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Unformatted text preview: Problem 1. Find the critical points of the function f ( x,y ) = 1 2 x 2 + 3 2 y 2 xy 3 and determine their nature. Solution: We have f x = x y 3 = 0 = x = y 3 f y = 3 y 3 xy 2 = 0 = y = xy 2 = y = y 5 = y = 0 ,y = 1 , or y = 1 . We find the critical points (0 , 0) , (1 , 1) ( 1 , 1) . We calculate f xx = 1 , f xy = 3 y 2 , f yy = 3 6 xy. At the critical point (0 , 0) the Hessian is H f = 1 0 0 3 which is positive definite, hence (0 , 0) is a local minimum. At ( x,y ) = (1 , 1) , the Hessian is H f = 1 3 3 3 which has negative determinant, hence it is definite, so (1 , 1) are both saddle points. Problem 2. Consider the function f ( x,y ) = y 2 x 4 . (i) Draw the level curves of f . (ii) At the point P (1 , 1), find the unit direction of steepest increase for f . (iiii) For the vector ~u = 4 ~ i +3 ~ j 5 , find the directional derivative of f at P in the direction ~u , and use it to estimate f ((1 , 1) + . 01 ~u ) . Solution: (i) The function f is undefined at x = 0 . The level curves of f are f ( x,y ) = c y 2 = cx 4 y = cx 2 . When c > , we get two arches of a parabola i.e. the parabola minus the point (0 , 0) . When c = 0 , we get the axis y = 0 minus the point (0 , 0) . (ii) We calculate the gradient f = 4 y 2 x 5 , 2 y x 4 which at P equals f ( P ) = ( 4 , 2) . The unit direction of steepest increase is ~u = f  f  = ( 4 , 2) 20 = ( 2 , 1) 5 . (iii) We have ~u f = f ~u = ( 4 , 2) 4 5 , 3 5 = 2 . We obtain f ((1 , 1) + . 01 ~u )) f (1 , 1) + . 01 ~u f = 1 . 02 = . 98 . Problem 3. Show that the function f : R 2 R 2 given by f ( x,y ) = ( e x + e 2 y ,e 3 y + e 2 x ) is locally invertible at every point. Solution: We calculate the Jacobian J f = e x 2 e 2 y 2 e 2 x 3 e 3 y which has determinant det J f = 3 e x +3 y 4 e 2 x +2 y 6 = 0 ....
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 Winter '11
 DragosOprea
 Critical Point

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