Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.6 Exercises - Page 969: 44

Answer

(a) $3x+2y+3z=10$ (b) $\dfrac{(x-1)}{3}=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{3}$

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$(1,2,1)$ $(x-1)(3)+(y-2)(2)+(z-1)(3)=0$ $3x-3+2y-2+3z-3=0$ $5(3x-3+2y-2+3z-3)=0$ This implies, $3x+2y+3z=10$ (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$(1,2,1)$ $\dfrac{(x-1)}{3}=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{3}$
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