Fall 2007 - Mengi's Class - Exam 2 (Version A)

Fall 2007 - Mengi's Class - Exam 2 (Version A) - Midterm 2,...

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Unformatted text preview: Midterm 2, Math 20C (Lecture C) November 28th, 2007 1. Consider the function f ( x, y ) = x 2 +4 y 2 4 xy ( x, y ) 6 = (0 , 0) ( x, y ) = (0 , 0) a) (3 points) Plot the contour diagrams of z = f ( x, y ) for z =- 1 , , 1. Setting z =- 1, we have 0 = x 2 + 4 xy + 4 y 2 = ( x + 2 y ) 2 . Setting z = 1, we have 0 = x 2- 4 xy + 4 y 2 = ( x- 2 y ) 2 . The function takes the value z = 0, only at ( x, y ) = (0 , 0). Therefore the level sets when z =- 1 and z = 1 are the lines x =- 2 y and x = 2 y , respectively. The level set z = 0 is just the origin.-4.8-4-3.2-2.4-1.6-0.8 0.8 1.6 2.4 3.2 4 4.8-2.4-1.6-0.8 0.8 1.6 2.4 z=-1 z=1 z=0 Answer for the other version: The level sets when z =- 2 and z = 2 are the lines y =- 2 x and y = 2 x , respectively. The level set z = 0 is just the origin.-4.8-4-3.2-2.4-1.6-0.8 0.8 1.6 2.4 3.2 4 4.8-2.4-1.6-0.8 0.8 1.6 2.4 z=-2 z=2 z=0 b) (3 points) Is f ( x, y ) continuous at ( x, y ) = (0 , 0)? Justify your answer. No, it is not continuous at (0 , 0), because lim ( x,y ) (0 , 0) f ( x, y ) does not exist. Suppose we approach the origin along the line y = kx . The function along this line is f ( x, kx ) = x 2 +4 k 2 x 2 4 kx 2 = 1+4 k 2 4 k . For instance along the line y = x ( k = 1), the function is constant and equal to 5 4 . But along the line y =- x ( k =- 1), the function takes the value- 5 4 . Since along different paths leading to (0 , 0) the function f ( x, y ) approaches different values, the limit as ( x, y ) (0 , 0) does not exist....
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Fall 2007 - Mengi's Class - Exam 2 (Version A) - Midterm 2,...

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