Answer
$\textbf{r}''(t) = \langle 90t^8, 0, -cos(t) \rangle$
Work Step by Step
$\textbf{r}(t) = \langle t^{10}, 8t, cos(t) \rangle$
$\textbf{r}'(t) = \langle 10t^9, 8, -sin(t) \rangle$
$\textbf{r}''(t) = \langle 90t^8, 0, -cos(t) \rangle$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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 - Page 814: 4
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