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: 55

Answer

No

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 $f(x,y,z)=(1,1,-1)$ Equation (1), becomes: Thus, $(x-x_0)(2x)+(y-y_0)(-2y)+(z-z_0)(-2z)=0$ There must exist a constant number $c$ when the planes are parallel. Therefore, $(\dfrac{c}{2})^2-(-\dfrac{c}{2})^2-(-\dfrac{c}{2})^2=-\dfrac{c^2}{4}$ This yields no solution. Thus, No point can be found on the hyperboloid.
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