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: 1


$$\mathbf{T} =\left(-\frac{2}{3} \sin t\right) \mathbf{i}+\left(\frac{2}{3} \cos t\right) \mathbf{j}+\frac{\sqrt{5}}{3} \mathbf{k}$$ $$\text { Length } =3\pi $$

Work Step by Step

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