Conservative Force: Work Done and Potential Energy Calculation

AI Thread Summary
The discussion focuses on determining the work done by the force F = ix^2y^3 + jx^3y^2 and verifying its conservativeness through the condition dFx/dy = dFy/dx. Participants clarify that the derivatives should be treated as partial derivatives, which is essential for correctly applying the conservative force criteria. The conversation also addresses the confusion surrounding the derivative technique, emphasizing that when taking partial derivatives, one variable is held constant while differentiating with respect to the other. Additionally, there are inquiries about the educational background required to understand these concepts, particularly regarding their inclusion in Calculus 2. Understanding these principles is crucial for solving the problem effectively.
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??
 
daivinhtran said:
Are those material in Calculus 2??
I don't know what education system you're in, let alone what's in what syllabus.
 
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