Arkhi Posted May 3, 2014 Share Posted May 3, 2014 Alternatively, this thread can be called "Ark hates graphs" thread.Anyway, the creation of this was probably only a matter of time. I am wondering if someone well-suited in the caliber of Calculus would be able to assist me.I need to convert the following polar equation into its rectangular equation:r=6/(2-3sinθ) Or, if you work visually, here's a picture (Number 25): Quote Link to comment Share on other sites More sharing options...
Yash Posted May 3, 2014 Share Posted May 3, 2014 (edited) 4x^2 + 3y^2 + 36y - 36 = 0 if you need the working.. just cross multiply 2-3sint with r.. replace rsint as y and r as sqrt(x^2 + y^2) and rearrange to get only the squareroot term on one side.. then square both sides.. and rearrange the terms.. Edited May 3, 2014 by Aquibex Quote Link to comment Share on other sites More sharing options...
Arkhi Posted May 3, 2014 Author Share Posted May 3, 2014 I derped my multiplication. Thank you. Additionally, I try to avoid roots where possible Quote Link to comment Share on other sites More sharing options...
Yash Posted May 3, 2014 Share Posted May 3, 2014 unfortunately in this case you dont have an option of avoiding the root.. incase you have any more problems im happy to help.. Quote Link to comment Share on other sites More sharing options...
Support Squad Felicity Posted May 3, 2014 Support Squad Share Posted May 3, 2014 OW Quote Link to comment Share on other sites More sharing options...
Arkhi Posted May 4, 2014 Author Share Posted May 4, 2014 4x^2 + 3y^2 + 36y - 36 = 0 if you need the working.. just cross multiply 2-3sint with r.. replace rsint as y and r as sqrt(x^2 + y^2) and rearrange to get only the squareroot term on one side.. then square both sides.. and rearrange the terms.. I notice one problem. If either side is being squared, that means that there cannot be a single y^1 unit. 2r - 3y = 6 2(sqrt(x^2 + y^2)) = 6 + 3y 4x^2 + 4y^2 = 36 + 9y^2 4x^2 - 5y^2 - 36 = 0 is what I ended with. Though, curiously enough, I received another solution through a different step (the step) 2r - 3y = 6 (Squaring everything immediately after this step) 4x^2 + 4y^2 + 9y^2 = 36 (I suppose this could be squared, but then everything is + or - and there are too many combinations of answers) 4x^2 + 13y^2 - 36 = 0 is another answer I received. Because of the property of square roots, there can be multiple different answers, right? One in which -3y is added and one in which -3y is squared. Either way, I never received a y^1 anywhere, so that has me confused. Quote Link to comment Share on other sites More sharing options...
Yash Posted May 6, 2014 Share Posted May 6, 2014 hey sorry about the really late reply.. my net was down for the past couple of days.. it seems i made a mistake as well.. here is the correct solution.. and you will get y^1.. the square of 6-3y becomes 38 - 36y + 9y^2.. putting an image for ur ref.. Quote Link to comment Share on other sites More sharing options...
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