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 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 27


$|r|$ is also constant.

Work Step by Step

As we are given that $r \dfrac{dr}{dt}=0 \\ \implies (2r) \dfrac{dr}{dt}=0 \\ \implies r \dfrac{dr}{dt}+r \dfrac{dr}{dt}=0$ Thus, we have $\dfrac{d(r\cdot r)}{dt}=0\\ \implies \dfrac{d(|r^2|)}{dt}=0$ This implies that $|r^2|$ is constant . Hence, $|r|$ is also constant.
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