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Fall 2007 - Cioaba's Class - Exam 2 (Version 1)

# Fall 2007 - Cioaba's Class - Exam 2 (Version 1) - Name PID...

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Unformatted text preview: Name: PID: Discussion Section- No: Time: TA’s name: Midterm 2, Math 20C - Lecture D (Fall 2007) Duration: 50 minutes Please close your books, turn off your phones and put them away. You can use one page of handwritten notes. To get full credit you should explain your answers. Don’t forget to write your name, ID number and your TA’s name. 1.a.(3 points) Find all second order partial derivatives of the function f ( x, y ) = radicalbig x + y 2 . Solution. We write f ( x, y ) = ( x + y 2 ) 1 2 . The first order partial derivatives are f x = ( x + y 2 ) − 1 2 2 f y = y ( x + y 2 ) − 1 2 The second order partial derivatives are f xx = 1 2 · − 1 2 ( x + y 2 ) − 3 2 = − ( x + y 2 ) − 3 2 4 f xy = 1 2 · − 1 2 ( x + y 2 ) − 3 2 2 y = − y ( x + y 2 ) − 3 2 2 f yy = ( x + y 2 ) − 1 2 + y · − 1 2 ( x + y 2 ) − 3 2 2 y = ( x + y 2 ) − 1 2 − y 2 ( x + y 2 ) − 3 2 b.(2 points) Find the linear approximation of the function g ( x, y ) = ln(2 x + y ) at (0 , 1) and use it to approximate f (0 . 1 , 1 . 1). Solution. The linear approximation at (0 , 1) is f ( x, y ) ∼ f (0 , 1) + f x (0 , 1)( x − 0) + f y (0 , 1)( y − 1) We have f x ( x, y ) = 2 2 x + y and f y ( x, y ) = 1 2 x + y which implies f x (0 , 1) = 2 and f y (0 , 1) = 1. Thus, the linear approximation near (0 , 1) is f ( x, y ) ∼ 0 + 2 x + y − 1 which implies that f (0 . 1 , 1 . 1) ∼ . 2 + 1 . 1 − 1 = 0 . 3 2. Suppose you are climbing a hill whose shape is given by the equation z = 1000 − . 01 x 2 − . 02 y 2 where x, y and z are measured in meters, and you are standing at a point with coordinates...
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Fall 2007 - Cioaba's Class - Exam 2 (Version 1) - Name PID...

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