lecture10-09[1] - Math 18.02(Spring 2009 Lecture 10 Second...

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Math 18.02 (Spring 2009): Lecture 10 Second derivative test. Global max/min: Boundaries and infinity. Level curves and lever surfaces February 26 Last time: Max/Min problems. Least squares approximation. Today: Second derivative test. Global max/min: Boundaries and infinity. Level curves and level surfaces. Reading Material: From Simmos 19.1, 19.7. From the Lecture Notes SD. 2 Second derivatives Given a function z = f ( x, y ) we denoted the first partial derivatives of f as f x = ∂f ∂x and f y = ∂f ∂y . Then it is simple to define the second partial derivatives: f xx = 2 nd derivative w.r.t. x = ∂f ∂x ∂x = 2 f ∂x 2 f xy = derivative w.r.t. y of derivative w.r.t. x = ( f x ) y = ∂f ∂x ∂y = 2 f ∂y dx f yy = 2 nd derivative w.r.t. y = ∂f ∂y ∂y = 2 f ∂y 2 . Remark. There is a very surprising theorem: if all second derivatives are continuous then f xy = f yx , and this in fact is going to be the case for all functions we will be dealing with in this class! 1
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3 2 nd derivative test Last time we learned that for a function z = f ( x, y ) local max/min occur on critical pints , that is on points ( x 0 , y 0 ) such that f x ( x 0 , y 0 ) = 0 f y ( x 0 , y 0 ) = 0 We also learned that not all critical points are either a local max or min, take for example the point (0 , 0) critical point for the function z = x 2 - y 2 which is a saddle point. Question: Is there a way to recognize if a critical point is a max a min or a saddle? The answer to this question is a “partial” YES . We first have to define the Discriminant D : Definition 1. Given a function z = f ( x, y ) we define D = f xx f yy - ( f xy ) 2 = f xx f xy f yx f yy note f xy = f yx . We have the following result: Theorem 1. Assume that ( x 0 , y 0 ) is a critical point for the function z = f ( x, y ) . Let D be the discriminant above evaluated at ( x 0 , y 0 ) . Then 1. if D < 0
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