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We are asked to find the curvature of the vector function \( \mathbf{r}(t) = \langle 5t, t^2, t^3 \rangle \) at the point \( (5, 1, 1) \).

### Steps to Solve:

1. **Parameter Identification**:  
   We first need to determine the value of \( t \) that corresponds to the point \( (5, 1, 1) \) on the curve:
   - From \( x(t) = 5t \), setting \( x = 5 \) gives \( t = 1 \).
   
   Therefore, we will calculate the curvature at \( t = 1 \).

2. **Curvature Formula**:  
   The curvature \( \kappa(t) \) of a curve given by a vector function \( \mathbf{r}(t) \) is:
   \[
   \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}
   \]
   where \( \mathbf{r}'(t) \) is the first derivative of the position vector, and \( \mathbf{r}''(t) \) is the second derivative.

3. **First Derivative \( \mathbf{r}'(t) \)**:
   \[
   \mathbf{r}'(t) = \langle 5, 2t, 3t^2 \rangle
   \]
   At \( t = 1 \), 
   \[
   \mathbf{r}'(1) = \langle 5, 2, 3 \rangle
   \]

4. **Second Derivative \( \mathbf{r}''(t) \)**:
   \[
   \mathbf{r}''(t) = \langle 0, 2, 6t \rangle
   \]
   At \( t = 1 \), 
   \[
   \mathbf{r}''(1) = \langle 0, 2, 6 \rangle
   \]

5. **Cross Product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \)**:
   We calculate the cross product \( \mathbf{r}'(1) \times \mathbf{r}''(1) \):
   \[
   \mathbf{r}'(1) \times \mathbf{r}''(1) = \langle 5, 2, 3 \rangle \times \langle 0, 2, 6 \rangle
   \]
   Using the determinant method for cross products:
   \[
   \mathbf{r}'(1) \times \mathbf{r}''(1) = \langle (2 \cdot 6 - 3 \cdot 2), -(5 \cdot 6 - 3 \cdot 0), 5 \cdot 2 - 2 \cdot 0 \rangle = \langle 12 - 6, -(30), 10 \rangle = \langle 6, -30, 10 \rangle
   \]

6. **Magnitude of the Cross Product**:
   \[
   |\mathbf{r}'(1) \times \mathbf{r}''(1)| = \sqrt{6^2 + (-30)^2 + 10^2} = \sqrt{36 + 900 + 100} = \sqrt{1036}
   \]

7. **Magnitude of \( \mathbf{r}'(t) \)**:
   \[
   |\mathbf{r}'(1)| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38}
   \]

8. **Curvature Calculation**:
   Finally, using the curvature formula:
   \[
   \kappa(1) = \frac{\sqrt{1036}}{(\sqrt{38})^3} = \frac{\sqrt{1036}}{38\sqrt{38}}
   \]

Now, let's compute the final value of the curvature.The curvature of the vector function \( \mathbf{r}(t) = \langle 5t, t^2, t^3 \rangle \) at the point \( (5, 1, 1) \) is approximately \( 0.1374 \).

Would you like further details or have any questions?

Here are 5 related questions to explore further:
1. How does the curvature change along different points of the curve?
2. What is the geometric significance of curvature in space curves?
3. How would the curvature formula differ for planar curves?
4. What role does the cross product play in determining the curvature?
5. Can you generalize the curvature formula to any vector function in \( \mathbb{R}^3 \)?

**Tip**: The magnitude of the curvature indicates how sharply the curve bends at a given point, with higher values representing sharper turns.

Find the curvature of r(t) = ⟨5t, t², t³⟩ at the point (5, 1, 1).

This solution explains how to calculate the curvature of the vector function r(t) = ⟨5t, t², t³⟩ at the point (5, 1, 1). It covers key vector calculus concepts, including the cross product, derivatives, and the curvature formula for space curves.

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