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Thread: 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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    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 $\displaystyle x = r*cos\theta$ and $\displaystyle y = r*sin\theta$.

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

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

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

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

    Your equation now reads: $\displaystyle (x^2+y^2)^{1/2}=3/(3-x/(x^2+y^2)^{1/2})$. 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:$\displaystyle ax^2+bx+cy^2+dy+e=0$. I found that it fits the form for an ellipse.)

    -Dan
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