15.5.Ex20-25

15.5.Ex20-25 - S E C T I O N 15.5 SOLUTION The Gradient and...

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SECTION 15.5 The Gradient and Directional Derivatives (ET Section 14.5) 685 SOLUTION By the Chain Rule for Paths we have d dt g ( c ( t )) =∇ g c ( t ) · c ± ( t ) (1) We compute the gradient and c ± ( t ) : g = ¿ g x , g y , g z À = D yz 1 , xz 1 , xyz 2 E c ± ( t ) = - e t , 1 , 2 t ® At the point t = 1wehave c ( 1 ) = ( e , 1 , 1 ) c ± ( 1 ) = h e , 1 , 2 i g c ( 1 ) g ( e , 1 , 1 ) = h 1 , e , e i Substituting the vectors in (1) gives the following derivative: d g ( c ( t )) ¯ ¯ ¯ ¯ t = 1 = h 1 , e , e i · h e , 1 , 2 i = e + e 2 e = 0 20. g ( x , y , z ,w) = x + 2 y + 3 z + 5 w , c ( t ) = ( t 2 , t 3 , t , t 2 ) , t = 1 We compute the gradient and c ± ( t ) : g = ¿ g x , g y , g z , g ∂w À = h 1 , 2 , 3 , 5 i c ± ( t ) = D 2 t , 3 t 2 , 1 , 1 E At the point t = 1 we have (notice that the gradient is a constant vector) g c ( 1 ) = h 1 , 2 , 3 , 5 i c ± ( 1 ) = h 2 , 3 , 1 , 1 i We now use the Chain Rule for Paths to obtain the following derivative: d g ( c ( t )) ¯ ¯ ¯ ¯ t = 1 g c ( 1 ) · c ± ( 1 ) = h 1 , 2 , 3 , 5 i · h 2 , 3 , 1 , 1 i = 2 + 6 + 3 + 5 = 16
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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15.5.Ex20-25 - S E C T I O N 15.5 SOLUTION The Gradient and...

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