Find the curvature of the curve rt=<-5t,-3t^4,-5t^5>
at the point t=-1

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

\[
\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-5 & -12t^3 & -25t^4 \\
0 & -36t^2 & -100t^3
\end{vmatrix}
\]

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.

Learn how to calculate the curvature of the curve given by the vector function rt=<-5t,-3t^4,-5t^5> at t=-1. Follow a step-by-step solution involving vector calculus and the curvature formula. Suitable for advanced undergraduate students studying vector calculus and curve properties.

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