Why must be that for curl vector in spherical coordinate?

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The discussion centers on the correct formulation of the curl vector in spherical coordinates, specifically debating why the second option is preferred over the first. Participants seek clarification on the derivation and the reasoning behind the correctness of the second formulation, which involves dividing by r²sinθ to yield a unit vector. There is a call for additional resources or links that explain this derivation in detail. The conversation highlights a need for more information to fully understand the distinctions between the two options. Overall, the focus is on clarifying the mathematical principles behind the curl vector in spherical coordinates.
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The correct one is 2nd, but why not first?
Please guide , or tell me any link that relate to this derivation. Thanks
 
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I'm sorry, but your post is lacking some details.
 
I know the second ( divided by r^2sinθ, we will get unit vector also). Why the first one is wrong?
Thanks
 

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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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