# Homework Help: Convert the polar equation into rectangular coordinates

1. Nov 17, 2008

### louie3006

1. The problem statement, all variables and given/known data
r^2= 2cos^2 θ+3sin^2θ

2. Relevant equations

X= r cosθ
y= r sinθ

3. The attempt at a solution
√r=√2cos^2 θ+3sin^2θ
r = 2 cos θ+ 3 sin θ
r = 2x + 2 y.
I doubt that i even got close to the correct answer so I like to ask anyone who knows how to deal with this type of equation, Did I make the right moves ( steps ) or did messed it all up ?and if i did, will you please help ( guide ) me to the right way to correct way of doing these type of equations. thanks in advance.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 17, 2008

### Staff: Mentor

$$\sqrt{a^2 + b^2} \neq a + b$$
I see what you did in your first equation, but the second one doesn't follow from the first, for the reason I gave above. The third doesn't follow from the second.

The relationship is x = rcos θ and y = rsin θ. How did you go from the second equation to the third?

Rather than take square root of both sides, start making replacements using the polar equivalents for x and y. Also, there is a useful formula involving x, y, and r.

3. Nov 17, 2008

### louie3006

i know that r =X^2+ Y^2
and what i basically did in the 3rd equation was that i plugged in x instead of Cos θ and Y for Sin θ.

4. Nov 17, 2008

### gabbagabbahey

But x isn't Cos θ : it's r*Cos θ or, in terms of x and y $$\sqrt{x^2+y^2}\cos \theta$$ and so $$\cos \theta=\frac{x}{\sqrt{x^2+y^2}}$$ ...how about Sin θ?

5. Nov 18, 2008

### Staff: Mentor

Then you know something that isn't true, so isn't really worth knowing.

6. Nov 18, 2008

### CrystalEyes

okay that was kinda harsh! But he's right, you're forgetting something in that r $$\neq$$ x2+y2 what does r equal? Or maybe, what does r2 equal?

7. Nov 19, 2008

### HallsofIvy

Multiplying on both sides by $r^2$ gives $r^4= 2r^2cos^2 \theta+ 3r^2 sin^2 \theta$ or $(r^2)^2= 2(r cos \theta)^2+ 3(r sin \theta)^2$.