Labware - MA35 Multivariable Calculus - Two Variable Calculus



Conservative Vector Fields


A two-dimensional vector field F = (p(x,y),q(x,y)) is conservative if there exists a function f(x,y) such that F = ∇f.

Note that the circulation of ∇f(x,y) along a curve C equals ab fx(x(t),y(t))x'(t) + fy(x(t),y(t))y'(t) dt = ∫abd/dt(f(x(t),y(t)) dt = f(x(t),y(t))|ab = f(x(b),y(b)) - f(x(a),y(a)). In particular, if F(x,y) is a conservative vector field then the circulation of F along a closed curve is 0.

If f exists, then it is called the potential function of F.

If a two-dimensional vector field F(p,q) is conservative, then py = qx.

Suppose that F is conservative and so there does exist a function f such that F = ∇f. Then p = fx and q = fy. Now consider the mixed partials fxy = py and fyx = qx. By the equality of mixed partials, py = fxy = fyx = qx.

If a two-dimensional vector field F(p,q) is conservative, then its curl is identically zero.



  • 1. For each of the following, use the demo to determine whether or not the vector field F is conservative. If it is conservative, find the potential function of F.
    • F = (x, y)
    • F = ( 1, y)
    • F = ( 0, -9.8)
    • F = (cos(x), sin(y))
    • F = (-y, x)
    • F = (x/(x2+y2), y/(x2+y2))
  • 2. Why won't the green curves necessarily lie on the graph of f(x, y) (the function whose gradient is F) if the blue and red paths have different starting points? (Assume F is conservative.)