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: 5

Answer

\begin{align*} \mathbf{T}&= (-\cos t) \mathbf{j}+(\sin t) \mathbf{k}\\ \text{Length }&= \frac{3}{2} \end{align*}

Work Step by Step

Since $\mathbf{r}=\left(\cos ^{3} t\right) \mathbf{j}+\left(\sin ^{3} t\right) \mathbf{k} $ Then $ \mathbf{v}=\left(-3 \cos ^{2} t \sin t\right) \mathbf{j}+\left(3 \sin ^{2} t \cos t\right) \mathbf{k}$ and \begin{align*} |v|&=\sqrt{\left(-3 \cos ^{2} t \sin t\right)^{2}+\left(3 \sin ^{2} t \cos t\right)^{2}}\\ &=\sqrt{\left(9 \cos ^{2} t \sin ^{2} t\right)\left(\cos ^{2} t+\sin ^{2} t\right)}\\ &=3|\cos t \sin t| \end{align*} Hence \begin{align*} \mathbf{T}&=\frac{\mathrm{v}}{|\mathrm{v}|}\\ &=\frac{-3 \cos ^{2} t \sin t}{3|\cos t \sin t|} \mathbf{j}+\frac{3 \sin ^{2} t \cos t}{3|\cos t \sin t|} \mathbf{k}\\ &=(-\cos t) \mathbf{j}+(\sin t) \mathbf{k}\\ \text{Length }&=\int_{0}^{\pi / 2} 3|\cos t \sin t| d t\\ &=\int_{0}^{\pi / 2} 3 \cos t \sin t d t\\ &=\int_{0}^{\pi / 2} \frac{3}{2} \sin 2 t d t\\ &=\left[-\frac{3}{4} \cos 2 t\right]_{0}^{\pi / 2}\\ &=\frac{3}{2} \end{align*}
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