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 13.6 - Velocity and Acceleration in Polar Coordinates - Exercises 13.6 - Page 775: 1


$v=3a \sin \theta \ u_r+3a (1-\cos \theta) \ u_{\theta}$ and, $a=9a (2\cos \theta -1)u_r+18a \sin \theta u_{\theta}$

Work Step by Step

The velocity and acceleration in terms of $u_r$ and $u_{\theta}$ can be expressed as: $v=\dfrac{dr}{dt}u_r+r \dfrac{d\theta}{dt}u_{\theta}$ and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$ Now, $v=\dfrac{d(a(1-\cos \theta))}{dt}u_r+3ru_{\theta}$ Now, we will use the chain rule. $v=\dfrac{d(a(1-\cos \theta))}{dt} \dfrac{d\theta}{dt} u_r+3ru_{\theta}$ or, $v=a \sin \theta (3 u_r)+3ru_{\theta}=3a \sin \theta u_r+3ru_{\theta}$ Substitute $r=1-\cos \theta$ So, $v=3a \sin \theta \ u_r+3a (1-\cos \theta) \ u_{\theta}$ and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$ So, $a=9a (2\cos \theta -1)u_r+18a \sin \theta u_{\theta}$
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