Calculus: Early Transcendentals (2nd Edition)

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

Chapter 12 - Functions of Several Veriables - 12.6 Directional Derivatives and the Gradient - 12.6 Exercises - Page 925: 14

Answer

$$\nabla {\text{ }}f\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\left\langle {3,2} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \left\langle {3,2} \right\rangle $$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = \sin \left( {3x + 2y} \right);\,\,\,\,\,\,P\left( {\pi ,3\pi /2} \right) \cr & {\text{compute the gradient of }}f\left( {x,y} \right),{\text{ }} \cr & {\text{first find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right).{\text{ then}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sin \left( {3x + 2y} \right)} \right] \cr & {\text{treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\frac{\partial }{{\partial x}}\left[ {\left( {3x + 2y} \right)} \right] \cr & {f_x}\left( {x,y} \right) = 3\cos \left( {3x + 2y} \right) \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sin \left( {3x + 2y} \right)} \right] \cr & {\text{treat }}x{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\frac{\partial }{{\partial y}}\left[ {\left( {3x + 2y} \right)} \right] \cr & {f_x}\left( {x,y} \right) = 2\cos \left( {3x + 2y} \right) \cr & {\text{using }}\nabla {\text{ }}f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr & {\text{derivatives}} \cr & \nabla {\text{ }}f\left( {x,y} \right) = 3\cos \left( {3x + 2y} \right){\bf{i}} + 2\cos \left( {3x + 2y} \right){\bf{j}} \cr & or \cr & \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {3\cos \left( {3x + 2y} \right),2\cos \left( {3x + 2y} \right)} \right\rangle \cr & \nabla {\text{ }}f\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\left\langle {3,2} \right\rangle \cr & {\text{evaluate }}\nabla {\text{ }}f\left( {x,y} \right){\text{ at the point }}P \cr & \nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \cos \left( {3\left( \pi \right) + 2\left( {\frac{{3\pi }}{2}} \right)} \right)\left\langle {3,2} \right\rangle \cr & \nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \cos \left( {6\pi } \right)\left\langle {3,2} \right\rangle \cr & \nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \left\langle {3,2} \right\rangle \cr & \cr & \nabla {\text{ }}f\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\left\langle {3,2} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \left\langle {3,2} \right\rangle \cr} $$
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