Convert the polar equation into rectangular coordinates

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Homework Help Overview

The discussion revolves around converting a polar equation, specifically r² = 2cos²(θ) + 3sin²(θ), into rectangular coordinates. Participants are exploring the relationships between polar and rectangular forms, particularly focusing on the equations x = r cos(θ) and y = r sin(θ).

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss various attempts to manipulate the polar equation, with some questioning the validity of steps taken, particularly regarding the transition from polar to rectangular coordinates. There is a focus on the correct application of relationships between r, x, and y.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's reasoning. Some have pointed out potential errors in the original poster's approach, while others suggest alternative methods for making the necessary substitutions. There is no clear consensus yet on the best way to proceed.

Contextual Notes

Participants are grappling with the definitions and relationships between polar and rectangular coordinates, particularly the correct interpretation of r in relation to x and y. There is also a mention of homework constraints that may limit the approaches discussed.

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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