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 958: 46

Answer

(a) $x+y+z=3$ (b) $x=y=z$

Work Step by Step

(a) Formula to calculate tangent plane equation is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ At point$(1,1,1)$ $(x-1)(-2)+(y-1)(-2)+(z-1)(-2)=0$ This implies, $x+y+z=3$ (b) Formula to 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$(1,1,1)$ $\dfrac{(x-1)}{-2}=\dfrac{(y-1)}{-2}=\dfrac{(z-1)}{-2}$ Hence, $x=y=z$
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