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: 29

Answer

$$\eqalign{ & \left( {\bf{a}} \right)L\left( {x,y} \right) = x + y \cr & \left( {\bf{b}} \right)L\left( {0.1, - 0.2} \right) = - 0.1 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = \ln \left( {1 + x + y} \right);\,\,\,\,\left( {0,0} \right)\,\,\,{\text{ and }}\,\,\,\left( {0.1, - 0.2} \right) \cr & f\left( {0,0} \right) = \ln \left( {1 + 0 + 0} \right) \cr & f\left( {0,0} \right) = 0 \cr & {\text{The partial derivatives are}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + x + y}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + x + y}} \cr & {\text{Evaluate the partial derivatives at }}\left( {0,0} \right) \cr & {f_x}\left( {0,0} \right) = 1 \cr & {f_y}\left( {0,0} \right) = 1 \cr & \cr & \left( a \right) \cr & {\text{The linear approximation to the function at }}\,\left( {0,0,0} \right){\text{ is}} \cr & L\left( {x,y} \right) = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr & L\left( {x,y} \right) = \left( {x - 0} \right) + \left( {y + 0} \right) + 0 \cr & L\left( {x,y} \right) = x + y \cr & \cr & \left( b \right){\text{Estimate }}\,\left( {0.1, - 0.2} \right) \cr & L\left( {0.1, - 0.2} \right) = 0.1 - 0.2 \cr & L\left( {0.1, - 0.2} \right) = - 0.1 \cr} $$
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