Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 999: 62

Answer

Volume of the pyramids cut off from the first octant by any tangent planes to the surface $xyz=1$ at points in the first octant is same.

Work Step by Step

Our aim is to determine the tangent plane equation. The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1) From the given data, we have $f(x,y,z)=(a,b,c)$ Equation (1), becomes: Thus, $(x-a)bc)+(y-b)ac+(z-c)ab=0$ or, $3abc=bcx+acy+abz$ Next step is to determine x-,y-,z- intercepts.Thus, x-intercept: 3a; y-intercept: 3b and z-intercept: 3c Volume of the pyramid,$V=xyz=abc=\dfrac{1}{3}(\dfrac{9ab}{2})(3c)=\dfrac{9abc}{2}$ This yields $V=\dfrac{9}{2}$ Hence, the result.The volume of the pyramids cut off from the first octant by any tangent planes to the surface $xyz=1$ at points in the first octant is same.
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