Solving a Difficult Math Problem

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In summary, When solving for the vorticity in cylindrical coordinates, the equation (1/r) ∂/∂r (r√(C/r))= is dependent on the variable r and can be solved using the rule for the derivative of powers. The resulting solution is ωz = C1/2/2rr1/2.
  • #1
alexmarison
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Hi, I'm doing some research on my own, but my math is pretty bad and I am stuck trying to find the solution to this problem:
(1/r) ∂/∂r (r√(C/r))=

I've seen these solutions of similar-looking problems:
(1/r) ∂/∂r (r^2 (Ω))=2Ω
and
(1/r) ∂/∂r (r K/r)=0, except when r=0, in which case the solution is infinite.

Thank you very much!
 
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  • #2
alexmarison said:
Hi, I'm doing some research on my own, but my math is pretty bad and I am stuck trying to find the solution to this problem:
(1/r) ∂/∂r (r√(C/r))=

Usually equations have something on both sides of the equal sign...:wink:
 
  • #3
gabbagabbahey said:
Usually equations have something on both sides of the equal sign...:wink:

Hi, sorry, guess the thing on the other side would matter. It happens to be vorticity or, in cylindrical coordinates, ω_z.
 
  • #4
alexmarison said:
Hi, sorry, guess the thing on the other side would matter. It happens to be vorticity or, in cylindrical coordinates, ω_z.

What coordinate variables does the vorticity depend on (i.e. does it depend on [itex]r[/itex],[itex]\phi[/itex] and/or[itex]z[/itex])?
 
  • #5
gabbagabbahey said:
What coordinate variables does the vorticity depend on (i.e. does it depend on [itex]r[/itex],[itex]\phi[/itex] and/or[itex]z[/itex])?

Out of the 3 cylindrical coordinates, there is only vortical motion wrt z, which motion is dependent on r, as the equation shows, I think.
 
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  • #6
I've been still looking at the problem too, and I think I noticed that in the example solution I gave:
ωz = (1/r) ∂/∂r (ruφ ) = (1/r) ∂/∂r (r2Ω) = 2Ω

it seems as though it was solved like a regular differential equation using only the rule for the derivative of powers:
If f(x)=xn, then f'(x)=nxn-1

Can I just do that to solve mine, too, in which case, I would get f'(x)=(1/2)r-1/2 to give me:
ωz = C1/2/2rr1/2?
 
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