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: 43

Answer

$\textbf{r}''(t) = \langle -9cos(3t), -16sin(4t), -36cos(6t)\rangle$ $\textbf{r}'''(t) = \langle 27sin(3t), -64cos(4t), 216sin(6t)\rangle$

Work Step by Step

To find the derivative, simply find the derivative of each component. $\textbf{r}(t) = \langle cos(3t), sin(4t), cos(6t)\rangle$ $\textbf{r}'(t) = \langle -3sin(3t), 4cos(4t), -6sin(6t)\rangle$ $\textbf{r}''(t) = \langle -9cos(3t), -16sin(4t), -36cos(6t)\rangle$ $\textbf{r}'''(t) = \langle 27sin(3t), -64cos(4t), 216sin(6t)\rangle$
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.