## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 998: 42

#### Answer

(a) $x-2y+2z=-1$ (b) $\dfrac{(x-3)}{-1}=\dfrac{(y-1)}{2}=\dfrac{(z+1)}{-2}$

#### Work Step by Step

a) 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 $f(x,y,z)=(3,1,-1)$ Equation (1), becomes: Thus, $(x-3)(-1)+(y-1)(2)-(z+1)(2)=0$ or, $-x+3+2y-2-2z-2=0$ Hence, $x-2y+2z=-1$ b) Our aim is to determine the normal line equation. The general form is: $\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ ...(2) From the given data, we have $f(x,y,z)=(3,1,-1)$ Equation (2), becomes: Hence, $\dfrac{(x-3)}{-1}=\dfrac{(y-1)}{2}=\dfrac{(z+1)}{-2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.