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: 12


$$\frac{3 \pi^{2}}{8}$$

Work Step by Step

Since $$\mathbf{r}=(\cos t+t \sin t) \mathbf{i}+(\sin t-t \cos t) \mathbf{j}$$Then \begin{align*} \mathbf{v}&=(-\sin t+\sin t+t \cos t) \mathbf{i}+(\cos t-\cos t+t \sin t) \mathbf{j}\\ &=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j} \\ |\mathbf{v}|&=\sqrt{(t \cos t)^{2}+(t \cos t)^{2}}\\ &=\sqrt{t^{2}}=t,\end{align*} Hence \begin{align*} \Rightarrow s(t)&=\int_{0}^{t} \tau d \tau\\ &=\frac{t^{2}}{2}\\ \text{Length }&=s(\pi)-s\left(\frac{\pi}{2}\right)\\ &=\frac{\pi^{2}}{2}-\frac{\left(\frac{\pi}{2}\right)^{2}}{2}\\ &=\frac{3 \pi^{2}}{8} \end{align*}
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