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Derivation of Phi-Hat wrt Phi in Spherical Unit Vectors

  1. Feb 11, 2014 #1
    1. The problem statement, all variables and given/known data

    I just want to know how to get from this: ∂ø^/∂ø = -x^cosø - y^sinø
    to this: = -(r^sinθ+θ^cosθ)

    2. Relevant equations

    All the equations found here in the Spherical Coordinates section: http://en.wikipedia.org/wiki/Unit_vector

    3. The attempt at a solution

    I've tried a bunch of ways of algebraically getting the answer but I seem to be getting nowhere. Maybe I'm missing an equation. I tried adding and subtracting z^cosθ to get -r^ but I'm still missing the other piece. Please help! Thanks so much.
     
  2. jcsd
  3. Feb 11, 2014 #2
    Have you tried going the other way? It is easier (at least more natural) to prove that statement that way, inserting the formulas for [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex].
     
  4. Feb 12, 2014 #3
    I have not, however, it seems like such a process would be much too long. If someone asked you to solve this for a test, how would one solve this without taking so long to do this? To be fair, if each unit vector derivation (d[itex]\hat{r}[/itex]/dt and d[itex]\hat{θ}[/itex]/dt) was incredibly short, why would this one partial derivative take 30 mins to do (if it does so)? I'll try, but I'd like to start from beginning to end to properly understand the process. I appreciate the reply though, thank you.
     
  5. Feb 12, 2014 #4
    On a test, I'd probably use a geometrical method - essentially drawing and trying to see how each unit axis would change. This particular derivative is probably the trickiest one even using that method though.

    However, you wanted an algebraic method. In one direction it looks like a trick to me, in the other way it comes out quite naturally (and that substitution doesn't take very long at all).
     
  6. Feb 21, 2014 #5
    Oh yes. You were right. I tried the derivation backwards and it only took about half a page. However I probably would never have thought about it, but there is a trick. The trick is to place (sin2θ + cos2θ) multiplying the cosø and the sinø in the equation. Once that's done you distribute out the sinθ in the x-hat and the cosθ in the y-hat and the negative one. Then add and subtract z-hat cosθsinθ and group it so that you get r-hat on one side and phi-hat on the other and you should get the answer.
     
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