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math237_a6

# math237_a6 - SOLUTIONS FOR ASSIGNMENT 6 Problem Set 5...

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SOLUTIONS FOR ASSIGNMENT 6 Problem Set 5: A1 (iii): a). Find and classify the critical points of f ( x, y ) = ( x 2 + y 2 1) y b). Sketch the level curves of f in a neighbourhood of each critical point. The Taylor polynomial P 2 ,vectora ( vectorx ) at each critical point will be helpful. Use symmetry to reduce the number of cases, if appropriate. a). To find the critical points of f we need to solve f ( x, y ) = (0 , 0) , i.e. we need to solve the system of equations: ∂ f ∂x ( x, y ) = 2 xy = 0 (1) ∂ f ∂y ( x, y ) = x 2 + 3 y 2 1 = 0 . (2) From equation (1) we have: either x = 0 and thus equation (2) gives y = ± 1 / 3 , or y = 0 and thus equation (2) gives x = ± 1 . Therefore, the critical points of f are: (0 , 1 / 3) , (0 , 1 / 3) , ( 1 , 0) , (1 , 0) . In order to classify these critical points we will try to make use of the Second Derivative Test. The Hessian matrix of f is: Hf ( x, y ) = parenleftBigg 2 y 2 x 2 x 6 y parenrightBigg . Then: Hf (0 , 1 / 3) = parenleftBigg 2 / 3 0 0 6 / 3 parenrightBigg which is positive definite (since its quadratic form Q ( u, v ) = 2 3 ( u 2 + 3 v 2 ) > 0 for all ( u, v ) negationslash = (0 , 0) ). Thus, by the Second Derivative Test the critical point (0 , 1 / 3) is a local minimum of f . Hf (0 , 1 / 3) = parenleftBigg 2 / 3 0 0 6 / 3 parenrightBigg is negative definite (since its quadratic form Q ( u, v ) = 2 3 u 2 6 3 v 2 < 0 for all ( u, v ) negationslash = (0 , 0) ). Thus, by the Second Derivative Test the critical point (0 , 1 / 3) is a local maximum of f . Finally, Hf ( ± 1 , 0) = parenleftBigg 0 ± 2 ± 2 0 parenrightBigg , which are indefinite matrices (since their quadratic forms Q ( u, v ) = ± 4 uv are both positive and negative at points ( u, v ) negationslash = (0 , 0) ). Thus by the Second Derivative Test the critical points ( ± 1 , 0) are saddle points of f . 1

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b). Since all the critical points of f are non-degenerate, the level curves of f can be approximated locally near the critical points by the level curves of the Taylor polynomials P 2 ,vectora ( x, y ) at these points. We have: P 2 , (0 , ± 1 3 ) ( x, y ) = 2 3 3 ± 1 3 x 2 + 3 parenleftBigg y 1 3 parenrightBigg 2 whose level curves P 2 , (0 , ± 1 3 ) ( x, y ) = C are ellipses with C < 0 near the local minimum point (0 , 1 3 ) and with C > 0 near the local maximum point (0 , 1 3 ) , and P 2 , ( ± 1 , 0) ( x, y ) = ± 2( x 1) y, whose level curves P 2 , ( ± 1 , 0) (
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math237_a6 - SOLUTIONS FOR ASSIGNMENT 6 Problem Set 5...

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