Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.6 - Directional Derivatives and the Gradient Vector - 14.6 Exercise - Page 1005: 8

Answer

a) $\nabla f\left( {x,y} \right) = 2x{e^y}{\bf{i}} + {x^2}{e^y}{\bf{j}}$ b) $6{\bf{i}} + 9{\bf{j}}$ c) $- \frac{{18}}{5}$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = {x^2}{e^y},{\text{ }}P\left( {3,0} \right),{\text{ }}{\bf{u}} = \frac{1}{5}\left( {3{\bf{i}} - 4{\bf{j}}} \right) \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2}{e^y}} \right] \cr & {f_x}\left( {x,y} \right) = {e^y}\left( {2x} \right) \cr & {f_x}\left( {x,y} \right) = 2x{e^y} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2}{e^y}} \right] \cr & {f_y}\left( {x,y} \right) = {x^2}{e^y} \cr & \cr & \left( {\text{a}} \right) \cr & {\text{The gradient of the function is given by }} \cr & \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr & {\text{Substituting the partial derivatives}}{\text{, we obtain}} \cr & \nabla f\left( {x,y} \right) = 2x{e^y}{\bf{i}} + {x^2}{e^y}{\bf{j}} \cr & \cr & \left( {\text{b}} \right) \cr & {\text{Evaluate the gradient at the given point }}P\left( {3,0} \right) \cr & \nabla f\left( {3,0} \right) = 2\left( 3 \right){e^0}{\bf{i}} + {\left( 3 \right)^2}{e^0}{\bf{j}} \cr & \nabla f\left( {3,0} \right) = 6{\bf{i}} + 9{\bf{j}} \cr & \cr & \left( {\text{c}} \right) \cr & {\text{The rate of change of }}f{\text{ at }}P{\text{ in the direction of the vector }}{\bf{u}}{\text{ is}} \cr & {D_u}f\left( {x,y} \right) = \nabla f\left( {x,y} \right) \cdot {\bf{u}} \cr & {\text{Therefore at the point }}\left( {3,0} \right) \cr & {D_u}f\left( {3,0} \right) = \nabla f\left( {3,0} \right) \cdot \left( {\frac{1}{5}\left( {3{\bf{i}} - 4{\bf{j}}} \right)} \right) \cr & {D_u}f\left( {3,0} \right) = \left( {6{\bf{i}} + 9{\bf{j}}} \right) \cdot \left( {\frac{1}{5}\left( {3{\bf{i}} - 4{\bf{j}}} \right)} \right) \cr & {D_u}f\left( {3,0} \right) = \left( 6 \right)\left( {\frac{3}{5}} \right) + 9\left( { - \frac{4}{5}} \right) \cr & {D_u}f\left( {3,0} \right) = - \frac{{18}}{5} \cr} $$
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