Converting a Cartesian equation to polar form

rdioface
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Homework Statement


Convert the following Cartesian equation to polar form.
x^2/9 + y^2/4 = 1

Homework Equations


r*cos(t)=x
r*sin(t)=y
r=Sqrt(x^2 + y^2)
y/x = Arctan(t)

The Attempt at a Solution


I get ugly looking things like r^2(cos^2(t)/9 + sin^2(t)/4) = 1 but being a simple ellipse (edit: duh) I expect a cleaner answer.
 
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rdioface said:

Homework Statement


Convert the following Cartesian equation to polar form.
x^2/9 + y^2/4 = 1

Homework Equations


r*cos(t)=x
r*sin(t)=y
r=Sqrt(x^2 + y^2)
y/x = Arctan(t)

The Attempt at a Solution


I get ugly looking things like r^2(cos^2(t)/9 + sin^2(t)/4) = 1 but being a simple hyperbola I expect a cleaner answer.
It's actually an ellipse, not a hyperbola.

Multiply both sides by 36 to get rid of the fractions. After that you get
4r2cos2(t) + 9r2sin2(t) = 36

You can break up the sin2 term into 4r2sin2(t) + 5r2sin2(t). Does that give you any ideas?
 
That gets me to r^2(4 + 5sin^2(t))=36. Are there any further steps to be done?
 
Alright thanks, it seemed like there might have been some crazy trig identity I was missing or something.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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