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R^2=9cos(3Ө) Convert to rectangular cordinates

  1. Aug 13, 2015 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution
    -In order for me to figure out this problem I had to reverse the equality to isolate the Ө on the left side making the new equation 9cos(3Ө)=r^2. The first thing I’m going to do is change cos(3Ө) in terms of cos and sin. That will make my new equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = r^2. Now since r^2 will just reduce to x^2 + y^2 I am now left with the equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = x^2 + y^2. After this I have no idea what comes next.
     
  2. jcsd
  3. Aug 13, 2015 #2

    HOI

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    I don't see any reason to "isolate the Ө". You are, however, wrong about cos(3Ө). You know, I presume, that cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) so that cos(2Ө)= cos^2(Ө)- sin^2(Ө) and sin(2Ө)= 2sin(Ө)cos(Ө). Then cos(3Ө)= cos(2Ө+ Ө)= cos(2Ө)cos(Ө)- sin(2Ө)sin(Ө)= (cos^2(Ө)- sin^2(Ө))cos(Ө)- 2sin(Ө)cos(Ө)(sin(Ө))= cos^3(Ө)- 3sin^2(Ө)cos(Ө).

    Your equation, r^2= 9cos(3Ө) is r^2= 9cos^3(Ө)- 9 sin^2(Ө) cos(Ө).

    I would mulriply both sides by r^3 to get r^5= 9r^3cos^3(Ө)- 9 (r^2 sin^2(\theta))(rcos(\theta). Now replace r sin(Ө) by y and r cos(Ө) by x:
    r^5= 9y^2x^3. Further, r= (x^2+ y^2)^(1/2) and r^5= (x^2+ y^2)^{5/2} so (x^2+ y^2)^{5/2}= 9x^3y^2. If you don't want the 1/2 power, square both sides:
    x^2+ y^2= 81x^6y^4.
     
  4. Aug 13, 2015 #3

    SammyS

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    Hello bubbly616. Welcome to PF !

    In the future:
    Please include all information regarding the statement of your problem in "1. The problem statement, all variables and given/known data", even if the information is in the title of the thread.
     
  5. Aug 13, 2015 #4

    SteamKing

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    I don't think you are going to find a simple algebraic expression for r in terms of x and y cartesian coordinates, especially when the RHS of the equation evaluates to a negative value.
     
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