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Homework Help: Gradient vector for polar coordinates

  1. Jul 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the gradient vector of:
    [tex]g(r, \theta) = e^{-r} sin \theta[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I know how to get gradients for Cartesian - partially derive the equation of the surface wrt each variable. But I have no idea how to do it for non-Cartesian coordinate systems.

    I tried using the chain rule to get [tex]\frac{{\partial g}{\partial x}}[/tex] and [tex]\frac{{\partial g}{\partial y}}[/tex] so I could plug them into [tex]\nabla g[/tex]. But then I'm ending up with tans and sines and cosines all over the place.
  2. jcsd
  3. Jul 1, 2009 #2
    What coordinate system do you have in mind? Different coordinate systems are set up differently with different basis vectors, for cylindrical/spherical coordinates refer to http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

    For clearer understanding I think it's best to work through the basis vectors, and then use the definition of gradient to work out the required expressions, which you can then finally (whew!) apply to the eqn you want.
  4. Jul 1, 2009 #3
    I want to convert del g into cartesians so I can find a directional derivative, considering the direction vector is in cartesians. I know how to calculate the directional derivative; I just don't know how to convert del g into cartesians from polars.
  5. Jul 1, 2009 #4


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    Why don't you just convert g into cartesians and take the grad there? It is a bit of a pain in the neck, but...
  6. Jul 2, 2009 #5
    because I don't know how, but I'll give it a shot.

    r = x/cos (theta) = y/sin(theta)
    theta = arccos (r/x) = arcsin (r/y)

    partial g/partial x = (partial g/partial r)(partial r/partial x) + (partial g/partial theta)(partial theta/partial x)

    = (-e^(-r))(1/cos(theta)) + (e^(-r) cos(theta))(-r ln x/sqrt(1-(r/x)^2))

    partial g/partial y = (-e^(-r))(1/sin(theta)) + (e^(-r) cos(theta))(r ln y/sqrt(1-(r/y)^2))
  7. Jul 2, 2009 #6
    I think you're making it even more complicated for yourself.

    What's r? sqrt of r^2, yes? Now what's r^2? And then what's sin (theta)?

    Draw it out if you don't see what I mean!
  8. Jul 2, 2009 #7
    ahh. of course.

    r^2 = x^2 + y^2, and sin theta = y/r, where r is the square root of r^2. I indeed was making it too complicated for myself. thanks everyone.
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