anonymous
  • anonymous
Transform each polar equation to an equation in rectangular coordinates and identify its shape. r = 6 r = 2cosθ
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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JoannaBlackwelder
  • JoannaBlackwelder
I think you will need these equations for the first one:|dw:1433136389230:dw|
JoannaBlackwelder
  • JoannaBlackwelder
Any ideas how to use those?
anonymous
  • anonymous
Sort of I plug r into the equations, no? so 6(cos)(theta)

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JoannaBlackwelder
  • JoannaBlackwelder
Well, I think it is easier to start with the bottom equation.
JoannaBlackwelder
  • JoannaBlackwelder
|dw:1433136768548:dw|
JoannaBlackwelder
  • JoannaBlackwelder
Do you see how I solved for costheta?
anonymous
  • anonymous
OH
anonymous
  • anonymous
Okay wait, but doesn't the botto equation just give you x/6 =costheta?
anonymous
  • anonymous
how do I find x?
Michele_Laino
  • Michele_Laino
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: \[\Large \sqrt {{x^2} + {y^2}} = 6\]
Michele_Laino
  • Michele_Laino
please square both sides of that equation, what do you get?
JoannaBlackwelder
  • JoannaBlackwelder
Yes, the bottom equation gives x/6=costheta, which makes cos^2theta = x^2/6^2
JoannaBlackwelder
  • JoannaBlackwelder
|dw:1433173120794:dw|
JoannaBlackwelder
  • JoannaBlackwelder
Can you do a similar process with y=rsintheta?
anonymous
  • anonymous
@JoannaBlackwelder To answer your first question you get \[x^2+y^2=36\]
anonymous
  • anonymous
Or that was Michele's.
anonymous
  • anonymous
Oh it's a circle with a radius of 6.
JoannaBlackwelder
  • JoannaBlackwelder
Yep :-)

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