# Thread: need help to sketch the curve

1. ## need help to sketch the curve

would you sketch the curve with the polar equation r = 2cos(2theta), and express the equation in rectangular coordinates, that is what must be represented.

2. Originally Posted by rcmango
would you sketch the curve with the polar equation r = 2cos(2theta), and express the equation in rectangular coordinates, that is what must be represented.
I've attached the graph of the curve.

As you easily can see you'll need 4 equations to describe this curve by the usual x,y-coordinates.

3. $r = 2 \cos2\theta$

$r^2 = x^2 + y^2$
...for $r>0~,~~r = + \sqrt{x^2+y^2}$
...for $r<0~,~~r = - \sqrt{x^2+y^2}$

$\tan\theta = \frac{y}{x}$
...for $-\frac{\pi}{2}<\theta<\frac{\pi}{2}~,~~\theta = \arctan(\frac{y}{x})$
...for $\frac{\pi}{2}<\theta<\frac{3\pi}{2}~,~~\theta = \arctan(\frac{y}{x})+\pi$

Well, if we look carefully at $\theta$..
$\theta = \arctan(\frac{y}{x})$
$\theta = \arctan(\frac{y}{x})+\pi$
$r = 2 \cos(2\theta)$
$r = 2 \cos(2(\arctan(\frac{y}{x})+\pi))$
$r = 2 \cos(2\arctan(\frac{y}{x})+2\pi))$
which is equal to $r = 2 \cos2(\arctan(\frac{y}{x}))$

That means, $\theta = \arctan(\frac{y}{x})$ and $\theta = \arctan(\frac{y}{x})+\pi$ are the same. Then, using one of them is enough.

So, we have
$r = 2 \cos2\theta$
...for $r>0~,~~r = + \sqrt{x^2+y^2}$
...for $r<0~,~~r = - \sqrt{x^2+y^2}$
$\theta = \arctan(\frac{y}{x})$

Placing them will give us two functions:
$\sqrt{x^2+y^2} = 2\cos(2\arctan(\frac{y}{x}))$
$-\sqrt{x^2+y^2} = 2\cos(2\arctan(\frac{y}{x}))$