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: 14

Answer

$\boldsymbol{r'}(t)=\langle2a\sin{at}\cos{at},e^{bt}+bte^{bt},-2c\cos{ct}\sin{ct}\rangle$.

Work Step by Step

$\boldsymbol{r}(t)=\langle \sin^2{at},te^{bt},\cos^2{ct}\rangle$ In order to compute $\boldsymbol{r'}(t)$ we simply take the derivative of each component with respect to t of $\boldsymbol{r}(t)$. $\boldsymbol{r'}(t)=\frac{d}{dt}\boldsymbol{r}(t)=\frac{d}{dt}\langle \sin^2{at},te^{bt},\cos^2{ct}\rangle=\langle \frac{d}{dt}\sin^2{at},\frac{d}{dt}te^{bt},\frac{d}{dt}\cos^2{ct}\rangle=\langle2a\sin{at}\cos{at},e^{bt}+bte^{bt},-2c\cos{ct}\sin{ct}\rangle$
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