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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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, Greens 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 i + x x 2 + y 2 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 doesnt apply. But this doesnt mean our vector field is not conservative. It just means we dont know, it may or may not be. Next, lets 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 ( y ) = 0 so g ( y ) is a constant, and we can choose it to be . 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, doesnt 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....
View Full Document

This note was uploaded on 04/12/2009 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online