How Do I Convert Polar Functions to Cartesian Functions?

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Lancelot59
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I'm having issues getting converting Polar functions to Cartesian functions. Take for example:

[tex]rcos(\theta)=1[/tex] I just figured that since it was going to always equal the same thing, and because [tex]x=rcos(\theta)[/tex] that the Cartesian equation was x=1, and I was right.

However logic fails here:[tex]r=3sin(\theta)[/tex]

Now I know I have the following tools to work with:
[tex]x=rcos(\theta)[/tex]
[tex]y=rsin(\theta)[/tex]
[tex]r^{2}=x^{2}+y^{2}[/tex]
[tex]tan(\theta)=\frac{y}{x}[/tex]I remember from an example in class that this form is a circle, but I want to be able to algebraically prove it. This looks simple compared to what's further down the page:
[tex]r=tan(\theta)sec(\theta)[/tex]
[tex]r=2sin(\theta)+2cos(\theta)[/tex]

I'm completely at a loss as to where I should begin. Is there some usual procedure for solving these problems?
 
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r=3sin(theta)

r^2 = 3rsin(theta) = 3y

x^2 + y^2 = 3y

Circle centered at (0,1.5) with diameter of 3
 
Lancelot59 said:
[tex]r=tan(\theta)sec(\theta)[/tex]
I'm completely at a loss as to where I should begin. Is there some usual procedure for solving these problems?

Just keep chopping them down as you see them, find patterns, exploit them and convert them

r=tan(theta)*sec(theta) is same as
r=tan(theta)*1/(cos(theta))
r*cos(theta) = tan(theta)
x= tan(theta) = y/x

x^2 = y A bloody parabola
 
Lancelot59 said:
[tex]r=2sin(\theta)+2cos(\theta)[/tex]

I'll throw you one more freebie, by now you should've gotten the clue that these problems are not hard at all, all you need is those 4 tools and some basic trig identities

r=2*sin(theta) + 2*cos(theta) multiply it out by r

r^2 = 2*r*sin(theta) + 2*r*cos(theta)

x^2 + y^2 = 2y + 2x

Its a circle centered at (1,1) with radius of sqrt(2)
 
cronxeh said:
x^2 = y A bloody parabola
I just had to do one substitution to finish that...

cronxeh said:
Circle centered at (0,1.5) with diameter of 3
I see...
cronxeh said:
I'll throw you one more freebie, by now you should've gotten the clue that these problems are not hard at all, all you need is those 4 tools and some basic trig identities

...

Its a circle centered at (1,1) with radius of sqrt(2)
It makes sense. I just needed to see the solutions to some, now I have a better idea of what to do. Thanks!
 
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