MHB Convert another equation x^2+y^2=4 to polar form

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SUMMARY

The equation \(x^2 + y^2 = 4\) represents a circle centered at the origin. To convert this Cartesian equation to polar form, the transformation involves substituting \(x\) and \(y\) with their polar equivalents: \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). This results in the equation \(r^2\cos^2(\theta) + r^2\sin^2(\theta) = 4\), which simplifies to \(r^2(\cos^2(\theta) + \sin^2(\theta)) = 4\). Utilizing the Pythagorean identity \(\cos^2(\theta) + \sin^2(\theta) = 1\), the final polar form is \(r = 2\).

PREREQUISITES
  • Understanding of polar coordinates and their relationship to Cartesian coordinates
  • Familiarity with trigonometric identities, specifically the Pythagorean identity
  • Basic algebraic manipulation skills
  • Knowledge of the geometric interpretation of circles in both Cartesian and polar forms
NEXT STEPS
  • Study the conversion of other Cartesian equations to polar form
  • Learn about the geometric properties of circles in polar coordinates
  • Explore trigonometric identities and their applications in coordinate transformations
  • Investigate the implications of polar equations in calculus, particularly in integration and area calculations
USEFUL FOR

Students and educators in mathematics, particularly those studying coordinate geometry and trigonometry, as well as anyone interested in understanding the conversion between Cartesian and polar forms of equations.

Elissa89
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x^2+y^2=4

I have so far:

(r^2)cos^(theta)+(r^2)sin(theta)=4

Idk what I'm supposed to do from here
 
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You need to also square the trig. functions:

$$x^2+y^2=4$$

$$(r\cos(\theta))^2+(r\sin(\theta))^2=4$$

$$r^2\cos^2(\theta)+r^2\sin^2(\theta)=4$$

Factor the LHS...what do you have...is there a trig. identity you can apply?
 
MarkFL said:
You need to also square the trig. functions:

$$x^2+y^2=4$$

$$(r\cos(\theta))^2+(r\sin(\theta))^2=4$$

$$r^2\cos^2(\theta)+r^2\sin^2(\theta)=4$$

Factor the LHS...what do you have...is there a trig. identity you can apply?

got it! thanks!
 
Elissa89 said:
got it! thanks!

We see we have a circle centered at the origin, and in polar coordinates, that's simply a constant value for \(r\). The Cartesian equation:

$$x^2+y^2=a^2$$ where \(0<a\)

Has the polar equation:

$$r=a$$
 

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