# R^2 -> R^2 transformation of unit circle into square.

1. Jul 16, 2008

### glog

I have a unit circle:
x^2+y^2 <= 1

And i'm asked to convert it to a square with verticies (0,0),(0,1),(1,0),(1,1).

Now obviously I have to do this in polar coordinates, so I've rewritten the equation as:

x = cos t
y = sin t

I'm sort of drawing a blank after setting up these equations. How can I convert this curve into straight lines. Any suggestions?

2. Jul 16, 2008

### epkid08

Those aren't really polar coordinates, they're parametric. Converting your unit circle equation into a polar equation would yield, $$r=\pm\sqrt{\frac{1}{2cos^2(\theta)}$$, to make this into a square, you'll have to convert this polar equation to a parametric equation.

3. Jul 16, 2008

### epkid08

Oops, I just noticed I did something wrong, I'll try to fix it.

4. Jul 16, 2008

### epkid08

Okay, I think I found my problem...
What you need to do is to convert the parametric equation you have into a polar parametric equation. The answer I got was:
$$r=\pm\sqrt{cos^4(T)+sin^2(T)}$$
$$\theta=tan^{-1}(tan(T)$$

This simplifies to four x-y parametric equations with constraints:

$$-1\leq T \leq1 for X_1,X_2,Y_3,Y_4$$

$$T=0 for Y_1,Y_2,X_3,X_4$$

$$X_1=tan^{-1}(tan(T))$$
$$Y_1=\sqrt{cos^4(T)+sin^2(T)}$$

$$X_2=tan^{-1}(tan(T))$$
$$Y_2=-\sqrt{cos^4(T)+sin^2(T)}$$

$$X_3=\sqrt{cos^4(T)+sin^2(T)}$$
$$Y_3=tan^{-1}(tan(T))$$

$$X_4=-\sqrt{cos^4(T)+sin^2(T)}$$
$$Y_4=tan^{-1}(tan(T))$$

Last edited: Jul 17, 2008