C4PracticeTest4Sol

C4PracticeTest4Sol - MA 242 Test 4 Solutions 1. [8 points]...

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Unformatted text preview: MA 242 Test 4 Solutions 1. [8 points] Given f ( x,y ) = xy- 2 x , find the gradient vector field of f . Solution: f = f x i + f y j = ( y- 2) i + x j 2. [15 points] Evaluate the line integral, Z C x 2 y zdz , where the curve C is given by the parametric equa- tions, x = t 3 , y = t , z = t 2 over the interval 0 t 1. Solution: Z C x 2 y zdz = Z 1 ( t 3 ) 2 ( t ) t 2 (2 tdt ) = Z 1 2 t 9 dt = [ t 10 5 ] 1 = 1 5 3. [20 points] Given- F ( x,y ) = x 3 y 4 i + x 4 y 3 j where C is the curve defined by- r ( t ) = t i + (1 + t 3 ) j on the interval 0 t 1. (a) Find a function f such that- F = f . (b) Use part (a) to evaluate the line integral, Z C- F d- r along the given curve C . (Use the Fundamental Theorem for Line Integrals). Solution: (a) f x ( x,y ) = x 3 y 4 implies f ( x,y ) = 1 4 x 4 y 4 + g ( y ) and f y ( x,y ) = x 4 y 3 + g ( y )....
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This note was uploaded on 03/31/2009 for the course MA 242 taught by Professor Bliss during the Spring '08 term at N.C. State.

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C4PracticeTest4Sol - MA 242 Test 4 Solutions 1. [8 points]...

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