7 Find arc length: r(t) = 3 cos ti + 3 sin tj + 5t²k Find K: r(t) =< 6, e-7t, 9te-5t >; t = 0
7 Find arc length: r(t) = 3 cos ti + 3 sin tj + 5t²k Find K: r(t) =< 6, e-7t, 9te-5t >; t = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please help find the arc length of the function provided in the photo below as well as finding K in the second problem
![### Calculus Problem: Arc Length and Curvature
#### Problem 1: Find Arc Length
Given the vector function:
\[ \mathbf{r}(t) = 3 \cos(t) \mathbf{i} + 3 \sin(t) \mathbf{j} + 5t^{\frac{7}{2}} \mathbf{k} \]
#### Problem 2: Find Curvature \(\mathbf{K}\)
Given the vector function:
\[ \mathbf{r}(t) = \langle 6, e^{-7t}, 9te^{-5t} \rangle; \ \ t = 0 \]
### Explanation:
1. **Finding Arc Length:**
To find the arc length \( L \) of a parametrized curve represented by \( \mathbf{r}(t) \) over an interval \([a, b]\), we use the formula:
\[
L = \int_a^b \| \mathbf{r}'(t) \| \, dt
\]
where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \), and \( \| \mathbf{r}'(t) \| \) is the magnitude of \( \mathbf{r}'(t) \).
2. **Finding Curvature \( K \):**
The curvature \( K \) of a vector function \( \mathbf{r}(t) \) at a point \( t = t_0 \) can be found using the formula:
\[
K = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}
\]
Here, \( \mathbf{r}'(t) \) is the first derivative, \( \mathbf{r}''(t) \) is the second derivative, and \( \times \) denotes the cross product. The magnitude \( \| \mathbf{r}'(t) \times \mathbf{r}''(t) \| \) gives the numerator, and the denominator is the cube of the magnitude of the first derivative. You need to evaluate this formula at \( t = 0 \).
These steps outline the fundamental process for determining the arc length and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fc33cc1-e822-44c8-a303-0b5fd2ff9f49%2F36f11c10-f672-4422-971f-ae7037097bb0%2Fmpace7i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Arc Length and Curvature
#### Problem 1: Find Arc Length
Given the vector function:
\[ \mathbf{r}(t) = 3 \cos(t) \mathbf{i} + 3 \sin(t) \mathbf{j} + 5t^{\frac{7}{2}} \mathbf{k} \]
#### Problem 2: Find Curvature \(\mathbf{K}\)
Given the vector function:
\[ \mathbf{r}(t) = \langle 6, e^{-7t}, 9te^{-5t} \rangle; \ \ t = 0 \]
### Explanation:
1. **Finding Arc Length:**
To find the arc length \( L \) of a parametrized curve represented by \( \mathbf{r}(t) \) over an interval \([a, b]\), we use the formula:
\[
L = \int_a^b \| \mathbf{r}'(t) \| \, dt
\]
where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \), and \( \| \mathbf{r}'(t) \| \) is the magnitude of \( \mathbf{r}'(t) \).
2. **Finding Curvature \( K \):**
The curvature \( K \) of a vector function \( \mathbf{r}(t) \) at a point \( t = t_0 \) can be found using the formula:
\[
K = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}
\]
Here, \( \mathbf{r}'(t) \) is the first derivative, \( \mathbf{r}''(t) \) is the second derivative, and \( \times \) denotes the cross product. The magnitude \( \| \mathbf{r}'(t) \times \mathbf{r}''(t) \| \) gives the numerator, and the denominator is the cube of the magnitude of the first derivative. You need to evaluate this formula at \( t = 0 \).
These steps outline the fundamental process for determining the arc length and
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