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: 3


$v=2a e^{a \theta}u_r+2e^{a \theta}u_{\theta}$ and $a=4e^{a \theta}(a^2 -1)u_r+8a e^{a \theta} u_{\theta}$

Work Step by Step

The velocity and acceleration in terms of $u_r$ and $u_{\theta}$ can be computed as: $v=r_t u_r+r \theta_t u_{\theta}$ and $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$ Now, $\dfrac{d \theta}{dt}=\theta^{.}=2$ and $\theta^{..}=0$ Thus, $v=2a e^{a \theta}u_r+2e^{a \theta}u_{\theta}$ Next, $a=(\dfrac{d^2r}{dt^2}-r\theta^{.2})u_r+(r\theta^{..} +2r^{.}\theta^{.})u_{\theta}$ So, $a=4e^{a \theta}(a^2 -1)u_r+8a e^{a \theta} u_{\theta}$
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