Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.6 Net Change as the Integral of a Rate of Change - Exercises - Page 268: 12

Answer

displacement $0$ meters distance $6$ meters

Work Step by Step

displacement = $\int_0^{3\pi}(cost)dt$ = $sint |_0^{3\pi}$ = $0$ meters distance $cost$ = $0$ $t$ = $\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},...$ = $\int_0^{\frac{\pi}{2}}(cost)dt$ + $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(cost)dt$ + $\int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}}(cost)dt$ + $\int_{\frac{5\pi}{2}}^{3\pi}(cost)dt$ = $sint |_0^{\frac{\pi}{2}}$ + $sint |_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$ + $sint |_{\frac{3\pi}{2}}^{\frac{5\pi}{2}}$ + $sint |_{\frac{5\pi}{2}}^{3\pi}$ = $6$ meters
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.