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 $\mathbf{r}'(t) \times \mathbf{r}''(t)$

\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.