To find the curvature $\kappa$ of a curve given by a vector function $\mathbf{r}(t)$, we use the formula:

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

Given $\mathbf{r}(t) = \langle -5t, -3t^4, -5t^5 \rangle$, we need to compute the first and second derivatives of $\mathbf{r}(t)$, and then use the curvature formula.

Step 1: Compute $\mathbf{r}'(t)$

$\mathbf{r}'(t) = \frac{d}{dt} \langle -5t, -3t^4, -5t^5 \rangle = \langle -5, -12t^3, -25t^4 \rangle$

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

$\mathbf{r}''(t) = \frac{d}{dt} \langle -5, -12t^3, -25t^4 \rangle = \langle 0, -36t^2, -100t^3 \rangle$

Step 3: Compute the cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & -12t^3 & -25t^4 \\ 0 & -36t^2 & -100t^3 \end{vmatrix}$$ Expanding the determinant: $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{i} \left((-12t^3)(-100t^3) - (-25t^4)(-36t^2)\right) - \mathbf{j} \left((-5)(-100t^3) - (0)(-25t^4)\right) + \mathbf{k} \left((-5)(-36t^2) - (0)(-12t^3)\right)$$ $$= \mathbf{i} \left(1200t^6 - 900t^6\right) - \mathbf{j} \left(500t^3\right) + \mathbf{k} \left(180t^2\right)$$ $$= \mathbf{i} \left(300t^6\right) - \mathbf{j} \left(500t^3\right) + \mathbf{k} \left(180t^2\right)$$ $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \langle 300t^6, -500t^3, 180t^2 \rangle$$ ### Step 4: Compute the magnitude $$|\mathbf{r}'(t) \times \mathbf{r}''(t)|$$ $$|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{(300t^6)^2 + (-500t^3)^2 + (180t^2)^2}$$ $$= \sqrt{90000t^{12} + 250000t^6 + 32400t^4}$$ ### Step 5: Compute the magnitude $$|\mathbf{r}'(t)|$$ $$|\mathbf{r}'(t)| = \sqrt{(-5)^2 + (-12t^3)^2 + (-25t^4)^2}$$ $$= \sqrt{25 + 144t^6 + 625t^8}$$ ### Step 6: Plug in $$t = -1$$ and compute $$\kappa(-1)$$ Let's calculate these expressions at $$t = -1$$. - For $$|\mathbf{r}'(-1)|$$: $$|\mathbf{r}'(-1)| = \sqrt{25 + 144(-1)^6 + 625(-1)^8} = \sqrt{25 + 144 + 625} = \sqrt{794}$$ - For $$|\mathbf{r}'(-1) \times \mathbf{r}''(-1)|$$: $$|\mathbf{r}'(-1) \times \mathbf{r}''(-1)| = \sqrt{90000(-1)^{12} + 250000(-1)^6 + 32400(-1)^4} = \sqrt{90000 + 250000 + 32400} = \sqrt{372400}$$ - $$\sqrt{372400} = 610$$ ### Step 7: Calculate the curvature $$\kappa(-1)$$ $$\kappa(-1) = \frac{610}{(794)^{3/2}}$$ Let me simplify the final expression.The curvature of the curve at the point where $$t = -1$$ is approximately $$\kappa(-1) \approx 0.0273$$. Would you like to know more details about the steps, or do you have any questions? Here are some related questions you might find interesting: 1. How does the curvature of a curve change with respect to the parameter $$t$$? 2. What is the geometric interpretation of curvature? 3. How is torsion related to the curvature of a space curve? 4. Can curvature be negative? If so, what does it signify? 5. How would you find the curvature of a plane curve rather than a space curve? **Tip:** The curvature of a curve gives a measure of how sharply it bends at a given point, with larger values indicating a tighter bend.