Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.5 Motion in 3-Space - Exercises - Page 744: 5


$$\left\langle\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\right\rangle$$ $$\left\langle-\frac{\sqrt{3}}{2},-\frac{1}{2}, 9\right\rangle$$

Work Step by Step

Given $$\mathbf{r}(\theta)=\langle\sin \theta, \cos \theta, \cos 3 \theta\rangle, \quad \theta=\frac{\pi}{3}$$ Velocity is given by \begin{aligned} \mathbf{v}(\theta) &=\mathbf{r}^{\prime}(\theta)\\ &=\langle\cos \theta,-\sin \theta,- 3 \sin 3 \theta\rangle \\ \mathbf{v}\left(\frac{\pi}{3}\right) &=\left\langle\cos \frac{\pi}{3},-\sin \frac{\pi}{3},-3 \sin \pi\right\rangle\\ &=\left\langle\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\right\rangle \end{aligned} and acceleration is given by \begin{aligned} \mathbf{a}(\theta) &=\mathbf{v}^{ \prime}(\theta)\\ &=\langle-\sin \theta,-\cos \theta,- 9 \cos 3 \theta\rangle \\ \mathbf{a}\left(\frac{\pi}{3}\right)&=\left\langle-\sin \frac{\pi}{3},-\cos \frac{\pi}{3},-9 \cos \pi\right\rangle\\ &=\left\langle-\frac{\sqrt{3}}{2},-\frac{1}{2}, 9\right\rangle \end{aligned}
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