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Need Help With Gradient (Spherical Coordinates)

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  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Find te gradient of the following function f(r) = rcos(##\theta##) in spherical coordinates.

    2. Relevant equations
    \begin{equation}
    \nabla f = \frac{\partial f}{\partial r} \hat{r} + (\frac{1}{r}) \frac{\partial f}{\partial \theta} \hat{\theta} + \frac{1}{rsin\theta} \frac{\partial f}{\partial \phi} \hat{\phi}
    \end{equation}

    3. The attempt at a solution
    I know that z = rcos##\theta##


    But I don't know where to go from there since I don't see any ##\hat{r}##, ##\hat{\theta}##, ##\hat{\phi}##
     
  2. jcsd
  3. Feb 1, 2017 #2

    ShayanJ

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    Gold Member

    Do you know how to calculate ## \frac{\partial f}{\partial r} ##,## \frac{\partial f}{\partial \theta} ## and ## \frac{\partial f}{\partial \phi} ##?
     
  4. Feb 1, 2017 #3
    if you're asking if I know how to take partial derivatives, then yes. The issue lies in I don't know where to begin since there is ##\hat{r}##, ##\hat{\theta}##, ##\hat{\phi}## in the equation.
     
  5. Feb 1, 2017 #4

    ShayanJ

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    Gold Member

    Its just like gradient in the Cartesian coordinates. You calculate ## \frac{\partial f}{\partial x} ##,etc. and assume they are components of a vector field. So you have ## \vec \nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})=\frac{\partial f}{\partial x}\hat x+\frac{\partial f}{\partial y}\hat y+\frac{\partial f}{\partial z}\hat z ##.
    What is so different about spherical coordinates?
     
  6. Feb 1, 2017 #5
    So would it be:

    \begin{equation}
    \begin{split}
    f(r) = rcos(\theta) \\
    \frac{\partial f}{\partial r} = cos\theta \\
    \frac{\partial f}{\partial \theta} = - rsin \theta \\
    \frac{\partial f}{\partial \phi} = 0
    \end{split}
    \end{equation}
    So the gradient is
    \begin{equation}
    cos\theta \ \hat{r} + -sin\theta \ \hat{\theta}
    \end{equation}

    Is this correct?
     
  7. Feb 1, 2017 #6

    ShayanJ

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    Gold Member

    Yes, that's correct.
     
  8. Feb 1, 2017 #7
    Thank you, very much! :)
     
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