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: 2


\begin{align*} \mathbf{T} =\left(\frac{12}{13} \cos 2 t\right) \mathbf{i}-\left(\frac{12}{13} \sin 2 t\right) \mathbf{j}+\frac{5}{13} \mathbf{k}\\ \text{ Length } =13 \pi \end{align*}

Work Step by Step

Since $\mathbf{r}=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k} $ Then $\mathbf{v}=(12 \cos 2 t) \mathbf{i}+(-12 \sin 2 t) \mathbf{j}+5 \mathbf{k}$ $ |\mathbf{v}|=\sqrt{(12 \cos 2 t)^{2}+(-12 \sin 2 t)^{2}+5^{2}}=13$ Hence \begin{align*} \mathbf{T}&=\frac{\mathbf{v}}{|\mathbf{v}|}\\ &=\left(\frac{12}{13} \cos 2 t\right) \mathbf{i}-\left(\frac{12}{13} \sin 2 t\right) \mathbf{j}+\frac{5}{13} \mathbf{k}\\ \text{ Length }&=\int_{0}^{\pi}|\mathbf{v}| d t\\ &=\int_{0}^{\pi} 13 d t=[13 t]_{0}^{\pi}\\ &=13 \pi \end{align*}
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