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Deriving a forumla for the gradient in cylindrical coordinates

  1. Sep 11, 2013 #1
    Problem:

    Starting from the gradient of a scalar function T(x,y,z) in cartesian coordinates find the formula for the gradient of T(s,ϕ,z) in cylindrical coordinates.

    Solution (so far):

    I know that the gradient is given by [itex] \nabla T = \frac{\partial T}{\partial x}\hat{x}+\frac{\partial T}{\partial y}\hat{y}+\frac{\partial T}{\partial z}\hat{z}[/itex]. We must use the chain rule, so we have [itex] \frac{\partial T}{\partial x}(\frac{\partial z}{\partial x})=\frac{\partial T}{\partial s}(\frac{\partial s}{\partial x})+\frac{\partial T}{\partial ϕ}(\frac{\partial ϕ}{\partial x})+\frac{\partial T}{\partial z}(\frac{\partial z}{\partial x}) [/itex], [itex] \frac{\partial T}{\partial y}=\frac{\partial T}{\partial s}(\frac{\partial s}{\partial y})+\frac{\partial T}{\partial ϕ}(\frac{\partial ϕ}{\partial y})+\frac{\partial T}{\partial z}(\frac{\partial z}{\partial y}) [/itex], and [itex] \frac{\partial T}{\partial x}=\frac{\partial T}{\partial s}(\frac{\partial s}{\partial z})+\frac{\partial T}{\partial ϕ}(\frac{\partial ϕ}{\partial z})+\frac{\partial T}{\partial z}(\frac{\partial z}{\partial z}) [/itex].

    So then I went and calculated ##\hat{x}## and ##\hat{y}## in terms of ##\hat{s}## and ##\hat{ϕ}##. I got ##\hat{x}=cosϕ\hat{s}-sinϕ\hat{ϕ}## and ##\hat{y}=sinϕ\hat{s}+cosϕ\hat{ϕ}##. Note that ##\hat{x} \cdot \hat{x} =1##, ##\hat{y} \cdot \hat{y}=1##, and ##\hat{x} \cdot \hat{y} =0## - as they should.

    Since I need partials of s and ϕ with respect to x and y, I found the inversion formulas ##s=\sqrt{x^2+y^2}## and ##ϕ=sin^{1}(\frac{y}{\sqrt{x^2+y^2}})##. (I used x=s cosϕ and y=s sinϕ)

    Obviously ##\frac{\partial ϕ}{\partial z}=0## and ##\frac{\partial s}{\partial z} =0##, but
    I'm concerned with how complicated ##\frac{\partial ϕ}{\partial x}## and ##\frac{\partial ϕ}{\partial y}## are going to turn out.

    Am I on the right track? Thanks in advance.
     
  2. jcsd
  3. Sep 11, 2013 #2
    Actually, after differentiating and putting it in terms of ϕ, I got ## \frac{\partial ϕ}{\partial x}=-\frac{1}{s}sinϕ## and ##\frac{\partial ϕ}{\partial y}=\frac{1}{s}cosϕ## - That's not too bad. I'd still like to know if this is the correct approach, though.
     
  4. Sep 11, 2013 #3
    Any input is appreciated!
     
  5. Sep 12, 2013 #4

    vela

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    Yup, keep going.
     
  6. Sep 12, 2013 #5
    It worked! Praise the Lord!
     
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