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anonymous

  • one year ago

Transform each polar equation to an equation in rectangular coordinates and identify its shape. r = 6 r = 2cosθ

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  1. JoannaBlackwelder
    • one year ago
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    I think you will need these equations for the first one:|dw:1433136389230:dw|

  2. JoannaBlackwelder
    • one year ago
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    Any ideas how to use those?

  3. anonymous
    • one year ago
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    Sort of I plug r into the equations, no? so 6(cos)(theta)

  4. JoannaBlackwelder
    • one year ago
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    Well, I think it is easier to start with the bottom equation.

  5. JoannaBlackwelder
    • one year ago
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    |dw:1433136768548:dw|

  6. JoannaBlackwelder
    • one year ago
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    Do you see how I solved for costheta?

  7. anonymous
    • one year ago
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    OH

  8. anonymous
    • one year ago
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    Okay wait, but doesn't the botto equation just give you x/6 =costheta?

  9. anonymous
    • one year ago
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    how do I find x?

  10. Michele_Laino
    • one year ago
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    hint: since the subsequent transformation equations holds: \[\Large \left\{ \begin{gathered} x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ \end{gathered} \right.\] then we can write: \[\Large \sqrt {{x^2} + {y^2}} = r\] Now, substituting into your first equation, namely r=6, we get: \[\Large \sqrt {{x^2} + {y^2}} = 6\]

  11. Michele_Laino
    • one year ago
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    please square both sides of that equation, what do you get?

  12. JoannaBlackwelder
    • one year ago
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    Yes, the bottom equation gives x/6=costheta, which makes cos^2theta = x^2/6^2

  13. JoannaBlackwelder
    • one year ago
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    |dw:1433173120794:dw|

  14. JoannaBlackwelder
    • one year ago
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    Can you do a similar process with y=rsintheta?

  15. anonymous
    • one year ago
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    @JoannaBlackwelder To answer your first question you get \[x^2+y^2=36\]

  16. anonymous
    • one year ago
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    Or that was Michele's.

  17. anonymous
    • one year ago
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    Oh it's a circle with a radius of 6.

  18. JoannaBlackwelder
    • one year ago
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    Yep :-)

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