R^2=9cos(3Ө) Convert to rectangular cordinates

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Homework Help Overview

The discussion revolves around converting the polar equation \( r^2 = 9 \cos(3\theta) \) into rectangular coordinates. The subject area includes polar coordinates and trigonometric identities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss isolating \( \theta \) and rewriting \( \cos(3\theta) \) in terms of \( \cos \) and \( \sin \). There are attempts to express the equation in rectangular form using \( r^2 = x^2 + y^2 \). Some participants question the necessity of isolating \( \theta \) and the correctness of the trigonometric expansion used.

Discussion Status

The discussion is ongoing, with various interpretations and methods being explored. Some participants have provided guidance on trigonometric identities and transformations, while others express uncertainty about the feasibility of finding a simple algebraic expression for \( r \) in terms of \( x \) and \( y \).

Contextual Notes

There are indications that the right-hand side of the equation may evaluate to a negative value, which raises concerns about the existence of a valid solution in rectangular coordinates. Participants are also reminded to include all relevant information in their problem statements.

bubbly616
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Homework Statement

Homework Equations

The Attempt at a Solution


-In order for me to figure out this problem I had to reverse the equality to isolate the Ө on the left side making the new equation 9cos(3Ө)=r^2. The first thing I’m going to do is change cos(3Ө) in terms of cos and sin. That will make my new equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = r^2. Now since r^2 will just reduce to x^2 + y^2 I am now left with the equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = x^2 + y^2. After this I have no idea what comes next.
 
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I don't see any reason to "isolate the Ө". You are, however, wrong about cos(3Ө). You know, I presume, that cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) so that cos(2Ө)= cos^2(Ө)- sin^2(Ө) and sin(2Ө)= 2sin(Ө)cos(Ө). Then cos(3Ө)= cos(2Ө+ Ө)= cos(2Ө)cos(Ө)- sin(2Ө)sin(Ө)= (cos^2(Ө)- sin^2(Ө))cos(Ө)- 2sin(Ө)cos(Ө)(sin(Ө))= cos^3(Ө)- 3sin^2(Ө)cos(Ө).

Your equation, r^2= 9cos(3Ө) is r^2= 9cos^3(Ө)- 9 sin^2(Ө) cos(Ө).

I would mulriply both sides by r^3 to get r^5= 9r^3cos^3(Ө)- 9 (r^2 sin^2(\theta))(rcos(\theta). Now replace r sin(Ө) by y and r cos(Ө) by x:
r^5= 9y^2x^3. Further, r= (x^2+ y^2)^(1/2) and r^5= (x^2+ y^2)^{5/2} so (x^2+ y^2)^{5/2}= 9x^3y^2. If you don't want the 1/2 power, square both sides:
x^2+ y^2= 81x^6y^4.
 
bubbly616 said:

Homework Statement



Homework Equations



The Attempt at a Solution


-In order for me to figure out this problem I had to reverse the equality to isolate the Ө on the left side making the new equation 9cos(3Ө)=r^2. The first thing I’m going to do is change cos(3Ө) in terms of cos and sin. That will make my new equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = r^2. Now since r^2 will just reduce to x^2 + y^2 I am now left with the equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = x^2 + y^2. After this I have no idea what comes next.
Hello bubbly616. Welcome to PF !

In the future:
Please include all information regarding the statement of your problem in "

Homework Statement

", even if the information is in the title of the thread.
 
bubbly616 said:

Homework Statement

Homework Equations

The Attempt at a Solution


-In order for me to figure out this problem I had to reverse the equality to isolate the Ө on the left side making the new equation 9cos(3Ө)=r^2. The first thing I’m going to do is change cos(3Ө) in terms of cos and sin. That will make my new equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = r^2. Now since r^2 will just reduce to x^2 + y^2 I am now left with the equation 9(2cos^3Ө – cosӨ – 2sin^2ӨcosӨ) = x^2 + y^2. After this I have no idea what comes next.
I don't think you are going to find a simple algebraic expression for r in terms of x and y cartesian coordinates, especially when the RHS of the equation evaluates to a negative value.
 

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