Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 23


a) $0.935^{\circ}C/ft$ b) $1.87^{\circ}C/sec$

Work Step by Step

a. Here, we have $u=(\dfrac{\sqrt 3}{2})i-(\dfrac{1}{2})j$; $\nabla T= \sin (\sqrt 3) i+ \cos (\sqrt 3) j$ Then, we get $D_uT=(\nabla T \cdot u)= [(\dfrac{\sqrt 3}{2}) \sin \sqrt 3-(\dfrac{1}{2}) \cos \sqrt 3] \approx 0.935^{\circ}C/ft$ b. Here, we have $u=(\dfrac{\sqrt 3}{2})i-(\dfrac{1}{2})j$ Then, we get $\dfrac{dT}{dt}=|v| (D_uT)$\ and $(2) (\dfrac{\sqrt 3}{2} \sin \sqrt 3-\dfrac{1}{2} \cos \sqrt 3)\approx 1.87^{\circ}C/sec$
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.