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 760: 21



Work Step by Step

Since \begin{align*} \mathbf{v}&=\frac{d}{d t}\left(x_{0}+t u_{1}\right) \mathbf{i}+\frac{d}{d t}\left(y_{0}+t u_{2}\right) \mathbf{j}+\frac{d}{d t}\left(z_{0}+t u_{3}\right) \mathbf{k}\\ &=u_{1} \mathbf{i}+u_{2} \mathbf{j}+u_{3} \mathbf{k}=\mathbf{u},\end{align*} Then \begin{align*} s(t)&=\int_{0}^{t}|\mathbf{v}| d t\\ &=\int_{0}^{t}|\mathbf{u}| d \tau\\ &=\int_{0}^{t} 1 d \tau\\ &=t \end{align*}
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