Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises: 44

Answer

$\textbf{r}''(t) = \langle 16e^{4t},32e^{-4t}, 2e^{-t}\rangle$ $\textbf{r}'''(t) = \langle 64e^{4t},-128e^{-4t}, -2e^{-t}\rangle$

Work Step by Step

To find the derivative, simply find the derivative of each component. $\textbf{r}(t) = \langle e^{4t},2e^{-4t}, 2e^{-t}\rangle$ $\textbf{r}'(t) = \langle 4e^{4t},-8e^{-4t}, -2e^{-t}\rangle$ $\textbf{r}''(t) = \langle 16e^{4t},32e^{-4t}, 2e^{-t}\rangle$ $\textbf{r}'''(t) = \langle 64e^{4t},-128e^{-4t}, -2e^{-t}\rangle$
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.