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anonymous
 3 years ago
If you have a function, f, in cylindrical coordinates, can you normalize the function (in a rigorous sense)?
\[f(r,\phi,z)=a*r^{"^"}+b*\phi ^{"^"}+c*z^{"^"}\]
You can find the magnitude by
\[\sqrt{a^2+c^2}\]
but can you truly normalize the function and apply that to a mathematical formula which requires the normalized vector. Or do you have to transform back to rectangular coordinates before normalizing? I have looked through several textbooks and the internet and cannot find anything about normalizing a vector in the cylindrical system.
anonymous
 3 years ago
If you have a function, f, in cylindrical coordinates, can you normalize the function (in a rigorous sense)? \[f(r,\phi,z)=a*r^{"^"}+b*\phi ^{"^"}+c*z^{"^"}\] You can find the magnitude by \[\sqrt{a^2+c^2}\] but can you truly normalize the function and apply that to a mathematical formula which requires the normalized vector. Or do you have to transform back to rectangular coordinates before normalizing? I have looked through several textbooks and the internet and cannot find anything about normalizing a vector in the cylindrical system.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What do you mean by "in a rigorous sense"? I would be interested to see what others think, but what you are describing seems okay to me: The norm of a vector, in the case of a 3D vector, like here, is its length. To normalize a vector, you divide the vector by its norm, creating a unit vector with the same direction as the original vector. So to normalize a 3D vector, you would divide the vector by its length. Thus it seems that \[g=f(r,\theta,z)/\sqrt{a^{2}+c^{2}}\] would be the normalized function, as you have suggested. Again, I do not know if this is "rigorous" enough.
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