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: 24


a) $\dfrac{736}{\sqrt {89}}^{\circ}C/m$ b) $736^{\circ}C/sec$

Work Step by Step

a. At the point $(8,6,-4)$ we have $D_uT=(\nabla T \cdot u) =[56 i+32 j-48 k \cdot \dfrac{8}{\sqrt 89}i+\dfrac{3}{\sqrt 89}j-\dfrac{4}{\sqrt 89}k]$ and $ \dfrac{8}{\sqrt {89}} \cdot (56)+(32) \cdot (3)-(48) \cdot (-4)=\dfrac{736}{\sqrt {89}}736^{\circ}C/m$ b. Now, $\dfrac{dT}{dt}=|v| (D_uT) =(\dfrac{8}{\sqrt 89}i+\dfrac{3}{\sqrt 89}j-\dfrac{4}{\sqrt 89}k) \cdot (\dfrac{736}{\sqrt {89}})$ and $ (\sqrt {89}) (\dfrac{736}{\sqrt {89}})=736^{\circ}C/sec$
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