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 - 13.4 Exercises - Page 905: 17

Answer

$$\eqalign{ & f\left( {x,y} \right) = {x^2}y \cr & dz = 0.44 \cr} $$

Work Step by Step

$$\eqalign{ & {\left( {2.01} \right)^2}\left( {9.02} \right) - {2^2} \cdot 9 \cr & {\text{Rewrite the expression}} \cr & = {\left( {2 + 0.01} \right)^2}\left( {9 + 0.02} \right) - {2^2} \cdot 9 \cr & z = \underbrace {{{\left( {2 + 0.01} \right)}^2}}_{{{\left( {x + \Delta x} \right)}^2}}\underbrace {\left( {9 + 0.02} \right)}_{\left( {y + \Delta y} \right)} - \underbrace {{2^2} \cdot 9}_{{x^2}y} \cr & {\text{Let the function }} \cr & z = f\left( {x,y} \right) = {x^2}y,{\text{ with }}x = 2{\text{ and }}y = 9,{\text{ }} \cr & \Delta x = 0.01{\text{ and }}\Delta y = 0.02 \cr & {\text{Therefore, the total differential is }} \cr & dz = \frac{\partial }{{\partial x}}\left[ {{x^2}y} \right]dx + \frac{\partial }{{\partial y}}\left[ {{x^2}y} \right]dy \cr & dz = 2xydx + {x^2}dy \cr & {\text{With }}dx = \Delta x{\text{ and }}dy = \Delta y \cr & dz = 2\left( 2 \right)\left( 9 \right)\left( {0.01} \right) + {\left( 2 \right)^2}\left( {0.02} \right) \cr & dz = 0.44 \cr & \cr & f\left( {x,y} \right) = {x^2}y \cr & dz = 0.44 \cr} $$
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