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# how to use wolfram to convert rectangular equations to polar equations??

Posted 9 years ago
 Trying to figure this out but havent been able to.Example of rectangular would be x^2-y+2=0 and it would need to be in some sort of "r=" form
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Posted 9 years ago
 In:= r Sin[\[Theta]] - r^2 Cos[\[Theta]]^2 - 2 == 0 Out= -2 - r^2 Cos[\[Theta]]^2 + r Sin[\[Theta]] == 0 In:= Solve[-2 - r^2 Cos[\[Theta]]^2 + r Sin[\[Theta]] == 0, {r, \[Theta]}] Out= {{r ->1/2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {r -> 1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {r -> 1/2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {r -> 1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}} In:= PolarPlot[{1/ 2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2, 1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2, 1/2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2, 1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[ 1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {\[Theta], -3, 3}, AspectRatio -> 1/2] Posted 9 years ago Posted 9 years ago
 Take x^2-y+2=a and later make a go to zero In:= x^2 - y^2 == a /. Thread[Rule[{x, y}, FromPolarCoordinates[{r, \[Theta]}]]]; Solve[%, r] Out= {{r -> -(Sqrt[a]/Sqrt[ Cos[\[Theta]]^2 - Sin[\[Theta]]^2])}, {r -> Sqrt[a]/Sqrt[ Cos[\[Theta]]^2 - Sin[\[Theta]]^2]}} You can see a approach zero in this manipulate: Manipulate[ PolarPlot[a/Sqrt[ Cos[\[Theta]]^2 - Sin[\[Theta]]^2], {\[Theta], 0, 2 \[Pi]}, PlotPoints -> 1000, PlotRange -> 5], {{a, 0.5}, 0.02, 1, Appearance -> "Labeled"}] Posted 9 years ago
 What specifically have you tried?
Posted 9 years ago
 Im a noob to this. I just typed in "x^2-y+2=0 Convert to polar" and its just popping up as a graph (helpful but not exactly what im looking for) Im wondering if it will somehow be able to give me an answer in r= some term of sin(theta)/cos(theta)
Posted 9 years ago
 Im a noob to this. I just typed in "x^2-y+2=0 Convert to polar" and its just popping up as a graph (helpful but not exactly what im looking for) Im wondering if it will somehow be able to give me an answer in r= some term of sin(theta)/cos(theta)