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: 57

Answer

Every tangent plane to the cone passes through the origin $(0,0,0)$.

Work Step by Step

Formula to calculate tangent plane equation for an sphere 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$(x_0,y_0,z_0)$ $(x-x_0)(2x_0)+(y-y_0)(2y_0)-(z-z_0)(2z_0)=0$ This implies, $xx_0+yy_0-zz_0=x_0^2+y_0^2-z_0^2$ When the tangent plane passes through the origin $(0,0,0)$: $xx_0+yy_0-zz_0=0$ Thus we have proved that every tangent plane to the cone passes through the origin $(0,0,0)$.
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