07spr-fsols

07spr-fsols - Math 51, Spring 2007 Final Exam Solutions...

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Unformatted text preview: Math 51, Spring 2007 Final Exam Solutions June 8, 2007 Page 1 of 17 1. (14 points) Let f ( x,y ) = 1 2 x 2 + 3 2 y 2- xy 3 . (a) Find all the critical points of f . For each, specify if it is a local maximum, a local minimum, or a saddle point, and briefly show how you know. Differentiating, we obtain f = f x f y = x- y 3 3 y- 3 xy 2 , and Hf = f xx f xy f yx f yy = 1- 3 y 2- 3 y 2 3- 6 xy . Thus, critical points are solutions to the system x- y 3 = 0 = 3 y (1- xy ) . We have x = y 3 from the first equation, and thus we may rewrite the second equation as 3 y (1- y 4 ) = 0. It follows that either y = 0 or y 4 = 1; the latter can only be true if y = 1 or y =- 1. Using x = y 3 , this gives us three possible critical points: (0 , 0), (1 , 1), and (- 1 ,- 1) . Since Hf is 2 2, we may use the signs of det Hf and of f xx to characterize each critical point: At (0 , 0), det Hf = 1 0 0 3 = 3 > 0, and f xx = 1 > 0, so f has a local minimum here. At (1 , 1), det Hf = 1- 3- 3- 3 =- 12 < 0, so f has a saddle point here. At (- 1 ,- 1), det Hf = 1- 3- 3- 3 < 0 as before, so f has a saddle point here. (b) Write the quadratic approximation (that is, the degree-2 Taylor polynomial) for f at the point ( x,y ) = (1 , 1). The approximation is Q ( x,y ) = f (1 , 1) + f x (1 , 1) ( x- 1) + f y (1 , 1) ( y- 1) + 1 2 f xx (1 , 1) ( x- 1) 2 + f xy (1 , 1) ( x- 1)( y- 1) + 1 2 f yy (1 , 1) ( y- 1) 2 = 1 + 1 2 ( x- 1) 2- 3( x- 1)( y- 1)- 3 2 ( y- 1) 2 . Math 51, Spring 2007 Final Exam Solutions June 8, 2007 Page 2 of 17 2. (12 points) Consider the function f ( x,y ) = p 50- x 2- y 2 . (a) Find an equation that defines the level set of f through the point ( x,y ) = (3 , 4). Sketch and label the curve and point on the axes below. (Be sure to include the scales on your axes.) At the given point, f (3 , 4) = 50- 3 2- 4 2 = 25 = 5, and so the level set has equation p 50- x 2- y 2 = 5. This simplifies to x 2 + y 2 = 25, a circle of radius 5 centered at (0 , 0).-5 5-5 5 f (3,4) x +y = 25 (3,4) (b) Calculate f , the gradient of f , at the point ( x,y ) = (3 , 4) and indicate it on your diagram above. We have f = f x f y =- x (50- x 2- y 2 )- 1 / 2- y (50- x 2- y 2 )- 1 / 2 , so f (3 , 4) =- 3 / 50- 9- 16- 4 / 50- 9- 16 =- 3 / 5- 4 / 5 . (c) Calculate the directional derivative of f at the point (3 , 4) in the direction of the vector (2 ,- 1). The unit vector in the direction of (2 ,- 1) is- u = 1 2 2 + 1 2 (2 ,- 1) = 2 5 ,- 1 5 . This means that the directional derivative is ( D- u f )(3 , 4) = f (3 , 4) - u =- 3 5 ,- 4 5 2 5 ,- 1 5 =- 6 + 4 5 5 =- 2 5 5 . Math 51, Spring 2007 Final Exam Solutions June 8, 2007 Page 3 of 17 3. (8 points) Suppose S is the surface in R 3 given by the equation xy + yz + xz = 1 ....
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07spr-fsols - Math 51, Spring 2007 Final Exam Solutions...

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