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: 12



Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.