Convert the polar equation into rectangular coordinates

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SUMMARY

The discussion centers on converting the polar equation \( r^2 = 2\cos^2 \theta + 3\sin^2 \theta \) into rectangular coordinates. The correct approach involves using the relationships \( x = r\cos \theta \) and \( y = r\sin \theta \) to replace \( r \), \( \cos \theta \), and \( \sin \theta \) with their rectangular equivalents. The final transformation leads to the equation \( r^2 = 2x^2 + 3y^2 \), which can be simplified further. Participants emphasized the importance of correctly applying these relationships to avoid errors in conversion.

PREREQUISITES
  • Understanding of polar coordinates and their relationships to rectangular coordinates
  • Familiarity with trigonometric identities and functions
  • Knowledge of algebraic manipulation and equation solving
  • Basic understanding of Cartesian coordinates
NEXT STEPS
  • Study the derivation of polar to rectangular coordinate transformations
  • Learn about the implications of using trigonometric identities in coordinate conversions
  • Explore examples of converting various polar equations to rectangular form
  • Investigate the geometric interpretations of polar and rectangular coordinates
USEFUL FOR

Students studying mathematics, particularly those focusing on calculus or analytical geometry, as well as educators teaching coordinate transformations.

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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