MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms V2 Gradient Fields and Exact Differentials 1 Criterion for gradient fields Let F M x y i N x y j be a two dimensional vector field where M and N are continuous functions There are three equivalent ways of saying that F is conservative i e a gradient field 1 F Vf H R F dr is path independent H F dr 0 for any closed C Unfortunately these equivalent formulations don t give us any effective way of deciding if a given field F is a conservative field or not However if we assume that F is not just continuous but is even continuously differentiable meaning M My N Ny all exist and are continuous then there is a simple and elegant criterion for deciding whether or not F is a gradient field in some region Criterion Let F M i N j be continuously differentiable in a region D Then in D F V f for some f x y 2 My N Proof Since F V f this means M f My fxy and and N fy N fy Therefore But since these two mixed partial derivatives are continuous since My and N are by hypothesis a standard 18 02 theorem says they are equal Thus My N This theorem may be expressed in a slightly different form if we define the scalar function called the two dimensional curl of F by curl F N M 3 Then 2 becomes This criterion allows us to test F to see if it is a gradient field Naturally we would also like to know that the converse is true if curl F 0 then F is a gradient field As we shall see however this requires some additional hypotheses on the region D For now we will assume D is the whole plane Then we have Converse t o Criterion Let F M i 4 My N for all x y N j be continuously differentiable for all x y F V f for some differentiable f and all x y The proof of 4 will be postponed until we have more technique For now we will illustrate the use of the criterion and its converse V2 GRADIENT FIELDS AND EXACT DIFFERENTIALS E x a m p l e 1 For which value s if any of the constants a b will axy i a gradient field 1 x2 by j be Solution The partial derivatives are continuous for all x y and My ax N 22 Thus by 2 and 4 F Vf H a 2 b is arbitrary E x a m p l e 2 Are the fields F xi x2 y j y2 G yi x j x2 y2 conservative Solution We have the second line follows from the first by interchanging x and y from this we see immediately that the two equations in the last line show respectively that F and G satisfy the criterion 2 However neither field is defined at O O so that the converse 4 is not applicable So the question cannot be decided just on the basis of 2 and 4 In fact it turns out that F is a gradient field since one can check that On the other hand G is not conservative since if C is the unit circle x cost y sin t we have We will return later on in these notes to this example 2 Finding the potential function a Example 2 above raises the question of how we found the function ln x2 Y More generally if we know that F Vf for example if curl F 0 in the whole xy plane how do we find the function f x y There are two methods some students prefer one some the other Method 1 Suppose F V f By the Fundamental Theorem for Line Integrals Read from left to right 5 gives us an easy way of finding the line integral in terms of f x y But read right to left it gives us a way of finding f x y by using the line integral V VECTOR INTEGRAL CALCULUS 2 Here c is an arbitrary constant of integration as 5 shows c f xo yo R e m a r k It is common to refer to f a y as the mathematical potential function The potential function used in physics is f a y The negative sign is used by physicists so that the potential difference will represent work done against the field F rather than work done by the field as the convention is in mathematics E x a m p l e 3 Let F x y2 i 2xy 3y2 j Verify that F satisfies the Criterion 2 and use method 1 above to find the potential function f a y d Y2 2y d 2xy Solution We verify 2 immediately dY dx We use 5 The point xo yo can be any convenient starting point 0 O is the usual choice if the integrand is defined there We will subscript the variables to avoid confusion with the variables of integration but you don t have to By 59 I Since the integral is path independent we can choose any path we like The usual choice is the one on the right as it simplifies the computations Most of what follows you can do mentally with a little practice On Cl we have y 0 dy 0 so the integral on C1 becomes On C2 we have x x l dx 0 so the integral is I IY1 2xly 1 2 x x dx 2 3y2 dy xlyf y Adding the integrals on C1 and C2 to get the integral along the entire path and dropping the subscripts we get by 6 and 5 1 f x y x2 xy2 y3 C 2 The constant of integration is added by 59 since the choice of starting point was arbitrary You should always confirm the answer by checking that V f F Method 2 Once again suppose F V f that is M i N j f i f y j It follows that f M and fy N 7 These are two equations involving partial derivatives which we can solve simultaneously by integration We illustrate using the previous example F x y2 2xy 3y2 Solution by M e t h o d 2 Using the first equation in 7 df dx 8 x y2 Hold x fixed integrate with respect to y 1 2 where g y is an arbitrary function of y f x2 y2x y d f two ways To find g y we calculate dY af 2yx gl y dY df 2xy 3y2 dY by 8 while from 7 second equation V2 GRADIENT FIELDS AND EXACT DIFFERENTIALS 3 Comparing these two expressions we see that gl y 3y2 so g y y3 c Putting it all together using 8 we get f x y x2 y2x Y3 C as before In the …
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