Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.3 Arc Length in Space - Exercises 13.3 - Page 759: 13


$$\frac{3 \sqrt{3}}{4}$$

Work Step by Step

Since $$\mathbf{r}=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+e^{t} \mathbf{k}$$Then \begin{align*} \mathbf{v}&=\left(e^{t} \cos t-e^{t} \sin t\right) \mathbf{i}+\left(e^{t} \sin t+e^{t} \cos t\right) \mathbf{j}+e^{t} \mathbf{k}\\ \Rightarrow|v|&=\sqrt{\left(e^{t} \cos t-e^{t} \sin t\right)^{2}+\left(e^{t} \sin t+e^{t} \cos t\right)^{2}+\left(e^{t}\right)^{2}}\\ &=\sqrt{3 e^{2 t}}\\ &=\sqrt{3} e^{t} \end{align*} Hence \begin{align*} s(t)&=\int_{0}^{t} \sqrt{3} e^{t} d \tau\\ &=\sqrt{3} e^{t}-\sqrt{3}\\ \text{ Length }&=s(0)-s(-\ln 4)\\ &=0-\left(\sqrt{3} e^{-\ln 4}-\sqrt{3}\right)\\ &=\frac{3 \sqrt{3}}{4}\end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.