Converting Polar Equations to Rectangular and vice versa...

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