Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 18

Answer

$$2x - 6y - 8z = 0$$

Work Step by Step

$$\eqalign{ & {x^2} + 2{z^2} = {y^2},{\text{ }}\left( {1,3, - 2} \right) \cr & {x^2} + 2{z^2} - {y^2} = 0 \cr & {\text{Considering }} \cr & F\left( {x,y,z} \right) = {x^2} + 2{z^2} - {y^2} \cr & {\text{Calculate the partial derivatives }} \cr & {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + 2{z^2} - {y^2}} \right] = 2x \cr & {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + 2{z^2} - {y^2}} \right] = - 2y \cr & {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{x^2} + 2{z^2} - {y^2}} \right] = 4z \cr & {\text{At the point }}\left( {1,3, - 2} \right){\text{ the partial derivatives are}} \cr & {F_x}\left( {1,3, - 2} \right) = 2\left( 1 \right) = 2 \cr & {F_y}\left( {1,3, - 2} \right) = - 2\left( 3 \right) = - 6 \cr & {F_z}\left( {1,3, - 2} \right) = 4\left( { - 2} \right) = - 8 \cr & {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr & {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr & + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr & {\text{Substituting}} \cr & 2\left( {x - 1} \right) - 6\left( {y - 3} \right) - 8\left( {z + 2} \right) = 0 \cr & {\text{Simplifying}} \cr & 2x - 2 - 6y + 18 - 8z - 16 = 0 \cr & 2x - 2 - 6y + 18 - 8z - 16 = 0 \cr & 2x - 6y - 8z = 0 \cr} $$
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