Convert the polar equation into rectangular coordinates

AI Thread Summary
The discussion focuses on converting the polar equation r^2 = 2cos^2(θ) + 3sin^2(θ) into rectangular coordinates. The initial attempts included incorrect steps, particularly in manipulating the equations and understanding the relationships between r, x, and y. Participants emphasize the importance of using the correct substitutions for x and y, specifically that x = r cos(θ) and y = r sin(θ). There is a consensus that the method of taking square roots was misapplied, leading to errors in the conversion process. The conversation highlights the need for clarity in the relationships between polar and rectangular coordinates to solve such equations correctly.
louie3006
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Homework Statement


r^2= 2cos^2 θ+3sin^2θ



Homework Equations



X= r cosθ
y= r sinθ

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.

 
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\sqrt{a^2 + b^2} \neq a + b
√r=√2cos^2 θ+3sin^2θ
r = 2 cos θ+ 3 sin θ
r = 2x + 2 y

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.
 
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.
 
louie3006 said:
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.

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?
 
louie3006 said:
i know that r =X^2+ Y^2
Then you know something that isn't true, so isn't really worth knowing.
 
Mark44 said:
Then you know something that isn't true, so isn't really worth knowing.

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? :wink:
 
louie3006 said:

Homework Statement


r^2= 2cos^2 θ+3sin^2θ
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.

Homework Equations



X= r cosθ
y= r sinθ

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.
 
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