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: 8

Answer

$${D_u}h\left( {0,0} \right) = 0$$

Work Step by Step

$$\eqalign{ & h\left( {x,y} \right) = {e^{ - \left( {{x^2} + {y^2}} \right)}},{\text{ }}P\left( {0,0} \right),{\text{ }}{\bf{v}} = {\bf{i}} + {\bf{j}} \cr & {\text{Calculate }}\nabla h\left( {x,y} \right) \cr & \nabla h\left( {x,y} \right) = {h_x}\left( {x,y} \right){\bf{i}} + {h_y}\left( {x,y} \right){\bf{j}} \cr & {h_x}\left( {x,y} \right) = - 2x{e^{ - \left( {{x^2} + {y^2}} \right)}} \cr & {h_y}\left( {x,y} \right) = - 2y{e^{ - \left( {{x^2} + {y^2}} \right)}} \cr & \nabla h\left( {x,y} \right) = - 2x{e^{ - \left( {{x^2} + {y^2}} \right)}}{\bf{i}} - 2y{e^{ - \left( {{x^2} + {y^2}} \right)}}{\bf{j}} \cr & {\text{Evaluate }}\nabla h\left( {0,0} \right) \cr & \nabla h\left( {0,0} \right) = - 2\left( 0 \right){e^{ - \left( 0 \right)}}{\bf{i}} - 2\left( 0 \right){e^{ - \left( 0 \right)}}{\bf{j}} \cr & \nabla h\left( {0,0} \right) = 0 \cr & {\text{Calculate }}\left| {\bf{v}} \right| \cr & \left| {\bf{v}} \right| = \left| {{\bf{i}} + {\bf{j}}} \right| = \sqrt 2 \cr & {\bf{v}}{\text{ is not a unit vector; the unit vector in the direction of }}{\bf{v}}{\text{ is:}} \cr & {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{{\bf{i}} + {\bf{j}}}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\text{The directional derivative at }}\left( {0,0} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr & {D_u}h\left( {x,y} \right) = \nabla h\left( {x,y} \right) \cdot {\bf{u}} \cr & {D_u}h\left( {0,0} \right) = \nabla h\left( {0,0} \right) \cdot \left( {\frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}}} \right) \cr & {D_u}h\left( {0,0} \right) = \left( 0 \right) \cdot \left( {\frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}}} \right) \cr & {D_u}h\left( {0,0} \right) = 0 \cr} $$
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