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: 62

Answer

The volume of the pyramid cut off from the first octant by any tangent plane to the surface $xyz=1$ at points in the first octant is the same.

Work Step by Step

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$(a,b,c)$ $(x-a)bc)+(y-b)ac+(z-c)ab=0$ $3abc=bcx+acy+abz$ We need to find the x-,y-,x- intercepts. x-intercept: 3a; y-intercept: 3b and z-intercept: 3c Volume of the pyramid is given as:$V=\dfrac{1}{3}(\dfrac{9ab}{2})(3c)=\dfrac{9abc}{2}$ Since, $xyz=1 \implies abc=1$ Thus, $V=\dfrac{9}{2}$ This has been proved. The volume of the pyramid cut off from the first octant by any tangent plane to the surface $xyz=1$ at points in the first octant is the same.
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