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 - 13.6 Exercises - Page 924: 16

Answer

$$\nabla z = - 6\sin \left( {25} \right){\bf{i}} - 8\sin \left( {25} \right){\bf{j}}$$

Work Step by Step

$$\eqalign{ & z = \cos \left( {{x^2} + {y^2}} \right),{\text{ }}\left( {3, - 4} \right) \cr & {\text{Calculate the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr & {\text{ }}{z_x} = \frac{\partial }{{\partial x}}\left[ {\cos \left( {{x^2} + {y^2}} \right)} \right] \cr & {\text{ }}{z_x} = - \sin \left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right] \cr & {\text{ }}{z_x} = - \sin \left( {{x^2} + {y^2}} \right)\left( {2x} \right) \cr & {\text{ }}{z_x} = - 2x\sin \left( {{x^2} + {y^2}} \right) \cr & and \cr & {z_y} = \frac{\partial }{{\partial y}}\left[ {\cos \left( {{x^2} + {y^2}} \right)} \right] \cr & {\text{ }}{z_y} = - \sin \left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right] \cr & {\text{ }}{z_y} = - \sin \left( {{x^2} + {y^2}} \right)\left( {2y} \right) \cr & {\text{ }}{z_y} = - 2y\sin \left( {{x^2} + {y^2}} \right) \cr & \cr & {\text{The gradient of }}z{\text{ is}} \cr & \nabla z = {z_x}{\bf{i}} + {z_y}{\bf{j}} \cr & \nabla z = - 2x\sin \left( {{x^2} + {y^2}} \right){\bf{i}} - 2y\sin \left( {{x^2} + {y^2}} \right){\bf{j}} \cr & {\text{At the point }}\left( {3, - 4} \right){\text{ the gradient is}} \cr & \nabla z = - 2\left( 3 \right)\sin \left( {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} \right){\bf{i}} - 2\left( 4 \right)\sin \left( {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} \right){\bf{j}} \cr & \nabla z = - 6\sin \left( {25} \right){\bf{i}} - 8\sin \left( {25} \right){\bf{j}} \cr} $$
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