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C4PracticeTest4Sol

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

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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 ˆ i + f y ˆ j = ( y - 2) ˆ i + x ˆ 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 0 ( t 3 ) 2 ( t ) t 2 (2 tdt ) = Z 1 0 2 t 9 dt = [ t 10 5 ] 1 0 = 1 5 3. [20 points] Given -→ F ( x, y ) = x 3 y 4 ˆ i + x 4 y 3 ˆ j where C is the curve defined by -→ r ( t ) = t ˆ i + (1 + t 3 ) ˆ 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 0 ( y ). But f y ( x, y ) = x 4 y 3 so g 0 ( y ) = 0 which means that g ( y ) = K . We can take K = 0, so f ( x, y ) = 1 4 x 4 y 4 . (b) The initial point of C is -→ r (0) = (0 , 1) and the terminal point is -→ r (1) = (1 , 2), so Z C -→ F d -→ r = f (1 , 2) - f (0 , 1) = 4 - 0 = 4

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

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