8 Notes 6

8 Notes 6 - MATH 20C Lecture 13 - Monday, October 25, 2010...

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MATH 20C Lecture 13 - Monday, October 25, 2010 Recall chain rule I: g = F ( u ) and u = u ( x,y ), then ∂g ∂x = dF du ∂u ∂x . Used this to compute the partial derivatives of g ( x,y,z ) = ln( x 2 + y 2 - xz ) . Get ∂g ∂x = 2 x - z x 2 + y 2 - xz , ∂g ∂y = 2 y x 2 + y 2 - xz , ∂g ∂z = - x x 2 + y 2 - xz . Higher order partial derivatives Are computed by taking successive partial derivatives. For instance 2 f ∂x 2 = ∂x ± ∂f ∂x ² and so on. Computed 2 g ∂z∂x = ∂x ∂z ³ ∂g ∂x ´ = ∂z ³ 2 x - z x 2 + y 2 - xz ´ = ( - 1)( x 2 + y 2 - xz ) - (2 x - z )( - x ) ( x 2 + y 2 - xz ) 2 2 g ∂x∂z = ∂x ³ ∂g ∂z ´ = ∂x ³ - x x 2 + y 2 - xz ´ = ( - 1)( x 2 + y 2 - xz ) - ( - x )(2 x - z ) ( x 2 + y 2 - xz ) 2 Notice that 2 g ∂z∂x = 2 g ∂x∂z . This is no coincidence. In general, 2 f ∂x∂y = ∂f ∂y∂x MATH 20C Lecture 14 - Wednesday, October 27, 2010 Recall that the gradient vector of f ( x,y,z ) is f = µ ∂f ∂x , ∂f ∂y , ∂f ∂z . Using this notation, the chain rule can be re-written as follows. Consider a function f ( x,y,z ) with x = x ( t ) ,y = y ( t ) ,z = z ( t ) . On the path described by ~ r ( t ) = h x ( t ) ,y ( t ) i , we have df dt = f x dx dt + f y dy dt + f z dz dt = f · µ dx dt , dy dt , dz dt . That is, df dt = f · d~ r dt = f · ~v where ~v is the velocity vector. Note:
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This note was uploaded on 04/28/2011 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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8 Notes 6 - MATH 20C Lecture 13 - Monday, October 25, 2010...

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