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Wolfram Language

The polar equation of the circle of R=1

Simon Cadrin

Simon Cadrin, autodidacte

Posted 10 years ago

In[1]:= r^2 - 6 r Cos[\[Theta]] - 4 r Sin[\[Theta]] == -12 Out[1]= r^2 - 6 r Cos[\[Theta]] - 4 r Sin[\[Theta]] == -12 In[2]:= Solve[r^2 - 6 r Cos[\[Theta]] - 4 r Sin[\[Theta]] == -12, r] Out[2]= {{r -> 1/2 (6 Cos[\[Theta]] + 4 Sin[\[Theta]] - Sqrt[2] Sqrt[-11 + 5 Cos[2 \[Theta]] + 12 Sin[2 \[Theta]]])}, {r -> 1/2 (6 Cos[\[Theta]] + 4 Sin[\[Theta]] + Sqrt[2] Sqrt[-11 + 5 Cos[2 \[Theta]] + 12 Sin[2 \[Theta]]])}} In[5]:= PolarPlot[ 1/2 (6 Cos[\[Theta]] + 4 Sin[\[Theta]] - Sqrt[2] Sqrt[-11 + 5 Cos[2 \[Theta]] + 12 Sin[2 \[Theta]]]), {\[Theta], 0, 2 Pi}]

In[1]:= r^2 - 6 r Cos[\[Theta]] - 4 r Sin[\[Theta]] == -12

Out[1]= r^2 - 6 r Cos[\[Theta]] - 4 r Sin[\[Theta]] == -12

In[2]:= Solve[r^2 - 6 r Cos[\[Theta]] - 4 r Sin[\[Theta]] == -12, r]

Out[2]= {{r -> 
   1/2 (6 Cos[\[Theta]] + 4 Sin[\[Theta]] - 
      Sqrt[2] Sqrt[-11 + 5 Cos[2 \[Theta]] + 
        12 Sin[2 \[Theta]]])}, {r -> 
   1/2 (6 Cos[\[Theta]] + 4 Sin[\[Theta]] + 
      Sqrt[2] Sqrt[-11 + 5 Cos[2 \[Theta]] + 12 Sin[2 \[Theta]]])}}

In[5]:= PolarPlot[
 1/2 (6 Cos[\[Theta]] + 4 Sin[\[Theta]] - 
    Sqrt[2] Sqrt[-11 + 5 Cos[2 \[Theta]] + 
      12 Sin[2 \[Theta]]]), {\[Theta], 0, 2 Pi}]

enter image description here

POSTED BY: Simon Cadrin

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Sean Clarke

Sean Clarke, Wolfram Research

Posted 10 years ago

I see that you are posting a lot of screen shots. I'm sorry. I'm not able to tell what the issue is from them. Can you please be explciit about what the issue is?