University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 12 - Section 12.1 - Curves in Space and Their Tangents - Exercises - Page 650: 30

Answer

$\dfrac{d(u \pm v)}{dt}=\dfrac{d}{dt}u \pm \dfrac{d}{dt}v $

Work Step by Step

Consider, $u=\lt u_x(t), u_y(t), u_z(t) \gt$ and $v=\lt v_x(t), v_y(t), v_z(t) \gt$ Apply product rule to get $\dfrac{d(u \pm v)}{dt}=\lt \dfrac{d}{dt}(u_xt \pm v_xt), \dfrac{d}{dt}(u_yt \pm v_yt),\dfrac{d}{dt}(u_zt \pm v_zt)$ or, $=\lt \dfrac{d}{dt}u_xt , \dfrac{d}{dt}u_y t,\dfrac{d}{dt}u_z t\gt \pm \lt \dfrac{d}{dt}v_xt , \dfrac{d}{dt}v_y t,\dfrac{d}{dt}v_z t\gt $ or, $= \dfrac{d}{dt}u \pm \dfrac{d}{dt}v $ Hence, $\dfrac{d(u \pm v)}{dt}=\dfrac{d}{dt}u \pm \dfrac{d}{dt}v $
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