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: 10


$$P(-\pi)=(12 \sin (-\pi),-12 \cos (-\pi),-5 \pi)=(0,12,-5 \pi)$$

Work Step by Step

Let $P\left(t_{0}\right)$ denote the point. Then $$\mathbf{v}=(12 \cos t) \mathbf{i}+(12 \sin t) \mathbf{j}+5 \mathbf{k}$$ and \begin{align*} -13 \pi&=\int_{0}^{t_{0}} \sqrt{144 \cos ^{2} t+144 \sin ^{2} t+25} d t\\ &=\int_{0}^{t_{0}} 13 d t\\ &=13 t_{0} \end{align*} Then $ t_{0}=-\pi,$ and the point is $$P(-\pi)=(12 \sin (-\pi),-12 \cos (-\pi),-5 \pi)=(0,12,-5 \pi)$$
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