hat{i} + e^t \, \hat{j} + e^{-t} \, \hat{k}$ At $P(0, 1, 1)$, this means the $i$-, $j$-, and $k$-components must match.

From the components:

• $\sqrt{5}t = 0$ implies $t = 0$,
• At $t = 0$, $e^t = 1$ and $e^{-t} = 1$, matching the point $P(0, 1, 1)$.

Thus, the curvature must be computed at $t = 0$.

Step 2: Compute $\mathbf{r}'(t)$ and $\mathbf{r}''(t)$

First, we calculate the first derivative of $\mathbf{r}(t)$:

= \sqrt{5} \, \hat{i} + e^t \, \hat{j} - e^{-t} \, \hat{k}$$ Next, we compute the second derivative: $$\mathbf{r}''(t) = \frac{d}{dt} \left( \sqrt{5} \, \hat{i} + e^t \, \hat{j} - e^{-t} \, \hat{k} \right) = 0 \, \hat{i} + e^t \, \hat{j} + e^{-t} \, \hat{k}$$ ### Step 3: Use the curvature formula The formula for curvature $$\kappa$$ is given by: $$\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}$$ First, find the magnitude of $$\mathbf{r}'(t)$$ at $$t = 0$$: $$\mathbf{r}'(0) = \sqrt{5} \, \hat{i} + 1 \, \hat{j} - 1 \, \hat{k}$$ $$|\mathbf{r}'(0)| = \sqrt{(\sqrt{5})^2 + 1^2 + (-1)^2} = \sqrt{5 + 1 + 1} = \sqrt{7}$$ Next, compute the cross product $$\mathbf{r}'(0) \times \mathbf{r}''(0)$$: $$\mathbf{r}''(0) = 0 \, \hat{i} + 1 \, \hat{j} + 1 \, \hat{k}$$ $$\mathbf{r}'(0) \times \mathbf{r}''(0) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \sqrt{5} & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} \sqrt{5} & -1 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} \sqrt{5} & 1 \\ 0 & 1 \end{vmatrix}$$ $$= \hat{i} (1 \times 1 - (-1) \times 1) - \hat{j} (\sqrt{5} \times 1 - (-1) \times 0) + \hat{k} (\sqrt{5} \times 1 - 1 \times 0)$$ $$= \hat{i}(2) - \hat{j}(\sqrt{5}) + \hat{k}(\sqrt{5})$$ $$\mathbf{r}'(0) \times \mathbf{r}''(0) = 2 \, \hat{i} - \sqrt{5} \, \hat{j} + \sqrt{5} \, \hat{k}$$ The magnitude of the cross product is: $$|\mathbf{r}'(0) \times \mathbf{r}''(0)| = \sqrt{2^2 + (-\sqrt{5})^2 + (\sqrt{5})^2} = \sqrt{4 + 5 + 5} = \sqrt{14}$$ ### Step 4: Compute the curvature Now, using the curvature formula: $$\kappa = \frac{\sqrt{14}}{(\sqrt{7})^3} = \frac{\sqrt{14}}{7\sqrt{7}} = \frac{\sqrt{2}}{7}$$ Thus, the curvature at the point $$P(0, 1, 1)$$ is: $$\kappa = \frac{\sqrt{2}}{7}$$ Would you like further details or explanations? Here are some related questions: 1. How does curvature change along different curves? 2. What is the geometric meaning of curvature for a 3D curve? 3. Can curvature be used to describe motion in space? 4. How is torsion related to curvature? 5. What role does the Frenet-Serret frame play in curvature analysis? **Tip:** Curvature describes how sharply a curve bends at a given point. The smaller the radius of curvature, the greater the bending.