Find the curvature of 
r(t) = 

9t, t2, t3

 at the point 
(9, 1, 1).

The curvature \(\kappa(t)\) of a vector-valued function \(\mathbf{r}(t)\) is given by the following formula:

\[
\kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}
\]

Where:
- \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \langle 9t, t^2, t^3 \rangle\),
- \(\mathbf{r}'(t)\) is the first derivative of \(\mathbf{r}(t)\),
- \(\mathbf{r}''(t)\) is the second derivative of \(\mathbf{r}(t)\).

### Step 1: Find the derivatives
\[
\mathbf{r}'(t) = \frac{d}{dt} \langle 9t, t^2, t^3 \rangle = \langle 9, 2t, 3t^2 \rangle
\]
\[
\mathbf{r}''(t) = \frac{d}{dt} \langle 9, 2t, 3t^2 \rangle = \langle 0, 2, 6t \rangle
\]

### Step 2: Find the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\)
\[
\mathbf{r}'(t) \times \mathbf{r}''(t) = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
9 & 2t & 3t^2 \\
0 & 2 & 6t
\end{vmatrix}
\]
This determinant evaluates as follows:
\[
\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{i}[(2t)(6t) - (3t^2)(2)] - \mathbf{j}[(9)(6t) - (3t^2)(0)] + \mathbf{k}[(9)(2) - (2t)(0)]
\]
Simplifying:
\[
= \mathbf{i}(12t^2 - 6t^2) - \mathbf{j}(54t) + \mathbf{k}(18)
\]
\[
= \mathbf{i}(6t^2) - \mathbf{j}(54t) + \mathbf{k}(18)
\]
Thus, 
\[
\mathbf{r}'(t) \times \mathbf{r}''(t) = \langle 6t^2, -54t, 18 \rangle
\]

### Step 3: Find the magnitude of the cross product
\[
|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{(6t^2)^2 + (-54t)^2 + 18^2}
\]
\[
= \sqrt{36t^4 + 2916t^2 + 324}
\]
This is the magnitude of the cross product.

### Step 4: Find the magnitude of \(\mathbf{r}'(t)\)
\[
|\mathbf{r}'(t)| = \sqrt{9^2 + (2t)^2 + (3t^2)^2} = \sqrt{81 + 4t^2 + 9t^4}
\]

### Step 5: Use the point (9, 1, 1) to find the specific value of \(t\)
From the coordinates \(\mathbf{r}(t) = (9t, t^2, t^3)\), we match:
\[
9t = 9 \quad \Rightarrow \quad t = 1
\]
At \(t = 1\):
\[
\mathbf{r}(1) = (9 \cdot 1, 1^2, 1^3) = (9, 1, 1)
\]
So, \(t = 1\) is the correct value.

### Step 6: Compute the curvature at \(t = 1\)
At \(t = 1\):
- \(\mathbf{r}'(1) = \langle 9, 2, 3 \rangle\),
- \(\mathbf{r}''(1) = \langle 0, 2, 6 \rangle\),
- \(\mathbf{r}'(1) \times \mathbf{r}''(1) = \langle 6, -54, 18 \rangle\).

Now, calculate the magnitudes:
\[
|\mathbf{r}'(1) \times \mathbf{r}''(1)| = \sqrt{6^2 + (-54)^2 + 18^2} = \sqrt{36 + 2916 + 324} = \sqrt{3276}
\]
\[
|\mathbf{r}'(1)| = \sqrt{9^2 + 2^2 + 3^2} = \sqrt{81 + 4 + 9} = \sqrt{94}
\]
Finally, the curvature is:
\[
\kappa(1) = \frac{\sqrt{3276}}{(\sqrt{94})^3} = \frac{\sqrt{3276}}{94^{3/2}}
\]

This gives the curvature at \(t = 1\).

Would you like to go into more detail on any step? Here are some related questions you might find useful:

1. How does the cross product affect the curvature formula?
2. Can curvature be negative? Why or why not?
3. How does the curvature relate to the radius of a circle in space curves?
4. What happens to the curvature when \(\mathbf{r}(t)\) is linear?
5. How can you interpret curvature geometrically?

**Tip:** Curvature is a measure of how sharply a curve bends. The larger the curvature, the tighter the bend.

Find the curvature of r(t) = <9t, t^2, t^3> at the point (9, 1, 1).

Learn how to calculate the curvature of the vector function r(t) = <9t, t^2, t^3> at the point (9, 1, 1). This problem explores key concepts in vector calculus, including derivatives, cross products, and the curvature formula for space curves. Suitable for college-level calculus students.

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