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Inverse Polar Graph

  1. Dec 8, 2005 #1
    I need to figure out what or if what I found is why it is.

    I need the inverse of a polar equation graphed.
    When I say inverse I mean switching the variables and then placing it in proper form again.

    I have the equation r=Cos(θ) which produces the following graph.

    What I need is r=ArcCos(θ), when put in Mathematica it gives me this.

    Between the left end of the line and the x and y axises multiple values of pheta exist but they are not ploted, I believe it is because they aren't real, but why? and can this be properly graphed?

    Mathematicas gives me these errors:
    {ArcCos[θ] Cos[θ],ArcCos[θ] Sin[θ]} does not evaluate to a pair of real numbers at θ = 1.0453537892798384
    {ArcCos[θ] Cos[θ],ArcCos[θ] Sin[θ]} does not evaluate to a pair of real numbers at θ = 1.011390375748485
    {ArcCos[θ] Cos[θ],ArcCos[θ] Sin[θ]} does not evaluate to a pair of real numbers at θ = 1.0036610320397505

    Any answers would be much appriciated
  2. jcsd
  3. Dec 10, 2005 #2


    Staff: Mentor

    Mathematica is correct (as it usually is). The problem is that the domain of ArcCos is [-1,1] (since the range of Cos is [-1,1]). In other words, there is no real number whose Cos is greater than 1. You can therefore only plot your equation from -1 <= theta <= 1.

  4. Dec 10, 2005 #3
    I didn't ask the right question sorry... I shouldn't have used cos... What I need to know is this...

    If the graph of r=θ is a spiral, what would the inverse of a spiral look like?
  5. Dec 11, 2005 #4


    Staff: Mentor

    To find inverses you switch variables and solve. E.g. to find the inverse of y=x^2 you switch variables x=y^2 and solve for y to get y=x^(1/2). Same thing here, you switch variables θ=r and solve for r to get r=θ. So, in polar coordinates, the spiral r=θ is its own inverse, just like the line y=x is its own inverse.

  6. Dec 12, 2005 #5
    You can use graphmatica ( <-feel lucky & google on ) for these quick-plot purposes
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