Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.6 - Directional Derivatives and the Gradient Vector - 14.6 Exercise - Page 959: 58

Answer

Every normal line to the sphere passes through the center of the sphere.

Work Step by Step

Formula to calculate normal line equation is: $\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ At point$(p,q,r)$ $\dfrac{(x-p)}{2p}=\dfrac{(y-q)}{2q}=\dfrac{(z-r)}{2r}$ This implies: $\dfrac{x}{p}=\dfrac{y}{q}=\dfrac{z}{r}$ Therefore, $x=y=z=0$ This means that: Every normal line to the sphere passes through the center of the sphere.
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