Calculus work problem help?

Calculus III problem, someone please help?

• Hello, I have a problem in Calculus that I'm not too sure how to work. If someone can offer some help, I'd be grateful. The problem has three parts and is as follows: ...show more

• Answer:

  Your professor is most likely attempting to demonstrate that certain line integrals are path independent (fundamental theorem of line integrals). That's why he's asking for a potential function f in part 3). 1) When y = x ∫(2 - 12xy) dx + (2y - 6x^2)dy = ∫(0 to 2) (2 + 2x - 18x^2) dx (because y = x) = (2x + x^2 - 6x^3) (from x = 0 to 2) = -40 2a) When y = 0 ∫(2 - 12xy) dx + (2y - 6x^2)dy = ∫(0 to 2) 2dx (because y = 0) = 2x (from x = 0 to 2) = 4 2b) When x = 0 ∫(2 - 12xy) dx + (2y - 6x^2)dy = ∫(0 to 2) 2ydy (because x = 0) = y^2 (from x = 0 to 2) = 4 3) You need f with df = Pdx + Qdx. Appealing to the chain rule you want df = (∂f/∂x)dx + (∂f/∂y)dy = (2 - 12xy)dx + (2y - 6x^2)dy ==> ∂f/∂x = 2 - 12xy & ∂f/∂y = 2y - 6x^2 ∂f/∂x = 2 - 12xy ==> f = ∫(2 - 12xy) dx = 2x - 6x^2y + h(y) Then ∂f/∂y = ∂/∂x(2x - 6x2y + h(y)) = -6x2 + h'(y) must = 2y - 6x^2 ==> h(y) = y^2 + C ==> f = 2x + y^2 - 6x^2y This means that regardless of path (see youtube vid below) ∫(2 - 12xy) dx + (2y - 6x^2)dy = ∫ <∂/∂x, ∂/∂y>(2x + y^2 - 6x^2y) .<dx,dy> = ∫ <∂(2x + y^2 - 6x^2y)/∂x, ∂(2x + y^2 - 6x^2y)/∂y>.<dx/dt,dydt> dt = ∫ {(∂(2x + y^2 - 6x^2y)/∂x)(dx/dt) + (∂(2x + y^2 - 6x^2y)/∂y)(dy/dt) }dt = ∫ d(2x + y^2 - 6x^2y)/dt} dt = 2x + y^2 - 6x^2y (regardless of path). So in 1) your answer is - 40, and f(2,2) - f(0,0) = -40, you can check this formula will work with your other line integrals in 2). http://en.wikipedia.org/wiki/Line_integral