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 998: 43

Answer

(a) $x+2y+6z=12$ (b) $(x-2)=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{6}$

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 $f(x,y,z)=(2,2,1)$ Equation (1), becomes: Thus, $(x-2)(4)+(y-2)(8)+(z-1)(24)=0$ or, $4x-8+8y-16+24z-24=0$ Hence, $4x+8y+24z=48$ or, $x+2y+6z=12$ 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 $f(x,y,z)=(2,2,1)$ Equation (2), becomes: Thus, $\dfrac{(x-2)}{4}=\dfrac{(y-2)}{8}=\dfrac{(z-1)}{24}$ Hence, $(x-2)=\dfrac{(y-2)}{2}=\dfrac{(z-1)}{6}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.