{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

15.5.Ex20-25

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online