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

[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?

 

[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.

 

[epkid08]

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

 

[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))