# Converting Polar To Cartesian

1. Sep 11, 2010

### Lancelot59

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

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

However logic fails here:$$r=3sin(\theta)$$

Now I know I have the following tools to work with:
$$x=rcos(\theta)$$
$$y=rsin(\theta)$$
$$r^{2}=x^{2}+y^{2}$$
$$tan(\theta)=\frac{y}{x}$$

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:
$$r=tan(\theta)sec(\theta)$$
$$r=2sin(\theta)+2cos(\theta)$$

I'm completely at a loss as to where I should begin. Is there some usual procedure for solving these problems?

Last edited: Sep 11, 2010
2. Sep 11, 2010

### cronxeh

r=3sin(theta)

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

x^2 + y^2 = 3y

Circle centered at (0,1.5) with diameter of 3

3. Sep 11, 2010

### cronxeh

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

4. Sep 11, 2010

### cronxeh

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)

5. Sep 11, 2010

### Lancelot59

I just had to do one substitution to finish that...

I see...
It makes sense. I just needed to see the solutions to some, now I have a better idea of what to do. Thanks!

Last edited: Sep 11, 2010