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: 50


$r'(2)=12i-29j+14k$ or, $r'(2)=\lt 12,-29,14 \gt$

Work Step by Step

As we are given that $r(t)=u(t) \times v(t), u(2)=\lt 1,2,-1 \gt,u'(2)=\lt 3,0,4 \gt$ and $v(t)=\lt t, t^2,t^3 \gt$ Use cross product rule. $r'(t)=u'(t) \times v(t)+u(t) \times v'(t)$ $v(t)=\lt t, t^2,t^3 \gt \implies v'(t)=\lt 1, 2t,3t^2 \gt$ and $v'(2)=\lt 1,4,12 \gt$ This yields, $r'(2)=\lt 3,0,4 \gt \times \lt 2,4,8 \gt+\lt 1,2,-1 \gt \times \lt 1,4,12\gt$ or,$=\begin{vmatrix}1&j&k \\3&0&4\\2&4&8 \end{vmatrix}+\begin{vmatrix}1&j&k \\1&2&-1\\1&4&12 \end{vmatrix}$ or, $=(-16i-16j+12k)+(28i-13j+2k)$ or, $=(-16+28)i+(-16-13)j+(12+2)k$ Hence, the result. $r'(2)=12i-29j+14k$ or,$r'(2)=\lt 12,-29,14 \gt$
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