The discussion revolves around determining the work done by a force defined as F = ix^2y^3 + jx^3y^2, establishing whether this force is conservative, and finding the potential energy U(x, y). The subject area includes concepts from vector calculus and physics, particularly relating to conservative forces and potential energy.
The discussion is ongoing, with participants providing insights into the nature of partial derivatives and their application in this context. Some guidance has been offered regarding the treatment of variables as constants during differentiation, but there is no explicit consensus on the methods or solutions being discussed.
Some participants express uncertainty about their educational background and whether the concepts discussed are covered in their curriculum, indicating a potential gap in foundational knowledge related to calculus and physics.
Should be partial derivatives: ∂Fx/∂y = ∂Fy/∂xdaivinhtran said:A force F is conservative when :
dFx/dy = dFy/dx
They will be when you change to partial derivatives. (∂y/∂x = ∂x/∂y = 0)dFx/dy = d(x^2y^3)/dy = (2xdx/dy)(y^3) + (x^2)(3y^2)
dFy/dx = d(x^3/y^2)dx = (3x^2)(y^2) + (x^3)(2ydy/dx)
haruspex said:Should be partial derivatives: ∂Fx/∂y = ∂Fy/∂x
They will be when you change to partial derivatives. (∂y/∂x = ∂x/∂y = 0)
haruspex said:If you have a number of independent variables (in this case, x and y), and a function f of these, the partial derivative of f wrt x (written ∂f/∂x) means the change in f as x changes slightly but y stays constant. So when performing a partial derivative wrt x, treat y as a constant: ∂y/∂x = 0. Likewise ∂x/∂y = 0.
Plug those into the equation you got.
I don't know what education system you're in, let alone what's in what syllabus.daivinhtran said:Are those material in Calculus 2??