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 860: 11

Answer

$\boldsymbol{r'}(t)=\langle2t,-2t\sin{t^2},2\sin{t}\cos{t}\rangle$.

Work Step by Step

$\boldsymbol{r}(t)=\langle t^2,\cos{t^2},\sin^2{t}\rangle$ In order to compute $\boldsymbol{r'}(t)$ we simply take the derivative of each component with respect to t of $\boldsymbol{r}(t)$. $\boldsymbol{r'}(t)=\frac{d}{dt}\boldsymbol{r}(t)=\frac{d}{dt}\langle t^2,\cos{t^2},\sin^2{t}\rangle=\langle \frac{d}{dt}t^2,\frac{d}{dt}\cos{t^2},\frac{d}{dt}\sin^2{t}\rangle=\langle2t,-2t\sin{t^2},2\sin{t}\cos{t}\rangle$
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