Find the curvature of 
r(t) = 

sprt5
ti + et j + e−t k
 at point 
P(0, 1, 1).

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 \( \mathbf{r}(t) \):
\[
\mathbf{r}'(t) = \frac{d}{dt} \left( \sqrt{5}t \, \hat{i} + e^t \, \hat{j} + e^{-t} \, \hat{k} \right)
= \sqrt{5} \, \hat{i} + e^t \, \hat{j} - e^{-t} \, \hat{k}
\]

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.

Find the curvature of r(t) = sqrt(5)t i + e^t j + e^{-t} k at point P(0, 1, 1).

Find the curvature of the 3D parametric curve r(t) = sqrt(5)t i + e^t j + e^{-t} k at point P(0, 1, 1). This detailed solution walks through finding derivatives, calculating the cross product, and using the curvature formula. Suitable for students studying multivariable calculus.

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