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: 43

Answer

$$\left( {2,2, - 4} \right)$$

Work Step by Step

$$\eqalign{ & z = {x^2} - xy + {y^2} - 2x - 2y \cr & {x^2} - xy + {y^2} - 2x - 2y - z = 0 \cr & {\text{Consider}} \cr & F\left( {x,y,z} \right) = {x^2} - xy + {y^2} - 2x - 2y - z \cr & {\text{Calculating the partial derivatives}} \cr & {F_x}\left( {x,y,z} \right) = 2x - y - 2 \cr & {F_y}\left( {x,y,z} \right) = - x + 2y - 2 \cr & {F_z}\left( {x,y,z} \right) = - 1 \cr & {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr & \nabla F\left( {x,y,z} \right) = \left( {2x - y - 2} \right){\bf{i}} + \left( { - x + 2y - 2} \right){\bf{j}} - {\bf{k}} \cr & {\text{Find the point}}\left( {\text{s}} \right){\text{ on the surface at which the tangent plane }} \cr & {\text{is horizontal}}. \cr & 2x - y - 2 = 0 \cr & - x + 2y - 2 = 0 \cr & {\text{Solving the equations simultaneosly we obtain}} \cr & x = 2,{\text{ }}y = 2 \cr & z = 3{x^2} + 2{y^2} - 3x + 4y - 5 \cr & z = {x^2} - xy + {y^2} - 2x - 2y \cr & z = {\left( 2 \right)^2} - \left( 2 \right)\left( 2 \right) + {\left( 2 \right)^2} - 2\left( 2 \right) - 2\left( 2 \right) \cr & z = - 4 \cr & {\text{The point is}} \cr & \left( {2,2, - 4} \right) \cr} $$
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