exam2sol_s07

exam2sol_s07 - MATH 2400 CALCULUS 3 MIDTERM 2 I have...

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Unformatted text preview: MATH 2400: CALCULUS 3 March 14, 2007 MIDTERM 2 I have neither given nor received aid on this exam. Name: 001 E. Kim . . . . . . . . . . . . . . . . (9am) 002 E. Angel . . . . . . . . . . . . .(10am) 003 I. Mishev . . . . . . . . . . . . (11am) 004 J. Boisvert . . . . . . . . . . (12am) 005 A. Gorokhovsky . . . . . . (1pm) If you have a question raise your hand and remain seated. In order to receive full credit your answer must be complete , legible and correct . Show all of your work, and give adequate explanations. DO NOT WRITE IN THIS BOX! Problem Points Score 1 18 pts 2 16 pts 3 10 pts 4 16 pts 5 10 pts 6 10 pts 7 10 pts 8 10 pts TOTAL 160 pts 1. Consider lim ( x,y ) → (0 , 0) 2 x 2 y x 4 + y 2 . (a) (6 pts) Compute the limit along the line y = x . lim ( x,y ) → (0 , 0) Along y =0 2 x 2 y x 4 + y 2 = lim ( x,y ) → (0 , 0) x 4 = 0 (b) (6 pts) Compute the limit along the parabola y = x 2 . lim ( x,y ) → (0 , 0) Along y = x 2 2 x 2 y x 4 + y 2 = lim ( x,y ) → (0 , 0) 2 x 4 x 4 + x 4 = lim ( x,y ) → (0 , 0) 1 = 1 (c) (6 pts) Based on parts (a) and (b), what is lim ( x,y ) → (0 , 0) 2 x 2 y x 4 + y 2 ? The limit does not exist because the limits along different paths to (0 , 0) are different. 2. Given a function f ( x, y ) = y cos( x ) and a point P (0 , 0), (a) (8 pts) Find the local linear approximation L to the function f ( x, y ) at the point P ....
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This note was uploaded on 04/13/2008 for the course MATH 1012 taught by Professor N/a during the Fall '07 term at Colorado.

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exam2sol_s07 - MATH 2400 CALCULUS 3 MIDTERM 2 I have...

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