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lec_week9

lec_week9 - MIT OpenCourseWare http/ocw.mit.edu 18.02...

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 200 7 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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18.02 Lecture 21. Tue, Oct 30, 2007 Test for gradient fields. Observe: if F = M ˆ ı + N ˆ j is a gradient field then N x = M y . Indeed, if F = f then M = f x , N = f y , so N x = f yx = f xy = M y . Claim: Conversely, if F is defined and differentiable at every point of the plane, and N x = M y , then F = M ˆ ı + N ˆ j is a gradient field. Example: F = y ˆ ı + x ˆ j : N x = 1, M y = 1, so F is not a gradient field. Example: for which value(s) of a is F = (4 x 2 + axy ı + (3 y 2 + 4 x 2 j a gradient field? Answer: N x = 8 x , M y = ax , so a = 8. Finding the potential : if above test says F is a gradient field, we have 2 methods to find the potential function f . Illustrated for the above example (taking a = 8): Method 1: using line integrals (FTC backwards): We know that if C starts at (0 , 0) and ends at ( x 1 , y 1 ) then f ( x 1 , y 1 ) f (0 , 0) = F d� r . Here C · f (0 , 0) is just an integration constant (if f is a potential then so is f + c ). Can also choose the simplest curve C from (0 , 0) to ( x 1 , y 1 ). Simplest choice: take C = portion of x -axis from (0 , 0) to ( x 1 , 0), then vertical segment from ( x 1 , 0) to ( x 1 , y 1 ) (picture drawn). Then F d� r = (4 x 2 + 8 xy ) dx + (3 y 2 + 4 x 2 ) dy : · C C 1 + C 2 x 1 4 x 1 4 Over C 1 , 0 x x 1 , y = 0, dy = 0: = (4 x 2 + 8 x 0) dx = x 3 = x 1 3 . · 3 3 C 1 0 0 y 1 Over C 2 , 0 y y 1 , x = x 1 , dx = 0: = (3 y 2 + 4 x 2 ) dy = y 3 + 4 x 1 2 y y 1 = y 3 + 4 x 1 2 y 1 . 1 0 1 C 2 0 4 So f ( x 1 , y 1 ) = x 3 1 + y 1 3 + 4 x 1 2 y 1 (+constant). 3 Method 2: using antiderivatives: We want f ( x, y ) such that (1) f x = 4 x 2 + 8 xy , (2) f y = 3 y 2 + 4 x 2 . Taking antiderivative of (1) w.r.t. x (treating y as a constant), we get f ( x, y ) = 3 4 x 3 + 4 x 2 y + integration constant (independent of x ). The integration constant still depends on y , call it g ( y ). So f ( x, y ) = 4 3 x 3 + 4 x 2 y + g ( y ). Take partial w.r.t. y , to get f y = 4 x 2 + g ( y ). Comparing this with (2), we get g ( y ) = 3 y 2 , so g ( y ) = y 3 + c . Plugging into above formula for f , we finally get f ( x, y ) = 3 4 x 3 + 4 x 2 y + y 3 + c . Curl. Now we have: N x = M y F is a gradient field F is conservative: C F · d� r = 0 for any closed curve.
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lec_week9 - MIT OpenCourseWare http/ocw.mit.edu 18.02...

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