Find the length of the portion of the arc given by r(t) = (e 'cos t) i + (e 'sin t) j + 5e ' k, -In 2 sts0 O 33 4 O 33 2.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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### Calculus - Arc Length of a Vector Function

#### Problem Statement:
Find the length of the portion of the arc given by \( \mathbf{r}(t) = (e^{t} \cos t) \mathbf{i} + (e^{t} \sin t) \mathbf{j} + 5e^{t} \mathbf{k} \), for \( \ln 2 \le t \le 0 \).

#### Multiple Choice Options:
- \( \frac{3 \sqrt{3}}{4} \)
- \( \frac{3 \sqrt{3}}{2} \)
- \( \frac{\sqrt{23}}{4} \)
- \( \frac{\sqrt{23}}{2} \) 

#### Solution Approach:
To find the length of the arc \( \mathbf{r}(t) \) over the given interval, we will need to use the arc length formula for a vector function:
\[ L = \int_{a}^{b} \| \mathbf{r'}(t) \| \, dt \]

Where:
\[ \mathbf{r'}(t) = \frac{d}{dt} \mathbf{r}(t) \]

Calculate the derivative \( \mathbf{r'}(t) \), find its magnitude, and then integrate over the given interval \([ \ln 2, 0 ]\).

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Transcribed Image Text:--- ### Calculus - Arc Length of a Vector Function #### Problem Statement: Find the length of the portion of the arc given by \( \mathbf{r}(t) = (e^{t} \cos t) \mathbf{i} + (e^{t} \sin t) \mathbf{j} + 5e^{t} \mathbf{k} \), for \( \ln 2 \le t \le 0 \). #### Multiple Choice Options: - \( \frac{3 \sqrt{3}}{4} \) - \( \frac{3 \sqrt{3}}{2} \) - \( \frac{\sqrt{23}}{4} \) - \( \frac{\sqrt{23}}{2} \) #### Solution Approach: To find the length of the arc \( \mathbf{r}(t) \) over the given interval, we will need to use the arc length formula for a vector function: \[ L = \int_{a}^{b} \| \mathbf{r'}(t) \| \, dt \] Where: \[ \mathbf{r'}(t) = \frac{d}{dt} \mathbf{r}(t) \] Calculate the derivative \( \mathbf{r'}(t) \), find its magnitude, and then integrate over the given interval \([ \ln 2, 0 ]\). ---
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