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: 11


$$\frac{5 \pi}{2}$$

Work Step by Step

Since $$\mathbf{r}=(4 \cos t) \mathbf{i}+(4 \sin t) \mathbf{j}+3 t \mathbf{k} $$ Then \begin{align*} \mathbf{v}&=(-4 \sin t) \mathbf{i}+(4 \cos t) \mathbf{j}+3 \mathbf{k} \\ |\mathbf{v}|&=\sqrt{(-4 \sin t)^{2}+(4 \cos t)^{2}+3^{2}}\\ &=\sqrt{25}=5 \end{align*} Hence \begin{align*} s(t)&=\int_{0}^{t} 5 d \tau\\ &=5 t \\ \text{ Length }&=s\left(\frac{\pi}{2}\right)\\ &=\frac{5 \pi}{2}\end{align*}
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