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8 Notes 6

# 8 Notes 6 - MATH 20C Lecture 13 Monday Recall chain rule I...

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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 and so on. Computed 2 g ∂z∂x = ∂x ∂z ∂g ∂x = ∂z 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 ∂g ∂z = ∂x - 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 = ∂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 · dx dt , dy dt , dz dt . That is, df dt = f · d~ r dt = f · ~v where ~v is the velocity vector. Note: f is a vector whose value depends on the point ( x, y ) where we evaluate f. Theorem : f is perpendicular to the level surfaces f = c. Proof: take a curve ~ r = ~ r ( t ) contained inside level surface f = c. Then velocity ~v = d~ r/dt is in the tangent plane, and by chain rule, dw/dt = f · vecv = 0 , so ~v ⊥ ∇ f.

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