Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.7 Tangent Planes and Linear Approximation - 12.7 Exercises - Page 936: 27

Answer

$$\eqalign{ & \left( {\bf{a}} \right)L\left( {x,y} \right) = - 6x - 4y + 7 \cr & \left( {\bf{b}} \right)L\left( {3.1, - 1.04} \right) = - 7.44 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = - {x^2} + 2{y^2};\,\,\,\,\left( {3, - 1} \right)\,\,\,{\text{ and }}\,\,\,\left( {3.1, - 1.04} \right) \cr & f\left( {3, - 1} \right) = - {\left( 3 \right)^2} + 2{\left( { - 1} \right)^2} \cr & f\left( {3, - 1} \right) = - 7 \cr & {\text{The partial derivatives are}} \cr & {f_x}\left( {x,y} \right) = - 2x \cr & {f_y}\left( {x,y} \right) = 4y \cr & {\text{Evaluate the partial derivatives at }}\left( {3, - 1} \right) \cr & {f_x}\left( {3, - 1} \right) = - 6 \cr & {f_y}\left( {3, - 1} \right) = - 4 \cr & \cr & \left( a \right) \cr & {\text{The linear approximation to the function at }}\,\left( {3, - 1, - 7} \right){\text{ is}} \cr & L\left( {x,y} \right) = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr & L\left( {x,y} \right) = - 6\left( {x - 3} \right) - 4\left( {y + 1} \right) - 7 \cr & L\left( {x,y} \right) = - 6x + 18 - 4y - 4 - 7 \cr & L\left( {x,y} \right) = - 6x - 4y + 7 \cr & \cr & \left( b \right){\text{Estimate }}\,\left( {3.1, - 1.04} \right) \cr & L\left( {3.1, - 1.04} \right) = - 6\left( {3.1} \right) - 4\left( { - 1.04} \right) + 7 \cr & L\left( {3.1, - 1.04} \right) = - 7.44 \cr} $$
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