Why is a conserved vector field a gradient of a certain func

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A conserved vector field is associated with a function whose gradient equals the vector field, which stems from the properties of conservative forces. The work done by a conservative force on a closed curve is zero, unlike non-conservative forces, due to the Fundamental Theorem of Calculus. This theorem states that if a function is differentiable, the integral of its derivative depends solely on the endpoints, not the path taken. Therefore, when evaluating a closed path, the endpoints coincide, resulting in a net work of zero. This relationship highlights the fundamental nature of conservative forces in physics.
Za Kh
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I know that if a vector field is conserved then there exits a function such that the gradient of this function is equal to the vector field but am just curious to know the reason of it.
 
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Hi ZA,

You of course googled conservative vector field but there must be something that isn't clear to you. What specifically ?
 
BvU said:
Hi ZA,

You of course googled conservative vector field but there must be something that isn't clear to you. What specifically ?
Hi! I've figured this out, by the stocks theorem and that the work done by a conservative force on a closed curve is zero but why it isn't true for the non conservative force?
 
That comes from the "Fundamental Theorem of Calculus", that if F is a differentiable function such that F'= f then \int_a^b= f(b)- f(a). A "conservative force" is the derivative of some "energy function". The integral depends only on the end points, not the path between the end points. The integral around a closed path, since the "end points" are the same point, is 0.
 
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