University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 216: 86

Answer

$2\sqrt{2}$ amperes.

Work Step by Step

We find critical points. $i'(t)=-2\sin t+2\cos t$ $ i'(t)=0\quad$ for $2\sin t=2\cos t$ $\tan t=1$ $t=\displaystyle \frac{\pi}{4}+k\pi,$ where k is a nonnegative integer. Evaluating $i(t)$ at $t=\displaystyle \frac{\pi}{4},$ $i(\displaystyle \frac{\pi}{4})=2\cos(\frac{\pi}{4})+2\sin\frac{\pi}{4}=2(\frac{\sqrt{2}}{2})+2(\frac{\sqrt{2}}{2})=2\sqrt{2}$ Evaluating $i(t)$ at $t=\displaystyle \frac{5\pi}{4},$ $i(\displaystyle \frac{\pi}{4})=2\cos(\frac{5\pi}{4})+2\sin\frac{5\pi}{4}=2(-\frac{\sqrt{2}}{2})+2(-\frac{\sqrt{2}}{2})=-2\sqrt{2}$ The greatest magnitude (absolute value) is $2\sqrt{2}$ amperes.
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.