Thread: Need help solving/proving how to convert a cartesian equation into a polar equation

1. Need help solving/proving how to convert a cartesian equation into a polar equation

I need help converting this cartesian equation:
0.2025=x2+(y+1.5)2
into this polar equation:
r=-sqrt(2.25(sin(theta)2)-2.05)-1.5sin(theta)

What steps do I need to take to convert the first equation to the second one? Please help explain this to me; a step by step walkthrough would be greatly appreciated. Thank you.

2. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Hey lienj.

When you convert from one system to another you need a set of equations to go from one space to the other.

In this case it is r(x,y) and theta(x,y) - Can you define these first?

3. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Originally Posted by lienj
I need help converting this cartesian equation:
0.2025=x2+(y+1.5)2
into this polar equation:
r=-sqrt(2.25(sin(theta)2)-2.05)-1.5sin(theta)

What steps do I need to take to convert the first equation to the second one? Please help explain this to me; a step by step walkthrough would be greatly appreciated. Thank you.
you simply need to make the substitutions

$x = r \cos(\theta)$

$y = r \sin(\theta)$

and wrangle through the resulting algebra.

Get it into a quadratic equation in $r$ and use the quadratic formula

4. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Chiro,
I'm pretty ignorant in this subject, but I believe they are both located at (0,-1.5) and have a radius of 0.45.

5. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Romsek,
I am aware of the conversions; I just don't know how to go about solving it from there. Could you elaborate on what steps are required? Thanks.

6. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Well, this is a circle with center at (0, -1.5) and radius 0.45.

Now, using $x= r cos(\theta)$ and $y= r sin(\theta)$, as romsek suggested.
$.2025= x^2+ (y+ 1.5)^2= x^2+ y^2+ 3.0y+ 2.25$ can be written as $x^2+ y^2+ 0.3y+ 2.0475= 0$. Of course, $x^2+ y^2= r^2cos^2(\theta)+ r^2 sin^2(\theta)= r^2(cos^2(\theta)+ sin^2(\theta))= r^2$ and $0.3y= 0.3r sin(\theta)$ so that becomes $r^2+ 0.3r sin\theta)+ 2.0475= 0$. You can use the quadratic formula solve for r. There are two such functions.

7. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Originally Posted by lienj
Romsek,
I am aware of the conversions; I just don't know how to go about solving it from there. Could you elaborate on what steps are required? Thanks.
You just want someone to slog through the algebra for you. Well that won't be me. Maybe someone else will.

8. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

I need help converting this cartesian equation:
0.2025=x^2+(y+1.5)^2
into this polar equation:
r=-sqrt(2.25(sin^2(theta))-2.05)-1.5sin(theta)
$0.2025 = \color{red}{x^2 + y^2} + 3\color{blue}{y} + 2.25$

$0.2025 = \color{red}{r^2} + 3\color{blue}{r\sin{\theta}} + 2.25$

$0 = r^2 + 3r\sin{\theta} + 2.0475$

quadratic formula to solve for $r$ ...

$r = \dfrac{-3\sin{\theta} \pm \sqrt{9\sin^2{\theta} - 8.19}}{2}$

I'll leave it for you to simplify.

9. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Skeeter,
Thank you very much for helping me. I am terribly sorry to inform you that I am still unable to simplify the equation any further. I know -3sin(theta)/2 can be reduced to -1.5sin(theta), but I do not know how to convert the radical divided by two to match the polar equation above. Sorry for everyone's trouble, but if anyone could, please help finish proving the equations. Thank you, I hope you understand that my math is terrible.

10. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

$r = \dfrac{-3\sin{\theta} \pm \sqrt{9\sin^2{\theta}-8.19}}{2}$

$r = \dfrac{-3\sin{\theta}}{2} \pm \dfrac{\sqrt{9\sin^2{\theta}-8.19}}{2}$

$r = \dfrac{-3\sin{\theta}}{2} \pm \dfrac{\sqrt{9\sin^2{\theta}-8.19}}{\sqrt{4}}$

$r = \dfrac{-3\sin{\theta}}{2} \pm \sqrt{\dfrac{9\sin^2{\theta}-8.19}{4}}$

$r = \dfrac{-3\sin{\theta}}{2} \pm \sqrt{\dfrac{9\sin^2{\theta}}{4}-\dfrac{8.19}{4}}$

$r = -1.5\sin{\theta} \pm \sqrt{2.25\sin^2{\theta}-2.0475}$

good enough there, Zeke?

11. Re: Need help solving/proving how to convert a cartesian equation into a polar equati

Thank you so much!