Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 999: 67


(a) $F_xG_x+F_yG_y+F_zG_z=0$ at $P$ (b) The two surfaces are orthogonal and their dot product is equal to zero.

Work Step by Step

a) Condition to be orthogonal: Two normal lines are known to be orthogonal at point P when $\nabla F \cdot \nabla G=0$ at point $P$ or, we can also say that their dot product must be zero. $\lt \dfrac{\partial F}{\partial x},\dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z} \gt \cdot \lt \dfrac{\partial G}{\partial x},\dfrac{\partial G}{\partial y},\dfrac{\partial G}{\partial z} \gt =0$ Thus, $\lt \dfrac{\partial F}{\partial x}\dfrac{\partial G}{\partial x},\dfrac{\partial F}{\partial y}\dfrac{\partial G}{\partial y},\dfrac{\partial F}{\partial z} \dfrac{\partial G}{\partial z} \gt =0$ Hence, $F_xG_x+F_yG_y+F_zG_z=0$ at $P$ b) From the given data in the question, we have $\nabla P(x,y,z) \cdot \nabla Q(x,y,z)=\lt 2x,2y,-2z \gt \times \lt 2x,2y,2z \gt=4(x^2+y^2-z^2)$ or, $\nabla P(x,y,z) \cdot \nabla Q(x,y,z)=0$ Hence, we can see that the two surfaces are orthogonal and their dot product is zero.
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.