Conservative Force: Work Done and Potential Energy Calculation

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daivinhtran
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Homework Statement



Find the work done by a force F = ix^2y^3 + jx^3y^2
Show that this
is a conservative force and find the potential energy U(x, y).

Homework Equations


A force F is conservative when :
dFx/dy = dFy/dx


The Attempt at a Solution



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)

They're not equivalent though
 
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Solution : dFx/dy = 3x^2y^2 = dFy/dx

Can someone explain me the derivative technique? I don't know they get these...
 
daivinhtran said:
A force F is conservative when :
dFx/dy = dFy/dx
Should be partial derivatives: ∂Fx/∂y = ∂Fy/∂x
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)
They will be when you change to partial derivatives. (∂y/∂x = ∂x/∂y = 0)
 
haruspex said:
Should be partial derivatives: ∂Fx/∂y = ∂Fy/∂x

They will be when you change to partial derivatives. (∂y/∂x = ∂x/∂y = 0)

WHat do you mean partial ??

Can you or anyone show me just a first few steps?
 
I'm self studying AP Physics C in high school. Am I expected to know it? Are there any simpler way to solve it? ( My way is to integrate in different paths, but I"m not sure)
 
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.
 
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.

Are those material in Calculus 2??