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: 59


$\dfrac{-5}{4}, \dfrac{-5}{4},\dfrac{25}{8}$

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$(1,1,2)$ $\dfrac{(x-1)}{2}=\dfrac{(y-1)}{2}=\dfrac{(z-2)}{-1}$ Consider $\dfrac{(x-1)}{2}=\dfrac{(y-1)}{2}=\dfrac{(z-2)}{-1}=k$ Therefore, $x=1+3k;y=1+2k,z=2-k$ After simplifications we get $t=-\dfrac{9}{8}$ Plug $t=-\dfrac{9}{8}$ in the equation of a paraboloid : $z=x^2+y^2$ The desired points are: $\dfrac{-5}{4}, \dfrac{-5}{4},\dfrac{25}{8}$
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