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Coordinate system transformation

  1. Sep 16, 2012 #1
    Can someone help me with the conversion of this equation to Cartesian coordinates:

    2cosθr + sinθθ

    (Due to formatting limitations, I just made the r_hat and theta_hat components bold-faced)

    I know the answer ought to be -(3y2)/[(x2+y2)+1] but I've tried every variation of the 3 main coordinate transformation eqns that I can think of and haven't gotten anywhere. Those 3 eqns I'm talking about are y=rsinθ, x=rcosθ, and r=sqrt(x2+y2).

    Any help would be great. (Not hw related, need to get this for some research I'm working on.)
     
  2. jcsd
  3. Sep 17, 2012 #2

    HallsofIvy

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    The first thing I would do is convert [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex] to i and j: for a point whose radial line makes angle [itex]\theta[/itex] with the positive x-axis, [itex]\hat{r}= cos(\theta)\hat{i}+ sin(\theta)\hat{j}[/itex] and [itex]\hat{\theta}= -sin(\theta)\hat{i}+ cos(\theta)\hat{j}[/itex].

    So [itex]2cos(\theta)\hat{r}+ sin(\theta)\hat(\theta)= 2cos(\theta)(cos(\theta)\hat{i}+ sin(\theta)\hat{j})+ sin(\theta)(sin(\theta)\hat{i}+ cos(\theta)\hat{j}[/itex][itex]= (2cos^2(\theta)- sin^2(\theta))\hat{i}+ 3cos(\theta)sin(\theta)\hat{j}[/itex]

    Now, convert those trig functions into x, y.
     
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