Inverse Polar Graph: Graphing r=ArcCos(θ)

[Mike AkA Don]

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.
http://images.iceogen.net/cospheta.gif

What I need is r=ArcCos(θ), when put in Mathematica it gives me this.
http://images.iceogen.net/inverse.gif

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

 

[Dale]

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.

-Dale

 

[Mike AkA Don]

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?

 

[Dale]

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.

-Dale

 

[inFinie]

You can use graphmatica ( <-feel lucky & google on ) for these quick-plot purposes