Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.2 - Derivatives and Integrals of Vector Functions - 13.2 Exercise - Page 861: 43



Work Step by Step

Our aim is to proove: $\dfrac{d}{dt}[u(t)+v(t)]=u'(t)+v'(t)$ ...(1) Suppose $u(t)=u_1(t)i+u_2(t)j+u_3(t)k$ and $v(t)=v_1(t)i+v_2(t)j+v_3(t)k$ Take the left side of the equation (1). $\dfrac{d}{dt}[u(t)+v(t)]=\dfrac{d}{dt}[(u_1(t)i+u_2(t)j+u_3(t)k)+(v_1(t)i+v_2(t)j+v_3(t)k)]$ $=\dfrac{d}{dt}[(u_1(t)+v_1(t))i+(u_2(t)+v_2(t))j+(u_3(t)+v_3(t))k$ $=(u_1'(t)+v_1'(t))i+(u_2'(t)+v_2'(t))j+(u_3'(t)+v_3'(t))k$ $=(u_1'(t)i+u_2'(t)j+u_3'(t)k)+(v_1'(t)i+v_2'(t)j+v_3'(t)k)$ $=u'(t)+v'(t)$ Hence, $\dfrac{d}{dt}[u(t)+v(t)]=u'(t)+v'(t)$
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