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: 9

Answer

$\boldsymbol{r'}(t)=\langle\frac{1}{2\sqrt{t-2}},0,\frac{-2}{t^3}\rangle$.

Work Step by Step

$\boldsymbol{r}(t)=\langle \sqrt{t-2},3,\frac{1}{t^2}\rangle$ In order to compute $\boldsymbol{r'}(t)$ we simply take the derivative of each component with respect to t of $\boldsymbol{r}(t)$. $\boldsymbol{r'}(t)=\frac{d}{dt}\boldsymbol{r}(t)=\frac{d}{dt}\langle \sqrt{t-2},3,\frac{1}{t^2}\rangle=\langle \frac{d}{dt}\sqrt{t-2},\frac{d}{dt}3,\frac{d}{dt}\frac{1}{t^2}\rangle=\langle\frac{1}{2\sqrt{t-2}},0,\frac{-2}{t^3}\rangle$
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