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MidtermReview - In this Midterm Review File I want to...

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In this Midterm Review File I want to answer a couple of subtle questions that have to do with the Fundamental Theorem of Line Integral, Green’s Theorem, and conservative vector fields. The main questions we want to address are the following: 1. Is the vector field ~ F = h- y, x i x 2 + y 2 = - y x 2 + y 2 ˆ i + x x 2 + y 2 ˆ j , conservative? Solution: First, check that ∂P ∂y = ∂Q ∂x since we know that every conservative vector field must satisfy this, so if it fails there is no way our vector field can be conservative. You can check yourself that indeed this holds for our vector field here. Now, you might recall Theorem 6 on page 1086 of the text which says that if this derivative condition holds and the domain of ~ F is an open, simply-connected region, then ~ F is necessarily conservative. But , since our ~ F is not defined at the origin, its domain is R - { (0 , 0) } and is not simply connected. So, this means Theorem 6 doesn’t apply. But this doesn’t mean our vector field is not conservative. It just means we don’t know, it may or may not be. Next, let’s try to find f and see if we can. To start, we can (partially) integrate P ( x, y ) = - y/ ( x 2 + y 2 ) with respect to x so that f ( x, y ) = Z - y x 2 + y 2 dx + g ( y ) = Z ( - y/x 2 ) 1 + ( y/x ) 2 dx + g ( y ) , where I just factored out an x 2 in the denominator and wrote it in the numerator. This looks just like 1 / (1 + u 2 ) whose integral we know is arctan( u ) . So, do a u substitution and let u = y/x so that du = - y/x 2 dx . Good, we have all these pieces above, so plugging in our u substitution we find that f ( x, y ) = Z du 1 + u 2 + g ( y ) = arctan( u ) + g ( y ) = arctan( y/x ) + g ( y ) . To finish, we take the derivative of f ( x, y ) above with respect to y and set it equal to Q = x/ ( x 2 + y 2 ) above. If you do this you will see that g 0 ( y ) = 0 so g ( y ) is a constant, and we can choose it to be 0 . That is, if we let f ( x, y ) = arctan( y/x ) then you can check directly that for our ~ F above we have that ~ F = ~ f . So, doesn’t this mean that ~ F is conservative? Almost. This is where this gets really subtle. You have to be careful about where ~ F = ~ arctan( y/x ) . If you notice, arctan( y/x ) is not defined when x = 0 . So, that means its gradient is definitely not defined there, even though when you calculate its partials you get something which is defined at points like (0 , y ) . The point is that even though when you calculate the partials of arctan( y/x ) you get something defined at points (0 , y ) for y 6 = 0 these are not the partials of arctan( y/x ) at those
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