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 V12 Gradient Fields in Space 1 The criterion for gradient fields The curl in space We seek now to generalize to space our earlier criterion Section V2 for gradient fields in the plane Criterion for a Gradient Field differentiable Then Let F Mi Nj P k be continuously Proof Since F V f when written out this says M d f dx dM dy N d f p df dy dz d2f d2f d N dydx dxdy dx therefore The two mixed partial derivatives are equal since they are continuous by the hypothesis that F is continuously differentiable The other two equalities in 1 are proved similarly Though the criterion looks more complicated to remember and to check than the one in two dimensions which involves just a single equation it is not difficult to learn and apply For theoretical purposes it can be expressed more elegantly by using the three dimensional vector curl F N j P k be differentiable We define curl F by N i M P j N My k Definition Let F M i 3 curl F P i M j k a a a N P d symbolic notation d etc dx d where V i dx V x F dydj dzdk The equation 3 is the definition The other two lines give symbolic ways of writing and of remembering the right side of 3 Neither the first nor second row of the determinant contains the sort of thing you are allowed to put into a determinant however if you evaluate it using the Laplace expansion by the first row what you get is the right side of 3 Similarly to evaluate the symbolic cross product in 311 we use the determinant 3 In d dM doing these by the product of and M we mean ax ax By using the vector field curl F our criterion 1 becomes V VECTOR INTEGRAL CALCLUS 2 In dealing with a plane vector field F M x y i N x y j we gave the name curl F to the scalar function N My whereas for a vector field F in space curl F is a vector function However if we think of the two dimensional field F as a field in space i e one with zero k component and not depending on z then using definition 3 you can compute that curl F N My k Thus curl F has only a k component so if we are dealing just with two dimensional fields it is natural to give the name curl F just to this k component This is not a universally accepted terminology however some call it the scalar curl others don t use any name at all for N My Naturally the question arises as to whether the converse of 1 is true if curl F 0 is F a gradient field As in two dimensionas this requires some sort of restriction on the domain and we will return to this point after we have studied Stokes theorem For now we will assume the domain is the whole three space in which case it is true Theorem If F is continuously differentiable for all x y z 4 curl F 0 F Vf for some differentiable f x y z We will prove this later If F is a gradient field we can calculate the corresponding mathematical potential function f x y z by the three dimensional analogue of either of the two methods described before Section V2 We illustrate with an example Example 1 For what value s if any of c will F y i x cyz j y2 z2 k be a conservative i e gradient field For each such c find a corresponding potential function f x1y1z Solution Using 1 and 4 we calculate the relevant partial derivatives Thus all three equations in 1 are satisfied H c 2 For this value of c we now find f x y z by two methods Method 1 We use the second fundamental theorem Section V11 l l taking O 0 0 as a convenient lower limit for the integral and using the subscript 1 on the upper limit to avoid confusion with the variables of integration This gives Since the integral is path independent for the choice c 2 we can use any path The usual choice is the path illustrated consisting of three line segments C1 C2 and C3 The parametrizations for them are don t write these out yourself we are only doing it here this first time to make it clear how the line integral is being calculated C1 C2 C3 x x y O z O x x l y y z O x x 1 y 91 Z Z thus dx dx dy 0 dz 0 thus d x O dy dy d z 0 thus d x O dy 0 dz d z V12 GRADIENT FIELDS IN SPACE 3 Using these we calculate the line integral 5 over each of the Ci in turn Dropping subscripts we have therefore by 5 where we have added an arbitrary constant of integration to compensate for our arbitary choice of O 0 0 as the lower limit of integration a different choice would have added a constant to the right side of 6 The work should always be checked from 6 one sees easily that V f F the field we started with Method 2 This requires no line integrals but the work must be carried out systematically otherwise you ll get lost in a mess of equations We are looking for an f x y z such that f f f equivalent to the three equations y x 2yz y2 z2 This is From the first equation integrating with respect to x holding y and z fixed we get 8 f x Y z xY g Y z g is an arbitrary function from 8 x 2yz 29 dY S Y z Y2z h z 9 f x y z xy Y2 from 7 second equation comparing Integrating with respect to y h is an arbitrary function thus y2z h z from the preceding and 8 z2 from 7 third equation comparing ht z z2 1 h z3 3 C 1 f x y z x y Y 2 z z3 C 3 finally by 9 as in Method 1 V VECTOR INTEGRAL CALCLUS 4 2 Exact differentials Just as we did in the two dimensional case we translate the previous ideas into the language of differentials The formal expression which appears as the integrand in our line integrals is called a differential If f x y z is a differentiable function then its total differential or just differential is defined to be The differential 10 is said to be exact in some domain D where M N and P are defined if it is the total differential of some differentiable function f x y z in this domain that is if there exists an …
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