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: 41

Answer

(a) $x+y+z=11$ (b) $(x-3)=(y-3)=(z-5)$

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$(3,3,5)$ $(x-3)(4)+(y-3)(4)+(z-5)(4)=0$ $4x-12+4y-12+4z-20=0$ This implies, $x+y+z=11$ (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$(3,3,5)$ $\dfrac{(x-3)}{4}=\dfrac{(y-3)}{4}=\dfrac{(z-5)}{4}$ or, $(x-3)=(y-3)=(z-5)$
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