Converting arbitrary Cartesian vector to cylindrical

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To convert a Cartesian vector V={Vx,Vy,Vz} to cylindrical coordinates, the process involves transforming the x and y components into polar coordinates while keeping the z component unchanged. Specifically, the conversion formulas are r = √(Vx² + Vy²) for the radial distance and θ = arctan(Vy/Vx) for the angular component. If Vz is zero, it remains zero in cylindrical coordinates, confirming that the z-axis does not affect the transformation. The discussion emphasizes understanding that cylindrical coordinates extend polar coordinates into three dimensions. This conversion is essential for applications involving cylindrical geometries.
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Hello PF, I have a problem to solve in the following form: Given a vector with Cartesian components, V={Vx,Vy,Vz}, find its components in circular cylindrical coordinate.

Given the actual vector components, it'd be very easy to convert. But I have no idea where to start on this. Any guide to where to start from will be much appreciated!
 
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If the z component were zero then what kind conversion would it be?
 
Well, its a cylindrical coordinate, so Vz wouldn't go over any transformation, right? So if Vz=0, then after the conversion it will still be 0. I am still very lost...
 
I was hoping you'd recognize it as a conversion from x,y to polar coordinates. Cylindrical are an extension of polar to 3D by adding the z component.
 

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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