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Convert the polar equation into rectangular coordinates

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    r^2= 2cos^2 θ+3sin^2θ



    2. Relevant equations

    X= r cosθ
    y= r sinθ

    3. The attempt at a solution
    √r=√2cos^2 θ+3sin^2θ
    r = 2 cos θ+ 3 sin θ
    r = 2x + 2 y.
    I doubt that i even got close to the correct answer so I like to ask anyone who knows how to deal with this type of equation, Did I make the right moves ( steps ) or did messed it all up ?and if i did, will you please help ( guide ) me to the right way to correct way of doing these type of equations. thanks in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 17, 2008 #2

    Mark44

    Staff: Mentor

    [tex]\sqrt{a^2 + b^2} \neq a + b[/tex]
    I see what you did in your first equation, but the second one doesn't follow from the first, for the reason I gave above. The third doesn't follow from the second.

    The relationship is x = rcos θ and y = rsin θ. How did you go from the second equation to the third?

    Rather than take square root of both sides, start making replacements using the polar equivalents for x and y. Also, there is a useful formula involving x, y, and r.
     
  4. Nov 17, 2008 #3
    i know that r =X^2+ Y^2
    and what i basically did in the 3rd equation was that i plugged in x instead of Cos θ and Y for Sin θ.
     
  5. Nov 17, 2008 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    But x isn't Cos θ : it's r*Cos θ or, in terms of x and y [tex]\sqrt{x^2+y^2}\cos \theta[/tex] and so [tex] \cos \theta=\frac{x}{\sqrt{x^2+y^2}}[/tex] ...how about Sin θ?
     
  6. Nov 18, 2008 #5

    Mark44

    Staff: Mentor

    Then you know something that isn't true, so isn't really worth knowing.
     
  7. Nov 18, 2008 #6
    okay that was kinda harsh! But he's right, you're forgetting something in that r [tex]\neq[/tex] x2+y2 what does r equal? Or maybe, what does r2 equal? :wink:
     
  8. Nov 19, 2008 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Multiplying on both sides by [itex]r^2[/itex] gives [itex]r^4= 2r^2cos^2 \theta+ 3r^2 sin^2 \theta[/itex] or [itex](r^2)^2= 2(r cos \theta)^2+ 3(r sin \theta)^2[/itex].

     
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