Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 901: 47


$\dfrac{d}{dt}[u(t) \cdot v(t))]=2t \cos t+2 \sin t -2 \cos t \sin t$

Work Step by Step

As we are given that $u(t)=\lt \sin t,\cos t, t \gt$ and $v(t)=\lt t, \cos t, \sin t \gt$ Need to prove $\dfrac{d}{dt}[u(t) \cdot v(t))]$. $u(t)=\lt \sin t,\cos t, t \gt \implies u'(t)=\lt \cos t,-\sin t, 1 \gt $ and $v(t)=\lt t, \cos t, \sin t \gt \implies v'(t)=\lt 1, -\sin t, \cos t \gt $ Use product formula: $\dfrac{d}{dt}[u(t) \cdot v(t)]=u'(t) \cdot v(t)+u(t) \cdot v'(t)$ We have $\dfrac{d}{dt}[u(t) \cdot v(t)]=\lt \cos t,-\sin t, 1 \gt \cdot \lt t, \cos t, \sin t \gt+\lt \sin t,\cos t, t \gt \cdot \lt 1, -\sin t, \cos t \gt$ or, $=2t \cos t+2 \sin t -2 \cos t \sin t$ Hence, the result. $\dfrac{d}{dt}[u(t) \cdot v(t))]=2t \cos t+2 \sin t -2 \cos t \sin t$
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