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32B(W08)_PracticeFinal_Solutions

# 32B(W08)_PracticeFinal_Solutions - MATH 32B Lecture 4...

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MATH 32B - Lecture 4 - Winter 2008 Solutions to Practice Final - March 12, 2008 NAME: STUDENT ID #: DISCUSSION SECTION: This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 9 problems for a total of 200 points. POINTS: 1. 2. 3. 4. 5. 6. 7. 8. 9. TOTAL: 1

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2 1. (20 points) Use Green’s theorem to find R C F d r where F ( x, y ) = x 2 ( y 2 - x 2 ) i + 2 3 x 3 y j , and C is a triangle with vertices at (1,1), (4,1), (3,5). Solution. Since D = Domain ( ~ F ) = R 2 , which is open and simply con- nected, and also ∂Q ∂x = 2 x 2 y = ∂P ∂y are continuous throughout D , ~ F is con- servative and thus the line integral of ~ F is independent of path. Therefore, Z C ~ F.d~ r = 0 for every closed path C in D . /
3 2. (20 points) Determine whether F ( x, y ) = [ln( y - x ) - x y - x ] i + x y - x j is conservative in the region where y > x . If so, find a potential function f with F = f . Solution. D = Domain ( ~ F ) = { ( x, y ) | y - x > 0 } P = ln( y - x ) - x y - x , Q = x y - x ∂P ∂y = 1 y - x - - x ( y - x ) 2 = y - x + x ( y - x ) 2 = y ( y - x ) 2 , and ∂Q ∂x = y - x - x ( - 1) ( y - x ) 2 = y ( y - x ) 2 Since ∂Q ∂x = ∂P ∂y and are continuous on D, ~ F is conservative. Thus, we can find f such that F = f. That is, F = ( P, Q ) = ( ∂f ∂x , ∂f ∂y ) . ∂f ∂x = ln( y - x ) - x y - x f = x ln( y - x ) + g ( y ) . Taking partial derivative with respect to y , we get ∂f ∂y = x y - x + g 0 ( y ) . Since we want ∂f ∂y = x y - x , g 0 ( y ) = 0, which means that g is constant. Setting this constant to be zero, we have, then, f = x ln( y - x ) . /

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4 3. (25 points)
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32B(W08)_PracticeFinal_Solutions - MATH 32B Lecture 4...

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