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Math Help - Converting Polar Equations to Rectangular and vice versa...

  1. #1
    telefl89
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    Converting Polar Equations to Rectangular and vice versa...

    I am having some difficulty with doing this. I am in a Pre-Calc class at my high school. My problem is, I was horrible at solving for identities, and now I am struggling with converting polar equations to rectangular and vice versa. An example of a polar equation I am trying to convert to x and y coordinates is r = 3/(3 - cos). I apologize, I don't know how to enter in the symbol for theta. An equation I am trying to convert to r and theta is 4(x^2)y = 1. Can someone please help explain this to me?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by telefl89
    I am having some difficulty with doing this. I am in a Pre-Calc class at my high school. My problem is, I was horrible at solving for identities, and now I am struggling with converting polar equations to rectangular and vice versa. An example of a polar equation I am trying to convert to x and y coordinates is r = 3/(3 - cos). I apologize, I don't know how to enter in the symbol for theta. An equation I am trying to convert to r and theta is 4(x^2)y = 1. Can someone please help explain this to me?
    Formally, to convert from rectangular to polar what you need to do is set x = r*cos\theta and y = r*sin\theta.

    So, given 4(x^2)y = 1 we get 4(r*cos\theta)^2*r*sin\theta=4r^3cos^2\theta sin\theta = 1.
    Thus, your polar equation is r=(1/4*cos^2\theta sin\theta)^{1/3}.

    Converting from polar to rectangular can be a tad messier. Using the transformations x = r*cos\theta and y = r*sin\theta we can solve for r and \theta:
    r=(x^2+y^2)^{1/2} and \theta=tan^{-1}(y/x).

    So, r = 3/(3 - cos\theta) becomes (x^2+y^2)^{1/2}=3/(3-cos[tan^{-1}(y/x)]).

    What is cos[tan^{-1}(y/x)]? Remember, the argument of cos[ ] is just an angle. Call it \alpha. \alpha is just the angle such that tan\alpha=y/x. What is the cosine of this angle? Well, it's just cos\alpha = x/(x^2+y^2)^{1/2}. So cos[tan^{-1}(y/x)]=x/(x^2+y^2)^{1/2}.

    Your equation now reads: (x^2+y^2)^{1/2}=3/(3-x/(x^2+y^2)^{1/2}). There's a bit of simplifying left to do, but it isn't hard, just busy. (Hint: You can't really come up with a "clean" solution for y, so put it in the form of a general conic section: ax^2+bx+cy^2+dy+e=0. I found that it fits the form for an ellipse.)

    -Dan
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