MHB Multivariable conservative field

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To find a conservative field F where the line integral ∫F ds equals zero without C being a closed path, one can consider scenarios like the force of gravity. For example, if an object is moved vertically and returns to the same height, the work done by gravity is zero, indicating that the potential energy remains unchanged. This illustrates that conservative fields can have zero work done even when the path is not closed. Such examples demonstrate the properties of conservative fields in physics. Understanding these principles is essential for applying concepts of work and energy in various contexts.
kenporock
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Goodnight,

How can I find a conservative field F, such that ∫F ds = 0 without C being a closed path
Can i have some examples ?
 
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kenporock said:
Goodnight,

How can I find a conservative field F, such that ∫F ds = 0 without C being a closed path
Can i have some examples ?

Hi kenporock, welcome to MHB! (Wave)

Consider the force of gravity.
Now move an object around, such that it ends up on the same height it had at the beginning.
The work done by gravity $∫\mathbf F\cdot d\mathbf s$ is then zero.
That is, the object has the same potential energy again.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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