# polar to cartesian

• Dec 9th 2009, 11:38 AM
osolage
polar to cartesian
im having problems figuring out how my teacher came up with this solution.

the problem is to convert the following polar to cartesian
r=2-cos(2theta)

we are in agreement for the following steps
r=2-[(cos^2 (theta))-(sin^2(theta))] using the double angle identity

then he has
√(x^2 + y^2) = 2 - [(x/r)^2] + [(y/r)^2]
where did the (x/r) and (y/r) come from and why are they squared??

he goes on to further solve the problem as follows:
√(x^2 + y^2) * (x^2 + y^2) = 2(x^2 + y^2) - (x^2) + (y^2)

final step
√(x^2 + y^2) * (x^2 + y^2) = (x^2) + 3(y^2)

this guy is know to make some mistake ... OFTEN. but i dont want to write this problem off. can someone explain to me what is happening here and a better way to arrive at his answer .. or in the least tell me if the solution is correct.

• Dec 9th 2009, 01:55 PM
Prove It
Quote:

Originally Posted by osolage
im having problems figuring out how my teacher came up with this solution.

the problem is to convert the following polar to cartesian
r=2-cos(2theta)

we are in agreement for the following steps
r=2-[(cos^2 (theta))-(sin^2(theta))] using the double angle identity

then he has
√(x^2 + y^2) = 2 - [(x/r)^2] + [(y/r)^2]
where did the (x/r) and (y/r) come from and why are they squared??

he goes on to further solve the problem as follows:
√(x^2 + y^2) * (x^2 + y^2) = 2(x^2 + y^2) - (x^2) + (y^2)

final step
√(x^2 + y^2) * (x^2 + y^2) = (x^2) + 3(y^2)

this guy is know to make some mistake ... OFTEN. but i dont want to write this problem off. can someone explain to me what is happening here and a better way to arrive at his answer .. or in the least tell me if the solution is correct.

I don't really understand your question...

Are you asking to change this equation

$r = 2 - \cos{(2\theta)}$

into Cartesians?

If so, recall that $r = \sqrt{x^2 + y^2}$.

Also, remember that $\cos{(2\theta)} = \cos^2{\theta} - \sin^2{\theta}$

Now since $x = r\cos{\theta}$ that means that $\cos{\theta} = \frac{x}{r}$

$= \frac{x}{\sqrt{x^2 + y^2}}$.

Also, $y = r\sin{\theta}$, which means that $\sin{\theta} = \frac{y}{r}$

$= \frac{y}{\sqrt{x^2 + y^2}}$.

So the equation becomes

$r = 2 - \cos{(2\theta)}$

$r = 2 - (\cos^2{\theta} - \sin^2{\theta})$

$r = 2 - \cos^2{\theta} + \sin^2{\theta}$

$\sqrt{x^2 + y^2} = 2 - \left(\frac{x}{\sqrt{x^2 + y^2}}\right)^2 + \left(\frac{y}{\sqrt{x^2 + y^2}}\right)^2$

$\sqrt{x^2 + y^2} = 2 - \frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}$

$\sqrt{x^2 + y^2} = 2 + \frac{y^2 - x^2}{x^2 + y^2}$

$\sqrt{x^2 + y^2} = \frac{2x^2 + 2y^2 + y^2 - x^2}{x^2 + y^2}$

$\sqrt{x^2 + y^2} = \frac{x^2 + 3y^2}{x^2 + y^2}$...

You may wish to simplify it further...