finalsol

# finalsol - Problem 1 Find the critical points of the...

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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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finalsol - Problem 1 Find the critical points of the...

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