Trig: from rectangular to polar

[Joe05]

equation: y=x^(2)-3

solve for R

given: R^(2)=x^(2)+y^(2)
x=RcosO
y=RsinO

problem: So i get about as far as plugging in the values for x and y so the equation is in terms of only R, sine, and cosine. my only problem is manipulating the problem so that it reads, "R=....."

 

[tkhunny]

Well, you may need to review your algebra. Show how far you have gotten on this one. You may need to come to grips with NOT solving for "R" explicitly.

 

[Joe05]

y=x^(2)-3
Plug in values for x and y
Rsin=(Rcos)^(2)-3
Square Rcos
Rsin=(R^(2)cos^(2))-3
cos^(2)=1-sin^(2)
Rsin=(R^(2))(1-sin^(2))-3
Distribute R^(2) into (1-sin^(2))
Rsin=R^(2)-(R^(2)sin^(2))-3
+(R^(2)sin^(2))+3 to both sides
(R^(2)sin^(2))+Rsin+3=R^(2)
...stuck
But the teacher did say it was solvable for R. But then again, it wouldn't be the first time a teacher was wrong

 

[tkhunny]

1) "cos" doesn't mean anything in this context.
2) "sin" doesn't mean anything in this context.
3) A little formality will go a long way. It will save you if you refuse to get lazy.
4) Like any other Quadratic-looking equation, why did you leave the R^2 on the right? That should be zero (0).
5) If you are willing to use that nasty +/- symbol, yes, you can solve.
6) What was the point of changing the cosine to sine?

Get to here and see if you can complete it.

\(\displaystyle R^{2}\cdot \cos^{2}(\theta) -R\cdot\sin(\theta)-3=0\)