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: 63



Work Step by Step

At points (-1,1,2): $\nabla P \times \nabla Q=\lt -2,2,-1 \gt \times \lt -8,2,4 \gt=\lt 10, 16, 12 \gt$ Formula to calculate tangent line equation is: $\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ This implies $\dfrac{(x+1)}{5}=\dfrac{(y-1}{8}=\dfrac{(z-2)}{6}$ Hence, the parametric line of equations are: $x=-1-10t,y=1-16t,z=2-12t$
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